Solution Manual for Advanced Mechanics of Materials and Applied Elasticity, 6th Edition by TBtutors

Type:

Solution Manual

Resource:

Advanced Mechanics of Materials and Applied Elasticity

Edition:

6th Edition

Author(s):

Ansel Ugural Saul Fenster

CONTENTS

Chapter 1

Analysis of Stress

Chapter 2

Strain and Material Properties

Chapter 3

Problem in Elasticity

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

NOTES TO THE INSTRUCTOR

The Solutions Manual to accompany the text Advanced Mechanics of Materials and Applied Elasticity 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 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 its 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

iii

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 = 50 : o

σ x ' = 700(10 3 ) = σ x cos 2 50o = 0.413σ x = 110.18P P = 6.35 kN

or and

3 ) σ x sin 50o = cos 50o 0.492 τ x ' y ' 560(10 σ x 131.2 P = = =

Solving

P = 4.27 kN = Pall

______________________________________________________________________________________ SOLUTION (1.2) Normal stress is

σ x=

125(103 ) 0.05×0.05

= 50 MPa

P A

( a ) Equations (1.11), with θ = 20 : o

cos 2 20o 44.15 MPa = σ x ' 50 =

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

( b ) Equations (1.11), with θ = 45 : o

= 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 y’ 25 MPa

25 MPa x’ 45 o x

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.3) From Eq. (1.11a),

σ x = cosσ θ = cos−7530 = −100 MPa x' 2

For θ = 50 , Eqs. (1.11) give then o

σ x' = −100 cos 2 50o = −41.32 MPa

τ x ' y ' = −( −100) sin 50o cos 50o = 49.24 MPa o Similarly, for θ = 140 : σ x' = −100 cos 2 140o = −58.68 MPa τ 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 σ y = 0 and τ xy = 0 , become or

σ x ' = 12 σ x + 12 σ x cos 2θ

and

τ x ' y ' = 12 σ x sin 2θ

20 = 2PA (1 + cos 2θ )

and

10 = 2PA 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

Thus,

θ = 26.56o

= 20

gives

P 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 π A 2 (50) 4 P τ max = = 38.2 MPa 2A

σ1 =

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.6) Shaded transverse area:

= A 2= at 2(10)(75) = 1.5(103 ) mm 2 Metal is capable of supporting the load

P =σ A =90(106 )(1.5 ×10−3 ) =135 kN

Apply Eqs. (1.11):

P (cos 2 55o ) , P = 114 kN −3 1.5(10 ) P τ x ' y ' = 12(106 ) = − sin 55o cos 55o , P = 38.3 kN −3 1.5(10 )

6 ) σ x ' 25(10 = =

Thus,

Pall = 38.3 kN

______________________________________________________________________________________ SOLUTION (1.7) Use Eqs. (1.11):

P (cos 2 40o ) , P = 51.1 kN −3 1.5(10 ) P sin 40o cos 40o , P = 24.4 kN τ x ' y ' = 8(106 ) = − 1.5(10−3 )

6 = σ x ' 20(10 = )

Thus,

Pall = 24.4 kN

______________________________________________________________________________________ SOLUTION (1.8)

A =15 × 30 = 450 mm 2 Apply Eqs. (1.11):

120(103 ) (cos 2 40o ) 156 MPa = 450 ×10−6 120(103 ) sin 40o cos 40o = τ x' y' = − −131 MPa 450 ×10−6

= σ x'

______________________________________________________________________________________ SOLUTION (1.9) We have A = 450(10

−6

) m2 .

Use Eqs. (1.11):

−100(10 ) (cos 2 60o ) = −55.6 MPa −6 450 ×10 −100(103 ) − sin 60o cos 60o = 96.2 MPa τ x' y' = −6 450 ×10

σ x' =

______________________________________________________________________________________ 3 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 (1.10)

θ = 40o + 90o = 130o 150(10 ) P σx = − π (0.08 = −31.83 MPa A = − 0.07 ) 3

Equations (1.11):

σ x' = −31.83cos 2 130o = −13.15 MPa

τ x ' y ' = 31.83sin130o cos130o = −15.67 MPa x’

13.15 MPa

130 o

15.67 MPa

x Plane of weld y’

