CI221 Finite Element Notes

8/3/2019 CI221 Finite Element Notes

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CI 221 Computational Methods

Introduction to Finite Element Analysis

Dr Andrew TM Phillips

While intended to give an overview of this part of the course these notes are not a complete

record and should not be treated as a substitute for attending lectures and tutorials.

Copyright 2010. All rights reserved.


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Aim of this part of the course:

The aim of this part of the course is to introduce the finite element method, and inves-

tigate its application in particular in structural engineering. Emphasis is placed on theuse of Matlab and GSA in achieving this.

No one believes the results of a computational model, except the person who put to-

gether the model. Everyone believes the results of an experimental test, except the person

who put together the experiment.

Recommended reading

The following suggested textbooks may be useful for students looking to do further read-

ing on the subjects introduced in this part of the course.

Kwon and Bang, The Finite Element Method Using MATLAB

Ferreira, MATLAB Codes for Finite Element Analysis

Zienkiewicz, Taylor and Zhu, The Finite Element Method: Its Basics and Funda-

mentals

Zienkiewicz and Morgan, Finite Elements and Approximation

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1 Introduction

The finite element method is a very powerful mathematical technique that has found

significant application in structural engineering, as well as other fields of science andengineering.

The first recognised use of the Finite Element Method as we know it is that of

Turner et al.1 who used it to numerically model aircraft wings using a mesh of triangular

elements. In introducing the finite element method we will first look at the Galerkin

method of weighted residuals for approximating functions, before moving on to look at

strong and weak function formulations.

1.1 Galerkin method of weighted residuals

Although the Galerkin method is not the only method of weighted residuals available for

approximating functions we will investigate it, as we will see later on that it has particular

relevance to the finite element method.

Let us first consider the following differential governing equation:

d2u

dx2 u + x = 0 (1.1)

In order to define the problem let us introduce a domain, 0 < x < 1, and let us also

define boundary conditions, u(0) = 0, and u(1) = 0. This information can be rewritten

in the form:

d2udx2 u + x = 0, 0 < x < 1u(0) = 0, and u(1) = 0 (1.2)The first step for any method of weighted residuals is then to determine a trial function

which contains unknown coefficients to be determined. Here we will select the trial

function:

1Turner et al., Stiffness and Deflection Analysis of Complex Structures, Journal of AeronauticalSciences 23(9):805-823,854, 1956

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u = ax(1 x) (1.3)

where u denotes an approximate solution for u, and a is the unknown coefficient. The trial

function, u suggested has been chosen as it satisfies the boundary conditions specified

for the governing differential equation (u(0) = 0, and u(1) = 0). It is observed that the

goodness of fit of the approximate solution to the exact solution is dependent on the trial

function. Having selected a a trial function we define a residual, R by substituting u into

the governing differential equation:

R =d2u

dx2 u + x = 2a ax(1 x) + x (1.4)

where because u will be different to the exact solution, the residual R will not be equal

to zero for all values of x within the domain. Having defined the residual, R we can now

combine this with a weighting function w in order to determine the value of the coefficient,

a such that our chosen trial function provides the best fit that can be achieved compared

to the exact solution. To do this we set the integration of the weighted residual over the

domain to zero:

10 wR dx = 10 w d2udx2 u + x dx=10

w(2a ax(1 x) + x) dx = 0(1.5)

For Galerkins method we set the weighting function based on the chosen trial function,

such that:

w =du

da= x(1 x) (1.6)

Hence for the chosen trial function, setting the integrated weighted residual over the

domain to zero we have:

10

x(1 x)(2a ax(1 x) + x) dx = 0 (1.7)

which can be solved to give a = 0.2272.

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Although not discussed here the governing differential function can be solved exactly

to give:

u = x1

(e1 e1)(ex

ex

) (1.8)

Hence we can plot the solution provided by our chosen trial function against our exact

solution, as shown in Figure 1.1.

0 0.2 0.4 0.6 0.8 10

0.01

0.02

0.03

0.04

0.05

0.06

x

u

(solid

red),u

Figure 1.1: u, the exact solution (solid red) and u, the approximate solution (dash-dotblue) found using the Galerkin weighted residual method

In order to improve the approximation we can change the trial function, and increase

the number of independent coefficients. Hence another trial function could be:

u = a1x(1

x) + a2x2(1

x) (1.9)

where a1 and a2 are independent coefficients. This trial functions gives the residual:

R =d2u

dx2 u + x = a1(2 x + x2) + a2(2 6x + x2 + x3) + x (1.10)

In order to find appropriate values for a1 and a2 we define the same number of weight-

ing functions as we have unknowns. For the Galerkin method these are defined:

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wi =u

ai(1.11)

So for the given trial function we have:

w1 = x(1 x), w2 = x2(1 x) (1.12)

Hence for the chosen trial function, we can define two weighted residuals, and set their

integrals to zero over the domain:

1

0

w1Rdx = 0,

1

0

w2Rdx = 0 (1.13)

giving simultaneous expressions, that in matrix form can be solved to give a1 = 0.1457

and a2 = 0.1628:

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a1

a2

=

1/121/20

(1.14)

Hence we can plot the solution provided by our 1st-order and 2nd-order chosen trialfunctions against our exact solution, as shown in Figure 1.2. It is observed that the

2nd-order chosen trial function has a very high goodness of fit for the exact solution.

