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# Solution Manual for Introduction to Finite Elements in Engineering 4th Edition

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1.1
CHAPTER 1 FUNDAMENTAL CONCEPTS
We use the first three steps of Eq. 1.11
x =x y z EEE
y =x +y z EEE
z =x y +z EEE
x +y +z =12(x +y +z)
Adding and subtracting x from the first equation, E
x =1+x (x +y +z) EE
Similar expressions can be obtained for y, and z. From the relationship for yz and Eq. 1.12,
= E etc. yz 2(1+) yz
Above relations can be written in the form = D
where D is the material property matrix defined in Eq. 1.15.
Note that u2(x) satisfies the zero slope boundary condition at the support.

E
1.2
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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1.3 Plane strain condition implies that
z =0=x y +z
which gives E E E z =(x +y)
Wehave, x =20000psi y =10000psi E=30106 psi =0.3. On substituting the values,
1.4 Displacement field
z =3000psi u=104(x2 +2y2 +6xy)
v =104 (3x + 6y y2 )
u =104(2x+6y) u =104(4y+6x)
x y
v =3104 x
v =104(6+2y) y
u x
= v y
u + v
y x at x = 1, y = 0
2 =104 6 9
1.5 On inspection, we note that the displacements u and v are given by u= 0.1y+4
v=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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x =u=0 x
y =v=0 y
xy =u+v=0.1 y x
1.6 The displacement field is given as
u= 1+3x+43 +6xy2
v = xy 72 (a) The strains are then given by
x =u=3+122 +6y2 x
y =v=x y
xy =u+v=12xy+y14x y x

(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 x 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:
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.

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
8
14
14
6
18
12
16
6
10
12
6
16
18
8
4
8
14
14
4
8
10
6
6
10
12
12
8 10
4
16
14
6
14
16
8
8
10
18
12
10
18
12
Contours of x
Contours of y and xy are obtained by changing Z in the script file. The numbers on
the contours show the function values.
(c) The maximum value of x is at any of the corners of the square region. The
maximum value is 21.
1.7

a) u=0.2y u0.2=y v0 = 1
u v u v
b) x=x=0y=y=0xy=y+x=0.2
(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.
1.8
x =40MPa y =20MPa z =30MPa
.
yz = 30MPa xz =15MPa
n=1 1 1T 2 2 2
From Eq.1.8 we get
Tx =xnx +xyny +xznz = 35.607 MPa
Ty =xynx +yny +yznz = 6.213 MPa
Tz =xznx +yzny +znz = 13.713 MPa
n =Txnx +Tyny +Tznz = 24.393 MPa
xy =10MPa
1.9 From the derivation made in P1.1, we have
= E [(1) + + ]

x (1+)(12) x y z which can be written in the form
= E [(12) + ] x (1+)(12) x v
and
=E yz 2(1+) yz
Lames constants and are defined in the expressions x =v +2x
yz =yz
On inspection,
= E (1+)(12)
=E 2(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.

is same as the shear modulus G.
1.10
=1.2105
T = 300 C
E = 200 GPa
=1210-6 /0 C
0 = T = 3.6104
= E( 0 )= 69.6 MPa
x =du=1+22 dx
Ldu 23L = 0 dxdx=x+3x 0
22 = L 1 + 3 L
1.11

1.12 Following the steps of Example 1.1, we have (80+40+50) 80q1=60
80 80 q2 50
Above matrix form is same as the set of equations: 170q1 80q2 = 60
80q1 +80q2 = 50 Solving for q1 and q2, we get
q1 = 1.222 mm q2 = 1.847 mm

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.

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

1.13
When the wall is smooth, x = 0 . 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 y = 0 . The generalized Hookes law yields the
equations
y
x = E+ T
y=y +T E
From the second equation, setting y = 0 , we get y = E T . x is then calculated
using the first equation as (1 )T .
b) When the block is very thick in the z direction, plain strain condition prevails. Now we
have z = 0 , in addition to y = 0 . z is not zero.
x
z
y EE

= + T
y
y= z +T0=
z
z EE
EE
y

= + + T= 0
From the last two equations, we get
ET 1+2
y= =z ET
1+ 1+
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
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.

1.14 For thin block, it is plane stress condition. Treating the nominal size as 1, we may set the initial strain 0 = T = 0.1 in part (a) of problem 1.13. Thus y = 0.1E .
1.15
1
The potential energy is given by
x=0
00
Consider the polynomial from Example 1.2,
u = a 3 ( 2 x + x 2 ) du=(2+2x)a3 =2(1+x)a3
g=1 E=1 A=1
x=2
dx
On substituting the above expressions and integrating, the first term of becomes 22
2a3 3 and the second term
2 2 x32 ugAdx=udx=a3x2 + 3 000
= 43 a 3 Thus
= 4 (a 2 + a )
33 =0 a3=1
a3
thisgives ux=1 =12(2+1)=0.5
2
3

1.16
E=1 A=1
x=0 f = x3
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 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 1 1 du 2 1
=2dx dxfudx 00
223
=11a (12x)dx1xax(1x)dx
211 00
2245 =11a (14x+4x)dx1a(x x)dx
211 00
124411 =2a1 12+3a156
=a2 a 11
6 30
=0 a11=0
a1 3 30 This yields, a1 = 0.1
Displacemen u = 0.1x(1 x)
Stress =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. u=a1 +a2x u=a3 +a4x
u1 0 200
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.
500

