ce220 reader for 2009

REPORT

UCB/SEMM

STRUCTURAL ENGINEERING MECHANICS AND MATERIALS

CE 220 - STRUCTURAL ANALYSIS,

THEORY AND APPLICATIONS

COURSE NOTES

by

FILIP C. FILIPPOU

FALL SEMESTER 2009 DEPARTMENT OF CIVIL ENGINEERING UNIVERSITY OF CALIFORNIA BERKELEY, CALIFORNIA

Notation

General:

• Upper case letters for matrices and vectors refer to variables of the entire structure, e.g. collection of forces, displacements, etc.

• Lower case letters for matrices and vectors refer to element variables. • A bold face font is used for matrices, a bold face italic font is used for vectors. An

italic font is used for scalars. • All variables refer to a global (structural) reference system under otherwise

indicated

General notation:

Static variables

P applied forces at global degrees of freedom (dofs) p element end forces q basic element forces (independent element end forces)

Q collection of all element basic forces for structural model

s section resultant force vector

σ stress tensor

Kinematic variables

U global dof displacement vector

u element end node displacements

v Element deformations

V collection of all element deformations for structural model

e section deformations ε strain tensor

Matrices relating static and kinematic variables

B static (equilibrium) matrix for structure

A kinematic (compatibility) matrix for structure

F structure flexibility matrix

f element flexibility matrix

K structure stiffness matrix

k element basic stiffness matrix

Page 1

Subscripts

Subscripts are used to specialize the variables on the preceding page often denoting a subset or subarray of the original variable. They have the following meaning:

Single subscript

Interpretation

b Boolean

c refers to the incipient collapse state for scalars and vectors denotes constraint for arrays

d refers to restrained dofs

e elastic ε refers to strain based deformations

f refers to free dofs

g global

h hinge

i refers to statically determinate primary system

p refers to the particular solution of a system of equations; denotes plastic in arrays or when used as second subscript

r refers to rotation for arrays denotes resisting when used with force vectors denotes residual when used with deformations in CE221

s collection of element matrices for structural model

u denotes unbalance when used with force vectors

x refers to force redundants

Double subscript

Interpretation

pl plastic capacity

ce elastic at incipient collapse

cp plastic at incipient collapse

Page 2

Variables with diacritical marks

B force influence matrix of structure

p element end forces in local (element) reference system

u element end displacements in local reference system

fA kinematic matrix of structure for constrained free dofs

fU independent free dofs of structural model under constraints

Scalar variables

N axial force

M bending moment

V shear force α coefficient of thermal expansion

aε normal strain at reference axis of frame element

θ rotation of element chord

κ curvature

λ load factor

Other variables or symbols

δ variation of (also denotes virtual)

Δ increment of

X, Y, Z global (structure) coordinates

x, y, z local (element) coordinates

Page 3

Variables by section

Statics or Equilibrium

P applied forces at global degrees of freedom (dofs)

fP applied forces at free global dofs

=dP R applied forces at restrained global dofs = support reactions

fwP equivalent nodal forces at free dofsdue to element loading

p element end forces in global (structure) reference system p element end forces in local (element) reference system q basic element forces (independent element end forces)

Q collection of all element basic forces for structural model

s section resultant force vector

σ stress tensor

N axial force

M bending moment

V shear force

b force transformation matrix from basic to local system

rb force transformation matrix from local to global system

gb force transformation matrix from basic to global system

B static (equilibrium) matrix for structure

fB static (equilibrium) matrix for free dofs of structure

dB static (equilibrium) matrix for restrained dofs of structure

iB submatrix of fB for primary basic forces

xB submatrix of fB for redundant basic forces

B force influence matrix of structure

iB force influence matrix of primary structure under applied nodal forces

xB force influence matrix of primary structure under redundant basic forces

pQ basic forces for particular solution of equilibrium equations

Page 4

iQ basic forces in primary structure

xQ redundant basic forces

plQ plastic capacities for basic forces; superscript + for positive, - for negative

refP reference applied force vector

λ load factor

cQ basic forces at incipient collapse

ceQ basic forces at incipient collapse at elastic locations

cpQ basic forces at incipient collapse at plastic hinge locations

eB submatrix of fB for elastic basic forces

pB submatrix of fB for plastic basic forces

peQ particular solution for elastic basic forces at incipient partial collapse

xeB force influence matrix of primary structure under redundant basic forces at incipient partial collapse

Kinematics or Compatibility

U displacements at global degrees of freedom (dofs)

fU displacements at free global dofs

dU displacements at restrained global dofs = imposed support displacements

u element end displacements in global (structure) reference system

u element end displacements in local (element) reference system

v element deformations

V collection of all element deformations for structural model

e section deformations ε strain tensor

a kinematic transformation matrix for extraction of rigid body modes from element end displacements in local reference system

ra kinematic transformation matrix from global to local system

ga kinematic transformation matrix from global to basic system

Page 5

A kinematic (compatibility) matrix relating the element deformations with the displacements at all dofs of the structure

fA submatrix of A relating the element deformations with the displacements at the free dofs of structure

dA submatrix of A relating the element deformations with the displacements at the restrained dofs of structure

dV element deformations due to imposed displacements at restrained dofs

α coefficient of thermal expansion

aε normal strain at reference axis of frame element

θ rotation of element chord

κ curvature

εv strain dependent element deformations

hv element deformations due to hinge (release) deformation

εV collection of all strain dependent element deformations for structure

hV collection of all hinge dependent element deformations for structure

cA kinematic matrix relating unconstrained with constrained dofs

fU independent free dofs of structural model under constraints

fA kinematic matrix of structure for constrained free dofs

iV subset of element deformations corresponding to basic forces of primary structure

xV subset of element deformations corresponding to redundant basic forces

eV subset of element deformations at elastic locations in the structure

iA submatrix of fA corresponding to element deformations iV

xA submatrix of fA corresponding to element deformations xV

eA submatrix of fA corresponding to element deformations eV

cpA kinematic matrix relating unconstrained dofs with single independent dof of the collapse mechanism

fpA kinematic (compatibility) matrix relating the element deformations with the single free independent dof of the collapse mechanism

pWΔ plastic work increment

Page 6

Force method

F structure flexibility matrix

f element flexibility matrix

E modulus of elasticity

A cross section area

I moment of inertia

sF block diagonal matrix of all element flexibility matrices

0v initial element deformations due to element loading

0V collection of initial element deformations due to element loading

w distributed element load

L element length

ixF xiF xxF intermediate flexibility matrices of force method

VB force influence matrix of structure under initial deformations

Displacement method

K structure stiffness matrix

k element basic stiffness matrix

sK block diagonal matrix of all element stiffness matrices

0q initial or fixed element forces

0Q collection of initial or fixed end basic element basic forces

fP applied forces at free global dofs

0P initial nodal force vector

Page 7

INTRODUCTION - STRUCTURAL MODELING

Objective: systematic description of structural model; ingredients; sign convention; structural variables; loading description

CE220-Theory of Structures Equilibrium © Prof. Filip C. Filippou, 2000

Page 8

Introduction; structural model

The structural model is an idealization of the real structure. It consists of nodes in space connected byelements. While the structure is a continuum, the structural model is discrete. The analyst has makesure to represent the actual structure as accurately as warranted by the demands of the analysis, and thentranslate the results of the discrete structural model back to the actual structure for its design and detailing.

In this course we limit ourselves to one-dimensional elements, whose properties depend on a singlevariable x. Moreover, we limit ourselves to illustrating the key concepts with straight, prismatic 2d trussand frame elements, but state these in general form, so as to allow the extension to any type of one-dimensional element (e.g. a circular frame element).

Limiting ourselves to straight, prismatic 2d elements means that the geometry of curvilinear structuresneeds to be approximated with as many elements as the accuracy of the results demands. We showtwo approximation examples below: one is a non-prismatic girder and the other a circular arch.

structure structural model

We distinguish one-dimensional, two-dimensional and three-dimensional elements in a structural model. The dimension refers to the number of variables for describing the element force-deformation properties.Thus, a truss element belongs in the one-dimensional element category, a plate element in the twodimensional element category, and a solid or brick element in the three-dimensional element category.Truss elements used in a 3d (three-dimensional) structural model are called 3d truss elements.

Another example of the approximation involved in the idealization of the actual structure are the structuraljoints: members of frame structures intersect at joints of finite size. In a structural model the nodes arepoints in space. It is not entirely clear how to represent the joint region with one-dimensional elements.We will mention this point in the latter part of the course.


Page 9

Structural model: Geometric description, degrees of freedom

X

Yglobal coordinate system

The geometry of the structural model is described by the location of the nodes in an orthogonal right-handed coordinate system. We call this the global or structural coordinate systemand identify its axes by upper case letters. If the global X and Y axis are in the plane of the paper or board, then the Z-axis points toward the viewer.

In a discrete structural model the response of the model is completely described by structural variablesassociated with the nodes. These are of two kinds: generalized forces and generalized displacements.We use the word "generalized" to indicate forces, in the conventional sense, and moments. "Generalized"displacements refers to translations and rotations of a node.

Since a node is a point in space, we can decompose its translation into 3 components along the global X,Y and Z-axes; we can also decompose the rotation into 3 components, i.e. a rotation about the X-axis, arotation about the Y-axis, and a rotation about the Z-axis. Note, however, that the decomposition of therotation into 3 components is only possible for very small (actually infinitesimal) rotations. In this coursewe limit ourselves to such very small rotations, but, moreover, plan to study mostly planar structures forwhich only the rotation about the Z-axis is relevant. With the right-handed coordinate system we usethe right hand grip rule to define positive rotations: with the Z-axis pointing toward the viewer in the abovefigure a counter clockwise rotation (CCW) is positive, and a clockwise (CW) rotation is negative.This is also the consistent convention for moments: a moment acting CCW is positive, and a momentacting CW is negative. Each node of the structural model thus experiences, in general, a translation and a rotation: theseare described by 3 components each. Each component is called a degree of freedom (dof) of the node.By decomposing the translation and rotation into components relative to the global coordinate system, weensure that the degrees of freedom (dofs) are independent.Thus, each node of a 3d structural model has 6 degrees of freedom. In a 2d or planar structural model,there are only two translation components and one rotation in the plane for each node. Thus, each nodehas only 3 degrees of freedom in a planar structural model.

We represent each node with a small black square in this course. This allows us to illustrate its rotation, which a point does not allow us to do. We depict translation dofs with a single arrowhead and rotation dofs with a double arrowhead. The same is true for forces and moments, respectively. In the plane the rotation dof is depicted the same way as the moment, as shown in the figure on the right.

Node dofs may be restrained by devices. We depict such devices with special symbols. Nodes with at least one dof restrained are known as the supports of the structural model. In the following figure of typical node restraint (support) symbols the restrained dofs are shown in gray, and the free dofs are shown in black; to each restrained dof there corresponds a restraining force, which is commonly known as support reaction.

A connectivity array is also required for the elements of themodel indicating which nodes each element connects.


Page 10

Structural model: Partition of global degrees of freedom

We collect the displacement values of all dofs of the structural model in a vector U. Vectors areunderstood in this course in the general sense as mathematical objects endowed with the operationsof addition and multiplication by a scalar and not as geometric objects. Vectors are shown in bold, italictypeface in this course.

We partition the degrees of freedom (dofs) of the structural model into two groups: the free dofs(subscript f), and the restrained dofs (subscript d). We reorder the entries of the displacement vector U, so that the free dofs appear in sequence as the first partition of the vector followed by the restrained dofs.In the structural model we achieve this by numbering the free dofs in sequence first followed by therestrained dofs. We show this in the following example.

Structural model with global coordinate systemand node numbering in arbitrary order

X


1

2

3

4

5

1

2

3

4

5

6

7

8

9

10

11

12

1314

15

Number degrees of freedom in sequence (automatic)1. Start with free dofs and proceed in node order

with translation in X, then Y, then Z (3d), thenrotation about X (3d), Y (3d) and Z

2. Number restrained dofs in the same fashion

is the free dof displacement vector

f

d

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠

UU

U

dof numbering

we use the abbreviated notation nf for the number of free dofs and nr for the number of restrained dofs;the total number of dofs then, nt, is equal to the sum of nf and nr, i.e. nt = nf + nr

with the above numbering the displacement vector is partitioned according to

fU

dU is the restrained dof displacement vector (note that restrained displacements are not necessarily zero;instead restrained displacements are of specific value at the start of the analysis with zero a special case; given support settlements thus fall into this category


Page 11

Structural models are subjected to generalized forces at the nodes. We collect the values of the appliedgeneralized forces in a vector P. It is logical to use the same numbering for the generalized forces as forthe generalized displacements, so that we can easily determine the external work by the inner productof the two vectors. The applied force vector is thus partitioned into the vector of the applied forces at thefree dofs of the structural model, and the vector of the applied forces at the restrained dofs. The latter arecommonly known as support reactions. The following example illustrates the applied force vector for aparticular load case. The applied force numbering is the same as the dof numbering on the preceding page.

Structural model: Generalized forces

10

20

15

200 1

2

3

4

5

6

7

8

9

10

11

12

1314

15

f

100002000

152000

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟=⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

P

is the vector of the applied forces at the free dofs

f

d

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠

PP

Pthe applied force vector is partitioned according to

fP

is the vector of the applied forces at the restrained dofs, commonly known as support reactions

f⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠

PP

Ror

R


Page 12

Structural model: Loading

Loading1. applied forces or moments at nodes grouped in generalized force vector Pf

2. imposed translations or rotations at nodes grouped in support displacement vector Ud

3. element loadinga. applied forces or moments along element axisb. initial deformations (thermal, shrinkage, creep)

element loading

this model has fewer nodes and elementsand thus a smaller number of dofs and equations;it is thus suitable for hand solution, but requirestreatment of concentrated forces as element loading

nodal loading

Distinction between nodal and element forces is a question of discretization

suitable for computer solution if element librarydoes not include concentrated element forces(more equations and dofs than solution to the right)

In this course we will consider loading type 1 and 2. For loading type 3 we limit ourselves to uniformlydistributed transverse forces and uniform initial deformations over the entire element span

We can avoid element loading by approximation

Actual model

requires element with distributed element loading

Approximation

no element loading required,but more computational effort


Page 13

EQUILIBRIUM (STATICS)

Objective: static variables for structure, element and section and their interrelationship through free body equilibrium equations; static matrix of structural model and its properties; degree of redundancy; basic force influence matrix; solution of equilibrium equations


Page 14

Because we deal exclusively with two node elements in this course, i.e. elements that connect to onlytwo nodes, the fictitious cuts separate each element from the structural model at its two ends.The cut reveals element force variables, i.e. a force and a moment at each cut. We decompose theseinto components relative to the global coordinate system and collect them in a generalized element forcevector p. We use the convention in this course that lower case letters refer to element variablesand upper case letters to structural variables. We identify the end points of each element with the lettersi and j, with i referring to the end point at the node with lower number and j to that with higher number.The end points of the element undergo generalized displacements, i.e. a translation and rotation at eachend. We decompose these into components relative to the global coordinate system and collect themin a generalized element displacement vector u. We number the components of the element force anddisplacement vector in sequence starting with the translation (or force) components in X, Y and Z at node i,continuing with the rotation (or moment) components about X, Y and Z at node i and then proceeding inthe same fashion with the translation (or force), and then rotation (or moment) components at node j. Fora two node element we end up with 6 generalized displacement (or force) components in a 2d model, and 12 components in a 3d model.

Structural model: Free bodies and element variablesGeneralized displacements and generalized forces are related by Newton's first law which states thatan object at rest remains at rest if it is acted upon by a set of balanced forces (the case of uniformvelocity is not of interest in this course). To formulate this law we isolate first the nodes of the structuralmodel by fictitious cuts into free bodies. These free bodies also isolate the elements of the model andreveal another important set of force and displacement variables, those belonging to each element.

node and element free bodies; node numbering as before; element numbering in arbitrary order denoted with lower case letters

1

2

3

4

5

a

b c

d

P

( )cp

a

b c

d

( )bp

element b

i

j

1

2

3

4

56 1

2

( ) 3

4

5

6

b

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟=⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

uuu

uuuu

1

2

( ) 3

4

5

6

b

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟=⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

ppp

pppp

generalized displacement andforce vector for 2d element b;the little white square are thereto indicate that the elementfree body extends from nodeto node, i.e. the full element length


Page 15

Structural model: Node equilibrium equationsWe are now in a position to apply Newton's first law to the node free bodies of the structural model.Instead of writing these equations in vectorial form, we write them in component form, i.e. we write oneequation in the direction of each dof. These are the node equilibrium equations.

X


1

2

3

4

5

Structural model

a

b c

d

1

2

3

4

5

6

7

8

9

10

11

12

1314

15

Node free bodies

Write equilibrium equations at free dofs in terms of the applied forces P and the element forces p(b)(a) 141 (b)(a) 22 5 (b)(a)3 3 (c)6 1(b)4 4 (c)

2(b)55 (c)

6 3(b)6

7

8

9

10

000

0000 00 00 00 0

⎛ ⎞⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟− − −⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎜ ⎟⎝ ⎠ ⎝ ⎠

ppPpP p

P pp pP ppP

pP p

pPPPP

(d)(c) 14 (d)(c) 25 (d)(c) 36 (d)

6

00

00

00

00

00

000000

0

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎛ ⎞⎜ ⎟

⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ − =⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

pp

pp

pp

p

which we can write in compact form as f fr− = 0P P

where Pfr is the resisting force vector at the free dofsof the structural model; we see that the resisting forcevector for all dofs is the sum of the element forcesand we write symbolically

( )r

el

el=∑P p with the understanding that each

element force term is assigned tothe appropriate dof equilibrium equation

The equilibrium equations for the restrained dofs result insimilar fashion dr− = 0R P

We can combine these two sets of equations in the expression of Newton's first law for the structural model

f fr

dr

⎛ ⎞ ⎛ ⎞ ⎛ ⎞− =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

00

P PR P

or, shorter r− = 0P P

if the applied and resisting forces are not in equilibrium, then we are dealing with Newton's second law which states that r

t− = MP P U

tUwhere is the dof acceleration vector and M is the mass matrix of the structural model (see CE225)


Page 16

Structural model: Element equilibrium equationsAfter writing the equilibrium equations for the node free bodies we investigate the element free bodies, whichare also separated by the fictitious cuts from the structural model. We note that there are again as manyequilibrium equations as dofs for the element free body: 3 for an element in a planar structural model, and 6 foran element in a 3d model. We write these equations for an element with general orientation in a planar modelunder the assumption that there are no loads acting along the element span. The presence of element loading complicates the equations without any benefit to the argument of comparing the number of available equationsto the number of force unknowns of the problem.

1 4

2 5

3 6 5 4

000X Y

+ =

+ =

+ + Δ − Δ =

p p

p p

p p p p

i

j

1

2

3

4

56

j iX X XΔ = −

j iY Y YΔ = −

the element equilibriumequations are

In a structural problem the applied forces at the free dofs of the structural model are given. Similarly, thedisplacements of the restrained dofs are also given. It is our task to determine the displacements at thefree dofs and the forces at the restrained dofs (support reactions). We write this in compact form

Given: and

Determine: and

fP dU

fU R

Structural Analysis Problem

Structural model: Degree of redundancy or static indeterminacySince the support reactions are unknown at the start of the analysis, we cannot make any use of the equilibrium equations at the restrained dofs of the structural model. We therefore focus our attention on the equilibrium equations at the free dofs only. There are nf available equations in 6 x ne unknowns where ne is the number of elements. However, the element equilibrium provides another 3 equations for each element, i.e. a total of 3 x ne equations without increasing the number of unknowns. Thus, in summary we have nf + 3 x ne equations in 6 x ne unknowns. The difference between number of force (or static) unknowns of the problem and number of available equations is known as the degree of redundancy or degree of static indeterminacy of the structural model. We use the shorthand NOS to denote the degree of redundancy of the model.

NOS = (3 x ne) - nf NOS = (6 x ne) - (nf + 3 x ne)

Note: instead of adding 3 x ne element equilibrium equations to the nf node equlibrium equations it is much more efficient to use the element equilibrium equations to express the element forces in terms of only three independent or basic element forces. This is what the boxed equation reflects: there are only 3 x ne unknown element forces and nf available node equilibrium equations. This assumes that none of the basic element forces are set to zero through special devices (releases). We will address this issue later on. Note: nothing changes to the above discussion by adding the nr equilibrium equations at the restrained dofs since these equations involve an additional nr unknown support reactions. Thus, it is convenient to leave the support reactions and the corresponding equilibrium equations out of consideration.


Page 17

Element variables: global, local and basic reference system

element coordinate system (e.g. for b)

x

y

i

jb

Definition of element specific or local coordinate system1. Local x-axis points from lower to higher numbered

node (we denote the former node i and latter node j)2. In 2d local y-axis lies in the plane and is normal to x;

in 3d local y-axis is specified by user (e.g. alongprincipal axis of cross section)

3. Local z-axis is normal to the plane formed by x-y

Instead of adding the element equations of equilibrium to the nodal equations of equilibrium it is more efficient and insightful to use the number of element equilibrium equations to reduce the number of unknown element forces*. For a 2-node, 2d element we have 6 element forces and 3 equilibrium equations. Thus, we have just 3 independent or basic element forces. We can use the element equilibrium equations to express the other element forces in terms of the basic forces. We denote the basic element forces by q.

*Note that adding the element equilibrium equations to the node equilibrium equations results in a very large system of equations with many 0 terms, since each element equilibrium equation involves only the forces of the particular element.In working out the relation between dependent and independent element forces it is convenient to workwith an element specific right handed coordinate system. This system is shown in the following figure. Itis called local or element coordinate system, and we use lower case letters to denote its axes.

The force vectors at the end points of the element can be expressed relative to the global or to the localcoordinate system. The former decomposition is shown on the left and the latter on the right of the following figure. We introduce a new symbol for denoting the components of the element force vector inthe local coordinate system.

In the local coordinate system the element equilibrium equations in the absence of element loading become

i

j

2 2L X Y= Δ + Δ1

2

3

456end forces in local system

i

j

1

2

3

4

56

j iX X XΔ = −

j iY Y YΔ = −

end forces in global system

element force vector in local coordinates

1 4

2 5

3 6 5

000L

+ =

+ =

+ + =

p p

p p

p p p

1

2

3

4

5

6

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟=⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

ppp

pppp


Page 18

There are some obvious choices for the basic element forces (note that we cannot pick any three!)

Another choice is this group

One common choice is this group

In this course we pick the second choice for the basic forces. We use then the element equilibrium equations to express theother three element forces in terms of the basic.

1 4

3 65

3 62 5

L

L

= −

+= −

+= − =

p p

p pp

p pp p

Because of their importance we introduce a new symbol for the basic element forces: q

with b the force transformation matrix from the basic to the local coordinate system. We call q1 the longitudinal basic force, q2 and q3 the flexural basic forces of the element. q2 acts at node i and q3 at node j

Element variables: basic element forces

We restate the element equilibrium equations in the absence of element loading

1 4

2 5

3 6 5

000L

+ =

+ =

+ + =

p p

p p

p p p

1 2 3p p p

4 3 6p p p

4 3 6p p p 1 2 3q q q

with this definition we have 4 1

3 2

6 3

1 1

2 35

2 32

L

L

=

=

=

= −

+= −

+=

p q

p q

p q

p q

q qp

q qp

or, in compact form

1

21

32

43

5

6

1 0 01 10

0 1 01 0 0

1 10

0 0 1

L L

L L

−⎡ ⎤⎢ ⎥⎛ ⎞⎢ ⎥⎜ ⎟⎢ ⎥⎜ ⎟ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟= = =⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎝ ⎠⎢ ⎥⎜ ⎟ − −⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎢ ⎥⎣ ⎦

b

pp

qp

p q qp

qpp

Summary: the relation between local components of the element force vector and basic element forces is

= bp q with

1 0 01 10

0 1 01 0 0

1 10

0 0 1

L L

L L

−⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎢ ⎥− −⎢ ⎥⎢ ⎥⎣ ⎦

b


Page 19

Force component transformation from local to global coordinate system

After having established a relation between the local and basic element forces, we need to relate the global element force components to the local. This is rather straightforward, if we note that the two coordinate systems are related by a rotation in the plane of the model. We use the direction cosines of the element orientation in the undeformed (original) position in lieu of trigonometric functions of the orientation angle.

X

Y

xy L

i

j

XΔ

YΔ

α

4p

5p4p

5p

α

α

4p

5p

5p

4p

The transformation for the forces at end i is identical. Note that the moments do not transform during the rotation. Thus, we can write for the relation between global components and local components of the element end forces

or, compactly

6 element end forces in global reference

6 element end forces in local reference

3 basic element end forces and3 dependent end forces

ip

jp

(a) ip

jp

(b)

1q

2q

3q

(c)

L

2 3

L+q q

2 3

L+q q

1q

global

local

basic

4 4 5 4 5

5 4 5 4 5

cos sin

sin cos

X YL LY XL L

α α

α α

Δ Δ= − = −

Δ Δ= + = +

p p p p p

p p p p p

4 4

5 5

X YL LY XL L

Δ Δ⎡ ⎤−⎢ ⎥⎛ ⎞ ⎛ ⎞⎢ ⎥=⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟Δ Δ⎢ ⎥⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

p pp p

1 1

2 2

3 3r

4 4

5 5

6 6

0 0 0 0

0 0 0 0

0 0 1 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0 0 1

X YL L

pY XpL LppX YpL L

Y X pL L

Δ Δ⎡ ⎤−⎢ ⎥⎢ ⎥⎛ ⎞ ⎛ ⎞Δ Δ⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟= = =⎢ ⎥Δ Δ⎜ ⎟ ⎜ ⎟⎢ ⎥−⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥Δ Δ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

b

ppp

p pppp


Page 20

combining with gives

with

Thus, finally the relation between element force components in the global coordinate system and the basic forces is

r= bp p

Element equilibrium: from basic forces to global element force components

= bp q r g= =b b bp q q

2 2

2 2

g r

2 2

2 2

0 0 0 0 1 0 01 10 0 0 0 0

0 0 1 0 0 0 0 1 0 0 1 01 0 00 0 0 0

1 100 0 0 0

0 0 10 0 0 0 0 1 0 0 1

X Y YX YL L LL LY X XY XLL L L L L L

X Y X Y YL L L L LY X Y X XL LL L L L L

Δ Δ Δ⎡Δ Δ⎡ ⎤ − − −⎢−⎢ ⎥ −⎡ ⎤⎢ ⎥ ⎢ ⎥ Δ Δ ΔΔ Δ⎢ ⎥ ⎢ ⎥ −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= = =⎢ ⎥ ⎢ ⎥Δ Δ Δ Δ Δ⎢ ⎥− ⎢ ⎥⎢ ⎥ ⎢ ⎥− −⎢ ⎥Δ Δ ⎢ ⎥ Δ Δ Δ⎢ ⎥ − −⎢ ⎥⎢ ⎥ ⎣ ⎦⎢ ⎥⎣ ⎦ ⎣

b b b

⎤⎥

⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎦

2 2

12 2

21

32 g

432 2

5

62 2

0 1 0

0 0 1

X Y YL L LY X XL L L

qX Y Y

qL L LY X XL L L

Δ Δ Δ⎡ ⎤− − −⎢ ⎥⎢ ⎥

⎛ ⎞ ⎢ ⎥Δ Δ Δ−⎜ ⎟ ⎢ ⎥

⎜ ⎟ ⎢ ⎥ ⎛ ⎞⎜ ⎟ ⎢ ⎥ ⎜ ⎟⎜ ⎟= = =⎢ ⎥ ⎜ ⎟Δ Δ Δ⎜ ⎟ ⎢ ⎥ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎢ ⎥⎜ ⎟ ⎢ ⎥⎜ ⎟ Δ Δ Δ⎢ ⎥⎝ ⎠ − −⎢ ⎥

⎢ ⎥⎢ ⎥⎣ ⎦

b

pp

qp

p qppp

with bg the force transformation matrix from the basic to the globall coordinate system; this matrix can also be regarded as the equilibrium or static matrix of the element.


Page 21

The basic forces of a 2d frame element are:the axial force q1 and the end moments q2 and q3

The end element forces in the global coordinate system can be expressed in terms of these

1q

1q

1XLΔ q

1YLΔ q

1XLΔ

− q

1YLΔ

− q

3q

3

Lq

3

Lq

32

YLΔ

− q

32

XLΔ q

32

XLΔ

− q

32

YLΔ q

3q

2q

2

Lq

2

Lq

22

YLΔ

− q

22

XLΔ q

22

XLΔ

− q

22

YLΔ q

2q

Element equilibrium: from basic forces to global element force components

The figures on the left show each basic force in turnalong with the dependent element force componentsin the local reference system satisfying equilibrium.

The figures on the right show the element forcecomponents in the global reference system for each basic force in turn; each group of force components corresponds to a column of the static matrix bg and satisfies element equilibrium


Page 22

end i

1q

1q2q

Special cases of basic element forces for 2d frame elements

We deal at first only with devices that release one or more of the element basic forces. We will returnlater to deal with more general internal force releases.

longitudinal basic force release

flexural basic force release

truss element: both flexural basicforces released; only one basic force

2d frame element; one flexural basicforce released; two basic forces only

2d frame element; one flexural basicforce released; two basic forces only 1q

2q

2d frame element; longitudinall basicforce released; two basic forces only

1q 2q

end j


Page 23

Relation between element and section static variablesIt is very important to distinguish between element forces and internal forces. The latter arise at a fictitious cut through the element at a distance x from end i, as shown in the following figure. The cut separates the element into two parts, each of which needs to satisfy the equations of equilibrium. For the 3 additional equilibrium equations there are 3 additional unknowns, the internal element forces at the cut at a distancex from end i. These internal forces are the resultants of the normal and shear stresses acting at the cut andare, therefore, known as stress resultants. They are also known as section forces, since they are the static variables of an infinitesimal slice of the frame element.

We collect the section forces in a vector s(x), which for the 2d element in the figure includes the normal force N(x), the shear force V(x) and the bending moment M(x). Note that the section or internal forces N, V, and M (there is another shear force, another bending moment, and a torsional moment in 3d) are continuous functions of x and not discrete variables as the basic element forces. As continous functions of x the internal forces can be differentiated with respect to x. Moreover, the sign convention for the internal forces needs to follow the sign convention for continuous functions: the forces at the face of the cut (section) closest to end i follow the axes orientation of the local coordinate system, while those at the opposite section act in the opposite direction. By contrast, the basic element forces are discrete static variables and follow the sign convention adopted for all structural and element variables.

2d frame element with moment release at xi j

1q2q 2

L xx−q

x L-x

We can now introduce releases of the internal forces at any section inside the element. These result in one constaint equation between the basic element forces for each release. Each release, therefore, reduces the number of basic element forces by one. Three common cases follow

2d frame element with shear force releasei j

1q2q

i j

1q 2q

2d frame element with normal force release

2q

( )M x

2q

2 3

L+q q

3q

3q

2 3

L+q q

2 3

L+q q 2 3

L+q q

( )V x

frame element

end jend i

x

y

+ 1q1q

( )N x

1qx

( )M x

( )V x

( )N x slice of length Δx


Page 24

Example 1: Equilibrium equations for gable frame, properties of static matrix After establishing the relation between basic element forces and element force components in the global coordinate system we return to the equilibrium equations of the node free bodies in the direction of the free dofs. Setting up these equilibrium equations in the undeformed configuration establishes a linear relation between applied forces and basic element forces. The coefficients of the basic element force can be collected in the static matrix of the structural model. We use example 1 to illustrate. The followingprovides a brief overview of process. More details are available in the example description.

X


1

2

3

4

5

Structural model

a

b c

d

1

2

3

4

5

6

7

8

9

10

11

12

1314

15

Node free bodies(b)(a)141(b)(a)22 5(b)(a)3 3 (c)6(b) 14 4 (c)(b) 255 (c)(b)6 36

7 4

8

9

10

000

0000 00 00 00 0

⎛ ⎞⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ = + +⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

ppPpP p

P pppP ppP p

P pp

P pPPP

(d)(c) 1

(d)(c) 25 (d)(c) 36 (d)

6

000000

0

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟+⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

p

ppppp

(a) (a) (a)g.41 g.42 g.43

1(a) (a) (a)

2 g.51 g.52 g.53

(a) (a) (a)3g.61 g.62 g.63

(a)41

52

63

7

8

9

10

0 0 00 0 00 0 00 0 00 0 00 0 00 0 0

⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ = ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠

⎝ ⎠

b b bPP b b bP b b bP

qP

qP

qPPPP

(b) (b) (b)g.11 g.12 g.13

(b) (b) (b)g.21 g.22 g.23

(b) (b) (b)g.31 g.32 g.33

(b) (b) (b)g.41 g.42 g.43

1(b) (b) (b)

2g.51 g.52 g.53

(b) (b) (b)g.61 g.62 g.63

0 0 00 0 00 0 00 0 0

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟+ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

b b b

b b b

b b b

b b b qqb b bqb b b

(c) (c) (c)g.11 g.12 g.13

(b) (c) (c) (c)g.21 g.22 g.23

(c) (c) (c)g.31 g.32 g.33

3 (c) (c) (c)g.41 g.42 g.43

(c) (c) (c)g.51 g.52 g.53

(c) (c) (c)g.61 g.62 g.63

0 0 00 0 00 0 0

0 0 0

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎛ ⎞

⎜ ⎟ ⎜ ⎟+⎜ ⎟ ⎜⎜ ⎟ ⎜⎝ ⎠ ⎜

⎜⎜⎜⎜⎜⎜⎝ ⎠

b b b

b b b

b b b

b b b

b b b

b b b

(c) (d)1 1

2 2(d) (d) (d)

g.11 g.12 g.133 3

(d) (d) (d)g.21 g.22 g.23

(d) (d) (d)g.31 g.32 g.33

(d) (d) (d)g.61 g.62 g.63

0 0 00 0 00 0 00 0 00 0 00 0 0

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟+⎜ ⎟ ⎜ ⎟⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎟ ⎜ ⎟⎟ ⎜ ⎟⎟ ⎜ ⎟⎟ ⎜ ⎟⎟ ⎜ ⎟⎟ ⎜ ⎟⎟ ⎝ ⎠

q qq q

b b bq q

b b b

b b b

b b b

Now, collect all basic element forces into a single vector Q for compact notation. We have:

(a)1

12

23

3(b)1 4

2 5

3 6(c)

71

82

93

10(d)1 11

2 12

3

⎛ ⎞⎛ ⎞⎜ ⎟ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ = =⎜ ⎟ ⎜ ⎟⎜ ⎟⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

Q

qQ

qQ

qQ

q Qq Qq Q

QqQqQqQ

q Qq Qq


Page 25

The equilibrium equations at the free dofs then become

P1Q2 Q3+

12Q4 0.8⋅−

Q5 Q6+

100.6⋅−

P2 Q1 0.6 Q4⋅−Q5 Q6+

100.8⋅+

P3 Q3 Q5+

P4 0.8 Q4⋅Q5 Q6+

100.6⋅+ 0.8 Q7⋅−

Q8 Q9+

100.6⋅+

P5 0.6 Q4⋅Q5 Q6+

100.8⋅− 0.6 Q7⋅+

Q8 Q9+

100.8⋅+

P6 Q6 Q8+

P7 0.8 Q7⋅Q8 Q9+

100.6⋅−

Q11 Q12+

12+

P8 0.6− Q7⋅Q8 Q9+

100.8⋅− Q10+

P9 Q9 Q11+

P10 Q12

Compare these with the equations set up one at a time in Example 1.

Bf

0

1

0

0

0

0

0

0

0

0

112

0

0

0

0

0

0

0

0

0

112

0

1

0

0

0

0

0

0

0

0.8−

0.6−

0

0.8

0.6

0

0

0

0

0

0.610

−

0.810

1

0.610

0.810

−

0

0

0

0

0

0.610

−

0.810

0

0.610

0.810

−

1

0

0

0

0

0

0

0

0.8−

0.6

0

0.8

0.6−

0

0

0

0

0

0.610

0.810

1

0.610

−

0.810

−

0

0

0

0

0

0.610

0.810

0

0.610

−

0.810

−

1

0

0

0

0

0

0

0

0

1

0

0

0

0

0

0

0

0

112

0

1

0

0

0

0

0

0

0

112

0

0

1

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

or, in compact form where is the static matrix of the structural model for the free dofsf f= BP Q fB

(a) (a) (a) (b) (b) (b)g.41 g.42 g.43 g.11 g.12 g.13

(a) (a) (a) (b) (b) (b)g.51 g.52 g.53 g.21 g.22 g.23

1(a) (a) (a) (b) (

2 g.61 g.62 g.63 g.31 g.32

3

4

5

6

7

8

9

10

0 0 0 0 0 0

0 0 0 0 0 0⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ =⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

b b b b b b

b b b b b bPP b b b b bPPPPPPPP

b) (b)g.33

(b) (b) (b) (c) (c) (c)g.41 g.42 g.43 g.11 g.12 g.13

(b) (b) (b) (c) (c) (c)g.51 g.52 g.53 g.21 g.22 g.23

(b) (b) (b) (c) (c) (c)g.61 g.62 g.63 g.31 g.32 g.33

(c) (c)g.41 g.42

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

b

b b b b b b

b b b b b b

b b b b b b

b b (c) (d) (d) (d)g.43 g.11 g.12 g.13

(c) (c) (c) (d) (d) (d)g.51 g.52 g.53 g.21 g.22 g.23

(c) (c) (c) (d) (d) (d)g.61 g.62 g.63 g.31 g.32 g.33

(d) (d) (d)g.61 g.62 g.63

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

⎛ ⎞⎜ ⎟⎜ ⎟⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝ ⎠

b b b b

b b b b b b

b b b b b b

b b b

1

2

3

4

5

6

7

8

9

10

11

12

⎛ ⎞⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎟⎜ ⎟⎜ ⎟⎟⎝ ⎠⎟⎟

QQQQQQQQQQQQ


Page 26

Static matrix- Stability and degree of redundancy (static indeterminacy) of structure

The equilibrium equations at the free dofs of the structural model including the effect of element loading take the following form

The degree of redundancy (static indeterminacy) of the structural model is given by the difference between the number of basic element forces and the number of linearly independent equations of equilibrium at the free dofs. For a stable structural model the degree of redundancy or static indeterminacy NOS is the difference between the number of columns and the number of rows of matrix Bf.

where Pfw are the equivalent nodal forces due to element loading

Degree of redundancy (static indeterminacy) NOS

STRUCTURE A

NOS = 18-12=6

STRUCTURE B

NOS = 15-11=4

STRUCTURE C

NOS = 15-11=4

NOS = 10-7=3

a b

Examples of statically indeterminate structures

NOS = 10 - 8 = 2

f f fw= +BP Q P

The above equation can be rewritten with the known loading terms on the left and the unknown basic element forces on the right. f fw f− = BP P Q

The solution of this linear system of equations depends on the properties of the static matrix Bf. In a stable structural model the number of linearly independent rows of static matrix Bf should be equal to to the number of free dofs. The number of linearly independent rows of a matrix is known as its rank. Thus, for a stable structural model the rank of the static matrix should be equal to the number of free dofs.


Page 27

CE220 - Theory of Structures Example 1 - Equilibrium for Gable Frame © Prof. Filip C. Filippou, 2000

Example 1 - Equilibrium Equations for Gable FrameObjectives: (a) set up the equilibrium equations for a frame structure with general orientation members(b) assess stability and degree of static indeterminacy of structural model(c) automate process of setting up equilibrium equations

X


1

2

3

4

5

Structural model

a

b c

d

1

2

3

4

5

6

7

8

9

10

11

12

1314

15

Node free bodies

8 8

12

6

We isolate node free bodies and number the equations of equilibrium (or, degrees of freedom dofs,according to Newton's second law). We number the free dofs first starting from the lowest numbered node,and number the support or restrained dof's after completing the numbering of the free dof's in the same way.

The equilibrium of the element free bodies is satisfied by selecting 3 basic forces as independent or basicforces and then expressing the element end forces in the global reference system in terms of them, so as tosatisfy the 3 equilibrium equations of the element. We obtain the relation

If we write the equilibrium in the undeformed configuration (first order or linear equilibrium)then the array bg depends only on the geometry of the element (length and direction cosines)p bg q⋅=

the element equilibrium transformation matrix to global coordinates bg has the form

ip

jp

(a) ip

jp

(b)

1q

2q

3q

(c)

L

2 3

L+q q

2 3

L+q q

1q

2 2

1

2 22

13

2 g4

2 2 35

62 2

0 1 0

0 0 1

X Y YL L LY X X

L L L

X Y YL L LY X X

L L L

pp

qp

p q b qp

qpp

⎡ ⎤− − −⎢ ⎥⎢ ⎥⎛ ⎞ ⎢ ⎥−⎜ ⎟ ⎢ ⎥⎜ ⎟ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥= = =⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎝ ⎠⎜ ⎟ ⎢ ⎥⎜ ⎟⎜ ⎟ ⎢ ⎥⎝ ⎠ − −⎢ ⎥⎢ ⎥⎣ ⎦

Δ Δ Δ

Δ Δ Δ

Δ Δ Δ

Δ Δ Δ

where ΔX/L and ΔY/L are the direction cosinesof the element and L is the element length

Page 28


When writing the equilibrium equations we proceed equation by equation and limit ourselves to theequations not involving support reactions, i.e. the equilibrium equations for the free dofs of the structuralmodel. We note in the following figure that the basic forces Q are numbered in sequence starting fromelement a through element d. The node i and node j correspondence for the elements of the model isshown next to the figure.

1

2

3

5

6

7

8

9

10

11

12

13 14

15

12Q

11Q

10Q

7Q

8Q9Q

4Q

5Q6Q

1Q

2Q

3Q

4 element node i node j

a 1 2

b 2 3

c 3 4

d 4 5

P1Q2 Q3+

12Q4 0.8⋅−

Q5 Q6+

100.6⋅−=

P2 Q1 0.6 Q4⋅−Q5 Q6+

100.8⋅+=

P3 Q3 Q5+=

P4 0.8 Q4⋅Q5 Q6+

100.6⋅+ 0.8 Q7⋅−

Q8 Q9+

100.6⋅+=

we can write this system ofequations in compact form Pf Bf Q⋅=P5 0.6 Q4⋅

Q5 Q6+

100.8⋅− 0.6 Q7⋅+

Q8 Q9+

100.8⋅+=

P6 Q6 Q8+=

P7 0.8 Q7⋅Q8 Q9+

100.6⋅−

Q11 Q12+

12+=

P8 0.6− Q7⋅Q8 Q9+

100.8⋅− Q10+=

P9 Q9 Q11+=

P10 Q12=

Page 29


Observations:

1. the basic longitudinal force is always assumed at node j2. we can use the appropriate terms of the element equilibrium matrix bg; the signs will be correct as long aswe remember that ΔX is measured by subtracting the X-coordinate of node i from that of node j and thesame is true for ΔY. Thus, for element c in the gable frame of the example, ΔY is negative because node j islocated lower than node i.

Thus, the bg matrix for element c is

bgc

810

−

6−10

−

0

810

6−10

0

6−

102−

8

102

1

6−

102

8

102−

0

6−

102−

8

102

0

6−

102

8

102−

1

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= bgc

0.8−

0.6

0

0.8

0.6−

0

0.06

0.08

1

0.06−

0.08−

0

0.06

0.08

0

0.06−

0.08−

1

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

=

The static matrix Bf for the freedofs of this structural model is

Bf

0

1

0

0

0

0

0

0

0

0

112

0

0

0

0

0

0

0

0

0

112

0

1

0

0

0

0

0

0

0

0.8−

0.6−

0

0.8

0.6

0

0

0

0

0

0.610

−

0.810

1

0.610

0.810

−

0

0

0

0

0

0.610

−

0.810

0

0.610

0.810

−

1

0

0

0

0

0

0

0

0.8−

0.6

0

0.8

0.6−

0

0

0

0

0

0.610

0.810

1

0.610

−

0.810

−

0

0

0

0

0

0.610

0.810

0

0.610

−

0.810

−

1

0

0

0

0

0

0

0

0

1

0

0

0

0

0

0

0

0

112

0

1

0

0

0

0

0

0

0

112

0

0

1

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

In the structure equilibrium matrix Bf we see the contribution of element c in columns 7-9. The terms of theequilibrium matrix bg for element c appear in rows 4 through 9. Before discussing the automatic assemblyof the equilibrium equations let us write down the equilibrium equations for the support reactions.

Page 30


For the restrained dofs the equilibrium equations become

1

2

3

5

6

7

8

9

10

11

12

13 14

15

12Q

11Q

10Q

7Q

8Q9Q

4Q

5Q6Q

1Q

2Q

3Q

4 P11Q2 Q3+

12−=

P12 Q1−=

P13 Q2=

P14Q11 Q12+

12−=

P15 Q10−=

we can write these equations in compact form as R Bd Q⋅=

with R

P11

P12

P13

P14

P15

⎛⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎠

= and Bd

0

1−

0

0

0

112

−

0

1

0

0

112

−

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

1−

0

0

0

112

−

0

0

0

0

112

−

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

we can see that this matrix contains a lot of zeros, since only one element connects to a support. Thus,the determination of the support reaction by the operation Bd Q⋅ is not very efficient. Instead, eachsupport reaction should be evaluated from the corresponding equation.

we can combine the equilibrium equations for the free and restrained dofs in a singleset by stacking the corresponding vector and matrices on top of each other. We get

Pf

R⎛⎜⎝

⎞⎟⎠

Bf

Bd

⎛⎜⎝

⎞⎟⎠

Q⋅=

the matrix BBf

Bd

⎛⎜⎝

⎞⎟⎠

= is the static matrix of the structural model for all dofs

Matrix B can be readily assembled automatically by considering the contribution of one element afteranother. The contribution of each element goes into 3 separate columns, so that the contribution of elementa goes into columns 1 to 3, element b into columns 4 to 6, and so on. The rows of each element equilibriummatrix bg go to the rows of matrix B according to the incidence relation of the dofs of this element. Thisincidence relation expresses the relation between element dof and global dof numbers. Since the 2-node, 2dframe element has 6 dofs, the incidence (id) array for this element has 6 rows. Each row contains the globaldof number that corresponds to the element dof number for that row. We illustrate this for the example

Page 31


Incidence matrices for element dofs

1

2

3

4

5

6

7

8

9

10

11

12

1314

15

Node free bodieselement a element b element c element d

11

12

13

1

2

3

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

1

2

3

4

5

6

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

4

5

6

7

8

9

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

7

8

9

14

15

10

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

this means that the first row of the equilibrium matrix of element c goes to row 4 of the structure matrix B,the fourth row to row 7, etc. In Matlab this reassignment of rows can be easily accomplished with arrayindexing as the FEDEASLab function B_matrix.m illustrates.

B stack Bf Bd,( ):=

B

0

1

0

0

0

0

0

0

0

0

0

1−

0

0

0

0.083

0

0

0

0

0

0

0

0

0

0.083−

0

1

0

0

0.083

0

1

0

0

0

0

0

0

0

0.083−

0

0

0

0

0.8−

0.6−

0

0.8

0.6

0

0

0

0

0

0

0

0

0

0

0.06−

0.08

1

0.06

0.08−

0

0

0

0

0

0

0

0

0

0

0.06−

0.08

0

0.06

0.08−

1

0

0

0

0

0

0

0

0

0

0

0

0

0.8−

0.6

0

0.8

0.6−

0

0

0

0

0

0

0

0

0

0

0.06

0.08

1

0.06−

0.08−

0

0

0

0

0

0

0

0

0

0

0.06

0.08

0

0.06−

0.08−

1

0

0

0

0

0

0

0

0

0

0

0

0

0

1

0

0

0

0

0

0

1−

0

0

0

0

0

0

0.083

0

1

0

0

0

0

0.083−

0

0

0

0

0

0

0

0.083

0

0

1

0

0

0

0.083−

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

Even though the function B_matrix.m assembles the static matrix B for all dofs of the structural model weuse only the static matrix Bf for the free dofs of the model for discussions of stability and staticindeterminacy. This is so because the addition of the support dofs does not change stability or staticindeterminacy considerations. In the case of static indeterminacy this is easy to see: there are as manyequilibrium equations at the restrained dofs as unknown support reactions.

Page 32


Moreover, the total set of equilibrium equations in the form Pf

R⎛⎜⎝

⎞⎟⎠

Bf

Bd

⎛⎜⎝

⎞⎟⎠

Q⋅= is not very useful

since it contains the unknown support reactions on the left hand side of the equations. We always workonly with the top portion of these equations, and determine the support reactions only after havingdetermined the basic element forces Q, as subsequent examples will illustrate.

We return now to deal with the equilibrium equations at the free dofs of the structure Pf Bf Q⋅=

It is the properties of the static matrix Bf for the free dofs of the structural model that determine whether asolution exists, and whether it is unique, or whether there are multiple solutions. Note that the specificationof a loading is not necessary for the task, except for the case of an unstable structure being stable for aspecific load, which is of no interest to us. Let us look at one important property of static matrix Bf :the rank of the matrix, i.e. the number of linearly independent rows

A necessary and sufficient condition for a structural model to be stable is that the rank of Bf be equal tothe number of rows, i.e. to the number of free dofs; such a matrix is known to have full rank

Let us check the present example: rank Bf( ) 10= which is equal to the number of rows rows Bf( ) 10=

A structure is called statically determinate, if it is stable and its static matrix Bf has the same number ofrows as columns (i.e. it is a square matrix). In such case the inverse of the structure equilibrium matrixexists and is called the force influence matrix. We give it the symbol Bbar.

The equilibrium equations have a unique solution for a stable, statically determinate structure. This is animportant class of structures, since the internal stress state does not depend on material properties.

A structure is called statically indeterminate, if it is stable and its structure equilibrium matrix has morecolumns than rows (i.e. more unknowns than available equations). The difference between number ofcolumns or unknowns and number of rows or available equations is called the degree of static indeterminacyand we use the shorthand NOS for this number.

In a stable, statically indeterminate structure the equilibrium equations have multiple solutions. Thesesolutions can be expressed as the sum of a primary basic force field (called primary or particular solution),and NOS self-stress basic force fields with NOS redundant basic forces as parameters (consult Strang, 3rdedition, pp 71-77). The selection of these redundant basic forces can be done automatically with theGauss-Jordan form of the structure equilibrium matrix. We will illustrate this solution process with examples.

The structure in this example has 12 unknown basic forces and 10 equations of equilibrium, i.e. NOS = 2.

In the above system of equilibrium equations, there is one that involves only one unknown basic force, whichcan be therefore determined immediately and independently from the rest, i.e.

P10 Q12= given the value for P10 at the beginning of the analysis we can immediately calculate Q12.Quite often the value for P10 is zero, resulting in a basic force with zero value and the finalresult 0 = 0 for the corresponding equilibrium equation. If the value of P10 is not zero, wecan still calculate Q12 and account for its effect on the remaining equilibrium equations (thiswould be equation 7 for this structure). To economize on the number of equations andbasic forces, it is advisable to identify such equations and basic forces at the beginning ofthe analysis and avoid numbering the corresponding dof and basic force. We should makesure to include the effect of a non-zero value for the corresponding basic force on theremaining equilibrium equations.

Page 33


An interesting case results with the insertion of a moment release at node j of element b, i.e. by releasingthe basic force Q6. Since Q6 = 0, equation 6 now involves a single unknown

P6 Q8= Thus, for this structure we should only set up 8 equations (i.e. eliminate #6 and #10 above)and then use only 9 unknown basic forces, because Q8 and Q12 can be determinedindependently at the start of the analysis from the corresponding equilibrium equation.Note that NOS = 1 in this case, since we have 8 equilibrium equations for 9 basic forces.

8 8

12

6

1

2

3

4

5

6

7

8

9

10

11

12

1314

15

12Q

11Q

10Q

7Q

8Q9Q

4Q

5Q

1Q

2Q

3Q

Page 34

Mat

lab

scri

pt fo

r E

xam

ple

1 in

CE

220

clas

s not

es

% Automatic assembly of equilibrium equations for gable frame with FEDEASLab function B_matrix

% Clear workspace memory and initialize global variables

CleanStart

Cre

ate

mod

el

% specify node coordinates (could only specify non-zero terms)

XYZ(1,:) = [ 0 0]; % first node

XYZ(2,:) = [ 0 12]; % second node, etc

XYZ(3,:) = [ 8 18]; %

XYZ(4,:) = [ 16 12]; %

XYZ(5,:) = [ 16 0]; %

% connectivity array

CON {1} = [ 1 2];

CON {2} = [ 2 3];

CON {3} = [ 3 4];

CON {4} = [ 4 5];

% boundary conditions (1 = restrained, 0 = free) (specify only restrained dof's)

BOUN(1,:) = [1 1 1];

BOUN(5,:) = [1 1 0];

% specify element type

[ElemName{1:4}] = deal('2dFrm'); % 2d frame element

Model = Create_SimpleModel(XYZ,CON,BOUN,ElemName);

Page 35

Equ

ilibr

ium

equ

atio

ns

% set up equilibrium equations for all dof's

B = B_matrix(Model);

% extract rows corresponding to free dof's to form matrix Bf

Bf = B(1:Model.nf,:);

% display equilibrium matrix

disp(Bf)

% check rank of matrix Bf

disp(['the rank of the structure equilibrium matrix is ' num2str(rank(Bf))])

% determine NOS by difference of no of columns and rows (use size command)

ncol = size(Bf,2);

disp(['the no of columns of the structure equilibrium matrix is ' num2str(ncol)])

nrow = size(Bf,1);

disp(['the no of rows of the structure equilibrium matrix is ' num2str(nrow)])

NOS = ncol-nrow;

disp(['the degree of static indeterminacy of the model is ' num2str(NOS)])

0 0.0833 0.0833 -0.8000 -0.0600 -0.0600 0 0 0 0 0 0

1.0000 0 0 -0.6000 0.0800 0.0800 0 0 0 0 0 0

0 0 1.0000 0 1.0000 0 0 0 0 0 0 0

0 0 0 0.8000 0.0600 0.0600 -0.8000 0.0600 0.0600 0 0 0

0 0 0 0.6000 -0.0800 -0.0800 0.6000 0.0800 0.0800 0 0 0

0 0 0 0 0 1.0000 0 1.0000 0 0 0 0

0 0 0 0 0 0 0.8000 -0.0600 -0.0600 0 0.0833 0.0833

0 0 0 0 0 0 -0.6000 -0.0800 -0.0800 1.0000 0 0

0 0 0 0 0 0 0 0 1.0000 0 1.0000 0

0 0 0 0 0 0 0 0 0 0 0 1.0000

the rank of the structure equilibrium matrix is 10

the no of columns of the structure equilibrium matrix is 12

the no of rows of the structure equilibrium matrix is 10

the degree of static indeterminacy of the model is 2

Page 36

When the structural model is stable and its static matrix Bf is square, the structure is statically determinate and a unique solution for the basic forces exists for any applied loading. This solution does not depend on the element properties. We can write this solution in compact form

In a statically determinate model the basic force influence matrix is the inverse of the static matrix for the free dofs. The coefficient (i,j) of the basic force influence matrix expresses the effect of a unit force at dof j on the basic force i of the structural model.

Once the basic forces are determined in a statically determinate structure, the support reactions can be obtained one by one from the equation of equilibrium of the corresponding restrained dof. Finally, the analyst should check global equilibrium between applied forces and support reactions for the free body of the entire structural model.

Consult examples 2, 3 and 4 for applications of the equilibrium equations to statically determinate structures.

Special case for equilibrium equations - Statically determinate structures

( )1f f fw−= −BQ P P ( )f fw= −BQ P P

is the basic force influence matrix of the structural modelB


Page 37

CE 220 - Theory of Structures Ex. 2 - Truss equilibrium © Prof. Filip C. Filippou, 2000

Example 2(r) - Equilibrium equations for statically determinate 2d truss

Objectives:

(a) set up and solution of the equilibrium equations for a statically determinate truss(b) determination of force influence matrix; physical interpretation(c) determination of support reactions and check of global equilibrium

In this example we will set up the equilibrium equations at the nodes of a statically determinate truss.The truss is given in Figure 1. The global coordinate system X-Y is also shown.

88

6

12

3

4

a b

c d e

X

Y

Fig. 1: Statically determinate truss

We number the nodes in arbitrary numerical sequence and identify the elements with lower case romanletters (could have also used numbers). The element numbering is also in arbitrary sequence.

Before embarking on the equilibrium equations we note the following: a truss is a structure made upexclusively of truss elements. These are straight elements (see Remark below) with a momentrelease at each end. Thus, each element has only one basic force q1. The following figure displaysall moment releases for the truss model.

1q

1q

truss element

It is quite clear from the above figure that no moments can be applied at the nodes of the truss. With thisassumption the moment equilibrium at the nodes is satisfied automatically, i.e. 0 = 0. Strictly speaking,however, this results in an unstable node. Thus, it is better to represent the model as shown in thefollowing figure and then postulate that the applied moments at the nodes are zero (this is what makessense in computer analysis).

Page 38


in this case there is one element with a rigid (moment) connection at each node so that the stability of thenodes against rotation is ensured. If no moments act at the nodes, there are no end moments in theelements. For convenience we simplify the display of these construction details with a single circle at thenode, as shown in Fig. 1. This indicates a node without any applied or resisting moments.

Remark: The longitudinal basic force q1 in each element should not be confused with the axial force N(x),which is the internal force acting parallel to the element axis at a section located a distance xfrom end i of the element. While N(x) is equal to q1 in the absence of longitudinal element loads,this is not the case in their presence. q1 and N(x) are also not equal if the "truss element" iscurved. It should then be considered as a frame element with end moment releases.

Based on the preceding discussion, there are no moment equilibrium equations at the nodes of a trussstructure. Starting with the first node we number the equilibrium equations at the node free bodies in thefollowing order: equilibrium equations without support reactions are numbered first, in the X-direction andthen Y-direction following the node sequence (we call these equilibrium equations for the free degrees offreedom or dofs). After we finish with the equations without support reactions, we number in the samefashion the equations with support reactions (we call these equilibrium equations for restrained dofs). Thenumbering in Figure 2 results.

12

3

4

67 8

Fig. 2: Dof or equilibrium equation numbering

Page 39


Length of elements La 8:= La 8=

Lb 8:= Lb 8=

Lc 62 82+:= Lc 10=

Ld 6:= Ld 6=

Le 62 82+:= Le 10=

We now set up the equilibrium equations at the free bodies of the nodes after separating these withmental cuts from the elements. Each truss element has a single basic force. Note that we keep theloading general, i.e. we assume that forces may be present at any degree of freedom. We have:

Equilibrium equations for free dofs

12

3

4

5

67 8

1Q 2Q

4Q3Q

5Q

P1 Q1− Q2+ 0=

P2 Q4+ 0=

P3 Q2− 0.8 Q5⋅− 0=

P4 0.8 Q3⋅− 0.8 Q5⋅+ 0=

P5 0.6 Q3⋅− Q4− 0.6 Q5⋅− 0=

moving the unknown basic forces to the right hand side of the equations gives the following equations

12

3

45

67 8

1Q 2Q

4Q3Q 5Q

P1 Q1 Q2−=

P2 Q4−=

P3 Q2 0.8 Q5⋅+=

P4 0.8 Q3⋅ 0.8 Q5⋅−=

P5 0.6 Q3⋅ Q4+ 0.6 Q5⋅+=

Conclusion: To determine the signs of the element forces Q directly on the right hand side of theabove equilibrium equations we can use the element force orientation at the element side of the cut ofthe corresponding node free body.

Page 40


The coefficients of the element forces Q (note that they range in absolute value from 0 to 1, since theyrepresent direction cosines) can be collected in static or equilibrium matrix of the structure for the free dofs.

We can then write the above equations in compact form: after denoting the applied forces at the free dofswith Pf the system of equilibrium equations for the free dofs of the structure becomes Pf = Bf * Q

Pf Bf Q⋅= with Pf

P1

P2

P3

P4

P5

⎛⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎠

= Bf

1

0

0

0

0

1−

0

1

0

0

0

0

0

0.8

0.6

0

1−

0

0

1

0

0

0.8

0.8−

0.6

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:= and Q

Q1

Q2

Q3

Q4

Q5

⎛⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎠

=

Static (equilibrium) matrix and force influence matrix Equation for dof 1

Equation for dof 2

The static matrix for the free dofs is Bf

1

0

0

0

0

1−

0

1

0

0

0

0

0

0.8

0.6

0

1−

0

0

1

0

0

0.8

0.8−

0.6

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

= Equation for dof 3

Equation for dof 4

Equation for dof 5

In the static matrix for the free dofs of a truss structure rows correspond to equilibriumequations and columns to the basic force of one element

Note that the static matrix for the free dofs Bf only depends on the geometry of the structural model; it isindependent of the loading. Thus, it can be set up for a given model geometry, without specification of theapplied forces, exactly as we have done in this example.If Bf is a square and nonsingular matrix, we talk of a statically determinate, stable structure.In such case we can invert the matrix to obtain the (basic) force influence matrix Bbar.

Bbar Bf1−:= Bbar

1

0

0

0

0

0.667−

0.667−

0.833

1−

0.833

1

1

0

0

0

0.5

0.5

0.625

0

0.625−

0.667−

0.667−

0.833

0

0.833

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

=

The rows of the basic force influence matrix correspond to the longitudinal basic force in a particularelement (e.g. the coefficients in row 3 correspond to element c), and the columns correspond to aparticular dof (e.g. the second column corresponds to the effect of a force at dof 2). Thus the term (i,j) ofthe basic force influence matrix shows the effect of a unit force at dof j on the basic force i of the structure.This basic force influence matrix coefficient is very useful in design optimization problems, since it showswhich member is most affected by the given loading.

Remark: the static matrix at the free dofs Bf and the basic force influence matrix Bbar depend only on themodel geometry for linear statics, in which case the equilibrium equations are formulated in the undeformed(original) configuration of the structural model. When the deformed shape deviates appreciably from theundeformed configuration, the equilibrium equations should be established in the deformed configuration.This is covered in Nonlinear Structural Analysis.

Page 41


Determination of basic forces for specific load case

For a specific load case the applied forces at the free dofs are given. It is then possible to solve the systemof equilibrium equations for the unknown element forces Q. In a stable, statically determinate structure forwhich Bf is square and nonsingular there is a unique set of element forces Q that equilibrate the givenloading (unique solution of equilibrium equations). Note that the solution does not depend on elementproperties, which means that the member sizes do not need to be known beforehand.

88

6

10

5

Pf

0

5−

0

10

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=

solve for unknown element forces by Gauss elimination (function lsolve in Mathcad, \ in Matlab)

basic forces in truss elements(+ : tension, -: compression)Q lsolve Bf Pf,( ):= Q

8.33

8.33

2.08

5

10.42−

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

=

Note that if we have spent the effort to determine the basic force influence matrix (getting the inverse of amatrix is more work than performing Gauss elimination!), then we can obtain the element forces also bymultiplying the second column of Bbar by (-5) and the fourth column by 10 and adding up the results.

Q Bbar 2⟨ ⟩ 5−( )⋅ Bbar 4⟨ ⟩ 10⋅+:= Q

8.33

8.33

2.08

5

10.42−

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

=

Thus, the end forces result from the linear combination of the 2nd and 5th column of the basic forceinfluence matrix after these are factored by the values of the force acting at the corresponding dof.

Page 42


Support reactions and global equilibriumWe can now determine the support reactions from the equilibrium equations at the restrained dofs

12

3

45

67 8

1Q 2Q

4Q3Q 5Q

Equilibrium equations at restrained dofs

P6 Q1− 0.8 Q3⋅−=

P7 0.6− Q3⋅=

P8 0.6− Q5⋅=

We do this directly for each support reaction (for convenience we can use the symbol R to denote thesupport reactions; in such case we number these independently).

R1 P6= Q1− 0.8 Q3⋅−= R1 Q1− 0.8 Q3⋅−:= R1 10−=

R2 P7= 0.6− Q3⋅= R2 0.6− Q3⋅:= R2 1.25−=

R3 P8= 0.6− Q5⋅= R3 0.6− Q5⋅:= R3 6.25=

and with these we can check the global equilibrium (a very important check indeed!)

88

6

10

5

10

1.25 6.25

In addition to the trivial sum of the forces in X and sum of the forces in Y (which are satisfied by inspection!)we check the sum of the moments about the left and right support (only one of the two is really necessary)

sum of moments about left support (CCW is +ve) 10− 6⋅ 5 8⋅− R3 16⋅+ 0=

sum of moments about right support (CCW is +ve) 10− 6⋅ 5 8⋅+ R2 16⋅− 0= Ok!

Page 43


If we are interested in the support reaction influence matrix, then we set up the static matrix for the supportdofs consisting of the coefficients of the basic forces in the equilibrium equations for the restrained dofs ofthe structure. We have from the earlier equations

R1 P6= Q1− 0.8 Q3⋅−= eq for dof 6

thus the complete staticmatrix for the support dofs isR2 P7= 0.6− Q3⋅= Bd

1−

0

0

0

0

0

0.8−

0.6−

0

0

0

0

0

0

0.6−

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

:= eq for dof 7

eq for dof 8R3 P8= 0.6− Q5⋅=

The support reaction influence matrix is the product of Bd and Bbar and contains the coefficients for theeffect of the applied forces at the free dofs on the support reactions.

Bd Bbar⋅

1−

0

0

0

0.5−

0.5−

1−

0

0

1−

0.375−

0.375

0

0.5−

0.5−

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

= e.g. a unit force at dof 4 produces R1 = -1, R2 = -0.375and R3 = 0.375

Note: if we had bothered setting up the matrix Bd , we could determine the support reactions from the basicelement forces with the following expression

support reactions R Bd Q⋅:= R

10−

1.25−

6.25

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

=

but the direct determination earlier is faster for hand calculations.

CONCLUSIONS:

1. In a stable, statically determinate structure there are as many equilibrium equations for the free dofs ofthe structural model as there are unknown basic element forces. The static matrix for the free dofs Bf is,therefore, square and invertible. Note that the inclusion of the support reactions does not change anythingto this discussion, since there are as many equilibrium equations at the support dofs as unknown supportreactions. Thus, it is better to work with the static matrix for the free dofs only.

2. The static matrix for the free dofs of the structure only depends on the geometry of the structural modelin its original (undeformed) configuration (linear equilibrium).

3. The inverse of the static equilibrium matrix for the free dofs Bf gives the basic force influence matrixBbar. This matrix is useful in many applications, such as design optimization, system identification, damageassessment.

4. For a particular loading, the equilibrium equations give a unique solution for the basic forces of thestructural model. The element properties are not required, thus the element sizes need not be available.

5. The support reactions are calculated after the determination of basic forces.

Page 44

Matlab script for Example 2 in CE220 class notes

% Solution for statically determinate truss

OPTION A: perform each step with Matlab functions

% clear memory clear all % specify static (equilibrium) matrix Bf Bf = [1 -1 0 0 0; 0 0 0 -1 0; 0 1 0 0 0.8; 0 0 0.8 0 -0.8; 0 0 0.6 1 0.6]; % determine and display force influence matrix Bbar Bbar = inv(Bf); disp('the force influence matrix Bbar is'); disp(Bbar); the force influence matrix Bbar is 1.0000 -0.6667 1.0000 0.5000 -0.6667 0 -0.6667 1.0000 0.5000 -0.6667 0 0.8333 0 0.6250 0.8333 0 -1.0000 0 0 0 0 0.8333 0 -0.6250 0.8333

specify applied force vector

Pf = [ 0; -5; 0; 10; 0]; % solve for basic forces Q (Note the use of the backslash or left matrix divide operator which computes the solution for a linear system of equations by Gauss elimination) Q = Bf\Pf; % display result format short disp('the basic forces are'); disp(Q); the basic forces are 8.3333 8.3333 2.0833 5.0000 -10.4167

OPTION B: use FEDEASLab functions

% clear memory; close any open windows CleanStart; % define model geometry XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 8 0]; % second node, etc XYZ(3,:) = [ 16 0]; %

Page 45

XYZ(4,:) = [ 8 6]; % % element connectivity array CON { 1} = [ 1 2]; CON { 2} = [ 2 3]; CON { 3} = [ 1 4]; CON { 4} = [ 2 4]; CON { 5} = [ 3 4]; % boundary conditions (1 = restrained, 0 = free) (specify only restrained dof's) BOUN(1,:) = [ 1 1]; BOUN(3,:) = [ 0 1]; % specify element type ne = length(CON); % number of elements [ElemName{1:ne}] = deal('Truss'); % truss element

create Model

Model = Create_SimpleModel (XYZ,CON,BOUN,ElemName); % plot and label model for checking (optional) Create_Window (0.80,0.80); % open figure window Plot_Model (Model); % plot model Label_Model (Model); % label model

form static (equilibrium) matrix B

B = B_matrix(Model); % extract submatrix for free dofs Bf = B(1:Model.nf,:);

specify loading

Pe(2,2) = -5; % force at node 2 in direction Y Pe(4,1) = 10; % force at node 4 in direction X % generate data object Loading Loading = Create_Loading(Model,Pe); % extract applied force vector Pf at free dofs from Loading field Pref Pf = Loading.Pref;

Page 46

solution for basic forces

solve for basic forces and display the result Q = Bf\Pf; disp('the basic forces are'); disp(Q); the basic forces are 8.3333 8.3333 2.0833 5.0000 -10.4167

determination of support reactions

% the product B*Q delivers all forces at the global dofs % the upper 5 should be equal to the applied forces, the lower 3 are the support reactions disp('B*Q gives'); disp(B*Q); B*Q gives 0 -5.0000 0 10.0000 0.0000 -10.0000 -1.2500 6.2500

post-processing of results

% plot axial force distribution Create_Window(0.80,0.80); Plot_Model(Model); Plot_AxialForces(Model,Q);

Page 47

CE 220 - Theory of Structures Ex. 3 - Beam equilibrium © Prof. Filip C. Filippou, 2000

Example 3(r) - Equilibrium equations for statically determinate 2d beam

Objectives:

(a) set up of equilibrium equations for a statically determinate beam(b) systematic reduction of equilibrium equations to exclude longitudinal basic forces for specific loading(c) systematic reduction of equilibrium equations involving a single unknown basic force(d) relation between basic forces and internal forces; bending moment and shear force diagrams(e) influence lines for moving loads

Geometry of structural model

a b c1

2 3 4

10 10 5

Case A: Complete set of equilibrium equations and basic element forces

12

3

4 56

7

8

9

a

b

c

1Q2Q 3Q

4Q5Q 6Q

7Q8Q 9Q

Complete set of equilibrium equations and basic element forces

X-direction Y-direction Moments about Z-axis

P1 Q2=

P2 Q1 Q4−= P3Q2 Q3+

10−

Q5 Q6+

10+= P4 Q3 Q5+=

P5 Q4 Q7−= P6 Q6 Q8+=

P7 Q7= P8Q8 Q9+

5−= P9 Q9=

Page 48


We distinguish 3 horizontal force equilibrium equations (equations 2, 5 and 7) involving 3 longitudinalbasic forces. We note that the longitudinal basic forces Q1, Q4 and Q7 do not appear in the otherequations. We can therefore uncouple these three equations from the rest and solve themindependently. We could have concluded this independence of equations with longitudinal basic forcesby inspection at the start of the analysis. If we are not interested in the longitudinal basic forces, we canset up only the equilibrium equations involving the flexural basic element forces. This is shown next

Case B: Complete set of equilibrium equations for flexural basic element forces

12

34

5

6

a b c1

2 3 4

10 10 5

a

b

c

1Q 2Q

3Q4Q

5Q 6Q

Y-direction Moments about Z-axis

P1 Q1=

P2Q1 Q2+

10−

Q3 Q4+

10+= P3 Q2 Q3+=

P4 Q4 Q5+=

P5Q5 Q6+

5−= P6 Q6=

Looking at these 6 equations and the unknown 6 basic element forces we conclude that equations 1 and 6involve a single unknown basic force each. We can therefore solve for the corresponding unknown basicforce independently from the rest. Assuming that the applied force at this degree of freedom is zero in mostcases, we can proceed to remove the two equations and the corresponding basic forces from the system ofequations. This process can be done by inspection at the start of the analysis once the loading is given. Ifthe applied forces at dofs 1 and 6 are not equal to zero, the corresponding basic forces will not be zero andwe should be careful to include the effect of the non-zero basic force on the remaining equilibriumequations. We will see later in this example how this is done.

Page 49


We renumber the equations and basic forces.

Case C: Reduced set of equilibrium equations for flexural basic element forces

a b c1

2 3 4

10 10 5

1

23 4

a

b

c

1Q

2Q 3Q

4Q

Y-direction Moments about Z-axis

P1Q1

10−

Q2 Q3+

10+= P2 Q1 Q2+=

P3 Q3 Q4+=P4

Q4

5−=

We note that equation 4 now involves only one unknown, which can be, therefore, determinedindependently from the other basic element forces. However, it is quite likely that a concentratedtransverse force acts at the tip of the cantilever, so that this basic force will not be zero in most cases. Forthis reason we defer further simplification until we determine the solution for a specific load case.

Page 50


Determination of basic forces for specific load case

The following loading is given. It is required to determine the basic forces for this loading.

15 5

a b c1

2 3 4

10 10 5

12

34

5

6

We start with the 6 equilibrium equations not involving axial forces, since the latter are zero by inspectionfor the given loading. We note that the applied forces are zero at dofs 1 and 6, so that the correspondingbasic forces are also zero. We can, therefore, set up the equilibrium equations in terms of the reduced setof 4 dofs with 4 basic forces corresponding to case C above. We restate the equilibrium equations

1

23 4

a

b

c

1Q

2Q 3Q

4Q

P1Q1

10−

Q2 Q3+

10+=

P2 Q1 Q2+=

P3 Q3 Q4+=

P4Q4

5−=

For the given loading we have: P1 15−= P2 0= P3 0= P4 5−=

and it is very straightforward to solve the equilibrium equations for the unknown basic forces. We get

Q4 25= from equation 4

Q3 25−= from equation 3

and then need to solve equation 1 realizing that Q1 = -Q2 from equation 2. We get 15−Q1

5− 2.5−=

and thus, Q1 62.5= and Q2 62.5−=

Page 51


We can also use Gauss elimination of the system of equilibrium equations. This approach is useful for largerstructures, which we do not wish to solve by hand.

The specified loading forms the applied nodal force vector Pf

15−

0

0

5−

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:=

The static or equilibrium matrix of the structure consists of the coefficientsof the basic forces Q in the equilibrium equations, namely Bf

110

−

1

0

0

110

1

0

0

110

0

1

0

0

0

1

15

−

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

We use Gauss elimination to solve for Q Q lsolve Bf Pf,( ):= Q

62.5

62.5−

25−

25

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

=

Bending moment diagram

Before determining and drawing the bending moment diagram we need to establish the relation betweenbasic forces and internal forces at a section located a distance x from end i of the frame element. Thefollowing figure is limited to the effect of the flexural basic forces. The corresponding internal forces are thebending moment M(x) and the shear force V(x) at a section x from the end i of the frame element.

2q

( )M x2q

2 3

L+q q

( )M x

2q 2 3( ) 1 x xM xL L

⎛ ⎞= − − +⎜ ⎟⎝ ⎠

q q

L

3q

3q

2 3

L+q q

2 3

L+q q 2 3

L+q q

3q

( )V x

2 3

L+q q -

( )V x

frame element

end jend i

x

y

+

the bending moment is positive when actingCCW at the face of the cut closest to the origin.Its reaction acts in the opposite direction at theopposite face of the cut; the shear force ispositive when acting in the positive y-direction atthe face of the cut closest to the origin; itsreaction acts in the opposite direction at theopposite face of the cut

we draw the bending moment "upside down" oron the tension side of the reference line, but thisis really a matter of taste; it is important to drawa happy or sad face so as to unambiguouslydefine the sense of flexure

we draw the shear normally, i.e. positive inpositive y. With this sign convention for theshear and moment diagram the slope of themoment diagram corresponds to the sign of theshear force in the graph, even thoughnumerically it is equal to the opposite

Page 52


With the above convention the bending moment diagram becomes

a b c

62.5 62.5 25 25

5 58.758.756.25 6.25

62.525

Once the basic forces are known we use element equilibrium to determine all dependent forces(in this case there are only shear forces). We then use the nodal equilibrium to determine the supportreactions. Finally, we check global equilibrium. This process is illustrated in the following figures.

(1) Element equilibrium: determine shear forces

a

b

c

62.5

62.5 25

25

5 5

8.758.75

6.25 6.25

(2) node equilibrium: determine support reactions and confirm correctness of solution at applied forces

5 5 5

13.75

6.25

15

6.25

6.25

8.75 8.75

(3) complete structure free body and global equilibrium

15 5

6.25 13.75

Page 53


It is instructive to display this 3-step process in a single figure of element, node and structure equilibrium

a

b

c

62.5

62.5 25

25

5 5

8.758.75

5 5 5

13.75

6.25 6.25

6.25

15

6.25

6.25

15 5

6.25 13.75

8.75 8.75

Page 54


Force influence matrix and influence line for moving loads

The inverse of the 4x4 static matrix for the free dofs of the structural model yields the basic force influencematrix Bbar. for transverse loading

Bbar Bf1−:= Bbar

5−

5

0

0

0.5

0.5

0

0

0.5

0.5−

1

0

2.5

2.5−

5

5−

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

=

The rows of the force influence matrix correspond to a particular basic force, e.g. the second rowcorresponds to Q2. The columns correspond to particular dofs. The terms of the force influence matrixare force influence coefficients that show the effect of a unit force at a certain dof on the particularbasic force. For example, a unit force at dof 1 gives rise to Q2 = 5, while a unit force at dof 4 gives riseto Q2 = -2.5 (recall that a positive force is directed upwards). Thinking that the unit vertical force maybe acting anywhere along the beam leads to the idea of influence line for Q2 The influence line for aparticular force is the graphical representation of the effect of a moving unit force on the basic force.The value of the diagram at a particular location represents the value of the basic force when a unitforce acts at that particular location of the structure.

a b c1

2 3 4

5

2.5

+

-

2Qinfluence line for under vertical force

moving unit force

CONCLUSIONS:

1. In beams the longitudinal basic forces and corresponding equilibrium equations are uncoupled from theother basic forces and the equilibrium equations. In the absence of applied longitudinal forces thelongitudinal basic forces are all zero and do not need to be considered.

2. The inverse of the static matrix of a statically determinate structure is called the force influence matrix.Rows of this matrix represent the effect of a unit force at any free dof on a particular basic force. Thecorresponding values can be used to construct influence lines for each basic force of interest. Theinfluence lines consist of straight line segments for statically determinate structures.

Page 55


Treatment of nodal force at tip of cantileverWe return to the earlier set of equilibrium equations for case C.

1

23 4

a

b

c

1Q

2Q 3Q

4Q

P1Q1

10−

Q2 Q3+

10+=

P2 Q1 Q2+=

P3 Q3 Q4+=

P4Q4

5−=

We note that equation 4 only involves one unknown basic force, Q4, and can be solved independently.For the general case we obtain

Q4 5− P4⋅= which says that the basic force at the base of the cantilever it equal to the force acting atthe tip times the lever arm.

after inserting this result into equation 3 we get P3 Q3 5 P4⋅−=

after moving the applied force to the left hand side of the equation we get P3 5 P4⋅+ Q3=

this is now another trivial equilibrium equation that can be solved for the basic force Q3 independently ofthe other equations.

The above equations simplify the number of equilibrium equations and basic forces for the given load caseto the following

15

a b1

2 3

10 10

1

2

a

b

1Q

2Q

25

25

5Force in Y-direction at node 2

P1Q1

10−

Q2 25−

10+=

Moments about Z-axis at node 2

P2 Q1 Q2+=

Conclusion: as far as equilibrium isconcerned we can replace the nodal forceat the tip of the cantilever by a force and amoment at the root of the cantilever. Thisreduces the total number of equations andbasic forces by two.

Page 56


This is an example of determining basic forces independently, one at a time, from as many equilibriumequations. We did this in this case first for Q4 from equation 4, and then for Q3 from equation 3. In eachcase we have accounted for the effect of the basic force on the remaining equilibrium equations. The effectof Q4 was accounted for in equation 3 where it appears. Then, the effect of Q3 was accounted for inequation 1 where it appears. The end result is the inclusion of Q3 = -25 in equation 1 leaving us with 2coupled equilibrium equations in two unknowns. The solution is very straightforward.

Page 57

Matlab script for Example 3 in CE220 class notes

% Solution for statically determinate beam with overhang

use FEDEASLab functions

% clear memory and close any open windows CleanStart; % define model geometry XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 10 0]; % second node, etc XYZ(3,:) = [ 20 0]; % XYZ(4,:) = [ 25 0]; % % element connectivity array CON { 1} = [ 1 2]; CON { 2} = [ 2 3]; CON { 3} = [ 3 4]; % boundary conditions (1 = restrained, 0 = free) BOUN(1,:) = [ 1 1]; BOUN(3,:) = [ 0 1]; % specify element type ne = length(CON); % number of elements [ElemName{1:ne}] = deal('2dFrm'); % 2d frame element

create Model

Model = Create_SimpleModel (XYZ,CON,BOUN,ElemName); % plot and label model for checking (optional) Create_Window (0.80,0.80); % open figure window Plot_Model (Model); % plot model Label_Model (Model); % label model

form static (equilibrium) matrix B

B = B_matrix(Model); % extract submatrix for free dofs Bf = B(1:Model.nf,:);

specify loading

Pe(2,2) = -15; % force at node 2 in direction Y Pe(4,2) = -5; % force at node 4 in direction Y % generate data object Loading Loading = Create_Loading(Model,Pe);

Page 58

% extract applied force vector Pf at free dofs from Loading field Pref Pf = Loading.Pref;

solve for basic forces

Q = Bf\Pf; % display the result disp('the basic forces are'); disp(Q); the basic forces are 0 0 62.5000 0 -62.5000 -25.0000 0 25.0000 0

determine support reactions

% the product B*Q delivers all forces at the global dofs the upper 9 should be equal to the applied forces, the lower 3 are the support reactions disp('B*Q gives'); disp(B*Q); B*Q gives 0 0 -15.0000 0 0 0 0 -5.0000 0 0 6.2500 13.7500

plotting

% open window and plot moment diagram Create_Window(0.80,0.80); Plot_Model (Model); % observe that the second argument is empty [] in the following function call Plot_2dMomntDistr (Model,[],Q);

Page 59

CE220 - Theory of Structures Ex. 4 - Equilibrium of portal frame with hinge © Prof. Filip C. Filippou, 2000

Example 4(r) - Equilibrium equations for statically determinate frame

Objectives: (a) set up reduced set of equilibrium equations for three-hinge frame(b) uncouple equilibrium equations for the longitudinal basic forces(c) solution for basic forces of primary importance, i.e. flexural basic forces of elements(d) determination of longitudinal basic forces one at a time from node equilibrium

12

8 8

10

1

2 3 4

5

a

b c

d

Reduced set of equilibrium equations and basic element forcesThe inspection of the structural model reveals that the moment equilibrium equations at nodes 1, 3 and 5involve only one basic force each. Assuming that there are no applied forces at these dofs thecorresponding basic forces will be zero. We, therefore, leave out these equilibrium equations andcorresponding basic forces and write directly the reduced set

1

23

4

5

6

7

81Q

2Q

3Q

4Q 5Q6Q

7Q

8Q

a

b c

d

Page 60


Equilibrium equations:

X-direction Y-direction Moments about Z

P1Q2

12Q3−= P2 Q1

Q4

8+= P3 Q2 Q4+=

P4 Q3 Q5−= P5Q4

8−

Q6

8+=

P6 Q5Q8

10+= P8 Q6 Q8+=P7

Q6

8− Q7+=

there are 8 equations in 8 unknown basic element forces and the structure is statically determinate

Suppose that we like to separate the equilibrium equations that involve the 4 longitudinal basic forces inthe frame elements, Q1, Q3, Q5 and Q7. To do so, we need to isolate 4 equations from the above 8 insuch a way that we are left with 4 equations that do not include the longitudinal basic forces.

This is easy to do for Q1 and Q7, since each appear in only one equilibrium equation, namely 2 and 7respectively.

We also note from the above equations that Q3, Q5 appear in only three equations, namely 1, 4 and 6.In order to isolate these two longitudinal basic forces from the remainder, we create a new equation bysumming up the three equations in the X-direction. We obtain:

P1 P4+ P6+Q2

12

Q8

10+= (*)

We note that the longitudinal basic forces do not appear in this equation. We can therefore set aside thethree equilibrium equations in the two longitudinal basic forces that we wish to isolate, as long as we insertthe sum of these 3 equations (*) in their place. We are left with the following two sets of equations:

Set A: equilibrium equations without longitudinal basic element forces


P1 P4+ P6+Q2

12

Q8

10+= P3 Q2 Q4+=

P5Q4

8−

Q6

8+=

P8 Q6 Q8+=

Page 61


Set B: equilibrium equations with longitudinal basic element forces

X-direction Y-direction

P1Q2

12Q3−= P2 Q1

Q4

8+=

P6 Q5Q8

10+= P7

Q6

8− Q7+=

Thus, by isolating the four longitudinal basic forces and an equal number of equilibrium equations fromthe four flexural basic forces, we have split the original set of 8 equations in 8 unknowns into two setsof 4 equations. The first set of 4 equations only involves 4 unknown flexural basic forces and can besolved independently. The second set, however, can only be solved after determining the value of theflexural basic forces Q2, Q4, Q6 and Q8 from the solution of the first set.

Reduced set of equilibrium equations without longitudinal basic element forces

We can simplify this process by numbering only the first set of equations and corresponding unknowns.This is the fastest of tackling the problem by hand (the preceding discussion serves only as explanation ofhow to get to this point systematically!). Basic forces and relevant dofs are shown in the following figure.

1

23

1Q

2Q

a

b c

d

3Q

4Q

4

Page 62


The corresponding equilibrium equations without longitudinal basic element forces are

P1Q1

12

Q4

10+=

P2 Q1 Q2+=

P3Q2

8−

Q3

8+=

P4 Q3 Q4+=

It is worth noting that the first equilibrium equation now physically corresponds to the force equilibriumin the global X-direction for the free body consisting of nodes 2, 3 and 4 and elements b and c, asshown in the following figure.

1

1Q

ad

4Q1

12Q

4

10Q

The process of setting up these equations directly results in a significant reduction of problem sizeand in better insight into the system response.

The static matrix of the frame without the longitudinal basic element forces is Bf

112

1

0

0

0

1

18

−

0

0

0

18

1

110

0

0

1

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

and the force influence matrix is Bbar Bf1−:= Bbar

5.45

5.45−

5.45−

5.45

0.55

0.45

0.45

0.45−

4.36

4.36−

3.64

3.64−

0.55−

0.55

0.55

0.45

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

=

Page 63


Solution of basic forces for specific load case including distributed element loading

We wish to solve for the basic forces under the given loading

55

12

8 8

10

1

2 3 4

5

a

b c

d

The nodal force vector is straightforward. It is Pf

5

0

0

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:=

How are we supposed to include the effect of the distributed element load?

We look at the equilibrium of a typical element

1q3qyw

2q

2 3

2yw L

Lq q+

+ 2 3

2yw L

Lq q+

− +

1q

1q3q2q

2 3

Lq q+

2 3

Lq q+

1q

+yw

2yw L

2yw L

0

homogeneous solution thatsatisfies the element free bodyequilibrium equations with 3unknown "parameters", the basicelement forces

particular solution for whichthe basic element forces are zero

Page 64


We note that the equilibrium solution of a typical element consists of two parts:(a) the homogeneous solution, i.e. the relation between basic and dependent end forces so as to satisfythe element equilibrium equations, and(b) the particular solution, i.e. the values of the dependent forces that satisfy the element equilibrium underthe applied element loads

We have already accounted for the homogeneous solution when writing the equilibrium equations in termsof the basic forces. Now it remains to include the particular solution for the element carrying a distributedload. This is element c in our case and the shear forces of the particular solution affect dof 5 and 7 of theoriginal equilibrium equations. We show this particular solution effect separately in the following figure.

1

23

4

5

6

7

8

5

c20 20

Equilibrium equations for given load case:


5Q2

12Q3−= 0 Q1

Q4

8+= 0 Q2 Q4+=

0 Q3 Q5−= 0Q4

8−

Q6

8+ 20+=

0 Q5Q8

10+= 0 Q6 Q8+=0

Q6

8− Q7+ 20+=

Page 65


and after separating the equations involving longitudinal basic element forces we get

Equilibrium equations without longitudinal basic forces


5Q2

12

Q8

10+= 0 Q2 Q4+=

0Q4

8−

Q6

8+ 20+=

0 Q6 Q8+=

Equilibrium equations with longitudinal basic forces

X-direction Y-direction

P1Q2

12Q3−= P2 Q1

Q4

8+=

P6 Q5Q8

10+= P7

Q6

8− Q7+ 20+=

After renumbering basic forces and equations, as before, to focus only on the effect of the flexural basicelement forces we get

5Q1

12

Q4

10+=

0 Q1 Q2+=

0Q2

8−

Q3

8+ 20+=

0 Q3 Q4+=

For hand calculations the solution of this problem can be simplified by realizing that equations 2 and 4essentially amount to Q2 = -Q1 and Q4=-Q3. This leaves us with two equations in two unknowns, an easytask.

Page 66


We can write the above equilibrium equations compactly, as Pf Bf Q⋅ Pfw+=

with Pf the vector of the forces applied directly at the nodes, and Pfw the vector of the equivalent nodalforces due to element loading. To solve this system we move all given terms to the left hand side, and solvefor the unknowns on the right hand side noting that we have as many unknowns as available equations.

Thus, we have to solve the following system of equations: Pf Pfw− Bf Q⋅=

For solving the equations we set up the vectors and matrices in Mathcad or Matlab and get

with the following vector of applied forces Pf

5

0

0

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

= and equivalent nodal forcesdue to element loading

Pfw

0

0

20

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:=

and with the static matrix for the free dofs defined earlier, i.e. Bf

112

1

0

0

0

1

18

−

0

0

0

18

1

110

0

0

1

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

we solve for the unknown element forces by Gauss elimination Q lsolve Bf Pf Pfw−,( ):= Q

60−

60

100−

100

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

=

As the equations reveal the element loading amounts to an equivalent nodal force vector of -Pfw

5

1

2 3 4

5

a

b c

d

20 20

5

equivalent nodal forces due to w = 5

Thus, it is possible to treat the element loading as an equivalent nodal force vector for the determination ofthe basic forces. For the later determination of the shear forces and bending moment distribution in elementc, however, we should not neglect to include the particular solution.

Page 67


With these results we return to the remaining equilibrium equations to obtain the longitudinal basic forces.Instead of doing this analytically with the equations we set up earlier, it is better to use the element andnode equilibrium directly to get the longitudinal basic element forces and dependent element end forces.This avoids the numbering of the longitudinal basic element forces and the corresponding equations.

The following figure provides all details. It is an excellent way for the analyst to check equilibriumand display applied and resisting forces. Note that we have selected to include the effect of element load inelement c as equivalent nodal forces. This results in the right values for all internal forces except for theshear forces and bending moments in element c for which the particular solution due to the elementloading needs to be added to the homogeneous solution. We will do this separately.

a

b c60

100

605

5

12.57.5

10

1012.5

32.5

d

10010

10

32.5

32.5

1010

7.5

1010

5

5 107.5

7.5

7.5

7.5

10

532.5

7.5

5

7.5

32.5

10

20

12.5

Node 1

Node 2

10

12.5

7.510

20

Node 3

Node 5

Node 4

only homogeneous solution

The steps in the figure above are:

Show the flexural basic element forces Q1 through Q4 with actual values and orientation.1.Use element equilibrium to get the shear forces in elements a through d; note that the shear forces in2.element c correspond to the homogeneous solution of element equilibrium only. The correct valuesresult only after including the particular solution under element loading, as will be shown subsequently.Use horizontal and vertical force equilibrium at nodes 2 and 4 to get the longitudinal basic forces (only3.4 equations are needed; we can use the vertical and horizontal force equilibrium at node 3 to check)Use the vertical and horizontal force equilibrium at nodes 1 and 5 to obtain the support reactions.4.

Remark: in an actual problem only the four flexural basic element forces would be numbered and thecorresponding equilibrium equations not involving longitudinal basic forces would be set up. Thus, there isno need to number the longitudinal basic element forces and corresponding equations. A computerprogram on the other hand sets up all equations and then solves them, since it is more work to considerspecial cases than to solve a system of equations, however large the size of it may be.

Page 68



For determining and drawing the bending moment distribution in all elements of the structural model we needto include the particular solution for element c under element loading. This is shown in the following figure.

100

100

60

60

2

8yw L

10 10100

12.5 12.5

homogeneous solution

05

20 20particular solution

+

Determination of maximum positive moment in element c w 5:=

use the shear force value at the left end Mmax7.52

2 w⋅:= Mmax 5.625=

use the shear force value at the right end(just for practice) Mmax 100−

32.52

2 w⋅+:= Mmax 5.625=

Page 69


Check of global equilibrium

Finally, it is very important to check global equilibrium for the free body of the entire structural model.This is shown in the following figure.

5

12

8 8

10

10

532.5

7.5

20 20

Sum of the forces in X and Y is satisfied by inspection. Sum of the moments about the right support yields:

5 2⋅ 5 10⋅− 7.5 16⋅− 40 4⋅+ 0= ok

Remark

Basic forces and support reactions are the same under the distributed element load or the equivalent nodalforces due to the element load (this is where the name equivalent comes from). The particular solution onlyplays a role in the determination of the shear force and bending moment distribution of the elementcarrying the distributed element load.

CONCLUSIONS:

1. In frame analysis longitudinal basic forces are of less significance than the flexural basic forces of theelements. Thus, we try to set up only the equilibrium equations involving the latter and set aside theequilibrium equations involving the longitudinal basic forces.

2. After solving for the basic forces of primary interest, we use the equilibrium equations that were setaside in the first place to determine the longitudinal basic forces. This is usually done with free bodydiagrams of the elements and nodes.

3. In determining the shear force and bending moment distribution of elements carrying the elementloads we should remember to include the particular solution of the element equilibrium equations underthe element loading.

Page 70

Matlab script for Example 4 in CE220 class notes % Solution for statically determinate portal frame under distributed loading

use FEDEASLab functions % clear memory and close any open windows CleanStart; % define model geometry XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 0 12]; % second node, etc XYZ(3,:) = [ 8 12]; % XYZ(4,:) = [ 16 12]; % XYZ(5,:) = [ 16 2]; % % element connectivity array CON { 1} = [ 1 2]; CON { 2} = [ 2 3]; CON { 3} = [ 3 4]; CON { 4} = [ 4 5]; % boundary conditions (1 = restrained, 0 = free) BOUN(1,:) = [ 1 1 0]; BOUN(5,:) = [ 1 1 0]; % specify element type ne = length(CON); % number of elements [ElemName{1:ne}] = deal('2dFrm'); % 2d frame element

create Model Model = Create_SimpleModel (XYZ,CON,BOUN,ElemName); % plot and label model for checking (optional) Create_Window (0.80,0.80); % open figure window Plot_Model (Model); % plot model Label_Model (Model); % label model

1

2 3 4

5

1

2 3

4

form static (equilibrium) matrix B B = B_matrix(Model); % extract submatrix for free dofs Bf = B(1:Model.nf,:);

Page 71

% insert hinge at left end of element c by removing corresponding column of Bf iq = setdiff(1:12,8); Bf = Bf(:,iq);

specify loading Pe(2,1) = 5; % force at node 2 in direction X % for vertical loading use equivalent nodal forces as determined in pp. 53-54 of class notes Pe(3,2) = -20; % force at node 2 in direction Y Pe(4,2) = -20; % force at node 4 in direction Y % generate data object Loading Loading = Create_Loading(Model,Pe); % extract applied force vector Pf at free dofs from Loading field Pref Pf = Loading.Pref;

solve for basic forces % make sure to re-insert zero at moment release Q = zeros(12,1); Q(iq) = Bf\Pf; % display the result for the basic forces disp('the basic forces are'); disp(Q); the basic forces are -7.5000 0 -60.0000 -10.0000 60.0000 0 -10.0000 0 -100.0000 -32.5000 100.0000 0

determine support reactions % the product B*Q delivers all forces at the global dofs; the upper 11 should be equal to the applied forces, the lower 4 are the support reactions

disp('B*Q gives');

disp(B*Q); B*Q gives 0 5.0000 0 0 0 -20.0000 0 0 -20.0000 0

Page 72

0 5.0000 7.5000 -10.0000 32.5000

plotting % open window and plot moment diagram first w/o distributed element load w Create_Window(0.80,0.80); Plot_Model (Model); % observe that the second argument is empty [] in the following function call % use scale factor of 2 for plot as last argument Plot_2dMomntDistr (Model,[],Q,2);

% to include the effect of the distributed load we need to specify it as part of the ElemData object, so that the element "knows" that it is carrying the load and plots the correct M-diagram % specify distributed load for element 3 ElemData{3}.w = [0;-5]; % open new window and plot new moment diagram with same scale factor Create_Window(0.80,0.80); Plot_Model (Model); Plot_2dMomntDistr (Model,ElemData,Q,2);

Page 73

Example for reduction of equilibrium equations for relevant basic forces

a

b

c

d

1

2 3

4

5

complete dofs

2Q

3Q1Q

4Q5Q6Q

7Q

8Q

9Q

10Q11Q 12Q

complete set of basic forces

NF = 10 NQ = 12

NOS=2

non-trivial dofs

1

2

34

5

67

89

10

1

23 4

56

NF = 6

4Q

2Q

3Q

1Q5Q

6Q

7Q

8Q

NQ = 8

dofs not involving longitudinal basic forces

1

32

NF = 3

3Q

1Q4Q

5Q2Q

corresponding basic forces

corresponding basic forces

NQ = 5

NQ = no of basic element forces

NF = no of equilibrium equations at relevant free dofs


Page 74

General solution for redundant (statically indeterminate) structures

start from equilibrium equations at free dofs

solve for the basic forces of the primary structure in terms of the applied forces (and element loading, if present) and the redundant basic forces according to

add the identity to recover the complete set of basic forces on the left hand side

combine

arrange vector of basic forces in original order

where are basic force influence matrices for the applied forces and for the redundant basic forces, respectively

In actual problems we are usually interested in the basic forces under a specific loading and thus determine the basic forces of the primary structure under the given loading directly without first setting up the force influence matrix Bbari We denote these basic forces with Qp

and thus the solution for a specific loading becomes

How do we select the redundant basic forces at the beginning of the solution process?(a) "by hand" we select them so that the primary structure is stable;we assess stability of structural models by kinematic criteria in the lectures on compatibility

(b) automatically we can select the redundant basic forces by a linear algebra procedure known as row-echelonform of the equilibrium matrix Bf (see Strang, 3rd edition, pp. 71-77). This is a built-in function in Mathcad and Matlab and is used in the FEDEASLab function ForceInfl_matrix to establish the force influence matrices. To select specific basic forces as redundants, one needs to re-order the columns of Bf and place the selected redundants last. If you are interested in this, consult the addendum to Example 6 and Matlab file Example_6.m.

The following is the general solution of the equilibrium equations for statically indeterminate structures.For hand calculations the process is shown in Examples 5 and 6.

f f fw= +BP Q P

select NOS redundant basic forces and collect them in vector call remaining basic forcesxQ iQ

the static matrix is partitioned accordingly into two submatrices andiB xB

is a square matrix that should not be singular; it corresponds to the static matrix of the primary structure(i.e. the statically determinate structure that results when the redundant basic force values are set to zero)

iB

with this split the equilibrium equations become f fw i i x x− = +B BP P Q Q

( )1i i f fw x x

− ⎡ ⎤= − −⎣ ⎦B BQ P P Q

x x=Q Q

( )1 1

i i i xf fw x

x

− −⎡ ⎤ ⎡ ⎤⎛ ⎞ −= − +⎢ ⎥ ⎢ ⎥⎜ ⎟⎜ ⎟ ⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎣ ⎦ ⎣ ⎦

B B B0 I

QP P Q

Q

( )i f fw x x= − +B BQ P P Q

we write ( )p i f fw= −BQ P P

i x,B B

p x x= +BQ Q Q


Page 75

CE220 - Theory of Structures Ex. 5 - Equilibrium for indeterminate truss © Prof. Filip C. Filippou, 2000

Example 5(r) - Equilibrium equations for statically indeterminate truss

Objectives:

(a) set up equilibrium equations for statically indeterminate truss(b) primary structure, basic force redundants, force influence matrices for primary structure(c) general solution of problem, superposition of homogeneous and particular solution

8

6

15

a

b c

d

e f

Geometry:

La 8:= Lb 6:=

Lc 6:= Ld 8:=

Le 10:= Lf 10:=

Dof and basic force numbering

8

6

1

23

4

5

1Q

2Q3Q

4Q

5Q6Q

Page 76


Step 1: Equilibrium equations- degree of static indeterminacy

P1 Q1 0.8 Q5⋅+=corresponding equilibrium matrix

P2 Q4− 0.8 Q5⋅−=

P3 Q2 0.6 Q5⋅+=Bf

1

0

0

0

0

0

0

1

0

0

0

0

0

0

1

0

1−

0

1

0

0.8

0.8−

0.6

0

0

0

0

0

0.8

0.6

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=P4 Q4 0.8 Q6⋅+=

P5 Q3 0.6 Q6⋅+=

There are 6 truss element forces and 5 independent equations of equilibrium at the free degrees offreedom. The structural model has a degree of static indeterminacy NOS of 6-5=1 (i.e. NOS=1).

