FEM Analysis Of Plane Beam Structure

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Lecture 16: FEM ANALYSIS OF PLANE BEAM STRUCTURE 162

TABLE OF CONTENTS

Page

16.1. A Plane Beam Structure 16316.2. Finite Element Solution 163

16.3. Analytical Solution By Discontinuity Functions 168

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163 16.2 FINITE ELEMENT SOLUTION

This is a supplementary Lecture on the Finite Element Method (FEM), designed to illustrate some

additional concepts: Rotational DOFs, Distributed Load Lumping, and FEM Convergence.

These fall beyond the basic exposition of the DSM in Lectures 1315.

Note: material not used in HWs or Midterm Exams, but it will be of use in the Airplane Shaker

Lab to be formalized on Fall 2011.

16.1. A Plane Beam Structure

In this Lecture we will analyze the plane beam shown in Figure 16.1, with a two-element FEM

discretization. The FEM displacement results will be compared with the exact analytical solution

obtained by discontinuity functions. Then the number of elements will be doubled to 4, 8, etc.,

usingMathematica. As the number of elements is increased, the FEM solution will be visually

shown to rapidly converge to the analytical solution.

The beam span is 2L . It has uniform cross section of elastic modulus Eand moment of inertia I

aboutz. The beam isfixed (clamped) atAand simply supported atB . It is loaded by a downward

uniform distributed force of magnitude w0 acting over the right half span L x 2L . Theproblem is statically indeterminate.

AB

y

;

;

;

; ;

x

2L

L L

0

0

wEI constantover entire beam

Numerical properties for examples:

L= 36, w = 16, EI = 1000000.C

Figure 16.1. Plane beam problem.

16.2. Finite Element Solution

To illustrate the use offinite elements for beam structures we will discretize the problem of Figure

16.1 using two plane-beamfinite elements as illustrated in Figure 16.2.

The equations of a generic plane beam element are worked out in detail in Chapter 12 of theIntroduction to FiniteElements course. For this course the development in that Chapter is overkill;

furthermore it requires graduate level mathematics. Only selected results from are reproduced here

for convenience.

This is posted as a PDFfile in

http://caswww.colorado.edu/courses.d/IFEM.d/IFEM.Ch12.d/IFEM.Ch12.index.html

click on Chapter 12 link (first one).

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y

x

0w

0-w L /20

-w L /2

1 2 3

3

(1)

(2)

1 2

(1) (2)

A

y

;

;

;

;

1 23(1) (2)

vv v1

1

2 3

2 3

L = L

(1)

L = L

(2)

transformed to node forces(c)

(b)

(a)E I same for both elements

Figure 16.2. Analysis of the problem of Figure 16.1 byfinite elements:(a) a two-element

FEM discretization, (b) the six degrees of freedom of the free-free

FEM model, (c) support conditions and applied forces.

Beam finite elements are obtained by subdividing beam members longitudinally. The simplest

plane beam element, depicted in Figure 3, has two end nodes,i and j , four node displacements

grouped in vectoru(e) and four node forces, grouped in vectorf(e):

u(e) =

v(e)i

(e)i

v(e)j

(e)j

, f(e) =

f(e)i

m(e)i

f(e)j

m(e)j

(16.1)

The stiffness relations of a generic prismatic plane beam element of length =L (e), elastic modulus

Eand moment of inertiaIare

E I

3

12 6 12 66 42 6 2212 6 12 6

6 22 6 42

v(e)

i

(e)i

vj

(e)j

=

f(e)i

m(e)if

(e)j

m(e)j

(16.2)

To form the master stiffness equations using the augment and add technique, write the previous

stiffness equations for elements (1) and (2) including all six degrees of freedom shown in Figure

16.2(b).

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i

i

j

(e)= L

x

E =E(e),I =I(e)

x, uj

y,v

vi

vj

Figure 16.3. The two-node plane beam element with four degrees of freedom.

Element (1), nodes 1-2, =L :

E I

L3

12 6L 12 6L 0 0

6L 4L2 6L 2L2 0 012 6L 12 6L 0 0

6L 2L2 6L 4L2 0 0

0 0 0 0 0 0

0 0 0 0 0 0

v(1)1

(1)

1

v(1)2

(1)2

v(1)3

(1)3

=

f(1)1

m(1)

1

f(1)2

m(1)2

f(1)3

m(1)3

(16.3)

Element (2), nodes 2-3, =L :

E I

L3

0 0 0 0 0 0

0 0 0 0 0 0

0 0 12 6L 12 6L

0 0 6L 4L2

6L 2L2

0 0 12 6L 12 6L

0 0 6L 2L2 6L 4L2

v(2)1

(2)1

v(2)2

(2)

2

v(2)3

(2)3

=

f(2)1

m(2)1

f(2)2

m(2)

2

f(2)3

m(2)3

(16.4)

Assemble by adding the matrices and suppressing element superscripts:

E I

L3

12 6L 12 6L 0 0

6L 4L2 6L 2L2 0 0

12 6L 24 0 12 6L

6L 2L2 0 8L2 6L 2L2

0 0 12 6L 12 6L

0 0 6L 2L2 6L 4L2

v11v22v33

=

f1m1f2m2f3m3

(16.5)

The applied uniform distributed load over element (2) is converted to the two node forces 12

w0L

on nodes 2 and 3, as shown in Figure 16.2(c).

