Principles of Computer-Aided Design and Manufacturing Second Edition 2004 ISBN 0-13-064631-8

Principles of Computer-Aided Design and Manufacturing

Second Edition 2004

ISBN 0-13-064631-8

Author: Prof. Farid. Amirouche

University of Illinois-Chicago

University of Illinois-Chicago

Chapter 7

Introduction to Finite Element Method

Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN 0-13-064631-8

Author: Prof. Farid. Amirouche, University of Illinois-Chicago

CHAPTER 7 7.1 Introduction

7.1 INTRODUCTION

Figure 7.1 A finite element mesh of hip prosthesis

FEM is a technique that discretizes a given physical or mathematical problem into smaller fundamental parts, called ‘elements’ . Then an analysis of the element is conducted using the required mathematics.



CHAPTER 7

Figure 7.2 A output of stress distribution after simulation

7.1 Introduction



CHAPTER 7

The basic steps in FEA (Finite Element Analysis) consists of 3 phases:

7.2 BASIC CONCEPTS IN THE FINITE-ELEMENT METHOD

a) Preprocessing Phase:

1) Create and discretize the solution domain into finite elements; that is ,subdivide the problem into nodes and elements.

2) Assume a shape function to represent the physical behavior of an element; that is, an approximate continuous function

is assumed to represent the solution of an element.

3) Develop equations for an element.

4) Assemble the elements to present the entire problem. Construct the global stiffness matrix.

5) Apply boundary conditions, initial conditions, and loading.

7.2 Basic Concepts in FEM



CHAPTER 7

b) Solution Phase

6) Solve a set of linear or nonlinear algebraic equations simultaneously to obtain nodal results,

such as displacement values at different nodes or temperature values at different nodes in a heat

transfer problem.

c) Post processor Phase

7) Obtain other important information. At this point, you may be interested in values of

principal stresses, heat fluxes, etc.




CHAPTER 7

Figure 7.3 A general discretization of a body into finite elements




CHAPTER 7

,eadd

eee fukf

,eee ukf

where ke (e being the element) is the local element stiffness matrix, fe is the external forces applied at each element, ue represents the nodal displacements for the element, and feadd represents the

additional forces. (7.1)

where e is the element stress matrix, and se is the constitutive relationship connecting ue and e .

e = seue

0eaddfAssuming Equation 7.1 reduces to

(7.2)

(7.3)

In terms of stresses and strains equation (7.2) becomes




CHAPTER 7

Example 7.1 A load of P =180 lbs act on the variable cross sectional bar in which one end is fixed and a compressive load act at the other end. Dimensions of the bar are shown in the figure 7.4. Calculate the deflection at various points along its length. Calculate the reaction force at the fixed end and compute stresses in each element.

1.18

1 i n

1.18

1 i n

0.39

37 in

Node 1

Node 3

Node 2

Element 3

Element 2

Element 1

Node 4

u3

u2

u4

u1

1.9685 in

1.7716 in

E1

E3

E2

L3

L1

L2

P

Figure 7.4(a) A variable cross-sectional bar subject to a compressive load




CHAPTER 7 7.2 Basic Concepts in FEM

Solution : From figure 7.4 we compute the cross-sectional areas , and assign values of the Young’s modulus of elasticity to each of the segment bars.

A1=2.3248 in2. E1=669400.12 lb/in2 .A2=0.7750 in2. E2=458 lb/in2 .A3=2.0922 in2. E3=669400.12 lb/in2 .P = 180 lb.The stiffness of each bar segment is derived from Hooke’sLaw where

lb/in. 253.1185909

lb/in. 575.901

lb/in. 986.1317714

3

333

2

222

1

111

L

EAk

L

EAk

L

EAk (7.4)

(7.5)

(7.6)



CHAPTER 7

Node 4

P

k3(u4-u3) k2(u3-u2)

Node 3

k1(u2-u1)

Node 2

R

Node 1

k3(u4-u3) k2(u3-u2) k1(u2-u1)

Figure 7.4(b) Free body diagram of the forces





0 )()( 343232 uuKuuK

0 PuuK )( 344

At node 1

At node 2

At node 3

At node 4

0 )( 1211 uuKR

0 )()( 232121 uuKuuK

(7.7)

(7.8)

(7.9)

(7.10)

P

R

u

u

u

u

kk

kkkk

kkkk

kk

0

0

00

0

0

00 1

4

3

2

1

44

3322

2211

11

The above equations can be written in matrix form as

(7.11)

Applying the I law of mechanics where F=0, at each node




Since u1 = 0 , this is equivalent to eliminating row 1 and column 4 ofthe stiffness matrix.