______________________________________________________________________________________ SOLUTION (1.11) Use Eqs. (1.14),

( 2 x ) + ( −2 xy ) + ( x ) + Fx = 0

( − y 2 ) + ( −2 yz + x ) + (0) + Fy = 0

( z − 4 xy ) + (0) + ( −2 z ) + Fz = 0 3

Solving, we have (in MN m ):

Fx = −3x + 2 xy

Fy =− x + y 2 + 2 yz

Fz = 4 xy + z (a)

Substituting x=-0.01 m, y=0.03 m, and z=0.06 m, Eqs. (a) yield the following values

= Fx 29.4 = Fy 14.5 = Fz 58.8 kN m3 kN m3 kN m3 Resultant body force is thus

Fx2 + Fy2 + Fz2 = 67.32 kN m3

______________________________________________________________________________________ SOLUTION (1.12) Equations (1.14):

− 2c1 y − 2c1 y + 0 + 0 = 0, 0 + c3 z + 0 + 0 = 0, 0+0+0+0=0

4c1 y ≠ 0 c3 z ≠ 0

No. Eqs. (1.14) are not satisfied. ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.13) ( a ) No. Eqs. (1.14) are not satisfied. ( b ) Yes. Eqs. (1.14) are satisfied. ______________________________________________________________________________________ SOLUTION (1.14) Eqs. (1.14) for the given stress field yield:

Fx = Fy = Fz = 0

______________________________________________________________________________________ SOLUTION (1.15) y

τ x ' y ' ∆A σ x ' ∆A

y’ 40 ∆A cos20o

50 ∆A cos20o

20o

x’

x 50 ∆A sin20o 60 ∆A sin20o

= ∑ Fx ' 0 :

σ x ' ∆A + 40 cos 2 20o − 60∆A sin 2 20o

−2(50∆A sin 20o cos 20o ) = 0 σ x' = −35.32 + 7.02 + 32.14 = 3.8 MPa F 0: τ ∑= y'

x' y'

∆A − 40∆A sin 20o cos 20o

−60∆A sin 20o cos 20o − 50∆A cos 2 20o +50∆A sin 2 20o = 0 τ x ' y ' = 12.86 + 19.28 + 44.15 − 5.85 = 70.4 MPa ______________________________________________________________________________________ SOLUTION (1.16) 50 ∆A cos25o y’

15 ∆A cos25o 15 ∆A sin25o

25o

90 ∆A sin25o

σ x ' ∆A x’

F 0 : σ ∆A + 50∆A cos 25 ∑= x'

τ x ' y ' ∆A

−90∆A sin 2 25o − 2(15∆A sin 25o cos 25o ) = 0 σ x' = −41.07 + 16.07 + 11.49 = −13.5 MPa (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.16 (CONT.)

F 0: τ ∑= y'

x' y'

∆A − 50∆A sin 25o cos 25o

0 −90∆A sin 25o cos 25o − 15∆A cos 2 25o +15∆A sin 2 25o = τ x ' y ' = 19.15 + 34.47 + 12.32 − 2.68 = 63.3 MPa ______________________________________________________________________________________ SOLUTION (1.17) y

τ x' y'

y’ 50 MPa 60 MPa

σ x' 20

x’ x

θ =20o

80 MPa

1 1 (−40 + 60) + (−40 − 60) cos 40o + 50sin 40o 2 2 = 10 − 38.3 + 32.1 = 3.8 MPa 1 τ x' y' = − (−40 − 60) sin 40o + 50 cos 40o 2 = 32.14 + 38.3 = 70.4 MPa

σ x' =

______________________________________________________________________________________ SOLUTION (1.18)

σ x'

τ x' y'

x’

θ y’

90 MPa 25

15 MPa

θ =115o

50 MPa

1 1 (90 − 50) + (90 + 50) cos 230o − 15sin 230o 2 2 = 20 − 45 + 11.5 = −13.5 MPa 1 τ x' y' = − (90 + 50) sin 230o − 15cos 230o 2 = 53.62 + 9.64 = 63.3 MPa

σ x '=

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.19) Transform from θ = 40 to θ = 0 . For convenience in computations, Let o

σx = −160 MPa, σy = −80 MPa, τ xy = 40 MPa and θ = −40o

Then

1 1 (σ x + σ y ) + (σ x − σ y ) cos 2θ + τ xy sin 2θ 2 2 1 1 = (−160 − 80) + (−160 + 80) cos(−80o ) + 40sin(−80o ) 2 2 = −166.3 MPa