0 x 200
u = 0 at x = 0 a1 = 0
u=u1 atx=200 a2 =u1/200 u = (u1/200)x du/dx = u1/200
200 x 500
u=0 atx=500 a3 +500a4 =0
u=u1 atx=200 a3 +200a4 =u1
a4 = u1/300 a3 = (5/3)u1
u = (5/3)u1 (u1/300)x du/dx = u1/200
1 200 du 2 1 500 du 2
= EA dx+ EA dx10000u
2 al 1 dx 2 st 2 dx 1
0
200
1u2 1u2
= EA 1 200+ EA 1 30010000u
2 al 1200 2 st 2 300 1
1EAEA2
= al 1 + st 2 u1 10000u1
2 200 300
EAEA
=0al 1+st 2u110000=0
u1 200 300
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 d EAdu+g=0
dx dx
Following the steps of Example 1.3, we get 2 dud2
00 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
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.

u=(2xx2)u1, and = (2 x x 2 ) 1
where u1 and 1 are the values of u and at x = 1 respectively,
u 2(12x)dx+2(2xx )dx=0 11
22 00
On integrating, we get
84
1 3 u 1 + 3 = 0
This is to be satisfied for every 1, which gives the solution
1.19 We use
u1 = 0.5
u=a1 +a2x+a3x2 +a4x3
u = 0 at x = 0
u = 0 at x = 2 This implies that
and
0 = a1
0=a1 +2a2 +43 +8a4
u=a3(x2 2x)+a4(x3 4x) du=2a3(x1)+a4(32 4)
dx
a3 and a4 are considered as independent variables in
1 2 2
=2 [2a3(x1)+a4(32 4)]dx2(a3 3a4)
0
on expanding and integrating the terms, we get
=1.333a 2 +12.8a 2 +8a a +2a +6a 343434
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.

=2.667a3 +8a4 +2=0 a3
=8a3 +25.6a4 +6=0 a4
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
200 =E and =du
On substitution, dx 1 60 du 2
0 0 30
1 60du2 30 60
=(6010) dx10xudx300udx 6
(b)
20dx0 30
Since u = 0 at x = 0 and x = 60, and u = a0 + a1x + a2x2, we have
u = a2 x(x 60) du = a2 (2x 60)
dx
On substituting and integrating,
=2161010a 2 +8775000a 22
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.

a2 = 2.03125106
= E du = 60.935(2x 60)
Plots of displacement and stress are given below:
2 x 10-3
1.8 1.6 1.4 1.2
1 0.8 0.6 0.4 0.2
00102030405060
dx
4000 3000 2000 1000
0 -1000 -2000 -3000
Displacement u
-40000 10 20 30 40 50 60
Stress
1.21 y = 20 at x = 60 implies that
20 = a0 + 60a1 + 3600a2 , which yields
a0 =20(13a1 180a2) Substituting for k, h, L, and a0 in I, we get
.
22 I=6010(a +2a x)dx+1(25)[20(13a 180a )800]
12212 0
I=6010(a2 +4xaa +4x2a2)dx+5000(3a +18a +39)2 112212
0
I = 45600a 2 + 612000a a + 45105 a 2 +117104 a + 702104 a + 7605000 112212
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.

dI =912000a1 +612000a2 +117104 =0 da1
dI =612000a1 +90105a2 +702104 =0 da2
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 13 du2
=2 0 EAdx dxPu Pu 2233
.
We have u2 = a1, u3 = 3a1, E = 1, A = 1, and du = a1 . Thus
dx = (a)dxa3a =a2 4a.
132 3 20111211
For stationary value, setting d = 0 , we get da1
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.

1.23 Use Galerkin approach with approximation u = a+ bx+ cx2 to solve du+3u=x 0 x1
dx u(0)=1
The week form is obtained by multiplying by satisfying (0) = 0 . 1du
dx + 3u x dx = 0 0
We now set u =1+ bx+ cx2 satisfying u(0)=1and = a1x into the above integral,
1(a1x+a2x2)(b+2cx+3+3bx+3cx2 x)dx=0 0
a2x+2 . On introducing these
a11(bx+3xx2 +3bx2 +2cx2 +3cx3)dx+a21(bx2 +32 x3 +3bx3 +2cx3 +3cx4)dx+=0 00
On integrating, we get
a b+31+b+2c+3c+a b+11+3b+c+3c=0
122334234425
a 3b+17c+7+a 13b+11c+3=0 12 12 6 212 10 4

This must be satisfied for every a1 and a2. Thus the equations to be solved are 3 b + 17 c + 7 = 0
2 12 6
13 b + 11 c + 3 = 0 12 10 4
The solution is b = 1.9157, c = 1.2048. Thus u =1 1.+9157x 1.20482 . 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.

111
The deflection and slope at a due to P are Pa3 and Pa2 . Using this the deflection
and slope at L due to load P1 are
3EI
2EI
Pa3 Pa2 (La) 11
v1 =
3EI 2EI
1 v= Pa2
1 2EI
The deflection and slope due to load P2 are
1.25
v = P L 2 2EI
We then get
3
2 3EI
2
2
v=v v+ 12
v=v v+ 12
(0,1)
v = PL 2

(0,0)
(1,0)
(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
u=a+ ax+ ay+ axy 1234
v=b+ bx+ by+ bxy 1234
The four constants can be evaluated using the known displacements
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.
(1,1)

At A (0, 0)
1
At B (1, 0)
12
a1 = 0 b=0
a1+a2= 0.1 b +b =0.1
a1 + a2 + a3 + a4 = 0 b+b+b+b=0
At C (1, 1) At D (0, 1)
The solution is
a1 =0,a2 =0.1,a3 =0,a4 =0.1
b1 =0,b2 =0.1,b3 =0,b4
This gives u = 0+.1x 0.1xy v = 0.1x 0.1xy (b) The shear strain at B is
= u + v = 0.1x +0.1 0.1y y x
B =0.1(1) 0.+1 0.1(0) 0=.2
1234 a1 + a3 = 0
b+b=0 13

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

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