MECH 420 Problems Problem Chap 5, Fall 2012

Chapter 5 5-1 MSS: σ1 − σ3 = Sy/n ⇒ n = Sy σ1 − σ3 DE: n = Sy σ ′ σ ′ = � σ 2 A − σAσB + σ 2 B �1/2 = � σ 2 x − σxσy + σ 2 y + 3τ 2 xy �1/2 (a) MSS: σ1 = 12, σ2 = 6, σ3 = 0 kpsi n = 50 12 = 4.17 Ans. DE: σ ′ = (122 − 6(12) + 62)1/2 = 10.39 kpsi, n = 50 10.39 = 4.81 Ans. (b) σA, σB = 12 2 ± ��12 2 �2 + (−8)2 = 16, −4 kpsi σ1 = 16, σ2 = 0, σ3 = −4 kpsi MSS: n = 50 16 − (−4) = 2.5 Ans. DE: σ ′ = (122 + 3(−82))1/2 = 18.33 kpsi, n = 50 18.33 = 2.73 Ans. (c) σA, σB = −6 − 10 2 ± ��−6 + 10 2 �2 + (−5)2 = −2.615, −13.385 kpsi σ1 = 0, σ2 = −2.615, σ3 = −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) σA, σB = 12 + 4 2 ± ��12 − 4 2 �2 + 12 = 12.123, 3.877 kpsi σ1 = 12.123, σ2 = 3.877, σ3 = 0 kpsi MSS: n = 50 12.123 − 0 = 4.12 Ans. DE: σ ′ = [122 − 12(4) + 42 + 3(12)]1/2 = 10.72 kpsi n = 50 10.72 = 4.66 Ans. �B �A 116 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 5-2 Sy = 50 kpsi MSS: σ1 − σ3 = Sy/n ⇒ n = Sy σ1 − σ3 DE: � σ 2 A − σAσB + σ 2 B �1/2 = Sy/n ⇒ n = Sy/ � σ 2 A − σAσB + σ 2 B �1/2 (a) MSS: σ1 = 12 kpsi, σ3 = 0, n = 50 12 − 0 = 4.17 Ans. DE: n = 50 [122 − (12)(12) + 122]1/2 = 4.17 Ans. (b) MSS: σ1 = 12 kpsi, σ3 = 0, n = 50 12 = 4.17 Ans. DE: n = 50 [122 − (12)(6) + 62]1/2 = 4.81 Ans. (c) MSS: σ1 = 12 kpsi, σ3 = −12 kpsi, n = 50 12 − (−12) = 2.08 Ans. DE: n = 50 [122 − (12)(−12) + (−12)2]1/3 = 2.41 Ans. (d) MSS: σ1 = 0, σ3 = −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 MPa MSS: σ1 − σ3 = Sy/n ⇒ n = Sy σ1 − σ3 DE: � σ 2 A − σAσB + σ 2 B �1/2 = Sy/n ⇒ n = Sy/ � σ 2 A − σAσB + σ 2 B �1/2 (a) MSS: σ1 = 180 MPa, σ3 = 0, n = 390 180 = 2.17 Ans. DE: n = 390 [1802 − 180(100) + 1002]1/2 = 2.50 Ans. (b) σA, σB = 180 2 ± ��180 2 �2 + 1002 = 224.5, −44.5 MPa = σ1, σ3 MSS: n = 390 224.5 − (−44.5) = 1.45 Ans. DE: n = 390 [1802 + 3(1002)]1/2 = 1.56 Ans. Chapter 5 117 (c) σA, σB = −160 2 ± �� −160 2 �2 + 1002 = 48.06, −208.06 MPa = σ1, σ3 MSS: n = 390 48.06 − (−208.06) = 1.52 Ans. DE: n = 390 [−1602 + 3(1002)]1/2 = 1.65 Ans. (d) σA, σB = 150, −150 MPa = σ1, σ3 MSS: n = 390 150 − (−150) = 1.30 Ans. DE: n = 390 [3(150)2]1/2 = 1.50 Ans. 5-4 Sy = 220 MPa (a) σ1 = 100, σ2 = 80, σ3 = 0 MPa MSS: n = 220 100 − 0 = 2.20 Ans. DET: σ ′ = [1002 − 100(80) + 802]1/2 = 91.65 MPa n = 220 91.65 = 2.40 Ans. (b) σ1 = 100, σ2 = 10, σ3 = 0 MPa MSS: n = 220 100 = 2.20 Ans. DET: σ ′ = [1002 − 100(10) + 102]1/2 = 95.39 MPa n = 220 95.39 = 2.31 Ans. (c) σ1 = 100, σ2 = 0, σ3 = −80 MPa MSS: n = 220 100 − (−80) = 1.22 Ans. DE: σ ′ = [1002 − 100(−80) + (−80)2]1/2 = 156.2 MPa n = 220 156.2 = 1.41 Ans. (d) σ1 = 0, σ2 = −80, σ3 = −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. 