σ x'=

1 2 1 = − (−160 + 80) sin(−80o ) + 40 cos(−80o ) 2 = −32.4 MPa

τ x' y' = − (σ x − σ y ) sin 2θ + τ xy cos 2θ

σ y' = σ x + σ y − σ x' = −160 − 80 + 166.3 = −73.7 MPa

For θ = 0 : o

73.7 MPa 32.4 MPa 166.3 MPa x

______________________________________________________________________________________ SOLUTION (1.20) −1 4 53.1o = θ tan = 3 45 + 90 45 − 90 cos106.2o = + σ x' 2 2 = 67.5 + 6.28 = 73.8 MPa 45 − 90 sin106.2o = 21.6 MPa − τ x' y' = 2

21.6 MPa y’

73.8 MPa x’

53.1o

x ______________________________________________________________________________________ 7 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 (1.21)

= τ xy 0= θ 70o (a)

σ − 60 sin140o τ x' y' = −30 = −

(b)

σ= 80 = x'

σ = 153.3 MPa

σ + 60 σ − 60 2

cos140o

σ = 231 MPa

______________________________________________________________________________________ SOLUTION (1.22) Equations(1.18) with θ = 60 , σ x = 110 MPa , σ y = 0 , τ xy = 50 MPa give o

o o 1 1 σ x' = 70.8 MPa 2 (110) + 2 (110) cos120 + 50sin120 =

τ x' y' = − 12 (110) sin120o + 50 cos120o = −72.6 MPa o o 1 1 σ y' = 39.2 MPa 2 (110) − 2 (110) cos120 − 50sin120 =

______________________________________________________________________________________ SOLUTION (1.23) Equations(1.18) with θ = 30 , σ x = 110 MPa , σ y = 0 , τ xy = 50 MPa result in o

o o 1 σ x' = 125.8 MPa 2 (110) + 55cos 60 + 50sin 60 =

τ x' y' = − 12 (110) sin 60o + 50 cos 60o = −22.6 MPa o o 1 σ y' = −15.8 MPa 2 (110) − 55cos 60 − 50sin 60 =

______________________________________________________________________________________ SOLUTION (1.24) We have

θ = 25 + 90 = 115o σ x = −10 MPa σ y = 30 MPa

τ xy = 0 (a)

x’

σ x'

τ x' y'

10 MPa 25

y’

30 MPa

1 1 σ x ' = (σ x + σ y ) + (σ x − σ y ) cos 2θ 2 2 1 1 = (−10 + 30) + (−10 − 30) cos 230o = 22.86 MPa 2 2

Thus,

σ= σ= 22.86 MPa w x'

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.24 (CONT.) (b)

1 2 1 = − (−10 − 30) sin 230o = −15.32 MPa 2

τ x' y' = − (σ x − σ y ) sin 2θ

τw

τ w = τ x ' y ' = −15.32 MPa

______________________________________________________________________________________ SOLUTION (1.25) (a)

(b)

0 + 50 2 τ = 49 MPa

80 =+ ( σ1 =

τ= max

(

0 − 50 2 ) +τ 2 2

−50 2 2 = ) + 49 55 MPa 2

50 = 25 MPa 2 0 − 50 2θ s =tan −1[− ] =27 o 2(49) 50 τ x ' y ' = sin 27o + 49 cos 27o = 55 MPa 2

σ='

Thus,

θ s ' = 13.5o y’

25 MPa 25 MPa x’ 13.5o x

55 MPa

______________________________________________________________________________________ SOLUTION (1.26) 80 MPa 40 MPa

θ =-30o x (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.26 (CONT.)

40 + 80 40 − 80 + cos(−60o ) = 60 − 10 = 50 MPa 2 2 σ y = 60 + 10 = 70 MPa

σx =

40 − 80 sin(−60o ) = −17.32 MPa 2

− τ xy =

70 MPa

70 MPa 50 MPa

50 MPa

17.32 MPa

20 MPa

2.68 MPa

50 + 70 50 − 70 2 2 ± ( ) + 2.68= 60 ± 10.35 2 2 = σ 1 70.35 = MPa σ 2 49.65 MPa 2(2.68) 2θ p = tan −1[ ] = −15o 50 − 70

σ 1,2 =

50 − 70 cos(−15o ) + 2.68sin(−15o ) 2 =60 − 9.66 − 0.694 =49.65 MPa

σ x ' =60 + Thus,

θ p " = −7.5o

x 7.5o 49.65 MPa

70.35 MPa ______________________________________________________________________________________ SOLUTION (1.27) τ (MPa)

60 MPa

40 MPa

(60, 50)

50 MPa O α

R C

σ (MPa)

(-40, -50) 40o (CONT.) x’ ______________________________________________________________________________________ 10 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.

