lecture 2 truss and beam fem

19/07/2007 1

Lecture 2 – Bar and Beam Elements

Yan ZhugeCIVE 3011 Structural Analysis and

Computer Applications

2

Element Types

3

Element Types

4

Element Types

5

Finite Element Analysis - The Steps

1. Define the physical problem2. Create the finite element model

– define geometry, nodes, and elements.– define material properties, loading and

boundary conditions.

6

Finite Element Analysis - The Steps

3. Perform the calculations. The software:– generate the stiffness matrix k of each element,– connect elements together, assemble the element k

matrices to obtain the "global" matrix K,– assemble loads into a global load vector R,– impose support conditions, and– solve the global equations KD=R for the vector D of

unknowns. In structural problem D contains displacement components of the nodes.

4. Postprocess the information contained in D. In stress analysis this means compute strains and stresses.

7

Spring ElementThe first step in FEM is the derivation of the element stiffness matrix ke. This is illustrated below by using a simple spring element

Two nodes: 1,2Nodal displacement: u1, u2 (m, mm)Nodal forces: F1, F2 (N)Spring constant (Stiffness) k (N/m, N/mm)

F1 F2

x

u1 u2k

1 2

8

Consider the equilibrium of forces for the spring. At node 1, we have:

Spring ElementWe have: F = kΔ with Δ = u2 - u1 (1)

21121 )( kukuuukFF −=−−=−= (2)

and at node 2,

21122 )( kukuuukFF +−=−== (3)In matrix form

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡−

−

2

1

2

1

FF

uu

kkkk

(4)

orku = F (5)

Where k = (element) stiffness matrixu = (element nodal) displacement vectorF = (element nodal) force vector

9

Bar ElementConsider a uniform prismatic bar:

L lengthA cross-sectional areaE elastic modulus

10

Stiffness matrix – Direct methodAssume that the displacement in the bar is δ, then we have:

AEFL /=δ (6)

and δ)/( LAEF = (7)

For the respective cases in the above figure, we have:

12111 uL

AEFF == and

22212 uL

AEFF == (8)

where Fij is the force at node i (i = 1,2) associated with displacement of node j (j = 1,2)

11

Stiffness matrix – Direct methodWritten the above expressions in matrix format:

(9)or

(10)

where F1 and F2 are the resultant forces applied to the bar at nodes 1 and 2

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡−

−

2

1

2221

1211

11

FF

FFFF

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧⎥⎦

⎤⎢⎣

⎡−

−

2

1

2

1

1111

FF

uu

LAE

From the above matrix, we can see that the bar is acting like a spring in this case, the element stiffness matrix for the bar is:

⎥⎦

⎤⎢⎣

⎡−

−=⎥

⎦

⎤⎢⎣

⎡−

−=

1111

LAE

kkkk

k

12

Stiffness matrix – A Formal Procedure

• We derive the same stiffness matrix for the bar using a formal procedure which can be applied to more complicated situations

• For different types of element, the variation of the displacement between the nodes must be assumed first (what kind of behaviour can be expected). This is called the displacement function and is usually a polynomial whose order depends on the number of degree of freedom (d.o.f.) in the element. Degree of freedom is number of components of the displacement vector at a node.

13


Bar elements are used to model truss structures and any other structures where axial effects predominate. Bar element consists of 2 d.o.f. per element and hence the displacement function can be assumed as:

xu 21 ββ += (11)

The FEM treats the nodal displacements as variables of an interpolation function, usually a polynomial, to give an analytical expression for displacement at any point inside the element.

14


Eq. (11) can be written by expressing the βi in terms of nodal displacements u1 and u2:

at x = 0 u = u1 = β1at x = L u = u2 = β1+ β2L

replace β1 and β2 into eq. (11), we have:

211 uLxu

Lxu +−= )( (12)

Define two linear shape functions as follows: ξξ −=1)(1N ξξ =)(2N (13)

where

Lx

=ξ 0 ≤ ξ ≤ 1

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Each shape function Ni describes how u varies with x when the corresponding d.o.f. ui is unity while the other is zero. Re-write eq. (12) in matrix form:

⎭⎬⎫

⎩⎨⎧⎥⎦⎤

⎢⎣⎡ −

=2

1

uu

Lx

LxLu or u = Nd (14)

where N is the shape function matrixd is the vector of element nodal d.o.f.

16


Axial strain εx is given by

BddNdxd

dxdu

=⎥⎦⎤

⎢⎣⎡==ε (15)

where B is the element strain-displacement matrix

⎥⎦⎤

⎢⎣⎡−=

LLB 11

(16)

Stress can be written as

σ = Eε = EBd (17)

17


(18)

Consider the strain energy stored in the bar

(20)

The work done by the two nodal forces is

For conservative system, we have

(19)

ddVEBBd

dVEBdBddVU

V

TT

V

TT

V

T

⎥⎦

⎤⎢⎣

⎡=

==

∫

∫∫

)(

)(

21

21

21 εσ

FduFuFW T

21

21

21

2211 =+=

U = W

18


(21)which gives

(23)

Then we have

where

(22) kd = F

FdddVEBBd T

V

TT

21

21

=⎥⎦

⎤⎢⎣

⎡∫ )(

FddVEBBV

T =⎥⎦

⎤⎢⎣

⎡∫ )( or

∫=V

T dVEBBk )( (24)

is the element stiffness matrix

Equation (24) is a general result which can be used for the construction of other types of elements.

