Capítulo 06 falhas resultantes de carregamento estático

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1. Chapter 6 6-1 MSS: 1 3 = Sy/n n = Sy 1 3 DE: n = Sy = 2 A AB + 2 B 1/2 = 2 x xy + 2 y + 32 xy 1/2 (a) MSS: 1 = 12, 2 = 6, 3 = 0 kpsi n = 50 12 = 4.17 Ans. DE: = (122 6(12) + 62 )1/2 = 10.39 kpsi, n = 50 10.39 = 4.81 Ans. (b) A, B = 12 2 12 2 2 + (8)2 = 16, 4 kpsi 1 = 16, 2 = 0, 3 = 4 kpsi MSS: n = 50 16 (4) = 2.5 Ans. DE: = (122 + 3(82 ))1/2 = 18.33 kpsi, n = 50 18.33 = 2.73 Ans. (c) A, B = 6 10 2 6 + 10 2 2 + (5)2 = 2.615, 13.385 kpsi 1 = 0, 2 = 2.615, 3 = 13.385 kpsi MSS: n = 50 0 (13.385) = 3.74 Ans. DE: = [(6)2 (6)(10) + (10)2 + 3(5)2 ]1/2 = 12.29 kpsi n = 50 12.29 = 4.07 Ans. (d) A, B = 12 + 4 2 12 4 2 2 + 12 = 12.123, 3.877 kpsi 1 = 12.123, 2 = 3.877, 3 = 0 kpsi MSS: n = 50 12.123 0 = 4.12 Ans. DE: = [122 12(4) + 42 + 3(12 )]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 2. 150 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design 6-2 Sy = 50 kpsi MSS: 1 3 = Sy/n n = Sy 1 3 DE: 2 A AB + 2 B 1/2 = Sy/n n = Sy/ 2 A AB + 2 B 1/2 (a) MSS: 1 = 12 kpsi, 3 = 0, n = 50 12 0 = 4.17 Ans. DE: n = 50 [122 (12)(12) + 122]1/2 = 4.17 Ans. (b) MSS: 1 = 12 kpsi, 3 = 0, n = 50 12 = 4.17 Ans. DE: n = 50 [122 (12)(6) + 62]1/2 = 4.81 Ans. (c) MSS: 1 = 12 kpsi, 3 = 12 kpsi, n = 50 12 (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 6-3 Sy = 390 MPa MSS: 1 3 = Sy/n n = Sy 1 3 DE: 2 A AB + 2 B 1/2 = Sy/n n = Sy/ 2 A AB + 2 B 1/2 (a) MSS: 1 = 180 MPa, 3 = 0, n = 390 180 = 2.17 Ans. DE: n = 390 [1802 180(100) + 1002]1/2 = 2.50 Ans. (b) A, B = 180 2 180 2 2 + 1002 = 224.5, 44.5 MPa = 1, 3 MSS: n = 390 224.5 (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 150 3. Chapter 6 151 (c) A, B = 160 2 160 2 2 + 1002 = 48.06, 208.06 MPa = 1, 3 MSS: n = 390 48.06 (208.06) = 1.52 Ans. DE: n = 390 [1602 + 3(1002)]1/2 = 1.65 Ans. (d) A, B = 150, 150 MPa = 1, 3 MSS: n = 380 150 (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 MPa MSS: n = 220 100 0 = 2.20 Ans. DET: = [1002 100(80) + 802 ]1/2 = 91.65 MPa n = 220 91.65 = 2.40 Ans. (b) 1 = 100, 2 = 10, 3 = 0 MPa MSS: n = 220 100 = 2.20 Ans. DET: = [1002 100(10) + 102 ]1/2 = 95.39 MPa n = 220 95.39 = 2.31 Ans. (c) 1 = 100, 2 = 0, 3 = 80 MPa MSS: n = 220 100 (80) = 1.22 Ans. DE: = [1002 100(80) + (80)2 ]1/2 = 156.2 MPa n = 220 156.2 = 1.41 Ans. (d) 1 = 0, 2 = 80, 3 = 100 MPa MSS: n = 220 0 (100) = 2.20 Ans. DE: = [(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 4. 