CHAPTER 1 FUNDAMENTAL CONCEPTStestbanksinstant.eu/samples/Solution...Finite-Elements...Chandrupatla.pdf · download full file at full file at . 1.3 Plane strain condition implies

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full file at http://testbankinstant.comCHAPTER 1

FUNDAMENTAL CONCEPTS 1.1 We use the first three steps of Eq. 1.11

EEE

EEE

EEE

zyxz

zyxy

zyxx

σ+

σν−

σν−=ε

σν−

σ+

σν−=ε

σν−

σν−

σ=ε

Adding the above, we get

( )zyxzyx Eσ+σ+σ

ν−=ε+ε+ε

21

Adding and subtracting E

xσν from the first equation,

( )zyxxx EEσ+σ+σ

ν−σ

ν+=ε

1

Similar expressions can be obtained for εy, and ε z. From the relationship for γyz and Eq. 1.12,

( ) yzyzE

γν+

=τ12

etc.

Above relations can be written in the form σ = Dε where D is the material property matrix defined in Eq. 1.15. 1.2 Note that u2(x) satisfies the zero slope boundary condition at the support.

Introduction to Finite Elements in Engineering, Fourth Edition, by T. R. Chandrupatla and A. D. Belegundu. ISBN 01-3-216274-1. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


full file at http://testbankinstant.com 1.3 Plane strain condition implies that

EEE

zyxz

σ+

σν−

σν−==ε 0

which gives ( )yxz σ+σν=σ

We have, 0.3 psi 1030 psi 10000 psi 20000 6 =ν×=−=σ=σ Eyx . On substituting the values, psi 3000=σ z 1.4 Displacement field

( )( )( ) ( )

( )

−=ε

==

∂∂

+∂∂∂∂∂∂

=ε

+=∂∂

×=∂∂

+=∂∂

+−=∂∂

−+=

++−=

−

−−

−−

−

−

962

10

0,1at

2610 103

6410 6210

63106210

4

44

44

24

224

yxxv

yu

yvxu

yyv

xv

xyyuyx

xu

yyxvxyyxu

1.5 On inspection, we note that the displacements u and v are given by u = 0.1 y + 4 v = 0 It is then easy to see that



full file at http://testbankinstant.com

1.0

0

0

=∂∂

+∂∂

=γ

=∂∂

=ε

=∂∂

=ε

xv

yu

yvxu

xy

y

x

1.6 The displacement field is given as u = 1 + 3x + 4x3 + 6xy2

v = xy − 7x2

(a) The strains are then given by

xyxyxv

yu

xyv

yxxu

xy

y

x

1412

6123 22

−+=∂∂

+∂∂

=γ

=∂∂

=ε

++=∂∂

=ε

(b) In order to draw the contours of the strain field using MATLAB, we need to create a

script file, which may be edited as a text file and save with “.m” extension. The file for plotting εx is given below

file “prob1p5b.m”

[X,Y] = meshgrid(-1:.1:1,-1:.1:1); Z = 3.+12.*X.^2+6.*Y.^2; [C,h] = contour(X,Y,Z);

clabel(C,h); On running the program, the contour map is shown as follows:




-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

4

4

4

6

6

6

6

6

6

8

88

8

8

8

810

10

1010

10

10

12

12

12

12

12

12

14

14

14

14

14

14

16

16

16

16

18

18

18

18

Contours of εx Contours of εy and γxy are obtained by changing Z in the script file. The numbers on the contours show the function values.

(c) The maximum value of εx is at any of the corners of the square region. The maximum value is 21.

1.7

a) 0.2 0.2 01

u y u y v= ⇒ = =

b) 0 0 0.2x y xyu v u vx y y x

ε ε γ∂ ∂ ∂ ∂= = = = = + =∂ ∂ ∂ ∂

(x, y) (u, v)



full file at http://testbankinstant.com 1.8

.