118 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 5-5 (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 F E D A B C Scale 1" � 200 MPa �B �A Chapter 5 119 5-6 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 120 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 5-7 Sut = 30 kpsi, Suc = 100 kpsi; σA = 20 kpsi, σB = 6 kpsi (a) MNS: Eq. (5-30a) n = Sut σx = 30 20 = 1.5 Ans. BCM: Eq. (5-31a) n = 30 20 = 1.5 Ans. MM: Eq. (5-32a) n = 30 20 = 1.5 Ans. (b) σx = 12 kpsi,τxy = −8 kpsi σA, σB = 12 2 ± ��12 2 �2 + (−8)2 = 16, −4 kpsi MNS: Eq. (5-30a) n = 30 16 = 1.88 Ans. BCM: Eq. (5-31b) 1 n = 16 30 − (−4) 100 ⇒ n = 1.74 Ans. MM: Eq. (5-32a) n = 30 16 = 1.88 Ans. (c) σx = −6 kpsi, σy = −10 kpsi,τxy = −5 kpsi σA, σB = −6 − 10 2 ± ��−6 + 10 2 �2 + (−5)2 = −2.61, −13.39 kpsi MNS: Eq. (5-30b) n = − 100 −13.39 = 7.47 Ans. BCM: Eq. (5-31c) n = − 100 −13.39 = 7.47 Ans. MM: Eq. (5-32c) n = − 100 −13.39 = 7.47 Ans. (d) σx = −12 kpsi,τxy = 8 kpsi σA, σB = −12 2 ± �� −12 2 �2 + 82 = 4, −16 kpsi MNS: Eq. (5-30b) n = −100 −16 = 6.25 Ans. 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 D E A B 1" � 20 kpsi �B �A 122 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 5-8 See Prob. 5-7 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 MM: n = O H OF = 3.3 0.82 = 4.0 5-9 Given: Sy = 42 kpsi, Sut = 66.2 kpsi, εf = 0.90. Since εf > 0.05, the material is ductile and thus we may follow convention by setting Syc = Syt. Use DE theory for analytical solution. For σ ′, use Eq. (5-13) or (5-15) for plane stress and Eq. (5-12) or (5-14) for general 3-D. (a) σ ′ = [92 − 9(−5) + (−5)2]1/2 = 12.29 kpsi n = 42 12.29 = 3.42 Ans. (b) σ ′ = [122 + 3(32)]1/2 = 13.08 kpsi n = 42 13.08 = 3.21 Ans. (c) σ ′ = [(−4)2 − (−4)(−9) + (−9)2 + 3(52)]1/2 = 11.66 kpsi n = 42 11.66 = 3.60 Ans. (d) σ ′ = [112 − (11)(4) + 42 + 3(12)]1/2 = 9.798 n = 42 9.798 = 4.29 Ans. Chapter 5 123 For graphical solution, plot load lines on DE envelope as shown. (a) σA = 9, σB = −5 kpsi n = O B O A = 3.5 1 = 3.5 Ans. (b) σA, σB = 12 2 ± ��12 2 �2 + 32 = 12.7, −0.708 kpsi n = O D OC = 4.2 1.3 = 3.23 (c) σA, σB = −4 − 9 2 ± ��4 − 9 2 �2 + 52 = −0.910, −12.09 kpsi n = OF OE = 4.5 1.25 = 3.6 Ans. (d) σA, σB = 11 + 4 2 ± ��11 − 4 2 �2 + 12 = 11.14, 3.86 kpsi n = O H OG = 5.0 1.15 = 4.35 Ans. 5-10 This heat-treated steel exhibits Syt = 235 kpsi, Syc = 275 kpsi and εf = 0.06. The steel is ductile (εf > 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 124 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design (a) σx = 90 kpsi, σy = −50 kpsi, σz = 0 � σA = 90 kpsi and σB = −50 kpsi. For the fourth quadrant, from Eq. (5-31b) n = 1 (σA/Syt) − (σB/Suc) = 1 (90/235) − (−50/275) = 1.77 Ans. (b) σx = 120 kpsi, τxy = −30 kpsi ccw. σA, σB = 127.1, −7.08 kpsi. For the fourth quadrant n = 1 (127.1/235) − (−7.08/275) = 1.76 Ans. (c) σx = −40 