How do we know that the 5 equilibrium equations are independent? The question we need to answer iswhether the structure is stable or not. For this structure this is easy to. For more complex structures with fewelements the best way is by the kinematic method, as we will discuss later in the course. Linear algebra,however, provides the standard method for any size structure. It has to do with the rank of the matrix, whichis equal to the number of linearly independent rows. Thus, when the rank of the static matrix is equal to thenumber of rows (which, in turn, are equal to the number of free dofs), then the structure is stable. Let uscheck this for the truss using two built-in functions of Mathcad (the corresponding Matlab functions are rankand size)

rank Bf( ) 5= rows Bf( ) 5= we conclude that the structure is stable

Step 2: Selection of redundant(s), primary structure

Select NOS basic forces as free parameters of the equilibrium equations (linear algebra terminology).These are known in structural analysis as redundant basic forces or simply redundants. Make sure thatthe structure without the NOS redundants is statically determinate and stable (we will learn later how todo this with kinematic methods). In the case at hand we select the force in member f Q6 as redundant.The structure that remains when the redundant is set equal to zero (or "removed") is known in structuralanalysis as primary structure.

Page 77


Review of Linear Algebra (Linear Algebra and Its Applications by G. Strang, pp. 71-77 of 3rd edition)

The general solution of a system of equations M x⋅ b=

can be obtained as the superposition of the particular solution which satisfies the system of equationsfor the given right hand side vector, and the homogeneous solution which satisfies the system ofequations M x⋅ 0=

denoting the homogeneous solution with xh and the particularsolution with xp we have the following conditions

M xp⋅ b= and M xh⋅ 0=

and we can see clearly that M xp xh+( )⋅ b=

If the matrix M is square and invertible, then the only homogeneous solution is the so-called trivial solution xh 0= . However, a matrix M with more columns than rows, has as many non-trivial homogeneous solutionsas the difference between the number of columns and the rank of the matrix M. These non-trivialhomogeneous solutions span the nullspace of matrix M.

Translated to the structural problem where M is the static matrix Bf of the structural model this means: thereare NOS non-trivial homogeneous solutions of the equilibrium equations for a stable structure with degree ofstatic indeterminacy of NOS.

What is the systematic way of finding the particular solution and the NOS homogeneous solutions?

In Linear Algebra we select the NOS free variables by a procedure known as row-echelon form. In Mathcadand Matlab there is a built-in function called rref for Reduced Row-Echelon Form, which goes even furtherby transforming the matrix M through row operations into a diagonal unit matrix and unreduced columns forthe free variables. Note that the diagonal unit matrix may not be contiguous as the following example shows

here the free variable is the third, since the row operationsresult in a diagonal unit matrix with columns 1, 2 and 4. If thiswere a static matrix, then Q3 would be the redundant that theprogram would select, because the primary structure with Q1, Q2and Q4 is stable (look at the rank of columns 1, 2 and 4). Bycontrast, the selection of Q4 as redundant results in an unstableprimary structure (the rank of columns 1, 2 and 3 is only 2).

rref

1

0

1

2

1

2

3

1

3

1

0

2

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

1

0

0

0

1

0

1

1

0

0

0

1

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

=

In structural analysis we select the free variables (i.e. the redundants) by inspection or by the kinematicmethod.

Once the free variables (i.e. redundants) are selected we set their value equal to 0 (primary structure)and solve the governing equations (equilibrium equations) for the remaining variables (basic forces). Thisgives the particular solution for the given vector (applied nodal forces) b.

To obtain linearly independent homogeneous solutions we set each free variable in turn equal to 1 (all otherfree variables are zero) and solve the system of equations M xh⋅ 0= . In structural analysis these solutionsare called self-stress states, since the structure is stressed without external forces. Such self-stress statesonly arise in statically indeterminate structures, since a square matrix M has only trivial homogeneoussolutions, as we have already stated.

Page 78


Step 3: Determine basic forces Q for given loading with redundants equal to zero

Set redundant equal to zero. Solve the equilibrium equations for the applied nodal force vector. This is theparticular solution of the equilibrium equations for the given load case; we denote the basic forces with Qp

Set redundant equal to zero, i.e. Q6 0=

We equilibrium equations for thegiven applied force vector become and their solution one by one is

0 Q1 0.8 Q5⋅+= Q1 15=

0 Q4− 0.8 Q5⋅−= Q5 18.75−=

0 Q2 0.6 Q5⋅+= Q2 11.25=

15 Q4= Q4 15=

0 Q3= Q3 0=

The particular solution therefore is: Qp

15

11.25

0

15

18.75−

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

We can also solve these equations in matrix form, but we leave this for Matlab (see Example_5.m).

With the basic forces we can obtain the support reactions from node equilibrium at the restrained dofs.

8

6

15

a

b c

d

e

f

6 0=Q

11.2511.2515

The final result for the basicforces and support reactions ofthe primary structure under theapplied force vector is

Page 79


Step 4: Determine the basic forces Q for a unit value of each redundant from the homogeneousequilibrium equations, i.e. without any applied force vector

Determine the truss element forces for Q6 1= This is a force influence vector and is denoted byBbarx. Because of the absence of applied forces thisrepresents a self-stress state

We solve the following equations:

0 Q1 0.8 Q5⋅+= Q1 0.8−=

0 Q4− 0.8 Q5⋅−= Q5 1=

0 Q2 0.6 Q5⋅+= Q2 0.6−=

0 Q4 0.8 1⋅+= Q4 0.8−=

0 Q3 0.6 1⋅+= Q3 0.6−=

We can also do this in matrix form, but it is easier to demonstrate in Matlab (see Example_5.m).

We can then obtain the support reactions from the equations at the restrained dofs.

The final result for the basic forces and support reactions becomes

b c

d

6 1=Q

000

8

6

a

e

f

Bbarx

0.8−

0.6−

0.6−

0.8−

1

1

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

Page 80


In reality, of course, the redundant basic force can assume any value (not just the value 1!). Consequently,the general solution of the equilibrium equations is a one parameter expression of the form

Q

15

11.25

0

15

18.75−

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

0.8−

0.6−

0.6−

0.8−

1

1

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

Q6⋅+=

in the more general case of a structural model with degree of static indeterminacy NOS, there are exactlyNOS redundant basic forces and we generalize the above expression as follows: the force influence vectorsfor each redundant get collected into a force influence matrix Bbarx and the redundant basic forces into avector of redundant basic forces Qx.

Q Qp Bbarx Qx⋅+=

We can see that the matrix Bbarx has NOS columns, one for each redundant basic force. We therefore saythat the final stress state is the linear superposition of a particular solution Qp for the given applied forces,and NOS self-stress states, each of which is expressed by a column of matrix Bbarx

The columns of matrix are linearly independent because we selected a unit value for each redundant in turn.

Solution with Linear Algebra functions (optional)

It is interesting to note that we can obtain the above solutions directly with the reduced row-echelon formfunction in Mathcad and Matlab. Let us see

The reduced row-echelon form of the static matrix Bf is

rref Bf( )

1

0

0

0

0

0

1

0

0

0

0

0

1

0

0

0

0

0

1

0

0

0

0

0

1

0.8

0.6

0.6

0.8

1−

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

=

the program selects the first 5 variables that give a unit matrix; in this case the first 5 do the job and, thus,the last variable, i.e. Q6, is selected as the free variable or redundant, exactly as we did above. We notethat the numerical values in the sixth column of the reduced row-echelon form of Bf are the first five valuesof the force influence matrix Bbarx with opposite sign. Recall that the sixth value for the redundant is 1.

Page 81


If we wish to obtain the particular solution automatically, then we need to perform the reduced row-echelonoperation on the augmented matrix consisting of the static matrix and the applied force vector. Wedemonstrate it below

with the appliedforce vector Pf

0

0

0

15

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:= we get rref augment Bf Pf,( )( )

1

0

0

0

0

0

1

0

0

0

0

0

1

0

0

0

0

0

1

0

0

0

0

0

1

0.8

0.6

0.6

0.8

1−

15

11.25

0

15

18.75−

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

=

Now, the last column contains the first 5 values of the particular solution. Recall that the sixth value is 0in this case.

CONCLUSIONS:

1. The general solution of the equilibrium equations for any stable structure is made up of the particularsolution for the given applied force vector (including element loading, if present) and NOS independentself-stress states, where NOS is the degree of static indeterminacy of the structure.

2. The final stress state of the structure under the given loading is then the linear superposition of theparticular solution with NOS independent self-stress states with free parameters. The independence of theself-stress states is achieved by selecting each redundant in turn equal to 1, with the other redundantsequal to zero. Other ways of selecting linearly independent self-stress states are possible, indeed they areideal for solution that take advantage of the symmetry of the structure. This will be discussed later in thecourse.

Page 82

Script for Example 5 in CE220 class notes

% Equilibrium equations for statically indeterminate truss % % specify each step of the process with Matlab functions clear all % structure equilibrium matrix for free dof's Bf=[ 1 0 0 0 0.8 0 ; 0 0 0 -1 -0.8 0 ; 0 1 0 0 0.6 0 ; 0 0 0 1 0 0.8; 0 0 1 0 0 0.6]; % select the redundant basic force Qx (index ix) ix = 6; % define basic forces of primary structure Qi (index ii) ii = 1:5; % extract matrix Bi (equilibrium matrix of primary structure) Bi = Bf(:,ii); % check if primary structure is stable disp(['the rank of the equilibrium matrix of the primary structure is ' num2str(rank(Bi))]) the rank of the equilibrium matrix of the primary structure is 5 % define loading Pf = [0 ; 0 ; 0 ; 15 ; 0]; % determine basic element forces under the applied loading: particular solution with subscript p Qp(ii,1) = Bi\Pf; Qp(ix,1) = 0; % extract matrix Bx Bx = Bf(:,ix); % determine force influence matrix for redundant basic forces % set up only one column since NOS=1 Bbarx(ii,1) = -Bi\Bx(:,1); Bbarx(ix,1) = 1;

Page 83

% display Qp, Bxbar Qp Bbarx Qp = 15.0000 11.2500 0 15.0000 -18.7500 0 Bbarx = -0.8000 -0.6000 -0.6000 -0.8000 1.0000 1.0000 % use FEDEASLab functions % after defining equilibrium matrix Bf use it as first argument to the function ForceInfl_matrix to obtain the force influence matrices of the primary structure for the applied forces and redundants [Bbari Bbarx] = ForceInfl_matrix (Bf); Qp = Bbari*Pf Bbarx Qp = 15.0000 11.2500 0 15.0000 -18.7500 0 Bbarx = -0.8000 -0.6000 -0.6000 -0.8000 1.0000 1.0000

Page 84

CE220 - Theory of Structures Ex 6 - Equilibrium for indeterminate frame © Prof. Filip C. Filippou, 2000

Example 6(r) - Equilibrium equations for statically indeterminate braced frame

Structural Model Geometry

a

b c

d

4 4

6

La 6:=

Lb 4:=

Lc 4:=

Ld 82 62+:=

Step 1: Equilibrium equations- degree of static indeterminacy

Dof and basic force numbering without axial forces and corresponding equations

dof 1

dof 2 dof 3dof 4

a

b c

d

4 4

6

Equilibrium equations at free dofs

P1Q1 Q2+

60.8 Q6⋅+=

P2 Q2 Q3+=

P3Q3 Q4+

4−

Q5

4+=

P4 Q4 Q5+=

1Q

2Q 5Q3Q4Q

6Q

static matrix Bf

Bf

16

0

0

0

16

1

0

0

0

1

14

−

0

0

0

14

−

1

0

0

14

1

0.8

0

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

Page 85


There are 6 basic element forces and 4 equations of equilibrium at free degrees of freedom.The structural model has a degree of static indeterminacy of 2 (NOS=2).

Step 2: Selection of redundant(s), primary structure

Select NOS basic forces as unknowns of the problem. These are known as redundant basic forces orsimply redundants. Make sure that the structure is statically determinate and stable without the NOSredundants. In the case at hand we select the basic forces Q1 and Q4 as redundants. The structure thatremains when the redundants are set equal to zero is known as primary structure.

Given load case

a

b c

d

4 4

6

3020

Step 3: Determine basic forces Q for given loading with redundant forces equal to zero

set Q1 0= Q4 0=

We solve the following equations by hand:

Q6300.8

804.8

+=30Q2

60.8 Q6⋅+=

0 Q2 Q3+= Q2 80−= Qp

0

80−

80

0

0

300.8

804.8

+

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

20−Q3

4−

Q5

4+= Q3 80=

0 Q5= Q5 0=

for automatic solution with rref function in Matlab see Example_6.m

check solution Bf Qp⋅

30

0

20−

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

= ok!

Page 86


Step 4: Determine basic forces Q for a unit value of each redundant without any applied forces.

set Q1 1= Q4 0=


Q61

4.8−=0

1 Q2+

60.8 Q6⋅+=

0 Q2 Q3+= Q2 0= Bbarx1⟨ ⟩

1

0

0

0

0

524

−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

0Q3

4−

Q5

4+= Q3 0=

0 Q5= Q5 0=

set Q1 0= Q4 1=


Q62

4.8−=0

Q2

60.8 Q6⋅+=

Bbarx2⟨ ⟩

0

2

2−

1

1−

512

−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=0 Q2 Q3+= Q2 2=

0Q3 1+

4−

Q5

4+= Q3 2−=

0 1 Q5+= Q5 1−=

check homogeneous solution Bf Bbarx⋅

0

0

0

0

0

0

0

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

=

thus, we have found two independent homogeneous solutions (self-stress) states which span what is knownin linear algebra as the nullspace of the static matrix Bf

In reality, of course, the redundants are free parameters that can assume any value (not just the value 1!).Consequently, the general solution of the equilibrium equations is a two parameter expression of the form

Q

0

80−

80

0

0

2604.8

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

1

0

0

0

0

524

−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

Q1⋅+

0

2

2−

1

1−

512

−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

Q4⋅+= or, in compact form Q Qp Bbarx Qx⋅+=

Page 87


We can see that the matrix Bbarx has NOS columns, one for each redundant. We therefore state that thefinal stress state is the linear superposition of a particular solution Qp for the given loading, and NOSself-stress states, each of which is expressed by a column of matrix Bbarx

It is important to be able to establish the particular and the homogeneous solutions of the equilibriumequations of a statically indeterminate structure by inspection. This is a much faster approach for mostsmall structures and yields significant insight into the behavior of the structure.

Consider first the applied loading on the primary structure. In the following figure we have inserted momentreleases at the locations of the redundants to form the primary structure. We have also replaced theelement loading by equivalent nodal forces, which give the same basic forces Q (see Example 3). Wedecompose the loading into the horizontal force and the vertical force, as shown.

30

4 4

6

20

applied loading on primary structure

4 4

6

2030

4 4

6

Decomposition of applied loading: (a) horizontal force (b) vertical force

Page 88


Equilibrium solution for applied loading (particular solution)

4 4

6

2080

80

a

b c80

80/6

80/6

80/6/0.8=16.67

30

4 4

630/0.8=37.5a

b c

no bending moments arise under horizontal force!

Why? The bending moment diagram is piecewiselinear in the absence of element loading. It changesslope ("kink") only when the shear force changes,which in turn requires that a concentrated force actsnormal to the beam axis. Looking at elements b andc no moment diagram is possible with a zero valueat the moment release. Thus, the moments are zero.If so, then the moments in column element a are alsozero and the solution is obtained.

there can be no shear transfer in element cbecause of zero moment at each end; thus,the applied force of 20 gets transferred toelement b producing a moment of 80 units atthe girder-column connection. The shear forcein the column element is equilibrated by atensile force in the brace.

Equilibrium solution for unit value of first redundant Equilibrium solution for unit value of second redundant

4 4

6

Q4 = 1

12

2

2/(6x0.8)=5/12

a

b c

4 4

6

Q1 = 1

1/(6x0.8)=5/24a

b c

gray circle = location with known basic force valuethe shear force needs to remain constant acrossgirder elements b and c, since no concentratedforce acts anywhere on the girder; this leads to avalue of 2 at the girder-column connection if thevalue at girder midspan is 1. The shear force incolumn element a is resisted by a compressiveforce in the brace element d

no bending moments in elements b and c forthe same reason as for particular solution; theshear force in column element a is resisted bya compressive force in the brace element d

Page 89


We can now use the remaining 3 equilibrium equations involving the axial forces in elements a, b and cto determine the axial forces in terms of the basic forces Q (i.e. eventually in terms of the applied loadingand the two redundant basic forces)

Basic axial force in element a (call it Q7)

1Q

2Q 5Q3Q4Q

6Q

0 Q7Q3 Q4+

4+=

i.e. Q7Q3 Q4+

4−=

Basic axial force in element b (call it Q8)

30Q1 Q2+

6Q8−=

i.e. Q8 30−Q1 Q2+

6+=

Support reactions and global equilibrium Basic axial force in element c (call it Q9)

a

b c

d

4 4

6

R1R2

R3

R4

3020 P7 Q8 Q9−= Q9 Q8=

Using the equilibrium equations at the restrained dof's we can determine the support reactions in termsof the basic forces Q (i.e. eventually in terms of the applied loading and the two redundant basic forces)

for ease of identification we call the support reactions by R and number them in sequence starting with 1

R1Q1 Q2+

6− 0.8 Q6⋅−=

bypassing the axial force in the columnand using the shear force in element bR2

Q3 Q4+

40.6 Q6⋅−=

R3 Q1=

R4Q5

4− 0.6 Q6⋅+=

Page 90


Remark: We can use the static matrix Bd for the restrained dofs to set up the direct relation between thesupport reactions and the applied loading plus the redundants. We demonstrate

Bd

16

−

0

1

0

16

−

0

0

0

0

14

0

0

0

14

0

0

0

0

0

14

−

0.8−

0.6−

0

0.6

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

the support reactions of the primary structure due to the applied loading are

Rp Bd Qp⋅:= Rp

30−

12.5−

0

32.5

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

=

the support reactions due to the redundants are Rx Bd Bbarx⋅ Qx⋅=

thus, with Bd Bbarx⋅

0

0.125

1

0.125−

0

0

0

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

= we get R

30−

12.5−

0

32.5

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

0

0.125

1

0.125−

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

Q1⋅+=

and it is easy to check global equilibrium

horizontal force equilibrium R1 30+ 0= ok

vertical force equilibrium R2 R4+ 20− 0= ok

moments about left support R3 30 6⋅− 20 4⋅− R4 8⋅+ 0= ok, because 30− 6⋅ 20 4⋅− 32.5 8⋅+ 0=

Remarks:

(a) this solution demonstrates the utility of matrix notation in establishing direct relationships betweenvariables; in this example the relation between the support reactions and the redundants of the structure

(b) it is interesting that there are no support reactions due to the redundant Q4; can you explain why?

Page 91

Script for Example 6 in CE220 class notes % Equilibrium equations for statically indeterminate braced frame

specify each step of the process with Matlab functions clear all % equilibrium matrix for non-trivial free dof's w/o axial forces in a-c Bf=[1/6 1/6 0 0 0 0.8; 0 1 1 0 0 0 ; 0 0 -1/4 -1/4 1/4 0 ; 0 0 0 1 1 0]; % select redundant basic forces Qx (index ix) ix=[1 4]; % define basic forces of primary structure Qi (index ii) ii=setdiff(1:6,ix); % extract matrix Bi (equilibrium matrix of primary structure) Bi=Bf(:,ii); % check if primary structure is stable disp(['the rank of the equilibrium matrix of the primary structure is ' num2str(rank(Bi))]) the rank of the equilibrium matrix of the primary structure is 4

define loading Pf=[30 ; 0 ; -20 ; 0]; % determine basic element forces of primary structure under the applied loading Qp(ii,1) = Bi\Pf; Qp(ix,1) = [0; 0]; % extract matrix Bx Bx=Bf(:,ix); % determine force influence matrix for redundant basic forces % set up 1st column Bbarx(ii,1) = -Bi\Bx(:,1); Bbarx(ix,1) = [1;0]; % set up 2nd column Bbarx(ii,2) =-Bi\Bx(:,2); Bbarx(ix,2) = [0;1];

Page 92

% display Qp, Bbarx Qp Bbarx Qp = 0 -80.0000 80.0000 0 0 54.1667 Bbarx = 1.0000 0 0 2.0000 0 -2.0000 0 1.0000 0 -1.0000 -0.2083 -0.4167

use FEDEASLab functions % after defining equilibrium matrix Bf use it as first argument to the function ForceInfl_matrix to obtain the force influence matrices of the primary structure for the applied forces and redundants; in order to use specific basic forces as redundants, these need to be last among the columns of the equilibrium matrix; the order is the second argument of ForceInfl_matrix (optional) [Bbari Bbarx] = ForceInfl_matrix (Bf,[ip ii]); Qp = Bbari*Pf Bbarx Qp = 0 -80.0000 80.0000 0 0 54.1667 Bbarx = 1.0000 0 0 2.0000 0 -2.0000 0 1.0000 0 -1.0000 -0.2083 -0.4167

Page 93

Collapse load factor of perfectly plastic structures

Lower bound theorem of plastic analysis: any equilibrium solution that satisfies the plastic conditionresults in a load factor λ which is a lower bound of the collapse load factor λc

Mathematically this is expressed as follows: the collapse load factor is the maximum load factor thatsatisfies the equilibrium equations and the plastic condition. This is a linear programming problem forwhich algorithms exist in numerical analysis toolboxes, like Matlab (optimization toolbox) and Mathcad.

Note that Pref, contains the applied forces to be factored and Pfu the applied forces that remain constant in the equilibrium equations.

We write the above statement in a single set of equations for the unknowns λ and Q.

with solution where c stands for collapse

FEDEASLab function LowerBound_PLAnalysis takes the static matrix for the free dofs, the plasticcapacity array, the reference load vector and the constant load vector and returns the collapse loadfactor and the basic element forces at collapse.

[lamdac,Qc] = LowerBound_PLAnalysis(Bf,Qpl,Pref,Pfu)

Examples 7, 8 and 9 discuss plastic analysis from different standpoints, including the case of partialcollapse.

ref f fu plmax for andcλ λ λ= = + ≤BP Q P Q Q

ref f fu

pl

pl

max 1λ

λ

λ+

−

⎛ ⎞⎡ ⎤ ⎜ ⎟⎣ ⎦

⎝ ⎠⎛ ⎞

⎡ ⎤− =⎜ ⎟⎣ ⎦⎝ ⎠

⎛ ⎞⎡ ⎤ ⎛ ⎞ ⎜ ⎟≤⎜ ⎟⎢ ⎥ ⎜ ⎟− ⎜ ⎟⎣ ⎦ ⎝ ⎠ ⎝ ⎠

0

B

0 I0 I

Q

P PQ

Q

Q Q

c

c

λ⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠Q


Page 94

Complete or partial collapse mode for structure

write the equilibrium equations at collapse in the form

isolating the contribution of the elastic and plastic basic forces on the right hand side, with subscript edenoting the elastic basic forces and subscript p the plastic. The static matrix is split into correspondingsubmatrices.

For perfectly plastic behavior of the structural elements the plastic basic forces are known, as is thecollapse load factor (trial and error or linear programming solution for locating the plastic hinges anddetermining the collapse load factor). We can then determine the elastic basic forces to satisfythe equilibrium equations

The properties of the static matrix Be determine whether a unique solution for the elastic basic forces is possible or not from the above system of equilibrium equations. It is useful to recall that the static matrix Bf has NOS more columns than rows. Thus, if NOS+1 plastic hinges form, meaning that we remove NOS+1 columns in Bp from Bf, then the remaining static matrix Be has a number of rows one greater than the number of columns. The rank of the matrix is now one less than the number of rows indicating that the structure with NOS+1 plastic hinges is unstable. We say that the structure has turned into a mechanism with a single independent dof . However, a unique solution for the elastic basic forces is possible, because the vector on the left hand side of the above equilibrium equations lies in the column space of the static matrix Be, i.e. it is a linear combination of the columns of Be with Qce being the combination factors (see Strang, 3rd edition, pp. 65-66). This case represents a complete collapse mode.

If plastic hinges form at less than NOS+1 locations, then the static matrix Be has more unknowns than available linearly independent equations (recall that the rank of the matrix is one less than the number of rows). In such case there are multiple solutions for the elastic basic forces and these can be obtained in systematic fashion as we did for statically indeterminate structures in Examples 5 and 6. We demonstrate this in Example 9. We speak of a partial collapse mode, for which there is no unique solution of the elastic basic forces.

ref e ce p cpcλ = +P Q QB B

ref p cp e cecλ − =P Q QB B


Page 95

CE220 - Theory of Structures Ex. 7 - Plastic analysis of simple truss © Prof. Filip C. Filippou, 2000

Example 7(r) - Plastic analysis of simple trussObjectives:

(a) use equilibrium equations to determine collapse load factor of simple truss by trial and error

In the following example we illustrate the application of the equilibrium equations to the plastic analysisand collapse load determination of a simple truss. The process is the same for any structure.

a

b c

λ(10)

8

6

λ(10)

The axial capacity in tension or compression of the truss elements of the structural model in the figure isgiven. It is equal to 10 units of force.

The loading here is specified more generally than what we have seen so far. The actual forces applied onthe structure consist of reference force values, which define the load pattern. These are arranged in areference force vector Pref. This force vector is then multiplied by a load factor λ to give the applied forcevector Pf.

For the example at hand this means that Pref10

10⎛⎜⎝

⎞⎟⎠

:= and Pf λ Pref⋅=

Note that the load factor is assumed to be greater than zero, i.e. load reversal is not allowed.

The equilibrium equations are straightforward in this case:

P1 Q1 0.8Q2+= λ 10⋅ Q1 0.8Q2+=

P2 0.6 Q2⋅ Q3+= λ 10⋅ 0.6 Q2⋅ Q3+=

There are two equilibrium equations and 3 unknown basic element forces. Thus, the degree of staticindeterminacy is NOS=1. With the load factor λ on the left hand side of the equations unknown thatthere are 4 unknowns for two available equations (the excess number of unknowns is thus NOS+1, i.e.the number of redundants of the structural model + the load factor).

We now ask the following question: what is the largest load factor λ that the structure can resist? Lookingat the above equilibrium equations it is easy to conclude that the largest load factor λ results when thebasic forces Q1 through Q3 are as large as possible. These basic forces cannot exceed, however, the axialcapacity of the element. Thus, there is a limit to how large λ can become.

Page 96


In fact, from the sign of the applied forces and the sign of the basic forces we conclude that the largest l willbe attained for positive (tensile) basic forces Q1 through Q3

Since there are 2 equilibrium equations in 4 unknowns we need to "guess" which two basic forces reach theaxial capacity. We can do this with insight, or, we can simply apply trial and error. Since there are only 3alternatives for the case at hand, we apply the latter method which yields some interesting information.

Our first guess is that the largest collapse load factor λ results when truss elements a and b reach theaxial capacity. With this assumption we get

set Q1 10= Q2 10=

from the first equation we determine λ 1.8=

substituting these into the second equation we obtain Q3 12=

Since no basic force can exceed the axial capacity, this is not a feasible solution.

The second guess is that the largest collapse load factor λ results when truss elements a and c reach theaxial capacity. With this assumption we get

set Q1 10= Q3 10=

λ 10⋅ 10 0.8Q2+=The two equations read

λ 10⋅ 0.6 Q2⋅ 10+=

by inspection we conclude that the solution is λ 1.0= Q2 0=

The third and final guess is that that the largest collapse load factor λ results when truss elements b and creach the axial capacity. With this assumption we get

set Q2 10= Q3 10=

from the second equation we determine λ 1.6=

substituting these values into the first equation we get Q1 8= which is smaller than the axial capacity

Definition:

(a) the plastic condition expresses the fact that the basic forces cannot exceed the axial capacity

(b) a feasible solution is any λ and set of Q's that satisfies the equilibrium equations and the plastic condition

Conclusion: There are two feasible solutions for the problem at hand. The largest load factor λcorresponds to the case when truss elements b and c reach the axial capacity. We call this value thecollapse load factor λc of the structure for the given loading. For this example it is equal to 1.6.

the corresponding axial forces are Q1 8= Q2 10= Q3 10= for λc 1.6:=

Page 97


We denote the basic forces at collapse with Qc and can confirm once more the equilibrium equations

Qc

8

10

10

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

:= Bf1

0

0.8

0.6

0

1⎛⎜⎝

⎞⎟⎠

:= Bf Qc⋅16

16⎛⎜⎝

⎞⎟⎠

= λc Pref⋅16

16⎛⎜⎝

⎞⎟⎠

= good!

In this problem we were able to check all possible solutions and then select the solution with the largestcollapse load factor among the feasible solutions. Thus, we are sure to have found the solution. In morecomplex structures it is not possible to check all feasible solutions. Thus, an additional condition becomesnecessary to ensure that we have reached the unique solution of the problem. We will discuss this in duetime. For now, we wish to formulate the problem that we solved in a mathematical way, so that we can usematrix methods to solve it for structural models of any size.

Problem formulation: determine the largest load factor λ and the corresponding set of basic forces Q thatsatisfy the equilibrium equations and the plastic condition. We realize that the problem involves themaximization of a function (the collapse load factor), under equality (equilibrium) and inequality constraints(plastic condition). The function and the constraints are linear in the unknowns of the problem. Thisconstitutes a linear programming problem for the solution of which there exist powerful algorithms. Mathcadand Matlab have such algorithms and we will demonstrate these in the following example.

For now, we give a name to the problem of finding the largest load factor λ that satisfies the equilibriumequations and the plastic condition: it is called the lower bound theorem of plastic analysis. The name stemsfrom the fact that any λ that satisfies the equilibrium equations and the plastic condition is a lower bound ofthe collapse load factor λc. Thus, the collapse load factor λc is the maximum λ that satisfies the equilibriumequations and the plastic condition, as we have already stated.

Note: we have not used any information about cross section or material properties for the solution of theproblem. In reality, the axial capacity is area * yield strength, so that both geometric and materialproperties are involved in its determination. Thus, the design of the structure is required before itscollapse load factor can be determined under a given loading.

Page 98

CE220 - Theory of Structures Example 8 - Plastic Analysis © Prof. Filip C. Filippou, 2000

Example 8 - Plastic Analysis of Portal Frame

Objectives:

(a) use equilibrium equations to determine collapse load factor of portal frame by hand calculations(b) determine plastic flexural hinge location at collapse (c) formulate the problem as linear programming problem to obtain solution by computer(d) consider the effect of factored or unfactored distributed element load (optional)

4 4

5

50

a

b c

d

1

2 34

5

30

Given:

(a) load pattern(b) plastic flexural capacity of girder, i.e.elements b, and c is 120 units(c) plastic flexural capacity of columns,i.e. elements a and c is 150 units(d) axial capacity of all elements is large(e) effect of axial force on flexuralcapacity is negligible

We avoid writing the equations of equilibrium involving longitudinal basic forces, so as to reduce theproblem size. We note that the collapse load factor is not affected, because of (d). Without the longitudinalbasic forces there are 8 basic forces (2 for each element) and 5 corresponding equilibrium equations.

4 4

5

12

34 5

1Q

2Q

3Q 4Q 5Q 6Q

7Q

8Q

The equilibrium equations are:

P1Q1 Q2+

5

Q7 Q8+

5+=

P2 Q2 Q3+=

P3Q3 Q4+

4−

Q5 Q6+

4+=

P4 Q4 Q5+=

P5 Q6 Q7+=

the reference applied force vector for the given loading is Pref

30

0

50−

0

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=

Page 99


a

b c

d150 150

120 120The following flexural capacities are givenunder the assumption that the positive andnegative capacity is the same.

We plan to solve this problem in trial and error fashion before using linear programming functions to getthe solution by computer. We note first that there are 5 equilibrium equations for 8 unknown basic forces,meaning that the degree of static indeterminacy NOS is three, i.e. NOS=3. We also note that there are 5possible plastic flexural hinge locations (we denote the locations by the corresponding subscript of thebasic force): at the base of each column, i.e. at 1 and 8, at the girder ends at the girder-columnconnection, i.e. at 3 and 6, and at the girder midspan, i.e. either at 4 or 5. Note that no plastic flexuralhinge is possible at the top of the columns, since their flexural capacity is larger than that of the girder("strong column" design). For each of these locations a positive or negative value is possible, i.e. there are5x2 = 10 possible outcomes, at least as long as we do not use structural insight to eliminate unlikelyoutcomes. For a complete collapse (more in the following example) we need NOS+1 = 4 plastic flexuralhinges for getting a solution to the problem, and so there are 210 possible combinations!!

combin 10 4,( ) 210= this is indeed a very large number for such a small problem, imagine what wouldhappen in a multi-story frame!

We can now use the signs of the terms in the equilibrium equations to reduce the number of possibleoutcomes to an extremely small number (many people call this process "insight into structural behavior").

we first specialize the equilibriumequations for the given loading λ 30⋅

Q1 Q2+

5

Q7 Q8+

5+= (1)

0 Q2 Q3+= (2)

λ 50−( )⋅Q3 Q4+

4−

Q5 Q6+

4+= (3)

0 Q4 Q5+= (4)

0 Q6 Q7+= (5)

and then take advantage of the fact that plastic hinges can occur at 1, 3, 4, 6 and 8 to retain only the basicforces at these locations in equations (1) and (2) after making use of moment equations (2), (4) and (5).

for largest λ Q1, and Q8 should be positive,Q3 and Q6 should be negative

We get: λ 30⋅Q1 Q3−

5

Q6− Q8+

5+= (1*)

for largest λ Q3, and Q4 should be positive,Q6 should be negative

λ− 50⋅Q3 Q4+

4−

Q4− Q6+

4+= (3*)

Page 100


Thus, we have reached some very specific conclusions regarding the signs of the basic forces at the likelyplastic hinge locations. The only contradiction regards Q3 which according to (1*) should be negative, andaccording to (3*) should be positive. This is a strong indication that no plastic flexural hinge is likely tooccur at 3, as long as neither applied force in the above equations dominates. Why? Because thepreceding discussion indicates the opposite effect that the horizontal force has on Q3 relative to thevertical force. Only when the horizontal force is very large a hinge will form at 3 with negative value. Onlywhen the vertical force at girder midspan is very large, a hinge will form at 3 with positive value. For theproblem at hand, we will assume first than neither force is very "large" (note that we have no criterion whatthis means, so we need to be cautious). Then, we will check the other possibilities just to be sure.

Note: studying the deformed shape of the structure gives good insight of what happens, but not as goodas the equilibrium equations. Let us see why. Here are the bending moments under each applied forceseparately (draw an approximate deformed shape for each load case)

We see that the moments at locations 6 and 8 act in the same sense under these two load cases. Thus,plastic hinges are a sure bet at these two locations with the indicated signs (-ve for 6 and +ve for 8). Thenext likely hinge location is at 4 and then it is not clear whether 1 or 3 is the most critical. This depends onthe relative magnitude of the horizontal and vertical force, as we have already discussed.

After this preliminary discussion we have one very strong candidate for the collapse load factor (instead of210!). It is the following case

plastic hinges at 1, 4, 6 and 8 Q1 150= Q4 120= Q6 120−= Q8 150=

We substitute these values into the equilibrium equations and obtain

λ 30⋅Q3

5−

4205

+=λ 30⋅150 Q3−

5120 150+

5+=

λ− 50⋅Q3 120+

4−

120− 120−4

+= λ 50⋅Q3

43604

+=

We use the Given Block in Mathcad to solve the two simultaneous equations

Page 101


with initial guess values λ 0:= Q3 0:=Given

λ 30⋅Q3

5−

4205

+=

λ 50⋅Q3

43604

+=

λ

Q3

⎛⎜⎝

⎞⎟⎠

Find λ Q3,( ):= λ 2.229= Q3 85.714=

we have determined a feasible solution, since Q3 is smaller than the flexural capacity at that location. Is thisthe largest load factor, i.e. is it the collapse load factor?

Let us see what happens if we assume that a plastic hinge arises at 3 either because the horizontal force is"large", i.e. using eq (1*), or because the vertical force is "large", i.e. using eq (3*).

With eq (1*) we assume Q1 150:= Q3 120−:= Q6 120−:= Q8 150:=

λ130

Q1 Q3−

5

Q6− Q8+

5+

⎛⎜⎝

⎞⎟⎠

⋅:= λ 3.6=

the load factor is now much larger; is this a feasible solution?

from eq (3*) we get Q4 2 λ 50⋅Q3

4−

Q6

4+

⎛⎜⎝

⎞⎟⎠

:= Q4 360= not feasible since larger than 120!

With eq (3*) we assume Q3 120:= Q4 120:= Q6 120−:= note that only 3 values are req'd

λ150−

Q3 Q4+

4−

Q4− Q6+

4+

⎛⎜⎝

⎞⎟⎠

⋅:= λ 2.4=

the load factor is again larger; is this a feasible solution?

we substitute in equation (1*) to obtain

λ 30⋅Q1 Q3−

5

Q6− Q8+

5+= Q1 Q8+ 150 λ⋅ Q3+ Q6+= i.e. Q1 Q8+ 360=

it is clear from the last equation that the solution is not feasible, since there is no way to select values for Q1and Q8 that satisfy the horizontal force equilibrium without exceeding the flexural capacity of the column.

Thus, it is clear that there can be no plastic hinge at 3. Is there another solution? It does not seem possiblein this case, since we do not seem to have further equilibrium solutions. Nonetheless, we will be able toanswer this definitively, once we look at the collapse mechanism later in the course.

Page 102


Automatic solution with linear programming

Let us now look at a computer solution. For this we need to write the static matrix of the structural model

so that the equilibrium equations become

Bf

0.2

0

0

0

0

0.2

1

0

0

0

0

1

0.25−

0

0

0

0

0.25−

1

0

0

0

0.25

1

0

0

0

0.25

0

1

0.2

0

0

0

1

0.2

0

0

0

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:= Pf Bf Q⋅=

with Pf λ Pref⋅=

We specify the plastic moment capacities of the elements in a vector with the same numbering as the basicforces vector. This way we can write the inequality in very compact form.

plastic moment capacities Qpl 150 150 120 120 120 120 150 150( )T:=

The determination of the collapse load factor of any structure can be formulated as a linearprogramming problem. To this end we define the function to maximize subject to given equality andinequality constraints. In a structural problem the unknowns are the load factor and the basic forces.The function to maximize is the load factor λ. The equality constraint are the equilibrium equations, andthe inequality constraints are the yield conditions stating that the basic forces cannot exceed the plasticcapacity of the member (positive or negative).

Mathcad has a built-in linear programming function called Maximize (there is also Minimize). We definethe function for our purposes and supply guess values for the start of the search.

function to maximize g λ Q,( ) λ:=

with initial guess values λ 0:= Q8 0:= note that this statement sets to zero the entire array

We then specify the equality and inequality constraints in a "Given" block, as follows

Given

λ Pref⋅ Bf Q⋅= equality constraint (equilibrium equations)

Q Qpl≤inequality constraints (plastic condition)

Q− Qpl≤

Note that we need to specify both negative and positive capacities,which affords the possibility of specifying different capacity values

We invoke the Maximize function with the name ofthe function and the variables. Mathcad returnsthe solution in a nested array, here called Sol.

Sol Maximize g λ, Q,( ):= Sol

2.229

150

85.714−

85.714

120

120−

120−

120

150

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

=

Page 103


We can extract the collapse load factor as the first element of the nested array Sol.

Collapse load factor is: λc Sol1:= λc 2.229=

and the basic forces at collapse as the second element of the nested array Sol.

Qc Sol2:=

QcT 150 85.714− 85.714 120 120− 120− 120 150( )=

We confirm that the plastic hinges have formed at 1, 4(5), 6 and 8. Substituting Q into the equilibriumequations we should be able to confirm the collapse load factor.

λc Pref⋅

66.857

0

111.429−

0

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

= Bf Qc⋅

66.857

0

111.429−

0

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

= confirmed

We will show later that the collapse mechanism of this structure under the given loading is

2.229*502.229*30

Note: the plastic hinge can be either at 4 or 5; this creates an ambiguity that is, however, not relevant

Page 104


After determining the flexural basic element forces we use element equilibrium to determine thetransverse (shear) forces and then use node equilibrium at the free dofs to determine the axial forces.It is important to recall that the applied forces should be multiplied by load factor λ.The following figure illustrates the determination of the shear forces.

4 4

5

150

85.71120

120

150

54

54

60 6051.43 51.4312.86

12.86

6051.43

2.229*505412.86

2.229*30

We conclude that the axial force in the left column is 51.43 (C), and in the right column it is 60 (C).The axial force in the girder is 54 (C).

The following figure shows the support reactions

4 4

5

2.229*502.229*30

5412.86150 150

6051.43

Check of global equilibrium

sum of forces in X

51.43 2.229 50⋅− 60+ 0.02−=

sum of forces in Y

2.229 30⋅ 12.86− 54− 0.01=

moment equilibrium about left support 150 2.229 30⋅ 5⋅− 2.229 50⋅ 4⋅− 150+ 60 8⋅+ 0.15−=

Page 105


Load Case: Factored distributed load in left half of girder (optional)

We change the loading to a distributed load in the left half of the girder. The magnitude of the distributedload is such that the equivalent vertical force at node 3 remains equal to 50. Consequently, the distributedelement load value is 50/2 = 25 units of load/per unit length. The force vectors in the equilibrium equationschange, but the end effect is the same, as long as it assumed that both the applied nodal force and thedistributed element load are factored. Consequently, the collapse load factor does not change, as thefollowing calculations reveal.

Pref

30

0

0

0

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

0

0

50

0

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

−:=distributed load effect moved from right hand sideof equilibrium equations

Given

λ Pref⋅ Bf Q⋅= equality constraint

Q Qpl≤inequality constraints

Q− Qpl≤

We invoke the Maximize function with the name of the function and the variables. Mathcad returnsthe solution in a nested array, here called Sol.

Sol Maximize f λ, Q,( ):= f Sol

2.229

150

85.714−

85.714

120

120−

120−

120

150

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

=


Upper bound load factor is: λu Sol1:= λu 2.229=

The same answer is also obtained with Matlab. See Example_8.m

Page 106


We show the flexural basic forces in element free bodies and determine the shear forces.

2.229*30 2.229*25=55.725

4 4

5

150

85.71120

120

150

54

54

6012.86

12.86

5412.86

60 60

6060

162.88

We determine the maximum moment at the left girder element. It is

602

2 λu⋅ 25⋅120+ 152.308=

This clearly exceeds the plastic moment capacity. Thus, the collapse load factor is not correct. If we scaledown, the left and the right hand side of the equilibrium equations, so that the maximum moment justreaches the plastic moment capacity we have a lower bound that satisfies equilibrium and the plasticcondition. It is

λ l λu120

152.308⋅:= λ l 1.756=

From this we can only conclude that the actual collapse load factor λc is between 1.756 and 2.229. Sincethe difference between the two values is significant, we need to repeat the analysis by inserting anothernode at the expected location of the maximum moment. Instead of the exact location of maximum moment(which will change after inserting the node!), we place the extra node at midspan of element b.We perform this analysis in Matlab (see Example_8.m)

Page 107


Load Case: Constant distributed load in left half of girder (optional)

How does the above result change if the distributed element load is not factored? In this case thesystem of equilibrium equations becomes

Pref

30

0

0

0

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:= Pfw

0

0

50

0

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=

Given

λ Pref⋅ Bf Q⋅ Pfw+= equality constraint


Q− Qpl≤



3.6

150

120

120−

100

100−

120−

120

150

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

=


Upper bound for load factor is: λu Sol1:= λu 3.6=

The result is now very different, of course and the plastic hinge locations have changed.

Page 108


We show the flexural basic forces in element free bodies and determine the shear forces.

3.6*30 25

4 4

5

150

120100

120

150

54

54

54

55 55

55

554554

54

54

55

We determine the maximum moment at the left girder element. It is

552

2 25⋅100+ 160.5=

This exceeds again the plastic moment capacity. Can you obtain an estimate of the lower bound?

A more exact analysis is obtained with Matlab, see Example_8.m

Page 109

Script for Example 8a in CE220 class notes % Plastic analysis of one story portal frame (consult pp. 84-94) solution with Matlab functions and FEDEASLab function LowerBound_PLAnalysis % Clear workspace memory and initialize global variables CleanStart

set up equilibrium matrix w/o axial forces in frame members Bf = [0.2 0.2 0 0 0 0 0.2 0.2; 0 1 1 0 0 0 0 0 ; 0 0 -0.25 -0.25 0.25 0.25 0 0 ; 0 0 0 1 1 0 0 0 ; 0 0 0 0 0 1 1 0 ];

specify applied force vector at free dof's for load case of nodal forces Pref = [ 30; 0; -50; 0; 0];

define plastic flexural capacity of elements in a column vector (2 entries per element) Qpl = [ 150 150 120 120 120 120 150 150]';

call function for lower bound plastic analysis in FEDEASLab [lambdac Qc] = LowerBound_PLAnalysis(Bf,Qpl,Pref); disp(' The collapse load factor for nodal force at girder midspan is'); disp(lambdac); disp(' The basic forces Q at collapse are'); disp(Qc); Optimization terminated. The collapse load factor for nodal force at girder midspan is 2.2286 The basic forces Q at collapse are 150.0000 -85.7143 85.7143 120.0000 -120.0000 -120.0000 120.0000 150.0000

Page 110

load case of distributed element load in left half of girder (factored) % consult pp. 91-92 in class notes: assume that a plastic hinge may form at midspan of element b % redefine equilibrium matrix, applied force vector, and plastic capacity vector Bf = [0.2 0.2 0 0 0 0 0 0 0.2 0.2; 0 1 1 0 0 0 0 0 0 0; 0 0 -0.5 -0.5 0.5 0.5 0 0 0 0; 0 0 0 1 1 0 0 0 0 0; 0 0 0 0 -0.5 -0.5 0.25 0.25 0 0; 0 0 0 0 0 1 1 0 0 0; 0 0 0 0 0 0 0 1 1 0]; Pref = [ 30; 0; -50; 0; -25; 0; 0]; Qpl = [ 150 150 120 120 120 120 120 120 150 150]';

call function for lower bound plastic analysis in FEDEASLab [lambdac Qc] = LowerBound_PLAnalysis(Bf,Qpl,Pref); disp(' The collapse load factor for factored distributed load is'); disp(lambdac); disp(' The basic forces Q at collapse are'); disp(Qc); Optimization terminated. The collapse load factor for factored distributed load is 2.1882 The basic forces Q at collapse are 150.0000 -91.7647 91.7647 120.0000 -120.0000 112.9412 -112.9412 -120.0000 120.0000 150.0000

Page 111

load case of distributed element load in left half of girder (not factored) % redefine applied force vector only Pref = [ 30; 0; 0; 0; 0; 0; 0]; Pfw =[ 0; 0; 50; 0; 25; 0; 0];

call function for lower bound plastic analysis in FEDEASLab [lambdac Qc] = LowerBound_PLAnalysis(Bf,Qpl,Pref,Pfw); disp(' The collapse load factor for unfactored distributed load is'); disp(lambdac); disp(' The basic forces Q at collapse are'); disp(Qc); Optimization terminated. The collapse load factor for unfactored distributed load is 3.2444 The basic forces Q at collapse are 150.0000 66.6667 -66.6667 120.0000 -120.0000 73.3333 -73.3333 -120.0000 120.0000 150.0000

Page 112

Script for Example 8b in CE220 class notes % Plastic analysis of one story portal frame (consult pp. 84-94) solution entirely with FEDEASLab functions

Clear workspace memory and initialize global variables CleanStart % define model geometry XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 0 5]; % second node, etc XYZ(3,:) = [ 4 5]; % XYZ(4,:) = [ 8 5]; % XYZ(5,:) = [ 8 0]; % % element connectivity array CON { 1} = [ 1 2]; CON { 2} = [ 2 3]; CON { 3} = [ 3 4]; CON { 4} = [ 4 5]; % boundary conditions (1 = restrained, 0 = free) BOUN(1,:) = [ 1 1 1]; BOUN(5,:) = [ 1 1 1]; % specify element type ne = length(CON); % number of elements [ElemName{1:ne}] = deal('2dFrm'); % 2d frame element


1

2 3 4

5

1

2 3

4

define plastic flexural capacity of elements in a column vector (3 entries per element) % axial force capacity is "very large" Qpl = [1e5 150 150 1e5 120 120 1e5 120 120 1e5 150 150]';

Page 113


specify loading Pe(2,1) = 30; % force at node 2 in direction X Pe(3,2) = -50; % force at node 3 in direction Y % generate data object Loading Loading = Create_Loading(Model,Pe); % extract applied force vector Pf at free dofs from Loading field Pref Pref = Loading.Pref;

call function for lower bound plastic analysis in FEDEASLab [lambdac Qc] = LowerBound_PLAnalysis(Bf,Qpl,Pref); disp(' The collapse load factor for nodal force at girder midspan is'); disp(lambdac); disp(' The basic forces Q at collapse are'); disp(Qc); % open window and plot moment diagram Create_Window(0.80,0.80); Plot_Model (Model); Plot_2dMomntDistr (Model,[],Qc);

% open window and show plastic hinge locations Create_Window(0.80,0.80); Plot_Model (Model); Plot_PlasticHinges (Model,Qpl,Qc);

Page 114

% we can also show the plastic hinge locations in the same window as the moment diagram Create_Window(0.80,0.80); Plot_Model (Model); Plot_2dMomntDistr (Model,[],Qc); Plot_PlasticHinges (Model,Qpl,Qc); Optimization terminated. The collapse load factor for nodal force at girder midspan is 2.2286 The basic forces Q at collapse are -51.4286 150.0000 -85.7143 -54.0000 85.7143 120.0000 -54.0000 -120.0000 -120.0000 -60.0000 120.0000 150.0000

Page 115

CE220 - Theory of Structures Example 9 - Partial Collapse © Prof. Filip C. Filippou, 2000

Example 9 - Plastic Analysis of Braced Frame with Partial Collapse

Objectives:

(a) use equilibrium equations to determine collapse load factor of portal frame by linear programming(b) study basic force distribution for partial collapse

a b

c

d e

12 3

4

f

56

40

30

6 6

8

Given:

(a) load pattern(b) plastic flexural capacity of horizontalelements a, b, d and e is 100 units(c) plastic flexural capacity of verticalelement c is 150 units(d) axial capacity of brace element f is 20units(e) elements a through e have very largeaxial capacity(f) effect of axial force on flexural capacityof any element is negligible

31

2

4

1Q 2Q 3Q4Q

5Q6Q

7Q

8Q

Global dof numbering (eq equations) without those affected by longitudinal basic forces in elements athrough e. The basic forces are also numbered accordingly.

The equilibrium equations are: P1Q1 Q2+

6−

Q3

6+

Q6

6−

Q7

6+ 0.8 Q8⋅+=

P2 Q2 Q3+ Q4+=

P3Q4 Q5+

80.6 Q8⋅−=

P4 Q5 Q6+ Q7+=

Page 116


which can also be written in matrix form as Pf Bf Q⋅= with

the applied force vector Pf resultsfrom a reference force vector Pref(load pattern) multiplied with aload factor λ

Bf

16

−

0

0

0

16

−

1

0

0

16

1

0

0

0

1

18

0

0

0

18

1

16

−

0

0

1

16

0

0

1

0.8

0

0.6−

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:= Pref

40−

0

30

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:=

Define the plastic moment capacities of the member in an array.

plastic moment capacities Qpl

100

100

100

150

150

100

100

20

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

The determination of the collapse load factor of any structure can be formulated as a linear programmingproblem. To this end we define the function to maximize subject to given equality and inequality constraints.In a structural problem the unknowns are the load factor and the basic forces. The equality constraint isthe set of equilibrium equations, and the inequality constraints are the so-called plastic conditions,which state that the basic forces should not exceed the plastic capacity of the member (positive or negative).

Mathcad has a built-in linear programming function called Maximize (there is also Minimize). We definethe function for our purposes and supply guess values for the start of the search.

function to maximize g λ Q,( ) λ:= with initial guess values λ 0:=

Q8 0:= note that this statementsets to zero the entire array


Given

λ Pref⋅ Bf Q⋅= equality constraint


Q− Qpl≤

Page 117




1.442

100

0

100−

100

150

50−

100−

20−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

=


Collapse load factor is: λc Sol1:= λc 1.442=

and the basic forces at collapse as the second element of the nested array Sol.

Qc Sol2:=

QcT 100 0 100− 100 150 50− 100− 20−( )=

Interestingly, Matlab's linear programming function returns a different set of basic forces for the solution.In both cases, however, the collapse load factor λc is the same. The collapse load factor is uniquewhile the solution for the basic forces is not! Why is this happening?

Let us first list the solution by Matlab: Qc.matlab 100 50− 100− 150 100 0 100− 20−( )T:=

a b

c

d e

12 3

4

f

56

40

30

6 6

8

Page 118


We note that both solutions agree at 1, 3, 7 and 8, and that the plastic capacities are reached at theselocations. We conclude, therefore, that these plastic hinge locations are unique. The rest of the basicforces is not unique. What does this mean?

(a) for a unique solution of the basic forces we require plastic hinges to form at NOS+1 locations,converting the equilibrium problem to a unique solution (NOS basic forces more than available equations +collapse load factor λc require NOS+1 values among the unknowns to be given; these values are thebasic forces at the plastic hinge locations).

(b) if plastic hinges form at less than NOS+1 locations, we speak of a partial collapse. In the case of theexample we have NOS=4 and 4 plastic hinges at collapse. We are therefore left with 5 unknowns, i.e. theother 4 basic forces + the collapse load factor. Since the values for these 5 unknowns are unique atcollapse, we are actually left with 3 independent equilibrium equations in 4 unknown basic forces (recallthat the structure becomes unstable at collapse). This is a statically indeterminate problem and its generalsolution has one parameter (one redundant basic force). We have studied such problems earlier.

How do we identify this static indeterminacy? By studying the rank of the equilibrium matrix withoutthe columns corresponding to the plastic hinges. We call this matrix Be

Be

16

−

1

0

0

0

1

18

0

0

0

18

1

16

−

0

0

1

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:= rank Be( ) 3= we note that the rank of the equilibrium matrix for theelastic portion of the structure is 3 and not 4.This means that only 3 columns of the matrix arelinearly independent. The mathematicians speakof the column space of matrix Be having dimension 3.

ref e ce p cpcλ = +P Q QB BSolving for the remaining elastic basic forces at incipient collapse

ref p cp e cecλ − =P Q QB Bsince the collapse load factor λc and the basic forces at the plastic hingelocations are known at this point from the linear programming problemwe solve the equilibrium equations for the remaining basic forces

Because matrix Be has rank 3 and not 4, such a system of equations does not generally have a solutionexcept for the very special case of the left hand side vector lying in the column space of the matrix Be. Thisis indeed the case at incipient collapse of a structure in a partial collapse mode. To see this we determinethe rank of the matrix that consists of the left hand side "load vector" next to the elastic equilibrium matrix Be

let us first form the left hand side in the above equation

Bp

16

−

0

0

0

16

1

0

0

16

0

0

1

0.8

0

0.6−

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:= Qcp

100

100−

100−

20−

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:= LHS λc Pref⋅ Bp Qcp⋅−:= LHS

8.333

100

31.25

100

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

=

Page 119


Here is the rank of the matrix that results from placing the left hand side load vector next to Be

rank augment Be LHS,( )( ) 3= indeed, the rank is still 3, which means that the column of the appliedload vector lies in the space spanned by the columns of Be, i.e. it alinear combination of these columns

In such case where the load vector lies in the space spanned by the columns of the equilibrium matrix, eventhough the rank of the matrix is less than the number of rows, and the matrix is thus singular, there is a m-nparameter solution for the elastic basic forces at incipient collapse Qce, with m being the number of unknownelastic basic forces and n the dimension of the column space or rank of matrix Be. In our example, m=4, n=3and there is a one parameter general solution which we now seek.

We use the method of Examples 5 and 6.

We select the last basic force in vector Qce (i.e. Q6) equal to zero and solve for the values of the other 3under the given left hand side load vector. We obtain the following particular solution (recall that we haveused the subscript p for the particular solution in Examples 5 and 6). In the following we drop the subscript cto avoid having to deal with 3 subscripts. It is clear from the context that we are referring to forces at collapse

Qep

50−

150

100

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:= a quick check confirms that Be Qep⋅

8.333

100

31.25

100

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

= is indeed equal to theleft hand side load vector

Then we set the last basic force in vector Qce (i.e. Q6) equal to 1 and determine the values of the other 3forces from the 4 equilibrium equations (this solution is known as the nullspace of Be).

We obtain Bbarxe

1−

1

1−

1

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:= a quick check confirms that Be Bbarxe⋅

0

0

0

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

=

The general solution for the elastic basic forcesat incipient collapse is

Qce Qep Bbarxe Q6⋅+=

The Mathcad solution above is obtained with Q6 50−:= Qep Bbarxe 50−( )⋅+

0

100

150

50−

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

=

and the Matlab solution with Q6 0:= Qep Bbarxe 0( )⋅+

50−

150

100

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

=

Page 120

Script for Example 9a in CE220 class notes % Plastic analysis with partial collapse mechanism % solution with Matlab functions and FEDEASLab function LowerBound_PLAnalysis % Clear workspace memory and initialize global variables CleanStart

set up equilibrium matrix w/o axial forces in frame members Bf = [-1/6 -1/6 1/6 0 0 -1/6 1/6 0.8 ; 0 1 1 1 0 0 0 0 ; 0 0 0 1/8 1/8 0 0 -0.6 ; 0 0 0 0 1 1 1 0 ];

specify applied force vector at free dof's for load case of nodal forces Pref = [-40; 0; 30; 0];

define plastic flexural capacity of elements in a column vector (2 entries per element) Qpl = [ 100 100 100 150 150 100 100 20]';

call function for lower bound plastic analysis in FEDEASLab [lambdac Qc] = LowerBound_PLAnalysis(Bf,Qpl,Pref); disp(' The collapse load factor for nodal force at girder midspan is'); disp(lambdac); disp(' The basic forces Q at collapse are'); disp(Qc); Optimization terminated. The collapse load factor for nodal force at girder midspan is 1.4417 The basic forces Q at collapse are 100 -50 -100 150 100 0 -100 -20

rank of equilibrium matrix for structure with plastic hinges at incipient collapse % basic force index at plastic hinge locations ip = [1 3 7 8]; % basic force index at non-plastic locations (subscript e for elastic) ie = setdiff(1:8,ip); % left hand side of equilibrium equations (consult pp. 101 in class notes) LHS = lambdac*Pref - Bf(:,ip)*Qc(ip);

Page 121

% remainder of equilibrium matrix (elastic part) Be = Bf(:,ie); disp('number of rows of elastic part of equilibrium matrix'); disp(size(Be,1)) disp('rank of elastic part of equilibrium matrix'); disp(rank(Be)) % check that the load vector lies in the column space of elastic equilibrium matrix disp('rank of elastic part of equilibrium matrix with load vector placed alongside') disp(rank([Be LHS])) number of rows of elastic part of equilibrium matrix 4 rank of elastic part of equilibrium matrix 3 rank of elastic part of equilibrium matrix with load vector placed alongside 3

general solution for elastic basic forces % particular solution for given left hand side (last elastic basic force = 0) Qep = zeros(4,1); Qep(1:3) = Be(1:3,1:3)\LHS(1:3); disp('the particular solution for the elastic basic forces is'); disp(Qep); % determine the homogeneous solution using the nullspace command of Matlab Bbarxe = null(Be,'r'); disp('the homogeneous solution for the elastic basic forces is'); disp(Bbarxe); the particular solution for the elastic basic forces is -50.0000 150.0000 100.0000 0 the homogeneous solution for the elastic basic forces is -1 1 -1 1

Page 122

Script for Example 9b in CE220 class notes % Plastic analysis with partial collapse mechanism % solution entirely with FEDEASLab functions % Clear workspace memory and initialize global variables CleanStart % define model geometry XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 6 0]; % second node, etc XYZ(3,:) = [ 12 0]; % XYZ(4,:) = [ 0 8]; % XYZ(5,:) = [ 6 8]; % XYZ(6,:) = [ 12 8]; % % element connectivity array CON { 1} = [ 1 2]; CON { 2} = [ 2 3]; CON { 3} = [ 2 5]; CON { 4} = [ 4 5]; CON { 5} = [ 5 6]; CON { 6} = [ 3 5]; % boundary conditions (1 = restrained, 0 = free) (specify only restrained dof's) BOUN(1,:) = [ 1 1 1]; BOUN(3,:) = [ 0 1 0]; BOUN(4,:) = [ 0 1 0]; BOUN(6,:) = [ 0 1 0]; % specify element type [ElemName{1:5}] = deal('2dFrm'); % 2d frame element ElemName{6} = 'Truss'; % truss element


1 2 3

4 5 6

1 2

3

4 5

6

Page 123

define plastic flexural capacity of elements in a column vector (3 entries per element) % axial force capacity is "very large" except for brace (truss) element Qpl = [ 1e5 100 100 1e5 100 100 1e5 150 150 1e5 100 100 1e5 100 100 20]';


specify loading Pe(4,1) = 30; % force at node 4 in direction X Pe(5,2) = -40; % force at node 5 in direction Y % generate data object Loading Loading = Create_Loading(Model,Pe); % extract applied force vector Pf at free dofs from Loading field Pref Pref = Loading.Pref;

call function for lower bound plastic analysis in FEDEASLab [lambdac Qc] = LowerBound_PLAnalysis(Bf,Qpl,Pref); disp(' The collapse load factor for nodal force at girder midspan is'); disp(lambdac); disp(' The basic forces Q at collapse are'); disp(Qc); Optimization terminated. The collapse load factor for nodal force at girder midspan is 1.4417 The basic forces Q at collapse are 43.2500 100.0000 -50.0000 12.0000 -100.0000 0 -25.0000 150.0000 100.0000 -43.2500 0 -0.0000 0 -100.0000 0 -20.0000 % open window and plot moment diagram Create_Window(0.80,0.80); Plot_Model (Model); Plot_2dMomntDistr (Model,[],Qc);

Page 124

% open window and show plastic hinge locations Create_Window(0.80,0.80); Plot_Model (Model); Plot_PlasticHinges (Model,Qpl,Qc);

Page 125

% we can also show the plastic hinge locations in the same window as the moment diagram Create_Window(0.80,0.80); Plot_Model (Model); Plot_2dMomntDistr (Model,[],Qc); Plot_PlasticHinges (Model,Qpl,Qc);

Page 126

Equilibrium Summary

f f fwP B Q P= +

degree of static indeterminacy NOS = no of columns - rank (usually no of rows) of matrix Bf

( )(1) (1)

f f

(2) (2) (1) (2) (1) (2)f f f f f

if

and then

=

= + = +

P B Q

P B Q P P B Q Q

superposition principle since Bf does not depend on Q or P for equilibrium in the original (undeformed) configuration of the structure

lower bound theorem of plastic analysis

1fB B −=for stable, statically determinate structures where is the force influence matrixB

for stable, statically indeterminate structures ( )f fw i i x x− = +P P B Q B Q

ref e ce p cp fucλ = + +P B Q B Q Pfor perfectly plastic response at collapse ( )ref fu p cp e cecλ − − =P P B Q B Q

if the rank of matrix Be is equal to the number of columns a complete collapse mode exists accompanied by a unique set of elastic basic forces Qce. If the rank is smaller than the number of columns, a partial collapse mode arises and difference between the number of columns and the rank of the matrix defines the number of free basic force parameters of the general solution.

ref f fu plmax for andcλ λ λ= = + ≤BP Q P Q Q


( )f fw= −BQ P P

static variables fP Rapplied forces at free dofs support reactions

element end forces p qbasic or free element forces

collect all basic element forces in vector Q

static matrix of structure for free dofs fB

basic force influence matrix of structure due to applied loads B

basic force influence matrix of primary structure due to applied loads iB

basic force influence matrix of primary structure due to redundants xB

static matrices representing equilibrium relations between static variables

select NOS redundant basic forces and collect them in a vector xQ


Page 127

KINEMATICS (COMPATIBILITY)

Objective: kinematic variables for structure, element and section and their interrelationship through compatibility equations; kinematic matrix of structural model and its properties; stability of structural model

CE220-Theory of Structures Kinematics © Prof. Filip C. Filippou, 2000

Page 128

Relation between structural and element displacementsRecall the numbering of the global dofs for the structural model in the following figure.

Recall also the numbering of the element dofs for a typical 2 node element in a 2d model (here we use element b for reference). The following figure shows the corresponding displacement components in the global reference

element b

i

j

1

2

3

4

56

1

2

( ) 3

4

5

6

b

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟=⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

uuu

uuuu

There is a one to one correspondence between element displacement dofs in the global reference and structural displacement dofs. The following figure shows an example for the dofs at the middle node of the gable frame.

a

b c

d

5U

c

2u1u

3u

b

5u4u6u

4U6U

We have(b) (c)4 4 1 4(b) (c)5 5 2 5(b) (c)6 6 3 6

= =

= =

= =

u U u U

u U u U

u U u U

and completing these one-to-onerelations for all elements and dofs(including the restrained dofs) we get

(a) (b) (c) (d)1 11 1 1 1 4 1 7(a) (b) (c) (d)2 12 2 2 2 5 2 8(a) (b) (c) (d)3 13 3 3 3 6 3 9(a) (b) (c) (d)4 1 4 4 4 7 4 14(a) (b) (c) (d)5 2 5 5 5 8 5 15(a) (b) (c) (d)6 3 6 6 6 9 6 10

= = = =

= = = =

= = = =

= = = =

= = = =

= = = =

u U u U u U u U

u U u U u U u U

u U u U u U u U

u U u U u U u U

u U u U u U u U

u U u U u U u U

1

2

3

4

5

6

7

8

9

10

11

12

1314

15

We introduce the incidence or id array of each element which provides the correspondence between element and structural dofs. We list this array for each of the elements in the gable frame.

(a) (b) (c) (d)

11 1 4 712 2 5 813 3 6 91 4 7 142 5 8 153 6 9 10

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟= = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

id id id id

With this array as index the correspondence of elementand structural displacements can be compactly written

(el)(el)i

=d

u U


Page 129

the relation between element and structural displacement dofs can also be written in terms of a matrix-vector multiplication by a Boolean compatibility matrix (Boolean is a matrix with terms of 1 and 0); this is not used in computing but furnishes a convenient way of writing the displacement correspondence relations

(el)(el)i

=d

u U

(el) (el)b

1 0 0 0 0 0 0 0 0 0 0 0 0 0 00 1 0 0 0 0 0 0 0 0 0 0 0 0 00 0 1 0 0 0 0 0 0 0 0 0 0 0 0

reorder columns according to0 0 0 1 0 0 0 0 0 0 0 0 0 0 00 0 0 0 1 0 0 0 0 0 0 0 0 0 00 0 0 0 0 1 0 0 0 0 0 0 0 0 0

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

A id

(el) (el)b= Au Ucan be written as

where

the size of the Boolean compatibility matrix is 6 x nt for a planar model with nt total degrees of freedom andfor the example it becomes 6 x 15 as shown.

for element a the Boolean compatibility matrix is (a)b

0 0 0 0 0 0 0 0 0 0 1 0 0 0 00 0 0 0 0 0 0 0 0 0 0 1 0 0 00 0 0 0 0 0 0 0 0 0 0 0 1 0 01 0 0 0 0 0 0 0 0 0 0 0 0 0 00 1 0 0 0 0 0 0 0 0 0 0 0 0 00 0 1 0 0 0 0 0 0 0 0 0 0 0 0

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

A

note that the transpose of the Boolean compatibility matrix arises from reordering the rows instead of the columns; we write here explicitly the transpose of the Boolean compatibility matrix for element a

(a)Tb

0 0 0 1 0 00 0 0 0 1 00 0 0 0 0 10 0 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 01 0 0 0 0 00 1 0 0 0 00 0 1 0 0 00 0 0 0 0 00 0 0 0 0 0

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

A


Page 130

Remark: Note that the equilibrium equations at the node free bodies can be written in compact form with the help of the id array or Boolean compatibility matrix. Recall the following equilibrium equations from Part I

we include now also the equations at the restraineddofs (support reactions in the force vector)

(a)1

11 (a)2

12 (a)3 (d)13

414 (d)

515

00 0 00 0 00 0

0 0 00 0 0

00

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟= + + +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

pP

pPpP

pP

pP

(b)(a) 141 (b)(a) 22 5 (b)(a)3 3 (c)6 1(b)4 4 (c)2(b)5

5 (c)6 3(b)

67

8

9

10

000

0000 00 00 00 0

⎛ ⎞⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟= + +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎜ ⎟⎝ ⎠ ⎝ ⎠

ppPpP p

P pp pP ppP

pP p

pPPPP

(d)(c) 14 (d)(c) 25 (d)(c) 36 (d)

6

000000

0

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ + ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

pp

pp

pp

p

with the Boolean compatibility matrix for all elements of the structure the above equations can be written in compact form

(a)T (a) (b)T (b) (c)T (c) (d)T (d)b b b b= + + +A A A AP p p p p

or, simplyne

(el)T (el)b

el 1== ∑ AP p

First instance of a contradient transformation

We observe that the element displacements of the example structure are

(a) (a)b

(b) (b)b

(c) (c)b

(d) (d)b

=

=

=

=

A

A

A

A

u U

u U

u U

u U

or, in compact form

(a)(a)b(b)(b)b

b b (c)(c)b(d)(d)b

where

⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟⎢ ⎥⎜ ⎟⎢ ⎥⎜ ⎟ = =⎢ ⎥⎜ ⎟⎢ ⎥⎜ ⎟

⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦

A

AA A

A

A

u

uU

u

u

then the corresponding forces obey the following relation

(a) (a)

(b) (b)(a)T (b)T (c)T (d)T Tb b b b b(c) (c)

(d) (d)

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎡ ⎤ ⎜ ⎟ ⎜ ⎟= =⎢ ⎥⎣ ⎦ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

A A A A A

p p

p pP

p p

p p(a) (a)

(b) (b)T

b b(c) (c)

(d) (d)

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟= =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

A A

u p

u pU P

u p

u p

The pair of relations are known as contragradient transformations. They will appear over and over in the next pages and are the direct result of the principle of virtual work, as will be shown later.


Page 131

Element displacements from global to local reference system

Similar to the transformation of the generalized element forces from the local to the global coordinate system in Part I, we transform the components of the displacement vector from the global to the local reference. The following figure shows the details for end node j; the same relations hold for node i; the rotation at node i and node j does not change during the transformation. We note again that the matrix relating the components is a rotation matrix

4 4 5 4 5

5 4 5 4 5

cos sin

sin cos

X Yu u u uL L

Y Xu u u uL L

α α

α α

Δ Δ= + = +

Δ Δ=− + = − +

u

u

or, compactly 4 4

5 5

X YL L

Y XL L

Δ Δ⎡ ⎤⎢ ⎥⎛ ⎞ ⎛ ⎞⎢ ⎥=⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟Δ Δ⎢ ⎥⎝ ⎠ ⎝ ⎠−⎢ ⎥⎣ ⎦

u uu u

The transfromation of all element displacement components becomes

1 1

2 2

3 3r

4 4

5 5

6 6

0 0 0 0

0 0 0 0

0 0 1 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0 0 1

X YL L

uY XuL LuuX YuL L

Y X uL L

Δ Δ⎡ ⎤⎢ ⎥⎢ ⎥⎛ ⎞ ⎛ ⎞Δ Δ⎢ ⎥⎜ ⎟ ⎜ ⎟−⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟= = =⎢ ⎥Δ Δ⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥Δ Δ⎜ ⎟ ⎜ ⎟

−⎝ ⎠ ⎝ ⎠⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

a

uuu

u uuuu

and we note that the transformationof the displacement components isthe transpose of the correspondingtransformation of the force components

It remains now to relate the element end displacement components in the local reference system to the relative element displacements or element deformations. We derive this relation on the following page.

Tr r=b a

Second instance of a contradient transformation

then the corresponding forces obey the following relation

The relation between local and global components of the element displacement vectors is r= au u

Tr= ap p

X

Y

xy L

i

j

XΔ

YΔ

α

4u 5u

5u

4u

α

α5u

5u

4u

4u


Page 132

From element end displacements in local coordinates to element deformations The following figure (a) shows two nodes i and j that will be connected by an element of any geometry. In any case the straight line connecting the nodes will be the element chord. This is the name we plan to use for this line from now on. Because the actual element geometry between nodes i and j is not important, the straight line connecting nodes i and j is dashed. The original distance between the nodes is L. The figure also shows the nodes in the displaced position involving translation and rotation at each end. The rotation is depicted by drawing a line emanating from the node that originally coincides with the element chord. We call this line the node tangent for the particular element chord. The angle of this line relative to the original orientation of the element chord is the end rotation of the element. Figures (b) and (c) decompose the displaced state of the nodes in two parts: (a) a rigid body translation in figure (b); we note that the node tangents rotate so as to continue coinciding with the element chord which rotates through the angle β relative to the original position in the process; (b) end rotations relative to the chord in figure (c). It is clear from figure (b) that the change in distance ΔL between the nodes depends only on the relative translation components between nodes i and j.

We have the following relations under large displacements.

change in distance between nodes i and j

angle between node tangent and chord in displaced position at node i

at node j

under large displacements we have the following relations for the change of distance ΔL and the chord rotation angle β

the change in distance between nodes i and j and the angle between node tangent and chord inthe displaced position at node i and j make up the components of the element deformation vector v

2 3 β= −v u

3 6 β= −v u

1 nL L L= Δ = −v

( ) ( )22x yL L LΔ = + Δ + Δ −u u

arctan y

xLβ

Δ=

+ Δ

u

u

i jx

y

β

3u2v

L

(c)

i jx

y

yΔu

β

position under rigid body translation

nL

LΔ

(b)

1u X

Y

xy

L

L

4 1xΔ = −u u u

i

j

5 2yΔ = −u u u

nL

(a)

β

3vyΔu

xΔu

xΔu

6u2u

3u

6u

5u

4u


Page 133

Element deformations under linear and nonlinear kinematics

We transform the expressions for the change in distance and the chord rotation angle using an expansionin power series that allows us to assess the relative magnitude of individual terms.

change of distance

where we made use of 21 11 1 higher order terms for 12 8

x x x x+ = + − +

finally, we obtain for a moderate relative transverse displacement ratio

00

and for a small relative transverse displacement ratio

noting that in structural engineering practice, typically

We can show that the relative transverse displacement ratio is approximately equal to the chord rotation angle. To this end we expand the trigonometric arctan function into a power series

a small relative transverse displacement ratio is

and a moderate relative transverse displacement ratio is

noting that in structural engineering practice, typically it follows that

We conclude that the chord rotation angle is approximately equal to the relative transverse displacementratio. This is known as tranverse "drift" in professional engineering practice, or, as lateral drift or interstorydrift for vertical members.

22

1 yxL L LL L

Δ⎛ ⎞Δ⎛ ⎞⎜ ⎟Δ = + + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

uu

( ) ( )22x yL L LΔ = + Δ + Δ −u u

22

1 2 yx xL L LL L L

Δ⎛ ⎞Δ Δ⎛ ⎞⎜ ⎟Δ = + + + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

uu u

22 221 1 11 higher order terms2 2 8

yx x xL L LL L L L

⎡ ⎤Δ⎛ ⎞Δ Δ Δ⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟Δ = + + + − + −⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠⎢ ⎥⎣ ⎦

uu u u

2 21 112 2

y yx xL L L LL L L L

⎡ ⎤ ⎡ ⎤Δ Δ⎛ ⎞ ⎛ ⎞Δ Δ⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟Δ ≈ + + − = +⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

u uu u

4 11 xxL L L

LΔ⎡ ⎤

Δ ≈ + − = Δ = −⎢ ⎥⎣ ⎦

uu u u

35 10xL

−Δ≤ ⋅

u

35 10y

L−Δ

≤ ⋅u

25 10y

L−Δ

≤ ⋅u

arctan arctan1

yy

xx

LL

L

β

ΔΔ

= =Δ+ Δ +

uu

uu

2

1 higher order termsy x xL L L

β⎡ ⎤Δ Δ Δ⎛ ⎞⎢ ⎥≈ − + −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

u u u

y

Lβ

Δ≈

u35 10xL

−Δ≤ ⋅

u


Page 134

Element deformations for linear kinematics

As long as and the element deformations become

or, in compact form

T=b awe note again that the relation between element deformations and the end displacement components in the local reference system is the transpose of the transformation of the element force components from the local to the basic reference system

We note that the relation between relative element displacements and end displacement components in the local reference system is linear under the small displacement assumption. We speak of linear kinematics. The matrix a in the above relation is known as the rigid body kinematic matrix, since it removes the three (3) rigid body components from the set of six (6) element end displacements leaving only three (3) relative displacements for the 2 node element in a planar model.

1 0 0 1 0 01 10 1 0 0

1 10 0 0 1

L L

L L

⎡ ⎤−⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥⎢ ⎥

−⎢ ⎥⎣ ⎦

It is worth pointing out at this stage that the rigid body kinematic matrix reveals the uncoupled nature between the relative translation in the direction of the chord, and the chord rotation angles at the element ends; this means that transverse translations and rotation do not cause a change of distance between the nodes, and that translations in the direction of the chord do not cause the chord to rotate under linear kinematic considerations..

35 10xL

−Δ≤ ⋅

u 35 10y

L−Δ

≤ ⋅u

1 4 1x= Δ = −v u u u

2 3y

L

Δ= −

uv u

3 6y

L

Δ= −

uv u

1

21

32

43

5

6

1 0 0 1 0 01 10 1 0 0

1 10 0 0 1

L L

L L

⎛ ⎞⎡ ⎤ ⎜ ⎟⎢ ⎥− ⎜ ⎟⎛ ⎞ ⎢ ⎥ ⎜ ⎟⎜ ⎟ ⎢ ⎥ ⎜ ⎟= = − =⎜ ⎟ ⎢ ⎥ ⎜ ⎟⎜ ⎟ ⎢ ⎥ ⎜ ⎟⎝ ⎠ ⎢ ⎥− ⎜ ⎟⎢ ⎥ ⎜ ⎟⎣ ⎦ ⎝ ⎠

a

uu

vu

v v uu

vuu

Fourth instance of a contradient transformation

then the relation between basic forces and global components of the element force vector is

The relation between global components of the element displacement vector and the element deformations is

Element kinematics: from element end displacement components to deformations

combining with gives= av u r= au u r g= =aa av u u where ( )TT T Tg r r r g= = = =b b b a a aa a

g= av u

Third instance of a contradient transformation

then the relation between basic forces and local components of the element force vector is

The relation between local components of the element displacement vector and the element deformations is

T= ap q

= av u

Tg= ap q


Page 135

g= av uThe relation between global components of the element displacement vector and the element deformations is

g 2 2 2 2

2 2 2 2

0 0

1 0

0 1

X Y X YL L L LY X Y X

L L L LY X Y X

L L L L

⎡ ⎤Δ Δ Δ Δ− −⎢ ⎥⎢ ⎥⎢ ⎥Δ Δ Δ Δ

= − −⎢ ⎥⎢ ⎥⎢ ⎥Δ Δ Δ Δ− −⎢ ⎥⎣ ⎦

awith

j

= j iX X XΔ −

= j iY Y YΔ −2 2L X Y= Δ + Δ

X

Y

i

5u

1u

4u

2u

1

XLΔ

YLΔ

i

ji

1

YLΔ

j

XLΔ

1

YLΔ

XLΔ

i

j

1

XLΔ

YLΔ

i

j

1 1=u deformations are 2

2

XLY

LY

L

⎛ ⎞Δ−⎜ ⎟⎜ ⎟⎜ ⎟Δ−⎜ ⎟

⎜ ⎟⎜ ⎟Δ−⎜ ⎟

⎝ ⎠

note that the angle is measured from the chordin the displaced position to the node tangent (note that CCW is positive and CW negative)


2

YLX

LX

L

⎛ ⎞Δ−⎜ ⎟⎜ ⎟⎜ ⎟Δ⎜ ⎟⎜ ⎟⎜ ⎟Δ⎜ ⎟⎝ ⎠


2

XLY

LY

L

⎛ ⎞Δ⎜ ⎟⎜ ⎟⎜ ⎟Δ⎜ ⎟⎜ ⎟⎜ ⎟Δ⎜ ⎟⎝ ⎠

2

2

YL

XLX

L

⎛ ⎞Δ⎜ ⎟⎜ ⎟⎜ ⎟Δ−⎜ ⎟⎜ ⎟⎜ ⎟Δ−⎜ ⎟⎝ ⎠

deformations are5 1=u


Page 136

Structure kinematics: from global dof displacements to all element deformations

1

2

3

4

5

6

7

8

9

10

11

12

1314

15

(a)(a)b(b)(b)b

b (c)(c)b(d)(d)b

⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟⎢ ⎥⎜ ⎟⎢ ⎥⎜ ⎟ =⎢ ⎥⎜ ⎟⎢ ⎥⎜ ⎟

⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦

A

AA

A

A

u

uU = U

u

u

we write g= av u

for all elements of the structuralmodel and then combine it with

to obtain

(a) (a) (a) (a)(a) g g b

(b) (b) (b) (b)(b)g g b

(c) (c) (c) (c) (c)g g b

(d) (d) (d) (d) (d)g g b

⎛ ⎞ ⎡ ⎤⎛ ⎞ ⎜ ⎟ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎢ ⎥⎜ ⎟ = ⎜ ⎟ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎢ ⎥⎝ ⎠ ⎣ ⎦

a a A

a a A

a a A

a a A

uv

uv= U

v u

v u

We collect the element deformations into the deformation vector V for the elements of the entire structure.Similarly, the compatibility matrices on the right hand side which are stacked one on top of the other arecollected to the kinematic or compatibility matrix A for the entire structure. We get

= AV U

(a)

(b)

(c)

(d)

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟=⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

v

vV

v

v

with

(a) (a)g b

(b) (b)g b

(c) (c)g b

(d) (d)g b

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

a A

a AA

a A

a A

=and

We partition the displacement vector into free and restrained dofs, as we have already discussed. Thekinematic or compatibility matrix of the structure gets partitioned accordingly into a set of columns that correspond to the free dofs and another set that corresponds to the restrained dofs. We write

f f d d f f d= + = +A A AV U U U V where d d d= AV U the element deformations due to displacementsat the restrained dofs (e.g. support settlements)are known at the start of the analysis

We express the equilibrium equations in terms of the basic forces of the structure and get(a)T (a)

(a) (a)g

(b)T (b)(b) (b)gT (a)T (b)T (c)T (d)T (a)T (a)T (b)T (b)T (c)T (c)T (d)T (d)T

b b b b b b g b g b g b g(c) (c)T (c) (c)g

(d) (d)(d)T (d)g

⎛ ⎞⎛ ⎞ ⎛⎜ ⎟⎜ ⎟ ⎜⎜ ⎟⎜ ⎟ ⎜ ⎟⎡ ⎤ ⎡ ⎤⎜ ⎟= = =⎜ ⎟⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝⎝ ⎠

a

aA A A A A A a A a A a A a

a

a

qp q

qp qP

p q q

p qq

⎞⎟

⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟

⎠

T= AP Qor, after introducing the structure kinematic matrix on the right hand side and collecting the basic forces into a vector Q for the entire structure we getWe partition the applied force vector into free and restrained dofs, as we have already discussed. Thetranspose of the kinematic or compatibility matrix of the structure gets partitioned accordingly into a set of that correspond to the free dofs and another set that corresponds to the restrained dofs. We write

TffTd

R

⎡ ⎤⎛ ⎞ ⎢ ⎥=⎜ ⎟⎢ ⎥⎝ ⎠ ⎣ ⎦

A

A

PQ

we have encountered the first equation already in the form f f= BP Q

we note that R are the forces at the restrained dofs or support reactions

we note that Tf f=B A


Page 137

Fifth instance of a contradient transformation

then the relation between basic forces and global components of the applied force vector is

The relation between global dof displacement components and the element deformations of the entire structure is

f f d d= +A AV U U

TffTd

R

⎡ ⎤⎛ ⎞ ⎢ ⎥=⎜ ⎟⎢ ⎥⎝ ⎠ ⎣ ⎦

A

A

PQ

Limiting ourselves to the free dofs only by assuming that the displacements at the restrained dofs are all zero, we set aside the support reactions and get the following important pair of relations

f f= AV U

Tf f= AP Q

kinematic or compatibility relation for structure

static or equilibrium relations for struct ure f f= BP Q or

The properties of the static matrix Bf that we have discussed in Part I are also properties of the transposeof the kinematic matrix. In particular, a statically determinate structure has a square kinematic matrix withas many element deformations as global dof displacements. Thus, any set of element deformations is compatible with a unique set of global dof displacement values, as we can see in Examples 10, 11, 12

A statically indeterminate structure has more element deformations than global dof displacements. This means that an arbitrary set of deformations will, in general, not be compatible with a set of global dof displacement values, since the compatibility relations have more equations than unknowns. The only way that such a system will have a unique solution is, if the deformation vector V lies in the column space of the kinematic matrix Af (Strang, pp. 66). For this to be the case, the deformation vector V needs to be orthogonal to the nullspace of the transpose of the kinematic matrix (Strang, pp. 135-139). Since the latter is equal to the static matrix, we have already determined the nullspace of the static matrix in Part I as theforce influence matrix for the redundant basic forces (see Examples 5 and 6). Thus, the necessary andsufficient condition for a set of given element deformations to be compatible with a unique set of global dof displacement values, is that

Tx =VB 0 compatibility condition for statically indeterminate structure

We use this condition in Examples 13 and 14


Page 138

CE220-Theory of Structures Ex. 10 - Compatibility for determinate truss © Prof. Filip C. Filippou, 2000

Example 10(r) - Compatibility relations for statically determinate 2d truss

Objectives:

(a) set up of the compatibility relations between the displacements at the free dofs of a staticallydeterminate truss and the element deformations

(b) determination of deformed shape for given element deformations

The truss is shown in Figure 1. The global coordinate system X-Y is also shown.

88

6

12

3

4

a b

c d e

X

Y

Figure 1: Statically determinate truss with global coordinate system

We focus on the number of independent ways that the nodes of the structure can displace. Since rotationsare of no interest in this case, there are two independent translations at each node, except for thetranslations restrained by supports. We call each independent translation a degree of freedom of thestructure. We number these in sequence starting with the node with lowest number exactly as we did for theequilibrium equations of Example 1. The one to one correspondence between degree of freedom (dof) andequilibrium equation is established by Newton's second law. The numbering is, therefore, the same for both.

8 8

6

1

23

4

5

a b

c d e

Figure 2: Free dof numbering

Page 139


If rotations at the nodes are of no significance, then the only "deformation" between nodes of interest hereis the change in the original distance. In the following we wish to establish a relation between this change ofdistance between nodes and the deformed element length. If there are no gaps in the structure afterdeformation, then the element deformation must be equal to the change in node distance. For a trusselement the only element deformation is the change of its length denoted by v1.

To set up the compatibility relation between free dof displacements and element deformations we proceedin the following systematic way: (1) set each dof in turn to a unit value, while holding the other dofs equalto zero, (2) determine the change in distance between the nodes in the deformed configuration assumingsmall displacements and deformations, (3) record this change of distance in a column vector with each rownumber corresponding to the number of the element that connects the corresponding nodes (recall thatthere is only one "deformation" of interest for each element, the change in distance of the nodes). Thecolumn number corresponds to the number of the dof we "activate". The corresponding array or matrix isknown as compatibility or kinematic matrix Af

Preliminary geometric calculation: original distance between nodes numbered according to thecorresponding element number.

La 8:= La 8=

Lb 8:= Lb 8=

Lc 62 82+:= Lc 10= original distance between nodes 1 and 4

Ld 6:= Ld 6=

Le 62 82+:= Le 10=

Unit value for degree of freedom (dof) 1.

for unit displacement at dof 1 (translation)dof 1dist change btw 1 and 2

a b

c d e

11

2 3

4dist change btw 2 and 3

Af1⟨ ⟩

1

1−

0

0

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=why??very small displacements

Page 140


for unit displacement at dof 2 (translation)dof 2

a b

c d e

12 3

4 why??very small displacements

Af2⟨ ⟩

0

0

0

1−

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=

dist change btw 2 and 4


a b

c d e

12 3

4dist change btw 2 and 3

Af3⟨ ⟩

0

1

0

0

8Le

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=


Page 141



a b

c d e

12 3

4

dist changebtw 1 and 4Af

4⟨ ⟩

0

0

8Lc

0

8Le

−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

dist changebtw 3 and 4


a b

c d e

12 3

4

dist change btw 1 and 4Af5⟨ ⟩

0

0

6Lc

1

6Le

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=



Page 142


The change in node distances when all dof's are "activated" is the linear combination of the individual dofeffects. Since the dof translation values are parameters to be determined, we collect them into thefree dof displacement vector Uf and write for the change in node distances the following general relation.We restate that the change in distance between the nodes is the linear combination of the columns ofthe kinematic matrix for the free dofs Af with the free dof displacement values serving as parameters.

change in node distances is Af Uf⋅ with Af

1

1−

0

0

0

0

0

0

1−

0

0

1

0

0

0.8

0

0

0.8

0

0.8−

0

0

0.6

1

0.6

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

=

we note that matrix Af is the transpose of the equilibrium matrix Bf for the same truss structure in Example 2.This is a consequence of the principle of virtual work, as we prove later.

Now let us look at the elements that are supposed to connect the nodes of the truss structure.In general, elements may have different lengths than those resulting on the basis of the original(undeformed) configuration of the structure. One reason is that the elements are fabricated longer orshorter than specified. This happens in prefabricated structural systems like most steel and timberstructures where the elements are fabricated in a shop or plant and delivered to the site for installation.Elements may be longer or shorter either intentionally (we will see why), or unintentionally due tounavoidable fabrication errors (fabrication tolerances are usually prescribed for a structure). Even if theelement length can be assumed as "perfect", another reason that it may not match the actual distancebetween the corresponding nodes during installation is a temperature change between fabrication andinstallation time, particularly in severe climate sites like underwater, arctic, desert etc. Finally, it is knownfrom earlier courses that elements change length due to stress (forces cause stress at every material pointwhich gives rise to strain which results in change of the element length). For now we will limit ourselves tothe first two causes for change of element length, the so called non-mechanical causes: intentional orunintentional fabrication "error", and temperature.

We denote the difference between the actual and the ideal or original element length the elementdeformation and denote it by the lower case letter v. A truss element has only this one deformation. We willsee in the following example that beam elements have two and frame elements have 3 deformationmeasures. It is, therefore, justified to use already a vector to contain the element deformations.

Exactly as we have done for the basic forces of the element we collect the element deformations for theentire structure into a single vector which we denote with the upper case letter V. It contains thedeformations of all elements in the structure numbered in element sequence, i.e. deformations of element astacked on top of deformations of element b and so on.

Page 143


Compatibility states that in order for the elements to fit perfectly in the space between the displaced nodesthe element deformations V need to be equal to the change in the distance between nodes. We write thisequality relation as follows:

V Af Uf⋅=let us write it out in full to appreciate itbetter for this small structure

V1 U1=

V2 U1− U3+=

V3 0.8 U4⋅ 0.6 U5⋅+=

V4 U2− U5+=

V5 0.8 U3⋅ 0.8 U4⋅− 0.6 U5⋅+=

Let us look at these compatibility relation in some more detail. First, we reiterate that the coefficientsmultiplying the displacements at the free dof's are the same as those of the equilibrium equations for thesame structure but in transpose form, i.e. the rows of the equilibrium matrix have now become columns.We will see that this relation holds for every structure in this course and we will prove it by the principle ofvirtual work in a later lecture. Secondly, we note that if the element deformations V are given on the lefthand side of the compatibility equations, then we have exactly as many unknown global dof displacementvalues on the right hand side as equations of compatibility. In such case if the matrix Af is non-singularthere is a unique solution. This is the case for a stable structure and we can confirm this by checking therank of the kinematic matrix Af. (recall a similar argument for the static matrix Bf, which is the transpose ofthe kinematic matrix).

rank Af( ) 5= this means that all equations are independent

Let us then solve an example. Let us assume that elements c, d and e are fabricated longer or shorter thanrequired by the specified geometry such that c is longer by 0.02 units, d is shorter by -0.006 units and e islonger by 0.01 units of length. We assume that elements a and b have ideal length according to theundeformed geometry of the structure. For this scenario the element deformation vector becomes

V

0

0

0.02

0.006−

0.01

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:= We can now solve for the displacementsthat are compatible with these deformationsfrom the compatibility equations

Uf lsolve Af V,( ):= Uf

0

0.031

0

0.006

0.025

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

=

We can easily solve the 5 compatibility equations by hand and gain more insight into how the nodesdisplace due to the element deformations, e.g.

from the first equation we conclude that U1 is zero, and then from the second that U3 is zero, if V1=V2=0.Then, we can use the third and fifth equation to determine U4 and U5 and locate the displaced position ofnode 4. In fact, because elements c and e have the same direction cosine values the solution is easy.

By addition of eq. (3) and (5) we get V3 V5+ 1.2U5=

and by subtraction V3 V5− 1.6U4=

Page 144


From eq (4) we conclude then that the translation U2 is equal to U2 U5 V4−=

The resulting deformed shape of the truss structure looks as follows (magnification factor is 50)

The geometrical relations we noted are clear in the figure (i.e. node 4 "at the intersection of elements cand e each pivoting about the supports that do not move; vertical translation of node 2 is the differencebetween vertical translation of node 4 and the deformation of element d)

CONCLUSIONS

In a statically determinate structure there are as many element deformations (rows of Af) as there are freeglobal dof displacements (columns of Af). Consequently, Af is a square matrix. In a stable structure therank of the matrix is equal to the number of columns (or rows for a determinate structure). This meansthat it is always possible to find a deformed configuration of the structure that matches any given elementdeformations. It also means that the only way to match the original length of the elements, i.e. a zerodeformation vector on the left hand side of the compatibility equations, is with zero displacement values atall free dof's, i.e. only the undeformed configuration results in no-element deformations. We will see laterthat the kinematic matrix of an unstable structure is "rank deficient", meaning that the rank is smaller thanthe number of columns. In such case it is possible to find deformed positions of the structure withoutdeforming any of the elements at least under the assumption of small displacements. This is clearly notdesirable for structures, unless we plan to "deploy a folded structure" in which case these considerationsneed to be extended to large displacements.

Corollary

In a statically determinate structure it is possible to have any element deformations (and thus elementlengths) and "fit them together" without "forcing" the elements. This is not possible in a staticallyindeterminate structure, as later examples will demonstrate.

The above structure is not subjected to any forces at the free global dof's. This means that there are nobasic element forces and, of course, no support reactions.The nodes displace due to the non-mechanical element deformations while the elements are stress-free!

Page 145

Matlab script for Example 10 in CE220 class notes % Deformations and deformed shape for statically determinate 2d truss

OPTION A: perform each step with Matlab functions % clear memory clear all % specify kinematic matrix Af (consult page 144 in Part I Notes) Af = [ 1 0 0 0 0; -1 0 1 0 0; 0 0 0 0.8 0.6; 0 -1 0 0 1; 0 0 0.8 -0.8 0.6]; % specify the element deformations V = [ 0; 0; 0.02; -0.006; 0.01]; % solve for free global dof displacements Uf = Af\V; % display result format short disp('the free global dof displacements are'); disp(Uf); the free global dof displacements are 0 0.0310 0 0.0063 0.0250

OPTION B: use FEDEASLab functions % clear memory; close any open windows CleanStart

Create model % specify node coordinates XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 8 0]; % second node, etc XYZ(3,:) = [ 16 0]; % XYZ(4,:) = [ 8 6]; % % connectivity array CON { 1} = [ 1 2]; CON { 2} = [ 2 3]; CON { 3} = [ 1 4]; CON { 4} = [ 2 4]; CON { 5} = [ 3 4]; % boundary conditions (1 = restrained, 0 = free) BOUN(1,:) = [ 1 1]; BOUN(3,:) = [ 0 1]; % specify element type ne = length(CON); % number of elements

Page 146

[ElemName{1:ne}] = deal('Truss'); % truss element Model = Create_SimpleModel (XYZ,CON,BOUN,ElemName);

Post-processing functions on Model (optional) Create_Window (0.80,0.80); % open figure window Plot_Model (Model); % plot model (optional) Label_Model (Model); % label model (optional)

specify element deformations V = zeros(Model.ne,1); V(3) = 0.02; V(4) = -0.006; V(5) = 0.01;

Kinematic (compatibility) relations % set up kinematic matrix for all dof's A = A_matrix(Model); % extract columns corresponding to free dof's to form matrix Af Af = A(:,1:Model.nf);

solution for free global dof displacements Uf = Af\V; % put free dof displacement values in complete displacement vector U = zeros(Model.nt,1); U(1:Model.nf) = Uf;

plot deformed shape of structural model MAGF = 50; % magnification factor for deformed configuration Create_Window(0.80,0.80); Plot_Model(Model); Plot_DeformedStructure(Model,[],U,V);

Page 147

Section kinematics and non-mechanical effects

x

reference axis h

dx

x L

i j

( )a x dxε

( )x dxκ

section x section x+ dx

Under the assumption that plane sections normal to the reference axis remain plane after deformationand normal to the deforrmed reference axis, there are two kinematic variables at a section x: the axialstrain at the reference axis and the curvature. We call these the section deformations e(x).

( )( )

( )a x

xx

εκ

⎛ ⎞= ⎜ ⎟⎝ ⎠

e

section deformations

section deformations may be caused by mechanical and non-mechanical effects; section forces, suchas the normal force, shear and bending moment, cause mechanical effects; we can quantify these onlyafter studying the material response in Part IV. For now we limit ourselves to non-mechanical effects. These effects arise from environmental causes, or from intentional camber or lack-of-fit considerations. As an example, we look at the effect of differential temperature in the following.

Assume a temperature difference between the time of fabrication of the element and the time of installation.Then, the axial strain at the reference axis is

where α is the thermal expansioncoefficient of the material and ΔTa

is the temperature difference

x

reference axis

temperature increment

h

ΔTt

Δx

aTΔ

ΔTb

ΔTt

ΔTb

If there is a different temperature difference at the top than at the bottom of the cross section, then there is not only an axial strain at the reference axis, but also a curvature. The latter is given by the following expression

a aTε α= Δ

( ) ( )b tT T Th h

α ακ

Δ − Δ Δ Δ= =

Shrinkage of a material produces similar effects to temperature. On the other hand, it is common to fabricate or build members with intentional camber or different length then specified so as to induce a desired level of prestressing in the structure. We will investigate this in later lectures.

In order to distinguish non-mechanical deformations from mechanical deformations later in the course wedenote the former with a subscript 0. We leave this out for the moment, since we focus exclusively on non-mechanical deformations and there is no danger of confusion.


Page 148

Element deformations and section kinematics

The change of element length is equal to the integral of the distribution of the axial strain at the reference axis, as long as the element does not contain axial force release devices.

1v( )a xε

Lx

we write

and for uniform axial reference strain

1 a Lε=v

The axial deformation of a force release device needs to be added to the above deformation. We will investigate this later, after first discussing the effect of the curvature on the deformations.

( )x dxκ 3v

0

( )

L

x x dxκ∫x L x−

( )0

( )

L

L x x dxκ−∫

2v

L

element chord

i

j

The following figure shows the element chord connecting nodes i and j and the deformed reference axis of the element. In determining the angle between the element chord and the tangent to the deformed axis of the element at the end nodes we make use of the height of the highlighted triangle and the total distance at node i and node j that it covers as x increases from 0 to L. For small deformations we have

( )2

0 0

3

0 0

1 ( ) 1 ( )

1 ( ) ( )

L L

L L

xL x x dx x dxL L

xx x dx x dxL L

κ κ

κ κ

⎛ ⎞⎛ ⎞⎜ ⎟= − − = − −⎜ ⎟⎜ ⎟ ⎝ ⎠⎜ ⎟

⎝ ⎠⎛ ⎞⎜ ⎟= =⎜ ⎟⎜ ⎟⎝ ⎠

∫ ∫

∫ ∫

v

v

for the deformation at node i

for the deformation at node j

Note: the deformed axis has a positive curvature, since the angle of the tangent to the horizontal increases with increasing x. For this case the deformation at node i is negative and that at node j is positive, as shown. Recall that the deformation angle is measured from the chord of the element to the tangent of the deformed reference axis at the end node.

