Materials for Ijima’s seminar：Section1.1＆1.2 May 17. 2020

1-1

1. Statically linear Analysis of frame structures Chapter 1 shows the displacement method for deriving displacement, member forces and member strains in frame structures subjected to loads. In the deformation of the structures, we assume that the members consisting of the structures are elastic, and that the displacement of the structures is very small. Therefore, the response of the structure is proportional to the load, and the relation is called “linear”. Pay careful attention to the three points of the followings. i) Be accustomed to equations expressed by using vectors and matrices, and seize the physical meaning of the equations. ii) Fully understand that forces acting on a member keep the balance as well as forces acting on a node keep the balance and forces acting on the whole of a structure keep the balance. iii) Understand the algorithm that is the full flow of the computations of the displacement method. 1.1 The outline of the displacement method The displacement method can be commonly applied to various structural analyses. The method can analyze structures without differentiating structurally determinate ones and indeterminate ones. The method cannot be only applied to statically linear analysis but also geometrically nonlinear analysis, materially inelastic analysis and dynamically nonlinear analysis. Though the displacement method may be included in the method of FEM, the displacement method is utterly different from the up-to-date popular FEM. The present FEM is multiply usable, but there is a disadvantage of inaccuracy in nonlinear structural analysis. Therefore, the text uses the word of the displacement method with more original concepts than that of FEM. The method divides a frame structure to finite members. The cut members are called to the elements, and a cut point is called to a node. The displacement method uses the four of physical quantities that are the nodal displacement, the end deformation of the member, the end force of the member and the nodal force The four quantities in the displacement method are shown in Fig.1.1. The figure shows that the load acts on the planes truss, and that the member deformation is caused by the nodal displacement．The coordinate axes of u, v are universally applied to the whole of the Fig.1.2 The nodal forces equilibrated

with the end forces

!Ui

!Vi

N

N

!Uj

!Vj

i

j

Fig. 1.1 The nodal displacement and the member deformation in the plane truss

u

v

The displacement vector at the node i

ll + !l

i (ui,v

i)

j (uj,v

j)

i (ui+ !u

i,v

i+ !v

i)

j (uj+ !u

j,v

j+ !v

j)

Load

The displacement vector at the node j

1-2

structure and define the nodal position vectors ui, u j in the plane.

The nodal position vectors are the following,

ui=

ui

vi

!"#

$%&

, uj=

uj

vj

!"#

$%&

. (1.1,1)

The nodal displacement vectors in Fig.1.1 are the following,

!ui=

!ui

!vi

"#$

%&'

, !uj=

!uj

!vj

"#$

%&'

. (1.1.2)

The end deformation of the axial member is only the elongation !l of the member. However, general members such like bending-axial-member have end deformation expressed by more than one physical quantity, so that the text uses the vector s as the end deformation. That is, s = !l . (1.1.3) The end deformation occurs by the end force acting on the member as shown in Fig. 1.2. The end deformation of the axial member is only the axial force N , but end force of general member is more than one. Therefore, the text uses the vector S as end forces of a member. That is, S = N . (1.1.4) The nodal forces equilibrated with the end force are defined as the vector !U

i, !Vj . That is,

!Ui=

!Ui

!Vi

"#$

%&'

, !Vj=

!Uj

!Vj

"#$

%&'

. (1.1.5)

The four of the physical quantities in the displacement method are gathered as follows, the nodal displacement Δu, the end deformation s, the end force S and the nodal force ΔU. The four of the physical quantities can compose Fig. 1.3 that conceptually shows the displacement method, Additionally, Fig. 1.3 expresses the relations between the four quantities. When it is assumed that the displacement is very small, the relation between the end deformation of a member and the displacement as well as the relation between the nodal force and the end force of the member becomes linear. Further, when the member is elastic, the relation between the end force and the end deformation becomes linear. The three relations are linear, and