Pu

u

u

kk

kkkk

kkk

0

0

0

0

4

3

2

33

3322

221

The matrix equation (7.11) reduces to

(7.12)

Plugging in the values for all the Ks and substituting the value of P :

180

0

0

253.1185909253.11859090

823.1186810823.118681057.901

057.901556.1318616

4

3

2

u

u

u(7.13)

Column 1




From the Equation (7.13) we get:

lb. 180 R forcereaction thecompute which wefrom R - .uk-

:obtain we(7.7)in 0u Putting

in. 0.19940 u

in. 0.199788 u

in. 0.0001366 u

1121

1

4

3

2

2

3

343

(3)

2

2

232

(2)

2

1

121

(1)

lb/in 113.351 L

uuEσ

lb/in 232.2589 L

uuEσ

lb/in 77.4259 L

uuEσ

:iprelationshstrain -stress theusing obtained are Stresses



CHAPTER 7

Example 7.2 This is a similar example to 7.1, however element z is replaced with two identical supports which forms element 2 and 3.A load of P = 180 lbs acts on the variable cross sectional bar in which one end is fixed and a compressive load acts at the other end. Dimensions of the bar are indicated in the figure 7.5. Calculate the deflection at various points along its length. Calculate the reaction force at the fixed end and the corresponding stresses in each element.

1.7716 in

1.9685 in

0.5905 in

Element 3

1.18

1 i n

u1

u2

1.18

1 i n

0.39

37 in

u3 E3

u4

E1

Element 1

Node 1

E4

E3

P

Element 4

Element 2

Node 3

Node 2

Node 4

Figure 7.6 A variable cross-section bar with identical supports in the middle.





21 04.3 inA 2

1 /12.669400 inlbE 2

32 2739. inAA 232 /458 inlbEE

24 456.2 inA

24 /12.669400 inlbE

lbP 180

lb/in. 221.17248871

111

L

EAk

The cross sectional areas are found to be

The load P is:

The Young’s modulus of elasticity for each segment

The element stiffness is found as:

lb/in. 64.3182

2232

L

EAkk

lb/in. 65.13971584

444

L

EAk




0 )())(( 3442332 uuKuukK

0 )( 343 PuuK

At node 1

At node 2

At node 3

At node 4

0 )( 1211 uuKR

0 ))(()( 2332121 uukKuuK

(7.14)

(7.15)

(7.16)

(7.17)

P

R

u

u

u

u

kk

kkkk)k(k

)k(kkkkk

kk

0

0

00

0

0

00 1

4

3

2

1

44

443232

323211

11

The above equations can be written in matrix form as

(7.18)




Since u1 = 0, Equation ( 7.18 ) can be reduced to:

P

0

0

u

u

u

kk0

kkkk)k(k

0)k(kkkk

4

3

2

44

443232

32421

(7.19)

180

0

0

65.139715865.13971580

65.1397158936.13977952856.637

02856.637507.1725524

4

3

2

u

u

u

Substituting the values of Ks and P we obtain

The solution of which is found to be

inu 0001.02 inu 2826.03 inu 2827.04

nodes. ofnt displaceme the tocorrespond and , 432 uuu



CHAPTER 7

675.56

1

121

)1(

l

uuE

63.328

2

232

)2(

l

uuE

675.56

3

344

)4(

l

uuE

lb/in2

lb/in2

lb/in2


lb. 172.3 R forcereaction thecompute which wefrom R - .uk-

:obtain we(7.18)in 0u Putting

1121

1

Stresses for each element are obtained as:



CHAPTER 7

Example 7.3 A tapering round bar is fixed at one end and a tensile load P=180 lbs is applied at the other end. The maximum and minimum radius of the bar is 20 in and 10 in respectively. The bar’s modulus of elasticity E = 669400.12 lb/in2. Consider the bar as a set of 4 elements of equal length and uniformly increasing diameter. Find the global stiffness matrix and displacements at each node and reaction force.