1.2 Weak formulation of the Galerkin method

When we were operating on the trial function previously we were using the strong form,

where the evaluation of the integrated weighted residual relied on the highest order of

derivative term in the residual having a non-zero value. This restricted our choice of trial

functions to those where:

2u

x2= 0 (1.15)

We can however apply integration by parts to the strong form, in order define the

weak form:

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0 0.2 0.4 0.6 0.8 10

0.01

0.02

0.03

0.04

0.05

0.06

x

u

(solid

red),

u

Figure 1.2: u, the exact solution (solid red), 1st-order approximate solution (dash-dot blue), and 2nd-order approximate solution (dotted black) found using the Galerkinweighted residual method

10

w

d2u

dx2 u + x

dx =

10

dw

dx

du

dx wu + wx

dx +

w

du

dx

10

= 0 (1.16)

If we refer back the our original trial function for the governing differential equation

it can be seen that we will arrive at the same value of a using the weak form as we did

for the strong form:

10

(wu wu + wx)dx + [wu]10 = 0

u = ax(1

x), u = a

2ax, w = x(1

x), w = 1

2x

a = 0.2272

(1.17)

While for the originally chosen trial function this is not seen to have any immediate

advantage, we can go on to introduce the concept of using piecewise continuous trial

functions.

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1.3 Piecewise continuous trial functions

It is observed that regardless of whether a strong or weak formulation is adopted the

accuracy of any approximate solutions depends on the chosen trial function. However

selecting a trial function is not always an easy task. The is particularly true if the exact

function varies widely over the problem domain, if the domain has a complex shape in

two or three dimensions, or if the domain, or problem has a complex set of boundary

conditions. The task can be made easier by selecting piecewise continuous trial functions.

Consider piecewise linear functions in a one dimensional domain:

i(x) = (x

xi1)/hi for xi1

x

xi

(xi+1 x)/hi+1 for xi x xi+10 for x < xi1, or x > xi+1

(1.18)

The piecewise linear functions can be represented graphically as shown in Figure 1.3.

Figure 1.3: Piecewise linear functions

Let us go on to define a piecewise linear trial function such that:

u = a11(x) + a22(x) (1.19)

where a1 and a2 are unknown coefficients and 1 and 2 are defined:

1(x) =

3x for 0 x 1/32 3x for 1/3 x 2/30 for 2/3 x 1

(1.20)

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2(x) =

0 for 0 x 1/33x 1 for 1/3 x 2/33 3x for 2/3 x 1

(1.21)

The trial function, u can then be rewritten as:

u =

a1(3x) for 0 x 1/3a1(2 3x) + a2(3x 1) for 1/3 x 2/3a2(3 3x) for 2/3 x 1

(1.22)

and graphically represented as shown in Figure 1.4.

Figure 1.4: Piecewise linear trial functions

The Galerkin method gives the following weighting functions:

w1 =

3x for 0 x 1/32 3x for 1/3 x 2/30 for 2/3 x 1

(1.23)

w2 =

0 for 0 x 1/33x 1 for 1/3 x 2/3

3 3x for 2/3 x 1

(1.24)

The integrated weighted residuals over the domain are given:

IwR1 =

10

(w1u w1u + w1x)dx = 0 (1.25)

IwR2 = 1

0

(

w1u

w1u + w1x)dx = 0 (1.26)

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where [wiu] is omitted recognising that w1(0) = w1(1) = w2(0) = w2(1) = 0.

We can go on to evaluate the weighted integrals:

IwR1 =1/30

[3(3a) 3x(a1(3x)) + (3x)x] dx+2/31/3

[3(3a1 + 3a2) (2 3x)(2a1 3a1x + 3a2x a2) + (2 3x)x] dx+12/3

0 dx = 0

(1.27)

IwR2 = 1/3

00 dx

+ 2/31/3 [3(3a1 + 3a2) (3x 1)(2a1 3a1x + 3a2x a2) + (3x 1)x] dx+12/3

[3(3a2) (3 3x)(3a2 3a2x) + (3 3x)x] dx = 0(1.28)

giving simultaneous equations, that in matrix form can be solved to give a1 = 0.0488 and

a2 = 0.0569:

6.2222 2.9444

2.9444 6.2222 a1

a2 = 0.1111

0.2222 (1.29)Hence we can plot the solution provided by our piecewise linear trial functions against

our exact solution, as shown in Figure 1.5. It is observed that although a good fit is not

achieved with just two piecewise linear trial functions, a good fit could be achieved with

the introduction of additional piecewise linear trial functions. While this will not be done

here, it is clear that computationally the process could quickly be automated.

Exercise: Download the Oasys GSA refresher tutorial from BlackBoard

and remind yourself of how to put together simple truss and frame structures.

GSA is based on the stiffness method and you should look in the help documents for

the stiffness matrices used for bar and beam elements respectively. When we move on to

the implementation of the finite element method for analysing structures we will first of

all remind ourselves of the matrix stiffness method, which is covered in the second year

structural mechanics course.

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0 0.2 0.4 0.6 0.8 10

0.01

0.02

0.03

0.04

0.05

0.06

x

u

(solid

red),

u

Figure 1.5: u, the exact solution (solid red), u approximate solution (dash-dot blue) usingpiecewise linear trial functions

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1.4 Galerkin Finite Element Formulation

Having introduced the concepts of the Galerkin method, and piecewise continuous trial

functions we can go on to introduce a more formal system of finite elements. Let us

consider an individual finite element as shown in Figure 1.6.

Figure 1.6: Individual finite element with nodes at xi and xi+1

The element shown has two nodes at xi and xi+1 with nodal variables ui and ui+1

at those nodes. We will assume that our unknown trial function is of the form:

u = c1x + c2 (1.30)

This trial function can be expressed in terms of the nodal variables (ui and ui+1),

replacing c1 and c2 with the nodal variables:

u(xi) = c1xi + c2 = ui

u(xi+1) = c1xi+1 + c2 = ui+1(1.31)

Thus solving simultaneously we obtain:

c1 =ui+1 uixi+1 xi

c2 =

uixi+1

ui+1xixi+1 xi

(1.32)

Substituting expressions for c1 and c2 back into our unknown trial function we have:

u = N1(x)ui + N2(x)ui+1 (1.33)

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where:

N1(x) =xi+1 x

hi, N2(x) =

x xihi

, hi = xi+1 xi (1.34)

and N1 and N2 are referred to as linear shape functions.