1

0

( )L

a x dxε= ∫v


Page 149

Element deformations

The expressions for the deformations v2 and v3 involve integrals of the type

0

( )L

x x dxκ∫ which is equal to x Aκ κ

where is the area under the curvature diagram, and is the distance of the centroid of the areaunder the curvature diagram from node i

Aκ xκ

Because of this interpretation of the integral and the fact that curvatures are proportional to bending moments for linear elastic material response, this geometric method for determining beam deformations is known in the literature as "moment area method". Unfortunately, this is not the most appropriate name; curvature-area method would be a lot better and completely general for small deformations. With this definition the deformation at node i and node j becomes

for the deformation at node i

for the deformation at node j

2L x x

A AL L

κ κκ κ

′−= − = −v

3x

ALκ

κ=v

xκ′with the distance of the centroid of the area under the curvature diagram from node j

For the case of a constant curvature with value κ the deformations become

( )212

Lκ= −v ( )312

Lκ=v

For the case of a linear curvature with value κi at node i the deformations become

22 1 13 2 3i iL Lκ κ⎛ ⎞= − = −⎜ ⎟⎝ ⎠

v 31 1 13 2 6i iL Lκ κ⎛ ⎞= =⎜ ⎟⎝ ⎠

v

and we note that the angle at node j is one half the value of the angle at node i, but of opposite sign

Lx

2v 3v

iκ

i j

Lx

( ) const=xκ κ=

2v 3v

i j


Page 150

Element deformations (cont'd)

For the case of a linear curvature with value κj at node j the deformations become

21 1 13 2 6j jL Lκ κ⎛ ⎞= − = −⎜ ⎟⎝ ⎠

v 32 1 13 2 3j iL Lκ κ⎛ ⎞= =⎜ ⎟⎝ ⎠

v

For the case of a parabolic curvature with value κm at midspan the deformations become

Lx

mκ

i j

2v 3v2

1 2 12 3 3m mL Lκ κ⎛ ⎞= − = −⎜ ⎟⎝ ⎠

v 31 2 12 3 3m mL Lκ κ⎛ ⎞= =⎜ ⎟⎝ ⎠

v

Other cases can be obtained with the help of integration tables, or numerical integration

Lx

2v3v

jκ

i j


Page 151

CE 220 - Theory of Structures Ex. 11 - Compatibility for determinate beam © Prof. Filip C. Filippou, 2000

Example 11(r) - Compatibility relations for statically determinate 2d beamObjectives:

(a) set up of the compatibility relations between the displacements at the free dofs of a staticallydeterminate beam and the element deformations

(b) determination of deformed shape for given element deformations(c) reduced set of degrees of freedom

In this example we will set up the compatibility relations between the displacements at the free dofs of astatically determinate beam with overhang and the deformations of each element in the model. The beamwith overhang is shown in Figure 1. It is clear from the figure that the horizontal translations andcorresponding deformations are independent from the transverse translations and rotations. We will notbother with the longitudinal translations of the nodes of the structural model and the corresponding changeof length in this problem, and instead concentrate on the transverse translations and node rotations.

a b c1

2 3 4

10 10 5

Figure 1: Statically determinate beam

Each node of the structural model has 3 independent ways of displacing in the plane. We call these thedegrees of freedom of a node. They are: translation in the global X-direction, translation in the globalY-direction, and rotation about the Z-axis or rotation in the plane of the structural model). As already stated,we limit our attention in this example to the node translations in Y and the rotations. The unrestrained dofsare numbered in sequence starting with the lowest number node. The free dofs are shown in Figure 2.

12

34 5

6

Figure 2: Free dof numbering

Length of elements

La 10:= Lb 10:= Lc 5:=

We want to set up a relation between the displacements (i.e. translations and rotations) of the nodesand the angle that forms between the node tangent and the node connector line for nodes connected byelements (this will be called a node connector line in the course). To display this angle without ambiguitywe make use of the following figure

the figure shows a line connecting node i and j implyingthat these nodes are connected in the structural model(note that the element connecting the nodes need not bea straight element!). It also shows a line emanating fromthe face of the square symbol for the node. This linepoints originally in the direction from node i to node j andfollows the node in its rotation. The angle is measuredfrom the line connecting the nodes to the line attached tothe square symbol for the node and is positive in thecounterclockwise (CCW) direction.

node i node j

Page 152


Dof displacements and corresponding deformations (angles)

To set up the relation between the free dof displacements and the angles between the node tangents andthe node connector lines in the structural model we proceed as follows: we activate each dof in turn andgive it a unit value; during activation of a particular dof the others remain "locked". We measure the anglebetween the node tangents and the node connector lines in the structural model. We number the anglessystematically starting with end i of element a, to end j of element a, end i of element b, etc. as shown in thefollowing figure for dof 1. We record the values of these angles for a unit displacement of the particular dofin a column of the kinematic matrix Af (one column for each dof with corresponding number). The followingfigure shows the concept and the angle numbering

3

2 4 5

a b c61

1 2 3 4 5 6dof 1

dof 2

dof 3

dof 4

dof 5

dof 6

Page 153


for unit displacement at dof 1 (rotation)

1 2 3 4 5 6dof 1 Af

1⟨ ⟩

1

0

0

0

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

for unit displacement at dof 2 (translation)

dof 2 Af2⟨ ⟩

1La

−

1La

−

1Lb

1Lb

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

Note: we limit ourselves to small displacements in this course (linear geometry). In such case the arctan ofthe angle is a very close approximation of the angle itself.


dof 3 Af3⟨ ⟩

0

1

1

0

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=


dof 4 Af4⟨ ⟩

0

0

0

1

1

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

Page 154


for unit displacement at dof 5 (translation)

dof 5 Af5⟨ ⟩

0

0

0

0

1Lc

−

1Lc

−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=


dof 6 Af6⟨ ⟩

0

0

0

0

0

1

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

After establishing the individual relations we can conclude that the combined effect of all free dofdisplacements on the angles between node tangents and node connector lines can be written in the form

Af Uf⋅ with Af

1

0

0

0

0

0

0.1−

0.1−

0.1

0.1

0

0

0

1

1

0

0

0

0

0

0

1

1

0

0

0

0

0

0.2−

0.2−

0

0

0

0

0

1

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

=

we note that matrix Af is the transpose of the equilibrium matrix Bf for the same beam structure in Example 3(case B without longitudinal basic forces).

Page 155



Now we look at the elements connecting the nodes. Elements that are supposed to be straight in idealconditions may end up curved either by intentional or unintentional fabrication, or by thermal effects.The following figure shows the effect of curvature on the element deformation. Note that curvature meansthe change of angle of the tangent to the deformed shape of the element at two infinitesimally adjacentsections (κ(x) dx in the figure). We see that the element curvature results in an angle between the tangentto the deformed shape of the element at each end and the line connecting the ends of the element. Thelatter is known as the element chord. These angles can be expressed by curvature integrals as shown inthe figure.

( )x dxκ 3v

0

( )

L

x x dxκ∫x L x−

( )0

( )

L

L x x dxκ−∫

2v

L

element chord

i

j

Recalling that the change of length of the element is counted as the first element deformation v1, the anglebetween tangent and chord at end i represents element deformation v2, and the angle between tangent andchord at end j represents element deformation v3. Noting that the angle is measured from the chord to thetangent, as shown in the figure and that CCW is positive we have

v21L

−0

LxL x−( ) κ x( )⋅

⌠⎮⌡

d⋅= and v31L 0

Lxx κ x( )⋅

⌠⎮⌡

d⋅=

after bringing the length L of the element into the integral we can write

v2

0

L

x1xL

−⎛⎜⎝

⎞⎟⎠

κ x( )⋅⌠⎮⎮⌡

d−= and v3

0

L

xxL

κ x( )⋅⌠⎮⎮⌡

d=

Page 156


We note that the above integrals represent the "moment" of the area under the curvature function aboutend i (for v2) and about end j (for v3) divided by the element length. This interpretation of the geometricfacts is, therefore, known in the history of structural analysis as "moment area" method.for the case of constant curvature, which we denote with κ0, we have

Lx

0 0( ) const=xκ κ=0

2Lκ0

2L

−κ

A constant curvature arises from a temperature field with a difference in the change of temperature betweenthe bottom and top fiber of the section as the following figure shows

x

reference axis

temperature increment

h

beam elementx

thermal field

L

i j

ΔTt

Δx

aTΔ

ΔTb

ΔTt

ΔTb

in this case

κ0 αΔTb ΔTt−

h⋅=

where h is the section depth

Page 157


Continuity requirement

Assuming now that the elements are not straight but deformed we state that the continuity of tangents atnodes with rigid or continuous connections requires that, when the element chord is placed so as to line upwith the line connecting the corresponding nodes, each element deformation should match thecorresponding angle between the node tangent and the node connector line in the structural model. Weshow this state of continuity in the following figure for each dof separately.

3

24 5

a b c61

dof 1

dof 2

dof 3

dof 4

dof 5

dof 6

We can see from the above figure that the final deformed shape of the structural model will be the linearcombination of the deformed shapes for each dof scaled by the displacement value of the corresponding dof

Page 158


Compatibility relations

To write the compatibility statement of angle equality in compact form we put the element deformations in asingle vector V starting with the deformation at end i for element a, then deformation at end j for element aand so forth, so that there is perfect agreement in numbering between the complete set of elementdeformations and the angles between node tangents and node connector lines in the structural model.

The compatibility or geometric continuity statement then reads V Af Uf⋅=

where Uf is the vector of free dof displacements

and where the deformation vector V is

deformation at end i of element a

( )

( )

( )

a

b

c

v

V v

v

⎛ ⎞⎜ ⎟⎜ ⎟=⎜ ⎟⎜ ⎟⎝ ⎠

deformation at end j of element a

deformation at end i of element bor, V

V1

V2

V3

V4

V5

V6

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

=deformation at end j of element b

deformation at end i of element c

deformation at end j of element c

We write the compatibility relations for the example explicitly

V1 U1U2

10−=

V2U2

10− U3+=

We note that there are as many compatibility equations as variables on the right handside of the equations V Af Uf⋅= .V3

U2

10U3+=

V4U2

10U4+=

V5 U4U5

5−=

V6U5

5− U6+=

Page 159


Numerical results

Let us assume that element a is deformed due to a uniform curvature field of κ0 1.2 10 3−⋅:=

while elements b and c are in their ideal condition, i.e. perfectly straight. What is the resulting deformedshape of the structure that ensures tangent continuity at all nodes?

Given V

0.006−

0.006

0

0

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:= note κ0La

2⋅ 0.006= we obtain Uf lsolve Af V,( ):= Uf

0.009−

0.03−

0.003

0.003

0.015

0.003

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

=

it is worth noting that the translation values are roughly an order of magnitude greater than the rotations.

The system of compatibility equations can be readily solved by hand allowing insight into the deformationbehavior of the beam. Since V3 through V6 are zero we conclude that

U6U5

5=

U4U5

5= and thus U4 U6=

U4U2

10−=

U3U2

10−= and thus U3 U4=

thus, in the absence of any deformations in elements b and c the rotations U3, U4 and U6 are equal.

finally, U2 5− V2⋅= and U1 V1U2

10+=

The deformed shape of the structure under a magnification factor of 50 is shown below. The geometricinterpretation of the above equations is clear from the figure.

10 10 5

U2 5− V2⋅

U1 V1U2

10+

Page 160


We repeat the conclusions and corollary of Example 10.

CONCLUSIONS

In a statically determinate structure there are as many element deformations (rows of Af) as there are freeglobal dof displacements (columns of Af). Consequently, Af is a square matrix. In a stable structure the rankof the matrix is equal to the number of columns (or rows for a determinate structure). This means that it isalways possible to find a deformed configuration of the structure that matches any given elementdeformations. It also means that the only way to match the undeformed ("ideal") state of all elements is byzero displacement values at all free dof's, i.e. by the undeformed configuration. We will see later that thekinematic matrix of an unstable structure is "rank deficient", meaning that the rank is smaller than thenumber of columns. In such case it is possible to find deformed positions of the structure without deformingany of the elements. This is the key criterion for judging the stability of a structural model.

There is a clear geometric interpretation of the compatibility relations. The solution of the compatibilityequations "by hand" affords insight into the deformed shape of the structure and the interplay of elementdeformations and global dof displacements.

Corollary

In a statically determinate structure it is possible to have any element deformations and "fit themtogether" without "forcing" the elements. This is not possible in a statically indeterminate structure.

The above structure is not subjected to any forces at the free global dof's. This means that there are nobasic element forces and, of course, no support reactions. The structure deforms but is stress-free!

Page 161


Reduced set of dofsFrom the compatibility equations, which we summarize again on theright, we can see that the displacement of dof 1 only affects thedeformation V1 and the displacement of dof 6 only affects thedeformation V6. We say that the corresponding dof does not have tonegotiate its value with more than one element. This is reflected by acolumn with a single non-zero entry in the compatibility matrix. In suchcase we can set aside the equation and the corresponding dof, andwork with a smaller set of free dofs and corresponding elementdeformations (smaller system of compatibility equations). If we areinterested in determining the displacement of a dof that has been setaside, we can do so after obtaining the values of the free dofdisplacements for the reduced set using the compatibility equationthat was set aside. We will demonstrate this in the following. We markthe equations that we set aside for later use with an asterisk (*). Notethat by setting aside two equations we also set aside thecorresponding dofs for the case at hand. We emphasize again that forthis to be possible it is necessary that the dof to be set aside affectsonly one compatibility relation and thus only one element deformation.

V1 U1U2

10−= (*)

V2U2

10− U3+=

V3U2

10U3+=

V4U2

10U4+=

V5 U4U5

5−=

V6U5

5− U6+= (*)

Because we intend to use this reduced set of dofs in future problems we number the reduced set of freedofs and the corresponding element deformations from scratch. We have

1

23 4

a b c

a b c

a b c

a b c

a b c

1 2 3 4

The kinematic matrix Af becomes

Af

1La

−

1Lb

1Lb

0

1

1

0

0

0

0

1

1

0

0

0

1Lc

−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

Af

0.1−

0.1

0.1

0

1

1

0

0

0

0

1

1

0

0

0

0.2−

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

=

note that the kinematic matrixAf is the transpose of the staticmatrix Bf for case C ofexample 3.

Page 162


It is worth noting in the preceding figure that the end nodes are free to rotate with the node connector lineemanating from them, since there is no rotation dof present at the end.

We restate that the kinematic matrix Af for the reduced set of dofs can be obtained directly from the matrixfor the full set of dofs by setting aside rows 1 and 6 and columns 1 and 6. With the renumbered dofs andelement deformations the compatibility relations become.

we now use the values for the element deformations to solvefor the displacements at the free dofsV1

U1

10− U2+=

deformation of node j of element aV2

U1

10U2+=

V

0.006

0

0

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:=V3

U1

10U3+= Uf lsolve Af V,( ):= Uf

0.03−

0.003

0.003

0.015

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

=

V4 U3U4

5−=

which, of course, match the earlier values for the same dofs. How can we calculate the rotationat node 1 and node 4? Quite simply, using the rows of the complete kinematic matrix that we have set aside.We renumber the corresponding deformation and dof: the rotation at node 1 is U5 and that at node 4 is U6.The deformation at end i of element a is V5 and that at end j of element c is now V6. We have the followingtwo compatibility relations that we set up earlier (see equations with asterisk):

V5 U5U1

10−= (*) solve for U5 U5 V5

U1

10+= U5 0.006−

Uf1

10+:= U5 0.009−=

V6U4

5− U6+= (*) solve for U6 U6 V6

U4

5+= U6 0

Uf4

5+:= U6 0.003=

the values are of course the same as before (compare with dofs 1 and 6 of the complete dof solution).

We can see the geometric interpretation of the above equations in the deformed shape of the structure.The rotation of node 1 is equal to the deformation of element a at the corresponding node added to therotation of the chord connecting node 1 and 2. Because element c does not deform, the rotation of node 4is equal to the rotation of the chord connecting nodes 3 and 4. Thus, for a quick determination in thefuture we can simply add the element deformations to the chord rotations at the node of interest.The deformed shape of the structure under a magnification factor of 50 is shown below.

10 10 5

U5 V5U1

10+ U1

Page 163


With the reduced set of dofs the deformed shape of the structural model will be the linear combination ofthe following deformed shapes for each independent dof after factoring these by the correspondingdisplacement value.

1

23 4

a b c

a b c

a b c

a b c

a b c

Page 164

Matlab script for Example 11 in CE220 class notes % Deformations and deformed shape for simply supported beam with overhang

use FEDEASLab functions % clear memory; close any open windows CleanStart % define model geometry XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 10 0]; % second node, etc XYZ(3,:) = [ 20 0]; % XYZ(4,:) = [ 25 0]; % % element connectivity array CON { 1} = [ 1 2]; CON { 2} = [ 2 3]; CON { 3} = [ 3 4]; % boundary conditions (1 = restrained, 0 = free) BOUN(1,:) = [ 1 1]; BOUN(3,:) = [ 0 1]; % specify element type ne = length(CON); % number of elements [ElemName{1:ne}] = deal('2dFrm'); % 2d frame element


specify element deformations (note that we also include the axial) V = zeros(3*Model.ne,1); V(2) = -0.006; V(3) = 0.006;


Page 165

solution for free global dof displacements Uf = Af\V; % display result format short disp('the free global dof displacements are'); disp(Uf); the free global dof displacements are -0.0090 0 -0.0300 0.0030 0 0.0030 0 0.0150 0.0030

plot deformed shape of structural model % put free dof displacement values in complete displacement vector U = zeros(Model.nt,1); U(1:Model.nf) = Uf; MAGF = 50; % magnification factor for deformed configuration Create_Window(0.80,0.80); Plot_Model(Model); Plot_DeformedStructure(Model,[],U,V);

Page 166

Force release deformations

we see that

Note: the presence of an axial device implies that the axialforce is zero only if the device "slips" under zero axial force.More typical are devices that "slip" at a certain axial force level

We investigate now the effect of force release devices on the element deformations. The devices can beof three types, as we have already mentioned in Part I, and can be located anywhere inside the element.

Normal force release or axial translation device

Bending moment release or rotational device

we see that

we obtain these formulas either directly from the figure or by using the Dirac delta function for the curvature κ(x) in the earlier integrals for the element deformations

If the moment release is located at node i or node j we obtain two special cases:

at node i

at node j

It is important to recall at this point that the element deformations are measured from the chord to the tangent at the node, which is shown in the above figures as a short line.

Shear force release or transverse translation device

we see that

Note that no relative rotation is allowed by thetransverse translation device, so that the two partsof the element are supposed to remain parallel.Thus, the end deformation angles are equal.

1h ah=v v

i j

ahv 1hv

i j

x L-xL

fhv

2hv3hv

2h fhL x

L−

= −v v

3h fhxL

=v v

2h fh= −v v 3 0h =v

2 0h =v 3h fh=v v

i j

fhv2hv

i j

fhv 3hv

2 3th

h h L= =−

vv v

x L-xL

thv 2hv 3hv


Page 167

Element deformationsWe generalize the relations on the preceding page by stating that the total element deformations v are made up of two contributions: the deformations resulting from the integration of continuous section deformations (we call these strain-dependent deformations in the following), and the deformations resulting from the effect of concentrated relative translations or rotations due to the presence of force release devices. We denote the former with vε and the latter with vh. With this notation we write for the element deformations

hε= +v v v

We collect each of the above contributions into a separate vector for all elements of the structural model and obtain hε= +V V V

the kinematic or compatibility relation for the structure then becomes f fhε= + = AV V V U

In a statically determinate structure there are as many free dof displacements on the right hand side of theabove kinematic relations, as there are strain-dependent element deformations and concentrated hinge deformations. Examples 10 and 11 provide good examples without a release, and Example 12 provides a good example with a release. In the absence of releases we use V instead of Vε for the strain-dependent element deformations, since Vh = 0.In a statically indeterminate structure there are more element deformations on the left hand side than there are free global dof displacements on the right hand side of the above kinematic relations. The element deformations will be compatible if they satisfy the compatibility constraint we have encountered earlier, i.e.

Tx =VB 0 compatibility condition for statically indeterminate structure

or, else NOS axial or rotational devices need to be introduced in the redundant elements of the structure,so as to increase the number of unknown hinge deformations on the left hand side of the above equationsfor a unique solution. Example 13 is a good case in point illustrating both cases.

i j

2εv

2v3v

h3v

3εv

here is an example with a moment release at end j


Page 168

CE220 - Theory of Structures Ex. 12 - Compatibility for portal frame with hinge © Prof. Filip C. Filippou, 2000

Example 12 - Compatibility relations for statically determinate 2d portal frame

Objectives:

(a) set up of the compatibility relations between the displacements at the reduced set of free dofs of astatically determinate portal frame (three hinge frame) and the element deformations

(b) determination of deformed shape for given element deformations(c) determination of hinge rotation at girder midspan of portal frame

We make the following assumption: the nodes of the frame model do not displace relative to each other inthe longitudinal (axial) direction of the element connecting two nodes, which also amounts to saying that thelongitudinal deformation of the frame elements (change of length) is negligibly small. We call such elementsinextensible. The statically determinate frame model is shown below.

12

8 8

10

1

2 3 4

5

a

b c

d

We make the following observations: the vertical translation of dofs 2 and 4 is zero (or restrained) onaccount of the assumption that elements a and d are inextensible. Similarly, the horizontal translation ofnodes 2, 3 and 4 is equal on account of the assumption about elements b and c being inextensible. Afterthese observations we realize that there are seven independent global dofs, as shown in the following figure

2

34

6

a

b c

d

1

5

7

Page 169


Length of elements

La 12:= Lb 8:= Lc 8:= Ld 10:=

Before setting up the relation between the free dof displacements and the angles that form between thenode tangents and the node connector lines in the structural model we note that dofs 1 and 7 affect onlyone angle at the corresponding node. We have already encountered this case in the preceding example forthe statically determinate beam. The same is also true for dof 5, because the angle at the left end ofelement c is of no interest for the lack of continuity at that location due to the rotational hinge. We will returnto this issue subsequently to determine the hinge rotation at the left end of element c. We, therefore, setaside these dofs and plan to determine their values separately at the end, just in case their are of interest.The new set of reduced dofs is shown in the following figure

23

4

a

b c

d

1

We proceed as follows: we activate each dof in turn and give it a unit value; during activation of a particulardof the others remain "locked". We measure the angle between the node tangents and the node connectorlines in the structural model. We number the angles systematically starting with end i of element a, to end jof element a, end i of element b, etc. as shown in the following figure for dof 1. We skip angles at locationswhere continuity will not be enforced or is of no interest, as is the case with the hinge and the supports. Werecord the angle values for a unit displacement of the particular dof in a column of the kinematic matrix Af(one column for each dof with corresponding number). The following figures show the concept and theangle numbering

dof 11

312

4

Af1⟨ ⟩

1La

0

0

1Ld

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

Page 170


1dof 2

31

24

Af2⟨ ⟩

1

1

0

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:=

dof 3

1

31

24

Af3⟨ ⟩

0

1Lb

−

1Lc

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

1dof 4

3

12

4

Af4⟨ ⟩

0

0

1

1

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:=

Page 171


After establishing the individual relations we can conclude that the combined effect of all free dofdisplacements on the angles between node tangents and node connector lines can be written in the form

Af Uf⋅ with Af

112

0

0

110

1

1

0

0

0

18

−

18

0

0

0

1

1

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

we note that matrix Af is the transpose of the static matrix Bf for the same frame model after setting asidethe equilibrium equations that involve axial forces in Example 4. In fact, we observe that it may be a littleeasier to set up the kinematic matrix under the assumption that the elements are inextensible than tocombine the equilibrium equations so as to eliminate the axial forces. We will make ample use ofthis fact through the principle of virtual displacements later on in the course.To write the compatibility statement of angle equality in compact form we put the element deformations in asingle vector V starting with the deformation at end i for element a, then deformation at end j for element aand so forth, so that there is perfect agreement in numbering between the set of element deformationsand the angles between nodes and lines connecting nodes in the structural model. We note again that inthis example we do not include in vector V the deformations at the element ends that are of no interest(like end i of element a, or end j or element d, and at the element end with no continuity because of thepresence of a rotational hinge, i.e. end i of element c

The compatibility or geometric continuity equations then read V Af Uf⋅=where Uf is the vector of free dof displacements

and where the deformation vector V is


deformation at end i of element bV

V1

V2

V3

V4

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

=deformation at end j of element c

deformation at end i of element d

We write the compatibility relations for the example explicitly

V1U1

12U2+=

V2 U2U3

8−=

We note that there are as many compatibility relations as variableson the right hand side of the equations V Af Uf⋅= .

V3U3

8U4+=

V4U1

10U4+=

Page 172


Numerical resultsLet us assume now that elements a and b are curved or cambered with uniform curvature κ0 (curvingoutward). The deformations are thus equal to

v κ0L2⋅= with the appropriate sign. We assume a curvature value of κ0 6 10 4−⋅:=

For the same κ0 the deformation is proportional to the length

κ0La

2⋅ 0.0036= for element a κ0

Lb

2⋅ 0.0024= for element b

leading to the following element deformation vector V 0.0036− 0.0024 0 0( )T:=

where we use the transpose on the right hand side to save some space in specifying the vector!

We can now solve for the displacements at the free dofs of the model

we observe that the translations (dofs 1 and 3) are anorder of magnitude larger than the rotations! This isassociated with the fact that the angle between nodetangent and element connector is equal to the transversetranslation value divided by the element length; thus, thedifference in magnitude between translations androtations is always on the order of the element length inconsistent units

Uf lsolve Af V,( ):= Uf

32.73−

0.87−

26.18−

3.27

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

10 3−=

Noting that V3 and V4 are zero, we can determine the free dof displacements easily by hand and thus gaininsight into the interplay between element deformations and global dof displacements.

from (1) and (2) we get V1 V2−U1

12

U3

8+=

from (3) and (4) we get 0U3

8

U1

10−= (*) combining we get V1 V2−

U1

12

U1

10+=

and we can solve for U1 and then substitute back to equation (*) to get U3 U3 0.8 U1⋅=

Finally, we can get the rotations U2 and U4 from the original equations namely: U4U1

10−=

U2 V2U3

8+=

adding the last two equations up and noting the earlier relation (*) we get U2 U4+ V2=

Finally, the rotations at the supports can be readily determined by adding the element deformation tothe rotation of the chord at the corresponding node, as we have already remarked for the precedingexample of the statically determinate beam.

Page 173


The deformed shape of the structure under a magnification factor of 50 is shown below

observe the chords andthe deformations forelements a and b !

The geometric interpretation of the earlier relations is left as an exercise. Note that elements c and ddisplace as a unit that pivots about the right support

Page 174


Determination of hinge rotation

We return to the compatibility equations with the intent to also include in the calculations the determinationof the hinge rotation at girder midspan.

12

8 8

10

1

2 3 4

5

a

b c

d

Structural system with node and element numbering

For determining the value of the hinge rotation we retain the rotation at node 3 as free dof. With this thereare five free global dofs as shown in the following figure.

1

23

5

a

b c

d

4

Page 175


The figures with the angles that result under a unit displacement of each free global dof do not change,with the exception of dof 3 and the new rotation dof. We insert these below and pay particular attention tothe fact that we also choose to record the angle between node 3 and the node connector line 2-3 as wellas the node connector line 3-4. Thus, the angles to be monitored are now 6, as labeled in the figures.

1

dof 31

2

3 4

56

13 5

612 4

dof 4

We write the structure kinematic matrix, i.e. the relation between the angles at the six locations in theabove figures and the global dof displacements (you may like to consult the earlier figures about dofs 1, 2,and 5). We obtain

Page 176


The element deformations are numbered accordingly and we have avector V that contains six values, as shown in the figure above.

Af

1La

0

0

0

0

1Ld

1

1

0

0

0

0

0

1Lb

−

1Lb

−

1Lc

1Lc

0

0

0

1

1

0

0

0

0

0

0

1

1

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= V

V1

V2

V3

V4

V5

V6

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

= Af

0.083

0

0

0

0

0.1

1

1

0

0

0

0

0

0.125−

0.125−

0.125

0.125

0

0

0

1

1

0

0

0

0

0

0

1

1

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

=

The compatibility relations now state that the angles need to match at five locations, namely 1, 2, 3, 5 and 6.

We write these compatibility relations explicitly,

V1U1

LaU2+= V4 V4ε V4h+=

U3

LcU4+= (**)

V2 U2U3

Lb−= V5

U3

LcU5+=

V3U3

Lb− U4+= V6

U1

LdU5+=

(**) at end i of element c, i.e. at number 4 we do not have continuity; instead we can only calculate thelack of continuity by subtracting the element deformation from the angle of the node relative to the lineconnecting nodes 3 and 4. Thus, instead of

V4U3

LcU4+= we have the relation V4h

U3

LcU4+

⎛⎜⎝

⎞⎟⎠

V4ε−=

where V4h is the hinge rotation at end i of element c.

We perform again the numerical calculations, but this time we also include a uniform curvature in element c.We assume the same uniform curvature value as for elements a and b (curving outward).

Page 177


Numerical resultsElements a, b and c are curved or cambered with uniform curvature κ0 (curving outward). Thedeformations are thus equal to

v κ0L2⋅= with the appropriate sign; we assume a curvature value of κ0 6 10 4−⋅:=

For the same κ0 the deformation is proportional to the length

κ0La

2⋅ 0.0036= for element a κ0

Lb

2⋅ 0.0024= for element b

κ0Lc

2⋅ 0.0024= for element c

this leads to the following element deformation vector


deformation at end i of element b

deformation at end j of element bVε

0.0036−

0.0024

0.0024−

0.0024

0.0024−

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=deformation at end i of element c

deformation at end j of element c

deformation at end i of element d

We need to satisfy continuity or compatibility relations at 1, 2, 3, 5 and 6. We set up the coefficient matrix forthe displacements at the free dofs and the corresponding left hand side deformation vector by extractingrows 1, 2, 3, 5 and 6 from Af and V, respectively. We have

Af

0.083

0

0

0

0

0.1

1

1

0

0

0

0

0

0.125−

0.125−

0.125

0.125

0

0

0

1

1

0

0

0

0

0

0

1

1

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

= Vε

3.6−

2.4

2.4−

2.4

2.4−

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

10 3−=

Coeff_Matrix

112

0

0

0

110

1

1

0

0

0

0

0.125−

0.125−

0.125

0

0

0

1

0

0

0

0

0

1

1

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= LHS

3.6−

2.4

2.4−

2.4−

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

10 3−:= only rows 1, 2, 3, 5 and 6

Page 178


We can now solve for the displacements at the free dofs of the model

Uf lsolve Coeff_Matrix LHS,( ):= UfT 19.64− 1.96− 34.91− 6.76− 1.96( ) 10 3−=

With these displacement values, which represent a unique solution of the compatibility equations for thegiven element deformations for the statically determinate frame we can determine the hinge rotations at alllocations of the structural model, i.e. 1 through 6. We have

continuity!! i.e. equality between Af Uf and V

Vh Af Uf⋅ Vε−:= Vh

0

0

0

13.53−

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

10 3−= discontinuity = hinge rotation at 4

i.e. inequality between Af Uf and V

hinge rotation

The figure shows that the hinge rotation is measured from the tangent of the deformed element at end ito the line following the node rotation. Since CCW is defined as positive for rotations and moments in thiscourse, the rotation as shown is negative, in perfect agreement with the numerical result.

CONCLUSIONIn a statically determinate structure there are as many deformation continuity conditions as independentfree global dofs. Including compatibility conditions at locations with rotational hinges permits thedetermination of the hinge rotations with the equation Vh Af Uf⋅ Vε−= where Vh are the hinge

rotations.The hinge rotations Vh are zero at locations of continuity between node and element. Thus, ingeneral the kinematic matrix may have more rows than columns, but continuity in a statically determinatestructure can be enforced only at as many locations as available free global dofs.

Page 179

Matlab script for Example 12 in CE220 class notes % Deformations and deformed shape for determinate portal frame (three hinge portal frame)

use FEDEASLab functions % clear memory and close any open windows CleanStart; % define model geometry XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 0 12]; % second node, etc XYZ(3,:) = [ 8 12]; % XYZ(4,:) = [ 16 12]; % XYZ(5,:) = [ 16 2]; % % element connectivity array CON { 1} = [ 1 2]; CON { 2} = [ 2 3]; CON { 3} = [ 3 4]; CON { 4} = [ 4 5]; % boundary conditions (1 = restrained, 0 = free) BOUN(1,:) = [ 1 1 0]; BOUN(5,:) = [ 1 1 0]; % specify element type ne = length(CON); % number of elements [ElemName{1:ne}] = deal('2dFrm'); % 2d frame element


1

2 3 4

5

1

2 3

4

specify element deformations (note that we also include the axial) V = zeros(3*Model.ne,1); V(2) = 0.0036; V(3) = -0.0036; V(5) = 0.0024; V(6) = -0.0024;

Page 180

V(8) = 0.0024; % with curvature in element c V(9) = -0.0024; % with curvature in element c


remove row of kinematic matrix that corresponds to release (i.e. row 8) ih = 8; ic = setdiff(1:3*Model.ne,ih); % indices of continuous element ends

solution for free global dof displacements Uf = Af(ic,:)\V(ic); % display result format short disp('the free global dof displacements are'); disp(Uf); the free global dof displacements are 0.0052 -0.0196 0 -0.0020 -0.0196 -0.0349 -0.0068 -0.0196 0 0.0020 0.0020

determine hinge rotation at left end of element c (consult pages 191-195 of reader) Vh = Af*Uf - V; disp('the hinge rotation is'); disp(Vh(ih)); the hinge rotation is -0.0135

plot deformed shape of structural model % put free dof displacement values in complete displacement vector U = zeros(Model.nt,1); U(1:Model.nf) = Uf; MAGF = 50; % magnification factor for deformed configuration Create_Window(0.80,0.80); Plot_Model(Model); % it is necessary to insert V in the argument list for the hinge discontinuity Plot_DeformedStructure(Model,[],U,V);

Page 181

CE220-Theory of Structures Ex. 13 - Compatibility for indeterminate truss © Prof. Filip C. Filippou, 2000

Example 13(r) - Compatibility relations for statically indeterminate 2d trussObjectives:

(a) set up of the compatibility relations between the displacements at the free dofs of a staticallyindeterminate truss and the element deformations

(b) determination of deformed shape for given element deformations(c) compatibility condition; hinge deformation at "redundant" element

The truss geometry is given in the following figure.

8

6

a

b c

d

e f

1 2

3 4

We focus on the number of independent ways that the nodes of the structure can displace. Since rotationsare of no interest in this case, there are two independent translations at each node, except for thetranslations that are restrained by supports. There are 5 independent free global dofs in this case.

8

6

a

b c

d

e f

1

23

4

5

Page 183


To set up the compatibility relation between free dof displacements and element deformations we proceed inthe following systematic way: (1) set each dof in turn to a unit value, while holding the other dofs equal tozero, (2) determine the change in distance between the nodes in the deformed configuration, (3) record thischange of distance in a column vector with each row number corresponding to the element number thatconnects the corresponding nodes. The column number corresponds to the number of the dof we "activate".The corresponding array or matrix is known as compatibility matrix Af

Preliminary geometric calculation: original distance between nodes numbered according to thecorresponding element number.

La 8:= Lb 6:= Lc 6:= Ld 8:= Le 62 82+:= Lf 62 82+:=

La 8= Lb 6= Lc 6= Ld 8= Le 10= Lf 10=

8

6

a

b c

d

e f

1

dof 2

8

6

a

b c

d

e f

1dof 1

8

6

a

b c

d

e f

1dof 3

8

6

a

b c

d

e f

dof 41

Page 184


8

6

a

b c

d

e f

1 dof 5

Af

1

0

0

0

8Le

0

0

0

0

1−

8Le

−

0

0

1

0

0

6Le

0

0

0

0

1

0

8Lf

0

0

1

0

0

6Lf

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

The change in node distances when all dof's are "activated" is the linear combination of the individual dofeffects. Since the dof translation values are parameters to be determined, we collect them into thefree dof displacement vector Uf and write for the change in node distances the following general relation

Af Uf⋅ with Af

1

0

0

0

0.8

0

0

0

0

1−

0.8−

0

0

1

0

0

0.6

0

0

0

0

1

0

0.8

0

0

1

0

0

0.6

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

=

we note that matrix Af is the transpose of the equilibrium matrix Bf for the same truss structure in Example 5.

There are now six element deformations for the five free global dofs and the compatibility matrix is 6 x 5.

Compatibility states that for the elements to fit perfectly in the space between the displaced nodes theelement deformations V need to be equal to the change in the distance between nodes. We write thisequality relation as follows:

V1 U1=V Af Uf⋅=

V2 U3=let us write it out in full to appreciate itbetter for this small structure

V3 U5=

V4 U2− U4+=

V5 0.8 U1⋅ 0.8 U2⋅− 0.6 U3⋅+=

V6 0.8 U4⋅ 0.6 U5⋅+=

Page 185


There are now more compatibility relations for the global free dof displacements to satisfy, than there areavailable free global dofs. Thus, if an arbitrary set of element deformations V were supplied, it is most likelythat a solution to all compatibility relations could not be found. The linear system of compatibility relationshas a unique solution only if the V's satisfy extra conditions, which we call deformation constraints orcompatibility conditions. The number of these conditions is equal to the difference between the number ofrows and the number of columns under the assumption that the structural model we are dealing with isstable. Recall from the equilibrium discussion that this means that the rank of matrix Af is equal to thenumber of columns, which amounts to saying that the rank of its transpose matrix Bf is equal to the numberof rows (recall discussion about equilibrium matrix in examples 1-6).

rank Af( ) 5= the rank of this truss structure is indeed 5, as is the number of columns; it is, therefore,stable and the element deformations need to satisfy a single constraint or compatibilitycondition. We embark on finding this constraint. To do this we proceed as follows:we select the first 5 compatibility relations and use them to express the free global dofdisplacements U1 through U5 in terms of element deformations V1-V5. Since this is asystem of five linear equations in five unknowns, it has a unique solution as long as therank of the coefficient matrix is equal to the number of equations. We try this and thenform the inverse symbolically.

Coeff_Matrix

1

0

0

0

0.8

0

0

0

1−

0.8−

0

1

0

0

0.6

0

0

0

1

0

0

0

1

0

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:= rank Coeff_Matrix( ) 5=

U1 V1=

U2 V1 0.75 V2⋅+ 1.25 V5⋅−=Coeff_Matrix 1−

1

1

0

1

0

0

0.75

1

0.75

0

0

0

0

0

1

0

0

0

1

0

0

1.25−

0

1.25−

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

= i.e.

U3 V2=

U4 V1 0.75 V2⋅+ V4+ 1.25 V5⋅−=

U5 V3=

of course, we note that the original 5 compatibility relations were simple enough that we could haveobtained the same answer "by hand". Can you try?

We take now the last two equations for U4 and U5 and substitute into the 6th compatibility relation aboveto obtain

V6 0.8 V1 0.75 V2⋅+ V4+ 1.25 V5⋅−( )⋅ 0.6 V3⋅+=

this is the condition between the element deformations to make sure that all compatibility relationscan be satisfied by a unique set of global dof displacements.

Page 186


Let us simplify the above condition and collect all terms on the left hand side. We get

0.8− V1⋅ 0.6 V2⋅− 0.6 V3⋅− 0.8 V4⋅− V5+ V6+ 0=

We write this relation as the vector product of two vectors and make an amazing discovery!

We have: 0.8− 0.6− 0.6− 0.8− 1 1( )

V1

V2

V3

V4

V5

V6

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

⋅ 0=

and we note that the vector multiplying the element deformations is the transpose of the force influencematrix Bbarx from example 5. Thus, the deformation constraint or compatibility condition that ensures that allcompatibility relations can be satisfied with a unique set of global dof displacements is

BbarxT V⋅ 0=

This is such a well known result in linear algebra that MIT Professor Gilbert Strang calls it thefundamental theorem of linear algebra, Part II. You can read about all this when you read Chapter 2 frompage 62 to page 99 and Chapter 3 from pages 132 to 139 of his outstanding book "Linear Algebra andits Applications", 3rd edition (on reserve in the Engineering Library).

We conclude our example by stating the following results:

In a statically indeterminate structure the compatibility matrix Af has more rows than columns. In fact, sincethe compatibility matrix is always the transpose of the equilibrium matrix for the same structural model (tobe proven later), the difference between number of rows and number of columns of the compatibility matrixis the degree of static indeterminacy or redundancy NOS. For a stable structure the rank of thecompatibility matrix is equal to the number of columns, which means that the rank of the equilibrium matrixis equal to the number of rows.

Since the compatibility matrix Af has more rows than columns in a statically indeterminate structure, there aremore compatibility relations than available free global dof displacements. The latter need to satisfy allcompatibility relations so that the deformed structure is continuous without gaps and angle discrepancies. Foran arbitrary set of element deformation values V this is not possible, unless NOS "hinges" are inserted andNOS redundant forces are released. If this is not the case, then the element deformation vector V satisfies the

deformation constraint or compatibility condition that BbarxT V⋅ 0=

Recall that for a degree of static indeterminacy NOS, there are NOS linearly independent self-stress statesthat make up the columns of matrix Bbarx. Thus, NOS independent compatibility constraints are imposed

on the element deformations through the compatibility condition that BbarxT V⋅ 0= . We will see this in the

following example 14.Linear algebra specialists (like Professor Strang) say that the column space of the compatibility matrix isalways orthogonal to the nullspace of the equilibrium matrix (which is its transpose). If the elementdeformations V lie in this column space then a unique solution of the compatibility relations exists.

Page 187


Let us see now what happens if an arbitrary set of element deformations is given. We assume thefollowing values

V 0.02 0.05 0.02− 0.05 0.02 0.02( )T:=

We use the first 5 compatibility equations to determine the set of global dof displacement values thatsatisfy these. We obtain

Ai

1

0

0

0

0.8

0

0

0

1−

0.8−

0

1

0

0

0.6

0

0

0

1

0

0

0

1

0

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:= Vi

0.02

0.05

0.02−

0.05

0.02

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:= Uf lsolve Ai Vi,( ):= Uf

0.02

0.033

0.05

0.083

0.02−

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

=

The distance between nodes 1 and 4 then can be obtained from the 6th row of the kinematic matrix Af

0 0 0 0.8 0.6( ) Uf⋅ 0.054=

or, quite simply with

U1

U2

U3

U4

U5

⎛⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎠

0.02

0.033

0.05

0.083

0.02−

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:= 0.8 U4⋅ 0.6 U5⋅+ 0.054=

since the corresponding element deformation, i.e. the deformation of element f, is only 0.02 an axialgap or slip of an axial translation device results. It is equal to 0.054-0.02. The formal way of expressingthis is that between nodes 1 and 4 we have the following relation

V6ε V6h+ 0.8 U4⋅ 0.6 U5⋅+= with V6ε 0.02:=

so that V6h 0.8 U4⋅ 0.6 U5⋅+ V6ε−:= V6h 0.034=

Page 188


The following figure provides the representation of the deformed shape of the structural model withthe gap in the in the axial translation device in element f.

0.034

Page 189

Matlab script for Example 13 in CE220 class notes % Deformations for statically indeterminate 2d truss

Clear workspace memory and initialize global variables CleanStart

Create model % specify node coordinates XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 8 0]; % second node, etc XYZ(3,:) = [ 0 6]; % XYZ(4,:) = [ 8 6]; % % connectivity array CON { 1} = [ 1 2]; CON { 2} = [ 1 3]; CON { 3} = [ 2 4]; CON { 4} = [ 3 4]; CON { 5} = [ 2 3]; CON { 6} = [ 1 4]; % boundary conditions (1 = restrained, 0 = free) BOUN(1,:) = [ 1 1]; BOUN(2,:) = [ 0 1]; % specify element type ne = length(CON); % number of elements [ElemName{1:ne}] = deal('Truss'); % truss element Model = Create_SimpleModel (XYZ,CON,BOUN,ElemName);


1 2

3 4

1

2 3

4

56

Compatibility relations kinematic matrix for all dofs in structural model

A = A_matrix (Model); % extract free dofs

Page 190

Af = A(:,1:Model.nf); % specify the element deformations V = [0.02;0.05;-0.02;0.05;0.02;0.02]; % extract nf rows of kinematic matrix Ai = Af(1:5,:); % extract corresponding deformations Vi = V(1:5);

solution for free global dof displacements Uf = Ai\Vi; % display result format short disp('the free global dof displacements are'); disp(Uf); the free global dof displacements are 0.0200 0.0325 0.0500 0.0825 -0.0200

determine incompatible ("hinge") deformation at NOS redundant elements Vh = Af*Uf - V; disp('the incompatible deformation is'); disp(Vh(6)); the incompatible deformation is 0.0340

plot deformed shape of structural model % put free dof displacement values in complete displacement vector U = zeros(Model.nt,1); U(1:Model.nf) = Uf; MAGF = 10; Create_Window(0.80,0.80); Plot_Model (Model); Plot_Model (Model,U);

Page 191

CE220 - Theory of Structures Ex 14 - Compatibility for indeterminate braced frame © Prof. Filip C. Filippou, 2000

Example 14 - Compatibility relations for statically indeterminate braced frameObjectives:

(a) set up of the compatibility relations between the displacements at the reduced set of free dofs of astatically indeterminate braced frame and the element deformations

(b) derive the necessary conditions for the element deformations so that a unique solution is possiblefor the displacements at the free global dofs (compatibility conditions for the deformations)

We assume that the elements of the braced frame in the figure are inextensible, so that the nodes cannotmove relative to each other in the longitudinal (axial) direction of the elements. Under this assumption thestructural model has only 4 free global dofs, as shown.

dof 1

dof 2 dof 3dof 4

a

b c

d

4 4

6

1

2 34

La 6:=

Lb 4:=

Lc 4:=

Ld 82 62+:=

We are interested in the angle between node tangent and node connector line at 5 locations. These aresystematically numbered in the following figures. We are also interested in the change of distance betweennode 1 and 4, since there is an axially deformable element at this location. The relation between global dofdisplacements and these deformations for the translation dofs can be inferred from the following figures

4 4

dof 3

6

1

2

34 5

6

4 4

6

dof 1

1

2

3 4 5

6

Page 192


The compatibility matrix is We recall the equilibrium equations from Example 6 andconfirm again that the compatibility matrix is the transpose ofthe equilibrium matrix.

Equilibrium equations from Example 6

P1Q1 Q2+

60.8 Q6⋅+=

P2 Q2 Q3+=Af

16

16

0

0

0

0.8

0

1

1

0

0

0

0

0

14

−

14

−

14

0

0

0

0

1

1

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=P3

Q3 Q4+

4−

Q5

4+=

P4 Q4 Q5+=

There are now 5 angles and one change of distance that reflect the effect of the free global displacements.Compatibility is satisfied if the element deformations match these 5 angles and the change of distance.

We can write V Af Uf⋅=

V1U1

6=

or, written out in full

V2U1

6U2+=

V3 U2U3

4−=

V4U3

4− U4+=

V5U3

4U4+=

V6 0.8 U1⋅=

There are now 6 equations in involving 4 variables, the free global dof displacement values. The differencebetween the number of compatibility equations and the variables is NOS, the degree of static indeterminacy,something to be expected, of course, since we noted that the kinematic matrix Af is the transpose of thestatic matrix Bf, which has NOS more columns than rows. If the left hand side vector is given, such a systemof equations does not have a solution, unless the vector satisfies certain conditions. We set out to determinethese conditions now. To do so, we select four linearly independent compatibility relations and solve for theglobal dof displacements in terms of the element deformations (we assume that the latter will be given). Inthis example we have quite a choice of which four equations to select, but because we need to prove acertain relation of the resulting equations with a result from Example 6, we select the compatibility equations2, 3, 5 and 6. We solve for the free global dof displacement values by hand and get

Page 193


from eq (6) U1 1.25 V6⋅=

from eq (2) U2 V2U1

6−= V2

524

V6⋅−=

from eq (3) U3 4 U2⋅ 4 V3⋅−= 4 V2⋅56

V6⋅− 4 V3⋅−=

from eq (5) U4 V5U3

4−= V5 V2−

524

V6⋅+ V3+=

To ensure that the two deformation values that we did not use above are compatible with the rest, wesubstitute the free global dof displacement values into the compatibility equations 1 and 4 and get

from eq (1) V1524

V6⋅= I.e. V1524

V6⋅− 0=

from eq (4) V4 V2−524

V6⋅+ V3+ V5+ V2−524

V6⋅+ V3+= 2− V2⋅ 2 V3⋅+ V5+512

V6⋅+=

i.e. 2 V2⋅ 2 V3⋅− V4+ V5−512

V6⋅− 0=

We write the two compatibility conditions in compact form

1

0

0

2

0

2−

0

1

0

1−

524

−

512

−

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

V1

V2

V3

V4

V5

V6

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

⋅0

0⎛⎜⎝⎞⎟⎠

= or, after checking with Example 6 BbarxT V⋅ 0=

Thus, we conclude that in order for a solution of the compatibility equations V Af Uf⋅= to exist for a givendeformation vector V, the element deformations V need to satisfy NOS conditions, which take the form

BbarxT V⋅ 0= . These conditions are known as compatibility conditions. There are no compatibility conditions

in a stable, statically determinate structure (NOS=0), which means that any given deformation vector iscompatible with a unique set of free global dof displacements.

Linear algebra deals with the solution of linear systems of the form V Af Uf⋅= extensively. If there are moreequations than unknowns, then it is stated that a solution exists only if the vector V lies in the column spaceof Af. (Strang, 3rd ed, pp. 66). To prove that a vector V lies in the column space of a matrix Af, it is ofteneasier to show that the vector is orthogonal to the so-called left nullspace of the matrix, i.e. the nullspace ofthe transpose of the matrix. Since Bf = transpose (Af) and the vectors of Bbarx span the nullspace of Bf, this

results in the condition that BbarxT V⋅ 0= , as stated earlier. This is the necessary and sufficient condition

for V to lie in the column space of the matrix Af, so that a solution exists to the compatibility equations V Af Uf⋅= (Strang, 3rd ed, pp. 138-139).

Page 194

Let us show the relationship between equilibrium equations and compatibility relations.

Relation between equilibrium and compatibility

Recall that

Recall also that

noting that

start with or

solve top equation since Ai square and non-singular

or

substitute solution into the second equation

noting that

and

f fw i i x x− = +B BP P Q Q

Tf f=A B

( )1i i f fw x x

− ⎡ ⎤= − −⎣ ⎦B BQ P P Q

x x=Q Q

( )1 1

i i i xf fw x

x

− −⎡ ⎤ ⎡ ⎤⎛ ⎞ −= − +⎢ ⎥ ⎢ ⎥⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎣ ⎦ ⎣ ⎦

B B B0 I

QP P Q

Q


f f= AV U i if

x x

⎛ ⎞ ⎡ ⎤=⎜ ⎟ ⎢ ⎥

⎝ ⎠ ⎣ ⎦

AA

VU

V

1f i i

−= AU V 1 if i

x

− ⎛ ⎞⎡ ⎤= ⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠A 0

VU

V

Tf i= BU V T

i i=A B

1x x i i

−= A AV V 1x i i x 0−− + =A A V V

1 ix i

x0− ⎛ ⎞⎡ ⎤− =⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠

A A IVV

1x i 0−⎡ ⎤− =⎢ ⎥⎣ ⎦

A A I V

Tx 0=B V T

x x=A B

( )T T T=CD D C

TifTx

0

⎛ ⎞⎛ ⎞ ⎜ ⎟=⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

B

B

UV


Page 195

A significant reduction in the number of independent free dof's can be achieved by considering approximations of the actual behavior of structural elements of the form: (a) axial deformations of frame elements can be treated as negligible in many cases, because they contribute very little to the global dof displacements of frame structures, and (b) certain elements of the structural model can be considered relatively inflexible to the other elements, because they possess a rather large flexural stiffness (e.g. shear wall elements). We impose these conditions in the form of constraint equations between the original free dof's. Using the constraint equations we can express all original dof displacements in terms of a smaller number of constrained dof's. Under small displacements the constraint relations are linear.

1. Case: inextensible element

recall that the axial deformation of a 2d frame element is given by

Thus, if v=0, a linear relation between the end translation dof's results.

2. Case: inflexible element

recall that the end deformations of a 2d frame element are given by

Linear constraints for inextensible and inflexible elements

j

= j iX X XΔ −

= j iY Y YΔ −2 2L X Y= Δ + Δ

X

Y

i

5u

1u

4u

2u

( ) ( )

1

2

3

4

5

6

4 1 5 2

0 0X Y X YL L L L

X YL L

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟Δ Δ Δ Δ⎛ ⎞⎜ ⎟= − −⎜ ⎟⎜ ⎟⎝ ⎠⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

Δ Δ= − + −

uuu

vuuu

v u u u u

( ) ( )4 1 5 20 X YL LΔ Δ

= − + −u u u u

1

22 2 2 2

2 3

3 42 2 2 2

5

6

1 0

0 1

Y X Y XL L L LY X Y X

L L L L

⎛ ⎞⎜ ⎟

Δ Δ Δ Δ⎛ ⎞⎜ ⎟− −⎜ ⎟⎜ ⎟⎛ ⎞ ⎜ ⎟⎜ ⎟=⎜ ⎟⎜ ⎟ ⎜ ⎟Δ Δ Δ Δ ⎜ ⎟⎝ ⎠ − −⎜ ⎟⎜ ⎟⎝ ⎠⎜ ⎟

⎜ ⎟⎝ ⎠

uu

v uv u

uu

For an inflexible element both deformations are zero, i.e.( ) ( )

( ) ( )

4 1 5 2 32 2

4 1 5 2 62 2

0

0

Y XL L

Y XL L

Δ Δ= − − − +

Δ Δ= − − − +

u u u u u

u u u u u

which can be rewritten in more compact form as ( ) ( )3 6 4 1 5 22 2Y X

L LΔ Δ

= =− − + −u u u u u u

the constraint equations for an inflexible element show that the end rotations are equal; furthermore, the relative end translations satisfy a specific linear relation

From the above two cases we conclude that an inextensible element introduces one constraint equation, while an inflexible element introduces two constraint equations. In general, in a structural model constraint equations are independent. In such case the number of indepedent free dof's is reduced by the number of available constraint equations. We can be more specific. An inextensibility condition involves only translation dof's. Thus, it reduces the number of independent translation dof's, but does not constrain the rotation dof's in any way. By contrast, the inflexibility constraint involves both rotation and translation dof's.


Page 196

CE220 - Theory of Structures Ex. 15 - Compatibility for portal frame with constraints © Prof. Filip C. Filippou, 2000

Example 15 - Compatibility relations for 2d portal frame with linear constraintsObjectives:

(a) set up formally the compatibility relations for a portal frame with inextensible elements

Start with full set of free global dofs for the case that axial deformations are not negligible.

4 4

5

1

2

a

b c

d

3 4

56 7

89

There are 3 deformations per elementfor a total of 12 deformations. Thecompatibility relations are

V Af Uf⋅= with

The numbering of the elementdeformations is shown in thefollowing figure in italic font.

3

1

2

45 6 78 9

10

11

12

Af

0

15

15

1−

0

0

0

0

0

0

0

0

1

0

0

0

14

14

0

0

0

0

0

0

0

0

1

0

1

0

0

0

0

0

0

0

0

0

0

1

0

0

1−

0

0

0

0

0

0

0

0

0

14

−

14

−

0

14

14

0

0

0

0

0

0

0

0

1

0

1

0

0

0

0

0

0

0

0

0

0

1

0

0

0

15

15

0

0

0

0

0

0

0

14

−

14

−

1

0

0

0

0

0

0

0

0

0

0

1

0

1

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

Page 197


We assume now that elements a, b, c and d are inextensible. This means that

V1 0= or, from the compatibility relations 0 U2=

V4 0= 0 U1− U4+= U1 U4=

V7 0= 0 U4− U7+= U4 U7=

V10 0= 0 U8=

These 4 linearly independent linear constraints establish 4 linearly independent linear relations betweenthe original set of 9 free global dofs. Consequently, only 5 independent free dofs remain. We select theoriginal dofs 1, 3, 5, 6 and 9 as the independent free global dofs and denote them with a new symbol Utild,which stands for constrained free global dofs (we cannot insert a tilde in Mathcad, so we need to resort tothe spelled out version). We have the following relations between the original and the constrained dof set

4 4

5

12

a

b c

d

1

34 15

U1 Utild1=which we can write compactly in matrix form as Uf Ac Utildf⋅=

U2 0=where Ac is the constraint compatibility matrix

U3 Utild2=

U4 Utild1=

U5 Utild3=

U6 Utild4=with Ac

1

0

0

1

0

0

1

0

0

0

0

1

0

0

0

0

0

0

0

0

0

0

1

0

0

0

0

0

0

0

0

0

1

0

0

0

0

0

0

0

0

0

0

0

1

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=U7 Utild1=

U8 0=

U9 Utild5=

Page 198


We can now obtain the following compatibility relations in terms of the constrained free global dofs

V Af Uf⋅= Af Ac⋅ Utildf⋅= Atildf Utildf⋅= with Atildf Af Ac⋅:=

we note, of course, that the deformations 1, 4, 7 and 10 areidentically equal to zero, as we have assumed at the beginning

deformations for dof 1

4 4

5

11

where Atildf

0

15

15

0

0

0

0

0

0

0

15

15

0

0

1

0

1

0

0

0

0

0

0

0

0

0

0

0

14

−

14

−

0

14

14

0

0

0

0

0

0

0

0

1

0

1

0

0

0

0

0

0

0

0

0

0

0

0

1

0

1

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

deformations for dof 3

4 4

5

1

3

Page 199


Since we know that the axial deformations are zero, we can write the compatibility matrix for the constrainedfree global dofs with only 8 deformation entries (note that the difference between deformations andconstrained free global dofs is NOS, i.e. 3) in this case. We have

Atildf

15

15

0

0

0

0

15

15

0

1

1

0

0

0

0

0

0

0

14

−

14

−

14

14

0

0

0

0

0

1

1

0

0

0

0

0

0

0

0

1

1

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= observe that this compatibility matrix is the transpose ofthe equilibrium matrix of the same portal frame from Example 8

P1Q1 Q2+

5

Q7 Q8+

5+=

P2 Q2 Q3+=

P3Q3 Q4+

4−

Q5 Q6+

4+=

P4 Q4 Q5+=

P5 Q6 Q7+=

Bf

0.2

0

0

0

0

0.2

1

0

0

0

0

1

0.25−

0

0

0

0

0.25−

1

0

0

0

0.25

1

0

0

0

0.25

0

1

0.2

0

0

0

1

0.2

0

0

0

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=

Bf

15

0

0

0

0

15

1

0

0

0

0

1

14

−

0

0

0

0

14

−

1

0

0

0

14

1

0

0

0

14

0

1

15

0

0

0

1

15

0

0

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

Page 200

1. Example: simple case of portal frame with inextensible elements

b

a c

6

1

23 4

5

b

a c

1 1

2 3

with constraints

becomes

Let us look at a few examples:

2

4 1

5

000

=

= −

=

U

U U

U

Note: only translation dofs are affected with inextensible constraints

2. Example: gable frame with inextensible elements

1

2

3

4

5

6

7

8

9

a

b c

d

8 8

12

6

with constraints

1 4 5 7U U U U( ) ( )

( ) ( )

2

4 1 5 2

7 4 8 5

8

08 60

10 108 60

10 100

=

= − + −

= − − −

=

U

U U U U

U U U U

U

= 0

Two constraint equations for 4 translation dofs. Thus, several choices for two independent translation dof's are possible

One choice is to select the original dof's1 and 4 as independent translation dof's

12

3 4

5

Another choice is to select theoriginal dof's 1 and 5 as independent

12

3

4

5


Page 201

3. Example: portal frame with inextensible and one inflexible element

a

b

c

1

24

3

5

67

H

with constraints

3

21 4

5 2

6

0

0

0

H

=

= = −

= −

=

U

UU U

U U

U

rigid

a

b

c

12

becomes

4. Example: dependent or inactive constraints

a

b

12

a

b

1

2

3

with constraints

becomes

a b

c

de

1

a b

c

d

e

1

23

with constraints

becomes

2

2

00=

=

U

U

1

2

0000

=

=

==

U

U

Thus, the number of independent free dofs is equal to the number of unconstrained free dofs minus the number of linearly independent constraint conditions. This number corresponds to the rank of the constraint matrix, as we will see subsequently.


Page 202

The rank of the constraint matrix ALC is equal to the number of independent constraints.Thus, the number of indepedent free dof's is equal to the number of original free dof's of the structural model (the number of unconstrained dof's) minus the rank of the constraint matrix.

Constraint compatibility matrix

The linear constraint conditions form a set of linear homogeneous equations in terms of the original dof displacements. We can write this system of equations symbolically

(1)

We can use equation (1) to select a set of independent dof's and express the remainder in terms of these. We have seen this process already in action with the equilibrium equations and the determination of the force influence matrix for the redundant basic forces, but we repeat it here for review's sake.

We select a set of independent dof's among the global dof displacement vector U and denote them with subscript i (forindependent). The remainder are denoted with subscript c (for constrained). From (1) we get

solving for Uc we get

We supplement the last equations with the identity

where I is the identity matrix; this way we recover the complete set of dof's on the left hand side

we get

for better identification we denote the independent free dof's under constraints with a new symbol

and write

Naturally, in the absence of any constraints the constraint compatibility matrix is the identity matrix.

for the relation between the original set of dof'sand the independent free dof's under constraints

we call Ac the constraint compatibility matrix

LC0 = A U

LCi i LCc c0 = +A AU U ( ) 1c LCc LCi i

−= − A AU U

i i= IU U

( ) 1c LCc LCi

f i c ii

reorder reorder−⎡ ⎤⎛ ⎞ −⎢ ⎥= = =⎜ ⎟⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦

A A AI

UU U U

U

fU

f c f= AU U

for application consult Examples 15 and 16


Page 203

Geometric determination of constraint compatibility matrix

1. Step: displacement of point R lying on plane that undergoes infinitesimal rotation about point 0

0

R

R'

RXΔ

RYΔ

X

Y

RXU

RYU

θ

RL

θ RLfrom similar triangles

2. Step: change of distance of two nodes rotating infinitesimally about common instantaneous center of rotation IC

for relative end displacement

Infinitesimal rotation of the line connecting two nodes is equivalent to inextensible response; thus, inextensible elements rotate about an instantaneous center of rotation; how to locate this center?

X

Y

IC

θθ

i

j

θ

4u

5u

1u

CYΔ

θ

CXΔ

XΔ

YΔ

2u

R RXRX R

R Rθ

θY

YL LΔ −

≅ ⇒ = − ΔU

U

R RYRY R

R Rθ

θX

XL LΔ

≅ ⇒ = ΔU

U

( )( )

1 C

2 C

4 C

5 C

θθ

θ

θ

YX

Y Y

X X

= − Δ

= Δ

= − Δ + Δ

= Δ + Δ

u

u

u

u

4 1

5 2

θθ

YX

− = − Δ

− = Δ

u u

u u

( ) ( )4 1 5 2 θ θ 0X Y Y X X Y− Δ + − Δ = − Δ Δ + Δ Δ =u u u u


Page 204

Location of instantaneous center of rotation

from relations on preceding page we have

for infinitesimal rotation angle

coordinates of IC

X

Y

IC

θθ

i

j

θ

4u

5u

1u

CYΔ

θ

CXΔ

XΔ

YΔ

2u

after subtitution of θ

Determination of rotation angle with given instantaneous center of rotation

for infinitesimal rotation angle

or, better or, or, or,

i.e. with the translation at either end of the element determine the rotation angle by dividing this translation with the distance normal to the translation direction from the point of translation to the instantaneous center of rotation

5 24 1θY X

−−= − =

Δ Δ

u uu u

1 2C Cθ θ

Y XΔ = − Δ =u u

1 2C C

4 1 5 2Y Y X XΔ = Δ Δ = Δ

− −

u u

u u u u

5 24 1θY X

−−= − =

Δ Δ

u uu u

1

Cθ

Y=−

Δ

u 2

Cθ

X=Δ

u 4

Cθ

Y Y=−

Δ + Δ

u5

Cθ

X X=Δ + Δ

u


Page 205

consider 4 special cases

5θX

=Δ

u

4u

5u

i

j

1 0=u

2 0=uIC

Case a

IC is located at end i of element; rotationcan be determined with either translation

4θY

=−Δ

u 5θX

=Δ

uor

IC is located somewhere on the line parallel to Xthrough node i; rotation can be determined fromhorizontal translation at node j

4θY

=−Δ

u

i

j

2u

1 0=u

4u

IC

Case b

5u

i

j

1u

IC Case c

4u

5u

2 0=u

IC is located somewhere on the line parallel to Y through node i; rotation can be determined fromvertical translation at node j

i

j

ICXΔ

YΔCase d

4u5 0=u

1 0=u

2u

IC is located at the intersection of the line parallelto X through node i and the line parallel to Ythrough node j; rotation can be determinedwith either translation

4θY

=−Δ

uor 2θ

X=−

Δ

u

for application consult Example 17


Page 206

We have encounted two very important matrices of a structural model: the equilibrium matrix Bf and the compatibility matrix Af. Since the compatibility matrix is the transpose of the equilibrium matrix the following remarks apply to both. We have mentioned repeatedly that the rank of either matrix needs to be equal to the number of independent free global dofs in the model for the structure to be stable. With a mathematical toolbox we can readily check the rank of a matrix. Is there a way to do it by "inspection"? To do this we need to consider what it means for the compatibility matrix not to have full rank, i.e. rank equal to the number of free global dofs. It means that the system of compatibility relations

Tranlated to our structural problem this means, that V=0 implies Uf = 0 in a stable structure (the undeformed configuration), while V=0 can result in non-zero vector Uf, which constitutes a collapse mechanism and is unacceptable. (Note that the stability of a structure at the undeformed configuration is a property of the compatibility matrix!).

To apply the stability check in practice we ask the following question:

The answer is no, if the IC locations contradict, in which case the structure is stable.

Is it possible for global dof displacements to take place without element deformations?

The answer is yes, if a set of consistent IC's of rotation for the different elements or blocks of the structural model exist.

Let us look at a few examples

Answering this question becomes very important for statically indeterminate structures, when we select the basic force redundants, hoping that we can solve the equilibrium equations for the primary basic forces. The system that remains when the redundants are set equal to zero is called the primary system, and it needs to be stable.

Stability of structural models

can have a non-zero solution for V=0.f f= AV U

NOS=4

example frame with NOS=4statically determinate system: unstable

Alternative A: unstable

A B C D

ICD

ICC

RCCD

ICC

RCBC

ICB

1

2

3

4

A B CD

ICC @ ∞


Page 207

A B C

D

ICA

ICC

A and C form a three hinge frame = stable

NOS=4

example frame with NOS=4

Alternative B: stable

A B C

ICA RCAB ICC

RCBC

ICBICB

ICB

statically determinate system: stable

Note the contradictory location of the instantaneous center for element B when viewed from A or C.

Note that we set the axial deformations to zero, since it is enough to make the argument with the flexural deformations.

stable

A statically indeterminate example (w/o hinges NOS=9)

A B C

D

stable

another example (originally NOS=4)

A B

CD

E

unstablestable


Page 208

CE220 - Theory of Structures Example 16 - Gable Frame with Constraints © Prof. Filip C. Filippou, 2000

Example 16 - Compatibility Relations for Gable Frame with Constraints - Analytic Method

In this example we set up the compatibility relations for a gable frame structure. We start will all free globaldofs and then constrain them by inextensibility and then inflexibility conditions.

The gable frame structure is the same as in Example 1. The following figure shows the geometry.

1

2

3

4

5

6

7

8

9

a

b c

d

10

12

6

8 8

XY

node 1

node 2

node 3

node 4

node 5

there are 10 free global dof's,as shown. Dof #10 only affectsthe angle at the base ofelement d and is put aside.

The resulting free global dof's without any constraints are:

1

2

3

4

5

6

7

8

9

a

b c

d 12

6

8 8

XY

Page 209


To set up the compatibility matrix Af we study a pair of nodes after translation and rotation. We note thatthere are the following measures of relative displacement: the change in the distance between the nodes,and the angles between the node tangent and the node connector line. We number these in order 1, 2 and3.

4u

5u

1u

2u

X

Y

xy

L

L

i

j

nL

3u

6u

1v

2v

3v

j

i

β

We have for the change in distance v1 Ln L−=

and limiting ourselves to small displacements v1ΔXL

u4 u1−( )⋅ΔYL

u5 u2−( )⋅+=

for the angle we have v2 u3 β−= and u3 u6 β−=

for small displacements β1L

ΔXL

u5 u2−( )⋅ΔYL

u4 u1−( )⋅−⎡⎢⎣

⎤⎥⎦

⋅=

Page 210


collecting all these relations into a matrix relation we get the compatibility matrix that expresses the relationbetween the change of distance and the angles between the node tangent and the node connector line onthe one hand and the end node displacements in the global reference on the other. We have

v ag u⋅= where ag

ΔXL

−

ΔY

L2−

ΔY

L2−

ΔYL

−

ΔX

L2

ΔX

L2

0

1

0

ΔXL

ΔY

L2

ΔY

L2

ΔYL

ΔX

L2−

ΔX

L2−

0

0

1

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

=

We apply this relation systematically to the above frame noting the following:

element a will be placed between nodes 1 and 2 ΔX 0:= ΔY 12:= L 12:=

element b will be placed between nodes 2 and 3 ΔX 8:= ΔY 6:= L 10:=

element c will be placed between nodes 3 and 4 ΔX 8:= ΔY 6−:= L 10:=

element b will be placed between nodes 4 and 5 ΔX 0:= ΔY 12−:= L 12:=

The resulting structure compatibility matrix Af is

there are 9 free global dof's represented bythe 9 columns of the matrix, one columnrepresenting the effect of a single dof

there 11 relative displacement entries, i.e. eitherchange of distance or angle between node andline connecting the nodes, represented by the11 rows of the matrix, one row for each relativedisplacement

Af

0

112

112

0.8−

0.610

−

0.610

−

0

0

0

0

0

1

0

0

0.6−

0.810

0.810

0

0

0

0

0

0

0

1

0

1

0

0

0

0

0

0

0

0

0

0.8

0.610

0.610

0.8−

0.610

0.610

0

0

0

0

0

0.6

0.810

−

0.810

−

0.6

0.810

0.810

0

0

0

0

0

0

0

1

0

1

0

0

0

0

0

0

0

0

0

0.8

0.610

−

0.610

−

0

112

0

0

0

0

0

0

0.6−

0.810

−

0.810

−

1

0

0

0

0

0

0

0

0

0

1

0

1

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

Page 211


We observe that the compatibility matrix is the transpose of the equilibrium matrix for the same structure withthe same numbering of equilibrium equations and basic forces. For a stable structure the rank of thecompatibility matrix should be equal to the number of columns, i.e. equal to the number of independent freeglobal dofs. Here we have

rank Af( ) 9= and indeed the structure is stable

The degree of static indeterminacy NOS of the structure is equal to the difference between the number ofcontinuity conditions and the number of independent free dofs, assuming that the latter is equal to the rank.Thus, if we enforce continuity in all 11 locations, as the detailing of the structural model suggests, then thedegree of static indeterminacy of this structure is NOS=11-9 = 2.

Inextensible element - Linear constraints

Let us now assume that the nodes are constrained to translate so that the distance between nodes withelements does not change. These introduces as many constraints as elements. These constraints are linearunder small displacements. For the structure in this example we have 4 constraints

0 U2= using the first two equations and simplifying we have

0810

U4 U1−( )⋅610

U5 U2−( )⋅+= 0810

U4 U1−( ) 610

U5⋅+=

0810

U7 U4−( )⋅610

U8 U5−( )⋅−= 0810

U7 U4−( )⋅610

U5⋅+=

0 U8=

thus, we are left with 2 linear equations for 4 translation dofs. These equations constrain the 4 translationdofs, so that only 2 are independent and the other 2 need to have values that satisfy the two equations(recall the discussion about basic and free variables in linear algebra, or the discussion about primaryand redundant basic forces for the equilibrium equations; the situation here is entirely analogous).

Let us pick the following independent free dof's and renumber them to be in sequence

13

4 6

91

2

3 4

5

Page 212


We solve first the two linear constraint equations:

set U1 1= and U4 0= from the first equation we get U543

U1⋅=43

=

and from the second U7 1− U1⋅= 1−=

set U1 0= and U4 1= from the first equation we get U543

− U4⋅=43

−=

and from the second U7 2 U4⋅= 2=

We are now ready to write the linear transformation relation between the original and the constrained setof free dof's. For the latter we use the symbol Utildf and the numbering on the right figure above. We have

U1 Utild1= U543

Utild1⋅43

Utild3⋅−= U9 Utild5=

U2 0= U6 Utild4=

U3 Utild2= U7 Utild1− 2 Utild3⋅+=

U4 Utild3= U8 0=

clearly, the rotation dofs are not affected by these constraints (just renumbered), but the translation dofs are

we can write the above relation between Uf and Utildf in compact from

Uf Ac Utildf⋅= with Ac

1

0

0

0

43

0

1−

0

0

0

0

1

0

0

0

0

0

0

0

0

0

1

43

−

0

2

0

0

0

0

0

0

0

1

0

0

0

0

0

0

0

0

0

0

0

1

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= this is called the constraint compatibilitymatrix in this course

since we plan to work with the constrained set of dofs in the future we like to express the structurecompatibility matrix in terms of the constrained set of free dofs. We call this matrix Atildf

in other words the changes in length and angles are expressed in terms of the displacements of theconstrained set of dofs by the relation

Atildf Utildf⋅ and since Af Uf⋅ Af Ac⋅ Utildf⋅= we conclude that Atildf Af Ac⋅=

Page 213


For the structure in this example the compatibility matrix Atildf becomes

change of length between nodes 1 and 2


Atildf Af Ac⋅:= Atildf

0

0.083

0.083

0

0.167−

0.167−

0

0.167

0.167

0

0.083−

0

0

1

0

1

0

0

0

0

0

0

0

0

0

0

0.167

0.167

0

0.167−

0.167−

0

0.167

0

0

0

0

0

1

0

1

0

0

0

0

0

0

0

0

0

0

0

1

0

1

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=



we confirm from the above that the distance between nodes 1-2, 2-3, 3-4 and 4-5 does not change.Thus, we plan to drop the corresponding rows from future uses of the structure compatibility matrix Atildf forthe constrained set of free dofs.

We will see later that the entries of the structure compatibility matrix Atildf for the constrained set of freedofs can be established by geometry in a much more straightforward way for small structures.

Inflexible element - Linear constraint

Let us now assume that element be is not only inextensible, but quite inflexible relative to the otherelements in the model. This means that the rotations of nodes 2 and 3 are constrained to be equal to theangle β as the following figure shows for an extensible, but inflexible element

4u

5u

1u

2u

X

Y

xy

L

L

i

j

nL

1vj

i

β3u

6u

the inflexibility constraints are

u3 β=ΔX

L2u5 u2−( )⋅

ΔY

L2u4 u1−( )⋅−=

u6 β=ΔX

L2u5 u2−( )⋅

ΔY

L2u4 u1−( )⋅−=

which implies, of course u6 u3=

Page 214


To introduce the inflexibility constraint between nodes 2 and 3 after having introduced the inextensibilityconstraints for the structure in this example we can proceed in one of two ways: (a) we can return to thelinear constraint conditions for inextensibility and add the inflexibility conditions, or, (b) we can introduce theinflexibility conditions on the already constrained set of free global dofs. Let us show both methods:

Method a: introduce inextensibility and inflexibility constraints in one step

We have 4 inextensibility and 2 inflexibility conditions. We use 2 inextensibility conditions to conclude thatU2 and U8 are equal to zero, as before. We have:

remaining inextensibility conditions inflexibility conditions

0810

U4 U1−( ) 610

U5⋅+= U38

10 10⋅U5⋅

610 10⋅

U4 U1−( )⋅−=

0810

U7 U4−( )⋅610

U5⋅+= U68

10 10⋅U5⋅

610 10⋅

U4 U1−( )⋅−=

set U1 1= and U4 0= from the first equation on the left we get U543

U1⋅=43

=

from the second equation on the left we get U7 1− U1⋅= 1−=

from the first and second equation on the right U316

U1⋅=16

=

U616

U1⋅=16

=

set U1 0= and U4 1= from the first equation on the left we get U543

− U4⋅=43

−=

from the second equation on the left we get U7 2 U4⋅= 2=

from the first and second equation on the right U316

− U4⋅=16

−=

U616

− U4⋅=16

−=

Page 215


The independent free dofs are now three, as shown in the following figure. We denote them with U'fTtodistinguish them from the Utildf in method (b) later. The above relations give

3

a

b c

d

8 8

12

6

2

1

U1 U'1=

U2 0=

U316

U'1⋅16

U'2⋅−=

U4 U'2=

U543

U'1⋅43

U'2⋅−=

U616

U'1⋅16

U'2⋅−=

U7 U'1− 2 U'2⋅+=

U8 0=

U9 U'3=The constraint compatibility matrix A'c now is

this is the matrix that relates the original set of nine independent free dofs tothe constrained set of 3 independent free dofs, which displace underconsideration of 4 inextensible and 1 inflexible element

A'c

1

0

16

0

43

16

1−

0

0

0

0

16

−

1

43

−

16

−

2

0

0

0

0

0

0

0

0

0

0

1

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

Page 216


The compatibility matrix Atild'f for the three independent free dofs becomes



angle between node 2 and chord at end i of b

Atild'f Af A'c⋅:= Atild'f

0

0.083

0.25

0

0

0

0

0.333

0.167

0

0.083−

0

0

0.167−

0

0

0

0

0.333−

0.167−

0

0.167

0

0

0

0

0

0

0

0

1

0

1

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

= angle between node 3 and chord at end j of b



We note that the change in distance between nodes 1-2, 2-3, 3-4 and 4-5 is now equal to zero, as isthe angle between nodes 2 and 3 and the chord connecting them. Thus, we plan to drop the correspondingrows from future uses of the structure compatibility matrix Atild'f for the constrained set of free dofs.

Method b: introduce inflexibility constraints after having introduced inextensibility constraints

in this case we plan to constrain the following set of 5 independent free dofs to 3

12

3 4

5

3

a

b c

d

2

1

To establish the relation between these two sets of dofs in the most direct way possible we look at the rowsof the structure compatibility matrix that correspond to the angles between nodes 2 and 3 and the chordconnecting them for the case of 5 independent free dofs (top of page 6). We have

16

− Utild1 Utild2+16

Utild3⋅+ for the angle between node 2 and the line connecting 2-3

16

− Utild116

Utild3⋅+ Utild4+ for the angle between node 3 and the line connecting 2-3

Page 217


For an inflexible element b these two angles need to be zero, i.e. there is no angle between the nodeand the chord connecting it to the next node, because the node is forces to rotate with the chord.

Thus, we get Utild216

Utild1⋅16

Utild3⋅−=and we conclude, of course that Utild2 Utild4=

Utild416

Utild1⋅16

Utild3⋅−=

After renumbering the new set of constrained dofs, as shown in the figure on the right and denoting themwith U'f we have:

Utild416

U'1⋅16

U'2⋅−=Utild1 U'1=

Utild216

U'1⋅16

U'2⋅−= Utild5 U'3=

Utild3 U'2=

this means that the constraint matrix from 5 to 3 dofs (call it A''c) is A''c

1

16

0

16

0

0

16

−

1

16

−

0

0

0

0

0

1

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

and we now have for the compatibility matrix Atild'f for the three independent free dofs



angle between node 2 and chord at end i of b

Atild'f Atildf A''c⋅:= Atild'f

0

0.083

0.25

0

0

0

0

0.333

0.167

0

0.083−

0

0

0.167−

0

0

0

0

0.333−

0.167−

0

0.167

0

0

0

0

0

0

0

0

1

0

1

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

= angle between node 3 and chord at end j of b



which, of course, is the same result as for method a. In the latter we went in a single step from the original 9independent free dofs to the three constrained free dofs. In method b we went in two steps: first from 9 to 5free dofs by invoking the inextensible element constraints, and then from 5 to 3 by invoking the inflexibleelement constraints.

Page 218

Script for Example 16 in CE220 class notes (a) gable frame with inextensible constraints (b) gable frame with inextensible and one inflexible constraint


Create model % specify node coordinates (could only specify non-zero terms) XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 0 12]; % second node, etc XYZ(3,:) = [ 8 18]; % XYZ(4,:) = [ 16 12]; XYZ(5,:) = [ 16 0]; % % connectivity array CON {1} = [ 1 2]; CON {2} = [ 2 3]; CON {3} = [ 3 4]; CON {4} = [ 4 5]; % boundary conditions (1 = restrained, 0 = free) (specify only restrained dof's) BOUN(1,:) = [1 1 1]; BOUN(5,:) = [1 1 0]; % specify element type [ElemName{1:4}] = deal('2dFrm'); % 2d beam element Model = Create_SimpleModel (XYZ,CON,BOUN,ElemName);

Compatibility analysis with inextensible elements % set up linear constraint equations for inextensible elements % (w/o specification, all elements are by default inextensible) LC_Eqs = LinConEqs(Model); % the size of array LC_Eqs is: % number of rows = no of inextensible elem + 2*no of inflexible elem + number of restrained dofs % number of columns = no of total dofs of unconstrained model % in this case the numbers are 9x15, the difference is the number of independent free dofs % form compatibility matrix Ac from linear constraint equations % place those translation dofs you wish to retain as independent last; % here we use the horizontal translations at node 2 and 3 pick_dof = [Model.DOF(2,1), Model.DOF(3,1)]; dof_reord = [setdiff(1:Model.nt,pick_dof) pick_dof]; Ac = Ac_matrix (LC_Eqs,dof_reord); % display independent free dofs magf = [2;1;2;1;1;1]; for k=1:size(Ac,2); Create_Window (0.80,0.80); Plot_Model (Model);

Page 219

title('Deformed shape for each independent free dof with inextensible elements') % extract global dof displacements for each "mode" U = Ac(:,k); % specify magnification factor for each "mode" MAGF = magf(k); % plot model in displaced position (chords) Plot_Model (Model,U); % plot deformed shape of elements Plot_DeformedStructure(Model,[],U); % pause and wait for user feedback % disp('press any key to continue...'), pause % get all figure elements and wipe them out % dum = get(gca,'Children'); % delete(dum); end clear Ac;

Page 220

Compatibility analysis with inextensible and one inflexible elements % set up linear constraint equations for inextensible elements % and specify inflexible element (b or #2 in this case) LC_Eqs = LinConEqs(Model,1:4,2); % in this case the size of LC_Eqs is 11x15 and there are 4 independent free dofs % form compatibility matrix Ac from linear constraint equations % place translation dofs to retain as independent last; here % we use the horizontal translations at node 2 and 3 pick_dof = [Model.DOF(2,1), Model.DOF(3,1)]; dof_reord = [setdiff(1:Model.nt,pick_dof) pick_dof]; Ac = Ac_matrix (LC_Eqs,dof_reord); % display independent free dofs magf = [2;2;1;1]; % magnification factor for each deformed shape for k=1:size(Ac,2); Create_Window (0.80,0.80); Plot_Model (Model); title('Deformed shape for each independent free dof with inextensible and one inflexible elements') % extract global dof displacements for each "mode" U = Ac(:,k); % specify magnification factor for each "mode" MAGF = magf(k); % plot model in displaced position (chords) Plot_Model (Model,U); % plot deformed shape of elements Plot_DeformedStructure(Model,[],U); % pause and wait for user feedback % disp('press any key to continue...'), pause % get all figure elements and wipe them out % dum = get(gca,'Children'); % delete(dum); end

Page 223


Example 17 - Compatibility for Gable Frame with Constraints - Graphical Method

The gable frame structure is the same as in Example 1 and 16. The following figure shows the geometry.

The resulting free global dof's without any constraints are:

1

2

3

4

5

6

7

8

9

a

b c

d 12

6

8 8

XY

With four inextensible elements a through d there are 4 independent linear constraint conditions betweenthe 6 translation dofs, while the rotation dofs are not affected. It is quite obvious that the displacementvalues at dofs 2 and 8 are zero. We can, therefore, select any two of the other 4 translation dofs as theindependent free global dofs for the structural model with inextensible elements.

Thus, we select the following 5 independent free dofs for the structural model with inextensible elements.

12

3 4

53

1

2

4 5

6

7

We are interested in the element deformations at thelocations numbered in the above figure. For the rotationdofs the compatibility matrix is straightforward.

Page 225


Let us look at the translation dofs

ICd

ICb

ICc

1

2

1.33

dof 3

For unit value of dof 1

1 1

1.33ICb ICc

ICdICa

IC = instantaneous center of rotation ofcorresponding element

1. Location of IC for elements a and d is obvious2. IC for element b is identical to case d, because

the right node cannot move horizontally, sinceit is an independent dof.

3. Similar consideration for IC of element c

1. Location of IC for element d is obvious2. IC for element b is at the left node, because it

cannot move horizontally, since it is anindependent dof.

3. IC of element c lies on the extension of element daxis and on the extension of element b axis.

Conclusion: The IC's of two adjoining elements and their common node lie on a straight line.

From the unit value at the particular dof determine the rotations of the element chords.

ICd

ICb

ICc

1

2

1.33

a

b

c

d

θ 0θ 1 6θ 1 6θ 1 6

== −=

= −

1 1

1.33ICb ICc

ICdICa

a

b

c

d

θ 1 12θ 1 6θ 1 6θ 1 12

= −

=

= −=

Page 226


1. From unit value of dof 1 and location of IC forelement a and b determine θa and θb.

2. From relative distance of node 3 from IC ofelement b determine vertical translation.

3. From vertical translation and distance of IC ofelement c determine θc.

4. From relative distance of node 4 from IC ofelement c determine horizontal translation.

5. From horizontal translation and distance of ICof element d determine θd.

1. From unit value of dof 3 and location of IC forelement b determine θb.

2. From unit value of dof 3 and distance of IC ofelement c determine θc.

4. From relative distance of node 4 from IC ofelement c determine horizontal translation.

5. From horizontal translation and distance of ICof element d determine θd.

The angles between the node tangent and the chord are equal to the opposite of the chord rotation, sincethe angles are measured from the chord to the tangent.

a

a

b1

bf

c

c

d

-θ 1 12-θ 1 12-θ 1 6-θ 1 6-θ 1 6-θ 1 6-θ 1 12

< >

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎜ ⎟ ⎜ ⎟

= = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠⎝ ⎠

A

a

a

b3

bf

c

c

d

-θ 0-θ 0-θ 1 6-θ 1 6-θ 1 6-θ 1 6-θ 1 6

< >

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

= =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎜ ⎟ ⎜ ⎟

−⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

A

The complete compatibility matrix for the 5 independent free dofs of the structural model with inextensibleelements becomes.

1

3

1

2

4 5

6

7

2

3

Atildf

112

112

16

−

16

−

16

16

112

−

0

1

1

0

0

0

0

0

0

16

16

16

−

16

−

16

0

0

0

1

1

0

0

0

0

0

0

0

1

1

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= 4

5

6

7

Page 227


Case for inflexible element bThe two rotations dofs #2 and #4 are no longer independent. As Example 16, shows the rotations can bederived from the translation dofs at the ends of the element, if it is inflexible. We have only threeindependent free global dofs, as shown in the figure on the right.

3

a

b c

d

8 8

12

6

2

1

3 4

a

b c

d

8 8

521

ICd

ICb

ICc

1

2

1.33

The translations remain the same as before, i.e.

1 1

1.33ICb ICc

ICdICa

1 1

1.33

rotations at node 2 and 3 constrained with translation

1

2

1.33

rotations at node 2 and 3 constrained with translation

Page 228


The figures on the preceding page show that the inflexible element b forces the nodes 2 and 3 to rotate withthe chord. This induces a rotation of the node tangent at the ends of elements a and c, as shown. Thecompatibility matrix for the two dofs becomes.

Recalling

ICd

ICb

ICc

1

2

1.33

a

b

c

d

θ 0θ 1 6θ 1 6θ 1 6

== −=

= −

1 1

1.33ICb ICc

ICdICa

a

b

c

d

θ 1 12θ 1 6θ 1 6θ 1 12

= −

== −

=

1 1

1.33 1

2

1.33

a

a b

1f

c b

c

d

-θ 1 12-θ θ 1 12 1 6

0 00 0

-θ θ 1 6 1 6-θ 1 6-θ 1 12

< >

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟+ +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

= =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟+ +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠⎝ ⎠

A

a

a b

2f

c b

c

d

-θ 0-θ θ 1 6

0 00 0

-θ θ 1 6 1 6-θ 1 6-θ 1 6

< >

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟+ −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

= =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟+ − −⎜ ⎟ ⎜ ⎟

−⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

A

Page 229

Kinematics of (plastic) collapse mechanismLet us now study the kinematics of an unstable structure. A typical case is an originally stable structure that turns into a mechanism when the collapse load factor is reached for the given loading. We study the kinematics of the structure past the point of incipient collapse, i.e. just after the last hinge has formed. Since this is an incremental motion we use the Greek letter Δ to denote displacements and deformations.: Past the point of incipient collapse the element deformation increments are zero, since there is no change in the element end forces after the collapse load is reached. Consequently, we can write

and thussince

with the hinge deformation increments

0εΔ =V hΔ = ΔV Vh f fΔ = ΔA UV

hΔV

H

L L

potential plastic hinge locations

1

23

4 5

H

L L

Example: consult detailed calculations in Example 18 and 19

The columns of the compatibility matrix describe the independent elementary collapse mechanisms.3

dof 3beam mechanism

dof 5joint mechanism

1

dof 1sidesway mechanism

The actual collapse mechanism is a linear combination of elementary collapse mechanisms.


Page 230

CE220 - Theory of Structures Example 18 - Collapse Mechanism © Prof. Filip C. Filippou, 2000

Example 18 - Full and Partial Collapse Mechanism - Analytical Method

We consider the portal frame that we investigated in example 8 in connection with the lower bound theoremof plastic analysis. Just before the last hinge forms there are 3 plastic hinges in the structure, which istherefore statically determinate. A similar case exists for a structure with three moment releases at thesame location. We study, therefore, either case, as the following figures show.

dof 1dof 2

dof 3 dof 5

4 4

5

dof 4

a

b c

d

23 4 5 6

7

81

dof 1dof 2

dof 3 dof 5

4 4

5

dof 4

a

b c

d

23 4 5 6

7

81

Portal frame of Example 8 before last plastic hingeat 1 forms. It is statically determinate (plastic hingesare indicated with gray circles)

Statically determinate portal frame with threemoment releases at 5, 6 and 8. Moment releasesor rotational hinges are indicated with white circles.

Assuming that all elements are inextensible there are 5 independent free global dofs, as shown.We determine the angles between the nodes and the lines connecting the nodes at all element ends.The compatibility matrix is then equal to the transpose of the equilibrium matrix of example 8. We haveencountered this matrix in Example 15

Af

15

15

0

0

0

0

15

15

0

1

1

0

0

0

0

0

0

0

14

−

14

−

14

14

0

0

0

0

0

1

1

0

0

0

0

0

0

0

0

1

1

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

Page 231


With hinges at locations 5, 6 and 8 whose rotations are unknown at this stage we use the compatibilityrelations at locations 1, 2, 3, 4 and 7 where no hinges are present and we are only dealing with straindependent element deformations. We extract the corresponding rows of the compatibility matrix and call it Ai

Ai

15

15

0

0

15

0

1

1

0

0

0

0

14

−

14

−

0

0

0

0

1

0

0

0

0

0

1

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= the rank of this compatibility matrix is rank Ai( ) 5=

and this means that the structure is stable, since the rank of the matrix isequal to the number of columns; furthermore, the number of columns is equalto the number of rows meaning that we are dealing with a staticallydeterminate structure. How can we "prove" that the structure is stable, if wedo not have access to the rank function?

We set Vi = 0 and seek to find a set of non-zero displacement that satisfyVi = Ai * Uf. If this is possible, the structure is unstable (instability means thatthe nodes of the structural model are able to displace infinitesimally withoutdeforming any elements, i.e. "without resistance"). If the only solution of theset of compatibility relation is the trivial solution Uf=0, then the structure isstable. Linear algebra says that the nullspace of kinematic matrix Ai consistsof the null vector only.

For the example at hand we write the compatibility equations out in full. We have:

V115

U1⋅=

V215

U1⋅ U2+=

V3 U214

U3⋅−=

V414

− U3⋅ U4+=

V715

U1⋅ U5+= recall that we eliminated rows 5, 6 and 8 of the compatibility matrix Af

Setting all V's equal to zero it is easy to see that all the U's are zero as well (proceed from the first to the lastequation). Thus, the structure is stable. Of course, we have an intuitive understanding that this is indeed thecase, but what we do with our "intuition" is precisely what this algebraic process illustrates: we try to see if itis possible to "move" the nodes without deforming the elements. If so, we call the structure unstable. We willsee soon that we can do this geometrically with the method of instantaneous centers of rotation.

Page 232


Let us now assume that an additional hinge is inserted at end 1. The structure now looks like this.

23 4 5 6

7

81

4 4

5a

b c

d

The compatibility matrix for the continuous element ends now has only four rows (it lost the first row)

Ae

15

0

0

15

1

1

0

0

0

14

−

14

−

0

0

0

1

0

0

0

0

1

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= the rank of this compatibility matrix is rank Ae( ) 4=

which is to be expected since there are now only four rows. Since the rankof the matrix is smaller than the number of free dof's, we have an unstablestructure. The question is: what are the displacement values that satisfy thecondition 0 = Ae * Uf ? We may recall that the solution of these homogeneouslinear equations is known as the nullspace of the compatibility matrix. Let usfind this solution.

Note that strain dependent deformation increments are zero for an unstable structure as well as for thecollapse mechanism of a fully plastic structure. Thus, the more accurate way of interpreting the aboveequation is that 0 = Ae * ΔUf in such case. The reason for the deformation increments being zero past thepoint at which the collapse load is reached is that the basic forces do not change for perfectly plasticbehavior.

V2 0=15

U1⋅ U2+= setting U1 1= we get U215

−=

V3 0= U214

U3⋅−= U3 4 U2⋅= U345

−=

V4 0=14

− U3⋅ U4+= U414

U3= U415

−=

V7 0=15

U1⋅ U5+= U515

−=

Page 233


Thus, any vector of displacement values of the form Uf

1

15

−

45

−

15

−

15

−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

U1⋅= satisfies the condition 0 = Ae * Uf

and forms the nullspace of thecompatibility matrix or, expresseddifferently, is the completehomogeneous solution of thecompatibility equations

This displacement vector represents the collapsemechanism of the unstable structure.

Note that U1 can assume any value.

We can interpret the above as follows: the free dof's are constrained so that only one independent free dofremains during the collapse mechanism. Under this aspect the vector of relative displacement values thatmultiplies U1 in the above relation represents a constraint matrix Ac. Since this constraint matrixcorresponds to the plastic collapse mechanism we add the subscript p. We can now return to thecompatibility relations in order to calculate the hinge rotations in terms of U1. But before we do this, let uslook at the depiction of the collapse mechanism.

The above vector of displacement values represents the collapse mechanism of the structure. A graphicalrepresentation of the collapse mechanism is shown in the following figure.

4 4

5

1

0.8

0.2 0.2

We note that the elements maintain continuity of tangents at the four locations 2, 3, 4 and 7 withoutdeforming. We also note that the collapse mechanism is the linear combination of the columns of thecompatibility matrix with factors 1, -1/5, -4/5, -1/5, -1/5.

Page 234


We return to the compatibility relations in order to calculate the hinge rotations in terms of U1

We have in general Vh Af Uf⋅ Vε−=

for the collapse mechanism Uf Acp U1⋅= with Acp

1

15

−

45

−

15

−

15

−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

since the element deformations in our case are all zero, we have Vε 0 0 0 0 0 0 0 0( )T:=

We also need the product Afp Af Acp⋅:= Afp

0.2

0

0

0

0.4−

0.4−

0

0.2

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

= the kinematic matrix of the collapsemechanism

With these terms the hinge rotations become Vh

0.2

0

0

0

0.4−

0.4−

0

0.2

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

U1⋅=

Page 235

Script for Example 18a in CE220 class notes % Collapse mechanism of one-story portal frame


Create model % specify node coordinates (could only specify non-zero terms) XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 0 5]; % second node, etc XYZ(3,:) = [ 4 5]; % third node, etc XYZ(4,:) = [ 8 5]; % XYZ(5,:) = [ 8 0]; % % connectivity array CON {1} = [ 1 2]; CON {2} = [ 2 3]; CON {3} = [ 3 4]; CON {4} = [ 4 5]; % boundary conditions (1 = restrained, 0 = free) BOUN(1,:) = ones(1,3); BOUN(5,:) = ones(1,3); [ElemName{1:4}] = deal ('2dFrm'); % 2d beam element Model = Create_SimpleModel (XYZ,CON,BOUN,ElemName);

Analysis % set up linear constraint equations for inextensible elements LC_Eqs = LinConEqs(Model); % form compatibility matrix Ac from linear constraint equations % re-order horizontal translation at node 2 and vertical translation at node 3 last pick_dof = [Model.DOF(2,1) Model.DOF(3,2)]; dof_reord = [setdiff(1:Model.nt,pick_dof) pick_dof]; Ac = Ac_matrix (LC_Eqs,dof_reord); % set up compatibility matrix A of structural model A = A_matrix(Model); % compatibility matrix for constrained independent free dofs Aftild (drop tild in the following) Af = A*Ac; % set up linear constraint equations from the deformations at the locations w/o hinge LC_Eqs = Af([3 5 6 11],:); % derive constrained collapse mechanism: put horizontal translation last for scaling dof_reord = [2:5 1]; Acp(dof_reord,:) = Ac_matrix(LC_Eqs(:,dof_reord)); Acp

Page 236

% displacement increments for collapse mechanism (horizontal translation dof scaled to 1) DU = Ac*Acp; % hinge deformation increments for collapse mechanism Afp = Af*Acp; DVh = Afp; DVh % display collapse mechanism; MAGF = 1; Create_Window (0.80,0.80); Plot_Model (Model); Plot_Model (Model,DU); Acp = 1.0000 -0.2000 -0.8000 -0.2000 -0.2000 DVh = 0 0.2000 0 0 0 0 0 -0.4000 -0.4000 0 0 0.2000

Page 237


Partial Collapse Mechanism

We turn our attention now to the structure of Example 9.

a b

c

d e

12 3

4

f

56

6 6

8

Under the assumption that elements a through e are inextensible the structure has only 4 independent freedofs, as shown in the following figure (the rotations at nodes 3, 4 and 6 are set aside). The locations forrecording the angles between nodes and the lines connecting them as well as the change of length betweennodes 3 and 5 are numbered in the figure.

dof 3

dof 1

dof 2

dof 4

1 2 34

56 7

8

Page 238


The compatibility matrix is

and we observe that it is the transpose of theequilibrium matrix Bf from Example 9.Af

16

−

16

−

16

0

0

16

−

16

0.8

0

1

1

1

0

0

0

0

0

0

0

18

18

0

0

0.6−

0

0

0

0

1

1

1

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

We now assume that the following releases are inserted: a circle indicates a flexural hinge or momentrelease, while a telescope connection indicates an axial hinge or axial force release.

a b

c

d e

12 3

4

f

56

6 6

8

For this structure the continuity conditions between element deformations and angles between nodes andlines connecting the nodes or changes of distance between nodes are:

V216

− U1⋅ U2+= V518

U3⋅ U4+= there seem to be four relations for 4 unknown free globaldofs, indicating that the structure may be stable. However,the rank of the compatibility matrix is not 4 but 3. Let us see

V4 U218

U3⋅+= V616

− U1⋅ U4+=

Page 239


Ae

16

−

0

0

16

−

1

1

0

0

0

18

18

0

0

0

1

1

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= rank Ae( ) 3= this means that only three compatibility relations arelinearly independent and that the fourth can be derivedfrom them. This is easy to see with the above equations:add V2 and V5 and subtract V4 and V6. We get

V2 V5+ V4− V6− 0= linear dependence!

Setting all V's equal to zero in the above equation, we can determine the vector of the displacement valuesfor the collapse mechanism. Setting U1 = 1 we get

016

− U1⋅ U2+= U216

=

0 U218

U3⋅+= U343

−= thus the collapse mechanism is represented by Acp U1⋅ with Acp

1

16

43

−

16

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

018

U3⋅ U4+= U416

=

and the release deformations can be determined from Vh Afp U1⋅= with Afp Af Acp⋅:=

The collapse mechanism is shown in the figure below.