______________________________________________________________________________________ 1.27 (CONT.) −1 50 45o α tan = = 50 1 2 R =(50 + 502 ) 2 =70.7

85o (70.7) 70.4 MPa τ x ' y ' sin = = 10 − cos85o (70.7) = 3.84 MPa σ x' = ______________________________________________________________________________________ SOLUTION (1.28) τ (MPa)

50 MPa

15 MPa 90 MPa

σ’=20

(-50, -15) x

62.1o

y’ (90, 15) α σ (MPa)

(σx’, τ x’y’) x’

−1 15 12.1o = α tan = 70 1 R =(152 + 702 ) 2 =71.6

13.5 MPa

y’

= = τ x ' y ' 71.6sin 62.1o 63.3 MPa

63.3 MPa 90 MPa

25o

σ x' = −71.6 cos 62.1o + 20

15 MPa

x’

= −13.5 MPa

50 MPa

______________________________________________________________________________________ SOLUTION (1.29) τ (MPa)

σ ' =67.5

x’ 45 MPa

R =22.5 R

53.1

90 MPa

C 106.2o

90 σ (MPa)

73.8o x’ (σx’, −τ x’ y’ )

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.29 (CONT.)

= τ x ' y ' 22.5sin = 73.8o 21.6 MPa

σ x' = 67.5 + 22.5cos 73.8o = 73.8 MPa

Sketch of results is as shown in solution of Prob. 1.20. ______________________________________________________________________________________ SOLUTION (1.30) τ (MPa) (a) x’ 1 60 MPa 140o

x’ 70

= R σ

(σ − 60)

σ (MPa)

σ − 60 −30 = τ x' y' = sin(−40o );

σ= 153.3 MPa

(b)

σ x ' =80 =60 +

σ − 60

σ = 231 MPa

[1 − cos(−40o )]

______________________________________________________________________________________ SOLUTION (1.31) ( a ) From Mohr’s circle, Fig. (a):

σ1 = 121 MPa σ2 = −71 MPa τ max = 96 MPa θ p ' = −19.3o θ s ' = 25.7 o τ (MPa) τmax

A(100,60) σ2

2θ p '

σ (MPa)

σ1

B Figure (a) By applying Eq. (1.20):

1 2

50  22,500 + 3600  =± σ 1,2 = 25 96 2 ± 4

σ 1 = 121 MPa

Using Eq. (1.19):

σ 2 = −71 MPa

tan 2θ p = − 12 15 = −0.8

θ p ' = −19.3o

θ s ' = 25.7 o

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.31 (CONT.) ( b ) From Mohr’s circle, Fig. (b):

σ1 = 200 MPa σ2 = τ max = 125 MPa −50 MPa

θ p ' = 26.55o

θ s ' = 71.55o

τ (MPa)

σ2

τmax

2θp’

σ (MPa)

σ1

A(150,-100) Figure (b) Through the use of Eq. (1.20),

σ 1, 2 = 75 ± [22,4500 + 10,000] = 75 ± 125 1 2

σ 1 = 200 MPa σ 2 = −50 MPa Using Eq. (1.19), tan 2θ p = 4 3 :

o = θ p ' 26.57 = θ s ' 71.57o

______________________________________________________________________________________ SOLUTION (1.32) Referring to Mohr’s circle, Fig. 1.15:

σ x ' = σ +2σ + σ −2σ cos 2θ 1

(a)

σ y ' = σ +2σ − σ −2σ cos 2θ 1

τ x ' y ' = σ −2σ sin 2θ 1

(b)

From Eqs. (a),

σ x' + σ y' = σ 1 + σ 2 2θ + sin 2 2θ = 1 , and Eqs. (a) and (b), we have σ x ' ⋅ σ y ' − τ x2' y ' = σ 1 ⋅ σ 2 = const .

