Finite Elements in Engineering Chandrupatla

8/10/2019 Finite Elements in Engineering Chandrupatla

1/17


2/17

1.3 Plane strain condition implies that

EEE

zyx

z

+

== 0

which gives

( )yxz += We have, 0.3psi1030psi10000psi20000 6 ==== Eyx .

On substituting the values,

psi3000=z

1.4 Displacement field

( )

( )( ) ( )

( )

=

==

+

=

+=

=

+=

+= +=

++=

9

6

2

10

0,1at

2610103

64106210

6310

6210

4

44

44

24

224

yx

x

v

y

uy

vx

u

yy

v

x

v

xyy

uyx

x

uyyxv

xyyxu

1.5 On inspection, we note that the displacements uand vare given by

u= 0.1y+ 4v= 0

It is then easy to see that

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

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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.


3/17

1.0

0

0

=+=

=

=

=

=

xv

yu

y

v

x

u

xy

y

x

1.6 The displacement field is given as

u= 1 + 3x+ 4x3+ 6xy

2

v=xy7x2

(a) The strains are then given by

xyxyx

v

y

u

xy

v

yxx

u

xy

y

x

1412

6123 22

+=

+

=

=

=

++=

=

(b)In order to draw the contours of the strain field using MATLAB, we need to create ascript file, which may be edited as a text file and save with .m extension. The file

for plotting xis given below

file prob1p5b.m[ X, Y] = meshgr i d( - 1: . 1: 1, - 1: . 1: 1) ;Z = 3. +12. *X. 2+6. *Y. 2;[ C, h] = cont our ( X, Y, Z) ;cl abel ( C, h) ;

On running the program, the contour map is shown as follows:






4/17

-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

8

8

8

8

8

8

10

10

10

10

10

10

12

12

12

12

12

12

14

14

14

14

14

14

16

16

16

16

18

18

18

18

Contours of x

Contours of yand xyare obtained by changing Z in the script file. The numbers onthe contours show the function values.

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

1.7

a)

0.2

0.2 01u y u y v= = =

b) 0 0 0.2x y xyu v u v

x y y x

= = = = = + =

(x,y) (u, v)






5/17

1.8

.

MPa393.24

MPa713.13

MPa213.6

MPa607.35

getwe1.8Eq.From

2

1

2

1

2

1

MPa10MPa15MPa30

MPa30MPa20MPa40

T

=

++=

=++=

=

++=

=

++=

=

===

===

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

+

=

++

=

+++

=

12

and

21211

formin thewrittenbecanwhich

1211

Lames constants and are defined in the expressions

( )( )

( )+=

+

=

=

+=

12

211

,inspectionOn

2

E

E

yzyz

xvx






6/17

is same as the shear modulus G.

1.10

( ) MPa6.69

106.3

/1012

GPa200

30

102.1

0

4

0

06-

0

5

==

==

=

==

=

E

T

C

E

CT

1.11

+=

+==

+==

2

0

3

0

2

3

21

3

2

21

LL

xxdxdx

du

xdx

du

LL

x

1.12 Following the steps of Example 1.1, we have

( )

=

++

50

60

8080

80504080

2

1

q

q

Above matrix form is same as the set of equations:

170 q1 80 q2 = 60

80 q1 + 80 q2 = 50

Solving for q1and q2, we get

q1= 1.222 mm

q2= 1.847 mm






7/17

1.13

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

a)

When the block is thin in thezdirection, it corresponds to plane stress condition. The

rigid walls in theydirection require 0y = . The generalized Hookes law yields the

equations

y

x

y

y

TE

TE

= +

= +

From the second equation, setting 0y = , we get y E T = . x is then calculated

using the first equation as ( )1 T .

b)

When the block is very thick in thezdirection, plain strain condition prevails. Now wehave 0z = , in addition to 0y = . z is not zero.

