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Chapter 2: Bars and Beams■ Static analysis: forces are constant in time or change very

slowly.

■ Linear analysis: deflections are small so that material behavior is elastic. No failure, no gaps that open or close.

■ Truss elements (bars, rods): pinned(hinged) at connectionpoints; resist axial forces only. Hence it has axial dofs only.

■ Frame elements (beams): welded (or, connected withmultiple fasteners) at connection points; resist axial and transverse forces and bending moments. Has axial, transverse and rotational dofs.

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Bar Element: Nodal displacements and nodal forces■ Nodal displacements and nodal forces of a finite element are related through the stiffness matrix of the element.

■ We’ll derive the stiffness matrix of a bar element now:

In Fig. (a), the left node is displaced while u2=0. In Fig. (b), it is the opposite. The forces to maintain these displacements:

F11= -F21=(AE/L)u1 ; -F12= F22=(AE/L)u2

• •

Lx

u1

F21F11

A, E

1 2

(a)

• •

Lx

F22F12

u2

A, E1 2

(b)

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Bar Element: Stiffness Matrix Derivation■ If both u1 and u2 are nonzero, the nodal forces are

F1=F11+F12=(AE/L)(u1-u2) ; F2=F21+F22=(AE/L)(u2-u1)

■ Writing these equations in matrix form:

where the coefficient matrix is called the element stiffness matrix.

=

−

−

2

1

2

11111

FF

uu

LAE

≡

≡=

−

−=

2

1

2

1 , with and 1111

FF

uu

LAE rdrkdk

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General Stiffness Matrix Formulation

■ The above is a direct method to compute the element stiffness matrix. This method is feasible for simple elements only. There is also a formal procedure which uses the following:

where B: strain-displacement matrix for the elementE: stress-strain matrixdV: volume element

∫= dVTEBBk

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Bar Element: Shape (interpolation) functions■ To derive B we interpolate axial displacement u of an arbitrary point on the bar between its nodal values u1 and u2:

L

u1u2

xu=N1u1+N2u2

where N1 and N2 are called the shape functions:

N2=x/L1

x

N1=(L-x)/L1

x

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Bar Element: Strain matrix B

■ Rewriting u:

Nd=

−

=2

1uu

Lx

LxLu

where N is the shape function matrix. Then,

−==

===

LLdxd

dxd

dxdu

x11 where)( BBddNNd

ε

Now, since there is only one stress component in an axial bar, σ=Eε and, therefore, the stress-strain matrix is just the elastic modulus E.

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Bar Element: Stiffness Matrix derivation

■ Substituting B and E into the integral expression for the element stifness matrix,

−

−=

−

−

= ∫ 111111

/1/1

0LAEAdx

LLE

LLL

k

which is the same matrix as before.

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Bar Element: When is element exact?NOTES:

■ The two-node bar element we’ve just derived can only represent a constant state of strain within a bar.

■ If ◆ axial forces are applied along the length of an actual bar

(instead of just the nodes), ◆ the bar is not uniform (non-uniform A and/or E)

then the bar element represents the actual bar only approximately. Then the actual bar can be divided into multiple bar elements and the exact results are approached as more and more elements are used to model the actual bar.

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Plane Beam Element: Nodal displacements and nodal forces

■ Resists transverse shear force and in-plane bending only

■The corresponding displacements are transverse displacement, v=v(x), and rotation, θ=θ(x), respectively.

■The element, therefore, has two degrees of freedom (d.o.f) at each node (i.e., at each end) : (v1, θ1) and (v2, θ2)

LM1 M2

F1 F2

xθ1 θ2

v1 v2

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Plane Beam Element: Stiffness Matrix Derivation

■ Similar to bar element, we can derive the entries of the stiffness matrix by making all nodal d.o.f. zero except one (seeFig. 2.3-1). This is the direct method.

■ There is again a formal procedure which this time uses

(EI: bending stiffness)

where B: curvature-displacement matrix for the element.