kpsi, σy = −90 kpsi, τxy = 50 kpsi. σA, σB = −9.10, −120.9 kpsi. Although no solution exists for the third quadrant, use n = −Syc σy = − 275 −120.9 = 2.27 Ans. (d) σx = 110 kpsi, σy = 40 kpsi, τxy = 10 kpsi cw. σA, σB = 111.4, 38.6 kpsi. For the first quadrant n = Syt σA = 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 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) σx = 9 kpsi, σy = −5 kpsi. σA, σB = 9, −5 kpsi. For the fourth quadrant, |σB σA | = 5 9 < 1, use Eq. (5-32a) n = Sut σA = 22 9 = 2.44 Ans. (b) σx = 12 kpsi, τxy = −3 kpsi ccw. σA, σB = 12.7, −0.708 kpsi. For the fourth quad- rant, |σB σA | = 0.708 12.7 < 1, n = Sut σA = 22 12.7 = 1.73 Ans. (c) σx = −4 kpsi, σy = −9 kpsi, τxy = 5 kpsi. σA, σB = −0.910, −12.09 kpsi. For the third quadrant, no solution exists; however, use Eq. (6-32c) n = −83 −12.09 = 6.87 Ans. (d) σx = 11kpsi, σy = 4kpsi,τxy = 1kpsi. σA, σB = 11.14, 3.86 kpsi. Forthefirstquadrant n = SA σA = Syt σA = 22 11.14 = 1.97 Ans. 30 30 Sut � 22 Sut � 83 �B �A –50 –90 (d ) (b) (a) (c) 126 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 5-12 Since εf < 0.05, the material is brittle. Thus, Sut .= Suc and we may use MM which is basically the same as MNS. (a) σA, σB = 9, −5 kpsi n = 35 9 = 3.89 Ans. (b) σA, σB = 12.7, −0.708 kpsi n = 35 12.7 = 2.76 Ans. (c) σA, σB = −0.910, −12.09 kpsi (3rd quadrant) n = 36 12.09 = 2.98 Ans. (d) σA, σB = 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. 5-13 Sut = 30 kpsi, Suc = 109 kpsi Use MM: (a) σA, σB = 20, 20 kpsi Eq. (5-32a): n = 30 20 = 1.5 Ans. (b) σA, σB = ± � (15)2 = 15, −15 kpsi Eq. (5-32a) n = 30 15 = 2 Ans. (c) σA, σB = −80, −80 kpsi For the 3rd quadrant, there is no solution but use Eq. (5-32c). Eq. (5-32c): 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 Chapter 5 127 (d) σA, σB = 15, −25 kpsi, |σB|σA| = 25/15 > 1, Eq. (5-32b): (109 − 30)15 109(30) − −25 109 = 1 n n = 1.69 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 = 4.9 2.9 = 1.69 5-14 Given: AISI 1006 CD steel, F = 0.55 N, P = 8.0 kN, and T = 30 N · m, applying the DE theory to stress elements A and B with Sy = 280 MPa A: σx = 32Fl πd3 + 4P πd2 = 32(0.55)(103)(0.1) π(0.0203) + 4(8)(103) π(0.0202) = 95.49(106) Pa = 95.49 MPa O (d) (b) (a) (c) E F C B A G D H 1 cm � 10 kpsi �B �A 128 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design τxy = 16T πd3 = 16(30) π(0.0203) = 19.10(106) Pa = 19.10 MPa σ ′ = � σ 2 x + 3τ 2 xy �1/2 = [95.492 + 3(19.1)2]1/2 = 101.1 MPa n = Sy σ ′ = 280 101.1 = 2.77 Ans. B: σx = 4P πd3 = 4(8)(103) π(0.0202) = 25.47(106) Pa = 25.47 MPa τxy = 16T πd3 + 4 3 V A = 16(30) π(0.0203) + 4 3 � 0.55(103) (π/4)(0.0202) � = 21.43(106) Pa = 21.43 MPa σ ′ = [25.472 + 3(21.432)]1/2 = 45.02 MPa n = 280 45.02 = 6.22 Ans. 5-15 Sy = 32 