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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
and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval
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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:
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.
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-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)
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.
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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
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.
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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
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.
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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.
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
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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.
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.
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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
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
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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
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
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( )( ) 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.
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.
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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
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.
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( )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
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.
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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.
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.
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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
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.
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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)
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.
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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
= + = +
= + =
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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