Full downloaIntroduction To Finite Elements In Engineering 4th Edition Chandrupatla Solutions Manual

Introduction To Finite Elements In Engineering 4th Edition Chandrupatla Solutions Manual Full Download: https://testbanklive.com/download/introduction-to-finite-elements-in-engineering-4th-edition-chandrupatla-solutio

Introduction to

Finite Elements in Engineering Tirupathi R. Chandrupatla Rowan University Glassboro, New Jersey

Ashok D. Belegundu The Pennsylvania State University University Park, Pennsylvania

Solutions Manual

Prentice Hall, Upper Saddle River, New Jersey 07458

Introduction to Finite Elements in Engineering, Fourth Edition, by T. R. Chandrupatla and A. D. Belegundu. ISBN 01-3-216274-1. Š 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 download all chapters instantly please go to Solutions Manual, Test Bank site: TestBankLive.com

CONTENTS

Preface Chapter 1

Fundamental Concepts

1

Chapter 2

Matrix Algebra and Gaussian Elimination

18

Chapter 3

One-Dimensional Problems

26

Chapter 4

Trusses

61

Chapter 5

Beams and Frames

86

Chapter 6

Two-Dimensional Problems Using Constant Strain Triangles

103

Chapter 7

Axisymmetric Solids Subjected to Axisymmetric Loading

151

Chapter 8

Two-Dimensional Isoparametric Elements and Numerical Integration

181

Chapter 9

Three-Dimensional Problems in Stress Analysis

207

Chapter 10

Scalar Field Problems

218

Chapter 11

Dynamic Considerations

264

Chapter 12

Preprocessing and Postprocessing

282

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

PREFACE This solutions manual serves as an aid to professors in teaching from the book Introduction to Finite Elements in Engineering, 4th Edition. The problems in the book fall into the following categories: 1. Simple problems to understand the concepts 2. Derivations and direct solutions 3. Solutions requiring computer runs 4. Solutions requiring program modifications Our basic philosophy in the development of this manual is to provide a complete guidance to the teacher in formulating, modeling, and solving the problems. Complete solutions are given for problems in all categories stated. For some larger problems such as those in three dimensional stress analysis, complete formulation and modeling aspects are discussed. The students should be able to proceed from the guidelines provided.

For problems involving distributed and other types of loading, the nodal loads are to be calculated for the input data. The programs do not generate the loads. This calculation and the boundary condition decisions enable the student to develop a physical sense for the problems.

The students may be encouraged to modify the programs to calculate the loads automatically.

The students should be introduced to the programs in Chapter 12 right from the point of solving problems in Chapter 6. This will enable the students to solve larger problems with ease. The input data file for each program has been provided. Data for a problem should follow this format. The best strategy is to copy the example file and edit it for the problem under consideration. The data from program MESHGEN will need some editing to complete the information on boundary conditions, loads, and material properties.

We thank you for your enthusiastic response to our first three editions of the book. We look forward to receive your feedback of your experiences, comments, and suggestions for making improvements to the book and this manual.

Tirupathi R. Chandrupatla P.E., CMfgE Department of Mechanical Engineering Rowan University, Glassboro, NJ 08028 e-mail: chandrupatla@rowan.edu Ashok D. Belegundu Department of Mechanical and Nuclear Engineering The Pennsylvania State University University Park, PA 16802 e-mail: adb3@psu.edu Introduction to Finite Elements in Engineering, Fourth Edition, by T. R. Chandrupatla and A. D. Belegundu. ISBN 01-3-216274-1. Š 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.

CHAPTER 1 FUNDAMENTAL CONCEPTS 1.1 We use the first three steps of Eq. 1.11 σ σx σ −ν y −ν z E E E σ σ σ ε y = −ν x + y − ν z E E E σ σ σ ε z = −ν x − ν y + z E E E

εx =

Adding the above, we get

εx + εy + εz = Adding and subtracting ν

1 − 2ν (σ x + σ y + σ z ) E

σx from the first equation, E

1+ ν ν σ x − (σ x + σ y + σ z ) E E Similar expressions can be obtained for εy, and ε z. εx =

From the relationship for γyz and Eq. 1.12, E γ yz etc.

2(1 + ν ) Above relations can be written in the form σ = Dε where D is the material property matrix defined in Eq. 1.15.  τ yz =

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.

1.3

Plane strain condition implies that σy σz σ ε z = 0 = −ν x − ν + E E E which gives σ z = ν(σ x + σ y ) We have, σ x = 20000 psi σ y = −10000 psi E = 30 × 10 6 psi ν = 0.3 .