That is,

∫=LT dxEI

0

BBk

Bd=2

2

dxvd

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Plane Beam Element: Transverse displacement function■ Transverse displacement v(x) is cubic in x for uniform prismatic beams loaded only at its ends (elementary beam theory).

■ So, the following function is used for a beam element:

v=v(x)=β1+ β2x+ β3x2+ β4x3

which is approximate for a beam element loaded along its length instead of just at its ends. (Use of a linear displacement function would lead to slope discontinuity at a common node of two beam elements.)

■ βi can be expressed in terms of nodal displacements vj and θj using boundary conditions:

v=v1 and θ=θ1 at x=0 ; v=v2 and θ=θ2 at x=L where θ=dv/dx

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Plane Beam Element: Shape functions■ Transverse displacement can then be written in terms of shape functions Ni and nodal dof:

■See Fig. 2.3-1 for other shape functions ■Curvature of the beam element is

[ ]1

11 2 3 4

2

2

2 3 2 3

1 22 3 2

3 2 31

v

v N N N Nv

x x x xN N xL L L L

θ

θ

= =

= − + = − +

N d

BddNNd=

== 2

2

2

2

2

2 )(dxd

dxd

dxvd

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Plane Beam Element: Stiffness matrix■ Curvature-displacement matrix is given by

■Substitution of B into the stiffness matrix expression yields

(SYMMETRIC)=2 LEIk

+−−+−+−= 232232

62 126 64 126Lx

LLx

LLx

LLx

LB

−

−

−

2

36

132

3636

2

22

LL

LLLL

L

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Plane Beam Element: Diagonal entries of a stiffness matrix■ The element stiffness matrix relates the end forces and

moments to the nodal d.o.f. in the following manner:

■ For example,

where, for instance,

■ If all d.o.f but θ1 were zero, M1=k22 θ1. Hence, k22>0 !!!

■ Similarly, all diagonal entries of a stiffness matrix are positive

=

2211

2211

θ

θv

v

MFMF

kL

M1 M2

F1 F2

xθ1 θ2

v1 v2

2242231221211 θθ kvkkvkM +++=

223 6 LEIk −=

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Plane Beam Element: Flexural stress

■ Normal stress on a cross-section located at x caused by a bending deformation is given by

where y is the distance from the neutral axis of the cross-section (σ is the value of the stress at y). M is the internalresultant bending moment at that cross-section and isrelated to the curvature:

NOTE: From B, the internal moment distribution M is linear in x.

IMyx == )(σσ

)( ; )( 2

2xEI

dxvdEIxMM BBBd ====

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General Plane Beam Element■ A finite element can resist a certain force or a moment if it has the corresponding dof at its nodes. The two-dof bar element cannot resist moment or transverse shear force while the plane beam element we’ve just seen cannot resist axial force.

■ General plane beam element (2D frame element) has three dof at each node and can resist axial force, transverse shear and bending in one plane.

■The 6x6 stiffness matrix is a combination of those of the bar element and the simple beam element (Eq. 2.3-9 in textbook)

LM1 M2

F1 F2

θ1 θ2

v1 v2

u1P1 P2

u2

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3D Beam Element■ A 3D beam (space beam) element can resist forces and moments in all directions. It thus has 6 dof at each end (3 displacements and 3 rotations)■ The stiffness matrix is derived in reference to axes directed along the beam element and along other suitable dimensions of the element (local axes x,y,z). That is what we did for the bar and plane elements also.■ A given structure to be modelled would have beams in arbitrary orientations. Stiffness matrix of each element is defined in its own local axes.■ A common axis system is needed for the structure. Global axes or structure axes X,Y,Z !!! (read pp. 24-25)■ Element stiffness matrices are transformed from local to globalcoordinates. (Local dof at a node are transformed to global dof.)