By using cos

______________________________________________________________________________________ SOLUTION (1.33) We have

2τ

tan 2θ p = σ x −xyσ y = 502(−(−−70) 190) = −0.583

2θ p = −30.24o

and θ p = −15.12

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.33 (CONT.) Equations (1.18):

σ x ' = 50−2190 + 50+2190 cos( −30.26o ) − 70 sin( −30.26o ) = −70 + 103.65 + 35.275 = σ1 68.93 MPa = σ y' = σ x + σ y − σ x' = −208.9 MPa = σ2 208.9 MPa 15.13o 68.92 MPa

______________________________________________________________________________________ SOLUTION (1.34)

τ max = (

σ x −σ y 2

) + τ xy2

Substituting the given values

140 2 = ( 60+2100 ) + τ xy2 2

τ xy ,max = 114.89 MPa

______________________________________________________________________________________ SOLUTION (1.35) Transform from θ = 60 to θ = 0 with σ x ' = −20 MPa, σ y ' = 60 MPa , o

τ x ' y ' = −22 MPa , and θ = −60o . Use Eqs. (1.18):

σ x = −202+ 60 + −202−60 cos 2(−60o ) − 22sin 2(−60o ) = 59 MPa σy = σ x' + σ y' − σ x = −19 MPa τ xy = −23.6 MPa y

19 MPa 23.6 MPa 59 MPa x

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.36) σy

σycos60o

14 MPa τxy

τxycos60o σxsin60o

30 MPa Area =1

30 MPa

τxysin60o Figure (b)

Figure (a) ( a ) Figure (a):

= σ y 14sin = 60o 12.12 MPa

= = τ xy 14 cos 60o 7 MPa Figure (b):

∑ F = 12.12 cos 60 − τ sin 60 = 0 o

τ xy = 7 MPa (as before)

∑ F = −σ sin 60 + 30 + 7 cos 60 = 0 x

σ x = 38.68 MPa

( b ) Equation (1.20) is therefore:

σ 1, 2 = 38.68+212.12 ± [( 38.682−12.12 ) 2 + 7 2 ]

1 2

or = σ 1 40.41 = σ 2 10.39 MPa MPa, Also,

θ p = 12 tan −1 38.682 (−712) .12 = 13.9 o

Note: Eq. (1.18a) gives, σ x ' = 40.41 MPa . Thus,

θ p ' = 13.9 o 10.39 MPa 40.41 MPa x’ θp’

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.37) σy τxy Figure (a)

σx

Figure (a):

= σ x 100 = cos 45o 70.7 MPa

45o 70.7 MPa = σ y 100sin = = τ xy 100 = cos 45o 70.7 MPa Now, Eqs. (1.18) give (Fig. b):

o 70.7 + 0 + 70.7 sin 240= 9.47 MPa σ= x'

τ x ' y ' =−0 + 70.7 cos 240o =−35.35 MPa o σ= 70.7 − 0 − 70.7 sin 240= 131.9 MPa y'

σy

σx’ τxy

x’ σx y’

τx’y’

Figure (b)

n 30o

σy’

______________________________________________________________________________________ SOLUTION (1.38)

−70sin 30o = −35 MPa σy =

cos 30o 60.6 MPa = τ xy 70 = ( a ) Figure (a):

∑ F = −150 + 0.5σ + 60.6(0.866) = 0 x

σ x = 195 MPa

35cos30o 150 MPa

60.6cos30o

30o Area=1

60.6sin30o σxsin30

Figure (a)

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.38 (CONT.) ( b ) Equation (1.20): 195 −35 σ 1,2 = ± ( 1952+35 ) 2 + 60.62  2

or Also,

σ 1 = 210 MPa

1 2

σ 2 = −50 MPa

2 ( 60.6 ) o θ p = 12 tan −1 195 + 35 = 13.89

Equation (1.18a): Thus,

σ x' = 80 + 115cos 2(13.89o ) + 60.6sin 2(13.89o ) = 210 MPa

θ p ' = 13.89 o

50 MPa 210 MPa x’ θp’

______________________________________________________________________________________ SOLUTION (1.39) For pure shear, σ 1 = −σ 2 : pr t

from which

σ 1 = pr t

= − 2prt + 2πPrt

σ 2 = 2prt − 2πPrt

P = 3πpr 2

______________________________________________________________________________________ SOLUTION (1.40) σθ y Table D.4:

σ +σa

A = 2πrt J = 2πr 3t

τ xy

Figure (a)

Stresses are (Fig. a):