19


(26)

Now we use equation (24) to re-calculate the element stiffness matrix for the bar element:

(25)

which is the same as we derived using the direct method

[ ] ⎥⎦

⎤⎢⎣

⎡−

−=−

⎭⎬⎫

⎩⎨⎧−

= ∫ 1111

1111

0 LAEAdxLLE

LL

kL

////

The strain energy can be written as

kddU T

21

=

20

Example: Find the stresses in the two bar assembly which is loaded with force P, and constrained at the two ends, as shown in the figure.

21

Notes to example

• In this case, the calculated stresses in elements 1 and 2 are exact within the linear theory for for 1D bar structures. It will not help if we further divide element 1 or 2 into smaller finite elements.

• For tapered bars, averaged values of the cross-sectional areas should be used for the elements.

• We need to find the displacements first in order to find the stresses, since we are using the displacement based FEM.

22

The beam element is used to model beams or frames where flexural effects (shear forces and bending moments) dominate. Beam element consists of 4 DOF per element and a cubic variation in displacement has to be assumed in the form:

Beam Element

34

2321 xxxv ββββ +++= (27)

I = moment of inertia of the cross-sectional area

E = elastic modulusv = v(x) lateral displacement

rotation about the z-axis

F = shear forceM = bending moment about z-axis

dxdv

=θ

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Beam ElementAxial effect can be added on if necessary

Similar to bar element, eq. (27) can be written by expressing the βi in terms of nodal d.o.f. with θz=dv/dx=β2+2β3x+3β4x2

for example: x = 0, β1 = v1, β2 = θz1

We can then derive the shape functions for beam element.

2324

33223

2322

33221

23

2

231

LxLxxN

LxLxxN

LxLxxxN

LxLxxN

//)(

//)(

//)(

//)(

+=

−=

+−=

+−=

(28)

24

Beam ElementThen, we can represent the displacement as

where strain-displacement matrix B is given by

Curvature of the beam element is

[ ] Ndv

v

NNNNv

z

z =

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

=

2

2

1

1

4321

θ

θ(29)

BddNdxd

dxvd

=⎥⎦

⎤⎢⎣

⎡= 2

2

2

2(30)

⎥⎦⎤

⎢⎣⎡ +−−+−+−= 232232

6212664126L

xLL

xLL

xLL

xL

B (31)

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Beam ElementStrain energy stored in the beam element is

Based on eq (32) the stiffness matrix for the beam element is:

Can you prove this??

(32)

(33)

dEIBdxBddVUL

TT

V

T⎥⎦

⎤⎢⎣

⎡== ∫∫

021

21 εσ

∫=L

T EIBdxBk0

26

Beam ElementApplying the result in (31) and carrying out the integration, the expression of k is as follows

Bending moment M is computed from curvature d2v/dx2, which in turn depends on nodal d.o.f. d.

(34)

(35)

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−

=

LEILEILEILEILEILEILEILEILEILEILEILEILEILEILEILEI

k

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

46266126122646612612

22

2323

22

2323

Stress σx = My/I

EIBddx

vdEIM == 2

2

27

2D Beam Element

Combining the axial stiffness

2

2

2

1

1

1

22

2323

22

2323

46026061206120

000026046061206120

0000

z

z

vu

vu

LEILEILEILEILEILEILEILEI

LAELAELEILEILEILEILEILEILEILEI

LAELAE

k

θ

θ

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−

−−−

−

=

////////

//////////

//

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Boundary conditions

• In order to prevent the finite element model from moving freely through space, each of the possible degrees of freedom must be constrained somewhere on the model. This can be done by imposing boundary conditions at appropriate nodal points on the model.

• In general, displacement boundary conditions simulate the actual supports of the structure. Same as general structural analysis, there are three basic types of supports: simply supported, fixed and roller supports.

29

Boundary Conditions

• The application of boundary conditions depends on the type of structure being analysed and also on the finite element program being used.

• If the program supports 2D elements, then the application of boundary conditions in the third dimension for a 2D analysis is unnecessary. On the other hand, programs that use 3D elements for 2D problems require that all displacements in the third dimension are zeroed.

30

Boundary Conditions

• Engineering judgement must be applied in determining what boundary conditions best simulate the behaviour of the actual structure. In some cases, it is necessary to try the analysis using different boundary conditions to determine which set of conditions produces the worst case results.

lecture 2 truss and beam fem

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