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Chapter 66-1MSS: 1 3 = Sy/n n = Sy1 3DE: n = Sy

=

2A AB + 2B

1/2=

2x xy + 2y +32xy

1/2(a) MSS: 1 = 12, 2 = 6, 3 = 0 kpsin =5012 = 4.17 Ans.DE:

= (1226(12) +62)1/2= 10.39 kpsi, n =5010.39 = 4.81 Ans.(b) A, B =122

122

2+(8)2= 16, 4 kpsi1 = 16, 2 = 0, 3 = 4 kpsiMSS: 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 +102

2+(5)2= 2.615, 13.385 kpsi1 = 0, 2 = 2.615, 3 = 13.385 kpsiMSS: n =500 (13.385) = 3.74 Ans.DE:

= [(6)2(6)(10) +(10)2+3(5)2]1/2= 12.29 kpsin =5012.29 = 4.07 Ans.(d) A, B =12 +42

12 42

2+12= 12.123, 3.877 kpsi1 = 12.123, 2 = 3.877, 3 = 0 kpsiMSS: n =5012.123 0 = 4.12 Ans.DE:

= [12212(4) +42+3(12)]1/2= 10.72 kpsin =5010.72 = 4.66 Ans.BAshi20396_ch06.qxd 8/18/03 12:22 PM Page 149150 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design6-2 Sy = 50 kpsiMSS: 1 3 = Sy/n n = Sy1 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 = 4.816-3 Sy = 390 MPaMSS: 1 3 = Sy/n n = Sy1 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[1802180(100) +1002]1/2 = 2.50 Ans.(b) A, B =1802

1802

2+1002= 224.5, 44.5 MPa = 1, 3MSS: n =390224.5 (44.5) = 1.45 Ans.DE: n =390[1802+3(1002)]1/2 = 1.56 Ans.shi20396_ch06.qxd 8/18/03 12:22 PM Page 150Chapter 6 151(c) A, B = 1602

1602

2+1002= 48.06, 208.06 MPa = 1, 3MSS: 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 =380150 (150) = 1.27 Ans.DE: n =390[3(150)2]1/2 = 1.50 Ans.6-4 Sy = 220 MPa(a) 1 = 100, 2 = 80, 3 = 0 MPaMSS: n =220100 0 = 2.20 Ans.DET:

= [1002100(80) +802]1/2= 91.65 MPan =22091.65 = 2.40 Ans.(b) 1 = 100, 2 = 10, 3 = 0 MPaMSS: n =220100 = 2.20 Ans.DET:

= [1002100(10) +102]1/2= 95.39 MPan =22095.39 = 2.31 Ans.(c) 1 = 100, 2 = 0, 3 = 80 MPaMSS: n =220100 (80) = 1.22 Ans.DE:

= [1002100(80) +(80)2]1/2= 156.2 MPan =220156.2 = 1.41 Ans.(d) 1 = 0, 2 = 80, 3 = 100 MPaMSS: n =2200 (100) = 2.20 Ans.DE:

= [(80)2(80)(100) +(100)2] = 91.65 MPan =22091.65 = 2.40 Ans.shi20396_ch06.qxd 8/18/03 12:22 PM Page 151152 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design6-5(a) MSS: n = OBOA =2.231.08 = 2.1DE: n = OCOA =2.561.08 = 2.4(b) MSS: n = OEOD =1.651.10 = 1.5DE: n = OFOD =1.81.1 = 1.6(c) MSS: n = OHOG =1.681.05 = 1.6DE: n = OIOG =1.851.05 = 1.8(d) MSS: n = OKOJ =1.381.05 = 1.3DE: n = OLOJ =1.621.05 = 1.5O(a)(b)(d)(c)HIGJKLFEDABCScale1" 200 MPaBAshi20396_ch06.qxd 8/18/03 12:22 PM Page 152Chapter 6 1536-6 Sy = 220 MPa(a) MSS: n = OBOA =2.821.3 = 2.2DE: n = OCOA =3.11.3 = 2.4(b) MSS: n = OEOD =2.21 = 2.2DE: n = OFOD =2.331 = 2.3(c) MSS: n = OHOG =1.551.3 = 1.2DE: n = OIOG =1.81.3 = 1.4(d) MSS: n = OKOJ =2.821.3 = 2.2DE: n = OLOJ =3.11.3 = 2.4BAO(a)(b)(c)(d)HGJKLIFEDABC1" 100 MPashi20396_ch06.qxd 8/18/03 12:22 PM Page 153154 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design6-7 Sut = 30 kpsi, Suc = 100 kpsi; A = 20 kpsi, B = 6 kpsi(a) MNS: Eq. (6-30a) n = Sutx=3020 = 1.5 Ans.BCM: Eq. (6-31a) n =3020 = 1.5 Ans.M1M: Eq. (6-32a) n =3020 = 1.5 Ans.M2M: Eq. (6-33a) n =3020 = 1.5 Ans.(b) x = 12 kpsi, xy = 8 kpsiA, B =122