3

6 6

8

1

4/3

with Afp

16

−

0

13

0

0

0

13

85

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

= Vh

16

−

0

13

0

0

0

13

85

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

U1⋅=

Page 240

Script for Example 18b in CE220 class notes % partial collapse mechanism of braced frame with two floors


Create model % specify node coordinates (could only specify non-zero terms) XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 6 0]; % second node, etc XYZ(3,:) = [ 12 0]; % third node, etc XYZ(4,:) = [ 0 8]; % XYZ(5,:) = [ 6 8]; % XYZ(6,:) = [ 12 8]; % % connectivity array CON {1} = [ 1 2]; CON {2} = [ 2 3]; CON {3} = [ 2 5]; CON {4} = [ 4 5]; CON {5} = [ 5 6]; CON {6} = [ 3 5]; % boundary conditions (1 = restrained, 0 = free) BOUN(1,:) = ones(1,3); BOUN(3,:) = [0 1 0]; BOUN(4,:) = [0 1 0]; BOUN(6,:) = [0 1 0]; [ElemName{1:5}] = deal ('2dFrm'); % 2d beam element ElemName{6} = 'Truss'; % truss element Model = Create_SimpleModel (XYZ,CON,BOUN,ElemName);

Analysis % set up linear constraint equations for inextensible elements LC_Eqs = LinConEqs(Model,1:5); % form compatibility matrix Ac from linear constraint equations % re-order horizontal translation at node 4 and vertical translation at 5 last pick_dof = [Model.DOF(4,1) Model.DOF(5,2)]; dof_reord = [setdiff(1:Model.nt,pick_dof) pick_dof]; Ac = Ac_matrix (LC_Eqs,dof_reord); % set up compatibility matrix A of structural model A = A_matrix(Model); % compatibility matrix for constrained independent free dofs Aftild Af = A*Ac; % set up linear constraint equations from deformations at locations w/o hinge LC_Eqs = Af([3 6 8 9 11 12 15],:); % derive collapse mechanism: put vertical translation last for unit scaling dof_reord = [1:4 6:7 5]; Acp(dof_reord,:) = Ac_matrix(LC_Eqs(:,dof_reord)); Acp

Page 241

% displacement increments for collapse mechanism (vertical translation dof scaled to 1) DU = Ac*Acp; % hinge deformation increments for collapse mechanism Afp = Af*Acp; DVh = Afp; DVh % display collapse mechanism; MAGF = 1; Create_Window (0.80,0.80); Plot_Model (Model); Plot_Model (Model,DU); Acp = 0.1667 -0.1667 -1.3333 0.1667 1.0000 0.1667 -0.1667 DVh = 0 -0.1667 0 0 0.3333 0 0 0 0 0 0 0 0 0.3333 0 1.6000

Page 242


Example 19 - Full and Partial Collapse Mechanism - Graphical Method

We return to the two structures that were investigated in Example 18 and present here the graphicalsolution, which is much faster for hand calculations. We treat the first as a collapse mechanism andinvestigate the incremental displacements past the point at which the collapse load is reached. We use theprefix Δ to indicate "increment of". The following figure shows that with four hinges, 3 rigid bodies result that can rotate relative to each otherand the supports. They are denoted with upper case letters, A, B and C in the following figure. Theinstantaneous centers of rotation of the three rigid bodies are shown in the figure. Using the horizontaltranslation dof as the reference value (which we set equal to 1 in the figure) we have the following rigidbody rotations

4 4

5

ICA

ICB

ICC

1

A

B

C

1

56

8

5

5

ΔθA15

−:= CW

ΔθB15

:= CCW

ΔθC15

−:= CW

The plastic hinge rotations are the deformations at 1, 5, 6 and 8. Note that ΔVε = 0, since the basic forcesdo not change past the point at which the collapse load is reached.

We have: ΔV1 ΔθA−:= ΔV1 0.2= CCW when measured from chord to node tangent

ΔV5 ΔθB− ΔθA+:= ΔV5 0.4−= CW when measured from chord to node tangent

ΔV6 ΔθB− ΔθC+:= ΔV6 0.4−=

ΔV8 ΔθC−:= ΔV8 0.2= compare with Afp "matrix" in Example 18

Page 243


We treat the second structure as a mechanism, i.e. we assume that the inserted hinges do not carry anyforces. There are again 3 rigid bodies that can rotate relative to each other and the supports, denoted withA, B and C. The instantaneous centers of rotation are shown in the following figure. Selecting the verticaltranslation as the reference value, we obtain the value of 4/3 for the horizontal translation, as shown.

3

6 6

8

1

4/3

ICA

ICB

ICC

A

B

C

1

3

7

8

8

We obtain the following rotationvalues for the rigid bodies A, B, C.

θA16

:=

θB16

−:=

θC16

−:=

The hinge rotations are the deformations at 1, 3 and 7. Moreover, we also have a hinge with axialdeformation at 8. We denote the hinge deformations with subscript h and note that there are no straindependent element deformations, since the model is a mechanism that is unable to resist any loading.

We have: ΔVh1 θA−:= ΔVh1 0.167−= CW when measured from chord to node tangent

ΔVh3 θC− θA+:= ΔVh3 0.333= CCW when measured from chord to node tangent

ΔVh7 θB− θA+:= ΔVh7 0.333=

ΔVh8 0.8 1( )⋅ 0.643⎛⎜⎝⎞⎟⎠

⋅+:= ΔVh8 1.6= compare with values in Example 18

Page 244

Equilibrium Compatibility

displacement or equilibrium method

NOS unknowns NOS conditions

force or compatibility method

nf unknownsnf equations

f f fw= +BP Q P f f d= +AV U V

( )i f fw x x= − +B BQ P P Q ( )Tx d 0− =B V V

Tf f=A B


Page 245

REAL AND VIRTUAL WORK

Objective: relation between static and kinematic variables through virtual work statements; incremental plastic work for upper bound theorem of plastic analysis

CE220-Theory of Structures Real and virtual work © Prof. Filip C. Filippou, 2000

Page 246

3. Virtual Work Principles

3.1 Principle of virtual work or virtual displacements

The principle of virtual work involves the equality between internal and external work. Work

is performed by a set of forces on an independent set of displacements and deformations.

Because of this causal independence we denote the set of displacements and the resulting work

as virtual. Consequently, the principle is also known as principle of virtual displacements. We

denote the virtual set of displacements and corresponding deformations by preceding the lower

case letter δ to the variables. Limiting ourselves at first to nodal forces we can write for the

external virtual work

T Tf f d deWδ δ δ= +U P U P

and the internal virtual work takes the form

TiWδ δ= V Q

The principle of virtual work states:

If a structure is in a state of equilibrium, then the external virtual work is equal to the internal

virtual work under any set of virtual displacements that satisfy the conditions of compatibility.

This is a necessary and sufficient condition meaning that the reverse is also true, i.e. if the

external work of the applied forces is equal to the internal work of some basic element forces for

all sets of virtual displacements and deformations that satisfy the conditions of compatibility,

then the applied forces are in equilibrium with the basic element forces.

In practice we use the principle of virtual work for determining individual force values and,

later on, individual stiffness coefficients in hand calculations (Note that a stiffness coefficient

ijK is the force at dof i due to a unit displacement at free dof j only). To this end we select a

virtual displacement field with a unit displacement value at the particular dof of interest and zero

values at all other dofs. The corresponding deformations can be determined with the methods of

the preceding section. In the process we simplify the compatibility problem as much as possible

by introducing constraints among the global dofs, so as to eliminate certain element forces from

the internal work. For example, in order to eliminate the contribution of axial forces we select a

virtual displacement field with zero virtual axial deformations.

Page 247

The external work of element loads can be obtained by integration of the inner product of

element forces with corresponding virtual displacements over the element length and summation

over all elements. Alternatively, we express the element loads through equivalent end forces that

get assembled into nodal forces fwP due to element loads. In the latter case we write for the

external virtual work

( )T Tf f fw d deWδ δ δ= − +U P P U P

3.2 Principle of complementary virtual work or virtual forces

The principle of complementary virtual work or virtual forces involves a set of displacements

and their compatible deformations and an independent set of applied forces and the basic element

forces in equilibrium with these. The forces represent the virtual set that has no causal relation

with the real displacement/deformation set. We call the work complementary and have for the

external complementary virtual work

T Tf f d deWδ δ δ= +P U P U

and for the internal complementary virtual work

TiWδ δ= Q V

The principle of complementary virtual work states:

If a structure is in a state of compatible deformation, then the external complementary virtual

work is equal to the internal complementary virtual work under any set of virtual forces that

satisfy equilibrium. This is a necessary and sufficient condition meaning that the reverse is also

true, i.e. if the external complementary virtual work of the virtual forces is equal to the internal

complementary virtual work of the corresponding virtual basic element forces for all sets of

virtual applied forces and corresponding element forces in equilibrium, then the actual

deformations and the global dof displacements are compatible.

In practice we use the principle of virtual forces for determining individual displacement

values and, later on, individual flexibility coefficients in hand calculations (Note that a flexibility

coefficient ijF is the displacement at dof i due to a unit force at free dof j only). To this end we

select a virtual force field with an applied unit force value at the particular dof of interest and

zero values at all other dofs. We select a set of basic element forces in equilibrium with the unit

Page 248

force. The simplest set of basic element forces (i.e. with the most zero values in basic force

vector Q ) is the most convenient choice, because it involves the least amount of arithmetic

operations for determining the internal virtual work. Because of the presence of a unit applied

force at the dof of interest, the principle of virtual forces is sometimes called the dummy unit

load method.

3.3 Plastic work increment and upper bound theorem of plastic analysis

Once the collapse load factor is reached and the structural model forms a complete or partial

collapse mechanism the plastic work increment of the external forces on the corresponding

displacement increments of the collapse mechanism is

( )Tf refpeW λΔ = ΔU P

The internal work increment of the basic element forces on the plastic deformation increments is

a bit harder to express. Because plQ represents the absolute values of the plastic capacity of the

elements, we need to distinguish between positive and negative plastic deformation increments

and change the sign of the latter. We define these as follows

pl pl pl plif 0 otherwise 0+ +Δ = Δ Δ > Δ =V V V V

pl pl pl plif 0 otherwise 0− −Δ = −Δ Δ < Δ =V V V V

With this definition the internal plastic work increment becomes

pl pl pl plpiW + + − −Δ = Δ + ΔQ V Q V

The superscript + or - denotes the location with a positive or negative plastic deformation

increment, respectively. The equality of work requires pe piW WΔ = Δ . The upper bound theorem

of plastic analysis states that the collapse load factor cλ is determined with the collapse

mechanism producing the smallest plastic work increment. In compact form we can write this as

( )T

f ref pl pl pl pl

pl pl f f pl pl

min for

and 0, 0

cλ λ λ + + − −

+ − + −

= Δ = Δ + Δ

Δ −Δ = Δ Δ ≥ Δ ≥

U P Q V Q V

V V A U V V

It turns out that this is the dual linear programming problem to the lower bound theorem. Thus,

both formulations produce the same answer, i.e. a unique collapse load factor for the structural

model under the given applied force pattern.

Page 249

CE220 - Theory of Structures Example 20 - Principle of Virtual Displacements © Prof. Filip C. Filippou, 2000

Example 20 - Applications of Principle of Virtual Displacements

In this example we demonstrate several uses of the principle of virtual displacements. This principle is usedfor determining individual force terms, such as applied forces, and later individual stiffness coefficients.

Example a

Given is the following structure with an applied force whose magnitude is unknown. Moreover, several basicforces are also supplied. These forces were obtained from an analysis of the structure under the appliedforce. Further details (e.g. linear or nonlinear material) are not supplied, other than that the equilibrium issatisfied in the original (undeformed) configuration. Req'd: find the magnitude of the applied force.

6

88

a b

c1

2

34

?

16.04

0.181

32.97 38.02

Q

16.04

0.18

0.18−

32.97

32.97−

38.02−

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

Page 250


To determine the magnitude of the applied force we select a virtual displacement field with the followingcharacteristics:

1. it has a unit value in the direction of the applied force (positive global Y-direction at node 3)2. it has zero displacements at the supports in the direction of unknown support reactions3. it does not involve axial deformations in elements a, b, and c, since the axial forces under the unknownapplied force are not given.

First choice

We select the virtual displacement field in the following figure. It meets all requirements above.

ICa=ICb

ICc

1

1

6

88

0.75 IC's as shownVirtual chord rotations

δθa18

:=

δθb18

:=

δθc18

−:=

the corresponding virtual deformations are δV Af δUf⋅=

for the virtual displacement field above we get δV

δθa−

δθa−

δθb−

δθb−

δθc−

δθc−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= δV

0.125−

0.125−

0.125−

0.125−

0.125

0.125

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

=

the principle of virtual displacements says that the external work equal the internal work

P1 δU1⋅ P1 1( )⋅= is the external work

δVT Q⋅ 15−= is the internal work consequently, the applied force has magnitude of 15 unitsand acts downward.

Page 251


Second choice


ICa=ICb

ICc

1

1

6

88


δθa18

:=

δθb18

:=

δθc18

−:=

the corresponding virtual deformations are δVh Af δUf⋅=

since the element deformations δV are equal to zero in this case

for the virtual displacement field above we get δVh

δθa−

δθa− δθb+

δθb− δθb+

δθb− δθc+

δθc− δθc+

δθc−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= δVh

18

−

0

0

14

−

0

18

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

Note that this virtual deformation field is a lot more convenient, since it has only 3 non-zero terms.


P1 δU1⋅ P1 1( )⋅= is the external work

δVhT Q⋅ 15−= is the internal work consequently, the applied force has magnitude of 15 units

and acts downward, as before.

Page 252


Example b

Given the same structure as above we want the vertical support reaction at node 1. It is not clear whether wecan depend or not on the magnitude of the applied force (after all we may have an error in the abovecalculations) and so we choose a virtual displacement field with zero translation at node 3, so as to avoidincluding the external work of the applied force of uncertain value.

6

88

a b

c1

2

34

?

Virtual displacement field requirements

1. it has a unit value in the direction of the unknown support reaction (positive global Y-direction at node 1)2. it has zero displacements at the supports in the direction of the other unknown support reactions3. it does not involve axial deformations in elements a, b, and c, since the axial forces under the unknownapplied force are not given.4. it has zero translation at node 3 so as to exclude the contribution of the unknown applied force to theexternal work.

First choice


ICa=ICb ICc

1

6

88

0.75

R

IC's as shownVirtual chord rotations

δθa18

−:=

δθb18

−:=

δθc 0:=

Page 253


the corresponding virtual deformations are δV Af δUf⋅=

for the virtual displacement field above we get δV

δθa−

δθa−

δθb−

δθb−

δθc−

δθc−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= δV

0.125

0.125

0.125

0.125

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

=

the principle of virtual displacements says that the external work equal the internal work,if all forces are in equilibrium.

R δU1⋅ R 1( )⋅= is the external work

δVT Q⋅ 6.126= is the internal work consequently, the vertical support reaction at node 1 isequal to 6.126 units and acts upward.

Second choice


ICa=ICb ICc

1

6

88

0.75

R

IC's as shownVirtual chord rotations

δθa18

−:=

δθb18

−:=

δθc 0:=




δθa−

δθa− δθb+

δθb− δθb+

δθb−

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= δVh

0.125

0

0

0.125

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

=

Page 254


Note that this virtual deformation field is a lot more convenient, since it has only 2 non-zero terms.


R δU1⋅ R 1( )⋅= is the external work

δVhT Q⋅ 6.126= is the internal work consequently, the vertical support reaction at node 1 is

equal to 6.126 units and acts upward, as before.

Third choice

We select the virtual displacement field in the following figure. It meets all requirements except #4 above.This means that we have to know the magnitude of the applied force in this case.

ICc

1

1

6

88

1

15R

δV

0

0

0

0

18

18

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

R δU1⋅ 15 1( )⋅− R 15−= is the external work

δVT Q⋅ 8.874−= is the internal work R 15 8.874−:= R 6.126=

consequently, the vertical support reaction at node 1 isequal to 6.126 units and acts upward, as before.

Page 255


Example c

Given the same structure as above under the effect of a uniformly distributed load in element a.

w=?

a b

c1

2

34

6

88

The corresponding basic forces are

56.94 59.6876.45

24.29

Q

76.45

24.29−

24.29

56.94

56.94−

59.68−

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

Page 256


We proceed exactly as for example a. Let us first look at the external work done by the uniformly distributedelement load. Assuming a linear virtual displacement field as in the following figure we have

w

1( ) xxL

uδ =

xL

external work δWe

δWe

0

L

xwxL⋅

⌠⎮⎮⌡

d=

since w is uniform, i.e constant

δWewL 0

Lxx2⌠

⎮⌡

d⋅=w L⋅

2=

We can think of the following interpretation of the above result: the external work is equal to the workof the resultant of the uniformly distributed load times the translation that its centroid undergoes (seefollowing figure)

wL

10.5δWe wL

12⎛⎜⎝⎞⎟⎠

⋅=

Consequently, we simplify the given distributed load as follows

ab

c1

2

34

6

88

10 w8 w

6 w

Page 257


Virtual displacement field requirements:1. It activates external work of the distributed element load by translation of at least one end of element a.2. it has zero displacements at the supports in the direction of unknown support reactions3. it does not involve axial deformations in elements a, b, and c, since the axial forces under the unknownapplied force are not given.


ICa=ICb

ICc

1

1

6

88


δθa18

:=

δθb18

:=

δθc18

−:=




δθa−

δθa− δθb+

δθb− δθb+

δθb− δθc+

δθc− δθc+

δθc−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= δVh

18

−

0

0

14

−

0

18

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

The centroid of the distributed element load moves horizontally δθa− 3⋅ 0.375−= units in negative X

and vertically δθa 4⋅ 0.5= units in positive Y

The components of the resultant of the uniformly distributed element load are

6w units in the positive X direction

8w− units in the negative direction

Page 258


Thus, the external work of the distributed element load is 6w 0.375−( )⋅ 8− w( ) 0.5⋅+ 6.250−( ) w⋅→

The internal work is δVhT Q⋅ 31.251−=

and from the equality of internal and external work we obtain

w31.251−

6.25−:= w 5= in the given direction!

Example d: determination of collapse load factor

Determine the collapse load factor λ for the structure in the following figure under the given loading. Thelocation of plastic hinges at incipient collapse is given. The plastic flexural capacities of the horizontal frameelements a, b, d and e is 100 units. The plastic flexural capacity of frame element c is 150 units. The plasticaxial capacity of element f is 20 units.

a b

c

d e

12 3

4

f

56

40

30

6 6

8

Page 259


Virtual displacement field requirements:1. it has zero displacements at the supports in the direction of unknown support reactions2. it does not involve axial deformations in any elements with unknown axial forces.3. It does not involve flexural deformations in any elements with unknown end moments.4. It involves translation under one or both applied forces, so that external work appears.

dof 3

dof 1

dof 2

dof 4

1 2 34

56 7

8

The inextensibility requirement for elements a through e results in the independent free dofs of the abovefigure. With these dofs we write the compatibility equations (consult Example 9). Setting the deformations atlocations 2, 4, 5 and 6 equal to zero gives us four equations for the four independent dofs. However, asexplained in Example 9, only three of these equations are linearly independent and there exists a non-zerodisplacement vector that satisfies these four compatibility relations. This vector is the collapse mechanism ofthe structural model. We pursue first the analytical solution, which is lengthier, and tackle the fastergeometric solution subsequently.

Example 9 shows that the collapse mechanism is given by the following displacement vector.

3

6 6

8

1

4/3

Uf

1

16

43

−

16

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

U1⋅=

we use the same vector witha unit value for U1 as thevirtual displacement vector

Page 260



with Af

16

−

16

−

16

0

0

16

−

16

0.8

0

1

1

1

0

0

0

0

0

0

0

18

18

0

0

0.6−

0

0

0

0

1

1

1

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= from example 9

δVh Af

1

16

43

−

16

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

⋅:= δVh

0.167−

0

0.333

0

0

0

0.333

1.6

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

We have to assume at this point that we know the sign of the basic forces at the plastic hinge locations. Wewill show in the following example how to get these from the consideration of real plastic work. The Q valuesat the other locations are of no significance, since the corresponding virtual deformations are zero and nocontribution to the internal work results. We set them equal to the value 99 in the vector Q

We have: Q

100

99

100−

99

99

99

100−

20−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= the internal work is δVhT Q⋅ 115.333−=

the external work is δUfT

λ Pref⋅( )⋅ with Pref

40−

0

30

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:= δUf

1

16

43

−

16

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

thus, the collapse load factor λ is λδVh

T Q⋅

δUfT Pref⋅

:= λ 1.442=

Page 261


We can also derive the mechanism by geometric considerations, as shown in the following figure. We are notworried about the scaling of the collapse mechanism and the direction.

3/4C

ICA

A B ICB

ICC

1

1

2

3

4

θA18

−:=

θB16

34⋅:= θB 0.125=

θC θB:=

δVh

θA−

0

θB− θA+

0

0

0

θC− θA+

1− 0.6⋅34

0.8⋅−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= δVh

0.125

0

0.25−

0

0

0

0.25−

1.2−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

= δUf

34

−

18

−

1

18

−

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

thus, the collapse load factor λ is λδVh

T Q⋅

δUfT Pref⋅

:= λ 1.442= as before

Page 262

CE220 - Theory of Structures Ex. 21 - Principle of virtual forces © Prof. Filip C. Filippou, 2000

Example 21 - Principle of virtual forces in lieu of compatibility relations

In this example we will demonstrate the principle of virtual forces and its use as a substitute for thecompatibility relations of a structural model.

8 8

6

a

b

c

de

12

3 4

We assume that the vector of element deformations is supplied. We take it for granted that the givendeformations are compatible and, thus, do not bother to check that they satisfy the compatibility conditionsbefore starting the problem. We will do so, at the end of this example.

The axial deformations are assumed negligible. Under this assumption there are 4 independent free dofs,as shown in the following figure. The supplied element deformations are ordered as shown in italics.

dof 1

dof 2

dof 3dof 4

1

2

34

56

V

0.002530

0.002846−

0.002846

0.002530−

0.02580−

0.02580

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

Page 263


The structure compatibility matrix Af for the given free dofs is (we show the deformations only forthe translation dofs 1 and 3):

8 8

6

dof 1

Af

18

−

0

0

18

0.6

0.6−

1

1

0

0

0

0

0

16

16

0

0.8

0.8

0

0

1

1

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

8 8

6

dof 3

written out in full the compatibility relations read

V118

− U1⋅ U2+= V5 0.6 U1⋅ 0.8 U3⋅+=

V6 0.6− U1⋅ 0.8 U3⋅+=V2 U216

U3⋅+=

V316

U3 U4+=

V418

U1⋅ U4+=

Page 264


We make the following general observation from these equations: rotation dofs do not affect the change ofdistance between nodes, or, conversely, truss element deformations only depend on translation dofs.

Thus, if we are interested in calculating the vertical translation of dof 1 under the given elementdeformations, we are well advised to use the last two compatibility relations which do not involve the rotationdofs. By subtracting equation 6 from equation 5 we get:

V5 V6− 1.2 U1⋅= and consequently U1

V5 V6−

1.2:= U1 43− 10 3−=

from the addition of these two equations we can obtain

V5 V6+ 1.6 U3⋅= U3

V5 V6+

1.6:= U3 0=

Interestingly, under the given element deformations the horizontal translation of dof 3 is equal to zero.

if we like to determine one of the rotation dofs, we can use any of the other 4 compatibility relations, e.g.for dof 2 we can use the first one.

U2 V118

U1⋅+:= U2 2.845− 10 3−=

We can use the second compatibility relation to check that the values for dofs 2 and 3 satisfy it as well.We have

V2 U2−16

U3⋅− 1− 10 6−= and it is indeed satisfied except for a very small round-off error.

We can use the third compatibility equation to determine the rotation of dof 4. We have

U4 V316

U3⋅−:= U4 2.84610 3−=

and after substituting into the fourth relation we confirm that it is also satisfied

V418

U1⋅− U4− 1− 10 6−=

Since the given six element deformations give a single solution for the global dof displacement values, theyare compatible.

Page 265


Principle of virtual forces

We use the following statement of the principle of virtual forces: If a structure is in a state of compatibledeformation, then the external complementary virtual work is equal to the internal complementary virtual workunder any set of virtual forces that satisfy equilibrium.

it is our job now to select the most convenient virtual force system that satisfies equilibrium and at the sametime allows us to calculate the variable of interest. Let us say that we wish to determine the verticaltranslation at dof 1. Then, we apply a unit virtual force in the direction of dof 1 and find the virtual basicforces that satisfy the equilibrium equations.

1

2

34

56

1 1δ =P

Since axial deformations are negligible, it is in our best interest to use the equilibrium equations that do notinclude axial forces. To this end we use the transpose of the above compatibility matrix. We have

P1Q1

8−

Q4

8+ 0.6 Q5⋅+ 0.6 Q6⋅−=

P2 Q1 Q2+=

P3Q2 Q3+

60.8 Q5⋅+ 0.8 Q6⋅+=

P4 Q3 Q4+=

for the given virtual force at dof 1 we have δP1 1= and the virtual basic forces δQ need to satisfy thefollowing equations

1δQ1

8−

δQ4

8+ 0.6 δQ5⋅+ 0.6 δQ6⋅−=

0 δQ1 δQ2+=

0δQ2 δQ3+

60.8 δQ5⋅+ 0.8 δQ6⋅+=

0 δQ3 δQ4+=

Page 266


What is the simplest choice of δQ that satisfies all these four equations?

We set δQ1 δQ2= δQ3= δQ4= 0= With this choice equations 2 and 4 are satisfied.

We determine δQ5 and δQ6 to satisfy equations 1 and 3. We have:

1 0.6 δQ5⋅ 0.6 δQ6⋅−=

0 0.8 δQ5⋅ 0.8 δQ6⋅+= from these two equations we get δQ556

= δQ656

−=

The vertical translation at dof 1 is obtained from the principle of virtual forces δPfT

Uf⋅ δQT

V⋅=

with the choice δPf δP1= 1= we ensured that Uf U1= is the only non-zero entry on the left hand side

With the basic virtual force values we get: U156

V5⋅56

V6⋅−:= U1 0.043−=

the negative sign signifies that dof 1 moves in the direction opposite to the applied virtual force for thegiven deformations V. Since we took care to apply the virtual force in the positive Y-direction, thenegative sign means that the translation is in the negative Y-direction, i.e. downward.

We see that the resulting equation for the translation of global dof 1 is the same as the one we obtainedby the combination of the last two compatibility relations.

Is there another choice of virtual force system in equilibrium with the virtual unit force at dof 1? Sure thereare a few. Here is another one

1δQ1

8−

δQ4

8+ 0.6 δQ5⋅+ 0.6 δQ6⋅−=

0 δQ1 δQ2+=

0δQ2 δQ3+

60.8 δQ5⋅+ 0.8 δQ6⋅+=

0 δQ3 δQ4+=

We set δQ5 δQ6= 0= from equations 2 and 4 we conclude that δQ1 δQ2−=

δQ3 δQ4−=

The first equation becomes: 1δQ2

8

δQ4

8+=

and the third becomes 0δQ2 δQ4−

6= from which δQ2 δQ4=

and from the first equation we conclude that δQ2 δQ4= 4=

Page 267


now the virtual basic force vector becomes δQ

4−

4

4−

4

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

and the desired displacement can be determined from U1 δQT V⋅:= U1 0.043−=

which gives, of course, the same result.

Thus, any virtual force system in equilibrium with the virtual unit force at the location and in the direction ofthe translation whose value we want to determine can be used. The fewer terms are non-zero, the easierthe calculation. Thus, the first choice is superior.

The second task is to determine the rotation of node 3. To this end we apply a virtual unit moment at node 3(dof 4) and determine the virtual basic forces that are in equilibrium with it.

4 1δ =P

We use the following equations

0δQ1

8−

δQ4

8+ 0.6 δQ5⋅+ 0.6 δQ6⋅−=

0 δQ1 δQ2+=

0δQ2 δQ3+

60.8 δQ5⋅+ 0.8 δQ6⋅+=

1 δQ3 δQ4+=

Page 268


We set δQ1 δQ2= δQ3= 0= and δQ4 1= with this choice equations 2 and 4 are satisfied.

We determine δQ5 and δQ6 to satisfy equations 1 and 3. We have:

018

0.6 δQ5⋅+ 0.6 δQ6⋅−=

0 0.8 δQ5⋅ 0.8 δQ6⋅+= from these two equations we get δQ51

9.6−= δQ6

19.6

=

The rotation at dof 4 can be obtained from the principle of virtual forces δPfT

Uf⋅ δQT

V⋅=

in this case we get: U4 1 V4⋅1

9.6V5⋅−

19.6

V6⋅+:= U4 2.84510 3−=

which is the same as the result we obtained from the compatibility relations.

In conclusion, the principle of virtual forces with an applied unit virtual force at the dof whose displacementis of interest can be used in lieu of the compatibility relations. To this end we need to find any set of virtualbasic forces that satisfy the equilibrium equations under the applied unit virtual force. The simplest choiceinvolves the smallest number of non-zero terms. This method is often referred to in the literature as"dummy unit load method".

Page 269

CE220 - Theory of Structures Example 22 - Upper Bound Theorem of Plastic Analysis © Prof. Filip C. Filippou, 2000

Example 22 - Upper Bound Theorem of Plastic Analysis

In this example we determine the collapse load factor of the simple portal frame that we have alreadystudied extensively in earlier examples from the equilibrium standpoint (Example 8), and for thedetermination of the collapse mechanism by kinematics (Examples 18 and 19). Here we pretend that wedo not know the location of the plastic hinges in the portal frame and proceed in a systematic way bycombination of elementary collapse mechanisms in the determination of the collapse load factor.

The following figure shows the structural model with the plastic flexural capacities of the elements. Theplastic axial capacities are assumed very large.

4 4

5

50

a

b c

d

1

2 34

5

30

150 150

120 120

Assuming that all elements are inextensible there are 5 independent free global dofs, as shown in thefollowing figure. The figure also shows the locations of the angles between node tangents and chords.

4 4

5

dof 2dof 3

dof 4dof 5

dof 1

23 4 5 6

7

81

Page 270


We determine the angles between the nodes and the lines connecting the nodes at all element ends.The compatibility matrix is then equal to the transpose of the equilibrium matrix of example 8. We have

Under the assumption that the elements do not experience deformationincrements during collapse, i.e. ΔVε=0, we have the following compatibilityrelation

ΔVh Af ΔUf⋅= with ΔUf Ac ΔUref⋅= the collapse displacement vector

and with ΔVh the plastic deformation increments (in this case only rotations)

With the above compatibility relations the columns of the compatibility matrixrepresent elementary collapse mechanisms as shown in the following figures.Af

15

15

0

0

0

0

15

15

0

1

1

0

0

0

0

0

0

0

14

−

14

−

14

14

0

0

0

0

0

1

1

0

0

0

0

0

0

0

0

1

1

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

Joint collapse mechanisms (columns 2, 4 and 5 of compatibility matrix)

4 4

5

11

Af1⟨ ⟩

15

15

0

0

0

0

15

15

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

sidesway collapse mechanism (first column of compatibility matrix)

Page 271


4 4

5

3

1

Af3⟨ ⟩

0

0

14

−

14

−

14

14

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

beam collapse mechanism (third column of compatibility matrix)

The upper bound theorem of plastic analysis now says: the load factor determined by the equation of theexternal with the internal plastic work increment for any collapse mechanism constitutes an upper boundof the actual collapse load factor. We are, therefore, interested in finding the collapse mechanism thatminimizes this load factor. Since the load factor is the ratio of internal to external plastic work, we seek acollapse mechanism that minimizes the internal work (fewer hinges with as small rotations as possible),while maximizing the external work (largest translations for applied forces).

Let us proceed systematically. We start with the determination of the collapse load factor for the elementarycollapse mechanisms. Since there are no applied moments at dofs 2, 4 and 5, the corresponding load factoris infinitely large and a definite upper bound. We look at the sidesway mechanism

for a unit displacement value at dof 1 the external plastic work increment is ΔWpe λ 30⋅ 1( )⋅=

the internal plastic work increment for this mechanism is ΔWpi 4 150⋅15⎛⎜⎝⎞⎟⎠

⋅=

note that the plastic deformation increments are all positive and the corresponding moments have alwaysthe same sign

The collapse load factor for the sidesway mechanism is therefore λ4 150⋅

15⎛⎜⎝⎞⎟⎠

⋅

30 1( )⋅:= λ 4=

Page 272


The next elementary mechanism is the beam mechanism. In this case we note that the applied force is in thenegative Y-direction. Thus, to induce positive external work (nothing else makes sense) we use a unitdisplacement in the negative direction of dof 3. We take then the negative values of the angles from columnthree of the compatibility matrix.

for a negative unit displacement value at dof 3 the external plastic work increment is ΔWpe λ 50−( )⋅ 1−( )⋅=

the internal plastic work increment for this mechanism is ΔWpi 2 120⋅14⎛⎜⎝⎞⎟⎠

⋅ 2 120−( )⋅14

−⎛⎜⎝

⎞⎟⎠

⋅+=

The collapse load factor for the beam mechanism is therefore λ2 120⋅

14⎛⎜⎝⎞⎟⎠

⋅ 2 120−( )⋅14

−⎛⎜⎝

⎞⎟⎠

⋅+

50−( ) 1−( )⋅:= λ 2.4=

We note that the beam mechanism is much more critical in this structure under the given loading. In tryingto reduce the collapse load factor we realize that we can use the sidesway mechanism with hinges in thebeam instead of the columns, since the plastic capacity there is only 120 units. We accomplish this by thefollowing elementary collapse mechanism combination = sidesway - 1/5 of joint mechanism at dof 2 - 1/5of joint mechanism at dof 5. Here are the corresponding columns of the compatibility matrix

Af1⟨ ⟩

15

15

0

0

0

0

15

15

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

= Af2⟨ ⟩

0

1

1

0

0

0

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

= Af5⟨ ⟩

0

0

0

0

0

1

1

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

= thus, Af1⟨ ⟩ 1

5Af

2⟨ ⟩⋅−15

Af5⟨ ⟩⋅−

15

0

15

−

0

0

15

−

0

15

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

we see that the plastic deformations now appear at locations 3 and 6, i.e. in the beam instead of thecolumns.

The external work of this combined mechanism is the same as for the sidesway mechanism

the internal work is now a bit smaller ΔWpi 2 150⋅15⎛⎜⎝⎞⎟⎠

⋅ 2 120−( )⋅15

−⎛⎜⎝

⎞⎟⎠

⋅+=

and the collapse load factor becomes λ2 150⋅

15⎛⎜⎝⎞⎟⎠

⋅ 2 120−( )⋅15

−⎛⎜⎝

⎞⎟⎠

⋅+

30 1( )⋅:= λ 3.6=

Page 273


i.e. a bit smaller than the sidesway mechanism, but still larger than the beam mechanism.

The question now is: is there a way of combining the beam mechanism with the last sidesway+jointmechanism so as to maximize the external work (both applied forces will do work), while minimizing theinternal work. Note that if we just added the two mechanisms without canceling any plastic hinge rotationswe would get a value between 2.4 and 3.6. Let us show this

beam mechanism innegative dof 3 direction sidesway + joint mechanism

the internal and external work now become

ΔWpi 2 150⋅15⎛⎜⎝⎞⎟⎠

⋅ 2 120( )⋅14⎛⎜⎝⎞⎟⎠

⋅+ 2 120−( )⋅14

−⎛⎜⎝

⎞⎟⎠

⋅+ 2 120−( )⋅15

−⎛⎜⎝

⎞⎟⎠

⋅+:=

Af3⟨ ⟩−

0

0

14

14

14

−

14

−

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

15

0

15

−

0

0

15

−

0

15

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

ΔWpe λ 50⋅ 1( )⋅ λ 30⋅ 1( )⋅+= λΔWpi

50 1( )⋅ 30 1( )⋅+:= λ 2.85=

It is clear that something strange is going on at location 3, where we have inserted the plastic capacity witha positive value for the beam mechanism and then with a negative value for the sidesway+joint mechanism.Obviously, we need to combine the two mechanisms so that the plastic deformation cancels at that location.For this to happen we add 4/5 x the beam mechanism to the sidesway+ joint mechanism. In this case we get

the internal and external work now become

ΔWpi 2 150⋅15⎛⎜⎝⎞⎟⎠

⋅ 120( )15⎛⎜⎝⎞⎟⎠

⋅+ 120−( )15

−⎛⎜⎝

⎞⎟⎠

⋅+ 120−( )25

−⎛⎜⎝

⎞⎟⎠

⋅+:=

45

0

0

14

14

14

−

14

−

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

⋅

15

0

15

−

0

0

15

−

0

15

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

+

15

0

0

15

15

−

25

−

0

15

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

ΔWpe λ 50⋅45⎛⎜⎝⎞⎟⎠

⋅ λ 30⋅ 1( )⋅+= λΔWpi

5045⎛⎜⎝⎞⎟⎠

⋅ 30 1( )⋅+:= λ 2.229=

Page 274


and we have managed to reduce the collapse load factor a little bit. There does not seem to be much elsethat we can do with this simple structure and we decide to call this the result. How can we be sure? Bydetermining the other basic forces and drawing the moment diagram of the structure. If the plastic flexuralcapacity is not exceeded anywhere in the structure, then we have also found a lower bound, which thenimplies that we have found the unique value of the collapse load factor. Fortunately, we have done this inexample 8 and we display the results one more time in the following figure

4 4

5

150

85.71

120

120

150

54

54

60 6051.43 51.4312.86

12.86

6051.43

2.229*50

5412.86

2.229*30

54

54

60

6051.43

51.43

12.86

2.229*30 54

We have concluded that the final collapse mechanism is the following combination of elementary collapsemechanisms

note that this combination shows a "double hinge"at location 4-5. To avoid confusion we add -1/5 x jointmechanism of dof 4 to get the following final result

Af1⟨ ⟩ 1

5Af

2⟨ ⟩⋅−45

Af3⟨ ⟩⋅−

15

Af5⟨ ⟩⋅−

15

0

0

15

15

−

25

−

0

15

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

Af1⟨ ⟩ 1

5Af

2⟨ ⟩⋅−45

Af3⟨ ⟩⋅−

15

Af4⟨ ⟩⋅−

15

Af5⟨ ⟩⋅−

15

0

0

0

25

−

25

−

0

15

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

Page 275


We can confirm this result with geometric considerations once we know the location of the plastic hinges(recall Example 19).

4 4

5

ICA

ICB

ICC

1

A

B

C

collapse displacementincrements

ΔUf

1

15

−

45

−

15

−

15

−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

ΔU1⋅=

collapse plasticdeformation increments

4 4

5

1

0.8

0.2 0.25030

0.4

0.4

ΔVh

15

0

0

0

25

−

25

−

0

15

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

ΔU1⋅=

Page 276


Upper bound theorem of plastic analysis as linear programming problem (optional)

from Lecture 15

T + T Tref f pl pl pl pl

+ +pl pl f f pl pl

min for

and 0, 0

cλ λ λ + − −

− −

= Δ = Δ + Δ

Δ − Δ = Δ Δ ≥ Δ ≥

P U Q V Q V

V V A U V V

We define the plastic moment capacities Qpl 150 150 120 120 120 120 150 150( )T:=

and the applied force vector Pref 30 0 50− 0 0( )T:=

positive and negative plastic moment capacities are the same Qplp Qpl:= Qpln Qpl:=

define function to minimize f ΔUf ΔVplp, ΔVpln,( ) QplpT ΔVplp⋅ Qpln

T ΔVpln⋅+:=

with initial guess values ΔUf50:= ΔVplp8

0:= ΔVpln80:=


Given

equality constraints PrefT ΔUf⋅ 1= and ΔVplp ΔVpln− Af ΔUf⋅=

inequality constraints ΔVplp 0≥ ΔVpln 0≥ solution Sol Minimize f ΔUf, ΔVplp, ΔVpln,( ):=

collapse load factor λ f Sol1 Sol2, Sol3,( ):= λ 2.229=

Displacement increments of collapse mechanisms (values can be scaled at will, one independent dof)

ΔUf Sol1:= ΔUfT 14.286 2.857− 11.429− 2.857− 2.857−( ) 10 3−=

scale values to unit translation (dof 1) ΔUfT 1 0.2− 0.8− 0.2− 0.2−( ) ΔUf1=

positive plastic rotation increments scaled to unit translation at dof 1

ΔVplp Sol2:= ΔVplpT 0.2 0 0 0 0 0 0 0.2( ) ΔUf1=

negative plastic rotation increments scaled to unit translation at dof 1

ΔVpln Sol3:= ΔVplnT 0 0 0 0 0.4 0.4 0 0( ) ΔUf1=

plastic deformation increments scaled to unit translation at dof 1

ΔVpl ΔVplp ΔVpln−:= ΔVplT 0.2 0 0 0 0.4− 0.4− 0 0.2( ) ΔUf1=

Page 277

Script for Example 22a in CE220 class notes % Upper bound theorem of plastic analysis - Portal frame example solution with Matlab functions and FEDEASLab function UpperBound_PLAnalysis

Clear workspace memory and initialize global variables CleanStart % set up compatibility matrix for free dofs Af = [1/5 0 0 0 0; 1/5 1 0 0 0; 0 1 -1/4 0 0; 0 0 -1/4 1 0; 0 0 1/4 1 0; 0 0 1/4 0 1; 1/5 0 0 0 1; 1/5 0 0 0 0]; % applied force vector Pref at free dof's Pref = [ 30; 0; -50; 0; 0]; % specify plastic flexural capacities Qpl = [ 150 150 120 120 120 120 150 150]';

Upper bound theorem of plastic analysis as linear programming problem [lambdac,DUf,DVpl] = UpperBound_PLAnalysis(Af,Qpl,Pref); % report the collapse load factor disp(['the collapse load factor is ',num2str(lambdac)]); % report the collapse displacement vector scaled by dof 1 disp( 'the collapse displacement vector is'); disp(DUf./DUf(1)) % report the plastic deformation increments scaled by dof 1 disp( 'the plastic deformation increments are'); disp(DVph./DUf(1)); Optimization terminated. the collapse load factor is 2.2286 the collapse displacement vector is 1.0000 -0.2000 -0.8000 0.2000 -0.2000 the plastic deformation increments are 0.2000 0 0 0.4000 0 -0.4000 0 0.2000

Page 278

Script for Example 22b in CE220 class notes % Upper bound theorem of plastic analysis - Portal frame example solution entirely with FEDEASLab functions

Clear workspace memory and initialize global variables CleanStart % define model geometry XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 0 5]; % second node, etc XYZ(3,:) = [ 4 5]; % XYZ(4,:) = [ 8 5]; % XYZ(5,:) = [ 8 0]; % % element connectivity array CON { 1} = [ 1 2]; CON { 2} = [ 2 3]; CON { 3} = [ 3 4]; CON { 4} = [ 4 5]; % boundary conditions (1 = restrained, 0 = free) (specify only restrained dof's) BOUN(1,:) = [ 1 1 1]; BOUN(5,:) = [ 1 1 1]; % specify element type ne = length(CON); % number of elements [ElemName{1:ne}] = deal('2dFrm'); % 2d frame element


1

2 3 4

5

1

2 3

4

define plastic flexural capacity of elements in a column vector % axial force capacity is "very large" Qpl = [1e5 150 150 1e5 120 120 1e5 120 120 1e5 150 150]';

Page 279

form kinematic (compatibility) matrix A A = A_matrix(Model); % extract submatrix for free dofs Af = A(:,1:Model.nf);

specify loading Pe(2,1) = 30; % force at node 2 in direction X Pe(3,2) = -50; % force at node 3 in direction Y % generate data object Loading Loading = Create_Loading(Model,Pe); % extract vector Pref at free dofs from Loading field Pref Pref = Loading.Pref;

Upper bound theorem of plastic analysis as linear programming problem [lambdac,DUf,DVph] = UpperBound_PLAnalysis(Af,Qpl,Pref); % report the collapse load factor disp(['the collapse load factor is ',num2str(lambdac)]); Optimization terminated. the collapse load factor is 2.2286

plot collapse mode Create_Window (0.80,0.80); Plot_Model (Model); % set displacemement increments to zero DU = zeros(Model.nt,1); % put the free dof displacement increments into the first nf components of DU DU(1:Model.nf) = DUf; MAGF = 50; Plot_Model (Model,DU); % use LowerBound_PLAnalysis function to find the basic forces at collapse % note that the static (equilibrium) matrix is the transpose of the kinematic matrix [lambdac Qc] = LowerBound_PLAnalysis(Af',Qpl,Pref); % display plastic hinge locations on collapse mechanism Plot_PlasticHinges (Model,Qpl,Qc,DU); Optimization terminated.

Page 280

CE220 - Theory of Structures Example 23 - Upper bound with partial collapse © Prof. Filip C. Filippou, 2000

Example 23 - Upper Bound Theorem of Plastic Analysis with Partial Collapse

In this example we determine the collapse load factor of the structural model in Example 9 that exhibits apartial collapse mechanism. The geometry, loading and plastic capacities are shown in the following figure.

a b

c

d e

12 3

4

f

56

40

30

6 6

820150

100 100

100 100

La 6:=

Lb 6:=

Lc 8:=

Ld 6:=

Le 6:=

Lf 62 82+:= Lf 10=

The independent free global dofs under the assumption that elements a through e are inextensible areshown in the following figure along with the numbering of the locations where relative displacements andelement deformations are determined for the compatibility relations.

dof 3

dof 1

dof 2

dof 4

1 2 34

56 7

8

Page 281


We write the compatibility relations and identify the elementary collapse mechanisms.

double beam mechanism - negative dof 1 - negative column 1 ofcompatibility matrix

40

30

6 6

8Af

1La

−

1La

−

1Lb

0

0

1Ld

−

1Le

8Lf

0

1

1

1

0

0

0

0

0

0

0

1Lc

1Lc

0

0

6Lf

−

0

0

0

0

1

1

1

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

sidesway mechanism - dof 3 - column 3 of compatibility matrix

40

30

6 6

8

Af

16

−

16

−

16

0

0

16

−

16

45

0

1

1

1

0

0

0

0

0

0

0

18

18

0

0

35

−

0

0

0

0

1

1

1

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

Page 282


elementary joint mechanisms= dofs 2 and 4 = columns 2 and 4of compatibility matrix

We start with the determination of the collapse load factor for the elementary collapse mechanisms. Sincethere are no applied moments at dofs 2, and 4, the corresponding load factor is infinitely large and adefinite upper bound. We look at the double beam mechanism

for a negative unit displacement at dof 1 the external plastic work increment is ΔWpe λ 40−( )⋅ 1−( )⋅=

the internal plastic work increment forthis mechanism is ΔWpi 3 100⋅

16⎛⎜⎝⎞⎟⎠

⋅ 2 100−( )⋅16

−⎛⎜⎝

⎞⎟⎠

⋅+ 20−( )45

−⎛⎜⎝

⎞⎟⎠

⋅+:=

note that the plastic deformation increments and the corresponding basic forces have the same sign

The collapse load factor for the double beam mechanism is λΔWpi

40−( ) 1−( )⋅:= λ 2.483=

We look at the sidesway mechanism

for a unit displacement at dof 3 the external plastic work increment is ΔWpe λ 30( )⋅ 1( )⋅=

the internal plastic work increment forthis mechanism is ΔWpi 2 150⋅

18⎛⎜⎝⎞⎟⎠

⋅ 20−( )35

−⎛⎜⎝

⎞⎟⎠

⋅+:=

note that the plastic deformation increments and the corresponding basic forces have the same sign

The collapse load factor for the sidesway mechanism is λΔWpi

30( ) 1( )⋅:= λ 1.65=

Clearly, the sidesway mechanism is more critical. The question now is: can we find a combination of thesetwo elementary collapse mechanisms that will reduce the collapse load factor by minimizing the internalplastic work increment and maximizing the external plastic work increment?

Page 283


We list the columns of the compatibility matrix for the two elementary mechanisms side by side

we note that a direct combination of these two elementarycollapse mechanisms does not close any plastic hinges. Wenote, however, that at node 2 there are 3 hinges at 2, 3 and4. Thus, by including the rotation dof 2 we should be able toclose some of these. Similarly, at node 5 there are threehinges open at 5, 6 and 7. Thus, by including the rotation ofdof 4 we should be able to close some of these. Either waywe note that we should scale the first dof by 3/4 to bring thehinge rotation values to the same magnitude as for dof 3.

Af1⟨ ⟩−

16

16

16

−

0

0

16

16

−

45

−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

= Af3⟨ ⟩

0

0

0

18

18

0

0

35

−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

With this insight we try the following elementary mechanismcombination: (-3/4 x dof 1) + dof 3 + (-1/8 x dof 2) + (-1/8 x dof 4)

34

−⎛⎜⎝

⎞⎟⎠

Af1⟨ ⟩⋅

18

−⎛⎜⎝

⎞⎟⎠

Af2⟨ ⟩⋅+ Af

3⟨ ⟩+18

−⎛⎜⎝

⎞⎟⎠

Af4⟨ ⟩⋅+

18

0

14

−

0

0

0

14

−

65

−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

= and indeed we were successful in closingthe plastic hinges at 2 and 4, as well as at5 and 6.

The external plastic work increment for this combined mechanism is: ΔWpe λ 40−( )⋅34

−⎛⎜⎝

⎞⎟⎠

⋅ λ 30( )⋅ 1( )⋅+=

The internal plastic work increment is: ΔWpi 10018⎛⎜⎝⎞⎟⎠

⋅ 2 100−( )⋅14

−⎛⎜⎝

⎞⎟⎠

⋅+ 20−( )65

−⎛⎜⎝

⎞⎟⎠

⋅+:=

Page 284


and the collapse load factor becomes λΔWpi

40−( )34

−⎛⎜⎝

⎞⎟⎠

⋅ 30( ) 1( )⋅+:= λ 1.442=

and this agrees with the answer of Example 9 obtained with the lower bound theorem of plastic analysis;we conclude, therefore, that this is the collapse load factor of the model under the given loading.The collapse mechanism is shown in the following figure.

3/4C

ICA

A B ICB

ICC

1

1

2

3

4

collapse mechanism

displacement incrementsof collapse mechanism

ΔUf

1

18

−

34

−

18

−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

ΔU1⋅=

plastic deformationincrements of collapsemechanism

ΔVh

18

0

14

−

0

0

0

14

−

65

−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

ΔU1⋅=

We recall that this is a partial collapse mechanism and that there is no unique set of basic forces inelements a, c and d when the collapse load factor is attained (consult Example 9). Can we still hope todetermine the displacements and deformations at incipient collapse? Not without a statically indeterminateanalysis with basic force Q6 as redundant (recall that we selected Q6 for the homogeneous solution inExample 9). We will return to this after presenting the force method of analysis.

Page 285

Script for Example 23 in CE220 class notes % Upper bound theorem of plastic analysis with partial collapse mechanism solution entirely with FEDEASLab functions using inextensible element constraints


Create model % specify node coordinates (could only specify non-zero terms) XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 6 0]; % second node, etc XYZ(3,:) = [ 12 0]; % XYZ(4,:) = [ 0 8]; % XYZ(5,:) = [ 6 8]; % XYZ(6,:) = [ 12 8]; % % connectivity array CON {1} = [ 1 2]; CON {2} = [ 2 3]; CON {3} = [ 2 5]; CON {4} = [ 4 5]; CON {5} = [ 5 6]; CON {6} = [ 3 5]; % boundary conditions (1 = restrained, 0 = free) BOUN(1,:) = [1 1 1]; BOUN(3,:) = [0 1 0]; BOUN(4,:) = [0 1 0]; BOUN(6,:) = [0 1 0]; % specify element type [ElemName{1:5}] = deal('2dFrm'); % 2d frame element ElemName{6} = 'Truss'; % truss element % create model Model = Create_SimpleModel (XYZ,CON,BOUN,ElemName);


1 2 3

4 5 6

1 2

3

4 5

6

Page 286

define plastic flexural capacity of elements in a column vector % axial force capacity is "very large" except for brace element Qpl = [ 1e5 100 100 1e5 100 100 1e5 150 150 1e5 100 100 1e5 100 100 20]';

form kinematic (compatibility) matrix A A = A_matrix(Model); % extract submatrix for free dofs Af = A(:,1:Model.nf);

define loading Pe(4,1) = 30; Pe(5,2) = -40; Loading = Create_Loading (Model,Pe); % extract vector Pref at free dofs from Loading field Pref Pref = Loading.Pref;

Upper bound theorem of plastic analysis as linear programming problem [lambdac,DUf,DVph] = UpperBound_PLAnalysis(Af,Qpl,Pref); % report the collapse load factor disp(['the collapse load factor is ',num2str(lambdac)]); Optimization terminated. the collapse load factor is 1.4417

plot collapse mode Create_Window (0.80,0.80); Plot_Model (Model); % set displacemement increments to zero DU = zeros(Model.nt,1); % put the free dof displacement increments into the first nf components of DU DU(1:Model.nf) = DUf; MAGF = 50; Plot_Model (Model,DU); % use LowerBound_PLAnalysis function to find the basic forces at collapse % note that the static matrix is the transpose of the kinematic matrix [lambdac Qc] = LowerBound_PLAnalysis(Af',Qpl,Pref); % display plastic hinge locations on collapse mechanism Plot_PlasticHinges (Model,Qpl,Qc,DU); Optimization terminated.

Page 287

DEFORMATION-FORCE RELATIONSFOR

MATERIAL, SECTION, ELEMENT AND STRUCTURE

Objective: cause and effect relation between static and kinematic variables for linear elastic material response; deformation force relation for material; force-deformation for section under Bernoulli assumption; deformation-force for basic element and collection of elements; displacement-force relation for statically determinate structures; force method of analysis and displacement-force relation for indeterminate structures

CE220-Theory of Structures Deformation-Force Relations © Prof. Filip C. Filippou, 2000

Page 288

Material strain-stress and stress-strain relation

in general, the behavior at a material point is described by a nonlinear relation between the stress and strain tensor ˆ ( , , )γ=σ σ ε ε

where we see that the stress tensor may depend not only on the strain, but also on the strain rate and on internal variables

Different types of constitutive models are available for the description of the behavior of specific materials: linear elastic, linear viscolelastic, perfectly plastic, viscoplastic, etc... Here we restrict ourselves to the relatively simple case of linear elastic material response. Moreover, we limit ourselves to uniaxial behavior. Thus, both stress and strain are scalars.

The most natural way of describing the linear elastic response of a material is in strain-stress form, where we distinguish the mechanical and non-mechanical contribution. An example of the latter is the thermal and shrinkage strain. We write

is the elastic modulus of the material and can be thought of as material stiffness

its inverse is known as material compliance or flexibility

Thus, the relation is the flexibility relation at the material level

with the mechanical strain

the non-mechanical strain

where is the material coefficient of thermal expansionthe change in reference temperature (e.g. from construction to service)

We can invert this relation to obtain the stiffness relation at the material level

The stiffness relation in this form shows clearly that the stress is proportional to the mechanical strain

Less clear is the following common form of the stiffness relation with

σ0 is the required initial stress to compensate the non-mechanical strain before the total strain is applied.

for a summary of material relations consult chapter 5 in S. Crandall, "An introduction to the Mechanics of Solids", 2nd edition

0mε ε ε= + m Eσε =

0ε

in the special case of thermal strain we have 0 Tε α= Δ

αTΔ

E

1 E

( ) 01 Eε σ ε= +

( )0Eσ ε ε= −

0Eσ ε σ= + 0 0Eσ ε= −


Page 289

Force-deformation behavior of sectionWhat is the force-deformation behavior of the section under the combined action of axial force and bending moment? The Euler-Bernoulli beam theory of small deformations (1750 AD) with the assumption that plane sections remain plane (Daniel Bernoulli's contribution) results in the following pair of simple relations under the assumption that the x-axis coincides with the centroid of the cross section and that the y-z axes pair coincides with the principal axes of the cross section.

Note that we have introduced non-mechanical section deformations by analogy with the stress-strain relation. Note also that the section force-deformation relation comes out of the Euler-Bernoulli theory in stiffness form with EA the axial and EI the flexural section stiffness. We can invert these relations to the flexibility form

The axial strain at the reference axis and the curvature are components of the section deformation vector e, which in more advanced applicationsincludes shearing and torsional deformations. This notation will be useful in CE221.

where A is the cross sectional area and I is the moment of inertia or second moment of area

is the axial strain at the reference axis (subscript a)

is the curvature of the beam which is defined as change of tangent rotation per unit length

with the axial section flexibility

the flexural section flexibility

for the thermal strain at the reference axis

for the thermal curvature

( )0a aN EA ε ε= −

( )0M EI κ κ= −

aε

κ

0a aNEA

ε ε= + 1 EA

1 EI0MEI

κ κ= +

We have now expressed the section deformations as the superposition of two effects: the mechanical effect which is proportional to the section (internal) forces, and the non-mechanical effect which we have already studied on page 13 of Part II. We recall the effect of a differential temperature field on the non-mechanical section deformations and introduce now the subscript 0 for the non-mechanical effect

0a aTε α= Δ

( )0

Th

ακ

Δ Δ=


Page 290

Deformation-force behavior of 2d frame element

After having established the force-deformation behavior of the section we use the equilibrium relations of Part I between the internal forces at a section x and the basic force of the frame element (page 18 of Part I) as well as the kinematic relations between the section and element deformations (page 14 of Part II) to derive the relation between basic forces and corresponding deformations for the 2d frame element.

from equilibrium

where Mp is the moment distribution due to element loading

1

2 3 p

( )

( ) 1 ( )

N x

x xM x M xL L

=

⎛ ⎞= − + +⎜ ⎟⎝ ⎠

q

q q

from kinematics 1

0

2

0

3

0

( )

1 ( )

( )

L

a

L

L

x dx

x x dxL

x x dxL

ε

κ

κ

=

⎛ ⎞= − −⎜ ⎟⎝ ⎠

=

∫

∫

∫

v

v

v

for linear elastic material response we use the relations on the preceding page 0

0

( )( ) ( )( )( )( ) ( )( )

a aN xx xEA xM xx xEI x

ε ε

κ κ

= +

= +

Substitute the section deformation-force into the kinematic relations and then substitute the internal forces in terms of the basic forces of the element and the element loading. We get

1 0 1 0

0 0 0

( ) ( ) ( )( ) ( )

L L L

a aN x dxx dx x dxEA x EA x

ε ε⎡ ⎤

= + = +⎢ ⎥⎣ ⎦∫ ∫ ∫v q

2

p2 0 2 3 0

0 0 0 0 0

1 1 ( )( )1 ( ) 1 ( ) 1( ) ( ) ( ) ( )

L L L L Lx x xM xx M x x xL L Lx dx dx dx x dx dx

L EI x EI x EI x L L EI xκ κ

⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎝ ⎠ ⎝ ⎠= − − + = − − − − −⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦∫ ∫ ∫ ∫ ∫v q q

2

p3 0 2 3 0

0 0 0 0 0

1 ( )( ) ( ) ( )( ) ( ) ( ) ( )

L L L L Lx x xM xx M x x xL L Lx dx dx dx x dx dx

L EI x EI x EI x L L EI xκ κ

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎡ ⎤ ⎝ ⎠ ⎝ ⎠= + = − + + +⎢ ⎥⎣ ⎦∫ ∫ ∫ ∫ ∫v q q

These relations express the deformation-force behavior of a 2d frame element. For a tapered element with variable section properties along its length (tapered element), the integrals need to be evaluated numerically. We do not pursue this further but write the deformation-force relation in compact form on the following page.

note that all element deformations on this and the following pages correspond to εv


Page 291

Flexibility matrix and initial deformations of 2d frame element

with the element flexibility matrix

and the initial deformations

non-mechanical effects

element load effects

The deformation-force relations can be written in compact form as follows

2

0 2

1 0 0( )

1 10

( ) ( )

10

( ) ( )

L

EA x

x x xL L L dx

EI x EI x

x x xL L L

EI x EI x

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

⎛ ⎞ ⎛ ⎞⎢ ⎥− −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠= −⎢ ⎥⎢ ⎥⎢ ⎥

⎛ ⎞ ⎛ ⎞⎢ ⎥−⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠−⎢ ⎥⎣ ⎦

∫f

0

0

p0 0

0 0

p0

0 0

( )0

( )1 ( ) 1

( )

( )( )

( )

L

a

L L

L L

x dx

M xx xx dx dxL L EI x

M xx xx dx dxL L EI x

ε

κ

κ

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= − − + − −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∫

∫ ∫

∫ ∫

v

For a prismatic 2d frame element (uniform properties) under uniform non-mechanical effects and undera uniformly distributed element load these expressions simplify to

element flexibility matrix

initial deformations

0 0

03 6

06 3

LEA

L LEI EIL LEI EI

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥= −⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦

f

0 3

0 0

30

0

12 24

12 24

ay

y

Lw L

LEI

w LLEI

ε

κ

κ

⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟= − + − +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

v …

other loading effects

0= +fv q v


Page 292

Summary of deformation-force behavior of prismatic 2d frame element

1q1q

x1q

( )N x( )N x

1q+

1q

L

end jend i

1

EAq

+

1L

EAq

1q1q

( )a xε for uniform EA

x

2q

( )M x

2q

2

Lq 2

Lq

( )M x

2q 2( ) 1 xM xL

q ⎛ ⎞= − −⎜ ⎟⎝ ⎠

2

Lq

2

Lq

L

2EIq

( )xκ for uniform EI

2q

2

Lq 2

Lq

23LEI

q 26LEI

q

x

3

Lq 3

Lq

( )M x3( ) xM x

Lq=

3

Lq

3q

3q

3

Lq

3q

L

( )M x

3EIq


33LEI

q36LEI

q3q

3

Lq3

Lq

1 1

2 2

3 3

0 0

03 6

06 3

LEA

L LEI EIL LEI EI

⎡ ⎤⎢ ⎥

⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥= −⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎢ ⎥−⎢ ⎥⎣ ⎦

v qv qv q


Page 293

Summary of deformation-force behavior of prismatic 2d frame element (con'd)

Lx

0 0( ) const=xκ κ=

20v 30v

10v

0 0( ) const=a axε ε=yw

x

( )M x

( )M x

2

8yw L

yw

2yw L

2yw L

2yw L

2yw L

L

end jend i

( )V x


2

8yw L

EI

3

24yw L

EI

3

24yw L

EI

parabola

uniform non-mechanical effects uniformly distributed element load

0 01

0 02

0 03

12

12

a L

L

L

ε

κ

κ

=

= −

=

v

v

v

012 3

02

2 3

03

0

13 8 24

13 8 24

y y

y y

w L w LL

EI EI

w L w LL

EI EI

=

⎛ ⎞⎜ ⎟= − = −⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟= =⎜ ⎟⎝ ⎠

v

v

v


Page 294

Deformation-force behavior of prismatic 2d frame element with end moment release

2q 2

Lq

end jend i

x

y

+

2

Lq

1q1q

L

element flexibility matrix

initial deformations

deformation-force relations in compact form

0

03

LEA

LEI

⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦

f

03

00

0

12 24

a

y

Lw LL

EI

ε

κ

⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟= + +⎜ ⎟⎜ ⎟−⎜ ⎟ −⎜ ⎟⎝ ⎠ ⎝ ⎠

v …

strain dependent deformation at release (for hinge rotation determination)3

03 26 2 24

yw LLLEI EI

κ= − + +v q

0= +fv q v


Page 295

Combine the deformation-force relations for all elements in the structural model

or, in compact form

with and we get where

block-diagonal

Deformation-force behavior for collection of elements in structural model

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

a a a a0

b b b b0

c c c c0

ne ne ne ne0

= +

= +

= +

= +

f

f

f

f

v q v

v q v

v q v

v q v

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

aa a a0

bb b b0

cc c c0

nene ne ne0

0 0 0 0

0 0 0 0

0 0 0 00 0 0 0

0 0 0 0

⎛ ⎞⎛ ⎞ ⎡ ⎤ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥= + ⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎜ ⎟⎝ ⎠ ⎣ ⎦ ⎝ ⎠ ⎝ ⎠

f

f

f

f

vv q

vv q

vv q

vv q

( )

( )

( )

( )

a

b

c

ne

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟=⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

v

vV v

v

( )

( )

( )

( )

a

b

c

ne

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟=⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

q

qQ q

q

s 0= +FV Q V

( )

( )

( )

( )

a

b

cs

ne

0 0 0 0

0 0 0 0

0 0 0 00 0 0 0

0 0 0 0

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

f

fF f

f

s 0= +FV Q V


Page 296

Displacement-force behavior of structures

Statically determinate structures

Equilibrium

for stable, statically determinate structures solve directly for Q

where 1fB B −=is the force influence matrixB

Compatibility f f= AV U

Tf = BU Vfor stable, statically determinate structures solve for Uf or since T

f f=A B

Deformation-force relation for linear elastic material response

Combine

( ) ( )T T Tf s f s fw 0= − +B F B B F B BU P P V

f f 0= +FU P U

with Ts=F B F B the structure flexibility matrix, and

T0 fw 0= − +F BU P V the initial displacement vector

f f fw= +BP Q P

( )f fw= −BQ P P

s 0= +FV Q V

{ }( ){ }

Tf

Ts 0

Ts f fw 0

=

= +

⎡ ⎤= − +⎣ ⎦

B

B F

B F B

U V

Q V

P P V

Consult examples 24, 25 and 26

1f f

−= AU V


Page 297

CE 220 - Theory of Structures Ex. 24 - Force-displacement of two beams © Prof. Filip C. Filippou, 2000

Example 24 - Force-displacement relations for two beams with overhang

A - Simply supported beam with cantileverWe consider a relatively simple structural model. The model is subjected to a concentrated force of 15 unitsat the tip of the cantilever.

15 5

15

1 2 3a b

The structure is statically determinate: there are 2 free global dofs (or equilibrium equations) and 2unknown basic forces, as shown in the following figure.

dof 2dof 1

1Q 2Q

The equilibrium equations are relatively simple: P1 Q1 Q2+= with P1 0=

P2Q2

5−= P2 15−=

We solve with ease and obtain Q2 75:= Q1 75−:=

The following figure shows node and element equilibrium and the bending moment diagram.

1515

15

755

5

5

5

75M(x)

5

20

Page 298


We wish to calculate the vertical translation under the applied force at the tip of the cantilever. To this endwe use the compatibility relations (we can use the transpose of the above equilibrium equations, or simplywrite them down directly). We have

V1 U1=

V2 U1U2

5−= from which we conclude that U2 5 V1 V2−( )⋅=

The following figure gives a geometric interpretation of this relation. Note that the deformation is measuredfrom the element chord to the tangent at each end.

15 5

1V

2V( )1 25 V V−

We determine the element deformations from the element deformation-force relation of each element, i.e.

v f q⋅= where fL

6 EI⋅

2

1−

1−

2⎛⎜⎝

⎞⎟⎠

⋅=

instead of worrying about specifying a stiffness value EI, we will assume that both elements have the samesection stiffness EI. We will then calculate the EI-fold element deformations. We have:

EIV1156

2 Q1⋅( )⋅:= EIV1 375−=

EIV256

2 Q2⋅( )⋅:= EIV2 125=

With the EI-fold deformation values we can get the EI-fold translation value.

EIU2 5 EIV1 EIV2−( )⋅:= EIU2 2500−= negative means downward (in negative Y-direction)

If we know the value of the stiffness EI, we can obtain the actual vertical translation value at the end.Since EI is a very large number (e.g. in the range of 20,000 to 100,000 for consistent units of force andlength), it is clearly better to leave this division to the end and not divide every element deformation by EI,since it is error prone dealing with very small numbers in hand calculations. The computer by contrast doesnot mind at all.

What would be a reasonable value of stiffness EI for this structure? The tip of the cantilever may deflect1/200 of the cantilever length. This means that the vertical translation is 0.025 in consistent units. For thisto be true, the stiffness EI should be 100,000 consistent units. We will see numbers in this range for theflexural stiffness EI in this course.

Page 299


Let us now try to determine the midspan deflection in element a under the given applied force of 15 units atthe tip of the cantilever. How do we proceed?

First, we draw the moment diagram in element a. Since we wish to determine the translation at midspan weinsert a node m into the middle of element a, as shown in the following figure. This splits the element into twoelements a1 and a2. We denote the degrees of freedom at the new node with m1 and m2.

7.5 5

15

7537.5

a b

a1 a2m

m1

m2

21

1 2 3

7.5

We write the compatibility relations

vja1 Um2Um1

7.5−=

via2 Um2Um1

7.5+=

where ja1 stands for end j of element a1and ia2 stands for end i of element a2.

We can see clearly from the abovetwo equations that the vertical translationcan be obtained by subtracting the firstequation from the second. We get

Um17.52

via2 vja1−( )⋅=

To determine via2 and vja1 we use the end moments in element a1 and a2. We obtain:

EIvia27.56

2 37.5⋅ 75−( )−[ ]⋅:= EIvja17.56

2 37.5−( )⋅[ ]⋅:=

With this we get: EIUm17.52

EIvia2 EIvja1−( )⋅:= EIUm1 1054.688=

and we observe that the translation is positive, which means that the point moves upward under the appliedforce of 15 units at the tip of the cantilever.

The following figure shows a different geometric interpretation for the determination of the translation Um1

iav

ia1v

7.5 57.5

via is the deformation atend i of element a

via1 is the deformationat end i of element a1

EIvia156

75−( )−[ ]⋅:= EIvia17.56

37.5−( )−[ ]⋅:=

According to this figure EIUm1 7.5 EIvia EIvia1−( )⋅:= EIUm1 1054.688= same as above!

Page 300


We move now to consider the case of a uniform distributed load.

5

w=10

15

1 2 3a b


P2Q2

5− 25+= P2 0=

We solve with ease and obtain Q2 125:= Q1 125−:=

The following figure shows node and element equilibrium and the bending moment diagram.

50125

83.33

125M(x)

218.75

5083.3366.67

66.67

2 8 281.25wL =

133.33

We can repeat the previous calculations for the vertical translation at the tip of the cantilever and themidspan of element a noting, however, the that force-deformation relation for each element needs toinclude the effect of the distributed load according to

v f q⋅ v0+= where fL

6 EI⋅

2

1−

1−

2⎛⎜⎝

⎞⎟⎠

⋅= and v0w L3⋅24 EI⋅

1−

1⎛⎜⎝

⎞⎟⎠

⋅= for a downward uniform load

We demonstrate for the tip of the cantilever.

EIV1156

2 Q1⋅( )⋅10 153⋅

24+:= EIV1 781.25=

EIV256

2 Q2⋅( )⋅10 53⋅

24−:= EIV2 156.25=

EIU2 5 EIV1 EIV2−( )⋅:= EIU2 3125= positive means upward (in positive Y-direction)

Page 301


The geometric interpretation of this calculation is shown in the following figure of the deformedshape of the beam

iav

ia1vja1v

ia2v1V

2V( )1 25 V V−

7.5 57.5

The vertical translation at midspan of element a can again be obtained from Um17.52

via2 vja1−( )⋅=

with EIvia27.56

2 218.75−( )⋅ 125−( )−[ ]⋅10 7.53⋅

24−:=

EIvja17.56

2 218.75⋅( )⋅10 7.53⋅

24+:=

With this we get: EIUm17.52

EIvia2 EIvja1−( )⋅:= EIUm1 4833.984−= negative, i.e. downward

Alternatively, by the direct geometric method we have used before

EIvia156

125−( )−[ ]⋅10 153⋅

24−:= EIvia1

7.56

218.75−( )⋅10 7.53⋅

24−:=

According to the figure EIUm1 7.5 EIvia EIvia1−( )⋅:= EIUm1 4833.984−= same result!!

Page 302


B - Propped cantilever

We consider a relatively simple structure, known as a propped cantilever beam. The structure is subjectedto a concentrated force of 5 units at the tip of the cantilever.

5

30 10

1 2 3a b

The structure is statically indeterminate. Can you tell why? There are 2 free global dofs (or equilibriumequations) and 3 unknown basic forces, as shown in the following figure.

1 2 3

dof 2dof 1

1Q 2Q 3Q

a bNOS=1


P2Q3

10−= P2 5−=

There is something very interesting going on here! There are two equations, and only two unknown basicforces appear in these. Thus, given the applied forces it is possible to solve for the two unknown basicforces and we get:

Q3 50:= and Q2 50−:=

We have seen a similar case in the reduction of the free dofs in Example 3 (see page 3-4 and 3-5).

How are we supposed to determine the basic force Q1? By looking at the deformations of element a.

Recall from the force-deformation relations of the beam element that

Page 303


V1 Q1L

3 EI⋅⋅ Q2

L6 EI⋅⋅−= since V1 is equal to zero at the fixed end of element a we conclude that

V1 0= Q1Q2

2:= Q1 25−=

Thus, the moment at the fixed-end of a prismatic beam element is equal to 1/2 the moment acting at theopposite end. This is known as carry-over moment and the factor 1/2 is known as carry-over factor(tapered elements have different carry-over factors). The following figure displays this fact

L

M0.5 M

We wish to calculate the vertical translation at the tip of the cantilever under the applied force of 5 units.Let us first calculate the element deformations from the basic forces. To this end we use the elementdeformation-force relation, i.e.

v f q⋅= where fL

6 EI⋅

2

1−

1−

2⎛⎜⎝

⎞⎟⎠

⋅=

instead of worrying about specifying a stiffness value EI, we will assume that both elements have the samesection stiffness EI. We will then calculate the EI-fold element deformations. We have:

EIV1306

2Q1 Q2−( )⋅:= EIV1 0= of course, we knew this so we should not have bothered

EIV2306

2Q2 Q1−( )⋅:= EIV2 375−=

EIV3106

2 Q3⋅( )⋅:= EIV3 166.67=

In order to calculate the vertical translation at the tip of the cantilever we write the compatibility relations(we can use the transpose of the above equilibrium equations, or simply write them down directly).

V2 U1=

V3 U1U2

10−=

thus, we can determine the translation U2 from the second compatibility relation after substituting U1 fromthe first into the second. We get

U2 10 V2 V3−( )=

Page 304


Since we have the EI-fold deformation values we can only get the EI-fold translation value.

EIU2 10 EIV2 EIV3−( )⋅:= EIU2 5416.67−=

If we know the value of the stiffness EI, we can obtain the actual vertical translation value at the end.Since EI is a very large number (e.g. in the range of 20,000 to 100,000 for consistent units of force andlength), it is clearly better to leave this division to the end and not divide every element deformation by EI,since it is error prone dealing with very small numbers in hand calculations. The computer by contrast doesnot mind at all.

There is some important information in the final result that we would like to bring to light. We havedetermined the vertical translation at the tip of the cantilever under a force of 5 units. We now ask thefollowing question: what is the vertical translation value under a unit force? It is easy to see that thebasic forces are proportional to the applied force and would therefore be equal to 1/5 the values above.The deformations would then also be equal to 1/5 the above values and the final translation would also beequal to 1/5 of the above value. Assuming that the unit force acts in the positive direction (i.e. upward),the translation would also be positive (i.e. upward) and have the value of 5416.67/5 = 1083.33/EI.

Thus, under a unit force at the tip of the cantilever the vertical translation is equal to 1083.33/EI. This isknown as the flexibility coefficient F22 of the propped cantilever. Why are there two subscripts for theflexibility coefficient? The first subscript denotes the dof where the effect is measured and the secondsubscript denotes the dof where the cause acts. Thus, F22 indicates the translation at dof 2 under a unitforce acting at dof 2. If this is the only dof where forces act in this structure (i.e. there is never aconcentrated moment acting at dof 1), then we will not be interested in any flexibility coefficients with secondsubscript equal to 1.

How about if we are interested in the effect of a force at dof 2 on the rotation at dof 1. We have this valuealready from the above analysis. It is equal to U1 for a force of -5 units.

EIU1 EIV2:= EIU1 375−=

thus, the rotation at dof 1 under a unit positive force acting at dof 2 will be equal to 375/5 = 75/EI.

Therefore, the flexibility coefficient F12 of this structure is 75/EI. Flexibility coefficients play an important rolein condition assessment of structures (think for example that the propped cantilever is a bridge, that youplace a truck at the tip and are able to measure the tip translation and the rotation at the free support; thisallows you to determine the stiffness of the bridge; by comparing with the ideal stiffness value you may beable to determine how much damage the bridge has sustained (note that stiffness reduction is a commondamage measure in structures).

Let us now think of another interestingapplication of the flexibility coefficient. Saythat we want to estimate the vibrationfrequency of the propped cantilever underthe assumption that the vibratory mass islocated at the tip. In this case we need thespring stiffness under the mass. This meansthat we only care about the cause-effect(force-displacement) relation at dof 2. Thisis supplied by flexibility coefficient F22. Theinverse of this coefficient is the springstiffness K. Given the mass M we may recallfrom Physics that the frequency of vibrationof this single dof spring-mass system is

30 10

single dof

K = EI/1083

M

EI EI

ωKM

=

Page 305


Let us now try to determine the midspan deflection in element a under the given applied force of 5 units atthe tip of the cantilever. How do we proceed?

First, we draw the moment diagram in element a. Since we wish to determine the translation at midspanwe insert a node m into the middle of element a, as shown in the following figure. This splits the elementinto two elements a1 and a2. We denote the degrees of freedom at the new node with m1 and m2.

5

30 10

1 2 3

30

50

25

a b

a1 a2 21

12.5

m

m1

m2

We write the compatibility relations

vja1 Um2Um1

15−=

via2 Um2Um1

15+=

where ja1 stands for end j of element a1and ia2 stands for end i of element a2.

We can see clearly from the abovetwo equations that the vertical translationcan be obtained by subtracting the firstequation from the second. We get

Um1152

via2 vja1−( )⋅=

To determine via2 and vja1 we usethe end moments in element a1 and a2.We obtain:

EIvia2156

2 12.5⋅ 50−( )−[ ]⋅:=

EIvja1156

2− 12.5⋅ 25−( )−[ ]⋅:=

With this we get: EIUm1152

EIvia2 EIvja1−( )⋅:= EIUm1 1406.25=

and we observe that the translation is positive, which means that the point moves upward under the appliedforce of 5 units at the tip of the cantilever. If we were interested in the flexibility coefficient between thevertical translation dof m1 and the applied force dof 2, we can obtain it from

we reiterate that this is the translation at dof m1 under a unit forceat dof 1; the negative sign means that the translation is downwardfor a unit force acting upward at dof 1

Fm1.21406.25

5 EI⋅−= 281.35

EI−=

Page 306

Script for Example 24 in CE220 class notes % statically determinate simply supported beam with overhang


Create model % specify node coordinates (could only specify non-zero terms) XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 7.5 0]; XYZ(3,:) = [ 15 0]; % XYZ(4,:) = [ 20 0]; % % connectivity array CON {1} = [ 1 2]; CON {2} = [ 2 3]; CON {3} = [ 3 4]; % boundary conditions (1 = restrained, 0 = free) BOUN(1,:) = [1 1 0]; BOUN(3,:) = [0 1 0]; % specify element type ne = length(CON); [ElemName{1:ne}] = deal('Lin2dFrm'); % 2d linear frame element % create model Model = Create_SimpleModel (XYZ,CON,BOUN,ElemName);


Define element properties ne = Model.ne; % number of elements in structural model % Modulus, area and moment of inertia for el=1:ne; ElemData{el}.E = 1000; ElemData{el}.A = 10; % value is not used, it is thus irrelevant ElemData{el}.I = 100; end

define loading % 1. Load case: vertical nodal force at tip of overhang Pe(4,2) = -15; Loading = Create_Loading (Model,Pe); % nodal forces at free dofs Pf = Loading.Pref;

Page 307

set up equilibrium matrix B = B_matrix(Model); % extract free dofs Bf = B(1:Model.nf,:); % determine force influence matrix Bbar = inv(Bf);

solution % determine basic forces from equilibrium equations Q = Bf\Pf; % display basic forces disp('the basic forces are') disp(Q) the basic forces are 0 0 -37.5000 0 37.5000 -75.0000 0 75.0000 0

plotting % open window and plot moment diagram Create_Window(0.80,0.80); Plot_Model (Model); % observe that the second argument is empty [] in the following function call Plot_2dMomntDistr (Model,[],Q);

set up collection of flexibility matrices Fs = Fs_matrix(Model,ElemData);

determine element deformations due to Q V = Fs*Q;

determine global dof displacements % free global dof displacements Uf = Bbar'*V; disp('the free dof displacements are') disp(Uf) the free dof displacements are 0.0019 0 0.0105 0.0005

Page 308

0 -0.0038 0 -0.0250 -0.0056

plot deformed shape of structural model % put free dof displacement values in complete displacement vector U = zeros(Model.nt,1); U(1:Model.nf) = Uf; MAGF = 50; % magnification factor for deformed configuration Create_Window(0.80,0.80); Plot_Model(Model); Plot_DeformedStructure(Model,[],U); clear Pe; % 2. Load case: uniformly distributed element load w=-10 % use equivalent nodal forces to determine the basic forces Pe(1,2) = -10*7.5/2; Pe(2,2) = -10*7.5; Pe(3,2) = -10*7.5/2 - 10*5/2; Pe(4,2) = -10*5/2; Loading = Create_Loading (Model,Pe); % nodal forces at free dofs Pf = Loading.Pref;

specify distributed load value in ElemData for el=1:ne; ElemData{el}.w = [0;-10]; end

solution % determine basic forces from equilibrium equations Q = Bf\Pf; % display basic forces disp('the basic forces under distributed element loading are') disp(Q) the basic forces under distributed element loading are 0 0 218.7500 0 -218.7500 -125.0000 0

Page 309

125.0000 0

plotting % open window and plot moment diagram Create_Window(0.80,0.80); Plot_Model (Model); % observe that the second argument is ElemData in this case Plot_2dMomntDistr (Model,ElemData,Q);

determine element deformations due to Q % set up first initial deformations due to w V0 = V0_vector(Model,ElemData); % element deformations V = Fs*Q + V0;

determine global dof displacements % free global dof displacements Uf = Bbar'*V; disp('the free dof displacements are') disp(Uf) the free dof displacements are -0.0109 0 -0.0483 0.0008 0 0.0078 0 0.0313 0.0057

plot deformed shape of structural model % put free dof displacement values in complete displacement vector U = zeros(Model.nt,1); U(1:Model.nf) = Uf; MAGF = 50; % magnification factor for deformed configuration Create_Window(0.80,0.80); Plot_Model(Model); Plot_DeformedStructure(Model,ElemData,U);

Page 310

CE220 - Theory of Structures Ex 25 - Force-displacement for determinate truss © Prof. Filip C. Filippou, 2000

Example 25- Force-displacement for statically determinate truss

In this example we will set up the relation between applied forces at the free dofs of a staticallydeterminate structural model and the corresponding displacements. To this end we will go through theentire process of structural analysis: equilibrium equations, compatibility relations, and force-deformationrelations for each element. The following figure shows the simple truss model with the numbering of thefree dofs and basic forces.

8

6

1

23

4

5

1Q

2Q3Q

4Q

5Q

8

6

15

a

b c

d

e

The axial stiffness of all truss elements is the same: EA 10000:=

Length of truss elements La 8:= Lb 6:= Lc 6:= Ld 8:= Le 10:=

Equilibrium equations

P1 Q1 0.8 Q5⋅+= corresponding equilibrium matrix

P2 Q4− 0.8 Q5⋅−=

P3 Q2 0.6 Q5⋅+= Bf

1

0

0

0

0

0

0

1

0

0

0

0

0

0

1

0

1−

0

1

0

0.8

0.8−

0.6

0

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=

P4 Q4=

P5 Q3=Generic solution of equilibrium equations (i.e. for a force at any dof) yields the force influence matrix Bbar

Bbar Bf1−:= Bbar

1

0

0

0

0

1

0.75

0

0

1.25−

0

1

0

0

0

1

0.75

0

1

1.25−

0

0

1

0

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

=

Page 311


The force influence matrix identifies the most highly stressed elements and plays important role inoptimization, system identification, damage detection, etc. We have already encountered this matrix in thefirst example of this course.

Typically, in homework problems we solve for a specific load case , while we may be interested in severalload cases in practice. In this example the applied loading consists of a force of 15 units at dof 4. Havingthe force influence matrix Bbar we can obtain the corresponding truss element forces by multiplication. Fora single load case it is more expedient to solve the equilibrium equations "by hand" and obtain Q directly.However, the force influence matrix plays an important role in the determination of the flexibility matrix forthe structure, so we keep it under consideration for a little longer.

for the applied force vector Pf

0

0

0

15

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:= we get the basic forces Q Bbar Pf⋅:= Q

15

11.25

0

15

18.75−

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

=

Note that the basic forces for this load case are equal to the 4th column of the force influence matrixmultiplied by 15. We reiterate that the columns of the force influence matrix represent the effect of aunit force at the corresponding dof on the basic forces of the structural model.


With all basic force values established we obtain the corresponding element deformations from theforce-deformation relation of each element. We can do this "long hand" or we can write it in compactform, as we will see in a moment. For reasons of computational convenience in hand calculations wedetermine the EA-fold value of the element deformations and delay the division by EA until the end.

Applying the force-deformation relation element by element means:

V1Q1

EALa⋅=

V2Q2

EALb⋅= if we have a matrix toolbox

at our disposal we organizethese relations in compactform as followsV3

Q3

EALc⋅= V

V1

V2

V3

V4

V5

⎛⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎠

=

La

EA

0

0

0

0

0

Lb

EA

0

0

0

0

0

Lc

EA

0

0

0

0

0

Ld

EA

0

0

0

0

0

Le

EA

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

Q1

Q2

Q3

Q4

Q5

⎛⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎠

⋅= Fs Q⋅=

V4Q4

EALd⋅=

V5Q5

EALe⋅=

Since each truss element deformation only depends on the corresponding basic force, the resulting matrixis diagonal. We call this matrix Fs. It represents the collection of element flexibilities f = L/EA for the trusselements of the entire structural model.

Page 312


We can see from the above relations that it is convenient to factor out EA from the Fs matrix and write it as

if the truss elements have different axial stiffness values, wecan still factor out a reference value (e.g. the highest value orthe one common to most elements), and then include the axialstiffness ratio in the diagonal terms. The advantage of this for handcalculations is significant. We are dealing with "normal" numericalvalues that we can handle easier.

Fs1

EA

La

0

0

0

0

0

Lb

0

0

0

0

0

Lc

0

0

0

0

0

Ld

0

0

0

0

0

Le

⎛⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎠

⋅=

This leads to the idea of determining the EA-fold element deformations, i.e. the EA-fold matrix Fs. We have:

EAV1 Q1 La⋅=

EAV2 Q2 Lb⋅=

EAV3 Q3 Lc⋅= or in compact form, EAV EAFs Q⋅= with EAFs

La

0

0

0

0

0

Lb

0

0

0

0

0

Lc

0

0

0

0

0

Ld

0

0

0

0

0

Le

⎛⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎠

=

EAV4 Q4 Ld⋅=

EAV5 Q5 Le⋅=

Let us get some numerical values.

Q1

Q2

Q3

Q4

Q5

⎛⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎠

Q:= EAV1 Q1 La⋅:= EAV1 120=

EAV2 Q2 Lb⋅:= EAV2 67.5=

EAV3 Q3 Lc⋅:= EAV3 0=

EAV4 Q4 Ld⋅:= EAV4 120=

EAV5 Q5 Le⋅:= EAV5 187.5−=

Note: Mathcad has a nice feature that allows us to express any numerical value in terms of any variable(just like units). If we have Mathcad around for all our calculations, we can therefore proceed as follows:

Page 313


Define the collection of element flexibility matrices for the entire structure

Fs

La

0

0

0

0

0

Lb

0

0

0

0

0

Lc

0

0

0

0

0

Ld

0

0

0

0

0

Le

⎛⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎠

1EA

⋅:=

the element deformations under a force of 15 units at dof 4 become V Fs Q⋅:=

we can express their numerical values directly V

12

6.75

0

12

18.75−

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

10 3−= note that we factored out 10-3

or, in terms of the axial stiffness EA V

120

67.5

0

120

187.5−

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

1EA

=

in which case we obtain the EA-fold values of the element deformations in the vector (compare with thevalues in the preceding page).

Page 314


Global dof displacements - Compatibility relations

We write the compatibility relations for the structural model. We can do this directly, or simply use thetranspose of the equilibrium relations. We have:

8

6

dof 1

dof 2dof 3

dof 4

dof 5

a

b c

d

e

V1 U1=

V2 U3=

V3 U5=

V4 U2− U4+=

V5 0.8 U1⋅ 0.8 U2⋅− 0.6 U3⋅+=

or, V Af Uf⋅= with

Af

1

0

0

0

0.8

0

0

0

1−

0.8−

0

1

0

0

0.6

0

0

0

1

0

0

0

1

0

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=

and we observe again that the compatibility matrix is the transpose of the equilibrium matrix

to solve for the displacements in terms of the element deformations we need to solve the system of thecompatibility relations; we do this by hand and in the process obtain the inverse of the compatibility matrix

U1 V1=

U3 V2=

U5 V3=

U2 U1 0.75 U3⋅+ 1.25 V5⋅−= V1 0.75 V2⋅+ 1.25 V5⋅−=

U4 U2 V4+= V1 0.75 V2⋅+ V4+ 1.25 V5⋅−=

Page 315


we compare with the inverse of the compatibility matrix

thus, Uf

1

1

0

1

0

0

0.75

1

0.75

0

0

0

0

0

1

0

0

0

1

0

0

1.25−

0

1.25−

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

V⋅:= Af1−

1

1

0

1

0

0

0.75

1

0.75

0

0

0

0

0

1

0

0

0

1

0

0

1.25−

0

1.25−

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

=

and we get, of course, the same answer, which means that we can compete with Mathcad for this smallproblem. For a larger one, we are unfortunately completely outmatched.

We also note that the inverse of the compatibility matrix is equal to the transpose of the force influencematrix Bbar.

this is to be expected of course, since Bbar is the inverse of theequilibrium matrix which is the transpose of the compatibility matrix.You may recall from linear algebra that the inverse and transposeoperations are interchangeable, i.e. we can take them in any order.

BbarT

1

1

0

1

0

0

0.75

1

0.75

0

0

0

0

0

1

0

0

0

1

0

0

1.25−

0

1.25−

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

=

We write therefore from now on: Uf BbarT V⋅=

and note that we have obtained this relations from the solution of the compatibility relations.

We can explain the relation Uf BbarT V⋅= directly with the principle of virtual forces.

Given the element deformations V we seek a system of virtual basic forces in equilibrium with a unit virtualforce at the dof whose displacement we wish to determine. For a unit force at dof 1 the virtual basic forcevalues are equal to the first column of the force influence matrix (after all we reiterate that the entries ofthat column represent the basic forces in the structure under a unit force at dof 1). For a unit force at dof 2we get the second column, etc, etc. If we are interested in the displacement of a particular dof only, thismay be for some a slightly faster way of getting the answer than solving the compatibility relations. Let usdemonstrate for dof 2.

The equilibrium equations to solve with a unit force at dof 2 are:

0 δQ1 0.8 δQ5⋅+= These equations are relatively easy to solve:the last two give immediately δQ4 and δQ3,then the second gives δQ5 and then the othertwo give directly the values for δQ2, δQ1. Thecomplete vector of virtual basic forces is

1 δQ4− 0.8 δQ5⋅−= δQ

1

0.75

0

0

1.25−

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=

0 δQ2 0.6 δQ5⋅+=

0 δQ4=

0 δQ3=

Page 316


Thus, the horizontal translation at dof 2 is by the principle of virtual forces

U2 δQT V⋅:= U2 40.510 3−= or in terms of EA U2 4051

EA=

for hand calculations we would use EAU2 EAV1 0.75 EAV2⋅+ 1.25 EAV5⋅−:=

EAU2 405= same as above

Since we have obtained the complete force influence matrix, we can determine all dof displacements. We get

Uf BbarT V⋅:= Uf

12

40.5

6.75

52.5

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

10 3−= or, in EA-fold form Uf

120

405

67.5

525

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

1EA

=

This leads to the following generalization: we have obtained the displacements at all dofs under a force of15 units at dof 4. If there was only a unit force acting at dof 4, then the displacements would be 15 timessmaller. Let us calculate these for later reference

we reiterate that these are the displacements at all five dofs under a unit force atdof 4. Let us summarize how we determined them: (a) a unit force at dof 4 givesbasic force values Q equal to the 4th column of the force influence matrix Bbar; (b)with these values we calculate the deformations in all elements by Fs*Q, (c) finally,

we obtain the displacements at all dofs by the relation BbarT V⋅ . Putting the wholeprocess together we conclude that the displacements under a unit force at dof 4are given by

115

Uf⋅

8

27

4.5

35

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

1EA

=

BbarT Fs⋅ Bbar 4⟨ ⟩⋅

8

27

4.5

35

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

1EA

= which gives the same result, of course

Repeating the process for a unit force at dof 1 will give another 5 displacement values, and another 5 for dof2 and so on. Thus, for a unit force acting at any one of the 5 free global dofs of the structural model we get

F BbarT Fs⋅ Bbar⋅:= F

0.8

0.8

0

0.8

0

0.8

2.7

0.45

2.7

0

0

0.45

0.6

0.45

0

0.8

2.7

0.45

3.5

0

0

0

0

0

0.6

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

10 3−= or, F

8

8

0

8

0

8

27

4.5

27

0

0

4.5

6

4.5

0

8

27

4.5

35

0

0

0

0

0

6

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

1EA

=

this matrix is known as the flexibility matrix of the structural model.

Page 317


Flexibility matrix of structural model

As the name suggests, the typical term (i,j) of the flexibility matrix of the structural model represents thedisplacement at dof i under a unit force acting at dof j. This matrix is, therefore, a representation of thedeformation-force behavior of the entire structure and can be considered a material property of thestructural model much like 1/E is the material flexibility or compliance and f is the element flexibility matrix.

The inverse of the flexibility matrix of the structural model gives the stiffness matrix of the structural model(we have seen this process in action at the element level between element flexibility and stiffness matrix).The entry (i,j) of the stiffness matrix represents the force at dof i under a unit displacement of dof j. Thismeans that only the particular dof j is activated with a unit value, while the other dofs are equal to zero. Toaccomplish this feat forces are typically required at the dofs, both the activated one and those held fixed.

If we are interested in the stiffness matrix for all dofs, we have

Complete stiffness matrix of structural model

K F 1−:= K

1.89

0.64−

0.48

0

0

0.64−

1.89

0.48−

1.25−

0

0.48

0.48−

2.027

0

0

0

1.25−

0

1.25

0

0

0

0

0

1.667

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

103=

The usefulness of the flexibility matrix comes into full play, however, if we are interested in the interactionbetween a few specific dofs among the many dofs of the structural model. Let us say, for example, that weare only interested in the interaction between dofs 2 and 3 (e.g. when forces are applied only at these dofsof the structural model).

To extract the corresponding flexibility matrix all we have to do is extract the terms at the intersectionof the second and third row and column. We have

flexibility matrix for dofs 2 and 3 F23

F2 2,

F3 2,

F2 3,

F3 3,

⎛⎜⎜⎝

⎞⎟⎟⎠

:= F232.7

0.45

0.45

0.6⎛⎜⎝

⎞⎟⎠

10 3−= F2327

4.5

4.5

6⎛⎜⎝

⎞⎟⎠

1EA

=

Now, the inverse of the flexibility matrix for dofs 2 and 3 is known as the condensed stiffness matrix of thestructural model for dofs 2 and 3. We obtain

Condensed stiffness matrix for specific dofs

condensed stiffness matrix for dofs 2 and 3 K23 F231−:= K23

0.423

0.317−

0.317−

1.905⎛⎜⎝

⎞⎟⎠

103=

We note that the stiffness coefficients are very different from those at the intersection of the second andthird row and column of the complete stiffness matrix. Why? Let us explain with the help of the entry at theintersection of the third row and third column. In the complete stiffness matrix this coefficient (2027)represents the force at dof 3 necessary to impose a unit displacement at the same dof 3, while all other dofsare held fixed. In the condensed stiffness matrix, this coefficient (1905) represents the force at dof 3necessary to impose a unit displacement at the same dof 3, while dof 2 is held fixed, but all other dofs arerelaxed to assume the values that satisfy force equilibrium at dofs 1, 4 and 5. Consequently, the stiffnesscoefficient on the diagonal of the condensed stiffness matrix is smaller.

Conclusion: Static condensation of a stiffness matrix is very conveniently performed with access to theflexibility matrix of the structural model.

Page 318

Script for Example 25 in CE220 class notes % Solution for statically determinate truss


Create model % specify node coordinates (could only specify non-zero terms) XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 8 0]; % second node, etc XYZ(3,:) = [ 0 6]; % XYZ(4,:) = [ 8 6]; % % connectivity array CON {1} = [ 1 2]; CON {2} = [ 1 3]; CON {3} = [ 2 4]; CON {4} = [ 3 4]; CON {5} = [ 2 3]; % boundary conditions (1 = restrained, 0 = free) BOUN(1,:) = [1 1 0]; BOUN(2,:) = [0 1 0]; % specify element type [ElemName{1:5}] = deal('Truss'); % linear truss element % create model Model = Create_SimpleModel (XYZ,CON,BOUN,ElemName);


1 2

3 4

1

2 3

4

5

Define element properties ne = Model.ne; % number of elements in structural model % Modulus and area for el=1:ne; ElemData{el}.E = 1000; ElemData{el}.A = 10; end

Page 319

define loading % horizontal force at node 4 Pe(4,1) = 15; Loading = Create_Loading (Model,Pe); % nodal forces at free dofs Pf = Loading.Pref;

set up equilibrium matrix B = B_matrix(Model); % extract free dofs Bf = B(1:Model.nf,:); % invert Bf to obtain Bbar Bbar = inv(Bf);

solution determine basic forces from equilibrium equations

Q = Bbar*Pf; % display result disp('truss forces Q under applied horizontal force') disp(Q); truss forces Q under applied horizontal force 15.0000 11.2500 0 15.0000 -18.7500

plotting open window and plot axial force diagram

Create_Window(0.80,0.80); Plot_Model (Model); Plot_AxialForces (Model,Q);

set up collection of flexibility matrices Fs = Fs_matrix(Model,ElemData); V = Fs*Q;

determine global dof displacements % free global dof displacements

Page 320

Uf = Bbar'*V; disp('the free dof displacements are') disp(Uf) the free dof displacements are 0.0120 0.0405 0.0067 0.0525 0

plot deformed shape of structural model put free dof displacement values in complete displacement vector

U = zeros(Model.nt,1); U(1:Model.nf) = Uf; MAGF = 20; % magnification factor for deformed configuration Create_Window(0.80,0.80); Plot_Model(Model); Plot_DeformedStructure(Model,[],U);

determine flexibility matrix (gives global dof displacement to applied force relation) F = Bbar'*Fs*Bbar; % display flexibility matrix disp('flexibility matrix of truss (x 1000)') disp(F.*1000); % determine displacements at free dofs for this particular loading (should be the same as above) Uf = F*Pf; % display displacements disp('free dof displacements Uf under applied nodal force') disp(Uf); flexibility matrix of truss (x 1000) 0.8000 0.8000 0 0.8000 0 0.8000 2.7000 0.4500 2.7000 0 0 0.4500 0.6000 0.4500 0 0.8000 2.7000 0.4500 3.5000 0 0 0 0 0 0.6000 free dof displacements Uf under applied nodal force 0.0120 0.0405 0.0067 0.0525 0

Page 321

CE 220 - Theory of Structures Ex 26 - Force-displacement for frame with hinge © Prof. Filip C. Filippou, 2000

Example 26 - Force-displacement relations for frame with hinge

In this example we investigate the force-displacement relations for the simple portal frame with hinge whoseequilibrium we studied in Example 4 and whose compatibility relations we studied in Example 12.

We seek to determine the translations and rotations at the nodes of the frame in the following figure underthe given loading.

55

12

8 8

10

1

2 3 4

5

a

b c

d

Length of elements

La 12:= Lb 8:= Lc 8:= Ld 10:=

Under the assumption that axial deformations can be neglected in all elements, we have the independentfree global dofs in the following figure.

2

34

6

a

b c

d

1

5

7

Page 322


The compatibility relations are (the numbering of the locations for these is shown in the following figure)

V1 U1U2

La+= (*)

a

b c

d

3 56

1

2 4

7

V2U2

LaU3+=

V3 U3U4

Lb−=

V4U4

Lb− U5+= (*)

V5U4

LcU6+=

V6U2

LdU6+=

V7U2

LdU7+= (*)

The structural compatibility matrix Af for these dofs and deformations is

Af

1

0

0

0

0

0

0

1La

1La

0

0

0

1Ld

1Ld

0

1

1

0

0

0

0

0

0

1Lb

−

1Lb

−

1Lc

0

0

0

0

0

1

0

0

0

0

0

0

0

1

1

0

0

0

0

0

0

0

1

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= Af

1

0

0

0

0

0

0

112

112

0

0

0

110

110

0

1

1

0

0

0

0

0

0

18

−

18

−

18

0

0

0

0

0

1

0

0

0

0

0

0

0

1

1

0

0

0

0

0

0

0

1

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

Page 323


The equilibrium matrix Bf for the corresponding dofs and basic forces is the transpose of the structuralcompatibility matrix Af.

Bf AfT:= Bf

1

112

0

0

0

0

0

0

112

1

0

0

0

0

0

0

1

18

−

0

0

0

0

0

0

18

−

1

0

0

0

0

0

18

0

1

0

0

110

0

0

0

1

0

0

110

0

0

0

0

1

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

and the equilibrium equations can be written out as follows

P1 Q1= for the given loading the applied force vector is

P2Q1

La

Q2

La+

Q6

Ld+

Q7

Ld+=

P3 Q2 Q3+=

P4Q3

Lb−

Q4

Lb−

Q5

Lc+= Pf

0

5

0

0

0

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= Pfw

0

0

0

20

0

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

P5 Q4=

P6 Q5 Q6+=

P7 Q7=

there are three uncoupled equations above, namely #1, #5 and #7, which we solved by inspection inExample 4, concluding that the corresponding basic forces are zero in the absence of applied moments atthe corresponding dofs. This led us in Example 4 to a system of 4 equations in 4 unknowns which could bereadily solved. Here we do not bother with this simplification and let Mathcad's lsolve function do the work.

