Solution Manual for Stresses in Beams Plates and Shells 3rd edition by Ugural pdf

Third Edition
Third Edition
Ansel C. Ugural
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PART I FUNDAMENTALS Chapter 1 BASIC CONCEPTS 1 Chapter 2 STRESSES IN SIMPLE STRUCTURAL MEMBERS 8 PART II PLATES Chapter 3 ELEMENTS OF PLATE - BENDING THEORY 16 Chapter 4 CIRCULAR PLATES 25 Chapter 5 RECTANGULAR PLATES 38 Chapter 6 PLATES OF VARIOUS GEOMETRICAL FORMS 57 Chapter 7 NUMERICAL METHODS 63 Chapter 8 ANISOTROPIC PLATES 79 Chapter 9 PLATES UNDER COMBINED LOADINGS 87 Chapter 10 LARGE DEFLECTIONS OF PLATES 94 Chapter 11 THERMAL STRESSES IN PLATES 97 PART III SHELLS Chapter 12 MEMBRANE STRESSES IN SHELLS 104 Chapter 13 BENDING STRESSES IN SHELLS 116 Chapter 14 APPLICATIONS TO PIPES, TANKS, AND PRESSURE VESSELS 122 Chapter 15 CYLINDRICAL SHELLS UNDER GENERAL LOADS 131 iii
NOTES TO THE INSTRUCTOR
The Solutions Manual to accompany the text Stresses in Beams, Plates and Shells 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. iv
CHAPTER 1 SOLUTION (1.1) ( a )
Entire Structure M A∑ = 0: 7 5 1 3 1 15. ( ) ( ) .+ = RBx
or R kNBx = 7
Fy∑ = 0: R kNAy = 10 5.
( b ) Member AD M A∑ = 0: Cy ( ) . ( );2 75 1= C kNy = 375.
Fy∑ = 0: A kNy = − =7 5 375 375. . .
Fx =∑ 0: C kNx = 7
( c ) Segment AE Fx∑ = 0: P kN= 7
Fy∑ = 0: V kN= − − =375 0 4175 6
10 3. .
2 3
10 3
1 2. ( ) ( ) ( )
= ⋅1528. kN m SOLUTION (1.2) Refer to Fig. P1.2: M A =∑ 0: R paB = 2
3
1 2
3
2 9
M pa a p a pa pa= − = − =2 3
3 2
1 2
3 2
V pa pa pa= − + =2 3
3 2
5 6
Fx∑ = 0: R kNBx = 70
Fy∑ = 0: R kNBy = 25 (CONT.)
B
1.5 m
1 2 RA
(1.3 CONT.) Member AC M A =∑ 0: R RCx Cy= 2
Fy∑ = 0: R kNCy = 35
Fx =∑ 0: R kNCx = 70
( b )
Fy∑ = 0: V kN= 35
MD =∑ 0: M kN m= ⋅35 SOLUTION (1.4) Link BD is a two-force member and hence the direction of BDF is known.
(a) Free body-Member ADE
BD BD
=
− + = = ↑ ∑
=
− + − = = ↓
∑
0 :y OF V P= = ↑∑
0 : 1.5O OM M Pa= =∑
SOLUTION (1.5) We have 115oθ = . Apply Eqs. ( 1.11 ): 1 1
n 2 2( 50 40) ( 50 40)cos 230 20sin 230o o xσ = − + + − − −
5 28.92 15.32 39.2MPa= − + + = 1
' ' 2 ( 50 40)sin 230 20cos 230 21.6o o x y MPaτ = − − − − = −
35
A
70
r = + =( . ) .50 37 5 62 52 2 1 2
Thus, σ1 62 5 62 5 125= + =. . MPa
σ2 0= ( b ) τmax .= =r MPa62 5 SOLUTION (1.7) ( a ) σ ' ( )= − + = −1
2 150 80 35 MPa
θp o" tan .= =−1
r = + =( ) .115 70 134 62 2 1 2
Thus, σ1 35 134 6 99 6= − + =. . MPa
σ2 169 6= − . MPa τmax .= =r MPa134 6 ( b ) SOLUTION (1.8) ( a ) (CONT.)
26 6. o
σ x
(1.8 CONT.) We haveσ y = 0 and τ xy MPa= 60 .
Fx =∑ 0: σ x o o osin cos cos40 60 40 60 40= +
or σ x MPa comp= 143 ( . ) Apply Eqs. (1.11) with θ = + =90 25 115o:
σ x
143 2 230 60 230 715
τ x y o o MPa' ' . sin cos .= + = −715 230 60 230 9334
( b )
τmax [( . ) ] .= − + =715 60 93342 2 1 2 MPa
σ ' .= −715 MPa It may be seen from a sketch of Mohr’s Circle that θp
o" tan .= =−1 2
o MPa= = =80 45 56 57cos .
τ xy o MPa= =80 45 56 57sin .
Apply Eqs. (1.11): σ x
o MPa' ( . . ) . sin .= + + + =1 2 56 57 56 57 0 56 57 60 105 6
σ y MPa' . . .= − =56 57 48 99 7 58
τ x y o MPa' ' . cos .= − + =0 56 57 60 28 29
SOLUTION (1.10) We have σ x = 0, σ y MPa= −50 , τxy MPa= 100 , θ = 70o
Apply Eqs. (1.11): σ x
o o ' cos sin= − + +50
2 50 2 140 100 140
= 2013. MPa σ y MPa' . . .= − + − = −25 1915 64 28 7013
τ x y o o
' ' sin cos= − +25 140 100 140
= −92 67. MPa