The shape functions are such that at node ni, the shape function Ni will have a value

of 1, while at all other nodes it will have a value of zero. Thus the linear shape functions,

N1 and N2 for a single element can be graphically represented as shown in Figure 1.7.

Figure 1.7: Linear shape functions for a single finite element (e1), with two nodes (n1and n2)

It is noted that in general:

ni=1

Ni(x) = 1 (1.35)

We can go on to look at a system of four finite elements as shown in Figure 1.8 which

we will use to find an approximation to the actual solution using the weak form of the

governing differential we looked at previously:

IwR =

ni=1

xi+1xi

(wu wu + wx)dx + [uw]10 = 0 (1.36)

For an individual element ei we maintain the relationship:

u = N1(x)ui + N2(x)ui+1 (1.37)

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Figure 1.8: Four finite elements (e1 to e4), with five nodes (n1 to n5) with linear shapefunctions

where from the Galerkin method the weighting functions w1 and w2 are based on the

shape functions N1 and N2:

w1 = N1(x)

w2 = N2(x)(1.38)

and the integrated weighted residual for an individual element ei is:

IwR(ei) =xi+1xi

N1N2

[N1N

2]

N1

N2

[N1N2]

dx

uiui+1

+ xi+1xi

N1

N2 x dx(1.39)

Hence if we have a domain between 0 and 1, with nodes at x = 0, x = 0.25, x =

0.5, x = 0.75 and x = 1, then we can derive the integrated weighted residual for each

element in the form:

IwR(ei) =

1hi

+ hi3

1hi

+ hi6

1hi

+ hi6

1hi

+ hi3

ui

ui+1

+

hi6

(xi+1 + 2xi)hi6

(2xi+1 + xi)

(1.40)

where hi = xi+1 xi

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We can then write the integrated weighted residual for each element in matrix form:

IwR(e1) = 4.08333 3.95833

3.95833 4.08333

u1

u2 + 0.0104167

0.0208333 (1.41)

IwR(e2) =

4.08333 3.958333.95833 4.08333

u2u3

+

0.0416667

0.0520833

(1.42)

IwR(e3) =

4.08333 3.958333.95833 4.08333

u3

u4

+

0.0729167

0.0833333

(1.43)

IwR(e4) = 4.08333 3.958333.95833 4.08333 u4u5 + 0.10416670.1145833 (1.44)

which can be combined to give:

IwR =

4.08333 3.95833 0 0 03.95833 8.16667 3.95833 0 0

0 3.95833 8.16667 3.95833 00 0 3.95833 8.16667 3.958330 0 0 3.95833 4.08333

u1

u2u3u4u5

+

0.0104167 u(0)0.0625

0.125

0.1875

0.1145833 + u(1)

= 0

(1.45)

where u(0) and u(1) are referred to as the Neumann boundary conditions. In practice we have

what are referred to as the Dirichlet boundary conditions:

u(0) = u1 = 0, u(1) = u5 = 0

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Thus we can alter the integrated weighted residual relationship with the known boundary

conditions:

1 0 0 0 0

3.95833 8.16667 3.95833 0 00 3.95833 8.16667 3.95833 00 0 3.95833 8.16667 3.958330 0 0 0 1

u1u2u3u4u5

=

0

0.06250.1250.1875

0

(1.46)

giving:

u1 = 0, u2 = 0.0352, u3 = 0.0569, u4 = 0.0505, u5 = 0

We can substitute back into our unknown trial function for each element to give:

u

N1(x)u1 + N2(x)u2 = 0.1408x for 0 x 0.25N1(x)u2 + N2(x)u3 = 0.0352 + 0.0868(x 0.25) for 0.25 x 0.5

N1(x)u3 + N2(x)u4 = 0.0569 0.0256(x 0.50) for 0.5 x 0.75N1(x)u4 + N2(x)u5 = 0.0505 0.2020(x 0.75) for 0.75 x 1

(1.47)

Hence we can plot the solution provided by the Galerkin finite element formulation against

our exact solution, as shown in Figure 1.9. It is observed that although a reasonable fit is

achieved with four elements, a better fit could be achieved with the introduction of additional

elements. While this will not be done here, it is clear that computationally the process could

quickly be automated. It should be noted in particular that although the residual (u u) iszero at the nodes, this is not the case in the length of the element.

So we have demonstrated the application of the Galerkin finite element formulation to a

particular governing differential relationship. It can be seen that the method is easily applied

to a variety of physical phenomena. We shall look at the finite element method as applied in

structural mechanics.

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0 0.2 0.4 0.6 0.8 10

0.01

0.02

0.03

0.04

0.05

0.06

x

u

(solid

red),

u

Figure 1.9: u, the exact solution (solid red), u approximate solution (dash-dot blue) usingthe Galerkin finite element formulation

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2 Structural Finite Elements

2.1 Axial bar element

Let us first consider a bar element, that is an element which only resists axial forces. At this

stage we will also introduce a local coordinate system onto the bar such that at the first node,

n1 we have = 1, and at the second node, n2 we have = +1, as shown in Figure 2.1.

Figure 2.1: 1D bar element and associated linear shape functions

If a force per unit length is applied to the bar as shown in Figure 2.2 we can go on to look

at the strain energy and external work associated with axial deformation of the element.

Figure 2.2: 1D bar element with an axial load per unit length, p applied

The total strain energy can be expressed as:

U =1

2 +a

axx A dx (2.1)

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where A is the cross-sectional area of the bar (assumed to be constant). Taking the elastic

modulus of the material to be E we have x = Ex and:

U =

1

2 +a

a EA

2

x dx (2.2)

with x =dudx we have:

U =1

2

+aa

EA

du

dx

2dx (2.3)

The external work associated with the axial deformation of the element is:

W =

+aa

p u dx (2.4)

In the relative coordinate system the axial displacement function can be expressed as:

u = N1()u1 + N2()u2 (2.5)

which can be expressed in matrix form:

u = N ue = [N1 N2]

u1

u2

(2.6)

The strain energy in terms of the relative coordinate system can be expressed:

U =1

2 +1

1EA

1

a2du

d2

a d (2.7)

dud is found from

= J

x where J is referred to as the Jacobian operator, or derivative

transformation matrix between the global and the local coordinate systems.

x= J1

(2.8)

In the case of a bar element J = x = a, and J1 = 1a In addition it is noted that x = a.

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Given dud =dN1d

dN2d

u1u2

= Nue the strain energy can be expressed:

U =1

2uTe

+1

1

EA

aNTN d ue (2.9)

giving:

U =1

2uTe Keue (2.10)

where Ke is the element stiffness matrix:

Ke = EAa+11

NTN d (2.11)

For the bar element dN1d = 12 and dN2d = +12 and the element stiffness matrix is:

Ke =EA

a

+11

12+12

[12 + 12 ]d =

EA

2a

+1 11 +1

=

EA

L

+1 11 +1

(2.12)

The external work in terms of the relative coordinate system can be expressed:

We =

+11

pua d = uTe a

+11

pNT d (2.13)

giving:

We = uTe fe (2.14)

where fe is the matrix of nodal forces equivalent to the applied distributed load:

fe = a

+11

pNT d =pL

2

1

1

(2.15)

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For a system of elements, as illustrated in Figure 2.3 we can express the strain energy and

work done by the external forces as:

U =

1

2 uT

n

i=1

Keu =

1

2 uT

Ku (2.16)

W = uTni=1

fe = uTf (2.17)

where K and f are the system stiffness and force matrices.

Hence we have a global system expressed in matrix form:

Ku = f (2.18)

Figure 2.3: System of 1D bar elements

2.1.1 Numerical integration

Integrals arising from the computational implementation of the finite element method may

be solved using numerical integration, for example Gauss quadrature. Considering a function

f(x), x [1, +1] the integral I = +11 f(x) dx may be represented by the weighted sum of the

values of f(x) at the Gauss points. The position and weightings of the Gauss points relate to

the chosen number, as described in Table 1.

Table 1: Gauss point positions and weightings

n xi Wi1 0 2.0

2 1/

3 1.03 0 8/9

3/5 5/9

The technique is exact for 2n1 polynomials where n is the chosen number of Gauss points.

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Exercise:

The MATLAB code over the page, uses the finite element method to analyse

the bar structure shown in Figure 2.4.

Make notes next to the Matlab code and draw a flowchart on the blank page following,

using pseudo-code to indicate your understanding of the Matlab code. Experiment with making

changes to the code. In particular develop your understanding of how the stiffness and force

matrices are constructed, and how the displacement matrix is derived.

The Matlab code can also be downloaded from BlackBoard.

Figure 2.4: Example Bar Structure

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1 % Axial bar problem using the Finite Element Method by Andrew Phillips

2 % adapted from Ferreira, MATLAB codes for Finite Element Analysis

3

4 clear all

5 close all

6

7 % E, elastic modulus (N/mm2)

8 % A, area of crosssection (mm2)9 % L, length of bar (mm)

10

11 E=2.05*105;

12 A=pi*102;

13 EA=E*A;

14 L=900;

15

16 % information on the structure will be kept in a structure array

17 s1 = struct();

18

19 % generating node coordinates and element connectivities

20

21 % setting the number of Elements

22 s1.numElements = 3;

23 % generating equally spaced nodes

24 s1.nodeCoords = 0:L/s1.numElements:L;

25 % setting coords in the x direction

26 s1.nodeXCoord = s1.nodeCoords;

27 % setting the number of Nodes

28 s1.numNodes = size(s1.nodeCoords,2);

29

30 % assigning Nodes to Elements

31 ii = 1:s1.numElements;

32 s1.elementNodes(:,1) = ii;

33 s1.elementNodes(:,2) = ii+1;

34

35 % degrees of freedom (1 at each node for a bar element)

36 s1.globalDOF = s1.numNodes;

37

38 % setting up matrices (K, u, f)

39 s1.stiffness = zeros(s1.globalDOF); % square matrix (n x n)

40 s1.displacements = zeros(s1.globalDOF,1); % column matrix (n x 1);

41 s1.force = zeros(s1.globalDOF,1); % column matrix (n x 1);

42

43 % applied loading (N)

44 s1.force(2) = 1200;

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45

46 % deriving the system stiffness matrix

47

48 % element degrees of freedom

49 for ee=1:s1.numElements

50 elementDOF = s1.elementNodes(ee,:);

51 nn = size(elementDOF,2);

52 lengthElement = s1.nodeXCoord(elementDOF(2))s1.nodeXCoord(elementDOF(1));53 detJacobian = lengthElement/2;

54 invJacobian = 1/detJacobian;

55

56 % using a single central Gauss point in each element (xi=0, W=2)

57 xi=0; W=2;

58 shape=struct();

59 shape.shape=[1

xi; 1+xi]/2;

60 shape.naturalDerivatives=[1; 1]/2;61 Xderivatives=shape.naturalDerivatives *invJacobian;

62

63 % B matrix (derivates of the shape functions)

64 B=zeros(1,nn); % single row

65 B(1:nn) = Xderivatives;

66

67 % assembling the system stiffness matrix

68 s1.stiffness(elementDOF,elementDOF)=s1.stiffness(elementDOF,elementDOF) ...

