Chapter 6 6-1 MSS: σ 1 − σ 3 = S y /n ⇒ n = S y σ 1 − σ 3 DE: n = S y σ σ = σ 2 A − σ A σ B + σ 2 B 1/2 = σ 2 x − σ x σ y + σ 2 y + 3τ 2 xy 1/2 (a) MSS: σ 1 = 12, σ 2 = 6, σ 3 = 0 kpsi n = 50 12 = 4.17 Ans. DE: σ = (12 2 − 6(12) + 6 2 ) 1/2 = 10.39 kpsi, n = 50 10.39 = 4.81 Ans. (b) σ A , σ B = 12 2 ± 12 2 2 + (−8) 2 = 16, −4 kpsi σ 1 = 16, σ 2 = 0, σ 3 =−4 kpsi MSS: n = 50 16 − (−4) = 2.5 Ans. DE: σ = (12 2 + 3(−8 2 )) 1/2 = 18.33 kpsi, n = 50 18.33 = 2.73 Ans. (c) σ A , σ B = −6 − 10 2 ± −6 + 10 2 2 + (−5) 2 =−2.615, −13.385 kpsi σ 1 = 0, σ 2 =−2.615, σ 3 =−13.385 kpsi MSS: n = 50 0 − (−13.385) = 3.74 Ans. DE: σ = [(−6) 2 − (−6)(−10) + (−10) 2 + 3(−5) 2 ] 1/2 = 12.29 kpsi n = 50 12.29 = 4.07 Ans. (d) σ A , σ B = 12 + 4 2 ± 12 − 4 2 2 + 1 2 = 12.123, 3.877 kpsi σ 1 = 12.123, σ 2 = 3.877, σ 3 = 0 kpsi MSS: n = 50 12.123 −0 = 4.12 Ans. DE: σ = [12 2 − 12(4) + 4 2 + 3(1 2 )] 1/2 = 10.72 kpsi n = 50 10.72 = 4.66 Ans. B A shi20396_ch06.qxd 8/18/03 12:22 PM Page 149 150 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 6-2 S y = 50 kpsi MSS: σ 1 − σ 3 = S y /n ⇒ n = S y σ 1 − σ 3 DE: σ 2 A − σ A σ B + σ 2 B 1/2 = S y /n ⇒ n = S y / σ 2 A − σ A σ B + σ 2 B 1/2 (a) MSS: σ 1 = 12 kpsi, σ 3 = 0, n = 50 12 − 0 = 4.17 Ans. DE: n = 50 [12 2 − (12)(12) + 12 2 ] 1/2 = 4.17 Ans. (b) MSS: σ 1 = 12 kpsi, σ 3 = 0 , n = 50 12 = 4.17 Ans. DE: n = 50 [12 2 − (12)(6) + 6 2 ] 1/2 = 4.81 Ans. (c) MSS: σ 1 = 12 kpsi, σ 3 =−12 kpsi , n = 50 12 − (−12) = 2.08 Ans. DE: n = 50 [12 2 − (12)(−12) + (−12) 2 ] 1/3 = 2.41 Ans. (d) MSS: σ 1 = 0, σ 3 =−12 kpsi, n = 50 −(−12) = 4.17 Ans. DE: n = 50 [(−6) 2 − (−6)(−12) + (−12) 2 ] 1/2 = 4.81 6-3 S y = 390 MPa MSS: σ 1 − σ 3 = S y /n ⇒ n = S y σ 1 − σ 3 DE: σ 2 A − σ A σ B + σ 2 B 1/2 = S y /n ⇒ n = S y / σ 2 A − σ A σ B + σ 2 B 1/2 (a) MSS: σ 1 = 180 MPa, σ 3 = 0, n = 390 180 = 2.17 Ans. DE: n = 390 [180 2 − 180(100) + 100 2 ] 1/2 = 2.50 Ans. (b) σ A , σ B = 180 2 ± 180 2 2 + 100 2 = 224.5, −44.5MPa= σ 1 , σ 3 MSS: n = 390 224.5 − (−44.5) = 1.45 Ans. DE: n = 390 [180 2 + 3(100 2 )] 1/2 = 1.56 Ans. shi20396_ch06.qxd 8/18/03 12:22 PM Page 150 Chapter 6 151 (c) σ A , σ B =− 160 2 ± − 160 2 2 + 100 2 = 48.06, −208.06 MPa = σ 1 , σ 3 MSS: n = 390 48.06 − (−208.06) = 1.52 Ans. DE: n = 390 [−160 2 + 3(100 2 )] 1/2 = 1.65 Ans. (d) σ A , σ B = 150, −150 MPa = σ 1 , σ 3 MSS: n = 380 150 − (−150) = 1.27 Ans. DE: n = 390 [3(150) 2 ] 1/2 = 1.50 Ans. 6-4 S y = 220 MPa (a) σ 1 = 100, σ 2 = 80, σ 3 = 0MPa MSS: n = 220 100 − 0 = 2.20 Ans. DET: σ = [100 2 − 100(80) + 