20 in

10 in

10

in

Figure 7.6 A Tapered round bar subject to a tensile load P




CHAPTER 7

element 3

element 2

element 1

element 4

Figure 7.7 A four element representation of the tapered round bar




CHAPTER 7

(7.20)

(7.21)

(7.22)

(7.23)

(7.24)

(7.25)


21 5398.78 inA 2

2 626.139 inA 23 166.218 inA 2

4 159.314 inA

210298251 K 52.373863552 K 5.584161803 K 841193004 K

The geometrical data and stiffness properties for each element are:

0)( 121 PuuK0)()( 121232 uuKuuK

0)()( 232343 uuKuuK0)()( 343454 uuKuuK

0)( 454 uuKR

R

P

u

u

u

u

u

kk

kkkk

kkkk

kkkk

kk

0

0

0

000

00

00

00

000

5

4

3

2

1

44

4433

3322

2211

11

At node 1

At node 2

At node 3

At node 4

At node 5

Since F=0 at each node therefore,



CHAPTER 7

(7.26)

(7.27)

(7.28)

(7.29)


0

0

0

00

0

0

00

4

3

2

1

433

3322

2211

11 P

u

u

u

u

kkk

kkkk

kkkk

kk

Applying the boundary conditions, that is u5 = 0 we get

A solution of which is found to be

The reaction force is obtained by writing the equation form, (7.25) where

)( 454 uukR

Substituting the values of u5 and u4 and k4, we obtain

lb 180.01 R

inu 41 101860.0

inu 42 101004.0

inu 43 100522.0

inu 44 100214.0



CHAPTER 7

7.3 POTENTIAL ENERGY FORMULATION

7.3 Potential Energy Formulation

• Finite element method is based on the minimization of the total potential energy formulation.

• When close form solution is not possible approximation methods such as Finite Element are the most commonly used in solid mechanics.

• To illustrate how finite element is formulated, we will consider an elastic body such as the one shown in Figure 7.9 where the body is subjected to a loading force which causes it to deform.



CHAPTER 7 7.3 Potential Energy Formulation

dy

FORCEElement

Figure 7.9 Elastic body



CHAPTER 7

,L

AEF

kyF

From Hooke’s Law, we write

where F represents the compressive force A represents area of the elastic body E represents the young’s modulus of elasticity

be the displacement in the y-direction, and L is the length of the segment. From Hooke’s Law the force displacement relationship is

,2

1 21

00

11

kykydyFdydyy

where represents the energy

,dxdzFA

Fyy

(7.30)


(7.32)




dVd 2

1

dxdydzdV

We can express the energy equation in terms of the stress-strain as:

Where represents the element strain energy and the volume of element isd

we can then write the strain energy for an element introducing the superscript e

,2

dVde

(7.35)

The stress can be substituted for by

,)( Ee (7.36)

Using the above relation in Equation (7.32 ), we obtain

.2

2

dVEe

(7.37)




It follows that a stable system requires that the potential energy be minimum at equilibrium

,0)(

ii

e

ii

uFuu

(7.39)

We know that strain is defined making use of the relative displacement between adjacent elements

l

uu ii )( 1 (7.40)

The total energy, which consists of the strain energy due to deformation of the body and the

work performed by the external forces can be expressed as function of combined energy as

,11

ii

i

n

e

e uF

(7.38) Where denotes displacement along the force direction Fi . iu



CHAPTER 7

),( 1

iiavg

i

e

uul

EA

u

),( 11

iiavg

i

e

uul

EA

u

,11

11

1

1

)(

)(

i

ieq

i

ei

e

u

uk

u

u

l

EAk avg

eq

)(where

(7.41)

(7.42)

(7.43)


Equation (7.39) has two components. Expression for the first term is

and

In matrix form



CHAPTER 7

iiii

FuFu

)(

1111

)(

iii

i

FuFu

)()()( eee ukF

1111

11

i

i

i

ieq F

F

u

ukor

and


Similarly , we take the second term in (7.39)

(7.44)

(7.45)

Combining Equations (7.43),(7.44), and (7.45) leads to elemental force-displacement relation

(7.46)

(7.47)

where k(e) is the element stiffness matrix, u(e) the element displacement associated with node i or i+1 ,

and F(e) denotes the external forces acting at the nodes.



CHAPTER 7

7.4 CLOSED FORM SOLUTION

7.4 Closed Loop Formulation

• The closed form solution is used when all the variableshave explicit mathematical forms that can be dealt with in terms of extracting a solution.

• Consider a continuous body subject to a compressive load P (Figure 7.10).The objective is to determine the displacement or deformation at any point.

•Let the continuous body have a variable cross-sectional area .



CHAPTER 7 7.4 Closed Loop Formulation

d y

L

r1

r2

P

cross-section

e

Figure 7.9 A non-uniform bar subjected to compressive load P.