69 +B'*

B*

W*

detJacobian*

EA;

70 end

71

72 % prescribed global degrees of freedom

73 s1.prescribedDOF = [1 4];

74

75 % solution

76 activeDOF=setdiff([1:s1.globalDOF]', [s1.prescribedDOF]);

77 U=s1.stiffness(activeDOF,activeDOF) \s1.force(activeDOF);78 s1.displacements(activeDOF)=U;

79

80 % output displacements and reactions

81 disp('Displacements')

82 [(1:s1.globalDOF)' s1.displacements]

83

84 F=s1.stiffness*s1.displacements;

85 reactions=F(s1.prescribedDOF);

86 disp('Reactions')

87 [s1.prescribedDOF' reactions]

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Blank page for pseudo-code flowchart for the bar structure example.

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2.1.2 Rotational transformation

In introducing the finite element formulation for an axial bar element we implicitly dealt with

transformation between a global (x) and a relative () coordinate system in one-dimension. If

we are to use bar elements in two-dimensional or three-dimensional problems we must also be

able to deal with transforming between the relative coordinate system, and a x, y or x,y ,z

coordinate system. We will limit ourselves here to transforming between the one-dimensional

relative () system and a two-dimensional global (x, y) system.

Referring to the global system of equations:

Ku = f (2.19)

We can recall the system of equations for an individual bar element:

Keue = fe EAL

+1 11 +1

u1

u2

=

f1

f2

(2.20)

It is noted that in the above equation u1 and u2 are one-dimensional deformation values at

nodes n1 and n2, and f1 and f2 are the associated applied forces. Going to a two-dimensional

coordinate system ue

and fe

will expand to 4

1 matrices and the element stiffness matrix Ke

will expand to a 4 4 matrix:

Keue = fe EAL

+1 0 1 00 0 0 0

1 0 +1 00 0 0 0

ux1

uy1

ux2

uy2

=

fx1

fy1

fx2

fy2

(2.21)

It is observed that Ke is similar to second order strain and stress tensors that we have

encountered before. As such it transforms in the same way as every second order tensor. That

is if we have an element at an angle in the x, y system as shown in Figure 2.5 we can transform

the element stiffness matrix in the relative system into the element stiffness matrix in the global

system, required in order to allow us for a structural system to combine all of the element

stiffness matrices to give us a system stiffness matrix.

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Figure 2.5: Bar element at an angle with respect to the x, y coordinate system

KGe = LTe K eLe (2.22)

K

Ge =

cos sin 0 0sin cos 0 0

0 0 cos

sin

0 0 sin cos

EA

Le

+1 0 1 00 0 0 0

1 0 +1 0

0 0 0 0

cos sin 0 0

sin cos 0 00 0 cos sin

0 0 sin cos

(2.23)

KGe = EALe

cos2 cos sin cos2 cos sin cos sin sin2 cos sin sin2 cos2 cos sin cos2 cos sin

cos sin sin2 cos sin sin2

(2.24)

where KGe is the element stiffness matrix in the global system, K e is the element stiffness matrixin the relative system, Le is the directional cosine matrix, and Le is the length.

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Truss Example

We can now adapt the Matlab code developed for our first bar element example

in order to analyse the truss system shown in Figure 2.6.

Figure 2.6: Example Truss Structure

You should check the results that you find using Oasys GSA. You should find a close matchbetween the stiffness method and the finite element method when using bar elements, as imple-

mented here.

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1 % Axial bar truss problem using the Finite Element Method by Andrew Phillips


3

4 clear all

5 close all

6


8 % A, area of crosssection (mm2)9 % L, length of bar (mm)

10

11 E=2.05*105;

12 A=pi*102;

13 EA=E*A;

14


16 s1 = struct();

17


19

20 % generating the node coordinates (6 nodes)

21 s1.nodeCoords = [0 0; 0 3000; 3000 0; 3000 3000; 6000 0; 6000 3000];

22 % setting coords in the x and y directions

23 s1.nodeXCoord = s1.nodeCoords(:,1);

24 s1.nodeYCoord = s1.nodeCoords(:,2);



27

28 % assigning Nodes to Elements (11 elements)

29 s1.elementNodes = [1 2; 1 3; 2 3; 2 4; 1 4; 3 4; 3 6; 4 5; 4 6; 3 5; 5 6];


31 s1.numElements = size(s1.elementNodes,1);

32

33 % degrees of freedom (2 at each node for a bar element in 2D space)

34 s1.globalDOF = s1.numNodes.*2;

35





40

41 % applied loading (N)

42 s1.force(4) = 50000; % node 2, DOF 243 s1.force(8) = 100000; % node 4, DOF 244 s1.force(12) = 50000; % node 6, DOF 2

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45


47



50 indice = s1.elementNodes(ee,:);

51 % assigns Global DOF to the element

52 elementDOF = [indice(1)*21 indice(1)*2 indice(2)*21 indice(2)*2];53 % length of each element in the x and y directions

54 xa = s1.nodeXCoord(indice(2))s1.nodeXCoord(indice(1));55 ya = s1.nodeYCoord(indice(2))s1.nodeYCoord(indice(1));56 lengthElement = sqrt(xa.2+ya.2);

57 thetaElement = atan(ya/xa);

58 % Ke in the relative system

59 Ke Relative = [1 0

1 0 ; 0 0 0 0 ;

1 0 1 0 ; 0 0 0 0 ] ;

60 % Setting up the transformation to the global system

61 C = cos(thetaElement);

62 S = sin(thetaElement);

63 L element = [C S 0 0; S C 0 0; 0 0 C S; 0 0 S C];64 % Ke in the global system

65 Ke Global = EA./lengthElement.*(L element'*Ke Relative*L element);


67 s1.stiffness(elementDOF,elementDOF) =...

