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Transcript of FIRST PAGES Chapter 5 - MechFamily | HU desing 1/solution manual/budynFIRST PAGES 116 Solutions...

• FIRST PAGES

Chapter 5

5-1

MSS: 1 3 = Sy/n n = Sy1 3

DE: n = Sy

= ( 2A AB + 2B)1/2 = ( 2x xy + 2y + 3 2xy)1/2(a) MSS: 1 = 12, 2 = 6, 3 = 0 kpsi

n = 5012

= 4.17 Ans.

DE: = (122 6(12) + 62)1/2 = 10.39 kpsi, n = 5010.39

= 4.81 Ans.

(b) A, B = 122 (

12

2

)2+ (8)2 = 16, 4 kpsi

1 = 16, 2 = 0, 3 = 4 kpsi

MSS: n = 5016 (4) = 2.5 Ans.

DE: = (122 + 3(82))1/2 = 18.33 kpsi, n = 5018.33

= 2.73 Ans.

(c) A, B = 6 102 (6 + 10

2

)2+ (5)2 = 2.615, 13.385 kpsi

1 = 0, 2 = 2.615, 3 = 13.385 kpsi

MSS: n = 500 (13.385) = 3.74 Ans.

DE: = [(6)2 (6)(10) + (10)2 + 3(5)2]1/2= 12.29 kpsi

n = 5012.29

= 4.07 Ans.

(d) A, B = 12 + 42 (

12 42

)2+ 12 = 12.123, 3.877 kpsi

1 = 12.123, 2 = 3.877, 3 = 0 kpsi

MSS: n = 5012.123 0 = 4.12 Ans.

DE: = [122 12(4) + 42 + 3(12)]1/2 = 10.72 kpsi

n = 5010.72

= 4.66 Ans.

B

A

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• FIRST PAGES

116 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design

5-2 Sy = 50 kpsiMSS: 1 3 = Sy/n n = Sy

1 3DE:

( 2A AB + 2B

)1/2 = Sy/n n = Sy/ ( 2A AB + 2B)1/2(a) MSS: 1 = 12 kpsi, 3 = 0, n = 5012 0 = 4.17 Ans.

DE: n = 50[122 (12)(12) + 122]1/2 = 4.17 Ans.

(b) MSS: 1 = 12 kpsi, 3 = 0, n = 5012 = 4.17 Ans.

DE: n = 50[122 (12)(6) + 62]1/2 = 4.81 Ans.

(c) MSS: 1 = 12 kpsi, 3 = 12 kpsi , n = 5012 (12) = 2.08 Ans.

DE: n = 50[122 (12)(12) + (12)2]1/3 = 2.41 Ans.

(d) MSS: 1 = 0, 3 = 12 kpsi, n = 50(12) = 4.17 Ans.

DE: n = 50[(6)2 (6)(12) + (12)2]1/2 = 4.81

5-3 Sy = 390 MPaMSS: 1 3 = Sy/n n = Sy

1 3DE:

( 2A AB + 2B

)1/2 = Sy/n n = Sy/ ( 2A AB + 2B)1/2(a) MSS: 1 = 180 MPa, 3 = 0, n = 390180 = 2.17 Ans.

DE: n = 390[1802 180(100) + 1002]1/2 = 2.50 Ans.

(b) A, B = 1802 (

180

2

)2+ 1002 = 224.5, 44.5 MPa = 1, 3

MSS: n = 390224.5 (44.5) = 1.45 Ans.

DE: n = 390[1802 + 3(1002)]1/2 = 1.56 Ans.

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• FIRST PAGES

Chapter 5 117

(c) A, B = 1602 (

1602

)2+ 1002 = 48.06, 208.06 MPa = 1, 3

MSS: n = 39048.06 (208.06) = 1.52 Ans.

DE: n = 390[1602 + 3(1002)]1/2 = 1.65 Ans.

(d) A, B = 150, 150 MPa = 1, 3MSS: n = 390

150 (150) = 1.30 Ans.

DE: n = 390[3(150)2]1/2

= 1.50 Ans.

5-4 Sy = 220 MPa(a) 1 = 100, 2 = 80, 3 = 0 MPa

MSS: n = 220100 0 = 2.20 Ans.