80 2 ] 1/2 = 91.65 MPa n = 220 91.65 = 2.40 Ans. (b) σ 1 = 100, σ 2 = 10, σ 3 = 0MPa MSS: n = 220 100 = 2.20 Ans. DET: σ = [100 2 − 100(10) + 10 2 ] 1/2 = 95.39 MPa n = 220 95.39 = 2.31 Ans. (c) σ 1 = 100, σ 2 = 0, σ 3 =−80 MPa MSS: n = 220 100 − (−80) = 1.22 Ans. DE: σ = [100 2 − 100(−80) + (−80) 2 ] 1/2 = 156.2MPa n = 220 156.2 = 1.41 Ans. (d) σ 1 = 0, σ 2 =−80, σ 3 =−100 MPa MSS: n = 220 0 − (−100) = 2.20 Ans. DE: σ = [(−80) 2 − (−80)(−100) + (−100) 2 ] = 91.65 MPa n = 220 91.65 = 2.40 Ans. shi20396_ch06.qxd 8/18/03 12:22 PM Page 151 152 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 6-5 (a) MSS: n = OB OA = 2.23 1.08 = 2.1 DE: n = OC OA = 2.56 1.08 = 2.4 (b) MSS: n = OE OD = 1.65 1.10 = 1.5 DE: n = OF OD = 1.8 1.1 = 1.6 (c) MSS: n = OH OG = 1.68 1.05 = 1.6 DE: n = OI OG = 1.85 1.05 = 1.8 (d) MSS: n = OK OJ = 1.38 1.05 = 1.3 DE: n = OL OJ = 1.62 1.05 = 1.5 O (a) (b) (d) (c) H I G J K L F E D A B C Scale 1" ϭ 200 MPa B A shi20396_ch06.qxd 8/18/03 12:22 PM Page 152 Chapter 6 153 6-6 S y = 220 MPa (a) MSS: n = OB OA = 2.82 1.3 = 2.2 DE: n = OC OA = 3.1 1.3 = 2.4 (b) MSS: n = OE OD = 2.2 1 = 2.2 DE: n = OF OD = 2.33 1 = 2.3 (c) MSS: n = OH OG = 1.55 1.3 = 1.2 DE: n = OI OG = 1.8 1.3 = 1.4 (d) MSS: n = OK OJ = 2.82 1.3 = 2.2 DE: n = OL OJ = 3.1 1.3 = 2.4 B A O (a) (b) (c) (d) H G J K L I F E D A B C 1" ϭ 100 MPa shi20396_ch06.qxd 8/18/03 12:22 PM Page 153 154 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 6-7 S ut = 30 kpsi, S uc = 100 kpsi; σ A = 20 kpsi, σ B = 6 kpsi (a) MNS: Eq. (6-30a) n = S ut σ x = 30 20 = 1.5 Ans. BCM: Eq. (6-31a) n = 30 20 = 1.5 Ans. M1M: Eq. (6-32a) n = 30 20 = 1.5 Ans. M2M: Eq. (6-33a) n = 30 20 = 1.5 Ans. (b) σ x = 12 kpsi,τ xy =−8 kpsi σ A , σ B = 12 2 ± 12 2 2 + (−8) 2 = 16, −4 kpsi MNS: Eq. (6-30a) n = 30 16 = 1.88 Ans. BCM: Eq. (6-31b) 1 n = 16 30 − (−4) 100 ⇒ n = 1.74 Ans. M1M: Eq. (6-32a) n = 30 16 = 1.88 Ans. M2M: Eq. (6-33a) n = 30 16 = 1.88 Ans. (c) σ x =−6 kpsi, σ y =−10 kpsi,τ xy =−5 kpsi σ A , σ B = −6 − 10 2 ± −6 + 10 2 2 + (−5) 2 =−2.61, −13.39 kpsi MNS: Eq. (6-30b) n =− 100 −13.39 = 7.47 Ans. BCM: Eq. (6-31c) n =− 100 −13.39 = 7.47 Ans. M1M: Eq. (6-32c) n =− 100 −13.39 = 7.47 Ans. M2M: Eq. (6-33c) n =− 100 −13.39 = 7.47 Ans. (d) σ x =−12 kpsi,τ xy = 8 kpsi σ A , σ B =− 12 2 ± − 12 2 2 + 8 2 = 4, −16 kpsi MNS: Eq. (6-30b) n = −100 −16 = 6.25 Ans. shi20396_ch06.qxd 8/18/03 12:22 PM Page 154 Chapter 6 155 BCM: Eq. (6-31b) 1 n = 4 30 − (−16) 100 ⇒ n = 3.41 Ans. M1M: Eq. (6-32b) 1 n = (100 −30)4 100(30) − −16 100 ⇒ n = 3.95 Ans. M2M: Eq. (6-33b) n 4 30 + n(−16) + 30 30 − 100 2 = 1 Reduces to n 2 − 1.1979n − 15.625 = 0 n = 1.1979 + 1.1979 2 + 4(15.625) 2 = 4.60 Ans. (c) L (d) J (b) (a) I H G K F O C D E A B 1" ϭ 20 kpsi B A shi20396_ch06.qxd 8/18/03 12:22 PM Page 155 156 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 6-8 See Prob. 6-7 for plot. (a) For all methods: n = OB OA = 1.55 1.03 = 1.5 (b) BCM: n = OD OC = 1.4 0.8 = 1.75 All other methods: n = OE OC = 1.55 0.8 = 1.9 (c) For all methods: n = OL OK = 5.2 0.68 = 7.6 (d) MNS: n = OJ OF = 5.12 0.82 = 6.2 BCM: n = OG OF = 2.85 0.82 = 3.5 M1M: n = OH OF = 3.3 0.82 = 4.0 M2M: n = OI OF = 3.82 0.82 = 4.7 6-9 Given: S y = 42 kpsi, S ut = 66.2 kpsi, ε f = 0.90. Since ε f > 0.05, the material is ductile and thus we may follow convention by setting S yc = S yt . Use DE theory for analytical solution. For σ , use Eq. (6-13) or (6-15) for plane stress and Eq. (6-12) or (6-14) for general 3-D. (a) σ = [9 2 − 9(−5) + (−5) 2 ] 1/2 = 12.29 kpsi n = 42 12.29 = 3.42 Ans. (b) σ = [12 2 + 3(3 2 )] 1/2 = 13.08 kpsi n = 42 13.08 = 3.21 Ans. (c) σ = [(−4) 2 − (−4)(−9) + (−9) 2 + 3(5 2 )] 1/2 = 11.66 kpsi n = 42 11.66 = 3.60 Ans. (d) σ = [11 2 − (11)(4) + 4 2 + 3(1 2 )] 1/2 = 9.798 n = 42 9.798 = 4.29 Ans. shi20396_ch06.qxd 8/18/03 12:22 PM Page 156 Chapter 6 157 For graphical solution, plot load lines on DE envelope as shown. (a) σ A = 9, σ B =−5 kpsi n = OB OA = 3.5 1 = 3.5 Ans. (b) σ A , σ B = 12 2 ± 12 2 2 + 3 2 = 12.7, −0.708 kpsi n = OD OC = 4.2 1.3 = 3.23 (c) σ A , σ B = −4 − 9 2 ± 4 − 9 2 2 + 5 2 =−0.910, −12.09 kpsi n = OF OE = 4.5 1.25 = 3.6 Ans. (d) σ A , σ B = 11 + 4 2 ± 11 − 4 2 2 + 1 2 = 11.14, 3.86 kpsi n = OH OG = 5.0 1.15 = 4.35 Ans. 6-10 This heat-treated steel exhibits S yt = 235 kpsi, S yc = 275 kpsi and ε f = 0.06. The steel is ductile (ε f > 0.05) but of unequal yield strengths. The Ductile Coulomb-Mohr hypothesis (DCM) of Fig. 6-27 applies — confine its use to first and fourth quadrants. (c) (a) (b) (d) E C G H D B A O F 1 cm ϭ 10 kpsi B A shi20396_ch06.qxd 8/18/03 12:22 PM Page 157 158 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design (a) σ x = 90 kpsi, σ y =−50 kpsi, σ z = 0 І σ A = 90 kpsi and σ B =−50 kpsi. For the fourth quadrant, from Eq. (6-13) n = 1 (σ A /S yt ) − (σ B /S uc ) = 1 (90/235) − (−50/275) = 1.77 Ans. (b) σ x = 120 kpsi, τ xy =−30 kpsi ccw. σ A , σ B = 127.1, −7.08 kpsi. For the fourth quadrant n = 1 (127.1/235) − (−7.08/275) = 1.76 Ans. (c) σ x =−40 kpsi, σ y =−90 kpsi, τ xy = 50 kpsi . σ A , σ B =−9.10, −120.9 kpsi. Although no solution exists for the third quadrant, use n =− S yc σ y =− 275 −120.9 = 2.27 Ans. (d) σ x = 110 kpsi, σ y = 40 kpsi, τ xy = 10 kpsi cw. σ A , σ B = 111.4, 38.6 kpsi. For the first quadrant n = S yt σ A = 235 111.4 = 2.11 Ans. Graphical Solution: (a) n = OB OA = 1.82 1.02 = 1.78 (b) n = OD OC = 2.24 1.28 = 1.75 (c) n = OF OE = 2.75 1.24 = 2.22 (d) n = OH OG = 2.46 1.18 = 2.08 O (d) (b) (a) (c) E F B D G C A H 1 in ϭ 100 kpsi B A shi20396_ch06.qxd 8/18/03 12:22 PM Page 158 [...]... = 67 6 .6 tan 20 = 2 46. 3 lbf y ROy = 193.7 lbf O ROz = 233.5 lbf 2 46. 3 lbf A C B 20" 16" 10" x RBy = 158.1 lbf 281.9 lbf O 67 6 .6 lbf A B 20" 16" z 102 .6 lbf C 10" RBz = 807.5 lbf xz plane xy plane M A = 20 193.72 + 233.52 = 60 68 lbf · in M B = 10 2 46. 32 + 67 6 .62 = 7200 lbf · in (maximum) σx = 32(7200) 73 340 = 3 πd d3 16( 3383) 17 230 = 3 πd d3 Sy 2 2 1/2 σ = σx + 3τx y = n τx y = 73 340 d3 d = 1 .66 5 in... 0.011 61 ω2 = 160 7 rad/s So the inner radius governs and n = 13 000 rev/min 6- 18 Ans For a thin-walled pressure vessel, di = 3.5 − 2(0. 065 ) = 3.37 in σt = p(di + t) 2t σt = 500(3.37 + 0. 065 ) = 13 212 psi 2(0. 065 ) σl = 500(3.37) pdi = = 64 81 psi 4t 4(0. 065 ) σr = − pi = −500 psi shi203 96_ ch 06. qxd 8/18/03 12:22 PM Page 165 165 Chapter 6 These are all principal stresses, thus, 1 σ = √ {(13 212 − 64 81) 2 + [64 81... T1 = 310 .6 N, T2 = 0.15(310 .6) = 46. 6 N (T1 + T2 ) cos 45 = 252 .6 N 107.0 N y 163 .4 N 252 .6 N 300 O A 89.2 N 400 300 150 400 z C B 252 .6 N 150 320 N 174.4 N xz plane xy plane M A = 0.3 163 .42 + 1072 = 58.59 N · m (maximum) M B = 0.15 89.22 + 174.42 = 29.38 N · m 32(58.59) 5 96. 8 = 3 πd d3 16( 33) 168 .1 = = 3 πd d3 σx = τx y σ = σx + 3τx y 2 2 1/2 5 96. 8 d3 = d = 17.5(10−3 ) m = 17.5 mm, 6- 26 2 168 .1 +3... ]1/2 Eq (6- 13) = 11 320 psi Ans (b) For a solid inner tube, p= 30(1 06 )(0.0005) (1.52 − 12 )(12 ) = 4 167 psi Ans 1 2(12 )(1.52 ) (σt ) i = − p = −4 167 psi, (σr ) i = −4 167 psi σi = [(−4 167 ) 2 − (−4 167 )(−4 167 ) + (−4 167 ) 2 ]1/2 = 4 167 psi Ans (σt ) o = 4 167 1.52 + 12 1.52 − 12 = 10 830 psi, (σr ) o = −4 167 psi σo = [10 8302 − 10 830(−4 167 ) + (−4 167 ) 2 ]1/2 = 13 410 psi Ans 6- 33 Using Eq (4 -60 ) with diametral... 17.5 mm, 6- 26 2 168 .1 +3 d3 2 1/2 66 4.0 370(1 06 ) = = d3 3.0 so use 18 mm Ans From Prob 6- 25, τmax = 5 96. 