68 s1.stiffness(elementDOF,elementDOF)+Ke Global;

69 end

70


72 s1.prescribedDOF = [1 2 10];

73

74 % solution



78

79 % output

80 fprintf('Solution\n\n')81


83 disp('Displacements:')

84 for n=1:s1.globalDOF

85 fprintf('Global DOF %02d, %+1.4e \n',n,s1.displacements(n));86 end

87 fprintf('\n');88

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91 disp('Reactions:')

92 for n=1:size(s1.prescribedDOF,2);

93 fprintf('Global DOF %02d, %+1.4e

\n',s1.prescribedDOF(n),reactions(n));

94 end

95 fprintf('\n');96

97 % output axial strains

98 for n=1:s1.numElements

99 indice = s1.elementNodes(n,:);

100 xa = s1.nodeXCoord(indice(2))s1.nodeXCoord(indice(1));101 ya = s1.nodeYCoord(indice(2))s1.nodeYCoord(indice(1));102 s1.originallengthElement(n) = sqrt(xa.2+ya.2);

103 xanew = (s1.nodeXCoord(indice(2)) + s1.displacements(indice(2)*2

1))...

104 (s1.nodeXCoord(indice(1))+ s1.displacements(indice(1)*21));105 yanew = (s1.nodeYCoord(indice(2)) + s1.displacements(indice(2)*2))...

106 (s1.nodeYCoord(indice(1))+ s1.displacements(indice(1)*2));107 s1.finallengthElement(n) = sqrt(xanew.2+yanew.2);

108 s1.strainsElement(n) = (s1.finallengthElement(n)s1.originallengthElement(n)) ...109 ./s1.originallengthElement(n);

110 end

111

112 disp('Strains:')


114 fprintf('Element %02d, %+1.4e \n',n,s1.strainsElement(n));115 end

116 fprintf('\n');117

118 % output axial stresses

119 s1.stressesElement = E.*s1.strainsElement;

120 disp('Stresses:')


122 fprintf('Element %02d, %+1.4e \n',n,s1.stressesElement(n));123 end

124 fprintf('\n');

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2.2 Euler-Bernoulli beam element

The Euler-Bernoulli beam element is based on standard or engineering beam bending theory.

The assumptions used in formulating standard beam bending theory are:

Plane sections remain plane after deformation.

Plane sections remain perpendicular to the neutral axis after deformation.

These assumptions are illustrated in Figure 2.7.

Figure 2.7: Euler-Bernoulli Beam Bending

Based on Figure 2.7 and the standard definition of strain we have the relationship:

11(x2) =du1dx1

=d

dx1

x2du2

dx1

= x2d

2u2dx21

= x2

Where is the curvature. The curvature of the beam is taken as:

=1

R=

d

ds=

d2u2dx21

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Where R is the radius of curvature, and ds is the distance along the curve as illustrated in

Figure 2.8, which for small deflections can be considered equal to dx1.

Figure 2.8: Beam curvature

With regards to a finite element formulation for a two noded element we can specify two

degrees of freedom at each node, as shown in Figure 2.9.

Figure 2.9: Degrees of freedom for a two noded Euler-Bernoulli beam element

We can define four shape functions as illustrated in Figure 2.10 and given:

N1() = +1

4(1 )2(2 + ) (2.25)

N2() = +1

8L(1 )2(1 + ) (2.26)

N3() = +1

4(1 + )2(2 ) (2.27)

N4() = 18

L(1 + )2(1 ) (2.28)

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1 0.5 0 0.5 10.4

0.2

0

0.2

0.4

0.6

0.8

1

N

N1

N2

N

3N

4

Figure 2.10: Hermite cubic shape functions for an Euler-Bernoulli beam element

The adopted hermite shape functions are said to have C1 continuity, whereas our original

linear shape functions had C0 continuity. That is to say that C1 shape functions are continuous

across nodes for the first derivative, while C0 shape functions are continuous for the value only.

It is noted that N2() and N4() do not reach values of 1 or zero at the relevant nodes. However

these shape functions relate to the angle of rotation, , or the slope at each of the nodes, = du2dx1 .

Thus the relationship is true for dN2d and dN4d as shown in Figure 2.11 and given:

dN2d

=1

4(1 2 + 32) (2.29)

dN4d

=1

4(1 + 2 + 32) (2.30)

The strain energy associated with bending of the element can be obtained as:

U =1

2

V

E211 dV (2.31)

Where dV = dAdx1, thus:

U =1

2

+aa

EI

2u2x21

2dx1 =

1

2

+11

EI

a4

2u2x21

2a d (2.32)

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1 0.5 0 0.5 10.4

0.2

0

0.2

0.4

0.6

0.8

1

dN

dN2

dN4

Figure 2.11: dN2 and dN4 for an Euler-Bernoulli beam element

From the chain rule it is found that d2u2d2

= 1a2

d2u2dx2

1

= 4L2

d2u2dx2

1

. In addition it is noted that

x = a = L2 .

Given d2u2d2

= d2Nd2

ue = Nue the strain energy can be expressed:

U =

1

2 uT

e

EI

a3 +11 NTN due (2.33)giving

U =1

2uTe Keue (2.34)

where Ke is the element stiffness matrix:

Ke =EA

a3

+11

NTN d (2.35)

and:

N =1

L

6

L3 1 6

L3 + 1

(2.36)

Using the Hermite shape functions the element stiffness matrix is found as:

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Ke =EI

L3

12 6L 12 6L6L 4L2 6L 2L212 6L 12 6L

6L 2L2

6L 4L2

(2.37)

It is observed that this is the same as the matrix derived using a virtual work approach,

therefore an exact match should be found between the stiffness method and the finite element

method when using Euler-Bernoulli beam elements as formulated here.

If we apply a uniform load to the element as shown in Figure 2.12 we can state the work

done by the external forces in deforming the element as:

W = +aa

pu dx = uTe a+11

pNT d (2.38)

Figure 2.12: Uniformly distributed loading for an Euler-Bernoulli beam element

The vector of nodal forces and moments is then found as:

fe = a

+11

pNT d = pL

1/2

L/12

1/2

L/12

(2.39)

It is observed that these have the same magnitude as the fixed end reactions and moments

for a uniformly loaded fixed end beam, which you will have encountered when using the stiffness

method.