With the system of linear equilibrium equations to be solved Pf Bf Q⋅ Pfw+=

we obtain the solution for the basic forces Q with Q lsolve Bf Pf Pfw−( ),[ ]:=

and we list the transpose of the basic force vector Q to save space

QT 0 60− 60 0 100− 100 0( )= which gives the same answers as Example 4, of course

Page 324


With these basic forces we can determine the corresponding deformations by the deformation-force relationof each element. We have under the assumption that the flexural stiffness is EI = 50,000

EI 50000:= w 5:=

V1La

6 EI⋅2 Q1⋅ Q2−( )⋅:= V1 120

1EI

=

V2La

6 EI⋅2 Q2⋅ Q1−( )⋅:= V2 240−

1EI

=

V3Lb

6 EI⋅2 Q3⋅ Q4−( )⋅:= V3 160

1EI

= V3 3.210 3−=

V4Lb

6 EI⋅2 Q4⋅ Q3−( )⋅:= V4 80−

1EI

=

V5Lc

6 EI⋅2 Q5⋅( )⋅

w Lc3⋅

24 EI⋅+:= V5 160−

1EI

=

V6Ld

6 EI⋅2 Q6⋅ Q7−( )⋅:= V6 333.33

1EI

=

V7Ld

6 EI⋅2 Q7⋅ Q6−( )⋅:= V7 166.67−

1EI

= or, in a vector V

V1

V2

V3

V4

V5

V6

V7

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

A couple of observations are in order:

1. In this example every element has one end with zero moment. In the absence of a distributed elementload the deformation at the end with no moment is -1/2 of the deformation value at the other element end.

2. It is very useful to factor out the stiffness EI and not bother dividing every deformation by this largevalue. In Mathcad this can be done by inserting the 1/EI term like a unit at the end of the result. If we areinterested to see the actual numerical value, we can leave this out, as shown for V3.

3. We should not forget to include the effect of the distributed element load on the element deformations.This is the case with V5, which belongs to end j of element c.

Page 325


With the element deformation values we can use the compatibility relation to determine the global dofdisplacement values. We can do this by hand, working carefully to isolate individual dofs by elimination ofthe others, but we will do this later when we work with the smaller set of dofs, and so here we just use thecapabilities of Mathcad to give us the result.

from V Af Uf⋅= we get Uf lsolve Af V,( ):= Uf

77.58

509.09

282.42−

3539.39−

522.42−

282.42

217.58−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

1EI

=

We can express the whole process in very compact form noting that we solved a system of linear equationsinvolving Bf or its transpose Af twice (once for the equilibrium equations and the second time for thecompatibility equations). Recalling from Example 2 that the inverse of the equilibrium matrix Bf is the forceinfluence matrix Bbar we can write the process in very compact form as follows.

for the solution of the equilibrium equations Q Bbar Pf Pfw−( )⋅=

for the deformation-force relations of all elements we write V Fs Q⋅ Vo+=

and for the solution of the compatibility relations we write

Uf Af1− V⋅= Bf

T( ) 1−V⋅= Bf

1−( )T V⋅= BbarT V⋅= Uf BbarT V⋅=

we combine these three steps and get Uf BbarT Fs⋅ Bbar⋅ Pf Pfw−( )⋅ BbarT Vo⋅+=

the expression BbarT Fs⋅ Bbar⋅ represents the flexibility matrix F of the structural model

F BbarT Fs⋅ Bbar⋅=

with it we can write compactly Uf F Pf Pfw−( )⋅ BbarT Vo⋅+=

So, let us see how this flexibility matrix looks. For this let us first get Bbar Bbar Bf1−:=

Page 326


We also need to set up the collection of element flexibility matrices Fs

Fs1EI

La

3

La

6−

0

0

0

0

0

La

6−

La

3

0

0

0

0

0

0

0

Lb

3

Lb

6−

0

0

0

0

0

Lb

6−

Lb

3

0

0

0

0

0

0

0

Lc

3

0

0

0

0

0

0

0

Ld

3

Ld

6−

0

0

0

0

0

Ld

6−

Ld

3

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

⋅:=

and here comes the flexibility matrix !

F BbarT Fs⋅ Bbar⋅:= F

8.44

42.31−

0.29−

12.03−

2.11−

2.72

4.99

42.31−

376.86

9.59−

39.67

12.23

19.5−

46.78−

0.29−

9.59−

2.98

14.15

1.16

0.56−

1.72

12.03−

39.67

14.15

206.28

31.6

16.09−

2.09

2.11−

12.23

1.16

31.6

7.34

2.74−

0.47−

2.72

19.5−

0.56−

16.09−

2.74−

3.47

1.19

4.99

46.78−

1.72

2.09

0.47−

1.19

8.92

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

1EI

=

What do all these numbers mean? A matrix is a doubly subscripted array. In the case of the flexibility matrixboth subscripts correspond to dof numbers: the first subscript corresponds to the displacement dof and thesecond subscript to the force dof. Thus, the flexibility coefficient Fij expresses the effect of a unit force at dofj on the displacement at dof i. The larger this number the "softer" is the corresponding interaction, i.e. thelarger the effect under a unit force at dof j.Studying the above matrix we note that the diagonal terms are positive and this is always the case. A unitforce at a certain dof causes a translation in the direction of the unit force. By contrast, off diagonal termscan be either positive or negative. We also note that a few terms are a lot larger than others, e.g. thesecond and 4th diagonal term. These correspond to the translation dofs and they are larger becausetranslations have units of length, while rotations are dimensionless (i.e. length/length). So, while rotationscan be scaled from a structure of one size to another, translations always carry the dimensions of thestructure which makes for the discrepancy in the magnitude of the flexibility coefficients. We have alreadynoted this difference in magnitude when commenting on the values of the free dof displacement vector.

Page 327


The most important property of the flexibility matrix is, of course, its symmetry! This means that a unit forceat dof i produces the same effect at dof j, as a unit force at dof j produces at dof i. This is known asMaxwell's Reciprocal Theorem (this is the same James Clerk Maxwell of electro-magnetic fame).

Another interesting property of the flexibility matrix is that we can isolate the effect of dofs of interest bysimply extracting them from the matrix. As an example let us say that we are not interested in the effect ofmoments on the portal frame (e.g. because there will not be such actions ever). In this case we are onlyinterested in the effect of forces at the translation dofs. We can set up a reduced force influence matrixBbar' that contains only the second and fourth column, since the other columns are irrelevant and willnever be of use. Let us see what this looks like

these are the forces Qdue a unit force at dof 2

these are the forces Qdue a unit force at dof 4Bbar 2⟨ ⟩

0

5.45

5.45−

0

5.45−

5.45

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

= Bbar 4⟨ ⟩

0

4.36

4.36−

0

3.64

3.64−

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

Thus, the reduced force influence matrix Bbar' becomes Bbar' 1⟨ ⟩ Bbar 2⟨ ⟩:=

Bbar' 2⟨ ⟩ Bbar 4⟨ ⟩:=

now we can calculated the flexibility matrix for the effects of interest. We denote this flexibility matrix F', sinceit does not involve all dofs of the structural model. We have

F' Bbar'T Fs⋅ Bbar'⋅:= F'376.86

39.67

39.67

206.28⎛⎜⎝

⎞⎟⎠

1EI

=

and we observe that it corresponds exactly to the coefficients (2,2), (2,4), (4,2) and (4,4) of the completeflexibility matrix. If had known from the beginning that applied moments at dofs 1, 3, 5, 6 and 7 were of nointerest, we would have set up just the flexibility matrix F' (and of course just the reduced force influencematrix Bbar').

Having focused attention on just the dofs of interest we can study the flexibility coefficients for dofs 2 and 4more carefully. We observe two things:

1. Dof 2 is more flexible than dof 4, i.e. a unit force a dof 2 causes a larger translation in that direction thanis the case with a unit force at dof 4 (in fact, a little more than 1.5 times larger as we can tell from theflexibility coefficients 376.86 and 206.28).

2. The "coupling" between dofs 2 and 4 is not particularly strong, i.e. a unit force at dof 2 causes a relativelysmall effect on dof 4 and vice versa. Interestingly, if the column on the right (i.e. element d) were as tall asthe column on the left, there would be no coupling on account of symmetry and anti-symmetry (more on thesubject later in the course).

Page 328


Solution with reduced number of degrees of freedom

We have done a lot of work with large vectors and matrices above without really taking advantage of the factthat the basic forces Q1, Q4 and Q7 are always equal to zero, when no forces are applied at dofs 1, 5 and 7.While this may not be important with matrix software, it is always more insightful, and also more economicwhen solving the problem "by hand". Let us therefore show the solution

The following figure shows the set of reduced free dofs (consult also Examples 4 and 12).

dof 1

dof 2 dof 3dof 4

a

b c

d

1Q2Q

3Q

4Q

The equilibrium equations are

P1Q1

La

Q4

Ld+=

P2 Q1 Q2+=with Pf

5

0

0

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:= Pfw

0

0

20

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:= for the given loading

P3Q2

Lb−

Q3

Lc+=

P4 Q3 Q4+=

Bf

1La

1

0

0

0

1

1Lb

−

0

0

0

1Lc

1

1Ld

0

0

1

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:= or, numerically Bf

112

1

0

0

0

1

18

−

0

0

0

18

1

110

0

0

1

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

=

Page 329


The compatibility equations are

V1U1

LaU2+= V3

U3

LcU4+=

V2 U2U3

Lb−= V4

U1

LdU4+=

and the compatibility matrix is

and we note, of course, that thetranspose of the compatibility matrixis equal to the equilibrium matrixAf

1La

0

0

1Ld

1

1

0

0

0

1Lb

−

1Lc

0

0

0

1

1

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= AfT

112

1

0

0

0

1

18

−

0

0

0

18

1

110

0

0

1

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

=

We determine the basic forces for the given loading. "By hand" we have

5Q1

12

Q4

10+=

0 Q1 Q2+=

0Q2

8−

Q3

8+ 20+= substitute Q2 and Q3 in terms of Q1 and Q4 20−

Q1

8

Q4

8−=

0 Q3 Q4+=

from equations #1 and #3 solve for Q1 and Q4 Q1 60−:= Q4 100:=

so that Q2 Q1−:= Q3 Q4−:=

and the basic force vector becomes Q

Q1

Q2

Q3

Q4

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=

Page 330


the corresponding element deformations now are

V1La

6 EI⋅2 Q1⋅( )⋅:= V1 240−

1EI

=

V2Lb

6 EI⋅2 Q2⋅( )⋅:= V2 160

1EI

=

V3Lc

6 EI⋅2 Q3⋅( )⋅

w Lc3⋅

24 EI⋅+:= V3 160−

1EI

=

V4Ld

6 EI⋅2 Q4⋅( )⋅:= V4 333.33

1EI

=

If we are interested in the horizontal translation at dof 1 and vertical translation at dof 3 we can obtainthese from the compatibility relations

V1U1

12U2+= V3

U3

8U4+=

V2 U2U3

8−= V4

U1

10U4+=

subtracting the first two and the last two we get V1 V2−U1

12

U3

8+=

V3 V4−U3

8

U1

10−=

and after subtracting the second from the first we get

V1 V2− V3 V4−( )−U1

12

U1

10+= and finally U1

12022

V1 V2− V3− V4+( )⋅:=

U1 509.091EI

= same as before

and we can obtain in similar fashion U3 or simpler U3 8 V1 V2−( )⋅812

U1⋅−:= U3 3539.394−1EI

=

Returning to the four original compatibility relations we can then obtain U2 and U4, if interested.

Page 331


How can we determine the node rotations that have not been included in the free dofs? By geometry

The following figure shows the deformed shape of the structural model under the given loading. We cansee that the rotation of an "unrestrained" node, i.e. a node with only one continuous element connectioncan always be determined from the chord rotation of the element with the continuous connection to thenode and the element deformation (i.e. the node rotation is equal to the tangent rotation of the element)

1

2

3

4

5

Thus, for rotation of node 1: chord rotation of element a + deformation at end i of element a

U1

La−

Q1−( ) La⋅

6 EI⋅+ 77.58

1EI

= compare with value on page 5

for rotation of node 3 chord rotation of element b + deformation at end j of element b

U3

Lb

Q2−( ) Lb⋅

6 EI⋅+ 522.42−

1EI


for rotation of node 5 chord rotation of element d+ deformation at end j of element d

U1

Ld−

Q4−( ) Ld⋅

6 EI⋅+ 217.58−

1EI


The hinge rotation at node 3 can be determined from the difference of the angles of the tangent to thedeformed shape of element c and the tangent to the deformed shape of element b (the latter is equal tothe rotation of node 3).

Page 332

Script for Example 26 in CE220 class notes %statically determinate portal frame (three hinge frame) % use input data of Example 4 Example_4_mod the basic forces are -7.5000 0 -60.0000 -10.0000 60.0000 0 -10.0000 0 -100.0000 -32.5000 100.0000 0

Define element properties ne = Model.ne; % number of elements in structural model % Modulus, area and moment of inertia for el=1:ne; ElemData{el}.E = 1000; ElemData{el}.A = 1e6; % very large value to exclude axial deformations ElemData{el}.I = 100; end

set up collection of flexibility matrices Fs = Fs_matrix(Model,ElemData); % set up first initial deformations due to w V0 = V0_vector(Model,ElemData); % determine element deformations V = Fs*Q + V0;

determine global dof displacements % free global dof displacements Bbar = inv(Bf); % use deformations at iq locations, since Bbar is set up for iq basic forces Uf = Bbar'*V(iq); disp('the free dof displacements are') disp(Uf) the free dof displacements are 0.0008 0.0051 -0.0000 -0.0028 0.0051 -0.0354 -0.0052 0.0051 -0.0000 0.0028 -0.0022

Page 333

plot deformed shape of structural model % put free dof displacement values in complete displacement vector U = zeros(Model.nt,1); U(1:Model.nf) = Uf; MAGF = 80; % magnification factor for deformed configuration Create_Window(0.80,0.80); Plot_Model(Model); % it is necessary to insert V in the argument list for the hinge discontinuity Plot_DeformedStructure(Model,ElemData,U,V);

determine flexibility matrix (gives global dof displacement to applied force relation) F = Bbar'*Fs(iq,iq)*Bbar; % display flexibility matrix disp('flexibility matrix of truss (x 1000)') disp(F.*1000); flexibility matrix of truss (x 1000) 0.0844 -0.4231 -0.0000 -0.0029 -0.4231 -0.1203 -0.0211 -0.4231 0.0000 0.0272 0.0499 -0.4231 3.7686 0.0000 -0.0959 3.7686 0.3967 0.1223 3.7686 -0.0000 -0.1950 -0.4678 -0.0000 0.0000 0.0000 -0.0000 0.0000 0.0000 -0.0000 0.0000 0.0000 -0.0000 -0.0000 -0.0029 -0.0959 -0.0000 0.0298 -0.0959 0.1415 0.0116 -0.0959 0.0000 -0.0056 0.0172 -0.4231 3.7686 0.0000 -0.0959 3.7686 0.3967 0.1223 3.7686 -0.0000 -0.1950 -0.4678 -0.1203 0.3967 0.0000 0.1415 0.3967 2.0628 0.3160 0.3967 0.0000 -0.1609 0.0209 -0.0211 0.1223 -0.0000 0.0116 0.1223 0.3160 0.0734 0.1223 0.0000 -0.0274 -0.0047 -0.4231 3.7686 0.0000 -0.0959 3.7686 0.3967 0.1223 3.7686 -0.0000 -0.1950 -0.4678 0.0000 -0.0000 0 0.0000 -0.0000 0.0000 0.0000 -0.0000 0.0000 0.0000 0.0000 0.0272 -0.1950 -0.0000 -0.0056 -0.1950 -0.1609 -0.0274 -0.1950 0.0000 0.0347 0.0119 0.0499 -0.4678 -0.0000 0.0172 -0.4678 0.0209 -0.0047 -0.4678 0.0000 0.0119 0.0892

Page 334

Displacements at incipient collapse

Case of full collapse mechanism

Just before the last hinge forms the structure is statically determinate if a full collapse mechanism forms.Thus, all basic forces can be determined from the equilibrium equations (page 26 of Part I). With all basicforces known we can determine the corresponding strain dependent element deformations. In determining the global dof displacements we can only use the compatibility equations in the locations without a plastic hinge, because at the NOS plactic hinge locations there are unknown plastic hinge deformations at the instant just before the last hinge forms.

Denoting the locations without plastic hinge including the last hinge to form with subscript i, as we havedone in Example 18, we have

i si i i0= +FV Q Vwith

solving for the displacements at the free dofs we obtain

i i f=V UA Tf i i= BU V T

i i=A B 1i i

−=B Bsince and

Since the location of the last hinge to form is not known from the lower or upper bound theorem of plastic analysis, trial and error is required: we guess the location of the last plastic hinge to form and then determine the plastic deformations and the free global dof displacements. The correct guess for the last plastic hinge to form corresponds to the case with no contradiction between the sign of the plastic deformations and the corresponding basic forces, which also results in the largest value for the translation dofs.

consult examples 27 and 28.

Case of partial collapse mechanism

In this case the structure is still statically indeterminate before the last plastic hinge forms. Recall that there are less than NOS+1 hinges at collapse, thus less than NOS hinges before the last plastic hinge forms. Let us call the number of plastic hinges at collapse NP. Before the last plastic hinge forms there are, therefore, NOS+1 - NP redundant basic forces and a statically indeterminate analysis, as discussed in the preceding page is required to determine the redundant values and then all basic forces just before the last plastic hinge forms. Following this we can follow the same procedure as for the case of the full collapse mechanism above to determine the displacements at incipient collapse.

consult example 33


Page 335

CE220 - Theory of Structures Ex. 27 - Displacements at incipient collapse of truss © Prof. Filip C. Filippou, 2000

Example 27 - Displacements at incipient collapse of simple truss

Objectives:

(a) determine the displacements at incipient collapse of truss: trial and error approach of guessing location of last hinge to form

a

b c

λ(10)

8

6

λ(10)

L1 8:=

L2 10:=

L3 6:=

We investigate the collapse load determination of this simple truss by the lower bound theorem of plasticanalysis in Example 7. Here we determine the displacements at incipient collapse by guessing the locationof the last hinge to form.

We state the equilibrium and compatibility relations

V1 U1=P1 Q1 0.8Q2+=

V2 0.8 U1⋅ 0.6 U2⋅+=P2 0.6 Q2⋅ Q3+=

V3 U2=

In example 7 we determined the collapse load factor as λc 1.6:=

and the corresponding axial forces as Q1 8:= Q2 10:= Q3 10:=

with hinges forming at elements b and c; we do not know which of the two is the last hinge to form, and weplan to determine this along with the corresponding global dof displacement values in this example.

We assume now that the behavior of the material is linear elastic-perfectly plastic and that the axial stiffnessof all truss elements is equal to EA = 10,000.

EA 10000:=

Page 336


We assume first that the last hinge to form is in element c. If this is the case, then element c is still elastic atthe instant the last hinge forms and we can use the deformation-force relations for elements a and c todetermine the deformations from the given forces.

We have:

V1ε Q1L1

EA⋅:= V1ε 6.410 3−= V1h 0:= V1 V1ε V1h+:=

V3ε Q3L3

EA⋅:= V3ε 610 3−= V3h 0:= V3 V3ε V3h+:=

we use the first and third compatibility relation to determine the global dof displacements

U1 V1:= U1 6.410 3−=

U2 V3:= U2 610 3−=

now we substitute into the second compatibility relation to determine the plastic deformation

first we determine the elastic deformation of element b V2ε Q2L2

EA⋅:= V2ε 1010 3−=

V2h 0.8 U1⋅ 0.6 U2⋅+( ) V2ε−:= V2h 1.28− 10 3−=

and the plastic deformation in element b is negative, which contradicts the fact that the corresponding basicforce in element b is positive (tension). We conclude that the guess of the last hinge forming in c is notphysically correct. We assume then that the last hinge forms in b. We use now the first two compatibilityrelations to determine the global dof displacements, since we know the deformations V1 and V2.

We have now:

V2h 0:= V2 V2ε V2h+:=

U1 V1:= U1 6.410 3−= from the first compatibility relation

U2V2 0.8 U1⋅−

0.6:= U2 8.13310 3−= from the second compatibility relation

substituting into the third compatibility relation we get the plastic deformation in c

V3h U2 V3ε−:= V3h 2.13310 3−= and it is positive, in agreement with the sign of the basic force

Since there are only two plastic hinges forming in this example, we know now the complete sequence:element c reaches its capacity first, followed by element b. Once element b reaches its capacity, the entirestructure has reached its capacity under the given loading.

Page 337


We like to demonstrate now that the relative axial stiffness of the elements affects the hinge sequence, whileit does not affect the collapse load factor, which obviously depends only on the plastic capacities (weassume here that we can change the stiffness of a member without changing its capacity).

We make element b twice as stiff axially as the other two, i.e. EAb 20000:=

Using the solution from the last case with the last hinge forming in element b we get

V2ε Q2L2

EAb⋅:= V2ε 510 3−=

V2h 0:= V2 V2ε V2h+:=

U1 V1:= U1 6.410 3−= from the first compatibility relation

U2V2 0.8 U1⋅−

0.6:= U2 0.2− 10 3−= from the second compatibility relation, which is already

suspect, since the point moves down rather than up

substituting into the third compatibility relation we get the plastic deformation in c

V3h U2 V3ε−:= V3h 6.2− 10 3−= and it contradicts the sign of the basic force in c

Thus, in this case the last hinge forms in c and not in b. We determine the global dof displacements and theplastic deformation in c. We get

V3h 0:= V3 V3ε V3h+:=

we use the first and third compatibility relation to determine the global dof displacements

U1 V1:= U1 6.410 3−=

U2 V3:= U2 610 3−= same answer as the wrong guess of the first case, of course

now we substitute into the second compatibility relation to determine the plastic deformation in b

V2h 0.8 U1⋅ 0.6 U2⋅+( ) V2ε−:= V2h 3.7210 3−=

and this time it is positive in agreement with the sign of the basic force in element b.

Conclusion: if we make element b stiffer, it "attracts" a larger portion of the applied loads and yieldsbefore element c.

Page 338

CE 220 - Theory of Structures Ex. 28 - Displacements at incipient collapse of portal frame © Prof. Filip C. Filippou, 2000

Example 28 - Displacements at Incipient Collapse of Portal Frame

this is the continuation of Example 22; instead of repeating the calculations we insert a link to the file

Reference:C:\Drive D\COURSES\CE220\Classnotes\Mathcad files\Example 22_Upper Bound Theorem.xmcd(R)

Determination of displacements and deformations at incipient collapse

With the collapse mechanism we can determine the displacement increments at all dofs of the structuralmodel in terms of a single independent dof (in this case ΔU1). Note that for perfectly plastic behavior thebasic forces do not change once the collapse mechanism has formed, so that there are no elementdeformation increments during the collapse "motion". The only deformation increments occur at the plastichinge locations.

So far, we have not made any assumptions about the material behavior other than that it needs to beperfectly plastic once it reaches its plastic capacity. We now ask the question: is it possible to determine thedeformed shape of the structural model at incipient collapse, i.e. just before the collapse mechanismforms? The answer is yes, as long as we know the material behavior before reaching its plastic capacity. Asan example we use the simplest material behavior for this: linear elastic. Thus, the material behavior for thefollowing calculations is assumed to be linear elastic, perfectly plastic (note that under "material" weunderstand the axial and flexural behavior; we will see in the Nonlinear Analysis course that a linear elastic,perfectly plastic material does not result in a linear elastic-perfectly plastic moment-curvature relation, butthat it can be approximated as such in many cases).

To determine the displacements at incipient collapse we make the following observations:

1. As the load factor λ increases the plastic hinges form one after another in the structural model.

2. We are not interested here in the plastic hinge formation sequence, but we need to know which is thelast plastic hinge to form. The reason for this is that just before the last plastic hinge forms, i.e. at incipientcollapse, the structure is statically determinate, if we assume a complete collapse mechanism (we willdiscuss a partial collapse mechanism later). This means that there are enough compatibility relations withknown deformations on the left hand side that a unique solution for the global dof displacements can beobtained. Note that the latter can be readily determined from the deformation-force relations, since weknow all basic forces at incipient collapse.

3. How do we "find" the last plastic hinge to form? By trial and error, as shown in the following.

Let us first supply a value for the flexural stiffness of the elements of the portal frame EI 50000:=

We have the following basic forces at incipient collapse: Q1

Q2

Q3

Q4

Q5

Q6

Q7

Q8

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

150

85.71−

85.71

120

120−

120−

120

150

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= Q

Q1

Q2

Q3

Q4

Q5

Q6

Q7

Q8

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

Page 339


Given are also the following element lengths La 5:= Lb 4:= Lc 4:= Ld 5:=

We have the following element deformations at incipient collapse

V1εLa

6 EI⋅2 Q1⋅ Q2−( )⋅:= V1ε 321.43

1EI

=

V2εLa

6 EI⋅2 Q2⋅ Q1−( )⋅:= V2ε 267.85−

1EI

=

V3εLb

6 EI⋅2 Q3⋅ Q4−( )⋅:= V3ε 34.28

1EI

=

V4εLb

6 EI⋅2 Q4⋅ Q3−( )⋅:= V4ε 102.86

1EI

=

V5εLc

6 EI⋅2 Q5⋅ Q6−( )⋅:= V5ε 80−

1EI

=

V6εLc

6 EI⋅2 Q6⋅ Q5−( )⋅:= V6ε 80−

1EI

=

V7εLd

6 EI⋅2 Q7⋅ Q8−( )⋅:= V7ε 75

1EI

=

V8εLd

6 EI⋅2 Q8⋅ Q7−( )⋅:= V8ε 150

1EI

=

We recall that the collapse mechanism involves plastic hinges at locations 1, 5, 6 and 8. Let us assume thatthe last hinge to form is at 6, meaning that there is no hinge deformation at this location. With thisassumption we have the following deformations

V2h 0:= V2 V2ε V2h+:=

V3h 0:= V3 V3ε V3h+:=

V4h 0:= V4 V4ε V4h+:=

V6h 0:= V6 V6ε V6h+:=

V7h 0:= V7 V7ε V7h+:=

Page 340


V215

U1⋅ U2+= these are 5 equations in the five unknown global dof displacements and we cansolve them relatively easily. Let us focus on the translations

V3 U214

U3⋅−= subtract equation #2 from #1 to get V2 V3−15

U1⋅14

U3⋅+=

V414

− U3⋅ U4+= subtract equation #4 from #5 to get V7 V6−15

U1⋅14

U3⋅−=

V614

U3⋅ U5+= then we obtain U1 2.5 V2 V3− V7+ V6−( )⋅:= U1 367.825−1EI

=

V715

U1⋅ U5+= which clearly does not make sense, since it moves opposite to the applied force;this is not a good guess therefore

replacing the equation #4 by the equation at location 5, i.e. by V514

U3⋅ U4+=

and the assumption that V5h 0:= V5 V5ε V5h+:=

gives us the following quick result for U3: V5 V4−12

U3⋅= U3 2 V5 V4−( )⋅:= U3 365.72−1EI

=

and then the following result for U1 U1 5 V2 V3−( )⋅54

U3⋅−:= U1 1053.5−1EI

=

again not possible because of the sign of U1 opposite to the applied force

We, therefore, concentrate now on the two easiest choices, i.e. 1 or 8 that we should have looked at firstfrom the beginning (they give us directly the displacement at dof 1). We have the following continuityrelations from before

V215

U1⋅ U2+=

V3 U214

U3⋅−= and we need to add either one of V115

U1⋅=

V414

− U3⋅ U4+= V815

U1⋅=

V715

U1⋅ U5+=

If we add the first one we get U1 5 V1⋅= with the assumption V1h 0:= V1 V1ε V1h+:=

if we add the second one we get U1 5 V8⋅= with the assumption V8h 0:= V8 V8ε V8h+:=

which value is the right one? the one resulting in a larger displacement for dof 1

Page 341


in this case the larger value is V1 and thus U1 5 V1⋅:= U1 1607.1251EI

=

why do we select the larger value? because in the opposite case, the hinge rotation at node 1 would benegative, while the corresponding moment is positive and this does not make sense.

with the dof 1 displacement determined we can use the other four continuity relations to get the other dofdisplacement values. We get

U2 V215

U1⋅−:= U2 589.275−1EI

=

U3 4 U2 V3−( )⋅:= U3 2494.22−1EI

=

U4 V414

U3⋅+:= U4 520.695−1EI

=

U5 V715

U1⋅−:= U5 246.425−1EI

=

With these values we have the following summary at incipient collapse:

element deformations Vε

V1ε

V2ε

V3ε

V4ε

V5ε

V6ε

V7ε

V8ε

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= global dof displacements Uf

U1

U2

U3

U4

U5

⎛⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎠

:= Uf

32.14

11.79−

49.88−

10.41−

4.93−

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

10 3−=

and the plastic deformations at incipient collapse are given by Vh Af Uf⋅ Vε−:=

Vh

0

0

0

0

21.29−

15.8−

0

3.43

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

10 3−= and we know that we have the right displacementsbecause the corresponding basic forces have thesame sign as the plastic deformations

Q

150

85.71−

85.71

120

120−

120−

120

150

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

Page 342


Remark: the quick solution with matrices gives (see Example 18 and page Part II-48)

compatibility matrix for deformations 1, 2, 3, 4 and 7 where no hinges have formed and correspondingdeformation vector

Ai

15

15

0

0

15

0

1

1

0

0

0

0

14

−

14

−

0

0

0

0

1

0

0

0

0

0

1

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= Vi

V1ε V1h+

V2ε V2h+

V3ε V3h+

V4ε V4h+

V7ε V7h+

⎛⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎠

:= solve for displacements

Uf lsolve Ai Vi,( ):= Uf

32.14

11.79−

49.88−

10.41−

4.93−

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

10 3−=

if we had selected the smaller displacement value for dof 1, we would have U1 5 V8⋅:= U1 7501EI

=

U2 V215

U1⋅−:= U2 417.85−1EI

=

U3 4 U2 V3−( )⋅:= U3 1808.52−1EI

=

U4 V414

U3⋅+:= U4 349.27−1EI

=

U5 V715

U1⋅−:= U5 75−1EI

=

U'f

U1

U2

U3

U4

U5

⎛⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎠

:= U'f

15

8.36−

36.17−

6.99−

1.5−

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

10 3−= V'h Af U'f⋅ Vε−:= V'h

3.43−

0

0

0

14.43−

8.94−

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

10 3−=

and this result does not make sense, because the first plastic deformation is of opposite sign relativeto the corresponding basic force at incipient collapse

Page 343


The following figure shows the deformed shape of the portal frame at incipient collapse.

it is worth observing that the hinge at 1 is about to form and that the element tangent is still vertical.It is also worth observing that the hinge rotations dominate the deformations of the rest of the structure sothat the element deformations are almost indistinguishable from the chord, even though the magnificationfactor is 25.

Determination of displacements and deformations of collapse mechanismAfter the last hinge forms the portal frame displaces incrementally with the collapse displacement vectorthat we have established earlier. During this process the lateral load remains constant under theassumption that displacements and deformations are small and that the equilibrium is satisfied in theoriginal, undeformed configuration. The latter is not true, of course, and as the lateral displacementsincrease it is necessary to consider the equilibrium in the deformed configuration, which leads to anonlinear geometric analysis (we will postpone this discussion until the Nonlinear Analysis course).

Let us say that we would like to determine the displacements and deformations of the portal frame at ahorizontal displacement of 0.1 units. How do we proceed? In two stages: first we know from above whathappens until incipient collapse. We restate

Uf

32.14

11.79−

49.88−

10.41−

4.93−

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

10 3−= Vh

0

0

0

0

21.29−

15.8−

0

3.43

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

10 3−=

Page 344


Then from the horizontal displacement of 0.032143 to the displacement of 0.1 the portal frame displacesaccording to the collapse displacement vector. We have

ΔU1 0.1 Uf1−:= ΔU1 67.858 10 3−=

the other free dof displacement increments areΔUf

1

15

−

45

−

15

−

15

−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

ΔU1⋅:= ΔUf

67.86

13.57−

54.29−

13.57−

13.57−

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

10 3−=

and the plastic deformation increments are

ΔVh

15

0

0

0

25

−

25

−

0

15

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

ΔU1⋅:= ΔVh

13.57

0

0

0

27.14−

27.14−

0

13.57

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

10 3−=

Thus, the total displacements and plastic deformations become

Ufinal Uf ΔUf+:= Ufinal

100

25.36−

104.17−

23.99−

18.5−

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

10 3−= and Vhfinal Af Ufinal⋅ Vε−:= Vhfinal

13.57

0

0

0

48.43−

42.94−

0

17

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

10 3−=

or, Vhfinal Vh ΔVh+:=

Page 345

Script for Example 28 in CE220 class notes % free dof displacements of portal frame at incipient collapse % use data of Example 22b with lower and upper bound solution of portal frame Example_22b Optimization terminated. the collapse load factor is 2.2286 Optimization terminated.

elastic element property data for el=1:Model.ne ElemData{el}.E = 1000; ElemData{el}.A = 1e6; % very high axial stiffness ElemData{el}.I = 50; end

form collection of element flexibility matrices Fs = Fs_matrix(Model,ElemData); % index of plastic hinge locations under the assumption that last hinge to form % is at base of left column, i.e. the latter it is still elastic ih = [8 9 12]; % index of plastic hinge locations ii = setdiff(1:12,ih); % index of locations with only elastic deformations at incipient collapse % determine elastic element deformations Ve = Fs*Qc;

free dof displacements at incipient collapse % extract rows for locations with only elastic deformations and determine free dof displacements at incipient collapse (statically determinate structure) Uf = Af(ii,:)\Ve(ii); disp('the free dof displacements at incipient collapse are') disp(Uf)

Page 346

the free dof displacements at incipient collapse are 0.0321 -0.0000 -0.0118 0.0321 -0.0499 -0.0104 0.0321 -0.0000 -0.0049

plastic deformations at incipient collapse V = Af*Uf; % element deformations Vh = Af*Uf - Ve; % hinge deformations disp('the plastic deformations at incipient collapse are') disp(Vh(ih)) the plastic deformations at incipient collapse are -0.0213 -0.0158 0.0034

plot deformed shape with plastic hinge locations MAGF = 20; % put free dof displacement values in complete displacement vector U = zeros(Model.nt,1); U(1:Model.nf) = Uf; Create_Window(0.80,0.80); Plot_Model(Model); % plot element chords for reference Plot_Model(Model,U); % it is necessary to insert Ve in the argument list for the plastic hinge discontinuities Plot_DeformedStructure(Model,[],U,Ve); Plot_PlasticHinges(Model,Qpl,Qc,U);

determine the plastic hinge rotations at a specific value of the horizontal translation

index of plastic hinge locations

ih = [2 8 9 12]; % index of plastic hinge locations ie = setdiff(1:12,ih); % extract the rows of the compatibility matrix with elastic deformations only Ae = Af(ie,:); % set displacement increment for horizontal translation to 1 and solve for the other displacements % (constrained motion of collapse mechanism) % dof index for horizontal translation ir = Model.DOF(2,1); % index for dependent dofs of collapse mechanism id = setdiff(1:Model.nf,1); % compatibility matrix (vector) of collapse mechanism Acp = zeros(Model.nf,1); Acp(ir) = 1; Acp(id) = Ae(:,id)\-Ae(:,ir);

Page 347

displacements and plastic deformations at specific value of horizontal translation % assume a specific value of the horizontal translation is given, say 0.1 units factor displacement increments by the difference between 0.1 and its value at incipient collapse DUf = Acp.*(0.1-Uf(ir)); % total displacements at a horizontal translation of 0.1 are Uft = Uf + DUf; % total deformations at a horizontal translation of 0.1 are DV = Af*Acp.*(0.1-Uf(ir)); Vt = V + DV; % plastic (hinge) deformations at a horizontal translation of 0.1 are DVh = Af*Acp.*(0.1-Uf(ir)); Vht = Vh + DVh; disp('the free dof displacements at a horizontal translation of 0.1 are') disp(Uft) disp('the plastic deformations at a horizontal translation of 0.1 are') disp(Vht(ih)) the free dof displacements at a horizontal translation of 0.1 are 0.1000 -0.0000 -0.0254 0.1000 -0.1042 -0.0240 0.1000 -0.0000 -0.0185 the plastic deformations at a horizontal translation of 0.1 are 0.0136 -0.0484 -0.0429 0.0170

plot deformed shape with plastic hinge locations at a horizontal translation of 0.1 % use the same plotting window for comparison put free dof displacement values in complete displacement vector U = zeros(Model.nt,1); U(1:Model.nf) = Uft; % it is necessary to insert Ve in the argument list for the plastic hinge discontinuities Plot_DeformedStructure(Model,[],U,Ve);

Page 348

Compatibility

f fV A U=

Equilibrium

f f fw= +P B Q P

( )i f fw x x= − +Q B P P B Q Tx 0=B VNOS redundants NOS equations

Tf f=B A

Element deformation-force

s 0V F Q V= +

gp xx x 0V F Q+ =

for general load case

FORCE OR COMPATIBILITY METHOD OF ANALYSIS

In the force method of analysis the unknowns of the problem are the redundant basic forces xQ

Equilibrium equations in undeformed configuration (linear) f f fw= +P B Q P

Compatibility relations for small displacements/deformations (linear) f fV A U=

(1)( )i f fw x xQ B P P B Q= − +

Element deformation-force for linear elastic material (3)s 0V F Q V= +

(2)Tx 0B V =

from (2) and (3*) ( )( )Tx s i f fw x x 0 0B F B P P QB V− + + =⎡ ⎤⎣ ⎦

( )T Tx s i f fw 0 x s x x 0B F B P P V B F B Q− + + =⎡ ⎤⎣ ⎦

from (1) and (3) (3*)( )s i xf fw x 0V F B P P B VQ= − + +⎡ ⎤⎣ ⎦

gp xx x 0V F Q+ = withT

xx x s xF B F B=

( )Tgp x s i f fw 0V B F B P P V= − +⎡ ⎤⎣ ⎦


solve for general load case

CE220 - Theory of Structures © Prof. Filip C. Filippou, 2000

Page 349

Compatibility

f f=V A U

Equilibrium

f f fw= +P B Q P

p x x= +Q Q B Q Tx 0=B VNOS redundants NOS equations

Tf f=B A

Element deformation-force

s 0V FQ V= +

gp xx x 0V F Q+ =


for specific load case

In the force method of analysis the unknowns of the problem are the redundant basic forces xQ

Equilibrium equations in undeformed configuration (linear) f f fw= +P B Q P

Compatibility relations for small displacements/deformations (linear) f f=V A U

(1)p x x= +Q Q B Q

Element deformation-force for linear elastic material (3)s 0V F Q V= +

(2)Tx 0B V =

from (2) and (3*) ( )Tx s p x 0x 0⎡ ⎤+ + =⎣ ⎦QB F Q B V

( )T Tx s p 0 x s x x 0+ + =B F Q V B QF B

from (1) and (3) (3*)( )s p x 0x= + +V F Q B VQ

gp xx x 0+ =V F Q withT

xx x s xF B F B=

( )Tgp x s p 0= +V B F Q V


solve for specific load case


Page 350

Solution steps for force method of analysis "by hand"

Step 1: write equilibrium relations and establish structure equilibrium matrix for free dof's

Step 3: solve equilibrium equations under the given loading with redundants set equal to zero --> particular solution

Step 2: identify redundants making sure that primary structure is stable (primary structure = structure with redundants equal to zero)

Step 4: solve equilibrium equations for general homogeneous solution

Step 6: solve compatibility conditions for redundant basic forces

Step 5: determine compatibility errors under particular and homogeneous solution

Step 7: determine basic forces under given loading by superposition of NOS+1 stress states

Step 8: with Q determine other element forces by equilibrium (forces in inextensible or inflexible elements, shear forces)

Step 9: with Q determine support reactions from equilibrium at restrained dof's

Step 10: with support reactions and applied forces check global equilibrium on complete structure free body

Step 11: (if necessary) determine element deformations

Step 12: (if necessary) determine displacements by compatibility or principle of virtual forces

orfB TfA with constraints

pQ

xB

and( )Tx s p 0+B F Q V T

xx x s x=F B F B

( )Tx s p 0 xx x 0+ + =B F Q V F Q

p x x= +Q Q B Q

s 0= +V F Q V

Tf i=U B V


Page 351

Displacement-force behavior of structuresStatically indeterminate structures

Equilibrium f f fw= +BP Q P

express basic forces in terms of applied loads and NOS redundants ( )i f fw x x= − +B BQ P P Q

Compatibility f f= AV U

element deformations need to be compatible; this leads to NOS compatibility conditions

Deformation-force relation for linear elastic material response s 0= +FV Q V

use compatibility conditions to obtain unique solution for redundant basic forces

{ }( )

TxTx s 0

Tx s i f fw x x 0

0 =

= +

⎡ ⎤⎡ ⎤= − + +⎢ ⎥⎣ ⎦⎣ ⎦

B

B F

B F B B

V

Q V

P P Q V

Tx0 = B V

( )T T Tx s i f fw x 0 x s x x0 = − + +B F B B B F BP P V Q

or, more compact ( ) Txi f fw x 0 xx x0 = − + +F B FP P V Q

with Txi x s i=F B F B T

xx x s x=F B F B

solve for redundant basic forces ( )1 Tx xx xi f fw x 0

− ⎡ ⎤= − − +⎢ ⎥⎣ ⎦F F BQ P P V

return to equilibrium equations and substitute for the redundant basic forces

( )( ) ( )

( )( )

i f fw x x

1 Ti f fw x xx xi f fw x 0

1 1 Ti x xx xi f fw x xx x 0

−

− −

= − +

⎡ ⎤= − − − +⎢ ⎥⎣ ⎦

= − − −

B B

B B F F B

B B F F B F B

Q P P Q

P P P P V

P P V

we recognize the force influence matrix for a statically indeterminate structure

and can express the basic forces in compact form in terms of the applied loading

( ) 1 Tf fw x xx x 0

−= − −B B F BQ P P V ( )f fw V 0= − +B BQ P P V

1i x xx xi

−= −B B B F F

1 TV x xx x

−= −B B F B


Page 352

Displacement-force behavior of structures (cont'd)

Tf = BU V

the displacements at the free dof's can be obtained from the same relation as for statically determinate structures (think of the principle of virtual forces!)

substitute the deformation-force relation for linear elastic material response and the basic force expression

{ }( ){ }

Tf

Ts 0

T 1 Ts f fw x xx x 0 0

−

=

= +

⎡ ⎤= − − +⎢ ⎥⎣ ⎦

B

B F

B F B B F B

U V

Q V

P P V V

after collecting terms we get ( ) ( )T 1 Tf f fw s x xx x 0

−= − + −F B I F B F BU P P V

with Ts=F B F B the structure flexibility matrix with 1

i x xx xi−= −B B B F F

( )T 1 T0 fw s x xx x 0

−= − + −F B I F B F BU P V the initial displacement vector

and

Consult examples 29, 30, 31 and 32

T 1 Ts x xx x 0− =B F B F B so that the last expression becomesNote that T

0 fw 0= − +U P VF B

which gives the following relation for the structure ( ) Tf f fw 0= − +U P P VF B

Proof: to show that T 1 Ts x xx x 0− =B F B F B

Recall that 1i x xx xi

−= −B B B F F consequently T T T T Ti xi xx x

−= −B B F F B

Thus, ( )T 1 T T T T T 1 Ts x xx x i xi xx x s x xx x

T 1 T T T T 1 Ti s x xx x xi xx x s x xx x

T 1 T T T 1 Txi xx x xi xx xx xx x

T 1 T T T Txi xx x xi xx x

0 q.e.d

− − −

− − −

− − −

− −

= −

= −

= −

= −

=

B F B F B B F F B F B F B

B F B F B F F B F B F B

F F B F F F F B

F F B F F B

where we have used the earlier definitions Txi x s i=F B F B and thus T T

xi i s x=F B F B

Txx x s x=F B F Band

as well as the symmetry of xxF and its inverse


Page 353

CE 220 - Theory of Structures Ex 29 - Force method: indeterminate truss © Prof. Filip C. Filippou, 2000

Example 29 - Force method: statically indeterminate truss

We return to complete the analysis of a structure whose equilibrium equations we studied in Example 5,and whose compatibility relations we studied in Example 13. After adding the linear deformation-forcerelation for the elements of the structural model, we will be able to obtain the unique solution of thestructural response that satisfies all three requirements: (a) equilibrium, (b) compatibility, and (c) linearelastic material response. For the solution in this example we will use the force method of analysis. In thismethod the unknowns of the problem are the redundant basic forces of the equilibrium equations. Usingthe compatibility condition on the element deformations we will be able to determine these redundant basicforces so as to satisfy these conditions.

The following figure shows the truss model with the numbering of the free dofs and basic forces.

8

6

1

23

4

5

1Q

2Q3Q

4Q

5Q6Q

8

6

15

a

b c

d

e f

1 2

34

The axial stiffness of all truss elements is the same: EA 10000:=

Length of truss elements La 8:= Lb 6:= Lc 6:= Ld 8:= Le 10:= Lf 10:=

Solution of equilibrium equations

Recall from Example 5 that there are 5 equilibrium equations and 6 unknown basic forces. The degree ofstatic indeterminacy of this structure is NOS=1. In such case, we can select one of the unknown basic forcesas redundant and express the general solution of the equilibrium equations as the sum of a particularsolution that satisfies equilibrium with the applied forces, and a homogeneous self-stress state that dependson the unknown redundant basic force value. The latter is, therefore, a free parameter as far as the solutionof the equilibrium equations is concerned.

We summarize the findings from Example 5 in the following:

Page 354


Particular solution of equilibrium equations Pf as given Q6 0=

8

6

15

a

b c

d

e

f

6 0=Q

11.2511.2515

Qp

15

11.25

0

15

18.75−

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

Homogeneous solution of equilibrium equations Q6 1= and Pf 0=

b c

d

6 1=Q

000

8

6

a

e

f

Bbarx

0.8−

0.6−

0.6−

0.8−

1

1

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

Page 355


The general solution of the equilibrium equations is a one parameter expression of the form

Q

15

11.25

0

15

18.75−

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

0.8−

0.6−

0.6−

0.8−

1

1

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

Q6⋅+= for equilibrium alone any value of Q6 is suitable

note that in this expression Q6 carries the units, while the force influencematrix (vector in this case) has units of force per unit of force

Compatibility relations

In a statically indeterminate structure there are more compatibility relations than available displacementsat the free global dofs of the structural model. As we have seen in Example 13, the compatibility relations only

have a unique solution when the element deformations satisfy the condition BbarxT V⋅ 0=

The element deformations are dependent on the basic element forces, and these depend for the case athand on the unknown value of the redundant basic force Q6. There is one unknown redundant force andone equation that needs to be satisfied for the element deformations. This completes the loop and allowsthe determination of the unique value of Q6 that satisfies the compatibility condition.

We write this condition out in detail

V Fs Q⋅= Fs Qp Bbarx Qx⋅+( )⋅= Fs

15

11.25

0

15

18.75−

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

0.8−

0.6−

0.6−

0.8−

1

1

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

Q6⋅+

⎡⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎦

⋅=

with Fs1

EA

La

0

0

0

0

0

0

Lb

0

0

0

0

0

0

Lc

0

0

0

0

0

0

Ld

0

0

0

0

0

0

Le

0

0

0

0

0

0

Lf

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

⋅:=

Page 356


then the compatibility condition gives

BbarxT V⋅ Bbarx

T Fs Qp Bbarx Qx⋅+( )⋅[ ]⋅= BbarxT Fs⋅ Qp⋅ Bbarx

T Fs⋅ Bbarx⋅ Qx⋅+= 0=

or, spelled out we get

0.8− 0.6− 0.6− 0.8− 1 1( ) Fs⋅

15

11.25

0

15

18.75−

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

0.8−

0.6−

0.6−

0.8−

1

1

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

Q6⋅+

⎡⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎦

⋅ 0=

we identify two contributions, one for the applied forces, and one for the redundant. We write them out

0.8− 0.6− 0.6− 0.8− 1 1( ) Fs⋅

15

11.25

0

15

18.75−

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

⋅ 0.8− 0.6− 0.6− 0.8− 1 1( ) Fs⋅

0.8−

0.6−

0.6−

0.8−

1

1

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

⋅ Q6⋅+ 0=

Because Fs is diagonal in this case, the evaluation of these expressions is relatively straightforward by hand:we multiply the six corresponding terms of the row and column and then add them up. We demonstrate

0.8−La

EA⋅ 15⋅ 0.6

Lb

EA⋅ 11.25⋅− 0.8

Ld

EA⋅ 15⋅− 1

Le

EA⋅ 18.75−( )⋅+ 420−

1EA

= for the first term

La

EA0.8−( )2⋅

Lb

EA0.6−( )2⋅+

Lc

EA0.6−( )2⋅+

Ld

EA0.8−( )2⋅+

Le

EA12⋅+

Lf

EA12⋅+ 34.56

1EA

= for thesecond

we note that all contributions of the second term are positive!

in compact form, we can also write with matrix software

BbarxT Fs⋅ Qp⋅ 420−

1EA

= for the first term, and BbarxT Fs⋅ Bbarx⋅ 34.56

1EA

= for the second

Page 357


The compatibility condition for the element deformations is a scalar equation for the case at hand

BbarxT Fs⋅ Qp⋅ Bbarx

T Fs⋅ Bbarx⋅ Q6⋅+ 0=

and solving for Q6 we get Q6Bbarx

T Fs⋅ Qp⋅

BbarxT Fs⋅ Bbarx⋅

−:= Q6 12.15=

with this result the final forces in the truss elements under the applied force of 15 units are

Q

15

11.25

0

15

18.75−

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

0.8−

0.6−

0.6−

0.8−

1

1

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

Q6⋅+:= Q

5.28

3.96

7.29−

5.28

6.6−

12.15

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

=

and the element deformations V are V Fs Q⋅:= V

42.22

23.75

43.75−

42.22

65.97−

121.53

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

1EA

=

we confirm that BbarxT V⋅ 0=

and we can use any five of the six compatibility relations to determine thedisplacements at the free global dofs

Recall e.g. from Example 13 that U1 V1:= U1 42.221

EA=

U2 V1 0.75 V2⋅+ 1.25 V5⋅−:= U2 142.51

EA=

U3 V2:= U3 23.751

EA=

U4 V1 0.75 V2⋅+ V4+ 1.25 V5⋅−:= U4 184.721

EA=

U5 V3:= U5 43.75−1

EA=

Note: the value of the redundant force Q6 does not depend on the axial stiffness value EA, since the latterappears both in the numerator and in the denominator of the expression for Q6. This is a general result forstatically indeterminate structures. The force distribution does not depend on the absolute, but only on therelative stiffness value, i.e. the relation of the axial stiffness values of the different elements. In this case theyare all the same and we could have left out EA from the intermediate terms leading to the determination ofQ6. We demonstrate this once more

Page 358


EAFs

La

0

0

0

0

0

0

Lb

0

0

0

0

0

0

Lc

0

0

0

0

0

0

Ld

0

0

0

0

0

0

Le

0

0

0

0

0

0

Lf

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= BbarxT EAFs⋅ Qp⋅ 420−= Bbarx

T EAFs⋅ Bbarx⋅ 34.56=

or, by hand

0.8− La⋅ 15⋅ 0.6 Lb⋅ 11.25⋅− 0.8 Ld⋅ 15⋅− 1 Le⋅ 18.75−( )⋅+ 420−= for the first term

La 0.8−( )2⋅ Lb 0.6−( )2⋅+ Lc 0.6−( )2⋅+ Ld 0.8−( )2⋅+ Le 12⋅+ Lf 12⋅+ 34.56= for the second

to solve for the redundant we divide the negative value of the first term by the second

Q6420−

34.56−:= Q6 12.15=

for hand calculations this approach is a lot more convenient, since we do not need to bother about divisionthrough a very large number like EA=10,000.

Physical interpretation of results

It remains to discuss the meaning of the results for the compatibility condition. We recall that the generalsolution of the equilibrium equations is a one parameter expression of the form

Q Qp Bbarx Qx⋅+=

15

11.25

0

15

18.75−

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

0.8−

0.6−

0.6−

0.8−

1

1

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

Q6⋅+= for equilibrium alone any value of Q6 is suitable

Let us investigate the effect of the forces Qp on the deformations V and then on compatibility. We have

V1 15La

EA⋅:= V2 11.25

Lb

EA⋅:= V3 0:= V4 15

Ld

EA⋅:= V5 18.75−( )

Le

EA⋅:= V6 0:=

Recall from Example 13 that after solving the first five compatibility relations for the unknown displacementsat the free global dofs we obtain

Page 359


U1 V1=

U2 V1 0.75 V2⋅+ 1.25 V5⋅−=

U3 V2=

U4 V1 0.75 V2⋅+ V4+ 1.25 V5⋅−=

U5 V3=

the sixth compatibility relation reads V6 0.8 U4⋅ 0.6 U5⋅+=

let us then determine the two displacement values needed for the sixth compatibility relation

U4 15La

EA⋅

⎛⎜⎝

⎞⎟⎠

0.75 11.25Lb

EA⋅

⎛⎜⎝

⎞⎟⎠

⋅+ 15Ld

EA⋅

⎛⎜⎝

⎞⎟⎠

+ 1.25 18.75−Le

EA⋅

⎛⎜⎝

⎞⎟⎠

−:= U4 5251

EA=

U5 0:=

thus, the right hand side of the sixth compatibility relation reads 0.8 U4⋅ 0.6 U5⋅+ 4201

EA=

this is the necessary deformation for element f so that it fits exactly between nodes 1 and 4. However,element f does not deform under the forces Qp and this creates a gap or release deformation V6h equal tothe difference between the distance of nodes 1 and 4 and the element deformation (consult pages Part I -

182-183 for Example 13). This release deformation is equal to 4201

EA under the applied force.

We conclude that BbarxT Fs⋅ Qp⋅ 420−

1EA

= represents the opposite of the release deformation orthe gap closing value under the forces Qp

1 2

3 4

1420EA

the state of deformation andthe global dof displacementsunder the forces Qp are shown in this figure

Page 360


Now we investigate the state of deformation and the global dof displacements under the self-stressstate that results under a unit value of the redundant basic force Q6

V1 0.8−( )La

EA⋅:= V2 0.6−( )

Lb

EA⋅:= V3 0.6−( )

Lc

EA⋅:=

V4 0.8−( )Ld

EA⋅:= V5 1( )

Le

EA⋅:= V6 1( )

Lf

EA⋅:=

We determine again the displacement values for the sixth compatibility relation

U4 0.8−( )La

EA⋅ 0.75 0.6−( )

Lb

EA⋅

⎡⎢⎣

⎤⎥⎦

⋅+ 0.8−( )Ld

EA⋅

⎡⎢⎣

⎤⎥⎦

+ 1.25 1( )Le

EA⋅

⎡⎢⎣

⎤⎥⎦

−:= U4 28−1

EA=

U5 0.6−( )Lc

EA⋅:=

thus, the right hand side of the sixth compatibility relation reads 0.8 U4⋅ 0.6 U5⋅+ 24.56−1

EA=

the left hand side of the equation is not equal to zero this time. We have V6 101

EA=

subtracting the deformation of element f from the distance between the nodes 1 and 4 gives again the gapbetween the end of element f and node 4.

0.8 U4⋅ 0.6 U5⋅+( ) V6− 34.56−1

EA=

represents the gap closing value under theself-stress state represented by vector Bbarx

We conclude that BbarxT Fs⋅ Bbarx⋅ 34.56

1EA

=

110EA

124.56EA

1 2

34

the state of deformation andthe global dof displacementsunder the self-stress state are shown in this figure

Page 361


We, therefore, conclude that the compatibility condition furnishes an equation for determining the value ofthe redundant basic force Q6 that will close the gap between the end of element f and node 4.

Note: in the compatibility equations we have used the sixth equation because the redundant of theproblem is Q6, but this is not really necessary, just convenient.

Support reactions and global equilibrium

Since the self-stress state does not produce any support reactions for the problem at hand, the supportreactions are those of the particular solution.

8

6

15

11.2511.2515

horizontal and vertical force equilibriumis obviously satisfied

sum of moments about left support

11.25 8⋅ 15 6⋅− 0= ok!

Page 362


Load Case 2: thermal expansion of element e

Assume now that the only loading on the structure is ofthermal type, i.e. assume that member e is heated upexperiencing a positive thermal strain of 0.005 (extension)

In this case the particular solution is zero, since thereare no applied forces at the free dofs of the structure(note that Q6 0= for the particular solution)

Qp

0

0

0

0

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

The general solution of the equilibrium equations is in this case

Q

0.8−

0.6−

0.6−

0.8−

1

1

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

Q6⋅=

Element deformations and compatibility relations

Under the particular solution (i.e. the applied loading) the only deformation present is the non-mechanicaldeformation of element e. We, therefore, have

V V0= with V0

0

0

0

0

0.005 Le⋅

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:= V0

0

0

0

0

0.05

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

=

The solution of the compatibility relations for this deformation vector gives the following gap betweenelement f and node 4 (see earlier discussion)

BbarxT V0⋅ 0.05= in fact, in this case the gap is equal to the amount of element e deformation !

can you explain why geometrically?

note that in this case the axial stiffness EA does not enter into the expression (it only appears formechanical deformations!). Thus, we need to be careful when calculating the redundant basic force!

Page 363


The self-stress state due to a unit value of the redundant basic force Q6 produces the same deformationsand the same gap between element f and node 4 as before. Thus, we determine the value of Q6 to satisfythe compatibility condition that there is no gap between element f and node 4, i.e.

BbarxT V0⋅ Bbarx

T Fs⋅ Bbarx⋅ Qx⋅+ 0=

solving the above scalar equation for Q6 we get Q6Bbarx

T V0⋅


−:= Q6 14.47−=

The final stress state in this structure under the effect of a thermal deformation of element e are

Q

0.8−

0.6−

0.6−

0.8−

1

1

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

Q6⋅:= Q

11.57

8.68

8.68

11.57

14.47−

14.47−

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

=

Support reactions and global equilibrium

Recall that the support reactions are zero for the self-stress state. Since this is the only stress state for thisloading case, the support reactions are zero and global equilibrium with the zero applied forces is, of course,satisfied.

Note that if we determined the EA-fold deformations for the denominator for computational convenience, weneed to remember to multiply the result by EA. We demonstrate

Q6Bbarx

T V0⋅

BbarxT EAFs⋅ Bbarx⋅

−:= Q6 14.47−1

EA=

This means that in a statically indeterminate structure the self-stress state induced by non-mechanicalelement deformations (i.e. thermal, shrinkage etc) is proportional to the stiffness of the elements.

It is worth noting that the final force in element f is a compression of 14.47 units. We will discuss this furtherin the following, but it is worth remembering that the deformation-force relation for any element is

v f q⋅ vo+= which implies that we may have a positive total deformation with a negative force in theelement; the latter is proportional to the mechanical deformation only!

Page 364


Commentary on prestressing and prestressed structures

The last load case consists of an initial deformation of 0.05 units in element f. This may arise fromnon-mechanical effects, like temperature, hydro-expansion, or from the fact that the element is fabricatedlonger than required by the undeformed geometry (thus, the fabricated element length would be 10.05 inthis case). How is the installation of this element supposed to take place in the field?

Let us assume that the rest of the structure is assembled, i.e. elements a through e are in place. It is worthnoting that the elements are all undeformed, so that the assembly looks like the undeformed configurationof the structural model without element f. Then, the process of installing element f can be described asfollows: the element f is pre-compressed to a value compensating the initial deformation, i.e. necessary tobring the element to its required length of 10 units according to the undeformed geometry; the element isthen installed into position and the jacks that pre-compress it are removed. The force in element f is thenreleased into the structure giving rise to the stress-state that we have calculated above.

What is the necessary pre-compression force?

set v = 0 in the above deformation-force relation and solve for q with given vo. You will get vo−EAL

⋅

For the case at hand this is 0.05−10000

Lf⋅ 50−= units of force

Upon release this force drops to -14.47 according to our analysis. For the example at hand this is a verylarge loss of initial prestressing force. How can we assess how much force we will loose during installation?Look at the contribution of the prestressed element and the rest of the structure to the gap caused by theself-stress state.

110EA

124.56EA

1 2

34

Element f contributes 101

EA⋅

The rest of the structure contributes

24.561

EA⋅ i.e. about 2.5 times as much

This explains the prestress drop to thevalue of

1010 24.56+

50−( )⋅ 14.47−=

Alternatively, we can say that theprestress loss is dependent on the ratioof the flexibility of the rest of the structureto the total flexibility. Can you apply thisconcept to a prestressed concrete beam?

prestress loss during installation24.56

10 24.56+50−( )⋅ 35.53−= i.e. very close to 70%

In typical structures the prestressing element is very flexible relative to the rest of the structure andprestress loss amounts to 3-5%.

Page 365

Script for Example 29 in CE220 class notes % Solution for statically indeterminate truss with Matlab functions

continuation of Example 5 Example_5 the rank of the equilibrium matrix of the primary structure is 5 Qp = 15.0000 11.2500 0 15.0000 -18.7500 0 Bbarx = -0.8000 -0.6000 -0.6000 -0.8000 1.0000 1.0000 % define element properties and set up Fs matrix for structure EA = 10000; % axial stiffness L = [8 6 6 8 10 10]; % element lengths for el=1:6 f{el} = L(el)/EA; end % set up Fs matrix Fs = blkdiag (f{:}); % determine indetermediate term for compatibility equation Fxx = Bbarx'*Fs*Bbarx;

First load case: applied horizontal forces of 15 units (Example 5) % solve for redundant basic force under the loading Qx = -Fxx\(Bbarx'*Fs*Qp); % display redundant basic forces disp('redundant force Qx under applied horizontal force') disp(Qx); % determine final forces in truss Q = Qp+Bbarx*Qx; % display final forces disp('truss forces Q under applied horizontal force') disp(Q); redundant force Qx under applied horizontal force 12.1528 truss forces Q under applied horizontal force 5.2778 3.9583 -7.2917 5.2778 -6.5972 12.1528

Page 366

Second load case: initial deformation % define initial deformation vector for second load case (prestress) V0 = zeros(6,1); V0(5) = 0.05; % solve for redundant basic force under the loading of Example 5 (concentrated force) Qx = -Fxx\(Bbarx'*V0); % display redundant basic forces disp('redundant force Qx under initial deformation') disp(Qx); % determine final forces in truss Q = Bbarx*Qx; % display final forces disp('truss forces Q under initial deformation') disp(Q); redundant force Qx under initial deformation -14.4676 truss forces Q under initial deformation 11.5741 8.6806 8.6806 11.5741 -14.4676 -14.4676

Page 367

Script for Example 29 in CE220 class notes % Solution for indeterminate truss with FEDEASLab functions (step-by-step)


Model and element property data from file Example_29_Data Example_29_Data


1 2

3 4

1

2 3

4

56

functions for force method of analysis % set up equilibrium matrix B = B_matrix(Model); % extract free dofs Bf = B(1:Model.nf,:); % determine force influence matrix [Bbari,Bbarx] = ForceInfl_matrix(Bf); % set up collection of element flexibility matrices Fs = Fs_matrix(Model,ElemData); % determine indetermediate term for compatibility equation Fxx = Bbarx'*Fs*Bbarx;

First load case: applied horizontal forces of 15 units (Example 5) % specify nodal forces Pe(4,1) = 15; Loading = Create_Loading (Model,Pe); Pf = Loading.Pref; % solve for redundant basic force under the loading Qx = -Fxx\(Bbarx'*Fs*Bbari*Pf); % display redundant basic forces disp('redundant force Qx under applied horizontal force') disp(Qx); % determine final forces in truss Q = Bbari*Pf+Bbarx*Qx; % display final forces disp('truss forces Q under applied horizontal force') disp(Q); redundant force Qx under applied horizontal force 12.1528

Page 368

truss forces Q under applied horizontal force 5.2778 3.9583 -7.2917 5.2778 -6.5972 12.1528

Second load case: initial deformation % define initial deformation vector for second load case (prestress) V0 = zeros(6,1); V0(5) = 0.05; % solve for redundant basic force under the loading of Example 5 (concentrated force) Qx = -Fxx\(Bbarx'*V0); % display redundant basic forces disp('redundant force Qx under initial deformation') disp(Qx); % determine final forces in truss Q = Bbarx*Qx; % display final forces disp('truss forces Q under initial deformation') disp(Q); redundant force Qx under initial deformation -14.4676 truss forces Q under initial deformation 11.5741 8.6806 8.6806 11.5741 -14.4676 -14.4676

Page 369

Script for Example 29 in CE220 class notes % Solution for indeterminate truss with FEDEASLab functions (one step)


Model and element property data from file Example_29_Data Example_29_Data


1 2

3 4

1

2 3

4

56

force method of analysis % set up equilibrium matrix B = B_matrix(Model); % extract free dofs Bf = B(1:Model.nf,:); % set up collection of element flexibility matrices Fs = Fs_matrix(Model,ElemData); % invoke single function for force method of analysis [Bbar,Bvbar,F] = ForceMethod(Bf,Fs);

First load case: applied horizontal forces of 15 units (Example 5) % specify nodal forces Pe(4,1) = 15; Loading = Create_Loading (Model,Pe); Pf = Loading.Pref; % determine final forces in truss Q = Bbar*Pf; % display final forces disp('truss forces Q under applied horizontal force') disp(Q); % determine displacements Uf = F*Pf; % display displacements disp('free dof displacements Uf under applied horizontal force') disp(Uf); truss forces Q under applied horizontal force 5.2778 3.9583 -7.2917 5.2778

Page 370

-6.5972 12.1528 free dof displacements Uf under applied horizontal force 0.0042 0.0143 0.0024 0.0185 -0.0044

plot axial force distribution Create_Window(0.80,0.80); Plot_Model(Model); Plot_AxialForces(Model,Q);

plot deformed shape Create_Window(0.80,0.80); Plot_Model(Model); % put free dof displacement values in complete displacement vector U = zeros(Model.nt,1); U(1:Model.nf) = Uf; Plot_Model(Model,U);

Second load case: initial deformation % define initial deformation vector for second load case (prestress) V0 = zeros(6,1); V0(5) = 0.05; % determine final forces in truss Q = Bvbar*V0;

Page 371

% display final forces disp('truss forces Q under initial deformation') disp(Q); % determine displacements Uf = Bbar'*V0; % display displacements disp('free dof displacements Uf under initial deformation') disp(Uf); truss forces Q under initial deformation 11.5741 8.6806 8.6806 11.5741 -14.4676 -14.4676 free dof displacements Uf under initial deformation 0.0093 -0.0313 0.0052 -0.0220 0.0052

plot axial force distribution Create_Window(0.80,0.80); Plot_Model(Model); Plot_AxialForces(Model,Q);

plot deformed shape Create_Window(0.80,0.80); Plot_Model(Model); % put free dof displacement values in complete displacement vector U = zeros(Model.nt,1); U(1:Model.nf) = Uf; Plot_Model(Model,U);

Page 372

CE 220 - Theory of Structures Ex 30 - Force method: indeterminate braced frame © Prof. Filip C. Filippou, 2000

Example 30 - Force method: indeterminate braced frame

In this example we analyze the braced frame that we have already studied from the equilibrium andcompatibility standpoint in Examples 6 and 14, respectively. We will summarize the key equations in thisanalysis, so that we collect the entire solution process in one place, particularly, since the loadinginvolves several cases. The following figure shows the geometry of the frame.

a

b c

d

4 4

6

La 6:=

Lb 4:=

Lc 4:=

Ld 82 62+:=

Under the assumption that elements a, b and c are inextensible, there are four independent free globaldofs, as shown in the following figure.

dof 1

dof 2 dof 3dof 4

a

b c

d

4 4

6

The equilibrium and compatibility equations are the same as in Examples 6 and 14, respectively.

Reference:C:\Drive D\COURSES\CE220\Classnotes\Mathcad files\Example 6_Equilibrium for Hyperstatic Braced Frame.xmcd(R)

Reference:C:\Drive D\COURSES\CE220\Classnotes\Mathcad files\Example 14_Compatibility 4 Hyperstatic Braced Frame.xmcd(R)

Page 373


Equilibrium equations at free dofs (from example 6) Compatibility relations (from example 14)

P1Q1 Q2+

60.8 Q6⋅+= V1

U1

6=

P2 Q2 Q3+= V2U1

6U2+=

P3Q3 Q4+

4−

Q5

4+=

V3 U2U3

4−=

P4 Q4 Q5+=

V4U3

4− U4+=

V5U3

4U4+=

V6 0.8 U1⋅=

We note that there are four equilibrium equations and 6 unknown basic forces. The degree of staticindeterminacy of the structural model is therefore NOS=2. Correspondingly, there are NOS morecompatibility relations to satisfy than available displacement values at the free dofs. Such a problem will nothave a solution unless NOS linearly independent constraints are imposed on the deformations. Thesefurnish NOS compatibility conditions for determining the NOS redundant basic forces.

We select the basic forces Q1 and Q4 as redundants. We establish three equilibrium states: one equilibriumstate is known as the particular solution and satisfies equilibrium with the applied forces while theredundants are equal to zero; the other two equilibrium states are homogeneous solutions with no appliedloading and each redundant set equal to a unit value in turn.

First loading case

a

b c

d

2030

4 4

6

We state here the results since the answers are the same as those of example 6 (note that the equivalentnodal forces at nodes 2 and 4 from the distributed load of example 6 do not affect the basic forces ofinterest).

Page 374


30

4 4

6

20Particular solution under given loading

set Q1 0= Q4 0=

We solve the following equilibrium equations

30Q2

60.8 Q6⋅+=

0 Q2 Q3+=Qp

0

80−

80

0

0

300.8

804.8

+

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=20−

Q3

4−

Q5

4+=

0 Q5=

Homogeneous solution for each redundant; no loading is present

set Q1 1= Q4 0=Bbarx

1

0

0

0

0

524

−

0

2

2−

1

1−

512

−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=and then separately set Q1 0= Q4 1=

The general homogeneous solution of the equilibrium equations is BbarxQ1

Q4

⎛⎜⎝

⎞⎟⎠

⋅

and we confirm that we have obtained the right self-stress states by Bf Bbarx⋅

0

0

0

0

0

0

0

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

=

thus, the general solution of the equilibrium equations consists of the linear combination of threestress-states that satisfy equilibrium: the particular solution and two homogeneous solutions with unknownparameters

We can write therefore Q Qp BbarxQ1

Q4

⎛⎜⎝

⎞⎟⎠

⋅+=

we denote the vector of redundant basic forces by Qxand can write more generally

Q Qp Bbarx Qx⋅+=

Page 375


For each of the equilibrium states we proceed to calculate the element deformations and try to check thecompatibility conditions. We assume that the flexural and axial stiffness values are given. Let us say

EI 50000:= for elements a, b and c and EA 10000:= for element d

Let us first proceed to solve the problem by matrix operations following Example 29 while providingcommentary on the physical meaning of the terms. We will return subsequently and repeat the process "byhand" so that we can see the terms one by one.

the collection of element flexibility matrices is Fs1EI

La

3

La

6−

0

0

0

0

La

6−

La

3

0

0

0

0

0

0

Lb

3

Lb

6−

0

0

0

0

Lb

6−

Lb

3

0

0

0

0

0

0

Lc

3

0

0

0

0

0

0

LdEIEA⋅

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

⋅:=

We note that we have factored out 1/EI on the right hand side, so that only relative stiffness values appear inthe matrix. This is quite convenient for hand calculations, as we have already discussed.

The deformations under the particular solution are Fs Qp⋅

80

160−

106.67

53.33−

0

2708.33

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

1EI

=

and the compatibility conditions are not satisfied, because BbarxT Fs⋅ Qp⋅

484.24−

1715.14−

⎛⎜⎝

⎞⎟⎠

1EI

=

In fact, the above expression represents the angle between node tangent and element tangent at locations1 and 4 in the structural model. This angles need to be zero for tangent continuity at locations 1 and 4.

Page 376


The deformations under each homogeneous solution are: Fs Bbarx⋅

2

1−

0

0

0

10.42−

2−

4

3.33−

2.67

1.33−

20.83−

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

1EI

=

note that we obtain two sets of deformations, of course, one for each homogeneous solution or self-stressstate. The first column represents the element deformations for Q1 = 1 and the second for Q4 = 1

each of these deformation states does not satisfy compatibility either as we can tell immediately from


4.17

2.34

2.34

27.347⎛⎜⎝

⎞⎟⎠

1EI

=

We now obtain a 2 x 2 matrix. The first column corresponds to the compatibility error at locations 1 and 4 forthe deformations caused by the first self-stress state (Q1 = 1), and the second column represents thecompatibility error at locations 1 and 4 for the deformations caused by the second self-stress state (Q4 = 1)

The above expression appears in every analysis by the force method and we give it a special name.

Fxx BbarxT Fs⋅ Bbarx⋅:= we note that it is a flexibility matrix, since it represents the compatibility error

at locations 1 and 4 under the self-stress states, i.e. a unit value of theredundant force. Like every flexibility matrix it is symmetric, as can be readilyseen from its definition. Thus, in hand calculations only the diagonal and theupper or lower half of the matrix needs to be determined.

The compatibility conditions require that there be no compatibility error at locations 1 and 4 under the finalstress state. This means that

BbarxT Fs⋅ Qp⋅ Fxx Qx⋅+ 0= and these are exactly two equations for the two unknown redundants

BbarxT Fs⋅ Qp⋅ the compatibility error under the particular solution

and Vhp BbarxT− Fs⋅ Qp⋅:= the hinge deformation if a hinge were inserted at 1 or 4

we get from the solution of the system of equations Qx lsolve Fxx Vhp,( ):= Qx85.01

55.44⎛⎜⎝

⎞⎟⎠

=

We can see from the above solution that the value of the redundant basic forces is selected so that iteliminates the release deformation at locations 1 and 4 under the given loading.

Note that the values for the redundants do not change if we change EI and EA, but keep the ratio the same.This has important ramifications in preliminary design. During this stage we may not know the exact sizes ofthe different structural elements, but we may have an idea about their relative sizes.

Page 377


We can approach this problem also in a more step by step fashion. To this end let us find the elementdeformations under the particular solution. We have

V1p

V2p

V3p

V4p

V5p

V6p

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

Fs Qp⋅:= V1p 801EI

= V3p 106.671EI

= V5p 01EI

=

V2p 160−1EI

= V4p 53.33−1EI

= V6p 2708.331EI

=

We recall the compatibility relations from above noting that these cannot be satisfied, in general, for whichreason we insert a hinge or release at locations 1 and 4.

V1U1

6= V3 U2

U3

4−= V5

U3

4U4+=

V2U1

6U2+= V4

U3

4− U4+= V6 0.8 U1⋅=

We solve the four compatibility equations without a release deformation, namely at 2, 3, 5 and 6

U1p 1.25 V6p⋅:= U1p 3385.421EI

=

U2p V2pU1p

6−:= U2p 724.24−

1EI

=

U3p 4 U2p V3p−( )⋅:= U3p 3323.61−1EI

=

U4p V5pU3p

4−:= U4p 830.9

1EI

=

We now determine the release deformations at 1 and 4 from the first and fourth compatibility relations.

V1hpU1p

6V1p−:= V1hp 484.24

1EI

=

V4hpU3p

4− U4p+⎛⎜⎝

⎞⎟⎠

V4p−:= V4hp 1715.141EI

=

When we compare the above answers with Vhp484.24

1715.14⎛⎜⎝

⎞⎟⎠

1EI

= we note that the values are the same

Thus, the compatibility conditions furnish the necessary values for the redundant basic forces to close thehinges at locations 1 and 4.

Page 378


The following figures show the bending moment distribution and the deformed shape of the structural modelfor the particular solution with the release deformations at 1 and 4.

80

80

moment diagram forparticular solution

deformed shape for particular solution (magnification factor = 15)

Page 379


We can continue this process for each homogeneous solution and obtain release deformations at locations1 and 4 for each self-stress state with unit redundant value. As we have mentioned earlier, these are thecolumns of the flexibility matrix Fxx. Finally, we scale the redundant values so that the release deformationsat 1 and 4 are zero, since there are physically no such releases at these locations. This means

BbarxT Fs⋅ Qp⋅ Fxx Qx⋅+ 0=

We can now determine the unique set of basic forces that satisfies equilibrium and compatibility under thegiven linear elastic deformation-force relation of the elements. With the given values for the redundant basicforces we obtain

Q Qp Bbarx Qx⋅+:= Q

85.01

30.89

30.89−

55.44

55.44−

13.36

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

=

and this results in the following moment diagram

55.44

30.89

85.01

30.89

Page 380


With the final basic forces we can determine the element deformations

V Fs Q⋅:= V

139.13

23.23−

78.14−

94.51

73.92−

667.8

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

1EI

=

for hand calculations it is better to proceed element by element, i.e.

Q1

Q2

Q3

Q4

Q5

Q6

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

Q:=V1

V2

⎛⎜⎝

⎞⎟⎠

La

6 EI⋅

2

1−

1−

2⎛⎜⎝

⎞⎟⎠

⋅Q1

Q2

⎛⎜⎝

⎞⎟⎠

⋅:= or, V1La

6 EI⋅2 Q1⋅ Q2−( )⋅:= V1 139.13

1EI

=

V2La

6 EI⋅2 Q2⋅ Q1−( )⋅:= V2 23.23−

1EI

=

V3

V4

⎛⎜⎝

⎞⎟⎠

Lb

6 EI⋅

2

1−

1−

2⎛⎜⎝

⎞⎟⎠

⋅Q3

Q4

⎛⎜⎝

⎞⎟⎠

⋅:= or, V3Lb

6 EI⋅2 Q3⋅ Q4−( )⋅:= V3 78.14−

1EI

=

V4Lb

6 EI⋅2 Q4⋅ Q3−( )⋅:= V4 94.51

1EI

=

V5Lc

3 EI⋅Q5⋅:= V5 73.92−

1EI

=

V6Ld

EAQ6⋅:= V6 667.8

1EI

=

It is worth noting that the deformation of the brace, which is a change of length is several times larger thanthe deformations of the beam elements, since the latter are angles, i.e. have units of length over length. Itis good to think of the element lengths as a rough estimate of this scaling.

Page 381


With the element deformations we can determine the global dof displacements as we have done earlier.We can use the same four compatibility relations as before

U1 1.25 V6⋅:= U1 834.751EI

=

U2 V2U1

6−:= U2 162.36−

1EI

=

U3 4 U2 V3−( )⋅:= U3 336.87−1EI

=

U4 V5U3

4−:= U4 10.3

1EI

=

and we can check the other two compatibility relations

U1

6V1− 0

1EI

=

U3

4− U4+⎛⎜⎝

⎞⎟⎠

V4− 01EI

= and conclude that they are, of course, satisfied!

The deformed shape of the structure with a magnification factor of 50 is shown in the following figure.

Page 382


How do we draw the deformed shape? In three steps: 1. locate the nodes with the determined translations(just show points, no black squares yet!). Make sure to respect the scale, e.g. in the drawing abovehorizontal translation should be a bit more than twice than vertical translation. 2. draw the chords andusing the element deformations sketch the deformed shape of each element noting that the angles at thenodes must be respected (e.g. right angle at node 2, continuous tangent at node 3, etc). 3. after step 2 itshould be easy to draw the little black squares according to the determined joint rotations (in an idealcase, the black squares should "fall in place" respecting the element tangents, but checking thedetermined rotation values (if we have bothered to calculate these!) is extra security.

Second load case: prestressing of brace

We would like to investigate now the case that the brace is prestressed or prestrained. We treat this as aninitial deformation problem in the force method of analysis, since it is much easier to understand it this way.We assume, thus, that the brace will be delivered longer or shorter than its required length of 10 units.Since we would like to customize the amount of prestressing so as to control either the final momentdistribution or the deformed shape of the structure under the earlier loading, we assume a unit value ofprestrain at the start.

The basic forces of the particular solution are all zero, since there is no external forces at the global dofs.

Qp

0

0

0

0

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:= but the deformations are not all zero, because of the prestrain of element d. We have

we use a value of 1/100 as a "unit" value for the initial prestrain, since a value of 1 wouldproduce extremely large forces and displacements.Vo

0

0

0

0

0

0.01

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

The release deformation at locations 1 and 4 on account of the initial deformation is

BbarxT− Fs Qp⋅ Vo+( )⋅

104.17

208.33⎛⎜⎝

⎞⎟⎠

1EI

= but, of course, it is easier to write BbarxT− Vo⋅

104.17

208.33⎛⎜⎝

⎞⎟⎠

1EI

=

Page 383


this result is very easy to interpret from the compatibility relations, since all element deformations except V6are zero. We recall

V1U1

6= V3 U2

U3

4−= V5

U3

4U4+=

V2U1

6U2+= V4

U3

4− U4+= V6 0.8 U1⋅=

and get

U1o 1.25 0.01⋅:= U1o 6251EI

=

U2oU1o

6−:= U2o 104.17−

1EI

=

U3o 4 U2o⋅:= U3o 416.67−1EI

=

U4oU3o

4−:= U4o 104.17

1EI

=

and the relations at locations 1 and 4 becomeU1o

6104.17

1EI

=

U3o

4− U4o+ 208.33

1EI

=

but we could have obtained these answers much faster from the geometry of the deformed shape

Page 384


We denote the release deformation due to prestressing (or, initial deformation) with Vho

Vho BbarxT− Vo⋅:= Vho

104.17

208.33⎛⎜⎝

⎞⎟⎠

1EI

=

In reality there is no release; to reduce the deformation at the release to zero we need the followingredundant force values

Vho− Fxx Qx⋅+ 0= Qx lsolve Fxx Vho,( ):= Qx21.75

5.76⎛⎜⎝

⎞⎟⎠

=

We conclude that initial deformations produce a self-stress state in a statically indeterminate structure. Thebasic force values of the structure for an initial deformation value of 0.01 in the brace are

Q Bbarx Qx⋅:= Q

21.75

11.51

11.51−

5.76

5.76−

6.93−

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

=

The bending moment diagram in the structure under an initial deformation value of 0.01 in the brace is

21.75

11.51

11.51 5.76

Note that support reactions arise under the load case of initial deformation. These are self-equilibrating.You should determine these and check global equilibrium for practice.It is worth noting that Fxx in the compatibility equation depends on the absolute value of the stiffness. Bycontrast, Vho does not ! Thus, the redundant basic forces and the resulting self-stress state are directlyproportional to the absolute value of the stiffness.

Conclusion: In a statically indeterminate structure self stress-states generated by initial deformations areproportional to the absolute value of the stiffness ! The element deformations, however, only depend on therelative stiffness and not on the absolute stiffness value. The same is then true for the displacements!

Page 385


There is another aspect that if of some interest in prestressed structures. Even though the initialdeformation is positive (0.01 units), the axial force in the brace is compressive. We should think of this asthe necessary force to "push back" the brace to fit into the frame, which at the same time gets "pushedout" at the location of the brace to meet the end of it. This is synergy at work!!

Interestingly the final deformation of the brace element is

V6Ld

EAQ6⋅ 0.01+:= V6 0.0031= i.e. only 31% of its initial value

this "recoiling" of the brace from its initial deformation value is known as "prestress loss during installation".It depends on the ratio EI/EA since this ratio controls the contribution of the frame vs. the brace in the gapclosing operation.

Calibration of prestressing (prestraining) valueLet us determine how much is the horizontal translation for the initial deformation case. This is ratherstraightforward.

from V645

U1⋅= we get U1 1.25 V6⋅:= U1 191.91EI

=

We recall that the horizontal translation for the loading of the first load case was 834.751EI⋅

this means that if we selected the prestressing (prestraining) value just right we could eliminate thehorizontal translation under the applied forces of 30 units at dof 4 and 20 units at dof 2. Let us summarizefirst

under the applied forces the horizontal translation is 834.751EI⋅

under an initial deformation value of 0.01 in the brace the horizontal translation is 191.91EI⋅

thus, if we select an initial deformation of834.75191.9

− 0.01⋅ 0.0435−=

we can cancel out the horizontal translation under the applied forces.

We determine the corresponding forces by superposition of the basic forces from the two load cases, i.e.

85.01

30.89

30.89−

55.44

55.44−

13.36

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

4.35−( )

21.75

11.51

11.51−

5.76

5.76−

6.93−

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

⋅+

9.6−

19.18−

19.18

30.38

30.38−

43.51

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

=

Page 386


The corresponding moment diagram is shown in the following figure

30.38

9.6

19.18

We observe now a better balance between positive and negative bending moment values in the girder(note that the maximum value is now 30.38 in the girder and 19.18 in the column, while it was 55.44 in thegirder and 85.01 in the column under the applied forces alone. Thus, prestressing allows us to fine tunethe moment distribution to our heart's content and achieve more economical values and, thus, smaller andlighter cross sections. We also noted that it helps control the displacements. Not only is the horizontaltranslation eliminated, but the vertical translation at girder midspan is -336.87 1/EI without prestressand -136.5 1/EI with prestress (value from computer analysis, since it was not obtained above).

Page 387


Load case of distributed load on girder

We would like to investigate the load case of a uniformly distributed element load acting on the girder of thebraced frame. The following figure shows the loading

a

b c

d

w=5

4 4

6

The following figure represents schematically the equivalent nodal forces for the distributed element load, aswell as the effect of the distributed element load on the element deformations.

a

b c

d

w=5

4 4

6

10 10 10 10

10 20 10

Equilibrium equations at free dofs

0Q1 Q2+

60.8 Q6⋅+=

0 Q2 Q3+=

0Q3 Q4+

4−

Q5

4+ 20+=

0 Q4 Q5+=

or,

0Q1 Q2+

60.8 Q6⋅+=

0 Q2 Q3+=

20−Q3 Q4+

4−

Q5

4+=

0 Q4 Q5+=

Page 388


The particular solution is now (compare also Example 6)

0 Q2 Q3+= Q2 80−=

20−Q3 0( )+

4−

Q5

4+= Q3 80= Qp

0

80−

80

0

0

1006

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

0 0( ) Q5+= Q5 0=

Q61006

=00( ) Q2+

60.8 Q6⋅+=

We check the solution: Bf Qp⋅

0

0

20−

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

= and it is correct.

the following figure shows the moment diagram under the particular solution

80

80

In determining the deformations under the particular solution we should make sure to include thedeformations under the distributed element load in vector Vo. This effect is indicated with the gray hatchedarea in the moment diagram above.

Page 389


the element deformations for the particular solution are

w 5:= Vo1EI

0

0

w Lb3⋅

24−

w Lb3⋅

24

w Lc3⋅

24−

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

⋅:= Vo

0

0

13.33−

13.33

13.33−

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

1EI

= and Fs Qp⋅ Vo+

80

160−

93.33

40−

13.33−

833.33

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

1EI

=

note that Vo is caused by the curvatures due to the gray hatched area in the moment diagram above

release deformations due to particular solution Bbarx− T Fs Qp⋅ Vo+( )⋅93.61

880.56⎛⎜⎝

⎞⎟⎠

1EI

=

these are shown in the following figure

Page 390


with Vhp Bbarx− T Fs Qp⋅ Vo+( )⋅:=

we get from the solution of the system of equations Qx lsolve Fxx Vhp,( ):= Qx4.6

31.81⎛⎜⎝

⎞⎟⎠

=

the final forces under the uniformly distributed element load are the superposition of NOS+1 stress states


4.6

16.39−

16.39

31.81

31.81−

2.46

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

=

the following figure shows the moment diagram under the uniformly distributed load

16.39

31.81

4.6

the maximum moment occurs in element c

the shear force at end j of element c is Vjw Lc⋅

2

Q5

Lc−:= Vj 17.95=

and the moment becomes maxMVj

2

2 w⋅:= maxM 32.23=

all that remains is to determine the support reactions and check global equilibrium. Take it as exercise!

Page 391

Script for Example 30a in CE220 class notes % frame with single brace under horizontal and vertical force with static matrix


force method of analysis La = 6; Lb = 4; Lc = 4; Ld = 10; % static (equilibrium) matrix (from Example 6) Bf = [1/La 1/La 0 0 0 (Lb+Lc)/Ld; 0 1 1 0 0 0 ; 0 0 -1/Lb -1/Lb 1/Lc 0 ; 0 0 0 1 1 0]; % set up collection of element flexibility matrices EI = 50000; EA = 10000; % for column and girder elements f{1} = La/(6*EI).*[2 -1; -1 2]; f{2} = Lb/(6*EI).*[2 -1; -1 2]; f{3} = Lc/(3*EI); % for brace element f{4} = Ld/EA; % collection of element flexibility matrices Fs = blkdiag (f{:}); % invoke single function for force method of analysis [Bbar,Bvbar,F] = ForceMethod(Bf,Fs);

First load case: applied nodal forces Pf = [30;0;-20;0]; % determine final forces Q = Bbar*Pf; % display final forces disp('basic forces Q under applied nodal forces') disp(Q); basic forces Q under applied nodal forces 8.5006e+001 3.0885e+001 -3.0885e+001 5.5443e+001 -5.5443e+001 1.3356e+001

determine displacements Uf = F*Pf; % display displacements disp('free dof displacements Uf under applied nodal forces') disp(Uf); % store displacement value at first dof for later use U1_1 = Uf(1); free dof displacements Uf under applied nodal forces 1.6695e-002 -3.2472e-003 -6.7375e-003 2.0590e-004

Page 392

Second load case: initial deformation % define initial deformation vector for second load case (prestress) V0 = zeros(6,1); V0(6) = 0.01; % determine final forces Q = Bvbar*V0; % display final forces disp('basic forces Q under initial deformation') disp(Q); basic forces Q under initial deformation 2.1748e+001 1.1514e+001 -1.1514e+001 5.7569e+000 -5.7569e+000 -6.9296e+000

determine displacements Uf = Bbar'*V0; % display displacements disp('free dof displacements Uf under initial deformation') disp(Uf); % select displacement value of first dof U1_2 = Uf(1); free dof displacements Uf under initial deformation 3.8380e-003 -6.1407e-004 -9.2111e-004 7.6759e-005

Third load case: applied nodal forces and prestressing % determine prestressing force so as to cancel horizontal translation under applied nodal forces V0(6) = -U1_1/U1_2*0.01; % determine final forces Q = Bbar*Pf+ Bvbar*V0; % display final forces disp('basic forces Q under applied nodal forces and prestressing') disp(Q); basic forces Q under applied nodal forces and prestressing -9.6000e+000 -1.9200e+001 1.9200e+001 3.0400e+001 -3.0400e+001 4.3500e+001

determine displacements Uf = F*Pf + Bbar'*V0; % display displacements disp('free dof displacements Uf under applied nodal forces and prestressing') disp(Uf);

Page 393

free dof displacements Uf under applied nodal forces and prestressing 0 -5.7600e-004 -2.7307e-003 -1.2800e-004

Fourth load case: distributed load on girder w = 5; V0 = [0;0;-w*Lb^3/(24*EI);w*Lb^3/(24*EI);-w*Lc^3/(24*EI);0]; Pf = zeros(4,1); Pfw = [0;0;20;0]; % determine final forces Q = Bbar*(Pf-Pfw)+ Bvbar*V0; % display final forces disp('basic forces Q under distributed girder load') disp(Q); % determine displacements Uf = F*(Pf-Pfw) + Bbar'*V0; % display displacements disp('free dof displacements Uf under distributed girder load') disp(Uf); basic forces Q under distributed girder load 4.5987e+000 -1.6389e+001 1.6389e+001 3.1806e+001 -3.1806e+001 2.4563e+000 free dof displacements Uf under distributed girder load 3.0704e-003 -1.2593e-003 -4.0222e-003 -1.0926e-004

Page 394

Script for Example 30b in CE220 class notes % frame with single brace under horizontal and vertical force with full set of basic forces


define model geometry and element types % specify node coordinates (could only specify non-zero terms) XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 0 6]; % second node, etc XYZ(3,:) = [ 4 6]; % XYZ(4,:) = [ 8 6]; % connectivity array CON {1} = [ 1 2]; CON {2} = [ 2 3]; CON {3} = [ 3 4]; CON {4} = [ 1 4]; % boundary conditions (1 = restrained, 0 = free) BOUN(1,:) = [1 1 1]; BOUN(4,:) = [0 1 0]; % specify element type [ElemName{1:3}] = deal('Lin2dFrm'); % 2d linear beam element ElemName{4} = 'LinTruss'; % linear truss element

create Model Model = Create_SimpleModel (XYZ,CON,BOUN,ElemName);

element properties for el=1:3 ElemData{el}.E = 1000; ElemData{el}.A = 1e6; % "inextensible" ElemData{el}.I = 50; end ElemData{4}.E = 1000; ElemData{4}.A = 10; % area for brace element

force method of analysis

static(equilibrium) matrix at free dofs

B = B_matrix(Model); Bf = B(1:Model.nf,:); % collection of element flexibility matrices Fs = Fs_matrix(Model,ElemData); % invoke single function for force method of analysis [Bbar,Bvbar,F] = ForceMethod(Bf,Fs);

First load case: applied nodal forces % specify nodal forces Pe(2,1) = 30; Pe(3,2) = -20;

Page 395

Loading = Create_Loading (Model,Pe); Pf = Loading.Pref; % determine final forces Q = Bbar*Pf; % display final forces disp('basic forces Q under applied nodal forces') disp(Q); basic forces Q under applied nodal forces -6.1393e+000 8.5006e+001 3.0885e+001 -1.0685e+001 -3.0885e+001 5.5443e+001 -1.0685e+001 -5.5443e+001 0 1.3356e+001

plot bending moment diagram Create_Window(0.80,0.80); Plot_Model (Model); Plot_2dMomntDistr (Model,[],Q,2);

determine displacements Uf = F*Pf; % display displacements disp('free dof displacements Uf under applied nodal forces') disp(Uf); free dof displacements Uf under applied nodal forces 1.6695e-002 -3.6836e-008 -3.2472e-003 1.6695e-002 -6.7375e-003 2.0591e-004 1.6695e-002 2.4236e-003

open window and plot chords and then deformed shape % put free dof displacement values in complete displacement vector

Page 396

U = zeros(Model.nt,1); U(1:Model.nf) = Uf; Create_Window(0.80,0.80); Plot_Model (Model); MAGF = 100; Plot_Model (Model,U); Plot_DeformedStructure (Model,[],U); % store displacement value at first dof for later use U1_1 = Uf(1);

Second load case: initial deformation

define initial deformation vector for second load case (prestress)

ElemData{4}.e0 = 1e-3; V0 = V0_vector(Model,ElemData); % determine final forces Q = Bvbar*V0; % display final forces disp('basic forces Q under initial deformation') disp(Q); basic forces Q under initial deformation 1.4392e+000 2.1748e+001 1.1514e+001 5.5437e+000 -1.1514e+001 5.7569e+000 5.5437e+000 -5.7569e+000 0 -6.9296e+000


Page 397

determine displacements Uf = Bbar'*V0; % display displacements disp('free dof displacements Uf under initial deformation') disp(Uf); % select displacement value of first dof U1_2 = Uf(1); free dof displacements Uf under initial deformation 3.8379e-003 8.6354e-009 -6.1407e-004 3.8380e-003 -9.2110e-004 7.6758e-005 3.8380e-003 3.0703e-004

open window and plot chords and then deformed shape % put free dof displacement values in complete displacement vector U = zeros(Model.nt,1); U(1:Model.nf) = Uf; Create_Window(0.80,0.80); Plot_Model (Model); Plot_Model (Model,U); Plot_DeformedStructure (Model,[],U);

Page 398

Third load case: applied nodal forces and prestressing

determine prestressing force so as to cancel horizontal translation under applied nodal forces

ElemData{4}.e0 = -U1_1/U1_2*1e-3; V0 = V0_vector(Model,ElemData); % determine final forces Q = Bbar*Pf+ Bvbar*V0; % display final forces disp('basic forces Q under applied nodal forces and prestressing') disp(Q); basic forces Q under applied nodal forces and prestressing -1.2400e+001 -9.5999e+000 -1.9200e+001 -3.4800e+001 1.9200e+001 3.0400e+001 -3.4800e+001 -3.0400e+001 0 4.3500e+001


determine displacements Uf = F*Pf + Bbar'*V0; disp('free dof displacements Uf under applied nodal forces and prestressing') disp(Uf); free dof displacements Uf under applied nodal forces and prestressing 0 -7.4400e-008 -5.7600e-004 -1.3920e-007 -2.7307e-003 -1.2799e-004 -2.7840e-007 1.0880e-003

Page 399

plot deformed shape % put free dof displacement values in complete displacement vector U = zeros(Model.nt,1); U(1:Model.nf) = Uf; % open window and plot chords and then deformed shape Create_Window(0.80,0.80); Plot_Model (Model); Plot_Model (Model,U); Plot_DeformedStructure (Model,[],U);

Fourth load case: distributed load on girder ElemData{4}.e0 = 0; % clear prestressing ElemData{2}.w = [0;-5]; ElemData{3}.w = [0;-5]; V0 = V0_vector(Model,ElemData); Pf = zeros(Model.nf,1); clear Pe; % specify equivalent nodal forces for distributed element loading Pe(2,2) = 10; Pe(3,2) = 20; Pe(4,2) = 10; Loading = Create_Loading(Model,Pe); Pfw = Loading.Pref; % determine final forces Q = Bbar*(Pf-Pfw)+ Bvbar*V0; % display final forces disp('basic forces Q under distributed girder load') disp(Q); basic forces Q under distributed girder load -2.2049e+001 4.5987e+000 -1.6389e+001 -1.9650e+000 1.6389e+001 3.1806e+001 -1.9650e+000 -3.1806e+001 0 2.4563e+000

plot bending moment diagram Create_Window(0.80,0.80); Plot_Model (Model);

Page 400

Plot_2dMomntDistr (Model,ElemData,Q,2);

determine displacements Uf = F*(Pf-Pfw) + Bbar'*V0; % display displacements disp('free dof displacements Uf under distributed girder load') disp(Uf); free dof displacements Uf under distributed girder load 3.0703e-003 -1.3229e-007 -1.2592e-003 3.0703e-003 -4.0223e-003 -1.0924e-004 3.0703e-003 1.6963e-003

plot deformed shape % put free dof displacement values in complete displacement vector U = zeros(Model.nt,1); U(1:Model.nf) = Uf; % open window and plot chords and then deformed shape Create_Window(0.80,0.80); Plot_Model (Model); Plot_Model (Model,U); Plot_DeformedStructure (Model,ElemData,U);

Page 401

CE220 - Theory of Structures Ex 31 - Force method for continuous beam on flexible support © Prof. Filip C. Filippou, 2000

Example 31 - Force Method for Continuous Beam on Flexible Support

We consider the following model of a continuous beam over two spans with flexible middle support.The beam carries a uniformly distributed load over both spans, as shown in the following figure.

a b1

2 3

18 15

4

w=10 w=10

Geometric and material properties are as follows: La 18:= Lb 15:=

beam flexural stiffness EI 100000:= middle support axial stiffness k 5000:=

There are two free dofs or equilibrium equations and three basic forces, as shown in the following figure

a b1 3

4

1Q 2Q

3Q

12

18 15

The degree of static indeterminacy is thus NOS=1. The equilibrium equations are

P1Q1

La−

Q2

Lb+ Q3+=

P2 Q1 Q2+=

We select Q1 as redundant basic force.

Page 402


Particular solution of equilibrium equations for primary system

We set Q1 = 0 and solve the equilibrium equations for the other basic forces under the given loading

The following figure represents schematically the equivalent nodal forces for the distributed element load, aswell as the effect of the distributed element load on the element deformations.

a b

18 15

w=10

9090 75

75165

00La

−Q2

Lb+ Q3+ 165+= 165−

Q2

LbQ3+=

Qp

0

0

165−

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

:=or

0 0( ) Q2+= 0 Q2=

In determining the deformations under the particular solution we should make sure to include thedeformations under the distributed element load in vector Vo.

w 10:= Vo1EI

w La3⋅

24

w Lb3⋅

24−

0

⎛⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎠

⋅:= Fs1EI

La

3

0

0

0

Lb

3

0

0

0

EIk

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

⋅:= V Fs Qp⋅ Vo+:= V

2430

1406.25−

3300−

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

1EI

=

The compatibility equations are V1U1

La− U2+=

V2U1

LbU2+=

V3 U1=

We use the last two to solve for the displacements and substitute into the first to find the compatibility erroror discontinuity under the applied loading. We get

Page 403


U1 V3:= U1 3300−1EI

=

U2 V2

U1

Lb−:= U2 1186.25−

1EI

=

and from the first compatibility relation we getU1

La− U2+ V1− 3432.92−

1EI

=

which is the hinge rotation at location 1. The compatibility error is the negative value of the hinge rotation(or the hinge closing value), and we will confirm this later. This hinge closing value is shown in the followingfigure which shows the moment diagram and the deformed shape of the continuous beam.