kpsi At A, M = 6(190) = 1 140 lbf·in, T = 4(190) = 760 lbf · in. σx = 32M πd3 = 32(1140) π(3/4)3 = 27 520 psi τzx = 16T πd3 = 16(760) π(3/4)3 = 9175 psi τmax = ��27 520 2 �2 + 91752 = 16 540 psi n = Sy 2τmax = 32 2(16.54) = 0.967 Ans. MSS predicts yielding 5-16 From Prob. 4-15, σx = 27.52 kpsi, τzx = 9.175 kpsi. For Eq. (5-15), adjusted for coordinates, σ ′ = � 27.522 + 3(9.175)2�1/2 = 31.78 kpsi n = Sy σ ′ = 32 31.78 = 1.01 Ans. DE predicts no yielding, but it is extremely close. Shaft size should be increased. Chapter 5 129 5-17 Design decisions required: • Material and condition • Design factor • Failure model • Diameter of pin Using F = 416 lbf from Ex. 5-3 σmax = 32M πd3 d = � 32M πσmax �1/3 Decision 1: Select the same material and condition of Ex. 5-3 (AISI 1035 steel, Sy = 81 000). Decision 2: Since we prefer the pin to yield, set nd a little larger than 1. Further explana- tion will follow. Decision 3: Use the Distortion Energy static failure theory. Decision 4: Initially set nd = 1 σmax = Sy nd = Sy 1 = 81 000 psi d = �32(416)(15) π(81 000) �1/3 = 0.922 in Choose preferred size of d = 1.000 in F = π(1)3(81 000) 32(15) = 530 lbf n = 530 416 = 1.274 Set design factor to nd = 1.274 Adequacy Assessment: σmax = Sy nd = 81 000 1.274 = 63 580 psi d = �32(416)(15) π(63 580) �1/3 = 1.000 in (OK ) F = π(1)3(81 000) 32(15) = 530 lbf n = 530 416 = 1.274 (OK) 130 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 5-18 For a thin walled cylinder made of AISI 1018 steel, Sy = 54 kpsi, Sut = 64 kpsi. The state of stress is σt = pd 4t = p(8) 4(0.05) = 40p, σl = pd 8t = 20p, σr = −p These three are all principal stresses. Therefore, σ ′ = 1 √ 2 [(σ1 − σ2)2 + (σ2 − σ3)2 + (σ3 − σ1)2]1/2 = 1 √ 2 [(40p − 20p)2 + (20p + p)2 + (−p − 40p)2] = 35.51p = 54 ⇒ p = 1.52 kpsi (for yield) Ans. For rupture, 35.51p .= 64 ⇒ p .= 1.80 kpsi Ans. 5-19 For hot-forged AISI steel w = 0.282 lbf/in3, Sy = 30 kpsi and ν = 0.292. Then ρ = w/g = 0.282/386 lbf · s2/in;ri = 3 in;ro = 5 in;r2 i = 9;r2 o = 25; 3 + ν = 3.292; 1 + 3ν =1.876. Eq. (3-55) for r = ri becomes σt = ρω2 �3 + ν 8 � � 2r2 o + r2 i � 1 − 1 + 3ν 3 + ν �� Rearranging and substituting the above values: Sy ω2 = 0.282 386 �3.292 8 � � 50 + 9 � 1 − 1.876 3.292 �� = 0.016 19 Setting the tangential stress equal to the yield stress, ω = � 30 000 0.016 19 �1/2 = 1361 rad/s or n = 60ω/2π = 60(1361)/(2π) = 13 000 rev/min Now check the stresses at r = (rori)1/2, or r = [5(3)]1/2 = 3.873 in σr = ρω2 �3 + ν 8 � (ro − ri)2 = 0.282ω2 386 �3.292 8 � (5 − 3)2 = 0.001 203ω2 Applying Eq. (3-55) for σt σt = ω2 �0.282 386 � �3.292 8 � � 9 + 25 + 9(25) 15 − 1.876(15) 3.292 � = 0.012 16ω2 Chapter 5 131 Using the Distortion-Energy theory σ ′ = � σ 2 t − σrσt + σ 2 r �1/2 = 0.011 61ω2 Solving