MPa 393.24

MPa 713.13

MPa 213.6

MPa 607.35

get we1.8 Eq. From2

121

21

MPa 10 MPa 15 MPa 30

MPa 30 MPa 20 MPa 40

T

=

++=σ=

σ+τ+τ=−=

τ+σ+τ==

τ+τ+σ=

=

=τ=τ−=τ

=σ=σ=σ

zzyyxxn

zzyyzxxzz

zyzyyxxyy

zxzyxyxxx

xyxzyz

zyx

nTnTnT

nnnT

nnnT

nnnT

n

1.9 From the derivation made in P1.1, we have

( )( ) ( )[ ]

( )( ) ( )[ ]

( ) yzyz

vxx

zyxx

E

E

E

γν+

=τ

νε+εν−ν−ν+

=σ

νε+νε+εν−ν−ν+

=σ

12

and

21211

form in the written becan which

1211

Lame’s constants λ and µ are defined in the expressions

( )( )

( )ν+=µ

ν−ν+ν

=λ

µγ=τµε+λε=σ

12

211

,inspectionOn

2

E

E

yzyz

xvx



full file at http://testbankinstant.comµ is same as the shear modulus G. 1.10

( ) MPa 6.69106.3

/1012GPa 200

30102.1

0

40

06-

0

5

−=ε−ε=σ×=∆α=ε

×=α

==∆

×=ε

−

−

ET

CE

CT

1.11

+=

+==δ

+==ε

∫2

0

3

0

2

321

32

21

LL

xxdxdxdu

xdxdu

LL

x

1.12 Following the steps of Example 1.1, we have

( )

=

−

−++5060

808080504080

2

1

qq

Above matrix form is same as the set of equations: 170 q1 − 80 q2 = 60 − 80 q1 + 80 q2 = 50 Solving for q1 and q2, we get q1 = 1.222 mm q2 = 1.847 mm



full file at http://testbankinstant.com 1.13

When the wall is smooth, 0xσ = . T∆ is the temperature rise.

a) When the block is thin in the z direction, it corresponds to plane stress condition. The rigid walls in the y direction require 0yε = . The generalized Hooke’s law yields the equations

yx

yy

TE

TE

σε ν α

σε α

= − + ∆

= + ∆

From the second equation, setting 0yε = , we get y E Tσ α= − ∆ . xε is then calculated

using the first equation as ( )1 Tν α− ∆ . b) When the block is very thick in the z direction, plain strain condition prevails. Now we

have 0zε = , in addition to 0yε = . zσ is not zero.

0

0

y zx

y zy

y zz

TE E

TE E

TE E

σ σε ν ν α

σ σε ν α

σ σε ν α

= − − + ∆

= − + ∆ =

= − + + ∆ =

From the last two equations, we get 1 2

1 1y zE T E Tα νσ σ α

ν ν− ∆ +

= = − ∆+ +

xε is now obtained from the first equation.



full file at http://testbankinstant.com1.14 For thin block, it is plane stress condition. Treating the nominal size as 1, we may set the

initial strain 00.11

Tε α= ∆ = in part (a) of problem 1.13. Thus 0.1y Eσ = − .

1.15

The potential energy Π is given by

∫∫ −

=Π

2

0

2

0

2

21 ugAdxdx

dxduEA

Consider the polynomial from Example 1.2,

( )( ) ( ) 33

23

1222

2

axaxdxdu

xxau

+−=+−=

+−=

On substituting the above expressions and integrating, the first term of becomes

322 2

3a

and the second term

3

2

0

2

0

32

3

2

0

34

3

a

xxaudxugAdx

−=

+−== ∫∫

Thus

( )

21 0

34

33

32

3

−=⇒=∂Π∂

+=Π

aa

aa

this gives ( ) 5.01221

1 =+−−==xu

1.16

x=0

x=2

g=1 E=1 A=1

f = x3

E = 1 A = 1

x=0 x=1 Introduction to Finite Elements in Engineering, Fourth Edition, by T. R. Chandrupatla and A. D. Belegundu. ISBN 01-3-216274-1. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.



We use the displacement field defined by u = a0 + a1x + a2x2. u = 0 at x = 0 ⇒ a0 = 0 u = 0 at x = 1 ⇒ a1 + a2 = 0 ⇒ a2 = − a1

We then have u = a1x(1 − x), and du/dx = a1(1 − x). The potential energy is now written as

( ) ( )

( ) ( )