On substituting the values, σ z = 3000 psi

1.4

Displacement field



( ) (3x + 6 y − y )

u = 10 −4 − x 2 + 2 y 2 + 6 xy −4

2 v = 10 ∂u = 10 − 4 (− 2 x + 6 y ) ∂x ∂v = 3 × 10 − 4 ∂x

∂u = 10 − 4 (4 y + 6 x ) ∂y ∂v = 10 − 4 (6 + 2 y ) ∂y

 ∂u     ∂x   ∂v  ε=   ∂y   ∂u + ∂v   ∂y ∂x    at x = 1, y = 0 − 2    ε = 10  6  9    −4



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

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 =0 ∂x ∂v εy = =0 ∂y ∂u ∂v γ xy = + = 0.1 ∂y ∂x εx =



1.6

The displacement field is given as u = 1 + 3x + 4x3 + 6xy2 v = xy − 7x2 (a) The strains are then given by ∂u εx = = 3 + 12 x 2 + 6 y 2 ∂x ∂v εy = =x ∂y ∂u ∂v γ xy = + = 12 xy + y − 14 x ∂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 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:

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 18

10 8

6

12

16

16

6

10 14

4

14

4 10

8

10 12

14 16

12

18

-1 -1

6

8

-0.8 -0.8

-0.6

14 16

4

6

-0.4

10

-0.4

8

-0.2

0

0.2

10 0.4

12

0.6

18

-0.2

12

6

8

0

-0.6

12

0.4 0.2

6 8

0.6

18

8

14

0.8

0.8

1

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

(x, y)

a)

= u

b)

εx =

0.2 y ⇒ = u 0.2 y 1

(u, v)

= v 0

∂u ∂v ∂u ∂v = 0 εy = = 0 γ xy = + = 0.2 ∂x ∂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 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

σ x = 40 MPa σ y = 20 MPa σ z = 30 MPa τ yz = −30 MPa τ xz = 15 MPa τ xy = 10 MPa T

1 1 1  n=  2 2 2 From Eq. 1.8 we get Tx = σ x n x + τ xy n y + τ xz n z .

= 35.607 MPa T y = τ xy n x + σ y n y + τ yz n z = −6.213 MPa Tz = τ xz n x + τ yz n y + σ z n z = 13.713 MPa σ n = Tx n x + T y n y + Tz n z = 24.393 MPa

1.9



From the derivation made in P1.1, we have

[

]

E (1 − ν )ε x + νε y + νε z (1 + ν )(1 − 2ν ) which can be written in the form E [(1 − 2ν )ε x + νε v ] σx = (1 + ν )(1 − 2ν ) and E τ yz = γ yz 2(1 + ν ) Lame’s constants λ and µ are defined in the expressions σx =

σ x = λε v + 2µε x τ yz = µγ yz On inspection, Eν λ= (1 + ν )(1 − 2ν ) E µ= 2(1 + ν )

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.

µ is same as the shear modulus G.



1.10 ε = 1.2 × 10 −5 ∆T = 30 0 C E = 200 GPa α = 12 × 10 -6 / 0 C ε 0 = α∆T = 3.6 × 10 − 4

σ = E (ε − ε 0 ) = −69.6 MPa 

1.11 du = 1 + 2x 2 dx L du 2   dx =  x + x 3  δ=∫ 0 dx 3  

εx =

L 0

 2  = L1 + L2   3  

1.12 Following the steps of Example 1.1, we have (80 + 40 + 50 ) − 80  q1  60  =   − 80 80  q 2  50  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



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.

1.13

When the wall is smooth, σ 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 ε y = 0 . The generalized Hooke’s law yields the equations

ε x =−ν ε y=

σy

σy E

+ α∆T

+ α∆T E From the second equation, setting ε y = 0 , we get σ y = − Eα∆T . ε x is then calculated using the first equation as (1 −ν ) α∆T . b) When the block is very thick in the z direction, plain strain condition prevails. Now we have ε z = 0 , in addition to ε y = 0 . σ z is not zero.

ε x =−ν ε y=

σy E

ε z =−ν

σy E −ν

σy

−ν

σz

σz

+ α∆T= 0

E

E

+ α∆T

σz

+ α∆T =0 E E From the last two equations, we get − Eα∆T 1 + 2ν − σy = σz = Eα∆T 1 +ν 1 +ν +

ε 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 0.1 in part (a) of problem 1.13. Thus σ y = −0.1E .  initial strain ε 0 = α∆T = 1 1.15 The potential energy Π is given by x=0 2