There is a more refined way of converting distributed forces to node forces calledconsistent node forcecomputation,

which is discussed in the IFEM Chapter referenced in the previous footnote. Consistent forces give generally better

results. The simpleforce lumpingused here is easier to interpret physically and does not require knowledge of energy

methods.

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ClearAll[EI,L,w0];

L=36; w0=16; EI=1000000;

K=(EI/L^3){{12,6*L,-12,6*L,0,0},

{6*L,4*L^2,-6*L,2*L^2,0,0},

{-12,-6*L,24,0,-12,6*L},

{0,0,0,8*L^2,-6*L,2*L^2},

{0,0,-12,-6*L,12,-6*L},

{0,0,6*L,2*L^2,-6*L,4*L^2}};

Kred=(EI/L^3)*{{24,0,6*L},{0,8*L^2,2*L^2},{6*L,2*L^2,4*L^2}};

fred={-(1/2)*w0*L,0,0};

ured=LinearSolve[Kred,fred];

Print["Unknown displacements ured=",N[ured]];

u={0,0,ured[[1]],ured[[2]],0,ured[[3]]};

Print["Complete displacement solution u=",N[u]];

Print["Deflection vC at midspan= ", N[u[[3]]] ];

f=K.u; Print["Node forces including reactions=",N[f]];

Print["Node forces including reactions=",N[f]];

Unknown displacements ured={-0.979776, -0.011664, 0.046656}

Complete displacement solution u={0, 0, -0.979776, -0.011664, 0, 0.046656}

Deflection vC at midspan= -0.979776

Node forces including reactions={198., 3888., -288., 0, 90., 0}

Figure 16.4. Upper box:Mathematicascript for FEM analysis

of two-element beam model. Lower Box: results.

Apply the known forces and displacements:

E I

L3

12 6L 12 6L 0 0

6L 4L2 6L 2L2 0 0

12 6L 24 0 12 6L

6L 2L2 0 8L2 6L 2L2

0 0 12 6L 12 6L

0 0 6L 2L2 6L 4L2

0

0

v220

3

=

f1m1

12

w0L

0

f30

(16.6)

The applied force f3 = 12

w0Lbecomes part of the reaction taken by the right-end support, since

v3 = 0. (In FEM, displacement BCs take precedence over force BCs.)

Reduce by removing rows and columns 1, 2 and 5, which pertain to the known node displacements:

E I

L3

24 0 6L

0 8L2 2L2

6L 2L2 4L2

v223

=

1

2w0L

0

0

(16.7)

The master and reduced equations were implemented inMathematicausing the script shown on

the top box of Figure 16.4. Executing this script underMathematicaproduces the output shown in

the lower box.

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(* Define singularity functions ^n for n=0,1,2,3,4 *)

SF0[x_,a_]:=If[x>a,1,0,0];

SF1[x_,a_]:=If[x>a,x-a,0,0];

SF2[x_,a_]:=If[x>a,(x-a)^2,0,0];

SF3[x_,a_]:=If[x>a,(x-a)^3,0,0];

SF4[x_,a_]:=If[x>a,(x-a)^4,0,0];

(* Express load q(x) in terms of SF, and integrate four times *)

ClearAll[L,EI,w0,x,C1,C2,C3,C4];

L=36; w0=16; EI=1000000;

q[x_]:= -w0*SF0[x,L];

Vx[x_]:= -w0*SF1[x,L] + C1;

Mx[x_]:= -w0*SF2[x,L]/2 + C1*x+C2;

EIthetax[x_]:= -w0*SF3[x,L]/6 + C1*x^2/2+C2*x+C3;

EIvx[x_]:= -w0*SF4[x,L]/24+ C1*x^3/6+C2*x^2/2+C3*x+C4;

(* Set the boundary conditions and solve for integration constants *)

BC= {Mx[2*L]==0,EIthetax[0]==0,EIvx[0]==0,EIvx[2*L]==0};

Print["Boundary conditions are: ",BC]

solBC=Solve[BC,{C1,C2,C3,C4}]; sol=Simplify[solBC[[1]] ];Print["Constants of integration: ",sol];

(* Replace C1,C2,C3,C4 into the indefinite integrals & print results *)

V[x_]:= Simplify[ Vx[x]/.sol ];

M[x_]:= Simplify[ Mx[x]/.sol ];

theta[x_]:= Simplify[ (EIthetax[x]/.sol)/EI ];

v[x_]:= Simplify[ (EIvx[x]/.sol)/EI ];

Print["Moment MC at midspan= ",M[L]," = ", N[M[L]] ];

Print["Deflection vC at midspan= ",v[L]," = ", N[v[L]] ];