306

61

51

34

241

21

44121

12121

21

12

1

12

1

1

0

1

0

541

221

1

0

1

01

3221

1

0

1

0

2

aa

aa

dxxxadxxxa

dxxxaxdxxa

fudxdxdxdu

−=

−−

+−=

−−+−=

−−−=

−

=Π

∫ ∫

∫ ∫

∫ ∫

0301

3 0 1

1

=−⇒=∂Π∂ aa

This yields, a1 = 0.1 Displacemen u = 0.1x(1 − x) Stress σ =E du/dx = 0.1(1 − x) 1.17 Let u1 be the displacement at x = 200 mm. Piecewise linear displacement that is

continuous in the interval 0 ≤ x ≤ 500 is represented as shown in the figure. 0 200 500

u1

u = a3 + a4x u = a1 + a2x



full file at http://testbankinstant.com 0 ≤ x ≤ 200 u = 0 at x = 0 ⇒ a1 = 0 u = u1 at x = 200 ⇒ a2 = u1/200 ⇒ u = (u1/200)x du/dx = u1/200 200 ≤ x ≤ 500 u = 0 at x = 500 ⇒ a3 + 500 a4 = 0 u = u1 at x = 200 ⇒ a3 + 200 a4 = u1

⇒ a4 = −u1/300 a3 = (5/3)u1 ⇒ u = (5/3)u1 − (u1/300)x du/dx = − u1/200

12

121

1

21

2

21

1

1

500

200

2

2

200

0

2

1

100003002002

1

100003003002

12002002

1

1000021

21

uuAEAE

uuAEuAE

udxdxduAEdx

dxduAE

stal

stal

stal

−

+=

−

−+

=Π

−

+

=Π ∫∫

010000300200

0 121

1

=−

+⇒=

∂Π∂ u

AEAEu

stal

Note that using the units MPa (N/mm2) for modulus of elasticity and mm2 for area and

mm for length will result in displacement in mm, and stress in MPa. Thus, Eal = 70000 MPa, Est = 200000, and A1 = 900 mm2, A2 = 1200 mm2. On

substituting these values into the above equation, we get u1 = 0.009 mm This is precisely the solution obtained from strength of materials approach 1.18 In the Galerkin method, we start from the equilibrium equation

0=+ gdxduEA

dxd

Following the steps of Example 1.3, we get

∫∫ φ+φ

−2

0

2

0

dxgdxdxd

dxduEA

Introducing




( )( ) 1

21

2

2

and ,2

φ−=φ

−=

xxuxxu

where u1 and φ1 are the values of u and φ at x = 1 respectively,

( ) ( ) 02212

0

2

0

2211 =

−+−−φ ∫ ∫ dxxxdxxu

On integrating, we get

034

38

11 =

+−φ u

This is to be satisfied for every φ1, which gives the solution u1 = 0.5 1.19 We use

2at 00at 0

34

2321

====

+++=

xuxu

xaxaxaau

This implies that

4321

1

84200

aaaa

+++==

and

( ) ( )

( ) ( )4312

42

243

34

23

−+−=

−+−=

xaxadxdu

xxaxxau

a3 and a4 are considered as independent variables in

( ) ( )[ ] ( )∫ −−−−+−=Π2

043

2243 324312

21 aadxxaxa

on expanding and integrating the terms, we get

43432

42

3 6288.12333.1 aaaaaa ++++=Π

We differentiate with respect to the variables and equate to zero.




066.258

028667.2

434

433

=++=∂Π∂

=++=∂Π∂

aaa

aaa

On solving, we get a3 = −0.74856 and a4 = −0.00045.

On substituting in the expression for u, at x = 1, u1= 0.749 This approximation is close to the value obtained in the example problem. 1.20

(a) ( )udxxTAdxLL

∫∫ −εσ=Π00

T

21

dxduE =εε=σ and

On substitution,

( ) udxudxxdxdxdu

udxTudxTdxdxduEA

∫∫∫

∫∫∫

−−

×=Π

−−

=Π

60

30

30

0

60

0

26

60

30

30

0

60

0

2

30010106021

21

(b) Since u = 0 at x = 0 and x = 60, and u = a0 + a1x + a2x2, we have

( )

( )602

60

2

2

−=

−=

xadxdu

xxau

On substituting and integrating, 2

22

10 877500010216 aa +×=Π Setting dΠ/da2 = 0 gives




( )602935.60

1003125.2 62

−−==σ

×−= −

xdxduE

a

Plots of displacement and stress are given below:

0 10 20 30 40 50 60

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2x 10

-3

Displacement u

0 10 20 30 40 50 60

-4000

-3000

-2000

-1000

0

1000

2000

3000

4000

Stress

. 1.21 y = 20 at x = 60 implies that

( )210

210

1803120yields which ,36006020

aaaaaa

−−=++=

Substituting for k, h, L, and a0 in I, we get

( ) ( ) ( )[ ]

( ) ( )

76050001070210117104561200045600

3918350004410

80018031202521210

24

142

25

212

1

60

0

221

22

221

21

60

0

221

221

+×+×+×++=

+++++=

−−−++=

∫

∫

aaaaaaI

aadxaxaxaaI

aadxxaaI




0107021090612000

010117612000912000

42

51

2

421

1

=×+×+=

=×++=

aadadI

aadadI

On solving, a2 = 0.1699 a1 = −13.969 Substituting into the expression for a0, we get a0 = 246.538

. 1.22 Since u = 0 at x = 0, the displacement satisfying the boundary condition is u = a1x. Also

the coordinates are x2 = 1, and x3 = 3.