152 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design 6-5 (a) MSS: n = O B O A = 2.23 1.08 = 2.1 DE: n = OC O A = 2.56 1.08 = 2.4 (b) MSS: n = OE O D = 1.65 1.10 = 1.5 DE: n = OF O D = 1.8 1.1 = 1.6 (c) MSS: n = O H OG = 1.68 1.05 = 1.6 DE: n = OI OG = 1.85 1.05 = 1.8 (d) MSS: n = OK O J = 1.38 1.05 = 1.3 DE: n = OL O J = 1.62 1.05 = 1.5 O (a) (b) (d) (c) H I G J K L FE D A B C Scale 1" 200 MPa B A shi20396_ch06.qxd 8/18/03 12:22 PM Page 152 5. Chapter 6 153 6-6 Sy = 220 MPa (a) MSS: n = O B O A = 2.82 1.3 = 2.2 DE: n = OC O A = 3.1 1.3 = 2.4 (b) MSS: n = OE O D = 2.2 1 = 2.2 DE: n = OF O D = 2.33 1 = 2.3 (c) MSS: n = O H OG = 1.55 1.3 = 1.2 DE: n = OI OG = 1.8 1.3 = 1.4 (d) MSS: n = OK O J = 2.82 1.3 = 2.2 DE: n = OL O J = 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" 100 MPa shi20396_ch06.qxd 8/18/03 12:22 PM Page 153 6. 154 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design 6-7 Sut = 30 kpsi, Suc = 100 kpsi; A = 20 kpsi, B = 6 kpsi (a) MNS: Eq. (6-30a) n = Sut 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) x = 12 kpsi,xy = 8 kpsi A, 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) x = 6 kpsi, y = 10 kpsi,xy = 5 kpsi A, 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) x = 12 kpsi,xy = 8 kpsi A, B = 12 2 12 2 2 + 82 = 4, 16 kpsi MNS: Eq. (6-30b) n = 100 16 = 6.25 Ans. shi20396_ch06.qxd 8/18/03 12:22 PM Page 154 7. 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 n2 1.1979n 15.625 = 0 n = 1.1979 + 1.19792 + 4(15.625) 2 = 4.60 Ans. (c) L (d) J (b) (a) I H G K F O C D E A B1" 20 kpsi B A shi20396_ch06.qxd 8/18/03 12:22 PM Page 155 8. 156 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design 6-8 See Prob. 6-7 for plot. (a) For all methods: n = O B O A = 1.55 1.03 = 1.5 (b) BCM: n = O D 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 = O J OF = 5.12 0.82 = 6.2 BCM: n = OG OF = 2.85 0.82 = 3.5 M1M: n = O H OF = 3.3 0.82 = 4.0 M2M: n = OI OF = 3.82 0.82 = 4.7 6-9 Given: Sy = 42 kpsi, Sut = 66.2 kpsi, f = 0.90. Sincef > 0.05, the material is ductile and thus 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 and Eq. (6-12) or (6-14) for general 3-D. (a) = [92 9(5) + (5)2 ]1/2 = 12.29 kpsi n = 42 12.29 = 3.42 Ans. (b) = [122 + 3(32 )]1/2 = 13.08 kpsi n = 42 13.08 = 3.21 Ans. (c) = [(4)2 (4)(9) + (9)2 + 3(52 )]1/2 = 11.66 kpsi n = 42 11.66 = 3.60 Ans. (d) = [112 (11)(4) + 42 + 3(12 )]1/2 = 9.798 n = 42 9.798 = 4.29 Ans. shi20396_ch06.qxd 8/18/03 12:22 PM Page 156 9. Chapter 6 157 For graphical solution, plot load lines on DE envelope as shown. (a) A = 9, B = 5 kpsi n = O B O A = 3.5 1 = 3.5 Ans. (b) A, B = 12 2 12 2 2 + 32 = 12.7, 0.708 kpsi n = O D OC = 4.2 1.3 = 3.23 (c) A, B = 4 9 2 4 9 2 2 + 52 = 0.910, 12.09 kpsi n = OF OE = 4.5 1.25 = 3.6 Ans. (d) A, B = 11 + 4 2 11 4 2 2 + 12 = 11.14, 3.86 kpsi n = O H OG = 5.0 1.15 = 4.35 Ans. 6-10 This heat-treated steel exhibits Syt = 235 kpsi, Syc = 275 kpsi and f = 0.06. The steel is ductile (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) E C G H D B A O F 1 cm 10 kpsi B A shi20396_ch06.qxd 8/18/03 12:22 PM Page 157 10. 