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Euler Bernoulli Beam Example

The Matlab code below can be used to model the simply supported and fixed

end beams shown in Figure 2.13 using Euler-Bernoulli beam elements.

Adapt the code for the different boundary conditions, and perform a convergence study

(increasing the number of elements, and recording the predicted deflection at midspan) to find

the number of Euler-Bernoulli beam elements required to find the analytical midspan deflections.

You will recall that these are

max = 5pL4

384EI

for a simply supported beam, and

max = pL4

384EI

for a fixed end beam.

Note the effect that using the nodal force vector has on the reported reaction values. Plot

the nodal displacements for 2, 4, 8, 16, 32 and 64 elements for both the simply supported and

fixed end beams and comment on what you think is a sensible mesh density with regards to

accurately reproducing the displaced shapes of the beams.

Figure 2.13: Example Truss Structure

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1 % EulerBernoulli beam problem using the Finite Element Method2 % by Andrew Phillips

3

4 clear all

5

6 % comment to show successive plots on the same graph

7 close all

8


10 % I, second moment of area of crosssection (mm4)11 % L, length of beam (mm)

12

13 E=2.05*105;

14 I=8.249*107;

15 EI=E*I;

16 L=10000;

17


19 s1 = struct();

20


22


24 % asking the user for the number of elements

25 s1.numElements = input('How many elements do you wish to use? ');







32





37

38 % degrees of freedom (2 at each node for an EB beam element)


40





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45

46 % applied loading

47 % load per unit length

48 % (kN/m to N/mm)

49 p =

5;

50


52





57 elementDOF = [indice(1)*21 indice(1)*2 indice(2)*21 indice(2)*2];58 % length of the element

59 lengthElement = s1.nodeXCoord(indice(2))

s1.nodeXCoord(indice(1));

60 Le = lengthElement;

61 % Ke

62 Ke = EI./(Le.3)*[...

63 12 6.*Le 12 6.*Le;...64 6.*Le 4.*Le.2 6.*Le 2.*Le2;...65 12 6.*Le 12 6.*Le;...66 6.*Le 2.*Le.2 6.*Le 4.*Le.2];67 % fe

68 fe = p.*[Le./2; Le.2./12; Le./2; Le.2./12];69 % assembling the system stiffness matrix

70 s1.stiffness(elementDOF,elementDOF) =...

71 s1.stiffness(elementDOF,elementDOF)+Ke;

72 % assembling the nodal force vector

73 s1.force(elementDOF)=s1.force(elementDOF)+fe;

74 end

75


77

78 % simply supported

79 s1.prescribedDOF = [1 s1.globalDOF

1];

80

81 % fixed end

82 % s1.prescribedDOF = [1 2 s1.globalDOF1 s1.globalDOF];83

84 % solution



88

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89 % output

90 fprintf('Solution\n\n')91





97 fprintf('\n');98





103 fprintf('Global DOF %02d, %+1.4e

\n',s1.prescribedDOF(n),reactions(n));

104 end

105 fprintf('\n');106

107 % plot the vertical displacements

108 fig1=figure(1);

109 plot(s1.nodeXCoord,s1.displacements(1:2:s1.globalDOF), 'x r');110 grid on

111 title('EulerBernoulli Beam Example');112 xlabel('Horizontal Distance');

113 ylabel('Vertial Deflection');

114 hold on

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2.3 Timoshenko beam element

Alternative beam bending theories to the standard beam bending theory exist, and the formu-

lation of the Timonshenko beam element is an example of the application of one of these.

Timoshenko beam bending theory can be considered as an extension of standard beam

bending theory, in which deflections due to shear strains are also taken into account. The

assumptions used in formulating Timoshenko beam theory are:

Plane sections remain in plane after deformation.

Plane sections need not remain perpendicular to the neutral axis.

The first of these assumptions in particular will be looked at with regards to compatibility.

The additional contribution to deformation due to considering shear strains is illustrated in

Figure 2.14. The contribution of shear strains to deformation of a beam is seen to increase as

the depth:span ratio increases.

Figure 2.14: Timoshenko Beam Bending

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It is seen in Figure 2.14 that the assumption of plane sections remaining in plane after

deformation doesnt fit with what we know about the shear distribution within a section. This

can to some extent be taken account of through the introduction of a shear correction coefficient,

or shape factor, in our shear deformation relationship:

12 =V

GA(2.40)

Where V is the shear force acting across the section in the x2 direction, G is the shear

modulus, normally taken as E/2(1 + ) where is Poissons ratio, and A is the area of the

cross-section. Various values of have been put forward for cross-sections of different shapes,

although 5/6 is often used.

The finite element formulation for a Timoshenko beam element is different to the formulationfor an Euler-Bernoulli beam element. Expressions for the curvature, , caused by bending

moments, and the angle, , caused by shear forces can be written in matrix form as:

=

0 ddx1

ddx1 1

u2

(2.41)

Expressions for the bending moment, M, and the shear force, V can be written in matrix

form as:

M

V

=

EI 0

0 GA

(2.42)

We can define a Timoshenko beam element using the same degrees of freedom at each node

as we had for the Euler-Bernoulli beam element, as shown in Figure 2.9. For the formulation

presented here we will use linear shape functions.

u2

=

12(1 ) 0 12(1 + ) 0

0 12(1 ) 0 12(1 + )

u2(n1)

(n1)

u2(n2)

(n2)

(2.43)

We can define the B matrix as:

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B = dN =

0 ddx1

ddx1 1

12(1 ) 0 12(1 + ) 0

0 12(1 ) 0 12(1 + )

(2.44)

The element stiffness matrix can then be defined as:

Ke =

V

BTDB dV (2.45)

where D =

EI 0

0 GA

While the equivalent nodal force vector can be defined as:

fe =

V

NTp dV (2.46)

where p is a uniformly distributed load.