Fig. 1.3 Conceptual figure of the linear analysis with the displacement method

Nodal displacement vector

Nodal force vector

End deformation vector

End force vector

Stiffness equation

Compatibility End force equation

Equilibratory equation

Solving the linear equation

!U

!u

s

S

!U = KO!u

s = aT!uS = ks

!U = aS

1-3

therefore, the relation between the nodal force and the nodal displacement is linear. The relation in the fourth quadrant is called “Compatibility”. The relation in the third quadrant is called “End force equation”. The relation in the second quadrant is called “Equilibratory equation”. The relation in the first quadrant is called “Stiffness equation”. Gothic letters appeared in the following equations as well as Fig. 1.3 express vectors or matrices. The normal letters are scalars. Fig. 1.3 also shows the computational procedure in the displacement method. If the nodal displacement is obtained, the compatibility gives the end deformations of members, s = aT!u , (1.1.6) where a is called the transformation matrix composed by the length of a member and the direction in the universal coordinates covered to the whole of a structure. The upper subscript T transposes the matrix a . The matrix a is generally rectangular, and therefore, the end deformation cannot directly bring the nodal displacement. Next, the end deformations of the member brings the end forces as the following, S = ks , (1.1.7) where k is the member stiffness matrix composed by the material stiffness of the member. The matrix is square. The end forces of the member bring the nodal force equilibrated with the end forces, as the following, !U = aS , (1.1.8) where a is the matrix transposing the transformation matrix in the equation (1.1.6). The nodal forces cannot directly bring the end forces of the member. The principle of virtual work can prove the relation of the transposition between the equation (1.1.6) and the equation (1.1.8). The equations of (1.1.6), (1.1.7) and (1.1.8) bring the stiffness equation in the first quadrant, namely, the equation (1.1.7) is substituted for the equation (1.1.8), and then the equation (1.1.6) is substituted for that. The result is the following, !U = aS = aks = aka

T!u = K

O!u . (1.1.9)

The matrix K

O is called the stiffness matrix and it is,

K

O! aka

T . (1.1.10) Finally, the stiffness equation of the relation between the nodal forces and the nodal displacement is expressed as the following, !U = K

O!u . (1.1.11)

The procedure for computing the nodal displacement of a structure, member forces and so on is the followings. Combining the compatibility of each member, the end forces equation and the equilibratory equation brings the stiffness matrix K

O of each member. Then assembling them

makes the stiffness matrix of the whole of the structure. Since the nodal forces of given load acting on the structure become the constant vector in the equation (1.1.11), solving the linear equation brings the nodal displacement. Next, the nodal displacement set as the starting point, and following the computational flow in Fig. 1.3 brings the end deformation of each member, the end forces and the nodal forces including the reaction forces at supporting points.

1-4

1.2 Statically linear analysis of a truss in a two dimensional plane The text applies the displacement method to analyzing statically linear behavior of a plane truss. The four of the physical quantities used in the analysis are the same as shown in the section 1.1. The compatibility, the end force equation of an axial member and the equilibratory equation are the following. (1) Compatibility of axial member in a plane The text derives the compatibility by using the axial member that both ends are connected to the nodes i, j. Fig. 1.4 shows the elongation of the member expressed by the nodal displacement assumed to be very small. The left figure displays the axial member before and after the displacement. The elongation is !l . The right figure displays the two members after translating the member after the displacement until matching the end i with the end i of the member before the displacement. The relative displacement of the end j is !uj

" !ui of the horizontal component and !vj " !v

i of the vertical component. The dotted lines are orthogonal to the member after the displacement. The equation (1.2.1) is the elongation of the member given by referring Fig. 1.4. Though the equation is not geometrically exact, If we assume that the nodal displacement is very small, the equation can be permissible. !l = (!uj " !ui )cos# + (!vj " !vi )sin# , (1.2.1) where ! is the incline angle, and the cosine of the angle and the sine are shown by the following equations that use the nodal positions,

cos! =uj " ui

l# $, sin! =

vj " vi

l# % . (1.2.2a,b)

The equations (1.2.2a, b) are called the direction cosine, and substituting the direction cosine defined by ! and ! for the equation (1.2.1) gives the following compatibility,

!l = (!uj " !ui )# + (!vj " !vi )$ = "# "$ # $( )

!ui!vi!uj

!vj

%

&''

(''

)

*''

+''

= aT

!ui!u j

%&(

)*+= a

T!u . (1.2.3)

Fig. 1.4 Geometric relation between the elongation and the nodal displacement

!

