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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

    budynas_SM_ch05.qxd 11/29/2006 15:00 Page 115

  • 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.

    budynas_SM_ch05.qxd 11/29/2006 15:00 Page 116

  • 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.

    budynas_SM_ch05.qxd 11/29/2006 15:00 Page 117

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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

    budynas_SM_ch05.qxd 11/29/2006 15:00 Page 118

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    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

    budynas_SM_ch05.qxd 11/29/2006 15:00 Page 119

  • FIRST PAGES

    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.

    budynas_SM_ch05.qxd 11/29/2006 15:00 Page 120

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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

    budynas_SM_ch05.qxd 11/29/2006 15:00 Page 121

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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.

    budynas_SM_ch05.qxd 11/29/2006 15:00 Page 122

  • 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

    budynas_SM_ch05.qxd 11/29/2006 15:00 Page 123

  • FIRST PAGES

    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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  • FIRST PAGES

    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