DET: = [1002 100(80) + 802]1/2 = 91.65 MPan = 220

91.65= 2.40 Ans.

(b) 1 = 100, 2 = 10, 3 = 0 MPa

MSS: n = 220100

= 2.20 Ans.

DET: = [1002 100(10) + 102]1/2 = 95.39 MPa

n = 22095.39

= 2.31 Ans.

(c) 1 = 100, 2 = 0, 3 = 80 MPa

MSS: n = 220100 (80) = 1.22 Ans.

DE: = [1002 100(80) + (80)2]1/2 = 156.2 MPan = 220

156.2= 1.41 Ans.

(d) 1 = 0, 2 = 80, 3 = 100 MPaMSS: n = 220

0 (100) = 2.20 Ans.DE: = [(80)2 (80)(100) + (100)2] = 91.65 MPa

n = 22091.65

= 2.40 Ans.

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118 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design

5-5

(a) MSS: n = O BO A

= 2.231.08

= 2.1

DE: n = OCO A

= 2.561.08

= 2.4

(b) MSS: n = O EO D

= 1.651.10

= 1.5

DE: n = O FO D

= 1.81.1

= 1.6

(c) MSS: n = O HOG

= 1.681.05

= 1.6

DE: n = O IOG

= 1.851.05

= 1.8

(d) MSS: n = O KO J

= 1.381.05

= 1.3

DE: n = O LO J

= 1.621.05

= 1.5

O

(a)

(b)

(d)

(c)

H

I

G

J

KL

FED

A

BC

Scale1" 200 MPa

B

A

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• FIRST PAGES

Chapter 5 119

5-6 Sy = 220 MPa

(a) MSS: n = O BO A

= 2.821.3

= 2.2

DE: n = OCO A

= 3.11.3

= 2.4

(b) MSS: n = O EO D

= 2.21

= 2.2

DE: n = O FO D

= 2.331

= 2.3

(c) MSS: n = O HOG

= 1.551.3

= 1.2

DE: n = O IOG

= 1.81.3

= 1.4

(d) MSS: n = O KO J

= 2.821.3

= 2.2

DE: n = O LO J

= 3.11.3

= 2.4

B

AO

(a)

(b)

(c)

(d)

H

G

J

K

L

I

FE

D

A

BC

1" 100 MPa

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120 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design

5-7 Sut = 30 kpsi, Suc = 100 kpsi; A = 20 kpsi, B = 6 kpsi

(a) MNS: Eq. (5-30a) n = Sutx

= 3020

= 1.5 Ans.

BCM: Eq. (5-31a) n = 3020

= 1.5 Ans.

MM: Eq. (5-32a) n = 3020

= 1.5 Ans.(b) x = 12 kpsi,xy = 8 kpsi

A, B = 122 (

12

2

)2+ (8)2 = 16, 4 kpsi

MNS: Eq. (5-30a) n = 3016

= 1.88 Ans.

BCM: Eq. (5-31b)1

n= 16

30 (4)

100 n = 1.74 Ans.

MM: Eq. (5-32a) n = 3016

= 1.88 Ans.

(c) x = 6 kpsi, y = 10 kpsi,xy = 5 kpsi

A, B = 6 102 (6 + 10

2

)2+ (5)2 = 2.61, 13.39 kpsi

MNS: Eq. (5-30b) n = 10013.39 = 7.47 Ans.

BCM: Eq. (5-31c) n = 10013.39 = 7.47 Ans.

MM: Eq. (5-32c) n = 10013.39 = 7.47 Ans.(d) x = 12 kpsi,xy = 8 kpsi

A, B = 122 (

122

)2+ 82 = 4, 16 kpsi

MNS: Eq. (5-30b) n = 10016 = 6.25 Ans.

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Chapter 5 121

BCM: Eq. (5-31b)1

n= 4

30 (16)

100 n = 3.41 Ans.

MM: Eq. (5-32b)1

n= (100 30)4

100(30) 16

100 n = 3.95 Ans.

(c)

L

(d)

J

(b)

(a)

H

G

K

F

O

C

DE

AB1" 20 kpsi

B

A

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122 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design

5-8 See Prob. 5-7 for plot.