0

0

y zx

y zy

y zz

TE E

TE E

TE E

= +

= + =

= + + =

From the last two equations, we get1 2

1 1y z

E TE T

+= =

+ +

x is now obtained from the first equation.






8/17

1.14 For thin block, it is plane stress condition. Treating the nominal size as 1, we may set the

initial strain 00.1

1T = = in part (a) of problem 1.13. Thus 0.1y E = .

1.15

The potential energy is given by

=2

0

2

0

2

2

1ugAdxdx

dx

duEA

Consider the polynomial from Example 1.2,

( )

( ) ( ) 33

2

3

1222

2

axaxdx

du

xxau

+=+=

+=

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

3

22

2

3a

and the second term

3

2

0

2

0

32

3

2

0

3

4

3

a

xxaudxugAdx

=

+==

Thus

( )

2

10

3

4

3

3

3

2

3

==

+=

aa

aa

this gives ( ) 5.0122

11 =+==xu

1.16

x=0

x=2

g=1E=1

A=1

f=x3

E= 1

A= 1

x=0 x=1Introduction to Finite Elements in Engineering, Fourth Edition, by T. R. Chandrupatla and A. D. Belegundu. ISBN 01-3-216274-1.





9/17

We use the displacement field defined by u= a0+ a1x + a2x2.

u= 0 atx= 0 a0= 0

u= 0 atx= 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

6

1

5

1

3

4

2

41

2

1

4412

1

1212

1

2

1

1

2

1

1

2

1

1

0

1

0

54

1

22

1

1

0

1

0

1

322

1

1

0

1

0

2

aa

aa

dxxxadxxxa

dxxxaxdxxa

fudxdxdx

du

=

+=

+=

=

=

030

1

301

1 ==

a

a

This yields, a1= 0.1

Displacemen u= 0.1x(1 x)

Stress =E du/dx= 0.1(1 x)

1.17 Let u1be the displacement atx= 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+a4xu= a1+a2x






10/17

0 x 200u= 0 atx= 0 a1= 0u= u1 atx= 200 a2= u1/200 u= (u1/200)x du/dx= u1/200

200 x 500u= 0 atx= 500 a3+ 500a4= 0u= u1 atx= 200 a3+ 200a4= u1

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

1

2

121

1

2

12

2

11

1

500

200

2

2

200

0

2

1

100003002002

1

100003003002

1

2002002

1

100002

1

2

1

uuAEAE

u

u

AE

u

AE

udxdx

duAEdx

dx

duAE

stal

stal

stal

+=

+

=

+

=

010000300200

0 121

1

=

+=

uAEAE

u

stal

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

2for area and

mm for length will result in displacement in mm, and stress in MPa.

Thus, Eal= 70000 MPa, Est= 200000, andA1= 900 mm2,A2= 1200 mm

2. 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

+

2

0

2

0

dxgdxdx

d

dx

duEA

Introducing






11/17

( )( ) 12

1

2

2

and,2

=

=

xx

uxxu

where u1and 1are the values of uand atx= 1 respectively,

( ) ( ) 02212

0

2

0

22

11 =

+ dxxxdxxu

On integrating, we get

03

4

3

811 =

+ u

This is to be satisfied for every 1, which gives the solution

u1= 0.5

1.19 We use

2at0

0at0

3

4

2

321

==

==

+++=

xu

xu

xaxaxaau

This implies that

4321

1

84200

aaaa

+++==

and

( ) ( )

( ) ( )4312

42

2

43

3

4

2

3

+=

+=

xaxadx

du

xxaxxau

a3and a4are considered as independent variables in

( ) ( )[ ] ( ) +=2

0

43

22

43 3243122

1aadxxaxa

on expanding and integrating the terms, we get

4343

2

4

2

3 6288.12333.1 aaaaaa ++++=

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






12/17

066.258

028667.2

43

4

43

3

=++=

=++=

aaa

aaa

On solving, we geta3= 0.74856 and a4= 0.00045.