M(x)

13432.9EI

Homogeneous solution of equilibrium equations for primary system (self-stress state)

Determine the basic element forces for Q1 1= without applied loading (homogeneous solution)

a b1

23

4

2Q

3Q

1 1=Q

a b1

2

3xxF

bending moment distribution

18 15

Page 404


The equilibrium equations are now: 0Q1

La−

Q2

Lb+ Q3+= Q3

1La

1Lb

+=

0 Q1 Q2+= Q2 1−=

and thus, Bbarx

1

1−

1La

1Lb

+

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:= Bbarx

1

1−

0.122

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

=

the compatibility error under the self-stress state is BbarxT Fs⋅ Bbarx⋅ 11.3

1EI

=

which is also easy to confirm from the compatibility relations with V Fs Bbarx⋅:=

U1 V3:= U1 2.4441EI

=

U2 V2

U1

Lb−:= U2 5.163−

1EI

=


La− U2+ V1− 11.3−

1EI

=

which gives again the hinge rotation; the compatibility error is the negative value of the hinge rotation

defining Fxx BbarxT Fs⋅ Bbarx⋅:=

we can solve the compatibility condition BbarxT Fs Qp⋅ Vo+( )⋅ Fxx Qx⋅+ 0= we obtain the value

of the redundant

QxBbarx

T Fs Qp⋅ Vo+( )⋅Fxx

−:= Qx 303.83−=

the final forces under the uniformly distributed element load are the superposition of NOS+1 stress states


303.83−

303.83

202.13−

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

=

and the final deformations are V Fs Q⋅ Vo+:= V

607.01

112.91

4042.7−

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

1EI

=

Page 405


with these we can obtain the displacements at the free dof's from the compatibility relations

U1 V3:= U1 4042.7−1EI

=

U2 V2

U1

Lb−:= U2 382.42

1EI

=


La− U2+ V1− 0

1EI

= which means that we havea compatible set ofdeformations, as expected!

the following figure shows the bending moment diagram and the deformed shape

IP IP

405

303.8

281.25

the moment at midspan can be obtained fromw La

2⋅

8303.83

2− 253.09= for span a

w Lb2⋅

8303.83

2− 129.34= for span b

the maximum moment, however, occurs closer to the outer supports and have the following value

shear at left support Vlw La⋅

2303.83

La−:= maxMa

Vl2

2 w⋅:= maxMa 267.33= @

Vl

w7.31=

shear at right support Vrw Lb⋅

2303.83

Lb−:= maxMb

Vr2

2 w⋅:= maxMb 149.85= @

Vr

w5.47=

Page 406


Determination of vertical deflection at midspan of element a

We wish to determine the vertical deflection at midspan of element a. To this end we insert a node at thatlocation in element a. We have the following compatibility relations from the following figure.

18 15

m1

m21

2a1 a2

vja1Um1

9− Um2+=

via2Um1

9Um2+

U1

9−= subtract the first equation from the second to get

via2 vja1− 2Um1

9⋅

U1

9−=

with vja1253.09( ) 9( )⋅

3 EI⋅w 93⋅24 EI⋅

+:= via2253.09−( ) 9( )⋅

3 EI⋅303.83−( ) 9( )⋅

6 EI⋅−

w 93⋅24 EI⋅

−:=

and finally Um192

via2 vja1−( )⋅U1

2+:= Um1 9537.68−

1EI

=

the effect of the flexible middle support on the midspan deflection is clear from the following figure

with the given flexural stiffness value the vertical deflection is Um1 95.38− 10 3−=

as compared with the displacement at dof 1 U1 40.43− 10 3−=

Page 407

Script for Example 31a in CE220 class notes % continuous beam on flexible middle support with static matrix


define span lengths La = 18; Lb = 15;

force method of analysis

static (equilibrium) matrix

Bf = [-1/La 1/Lb 1; 1 1 0]; % set up collection of element flexibility matrices EI = 100000; k = 5000; % for beam elements f{1} = La/3/EI; f{2} = Lb/3/EI; % for spring element f{3} = 1/k; % collection of element flexibility matrices Fs = blkdiag (f{:}); % invoke single function for force method of analysis [Bbar,Bvbar,F] = ForceMethod(Bf,Fs);

Load case: distributed load on girder w = 10; V0 = [w*La^3/(24*EI);-w*Lb^3/(24*EI);0]; Pf = zeros(2,1); Pfw = [165;0];

determine final forces Q = Bbar*(Pf-Pfw)+ Bvbar*V0; % display final forces disp('basic forces Q under distributed load') disp(Q); basic forces Q under distributed load -3.0383e+002 3.0383e+002 -2.0213e+002

determine displacements Uf = F*(Pf-Pfw) + Bbar'*V0; % display displacements disp('free dof displacements Uf under distributed load') disp(Uf); free dof displacements Uf under distributed load -4.0427e-002 3.8242e-003

Page 408

Script for Example 31b in CE220 class notes % continuous beam on flexible middle support (with full set of basic forces)


define model geometry and element types La = 18; Lb = 15; % specify node coordinates (could only specify non-zero terms) XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ La 0]; % second node, etc XYZ(3,:) = [ La+Lb 0]; % XYZ(4,:) = [ La -1]; % connectivity array CON {1} = [ 1 2]; CON {2} = [ 2 3]; CON {3} = [ 2 4]; % boundary conditions (1 = restrained, 0 = free) BOUN(1,:) = [1 1 0]; BOUN(3,:) = [0 1 0]; BOUN(4,:) = [1 1 1]; % specify element type [ElemName{1:2}] = deal('Lin2dFrm'); % 2d linear beam element ElemName{3} = 'LinTruss'; % linear truss element


element properties for el=1:2 ElemData{el}.E = 1000; ElemData{el}.A = 1e6; % "inextensible" ElemData{el}.I = 100; end ElemData{3}.E = 1000; ElemData{3}.A = 5; % "area" for spring element to get k=EA

force method of analysis % static(equilibrium) matrix at free dofs B = B_matrix(Model); Bf = B(1:Model.nf,:); % collection of element flexibility matrices Fs = Fs_matrix(Model,ElemData); % invoke single function for force method of analysis [Bbar,Bvbar,F] = ForceMethod(Bf,Fs);

Load case: distributed load on continuous beam ElemData{1}.w = [0;-10]; ElemData{2}.w = [0;-10]; V0 = V0_vector(Model,ElemData); Pf = zeros(Model.nf,1);

Page 409

% specify equivalent nodal forces for distributed element loading Pe(1,2) = 90; Pe(2,2) = 165; Pe(3,2) = 75; Loading = Create_Loading(Model,Pe); Pfw = Loading.Pref; % determine final forces Q = Bbar*(Pf-Pfw)+ Bvbar*V0; % display final forces disp('basic forces Q under distributed girder load') disp(Q); basic forces Q under distributed girder load 0 0 -3.0383e+002 0 3.0383e+002 0 -2.0213e+002

plot bending moment diagram Create_Window(0.80,0.80); Plot_Model (Model); Plot_2dMomntDistr (Model,ElemData,Q);

determine displacements Uf = F*(Pf-Pfw) + Bbar'*V0; % display displacements disp('free dof displacements Uf under distributed girder load') disp(Uf); free dof displacements Uf under distributed girder load -1.7431e-002 0 -4.0427e-002 3.8242e-003 0 9.1619e-003

Page 410

plot deformed shape % put free dof displacement values in complete displacement vector U = zeros(Model.nt,1); U(1:Model.nf) = Uf; % open window and plot chords and then deformed shape MAGF = 30; Create_Window(0.80,0.80); Plot_Model (Model); Plot_Model (Model,U); Plot_DeformedStructure (Model,ElemData,U);

Page 411

CE220 - Theory of Structures Example 32 - Frame with inextensible elements © Prof. Filip C. Filippou, 2000

Example 32 - Force method for frame with inextensible elements and brace

Problem description

In this example we will analyze the frame with inextensible elements and a brace in the following figure. It isassumed that elements a through d are inextensible and have flexural stiffness EI, while the brace elemente has axial stiffness EA.

12

3

4a

b c

d

5

6 8 8

10

6

e

La 6:=

Lb 10:=

Lc 10:=

Ld 10:=

Le 16:=

Under the assumption that the elements a through d are inextensible, their axial flexibility is zero. Thismeans that axial forces do not cause axial deformations, so that the latter have no contribution to thecompatibility conditions of the force method. It is, therefore, advisable to avoid solving for the axial forceswith the equilibrium equations. To this end we make use of the principle of virtual displacements: we usevirtual displacement fields with inextensible elements a through d in setting up the equilibrium equations.We have four equilibrium equations in 6 unknown basic forces as shown in the following figures.

1

2

3

41Q 2Q 3Q

5Q

4Q6Q

Page 412


for the translation dofs we have the following relation between global dof displacements, chord rotationsand relative displacements between nodes or between node tangents and chords.

a

b c

d

ICbICa ICc

6 8 8

10

6

dof 2

1

2

θa26

:=

θb18

−:=

θc18

−:=

θd 0:=

Af2⟨ ⟩

θa−

θb−

θc−

θd−

θd−

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

ICd

ICb

ICa

ICc

b

d

a

c

6 8 8

10

6

dof 3

1

4/3

θa4

3 6⋅:=

θb16

−:=

θc 0:=

θd1−

10:=

Af3⟨ ⟩

θa−

θb−

θc−

θd−

θd−

1

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

Page 413


with the following values of the compatibility matrix for the translation dofswe can write all equilibrium equations

1Q 2Q 3Q

5Q

4Q6Q

Af2⟨ ⟩

13

−

18

18

0

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

= Af3⟨ ⟩

29

−

16

0

110

110

1

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

P1 Q1 Q2+=

P2Q1

3−

Q2

8+

Q3

8+=

P32Q1

9−

Q2

6+

Q4

10+

Q5

10+ Q6+=

P4 Q3 Q4+=

We select Q2 and Q5 as redundant basic forces and obtain first the homogeneous solution (thus makingalso sure that the selection results in a primary structure that is stable)

12

3

4a

b c

d

5

e

1xQ

2xQ

Page 414


Homogeneous solution of equilibrium equations (self-stress states)

we solve P 0= Q2 1= Q5 0=

0 Q1 1( )+= Q1 1−=

0Q1

3−

18

+Q3

8+= Q3

113

−=

02Q1

9−

16

+Q4

10+

010

+ Q6+= Q63445

−=

Q4113

=0 Q3 Q4+=

then we solve P 0= Q2 0= Q5 1=

0 Q1 0( )+= Q1 0=

0Q1

3−

08

+Q3

8+= Q3 0=

02Q1

9−

06

+Q4

10+

110

+ Q6+= Q6110

−=

Q4 0=0 Q3 Q4+=

The general homogeneous solution of the equilibrium equations is therefore Bbarx

1−

1

113

−

113

0

3445

−

0

0

0

0

1

110

−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

with the equilibrium matrix we confirm the answer

Bf

1

13

−

29

−

0

1

18

16

0

0

18

0

1

0

0

110

1

0

0

110

0

0

0

1

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:= Bf Bbarx⋅

0

0

0

0

0

0

0

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

= good!

Page 415


We will now solve for two load cases by determining the particular solution of the equilibrium equations andsatisfying the compatibility conditions for the element deformations.

1. Load case: concentrated force

12

3

4a

b c

d

5

6 8 8

10

6

e

20

particular solution for applied loading

Q2 0= Q5 0=

0 Q1 0( )+= Q1 0=

20−Q1

3−

08

+Q3

8+= Q3 160−=

Qp

0

0

160−

160

0

16−

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

02Q1

9−

06

+Q4

10+

010

+ Q6+= Q6 16−=

0 Q3 Q4+= Q4 160=

with this we are ready to determine the element deformations and then set up the compatibility conditions todetermine the values of the redundant basic forces. We have

EI 40000:= EA 10000:=

Page 416


set up collection of element flexibility matrices Fs1EI

La

3

0

0

0

0

0

0

Lb

3

0

0

0

0

0

0

Lc

3

0

0

0

0

0

0

Ld

3

Ld

6−

0

0

0

0

Ld

6−

Ld

3

0

0

0

0

0

0

LeEIEA⋅

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

⋅:=

The deformations under the particular solution are Fs Qp⋅

0

0

533.33−

533.33

266.67−

1024−

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

1EI

=

and the compatibility conditions are not satisfied, because BbarxT Fs⋅ Qp⋅

4684.8

164.27−

⎛⎜⎝

⎞⎟⎠

1EI

=

The deformations under each homogeneous solution are: Fs Bbarx⋅

2−

3.33

12.22−

12.22

6.11−

48.36−

0

0

0

1.67−

3.33

6.4−

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

1EI

=

each of these deformation states does not satisfy compatibility either as we can tell immediately from


131.498

1.276−

1.276−

3.973⎛⎜⎝

⎞⎟⎠

1EI

=

Page 417


with the definitions Vhp BbarxT− Fs⋅ Qp⋅:= Fxx Bbarx

T Fs⋅ Bbarx⋅:=

we get from the solution of the system of equations Qx lsolve Fxx Vhp,( ):= Qx35.34−

30⎛⎜⎝

⎞⎟⎠

=

the final stress state is the superposition of NOS+1 stress states (one particular and NOS homogeneous)


35.34

35.34−

30.44−

30.44

30

7.7

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

=

the following figure shows the resulting moment diagram under the applied force of 20 units

35.34

30.44

30

Page 418


the element deformations are V Fs Q⋅:= V

70.67

117.78−

101.46−

51.46

49.27

492.67

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

1EI

=

and we can use the compatibility relations to determine the two translations and then draw the deformedshape of the structural model under the applied force of 20 units. We have

V1 U1U2

3− 2

U3

9⋅−=

V2 U1U2

8+

U3

6+=

V3U2

8U4+=

BfT

1

1

0

0

0

0

13

−

18

18

0

0

0

29

−

16

0

110

110

1

0

0

1

1

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

V4U3

10U4+=

V5U3

10=

V6 U3=

from the above relations it is very straightforward to conclude that U3 V6:= U3 492.671EI

=

and from equations 3 and 4 V3 V4−U2

8

U3

10−= U2 8 V3 V4−( )⋅ 0.8 U3⋅+:= U2 829.2−

1EI

=

and with this information and the given element deformations we should be able to draw the deformed shapeof the structure.

Note: the horizontal translation of node 3 can be found from U2 and U3 according to the earlier derivationsfor the compatibility matrix

Uhor34

U2⋅ U3+:= Uhor 129.23−1EI

= it moves about 1/6.5 of the vertical translation to the left!

the vertical translation of node 2 is Uver 2 U2⋅43

U3⋅+:= Uver 1001.5−1EI

= downward

Page 419


the deformed shape is shown in the following figure; it is clear from the tangent continuity that node 2rotates clockwise (CW) and node 3 counter clockwise (CCW). The rotation of node 4 is so small that it isnot possible to make out from the deformed shape. Indeed, numerical calculations reveal that it is roughly50 times smaller than the rotation at the other nodes and thus, insignificant in the scale of the figure.

CW

CCW

?

Page 420


2. Load case: uniformly distributed load in element b

12

3

4a

b c

d

5

6 8 8

10

6

e

w=5

w 5:=

Equivalent nodal forces (on the left hand side of equilibrium equations)

12

3

4a

b c

d

5

6 8 8

10

6

e

1520

1520

equivalent nodal forces forconstrained translation dofs

dof 2:

20−( ) 2⋅ 20−( ) 1⋅+ 1534⋅+ 48.75−=

dof 3:

20−( )43⋅ 15 1⋅+ 11.67−=

thus,

Pf

0

48.75−

11.667−

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:=

Page 421


particular solution for applied loading

Q2 0= Q5 0=

0 Q1 0( )+= Q1 0=

48.75−Q1

3−

08

+Q3

8+= Q3 390−=

Qp

0

0

390−

390

0

50.667−

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

11.67−2Q1

9−

06

+Q4

10+

010

+ Q6+= Q6 50.667−=

0 Q3 Q4+= Q4 390=

The deformations underthe particular solution are Vo

0

w Lb3⋅

24 EI⋅−

0

0

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= V Fs Qp⋅ Vo+:= V

0

208.33−

1300−

1300

650−

3242.69−

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

1EI

=

with the definitions Vhp Bbarx− T Fs Qp⋅ Vo+( )⋅:= Fxx BbarxT Fs⋅ Bbarx⋅:=

we get from the solution of the system of equations Qx lsolve Fxx Vhp,( ):= Qx89.03−

53.4⎛⎜⎝

⎞⎟⎠

=

the final stress state is the superposition of NOS+1 stress states (one particular and NOS homogeneous)


89.03

89.03−

63.57−

63.57

53.4

11.26

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

=

the element deformations are V Fs Q⋅ Vo+:= V

178.05

505.09−

211.89−

122.89

72.05

720.52

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

1EI

=

Page 422


the following figure shows the moment diagram; can you determine the maximum moment in element b?

89.03

63.57

53.4

Page 423


We determine again the values for the translation dofs from the compatibility relations

U3 V6:= U3 720.521EI

=

U2 8 V3 V4−( )⋅ 0.8 U3⋅+:= U2 2101.84−1EI

=

and derive the other translation values from the constraints used in setting up the compatibility matrix

for the horizontal translation of node 3 we have Uhor34

U2⋅ U3+:= Uhor 855.87−1EI

=

the vertical translation of node 2 is Uver 2 U2⋅43

U3⋅+:= Uver 3243−1EI

= downward

CW

CCW

CCW

the effect of the distributed load on the deformation V2 and the deformed shape of element b is verypronounced.

can you determine the deflection (horizontal and vertical translation) of element b at its midspan?

Page 424

Script for Example 32a in CE220 class notes % frame with inextensible elements and brace using constraint transformation for inextensibility


define model geometry and element types % specify node coordinates (could only specify non-zero terms) XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 6 0]; % second node, etc XYZ(3,:) = [ 14 6]; % XYZ(4,:) = [ 22 0]; XYZ(5,:) = [ 22 -10]; % connectivity array CON {1} = [ 1 2]; CON {2} = [ 2 3]; CON {3} = [ 3 4]; CON {4} = [ 4 5]; CON {5} = [ 2 4]; % boundary conditions (1 = restrained, 0 = free) BOUN(1,:) = [1 1 0]; BOUN(5,:) = [1 1 1]; % specify element type [ElemName{1:4}] = deal('Lin2dFrm'); % 2d linear beam element ElemName{5} = 'LinTruss'; % linear truss element


element properties for el=1:4 ElemData{el}.E = 1000; ElemData{el}.A = 1; % any value, since it will be constrained ElemData{el}.I = 40; end ElemData{5}.E = 1000; ElemData{5}.A = 10; % area for brace element

force method of analysis with inextensible constraints % set up compatibility matrix A = A_matrix(Model); % set up constraint conditions LC_Eqs = LinConEqs(Model,1:4); % re-order dofs to select translation dofs in specific order pick_dof = [Model.DOF(3,2), Model.DOF(4,1)]; dof_reord = [setdiff(1:Model.nt,pick_dof) pick_dof]; Ac = Ac_matrix (LC_Eqs,dof_reord); % compatibility matrix under inextensible constraints Aftild = A*Ac; % specify release location for subsequent index of deformations ElemData{2}.Release = ['n';'y']; % select deformation modes only

Page 425

Id = Idef_index(Model,ElemData,1:4); % retain only deformation modes in compatibility matrix Aftild = Aftild(Id,:); % equilibrium matrix w/o axial forces is the transpose of the compatibility matrix Bf = Aftild'; % remove Release field from ElemData before setting up flexibility matrix ElemData{2} = rmfield(ElemData{2},'Release'); % collection of element flexibility matrices Fs = Fs_matrix(Model,ElemData); Fs = Fs(Id,Id); % invoke single function for force method of analysis [Bbar,Bvbar,F] = ForceMethod(Bf,Fs);

First load case: applied nodal forces % specify nodal forces Pe(3,2) = -20; Loading = Create_Loading (Model,Pe); % make sure to define applied force vector at constrained dofs (remove restrained dofs from Ac matrix) Pf = Ac(1:Model.nf,:)'*Loading.Pref; % determine final forces Q = Bbar*Pf; % display final forces disp('basic forces Q under applied nodal forces') disp(Q); basic forces Q under applied nodal forces 0 3.5335e+001 -3.5335e+001 0 -3.0437e+001 3.0437e+001 2.9999e+001 7.6979e+000

determine displacements Uf = F*Pf; % display displacements disp('free dof displacements Uf under applied nodal forces') disp(Uf); free dof displacements Uf under applied nodal forces -5.0563e-003 -2.4062e-003 -2.0730e-002 3.8595e-003 1.2317e-002 5.4813e-005

Second load case: distributed load on girder ElemData{2}.w = [0;-5]; V0 = V0_vector(Model,ElemData); V0 = V0(Id); Pf = zeros(Model.nf,1); clear Pe; % specify equivalent nodal forces for distributed element loading Pe(2,1) = -15;

Page 426

Pe(2,2) = 20; Pe(3,1) = -15; Pe(3,2) = 20; Loading = Create_Loading(Model,Pe); Pfw = Loading.Pref; % determine final forces Q = Bbar*(Ac(1:Model.nf,:)'*(Pf-Pfw))+ Bvbar*V0; % display final forces disp('basic forces Q under distributed girder load') disp(Q); % determine displacements Uf = F*(Ac(1:Model.nf,:)'*(Pf-Pfw)) + Bbar'*V0; % display displacements disp('free dof displacements Uf under distributed girder load') disp(Uf); basic forces Q under distributed girder load 0 8.9027e+001 -8.9027e+001 0 -6.3568e+001 6.3568e+001 5.3400e+001 1.1258e+001 free dof displacements Uf under distributed girder load -1.5738e-002 -9.0612e-003 -5.2546e-002 9.2169e-003 1.8013e-002 1.2710e-003

Page 427

Script for Example 32b in CE220 class notes % frame with inextensible elements and brace using large axial stiffness for inextensibility


define model geometry and element types % specify node coordinates (could only specify non-zero terms) XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 6 0]; % second node, etc XYZ(3,:) = [ 14 6]; % XYZ(4,:) = [ 22 0]; XYZ(5,:) = [ 22 -10]; % connectivity array CON {1} = [ 1 2]; CON {2} = [ 2 3]; CON {3} = [ 3 4]; CON {4} = [ 4 5]; CON {5} = [ 2 4]; % boundary conditions (1 = restrained, 0 = free) BOUN(1,:) = [1 1 0]; BOUN(5,:) = [1 1 1]; % specify element type [ElemName{1:4}] = deal('Lin2dFrm'); % 2d linear beam element ElemName{5} = 'LinTruss'; % linear truss element


element properties for el=1:4 ElemData{el}.E = 1000; ElemData{el}.A = 1e6; % "inextensible" ElemData{el}.I = 40; end ElemData{5}.E = 1000; ElemData{5}.A = 10; % area for brace element ElemData{2}.Release = ['n';'y'];

force method of analysis % static(equilibrium) matrix at free dofs B = B_matrix(Model); % extract hinge locations from static matrix Id = Idef_index(Model,ElemData); Bf = B(1:Model.nf,Id); % collection of element flexibility matrices Fs = Fs_matrix(Model,ElemData); % invoke single function for force method of analysis [Bbar,Bvbar,F] = ForceMethod(Bf,Fs);

First load case: applied nodal forces % specify nodal forces Pe(3,2) = -20; Loading = Create_Loading (Model,Pe);

Page 428

Pf = Loading.Pref; % determine final forces Q = Bbar*Pf; % display final forces disp('basic forces Q under applied nodal forces') disp(Q); basic forces Q under applied nodal forces -6.0436e+000 0 3.5335e+001 -1.4527e+001 -3.5335e+001 -1.9459e+001 0 -3.0437e+001 -1.4111e+001 3.0437e+001 2.9999e+001 7.6979e+000

plot bending moment diagram Create_Window(0.80,0.80); Plot_Model (Model); % put basic forces back into complete vector that includes hinge Id index Qf = zeros(sum(Model.nq)); Qf(Id) = Q; Plot_2dMomntDistr (Model,[],Qf,0.75);

determine displacements Uf = F*Pf; % display displacements disp('free dof displacements Uf under applied nodal forces') disp(Uf); free dof displacements Uf under applied nodal forces -5.0563e-003 -3.6261e-008 -2.5038e-002 -2.4062e-003 -3.2307e-003 -2.0730e-002 3.8595e-003 1.2317e-002 -1.4111e-007 5.4825e-005

Page 429

open window and plot chords and then deformed shape % put free dof displacement values in complete displacement vector U = zeros(Model.nt,1); U(1:Model.nf) = Uf; Create_Window(0.80,0.80); Plot_Model (Model); MAGF = 60; Plot_Model (Model,U); Plot_DeformedStructure (Model,[],U);

Second load case: distributed load on element b ElemData{2}.w = [0;-5]; V0 = V0_vector(Model,ElemData); Pf = zeros(Model.nf,1); clear Pe; % specify equivalent nodal forces for distributed element loading Pe(2,1) = -15; Pe(2,2) = 20; Pe(3,1) = -15; Pe(3,2) = 20; Loading = Create_Loading(Model,Pe); Pfw = Loading.Pref; % determine final forces Q = Bbar*(Pf-Pfw)+ Bvbar*V0; % display final forces disp('basic forces Q under distributed girder load') disp(Q); basic forces Q under distributed girder load 1.8303e+001 0 8.9027e+001 -3.2672e+000 -8.9027e+001 -3.3461e+001 0 -6.3568e+001 -2.5162e+001 6.3568e+001 5.3400e+001 1.1258e+001

plot bending moment diagram Create_Window(0.80,0.80); Plot_Model (Model);

Page 430

% put basic forces back into complete vector that includes hinge Id index Qf = zeros(sum(Model.nq)); Qf(Id) = Q; Plot_2dMomntDistr (Model,ElemData,Qf,0.75);

determine displacements Uf = F*(Pf-Pfw) + Bbar'*V0; % display displacements disp('free dof displacements Uf under distributed girder load') disp(Uf); free dof displacements Uf under distributed girder load -1.5738e-002 1.0982e-007 -8.1075e-002 -9.0612e-003 -2.1396e-002 -5.2547e-002 9.2169e-003 1.8013e-002 -2.5162e-007 1.2710e-003

plot deformed shape % put free dof displacement values in complete displacement vector U = zeros(Model.nt,1); U(1:Model.nf) = Uf; MAGF = 30; Create_Window(0.80,0.80); Plot_Model (Model); Plot_Model (Model,U); Plot_DeformedStructure (Model,ElemData,U);

Page 431

Displacements at incipient collapse

Case of full collapse mechanism

Just before the last hinge forms the structure is statically determinate if a full collapse mechanism forms.Thus, all basic forces can be determined from the equilibrium equations (page 26 of Part I). With all basicforces known we can determine the corresponding strain dependent element deformations. In determining the global dof displacements we can only use the compatibility equations in the locations without a plastic hinge, because at the NOS plactic hinge locations there are unknown plastic hinge deformations at the instant just before the last hinge forms.

Denoting the locations without plastic hinge including the last hinge to form with subscript i, as we havedone in Example 18, we have

i si i i0= +FV Q Vwith

solving for the displacements at the free dofs we obtain

i i f=V UA Tf i i= BU V T

i i=A B 1i i

−=B Bsince and

Since the location of the last hinge to form is not known from the lower or upper bound theorem of plastic analysis, trial and error is required: we guess the location of the last plastic hinge to form and then determine the plastic deformations and the free global dof displacements. The correct guess for the last plastic hinge to form corresponds to the case with no contradiction between the sign of the plastic deformations and the corresponding basic forces, which also results in the largest value for the translation dofs.

consult examples 27 and 28.

Case of partial collapse mechanism

In this case the structure is still statically indeterminate before the last plastic hinge forms. Recall that there are less than NOS+1 hinges at collapse, thus less than NOS hinges before the last plastic hinge forms. Let us call the number of plastic hinges at collapse NP. Before the last plastic hinge forms there are, therefore, NOS+1 - NP redundant basic forces and a statically indeterminate analysis, as discussed in the preceding page is required to determine the redundant values and then all basic forces just before the last plastic hinge forms. Following this we can follow the same procedure as for the case of the full collapse mechanism above to determine the displacements at incipient collapse.

consult example 33


Page 432

CE220 - Theory of Structures Example 33 - Force method for partial collapse © Prof. Filip C. Filippou, 2000

Example 33 - Force method for partial collapse mechanism

We tackle now a problem that we addressed partially in Examples 9 and 18, the partial collapse mechanismof the structural model in the following figures

La 6:=

Lb 6:=

Lc 8:=

Ld 6:=

Le 6:=

Lf 10:=

31

2

4

1Q 2Q 3Q4Q

5Q6Q

7Q

8Q

We recall from Example 9 that we obtained the following basic forces at incipient collapse

Q2

Q4

Q5

Q6

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

Qei Bbarxe Q6⋅+= with Qei

50−

150

100

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:= and Bbarxe

1−

1

1−

1

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:=

A unique solution was not possible, because there were four unknown basic forces with three independentequilibrium equations, as is typical of a partial collapse mechanism (review discussion of Example 9).

Page 433


A unique solution of this problem is only possible if we introduce a deformation-force relation for theelements and then seek to satisfy the compatibility relations. Let us assume the following stiffness values

elements a through e are inextensible and have flexural stiffness EI EI 50000:=

the brace element has axial stiffness EA EA 10000:=

we need the element deformations at locations 2, 4, 5 and 6; we note that the element deformation at 2 alsodepends on the basic force Q1

We recall from Example 9 that Q1 100:=

we introduce the collection of flexibility matrices for elements a, c, and d only (note that we are not interestedin the deformation at 1, which is why the flexibility matrix for element a has only one row)

and the element deformations at the locations ofinterest for the particular solution are (do not forgetto include the deformation at 2 due to Q1 = 100)

@2Fse

La

3 EI⋅

0

0

0

0

Lc

3 EI⋅

Lc

6 EI⋅−

0

0

Lc

6 EI⋅−

Lc

3 EI⋅

0

0

0

0

Ld

3 EI⋅

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=@4Vei Fse

50−

150

100

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

⋅

100 La⋅

6 EI⋅−

0

0

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

+:= Vei

200−

266.67

66.67

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

1EI

=@5

@6

the compatibility error at location 6 due to the particular solution is BbarxeT Vei⋅ 400

1EI

=

we determine the discontinuity at location 6 due to the homogeneous solution (i.e. Q6 = 1)

BbarxeT Fse⋅ Bbarxe⋅ 12

1EI

= thus, to satisfy the compatibilityrelation at 6 we need to select thevalue of Q6 from the relation

4001EI⋅ 12

1EI⋅ Q6⋅+ 0=

and we get Q640012

−:= Q6 33.333−=

and the basic forces at the elastic locations become

Q2

Q4

Q5

Q6

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

Qei Bbarxe Q6⋅+:=

Q2

Q4

Q5

Q6

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

16.67−

116.67

133.33

33.33−

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

=

Page 434


and we note that the forces satisfy the yield condition since Q2 and Q6 do not exceed 100and Q4 and Q5 do not exceed 150.

We have, therefore, obtained a unique solution under the assumption that the material is linear elastic.It turns out that the value of the flexural stiffness EI does not affect the result, since it appears in both termsof the compatibility condition (note that the axial stiffness does not appear because the brace has yielded).

With the unique solution for the basic forces, we can now determine the element deformations and thenproceed to determine the displacements at incipient collapse. For the given stiffness values the last hingeforms at location 1 in element a, as an event-to-event analysis reveals (note that the sequence of plastichinge formation depends on the relative stiffness value EI/EA, while the plastic hinge location and collapseload factor does not; to prove this use the upper and lower bound theorem of plastic analysis).

Recalling the compatibility matrix from Example 18 we write and there are no hinge deformations atlocations 1, 2, 4, 5 and 6

V1U1

6−=

V2U1

6− U2+=

V4 U2U3

8+=

Af

1La

−

1La

−

1Lb

0

0

1Ld

−

1Le

8Lf

0

1

1

1

0

0

0

0

0

0

0

1Lc

1Lc

0

0

6Lf

−

0

0

0

0

1

1

1

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= Af

16

−

16

−

16

0

0

16

−

16

45

0

1

1

1

0

0

0

0

0

0

0

18

18

0

0

35

−

0

0

0

0

1

1

1

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=V5

U3

8U4+=

V6U1

6− U4+=

we know, of course, that even though we have 5 compatibility relations only 4 are linearly independent. Wecan, therefore, obtain a unique solution for the global dof displacements from these equations. We have

from 1 U1 6− V1⋅=

from 2 U2 V2U1

6+= V2 V1−=

from 6 U4 V6U1

6+= V6 V1−=

Page 435


from 5 U3 8 V5 U4−( )⋅= 8 V5 V6− V1+( )⋅=

and we can check that 4 is satisfied V4 V2 V1− V5+ V6− V1+=

which gives V4 V2− V5− V6+ 0= to be satisfied; but it is satisfied if we recall that Bbarxe

1−

1

1−

1

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:=

therefore, the given continuity relations result in a unique set of global dof displacements that we nowdetermine; to this end we need to determine the deformations

recall that Q1 100:= V1εLa

6 EI⋅2 Q1⋅ Q2−( )⋅:= V1ε 216.67

1EI

=

Q2 16.67−= V2εLa

6 EI⋅2 Q2⋅ Q1−( )⋅:= V2ε 133.33−

1EI

=

Q4 116.67= V4εLc

6 EI⋅2 Q4⋅ Q5−( )⋅:= V4ε 133.33

1EI

=

Q5 133.33= V5εLc

6 EI⋅2 Q5⋅ Q4−( )⋅:= V5ε 200

1EI

=

Q6 33.33−= V6εLd

3 EI⋅Q6⋅:= V6ε 66.67−

1EI

=

the global dof displacements at incipient collapse become

U1 6− V1ε⋅:= U1 1300−1EI

= U1 0.026−=

U2 V2ε V1ε−:= U2 350−1EI

= U2 7− 10 3−=

U3 8 V5ε V6ε− V1ε+( )⋅:= U3 3866.671EI

= U3 77.3310 3−=

U4 V6ε V1ε−:= U4 283.33−1EI

= U4 5.667− 10 3−=

Page 436


and we can determine the plastic deformations at incipient collapse from the relation Vh Af Uf⋅ Vε−=

relevant are only the deformations at 3, 7 and 8 (since the other locations are all continuous)

recall from Example 9 Q3 100−:= V3εQ3 Lb⋅

3 EI⋅:= V3ε 200−

1EI

=

Q7 100−:= V7εQ7 Le⋅

3 EI⋅:= V7ε 200−

1EI

=

Q8 20−:= V8εQ8 Lf⋅

EA:= V8ε 1000−

1EI

=

@3

then Vh

16

16

45

1

0

0

0

0

35

−

0

1

0

⎛⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎠

U1

U2

U3

U4

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

⋅

V3ε

V7ε

V8ε

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

−:= @7 Vh

7.333−

6−

47.2−

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

10 3−= or, Vh

366.67−

300−

2360−

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

1EI

=

@8

extract rows 3, 7 and 8 of Af

Page 437

FORCE-DEFORMATION RELATIONSFOR

ELEMENT AND STRUCTURE

Objective: force-deformation relation of frame elements; displacement method of analysis; constraints

CE220-Theory of Structures Displacement Method © Prof. Filip C. Filippou, 2000

Page 438

L

2 1v =

3 1v =

4 EIL

4 EIL

2 EIL

2 EIL

i j

1 1v =

EAL

1

0 0

4 20

2 40

EAL

EI EIL LEI EIL L

−

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥= = ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

k f

Force-deformation behavior of prismatic 2d frame element

Invert deformation-force relation for the same element from Part IV

i

L

2 1q =3

LEI

6LEI

3 1q =6LEI 3

LEI

1 1q =

LEA

j1

1L 1

L

1L 1

L

recall from Part IV

with

0 0

03 6

06 3

LEA

L LEI EIL LEI EI

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥= −⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦

f

= fv q

1−= =f kq v v= fv q


Page 439

Initial forces (or fixed-end forces) due to uniformly distributed element load

yw2

12yw L2

12yw L

2yw L

2yw L

L

0 0= −kq v 3 2

0

3 2

0 00 0

4 2024 12

2 4024 12

y y

y y

EAL

w L w LEI EIL L EIEI EI w L w LL L EI

⎛ ⎞ ⎛ ⎞⎡ ⎤ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎜ ⎟ ⎜ ⎟= − − =⎢ ⎥ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎜ ⎟ ⎜ ⎟⎢ ⎥ −⎜ ⎟ ⎜ ⎟⎣ ⎦ ⎝ ⎠ ⎝ ⎠

q

0 0( ) const=xκ κ= 0EIκ

0 0( ) const=a axε ε=0aEAε 0aEAε

L

0EIκ

Initial forces (or fixed-end forces) due to non-mechanical effects

0 0= −kq v

00

0 0 0

00

0 0

4 2 102

2 4 102

aa

EALL EA

EI EI L EIL L

EIEI EI LL L

ε εκ κ

κκ

⎡ ⎤ ⎛ ⎞⎢ ⎥ ⎜ ⎟

⎛ ⎞−⎢ ⎥ ⎜ ⎟⎜ ⎟⎢ ⎥ ⎜ ⎟= − − = ⎜ ⎟⎢ ⎥ ⎜ ⎟⎜ ⎟⎢ ⎥ ⎜ ⎟ −⎝ ⎠⎢ ⎥ ⎜ ⎟⎜ ⎟⎢ ⎥ ⎝ ⎠⎣ ⎦

q

no shear forces

Initial forces (or fixed-end forces)

0= +fv q v ( ) ( )10 0 0

−= − = − = +f k kq v v v v v q with 0 0= −kq v


Page 440

Force-deformation behavior of prismatic 2d frame element with moment release

Invert deformation-force relation for the same element from Part IV

recall from Part IV

with

= fv q

L

2 1q =3

LEI 6

LEI

1 1q =

LEA

i j

1

1L 1

L

0

03

LEA

LEI

⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦

f

3 212

= −v vand

1−= =f kq v v= fv q

L

i j

312

= −v2 1v =

3 EIL

1 1v =

EAL

0

30

EAL

EIL

⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦

kwith


Page 441

Initial forces (or fixed-end forces) due to uniformly distributed element load

Initial forces (or fixed-end forces) for element with moment release at node j

0= +fv q v ( ) ( )10 0 0

−= − = − = +f k kq v v v v v q with 0 0= −kq v

0 0= −kq v 230

000

30 824yy

EAL

w Lw LEIEIL

⎡ ⎤ ⎛ ⎞⎛ ⎞⎢ ⎥ ⎜ ⎟⎜ ⎟⎢ ⎥= − = ⎜ ⎟⎜ ⎟⎢ ⎥ − ⎜ ⎟⎜ ⎟⎢ ⎥ ⎝ ⎠ ⎝ ⎠⎣ ⎦

q

yw

2

8yw L

2yw L

L

2yw L

8yw L

8yw L

yw

2

8yw L

2yw L

2yw L

8yw L

8yw Lnote additional shear forces

due to fixed-end moment

Initial forces (or fixed-end forces) due to non-mechanical effects

0 0

00 0

01 330 2 2

a aEA L EAL

EI L EIL

ε ε

κ κ

⎡ ⎤ −⎛ ⎞ ⎛ ⎞⎢ ⎥ ⎜ ⎟ ⎜ ⎟⎢ ⎥= − =⎜ ⎟ ⎜ ⎟−⎢ ⎥ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

q0 0= −kq v

note shear forces due tofixed-end moment

01.5 EIκ

01.5EILκ 01.5EI

Lκ

0 0( ) const=xκ κ=


Page 442

Element stiffness matrix of 2d prismatic beam element in local coordinates

from

we derive

4 22 12

2 4 1 2

EI EIEIL L

EI EI LL L

⎡ ⎤⎢ ⎥ ⎡ ⎤⎢ ⎥= = ⎢ ⎥⎢ ⎥ ⎣ ⎦⎢ ⎥⎣ ⎦

q v v

L

26 EIL

26 EIL

312 EIL 312 EI

L

4 EIL 2 EI

L26 EI

L

26 EIL

312 EIL312 EI

L

26 EIL

26 EIL

4 EIL

2 EIL

26 EIL

26 EIL

1

1

1

1

1

1L

L

⎛ ⎞⎜ ⎟⎜ ⎟=⎜ ⎟⎜ ⎟⎝ ⎠

v2

2

6

6

EILEIL

⎛ ⎞⎜ ⎟⎜ ⎟=⎜ ⎟⎜ ⎟⎝ ⎠

q

10⎛ ⎞

= ⎜ ⎟⎝ ⎠

v

4

2

EILEIL

⎛ ⎞⎜ ⎟⎜ ⎟=⎜ ⎟⎜ ⎟⎝ ⎠

q

1

1L

L

⎛ ⎞−⎜ ⎟⎜ ⎟=⎜ ⎟−⎜ ⎟⎝ ⎠

v2

2

6

6

EILEIL

⎛ ⎞−⎜ ⎟⎜ ⎟=⎜ ⎟−⎜ ⎟⎝ ⎠

q

01⎛ ⎞

= ⎜ ⎟⎝ ⎠

v

2

4

EILEIL

⎛ ⎞⎜ ⎟⎜ ⎟=⎜ ⎟⎜ ⎟⎝ ⎠

q

3 2 3 2

2 2T

3 2 3 2

2 2

12 6 12 61 1

4 2 1 1 6 4 6 21 01 01 1 2 4 1 10 1 12 6 12 6

0 16 2 6 4

EI EI EI EIL L L L

EI EI EI EIL L EI EIL LL L L L L L

EI EI EI EI EI EIL L L L L L L L L L

EI EI EI EIL LL L

⎡ ⎤−⎢ ⎥⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎡ ⎤ ⎡ ⎤⎢ ⎥ −− ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥− − − − − −⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎢ ⎥⎢ ⎥ ⎢⎣ ⎦ −⎢⎣ ⎦

a ka

⎥⎥

or, in one step

gives the beam stiffness matrix in local coordinates (axial effects are not present)


Page 443

Element stiffness matrix of 2d prismatic beam element with moment releasein local coordinates

from

we derive

3EIL

⎡ ⎤= ⎢ ⎥⎣ ⎦

q v

L

3 EIL

23 EIL

23 EIL

23 EIL

33 EIL 33 EI

L

33 EIL33 EI

L

23 EIL

1

1

1

1L

⎛ ⎞= ⎜ ⎟⎝ ⎠

v 23EIL

⎛ ⎞= ⎜ ⎟⎝ ⎠

q

( )1=v3EI

L⎛ ⎞= ⎜ ⎟⎝ ⎠

q

1L

⎛ ⎞= −⎜ ⎟⎝ ⎠

v 23EIL

⎛ ⎞= −⎜ ⎟⎝ ⎠

q

or, in one step

gives the beam stiffness matrix in local coordinates (axial effects are not present)

3 2 3

T 2 2

3 2 3

3 3 3 01

3 3 3 01 3 1 11 01

3 3 3 00

0 0 0 0

EI EI EIL L L

L EI EI EIEI

LL LL L LEI EI EI

LL L L

⎡ ⎤−⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥

−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= − =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎢ ⎥ ⎢ ⎥−− −⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎢ ⎥⎣ ⎦

a ka


Page 444

Basic force-deformation relation for collection of elements

from the preceding pages we know that the basic force-deformation relation for a single 2d frame element is given by the following relation 0= +kq v q

with

( )

( )

( )

( )

a

b

c

ne

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟=⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

v

vV v

v

( )

( )

( )

( )

a

b

c

ne

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟=⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

q

qQ q

q

and

( )

( )

( )

( )

a

b

cs

ne

0 0 0 0

0 0 0 0

0 0 0 00 0 0 0

0 0 0 0

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

k

kK k

k

( )

( )

( )

( )

a0

b0

c0 0

ne0

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟= ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

q

q

Q q

q

Combine the deformation-force relations for all elements in the structural model( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

a a a a0

b b b b0

c c c c0

ne ne ne ne0

= +

= +

= +

= +

k

k

k

k

q v q

q v q

q v q

q v q

or, in compact form

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

aa a a0

bb b b0

cc c c0

nene ne ne0

0 0 0 0

0 0 0 0

0 0 0 00 0 0 0

0 0 0 0

⎛ ⎞⎛ ⎞ ⎡ ⎤ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥= + ⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎜ ⎟⎝ ⎠ ⎣ ⎦ ⎝ ⎠ ⎝ ⎠

k

k

k

k

qq v

qq v

qq v

qq v

block-diagonal

we get s 0= +KQ V Q


Page 445

CompatibilityEquilibrium

Element force-deformation

or

nf equations in nf unknowns

DISPLACEMENT OR STIFFNESS METHOD OF ANALYSIS

nf = number of free degrees of freedom

or with fewer subscripts

f f fw= +BP Q P f df= +A UV V

s 0= +KQ V Q

T T Tf f s f fw f df f 0 s= + + +A K A A A KP P Q VU

f f0ff= +K UP P

0f= +KP U P

Tf fB A=note

In the displacement method of analysis the unknowns of the analysis problem are the displacements at free dof's fU

Equilibrium equations in undeformed configuration (linear) f f fw= +P B Q P (1)

Compatibility relations for small displacements/deformations (linear) f f=V A U (2)

Element force-deformation for linear elastic material s 0= +Q K V Q (3)

from (2) and (3) ( )s f f 0= +UQ K A Q (3*)

from (1) and (3*) ( )[ ]Tf f s f f 0 fw= + +UP A K A Q P

fT T

f f s f fw f 0= + +P PUA K A A Q

ff ff f0= +UP K P

for brevity f 0= +P KU P

the initial nodal force vector arises from two contributions:(1) end shear forces due to element loading, (2) fixed end or initial basic forces

with the structure stiffness matrix at free dof'sTf s f=K A K A

the initial nodal force vector at free dof'sT0 fw f 0= +P P A Q

DISPLACEMENT OR STIFFNESS METHOD OF ANALYSIS


Page 446

Displacement method of analysis

In the absence of constraints we have the following relations:

note

In the displacement method of analysis the unknowns of the analysis problem are the displacements at free dof's

Equilibrium equations in undeformed configuration (linear) (1)

Compatibility relations for small displacements/deformations (linear) (2)

Element force-deformation for linear elastic material (3)

from (2) and (3) (3*)

from (1) and (3*)

the structure stiffness matrix at free dofs

for brevity

the initial nodal force vector arises from 3 contributions:(1) end shear forces due to element loading,(2) fixed end or initial basic forces, and(3) element deformations due to support displacements

the initial nodal force vector at free dofs

with

f f fw= +BP Q P Tf f=B A

f f d= +AV U V

s 0= +KQ V Q

fU

where dV are the element deformations due to imposed displacements at the restrained dofs

( )fs f d 0= + +K AQ V QU

( )fT

f f s f d 0 fw⎡ ⎤= + + +⎣ ⎦A K AP V Q PU

T T Tf f s f f f s d f 0 fw= + + +A K A A K AP V Q PU

f f 0ff f= +K UP P

f 0f= +KUP P

Tf s f=K A K A

T T0 fw f 0 f s d= + +A A KP P Q V


Page 447

Solution steps for displacement method of analysis

Step 1: write compatibility relations and establish structure compatibility matrix for free dofs

Step 2: form structure stiffness matrix and initial force vector

Step 3: solve system of equilibrium equations for the unknown displacements at free dofs

Step 4: determine element deformations from global dof displacements

Step 5: determine basic element forces from element deformations

Step 6: with Q determine other element forces by equilibrium

Step 8: with support reactions and applied forces check global equilibrium on complete structure free body

fA

Tf s f=K A K A T T

0 fw f 0 f s d= + +A A KP P Q V

Step 7: with Q determine support reactions from equilibrium at restrained dof's

f 0 f− =KP P U fU

f df= +A UV V

(a) by computer: ( )s 0 s 0= − = +K KQ V V V Q

(b) by hand: element by element ( )0 0= − = +k kq v v v q

(c) by hand: faster if intermediate result availablefrom stiffness determination

(shear forces from element equilibrium, axial forces from node equilibrium)

one support reaction at a time from equilibrium of corresponding restrained dof

d dw= +BR Q P(a) by computer:

(b) by hand:

(a) by computer:

(b) term by term through physical interpretation

( )Ts ff=K A K A ( )T T

0 fw f 0 f s d= + +A A KQP VP

( ) fs f 0 s d= + +K A KQ QU V

consult Examples 34, 35 and 36


Page 448

CE220 - Theory of Structures Ex 34 - Displacement method for truss © Prof. Filip C. Filippou, 2000

Example 34 - Displacement method: simple truss

We concoct a somewhat unusual structure in an effort to reduce the number of independent free globaldofs to 2. Truss structures have many independent free global dofs, and this is the reason why the forcemethod of analysis gained the upper hand in the late 19th century for the analysis of truss bridges. Thedisplacement method received a boost in the 1930s with the introduction of the iterative method of HardyCross for continuous beams and frames. Finally, the advent of computer analysis methods in the 1960s ledto the exclusive use of the displacement method in computer programs.

This unusual truss has only two independent free global dofs.

1 2

3

45

a

b

c

d1U

2U

8

4

4

5 4

The properties of the truss elements are:

EAa 30000:= EAb 10000:= EAc 10000:= EAd 20000:=

Geometric information

La 52 82+:= La 9.43=

Lb 42 42+:= Lb 5.66=

Lc 42 42+:= Lc 5.66=

Ld 8:=

Page 449


Step 1: Set up the compatibility relations between independent free global dofs and element deformations.

dof 1 corresponds to first column of the compatibility matrix Af

a

b

c

d

4

4

5 4

81 1=U

Af1⟨ ⟩

5La

0

4Lc

−

1

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

dof 2 corresponds to the second column of the compatibility matrix Af

a

b

c

d

4

4

8

2 1=U

5 4

Af2⟨ ⟩

0

4Lb

−

4Lc

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

Page 450


1 2

3

45

1Q

1Q

2Q

2Q

3Q

3Q

4Q 4Q 1P

2P

a

b

d

c

4

4

8

5 4Equilibrium equations just for checking(transpose of compatibility relations)

For free dof's P1 Q15La⋅ Q3

4Lc⋅− Q4+=

thus Bf

5La

0

0

4Lb

−

4Lc

−

4Lc

1

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:= we note Bf AfT=

P2 Q2−4Lb⋅ Q3

4Lc⋅+=

We set up the collection of element stiffness matrices. It consists of the individual basic element stiffnessmatrices arranged on the diagonal (block diagonal form, in the case of the truss diagonal)

Collection of element stiffness matrices Ks

with kaEAa

La:= kb

EAb

Lb:= kc

EAc

Lc:= kd

EAd

Ld:= we have Ks

ka

0

0

0

0

kb

0

0

0

0

kc

0

0

0

0

kd

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=

for hand calculations it is better to factor out a reference stiffness and then include only relative stiffnessterms in Ks. We demonstrate

KsEAa

La

1

0

0

0

0

EAb

EAa

La

Lb⋅

0

0

0

0

EAc

EAa

La

Lc⋅

0

0

0

0

EAd

EAa

La

Ld⋅

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

⋅:= Ks

1

0

0

0

0

0.556

0

0

0

0

0.556

0

0

0

0

0.786

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

EAa

La=

Page 451


Step 2: Form the structure stiffness matrix :

Method A: K AfT Ks⋅ Af⋅:= K

4.28

0.88−

0.88−

1.77⎛⎜⎝

⎞⎟⎠

103= or, better K1.35

0.28−

0.28−

0.56⎛⎜⎝

⎞⎟⎠

EAa

La=

Method B: Let us identify the contribution of each element to the stiffness matrix. We do this asfollows: we isolate the deformation of each element in the compatibility matrix.

Afa5La

0⎛⎜⎝

⎞⎟⎠

:= Afb 04Lb

−⎛⎜⎝

⎞⎟⎠

:= Afc4Lc

−4Lc

⎛⎜⎝

⎞⎟⎠

:= Afd 1 0( ):=

The operation AfT Ks⋅ Af⋅ now breaks down in the following element contributions:

element a Ka AfaT ka⋅ Afa⋅:= Ka

0.89

0

0

0⎛⎜⎝

⎞⎟⎠

103= or, Ka0.28

0

0

0⎛⎜⎝

⎞⎟⎠

EAa

La=

element b Kb AfbT kb⋅ Afb⋅:= Kb

0

0

0

0.88⎛⎜⎝

⎞⎟⎠

103= or, Kb0

0

0

0.28⎛⎜⎝

⎞⎟⎠

EAa

La=

element c Kc AfcT kc⋅ Afc⋅:= Kc

0.88

0.88−

0.88−

0.88⎛⎜⎝

⎞⎟⎠

103= or, Kc0.28

0.28−

0.28−

0.28⎛⎜⎝

⎞⎟⎠

EAa

La=

element d Kd AfdT kd⋅ Afd⋅:= Kd

2.5

0

0

0⎛⎜⎝

⎞⎟⎠

103= or, Kd0.79

0

0

0⎛⎜⎝

⎞⎟⎠

EAa

La=

Then the stiffness matrix is the sum of the element contributions: K Ka Kb+ Kc+ Kd+:=

Although Method A of determining the structure stiffness matrix K is compact and convenient bycomputer, it is not advisable for "hand" calculations. Method B is faster for "hand" calculations.For determining individual stiffness coefficients the most suitable method will be presented later underthe section "physical interpretation of stiffness coefficients".

Page 452


Illustration of Method B: element contributions to structure stiffness matrix

( )

( )

( )

( )

5 0

5 4 0 0 0 40 1 00 0 0

4 4 4 40 0 00 00 0 0

1 0

aa

ba c b

c

db c c c

L

L L L

L L L L

⎡ ⎤⎢ ⎥⎢ ⎥

⎡ ⎤ ⎢ ⎥⎡ ⎤− −⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥

− −⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎢ ⎥

⎢ ⎥⎢ ⎥⎣ ⎦

kk

Kk

k

[ ]( ) ( ) ( ) ( )

5 4015 4 4 40 0 1 0

4 4 00

a ca b c d

a b c c

b c

L LL L L L

L L

⎡ ⎤ ⎡ ⎤ ⎡ ⎤−⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= + − + − +⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦−⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

K k k k k

( )

( )

( )

( )

5 0

5 4 0 0 0 40 1 00 0 0

4 4 4 40 0 00 00 0 0

1 0

aa

ba c b

c

db c c c

L

L L L

L L L L

⎡ ⎤⎢ ⎥⎢ ⎥

⎡ ⎤ ⎢ ⎥⎡ ⎤− −⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥

− −⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎢ ⎥

⎢ ⎥⎢ ⎥⎣ ⎦

kk

Kk

k

( )

( )

( )

( )

5 0

5 4 0 0 0 40 1 00 0 0

4 4 4 40 0 00 00 0 0

1 0

aa

ba c b

c

db c c c

L

L L L

L L L L

⎡ ⎤⎢ ⎥⎢ ⎥

⎡ ⎤ ⎢ ⎥⎡ ⎤− −⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥

− −⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎢ ⎥

⎢ ⎥⎢ ⎥⎣ ⎦

kk

Kk

k

( )

( )

( )

( )

5 0

5 4 0 0 0 40 1 00 0 0

4 4 4 40 0 00 00 0 0

1 0

aa

ba c b

c

db c c c

L

L L L

L L L L

⎡ ⎤⎢ ⎥⎢ ⎥

⎡ ⎤ ⎢ ⎥⎡ ⎤− −⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥

− −⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎢ ⎥

⎢ ⎥⎢ ⎥⎣ ⎦

kk

Kk

k

( )

( )

( )

( )

5 0

5 4 0 0 0 40 1 00 0 0

4 4 4 40 0 00 00 0 0

1 0

aa

ba c b

c

db c c c

L

L L L

L L L L

⎡ ⎤⎢ ⎥⎢ ⎥

⎡ ⎤ ⎢ ⎥⎡ ⎤− −⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎢ ⎥⎢ ⎥

− −⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎢ ⎥

⎢ ⎥⎢ ⎥⎣ ⎦

kk

Kk

k

[ ]( ) ( ) ( ) ( )

5 4015 4 4 40 0 1 0

4 4 00

a ca b c d

a b c c

b c

L LL L L L

L L

⎡ ⎤ ⎡ ⎤ ⎡ ⎤−⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= + − + − +⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦−⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

K k k k k [ ]( ) ( ) ( ) ( )

5 4015 4 4 40 0 1 0

4 4 00

a ca b c d

a b c c

b c

L LL L L L

L L

⎡ ⎤ ⎡ ⎤ ⎡ ⎤−⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= + − + − +⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦−⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

K k k k k [ ]( ) ( ) ( ) ( )

5 4015 4 4 40 0 1 0

4 4 00

a ca b c d

a b c c

b c

L LL L L L

L L

⎡ ⎤ ⎡ ⎤ ⎡ ⎤−⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= + − + − +⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦−⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

K k k k k [ ]( ) ( ) ( ) ( )

5 4015 4 4 40 0 1 0

4 4 00

a ca b c d

a b c c

b c

L LL L L L

L L

⎡ ⎤ ⎡ ⎤ ⎡ ⎤−⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= + − + − +⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦−⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

K k k k k

Step 3: Set up the applied nodal force vector; for load case 1 we have Pf5

8⎛⎜⎝⎞⎟⎠

:=

Solve for the unknown displacements of the global dof's (use Gauss elimination instead of inverse!)

Uf lsolve K Pf,( ):= Uf2.35

5.7⎛⎜⎝

⎞⎟⎠

10 3−=

if we do not include the term EAa/La in the stiffness we get global dof displacements with EAa/La factor

Uf7.46

18.12⎛⎜⎝

⎞⎟⎠

La

EAa= such numbers are easier to handle

Note: we use Gauss elimination or another suitable solution method for linear system of equations.We do not use multiplication by the inverse of the stiffness matrix !! (even though the latter approach isnot a crime for very small matrices)

Page 453


Step 4: Determine the deformations of all elements from the compatibility relations

V Af Uf⋅:= V

1.24

4.03−

2.37

2.35

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

10 3−= or,

va

vb

vc

vd

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

Af Uf⋅:=

va

vb

vc

vd

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

3.955

12.814−

7.538

7.462

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

La

EAa=

Step 5: Determine the basic force of all elements

Q Ks V⋅:= Q

3.96

7.12−

4.19

5.87

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

=

Even though this way of determining is compact and convenient for computer use, particularly,since we have already established Ks, it is strongly advisable to use the following element-by elementdetermination of the basic force in "hand" calculations.

qa ka va⋅:= qa 3.96=

qb kb vb⋅:= qb 7.12−=

qc kc vc⋅:= qc 4.19=

qd kd vd⋅:= qd 5.87=

If we had not included the term EAa/La in the stiffness we get the element forces by multiplication with therelative stiffness only. We demonstrate

qa 1( ) 3.96⋅:= qa 3.96=

qb 0.556 12.814−( )⋅:= qb 7.12−=

qc 0.556 7.538( )⋅:= qc 4.19=

qd 0.786 7.462( )⋅:= qd 5.87=

It is clear that basic element forces do not depend on the reference stiffness value, but only on the relativestiffness ratios. Global dof displacements and element deformations on the other hand, are proportional tothe reference stiffness value, as the above results clearly demonstrate.

Step 6: not applicable since all basic forces have been determined

Page 454


Step 7: Determine support reactions

1 2

3

45

a

b

c

d

1Q

1Q

2Q

2Q

3Q

3Q

4Q 4Q 1P

2P

1R

2R 3R

4R

6R5R7R 8R

With all basic element forces known it is now straightforward to determine the support reactions. Each oneof these appears in a separate equilibrium equation. We can thus determine them one at a time.

In Mathcad it may be convenient to use matrix multiplication with anequilibrium matrix BdR1 Q4−=

R2 0=

R3 Q1−8La⋅ Q3

4Lc⋅−=

R4 Q24Lb⋅ Q3

4Lc⋅+=

R5 Q2−4Lb⋅=

Bd

0

0

8La

−

0

0

0

5La

−

8La

0

0

0

4Lb

4Lb

−

4Lb

0

0

0

0

4Lc

−

4Lc

0

0

0

0

1−

0

0

0

0

0

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= R Bd Q⋅:= R

5.87−

0

6.32−

2.07−

5.04

5.04−

2.1−

3.35

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=R6 Q2

4Lb⋅=

R7 Q1−5La⋅=

R8 Q18La⋅=

Page 455


1 2

3

45

a

b

c

d 5

8

3.955

5.8675.8675.867

4.19

4.19

7.124

3.955 7.124

6.317

2.074

2.096

3.354 5.037

5.037

1 2

3

45

4

4

5 4

8

a

b

c

d5

8

1R

2R3R

4R

6R5R7R 8R

Step 8: check global equilibrium

force equilibrium in X R1 5+ R4+ R5+ R7+ 0=

force equilibrium in Y R2 8+ R3+ R6+ R8+ 0=

moment equilibrium about 4 R1 8⋅ 5 8⋅+ 8 4⋅+ R4 4⋅+ R8 5⋅− 0=

Page 456


Second load case: thermal heating of members a and b

Steps 1 and 2 are the same (in fact, the compatibility matrix, the element and the structure stiffness matrixare only geometry and property dependent and should be set up only once).

Step 3: Set up applied nodal force vector: for thermal heating there are no applied forces at global dofs.

Pf0

0⎛⎜⎝⎞⎟⎠

:=

We note, however, that the elements undergo initial deformations due to non-mechanical effects. These are

V0

0.001 La⋅

0.001 Lb⋅

0

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:=

in the displacement method of analysis, these initial deformations result in initial element forces (also knownas "fixed end forces", because they arise for the case that the global dofs are "held fixed" so that the globaldof displacements are zero, while the non-mechanical effects are applied). These initial element forces arecollected in a vector Q0, which can be obtained either by the operation

Q0 Ks− V0⋅:= Q0T 30− 10− 0 0( )=

or, more conveniently element by element in which case q0 EA− ε0⋅=

Collecting the initial element forces into a vector we get Q0

EAa− 0.001⋅

EAb− 0.001⋅

0

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:= Q0

30−

10−

0

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

=

Initial force at global dof's P0 AfT Q0⋅:= P0

15.9−

7.07⎛⎜⎝

⎞⎟⎠

=

even though this way of determining the initial force vector Pf0 is compact and convenient bycomputer, it is not advisable for "hand" calculations. The method that is suitable for hand calculationswill be shown later under the section "physical interpretation of stiffness coefficients and initial force vector"

Solve for the unknown displacements of the global dof's

Uf lsolve K Pf P0−( ),[ ]:= Uf3.22

2.39−

⎛⎜⎝

⎞⎟⎠

10 3−= Uf10.25

7.59−

⎛⎜⎝

⎞⎟⎠

La

EAa=

Page 457


Step 4: Determine the deformations of all elements

V Af Uf⋅:= V

1.71

1.69

3.97−

3.22

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

10 3−= or,

va

vb

vc

vd

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

Af Uf⋅:=

va

vb

vc

vd

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

5.434

5.37

12.619−

10.252

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

La

EAa=

Step 5: Determine the basic force of all elements. This can be done in a number of ways, each with itsadvantages and drawbacks. Let us look at the methods.

(a) From the physical understanding point of view the best approach is to realize that the element forcesare proportional to the mechanical deformations, not the total deformations. Thus, we subtract the initialdeformations from the total deformations and then multiply by the element stiffness matrix. The mostconvenient computer solution is by multiplication with Ks. By "hand" it is advisable to proceed elementby element.

computer solution Q Ks V V0−( )⋅:= QT 24.57− 7.01− 7.01− 8.06( )=

hand solution, element by element

qa ka va 0.001 La⋅−( )⋅:= qa 24.57−= or, qa 1( ) 5.434 0.001 La⋅EAa

La⋅−

⎛⎜⎝

⎞⎟⎠

⋅:= qa 24.57−=

qb kb vb 0.001 Lb⋅−( )⋅:= qb 7.01−= qb 0.556( ) 5.37 0.001 Lb⋅EAa

La⋅−

⎛⎜⎝

⎞⎟⎠

⋅:= qb 7.02−=

qc kc vc⋅:= qc 7.01−= qc 0.556( ) 12.619−( )⋅:= qc 7.02−=

qd kd vd⋅:= qd 8.06= qd 0.786( ) 10.252( )⋅:= qd 8.06=

(b) If we have already set up the element initial force vector for determining the structure initial forcevector, then we can simply write

Q Ks V⋅ Q0+:= Q

24.57−

7.01−

7.01−

8.06

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

=

In this case we can think of the final element forces as the linear superposition of the initial forcesthat we obtained without any global dof displacements (remember the name "fixed end forces"!) andthe additional forces that result from the deformations due to global dof displacements.

Step 6: not applicable since all basic forces have been determined

Page 458


Step 7: With all element forces known the determination of support reactions is the same as for the firstload case with the interesting footnote that there are no applied forces at the applied dofs, but, thesupport reactions are not zero, if there are more than 3 support forces!

1 2

3

45

a

b

c

d

8.06 8.068.06

24.566

24.566 7.015

7.015

7.015

7.015

7.015

25.793

9.921

4.96

4.96

13.02

20.832

1 2

3

45

4

4

5 4

8

a

b

c

d5

8

1R

2R3R

4R

6R5R7R 8R

R Bd Q⋅:= R

8.06−

0

25.79

9.92−

4.96

4.96−

13.02

20.83−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=


force equilibrium in X R1 R4+ R5+ R7+ 0=

force equilibrium in Y R2 R3+ R6+ R8+ 0=

moment equilibrium about 4 R1 8⋅ R4 4⋅+ R8 5⋅− 0=

Page 459


Physical interpretation of stiffness coefficients and initial forces

Impose a unit value at displacement dof 1 and determine element deformations and forces

a

b

c

d11K

1 1=U

21K

4

c c

EAL L

−

5

a a

EAL L

0

1d

EAL

The force at dof 1 due to a unit displacement at dof 1 is stiffness coefficient K11. Obtain it by equilibrium

K11 1( )EAd

Ld⋅ 1( )⋅

5La

⎛⎜⎝

⎞⎟⎠

EAa

La⋅

5La

⎛⎜⎝

⎞⎟⎠

⋅+4Lc

−⎛⎜⎝

⎞⎟⎠

EAc

Lc⋅

4Lc

−⎛⎜⎝

⎞⎟⎠

⋅+:= K11 4.28103= as above


K214Lc

⎛⎜⎝

⎞⎟⎠

EAc

Lc⋅

4Lc

−⎛⎜⎝

⎞⎟⎠

⋅:= K21 0.88− 103= as above

Page 460


Impose a unit value at displacement dof 2 and determine element deformations and forces

a b

c

d12K

2 1=U22K

4

c c

EAL L

0

0

4

b b

EAL L

−


K224Lc

⎛⎜⎝

⎞⎟⎠

EAc

Lc⋅

4Lc

⎛⎜⎝

⎞⎟⎠

⋅4Lb

−⎛⎜⎝

⎞⎟⎠

EAb

Lb⋅

4Lb

−⎛⎜⎝

⎞⎟⎠

⋅+:= K22 1.77103= as above


K124Lc

−⎛⎜⎝

⎞⎟⎠

EAc

Lc⋅

4Lc

⎛⎜⎝

⎞⎟⎠

⋅:= K21 0.88− 103= as above

Clearly for hand calculations this is the most direct way to set up the coefficients of the stiffness matrix.

All free displacement dofs are set to zero ('locked') and the members are heated up. The forces at dofs 1and 2 are the initial (or fixed end) forces P0. They are determined by equilibrium

a

b

c

d

1 0=U

2 0=U

010P

20P

0

0

aEA Tα− Δ

bEA Tα− Δ αΔT 0.001:=

P105La

EAa− αΔT⋅( )⋅:= P10 15.9−=

as above

P204Lb

−⎛⎜⎝

⎞⎟⎠

EAb− αΔT⋅( )⋅:= P20 7.07=

as above

Clearly for hand calculations this is the most direct way to set up the initial force vector.

Page 461

Script for Example 34a in CE220 class notes %truss with only 2 free global dofs: solution with separate FEDEASLab functions


define model geometry and element types % specify node coordinates (could only specify non-zero terms) XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 8 0]; % second node, etc XYZ(3,:) = [ 12 4]; % XYZ(4,:) = [ 8 8]; XYZ(5,:) = [ 3 8]; % connectivity array CON {1} = [ 2 5]; CON {2} = [ 3 4]; CON {3} = [ 2 3]; CON {4} = [ 1 2]; % boundary conditions (1 = restrained, 0 = free) BOUN(1,:) = [1 1 0]; BOUN(2,:) = [0 1 0]; BOUN(3,:) = [1 0 0]; BOUN(4,:) = [1 1 0]; BOUN(5,:) = [1 1 0]; % specify element type [ElemName{1:4}] = deal('LinTruss'); % linear truss element

create Model Model = Create_SimpleModel (XYZ,CON,BOUN,ElemName); Create_Window(0.80,0.80); Plot_Model(Model); Label_Model(Model);

element properties in ElemData ElemData{1}.E = 1000; ElemData{1}.A = 30;

Page 462

for el=[2:3] ElemData{el}.E = 1000; ElemData{el}.A = 10; end ElemData{4}.E = 1000; ElemData{4}.A = 20;

set up collection of element stiffness matrices Ks = Ks_matrix(Model,ElemData);

form compatibility matrix A = A_matrix(Model); Af = A(:,1:Model.nf);

form stiffness matrix % either by Af'Ks*Af Kf = Af'*Ks*Af; disp('The stiffness matrix by the displacement method is'); disp(Kf); % or better by direct assembly Kf = Kf_matrix(Model,ElemData); disp('The stiffness matrix by direct assembly is'); disp(Kf); The stiffness matrix by the displacement method is 4.2771e+003 -8.8388e+002 -8.8388e+002 1.7678e+003 The stiffness matrix by direct assembly is 4.2771e+003 -8.8388e+002 -8.8388e+002 1.7678e+003

1. Load case: applied nodal forces Pf = [5;8]; % displacement method of analysis Uf = Kf\Pf; % display displacements disp('free dof displacements Uf under applied nodal forces') disp(Uf); free dof displacements Uf under applied nodal forces 2.3467e-003 5.6988e-003

determine element forces Q Q = Ks*Af*Uf; disp('The basic forces under the applied nodal forces are') disp(Q); The basic forces under the applied nodal forces are 3.9551e+000 -7.1235e+000 4.1902e+000 5.8667e+000

determine support reactions Pr = A'*Q; % resisting force vector % check equilibrium at free dofs Pu = Pf - Pr(1:Model.nf); % unbalance force vector Res = sqrt(Pu'*Pu);

Page 463

disp('The error of the equilibrium equations is'); disp(Res); % support reactions R = Pr(Model.nf+1:end); disp('The support reactions under the applied nodal forces are') disp(R) The error of the equilibrium equations is 1.7764e-015 The support reactions under the applied nodal forces are -5.8667e+000 0 -6.3168e+000 -2.0742e+000 5.0371e+000 -5.0371e+000 -2.0962e+000 3.3539e+000

2. Load case: thermal heating of elements a and b Pf = [0;0]; ElemData{1}.e0 = 0.001; ElemData{2}.e0 = 0.001; % set up initial deformation vector V0 = V0_vector(Model,ElemData); % set up initial force vector P0 = Af'*(-Ks*V0); % displacement method of analysis Uf = Kf\(Pf-P0); % display displacements disp('free dof displacements Uf under thermal heating of elements a and b') disp(Uf); free dof displacements Uf under thermal heating of elements a and b 3.2239e-003 -2.3880e-003

determine element forces Q Q = Ks*(Af*Uf-V0); disp('The basic forces under thermal heating of elements a and b are') disp(Q); The basic forces under thermal heating of elements a and b are -2.4566e+001 -7.0150e+000 -7.0150e+000 8.0598e+000

determine support reactions Pr = A'*Q; % resisting force vector % check equilibrium at free dofs Pu = Pf - Pr(1:Model.nf); % unbalance force vector Res = sqrt(Pu'*Pu); disp('The error of the equilibrium equations is'); disp(Res); % support reactions R = Pr(Model.nf+1:end); disp('The support reactions under thermal heating of elements a and b are') disp(R) The error of the equilibrium equations is

Page 464

2.5121e-015 The support reactions under thermal heating of elements a and b are -8.0598e+000 0 2.5793e+001 -9.9207e+000 4.9603e+000 -4.9603e+000 1.3020e+001 -2.0832e+001

Page 465

Script for Example 34b in CE220 class notes % truss with only 2 free global dofs: solution with direct stiffness method


define model geometry and element types % specify node coordinates (could only specify non-zero terms) XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 8 0]; % second node, etc XYZ(3,:) = [ 12 4]; % XYZ(4,:) = [ 8 8]; XYZ(5,:) = [ 3 8]; % connectivity array CON {1} = [ 2 5]; CON {2} = [ 3 4]; CON {3} = [ 2 3]; CON {4} = [ 1 2]; % boundary conditions (1 = restrained, 0 = free) BOUN(1,:) = [1 1 0]; BOUN(2,:) = [0 1 0]; BOUN(3,:) = [1 0 0]; BOUN(4,:) = [1 1 0]; BOUN(5,:) = [1 1 0]; % specify element type [ElemName{1:4}] = deal('LinTruss'); % linear truss element


element properties in ElemData ElemData{1}.E = 1000; ElemData{1}.A = 30; for el=[2:3] ElemData{el}.E = 1000; ElemData{el}.A = 10; end ElemData{4}.E = 1000;

Page 466

ElemData{4}.A = 20;

1. Load case: applied nodal forces Pe(2,1) = 5; Pe(3,2) = 8; Loading = Create_Loading(Model,Pe);

linear solution with direct stiffness method of analysis S_LinearStep Norm of equilibrium error = 1.776357e-015

plot deformed shape Create_Window(0.80,0.80); MAGF = 200; % magnification factor Plot_Model(Model); % original configuration Plot_Model(Model,U); % deformed configuration (chords only)

plot axial forces Create_Window(0.80,0.80); Plot_Model (Model); Plot_AxialForces (Model,Post); % store the results for later use (if necessary) Uf_LC1 = Uf; Post_LC1 = Post;

Page 467

2. Load case: thermal heating of elements a and b % no nodal forces for this load case Loading = Create_Loading(Model); ElemData{1}.q0 = -30; ElemData{2}.q0 = -10;


plot deformed shape Create_Window(0.80,0.80); Plot_Model(Model); % original configuration Plot_Model(Model,U); % deformed configuration (chords only)

plot axial forces Create_Window(0.80,0.80); Plot_Model (Model); Plot_AxialForces (Model,Post); % store the results for later use (if necessary) Uf_LC2 = Uf; Post_LC2 = Post;

Page 468

CE220 - Theory of Structures Ex 35 - Displacement method for continuous beam © Prof. Filip C. Filippou, 2000

Example 35 - Displacement method: a three span continuous beam

1. Load Case: nodal forces

We analyze the 3-span continuous beam in the following figure with the displacement method.

20 15 15

1 2 3 4

a b c80 100

There are only two free global dofs as shown in the following figure

a b c1 2

We can proceed as in earlier examples and set up the compatibility matrix Af for the free dofs.The result is:

a

bc

1 1=U

a bc

2 1=U

Af

0

1

1

0

0

0

0

0

1

1

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=

Page 469


With the element properties we obtain: La 20:= Lb 15:= Lc 15:=

EI 60000:=

Ks

4 EI⋅La

2 EI⋅La

0

0

0

2 EI⋅La

4 EI⋅La

0

0

0

0

0

4 EI⋅Lb

2 EI⋅Lb

0

0

0

2 EI⋅Lb

4 EI⋅Lb

0

0

0

0

0

3 EI⋅Lc

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= or, factor out EI/Lb KsEILb

4Lb

La⋅

2Lb

La⋅

0

0

0

2Lb

La⋅

4Lb

La⋅

0

0

0

0

0

4

2

0

0

0

2

4

0

0

0

0

0

3Lb

Lc⋅

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

⋅:=

and the structure stiffness matrix becomes K AfT Ks⋅ Af⋅:= K

28000

8000

8000

28000⎛⎜⎝

⎞⎟⎠

=

It is much more straightforward to set up the stiffness matrix directly from the following figure

a

bc

a bc

11K21K

12K 22K

4

b

EIL

4

a

EIL

2

b

EIL

2

b

EIL

3

c

EIL

2

a

EIL

K114 EI⋅La

4 EI⋅Lb

+:= K11 28000= K122 EI⋅Lb

:= K12 8000=

K212 EI⋅Lb

:= K21 8000= K224 EI⋅Lb

3 EI⋅Lc

+:= K22 28000=

if we factor out EI/Lb K11EILb

4Lb

La⋅ 4+

⎛⎜⎝

⎞⎟⎠

⋅:= K12EILb

2( )⋅:=

K21EILb

2( )⋅:= K22EILb

4 3Lb

Lc⋅+

⎛⎜⎝

⎞⎟⎠

⋅:=

Page 470


We can set up the equations of equilibrium at the free dofs in the form: Pf K Uf⋅ P0+=

and solve for the displacement values at the free dofs. The only difficulty arises with the moment of 100units acting at the end of the beam.

According to the following figure the fixed-end moment at the left end of element c is one half theapplied end moment of 100 units. How did we obtain this?

L

1006

LEI 100

10050

First we determined the initial deformation v0 for the completely unrestrained beam in the upper figure.

it isL

6 EI⋅times the value of the end moment and of opposite sign.

Then the fixed-end moment of the beam with one end unrestrained becomes q0 k− v0⋅=

with k3 EI⋅

L:= we obtain q0

3 EI⋅L

L6 EI⋅⋅ Mend⋅=

Mend

2=

Thus, one half of the moment at the unrestrained end "carries over" to the restrained end. 1/2 is knownas the carry-over factor of the prismatic beam.

Page 471


Thus, we have for the applied loading: Pf80

0⎛⎜⎝

⎞⎟⎠

:= and P00

50⎛⎜⎝

⎞⎟⎠

:=

1

a b c10020P

50

We solve the linear system of equilibrium equations for the unknown displacement values

U lsolve K Pf P0−,( ):= U3.67

2.83−

⎛⎜⎝

⎞⎟⎠

10 3−=

We can determine the end moments of members a, b and c directly from the stiffness coefficients and theinitial (fixed-end) moments (superposition of different compatible states)

1Q 2Q 3Q 5Q4Q

a

bc

a bc

11K21K

12K 22K

4

b

EIL

4

a

EIL

2

b

EIL

2

b

EIL

4

b

EIL

3

c

EIL

2

a

EIL

1

a b c10020P

50

Q12 EI⋅La

U1⋅:= Q1 22= Q34 EI⋅Lb

U1⋅2 EI⋅Lb

U2⋅+:= Q3 36=

Q24 EI⋅La

U1⋅:= Q2 44= Q42 EI⋅Lb

U1⋅4 EI⋅Lb

U2⋅+:= Q4 16−=

Q53 EI⋅Lc

U2⋅ 50+:= Q5 16=

Page 472


calculation with relative stiffness coefficients

if we factor out EI/Lb K11EILb

4Lb

La⋅ 4+

⎛⎜⎝

⎞⎟⎠

⋅:= K12EILb

2( )⋅:=

K21EILb

2( )⋅:= K22EILb

4 3Lb

Lc⋅+

⎛⎜⎝

⎞⎟⎠

⋅:=

K'

4Lb

La⋅ 4+

2

2

4 3Lb

Lc⋅+

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:= K'7

2

2

7⎛⎜⎝

⎞⎟⎠

=

U' lsolve K' Pf P0−,( ):= U'14.67

11.33−

⎛⎜⎝

⎞⎟⎠

=

Q'1 2Lb

La⋅ U'1⋅:= Q'1 22=

Q'2 4Lb

La⋅ U'1⋅:= Q'2 44=

Q'3 4 U'1⋅ 2 U'2⋅+:= Q'3 36=

Q'4 2 U'1⋅ 4 U'2⋅+:= Q'4 16−=

Q'5 3Lb

Lc⋅ U'2⋅ 50+:= Q'5 16=

Page 473


Bending moment diagram and deformed shape

22

44

3616

100

IP IP

shear forces and support reactions

22 44 36 16 100

80

20 15 15

3.3 3.3 1.33 1.33 7.73 7.73

6.401.97

22

3.3

22

80 100

6.40 7.731.973.3

22

global equilibrium 3.3 1.97− 6.40+ 7.73− 0= ok

22 80+ 100+ 1.97 20⋅− 6.40 35⋅+ 7.73 50⋅− 0.1= ok

Page 474


2. Load Case: uniform distributed load on members a and c

20 15 15

1 2 3 4a b c

6 6

Set up the applied force vectors directly. There are no nodal forces, therefore, Pf0

0⎛⎜⎝⎞⎟⎠

:=

the initial (fixed-end) forces can be obtained from the following figure

1 ab

c

20P

168.75

10P

200 200

w 6:=w Lc

2⋅

8168.75=

w La2⋅

12200= P0

200−

168.75⎛⎜⎝

⎞⎟⎠

:=



8.78−

⎛⎜⎝

⎞⎟⎠

10 3−=

We can determine the end moments of members a, b and c directly from the stiffness coefficients and theinitial (fixed-end) moments (superposition of different compatible states):

1Q 2Q 3Q 5Q4Q

a

bc

a bc

11K21K

12K 22K

4

b

EIL

4

a

EIL

2

b

EIL

2

b

EIL

4

b

EIL

3

c

EIL

2

a

EIL

Page 475


1

a bc

20P

168.75

10P200 200

Q12 EI⋅La

U1⋅ 200+:= Q1 257.92=

Q24 EI⋅La

U1⋅ 200−:= Q2 84.17−=

Q34 EI⋅Lb

U1⋅2 EI⋅Lb

U2⋅+:= Q3 84.17=

Q42 EI⋅Lb

U1⋅4 EI⋅Lb

U2⋅+:= Q4 63.33−=

Q53 EI⋅Lc

U2⋅ 168.75+:= Q5 63.33=


U' lsolve K' Pf P0−,( ):= U'38.61

35.14−

⎛⎜⎝

⎞⎟⎠

=

Q'1 2Lb

La⋅ U'1⋅ 200+:= Q'1 257.92=

Q'2 4Lb

La⋅ U'1⋅ 200−:= Q'2 84.17−=

Q'3 4 U'1⋅ 2 U'2⋅+:= Q'3 84.17=

Q'4 2 U'1⋅ 4 U'2⋅+:= Q'4 63.33−=

Q'5 3Lb

Lc⋅ U'2⋅ 168.75+:= Q'5 63.33=

Page 476



257.9 84.1 63.3

IP IP IP


20 15 15

68.71.39 49.2 40.8

257.9 84.1 63.3

68.7

68.7

257.9257.9

257.9

51.3 1.39

52.7

52.7

w=6 w=6

w=6 w=6

47.8 40.8

40.847.8

global equilibrium 68.7 6 20⋅− 52.7+ 47.8+ 6 15⋅− 40.8+ 0=

257.9 6 20⋅ 10⋅− 52.7 20⋅+ 47.8 35⋅+ 6 15⋅ 42.5⋅− 40.8 50⋅+ 0.1−= OK!!

Page 477


3. Load Case: differential heating of members a and c

20 15 15

1 2 3 4a b c0κ 0κ

with κ0 2 10 3−⋅:=

Set up the applied force vectors directly. There are no nodal forces, therefore, Pf0

0⎛⎜⎝⎞⎟⎠

:=

the initial (fixed-end) forces can be obtained from the following figure

1 ab

c

20P

180

10P

120 120

EI κ0⋅ 120= 1.5 EI⋅ κ0⋅ 180= P0120−

180⎛⎜⎝

⎞⎟⎠

:=



8.33−

⎛⎜⎝

⎞⎟⎠

10 3−=

We determine the basic forces (bending moments) by superposition of the different compatible states.

1Q 2Q 3Q 5Q4Q

a

bc

a bc

11K21K

12K 22K

4

b

EIL

4

a

EIL

2

b

EIL

2

b

EIL

4

b

EIL

3

c

EIL

2

a

EIL

Page 478


1 ab

c

20P

180

10P

120 120

Q12 EI⋅La

U1⋅ 120+:= Q1 160=

Q24 EI⋅La

U1⋅ 120−:= Q2 40−=

Q34 EI⋅Lb

U1⋅2 EI⋅Lb

U2⋅+:= Q3 40=

Q42 EI⋅Lb

U1⋅4 EI⋅Lb

U2⋅+:= Q4 80−=

Q53 EI⋅Lc

U2⋅ 180+:= Q5 80=


U' lsolve K' Pf P0−,( ):= U'26.67

33.33−

⎛⎜⎝

⎞⎟⎠

=

Q'1 2Lb

La⋅ U'1⋅ 120+:= Q'1 160=

Q'2 4Lb

La⋅ U'1⋅ 120−:= Q'2 40−=

Q'3 4 U'1⋅ 2 U'2⋅+:= Q'3 40=

Q'4 2 U'1⋅ 4 U'2⋅+:= Q'4 80−=

Q'5 3Lb

Lc⋅ U'2⋅ 180+:= Q'5 80=

Page 479


Bending moment diagram, curvature distribution and deformed shape

16040

80

80/EI40/EI

80/EI

40/EI

120/EI

IP


20 15 15

62.67 5.33

160 40 80

6

8.67160160

6

2.67 5.33

5.33

5.33

8

8

6

8.67

160

global equilibrium 6 8.67− 8+ 5.33− 0=

160 8.67 20⋅− 8 35⋅+ 5.33 50⋅− 0.1= OK!!

Page 480


4. Load Case: support settlement at node 2

Assume now that the support at node 2 settles downward by 0.05 units

In this case we have again an initial force vector, as shown in the following figure (impose supportdisplacement while "holding fixed" the free global dofs and determine the forces at free dofs).

1

2

3

4a b c20P

10P

2

60.05b

EIL

2

60.05b

EIL

2

60.05a

EIL

2

60.05a

EIL

There are again no nodal forces, therefore, Pf0

0⎛⎜⎝⎞⎟⎠

:=

the initial (fixed-end) forces can be determined from the figure above

P0

0.056 EI⋅

La2

⋅ 0.056 EI⋅

Lb2

⋅−

0.05−6 EI⋅

Lb2

⋅

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:= P035−

80−

⎛⎜⎝

⎞⎟⎠

=



2.72⎛⎜⎝

⎞⎟⎠

10 3−=

We determine the basic forces (bending moments) by superposition of the different compatible states.

1Q 2Q 3Q 5Q4Q

a

bc

a bc

11K21K

12K 22K

4

b

EIL

4

a

EIL

2

b

EIL

2

b

EIL

4

b

EIL

3

c

EIL

2

a

EIL

Page 481


1

2

3

4a b c20P

10P

2

60.05b

EIL

2

60.05b

EIL

2

60.05a

EIL

2

60.05a

EIL

Q12 EI⋅La

U1⋅ 0.056 EI⋅

La2

⋅+:= Q1 47.83=

Q24 EI⋅La

U1⋅ 0.056 EI⋅

La2

⋅+:= Q2 50.67=

Q34 EI⋅Lb

U1⋅2 EI⋅Lb

U2⋅+ 0.056 EI⋅

Lb2

⋅−:= Q3 50.67−=

Q42 EI⋅Lb

U1⋅4 EI⋅Lb

U2⋅+ 0.056 EI⋅

Lb2

⋅−:= Q4 32.67−=

Q53 EI⋅Lc

U2⋅:= Q5 32.67=


U' lsolve K' Pf P0−,( ):= U'1.89

10.89⎛⎜⎝

⎞⎟⎠

=

Q'1 2Lb

La⋅ U'1⋅ 0.05

6 EI⋅

La2

⋅+:= Q'1 47.83=

Q'2 4Lb

La⋅ U'1⋅ 0.05

6 EI⋅

La2

⋅+:= Q'2 50.67=

Q'3 4 U'1⋅ 2 U'2⋅+ 0.056 EI⋅

Lb2

⋅−:= Q'3 50.67−=

Q'4 2 U'1⋅ 4 U'2⋅+:= Q'4 47.33=

Q'5 3Lb

Lc⋅ U'2⋅:= Q'5 32.67=

Page 482



47.8332.67

50.67

IP IP0.05


20 15 15

4.935.56 2.18

10.49 7.74

47.83 32.6750.67

47.8347.83

47.83

4.93

4.93

5.564.93 2.18

2.18

2.187.7410.49

global equilibrium 4.93 10.49− 7.74+ 2.18− 0=

47.83 10.49 20⋅− 7.74 35⋅+ 2.18 50⋅− 0.07−= OK!!

Page 483

Script for Example 35a in CE220 class notes % continuous beam over three spans: solution with separate FEDEASLab functions


define model geometry and element types % specify node coordinates (could only specify non-zero terms) XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 20 0]; % second node, etc XYZ(3,:) = [ 35 0]; % XYZ(4,:) = [ 50 0]; % connectivity array CON {1} = [ 1 2]; CON {2} = [ 2 3]; CON {3} = [ 3 4]; % boundary conditions (1 = restrained, 0 = free) BOUN(1,:) = [1 1 1]; BOUN(2,:) = [0 1 0]; BOUN(3,:) = [0 1 0]; BOUN(4,:) = [0 1 0]; % specify element type [ElemName{1:3}] = deal('Lin2dFrm'); % linear 2d frame element


set up compatibility matrix by hand (note that we include the trivial dof!) Af = [0 1 1 0 0 0; 0 0 0 1 1 0; 0 0 0 0 0 1]';

set up collection of element stiffness matrices by hand EI = 60000; for el=1:Model.ne xyz = Localize(Model,el);

Page 484

L = ElmLenOr(xyz); k{el} = 2*EI/L*[2 1;1 2]; end Ks = blkdiag(k{1:Model.ne});

form stiffness matrix % by Af'Ks*Af Kf = Af'*Ks*Af; disp('The stiffness matrix at the free dofs is'); disp(Kf); The stiffness matrix at the free dofs is 28000 8000 0 8000 32000 8000 0 8000 16000

1. Load case: applied nodal forces Pf = [80;0;100]; % displacement method of analysis Uf = Kf\Pf; % determine element forces Q Q = Ks*Af*Uf; disp('The basic forces under the applied nodal forces are'); disp(Q); The basic forces under the applied nodal forces are 2.2000e+001 4.4000e+001 3.6000e+001 -1.6000e+001 1.6000e+001 1.0000e+002

2. Load case: uniformly distributed load in elements a and c % set up initial deformation vector w = 6; for el=[1 3] xyz = Localize(Model,el); L = ElmLenOr(xyz); v0{el} = w*L^3/24/EI*[-1 1]; end v0{2} = [0 0]; V0 = [v0{:}]'; % set up initial force vector P0 = Af'*(-Ks*V0); % displacement method of analysis Uf = Kf\(-P0); % determine element forces Q Q = Ks*(Af*Uf-V0); disp('The basic forces under the uniformly distributed load in elements a and c are'); disp(Q); % note that for the support reactions we should include the equivalent nodal forces! The basic forces under the uniformly distributed load in elements a and c are 2.5792e+002 -8.4167e+001 8.4167e+001 -6.3333e+001 6.3333e+001

Page 485

-1.4211e-014

3. Load case: differential heating of elements a and c % set up initial deformation vector kap0 = 2e-3; for el=[1 3] xyz = Localize(Model,el); L = ElmLenOr(xyz); v0{el} = kap0*L/2*[-1 1]; end v0{2} = [0 0]; V0 = [v0{:}]'; % set up initial force vector P0 = Af'*(-Ks*V0); % displacement method of analysis Uf = Kf\(-P0); % determine element forces Q Q = Ks*(Af*Uf-V0); disp('The basic forces under differential heating of elements a and c are'); disp(Q); The basic forces under differential heating of elements a and c are 1.6000e+002 -4.0000e+001 4.0000e+001 -8.0000e+001 8.0000e+001 4.2633e-014

4. Load case: support settlement at node 2 % set up deformation vector due to support settlement Vd = 0.05*[1/20 1/20 -1/15 -1/15 0 0]'; % set up initial force vector P0 = Af'*(Ks*Vd); % displacement method of analysis Uf = Kf\(-P0); % determine element forces Q Q = Ks*(Af*Uf+Vd); disp('The basic forces under support settlement are'); disp(Q); The basic forces under support settlement are 4.7833e+001 5.0667e+001 -5.0667e+001 -3.2667e+001 3.2667e+001 -3.5527e-015

Page 486

Script for Example 35b in CE220 class notes % continuous beam over three spans: solution with direct stiffness method


define model geometry and element types % specify node coordinates (could only specify non-zero terms) XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 20 0]; % second node, etc XYZ(3,:) = [ 35 0]; % XYZ(4,:) = [ 50 0]; % connectivity array CON {1} = [ 1 2]; CON {2} = [ 2 3]; CON {3} = [ 3 4]; % boundary conditions (1 = restrained, 0 = free) BOUN(1,:) = [1 1 1]; BOUN(2,:) = [0 1 0]; BOUN(3,:) = [0 1 0]; BOUN(4,:) = [0 1 0]; % specify element type [ElemName{1:3}] = deal('Lin2dFrm'); % linear 2d frame element


element properties in ElemData for el=1:Model.ne ElemData{el}.E = 1000; ElemData{el}.A = 1e6; ElemData{el}.I = 60; end

1. Load case: applied nodal forces Pe(2,3) = 80; Pe(4,3) = 100; Loading = Create_Loading(Model,Pe);

linear solution with direct stiffness method of analysis S_LinearStep

Page 487

Norm of equilibrium error = 3.552714e-015

plot deformed shape Create_Window(0.80,0.80); MAGF = 100; % magnification factor Plot_Model(Model); % original configuration Plot_DeformedStructure(Model,ElemData,U);

plot bending moment distribution Create_Window(0.80,0.80); Plot_Model (Model); Plot_2dMomntDistr (Model,ElemData,Post,2); % store the results for later use (if necessary) Uf_LC1 = Uf; Post_LC1 = Post;

2. Load case: uniformly distributed load in elements a and c % no nodal forces Loading = Create_Loading(Model); % specify uniformly distributed load value in ElemData ElemData{1}.w = [0;-6]; ElemData{3}.w = [0;-6];



Page 488

plot bending moment distribution Create_Window(0.80,0.80); Plot_Model (Model); Plot_2dMomntDistr (Model,ElemData,Post); % store the results for later use (if necessary) Uf_LC2 = Uf; Post_LC2 = Post;

3. Load case: differential heating of elements a and c % no nodal forces Loading = Create_Loading(Model); % clear uniformly distributed load value in ElemData ElemData{1}.w = [0;0]; ElemData{3}.w = [0;0]; % specify initial deformation due to heating in ElemData ElemData{1}.e0 = [0;2e-3]; ElemData{3}.e0 = [0;2e-3];



plot bending moment distribution Create_Window(0.80,0.80); Plot_Model (Model); Plot_2dMomntDistr (Model,ElemData,Post);

Page 489

% store the results for later use (if necessary) Uf_LC3 = Uf; Post_LC3 = Post;

4. Load case: support settlement at node 2 % no nodal forces Pe = []; % specify support displacement Ue(2,2) = -0.05; Loading = Create_Loading(Model,Pe,Ue); % clear initial deformation due to heating in ElemData ElemData{1}.e0 = [0;0]; ElemData{3}.e0 = [0;0];


plot deformed shape Create_Window(0.80,0.80); MAGF = 60; % magnification factor Plot_Model(Model); % original configuration Plot_Model(Model,U); % deformed configuration (chords only) Plot_DeformedStructure(Model,ElemData,U);


Page 490

CE220 - Theory of Structures Ex 36 - Displacement method for portal frame © Prof. Filip C. Filippou, 2000

Example 36 - Displacement method: portal frame w/o axial deformations

Geometry-dimensions Geometric properties: La 8:=

b

d

a

c

2@10 = 20

12

8

Lb 10:=

Lc 10:=

Ld 12:=

If all elements are assumed to be inextensible, there are five free independent global dofs as shown

12

34 5

b

d

a

cMaterial properties EI 80000:=

Step 1: Set up compatibility matrix for independent free global dofsdof 1 to determine the first column of the compatibility matrix Af

1 1U =

b

d

a

c

Af1⟨ ⟩

1La

1La

0

0

0

0

1Ld

1Ld

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

Page 491


dof 2 to determine the second column of the compatibility matrix Af

b

d

a

c2 1U =

Af2⟨ ⟩

0

1

1

0

0

0

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

dof 3 to determine the third column of the compatibility matrix Af

3 1U =

b

d

a

c

Af3⟨ ⟩

0

0

1Lb

−

1Lb

−

1Lc

1Lc

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

dof 4 to determine the fourth column of the compatibility matrix Af

b

d

a

c

4 1U =

Af4⟨ ⟩

0

0

0

1

1

0

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

Page 492


dof 5 to determine the fifth column of the compatibility matrix Af

5 1U =

b

d

a

c

Af5⟨ ⟩

0

0

0

0

0

1

1

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

b

d

a

c

1Q

2Q

3Q 5Q

7Q

4Q 6Q

8Q

It is interesting to look at the equilibrium equations thatdo not involve axial forces: the numbering of basicforces without axial forces is shown in the figure

The free body for the first equilibrium equation is shown in the following figure

b

d

a

c

1Q

2Q

3Q 5Q

7Q

4Q 6Q

8Q

1

1 2

8+Q Q

7 8

12+Q Q

Thus, the equilibrium matrix Bf becomesWe have: P1

Q1 Q2+

8

Q7 Q8+

12+=

P2 Q2 Q3+=

P3Q3 Q4+

10−

Q5 Q6+

10+= Bf

18

0

0

0

0

18

1

0

0

0

0

1

110

−

0

0

0

0

110

−

1

0

0

0

110

1

0

0

0

110

0

1

112

0

0

0

1

112

0

0

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

P4 Q4 Q5+=

P5 Q6 Q7+=

Page 493


The equilibrium matrix is, therefore The transpose of the compatibility matrix is:

Bf

18

0

0

0

0

18

1

0

0

0

0

1

110

−

0

0

0

0

110

−

1

0

0

0

110

1

0

0

0

110

0

1

112

0

0

0

1

112

0

0

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

= AfT

18

0

0

0

0

18

1

0

0

0

0

1

110

−

0

0

0

0

110

−

1

0

0

0

110

1

0

0

0

110

0

1

112

0

0

0

1

112

0

0

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

and we observe that the two are identical. Thus, equilibrium equations can also be written as

Pf AfT Q⋅= which is in fact an application of the principle of virtual displacements

Step 2: Set up the structure stiffness matrix :

Collection of element stiffness matrices for structural model

Ks

4EILa⋅

2EILa⋅

0

0

0

0

0

0

2EILa⋅

4EILa⋅

0

0

0

0

0

0

0

0

4EILb⋅

2EILb⋅

0

0

0

0

0

0

2EILb⋅

4EILb⋅

0

0

0

0

0

0

0

0

4EILc⋅

2EILc⋅

0

0

0

0

0

0

2EILc⋅

4EILc⋅

0

0

0

0

0

0

0

0

4EILd⋅

2EILd⋅

0

0

0

0

0

0

2EILd⋅

4EILd⋅

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

Page 494


Method 2(a): by computer in a single operation

K AfT Ks⋅ Af⋅:= K

2.43

7.5

0

0

3.33

7.5

72

4.8−

16

0

0

4.8−

1.92

0

4.8

0

16

0

64

16

3.33

0

4.8

16

58.67

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

103=

Method 2(a): express element stiffness matrix in terms of global dof displacements and add up

Determination of individual element stiffness contributions

kaEILa

4

2

2

4⎛⎜⎝

⎞⎟⎠

⋅:= kbEILb

4

2

2

4⎛⎜⎝

⎞⎟⎠

⋅:= kcEILc

4

2

2

4⎛⎜⎝

⎞⎟⎠

⋅:= kdEILd

4

2

2

4⎛⎜⎝

⎞⎟⎠

⋅:=

Afa

1La

1La

0

1

0

0

0

0

0

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:= Afb

0

0

1

0

1Lb

−

1Lb

−

0

1

0

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:=

Afc

0

0

0

0

1Lc

1Lc

1

0

0

1

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:= Afd

1Ld

1Ld

0

0

0

0

0

0

1

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:=


1.88

7.5

0

0

0

7.5

40

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

103=


0

0

0

0

0

0

32

4.8−

16

0

0

4.8−

0.96

4.8−

0

0

16

4.8−

32

0

0

0

0

0

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

103=

Page 495



0

0

0

0

0

0

0

0

0

0

0

0

0.96

4.8

4.8

0

0

4.8

32

16

0

0

4.8

16

32

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

103=


0.56

0

0

0

3.33

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

3.33

0

0

0

26.67

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

103=


Illustration of element stiffness contribution to structure stiffness matrix

3 22

3 2 22

2 3 2

2

0 0 0 0 0 0 0 0 0 012 6 0 0 0 4 6 2 0 0 0 0 00 0

12 6 60 06 4 0 0 0 6 12 60 060 00 0 0 0 0

2 6 40 00 0 0 0 00 0 0 0 0

0 0 0 0 0

a ab b b

c c ca a

b b b

b b b

EI EIEI EI EIL LL L L EI EI EIEI EI

L L LEI EI EIL LL L LEI EI EIL L L

K

⎡ ⎤⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− −= + +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦

⎣ ⎦

3 2

2

22

12 60 0 0

0 0 0 0 00 0 0 0 0

4 20 0 0 0 0

6 40 0 06 2 40 0

d d

c c c

d dc c c

EI EIL L

EI EI EIL L L

EI EIEI EI EI

L LL L L

⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ + ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ ⎣ ⎦

⎣ ⎦

( ) ( ) ( )

1 1 0 0 0 0 1 10 1 1 0 0 0 0 0

1 0 0 0 0 0 1 1 0 0 0 0 1 1 00 0 1 1 1 1 0 0

1 1 0 0 0 0 0 1 1 0 0 0 1 0 10 0 0 1 1 0 0 00 0 0 0 0 1 1 0

a a d d

a b ca b cb b c c

a b c

L L L L

L L LL L L L

L L LK k k k

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= + − − + +⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

( ) 1 0 0 0 11 0 0 0 0

dd

d

LL

k⎡ ⎤⎢ ⎥⎣ ⎦

⎥

1 0 0 0 01 1 0 0 0

1 1 0 0 0 0 1 10 0 0 0 1 1 0 0

0 1 1 0 0 0 0 00 0 0 0 0 1 1 0

0 0 1 1 1 1 0 00 0 0 0 0 1 1 0

0 0 0 1 1 0 0 00 0 0 0 0 1 0 1

0 0 0 0 0 1 1 01 0 0 0 11 0 0 0 0

a

aa a d d

a b

b bb b c c

c c

d c

d

d

LL

L L L LLL

L L L LLL

LL

kk

Kk

k

⎡ ⎤⎢ ⎥⎢ ⎥⎡ ⎤ ⎢ ⎥−⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ −⎢ ⎥⎢ ⎥⎢ ⎥= − − ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦ ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

2 2 2

2 2

1 1 0 0 0 06 2 4 6 2 6 4 20 0 0 0 0 0 00 1 1 0 0 0

0 0 1 1 1 16 4 2 6 40 0 0 0 00 0 0 1 1 0

0 0 0 0 0 1

a a

a a b b b c cb b c c

a a b b b

L LEI EI EI EI EI EI EI

L L L L L L LL L L L

EI EI EI EI EIL L L L L

K

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥= + − − +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

2

2 2

1 16 40 0 00 0

0 06 2 4 6 20 0 0 0 00 0

1 0

d d

c d d

c c c d d

L LEI EI EIL L L


⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥+⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

Page 496


Method 2(b): Derivation of all stiffness coefficients by direct methodImpose a unit value at displacement dof 1 and determine element deformations and then forces

The force at dof 1 necessary to equilibrate all element forces due to a unit displacement at dof 1 is stiffnesscoefficient K11 and so on. Obtain the necessary forces at all free dof's from the corresponding equilibriumequations.