158 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 the fourth 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 fourth quadrant 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 = Syc y = 275 120.9 = 2.27 Ans. (d) x = 110 kpsi, y = 40 kpsi, xy = 10 kpsi cw. A, B = 111.4, 38.6 kpsi. For the rst quadrant n = Syt A = 235 111.4 = 2.11 Ans. Graphical Solution: (a) n = O B O A = 1.82 1.02 = 1.78 (b) n = O D OC = 2.24 1.28 = 1.75 (c) n = OF OE = 2.75 1.24 = 2.22 (d) n = O H OG = 2.46 1.18 = 2.08 O (d) (b) (a) (c) E F B D G C A H 1 in 100 kpsi B A shi20396_ch06.qxd 8/18/03 12:22 PM Page 158 11. Chapter 6 159 6-11 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 fourth quadrants. Sut = 22 kpsi, Suc = 83 kpsi Parabolic failure segment: SA = 22 1 SB + 22 22 83 2 SB SA SB SA 22 22.0 60 13.5 30 21.6 70 8.4 40 20.1 80 2.3 50 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 = Sut A = 22 9 = 2.44 Ans. (b) x = 12 kpsi, xy = 3 kpsi ccw. A, B = 12.7, 0.708 kpsi. For the rst quadrant, n = Sut A = 22 12.7 = 1.73 Ans. (c) x = 4 kpsi, y = 9 kpsi, xy = 5 kpsi. A, B = 0.910, 12.09 kpsi. For the third quadrant, no solution exists; however, use Eq. (6-33c) n = 83 12.09 = 6.87 Ans. (d) x = 11 kpsi,y = 4 kpsi,xy = 1 kpsi.A, B = 11.14, 3.86 kpsi. Fortherstquadrant n = SA A = Syt A = 22 11.14 = 1.97 Ans. 30 30 Sut 22 Sut 83 B A 50 90 shi20396_ch06.qxd 8/18/03 12:22 PM Page 159 12. 160 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design 6-12 Since f < 0.05, the material is brittle. Thus, Sut . = Suc and we may use M2M which is basically the same as MNS. (a) A, B = 9, 5 kpsi n = 35 9 = 3.89 Ans. (b) A, B = 12.7, 0.708 kpsi n = 35 12.7 = 2.76 Ans. (c) A, B = 0.910, 12.09 kpsi (3rd quadrant) n = 36 12.09 = 2.98 Ans. (d) A, B = 11.14, 3.86 kpsi n = 35 11.14 = 3.14 Ans. Graphical Solution: (a) n = O B O A = 4 1 = 4.0 Ans. (b) n = O D OC = 3.45 1.28 = 2.70 Ans. (c) n = OF OE = 3.7 1.3 = 2.85 Ans. (3rd quadrant) (d) n = O H OG = 3.6 1.15 = 3.13 Ans. 6-13 Sut = 30 kpsi, Suc = 109 kpsi Use M2M: (a) A, B = 20, 20 kpsi Eq. (6-33a): n = 30 20 = 1.5 Ans. (b) A, B = (15)2 = 15, 15 kpsi Eq. (6-33a) n = 30 15 = 2 Ans. (c) A, B = 80, 80 kpsi For the 3rd quadrant, there is no solution but use Eq. (6-33c). Eq. (6-33c): n = 109 80 = 1.36 Ans. O G C D A B E F H (a) (c) (b) (d) 1 cm 10 kpsi B A shi20396_ch06.qxd 8/18/03 12:22 PM Page 160 13. Chapter 6 161 (d) A, B = 15, 25 kpsi Eq. (6-33b): n(15) 30 + 25n + 30 30 109 2 = 1 n = 1.90 Ans. (a) n = O B O A = 4.25 2.83 = 1.50 (b) n = O D OC = 4.24 2.12 = 2.00 (c) n = OF OE = 15.5 11.3 = 1.37 (3rd quadrant) (d) n = O H OG = 5.3 2.9 = 1.83 6-14 Given: AISI 1006 CD steel, F = 0.55 N, P = 8.0 kN