(* Plot q,V,M,theta and v over beam *)

Plot[q[x],{x,0,2*L},AxesLabel->{"x","q"},PlotLabel->"Load"];

Plot[V[x],{x,0,2*L},AxesLabel->{"x","V"},PlotLabel->"Shear force"];Plot[M[x],{x,0,2*L},AxesLabel->{"x","M"},PlotLabel->"Bending moment"];

Plot[theta[x],{x,0,2*L},AxesLabel->{"x","theta"},PlotLabel->"Rotation"];

Plot[v[x],{x,0,2*L},AxesLabel->{"x","v"},PlotLabel->"Deflection"];

Boundary conditions are: {-10368 + 72 C1 + C2 == 0, C3 == 0, C4 == 0,

-1119744 + 62208 C1 + 2592 C2 + 72 C3 + C4 == 0}

Constants of integration: {C1 -> 207, C2 -> -4536, C3 -> 0, C4 -> 0}

Moment MC at midspan= 2916 = 2916.

41553

Deflection vC at midspan= -(-----) = -1.3297 31250

Figure 16.4. Upper box:Mathematicascript for exact analysis of beam structure using

Discontinuity Functions and 4th order method. Lower box: results.

The beam deflections given by the FEM model are compared in Figure 16.6 against the exact

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0 10 20 30 40 50 60 70

1.4

1.2

1

0.8

0.6

0.4

0.2

0

x

Deflectionv(x)

FEM, 2 elements

Exact

Figure 16.6. Deflectionv(x)of 2-element FEM model versus exact solution.

solution obtained by discontinuity functions below.

Note that the center deflection given by the FEM model is v2 = v(L) = 0.979776, whereas the

analytical solution gives v (L) = 1.3297. The difference is approximately 30%. FEM analysis

generally produces only approximations to the analytical solution of the mathematical model.

The approximation can be improved by using more elements over the beam span. This will be

demonstrated in class from a laptop runningMathematica.

16.3. Analytical Solution By Discontinuity Functions

Recall that the discontinuity function, also known as Singularity Functions or SFs, x an for

n 0 is defined as

x an = (x a)n if x >a

0 if x a

(16.8)

Here . . . are the so-called MacAuley brackets. The SF forn= 1, which isnotincluded in the

definition (16.8), is Diracs delta function atx = a, denoted as x a1 in Vables book. The SFs

forn =0 andn =1 receive the namesunit step function(or Heaviside function) andunit ramp

function, respectively, in electrical engineering.

We start by writing the applied loadq(x)in terms of the SF x L0, which is a unit step function:

q(x) = w0 x L0 (16.9)

Integrate four times inx :

V(x) = w0 x L1 + C1 (16.10)

M(x) = 12

w0 x L2 + C1x + C2 (16.11)

E I (x) =E Iv (x) = 16

w0 x L3 + 1

2C1x

2 + C2x + C3 (16.12)

E Iv (x) = 124

w0 x L4 + 1

6C1x

3 + 12C2x

2 + C3x + C4 (16.13)

One notable exception are pin-jointed truss models in 2D or 3D. The FEM solution is then exact and cannot be improved

by using more elements per member.

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169 16.3 ANALYTICAL SOLUTION BY DISCONTINUITY FUNCTIONS

Here (x)is the rotation of the beam cross section, which is same as the slopev (x) = dv(x)/dx

in the beam theory used in this course.

The four boundary conditions are

M(2L) = 0, (0) = v(0) = 0, v(0) = 0, v(2L) = 0 (16.14)

Applying these conditions to expressions (11), (12) and (13) gives four linear equations from which

the four constants of integrationC1throughC4can be obtained. Once this is done, substituting

these constants into (10) to (13) gives the expressions for the shear force V(x), bending moment

M(x), cross section rotation (x) = v(x)and cross section deflection v(x)as functions ofx .

These functions may be plotted over 0 x 2L .

To carryout the analysis for the specific values w0 = 16,L = 36,E I = 106 weuse theMathematica

program listed in the top box of Figure 16.5.

Executing this script underMathematica produces the print output shown in the lower box of Figure

16.5, and also the plots ofq(x),V(x),M(x), (x) = v(x), and v(x)over the beam span, shown

in Figure 16.7.

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10 20 30 40 50 60 70

x

Load

-15

-12.5

-10

-7.5

-5

-2.5

q

10 20 30 40 50 60 70

x

Shear force

-300

-200

-100

100

200

V

10 20 30 40 50 60 70x

Bending moment

-4000

-2000

2000

4000

M

10 20 30 40 50 60 70

x

Rotation

-0.04

-0.02

0.02

0.04

0.06

0.08

theta

10 20 30 40 50 60 70

x

Deflection

-1.4

-1.2

-1

-0.8

-0.6

-0.4

-0.2

v

Figure 16.6. Plots ofq(x),V(x),M(x), (x) = v(x), andv(x)

over the beam span, given by DF solution.

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FEM Analysis Of Plane Beam Structure

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