σ=

σ= a

−P A

pr 2t

−30(103 )π 2π (0.12)(0.005) 6

− 25 MPa

4(10 )120 2(5)

= 48 MPa

= σ θ 2= σ a 96 MPa 3

−10π (10 ) τ xy = −JTr = 2π (0.12 = −69.4 MPa )(0.005) 2

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.40 (CONT.) Hence,

σ x = 48 − 25 = 23 MPa

σ y = 96 MPa

Therefore, we have

1 2

τ max = ± ( 23−296 ) 2 + 69.42  = ±78.4 MPa Also

σ '=

and

1 2

y’

(23 + 96) = 59.5 MPa

−96 − 12 tan −1 2(23−69.4) = −13.87o θs =

θs ' '

59.4 MPa

78.4 MPa

Equation (1.18b) with θ s = −13.87 : o

−16.99 − 61.42 = −78.4 MPa τ x' y' = Thus,

x x’

Figure (b)

θ s ' ' = 13.87o

______________________________________________________________________________________ SOLUTION (1.41)

= A 2= π rt 2π (60)(4) = 1508 mm 2 3 J 2π r 3t 2π (60)= (4) 5.429 ×106 mm 4 = =

σ= y

pr t

5(60) 4

= 75 MPa

σy

( 0.05 ) τ xy = − TrJ = − 5600 .429 (10 ) −6

= −5.526 MPa P σ x = 2prt + PA = 37.5 + 1508 ( P in newtons )

σx τ xy

Thus

σ1 =

σ x +σ y 2

σ −σ

+ ( x 2 y ) 2 + τ xy2

Substituting the numerical values gives

[

]

80 = 56.3 + 331.6 × 10 −6 P + ( −18.75 + 331.6 × 10 −6 P ) 2 + ( −5.526) 2 2 Solving,

P = 64.01 kN

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.42) At point A, T = 8 kN and P = 400 kN

P A

− = − σx =

τ xy

4(400 ×103 ) = −50.9 MPa π (0.1) 2

16T 16(8 ×10 ) = = 40.7 MPa τ xy = πd3 π (0.1)3

σx

Hence

= τ max

(

σx 2 2

) + τ xy2

= (−25.45) 2 + (40.7) 2 = 48 = MPa R

σ avg =

τ y

σx

= −25.45 MPa

tan 2θ p " =

40.7 , 25.45

σ 2 2θp"

θ p " = 29o

O σavg

σ1

______________________________________________________________________________________ SOLUTION (1.43)

θ =α + 90 =50 + 90 =140o 120 ×103 = 95.5 MPa π P (0.04) 2 σx = θ 4 A x 16(1.5 ×10−3 ) 16T = = 119.4 MPa τ τ xy = 3 π (0.04)3 πd = σx

y' Equations(1.18):

95.5 2

cos 280o + 119.4sin 280o = + −61.5 MPa σw = σ x' = and

95.5 2

sin 280o − 119.4 cos 280o = 47.02 − 20.73 = 26.3 MPa τw = τ x' y' = − 61.5 MPa

x’

140o x

26.3 MPa

y’ ______________________________________________________________________________________ 19 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 (1.44)

A = π4 (1802 − 1202 ) = 14.14 ×103 mm 2 π J = 32 (1804 − 1204 ) = 82.70 ×106 mm 4 700 σ x =− PA =− 14.14 =−49.5 MPa,

τ xy=

Tr J

20(90)

σ y =0

= 21.77 MPa

82.70×10−6

Equation (1.20) is therefore 2 49.5 2 σ max = σ1 = − 49.5 2 + ( 2 ) + (21.76)

= 8.205 MPa ______________________________________________________________________________________ SOLUTION (1.45)

P = τ 0 Lt M = τ 0 Lth A = 2ht I = 121 t ( 2h ) 3 = 23 th 3

τ 0 Lt A

P M

τ L

Axial stress: σ a = PA = 20h

Bending stress: σ b = Mc I = Point A

3τ 0 L 2h

τ0

σ a + σ b = τ2 hL

From Eqs. (1.20) and (1.22), we obtain

σ 1, 2 = τ hL ± τ 0 ( Lh ) 2 + 1 0

τ max = τ 0 ( Lh ) 2 + 1 Point B

σ b − σ a = τ hL

B Hence

σ 1 = τ hL 0

τ max = 2τ h

σ2 = 0

______________________________________________________________________________________