Solutions Manual For Advanced Mechanics of Materials and Applied Elasticity 6th Edition By Ansel Ugural, Saul Fenster (All Chapters, 100% Original Verified, A+ Grade)

Solutions Manual
For

Advanced Mechanics of
Materials and Applied Elasticity
6th Edition

By

Ansel Ugural
Saul Fenster




Part 1: SOLUTIONS MANUAL: Page 1-352
Part 2: MATLAB SOLUTIONS: Page 353-414

, Part 1
CONTENTS




Chapter 1 Analysis of Stress 1

Chapter 2 Strain and Material Properties 48

Chapter 3 Problem in Elasticity 83

Chapter 4 Failure Criteria 111

Chapter 5 Bending of Beams 133

Chapter 6 Torsion of Prismatic Bars 166

Chapter 7 Numerical Methods 186

Chapter 8 Thick-Walled Cylinders and Rotating Disks 227

Chapter 9 Beams on Elastic Foundations 248

Chapter 10 Applications of Energy Methods 259

Chapter 11 Stability of Columns 284

Chapter 12 Plastic Behavior of Materials 309

Chapter 13 Stresses in Plates and Shells 335




ii




From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886)
Copyright © 2020 Pearson Education, Inc. All rights reserved.

, CHAPTER 1
SOLUTION (1.1)

We have
A = 50 × 75 = 3.75(10−3 ) m 2 , θ = 50o , and σ x = P A .
Equations (1.11), with θ = 50o :
σ x ' = 700(10 3 ) = σ x cos 2 50o = 0.413σ x = 110.18P
or P = 6.35 kN
and
τ x ' y ' 560(10
= = 3
) σ x sin 50o =
cos 50o 0.492
= σ x 131.2 P
Solving
P = 4.27 kN = Pall
______________________________________________________________________________________
SOLUTION (1.2)

Normal stress is
125(103 )
σ x= P
A= = 50 MPa
0.05×0.05

( a ) Equations (1.11), with θ = 20o :
=σ x ' 50
= cos 2 20o 44.15 MPa
τ x' y' =
−50sin 20o cos 20o =
−16.08 MPa
σ y ' 50 cos 2 (20o +=
= 90o ) 5.849 MPa

5.849 MPa
y’
44.15 MPa
16.08 MPa x’
20 o
x


θ = 45o :
( b ) Equations (1.11), with
=σ x ' 50
= cos 2 45o 25 MPa
τ x' y' =
−50sin 45o cos 45o =
−25 MPa
σ y ' 50 cos 2 (45o +=
= 90o ) 25 MPa

25 MPa
25 MPa
x’
y’
25 MPa
45 o
x
______________________________________________________________________________________


1
From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886)
Copyright © 2020 Pearson Education, Inc. All rights reserved.

, ______________________________________________________________________________________
SOLUTION (1.3)

From Eq. (1.11a),
σx'
σx = cos 2 θ
= −75
cos 2 30o
= −100 MPa
For θ = 50 , Eqs. (1.11) give then
o


σ x' =−100 cos 2 50o = −41.32 MPa
41.32 MPa
τ x ' y ' = −( −100) sin 50o cos 50o 58.68 MPa
= 49.24 MPa 50 o
Similarly, for θ = 140 :
o


σ x' = −100 cos 2 140o =
−58.68 MPa
τ x ' y ' = −49.24 MPa 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 σ y = 0 and τ xy = 0 , become

σ x ' = 12 σ x + 12 σ x cos 2θ and τ x ' y ' = 12 σ x sin 2θ
or
20 = P
2A (1 + cos 2θ ) and 10 = P
2A sin 2θ

The foregoing lead to
2 sin 2θ − cos 2θ = 1 (a)

By introducing trigonometric identities, Eq. (a) becomes
4 sin θ cos θ − 2 cos 2 θ = 0 or tan θ = 1 2 . Hence
θ = 26.56o
Thus,
=20 P
2(1300) (1 + 0.6)
gives
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 )
σ1 = = = −76.4 MPa
A π 2
(50)
4
P
τ max
= = 38.2 MPa
2A

______________________________________________________________________________________



2
From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886)
Copyright © 2020 Pearson Education, Inc. All rights reserved.