122

2+(8)2= 16, 4 kpsiMNS: Eq. (6-30a) n =3016 = 1.88 Ans.BCM: Eq. (6-31b)1n =1630 (4)100 n = 1.74 Ans.M1M: Eq. (6-32a) n =3016 = 1.88 Ans.M2M: Eq. (6-33a) n =3016 = 1.88 Ans.(c) x = 6 kpsi, y = 10 kpsi, xy = 5 kpsiA, B = 6 102

6 +102

2+(5)2= 2.61, 13.39 kpsiMNS: Eq. (6-30b) n = 10013.39 = 7.47 Ans.BCM: Eq. (6-31c) n = 10013.39 = 7.47 Ans.M1M: Eq. (6-32c) n = 10013.39 = 7.47 Ans.M2M: Eq. (6-33c) n = 10013.39 = 7.47 Ans.(d) x = 12 kpsi, xy = 8 kpsiA, B = 122

122

2+82= 4, 16 kpsiMNS: Eq. (6-30b) n = 10016 = 6.25 Ans.shi20396_ch06.qxd 8/18/03 12:22 PM Page 154Chapter 6 155BCM: Eq. (6-31b)1n =430 (16)100 n = 3.41 Ans.M1M: Eq. (6-32b)1n =(100 30)4100(30) 16100 n = 3.95 Ans.M2M: Eq. (6-33b) n430 +n(16) +3030 1002= 1Reduces to n21.1979n 15.625 = 0n =1.1979 +

1.19792+4(15.625)2 = 4.60 Ans.(c)L(d)J(b)(a)IHGKFOCDEAB1" 20 kpsiBAshi20396_ch06.qxd 8/18/03 12:22 PM Page 155156 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design6-8 See Prob. 6-7 for plot.(a) For all methods: n = OBOA =1.551.03 = 1.5(b) BCM: n = ODOC =1.40.8 = 1.75All other methods: n = OEOC =1.550.8 = 1.9(c) For all methods: n = OLOK =5.20.68 = 7.6(d) MNS: n = OJOF =5.120.82 = 6.2BCM: n = OGOF =2.850.82 = 3.5M1M: n = OHOF =3.30.82 = 4.0M2M: n = OIOF =3.820.82 = 4.76-9 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. (6-13) or (6-15) for plane stress andEq. (6-12) or (6-14) for general 3-D.(a)

= [929(5) +(5)2]1/2= 12.29 kpsin =4212.29 = 3.42 Ans.(b)

= [122+3(32)]1/2= 13.08 kpsin =4213.08 = 3.21 Ans.(c)

= [(4)2(4)(9) +(9)2+3(52)]1/2= 11.66 kpsin =4211.66 = 3.60 Ans.(d)

= [112(11)(4) +42+3(12)]1/2= 9.798n =429.798 = 4.29 Ans.shi20396_ch06.qxd 8/18/03 12:22 PM Page 156Chapter 6 157For graphical solution, plot load lines on DE envelope as shown.(a) A = 9, B = 5 kpsin = OBOA =3.51 = 3.5 Ans.(b) A, B =122