On substituting in the expression for u, atx= 1,

u1= 0.749

This approximation is close to the value obtained in the example problem.

1.20

(a) ( )udxxTAdxLL

=00

T

21

dx

duE == and

On substitution,

( ) udxudxxdxdxdu

udxTudxTdxdx

duEA

=

=

60

30

30

0

60

0

2

6

60

30

30

0

60

0

2

30010106021

2

1

(b)Since u= 0 atx= 0 andx= 60, and u= a0+ a1x+ a2x

2, we have

( )

( )602

60

2

2

=

=

xadx

du

xxau

On substituting and integrating,

2

2

2

10 877500010216 aa +=

Setting d/da2= 0 gives






13/17

( )602935.60

1003125.2 62

==

=

xdx

duE

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

1803120

yields which,36006020

aaa

aaa

=

++=

Substituting for k, h,L, and a0inI, we get

( ) ( ) ( )[ ]

( ) ( )

76050001070210117104561200045600

3918350004410

80018031202521210

2

4

1

42

2

5

21

2

1

60

0

2

21

2

2

2

21

2

1

60

0

221

221

+++++=

+++++=

++=

aaaaaaI

aadxaxaxaaI

aadxxaaI






14/17

0107021090612000

010117612000912000

4

2

5

1

2

4

21

1

=++=

=++=

aada

dI

aada

dI

On solving,a2= 0.1699

a1= 13.969

Substituting into the expression for a0, we get

a0= 246.538

.

1.22 Since u= 0 atx= 0, the displacement satisfying the boundary condition is u= a1x. Alsothe coordinates arex2= 1, andx3= 3.

The potential energy for the problem is

23

2 2 3 30

1

2

duEA dx P u Pudx

=

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

adx

= . Thus

( )3 2 2

1 1 1 1 10

1 33 4

2 2a dx a a a a= = .

For stationary value, setting1

0d

da

= , we get

3a1 4 = 0, which gives a1= 0.75.

The approximate solution is u= 0.75x.






15/17

1.23 Use Galerkin approach with approximation 2u a bx cx= + + to solve

( )

3 0 1

0 1

duu x x

dx

u

+ =

=

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

0

3 0du

u x dxdx

+ =

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

into the above integral,

( )( )

( ) ( )

12 2

1 20

1 12 2 2 3 2 2 3 3 3 4

1 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 0

2 2 3 3 4 3 4 4 2 5

3 17 7 13 11 30

2 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 a1and a2. Thus the equations to be solved are

3 17 70

2 12 6

13 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






16/17

The deflection and slope at adue toP1are3

1

3

Pa

EI and

2

1

2

Pa

EI . Using this the deflection

and slope atLdue to loadP1are

( )23 111

2

11

3 2

2

Pa L aPa

v EI EI

Pav

EI

=

=

The deflection and slope due to loadP2are3

22

2

22

3

2

P Lv

EI

P Lv

EI

=

=

We then get1 2

1 2

v v v

v v v

= +

= +

1.25

(a) The displacement ofBis given by (0.1, 0.1) andA, C, andDremain in their originalposition. Consider a displacement field of the type

1 2 3 4

1 2 3 4

u a a x a y a xy

v 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)






17/17

AtA (0, 0)1

1

0

0

a

b

=

=

AtB (1, 0)1 2

1 2

0.1

0.1

a a

b b

+ =

+ =

At C (1, 1)1 2 3 4

1 2 3 4

00

a a a ab b b b

+ + + =+ + + =

AtD (0, 1)1 3

1 3

0

0

a a

b 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 0.10.1 0.1

u x xyv x xy

= +=

(b) The shear strain atBis

( ) ( )

0.1 0.1 0.1

0.1 1 0.1 0.1 0 0.2B

u vx y

y x

= + = +

= + =



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Finite Elements in Engineering Chandrupatla