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

Name: Solution Manual for Introduction to Finite Elements in Engineering, 4th Edition
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Solution Manual for Introduction to Finite Elements in Engineering, 4th Edition helps you reinforce learning with in-depth, accurate solutions.

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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

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 103
Strain Triangles
Chapter 7 Axisymmetric Solids Subjected to 151
Axisymmetric Loading
Chapter 8 Two-Dimensional Isoparametric Elements 181
and Numerical Integration
Chapter 9 Three-Dimensional Problems in Stress 207
Analysis
Chapter 10 Scalar Field Problems 218
Chapter 11 Dynamic Considerations 264
Chapter 12 Preprocessing and Postprocessing 282

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

1
CHAPTER 1
FUNDAMENTAL CONCEPTS
1.1 We use the first three steps of Eq. 1.11
EEE
EEE
EEE
zyx
z
zyx
y
zyx
x


















Adding the above, we get
 zyxzyx E 

 21
Adding and subtracting E
x
 from the first equation,
 zyxxx EE 



 1
Similar expressions can be obtained for y, and  z.
From the relationship for yz and Eq. 1.12,
  yzyz
E 

 12 etc.
Above relations can be written in the form
 = D
where D is the material property matrix defined in Eq. 1.15. 
1.2 Note that u2(x) satisfies the zero slope boundary condition at the support.


2
1.3 Plane strain condition implies that
0 yx z
z E E E
   




Which gives
( )z x y 
   
On substitution of the given values
z
 = 0.3 (3000 – 15000) = 4500 psi
1.4 u = (x2 + 4y2 – 16xy) × 10-4
2 4( 5 8 ) 10v y x y    
4
10
u
x
 
 (2x – 16y)
4
(8 16 ) 10
u y x
y
   

4
5 10
u
y
   

u
y

 = (2y + 8) × 10-4
u
x
u
y
u u
y x

 
 
 
 
  
 
  
 
  
At x = 1, y = 0
4
2
8 10
21
x
y
xy





     
     
  

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3
1.5 From the figure P 1.5, it is revealed that the displacements u and v are given by
u (x) = 0.2 y + 6
v (y) = 0
 u1 = 0.2 × 0 + 6 = 6
v1 = 0
u2 = u1 + [0.2 × 0 + 6] = 12
v2 = 0
u4 = 0.2 × 15 + 6 = 9
u3 = u4 + [0.2 × 15 + 6] = 15
The strain values are computed as:
εx = u
x

 = 0
εy = v
y

 = 0
xy
u v
x y
  
 
 
= 0.2

1.6 u = 2 + 2x + 4x2 + 3xy2
v = xy = 8x2
a) Expressions for εx, εy and γxy
εx = u
x

 = 2 + 8x + 3y2
εy = v
y

 = x
xy
u v
x y
  
 
  = 6xy + y – 16x

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4
(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 εy and γxy is given below
b) MATLAB code for εx, εy, γxy:
[X,Y] = meshgrid(-1:.1:1,-1:.1:1);
Z = 2.+8.*X + 3.*Y.^2;
k = X.;
L = 6.*X.*Y+Y.-16.*X;
[C,h]= contour(X,Y,Z);
clabel(C,h);
On running the program, the contour map is shown as follows:

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7
  
 






12
211
,inspectionOn
2
E
E
yzyz
xvx
 is same as the shear modulus G. 
1.10 5
2 10
 

T = 45°C
E = 123GPa
6
16.2 10 /
 
  °C
From Eq. 1.20 and 1.21,
4
7.29 10o T
  
   
 = E (ε – εo) = – 87.2MPa
1.11
The strain at any point
εx = 5 + 4x2 = du
dx
δ= 34
5 3
LL
OO
dudx x x
dx
 
   

24
5 3
L L

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8
1
2
30 25 50 50 45
50 50 30
q
q
       
        
Solving
105q1 + 50q2 = 45 –(i)
–50q1 + 50q2 = 30 –(ii)
Adding (i) and (ii)
55q1 = 75
 q1= 1.363 mm and on substitution of q1,
q2= 1.963 mm
1.13
When the wall is smooth, 0x
  . 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 0y
  . The generalized Hooke’s law yields the
equations
y
x
y
y
T
E
T
E

  

 
   
  
From the second equation, setting 0y
 

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