1  du  EA  dx − ∫ ugAdx ∫ 2 0  dx  0 2

Π=

2

Consider the polynomial from Example 1.2,

(

u = a3 − 2 x + x 2

g=1 E=1 A=1

)

du = (− 2 + 2 x )a3 = 2(− 1 + x )a3 dx

x=2

On substituting the above expressions and integrating, the first term of becomes 2 2 2a 3    3 and the second term

 2 x3    ugAdx = udx = a 3− x + ∫0 ∫0 3   4 = − a3 3 2

2

Thus Π=

(

4 2 a3 + a3 3

∂Π =0⇒ ∂a3 this gives

u x =1 = −

2

0

)

a3 = −

1 2

1 (− 2 + 1) = 0.5 2



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 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 2

1  du    dx − ∫ fudx 2 ∫0  dx  0 1

Π=

1

1

1

1 2 2 = ∫ a1 (1 − 2 x ) dx − ∫ x 3 a1 x(1 − x )dx 20 0 1

(

)

1

(

)

=

1 2 a1 1 − 4 x + 4 x 2 dx − ∫ a1 x 4 − x 5 dx ∫ 20 0

=

1 2 4 4 1 1 a1 1 − +  − a1  −  2  2 3 5 6 2

a a = 1 − 1 6 30 ∂Π =0 ⇒ ∂a1

a1 1 − =0 3 30

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.

u = a3 + a4x

u = a1 + a2x u1

0

200

500

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.

0 ≤ x ≤ 200 u = 0 at x = 0 ⇒ a1 = 0 u = u1 at x = 200 ⇒ a2 = u1/200 du/dx = u1/200 ⇒ u = (u1/200)x 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 1 Π= 2

200

∫ 0

2

500

2

Π= =

2

1  du   du  E al A1   dx + ∫ E st A2   dx − 10000u1 2 200  dx   dx  2

1 1  u   u  E al A1  1  200 + E st A2  − 1  300 − 10000u1 2 2  300   200 

1  E al A1 E st A2  2 +  u1 − 10000u1 2  200 300 

∂Π E A E A  = 0 ⇒  al 1 + st 2 u1 − 10000 = 0 ∂u1 300   200 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 du EA +g =0 dx dx Following the steps of Example 1.3, we get 2

∫ − EA 0

du dφ dx + ∫ gφdx dx dx 0 2

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.

( ) φ = (2 x − x )φ

u = 2 x − x 2 u1 , and 2

1

where u1 and φ1 are the values of u and φ at x = 1 respectively, 2 2   2 φ1 − u1 ∫ (1 − 2 x ) dx + ∫ 2 x − x 2 dx  = 0 0 0  

(

)

On integrating, we get  8 φ1  − u1 +  3

4 =0 3

This is to be satisfied for every φ1, which gives the solution 

u1 = 0.5 1.19

We use

u = a1 + a 2 x + a3 x 2 + a 4 x 3 u = 0 at x = 0 u = 0 at x = 2 This implies that 0 = a1 0 = a1 + 2a 2 + 4 3 + 8a 4 and

(

)

( du = 2a (x − 1) + a (3 x dx

) − 4)

u = a3 x 2 − 2 x + a 4 x 3 − 4 x 3

2

4

a3 and a4 are considered as independent variables in 2

[

(

)]

2 1 Π = ∫ 2a3 ( x − 1) + a 4 3 x 2 − 4 dx − 2(− a3 − 3a 4 ) 20 on expanding and integrating the terms, we get

Π = 1.333a3 + 12.8a 4 + 8a3 a 4 + 2a3 + 6a 4 2

2

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.

Introduction To Finite Elements In Engineering 4th Edition Chandrupatla Solutions Manual Full Download: https://testbanklive.com/download/introduction-to-finite-elements-in-engineering-4th-edition-chandrupatla-solutio

∂Π = 2.667 a3 + 8a 4 + 2 = 0 ∂a3 ∂Π = 8a3 + 25.6a 4 + 6 = 0 ∂a 4 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 L

(a)

L

1 Π = ∫ σ T εAdx − ∫ T ( x )udx 20 0 σ = Eε and ε =

du dx

On substitution, 2

1  du  Π = ∫ EA  dx − ∫ T udx − ∫ T udx 2 0  dx  0 30 60

Π=

(

1 60 × 10 6 2

30

60

2

 )∫  du  dx − ∫ 10 xudx − ∫ 300udx dx 60

0





30

60

0

30

(b) Since u = 0 at x = 0 and x = 60, and u = a0 + a1x + a2x2, we have u = a 2 x( x − 60 ) du = a 2 (2 x − 60 ) dx On substituting and integrating, Π = 216 × 1010 a 2 + 8775000a 2 2

Setting dΠ/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.

Full download all chapters instantly please go to Solutions Manual, Test Bank site: TestBankLive.com