CHAPTER 7

,0))(( yAP avg .Eavgwhere(7.48) (7.49)


The equilibrium equation at any cross sectional area cab be written as

Substituting the above equation into (7.48) we get

.0)( yAEP (7.50)

Recall that strain is defined as

.dy

du (7.51)

Substituting the above equation into (7.50) we get

,0)( dy

duyEAP (7.52)

.)(yEA

Pdydu (7.53)



CHAPTER 7

u l

yEA

Pdyduu

0 0 )(

112 ry

L

rryA

L

ryL

rrE

Pdyu

01

12

,0 1

L

ray

dy

E

Puor .12 const

L

rra

where

(7.54)

(7.55)


Integrating the above equation will lead to exact solution of displacement.

If we assume the load is constant and that

Equation (7.48) becomes

( 7.54 )

(7.56)

(7.57) (7.58)



CHAPTER 7 7.4 Closed Loop Formulation

adydzrayz 1

czEa

P

z

dz

Ea

Pu ln

Let Hence,

(7.59)

0)0( u1ln r

Ea

PC Let , then

11 lnln rrayEa

Pu (7.60)

1

1lnr

ray

Ea

Puor (7.61)



CHAPTER 7

21 rraL

12 lnln rrEa

Pu

1r 2rThus, the solution depends upon and which we should know first hand.

(7.62)


The above solution can be used to find displacement at various points Along the length . From (7.58) we have

And the displacement is



CHAPTER 7

7.5 WEIGHTED RESIDUAL METHOD

7.5 Weighted Residual Method

•The WRM assumes an approximate solution to the governing differential equations.

•The solution criterion is one where the boundary conditions and initial conditions of the problem are satisfied.

• It is evident that the approximate solution leads to some marginal errors.

• If we require that the errors vanish over a given interval or at some given points then we will force the approximate solution to converge to an accurate solution.



CHAPTER 7

.0)0(

,0)(

u

with

Pdy

duEyA

,)( 2210 yCyCCyu

PyCCEryL

rr)2( 211

12

2 10 , CandCCwhere are unknown coefficients

where stands for residual.

(7.63)

(7.65)

(7.64)


Consider the differential equation discussed in previous example where

Let us choose a displacement field u to approximate the solution, that is let

if we substitute u(y) & A(y) into the differential equation we get



CHAPTER 7

02

Ly

0Ly

and

,)(

)3(

122

121 rrr

rr

E

PC

)(

)(

122

122 rrr

rr

LE

PC

00 C 0)0( ufor

(7.66)


The equation above has two constants C1 & C2. If we require that ε vanishes at two points we will get two equation which can be used to solve for C1 & C2.Solving Equation (7.66) for these conditions , we get

The final solution for the displacement field is

21212

122

3)(

)( yL

rryrr

rrrE

Pyu (7.67)

(7.65)



CHAPTER 7

7.6 GALERKIN METHOD

,0

dyi (7.68)

7.6 Galerkin Method

The Galerkin method requires the integer of the error function over some selected interval to be forced to zero and that the error be orthogonal to some weighting functions i, according to the integral

i s are selected as part of the approximate solution. This is simply done by assigning the function to the terms to multiply the coefficients.

Because we assume as defined in (7.63) then 1= y and 2 = y2.

,)( 2210 yCyCCyu

(7.64)

Now we use (7.67) and substitute 1= y and the residual ε from (7.65). Furthermore let the values of

r1 and r2 be given from Example 7.3 then we obtain,

Where = [ 0 , L ] for i = 1,….., n.



CHAPTER 7

0)2)(10(0

21

L

dyE

PyCCyy

0)2)(10(0

212

L

dyE

PyCCyy

)()( 221 yCCyu

41 102314. C 6

2 1050. C

Solving the above two equations we get

&

The solution is then approximated by

C1y+C2y2

(7.71)

(7.70)

(7.69)

7.6 Galerkin Method

P = 180 lb from Example 7.3



CHAPTER 7

Location of the point along the length of the object (in inch)

Results from the Approximation method

(Ex.7.3) (x10-4)

Results from the Exact Solution Method (x10-

4)

Results from Weighted residual Methods (X10-4)

Results from Galerkin Method(X10-4)

Y=0 0 0 0 0

Y=2.5 0.0214 .6000 .532 .547

Y=5 0.0522 1.090 1.008 1.032

Y=7.5 0.1004 1.504 1.428 1.454

Y=10 0.1860 1.863 1.792 1.814

Table 7.1 Comparison of displacement values by different methods

7.6 Galerkin Method

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