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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

Adding the above, we get

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

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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

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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

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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

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

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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

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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

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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 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

12 du2 2 =2EAdx dxugAdx

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

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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 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

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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

EAdx dxdx+gdx

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

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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

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=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

(a) =1L AdxLT(x)udx T

200 =E and =du

On substitution, dx 1 60 du 2

30 60 =2EAdx dxTudxTudx

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

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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

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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

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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

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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

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(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

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