ω = � 30 000 0.011 61 �1/2 = 1607 rad/s So the inner radius governs and n = 13 000 rev/min Ans. 5-20 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) 2(0.065) = 13 212 psi σl = pdi 4t = 500(3.37) 4(0.065) = 6481 psi σr = −pi = −500 psi These are all principal stresses, thus, σ ′ = 1 √ 2 {(13 212 − 6481)2 + [6481 − (−500)]2 + (−500 − 13 212)2}1/2 σ ′ = 11 876 psi n = Sy σ ′ = 46 000 σ ′ = 46 000 11 876 = 3.87 Ans. 5-21 Table A-20 gives Sy as 320 MPa. The maximum significant stress condition occurs at ri where σ1 = σr = 0, σ2 = 0, and σ3 = σt. From Eq. (3-49) for r = ri , pi = 0, σt = − 2r2 o po r2o − r2 i = − 2(1502) po 1502 − 1002 = −3.6po σ ′ = 3.6po = Sy = 320 po = 320 3.6 = 88.9 MPa Ans. 5-22 Sut = 30 kpsi, w = 0.260 lbf/in3, ν = 0.211, 3 + ν = 3.211, 1 + 3ν = 1.633. At the inner radius, from Prob. 5-19 σt ω2 = ρ �3 + ν 8 � � 2r2 o + r2 i − 1 + 3ν 3 + ν r2 i � 132 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Here r2 o = 25, r2 i = 9, and so σt ω2 = 0.260 386 �3.211 8 � � 50 + 9 − 1.633(9) 3.211 � = 0.0147 Since σr is of the same sign, we use M2M failure criteria in the first quadrant. From Table A-24, Sut = 31 kpsi, thus, ω = � 31 000 0.0147 �1/2 = 1452 rad/s rpm = 60ω/(2π) = 60(1452)/(2π) = 13 866 rev/min Using the grade number of 30 for Sut = 30 000 kpsi gives a bursting speed of 13640 rev/min. 5-23 TC = (360 − 27)(3) = 1000 lbf · in, TB = (300 − 50)(4) = 1000 lbf · in In xy plane, MB = 223(8) = 1784 lbf · in and MC = 127(6) = 762 lbf · in. In the xz plane, MB = 848 lbf · in and MC = 1686 lbf · in. The resultants are MB = [(1784)2 + (848)2]1/2 = 1975 lbf · in MC = [(1686)2 + (762)2]1/2 = 1850 lbf · in So point B governs and the stresses are τxy = 16T πd3 = 16(1000) πd3 = 5093 d3 psi σx = 32MB πd3 = 32(1975) πd3 = 20 120 d3 psi Then σA, σB = σx 2 ± ��σx 2 �2 + τ 2 xy �1/2 σA, σB = 1 d3    20.12 2 ± ��20.12 2 �2 + (5.09)2 �1/2   = (10.06 ± 11.27) d3 kpsi · in3 B A D C xz plane 106 lbf 8" 8" 6" 281 lbf 387 lbf B A D C 223 lbf 8" 8" 6" 350 lbf 127 lbf xy plane y Chapter 5 133 Then σA = 10.06 + 11.27 d3 = 21.33 d3 kpsi and σB = 10.06 − 11.27 d3 = −1.21 d3 kpsi For this state of stress, use the Brittle-Coulomb-Mohr theory for illustration. Here we use Sut(min) = 25 kpsi, Suc(min) = 97 kpsi, and Eq. (5-31b) to arrive at 21.33 25d3 − −1.21 97d3 = 1 2.8 Solving gives d = 1.34 in. So use d = 1 3/8 in Ans. Note that this has been solved as a statics problem. Fatigue will be considered in the next chapter. 