Solutions Manual for
Advanced Mechanics of Materials and Applied Elasticity
Fifth Edition
Ansel C. Ugural Saul K. Fenster
CONTENTS
Chapter 1
Analysis of Stress
1–1
Chapter 2
Strain and Material Properties
2–1
Chapter 3
Problem in Elasticity
3–1
Chapter 4
Failure Criteria
4–1
Chapter 5
Bending of Beams
5–1
Chapter 6
Torsion of Prismatic Bars
6–1
Chapter 7
Numerical Methods
7–1
Chapter 8
Axisymmetrically Loaded Members
8–1
Chapter 9
Beams on Elastic Foundations
9–1
Chapter 10
Applications of Energy Methods
10–1
Chapter 11
Stability of Columns
11–1
Chapter 12
Plastic Behavior of Materials
12–1
Chapter 13
Plates and Shells
13–1
NOTES TO THE INSTRUCTOR
The Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition supplements the study of stress and deformation analyses developed in the book. The main objective of the manual is to provide efficient solutions for problems dealing with variously loaded members. This manual can also serve to guide the instructor in the assignments of problems, in grading these problems, and in preparing lecture materials as well as examination questions. Every effort has been made to have a solutions manual that can cut through the clutter and is as self-explanatory as possible, thus reducing the work on the instructor. It is written and class-tested by the author, Ansel Ugural. As indicated in the book’s Preface, the text is designed for the senior and/or first year graduate level courses in stress analysis. In order to accommodate courses of varying emphasis, considerably more material has been presented in the book than can be covered effectively in a single three-credit course. The instructor has the choice of assigning a variety of problems in each chapter. Answers to selected problems are given at the end of the text. A description of the topics covered is given in the introduction of each chapter throughout the text. It is hoped that the foregoing materials will help instructor in organizing his or her course to best fit the needs of his or her students. Ansel C. Ugural Holmdel, NJ
CHAPTER 1 SOLUTION (1.1) We have
A = 50 " 75 = 3.75(10 !3 ) m 2 , ! = 90o " 40o = 50o , and ! x = P A . o
Equations (1.8), with ! = 50 :
! x ' = 700(10 3 ) = ! x cos 2 50o = 0.413! x = 110.18P
or
P = 6.35 kN
" x ' y ' = 560(10 3 )! x sin 50o cos 50o = 0.492! x = 131.2 P Solving
P = 4.27 kN = Pall
______________________________________________________________________________________ SOLUTION (1.2) Normal stress is 3
(10 ) " x = PA = 0125 .05!0.05 = 50 MPa
" 70o = 20o : ! x ' = 50 cos 2 20o = 44.15 MPa
( a ) Equations (1.11), with ! = 90
o
" x ' y ' = !50 sin 20o cos 20o = !16.08 MPa
! y ' = 50 cos 2 ( 20o + 90o ) = 5.849 MPa 5.849 MPa y’
44.15 MPa
16.08 MPa
x’ 20 o x o
( b ) Equations (1.11), with ! = 45 :
! x ' = 50 cos 2 45o = 25 MPa
" x ' y ' = !50 sin 45o cos 45o = !25 MPa ! y ' = 50 cos 2 ( 45o + 90o ) = 25 MPa 25 MPa y’ 25 MPa
25 MPa x’ 45 o x
______________________________________________________________________________________
Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 1–1
______________________________________________________________________________________ SOLUTION (1.3) From Eq. (1.11a),
# x = cos# x2'" = cos!27530o = !100 MPa o
For ! = 50 , Eqs. (1.11) give then
" x ' = !100 cos 2 50o = !41.32 MPa
" x ' y ' = !( !100) sin 50o cos 50o = 49.24 MPa o Similarly, for ! = 140 : " x ' = !100 cos 2 140o = !58.68 MPa " x ' y ' = !49.24 MPa
58.68 MPa
41.32 MPa 50 o
49.24 MPa
______________________________________________________________________________________ SOLUTION (1.4) Refer to Fig. 1.6c. Equations (1.11) by substituting the double angle-trigonometric relations, or Eqs. (1.18) with ! y = 0 and ! xy = 0 , become or
" x ' = 12 " x + 12 " x cos 2!
and
# x ' y ' = 12 " x sin 2!
20 = 2PA (1 + cos 2! )
and
10 = 2PA sin 2!
The foregoing lead to
2 sin 2" ! cos 2" = 1
(a)
By introducing trigonometric identities, Eq. (a) becomes
Thus, gives
4 sin " cos " ! 2 cos 2 " = 0 or tan ! = 1 2 . Hence ! = 26.56o P 20 = 2 (1300 ) = (1 + 0.6)
P = 32.5 kN
It can be shown that use of Mohr’s circle yields readily the same result. ______________________________________________________________________________________ SOLUTION (1.5) Equations (1.12):
P #150(103 ) = = #76.4 MPa " A 2 (50) 4 P ! max = = 38.2 MPa 2A
!1 =
______________________________________________________________________________________
Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 1–2