122

2+32= 12.7, 0.708 kpsin = ODOC =4.21.3 = 3.23(c) A, B = 4 92

4 92

2+52= 0.910, 12.09 kpsin = OFOE =4.51.25 = 3.6 Ans.(d) A, B =11 +42

11 42

2+12= 11.14, 3.86 kpsin = OHOG =5.01.15 = 4.35 Ans.6-10 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. 6-27 applies conne its use to rst and fourth quadrants.(c)(a)(b)(d)ECGHDBAOF1 cm 10 kpsiBAshi20396_ch06.qxd 8/18/03 12:22 PM Page 157158 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. (6-13)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 fourthquadrantn =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, usen = 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 therst quadrantn = SytA=235111.4 = 2.11 Ans.Graphical Solution:(a) n = OBOA =1.821.02 = 1.78(b) n = ODOC =2.241.28 = 1.75(c) n = OFOE =2.751.24 = 2.22(d) n = OHOG =2.461.18 = 2.08O(d)(b)(a)(c)EFBDGCAH1 in 100 kpsiBAshi20396_ch06.qxd 8/18/03 12:22 PM Page 158Chapter 6 1596-11 The material is brittle and exhibits unequal tensile and compressive strengths. Decision:Use the Modied II-Mohr theory as shown in Fig. 6-28 which is limited to rst and fourthquadrants. Sut = 22 kpsi, Suc = 83 kpsiParabolic failure segment:SA = 221

SB +2222 83

2

SB SA SB SA22 22.0 60 13.530 21.6 70 8.440 20.1 80 2.350 17.4 83 0(a) x = 9 kpsi, y = 5 kpsi. A, B = 9, 5 kpsi. For the fourth quadrant, use Eq. (6-33a)n = SutA=229 = 2.44 Ans.(b) x = 12 kpsi, xy = 3 kpsi ccw. A, B = 12.7, 0.708 kpsi. For the rst quadrant,n = SutA=2212.7 = 1.73 Ans.(c) x = 4 kpsi, y = 9 kpsi, xy = 5 kpsi . A, B = 0.910, 12.09 kpsi. For thethird quadrant, no solution exists; however, use Eq. (6-33c)n = 8312.09 = 6.87 Ans.(d) x = 11 kpsi,y = 4 kpsi,xy = 1 kpsi. A, B = 11.14, 3.86 kpsi. For therst quadrantn = SAA= SytA=2211.14 = 1.97 Ans.3030Sut 22Sut 83BA5090shi20396_ch06.qxd 8/18/03 12:22 PM Page 159160 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design6-12 Since f < 0.05, the material is brittle. Thus, Sut.= Sucand we may use M2M which isbasically the same as MNS.(a) A, B = 9, 5 kpsin =359 = 3.89 Ans.(b) A, B = 12.7, 0.708 kpsin =3512.7 = 2.76 Ans.(c) A, B = 0.910, 12.09 kpsi (3rd quadrant)n =3612.09 = 2.98 Ans.(d) A, B = 11.14, 3.86 kpsin =3511.14 = 3.14 Ans.Graphical Solution:(a) n = OBOA =41 = 4.0 Ans.(b) n = ODOC =3.451.28 = 2.70 Ans.(c) n = OFOE =3.71.3 = 2.85 Ans. (3rd quadrant)(d) n = OHOG =3.61.15 = 3.13 Ans.6-13 Sut = 30 kpsi, Suc = 109 kpsiUse M2M:(a) A, B = 20, 20 kpsiEq. (6-33a): n =3020 = 1.5 Ans.(b) A, B =

(15)2= 15, 15 kpsiEq. (6-33a) n =3015 = 2 Ans.(c) A, B = 80, 80 kpsiFor the 3rd quadrant, there is no solution but use Eq. (6-33c).Eq. (6-33c): n = 10980 = 1.36 Ans.OGCDABEFH(a)(c)(b)(d)1 cm 10 kpsiBAshi20396_ch06.qxd 8/18/03 12:22 PM Page 160Chapter 6 161(d) A, B = 15, 25 kpsiEq. (6-33b): n(15)30 +

25n +3030 109

2= 1n = 1.90 Ans.(a) n = OBOA =4.252.83 = 1.50(b) n = ODOC =4.242.12 = 2.00(c) n = OFOE =15.511.3 = 1.37 (3r