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Element Stiffness Matrix: Singularity

■ The element stiffness matrices we derived for a bar and a planebeam are singular as such because all dof are free there.

■ A bar with that stiffness matrix, for example, can have the rigid-body motion u1=u2=c. In this motion, the bar moves along its own axis with no strain induced.

■ This rigid-body motion is prevented if u1 or u2 is prescribed as zero or a nonzero value in which case the corresponding dof drops out of the stiffness matrix and the matrix becomes nonsingular.

■ Similarly, a plane beam can translate or rotate freely unless supported properly (which means proper dof are restrained).

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Global Stiffness Matrix■ A structure to be modelled is divided into a number of finite elements such as bar and beam elements and other types.

■ These elements either represent the same structural member (a beam, for instance, being modelled by three beam elements) or are connected to elements representing different structural members.

■ In either case, the interface between adjacent finite elements is provided by the nodes and elements sharing the same node have the same dof, that is, the same displacements and rotations at that node.

■ All element stiffness matrices, after being transformed to global coordinates, are assembled together to form the global stiffness matrix. (In the process, entries of element stiffness matrices corresponding to the same nodal dof are added directly.)

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Global Stiffness Matrix: Singularity■ A global stiffness matrix relates the nodal dof and the external forces and moments applied to the nodes:

where D is the vector of all the nodal dof for the wholestructure.

■ If the external loads are not just applied at the nodes of the actual structure, the distributed loads are first converted to equivalent nodal loads. Commercial codes usually do this for theuser for simple distributed loads.

■ K is singular if the structure is unsupported or inadequately supported. To prevent singularity, supports must be sufficient to prevent all possible rigid-body motions. (Read Section 2.4)

RKD =

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Loads

■ If the external loads are distributed on the actual structure and not just applied at the nodes, these are first converted to equivalent nodal loads. Commercial codes usually do this for theuser for simple distributed loads such as uniform or linearly varying loads.

■ Example to distributed contact loads: pressure on a wing surface, internal pressure in a tank, friction between two surfaces, line load along a beam.

■ Another form of distributed load is body forces: gravity, inertia (arising from acceleration), magnetic forces, thermal loads. These act at every material point in the body and not just on the surface of the body.

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Equivalent Nodal Loads For a Bar Element■ Shown below are a bar element and a bar structure with uniformly distributed axial force of intensity q (N/m, lb/ft). q may be a contact force or a body force. Below them are given the equivalent nodal loads.

■ The total force qL on an element is evenly divided between the two nodes. These nodal forces are statically equivalent to the actual loading.

■ Adjacent elements contribute to nodal forces on the structure.

LT

q

• •• •

3@L=LT

qL qL/2qL

Structure

qL/2

Element

• •1 2q

L

q• •1 2

LqL/2

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Consistent loads

■ Element equilibrium kd=r■ Work done by element loads W=dTr■ Should be equal to work of distributed loads

■ Therefore■ For example, with q=x2

■ when loads are concentrated by intuition, they are called lumped loads

1 1 2 2( )W qudx N u N u dx= = +∫ ∫1 1 2 2

0 0

L L

r qNdx r qN dx= =∫ ∫

2 21 2

0 0

(1 / ) /12 ( / ) / 4L L

r x L x dx L r x L x dx L= − = = =∫ ∫

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Accuracy of Computed Displacements and Stresses■ A FE program computes nodal dof first, then uses them to compute strains and/or stresses.

■ The computed nodal dof are mostly not exact. When statically equivalent nodal loads are used for bar and beam elements, the computed nodal dof are exact. But the displacements between the nodes may still be inexact even though usually highly accurate.

■ The strains (and stresses), obtained by differentiating displacements, are less accurate than displacements.

(Differentiation amplifies inaccuracy!!!)

■ Stresses are usually most accurately computed at the element center. Averaging nodal stresses increases accuracy. But stress computed at a boundary, which is usually the highest and, therefore, is of most interest, is usually less accurate.