5-24 As in Prob. 5-23, we will assume this to be statics problem. Since the proportions are un- changed, the bearing reactions will be the same as in Prob. 5-23. Thus xy plane: MB = 223(4) = 892 lbf · in xz plane: MB = 106(4) = 424 lbf · in So Mmax = [(892)2 + (424)2]1/2 = 988 lbf · in σx = 32MB πd3 = 32(988) πd3 = 10 060 d3 psi Since the torsional stress is unchanged, τxz = 5.09/d3 kpsi σA, σB = 1 d3    �10.06 2 � ± ��10.06 2 �2 + (5.09)2 �1/2   σA = 12.19/d3 and σB = −2.13/d3 Using the Brittle-Coulomb-Mohr, as was used in Prob. 5-23, gives 12.19 25d3 − −2.13 97d3 = 1 2.8 Solving gives d = 1 1/8 in. Ans. 5-25 (FA)t = 300 cos 20 = 281.9 lbf, (FA)r = 300 sin 20 = 102.6 lbf T = 281.9(12) = 3383 lbf · in, (FC)t = 3383 5 = 676.6 lbf (FC)r = 676.6 tan 20 = 246.3 lbf 134 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design MA = 20 � 193.72 + 233.52 = 6068 lbf · in MB = 10 � 246.32 + 676.62 = 7200 lbf · in (maximum) σx = 32(7200) πd3 = 73 340 d3 τxy = 16(3383) πd3 = 17 230 d3 σ ′ = � σ 2 x + 3τ 2 xy �1/2 = Sy n ��73 340 d3 �2 + 3 �17 230 d3 �2�1/2 = 79 180 d3 = 60 000 3.5 d = 1.665 in so use a standard diameter size of 1.75 in Ans. 5-26 From Prob. 5-25, τmax = ��σx 2 �2 + τ 2 xy �1/2 = Sy 2n ��73 340 2d3 �2 + �17 230 d3 �2�1/2 = 40 516 d3 = 60 000 2(3.5) d = 1.678 in so use 1.75 in Ans. 5-27 T = (270 − 50)(0.150) = 33 N · m, Sy = 370 MPa (T1 − 0.15T1)(0.125) = 33 ⇒ T1 = 310.6 N, T2 = 0.15(310.6) = 46.6 N (T1 + T2) cos 45 = 252.6 N xz plane z 107.0 N 174.4 N 252.6 N 320 N 300 400 150 y 163.4 N 89.2 N 252.6 N 300 400 150 xy plane A B C O xy plane x y A B C ROy = 193.7 lbf RBy = 158.1 lbf 281.9 lbf 20" 16" 10" 246.3 lbf O xz plane x z A B C ROz = 233.5 lbf RBz = 807.5 lbf O 102.6 lbf 20" 16" 10" 676.6 lbf Chapter 5 135 MA = 0.3 � 163.42 + 1072 = 58.59 N · m (maximum) MB = 0.15 � 89.22 + 174.42 = 29.38 N · m σx = 32(58.59) πd3 = 596.8 d3 τxy = 16(33) πd3 = 168.1 d3 σ ′ = � σ 2 x + 3τ 2 xy �1/2 = ��596.8 d3 �2 + 3 �168.1 d3 �2�1/2 = 664.0 d3 = 370(106) 3.0 d = 17.5(10−3) m = 17.5 mm, so use 18 mm Ans. 5-28 From Prob. 5-27, τmax = ��σx 2 �2 + τ 2 xy �1/2 = Sy 2n ��596.8 2d3 �2 + �168.1 d3 �2�1/2 = 342.5 d3 = 370(106) 2(3.0) d = 17.7(10−3) m = 17.7 mm, so use 18 mm Ans. 5-29 For the loading scheme shown in Figure (c), Mmax = F 2 �a 2 + b 4 � = 4.4 2 (6 + 4.5) = 23.1 N · m For a stress element at A: σx = 32M πd3 = 32(23.1)(103) π(12)3 = 136.2 MPa The shear at C is τxy = 4(F/2) 3πd2/4 = 4(4.4/2)(103) 3π(12)2/4 = 25.94 MPa τmax = ��136.2 2 �2�1/2 = 68.1 MPa Since Sy = 220 MPa, Ssy = 220/2 = 110 MPa, and n = Ssy τmax = 110 68.1 = 1.62 Ans. x y A B V M C 136 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design For the loading scheme depicted in Figure (d) Mmax = F 2 �a + b 2 � − F 2 �1 2 � �b 2 �2 = F 2 �a 2 + b 4 � This result is the same as that obtained for Figure (c). At point B, we also have a surface compression of σy = −F A = −F bd − −4.4(103) 18(12) = −20.4 MPa With σx = −136.2 MPa. From a Mohrs circle diagram, τmax = 136.2/2 = 68.1 MPa. n = 110 68.1 = 1.62 MPa Ans. 5-30 Based