8 2d 3 2 d = 17.7(10−3 ) m = 17.7 mm, σx 2 168 .1 + d3 1/2 2 + τx y = 2 2 1/2 = Sy 2n 342.5 370(1 06 ) = d3 2(3.0) so use 18 mm Ans shi203 96_ ch 06. qxd 8/18/03 12:22 PM Page 169 169 Chapter 6 6-27 For the loading scheme shown in Figure (c), Mmax F = 2 a b + 2 4 V 4.4 = (6 + 4.5) 2 M = 23.1 N · m y For... With σx = −1 36. 2 MPa From a Mohrs circle diagram, τmax = 1 36. 2/2 = 68 .1 MPa n= 6- 28 110 = 1 .62 MPa 68 .1 Ans Based on Figure (c) and using Eq (6- 15) 2 σ = σx 1/2 = (1 36. 22 ) 1/2 = 1 36. 2 MPa n= Sy 220 = 1 .62 Ans = σ 1 36. 2 shi203 96_ ch 06. qxd 170 8/18/03 12:22 PM Page 170 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Based on Figure (d) and using Eq (6- 15) and the... speed of 1 364 0 rev/min 6- 21 TC = ( 360 − 27)(3) = 1000 lbf · in , TB = (300 − 50)(4) = 1000 lbf · in y 223 lbf A 127 lbf B 8" C 8" 350 lbf xy plane 6" D shi203 96_ ch 06. qxd 166 8/27/03 4:38 PM Page 166 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design In x y plane, M B = 223(8) = 1784 lbf · in and MC = 127 (6) = 762 lbf · in 387 lbf 8" A 8" B 6" D C 1 06 lbf 281 lbf... 2 ¯2 ¯ ro − R 2 ¯2 = 18.70(1 06 ) δ 0. 562 52 + 0.37552 0. 562 52 − 0.37552 = 48. 76( 1 06 ) δ psi σot = 48. 76( 1 06 )(0.001) = 48. 76( 103 ) psi ¯ σσot = Cδ σot = 0.0707(48. 76) (103 ) = 34.45 psi ˆ ¯ І σot = N(48 760 , 3445) psi Ans 6- 42 From Prob 6- 41, at the fit surface σot = N(48.8, 3.45) kpsi The radial stress is the fit pressure which was found to be p = 18.70(1 06 ) δ p = 18.70(1 06 )(0.001) = 18.7(103 ) psi ¯... = 35 16 psi Ans 1 2(12 )(1.52 − 0.52 ) Inner member: (σt ) i = − p Eq (4-57) R 2 + ri2 12 + 0.52 = −35 16 2 1 − 0.52 R 2 − ri2 = −5 860 psi (σr ) i = − p = −35 16 psi 2 2 σi = σ A − σ A σ B + σ B Eq (6- 13) 1/2 = [(−5 860 ) 2 − (−5 860 )(−35 16) + (−35 16) 2 ]1/2 = 5110 psi Ans Outer member: (σt ) o = 35 16 Eq (4-58) 1.52 + 12 1.52 − 12 = 9142 psi (σr ) o = − p = −35 16 psi σo = [91422 − 9142(−35 16) + (−35 16) 2... kpsi For the 3rd quadrant, there is no solution but use Eq (6- 33c) Eq (6- 33c): C A OD 3.45 = = 2.70 Ans OC 1.28 Eq (6- 33a) (b) B O (b) n = Eq (6- 33a): D G OB 4 = = 4.0 Ans (a) n = OA 1 6- 13 (d) n=− 109 = 1. 36 Ans −80 A shi203 96_ ch 06. qxd 8/18/03 12:22 PM Page 161 161 Chapter 6 (d) σ A , σ B = 15, −25 kpsi n(15) −25n + 30 + 30 30 − 109 Eq (6- 33b): 2 =1 n = 1.90 Ans OB 4.25 = = 1.50 OA 2.83 OD 4.24 = . 4, − 16 kpsi MNS: Eq. (6- 30b) n = −100 − 16 = 6. 25 Ans. shi203 96_ ch 06. qxd 8/18/03 12:22 PM Page 154 Chapter 6 155 BCM: Eq. (6- 31b) 1 n = 4 30 − (− 16) 100 ⇒. lbf 20" 16& quot; 10" 2 46. 3 lbf O xz plane x z ABC R Oz = 233.5 lbf R Bz = 807.5 lbf O 102 .6 lbf 20" 16& quot; 10" 67 6 .6 lbf shi203 96_ ch 06. qxd