The potential energy for the problem is

2

3

2 2 3 30

12

duEA dx P u Pudx

π = − − ∫

We have u2 = a1, u3 = 3a1, E = 1, A = 1, and 1du adx

= . Thus

( )3 2 2

1 1 1 1 10

1 33 42 2

a dx a a a aπ = − − = −∫ .

For stationary value, setting 1

0ddaπ= , we get

3a1 – 4 = 0, which gives a1 = 0.75.

The approximate solution is u = 0.75x.



full file at http://testbankinstant.com 1.23 Use Galerkin approach with approximation 2u a bx cx= + + to solve

( )

3 0 1

0 1

du u x xdx

u

+ = ≤ ≤

=

The week form is obtained by multiplying by φ satisfying ( )0 0φ = . 1

03 0du u x dx

dxφ + − = ∫

We now set 21u bx cx= + + satisfying ( )0 1u = and 21 2a x a xφ = + . On introducing these

into the above integral,

( )( )( ) ( )

1 2 21 20

1 12 2 2 3 2 2 3 3 3 41 20 0

2 3 3 3 0

3 3 2 3 3 3 2 3 0

a x a x b cx bx cx x dx

a bx x x bx cx cx dx a bx x x bx cx cx dx

+ + + + + − =

+ − + + + + + − + + + + =

∫∫ ∫

On integrating, we get

1 2

1 2

3 1 2 3 1 3 31 02 2 3 3 4 3 4 4 2 5

3 17 7 13 11 3 02 12 6 12 10 4

b c c b b c ca b a

a b c a b c

+ − + + + + + − + + + =

+ + + + + =

This must be satisfied for every a1 and a2. Thus the equations to be solved are 3 17 7 02 12 613 11 3 012 10 4

b c

b c

+ + =

+ + =

The solution is b = –1.9157, c = 1.2048. Thus 21 1.9157 1.2048u x x= − + . 1.24



full file at http://testbankinstant.com The deflection and slope at a due to P1 are

31

3PaEI

− and 2

1

2Pa

EI− . Using this the deflection

and slope at L due to load P1 are

( )2311

1

21

1

3 2

2

Pa L aPavEI EI

PavEI

−= − −

′ = −

The deflection and slope due to load P2 are

32

2

22

2

3

2

P LvEI

P LvEI

= −

′ = −

We then get

1 2

1 2

v v vv v v= +′ ′ ′= +

1.25

(a) The displacement of B is given by (–0.1, 0.1) and A, C, and D remain in their original position. Consider a displacement field of the type

1 2 3 4

1 2 3 4

u a a x a y a xyv b b x b y b xy= + + += + + +

The four constants can be evaluated using the known displacements

(0,0) (1,0)

(1,1) (0,1)



full file at http://testbankinstant.comAt A (0, 0) 1

1

00

ab==

At B (1, 0) 1 2

1 2

0.10.1

a ab b+ = −+ =

At C (1, 1) 1 2 3 4

1 2 3 4

00

a a a ab b b b+ + + =+ + + =

At D (0, 1) 1 3

1 3

00

a ab b+ =+ =

The solution is

a1 = 0, a2 = –0.1, a3 = 0, a4 = 0.1 b1 = 0, b2 = 0.1, b3 = 0, b4

This gives 0.1 0.1

0.1 0.1u x xyv x xy= − += −

(b) The shear strain at B is

( ) ( )

0.1 0.1 0.1

0.1 1 0.1 0.1 0 0.2B

u v x yy x

γ

γ

∂ ∂= + = + −∂ ∂

= + − =


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CHAPTER 1 FUNDAMENTAL CONCEPTStestbanksinstant.eu/samples/Solution...Finite-Elements...Chandrupatla.pdf · download full file at full file at . 1.3 Plane strain condition implies