1 1U =51K

26

d

EIL

31K41K21K

11K

2

6

a

EIL

3

12

a

EIL

3

12

d

EIL

K1112 EI⋅

La3

12 EI⋅

Ld3

+:= K11 2.43103=

K216 EI⋅

La2

:= K21 7.5103=

K31 0:=

K41 0:=

K516 EI⋅

Ld2

:= K51 3.33103=

for dof 2 K126 EI⋅

La2

:= K12 7.5103=

2 1U =22K 52K

42K

32K

4

a

EIL

2

a

EIL

2

6

b

EIL

12K K224 EI⋅La

4 EI⋅Lb

+:= K22 72103=

K326 EI⋅

Lb2

−:= K32 4.8− 103=

K422 EI⋅Lb

:= K42 16103=

K52 0:=

Page 497


for dof 3

3 1U =

53K

33K

2

6

b

EIL

3

12

b

EIL

2

6

c

EIL

2

6

c

EIL

2

6

b

EIL

3

12

c

EIL

23K13K

43KK13 0:= K12 7.5103=

K236 EI⋅

Lb2

−:= K23 4.8− 103=

K3312 EI⋅

Lb2

12 EI⋅

Lc2

+:= K33 19.2103=

K436 EI⋅

Lb2

−6 EI⋅

Lc2

+:= K43 0103=

K536 EI⋅

Lc2

:= K53 4.8103=

for dof 4

4 1U =54K

34K

24K14K

44K

4

b

EIL

2

b

EIL

2

6

b

EIL

2

6

c

EIL 2

c

EIL

4

c

EIL

K14 0:= K14 0103=

K242 EI⋅Lb

:= K24 16103=

K346 EI⋅

Lb2

−6 EI⋅

Lc2

+:= K34 0103=

K444 EI⋅Lb

4 EI⋅Lc

+:= K44 64103=

K542 EI⋅Lc

:= K54 16103=

for dof 5

5 1U =15K

4

d

EIL

2

d

EIL

26

d

EIL

35K55K

45K25K

4

c

EIL

2

c

EIL

2

6

c

EIL

K156 EI⋅

Ld2

:= K15 3.33103=

K25 0:=

K356 EI⋅

Lc2

:= K35 4.8103=

K452 EI⋅Lc

:= K45 16103=

K554 EI⋅Lc

4 EI⋅Ld

+:= K55 58.67103=

Page 498


Step 3: Set up the applied nodal force vector and solve for global dof displacements

Load Case: Nodal Forces

1

3 4b

d

2@10 = 20

12

2

8 a

c

5

2520

Pf

20

0

25−

0

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=

Solve for the unknown displacement values at the free global dof's

Uf lsolve K Pf,( ):= UfT 18.79 3.73− 24.09− 0.76 0.7( ) 10 3−=


V Af Uf⋅:= VT 2.35 1.38− 1.32− 3.17 1.65− 1.71− 2.26 1.57( ) 10 3−=

Step 5a: Determine the basic forces of all elements with a single operation

Q Ks V⋅:= QT 66.28 8.36− 8.36 80.21 80.21− 81.21− 81.21 71.92( )=

Step 5b: Determine the basic forces element by elementEven though this way of determining is compact and convenient for computer use, particularly,since we have already established Ks, it is strongly advisable to use the following element-byelementdetermination of the basic force in "hand" calculations.

va

V1

V2

⎛⎜⎜⎝

⎞⎟⎟⎠

:= va2.35

1.38−

⎛⎜⎝

⎞⎟⎠

10 3−= ka2 EI⋅La

2

1

1

2⎛⎜⎝

⎞⎟⎠

⋅:= qa ka va⋅:= qa66.28

8.36−

⎛⎜⎝

⎞⎟⎠

=

vb

V3

V4

⎛⎜⎜⎝

⎞⎟⎟⎠

:= vb1.32−

3.17⎛⎜⎝

⎞⎟⎠

10 3−= kb2 EI⋅Lb

2

1

1

2⎛⎜⎝

⎞⎟⎠

⋅:= qb kb vb⋅:= qb8.36

80.21⎛⎜⎝

⎞⎟⎠

=

vc

V5

V6

⎛⎜⎜⎝

⎞⎟⎟⎠

:= vc1.65−

1.71−

⎛⎜⎝

⎞⎟⎠

10 3−= kc2 EI⋅Lc

2

1

1

2⎛⎜⎝

⎞⎟⎠

⋅:= qc kc vc⋅:= qc80.21−

81.21−

⎛⎜⎝

⎞⎟⎠

=

vd

V7

V8

⎛⎜⎜⎝

⎞⎟⎟⎠

:= vd2.26

1.57⎛⎜⎝

⎞⎟⎠

10 3−= kd2 EI⋅Ld

2

1

1

2⎛⎜⎝

⎞⎟⎠

⋅:= qd kd vd⋅:= qd81.21

71.92⎛⎜⎝

⎞⎟⎠

=

Page 499


Step 5c: using stiffness coefficients

An even more convenient way of obtaining the basic forces if we have established the individual stiffnesscoefficients is by superposition of the forces of the unit displacement states. We obtain

simplify notation for free dof displacement vector U Uf:=

Q16 EI⋅

La2

U1⋅2 EI⋅La

U2⋅+:= Q1 66.28=

Q26 EI⋅

La2

U1⋅4 EI⋅La

U2⋅+:= Q2 8.36−=

Q34 EI⋅Lb

U2⋅6 EI⋅

Lb2

U3⋅−2 EI⋅Lb

U4⋅+:= Q3 8.36=

Q42 EI⋅Lb

U2⋅6 EI⋅

Lb2

U3⋅−4 EI⋅Lb

U4⋅+:= Q4 80.21=

Q56 EI⋅

Lc2

U3⋅4 EI⋅Lc

U4⋅+2 EI⋅Lc

U5⋅+:= Q5 80.21−=

Q66 EI⋅

Lc2

U3⋅2 EI⋅Lc

U4⋅+4 EI⋅Lc

U5⋅+:= Q6 81.21−=

Q76 EI⋅

Ld2

U1⋅4 EI⋅Ld

U5⋅+:= Q7 81.21=

Q86 EI⋅

Ld2

U1⋅2 EI⋅Ld

U5⋅+:= Q8 71.92=

Page 500


Step 6: Determine shear forces by element equilibrium and then use the equilibrium equations thatinvolve the axial forces to determine the latter. Note that the axial forces cannot be determined from theforce-deformation relations like the end moments, because the elements are inextensible and thus theiraxial deformations are zero.

2@10 = 20

12

8

66.28

80.22

81.21

71.92

8.36

7.24

7.24

2520

12.76

12.76

16.14

16.148.86

8.86

12.76

207.24

16.14

16.14

7.24

25

7.247.24

7.24

8.86

8.8616.148.86

Page 501


Draw bending moment diagram and deformed shape of the structure(observe element deformations as the angle between the tangent and the chord)

66.28

80.22

81.21

71.92

8.36

deformed shape

Page 502


Step 7: Determine support reactions (trivial from node free bodies)

2@10 = 20

12

8

66.28

80.22

81.21

71.92

8.36

7.24

7.24

2520

12.76

12.76

16.14

16.148.86

8.86

12.76

207.24

16.14

16.14

7.24

25

7.247.24

7.24

8.86

8.8616.148.86


b

d

2@10 = 20

12

8 a

c

2520

8.86

16.14

7.24

12.76

66.28

71.92

sum of forces in X: 20 7.24− 12.76− 0=

sum of forces in Y: 8.86 16.14+ 25− 0=

sum of moments about left support: 66.28 25 10⋅− 20 8⋅− 16.14 20⋅+ 71.92+ 12.76 4⋅− 0.04−=

Page 503


Second load case: thermal deformation of element d

Set up the applied nodal force vector: for load case 2 of thermal heating there are nodirectly applied forces at the global dofs.

Pf

0

0

0

0

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=

We note, however, that the elements undergo initial deformations due to non-mechanical effects.

A positive non-mechanical curvature causes element d to deform inwards κ0 2 10 3−⋅:=

noting that the top of the element d is deformation 1 we have:

V0

0

0

0

0

0

0

0.5− κ0⋅ Ld⋅

0.5 κ0⋅ Ld⋅

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=


Q0 Ks− V0⋅:= Q0T 0 0 0 0 0 0 160 160−( )=

or, more conveniently element by element in which case q0

EI κ0⋅

EI− κ0⋅

⎛⎜⎜⎝

⎞⎟⎟⎠

=

Collecting the initial element forces into a vector we get Q0

0

0

0

0

0

0

EI κ0⋅

EI− κ0⋅

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

Page 504


1

3 4b

d

2

a

c

5

0EIκ

0EIκ

12

34 5

b

d

a

cinitial forces at global dofs

P0 AfT Q0⋅:= P0

0

0

0

0

160

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

= or directly

Step 3: Solve for the unknown displacements of the global dof's

Uf lsolve K Pf P0−,( ):= UfT 6.69 0.29− 9.81 1.13 4.22−( ) 10 3−=


V Af Uf⋅:= VT 0.84 0.54 1.27− 0.15 2.11 3.24− 3.66− 0.56( ) 10 3−=


Q Ks V⋅ Q0+:= QT 44.3 38.44 38.44− 15.7− 15.7 69.83− 69.83 193.94−( )=

Step 5b: Determine the basic forces element by elementEven though this way of determining is compact and convenient for computer use, particularly,since we have already established Ks, it is strongly advisable to use the following element-byelementdetermination of the basic force in "hand" calculations.

va

V1

V2

⎛⎜⎜⎝

⎞⎟⎟⎠

:= va0.84

0.54⎛⎜⎝

⎞⎟⎠

10 3−= ka2 EI⋅La

2

1

1

2⎛⎜⎝

⎞⎟⎠

⋅:= qa ka va⋅:= qa44.3

38.44⎛⎜⎝

⎞⎟⎠

=

vb

V3

V4

⎛⎜⎜⎝

⎞⎟⎟⎠

:= vb1.27−

0.15⎛⎜⎝

⎞⎟⎠

10 3−= kb2 EI⋅Lb

2

1

1

2⎛⎜⎝

⎞⎟⎠

⋅:= qb kb vb⋅:= qb38.44−

15.7−

⎛⎜⎝

⎞⎟⎠

=

vc

V5

V6

⎛⎜⎜⎝

⎞⎟⎟⎠

:= vc2.11

3.24−

⎛⎜⎝

⎞⎟⎠

10 3−= kc2 EI⋅Lc

2

1

1

2⎛⎜⎝

⎞⎟⎠

⋅:= qc kc vc⋅:= qc15.7

69.83−

⎛⎜⎝

⎞⎟⎠

=

vd

V7

V8

⎛⎜⎜⎝

⎞⎟⎟⎠

:= vd3.66−

0.56⎛⎜⎝

⎞⎟⎠

10 3−= kd2 EI⋅Ld

2

1

1

2⎛⎜⎝

⎞⎟⎠

⋅:=

Page 505


vd0

0.5− κ0⋅ Ld⋅

0.5 κ0⋅ Ld⋅

⎛⎜⎜⎝

⎞⎟⎟⎠

:= qd kd vd vd0−( )⋅:= qd69.83

193.94−

⎛⎜⎝

⎞⎟⎠

=

or, qd0

EI κ0⋅

EI− κ0⋅

⎛⎜⎜⎝

⎞⎟⎟⎠

:= qd kd vd⋅ qd0+:= qd69.83

193.94−

⎛⎜⎝

⎞⎟⎠

=

Step 5c: using stiffness coefficients

An even more convenient way of obtaining the basic forces if we have established the individual stiffnesscoefficients is by superposition of the forces of the unit displacement states. Do not forget to add the basicforces due to the initial state!! (for which all dof's are "locked"). We obtain


Q16 EI⋅

La2

U1⋅2 EI⋅La

U2⋅+:= Q1 44.3=

Q26 EI⋅

La2

U1⋅4 EI⋅La

U2⋅+:= Q2 38.44=

Q34 EI⋅Lb

U2⋅6 EI⋅

Lb2

U3⋅−2 EI⋅Lb

U4⋅+:= Q3 38.44−=

Q42 EI⋅Lb

U2⋅6 EI⋅

Lb2

U3⋅−4 EI⋅Lb

U4⋅+:= Q4 15.7−=

Q56 EI⋅

Lc2

U3⋅4 EI⋅Lc

U4⋅+2 EI⋅Lc

U5⋅+:= Q5 15.7=

Q66 EI⋅

Lc2

U3⋅2 EI⋅Lc

U4⋅+4 EI⋅Lc

U5⋅+:= Q6 69.83−=

Q76 EI⋅

Ld2

U1⋅4 EI⋅Ld

U5⋅+ EI κ0⋅+:= Q7 69.83=

Q86 EI⋅

Ld2

U1⋅2 EI⋅Ld

U5⋅+ EI κ0⋅−:= Q8 193.94−=

Page 506


Step 6: Determine shear forces by element equilibrium and then use the equilibrium equations that involvethe axial forces to determine the latter. Note that the axial forces cannot be determined from the force-deformation relations like the end moments, because the elements are inextensible and thus their axialdeformations are zero.

2@10 = 20

12

8

193.94

69.8338.44

44.3

10.34

10.34

10.3410.34 10.34

10.34

10.34

10.34

5.41 5.41

5.41

5.41

5.41

5.41

Page 507


moment distribution

44.3

69.83

193.94

38.44

Curvature distribution

Page 508


Curvature distribution

deformed shape

Note the deformed shape follows the curvature !

Page 509


Step 7: Determine support reactions (trivial from node free bodies)

2@10 = 20

12

8

193.94

69.8338.44

44.3

10.34

10.34

10.3410.34 10.34

10.34

10.34

10.34

5.41 5.41

5.41

5.41

5.41

5.41


b

d

20

12

8 a

c

5.41

5.41

10.34

44.3

193.94

10.34

sum of moments about left support: 44.3 5.41 20⋅+ 193.94− 10.34 4⋅+ 0.08−=

Page 510


Third load case: distributed uniform load in element b

1

3 4b

d

2@10 = 20

12

2

8 a

c

5

4

Step 3: Set up the applied nodal force vector

applied nodal forces equivalent nodal forces due to distributed loads (from equilibrium)

Pf

0

0

0

0

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:= Pfw

0

0

20

0

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=

In addition to these forces that arise at the global dofs by equilibrium, the elements suffer initialdeformations due to distributed element loads acting over their span (in this case only element b).We denote these with vw and obtain with uniform transverse element load w (downward)

w 4:= vw

w Lb3⋅

24 EI⋅−

w Lb3⋅

24 EI⋅

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=

We collect these initial deformations in initial deformation vector V0 V0

0

0vw1

vw2

0

0

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

Page 511



Q0 Ks− V0⋅:= Q0T 0 0 33.33 33.33− 0 0 0 0( )=

or, more conveniently element by element in which case q0

w Lb2⋅

12

w Lb2⋅

12−

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=

Collecting the initial element forces into a vector we get for all Initial (fixed-end) forces Q0

0

0q01

q02

0

0

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

initial forces at global dofs

P0 Pfw AfT Q0⋅+:= P0

0

33.33

20

33.33−

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

=

or, better directly by keeping all free dofs "locked" and determining the forces at these dofs

4

2020

33.3333.331

23

4 5

P0

0

33.33

20

33.33−

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

=

Page 512



Uf lsolve K Pf P0−,( ):= UfT 7.13 2.73− 19.56− 0.97 0.93( ) 10 3−=


V Af Uf⋅:= VT 0.89 1.83− 0.77− 2.93 0.99− 1.03− 1.52 0.59( ) 10 3−=


Q Ks V⋅ Q0+:= QT 1− 55.51− 55.51 47.95 47.95− 48.58− 48.58 36.18( )=

Step 5b: Determine the basic forces element by element

Even though this way of determining is compact and convenient for computer use, particularly,since we have already established Ks, it is strongly advisable to use the following element-byelementdetermination of the basic force in "hand" calculations.

va

V1

V2

⎛⎜⎜⎝

⎞⎟⎟⎠

:= va0.89

1.83−

⎛⎜⎝

⎞⎟⎠

10 3−= ka2 EI⋅La

2

1

1

2⎛⎜⎝

⎞⎟⎠

⋅:= qa ka va⋅:= qa1−

55.51−

⎛⎜⎝

⎞⎟⎠

=

vb

V3

V4

⎛⎜⎜⎝

⎞⎟⎟⎠

:= vb0.77−

2.93⎛⎜⎝

⎞⎟⎠

10 3−= kb2 EI⋅Lb

2

1

1

2⎛⎜⎝

⎞⎟⎠

⋅:=

vb0

vw1

vw2

⎛⎜⎜⎝

⎞⎟⎟⎠

:= qb kb vb vb0−( )⋅:= qb55.51

47.95⎛⎜⎝

⎞⎟⎠

=

or, qb0

q01

q02

⎛⎜⎜⎝

⎞⎟⎟⎠

:= qb kb vb⋅ qb0+:= qb55.51

47.95⎛⎜⎝

⎞⎟⎠

=

vc

V5

V6

⎛⎜⎜⎝

⎞⎟⎟⎠

:= vc0.99−

1.03−

⎛⎜⎝

⎞⎟⎠

10 3−= kc2 EI⋅Lc

2

1

1

2⎛⎜⎝

⎞⎟⎠

⋅:= qc kc vc⋅:= qc47.95−

48.58−

⎛⎜⎝

⎞⎟⎠

=

vd

V7

V8

⎛⎜⎜⎝

⎞⎟⎟⎠

:= vd1.52

0.59⎛⎜⎝

⎞⎟⎠

10 3−= kd2 EI⋅Ld

2

1

1

2⎛⎜⎝

⎞⎟⎠

⋅:= qd kd vd⋅:= qd48.58

36.18⎛⎜⎝

⎞⎟⎠

=

Step 5c: using stiffness coefficientsAn even more convenient way of obtaining the basic forces if we have established the individual stiffnesscoefficients is by superposition of the forces of the unit displacement states. Do not forget to add the basicforces due to the initial state!! (for which all dof's are "locked"). We obtain


Q16 EI⋅

La2

U1⋅2 EI⋅La

U2⋅+:= Q1 1−=

Page 513


Q26 EI⋅

La2

U1⋅4 EI⋅La

U2⋅+:= Q2 55.51−=

Q34 EI⋅Lb

U2⋅6 EI⋅

Lb2

U3⋅−2 EI⋅Lb

U4⋅+w Lb

2⋅

12+:= Q3 55.51=

Q42 EI⋅Lb

U2⋅6 EI⋅

Lb2

U3⋅−4 EI⋅Lb

U4⋅+w Lb

2⋅

12−:= Q4 47.95=

Q56 EI⋅

Lc2

U3⋅4 EI⋅Lc

U4⋅+2 EI⋅Lc

U5⋅+:= Q5 47.95−=

Q66 EI⋅

Lc2

U3⋅2 EI⋅Lc

U4⋅+4 EI⋅Lc

U5⋅+:= Q6 48.58−=

Q76 EI⋅

Ld2

U1⋅4 EI⋅Ld

U5⋅+:= Q7 48.58=

Q86 EI⋅

Ld2

U1⋅2 EI⋅Ld

U5⋅+:= Q8 36.18=

Step 6: Determine shear forces by element equilibrium and then use the equilibrium equations that involvethe axial forces to determine the latter. Note that the axial forces cannot be determined from the force-deformation relations like the end moments, because the elements are inextensible and thus their axialdeformations are zero.

2@10 = 20

12

8

55.51 48.58

36.18

1 7.06

7.06

7.067.06

47.95

430.35

30.35

30.35

7.06 7.06 7.06

9.65 9.65

9.65

9.65

9.65

7.06

7.06

7.069.65 9.65

Page 514


moment diagram

max M

55.51

47.95

48.58

36.18

1

determination of maximum moment:

shear at right end of element b Vr4 10⋅

247.95 55.51+

10−:= Vr 9.65=

maximum moment Mmax 47.95Vr

2

2 4⋅+:= Mmax 59.6=

Deformed shape:

Page 515


Step 7: Determine support reactions (trivial from nodal free bodies as before)

2@10 = 20

12

8

55.51 48.58

36.18

1 7.06

7.06

7.067.06

47.95

430.35

30.35

30.35

7.06 7.06 7.06

9.65 9.65

9.65

9.65

9.65

7.06

7.06

7.069.65 9.65


b

d

2@10 = 20

12

8 a

c

30.35

9.65

7.06

7.0636.18

4

1

sum of moments about left support: 1− 4 10⋅ 5⋅− 36.18+ 7.06 4⋅− 9.65 20⋅+ 0.06−= OK!

Page 516

Displacement method of analysis with constraints

Equilibrium equations in undeformed configuration (linear) (1)

Compatibility relations for small displacements/deformations (linear) (2)

Element force-deformation for linear elastic material (3)

f f fw= +BP Q P

f f d= +AV U V

s 0= +KQ V Q

note Tf f=B A

Introduce relation between unconstrained and constrained dofs from Part II, pp. 23 f c f= AU U

The compatibility relations become f c f d f f d= + = +A A AV U V U V

Pre-multiply both sides of the equilibrium equations by the transpose of the constraint matrix Ac

(recall Lecture 14, pp. 7-10) to derive the equilibrium equations for the constrained dofs by PVD

( )T T Tc f c f fw= +A A AP Q P

T Tf f c fw= +A AP Q P with

with f f c=A A A

Tf c f= AP P

Proceeding as before and substituting (2*) into (3) and then into (1*) we obtain

(2*)

(1*)

f 0f= +KP PU

T T T Tf f s f f s d f c fwf 0= + + +A K A A K A AP V Q PU

the structure stiffness matrix at the constrained free dofs

the initial nodal force vector at the constrained free dofs

with Tf s f=K A K A

T T T0 c fw f 0 f s d= + +A A A KP P Q V

consult Examples 37 and 38


Page 517

Script for Example 36a in CE220 class notes % portal frame with inextensible elements: solution with separate FEDEASLab functions


define model geometry and element types % specify node coordinates (could only specify non-zero terms) XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 0 8]; % second node, etc XYZ(3,:) = [ 10 8]; % XYZ(4,:) = [ 20 8]; XYZ(5,:) = [ 20 -4]; % connectivity array CON {1} = [ 1 2]; CON {2} = [ 2 3]; CON {3} = [ 3 4]; CON {4} = [ 4 5]; % boundary conditions (1 = restrained, 0 = free) BOUN(1,:) = [1 1 1]; BOUN(5,:) = [1 1 1]; % specify element type [ElemName{1:4}] = deal('Lin2dFrm'); % linear 2d frame element


1

2 3 4

5

1

2 3

4

element properties in ElemData for el=1:Model.ne ElemData{el}.E = 1000; ElemData{el}.A = 1; % irrelevant ElemData{el}.I = 80; end

Page 518

collection of element stiffness matrices Ks = Ks_matrix(Model,ElemData);

identify deformation modes with inextensible elements Idef = Idef_index(Model,ElemData,1:Model.ne);

form stiffness matrix % form stiffness matrix at the original free dofs by direct assembly Kf = Kf_matrix(Model,ElemData); % set up constraint matrix Ac LC = LinConEqs(Model); % reorder dofs to have horizontal translation at node 2 and vertical translation at node 3 pick_dof = [Model.DOF(2,1), Model.DOF(3,2)]; dof_reord = [setdiff(1:Model.nt,pick_dof) pick_dof]; % constraint matrix Ac = Ac_matrix(LC,dof_reord); % extract free dofs of constraint matrix Acf = Ac(1:Model.nf,:); % compatibility matrix for unconstrained dofs A = A_matrix (Model); % compatibiliy matrix for constrained dofs Aftild = A*Ac; % Method A: stiffness matrix for constrained free dofs (transformation after direct assembly) Kftild = Acf'*Kf*Acf; disp('The stiffness matrix at the constrained free dofs is'); disp(Kftild); % Method B: stiffness matrix for constrained dofs ("hand calculation" if Aftild available) Kftild = Aftild'*Ks*Aftild; disp('The stiffness matrix at the constrained free dofs is'); disp(Kftild); The stiffness matrix at the constrained free dofs is 2.4306e+003 7.5000e+003 0 0 3.3333e+003 7.5000e+003 7.2000e+004 -4.8000e+003 1.6000e+004 0 0 -4.8000e+003 1.9200e+003 0 4.8000e+003 0 1.6000e+004 0 6.4000e+004 1.6000e+004 3.3333e+003 0 4.8000e+003 1.6000e+004 5.8667e+004 The stiffness matrix at the constrained free dofs is 2.4306e+003 7.5000e+003 0 0 3.3333e+003 7.5000e+003 7.2000e+004 -4.8000e+003 1.6000e+004 0 0 -4.8000e+003 1.9200e+003 0 4.8000e+003 0 1.6000e+004 0 6.4000e+004 1.6000e+004 3.3333e+003 0 4.8000e+003 1.6000e+004 5.8667e+004

1. Load case: applied nodal forces Pe(3,2) = -25; Pe(4,1) = 20; Loading = Create_Loading(Model,Pe); % transform applied nodal forces to constrained dofs Pftild = Acf'*Loading.Pref;

Page 519

% displacement method of analysis Uftild = Kftild\Pftild; % display result for constrained free dof displacements disp('constrained free dof displacements under the applied nodal forces'); disp(Uftild); % transform to complete displacement vector Uf = Acf*Uftild; % display result for free dof displacements only disp('complete free dof displacements under the applied nodal forces'); disp(Uf); constrained free dof displacements under the applied nodal forces 1.8789e-002 -3.7320e-003 -2.4093e-002 7.5882e-004 6.9671e-004 complete free dof displacements under the applied nodal forces 1.8789e-002 0 -3.7320e-003 1.8789e-002 -2.4093e-002 7.5882e-004 1.8789e-002 0 6.9671e-004

determine the basic forces % determine element deformations V = Aftild*Uftild; % determine basic forces from force-deformation relation (only deformable) Q = Ks(Idef,Idef)*V(Idef); disp('the basic forces under the applied nodal forces are'); disp(Q); the basic forces under the applied nodal forces are 6.6277e+001 -8.3622e+000 8.3622e+000 8.0215e+001 -8.0215e+001 -8.1208e+001 8.1208e+001 7.1919e+001

2. Load case: thermal deformation of element d % initial deformations ElemData{4}.e0 = [0;0.002]; % initial deformation vector V0 = V0_vector(Model,ElemData); % initial basic force vector (only for deformable elements) Q0 = -Ks(Idef,Idef)*V0(Idef); % initial force vector at constrained dofs P0tild = Aftild(Idef,:)'*Q0; % displacement method of analysis Uftild = Kftild\(-P0tild); % display result for constrained free dof displacements

Page 520

disp('constrained free dof displacements under thermal loading of element d'); disp(Uftild); % transform to complete displacement vector Uf = Acf*Uftild; % display result for free dof displacements only disp('complete free dof displacements under thermal loading of element d'); disp(Uf); constrained free dof displacements under thermal loading of element d 6.6895e-003 -2.9340e-004 9.8105e-003 1.1278e-003 -4.2176e-003 complete free dof displacements under thermal loading of element d 6.6895e-003 0 -2.9340e-004 6.6895e-003 9.8105e-003 1.1278e-003 6.6895e-003 0 -4.2176e-003

determine the basic forces % determine element deformations V = Aftild*Uftild; % determine basic forces from force-deformation relation (only deformable) Q = Ks(Idef,Idef)*V(Idef) + Q0; disp('the basic forces under thermal loading of element d are'); disp(Q); the basic forces under thermal loading of element d are 4.4303e+001 3.8435e+001 -3.8435e+001 -1.5697e+001 1.5697e+001 -6.9829e+001 6.9829e+001 -1.9394e+002

3. Load case: uniformly distributed load in element b % clear previously defined applied force vector (always a good idea!) clear Pe; % equivalent nodal forces Pe(2,2) = -20; Pe(3,2) = -20; Loading = Create_Loading(Model,Pe); % erase initial deformation of element d ElemData{4}.e0 = [0;0]; % specify uniformly distributed load in element b ElemData{2}.w = [0;-4]; % initial deformation vector V0 = V0_vector(Model,ElemData); % initial basic force vector (only for deformable elements) Q0 = -Ks(Idef,Idef)*V0(Idef);

Page 521

% initial force vector at constrained dofs P0tild = Aftild(Idef,:)'*Q0; % transform applied nodal forces to constrained dofs Pftild = Acf'*Loading.Pref; % displacement method of analysis Uftild = Kftild\(Pftild-P0tild); % display result for constrained free dof displacements disp('constrained free dof displacements under uniform distributed load in element b'); disp(Uftild); % transform to complete displacement vector Uf = Acf*Uftild; % display result for free dof displacements only disp('complete free dof displacements under uniform load in element b'); disp(Uf); constrained free dof displacements under uniform distributed load in element b 7.1333e-003 -2.7251e-003 -1.9555e-002 9.6956e-004 9.3024e-004 complete free dof displacements under uniform distributed load in element b 7.1333e-003 0 -2.7251e-003 7.1333e-003 -1.9555e-002 9.6956e-004 7.1333e-003 0 9.3024e-004

determine the basic forces % determine element deformations V = Aftild*Uftild; % determine basic forces from force-deformation relation (only deformable) Q = Ks(Idef,Idef)*V(Idef) + Q0; disp('the basic forces under the uniform distributed load in element b are'); disp(Q); the basic forces under the uniform distributed load in element b are -1.0035e+000 -5.5506e+001 5.5506e+001 4.7955e+001 -4.7955e+001 -4.8584e+001 4.8584e+001 3.6181e+001

Page 522

Script for Example 36b in CE220 class notes % portal frame with inextensible elements: solution with direct stiffness method


define model geometry and element types % specify node coordinates (could only specify non-zero terms) XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 0 8]; % second node, etc XYZ(3,:) = [ 10 8]; % XYZ(4,:) = [ 20 8]; XYZ(5,:) = [ 20 -4]; % connectivity array CON {1} = [ 1 2]; CON {2} = [ 2 3]; CON {3} = [ 3 4]; CON {4} = [ 4 5]; % boundary conditions (1 = restrained, 0 = free) BOUN(1,:) = [1 1 1]; BOUN(5,:) = [1 1 1]; % specify element type [ElemName{1:4}] = deal('Lin2dFrm'); % linear 2d frame element


1

2 3 4

5

1

2 3

4

element properties in ElemData for el=1:Model.ne ElemData{el}.E = 1000; ElemData{el}.A = 1e6; % inextensible ElemData{el}.I = 80; end

1. Load case: applied nodal forces Pe(3,2) = -25; Pe(4,1) = 20; Loading = Create_Loading(Model,Pe);

Page 523


plot deformed shape Create_Window(0.80,0.80); MAGF = 100; % magnification factor Plot_Model(Model); % original configuration Plot_Model (Model,U); % deformed configuration (chords only) Plot_DeformedStructure(Model,ElemData,U);


2. Load case: thermal deformation of element d % no nodal forces Loading = Create_Loading(Model); ElemData{4}.e0 = [0;0.002];


Page 524


plot bending moment distribution Create_Window(0.80,0.80); Plot_Model (Model); Plot_2dMomntDistr (Model,ElemData,Post);

plot curvature distribution Create_Window(0.80,0.80); Plot_Model (Model); Plot_2dCurvDistr (Model,ElemData,Post,0.75); % store the results for later use (if necessary) Uf_LC2 = Uf; Post_LC2 = Post;

Page 525

3. Load case: uniformly distributed load in element b % no nodal forces Loading = Create_Loading(Model); % clear initial deformation due to heating in ElemData ElemData{4}.e0 = [0;0]; % specify uniformly distributed load in element b ElemData{2}.w = [0;-4];




Page 526

CE220 - Theory of Structures Ex 37 - Displacement method for gable frame with constraints © Prof. Filip C. Filippou, 2000

Example 37 - Displacement method for gable frame with inextensible constraints

We have dealt with this problem in Part II from the standpoint of the kinematic relation between free globaldof displacements and element deformations (see Example 17). We use these relations in the following forthe solution of the element response under different loadings with the displacement method of analysis.

The geometry of the portal frame is given. Without trivial dofs, there are 9 independent free global dofs.

1

2

3

4

5

6

7

8

9

a

b c

d

8 8

12

6

Geometric properties: La 12:=

Lb 10:=

Lc 10:=

Ld 12:=

Under the assumption that all elements are inextensible, there are five free independent globaldegrees of freedom. We select these as shown in the following figure

12

3 4

5Material properties EI 80000:=

Page 527


Step 1: Set up compatibility matrix for independent free global dofs

dof 1 to determine the first column of the compatibility matrix Af consult Example 17

1 1

1.33ICb ICc

ICdICa

θa112

−:= θb16

:=

θc16

−:= θd112

:=

Af1⟨ ⟩

θa−

θa−

θb−

θb−

θc−

θc−

θd−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

dof 2 to determine the second column of the compatibility matrix Af

1

Af2⟨ ⟩

0

1

1

0

0

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

Page 528


dof 3 to determine the third column of the compatibility matrix Af consult Example 17

ICd

ICb

ICc

1

2

1.33

dof 3

θa 0:= θb16

−:=

θc16

:= θd16

−:=

Af3⟨ ⟩

θa−

θa−

θb−

θb−

θc−

θc−

θd−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

dof 4 to determine the fourth column of the compatibility matrix Af

1

Af4⟨ ⟩

0

0

0

1

1

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

dof 5 to determine the fifth column of the compatibility matrix Af

1

Af5⟨ ⟩

0

0

0

0

0

1

1

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

Page 529


It is interesting to look at the equilibrium equations that do not involve axial forces.To this end we use the principle of virtual displacements. Pf Af

T Q⋅=

the transpose of the kinematic matrix is

AfT

112

0

0

0

0

112

1

0

0

0

16

−

1

16

0

0

16

−

0

16

1

0

16

0

16

−

1

0

16

0

16

−

0

1

112

−

0

16

0

1

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

the numbering of basic forces without axial forces is shown in the figure

1Q

7Q

6Q

2Q

3Q

4Q 5Q The equilibrium equations are

P1Q1 Q2+

12

Q3 Q4+

6−

Q5 Q6+

6+

Q7

12−=

P2 Q2 Q3+=

P3Q3 Q4+

6

Q5 Q6+

6−

Q7

6+=

P4 Q4 Q5+=

P5 Q6 Q7+=

Page 530


Step 2: Set up the structure stiffness matrix:


KsEILa

4

2

0

0

0

0

0

2

4

0

0

0

0

0

0

0

4La

Lb⋅

2La

Lb⋅

0

0

0

0

0

2La

Lb⋅

4La

Lb⋅

0

0

0

0

0

0

0

4La

Lc⋅

2La

Lc⋅

0

0

0

0

0

2La

Lc⋅

4La

Lc⋅

0

0

0

0

0

0

0

3La

Ld⋅

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

⋅:=



6.03

4.67−

5.61−

0

6.33

4.67−

58.67

8

16

0

5.61−

8

5.89

0

4.67−

0

16

0

64

16

6.33

0

4.67−

16

52

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

103=

Method 2(a): express element stiffness matrix in terms of global dof displacements and add upDetermination of individual element stiffness contributions

kaEILa

4

2

2

4⎛⎜⎝

⎞⎟⎠

⋅:= kbEILb

4

2

2

4⎛⎜⎝

⎞⎟⎠

⋅:= kcEILc

4

2

2

4⎛⎜⎝

⎞⎟⎠

⋅:= kd3EILd

:=

Afa

112

112

0

1

0

0

0

0

0

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:= Afb

16

−

16

−

1

0

16

16

0

1

0

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:= Afc

16

16

0

0

16

−

16

−

1

0

0

1

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:=

Afd112

− 016

0 1⎛⎜⎝

⎞⎟⎠

:=

Page 531



0.56

3.33

0

0

0

3.33

26.67

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

103=


2.67

8−

2.67−

8−

0

8−

32

8

16

0

2.67−

8

2.67

8

0

8−

16

8

32

0

0

0

0

0

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

103=


2.67

0

2.67−

8

8

0

0

0

0

0

2.67−

0

2.67

8−

8−

8

0

8−

32

16

8

0

8−

16

32

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

103=


0.14

0

0.28−

0

1.67−

0

0

0

0

0

0.28−

0

0.56

0

3.33

0

0

0

0

0

1.67−

0

3.33

0

20

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

103=


K

6.03

4.67−

5.61−

0

6.33

4.67−

58.67

8

16

0

5.61−

8

5.89

0

4.67−

0

16

0

64

16

6.33

0

4.67−

16

52

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

103= same of course

Page 532


Illustration of element stiffness contributions to structure stiffness matrix

1 12 0 0 0 01/12 1/12 1 6 1 6 1 6 1 6 1 12 1 12 1 0 0 0

0 0 00 1 1 0 0 0 0 1 6 1 1 6 0 0

0 0 00 0 1 6 1 6 1 6 1 6 1 6 1 6 0 1 6 1 0

0 0 00 0 0 1 1 0 0 1 6 0 1 6 1 0

0 0 00 0 0 0 0 1 1 1 6 0 1 6 0 1

1 12 0 1 6 0 1

a

b

c

d

⎡ ⎤⎢ ⎥− − −⎡ ⎤ ⎢ ⎥⎡ ⎤⎢ ⎥ ⎢ ⎥−⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥= − − −⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥−⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦⎢ ⎥ −⎣ ⎦ ⎢ ⎥⎢ ⎥−⎣ ⎦

kk

Kk

k

( ) ( ) ( )

1/12 1/12 1 6 1 6 1 6 1 6 1 120 1 1 0 0 0 0

1/12 0 0 0 0 1 6 1 1 6 0 0 1 6 0 1 6 1 00 0 1 6 1 6 1 6 1 6 1 6

1/12 1 0 0 0 1 6 0 1 6 1 0 1 6 0 1 6 0 10 0 0 1 1 0 00 0 0 0 0 1 1

a b c

− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢− −⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢= + + +− −⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

K k k k [ ]( ) 1 12 0 1 6 0 1d⎥⎥ −⎥⎥⎥

k

1/12 1/12 1 6 1 6 1 6 1 66 1 2 6 1 4 6 1 20 0 0 00 1 1 0 0 012 6 6

0 0 1 6 1 6 1 6 1 66 1 4 6 1 2 6 1 40 0 0 00 0 0 1 1 012 6 6

0 0 0 0 0 1

a a b b b b

a a b b b b

EI EI EI EI EI EIL L L L L LEI EI EI EI EI EIL L L L L L

− −⎡ ⎤ ⎡ ⎤ ⎡⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢−⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢= + + − −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢−⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢⎣ ⎦ ⎣ ⎦⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣

K

1 126 1 6 1 4 20 06 6 3 1 3 1 30 01 66 1 6 1 2 4 12 60 06 6

1

c c c c

d d d

c c c c

EI EI EI EIL L L L EI EI EIEI EI EI EI L L LL L L L

−⎤ ⎡ ⎤⎡ ⎤⎥ ⎢ ⎥−⎢ ⎥⎥ ⎢ ⎥ ⎡ ⎤⎢ ⎥⎥ ⎢ ⎥+ −⎢ ⎥⎢ ⎥⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦−⎢ ⎥⎥ ⎢ ⎥⎣ ⎦⎢ ⎥ ⎢ ⎥⎦ ⎣ ⎦

2 2

2 2

12 1 12 1 6 1 6 106 6 6 6

0 0 0 0 0

12 1 12 1 6 1 6 106 6 6 6

6 1 6 1 4 206 6

6 1 6 1 2 406 6

c c c c

c c c c

c c c c

c c c c

EI EI EI EIL L L L

EI EI EI EIL L L LEI EI EI EIL L L LEI EI EI EIL L L L

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

+

2 2

2 2

12 1 12 1 6 1 6 106 6 6 6

0 0 0 0 0

12 1 12 1 6 1 6 106 6 6 6

6 1 6 1 4 206 6

6 1 6 1 2 406 6

c c c c

c c c c

c c c c

c c c c

EI EI EI EIL L L L

EI EI EI EIL L L LEI EI EI EIL L L LEI EI EI EIL L L L

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

2

2

3 1 3 1 1 3 10 012 12 6 12

0 0 0 0 0

3 1 1 3 1 3 10 012 6 6 60 0 0 0 0

3 1 3 1 30 012 6

d d d

d d d

d d d

EI EI EIL L L

EI EI EIL L L

EI EI EIL L L

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− −⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ −⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎢ ⎥⎢ ⎥

⎛ ⎞ ⎛ ⎞⎢ ⎥− ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

2 2

2

12 1 6 1 12 1 6 1 06 6 6 612 1 6 1 0 0 0

6 1 4 6 1 212 12 06 66 1 4 0 0 0

12 120 0 0 0 00 0 0 0 00 0 0 0 0

b b b b

a a

b b b b

a a

EI EI EI EIL L L LEI EI

EI EI EI EIL LL L L LEI EI

L L E

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎡ ⎤ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎛ ⎞ ⎛ ⎞⎝ ⎠ ⎝ ⎠⎢ ⎥ − ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎝ ⎠ ⎝ ⎠⎛ ⎞⎢ ⎥⎜ ⎟= +⎢ ⎥⎝ ⎠

−⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

K 2 21 6 1 12 1 6 1 06 6 6 6

6 1 2 6 1 4 06 6

0 0 0 0 0

b b b b

b b b b

I EI EI EIL L L L

EI EI EI EIL L L L

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎢ ⎥⎛ ⎞ ⎛ ⎞−⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥⎢ ⎥⎣ ⎦

Page 533


Method 2(b): Derivation of stiffness coefficients by direct method (PVD)

We confirm only the stiffness coefficients for translation dof 1 with the principle of virtual displacements

Impose a unit value at displacement dof 1 and determine element deformations and forces, as shown in thefollowing figure. The stiffness coefficients are the forces at the free global dofs due to the element forcescaused by a unit displacement at global dof 1. In steps we can write:

51K

31K 41K

21K11K

6 112a

EIL

⎛ ⎞⎜ ⎟⎝ ⎠

6 112a

EIL

⎛ ⎞⎜ ⎟⎝ ⎠

6 16b

EIL

⎛ ⎞⎜ ⎟⎝ ⎠

6 16b

EIL

⎛ ⎞⎜ ⎟⎝ ⎠

6 16c

EIL

⎛ ⎞⎜ ⎟⎝ ⎠

6 16c

EIL

⎛ ⎞⎜ ⎟⎝ ⎠

3 112d

EIL

⎛ ⎞⎜ ⎟⎝ ⎠

We denote with Q` the forces due a unit displacement at dof 1

K11Q'1 Q'2+

12

Q'3 Q'4+

6−

Q'5 Q'6+

6+

Q'712

−=

K21 Q'2 Q'3+=

K31Q'3 Q'4+

6

Q'5 Q'6+

6−

Q'76

+=

K41 Q'4 Q'5+=

K51 Q'6 Q'7+=

We demonstrate the terms one by one

111 6 1 1 6 1 1 6 1 1 3 12 2 2

12 12 6 6 6 6 12 12a b c d

EI EI EI EIL L L L

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + − − + + − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠K

216 1 6 1

12 6a b

EI EIL L

⎛ ⎞ ⎛ ⎞= + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

K

Page 534


311 6 1 1 6 1 1 3 12 26 6 6 6 6 12b c d

EI EI EIL L L

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠K

416 1 6 1

6 6b c

EI EIL L

⎛ ⎞ ⎛ ⎞= − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

K

516 1 3 1

6 12c d

EI EIL L

⎛ ⎞ ⎛ ⎞= + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

K

Combining terms and evaluating we obtain the first column of the 5x5 structure stiffness matrix

K1112 EI⋅

La

112⎛⎜⎝

⎞⎟⎠

2⋅

12 EI⋅Lb

16⎛⎜⎝⎞⎟⎠

2⋅+

12 EI⋅Lc

16⎛⎜⎝⎞⎟⎠

2⋅+

3 EI⋅Ld

112⎛⎜⎝

⎞⎟⎠

2⋅+:= K11 6.03103=

K216 EI⋅La

112⎛⎜⎝

⎞⎟⎠

⋅6EILb

16

−⎛⎜⎝

⎞⎟⎠

⋅+:= K21 4.67− 103=

K3112EILb

16

−⎛⎜⎝

⎞⎟⎠

⋅16⋅

12 EI⋅Lc

16⎛⎜⎝⎞⎟⎠

⋅16⎛⎜⎝⎞⎟⎠

⋅−3 EI⋅Ld

112

−⎛⎜⎝

⎞⎟⎠

⋅16⋅+:= K31 5.61− 103=

K416 EI⋅Lb

16

−⎛⎜⎝

⎞⎟⎠

⋅6 EI⋅Lc

16⎛⎜⎝⎞⎟⎠

⋅+:= K41 0103=

K516 EI⋅Lc

16⎛⎜⎝⎞⎟⎠

⋅3 EI⋅Ld

112

−⎛⎜⎝

⎞⎟⎠

⋅+:= K51 6.33103=

Page 535



Load Case: Nodal Forces30

100

20

consult Lecture 14, pp. 7-10

Tf c f= AP Pfrom

( ) ( )1430 20 1 603

= − + − = −P

2 0=P

( ) ( )3430 20 2 803

⎛ ⎞= − − + =⎜ ⎟⎝ ⎠

P

4 100=P

5 0=P

Pf

60−

0

80

100

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=


Uf lsolve K Pf,( ):= UfT 56.66 6.81− 75.88 3.56 1.19−( ) 10 3−=


V Af Uf⋅:= VT 4.72 2.09− 3.61− 6.77 0.36 4.39− 6.74( ) 10 3−=


Q Ks V⋅:= QT 98.02 7.19 7.19− 158.81 58.81− 134.79− 134.79( )=

Even though this way of determining is compact and convenient for computer use, particularly,since we have already established Ks, it is strongly advisable to use the following element-by elementdetermination of the basic force in "hand" calculations.

va

V1

V2

⎛⎜⎜⎝

⎞⎟⎟⎠

:= va4.72

2.09−

⎛⎜⎝

⎞⎟⎠

10 3−= ka2 EI⋅La

2

1

1

2⎛⎜⎝

⎞⎟⎠

⋅:= qa ka va⋅:= qa98.02

7.19⎛⎜⎝

⎞⎟⎠

=

vb

V3

V4

⎛⎜⎜⎝

⎞⎟⎟⎠

:= vb3.61−

6.77⎛⎜⎝

⎞⎟⎠

10 3−= kb2 EI⋅Lb

2

1

1

2⎛⎜⎝

⎞⎟⎠

⋅:= qb kb vb⋅:= qb7.19−

158.81⎛⎜⎝

⎞⎟⎠

=

vc

V5

V6

⎛⎜⎜⎝

⎞⎟⎟⎠

:= vc0.36

4.39−

⎛⎜⎝

⎞⎟⎠

10 3−= kc2 EI⋅Lc

2

1

1

2⎛⎜⎝

⎞⎟⎠

⋅:= qc kc vc⋅:= qc58.81−

134.79−

⎛⎜⎝

⎞⎟⎠

=

vd V7( ):= vd 6.74( ) 10 3−= kd3 EI⋅Ld

:= qd kd vd⋅:= qd 134.79( )=

Page 536


Step 6: Determine shear forces by element equilibrium and then use the equilibrium equations that involvethe axial forces to determine the latter. Note that the axial forces cannot be determined from the force-deformation relations like the end moments, because the elements are inextensible and thus their axialdeformations are zero (this is left as an exercise)


98.02

7.21

158.81

58.81

134.79

deformed shape

V

4.72

2.09−

3.61−

6.77

0.36

4.39−

6.74

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

10 3−=

(observe element deformationsas the angle between the tangentand the chord)

Page 537


Step 7: Determine support reactions (left as exercise)

Step 8: check global equilibrium (left as exercise)

Second load case: distributed uniform load in element b

w=5

2520

1525

20

15

8 8

12

6

25

w=5

initial state

500/12

500/12

for the effect of end shear forces in element b review again Lecture 14, pp. 7-10

( )Tfw c fw− = −AP Pfrom

( )1w3 430 40 11.676 6

⎛ ⎞ ⎛ ⎞− = ⋅ + − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

P

2w 0− =P

( )3w3 430 40 41.676 6

⎛ ⎞ ⎛ ⎞− = ⋅ + − − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

P

4w 0− =P

5w 0− =P

equivalent nodal forces due to distributed loads Pfw

11.67

0

41.67−

0

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=

initial forces (fixed-end moments) due to distributed load in element b w 5:=

Q0 0 0w Lb

2⋅

12

w Lb2⋅

12− 0 0 0

⎛⎜⎝

⎞⎟⎠

T

:=

Page 538




11.67

41.67

41.67−

41.67−

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

= there are no nodal forces in this case Pf

0

0

0

0

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=

or, better directly by keeping all free dofs "locked" and determining the forces at the free global dofs

2520

15

20

15

initial state P0−

11.67−

41.67−

41.67

41.67

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

=


Uf lsolve K Pf P0−,( ):= UfT 82.71 7.93− 94.6 3.28 2.59−( ) 10 3−=


V Af Uf⋅:= VT 6.89 1.03− 5.94− 5.26 1.3 4.57− 6.28( ) 10 3−=


Q Ks V⋅ Q0+:= QT 170.04 64.37 64.37− 31.6 31.6− 125.59− 125.59( )=



va

V1

V2

⎛⎜⎜⎝

⎞⎟⎟⎠

:= va6.89

1.03−

⎛⎜⎝

⎞⎟⎠

10 3−= ka2 EI⋅La

2

1

1

2⎛⎜⎝

⎞⎟⎠

⋅:= qa ka va⋅:= qa170.04

64.37⎛⎜⎝

⎞⎟⎠

=

Page 539


vb

V3

V4

⎛⎜⎜⎝

⎞⎟⎟⎠

:= vb5.94−

5.26⎛⎜⎝

⎞⎟⎠

10 3−= kb2 EI⋅Lb

2

1

1

2⎛⎜⎝

⎞⎟⎠

⋅:=

initial deformations vw

w Lb3⋅

24 EI⋅−

w Lb3⋅

24 EI⋅

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:= vb0

vw1

vw2

⎛⎜⎜⎝

⎞⎟⎟⎠

:= qb kb vb vb0−( )⋅:= qb64.37−

31.6⎛⎜⎝

⎞⎟⎠

=

or with initial forces, q0

w Lb2⋅

12

w Lb2⋅

12−

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:= qb0

q01

q02

⎛⎜⎜⎝

⎞⎟⎟⎠

:= qb kb vb⋅ qb0+:= qb64.37−

31.6⎛⎜⎝

⎞⎟⎠

=

vc

V5

V6

⎛⎜⎜⎝

⎞⎟⎟⎠

:= vc1.3

4.57−

⎛⎜⎝

⎞⎟⎠

10 3−= kc2 EI⋅Lc

2

1

1

2⎛⎜⎝

⎞⎟⎠

⋅:= qc kc vc⋅:= qc31.6−

125.59−

⎛⎜⎝

⎞⎟⎠

=

vd V7( ):= vd 6.28( ) 10 3−= kd3 EI⋅Ld

:= qd kd vd⋅:= qd 125.59( )=


170.04

64.37

31.6

125.59

Page 540


deformed shape

V

6.89

1.03−

5.94−

5.26

1.3

4.57−

6.28

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

10 3−=


Page 541


Third load case: vertical translation of 0.1 units at right support

ICd

ICc

12

6

8 8 16

0.1

0,5P

0,4P

0,3P

0,1P

6 0.18c

EIL

⎛ ⎞⎜ ⎟⎝ ⎠

3 0.116d

EIL

⎛ ⎞⎜ ⎟⎝ ⎠

6 0.18c

EIL

⎛ ⎞⎜ ⎟⎝ ⎠

Initial force vector at free global dofs due to support displacement

(a) by computer Vd 0 0 0 018

−18

−116

⎛⎜⎝

⎞⎟⎠

T0.1⋅:=

P0 AfT Ks⋅ Vd⋅:= P0

210.42−

0

220.83

600−

475−

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

=

(b) directly from the basic element forces caused by the support displacement (see above figure)


P01Q'1 Q'2+

12

Q'3 Q'4+

6−

Q'5 Q'6+

6+

Q'712

−=

P02 Q'2 Q'3+=

P03Q'3 Q'4+

6

Q'5 Q'6+

6−

Q'76

+=

P04 Q'4 Q'5+=

P05 Q'6 Q'7+=

Page 542


we evaluate

P0112 EI⋅

Lc

0.18

−⎛⎜⎝

⎞⎟⎠

⋅16⋅

3 EI⋅Ld

0.116

⎛⎜⎝

⎞⎟⎠

⋅112⋅−:= P01 210.42−=

P02 0:= P02 0=

P0312 EI⋅

Lc−

0.18

−⎛⎜⎝

⎞⎟⎠

⋅16⋅

3 EI⋅Ld

0.116

⎛⎜⎝

⎞⎟⎠

⋅16⋅+:= P03 220.83=

P046 EI⋅Lc

0.18

−⎛⎜⎝

⎞⎟⎠

⋅:= P04 600−=

P056 EI⋅Lc

0.18

−⎛⎜⎝

⎞⎟⎠

⋅3 EI⋅Ld

0.116

⎛⎜⎝

⎞⎟⎠

⋅+:= P05 475−= as above

there are no nodal forces in this case Pf

0

0

0

0

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=


Uf lsolve K Pf P0−,( ):= UfT 37.86− 5.73 77.63− 6.77 4.7( ) 10 3−=


V Af Uf⋅ Vd+:= VT 3.15− 2.57 0.9− 0.14 0.9 1.18− 1.16( ) 10 3−=


Q Ks V⋅:= QT 49.81− 26.58 26.58− 9.94− 9.94 23.23− 23.23( )=



va

V1

V2

⎛⎜⎜⎝

⎞⎟⎟⎠

:= va3.15−

2.57⎛⎜⎝

⎞⎟⎠

10 3−= ka2 EI⋅La

2

1

1

2⎛⎜⎝

⎞⎟⎠

⋅:= qa ka va⋅:= qa49.81−

26.58⎛⎜⎝

⎞⎟⎠

=

vb

V3

V4

⎛⎜⎜⎝

⎞⎟⎟⎠

:= vb0.9−

0.14⎛⎜⎝

⎞⎟⎠

10 3−= kb2 EI⋅Lb

2

1

1

2⎛⎜⎝

⎞⎟⎠

⋅:= qb kb vb⋅:= qb26.58−

9.94−

⎛⎜⎝

⎞⎟⎠

=

Page 543


vc

V5

V6

⎛⎜⎜⎝

⎞⎟⎟⎠

:= vc0.9

1.18−

⎛⎜⎝

⎞⎟⎠

10 3−= kc2 EI⋅Lc

2

1

1

2⎛⎜⎝

⎞⎟⎠

⋅:= qc kc vc⋅:= qc9.94

23.23−

⎛⎜⎝

⎞⎟⎠

=

vd V7( ):= vd 1.16( ) 10 3−= kd3 EI⋅Ld

:= qd kd vd⋅:= qd 23.23( )=

49.91

26.58

9.94

23.23


deformed shape

V

3.15−

2.57

0.9−

0.14

0.9

1.18−

1.16

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

10 3−=


Page 544

Script for Example 37a in CE220 class notes % gable frame with inextensible elements: solution with separate FEDEASLab functions


define model geometry and element types % specify node coordinates (could only specify non-zero terms) XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 0 12]; % second node, etc XYZ(3,:) = [ 8 18]; % XYZ(4,:) = [ 16 12]; % XYZ(5,:) = [ 16 0]; % % connectivity array CON {1} = [ 1 2]; CON {2} = [ 2 3]; CON {3} = [ 3 4]; CON {4} = [ 4 5]; % boundary conditions (1 = restrained, 0 = free) BOUN(1,:) = [1 1 1]; BOUN(5,:) = [1 1 0]; % specify element type [ElemName{1:4}] = deal('Lin2dFrm'); % 2d linear frame element Model = Create_SimpleModel (XYZ,CON,BOUN,ElemName);

Post-processing functions on Model (optional) Create_Window (0.80,0.80); % open figure window Plot_Model (Model); % plot model (optional) Label_Model (Model);

element properties in ElemData ne = Model.ne; % number of elements in structural model % Modulus, area and moment of inertia for el=1:ne; ElemData{el}.E = 1000; ElemData{el}.A = 1; % irrelevant value, since it will be constrained ElemData{el}.I = 80; end

Page 545

collection of element stiffness matrices Ks = Ks_matrix(Model,ElemData);

identify deformation modes with inextensible elements Idef = Idef_index(Model,ElemData,1:Model.ne);

form stiffness matrix % form stiffness matrix at the original free dofs by direct assembly Kf = Kf_matrix(Model,ElemData); % set up constraint matrix Ac LC = LinConEqs(Model); % reorder dofs to have horizontal translation at node 2 and vertical translation at node 3 pick_dof = [Model.DOF(2,1), Model.DOF(3,1)]; dof_reord = [setdiff(1:Model.nt,pick_dof) pick_dof]; % constraint matrix Ac = Ac_matrix(LC,dof_reord); % extract free dofs of constraint matrix Acf = Ac(1:Model.nf,:); % compatibility matrix for unconstrained dofs A = A_matrix (Model); % compatibiliy matrix for constrained dofs Aftild = A*Ac; % Method A: stiffness matrix for constrained free dofs (transformation after direct assembly) Kftild = Acf'*Kf*Acf; disp('The stiffness matrix at the constrained free dofs is'); disp(Kftild); % Method B: stiffness matrix for constrained dofs ("hand calculation" if Aftild available) Kftild = Aftild'*Ks*Aftild; disp('The stiffness matrix at the constrained free dofs is'); disp(Kftild); The stiffness matrix at the constrained free dofs is 6.4444e+003 -4.6667e+003 -6.4444e+003 0 4.6667e+003 -3.3333e+003 -4.6667e+003 5.8667e+004 8.0000e+003 1.6000e+004 0 0 -6.4444e+003 8.0000e+003 7.5556e+003 0 -1.3333e+003 6.6667e+003 0 1.6000e+004 0 6.4000e+004 1.6000e+004 0 4.6667e+003 0 -1.3333e+003 1.6000e+004 5.8667e+004 1.3333e+004 -3.3333e+003 0 6.6667e+003 0 1.3333e+004 2.6667e+004 The stiffness matrix at the constrained free dofs is 6.4444e+003 -4.6667e+003 -6.4444e+003 0 4.6667e+003 -3.3333e+003 -4.6667e+003 5.8667e+004 8.0000e+003 1.6000e+004 0 0 -6.4444e+003 8.0000e+003 7.5556e+003 0 -1.3333e+003 6.6667e+003 -4.5475e-013 1.6000e+004 4.5475e-013 6.4000e+004 1.6000e+004 0 4.6667e+003 0 -1.3333e+003 1.6000e+004 5.8667e+004 1.3333e+004 -3.3333e+003 0 6.6667e+003 0 1.3333e+004 2.6667e+004

1. Load case: applied nodal forces Pe(3,2) = -30; Pe(3,3) = 100; Pe(4,1) = 20; Loading = Create_Loading(Model,Pe);

Page 546

% transform applied nodal forces to constrained dofs Pftild = Acf'*Loading.Pref; % displacement method of analysis Uftild = Kftild\Pftild; % display result for constrained free dof displacements disp('constrained free dof displacements under the applied nodal forces'); disp(Uftild); % transform to complete displacement vector Uf = Acf*Uftild; % display result for free dof displacements only disp('complete free dof displacements under the applied nodal forces'); disp(Uf); constrained free dof displacements under the applied nodal forces 5.6657e-002 -6.8125e-003 7.5884e-002 3.5623e-003 -1.1865e-003 -1.1296e-002 complete free dof displacements under the applied nodal forces 5.6657e-002 0 -6.8125e-003 7.5884e-002 -2.5636e-002 3.5623e-003 9.5111e-002 0 -1.1865e-003 -1.1296e-002

determine the basic forces % determine element deformations V = Aftild*Uftild; % determine basic forces from force-deformation relation (only deformable) Q = Ks(Idef,Idef)*V(Idef); disp('the basic forces under the applied nodal forces are'); disp(Q); the basic forces under the applied nodal forces are 9.8023e+001 7.1894e+000 -7.1894e+000 1.5881e+002 -5.8807e+001 -1.3479e+002 1.3479e+002 4.2633e-014

2. Load case: uniformly distributed load in element b % clear previously defined applied force vector (always a good idea!) clear Pe; % equivalent nodal forces Pe(2,1) = 15; Pe(2,2) = -20; Pe(3,1) = 15; Pe(3,2) = -20; Loading = Create_Loading(Model,Pe);

Page 547

% specify uniformly distributed load in element b ElemData{2}.w = [0;-5]; % initial deformation vector V0 = V0_vector(Model,ElemData); % initial basic force vector (only for deformable elements) Q0 = -Ks(Idef,Idef)*V0(Idef); % initial force vector at constrained dofs P0tild = Aftild(Idef,:)'*Q0; % transform applied nodal forces to constrained dofs Pftild = Acf'*Loading.Pref; % displacement method of analysis Uftild = Kftild\(Pftild-P0tild); % display result for constrained free dof displacements disp('constrained free dof displacements under distributed load in element b'); disp(Uftild); % transform to complete displacement vector Uf = Acf*Uftild; % display result for free dof displacements only disp('complete free dof displacements under uniform distributed load in element b'); disp(Uf); constrained free dof displacements under uniform distributed load in element b 8.2713e-002 -7.9252e-003 9.4597e-002 3.2808e-003 -2.5940e-003 -1.2013e-002 complete free dof displacements under uniform distributed load in element b 8.2713e-002 0 -7.9252e-003 9.4597e-002 -1.5846e-002 3.2808e-003 1.0648e-001 0 -2.5940e-003 -1.2013e-002

determine the basic forces % determine element deformations V = Aftild*Uftild; % determine basic forces from force-deformation relation (only deformable) Q = Ks(Idef,Idef)*V(Idef) + Q0; disp('the basic forces under the uniform distributed load in element b are'); disp(Q); the basic forces under the uniform distributed load in element b are 1.7004e+002 6.4370e+001 -6.4370e+001 3.1594e+001 -3.1594e+001 -1.2559e+002 1.2559e+002 9.9476e-014

Page 548

3. Load case: vertical translation of 0.1 units at right support % no nodal forces element deformations due to support translation (by hand) Vd = [0 0 0 0 -1/8 -1/8 1/16 1/16]'.*0.1; % initial basic force vector (only for deformable elements) Q0 = Ks(Idef,Idef)*Vd; % initial force vector at constrained dofs P0tild = Aftild(Idef,:)'*Q0; % displacement method of analysis Uftild = Kftild\(-P0tild); % display result for constrained free dof displacements disp('constrained free dof displacements under upward support movement of node 5'); disp(Uftild); % transform to complete displacement vector Uf = Acf*Uftild; % display result for free dof displacements only disp('complete free dof displacements under upward support movement of node 5'); disp(Uf); constrained free dof displacements under upward support movement of node 5 -3.7858e-002 5.7290e-003 -7.7634e-002 6.7688e-003 4.6957e-003 2.9534e-003 complete free dof displacements under upward support movement of node 5 -3.7858e-002 0 5.7290e-003 -7.7634e-002 5.3035e-002 6.7688e-003 -1.1741e-001 0 4.6957e-003 2.9534e-003

determine the basic forces % determine element deformations V = Aftild*Uftild; V(Idef) = V(Idef) + Vd; % determine basic forces from force-deformation relation (only deformable) Q = Ks(Idef,Idef)*V(Idef); disp('the basic forces under upward support movement of node 5 are'); disp(Q); the basic forces under upward support movement of node 5 are -4.9808e+001 2.6578e+001 -2.6578e+001 -9.9407e+000 9.9407e+000 -2.3230e+001 2.3230e+001 -6.9278e-014

Page 549

Script for Example 37b in CE220 class notes % gable frame with inextensible elements: solution with direct stiffness


define model geometry and element types % specify node coordinates (could only specify non-zero terms) XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 0 12]; % second node, etc XYZ(3,:) = [ 8 18]; % XYZ(4,:) = [ 16 12]; % XYZ(5,:) = [ 16 0]; % % connectivity array CON {1} = [ 1 2]; CON {2} = [ 2 3]; CON {3} = [ 3 4]; CON {4} = [ 4 5]; % boundary conditions (1 = restrained, 0 = free) BOUN(1,:) = [1 1 1]; BOUN(5,:) = [1 1 0]; % specify element type [ElemName{1:4}] = deal('Lin2dFrm'); % 2d linear frame element Model = Create_SimpleModel (XYZ,CON,BOUN,ElemName);

Post-processing functions on Model (optional) Create_Window (0.80,0.80); % open figure window Plot_Model (Model); % plot model (optional) Label_Model (Model);

element properties in ElemData ne = Model.ne; % number of elements in structural model % Modulus, area and moment of inertia for el=1:ne; ElemData{el}.E = 1000; ElemData{el}.A = 1e6; % inextensible ElemData{el}.I = 80; end

1. Load case: applied nodal forces Pe(3,2) = -30;

Page 550

Pe(3,3) = 100; Pe(4,1) = 20; Loading = Create_Loading(Model,Pe);




2. Load case: uniformly distributed load in element b % no nodal forces Loading = Create_Loading(Model);

Page 551

% specify uniformly distributed load in element b ElemData{2}.w = [0;-5];




3. Load case: vertical translation of 0.1 units at right support % no nodal forces Pe = []; % specify support displacement

Page 552

Ue(5,2) = 0.1; Loading = Create_Loading(Model,Pe,Ue); % erase uniformly distributed load in element b ElemData{2}.w = [0;0];


plot deformed shape Create_Window(0.80,0.80); MAGF = 40; % magnification factor Plot_Model(Model); % original configuration Plot_Model(Model,U); % deformed configuration (chords only) Plot_DeformedStructure(Model,ElemData,U);


Page 553

CE220 - Theory of Structures Ex 38 - Displacement method for gable frame with inflexible element © Prof. Filip C. Filippou, 2000

Example 38 - Displacement method for gable frame with inflexible constraint

The geometry of the portal frame is given. Without trivial dofs, there are 9 independent free global dofs.

1

2

3

4

5

6

7

8

9

a

b c

d

8 8

12

6

Geometric properties: La 12:=

Lb 10:=

Lc 10:=

Ld 12:=

Material properties EI 80000:=

Under the assumption that all elements are inextensible and element b is inflexible, there are threeindependent free global degrees of freedom. We select these as shown in the following figure

Page 554


Step 1: Set up compatibility matrix for independent free global dofs

dof 1 to determine the first column of the compatibility matrix Af consult Example 17

1 1

1.33

θa112

−:= θb16

:= θc16

−:= θd112

:=

Af1⟨ ⟩

θa−

θa− θb+

θc− θb+

θc−

θd−

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:= Af1⟨ ⟩

112

14

13

16

112

−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

dof 2 to determine the second column of the compatibility matrix Af consult Example 17

θa 0:= θb16

−:= θc16

:= θd16

−:=

1

2

1.33

Af2⟨ ⟩

θa−

θa− θb+

θc− θb+

θc−

θd−

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:= Af1⟨ ⟩

112

14

13

16

112

−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

Page 555


dof 3 to determine the third column of the compatibility matrix Af

1

Af3⟨ ⟩

0

0

0

1

1

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=

It is interesting to look at the equilibrium equations that do not involve axial forces.To this end we use the principle of virtual displacements. Pf Af

T Q⋅=

the transpose of the kinematic matrix is

AfT

112

0

0

14

16

−

0

13

13

−

0

16

16

−

1

112

−

16

1

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

=

the numbering of basic forces without axial forces and basic forces in element b is shown in the figure

1Q

2Q

3Q

4Q

5Q

The equilibrium equations are

P1Q1

12

Q2

4+

Q3

3+

Q4

6+

Q5

12−=

P2Q2

6−

Q3

3−

Q4

6−

Q5

6+=

P3 Q4 Q5+=

Page 556


Step 2: Set up the structure stiffness matrix:


KsEILa

4

2

0

0

0

2

4

0

0

0

0

0

4La

Lc⋅

2La

Lc⋅

0

0

0

2La

Lc⋅

4La

Lc⋅

0

0

0

0

0

3La

Ld⋅

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

⋅:=



8.77

7.8−

9

7.8−

7.52

7.33−

9

7.33−

52

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

103=

Method 2(a): express element stiffness matrix in terms of global dof displacements and add upDetermination of individual element stiffness contributions

kaEILa

4

2

2

4⎛⎜⎝

⎞⎟⎠

⋅:= kcEILc

4

2

2

4⎛⎜⎝

⎞⎟⎠

⋅:= kd3EILd

:=

Afa

112

14

0

16

−

0

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:= Afc

13

16

13

−

16

−

0

1

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:= Afd112

−16

1⎛⎜⎝

⎞⎟⎠

:=


2.41

1.3−

0

1.3−

0.74

0

0

0

0

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

103=


6.22

6.22−

10.67

6.22−

6.22

10.67−

10.67

10.67−

32

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

103=

Page 557



0.14

0.28−

1.67−

0.28−

0.56

3.33

1.67−

3.33

20

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

103=

Then the stiffness matrix is the sum of the element contributions: K Ka Kc+ Kd+:=

K

8.77

7.8−

9

7.8−

7.52

7.33−

9

7.33−

52

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

103= same of course

Method 2(b): Derivation of stiffness coefficients by direct method (PVD)

We confirm only the stiffness coefficients for translation dof 1 with the principle of virtual displacements

Impose a unit value at displacement dof 1 and determine element deformations and forces, as shown in thefollowing figure. The stiffness coefficients are the forces at the free global dofs due to the element forcescaused by a unit displacement at global dof 1. In steps we can write:

1 1

1.33

6 1 4 112 6a a

EI EIL L

⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

31K

21K

11K

6 1 2 112 6a a

EI EIL L

⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

6 1 4 16 6c c

EI EIL L

⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

3 112d

EIL

⎛ ⎞⎜ ⎟⎝ ⎠

6 1 2 16 6c c

EI EIL L

⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠


K11Q'112

Q'24

+Q'33

+Q'46

+Q'512

−=

K21Q'26

−Q'33

−Q'46

−Q'56

+=

K31 Q'4 Q'5+=

Page 558


We demonstrate only for the first term

111 6 1 2 1 1 6 1 4 1

12 12 6 4 12 6

1 6 1 4 1 1 6 1 2 1 1 3 13 6 6 6 6 6 12 12

a a a a

c c c c d

EI EI EI EIL L L L


⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + + + − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

K

⎟

etc. etc....


Load Case: Nodal Forces

30

100

20

consult Lecture 14, pp. 7-10

Tf c f= AP Pfrom

( ) ( )14 130 20 1 100 43.333 6

⎛ ⎞= − + − + = −⎜ ⎟⎝ ⎠

P

( ) ( )24 130 20 2 100 63.333 6

⎛ ⎞ ⎛ ⎞= − − + + − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

P

3 0=P

Pf

43.33−

63.33

0

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

:=


Uf lsolve K Pf,( ):= UfT 31.87 41.84 0.39( ) 10 3−=


V Af Uf⋅:= VT 2.66 0.99 3.33− 1.28− 4.7( ) 10 3−=


Q Ks V⋅:= QT 84.05 61.88 126.85− 94.08− 94.08( )=

Page 559



va

V1

V2

⎛⎜⎜⎝

⎞⎟⎟⎠

:= va2.66

0.99⎛⎜⎝

⎞⎟⎠

10 3−= ka2 EI⋅La

2

1

1

2⎛⎜⎝

⎞⎟⎠

⋅:= qa ka va⋅:= qa84.05

61.88⎛⎜⎝

⎞⎟⎠

=

vc

V3

V4

⎛⎜⎜⎝

⎞⎟⎟⎠

:= vc3.33−

1.28−

⎛⎜⎝

⎞⎟⎠

10 3−= kc2 EI⋅Lc

2

1

1

2⎛⎜⎝

⎞⎟⎠

⋅:= qc kc vc⋅:= qc126.85−

94.08−

⎛⎜⎝

⎞⎟⎠

=

vd V5( ):= vd 4.7( ) 10 3−= kd3 EI⋅Ld

:= qd kd vd⋅:= qd 94.08( )=

Step 6: Determine shear forces by element equilibrium and then use the equilibrium equations that involvethe axial forces to determine the latter. Note that the axial forces cannot be determined from the force-deformation relations like the end moments, because the elements are inextensible and thus their axialdeformations are zero (this is left as an exercise)


84.05

65.88

126.85

94.0865.88

b

226.85

Note: the basic end moments of element b are obtained from node equilibrium at nodes 2 and 3;to this end we use the basic forces of elements a and c and the applied load values.

Page 560


deformed shape

V

2.66

0.99

3.33−

1.28−

4.7

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

10 3−=


Step 7: Determine support reactions (left as exercise)

Step 8: check global equilibrium (left as exercise)

Page 561


Second load case: distributed uniform load in element b

w=5

2520

1525

20

15

8 8

12

6

25

w=5

initial state

equivalent nodal forces due to distributed loads (same as Example 37)

Pfw

11.67

41.67−

0

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

:=

initial forces (fixed-end moments) due to distributed load in element b w 5:=

Q0 0 0 0 0 0( )T:=



11.67

41.67−

0

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

= there are no nodal forces in this case Pf

0

0

0

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

:=


Uf lsolve K Pf P0−,( ):= UfT 47.25 53.99 0.56−( ) 10 3−=


V Af Uf⋅:= VT 3.94 2.81 2.25− 1.69− 4.5( ) 10 3−=

Page 562



Q Ks V⋅ Q0+:= QT 142.52 127.55 98.86− 89.92− 89.92( )=



va

V1

V2

⎛⎜⎜⎝

⎞⎟⎟⎠

:= va3.94

2.81⎛⎜⎝

⎞⎟⎠

10 3−= ka2 EI⋅La

2

1

1

2⎛⎜⎝

⎞⎟⎠

⋅:= qa ka va⋅:= qa142.52

127.55⎛⎜⎝

⎞⎟⎠

=

vc

V3

V4

⎛⎜⎜⎝

⎞⎟⎟⎠

:= vc2.25−

1.69−

⎛⎜⎝

⎞⎟⎠

10 3−= kc2 EI⋅Lc

2

1

1

2⎛⎜⎝

⎞⎟⎠

⋅:= qc kc vc⋅:= qc98.86−

89.92−

⎛⎜⎝

⎞⎟⎠

=

vd V5( ):= vd 4.5( ) 10 3−= kd3 EI⋅Ld

:= qd kd vd⋅:= qd 89.92( )=


142.52

127.55

98.86

89.92

b

Note: the basic end moments of element b are obtained from node equilibrium at nodes 2 and 3;to this end we use the basic forces of elements a and c and the applied load values.

Page 563


deformed shape

V

3.94

2.81

2.25−

1.69−

4.5

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

10 3−=


Page 564

Direct stiffness implementation of displacement method

ne(el)T (el)b

el 1== ∑ AP por

with 0= +kq v q g= av u

(el) (el)b= Au U

(a)T (a) (b)T (b) (el)T (el)b b b= + + +A A AP p p p

we get

which can be written in compact form as e 0= +kp u p

where Te g g=k a ka is the element stiffness matrix in the global reference system

andT T

0 g 0 r w= +a ap q p

T Tg r w= +a ap q p

is the initial element force vector (fixed-end forces) in the global reference

(1)

(2)

Substituting (2) into (1) we getne

(el)T (el) (el) (el)b e 0

el 1=

⎡ ⎤= +⎢ ⎥⎣ ⎦∑ A kP u p

We recall the relation between element end displacements and global dof displacements

and substituting into (3) we get

(3)

ne(el)T (el) (el) (el)b e b 0

el 1ne ne

(el)T (el) (el) (el)T (el)b e b b 0

el 1 el 1

=

= =

⎡ ⎤= +⎢ ⎥⎣ ⎦

⎡ ⎤= +⎢ ⎥⎢ ⎥⎣ ⎦

∑

∑ ∑

A k A

A k A A

P U p

U p

which can be written compactly 0= +KP U P

wherene

(el)T (el) (el)b e b

el 1== ∑K A k A is the structure stiffness matrix

andne

(el)T (el)0 b 0

el 1== ∑ AP p is the initial force vector for the structure

The last two expressions reveal that the structure stiffness matrix and the initial force vector can be obtained by direct summation of the element stiffness matrix and the initial force vector after the latter have been transformed from the local to the global degrees of freedom by the Boolean matrix of the element. This process is called "direct assembly", because in the computer implementation each term of the structure stiffness matrix is updated by addition of the corresponding element stiffness term. The same is true for the initial force vector. Before assembly the element stiffness matrix and initial force vector are expressed relative to the global reference system according to the relations.

Te g g=k a ka T T

0 g 0 r w= +a ap q p

These operations can be done element by element, which makes the process modular and very effective.Because of this fact, the direct stiffness implementation of the displacement method of analysis is the exclusive method of analysis used in commercial software in structural engineering.

In Matlab this becomesK (id,id) = K (id,id) + ke;

( ) ( )T T T T T T T T Tg r g 0 r g g 0 r g g g 0 rw w w w= + = + + = + + = + +a a a k a a ka a a ka a ap q p v q p u q p u q p


Page 565

The stiffness matrix of a 2d frame element in global coordinates is

Element stiffness matrix in global coordinates

2 2

2 2

Te g g 2 2 2 2

2 2

2 2 2 2

2 2

0 0 0 0

0 1 0 4 20 1 0

2 40 0 1

0 0 1

X Y YL L LY X X EA X Y X YL L L L L L L L

EI EI Y X Y XX Y Y L L L L L L

EI EIL Y X Y XL LL LY X X L L L L

L L L

Δ Δ Δ⎡ ⎤− − −⎢ ⎥⎢ ⎥⎢ ⎥Δ Δ Δ ⎛ Δ Δ Δ Δ⎡ ⎤− − −⎢ ⎥ ⎜⎢ ⎥⎢ ⎥ ⎜⎢ ⎥⎢ ⎥ ⎜ Δ Δ Δ Δ⎢ ⎥= = − −⎢ ⎥ ⎢ ⎥Δ Δ Δ⎢ ⎥ ⎢ ⎥⎢ ⎥ Δ Δ Δ Δ⎢ ⎥ − −⎢ ⎥ ⎢ ⎥Δ Δ Δ ⎣ ⎦⎢ ⎥ ⎝− −⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

k a k a

⎞⎟⎟⎟

⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟

⎠

The special cases of a 2d truss and a 2d frame element with an end moment release can be obtained by suitable modification. The elegant way of accomplishing this is described on the following page.

2d truss

e0

0 0

0

XLYL

EA X Y X YX L L L L LLYL

Δ⎡ ⎤−⎢ ⎥⎢ ⎥

Δ⎢ ⎥−⎢ ⎥⎢ ⎥

Δ Δ Δ Δ⎡ ⎤ ⎛ ⎞⎢ ⎥= − −⎜ ⎟⎢ ⎥⎢ ⎥Δ ⎣ ⎦ ⎝ ⎠⎢ ⎥⎢ ⎥⎢ ⎥Δ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

k

2d frame element with an end moment release at node j

2

2

e

2 2 2 22

2

0 000 13 1 00

0 0

X YL LY XL X Y X YL EA

L L L LLY X Y XX Y EI

L L L L L LLY XL L

Δ Δ⎡ ⎤− −⎢ ⎥⎢ ⎥⎢ ⎥Δ Δ−⎢ ⎥

Δ Δ Δ Δ⎛ ⎞⎡ ⎤⎢ ⎥ − −⎜ ⎟⎢ ⎥⎢ ⎥⎜ ⎟⎢ ⎥= ⎢ ⎥ Δ Δ Δ Δ⎜ ⎟Δ Δ ⎢ ⎥⎢ ⎥ − −⎜ ⎟⎢ ⎥⎢ ⎥ ⎣ ⎦ ⎝ ⎠⎢ ⎥

Δ Δ⎢ ⎥−⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

k


Page 566

Basic force-deformation for 2d frame with or w/o end moment releases

It is possible to describe the force-deformation behavior of a 2d frame element with one expression by introducing the concept of a release switch. We demonstrate this for end moment releases.

iL

j(a)

(b)

0 or 10 or 1

i

j

mrmr

==

Release? n y

for hinge rotation: measured from tangent inside to tangent outside

2εv 3εv

3εv

2εv 3εv

2 2ε =v v

3 3ε =v v

or,

The following figure shows how the "strain" related element deformations vε depend on the element deformations v. It is the "strain" related element deformations that are related to the basic forces q.

Generalize and include axial deformation (without release)

2 2 2 2

3 3 3 3

1 0 0for hinge at end j for hinge at end i

0

12

112 0

ε ε

ε ε

⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎢ ⎥ ⎢ ⎥= =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣−

⎣ ⎦

−

⎦

v v v vv v v v

chord to node tangentchord to element tangentε

vv

( )( )

1 1

2 2

3 3

1 0 010 1 12

10 1 12

i j i

i j j

mr mr mr

mr mr mr

ε

ε

ε

⎡ ⎤⎢ ⎥

⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥= − − −⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎢ ⎥− − −

⎢ ⎥⎣ ⎦

v vv vv v

hε = av v

( )h hε= − = −I av v v v

With the relations hε = av vand we get Th εaq= q by PVD

Combining all these we get

0ε ε ε ε+kq = v q

T Th h h 0 0ε ε+ = +a k a a kq= v q v q with T

h hεk a k a=T

0 h 0εaq = qand

Without any end releases present h =a I the 3x3 identity matrix, so that εk k= 0 0εq =q

With a release at end node i we get 1 0i jmr mr= =

which shows that the "strain" related stiffness and intial forces are those we defined in pages 3-6

h

1 0 010 02

0 0 1

⎡ ⎤⎢ ⎥⎢ ⎥= −⎢ ⎥⎢ ⎥⎣ ⎦

aand thus

we perform the calculations on the following page


Page 567

With a release at end node i we get 1 0i jmr mr= = h

1 0 010 02

0 0 1

⎡ ⎤⎢ ⎥⎢ ⎥= −⎢ ⎥⎢ ⎥⎣ ⎦

aand thus

con'd

0 0 1 0 0 0 0 0 01 0 0 1 0 04 2 10 0 0 0 0 0 0 0 0 0 0 0 0 0 0

21 1 3 32 4 0 0 10 1 0 1 0 0 0 002 2

EA EA EAL L LEI EI

L LEI EIEI EIL LL L

⎡ ⎤⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= − = =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎣ ⎦ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

k

for the stiffness matrix we get

same as on page 5

for the initial force vector due to uniformly distributed element load

similar to the result on page 62

02

2

01 0 0 00 0 0 0

1210 12 812

y

yy

w L

w Lw L

⎛ ⎞ ⎛ ⎞⎜ ⎟⎡ ⎤ ⎜ ⎟⎜ ⎟⎢ ⎥ ⎜ ⎟⎜ ⎟⎢ ⎥ ⎜ ⎟⎜ ⎟= =⎢ ⎥ ⎜ ⎟⎜ ⎟⎢ ⎥ ⎜ ⎟⎜ ⎟⎢ ⎥− ⎜ ⎟⎜ ⎟ −⎜ ⎟⎣ ⎦ −⎜ ⎟ ⎝ ⎠⎝ ⎠

q

and the same is true for all other load cases


Page 568

Computer implementation of displacement method - Assembly of stiffness matrix

ne = Model.ne; % number of elements in structural modelnf = Model.nf; % number of free dofs in structural model nt = Model.nt; % number of total dofs in structural model K = zeros(nt,nt); % initialize structure stiffness matrix for el=1:ne; % locate element in Model and return end coordinates and id array [xyz id] = Localize(Model,el); % form element stiffness matrix ke in global reference system ke = ke_matrix(xyz,Model.ElemName{el},ElemData{el}); % assemble element stiffness matrix ke into structure stiffness matrix K K (id,id) = K (id,id) + ke; end Kf = K(1:nf,1:nf); % extract stiffness matrix of free dof's

Typical syntax is: Kf = Kf_matrix(Model,ElemData)

Matlab command for assembly of stiffness matrix

To illustrate the compact nature of the direct assembly process we use FEDEASLab function Kf_matrix which sets up the stiffness matrix at the free dof's for 2d/3d truss elements and 2d frame elements with end moment release(s)

The stiffness matrix of each element in global coordinates ke is supplied by the local function ke_matrix; the function uses the end coordinates of the element, its name (or type) and the element properties as input, as shown in the following page. This function separation illustrates the way of breaking down a process into modules for "modular programming".


Page 569

% determine element length and orientation[L dcx] = ElmLenOr(xyz); % direction cosines of element x-axis dXL = dcx(1); dYL = dcx(2); E = ElemData.E; A = ElemData.A; EA = E * A; switch ElemName case 'LinTruss' % linear truss element % transformation matrix from basic system to complete system in global reference ag = [-dcx' dcx']; k = EA/L; ke = ag'*k*ag; case 'Lin2dFrm' % 2d linear frame element with or w/o release % moment release indices MR: 0 indicates no hinge, 1 indicates hinge MRi = 0; MRj = 0; if (isfield(ElemData,'Release')) if (strncmpi(ElemData.Release(1),'y',1)), MRi = 1; end if (strncmpi(ElemData.Release(2),'y',1)), MRj = 1; end end % transformation matrix from basic system to complete system in global reference ag = [-dXL -dYL 0 dXL dYL 0; -dYL/L dXL/L 1 dYL/L -dXL/L 0; -dYL/L dXL/L 0 dYL/L -dXL/L 1]; I = ElemData.I; EI = E * I; k = [ EA/L 0 0; 0 4*EI/L 2*EI/L; 0 2*EI/L 4*EI/L]; % compatibility matrix in the presence of moment release(s) ah = [ 1 0 0; 0 1-MRi -0.5*(1-MRj)*MRi; 0 -0.5*(1-MRi)*MRj 1-MRj ]; % transform basic stiffness matrix for moment release(s) k = ah'*k*ah; % transform basic stiffness to global reference ke = ag'*k*ag; otherwise error ('element name not found in stiffness options') end

gX Y X YL L L LΔ Δ Δ Δ⎛ ⎞= − −⎜ ⎟

⎝ ⎠a for 2d

gX Y Z X Y ZL L L L L LΔ Δ Δ Δ Δ Δ⎛ ⎞= − − −⎜ ⎟

⎝ ⎠a for 3d

g 2 2 2 2

2 2 2 2

0 0

1 0

0 1

X Y X YL L L LY X Y X

L L L LY X Y X

L L L L

⎛ ⎞Δ Δ Δ Δ− −⎜ ⎟⎜ ⎟

Δ Δ Δ Δ⎜ ⎟= − −⎜ ⎟⎜ ⎟

Δ Δ Δ Δ⎜ ⎟− −⎜ ⎟⎝ ⎠

a for 2d

2 2 2L X Y ZΔ Δ Δ= + +

X Y ZL L L

Δ Δ Δdirection cosines

Te g gk a k a=

function ke_matrix

Computer implementation of displacement method - Element stiffness matrix

( )( )

h

1 0 010 1 12

10 1 12

i j i

i j j

mr mr mr

mr mr mr

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥= − − −⎢ ⎥⎢ ⎥⎢ ⎥− − −⎢ ⎥⎣ ⎦

a

Te g gk a k a=

Th hεk a k a=


Page 570

SUBSTRUCTURESSYMMETRY

Objective: force-deformation relation of substructures; condensation of internal dofs; generalization of dof definition; symmetry; structural theorems

CE220-Theory of Structures Substructures - Symmetry © Prof. Filip C. Filippou, 2000

Page 571

Substructure solution - there are 3 approaches

1. Form flexibility matrix (most logical, but not available in commercial software)

Procedure: divide structural model dofs into two groups:(a) those that interact with other elements or substructures; we call these "external" but use the subscript r for retain(b) those that are internal to the substructure and therefore do not interact with other elements or substructures; we use the subscript c for these which stands for condense.

Assumption: forces at "internal dofs" are given (subscript c); forces at external dofs are not given since these will beobtained from the interaction with other substructures; this means that we seek to establish a force-deformation relationfor the "external" dofs.

rr rcr r

cr ccc c

⎡ ⎤⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

F FF F

U PU P

we are only interested in Ur r rr r rc c= +F FU P P

since Pc is given

r rr r r0= +FU P U0= +fv q vdeformation-force for substructure

e 0= +kp u p for substructure

( )10 0

−= − = +f kq v v v qadd rigid body modes, if necessary, but relation is alreadyestablished in the global reference system


Page 572

2. Static condensation (logical, available in more advanced commercial software, e.g. aerospace, automobile industry)

rr rcr r

cr ccc c

⎡ ⎤⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

K KK K

P UP U

solve lower group of equations

( )1c cc c cr r

−= −K KU P U

0= +kq v q

with 1rr rc cc cr

−= −k K K K K the condensed stiffness matrix of the substructure

10 rc cc c

−= K Kq P the initial force vector for the substructure

since this relation is already established in the global coordinate system

e 0= +kp u p for substructure

( )1 1r rr rc cc cr r rc cc c

− −= − +K K K K K KP U P

and substitute into upper

support dofs can be included


Page 573

1

15F

25F

35F

45F

55F

1

14F

24F

34F

44F54F

1

13F

23F

33F

43F

53F

1

12F

22F

32F

42F

52F

111F

21F

31F

41F

51F

3. Set up flexibility matrix by typical structural engineering software program (can be a lot of work)

Apply a unit force in turn at each dof to be retained, determine the displacements at all other dofs to retain and set up flexibilibility matrix column by column; determine the displacements due to applied forces at internal dofs to set up the initial displacement vector U0 (not shown below)

r rr r r0= +FU P U

proceed as in case 1


Page 574

CE 220 - Theory of Structures Ex. 39 - Frame element with semi-rigid connectors © Prof. Filip C. Filippou, 2000

Example 39 - Frame element with semi-rigid connectors

Element with axial connector

We consider first the axial force-deformation behavior of a truss element with a semi-rigid connector (orflexible expansion joint) with spring stiffness ka. The location of the connector is not relevant and we show itat the end of the element. The deformable portion of the element has axial stiffness EA and length L.

L1q

axial spring ka

1

2free dofs

basic forces

elementEA

1q

1Q2Q2Q

viewed as a two-node element

viewed as a substructure

basic forces of substructure

From the "outside" the element is a typical two-node truss element with a special force-deformation relation.From the inside it is a substructure with 3 nodes and 2 elements. The semi-rigid connector is a zero lengthor point element, i.e. its end nodes have the same coordinates resulting in what is known in computermodels as a double node. The force-deformation behavior of the spring is described by its axial stiffness.The axial connector is assumed rigid in the transverse direction, so that no relative displacement normal tothe element axis is allowed.

The substructure is supported so that rigid body modes are eliminated. These can be added with theprocedure in page Part I-136. The substructure has, thus, only two free dofs, as shown. Dof 1 is critical forthe interaction of the element with other elements in the structural model ("the outside world"). Dof 2 isinternal to the element. We are interested in establishing the force-deformation relation at dof 1 only. Thiscan be done in one of two ways: first, by applying a unit force at dof 1 and determining the correspondingdisplacement (i.e. establishing the flexibility of the substructure only for the relevant dof); secondly, bysetting up the stiffness matrix for the substructure and using condensation of the internal dof.

The substructure has only two free dofs and two basic forces, if we do not include for now flexural effects(more on this later). The substructure is, therefore, statically determinate. The equilibrium equations are

P1 Q2=

P2 Q2− Q1+=

Assuming that there is no nodal force at the "internal" dof 2, we have P2 0=

we obtain Q1 Q2= P1= which of course we could have obtained by inspection a lot faster

Page 575


There is only one dof to be retained in this case, i.e. dof 1. We determine the flexibility for this dof anddenote it with subscript r. To do so, we apply a unit force at dof 1 and determine the correspondingdisplacement. Setting it up in general form we need the force influence matrix which expresses the effectof a unit force at the global dof on the basic element forces. We have from above

Bbar1

1⎛⎜⎝⎞⎟⎠

:= Fs

LEA

0

0

1ka

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:=EA

Frr BbarT Fs⋅ Bbar⋅:= Fs FrrL

EA1ka

+→

We observe that the substructure (element) flexibility is equal to the sum of the component flexibilities. Suchan element is known as a series element.

Conclusion: in a series element (i.e. an element with springs under constant axial force) the flexibility isequal to the sum of the component flexibilities.

Let us try the stiffness method with condensation of the internal dof, which is more time consuming for astatically determinate structure. There are two independent free global dofs, one to be retained (dof 1) andone to be condensed out (dof 2). Let us write the stiffness matrix of the substructure.

with Af0

1

1

1−⎛⎜⎝

⎞⎟⎠

:= Ks

EAL

0

0

ka

⎛⎜⎜⎝

⎞⎟⎟⎠

:=

EA

K AfT Ks⋅ Af⋅:= Ks K

ka

ka−

ka−

EAL

ka+

⎛⎜⎜⎝

⎞⎟⎟⎠

→

Since there is no applied force at the "internal dof" 2 we have the following relation

P2 0= K21 U1⋅ K22 U2⋅+= we solve for the displacement at dof 2 and obtain

U2 U1ka

EAL

ka+⋅= or, denoting the ratio

EALka

α= we get U2 U11

1 α+⋅=

substituting into the first equilibrium equation of the displacement method, i.e. P1 K11 U1⋅ K12 U2⋅+=

we get P1 K11 U1⋅ K121

1 α+⋅ U1⋅+= ka

α

1 α+⋅ U1⋅= EA

L1

1 α+⋅ U1⋅=

Thus, we have condensed out the internal dof by satisfying equilibrium and have obtained the condensedstiffness matrix for dof 1. It is a scalar in this case. Denoting the condensed stiffness matrix with a star wecan write

K*rrEAL

11 α+⋅= note that r refers to dof 1 in this case, as already stated

It is obvious that the condensed stiffness matrix is not the same as the stiffness coefficient K11 of theoriginal stiffness matrix of the substructure. Why? The stiffness coefficient K11 of the original stiffnessmatrix represents the force at dof 1 under a unit displacement of dof 1 with all other dof displacementvalues equal to zero. By contrast, the condensed stiffness matrix coefficient K*11 represents the force atdof 1 under a unit displacement at dof 1 with the value of dof 2 adjusted to satisfy equilibrium at that dof.

Page 576


Observe that the result from the two approaches is, of course, the same.

The earlier expression for the flexibility matrix can be written as FrrL

EA1ka

+= LEA

1 α+( )⋅=

and it is equal to the inverse of the condensed stiffness matrix K*11

Element with flexural connectors

After establishing the effect of a semi-rigid connector on the axial response, we look now at the sameproblem in flexure. We can then combine the axial and flexural relations to establish the force-deformationof a frame element with axial and flexural connectors. For a beam the location of the semi-rigid connectorsis important, since the bending moment is not uniform over the span. We assume at first that there are twosemi-rigid connectors, one at each end.

L

free dofs1

23

4

basic forces

EIrotational spring kr

3q2q

1Q

2Q

4Q

3Q

viewed as a two-node element

viewed as a substructure

basic forces of substructure

The substructure now consists of three components, a flexible beam element with length L and flexuralsection stiffness EI, and two semi-rigid connectors with different stiffness kri and krj in the general case.Noting that the connectors do not allow relative axial and transverse displacement there are four basicforces for the four free dofs of the substructure and the substructure is again statically determinate.It is also again supported so that rigid body modes are restrained; these can be added with the procedurein page Part I-136.

The 4 equilibrium equations (dofs) for 4 basic forces Q are:

P1 Q1−= note the definition of the basic force Q1 and Q4 for a point element!

P2 Q4=

P3 Q1 Q2+=

P4 Q3 Q4−=

Page 577


We need to retain dofs 1 and 2, with dofs 3 and 4 to be condensed out. We assume that there are noforces acting at the internal dofs 3 and 4. Thus, we have:

P3 0= Q2 Q1−=

P4 0= Q3 Q4=

we obtain therefore for P1 1= and P2 0= Q1 1−= Q2 1= Q4 0= Q3 0=

and for P1 0= and P2 1= Q1 0= Q2 0= Q4 1= Q3 1=

the force influence matrix and the collection of flexibility matrices are

the flexibility matrix of the substructurethen becomes

Bbar

1−

1

0

0

0

0

1

1

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:= Fs

1kri

0

0

0

0

L3 EI⋅

L6 EI⋅

−

0

0

L6 EI⋅

−

L3 EI⋅

0

0

0

0

1krj

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

kri

Frr BbarT Fs⋅ Bbar⋅:= Fs

We proceed to establish the contribution of each component separately:

left connector1−

0⎛⎜⎝

⎞⎟⎠

1kri⋅ 1− 0( )⋅

1kri

0

0

0

⎛⎜⎜⎝

⎞⎟⎟⎠

→

observe that the forceinfluence matrix is theidentity matrix; thus, theflexibility contribution tothe substructureflexibility is equal to thecomponent flexibility

flexible beam element1

0

0

1⎛⎜⎝

⎞⎟⎠

L6 EI⋅⋅

2

1−

1−

2⎛⎜⎝

⎞⎟⎠

⋅1

0

0

1⎛⎜⎝

⎞⎟⎠

⋅

13

LEI⋅

1−6

LEI⋅

1−6

LEI⋅

13

LEI⋅

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

→

right connector0

1⎛⎜⎝⎞⎟⎠

1krj⋅ 0 1( )⋅

0

0

0

1krj

⎛⎜⎜⎝

⎞⎟⎟⎠

→

The flexibility of the substructure is the sum of the component flexibilities "weighted" by the force influencecoefficients of each component. We obtain

Page 578


which can also be written FrrL

6 EI⋅

2

1−

1−

2⎛⎜⎝

⎞⎟⎠

⋅

1kri

0

0

1krj

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

+=Frr

1kri

13

LEI⋅+

1−6

LEI⋅

1−6

LEI⋅

13

LEI⋅

1krj

+

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

→

We see from the last expression that the substructure flexibility is equal to the beam element flexibility plusthe flexibility of the connectors; the latter is a diagonal matrix, since each connector is located at thecorresponding end node of the substructure and does not affect the deformation at the opposite node.

Let us derive an expression for the stiffness matrix of this substructure as the inverse of the flexibility matrix;for simplicity's sake we assume that the end connectors have the same stiffness.

After introducing the following stiffness ratio η6EIL kr⋅

=

we have fL

6 EI⋅

2 η+

1−

1−

2 η+

⎛⎜⎝

⎞⎟⎠

⋅= Note: the flexibility f of the element is equal to F of the substructure

then k f 1−= k6 EI⋅

L1

3 4 η⋅+ η2

+⋅

2 η+

1

1

2 η+

⎛⎜⎝

⎞⎟⎠

⋅=

we make a quick check by setting η=0 (infinitely stiff connectors). In this case we get k2 EI⋅

L

2

1

1

2⎛⎜⎝

⎞⎟⎠

⋅=

We can see clearly that the semi-rigid connectors reduce the stiffness of the substructure, since thedenominator is growing faster with increasing positive η than the diagonal terms.

We pursue again the static condensation approach, which is more complicated in this case. We do this forthe case of connectors with equal stiffness only.

The compatibility matrix for the 4 free dofs of the substructure is the transpose of the equilibrium matrix(coefficient matrix of Q's in earlier equilibrium equations). It is

Af

1−

0

0

0

0

0

0

1

1

1

0

0

0

0

1

1−

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:= Ks

kr

0

0

0

0

4 EI⋅L

2 EI⋅L

0

0

2 EI⋅L

4 EI⋅L

0

0

0

0

kr

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

kr

K AfT Ks⋅ Af⋅:= Ks

K

kr

0

kr−

0

0

kr

0

kr−

kr−

0

kr 4EIL

⋅+

2EIL

⋅

0

kr−

2EIL

⋅

kr 4EIL

⋅+

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

→

Page 579


We could have established the stiffness matrix of the substructure much faster by physical considerations,i.e. by getting the stiffness coefficients as forces at dof i under a unit displacement at dof j.