(a) For all methods: n = O BO A

= 1.551.03

= 1.5

(b) BCM: n = O DOC

= 1.40.8

= 1.75

All other methods: n = O EOC

= 1.550.8

= 1.9

(c) For all methods: n = O LO K

= 5.20.68

= 7.6

(d) MNS: n = O JO F

= 5.120.82

= 6.2

BCM: n = OGO F

= 2.850.82

= 3.5

MM: n = O HO F

= 3.30.82

= 4.0

5-9 Given: Sy = 42 kpsi, Sut = 66.2 kpsi, f = 0.90. Since f > 0.05, the material is ductile andthus we may follow convention by setting Syc = Syt .

Use DE theory for analytical solution. For , use Eq. (5-13) or (5-15) for plane stress andEq. (5-12) or (5-14) for general 3-D.

(a) = [92 9(5) + (5)2]1/2 = 12.29 kpsi

n = 4212.29

= 3.42 Ans.

(b) = [122 + 3(32)]1/2 = 13.08 kpsi

n = 4213.08

= 3.21 Ans.

(c) = [(4)2 (4)(9) + (9)2 + 3(52)]1/2 = 11.66 kpsi

n = 4211.66

= 3.60 Ans.

(d) = [112 (11)(4) + 42 + 3(12)]1/2 = 9.798

n = 429.798

= 4.29 Ans.

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• FIRST PAGES

Chapter 5 123

For graphical solution, plot load lines on DE envelope as shown.

(a) A = 9, B = 5 kpsin = O B

O A= 3.5

1= 3.5 Ans.

(b) A, B = 122 (

12

2

)2+ 32 = 12.7, 0.708 kpsi

n = O DOC

= 4.21.3

= 3.23

(c) A, B = 4 92 (

4 92

)2+ 52 = 0.910, 12.09 kpsi

n = O FO E

= 4.51.25

= 3.6 Ans.

(d) A, B = 11 + 42 (

11 42

)2+ 12 = 11.14, 3.86 kpsi

n = O HOG

= 5.01.15

= 4.35 Ans.

5-10 This heat-treated steel exhibits Syt = 235 kpsi, Syc = 275 kpsi and f = 0.06. The steel isductile ( f > 0.05) but of unequal yield strengths. The Ductile Coulomb-Mohr hypothesis(DCM) of Fig. 5-19 applies confine its use to first and fourth quadrants.

(c)

(a)

(b)

(d)

E

C

G

H

D

B

A

O

F

1 cm 10 kpsi

B

A

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124 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design

(a) x = 90 kpsi, y = 50 kpsi, z = 0 A = 90 kpsi and B = 50 kpsi. For thefourth quadrant, from Eq. (5-31b)

n = 1(A/Syt ) (B/Suc) =

1

(90/235) (50/275) = 1.77 Ans.

(b) x = 120 kpsi, xy = 30 kpsi ccw. A, B = 127.1, 7.08 kpsi. For the fourthquadrant

n = 1(127.1/235) (7.08/275) = 1.76 Ans.

(c) x = 40 kpsi, y = 90 kpsi, xy = 50 kpsi . A, B = 9.10, 120.9 kpsi.Although no solution exists for the third quadrant, use

n = Sycy

= 275120.9 = 2.27 Ans.

(d) x = 110 kpsi, y = 40 kpsi, xy = 10 kpsi cw. A, B = 111.4, 38.6 kpsi. For thefirst quadrant

n = SytA

= 235111.4

= 2.11 Ans.Graphical Solution:

(a) n = O BO A

= 1.821.02

= 1.78

(b) n = O DOC

= 2.241.28

= 1.75

(c) n = O FO E

= 2.751.24

= 2.22

(d) n = O HOG

= 2.461.18

= 2.08

O

(d)

(b)

(a)

(c)

E

F

B

D

G

C

A

H

1 in 100 kpsi

B

A

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Chapter 5 125

5-11 The material is brittle and exhibits unequal tensile and compressive strengths. Decision:Use the Modified Mohr theory.

Sut = 22 kpsi, Suc = 83 kpsi(a) x = 9 kpsi, y = 5 kpsi. A, B = 9, 5 kpsi. For the fourth quadrant,

|BA