!uj" !u

i

!vj" !v

i!

(!uj" !u

i)cos#

(!vj" !v

i)sin#

(ui,v

i)

(uj,v

j)

!ui

!vi

!uj

!vj

Length l before the displacement

Length l + !l after the displacement

1-5

(2) End force of axial member Fig. 1.5 shows the end force of the axial member. As shown in the figure, the end force is not the axial force of the internal force but the external force acting on the end of the member. The end force of the axial member, however, equals to the axial force of the internal force, and therefore, the end force is,

N =EA

l0

!l " F!l , (1.2.4)

where E: the Young’s modulus，A: the area of the cross section orthogonal to the member axis， l

0: the length of the unstressed member. In the linear analysis,

the relative stiffness F = EA / l0

is constant. (3) Equilibratory equation of axial member in a plane Fig. 1.6 shows the equilibration at the node i that the nodal forces !U

i, !V

i as well as the

reaction of the end forces act on. The equilibration means that the sum total of the two components given by decomposing the reaction of the end force acting on each member connected to the node i equal to the nodal forces !U

i, !V

i respectively.

Fig. 1.7 focuses on the member p in the members connected to the node i. The subscript p is omitted in Fig. 1.7. The direction cosine (!,") can decompose the reaction of the end force N , and then the equilibration gives the following equation,

!U =!U

i

!Uj

"#$

%&'=

!Ui

!Vi

!Uj

!Vj

"

#((

$((

%

&((

'((

=

)*

)+

*

+

"

#((

$((

%

&((

'((

N = aN . (1.2.5)

The equation (1.2.5) brings the nodal forces equilibratory to the end force.

N

l + !l

EA

l0

i

j

Fig. 1.5 End force of the axial member

!"#$ %& i '(')*+,

U

V N

q

Nq

Nr

Nr

Np

Np

i

i

i

i

l

k

j p

q

r

!Ui =

!Uip+ !Ui

q+ !Ui

r

!Vi = !Vi

p+ !Vi

q+ !Vi

r

Np

Np

p (ui ,vi )

(u j ,v j ) l

!Ui

p

!Vi

p

!U j

p

!V j

p

!"#- ./ p 01'(')*+,

Fig. 1.7 The nodal forces equilibratory to the end forces of the member p

!Ui

!Vi

N

N

!Uj

!Vj

i

j

Fig. 1.6 Equilibration of forces acting on the node i

!

! 1

1-6

(4) The stiffness equation of the axial member in a plane The text derives the stiffness equation of the member p in Fig. 1.7. The stiffness equation becomes the following from referring the equation (1.1.9). !U = aN = aF!l = Faa

T!u = K

O!u . (1.2.6)

The concrete computation of the equation (1.2.6) gives the following,

!Ui

!Vi

!Uj

!Vj

"

#$$

%$$

&

'$$

($$

== F

)*

)+

*

+

"

#$$

%$$

&

'$$

($$

)* )+ * +{ }

!ui

!vi

!uj

!vj

"

#$$

%$$

&

'$$

($$

= F

* 2 *+ )* 2 )*+

*+ + 2 )*+ )+ 2

)* 2 )*+ * 2 *+

)*+ )+ 2 *+ + 2

,

-

.

.

.

.