There are no applied forces at dofs 3 and 4, which means that we can use these two equations to solve forthe displacements of the dofs to be condensed in terms of the displacements of the dofs to be retained.The formula for this is available on page Part II-338.

rr rcr r

cr ccc c

⎡ ⎤⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

K KK K

P UP U ( )1

c cc c cr r−= −K KU P U

( )1 1r rr rc cc cr r rc cc c

− −= − +K K K K K KP U P

in our case r refers to dofs 1 and 2, c refers to dofs 3 and 4, Pc = 0

Krrkr

0

0

kr

⎛⎜⎝

⎞⎟⎠

:=kr

Krckr−

0

0

kr−

⎛⎜⎝

⎞⎟⎠

:=kr

Kcrkr−

0

0

kr−

⎛⎜⎝

⎞⎟⎠

:=kr

Kcc kr

123η⋅+

13η⋅

13η⋅

123η⋅+

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:= kr

with η as defined previously, i.e. η6EIL kr⋅

= the condensed stiffness matrix becomes

K*rr Krr Krc Kcc1−⋅ Kcr⋅−:= Krr K*rr factor

kr η2 η+

η 3+( ) η 1+( )⋅⋅⋅

ηkr

η 3+( ) η 1+( )⋅⋅

ηkr

η 3+( ) η 1+( )⋅⋅

kr η2 η+

η 3+( ) η 1+( )⋅⋅⋅

⎡⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎦

→

Noting that η 3+( ) η 1+( )⋅ expand 3 4 η⋅+ η2

+→ is the same as the earlier denominator ofthe stiffness matrix

and kr η⋅6 EI⋅

L= we have obtained the same result as with the inversion of the flexibility matrix of

the substructure but with more effort

Page 580


Basic frame element stiffness matrix with semi-rigid axial and flexural connectors

We know that as long as we use the centroidal axis of the element as reference axis the axial and flexuralresponses are uncoupled. Thus, the basic frame element stiffness matrix with semi-rigid axial and flexuralconnectors becomes

with αEAL ka⋅

=

k

EAL

11 α+⋅

0

0

0

6 EI⋅L

2 η+

1 η+( ) 3 η+( )⋅⋅

6 EI⋅L

11 η+( ) 3 η+( )⋅

⋅

0

6 EI⋅L

11 η+( ) 3 η+( )⋅

⋅

6 EI⋅L

2 η+

1 η+( ) 3 η+( )⋅⋅

⎡⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎦

= and η6EIL kr⋅

=

Beam with connectors offset from the end nodesWe would like to study now the case that the connectors are offset from the end nodes. This case isdepicted in the following figure. Formally, there are now 8 free dofs and 8 basic element forces requiring usto divide the flexible beam components into 3 elements as shown in the figure.

free dofs

1

2

basic forces

EI

0.25L 0.25L0.5L

3

4

5 6

7

8

unit force

1

1/L 1/L

unit force

1

1/L

1/L 0.750.25

0.750.25

rotational spring kr

1Q 2Q

4Q3Q

3q2q

Page 581


Instead of setting up 8 equilibrium equations in 8 unknown basic forces and solving them, we do this byinspection, since we are dealing with a simply supported beam under unit end moments. With this approachwe do not need to subdivide the flexible beam component into 3 elements, but can directly add the flexibilitymatrices of the 3 components of the substructure after "weighing" them by the force influence coefficients,as we have already discussed. The contribution of the flexible beam element remains the same as before.The force influence coefficients for the connectors are now

for the left connector the force influence coefficients are34

−14

⎛⎜⎝

⎞⎟⎠

for the right connector the force influence coefficients are14

−34

⎛⎜⎝

⎞⎟⎠

The contribution of the flexibility of the left connector to the substructure flexibility becomes

34

−

14

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

1kri⋅

34

−14

⎛⎜⎝

⎞⎟⎠

⋅

916 kri⋅

3−16 kri⋅

3−16 kri⋅

116 kri⋅

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

→

The contribution of the flexibility of the right connector to the substructure flexibility becomes

14

−

34

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

1krj⋅

14

−34

⎛⎜⎝

⎞⎟⎠

⋅

116 krj⋅

3−16 krj⋅

3−16 krj⋅

916 krj⋅

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

→

under the assumption that both end connector stiffness values are the same we get for the substructureflexibility

FrrL

6 EI⋅

2

1−

1−

2⎛⎜⎝

⎞⎟⎠

⋅1kr

58

38

−

38

−

58

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

⋅+=

Page 582

CE 220 - Theory of Structures Ex. 40 - Beam with semi-rigid connection and rigid zone © Prof. Filip C. Filippou, 2000

Example 40 - Beam with SR connection at midspan and rigid link

We study now a slightly more complicated case than Example 39. In this case the substructure consists of 4subelements including an inflexible segment. We also investigate the effect of a distributed element load onthe substructure force-deformation relation. The following figure shows the substructure as a 2-node beamelement whose force-deformation behavior is to be established. Since axial deformations are of no interest,the basic force vector consists of q2 and q3 only.

L/2 0 L/4 L/4

EI EIrigid

w

3q2q

Properties L 10:= EI 10000:= kr 3000:= (for reference only)

The following figure shows the substructure without rigid body modes. Nodes separate the substructureinto 4 elements. With the inflexible segment and without the rigid body modes the substructure has 5independent free global dofs. There are also 5 basic forces, as shown, and the substructure is staticallydeterminate. Note that we have not included any basic forces for the inflexible segment after introducingthe inflexible segment constraint.

34

51

2

1Q 2Q 4Q3Q

5Q

It is worth noting that the 5 independent free global dofs above result from the original set of 7 independentfree dofs, shown in the following figure, as a consequence of the constraint of the inflexible segment.

34

51

26 7 With the inflexible segment dofs 6 and 7depend on the translation dof 5.

We write the equilibrium equations for translation dofs 3 and 5 with the principle of virtual displacements

Before doing so, we write the moment equilibrium equations for dofs 1, 2 and 4 directly.

P1 Q1=

P2 Q5=

P4 Q2 Q3−=

Page 583


Equilibrium equations for dof 3

1

L/2 0 L/4 L/4

δVT 10.5 L⋅

−1

0.5 L⋅−

10.25 L⋅

−1

0.25 L⋅− 0⎛⎜

⎝⎞⎟⎠

=

the equation for dof 3 then reads

P3Q1 Q2+

0.5 L⋅−

Q3

0.25 L⋅−

Q4

0.25 L⋅−= or, simpler P3 0.25⋅ L⋅

Q1 Q2+

2− Q3− Q4−=

Equilibrium equations for dof 5

L/2 0 L/4 L/4

1δVT 0 0

10.25 L⋅

20.25 L⋅

10.25 L⋅

⎛⎜⎝

⎞⎟⎠

=

the equation for dof 5 then reads

P5Q3

0.25 L⋅

2 Q4⋅

0.25 L⋅+

Q5

0.25 L⋅+= or, simpler P5 0.25⋅ L⋅ Q3 2 Q4⋅+ Q5+=

Let us set up the equilibrium matrix for later use in checking the answers.

Bf

1

0

0.5−

0

0

0

0

0.5−

1

0

0

0

1−

1−

1

0

0

1−

0

2

0

1

0

0

1

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=

solving the equations for the case P1 1= P2 0= P3 P4= P5= 0=

and then separately for the case P1 0= P2 1= P3 P4= P5= 0=

Page 584


we get the following force influence matrix Bbar

Bbar

1

0.5−

0.5−

0.25

0

0

0.5

0.5

0.75−

1

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:= and we can check Bf Bbar⋅

1

0

0

0

0

0

1

0

0

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

=

However, for this simply supported beam it is much more straightforward to set up the force influence matrixby inspection, i.e. draw the moment diagram for each unit end force (moment) and read off the values at theends of the segments. The following figure shows the well known linear moment diagrams.

unit force

unit force

1

0.50

0.50 0.75

0.25

1Q 2Q 4Q3Q

5Q

1/L

1/L

1/L

1/L

1

The collection of element flexibility matrices is FsL4

16 EI⋅⋅

4

2−

0

0

0

2−

4

0

0

0

0

0

24L

EIkr

0

0

0

0

0

2

1−

0

0

0

1−

2

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

⋅:=

We can obtain the flexibility matrix of the substructure for the two dofs to be retained

Frr BbarT Fs⋅ Bbar⋅:= Frr

73192

37192

−

37192

−

61192

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

LEI

=

the element flexibility thus is f Frr:= and the element stiffness is k f 1−:= k3.8

2.3

2.3

4.54⎛⎜⎝

⎞⎟⎠

EIL

=

and we can see that the semi-rigid connector reduces the stiffness coefficients, while the rigid segmentincreases the stiffness coefficients; the increase is higher for node j, since the rigid segment is "off center"

Page 585


We can also show the component-by-component contributions to the substructure flexibility so as toillustrate the fact that it is the summation of the component flexibilities "weighted" by the force influencecoefficients. In the process we note, of course, that the rigid element does not contribute (zero flexibility)

left segment Fa

1

12

−

0

12

⎛⎜⎜⎝

⎞⎟⎟⎠

T

L24 EI⋅⋅

4

2−

2−

4⎛⎜⎝

⎞⎟⎠

⋅

1

12

−

0

12

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅:= Fa

724

112

−

112

−

124

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

LEI

=

connector Fb12

−12

⎛⎜⎝

⎞⎟⎠

T 1kr⋅

12

−12

⎛⎜⎝

⎞⎟⎠

⋅:= Fb

112

112

−

112

−

112

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

LEI

=

right segment Fc

14

0

34

−

1

⎛⎜⎜⎝

⎞⎟⎟⎠

T

L24 EI⋅⋅

2

1−

1−

2⎛⎜⎝

⎞⎟⎠

⋅

14

0

34

−

1

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅:= Fc

1192

5192

−

5192

−

37192

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

LEI

=

flexibility matrix of substructure Frr Fa Fb+ Fc+:= Frr

73192

37192

−

37192

−

61192

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

LEI

= as before

Page 586


Effect of distributed load wDistributed element loads acting over elements of the substructure cause displacements at the dofs to beretained. These correspond to initial deformations when the substructure is viewed as a single element. Todetermine these displacements we determine first the basic element forces under the distributed elementloads, then the deformations of each element in the substructure from the deformation-force relation, andfinally the displacements at the dofs to be retained by compatibility or by the principle of virtual forces.

For the given distributed load we can establish the basic element forces by inspection

3wL/8 wL/8

w

wL/4 wL/4

wL /16

0 L/4

2 wL /322

with w 20:=

the basic forces are Qp

0

116

116

132

−

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

w⋅ L2⋅:=

we check the answer with the equilibrium equations

Bf Qp⋅

0

0

116

−

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

w L2⋅= which agrees with P3 0.25⋅ L⋅Q1 Q2+

2− Q3− Q4−=

for P3wL4

−⎛⎜⎝

⎞⎟⎠

=

In addition to the basic forces distributed element loads also causeinitial deformations in the element(s) in which they act; in this casethis is limited to the left segment of the substructure

Vo

18

−

18

0

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

w L3⋅24 EI⋅⋅:=

The displacements at the dofs to be retained due to the element deformations can be established bycompatibility or by the principle of virtual forces.

Tf = BU VRecall from Part IV, page 11 that

Page 587


We do first the calculations element by element from left to right

left segment

Uoa

1

12

−

0

12

⎛⎜⎜⎝

⎞⎟⎟⎠

T

L24 EI⋅⋅

4

2−

2−

4⎛⎜⎝

⎞⎟⎠

⋅

0

116

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅ w⋅ L2⋅

1

12

−

0

12

⎛⎜⎜⎝

⎞⎟⎟⎠

T 18

−

18

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

⋅w L3⋅24 EI⋅⋅+:= Uoa

7384

−

1128

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

w L3⋅EI

=

semi-rigid connector

Uob12

−12

⎛⎜⎝

⎞⎟⎠

T 1kr⋅

116⋅ w⋅ L2⋅:= Uob

196

−

196

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

w L3⋅EI

=

right segment

Uoc

14

0

34

−

1

⎛⎜⎜⎝

⎞⎟⎟⎠

T

L24 EI⋅⋅

2

1−

1−

2⎛⎜⎝

⎞⎟⎠

⋅

132

−

0

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅ w⋅ L2⋅:= Uoc

11536

−

51536

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

w L3⋅EI

=

The displacements of substructure at the dofsto be retained are therefore

Uro Uoa Uob+ Uoc+:= Uro

15512

−

11512

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

w L3⋅EI

=

This can be obtained, of course, with a single operation Uro BbarT Fs Qp⋅ Vo+( )⋅:= Uro

15512

−

11512

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

w L3⋅EI

=

The displacements of the substructure at the dofs to be retained under elementloading represent the initial deformations of the two node element, i.e. vo Uro:=

and the corresponding fixed-end forces for use in thedisplacement method of analysis can be established from qo f 1−− vo⋅:= qo

0.741

0.362−

⎛⎜⎝

⎞⎟⎠

w L2⋅12

=

Page 588

CE 220 - Theory of Structures Ex. 41 - Static condensation for dynamic analysis © Prof. Filip C. Filippou, 2000

Example 41 - Static condensation for dynamic analysis

In this example we derive the condensed stiffness matrix of a cantilever model. The objective is to determinethe periods of vibration of this model under the assumption that the uniformly distributed mass is discretizedinto a lumped value at each node of the model. With more nodes the eigenvalue analysis approaches theresults of the actual model which has uniformly distributed mass with specific density ρ. In the following weinvestigate the case of 3 lumped mass terms. We are only interested in translatory effects and rotary masseffects are neglected. The following figure shows the model and the mass discretization.

5

5

5

m

m

m 1

2

3

EI

EI

EI

EI 120000:= L 5:=

The structural model has 5 free dofs asshown in the following figure

1

2

3

4

5

We are interested, however, only in the translation dofs (first three in the selected numbering scheme).We can derive the stiffness matrix at these three dofs either directly with the force method (note that themodel is statically determinate), or with the stiffness method after condensation of the rotation dofs.

Page 589


1. Approach: Force method (great for statically determinate structure)

Apply a unit force at each dof to be retained and determine the displacements at all dofs to be retained.This establishes the flexibility matrix of the model for the dofs to be retained. We do this long hand and incompact form in the following.

1

2

3

unit force

unit force

5

5

5

unit force

Page 590


The basic forces are numbered as shown in the following figure

1Q

2Q

3Q

4Q

5Q

The basic force values under a unit force at each dof to be retained are (note thatthese are actually force influence coefficients)

unit force at dof 1 unit force at dof 2 unit force at dof 3

Q1 5= Q1 10= Q1 15=

Q2 0= Q2 5−= Q2 10−=

Q3 0= Q3 5= Q3 10=

Q4 0= Q4 0= Q4 5−=

Q5 0= Q5 0= Q5 5=

The compatibility equations are:

V1U1

5= it follows U1 5 V1⋅=

V2U1

5U4+=

V3U2 U1−

5U4+= U2 5 V3 V2−( )⋅ 2 U1⋅+= 5 V3 V2−( )⋅ 10 V1⋅+=

V4U2 U1−

5U5+=

V5U3 U2−

5U5+= U3 5 V5 V4−( )⋅ 2 U2⋅+ U1−=

U3 5 V5 V4−( )⋅ 10 V3 V2−( )⋅+ 15 V1⋅+=

observe that the coefficients of the deformations V for each dof are equal to the valuesof the transpose of the corresponding column of the force influence matrix (we did nothave to prove this, but just in case!). This agrees with Part IV, page 11, i.e. T

f = BU V

Page 591


We do the calculations in compact form, even though hand calculations are only slightly more timeconsuming (take these as an exercise)

Bbar

5

0

0

0

0

10

5−

5

0

0

15

10−

10

5−

5

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:= FsL

6EI

2

1−

0

0

0

1−

2

0

0

0

0

0

2

1−

0

0

0

1−

2

0

0

0

0

0

2

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

⋅:=

Frr BbarT Fs⋅ Bbar⋅:= Frr

8.33

20.83

33.33

20.83

66.67

116.67

33.33

116.67

225

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

LEI

=

the condensed stiffness matrix for the three translation dofs can then be obtained as the inverse of F

K*rr Frr1−:= K*rr

0.738

0.425−

0.111

0.425−

0.406

0.148−

0.111

0.148−

0.065

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

EIL

=

2. Approach: Condensation of stiffness matrix

We have already established the compatibility matrix for all free dofs. We restate it in matrix form

V1U1

5=

V2U1

5U4+=

V3U2 U1−

5U4+= Af

1L

1L

1L

−

1L

−

0

0

0

1L

1L

1L

−

0

0

0

0

1L

0

1

1

0

0

0

0

0

1

1

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

V4U2 U1−

5U5+=

V5U3 U2−

5U5+=

Page 592


The collection of element stiffness matrices is

Ks2EIL

2

1

0

0

0

1

2

0

0

0

0

0

2

1

0

0

0

1

2

0

0

0

0

0

1.5

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

⋅:=

the structure stiffness matrix becomes K AfT Ks⋅ Af⋅:= K

0.96

0.48−

0

0

1.2−

0.48−

0.6

0.12−

1.2

0.6

0

0.12−

0.12

0

0.6

0

1.2

0

8

2

1.2−

0.6

0.6

2

7

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

EIL

=

we extract the necessary submatrices of the stiffness matrix and perform the condensation

Krr submatrix K 1, 3, 1, 3,( ):= Krr

0.96

0.48−

0

0.48−

0.6

0.12−

0

0.12−

0.12

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

EIL

=

Krc submatrix K 1, 3, 4, 5,( ):= Krc

0

1.2

0

1.2−

0.6

0.6

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

EIL

=

Kcr submatrix K 4, 5, 1, 3,( ):= Kcr0

1.2−

1.2

0.6

0

0.6⎛⎜⎝

⎞⎟⎠

EIL

=

Kcc submatrix K 4, 5, 4, 5,( ):= Kcc8

2

2

7⎛⎜⎝

⎞⎟⎠

EIL

=

K*rr Krr Krc Kcc1−⋅ Kcr⋅−:= K*rr

0.738

0.425−

0.111

0.425−

0.406

0.148−

0.111

0.148−

0.065

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

EIL

= same as with the force method

The same result for the condensed stiffness matrix can be obtained with function Condense_Kf ofFEDEASLab (consult Example_41a.m). It is interesting to run a parametric analysis of the eigenfrequenciesof the problem for different model discretizations. The results are:

Period of first mode Period of second mode Period of third mode

4 elements 2.2270 (sec) 0.3945 (sec) 0.1584 (sec)



Page 593

Script for Example 41a in CE220 class notes % eigenmodes and frequencies of flexible cantilever column (step by step)


Create model % specify node coordinates (could only specify non-zero terms) nn = 4; ne = nn-1; XYZ(:,1:2) = [ zeros(nn,1) linspace(0,15,nn)']; % connectivity array CON (1:ne) = num2cell ([ [1:(nn-1)]' [ 2:nn]'],2); % boundary conditions (1 = restrained, 0 = free) (specify only restrained dof's) BOUN(1,:) = [1 1 1]; % specify element type [ElemName{1:ne}] = deal('Lin2dFrm'); % 2d linear elastic frame element % create Model Model = Create_SimpleModel (XYZ,CON,BOUN,ElemName);


Define element properties % Modulus, area, moment of inertia and loading of beams (elements 1 through 3) for el=1:ne; ElemData{el}.E = 1000; ElemData{el}.A = 4*10^5; ElemData{el}.I = 120; end

form stiffness matrix Kf = Kf_matrix(Model,ElemData);

specify nodal mass (not required for CE220) % define distributed mass m m = 5;

Page 594

l = 10/ne; Me(2:nn,1) = (m*l).*ones(ne,1); Me( nn,1) = m*l/2; % create nodal mass vector Ml = Create_LumpedMass(Model,Me);

solve dynamic eigenvalue problem (not required for CE220) % condense stiffness matrix to dofs with non-zero mass idm = find(Ml); % find dofs with non-zero terms in lumped mass vector Kmm = Condense_MV(Kf,idm); % create diagonal mass matrix from lumped mass vector Md = diag(Ml(idm)); % solve eigenvalue problem [phi omega] = eig (Kmm,Md); % sort eigenvalues in ascending order [omega io] = sort(diag(omega)); % number of eigenmodes nmode = length(omega); % put eigenmodes in the same order as the eigenvalues phi = phi (:,io(1:nmode)); % eigenfrequencies are equal to square root of eigenvalues omega = sqrt(omega(1:nmode))'; % determine eigenmode values at all free dofs Um = zeros(Model.nf,nmode); Um(idm,:) = phi; Ueig = Kf\(repmat(omega.^2,length(Ml),1).*repmat(Ml,1,nmode).*Um); % echo eigenmode periods disp('The three lowest eigenmode periods are'); T = 2*pi./omega; disp(T(1:3)); The three lowest eigenmode periods are 2.2270e+000 3.9452e-001 1.5844e-001

plot 3 mode shapes with lowest eigenvalues Create_Window (0.80,0.80); Plot_Model (Model); MAGF = 2; U = zeros(Model.nt,1); % plot for i=1:min(3,length(omega)); U(1:Model.nf,1) = Ueig(:,i); Plot_DeformedStructure(Model,[],U); end

Page 595

Script for Example 41b in CE220 class notes % eigenmodes and frequencies of flexible cantilever column in single step


Create model % specify node coordinates (could only specify non-zero terms) nn = 4; ne = nn-1; XYZ(:,1:2) = [ zeros(nn,1) linspace(0,15,nn)']; % connectivity array CON (1:ne) = num2cell ([ [1:(nn-1)]' [ 2:nn]'],2); % boundary conditions (1 = restrained, 0 = free) (specify only restrained dof's) BOUN(1,:) = [1 1 1]; % specify element type [ElemName{1:ne}] = deal('Lin2dFrm'); % 2d linear elastic frame element % create Model Model = Create_SimpleModel (XYZ,CON,BOUN,ElemName);


Define element properties

Modulus, area, moment of inertia and loading of beams (elements 1 through 3)

for el=1:ne; ElemData{el}.E = 1000; ElemData{el}.A = 4*10^5; ElemData{el}.I = 120; end

Page 596


specify nodal mass % define distributed mass m m = 5; l = 10/ne; Me(2:nn,1) = (m*l).*ones(ne,1); Me( nn,1) = m*l/2; % create nodal mass vector Ml = Create_LumpedMass(Model,Me);

solve dynamic eigenvalue problem with FEDEASLab function EigenMode [omega Ueig] = EigenMode4LumpedMass (Kf,Ml,3); % echo eigenmode periods disp('The three lowest eigenmode periods are'); T = 2*pi./omega; disp(T(1:3)); The three lowest eigenmode periods are 2.2270e+000 3.9452e-001 1.5844e-001

plot 3 mode shapes with lowest eigenvalues Create_Window (0.80,0.80); Plot_Model (Model); MAGF = 2; U = zeros(Model.nt,1); for i=1:min(3,length(omega)); U(1:Model.nf,1) = Ueig(:,i); Plot_DeformedStructure(Model,[],U); end

Page 597

CE 220 - Theory of Structures Ex. 42 - Substructuring © Prof. Filip C. Filippou, 2000

Example 42 - Substructuring

In this example we will analyze the following structure under the given loading. We intend to demonstratethe substructuring concept and plan to analyze the structure in two parts, assuming that two separateteams are occupied with the analysis of each part. The following figure shows the complete structure.

8 8

6

2010

20

6

4 4

All elements are inextensible and the flexural section stiffness is EI 50000:=The two parts of the structure are shown in the following figure.

8 8

6

detail of double node

a

b

c d

20

dof 1

dof 2

2010

Vierendeel truss

Page 598


It is clear that the two substructures interact only at one node with 2 dofs. The vertical translation of thenode is zero on account of the inextensible element assumption. The process for the analysis of the twoparts is as follows.

Team A is responsible for the part of the structure that consists of a Vierendeel truss. This team is supposedto generate a condensed stiffness matrix and an initial force vector at dofs 1 and 2 in the above figure. Thisis done with FEDEASLab functions in the file Example_42b. The condensed stiffness matrix and initial forcevector are:

ka4.4826 103⋅

4.9044 103⋅

4.9044 103⋅

4.8347 104⋅

⎛⎜⎜⎝

⎞⎟⎟⎠

:= qa01.744−

17.192−

⎛⎜⎝

⎞⎟⎠

:=

Team B is responsible for the second substructure. The free dofs and basic forces of the substructure areshown in the following figure

1

2

34

56

1Q

2Q 3Q

4Q5Q

6Q 7Q

4 4

6

There are 6 independent free dofs or equilibrium equations and 7unknown basic forces. Thus, the degree of static indeterminacy isNOS=1 and it would be much, much better to solve the substructureby the force method. We will do this later. We solve now by thedisplacement method, which is only advisable because we haveaccess to a computer. The compatibility matrix is:

Af

1

0

16

−

16

−

0

0

0

0

1

1

0

0

0

0

0

0

16

16

0

0

0

0

0

0

1

1

0

0

0

0

0

0

14

−

14

−

14

0

0

0

0

0

1

1

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

Page 599


The collection of element stiffness matrices is:

Ks

4.4826 103⋅

4.9044 103⋅

0

0

0

0

0

4.9044 103⋅

4.8347 104⋅

0

0

0

0

0

0

0

4 EI⋅6

2 EI⋅6

0

0

0

0

0

2 EI⋅6

4 EI⋅6

0

0

0

0

0

0

0

4 EI⋅4

2 EI⋅4

0

0

0

0

0

2 EI⋅4

4 EI⋅4

0

0

0

0

0

0

0

3 EI⋅4

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

the stiffness matrix at the free dofs of the model becomes

Kf AfT Ks⋅ Af⋅:= Kf

7.26

3.429−

2.778−

8.333−

0

0

3.429−

81.68

8.333

16.667

0

0

2.778−

8.333

2.778

8.333

0

0

8.333−

16.667

8.333

83.333

18.75−

25

0

0

0

18.75−

11.719

9.375−

0

0

0

25

9.375−

87.5

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

103=

The initial force vector can be determined from the initial forces of the elements (fixed-end forces)

with Q0 1.744− 17.192− 0 0 0 0 0( )T:= we get P0 AfT Q0⋅:=

and the applied force vector for the given loading is Pf 0 0 10− 0 20− 0( )T:=

we solve for the displacements at the free dofs of the substructure

Uf lsolve Kf Pf P0−,( ):= Uf

3.104−

1.153

11.228−

0.355

1.334−

0.244−

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

10 3−=

Page 600


and determine the deformations and basic forces of the substructure

V Af Uf⋅:= VT 3.104− 1.153 0.201− 0.999− 0.688 0.089 0.578−( ) 10 3−=

Q Ks V⋅ Q0+:= QT 10− 23.35 23.35− 36.65− 36.65 21.67 21.67−( )=

any other information, e.g. support reactions, global equilibrium, deformed shape of the substructure, etc.can be obtained as usually.

Team B now passes to Team A the forces that correspond to dofs 1 and 2. Team A then proceeds toanalyze substructure A with this information. This is done in the file Example_42c. One can compare theabove results for substructure B and the results for substructure A from Example_42c with the results of theentire structure under the given loading analyzed in Example_42a. The two sets of results are, of course,identical.

We illustrate briefly the solution of substructure B by the force method. There are 6 equilibrium equations in7 unknown basic forces and NOS=1. The equilibrium equations are:

P1 Q1Q3 Q4+

6−=

P2 Q2 Q3+=

P3Q3 Q4+

6= Af

T

1

0

0

0

0

0

0

1

0

0

0

0

16

−

1

16

0

0

0

16

−

0

16

1

0

0

0

0

0

1

14

−

0

0

0

0

0

14

−

1

0

0

0

0

14

1

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

P4 Q4 Q5+=

P5Q5 Q6+

4−

Q7

4+=

P6 Q6 Q7+=

Select Q2 as redundant basic force.

The particular solution with Q2 = 0 gives Q3 0= Q4 60−= Q1 10−= Q7 10−=

Q5 60= Q6 10=

Qp 10− 0 0 60− 60 10 10−( )T:=

could do by inspection (faster!)

Page 601


check the answer AfT Qp⋅

0

0

10−

0

20−

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

= good!

Homogeneous solution with Q2 = 1 gives Q3 1−= Q4 1= Q1 0= Q7 0.5−=

Q5 1−= Q6 0.5=

Bbarx 0 1 1− 1 1− 0.5 0.5−( )T:= could do by inspection (faster!)

check the answer AfT Bbarx⋅

0

0

0

0

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

= good!

invert the stiffness of substructure A and the initial force vector to get the flexibility and initial deformationvector for the substructure A dofs 1 and 2.

fa ka1−:= fa

12.547

1.273−

1.273−

1.163⎛⎜⎝

⎞⎟⎠

1EI

= va0 fa− qa0⋅:= va00

17.78⎛⎜⎝

⎞⎟⎠

1EI

=

Fs1EI

12.547

1.273−

0

0

0

0

0

1.273−

1.163

0

0

0

0

0

0

0

63

66

−

0

0

0

0

0

66

−

63

0

0

0

0

0

0

0

43

46

−

0

0

0

0

0

46

−

43

0

0

0

0

0

0

0

43

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

⋅:= V0

0

17.78

0

0

0

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

1EI⋅:=

Page 602


compatibility condition QxBbarx

T Fs Qp⋅ V0+( )⋅


−:= Qx 23.347=

Q Qp Bbarx Qx⋅+:= QT 10− 23.35 23.35− 36.65− 36.65 21.67 21.67−( )=

same basic forces as with displacement method

Team B now passes to Team A the forces that correspond to dofs 1 and 2. Team A then proceeds toanalyze substructure A with this information. This is done in the file Example_42c.

Page 603

Script for Example 42a in CE220 class notes % substructure analysis: solution of complete structure by single analysis


Create model % specify node coordinates (could only specify non-zero terms) XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 8 0]; % second node, etc XYZ(3,:) = [ 16 0]; % XYZ(4,:) = [ 0 6]; % XYZ(5,:) = [ 8 6]; % XYZ(6,:) = [ 16 6]; % XYZ(7,:) = [ 16 12]; % XYZ(8,:) = [ 20 12]; % XYZ(9,:) = [ 24 12]; % % connectivity array CON { 1} = [ 1 2]; CON { 2} = [ 2 3]; CON { 3} = [ 1 4]; CON { 4} = [ 2 5]; CON { 5} = [ 3 6]; CON { 6} = [ 4 5]; CON { 7} = [ 5 6]; CON { 8} = [ 6 7]; CON { 9} = [ 7 8]; CON {10} = [ 8 9]; % boundary conditions (1 = restrained, 0 = free) BOUN(1,:) = [1 1 0]; BOUN(3,:) = [0 1 0]; BOUN(9,:) = [0 1 0]; % specify element type ne = length(CON); [ElemName{1:ne}] = deal('Lin2dFrm'); % 2d linear elastic frame element % create Model Model = Create_SimpleModel (XYZ,CON,BOUN,ElemName);


Page 604


applied nodal forces Pe(5,2) = -20; Pe(7,1) = -10; Pe(8,2) = -20; Loading = Create_Loading(Model,Pe);



plot bending moment distribution Create_Window(0.80,0.80); Plot_Model (Model); Plot_2dMomntDistr (Model,ElemData,Post,1.25);

Page 605

Script for Example 42b in CE220 class notes % substructure A stiffness and initial force vector


Create model % specify node coordinates (could only specify non-zero terms) XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 8 0]; % second node, etc XYZ(3,:) = [ 16 0]; % XYZ(4,:) = [ 0 6]; % XYZ(5,:) = [ 8 6]; % XYZ(6,:) = [ 16 6]; % % connectivity array CON { 1} = [ 1 2]; CON { 2} = [ 2 3]; CON { 3} = [ 1 4]; CON { 4} = [ 2 5]; CON { 5} = [ 3 6]; CON { 6} = [ 4 5]; CON { 7} = [ 5 6]; % boundary conditions (1 = restrained, 0 = free) (specify only restrained dof's) BOUN(1,:) = [1 1 0]; BOUN(3,:) = [0 1 0]; % specify element type ne = length(CON); [ElemName{1:ne}] = deal('Lin2dFrm'); % 2d linear elastic frame element % create Model Model = Create_SimpleModel (XYZ,CON,BOUN,ElemName);


1 2 3

4 5 6

1 2

3 4 5

6 7

Page 606



select dofs to retain ir = [Model.DOF(6,1) Model.DOF(6,3)];

applied nodal forces Pe(5,2) = -20; Loading = Create_Loading(Model,Pe); Pf = Loading.Pref;


determine condensed stiffness matrix and initial force vector for substructure A with static condensation Kf = Kf_matrix(Model,ElemData); [Kfr Pf0] = Condense_MV (Kf,ir,Pf); format short e disp('the condensed stiffness matrix is') disp(Kfr); disp('the initial force vector is') disp(-Pf0); the condensed stiffness matrix is 4.4826e+003 4.9044e+003 4.9044e+003 4.8347e+004 the initial force vector is -1.7440e+000 -1.7192e+001

determine condensed stiffness matrix by inverse of flexibility matrix B = B_matrix(Model); Bf = B(1:Model.nf,:); Fs = Fs_matrix(Model,ElemData); [Bbar Bvbar F] = ForceMethod(Bf,Fs); disp('the inverse of the flexibility matrix at the horizontal translation dofs is') disp(inv(F(ir,ir))); the inverse of the flexibility matrix at the horizontal translation dofs is 4.4826e+003 4.9044e+003 4.9044e+003 4.8347e+004

Page 607

Script for Example 42c in CE220 class notes % substructure A analysis with results from Substructure B


Create model % specify node coordinates (could only specify non-zero terms) XYZ(1,:) = [ 0 0]; % first node XYZ(2,:) = [ 8 0]; % second node, etc XYZ(3,:) = [ 16 0]; % XYZ(4,:) = [ 0 6]; % XYZ(5,:) = [ 8 6]; % XYZ(6,:) = [ 16 6]; % % connectivity array CON { 1} = [ 1 2]; CON { 2} = [ 2 3]; CON { 3} = [ 1 4]; CON { 4} = [ 2 5]; CON { 5} = [ 3 6]; CON { 6} = [ 4 5]; CON { 7} = [ 5 6]; % boundary conditions (1 = restrained, 0 = free) (specify only restrained dof's) BOUN(1,:) = [1 1 0]; BOUN(3,:) = [0 1 0]; % specify element type ne = length(CON); [ElemName{1:ne}] = deal('Lin2dFrm'); % 2d linear elastic frame element % create Model Model = Create_SimpleModel (XYZ,CON,BOUN,ElemName);


1 2 3

4 5 6

1 2

3 4 5

6 7

Page 608


applied nodal forces (forces at node 6 from substructure B analysis in Mcad example) Pe(5,2) = -20; Pe(6,1) = -10; Pe(6,3) = 23.35; Loading = Create_Loading(Model,Pe);



plot bending moment distribution Create_Window(0.80,0.80); Plot_Model (Model); Plot_2dMomntDistr (Model,ElemData,Post,1.25);

Page 609

General case of symmetry in 3d structural models

For plane symmetry the structural geometry and structural properties satisfy the condition g(X,Y,Z) = g(X',Y',Z')

for a coordinate system transformation such that

X' = - X Y' = Y Z' = Z

Note that the prime coordinate system is this time left-handed,since only one axis reverses sign (-1)1. Consequently, momentsand rotations need to obey the left-hand rule.

We can identify the plane of symmetry by noting that the right half of thestructural model is obtained by reflection of the left half of the modelabout the Y-Z plane passing through the point of plane symmetry (shownin gray in the figure). Note that the X-Y and X-Z planes can also serveas planes of symmetry, as can any plane in space for complex models.

plane of symmetry

X'

Y'

Z'

X

Y

Z

Plane symmetry

axis of symmetry

X

Y

Z

X'

Y'Z'

For axis symmetry the structural geometry and structural properties satisfy the condition g(X,Y,Z) = g(X',Y',Z')

for a coordinate system transformation such thatX' = - X Y' = - Y Z' = Z

Note that the prime coordinate system is also right-handed,since two axes reverse sign (-1)2

We can identify the axis of symmetry by noting that the right half of the structural model is obtained by rotation of 180o about the Z-axisat the point of symmetry (dashed line). Note that the X- and Y-axescan also serve as axes of symmetry, as can any line in space forcomplex structural models.

Axis symmetry

CE220 - Theory of Structures Filip C. Filippou

Page 610

X

Y

Z

point of symmetry

X'

Y'

Z'

For point symmetry the structural geometry and structural properties satisfy the condition g(X,Y,Z) = g(X',Y',Z')

for a coordinate system transformation with X' = - X Y' = - Y Z' = - Z

Note that the prime coordinate system is left-handed, since three axes reverse sign, i.e. (-1)3

Moments and rotations obey a left-hand rule in a left-hand coordinate system!

The origin of the coordinate system X,Y,Z is, of course, arbitrary,but once selected it fixes the origin of X',Y',Z'.

Point symmetry

Restricting ourselves to 2d structural models we identify only 2 types of symmetry:(1) point symmetry or symmetry about axis normal to the plane(2) plane symmetry or symmetry about axis in the plane (reflection)

Z and Z' axis towards viewer (reflection)

X

Y

trace of symmetry plane

reflection

X'

Y'

point of symmetry

X

Y

X'

Y'

Z - axis towards viewer

Z' - axis away from viewer

Symmetry in 2d structural models


Page 611

Z and Z' axis towards viewer (reflection)

X

Y

trace of symmetry plane

reflection

X'

Y'

Structure geometry and structural properties satisfy the condition g(X,Y) = g(X',Y')

Under reflection about the symmetry plane both Z and Z' axes are pointing towards the viewer but the prime coordinate system is left handed. Thus, we conclude:

X-Y-Z is right-handed CCW moments and rotations are positive

X'-Y'-Z' is left-handedwith Z' towards viewer

CW moments and rotations are positive

Sign convention for plane symmetry in a 2d structural model

point of symmetry

X

Y

X'

Y'

Structure geometry and structural properties satisfy the condition g(X,Y) = g(X',Y')

Under point symmetry the Z-axis points towards the viewer but the Z'-axis points away from the viewer. The prime coordinate system is left handed.Thus, we conclude:

X-Y-Z is right-handed CCW moments and rotations are positive

X'-Y'-Z' is left-handedwith Z' away from viewer

CCW moments and rotations are positive

Sign convention for point symmetry in a 2d structural model


Page 612

This leads to the following general definition of a "degree of freedom" or dof

DOF = a set of node displacement values that is linearly independent from all other dofs of the model(in the end there should be nf linearly independent displacement vectors)

Similarly, we select redundant force fields such that they are linearly independent from all other redundant force fields(in the end there should be NOS linearly indepedent basic force fields or combinations of basic forces)

Take advantage of the symmetry in the geometry of the structural model by

collecting the original set of independent global free dofs into symmetric and anti-symmetric sets for the displacement method of analysis

grouping basic forces in the same fashion for the force method of analysis; we call each such group a "force field"

The loading does not need to be symmetric or anti-symmetric. Under any loading the structural model response separates into symmetric and anti-symmetric responses with a smaller number of independent free dofs, or smaller number of redundants each. Note, however, that the total number of independent free dofs or the NOS of the structure is not affected by thedecomposition!

Under symmetric or anti-symmetric loading only the corresponding set of dofs gets "excited", so that only a solutionof the symmetric or anti-symmetric response is required and the problem solution is substantially shortened.

s

s

X

Y

+X'

Y'

+

Symmetric independent free global dofs

Taking advantage of symmetric structural models in "hand calculations"

Example 44 - Single bridge span with external prestressing

12 3

4

5 630 40 30

10a

b c

d

e

fg

h

s

s

Geometry

12

3

4

12

3

4

Original set of independent free global dofs

1 32 4

57

86

s

s

nf = 8

nfs = 4

X

Y

X'Y's

s

Anti-symmetric independent free global dofs

nfa = 4

5

6

78

56

7

8


Page 613

s

s

a

b c

d

e

fg

h dof 1

s

a

b c

d

e

fg

h dof 2

Deformations under symmetric independent free global dofs

Deformations under anti-symmetric independent free global dofs

s

s

a

b c

d

ef g

h dof 5

K AfT Ks⋅ Af⋅:=

K

666.67

2000−

0

600−

0

0

0

0

2000−

90000

0

0

0

0

0

0

0

0

470.76

56.92−

0

0

0

0

600−

0

56.92−

618.97

0

0

0

0

0

0

0

0

891.67

2500

0

600−

0

0

0

0

2500

150000

0

0

0

0

0

0

0

0

170.76

56.92−

0

0

0

0

600−

0

56.92−

618.97

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

103=

The stiffness matrix becomes

fs fs fs 0s

fa fa fa 0a

00

⎛ ⎞ ⎡ ⎤ ⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥

⎝ ⎠ ⎣ ⎦ ⎝ ⎠ ⎝ ⎠

KK

P U PP U P

12 3

4

5 630 40 30

10a

b c

d

e

fg

h

50s

s

if the loading is symmetric, then only symmetric responsefs fs fs 0s

fa fa fa 0a

00

⎛ ⎞ ⎡ ⎤ ⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥

⎝ ⎠ ⎣ ⎦ ⎝ ⎠ ⎝ ⎠

KK

P U PP U P= 0

if the loading not symmetric, solve symmetric and anti-symmetric response separately and superimpose results

12 3

4

5 630 40 30

10a

b c

d

e

fg

h

50 s

s

fs fs fs 0s

fa fa fa 0a

00

⎛ ⎞ ⎡ ⎤ ⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥

⎝ ⎠ ⎣ ⎦ ⎝ ⎠ ⎝ ⎠

KK

P U PP U P

fsU

faU


Page 614

Another example

15

34

2

s

s

geometry and element properties are symmetric about plane

1

2

1

2X

Y

+X'

Y'

+

s

s

symmetric dofs

X

Y

+X'

Y'

+

34

55

s

s

antisymmetric dofs

3 1U =

0.750.75

ICb ICcICa

ICd

4

a

b

c

X

Y

+

X'

Y'

+

4

a

b

c

X

Y

+

X'

Y'

+

0

0

20 20

20

1 2

3

4

a

b

c

point symmetry

X

Y

+

X'

Y'

+

4

a

b

c

X

Y

+

X'

Y'

+

1

2

3

independent free dofs for inextensible elements

Example 43 - Frame with point symmetry

1

1 3

2

2

3


Page 615

X

Y

+

X'

Y'

+

2Q′

1Q′

2Q′1Q′

3Q′3Q′

X

Y

+

X'

Y'

+

1Q′′2Q′′

3Q′′3Q′′

1Q′′

2Q′′

symmetric dof's

2

X

Y

+

X'

Y'

+

0

0

2 1

anti-symmetric dof's

4

X

Y

+

X'

Y'

+

3

3

4

P1Q'2 Q'2+

60.8 Q'3 Q'3+( )⋅+

Q'23

1.6 Q'3⋅+

P2 Q'1 Q'2+ Q'2+ Q'1+ 2 Q'1⋅ 2 Q'2⋅+

P3Q"1 Q"1+

8− 0.6 Q"3 Q"3+( )⋅+

Q"1

4− 1.2 Q"3⋅+

P4 Q"1 Q"2+ Q"2+ Q"1+ 2 Q"1⋅ 2 Q"2⋅+

Example 45 - Frame with point symmetry - Solution by force method

1 more unknown than equations 1 more unknown than equations

NOS = 2

Introduce special boundary conditions at plane or point of symmetry and analyze only "sample" portion of structureFor general loading need to decompose it into symmetric and anti-symmetric components (see end of presentation)Need to conduct separate analyses for symmetric and anti-symmetric loading, since boundary conditions change in each case

Under symmetric loading -> response is symmetricAt symmetry plane displacement in X and rotation about Z are zero.

Equivalent boundary condition:

symmetry plane or axis


X

Y

+X'

Y'

+

symmetric loading

Under anti-symmetric loading -> response is anti-symmetricAt symmetry plane displacement in Y is zero




X

Y

+X'

Y'

+

anti-symmetric loading

Analysis of plane symmetry by computer

Taking advantage of symmetric structural models in computer analysis


Page 616

Under anti-symmetric loading -> response is anti-symmetricAt point of symmetry rotation about Z is zero


symmetry point or axis

point of symmetry

free to translate, no rotation

X'

Y'

+

X

Y

+

Under symmetric loading -> response is symmetricAt point of symmetry displacements in X and Y are zero


symmetry point or axis

point of symmetry

X

Y

+

X'

Y'

+

Analysis of point symmetry by computer

symmetric loading

anti-symmetric loading

+

1610

8 85 5

8 85 5

30

X

Y

+X'

Y'

+

15 15 1515

symmetric load case anti-symmetric load case

Decomposition of loading for computer analysis


Page 617

CE 220 - Theory of Structures Example 43 - Frame with point symmetry © Prof. Filip C. Filippou, 2000

Example 43 - Frame with point symmetry

The geometry is shown in the following figure. The element properties are

EIa 106:= EIb 3 106⋅:= EIc 1 106⋅:=

La 20:= Lb 20:= Lc 20:=

20 20

20

1 2

3

4

a

b

c

point symmetry

X

Y

+

X'

Y'

+

With this geometry and the given element properties the structural model is point symmetric about themidpoint of element b, as shown in the figure above.

If axial deformations are neglected and no advantage is taken of the point symmetry thestructural model has 3 independent free global dofs, as shown in the following figure.

4

a

b

c

X

Y

+

X'

Y'

+

1

2

3

Page 618


To take advantage of the point symmetry of the structural model we group the independent free global dofsinto a symmetric and antisymmetric set. We do this by looking at the original dofs and considering theircounterparts in the prime coordinate system. The total number of dofs does not change in the process.

symmetric dof

4

a

b

c

X

Y

+

X'

Y'

+

1

1

0

0

symmetric translation of the original dof 1 isnot possible, if element b is inextensible

the compatibility matrix becomesantisymmetric dofs

4

a

b

c

X

Y

+

X'

Y'

+

2

3

2

3

Af

0

1

1

1

1

0

1La

−

1La

−

0

0

1La

1La

0

1

1

1−

1−

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

The first column of the compatibility matrix represents the symmetric dof and the other two the antisymmetricdofs. Observe that the entries of the lower half of the compatibility matrix are the "reflection" of the entries ofthe upper half: for the symmetric dof the signs are equal, for the antisymmetric dofs the signs are opposite.

Page 619


Alternatively, the compatibility matrix for the symmetric and antisymmetric dofs can be set up directly.

We apply each global dof in turn, while holding the others equal to zero. We determine the elementdeformations and write them in the compatibility matrix Af

degree of freedom 1

20 20

20

1 1U =

1 1U =

Af1⟨ ⟩

0

1

1

1

1

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

degree of freedom 2

20 20

20

1 1=U

Af2⟨ ⟩

1La

−

1La

−

0

0

1Lc

1Lc

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

Page 620


degree of freedom 3

20 20

20

3 1U =

3 1U =

Af3⟨ ⟩

0

1

1

1−

1−

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

complete compatibility matrix Af

0

1

1

1

1

0

120

−

120

−

0

0

120

120

0

1

1

1−

1−

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

= as before

Page 621


Form the structure stiffness matrix

set up collection of element stiffness matrices

Ks

4 EIa⋅

La

2 EIa⋅

La

0

0

0

0

2 EIa⋅

La

4 EIa⋅

La

0

0

0

0

0

0

4 EIb⋅

Lb

2 EIb⋅

Lb

0

0

0

0

2 EIb⋅

Lb

4 EIb⋅

Lb

0

0

0

0

0

0

4 EIc⋅

Lc

2 EIc⋅

Lc

0

0

0

0

2 EIc⋅

Lc

4 EIc⋅

Lc

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

The stiffness matrix of the structural model becomes Kf AfT Ks⋅ Af⋅:= Kf

2200

0

0

0

3

30−

0

30−

1000

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

103=

We observe that the stiffness matrix is now composed of two uncoupled submatrices, a 1x1 matrix thatcorresponds to the single symmetric dof and a 2x2 submatrix that corresponds to the antisymmetric dofs.This happens because the column of the compatibility matrix that corresponds to the symmetric dof isorthogonal with respect to the stiffness to the columns of the compatibility matrix that correspond to theantisymmetric dofs

Page 622


Structure stiffness matrix "by hand"

Set up the basic forces for each independent free dof and use the principle of virtual displacements todetermine the forces at the global dofs due to these forces (note that the basic forces correspond to thedeformations caused by a unit displacement at each global dof). The forces at the global dofs are bydefinition the stiffness coefficients. We demonstrate

4 a

a

EIL

2 a

a

EIL

6 b

b

EIL

6 b

b

EIL

4 c

c

EIL 2 c

c

EIL

basic forces due to unitdisplacement at dof 1

observe the deformed shape of element b;it corresponds to a symmetric deformationand results in symmetric basic forces

2EIL

2

1

1

2⎛⎜⎝

⎞⎟⎠

⋅1

1⎛⎜⎝⎞⎟⎠

⋅

6EIL

⋅

6EIL

⋅

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

→

26 a

a

EIL

26 a

a

EIL

26 c

c

EIL

26 c

c

EIL


2 b

b

EIL

4 c

c

EIL 2 c

c

EIL

4 a

a

EIL

2 a

a

EIL

2 b

b

EIL


observe the deformed shape of element b;it corresponds to an antisymmetric deformationand results in anti-symmetric basic forces

2EIL

2

1

1

2⎛⎜⎝

⎞⎟⎠

⋅1

1−⎛⎜⎝

⎞⎟⎠

⋅

2EIL

⋅

2−( )EIL

⋅

⎡⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎦

→

Page 623


In setting up the stiffness coefficients we look at the "upper half" of the structure and then double the result.

K11 2 4EIa

La6

EIb

Lb⋅+

⎛⎜⎝

⎞⎟⎠

⋅:= K11 2200103=

K21 0:= note that the virtual displacement for dof 2 is "orthogonal" to the forces due to dof 1

K31 0:=

K22 26EIa

La2

−⎛⎜⎜⎝

⎞⎟⎟⎠

1La

−⎛⎜⎝

⎞⎟⎠

⋅6EIa

La2

−⎛⎜⎜⎝

⎞⎟⎟⎠

1La

−⎛⎜⎝

⎞⎟⎠

⋅+⎡⎢⎢⎣

⎤⎥⎥⎦

⋅:= K22 3103=

K32 26EIa

La2

−⎛⎜⎜⎝

⎞⎟⎟⎠

1⋅⎡⎢⎢⎣

⎤⎥⎥⎦

⋅:= K32 30− 103=

K33 2 4EIa

La⋅ 2

EIb

Lb⋅+

⎛⎜⎝

⎞⎟⎠

⋅:= K33 1000103=

Loading

if the loading is symmetric or anti-symmetric, then it excites only the corresponding set of dofs and we cansolve a smaller problem. We demonstrate

100

20 20

20

1 2

3

4

a

b

c

point symmetry

X

Y

+

X'

Y'

+

the nodal force vector is

Pf

0

100−

0

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

:=

and we can see clearly that the symmetricdof is "not excited", i.e. its value is zerounder this nodal force vector

Page 624


We can therefore solve the smaller problem with only two independent free global dofs (dof #1 is zero).

we obtain with the anti-symmetric submatrix of the stiffness matrix denoted by Kfa

Kfa3

30−

30−

1000⎛⎜⎝

⎞⎟⎠

103⋅:= Pfa100−

0⎛⎜⎝

⎞⎟⎠

:= Ufa lsolve Kfa Pfa,( ):= Ufa47.619−

1.429−

⎛⎜⎝

⎞⎟⎠

10 3−=

with the columns of the compatibility matrix corresponding to the anti-symmetric dofs we have

observe the anti-symmetricforce distributionAfa

1La

−

1La

−

0

0

1La

1La

0

1

1

1−

1−

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= V Afa Ufa⋅:= Q Ks V⋅:= Q

571.43

428.57

428.57−

428.57

428.57−

571.43−

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

=

If the loading is general, then there are non-zero nodal force terms for the symmetric and the anti-symmetricdofs. In such case, we need to solve two smaller problems and add up the answers. We demonstrate for thecase of uniformly distributed load in element c.

Element loadingSet up the initial force vector due to distributed load w 12:=

03P

2

12cwL 2

12cwL

2cwL

2cwL

2cwL

01P Pfw

0

w Lc⋅

2

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:=

Set up the initial element force vector Q0

Q0

0

0

0

0

w Lc2⋅

12

w Lc2⋅

12−

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

Page 625


The initial force vector at the global dof's P0 becomes P0 AfT Q0⋅ Pfw+:= P0

400

120

400−

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

=

The nodal force vector P is equal to zero for this loading Pf

0

0

0

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

:=

We solve for the symmetric and anti-symmetric dofs separately. We have for the symmetric dof

Pfs 0:= P0s 400:= Kfs 2200 103⋅:= thus UfsPfs P0s−

Kfs:= Ufs 0.182− 10 3−=

and for the anti-symmetric dofs

Pfa0

0⎛⎜⎝⎞⎟⎠

:= P0a120

400−

⎛⎜⎝

⎞⎟⎠

:= Ufa lsolve Kfa Pfa P0a−,( ):= Ufa51.429−

1.143−

⎛⎜⎝

⎞⎟⎠

10 3−=

the final result is the superposition of the two solutions; in this case the complete free dof displacementvector is the combination of the symmetric and anti-symmetric dofs

Uf stack Ufs Ufa,( ):= Uf

0.182−

51.429−

1.143−

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

10 3−=

with the free dof displacement values established we can determine element deformations and basic forcesin the usual way

V Af Uf⋅:= Q Ks V⋅ Q0+:= Q

638.96

506.49

506.49−

179.22

179.22−

1075.32−

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

=

Page 626

CE220 - Theory of Structures Example 44 - Bridge with external prestressing © Prof. Filip C. Filippou, 2000

Example 44 - Bridge model with symmetry plane and external prestressing

Geometry and loading

1 2 34

5 630 40 30

10a

b c

d

e

fg

h

50

s

s

With the given geometry and element properties we have

EIa 3 108⋅:= EId 3 108⋅:= EIh 3 108⋅:= EA 3 106⋅:= w 50:=

La 30:= Ld 40:= Lh 30:=

Lb 302 102+:= Lb 31.623= Le 40:= Lg Lb:= Lc 10:= Lf 10:=

Dof numberingThe structure has 8 free degrees of freedom, as indicated in the following figure.

1 32 4

57

86

s

s

Page 627


We note that the geometry and element properties of the structure are symmetric about plane s-s. Theloading also happens to be symmetric. To take advantage of the symmetric characteristics of the structurewe group the free dofs in symmetric and antisymmetric generalized dofs, as shown in the following figures.Positive orientation is based on a right handed system for the left half of the structure (note theorientation of the reference system for the right half of the structure).

Symmetric generalized dof's

1 12 2

3 3

44

s

s

X

Y

X'Y'

Antisymmetric generalized dofs

5 56 6

77

88

s

s

X

Y

X'Y'

Page 628


Solution under given symmetric loading with symmetric dofs and 1/2 the structural model

We set up the structure compatibility matrix for the symmetric generalized dofs. We only list deformationsfor the left half of the structure. We need to be careful and include a deformation for elements that areintersected by the symmetry axis, like element d (only one deformation at the left hand).

For dofs 1 and 2 we obtain:

s

s

a

b c

d

e

fg

h

s

a

b c

d

e

fg

h

rotation at right end of member a

change of length of member bAfs

1La

−

0

1

0

0

1

0

0

1

0

0

30Lb

0

0

1−

0

10Lb

−

1−

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=change of length of member c

rotation at left end of member d

change of length of member e for left half of its length!

Page 629


For the element force-deformation matrix we need to be careful with member d and make sure to includethe effect of the symmetric mode of deformation. From the following figure we conclude:

s

11 2EIL

2EIL

L

We have: Ks

3 EIa⋅

La

0

0

0

0

0

EALb

0

0

0

0

0

EALc

0

0

0

0

0

EId

Ld2

0

0

0

0

0

EALe2

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

Form the structure stiffness matrix for the free dofs•

Kfs AfsT Ks⋅ Afs⋅:=

Kfs

333.33

1000−

0

300−

1000−

45000

0

0

0

0

235.38

28.46−

300−

0

28.46−

309.49

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

103=

Set up the applied nodal force vector•

applied nodal forces at symmetric dofs Pfs

0

0

0

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:=

equivalent nodal forces due to distributed loads (from direct equilibrium) Pfw

w La⋅

2

w Ld⋅

2+

0

0

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=

Page 630


initial force vector (element fixed-end forces) Q0

w−La

2

8⋅

0

0

w Ld2⋅

12

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= Q0

5625−

0

0

6666.67

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

=

complete initial force vector at free symmetric dofs

P0s AfsT Q0⋅ Pfw+:= P0s

1937.5

1041.7

0

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

=

We need to solve: Pfs Kfs Ufs⋅ P0s+=

Solve for the unknown displacements of the symmetric global dofs•

Ufs lsolve Kfs Pfs P0s−,( ):= UfsT 11.508− 0.258− 1.364− 11.281−( ) 10 2−=

Determine the deformations of all elements•

Vs Afs Ufs⋅:= VsT 1.26 22.73 2.27− 2.58− 13.64( ) 10 3−=

Determine the basic element forces for the symmetric dofs• IMPORTANT: add initial forces

Q Ks Vs⋅ Q0+:= QT 32.04 2.16 0.68− 32.04− 2.05( ) 103=

Page 631


Internal forces and joint equilibrium

1818

3204050

318 1000 318 1818

682

1000

318

682

1000

20462156

1818

2156

2046

2500

32040

50 50

Effect of external prestressing

For the effect of prestress the initial force vector Q0 is modified to include both element load and prestress

initial force vector due to prestress (element fixed-end forces) Q0

w−La

2

8⋅

0

0

w Ld2⋅

12

16000

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

<<< prestressing forcecomplete initial force vector at free symmetric dofs

P0s AfsT Q0⋅ Pfw+:= P0s

1937.5

1041.7

16000−

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

=

We need to solve: Pfs Kfs Ufs⋅ P0s+=

Solve for the unknown displacements of the symmetric global dofs•

Ufs lsolve Kfs Pfs P0s−,( ):= UfsT 0.377− 0.011− 6.829 0.262( ) 10 2−=

Page 632


Determine the deformations of all elements•

Vs Afs Ufs⋅:= VsT 0.02 63.96 6.4− 0.11− 68.29−( ) 10 3−=

Determine the basic element forces for the symmetric dofs ( IMPORTANT: add initial forces )•

Q Ks Vs⋅ Q0+:= QT 5.062− 6.068 1.919− 5.062 5.756( ) 103=

Internal Forces and joint equilibrium

506250

919 1000 1000

1919

1000

57566068

2500

919

919

5062

581 581

1919

6068

5815756

50 50

It is worth noting that the axial forces in the truss members increased about threefold after theprestressing, but the moments in the girder dropped about six fold (in fact, if one were to calculate themaximum positive moment in both cases a large reduction would be observed). It is also worth noting thatthe vertical displacement at joint 2 reduced 30-fold by the prestressing. Since the prestressing operationis performed in the field by stressing against the existing structure, the initial force of 16,000 will not beexperienced by member e. In fact, during prestressing the tensile force in member e will gradually rise toits final value of 5756.2 and the girder will be pushed upward.

Page 633


Solution for general loading case

In this case we need to consider both symmetric and anti-symmetric dofs.

global free dofs without consideration of symmetry

1 32 4

57

86

s

s

symmetric dofs

1 12 2

3 3

44

s

s

X

Y

+X'

Y'

+

antisymmetric dofs

5 56 6

77

88

X

Y

X'Y's

s

Page 634


Compatibility matrix

s

s

a

b c

d

e

fg

h dof 1

s

a

b c

d

e

fg

h dof 2

s

s

a

b c

d

ef g

h dof 5

Af

1La

−

0

1

0

0

0

1

0

1Lh

1

0

0

1

0

1−

0

0

1−

0

30Lb

0

0

2−

0

0

30Lg

0

0

10Lb

−

1−

0

0

0

1−

10Lg

−

0

1La

−

0

1

2Ld

0

2Ld

1−

0

1Lh

−

1

0

0

1

0

1

0

0

1

0

30Lb

0

0

0

0

0

30Lg

−

0

0

10Lb

−

1−

0

0

0

1

10Lb

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

Page 635


collection of element stiffness matrices

Ks

3 EIa⋅

La

0

0

0

0

0

0

0

0

0

EALb

0

0

0

0

0

0

0

0

0

EALc

0

0

0

0

0

0

0

0

0

4 EId⋅

Ld

0

2 EId⋅

Ld

0

0

0

0

0

0

0

EALe

0

0

0

0

0

0

0

2 EId⋅

Ld

0

4 EId⋅

Ld

0

0

0

0

0

0

0

0

0

EALf

0

0

0

0

0

0

0

0

0

EALg

0

0

0

0

0

0

0

0

0

3 EIh⋅

Lh

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

Form the structure stiffness matrix for free dofs•

K AfT Ks⋅ Af⋅:=

K

666.667

2000−

0

600−

0

0

0

0

2000−

90000

0

0

0

0

0

0

0

0

470.763

56.921−

0

0

0

0

600−

0

56.921−

618.974

0

0

0

0

0

0

0

0

891.667

2500

0

600−

0

0

0

0

2500

150000

0

0

0

0

0

0

0

0

170.763

56.921−

0

0

0

0

600−

0

56.921−

618.974

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

103=

Page 636


We note that the structure stiffness matrix consists of two uncoupled 4x4 submatrices, one for the set ofsymmetric dofs, and the other for the set of antisymmetric dofs. Interestingly, the only coupling betweengirder and external prestressing cables are the vertical posts c and f. This coupling arises through theinteraction of dofs 1 and 4, and 5 and 8.

General load case

1 2 34

5 630 40 30

10a

b c

d

e

fg

h

50 s

s

Q0

w−La

2

8⋅

0

0

0

0

0

0

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= Q0

5625−

0

0

0

0

0

0

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

= Pfw

w La⋅

2

0

0

0

w La⋅

2

0

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:= Pfw

750

0

0

0

750

0

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

complete initial force vector Pf0 AfT Q0⋅ Pfw+:= Pf0

937.5

5625−

0

0

937.5

5625−

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

We can now solve two 4x4 problems and then superpose the results, as we did in Example 43.

Page 637

CE 220 - Theory of Structures Example 45 - Force method for model with point symmetry © Prof. Filip C. Filippou, 2000

Example 45 - Force method for structural model with point symmetry

With the given geometry and element properties we have

EI 50000:= EA 10000:=

La 8:= Lb 6:= Lc 8:= Ld 82 62+:= Le 82 62+:=

8 8

6

a

b

c

d e

12

34

point symmetry

20

30

The structure has four independent global dofs, if elements a, b and c are assumed inextensible. The dofsand corresponding deformations and basic element forces are numbered in the following figures.

1

2

3

4

1

2

3

4

5

6

1Q2Q

3Q4Q

5Q6Q

Page 638


The symmetric dofs are two. These are shown in the following figure

2

X

Y

+

X'

Y'

+

0

0

2 1

point symmetry

The corresponding basic forces are shown in the following figure

X

Y

+

X'

Y'

+

2Q′

1Q′

2Q′1Q′

3Q′3Q′

The equilibrium equations for the symmetric dofs are (using PVD!)

P1Q'2 Q'2+

60.8 Q'3 Q'3+( )⋅+=

Q'23

1.6 Q'3⋅+=

P2 Q'1 Q'2+ Q'2+ Q'1+= 2 Q'1⋅ 2 Q'2⋅+=

We have two equations in three unknown basic forces; thus, the degree of static indeterminacy is NOS=1.

We select Q`2 as the redundant basic force

the particular solution is 20 1.6 Q'3⋅= thus,Q'p

0

0

12.5

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

:=0 2 Q'1⋅=

Page 639


The homogeneous solution is: 013

1.6 Q'3⋅+=B'barx

1−

1

14.8

−

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:=0 2 Q'1⋅ 2 1( )⋅+=

We set up the collection of element flexibility matrices for only half the structure

Fs

La

3

0

0

0

Lb

6

0

0

0

LdEIEA⋅

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

1EI⋅:=

the compatibility equation gives Q'xB'barx

T Fs⋅ Q'p⋅

B'barxT Fs⋅ B'barx⋅

−:= Q'x 22.31=

and the final symmetric solution is Q' Q'p B'barx Q'x⋅+:= Q'

22.31−

22.31

7.85

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

=

Page 640


The antisymmetric dofs are also two. These are shown in the following figure.

4

X

Y

+

X'

Y'

+

3

3

4

point symmetry

The corresponding basic forces are shown in the following figure

X

Y

+

X'

Y'

+

1Q′′2Q′′

3Q′′3Q′′

1Q′′

2Q′′

The equilibrium equations for the antisymmetric dofs are (using PVD!)

P3Q"1 Q"1+

8− 0.6 Q"3 Q"3+( )⋅+=

Q"1

4− 1.2 Q"3⋅+=

P4 Q"1 Q"2+ Q"2+ Q"1+= 2 Q"1⋅ 2 Q"2⋅+=

We have two equations in three unknown basic forces; thus, the degree of static indeterminacy is NOS=1.

We select Q"1 as the redundant basic force

the particular solution is 30− 1.2 Q"3⋅= thus,Q"p

0

0

25−

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

:=0 2 Q"2⋅=

Page 641


The homogeneous solution is: 014

− 1.2 Q"3⋅+=B"barx

1

1−

14.8

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:=0 2 1( )⋅ 2 Q"2⋅+=

We set up the collection of element flexibility matrices for only half the structure

Fs

La

3

0

0

0

Lb

2

0

0

0

LdEIEA⋅

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

1EI⋅:=

the compatibility equation gives Q"xB"barx

T Fs⋅ Q"p⋅

B"barxT Fs⋅ B"barx⋅

−:= Q"x 33.23=

and the final symmetric solution is Q" Q"p B"barx Q"x⋅+:= Q"

33.23

33.23−

18.08−

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

=

The final solution is obtained by superposition of the symmetric and antisymmetric solution.In this regard we note the following relations between the original basic forces on the one hand, andthe symmetric and antisymmetric basic forces on the other.

Q1 Q'1 Q"1+= Q1

Q2

Q5

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

Q' Q"+:=Q2 Q'2 Q"2+=

Q3 Q'1 Q"1−= thus,

Q4 Q'2 Q"2−=

Q5 Q'3 Q"3+=

Q3

Q4

Q6

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

Q' Q"−:= finally,

Q1

Q2

Q3

Q4

Q5

Q6

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

10.92

10.92−

55.54−

55.54

10.22−

25.93

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

=Q6 Q'3 Q"3−=

Page 642

CE 220 - Theory of Structures Example 46 - Braced Frame with Symmetry © Prof. Filip C. Filippou, 2000

Example 46- Force Method for Braced Frame with Symmetry

6 6

8

25

a

b c

dfe

1

23

4

5

In the general case there are 14 basic element forces and 9 available equations of equilibrium.The degree of static indeterminacy NOS is therefore 14-9=5.

1Q

2Q

3Q

4Q5Q 6Q7Q8Q

10Q

9Q

11Q

12Q

13Q 14Q

1

2

3 4

56

7

89

If we can neglect axial deformations in members a, b, c and d, then we are not interested in thecontribution of basic forces Q1, Q4, Q7 and Q10 to virtual work. We can, therefore, leave them aside,putting at the same time aside 4 equilibrium equations that involve these forces. The degree of staticindeterminacy is obviously not affected by this process.

The relevant basic forces are shown in the following figure. The relevant equilibrium equations arealso indicated by renumbered dofs. We note that the horizontal force equilibrium involves thecombination of original equations 1, 4 and 7 into a single equation with dof number 1.

Page 643


1Q

2Q

3Q 4Q 5Q 6Q

7Q

8Q

10Q9Q

1 1 1

23

4 5

Element properties: La 8:= Lb 6:= Lc 6:= Ld 8:= Le 62 82+:= Le 10=

Lf Le:= EI 40000:= EA 5000:=

On account of the symmetry in geometry, properties and loading we renumber the internal forces, notingthat moments and axial forces are symmetric about the symmetry axis under symmetric loading. We note thatthere are now only 5 pairs of unknown internal forces with 2 available equations of equilibrium at thesymmetric global dof's: sum of vertical forces at node 3 and sum of moments at node 4. The degree of staticindeterminacy of the symmetric problem is therefore 3 (there is an uncoupled set of equations forthe anti-symmetric loading with degree of static indeterminacy of 2; thus, the combined system of equationshas the same degree of static indeterminacy as the original problem)

The following figure shows the numbering of internal forces for the right half of the structure. The orientationof internal and external forces that obeys symmetry in the left half of the structure is indicated.

1Q 2Q

3Q

4Q

5Q

symmetry axis

1P2P

34

2P

Page 644


In the following we propose to work only with the right half of the structure. Note the appropriateboundary condition on the axis of symmetry.

1Q 2Q

3Q

4Q

5Q

symmetry axis

2P1P


P1Q1 Q2+

LcQ5

ΔYLf

⎛⎜⎝

⎞⎟⎠

⋅+=

P2 Q2 Q3+=

1. We select basic element forces Q1, Q2 and Q4 as redundant forces of the problem Qx.

2. Determine the remaining basic element forces for unit forces at the symmetric free dof's and Q1=0,Q2=0 and Q4=0. The unit forces at the symmetric free dof's are applied each one in turn, with theothers equal to zero, and their influence on the basic element forces is recorded in a force influencematrix denoted by Bi

3Q

5Q

symmetry axis

4 0=Q

2 0=Q2P

1P

1 0Q =The equilibrium equations are:

P1Q1 Q2+

LcQ5

ΔYLf

⎛⎜⎝

⎞⎟⎠

⋅+=

P2 Q2 Q3+=

For P1 1= and P2 0= and Q1 0= Q2 0= Q4 0= we have: Q3 0= Q5Lf

ΔY=

For P1 0= and P2 1= and Q1 0= Q2 0= Q4 0= we have: Q3 1= Q5 0=

Page 645


We have the following values for the force influence matrix Bi

Bi

0

0

0

0

54

0

0

1

0

0

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:= with the given loading Pf

252

−

0

⎛⎜⎜⎝

⎞⎟⎟⎠

:=

3. Determine the remaining basic element forces for the following combinations: Q1=1, Q2=0 and Q4=0;Q1=0, Q2=1 and Q4=0; Q1=0, Q2=0 and Q4=1. This establishes the three columns of the force influencematrix denoted by Bx

3Q

5Q

symmetry axis

1 1=Q

4 0=Q

2 0=Q


0Q1 Q2+

LcQ5

ΔYLf

⎛⎜⎝

⎞⎟⎠

⋅+=

0 Q2 Q3+=

for

Q1 1=

Q2 0= Bx1⟨ ⟩

1

0

0

0

524

−

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

Q4 0=

3Q

5Q

symmetry axis

2 1=Q

4 0=Q

1 0Q =


0Q1 Q2+

LcQ5

ΔYLf

⎛⎜⎝

⎞⎟⎠

⋅+=

0 Q2 Q3+=

for

Q1 0=Bx

2⟨ ⟩

0

1

1−

0

524

−

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=Q2 1=

Q4 0=

Page 646


3Q

5Q

symmetry axis

4 1=Q

2 0=Q1 0=Q


0Q1 Q2+

LcQ5

ΔYLf

⎛⎜⎝

⎞⎟⎠

⋅+=

0 Q2 Q3+=

for

Q1 0= Bx3⟨ ⟩

0

0

0

1

0

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=

Q2 0=

Q4 1=

With the unknown basic forces denoted by Qx we can write Q Bi Pf⋅ Bx Qx⋅+= . In words this reflects the

fact that the basic element forces will result from the superposition of the forces under the applied forcesand the forces under the redundant forces.

4. The basic element deformations under the basic forces Q are simply: V Fs Q⋅= in the absence of initialdeformations.

5. To determine the value of the unknown basic forces Qx we write the compatibility conditions. BxT V⋅ 0=

After substituting the expression for V in terms of Q and then Q in terms of Pf and Qx we obtain

BxT Fs⋅ B0 Pf⋅ Bx Qx⋅+( )⋅ 0=

where Fs in the collection of element flexibility matrices for the right half of the structure

Fs

Lc

3 EI⋅

Lc

6 EI⋅−

0

0

0

Lc

6 EI⋅−

Lc

3 EI⋅

0

0

0

0

0

Ld

3 EI⋅

Ld

6 EI⋅−

0

0

0

Ld

6 EI⋅−

Ld

3 EI⋅

0

0

0

0

0

Lf

EA

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

Page 647


We define the following terms: Fxi BxT Fs⋅ Bi⋅:= Fxi

0.521−

0.521−

0

0

0.067−

0.033−

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

10 3−⋅=

Fxx BxT Fs⋅ Bx⋅:= Fxx

0.137

0.062

0

0.062

0.203

0.033

0

0.033

0.067

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

10 3−⋅=

With these terms the compatibility condition becomes: Fxi Pf⋅ Fxx Qx⋅+ 0=

and we can solve for the unknown Qx Qx Fxx1−− Fxi⋅ Pf⋅=

We obtain Qx Fxx1−− Fxi⋅ Pf⋅:= Qx

37.44−

22.46−

11.23

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

=

The basic element forces become Q Bi Pf⋅ Bx Qx⋅+:=

QT 37.44− 22.46− 22.46 11.23 3.14−( )=

Note that it is easier from the standpoint of hand calculations to factor some stiffness value, e.g. EI, outof the composite flexibility matrix. In this case we have:

EIFs

Lc

3

Lc

6−

0

0

0

Lc

6−

Lc

3

0

0

0

0

0

Ld

3

Ld

6−

0

0

0

Ld

6−

Ld

3

0

0

0

0

0

Lf

EAEI⋅

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

We define the following terms: EIFxi BxT EIFs⋅ Bi⋅:= EIFxi

20.83−

20.83−

0

0

2.67−

1.33−

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

=

EIFxx BxT EIFs⋅ Bx⋅:= EIFxx

5.47

2.47

0

2.47

8.14

1.33

0

1.33

2.67

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

=

Page 648


We note that the values are much more manageable from the standpoint of hand calculations. Naturally,the value of Qx does not change when we determine it from

Qx EIFxx1−− EIFxi⋅ Pf⋅:= Qx

37.44−

22.46−

11.23

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

=

6. Determination of displacements at global dof's Uf BiT V⋅=

After substituting the expression for V in terms of Q and then Q in terms of Pf and Qx we obtain

Uf B0T Fs⋅ Bi Pf⋅ Bx Qx⋅+( )⋅=

and with the definitions Fii BiT Fs⋅ Bi⋅:= Fix Bi

T Fs⋅ Bx⋅:=

we obtain Uf Fii Pf⋅ Fix Qx⋅+:=

The result is Uf7.862−

1.123⎛⎜⎝

⎞⎟⎠

10 3−⋅=

It is worth noting that Fix FxiT= since Fs is symmetric

For a general result we substitute the expression of Qx into the global displacement equation and have

Uf Fii Pf⋅ Fix Qx⋅+= Fii Fix Fxx1−⋅ Fxi⋅−( ) Pf⋅= F Pf⋅=

F is the structure flexibility matrix;its coefficient Fij corresponds to the displacement at global dof i due to a unit force at dof j.

In our case we have:

F Fii Fix Fxx1−⋅ Fxi⋅−:= F

0.629

0.09−

0.09−

0.034⎛⎜⎝

⎞⎟⎠

10 3−⋅=

The inverse of the system flexibility matrix F is the system stiffness matrix K(also denoted as Kf in the course)

Page 649


Anti-symmetric Loading Case

We now would like to solve the following load case with anti-symmetric loading.

6 6

8

30

a

b c

dfe

1

23 4

5

which is identical to the following case if axial deformations can be neglected in members b and c

6 6

8

15

a

b c

dfe

1

23 4

5

15

Page 650


We note that there are initially 5 pairs of unknown internal forces with asymmetric values as shown in thefollowing figure. There are 3 available equations of equilibrium at the asymmetric global dof's: sum ofhorizontal forces at node 4, sum of moments at node 3 and sum of moments at node 4. The degree of staticindeterminacy for this problem is therefore 2.

1Q 2Q

3Q

4Q

5Q

symmetry axis

1P2P

34

3P

1P2P

Because the external moment at dof 3 is zero, and the internal forces are anti-symmetric we conclude thatQ1 is zero. We therefore re-number internal forces and relevant dofs for the right half of the structure inthe following figure. The orientation of internal and external forces that obeys asymmetry in the left half ofthe structure is indicated.

1Q

2Q

3Q

4Q

symmetry axis

2P

3

1P2P

1PThe equilibrium equations are:

P1Q2 Q3+

LdQ5

ΔXLf

⎛⎜⎝

⎞⎟⎠

⋅−=

P2 Q1 Q2+=

Page 651


In the following we propose to work only with the right half of the structure. Note the appropriateboundary condition on the axis of symmetry.

1Q

2Q

3Q

4Q

symmetry axis

3

2P1P

1. We select basic element forces Q1 and Q3 as the redundants of the problem Qx.

2. Determine the remaining basic element forces for the given loading at the asymmetric free dof's andQ1=0 and Q3=0. This produces basic force vector QP.

symmetry axis

3 0Q =

1 0Q =

4Q

2Q

15The equilibrium equations are:

15Q2 Q3+

LdQ4

ΔXLf

⎛⎜⎝

⎞⎟⎠

⋅−=

0 Q1 Q2+=

We have the following values for the forces under the applied nodal forces

with Q1 0= Q3 0= we have: Q2 0= Q4 15−Lf

ΔX⋅= Qp

0

0

0

25−

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:=

Page 652


3. Determine the remaining basic element forces for the following combinations: Q1=1 and Q3=0; Q1=0and Q3=1. This establishes the two columns of the force influence matrix denoted by Bx


symmetry axis

34

3 0Q =

1 1Q =

4Q

2Q

0Q2 Q3+

LdQ4

ΔXLf

⎛⎜⎝

⎞⎟⎠

⋅−=

0 Q1 Q2+=

for

Q1 1=Bx

1⟨ ⟩

1

1−

0

524

−

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=Q3 0=

symmetry axis

1 0Q =

3 1Q =

4Q

2Q


0Q2 Q3+

LdQ4

ΔXLf

⎛⎜⎝

⎞⎟⎠

⋅−=

0 Q1 Q2+=

for

Q1 0= Bx2⟨ ⟩

0

0

1

524

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=Q3 1=

Thus Bx

1

1−

0

524

−

0

0

1

524

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:= we need to make sure to redefine the Bx matrix completely,so as to avoid confusion with the symmetric case

With the unknown basic forces denoted by Qx we can write Q Qp Bx Qx⋅+= . In words this reflects the fact

that the basic element forces will result from the superposition of the forces under the applied forces andthe forces under the unknown basic forces.

Page 653


4. The basic element deformations under the basic forces Q are simply: V Fs Q⋅= in the absence of initialdeformations.

5. To determine the value of the unknown basic forces Qx we write the compatibility conditions BxT V⋅ 0=

After substituting the expression for V in terms of Q and then Q in terms of QP and Qx we obtain

BxT Fs⋅ Qp Bx Qx⋅+( )⋅ 0=

where Fs in the collection of the element flexibility matrices for the right half of the structure

Fs

Lc

3 EI⋅

0

0

0

0

Ld

3 EI⋅

Ld

6 EI⋅−

0

0

Ld

6 EI⋅−

Ld

3 EI⋅

0

0

0

0

Lf

EA

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

:=

We define the following term: Fxx BxT Fs⋅ Bx⋅:= Fxx

0.203

0.053−

0.053−

0.153⎛⎜⎝

⎞⎟⎠

10 3−⋅=

With these term the compatibility condition becomes: BxT Fs⋅ Qp⋅ Fxx Qx⋅+ 0=

and we can solve for the unknown Qx Qx Fxx1−− Bx

T⋅ Fs⋅ Qp⋅=

We obtain Qx Fxx1−− Bx

T⋅ Fs⋅ Qp⋅:= Qx36.72−

55.08⎛⎜⎝

⎞⎟⎠

=

The basic element forces become Q Qp Bx Qx⋅+:= QT 36.72− 36.72 55.08 5.88−( )=

Page 654

CE 220 - Theory of Structures Ex. 47 - Vierendeel Truss with double symmetry © Prof. Filip C. Filippou, 2000

Example 47 - Vierendeel Truss with Double Symmetry

We investigate the following Vierendeel truss with double symmetry.

1 2 3

45

6

a b

c d

ef g

8 8

6

s1

s1

s2 s2

La 8:= Lb 8:= Lc 8:=

Ld 8:=

Le 6:= Lf 6:= Lg 6:=

EI 50000:=

The planes of symmetry as s1 and s2, as shown in the figure.

Under the assumption that all elements are inextensible the structural model has 8 independent free globaldofs, as shown in the following figure.

1 2 3

4

5

67

8

We reorganize these dofs now into 4 groups: one group of dofs that are symmetric about both planes (wedenote these with ss), one group of dofs that is symmetric about the first and anti-symmetric about thesecond (sa), one group of dofs that is anti-symmetric about the first and symmetric about the second (as),and one group of dofs that is anti-symmetric about both planes (aa). We have:

1 1

11

ss - dofonly one ss dof

Page 655


3

2

2

3

3 3

sa - dofstwo sa dofs

4 5 4

45

as - dof's

4

two as dofs

8

6

7

aa - dof's

687 7

7

three aa dofs

By selecting these 8 dofs instead of the original 8 the structure stiffness matrix separates into 4 uncoupledblock-diagonal stiffness matrices: one submatrix is 1x1 and corresponds to the ss dof, the second and thirdsubmatrix are 2x2 and correspond to the sa and as dofs, respectively, and the final submatrix is 3x3 andcorresponds to the aa dofs. In the following we solve this problem only for a particular loading. In thisregard we would like to find out to which dofs the loading contributes non-zero terms. Let us see.

Page 656


Symmetric - symmetric dofs

w=10

w=10

w 10:=

1 1

11

ss - dof ss dof clearly not excited by this loading

P0ss 0=

Just in case the stiffness coefficient is:

K11 4 4EILa⋅⎛

⎜⎝

⎞⎟⎠

⋅ 4 2EILe⋅⎛

⎜⎝

⎞⎟⎠

⋅+:= K11 166.67103=

Page 657


Symmetric - anti-symmetric dofs

w=10

w=10

2

2

sa - dof'ssa dofs clearly excited by this loading

we have

P0sa

w La⋅

2

w Ld⋅

2+

w La2⋅

12

w Ld2⋅

12+

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

:=

sa - dof's

33

3 3

P0sa80

106.67⎛⎜⎝

⎞⎟⎠

=

The stiffness coefficients are:

K22 4 12EI

La3

⋅⎛⎜⎝

⎞⎟⎠

⋅:= K33 4 4EILa⋅⎛

⎜⎝

⎞⎟⎠

⋅ 4 6EILe⋅⎛

⎜⎝

⎞⎟⎠

⋅+:= K23 4 6−EI

La2

⋅⎛⎜⎝

⎞⎟⎠

⋅:= K32 K23:=

KsaK22

K32

K23

K33

⎛⎜⎝

⎞⎟⎠

:= free dof displacements Usa lsolve Ksa P0sa−,( ):= Usa24.652−

1.896−

⎛⎜⎝

⎞⎟⎠

10 3−=

Page 658


Anti-symmetric - symmetric dofs

w=10

w=10

as dofs clearly excited by this loading

we have

as - dof's

4 4

4 4P0as

w La2⋅

12−

w Ld2⋅

12−

w La2⋅

12

w Ld2⋅

12+

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=

P0as106.67−

106.67⎛⎜⎝

⎞⎟⎠

=

5

5

as - dof's

The stiffness coefficients are:

K44 4 4EILa⋅⎛

⎜⎝

⎞⎟⎠

⋅ 4 2EILe

⎛⎜⎝

⎞⎟⎠

⋅+:= K55 4 4EILa⋅⎛

⎜⎝

⎞⎟⎠

⋅ 2 2EILe⋅⎛

⎜⎝

⎞⎟⎠

⋅+:= K45 4 2EILa⋅⎛

⎜⎝

⎞⎟⎠

⋅:= K54 K45:=

KasK44

K54

K45

K55

⎛⎜⎝

⎞⎟⎠

:= free dof displacements Uas lsolve Kas P0as−,( ):= Uas0.992

1.172−

⎛⎜⎝

⎞⎟⎠

10 3−=

Page 659


Anti-symmetric - anti-symmetric dofs

w=10

w=10

6

aa - dof's

6

7

aa - dof's

7 7

7

aa dofs clearly not excited by this loading

P0aa

0

0

0

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

=

aa - dof's

8

8

Page 660


Combination of solutions

We are interested in the displacement values at the original set of 8 independent free dofs, i.e.

1 2 3

4

5

67

8

3

2

2

3

3 3

sa - dofs

we denote these dofs with a prime andinclude only those who are non-zero, i.e.the sa and as dofs

we have U1 U'3 U'4+=

U2 U'5=

U3 U'3− U'4+=

U4 U'3 U'4−=

4 5 4

45

as - dof's

4

U5 U'2=

U6 U'5−=

U7 0=

U8 U'3− U'4−=

U'2

U'3

U'4

U'5

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

stack Usa Uas,( ):=

U'2

U'3

U'4

U'5

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

24.652−

1.896−

0.992

1.172−

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

10 3−=

Page 661


thus, the original free dof displacements become

U1 U'3 U'4+:= U1 0.905− 10 3−= U5 U'2:= U5 24.652− 10 3−=

U2 U'5:= U2 1.172− 10 3−= U6 U'5−:= U6 1.17210 3−=

U3 U'3− U'4+:= U3 2.88810 3−= U7 0:= U7 0=

U4 U'3 U'4−:= U4 2.888− 10 3−= U8 U'3− U'4−:= U8 0.90510 3−=

You can compare the answers with FEDEASLab file Example_47m. The moment distribution has pointsymmetry, as can be observed from the following figure.

78.3

78.3

75

75

78.3

111.3

55.4

55.4

111.319.5

19.5

deformed shape with magnification factor of 50 shows clearly the dominance of sa dof 2.

Page 662


Structural model with infinitely rigid vertical elements

This is an interesting simplification of the earlier problem. there are now only 4 independent freeglobal dofs, i.e. 4 = 8 - 2x2

1

2

34

ss - dof

0

0

0

0

in such case there is no ss dof

1

1

sa - dofs

there is only one sa dof

2

2

as - dof's

there is only as dof

Page 663


4

aa - dofs

3

3

4

there are 2 aa dofs

w=10

w=10

EI 50000=

w 10=

with the given loading we have to solve now two problems with one unknown displacement value each

symmetric - anti-symmetric dof case

Ksa 412EI

La3

⎛⎜⎝

⎞⎟⎠

⋅:= same as K22 earlier K22 4687.5= Ksa 4687.5=

P0saw La⋅

2

w Ld⋅

2+:= P0sa 80= Usa

P0sa

Ksa−:= Usa 17.07− 10 3−=

thus the vertical midspan translation reduces now to 0.017 from the earlier value of U5 24.65− 10 3−=

anti-symmetric - symmetric dof case

Kas 4 4EILa⋅⎛

⎜⎝

⎞⎟⎠

⋅ 2 2EILe⋅⎛

⎜⎝

⎞⎟⎠

⋅+:= same as K55 earlier K55 133.33103= Kas 133.33103=

P0asw La

2⋅

12

w Ld2⋅

12+:= P0as 106.67= Uas

P0as

Kas−:= Uas 0.8− 10 3−=

Page 664

CE 220 - Theory of Structures Example 48 - Substructuring with symmetry © Prof. Filip C. Filippou, 2000

Example 48 - Substructuring with Symmetry

Problem Statement

Determine the vertical displacement at the point of application of the vertical force of 20 unitsfor the structure in the figure. The flexural stiffness EI of the sloping members is 20,000 unitsand axial deformations can be neglected throughout.

20 208 8

12

6

20

A separate analysis of the portal frame in the following figure yielded the following relationbetween the horizontal force and moment at the upper right node and the correspondingdisplacements

3 5

5 5

2.340 10 9.000 109.000 10 3.643 10

− −

− −

⎡ ⎤⋅ − ⋅= ⎢ ⎥− ⋅ ⋅⎣ ⎦

F

20

12

12

Page 665


We demonstrate the solution process thoroughly but succinctly.

Identification of independent free global dof's under symmetry considerations

Without regard for inextensible elements and with the portal frame as substructure (i.e. an element with itsown force-deformation relation) the structural model has the following dof's (i.e. 3 dof's/node).

Accounting for the inextensibility of all frame elements and separating dof's into a symmetric andanti-symmetric set we obtain the following independent free symmetric dof's.

1

2

Page 666


The corresponding basic forces Q are shown in the following figure

1

2

1Q

2Q3Q

4Q

Equilibrium equations, equilibrium and compatibility matrix

We write the equilibrium equations by application of the principle of virtual displacements. We have:

1IC

0.75 0.75

P118

− Q1 Q2+( )⋅ 0.75 Q3⋅+= The degree of static indeterminacy is NOS=2.Under these conditions, the displacement method seemsto be the most straightforward choice (NFD = 2).

P2 Q2 Q4+=

The compatibility matrix is the transpose of the equilibrium matrix:

Af

18

−

18

−

0.75

0

0

1

0

1

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:=

Page 667


Solution by displacement method

EI 20000:= L 10:=

In forming the stiffness matrix we need the stiffness matrix of the subassembly. To this end weneed to invert the given flexibility matrix. We should factor out 1/EI in this calculation

F*234.0

9−

9−

3.643⎛⎜⎝

⎞⎟⎠

10 5−⋅:= F*46.8

1.8−

1.8−

0.729⎛⎜⎝

⎞⎟⎠

1EI

= k* F* 1−:= k*0.024

0.058

0.058

1.517⎛⎜⎝

⎞⎟⎠

EI=

Now we have for the collection of element stiffness matrices Ks

Ks

4 EI⋅L

2 EI⋅L

0

0

2 EI⋅L

4 EI⋅L

0

0

0

0

k*1 1,

k*2 1,

0

0

k*1 2,

k*2 2,

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:= Ks

0.4

0.2

0

0

0.2

0.4

0

0

0

0

0.024

0.058

0

0

0.058

1.517

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

EI=

The stiffness matrix for the left half of the structural model is K AfT Ks⋅ Af⋅:= K

0.032

0.031−

0.031−

1.917⎛⎜⎝

⎞⎟⎠

EI=

The loading for half the structure is Pf10−

0⎛⎜⎝

⎞⎟⎠

:=

The resulting displacements are: Uf lsolve K Pf,( ):= Uf317.24−

5.17−

⎛⎜⎝

⎞⎟⎠

1EI

=

We can now determine the basic forces from the element deformations

Q Ks Af⋅ Uf⋅:= QT 22.759 21.724 5.92− 21.724−( )=To obtain the forces in the substructure we need to perform an analysis of it under the following loading

20

12

5.92

21.724

Page 668


Solution by force methodThe solution by the force method of analysis is equally straightforward. We start from the equilibriumequations and select the basic forces of the substructure as redundants Qx. We solve the equilibriumequations for the three equilibrium states: (1) equilibrium state under applied loading with bothredundants set to zero (particular solution), (2) equilibrium state under a unit value of the first redundantwithout applied loads, and (3) equilibrium state under a unit value of the second redundant withoutapplied loads

Equilibrium equations for given loading (half the structure).

10−18

− Q1 Q2+( )⋅ 0.75 Q3⋅+=

0 Q2 Q4+=

First equilibrium state: Q3 0:= Q4 0:=

from the equations it follows: Q2 0:= Q1 80:= Qp

Q1

Q2

Q3

Q4

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:= Qp

80

0

0

0

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

=

Second equilibrium state: Q3 1:= Q4 0:=

from the equations without any loading on the left hand side

018

− Q1 Q2+( )⋅ 0.75 Q3⋅+=it follows Q2 0:= Q1 6:= Bbarx

1⟨ ⟩

Q1

Q2

Q3

Q4

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=0 Q2 Q4+=

Third equilibrium state: Q3 0:= Q4 1:=

from the equations without any loading on the left hand side

018

− Q1 Q2+( )⋅ 0.75 Q3⋅+=it follows Q2 1−:= Q1 1:= Bbarx

2⟨ ⟩

Q1

Q2

Q3

Q4

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

:=0 Q2 Q4+=

Thus: Bbarx

6

0

1

0

1

1−

0

1

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

=

Page 669


Collection of element flexibility matrices

Fs

L3 EI⋅

L6 EI⋅

−

0

0

L6 EI⋅

−

L3 EI⋅

0

0

0

0

F*1 1,

F*2 1,

0

0

F*1 2,

F*2 2,

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

:= Fs

3.33

1.67−

0

0

1.67−

3.33

0

0

0

0

46.8

1.8−

0

0

1.8−

0.73

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

1EI

=

Compatibility condition for redundant basic forces Fxx BbarxT Fs⋅ Bbarx⋅:= Fxx

166.8

28.2

28.2

10.73⎛⎜⎝

⎞⎟⎠

1EI

=

Vhp BbarxT− Fs⋅ Qp⋅:= Vhp

1600−

400−

⎛⎜⎝

⎞⎟⎠

1EI

=

Qx Fxx1− Vhp⋅:= Qx

5.92−

21.72−

⎛⎜⎝

⎞⎟⎠

=

Final basic forces are obtained by superposition of three equilibrium states:

Q Qp Bbarx Qx⋅+:= QT 22.76 21.72 5.92− 21.72−( )= same as before

To determine the vertical displacement at dof 1 with this solution approach, we need to use eithergeometry (i.e. the compatibility equations) or the principle of virtual forces. Let us try PVF first.

δP118

− δQ1 δQ2+( )⋅ 0.75 δQ3⋅+= set δP1 1= δQ3 0:=

δQ4 0:=0 δQ2 δQ4+=

then it follows that δQ2 0:= δQ1 8−:=

Thus, the vertical displacement is U1 8− V1⋅= with V1L

6 EI⋅2 Q1⋅ Q2−( )⋅:= V1 1.98310 3−=

U1 8− V1⋅:= U1 1.586− 10 2−= it agrees with displ. method

A faster solution than that by the principle of virtual forces is obtained by "geometry", i.e. by thecompatibility equations directly. We note that

V Af Uf⋅= where Af

0.125−

0.125−

0.75

0

0

1

0

1

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

= from the first row we can see clearly that

U1 8− V1⋅=

Page 670

Betti's Theorem was published in 1872. It deals with the effect of one system of forces on the displacements of another, independent system of forces under linear elastic material behavior.

Consider a structural model and two independent sets of forces and the corresponding set of displacements.

1P 2P

3U

4U

First set of forces and corresponding displacements (subscript m)

mP mU

The figure singles out two dofs with applied forcesand two separate dof's with resulting displacements.

Second set of forces and corresponding displacements (subscript n)

3P

4P

1U

2U

nP nU

The figure singles out two dofs with applied forcesand two separate dof's with resulting displacements.

T Tn m m nP U P U=

Betti's Theorem states that the work done by the first set of forces on the displacements caused by the second set is equal to the workdone by the second set of forces on the displacements caused by the first set. In compact mathematical notation this is stated as

If the first set of forces only included P1 and P2 and the second set of forces only P3 and P4 in the above figures, then this statementspecializes to

1 1 2 2 3 3 4 4P U P U P U P U+ = +

note, however, carefully the meaning of the terms: displacements on the left hand side of equation are caused by force set on the right handside of equation, and displacements on the right hand side by force set on the left hand side, as the above figures illustrate.

Another proof of Betti's theorem is instructive. We show it on the following page.

Proof: apply first set of forces followed by the second set and calculate the work done in the process: T T Tn n n m m m

1 12 2

W P U P U P U= + +

Reverse order of application and calculate work: T T Tm m m n n n

1 12 2

W P U P U P U= + + From conservation of energy (work) follows Betti's theorem.

Another proof of Betti's theoremT Tm n m nP U Q V=Start with equality of external and internal work

between forces of set m on displacements ordeformations of set n

(1) Note: by limiting the requirement on P and Q to equilibriumonly we have the principle of virtual forces.

Use the linear elastic force-deformation relationto express the deformations of set n in terms ofthe corresponding basic forces

(2) T Tm n m s nQ V Q F Q=

Make use of the symmetry ofthe element flexibility matrix

(3)T Tm s n n s mQ F Q Q F Q=

Use the linear elastic force-deformation relationto deformations of set m resulting from the actionsof the corresponding basic forces

(4) T Tn s m n mQ F Q Q V=

Use the equality of internal and external workbetween forces of set n on displacements ordeformations of set m this time

(5)T Tn m n mQ V P U=

Tn mP UT

m nP U = Q.E.D.

A special case of Betti's theorem is Maxwell's reciprocal theorem which was enunciated a few years earlier (1864) independently. It statesthat the displacement of a particular dof under a unit force at a second dof is equal to the displacement of the second dof under a unit forceat the first dof. Note that one displacement resp. force can be a rotation, resp. moment. In the following figure this means

1

1U 1

2U

1 2U U=

This directly implies the symmetry of the flexibility matrix F (recall definition of flexibility coefficient) and, consequently, the structural stiffnessmatrix K as its inverse.


Page 671

EIm n

Example: application of Maxwell's theorem

EI

unit force

given Um and Un under a unit force at n, and Um under a unit force at m determine the vertical translations at m and nunder the loading given below

EIm nEI

2030

i.e. given Fmn , Fnn , Fmm Maxwell's theorem states that Fmn = Fnm

Thus, ( ) ( )n nn nm20 30= +F FU

( ) ( )m mn mm20 30= +F FU

unit force

1P 2P

3U

4U

First set of forces and corresponding displacements (subscript m)

mP mU

Second set of forces and corresponding displacements (subscript n)

3P

4P

1U

2U

nP nU

T Tn m m nP U P U=

Betti's Theorem states that the work done by the first set of forces on the displacements caused by the second set is equal to the workdone by the second set of forces on the displacements caused by the first set. In compact mathematical notation this is stated as

Betti's theorem

1

1U 1

2U

1 2U U=

Maxwell's theorem


Page 672

Müller-Breslau Theorem (1885) - Generation of influence lines for moving loads

Let us first look at the generation of influence lines for displacements, since the procedure results from a straightforward extension of Maxwell's theorem. We use the earlier cable-stayed bridge structural model and assume that a unit force is moving over the bridge deck (a number of nodes have been inserted for this purpose). Two locations of the unit force have been singled out and identified with letters m and n. We are interested in effect of the moving force on the translation at the top of pylon. The latter is now denoted with a double subscript (first for the location of the displacement and the second for the location of the force causing the displacement).

What is an influence line? It is the graphical representation of the effect of a unit moving force on a displacement or internal force. Theabscissa of the graph represents the location of the moving force and the ordinate its quantitative influence on the displacement or internal force of interest.

1

tmU

m

1

tnU

n

by Maxwelltm mtU U=

1

influence coefficient mη

1

mtU

m

1

ntU

n

by Maxwelltn ntU U=

tU

moving unit force

m

influence line = deformed shape under unit force

sign convention: positive in direction of force (work!)

Influence line of horizontal translation at top of the pylon


influence line for displacement Ut at top of the pylon = deformed shape under unit force

1

tU

moving unit force

m

t

30 30 30

20

20


Page 673

Influence lines for support reactions

We return to Betti's theorem and specialize it to the case that the first set of forces consists of a unit force at a free dof and anotherforce at a restrained (support) dof. We demonstrate again with the cable-stayed structural model

1

m

1P

first set of forces and corresponding displacements

1 1= −U

mU

second set of forces and corresponding displacements

Apply Betti's theorem: ( ) m 1 11 0U P U+ = Note that there are no forces of the second set doing work on the displacements of the first

selecting ( ) ( )m 11 1 0U P+ − =1 1U = − we get[ ][ ]1 m

1 force unit1 displ. unit

P U=

if the unit force is moving, then Um represents the displacement at the point of load application, under a unit support displacementin the negative direction. This gives the procedure for determining the influence line for a support reaction.

influence coefficient m mη = Uinfluence line = deformed shape under unit displacement

Influence line of vertical support reaction under the pylon

unit vertical translation 1 1= −U


30 30 30

moving unit force

m

20

20


Page 674

0 10 20 30 40 50 60 70 80 90-1.8

-1.6

-1.4

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0critical force location

support reaction is 1.6 times applied force value

distance from left support

m

influence coefficient

Influence lines for axial force N(x), shear force V(x), and bending moment M(x)

In case that we are interested in the influence line for an internal force, then we introduce a release at the particular location and study the case that we impose a unit release deformation, so that only the particular internal force does work. We have studied this in Part II and we summarize below the findings

Normal force release or axial translation device

1h ah=v vi j

ahv 1hv

fixed-end force0 1=vfor unit 1hv

0EAL

= −q

Shear force release or transverse translation device

2 3th

h h L= =−

vv v

for unit thv

0

1

1L

L

⎛ ⎞−⎜ ⎟⎜ ⎟=⎜ ⎟−⎜ ⎟⎝ ⎠

v

x L-xL

thv 2hv 3hv fixed-end force 0 2161

EIL

⎛ ⎞= ⎜ ⎟

⎝ ⎠q

Bending moment release or rotational device

i j

x L-xL

fhv

2hv3hv 2h fh

L xL−

= −v v

3h fhxL

=v v

for unit fhv

0

1xLxL

⎛ ⎞−⎜ ⎟⎜ ⎟=⎜ ⎟⎜ ⎟⎝ ⎠

v fixed-end force

01 26 21 1

EI x EIL L L

⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠q


Page 675

It is important to note the difference between internal and external work at the release. We show it for the shear force and moment release

thv

positive internal work

negative external work

i j

positive internal work

negative external work

From Betti's theorem we have

moving unit force

m

M

first set of forces and corresponding displacements

1mU

second set of forces and corresponding displacements

( ) ( )m1 1 0M− =UApply Betti's theorem: Note that there are no forces of the second set doing work on the displacements of the first

[ ][ ]m1 force unit

1 rotation unitM = U

moving unit force

m

M

1mU

Influence line for bending moment M at specific location


Page 676

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