ALL 15 CHAPTERS COVERED
, CONTENTS
PART I FUNDAMENTALS
Chapter 1 BASIC CONCEPTS 1
Chapter 2 SIMPLE STRUCTURAL MEMBERS 11
PART II PLATES
Chapter 3 PLATE BENDING THEORY 25
Chapter 4 CIRCULAR PLATES 34
Chapter 5 RECTANGULAR PLATES 48
Chapter 6 PLATES OF VARIOUS GEOMETRICAL FORMS 67
Chapter 7 NUMERICAL METHODS 73
Chapter 8 ANISOTROPIC PLATES 94
Chapter 9 PLATES UNDER COMBINED LOADINGS 102
Chapter 10 LARGE DEFLECTIONS OF PLATES 110
Chapter 11 THERMAL STRESSES IN PLATES 113
PART III SHELLS
Chapter 12 MEMBRANE STRESSES IN SHELLS 120
Chapter 13 BENDING STRESSES IN SHELLS 132
Chapter 14 APPLICATIONS TO PIPES, TANKS, AND PRESSURE VESSELS 138
Chapter 15 CYLINDRICAL SHELLS UNDER GENERAL LOADS 148
APPENDIX C INTRODUCTION TO FINITE ELEMENT ANALYSIS 152
APPENDIX D INTRODUCTION TO MATLAB …
, NOTES TO THE INSTRUCTOR
The Solutions Manual to accompany the text PLATES AND SHELLS: Theory and
Analysis 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 structural 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 self-explanatory
as possible thus reducing the work on the instructor. It is written and class tested by
the author.
As indicated in its preface, the text is designed for the senior and/or first year
graduate level courses in the analysis of beams, pates and shells, stress analysis, pressure
vessels, advanced statics, or special topics in solid and structural mechanics. 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 course to best fit the needs of
his students.
Ansel C. Ugural
Holmdel, N.J.
, CHAPTER 1
SOLUTION (1.1)
(a) RAy 7.5 kN
1m Entire Structure
RAx
A C D M A = 0: 75
. (1) + 3(1) = 15
. RBx
1.5 m or RBx = 7 kN
RBx
B
3 kN
F x = 0: RAx = −7 kN
RBy = 0 F y = 0: R Ay = 10 .5 kN
(b)
7.5 kN Member AD
7 kN
1m 1m Cx
M A = 0: Cy (2) = 75
. (1);
A C C y = 3.75 kN
Ay Cy F y = 0: Ay = 7.5 − 3.75 = 3.75 kN
F x = 0: C x = 7 kN
(c)
5/6 Segment AE
10/3
5/3 10/3
M F = 0: P = 7 kN
x
7 kN
A E P
F = 0: V = 3.75 − −
y
5
6
10
3 = 0.417 kN
3.75 kN 1/3
1/2 1/2
V
M = 0: M = 375
E . (1) − ( 23 ) − 103 ( 12 )
5
6
= 1.528 kN m
SOLUTION (1.2)
Refer to Fig. P1.2: M A = 0: RB = 23 pa
(a)
p
M 0 = 0: M = 23 pax − 12 px 2 (a)
Thus
M B
V 0 = 0: 23 pa − px = 0;
dM
dx x = 23 a
x 2
pa
3
Equation (a), for x = 23 a; M max = 2
9 pa 2
( b ) Equation (a), for x = 3
2 a:
M = 23 pa( 23 a ) − 12 p( 23 a )2 = − 18 pa 2 = 18 pa 2
Shear force at x = 23 a:
V = − 23 pa + 23 pa = 56 pa
SOLUTION (1.3)
Free-Body: Piston
53
7 53
7
2
F x = 0: P −
53
FAB = 0 FAB =
7
P (a)
A FAB
P
RA (CONT.)
(1.3 CONT.)
1
, CONTENTS
PART I FUNDAMENTALS
Chapter 1 BASIC CONCEPTS 1
Chapter 2 SIMPLE STRUCTURAL MEMBERS 11
PART II PLATES
Chapter 3 PLATE BENDING THEORY 25
Chapter 4 CIRCULAR PLATES 34
Chapter 5 RECTANGULAR PLATES 48
Chapter 6 PLATES OF VARIOUS GEOMETRICAL FORMS 67
Chapter 7 NUMERICAL METHODS 73
Chapter 8 ANISOTROPIC PLATES 94
Chapter 9 PLATES UNDER COMBINED LOADINGS 102
Chapter 10 LARGE DEFLECTIONS OF PLATES 110
Chapter 11 THERMAL STRESSES IN PLATES 113
PART III SHELLS
Chapter 12 MEMBRANE STRESSES IN SHELLS 120
Chapter 13 BENDING STRESSES IN SHELLS 132
Chapter 14 APPLICATIONS TO PIPES, TANKS, AND PRESSURE VESSELS 138
Chapter 15 CYLINDRICAL SHELLS UNDER GENERAL LOADS 148
APPENDIX C INTRODUCTION TO FINITE ELEMENT ANALYSIS 152
APPENDIX D INTRODUCTION TO MATLAB …
, NOTES TO THE INSTRUCTOR
The Solutions Manual to accompany the text PLATES AND SHELLS: Theory and
Analysis 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 structural 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 self-explanatory
as possible thus reducing the work on the instructor. It is written and class tested by
the author.
As indicated in its preface, the text is designed for the senior and/or first year
graduate level courses in the analysis of beams, pates and shells, stress analysis, pressure
vessels, advanced statics, or special topics in solid and structural mechanics. 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 course to best fit the needs of
his students.
Ansel C. Ugural
Holmdel, N.J.
, CHAPTER 1
SOLUTION (1.1)
(a) RAy 7.5 kN
1m Entire Structure
RAx
A C D M A = 0: 75
. (1) + 3(1) = 15
. RBx
1.5 m or RBx = 7 kN
RBx
B
3 kN
F x = 0: RAx = −7 kN
RBy = 0 F y = 0: R Ay = 10 .5 kN
(b)
7.5 kN Member AD
7 kN
1m 1m Cx
M A = 0: Cy (2) = 75
. (1);
A C C y = 3.75 kN
Ay Cy F y = 0: Ay = 7.5 − 3.75 = 3.75 kN
F x = 0: C x = 7 kN
(c)
5/6 Segment AE
10/3
5/3 10/3
M F = 0: P = 7 kN
x
7 kN
A E P
F = 0: V = 3.75 − −
y
5
6
10
3 = 0.417 kN
3.75 kN 1/3
1/2 1/2
V
M = 0: M = 375
E . (1) − ( 23 ) − 103 ( 12 )
5
6
= 1.528 kN m
SOLUTION (1.2)
Refer to Fig. P1.2: M A = 0: RB = 23 pa
(a)
p
M 0 = 0: M = 23 pax − 12 px 2 (a)
Thus
M B
V 0 = 0: 23 pa − px = 0;
dM
dx x = 23 a
x 2
pa
3
Equation (a), for x = 23 a; M max = 2
9 pa 2
( b ) Equation (a), for x = 3
2 a:
M = 23 pa( 23 a ) − 12 p( 23 a )2 = − 18 pa 2 = 18 pa 2
Shear force at x = 23 a:
V = − 23 pa + 23 pa = 56 pa
SOLUTION (1.3)
Free-Body: Piston
53
7 53
7
2
F x = 0: P −
53
FAB = 0 FAB =
7
P (a)
A FAB
P
RA (CONT.)
(1.3 CONT.)
1