/

0

1111

!ui

!vi

!uj

!vj

"

#$$

%$$

&

'$$

($$

. (1.2.7)

Further, the nodal forces as well as the nodal displacement in the equation (1.2.7) are separated into the two vectors according to the node, in order to make the stiffness equation of the whole of the structure. Then, the stiffness matrix are divided into the four matrices,

!U

i

!Uj

"#$

%&'=kiikij

kjikjj

(

)*

+

,-

!ui

!uj

"#$

%&'

, (1.2.8)

where,

!U

i=

!Ui

!Vi

"#$

%&'

, !Uj=

!Uj

!Vj

"#$

%&'

, !ui=

!ui

!vi

"#$

%&'

, !uj=

!uj

!vj

"#$

%&'

,

kii=

( 2 ()

() ) 2

*

+,

-

./, k

ij= k

ji

T=

0( 2 0()

0() 0) 2

*

+,

-

./, k

jj=

( 2 ()

() ) 2

*

+,

-

./.

(1.2.9a~g)

(5) Composing the stiffness equation of the whole of the structure The text composes the stiffness equation of the whole of the structure from focusing the node i in Fig. 1.6. The three members of p, q and r are connected to the node i, as shown in Fig. 1.6. The stiffness equation with the matrix divided into the four matrices like the equation (1.2.8) can be applied to each member. The nodal force vectors equilibratory to the end forces of the three members respectively are defined as !Ui

p, !Ui

q, !Ui

r . Then, the nodal force vector at the node i !U

i can be expressed as the following,

!Ui = !Ui

p+ !Ui

q+ !Ui

r= (kii

p!ui + kij

p!u j ) + (kii

q!ui + kik

q!uk ) + (kii

r!ui + kil

r!ul )

= (kiip+ kii

q+ kii

r)!ui + kij

p!u j + kik

q!uk + kil

r!ul (1.2.10)

Putting the equation (1.2.10) into the stiffness equation composed by the whole of the nodal forces and the whole of the nodal displacement gives the following,

1-7

!

!Ui

!

!U j

!

!Uk

!

!Ul

!

"

#

$$$$$$

%

$$$$$$

&

'

$$$$$$

(

$$$$$$

=

" ! ! ! !

# kiip+ kii

q+ kii

r# kij

p# kik

q# kil

r#

! " ! ! !

# k ji

p# k jj

p# # # # #

! ! " ! !

# kkiq

# # # kkkq# # #

! ! ! " !

# klir

# # # # # kllr#

! ! ! ! "

)

*

++++++++++++

,

-

.

.

.

.

.

.

.

.

.

.

.

.

!

!ui!

!u j

!

!uk!

!ul!

"

#

$$$$$$

%

$$$$$$

&

'

$$$$$$

(

$$$$$$

. (1.2.11)

When the nodal forces are given, the equation (1.2.11) becomes the linear equation that the nodal displacement is unknown. The stiffness matrix in the equation (11.2.11), however, is singular at this stage, so that the linear equation cannot be solved. That is to say, since supporting conditions are not considered in the equation, the structure is unstable at this stage. Mathematically speaking, the determinant of the stiffness matrix is zero, so that there exists no inverse matrix of the stiffness matrix. (6) Applying supporting conditions to the nodal force vector and the stiffness matrix The nodal force vector and the stiffness matrix are changed in order that the nodal displacement at the supporting point may satisfy the supporting conditions. For example, the text uses a plane truss with the supporting conditions as shown by the equation (1.2.12). !uj = 0, !vj = 0, !vk = 0 (1.2.12) The supporting conditions are that the node j is fixed in the two directions of u and v, and that the node k is fixed in the direction of v only. The equation (1.2.13) is derived from applying the supporting conditions to the equation (1.2.11). , (1.2.13)

1 i uj vj uk vk l

!

!Ui

!

!Uj = 0

!Vj = 0

!

!Uk

!Vk = 0

!

!Ul

!

"

#

$$$$$$$

%

$$$$$$$

&

'

$$$$$$$

(

$$$$$$$

=

" ! 0c 0c ! 0c !