It is observed that the expressions for Ke and fe are general relationships, such that we

could have used them in the development of the axial bar element, and the Euler-Bernoulli

beam element formulations. They are easy to implement using short blocks of code within a

numerical program such as Matlab, and also form the basis of other structural and continuum

finite element formulations that you will investigate in future years.

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Timoshenko Beam Example

The Matlab code below can be used to model the simply supported and fixed

end beams shown in Figure 2.13 (for the Euler-Bernoulli beam element example)

using Timoshenko beam elements.

Instead of using the previous section you should calculate I and A values for a cross-section

300mm in depth and 30mm in width. You may take as 5/6.

As you did for the Euler-Bernoulli beam example adapt the code for the different boundary

conditions, and perform a convergence study (increasing the number of elements, and recording

the predicted deflection at midspan) to find the minimum number of Timoshenko beam elements

required to find consistent values for the midspan deflection. You will see that the code makes

use of numerical integration, with two gauss points per element used for the integration related

to curvature, and a single gauss point used per element for the integration related to shear.

Repeat the convergence study for beam depths of 100mm and 900mm.

The analytical values for midspan deflection using the Timoshenko formulation are:

max = 5pL4

384EI pL

2

8GA(2.47)

for a simply supported beam, and

max = pL4

384EI pL

2

8GA(2.48)

for a fixed end beam.

Repeat the exercise using Euler-Bernoulli beam elements for the new cross-sections and find

the percentage increase in the midspan deflections found using Timoshenko beam elements.

Discuss when you think the use of Timoshenko, as opposed to Euler-Bernoulli beam elements

could be important.

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1 % Axial bar problem using the Finite Element Method by Andrew Phillips


3

4 clear all

5 close all

6


8 % A, area of crosssection (mm2)9 % I, second moment of area (mm4)

10 % L, length of bar (mm)

11

12 E=2.05*105; %elastic (Young's) modulus

13 G=E/(2*(1+0.3)); %shear modulus

14 k=5/6; % shear correction factor%

15 b=30;

16 d=300;

17 A=b*d;

18 I=b*d3/12;

19 EI=E*I; %curvature%

20 kGA=k*G*A; %shear%

21 L=10000;

22

23 % D matrix

24 D=[EI 0; 0 kGA];

25


27 s1 = struct();

28


30


32 % asking the user for the number of elements

33 s1.numElements = input('How many elements do you wish to use? ');







40





44


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45

46 % degrees of freedom (2 at each node for a T beam element)


48





53

54 % applied loading

55 % load per unit length

56 % (kN/m to N/mm)

57 p = 5;58


60





65 elementDOF = [indice(1)*21 indice(1)*2 indice(2)*21 indice(2)*2];66 % length of the element

67 lengthElement = s1.nodeXCoord(indice(2))s1.nodeXCoord(indice(1));68 Le = lengthElement;

69 detJacobian = Le/2;

70 invJacobian = 1/detJacobian;

71

72 % two point gauss integration

73 xi=[+1/sqrt(3) 1/sqrt(3)]; W=[1 1];74

75 for n=1:size(W,2)

76

77 shape=struct();

78 shape.shape=[1xi(n); 1+xi(n)]/2;79 shape.naturalDerivatives=[

1; 1]/2;

80 Xderivatives=shape.naturalDerivatives *invJacobian;

81

82 %%%% theta contribution

83 % B matrix (derivates of the shape functions)

84 B=zeros(2,4);

85 B(1,2) = Xderivatives(1);

86 B(1,4) = Xderivatives(2);



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89 +B'*B.*W(n).*detJacobian.*D(1,1);

90 %%%%

91

92 % assembling the nodal force vector

93 s1.force(elementDOF)=s1.force(elementDOF)+ ...

94 p*[shape.shape(1) 0 shape.shape(2) 0]'.*W(n).*detJacobian;

95

96 end

97

98 % avoiding a problem described as 'shear locking'

99 % one point gauss integration when looking at shear

100 xi=[0]; W=[2];

101 for n=1:size(W,2)

102

103 shape=struct();

104 shape.shape=[1xi(n); 1+xi(n)]/2;105 shape.naturalDerivatives=[1; 1]/2;106 Xderivatives=shape.naturalDerivatives *invJacobian;

107

108 %%%% gamma contribution

109 % B matrix (derivatives of shape functions and shape functions)

110 B=zeros(2,4);

111 B(2,1) = Xderivatives(1);112 B(2,3) = Xderivatives(2);113 B(2,2) = shape.shape(1);

114 B(2,4) = shape.shape(2);



117 +B'*B.*W(n).*detJacobian.*D(2,2);

118 %%%%

119

120 end

121 end

122


124

125 % simply supported

126 s1.prescribedDOF = [1 s1.globalDOF1];127

128 % fixed end

129 % s1.prescribedDOF = [1 2 s1.globalDOF1 s1.globalDOF];130

131 % solution


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135

136 % output

137 fprintf('Solution

\n

\n')

138





144 fprintf('\n');145

146 % midspan displacements

147 disp('Midspan Displacement:')

148 fprintf('%+1.4e \n\n',s1.displacements(s1.globalDOF/2));149





154 fprintf('Global DOF %02d, %+1.4e \n',s1.prescribedDOF(n),reactions(n));155 end

156 fprintf('\n');157

158 % plot the vertical displacements

159 fig1=figure(1);

160 plot(s1.nodeXCoord,s1.displacements(1:2:s1.globalDOF), 'x r');161 grid on

162 title('Timoshenko Beam Example');

163 xlabel('Horizontal Distance');

164 ylabel('Vertial Deflection');

165 hold on

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Documents

CI221 Finite Element Notes