# kiip+ kii

q+ kii

r# 0c 0c # kik

q *0c # kil

r#

! " 0c 0c ! 0c !

0r

0r

0r

0r

0r

0r

1 0

0 1

0r

0r

0 0

0 0

0r

0r

0r

0r

0r

0r

! 0c 0c " ! 0c !

#

0r

kkiq *

0r

#

0r

0

0

0

0

#

0r

Fq() q

)2

0

0 1

#

0r

#

0r

#

0r

! ! 0c 0c ! 0c " !

# klir

# 0c 0c # ! 0c # kllr#

! 0c 0c ! 0c ! "

*

+

,,,,,,,,,,,,,,,

-

.

///////////////

!

!ui!

!uj

!vj

!

!uk!vk!

!ul!

"

#

$$$$$$$

%

$$$$$$$

&

'

$$$$$$$

(

$$$$$$$

1-8

where,

0r= 0 0{ }, 0

l=

0

0

!"#

$%&

(1.2.14a.b)

kkiq *

= Fq !(" q

)2 !" q# q{ } (1.2.15a)

kikq *

= kkiq *{ }

T

(1.2.15b) and F

q, (! q

, " q) are the relative

stiffness in the elongation of the member q, the direction cosine of the member q, respectively. The equation (1.2.13) is to specify the two components of the row vectors as well as the column vectors corresponding to the node j and the node k in the equation (1.2.11). First, in the nodal force vector, the elements corresponding to the degrees of the freedom fixed change into zero as shown in the equation (1.2.13), namely, the elements corresponding to both components of the node i as well as the v component of the node k are zeros. Next, in the stiffness matrix, the diagonal elements in the row vectors corresponding to the degree of the freedom fixed change into ones and the off-diagonal elements into zeros. As a result, the nodal displacement with the fixed condition is zeros in the solution. The off-diagonal elements in the column vectors corresponding to the degree of the freedom fixed also change into zeros. The diagonal elements in the column vectors keep ones. This is because the zero displacement due to the fixed conditions does not affect to the nodal forces. (7) Computational procedure of statically linear analysis by the displacement method Fig. 1.8 shows the computational procedure of statically linear analysis by the displacement method. First, the procedure inputs the data of a structure and the data of loads. The structural data is the specifications of the structure as shown in Fig. 1.8. Next, the procedure is the making of the nodal force vector by using the data of the load and the making of the stiffness matrix by using the data of the structure.

Fig. 1.8 Computational procedure of statically linear analysis by the displacement method

Managing the supporting conditions to the nodal vector!U and the stiffness matrixK

O

Computing the reaction forces at the supporting points and confirming the nodal equilibration, !U " aN#

Output the computational result

Input the load data

Making the nodal force vector!U

Computing the end deformation with the compatibility !l = aT!u

Computing the end force by the equation N = F!l

Making the stiffness matrix KO

Solving the linear equation !U = KO!u

Input the structural data: the number of all of the nodes, the number of all of the members, the supporting conditions, the nodal positions, both nodal numbers of members, member stiffness, and so on.

①

②

③

④

⑤

⑥

⑦

⑧

⑨

⑩

1-9

Next, the procedure is to apply the supporting conditions to the nodal force vector and the stiffness matrix. Then, solving the linear equation of the stiffness equation gives the nodal displacement. Next, the following procedure brings member forces, reaction forces at supporting points. First, the nodal displacement brings the elongation of the member through the compatibility, and then the elongation gives the axial end force through the end force equation. Finally, deducting the sum of aN of all of the members from the nodal force vector of all of the nodes gives the minus values of the reaction forces at the supporting nodes. The resultants at the nodes except the supporting nodes become zeros because all of the forces acting on the node without restriction are equilibratory.. Besides the static analysis of structures subjected by loads, there are the static analysis of structural deformation caused by supporting points moving, the static analysis of structural deformation caused by members expanding and contracting and so on. We do not treat them here.