on Figure (c) and using Eq. (5-15) σ ′ = � σ 2 x �1/2 = (136.22)1/2 = 136.2 MPa n = Sy σ ′ = 220 136.2 = 1.62 Ans. Based on Figure (d) and using Eq. (5-15) and the solution of Prob. 5-29, σ ′ = � σ 2 x − σxσy + σ 2 y �1/2 = [(−136.2)2 − (−136.2)(−20.4) + (−20.4)2]1/2 = 127.2 MPa n = Sy σ ′ = 220 127.2 = 1.73 Ans. 5-31 When the ring is set, the hoop tension in the ring is equal to the screw tension. σt = r2 i pi r2o − r2 i � 1 + r2 o r2 � We have the hoop tension at any radius. The differential hoop tension dF is dF = wσt dr F = � ro ri wσt dr = wr2 i pi r2o − r2 i � ro ri � 1 + r2 o r2 � dr = wri pi (1) dF r w Chapter 5 137 The screw equation is Fi = T 0.2d (2) From Eqs. (1) and (2) pi = F wri = T 0.2dwri dFx = f piri dθ Fx = � 2π o f piwri dθ = f Tw 0.2dwri ri � 2π o dθ = 2π f T 0.2d Ans. 5-32 (a) From Prob. 5-31, T = 0.2Fid Fi = T 0.2d = 190 0.2(0.25) = 3800 lbf Ans. (b) From Prob. 5-31, F = wri pi pi = F wri = Fi wri = 3800 0.5(0.5) = 15 200 psi Ans. (c) σt = r2 i pi r2o − r2 i � 1 + r2 o r � r=ri = pi � r2 i + r2 o � r2o − r2 i = 15 200(0.52 + 12) 12 − 0.52 = 25 333 psi Ans. σr = −pi = −15 200 psi (d) τmax = σ1 − σ3 2 = σt − σr 2 = 25 333 − (−15 200) 2 = 20 267 psi Ans. σ ′ = � σ 2 A + σ 2 B − σAσB �1/2 = [25 3332 + (−15 200)2 − 25 333(−15 200)]1/2 = 35 466 psi Ans. (e) Maximum Shear hypothesis n = Ssy τmax = 0.5Sy τmax = 0.5(63) 20.267 = 1.55 Ans. Distortion Energy theory n = Sy σ ′ = 63 35 466 = 1.78 Ans. dFx pirid� 138 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 5-33 The moment about the center caused by force F is Fre where re is the effective radius. This is balanced by the moment about the center caused by the tangential (hoop) stress. Fre = � ro ri rσtw dr = wpir2 i r2o − r2 i � ro ri � r + r2 o r � dr re = wpir2 i F � r2o − r2 i � � r2 o − r2 i 2 + r2 o ln ro ri � From Prob. 5-31, F = wri pi. Therefore, re = ri r2o − r2 i � r2 o − r2 i 2 + r2 o ln ro ri � For the conditions of Prob. 5-31, ri = 0.5 and ro = 1 in re = 0.5 12 − 0.52 �12 − 0.52 2 + 12 ln 1 0.5 � = 0.712 in 5-34 δnom = 0.0005 in (a) From Eq. (3-57) p = 30(106)(0.0005) (13) �(1.52 − 12)(12 − 0.52) 2(1.52 − 0.52) � = 3516 psi Ans. Inner member: Eq. (3-58) (σt)i = −p R2 + r2 i R2 − r2 i = −3516 �12 + 0.52 12 − 0.52 � = −5860 psi (σr)i = −p = −3516 psi Eq. (5-13) σ ′ i = � σ 2 A − σAσB + σ 2 B �1/2 = [(−5860)2 − (−5860)(−3516) + (−3516)2]1/2 = 5110 psi Ans. Outer member: Eq. (3-59) (σt)o = 3516 �1.52 + 12 1.52 − 12 � = 9142 psi (σr)o = −p = −3516 psi Eq. (5-13) σ ′ o = [91422 − 9142(−3516) + (−3516)2]1/2 = 11 320 psi Ans. R �t 1 2 " 1"R re r Chapter 5 139 (b) For a solid inner tube, p = 30(106)(0.0005) 1 �(1.52 − 12)(12) 2(12)(1.52) � = 4167 psi Ans. (σt)i = −p = −4167 psi, (σr)i = −4167 psi σ ′ i = [(−4167)2 − (−4167)(−4167) + (−4167)2]1/2 = 4167 psi Ans. (σt)o = 4167 �1.52 + 12 1.52 − 12 � = 10 830 psi, (σr)o = −4167 psi σ ′ o = [10 8302 − 10 830(−4167) + (−4167)2]1/2 = 13 410 psi Ans. 5-35 Using Eq. (3-57) with diametral values, p = 207(103)(0.02) (503) �(752 − 502)(502 − 252) 2(752 − 252) � = 19.41 MPa Ans. Eq. (3-58) (σt)i = −19.41 �502 + 252 502 − 252 � = −32.35 MPa (σr)i = −19.41 MPa Eq. (5-13) σ ′ i = [(−32.35)2 − (−32.35)(−19.41) + (−19.41)2]1/2 = 28.20 MPa Ans. Eq. (3-59) (σt)o = 19.41 �752 + 502 752 − 502 � = 50.47 MPa, (σr)o = −19.41 MPa σ ′ o = [50.472 − 50.47(−19.41) + (−19.41)2]1/2 = 62.48 MPa Ans. 5-36 Max. shrink-fit conditions: Diametral interference δd = 50.01 − 49.97 = 0.04 mm. Equa- tion (3-57) using diametral values: p = 207(103)0.04 503 �(752 − 502)(502 − 252) 2(752 − 252) � = 38.81 MPa Ans. Eq. (3-58): (σt)i = −38.81 �502 + 252 502 − 252 � = −64.68 MPa (σr)i = −38.81 MPa Eq. (5-13): σ ′ i = � (−64.68)2 − (−64.68)(−38.81) + (−38.81)2�1/2 = 56.39 MPa Ans. 140 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 5-37 δ = 1.9998 2 − 1.999 2 = 0.0004 in Eq. (3-56) 0.0004 = p(1) 14.5(106) �22 + 12 22 − 12 + 0.211 � + p(1) 30(106) �12 + 0 12 − 0 − 0.292 � p = 2613 psi Applying Eq. (4-58) at R, (σt)o = 2613 �22 + 12 22 − 12 � = 4355 psi (σr)o = −2613 psi, Sut = 20 kpsi, Suc = 83 kpsi ���� σo σA ���� = 2613 4355 < 1, ∴ use Eq. (5-32a) h = Sut/σA = 20/4.355 = 4.59 Ans. 5-38 E = 30(106) psi, ν = 0.292, I = (π/64)(24 − 1.54) = 0.5369 in4 Eq. (3-57) can be written in terms of diameters, p = Eδd D �� d2 o − D2� � D2 − d2 i � 2D2� d2o − d2 i � � = 30(106) 1.75 (0.002 46) �(22 − 1.752)(1.752 − 1.52) 2(1.752)(22 − 1.52) � = 2997 psi = 2.997 kpsi Outer member: Outer radius: (σt)o = 1.752(2.997) 22 − 1.752 (2) = 19.58 kpsi, (σr)o = 0 Inner radius: (σt)i = 1.752(2.997) 22 − 1.752 � 1 + 22 1.752 � = 22.58 kpsi, (σr)i = −2.997 kpsi Bending: ro: (σx)o = 6.000(2/2) 0.5369 = 11.18 kpsi ri: (σx)i = 6.000(1.75/2) 0.5369 = 9.78 kpsi Torsion: J = 2I = 1.0738 in4 ro: (τxy)o = 8.000(2/2) 1.0738 = 7.45 kpsi ri: (τxy)i = 8.000(1.75/2) 1.0738 = 6.52 kpsi Chapter 5 141 Outer radius is plane stress σx = 11.18 kpsi, σy = 19.58 kpsi, τxy = 7.45 kpsi Eq. (5-15) σ ′ = [11.182 − (11.18)(19.58) + 19.582 + 3(7.452)]1/2 = Sy no = 60 no 21.35 = 60 no ⇒ no = 2.81 Ans. Inner radius, 3D state of stress From Eq. (5-14) with τyz = τzx = 0 σ ′ = 1 √ 2 [(9.78 − 22.58)2 + (22.58 + 2.997)2 + (−2.997 − 9.78)2 + 6(6.52)2]1/2 = 60 ni 24.86 = 60 ni ⇒ ni = 2.41 Ans. 5-39 From Prob. 5-38: p = 2.997 kpsi, I = 0.5369 in4, J = 1.0738 in4 Inner member: Outer radius: (σt)o = −2.997 �(0.8752 + 0.752) (0.8752 − 0.752) � = −19.60 kpsi (σr)o = −2.997 kpsi Inner radius: (σt)i = −2(2.997)(0.8752) 0.8752 − 0.752 = −22.59 kpsi (σr)i = 0 Bending: ro: (σx)o = 6(0.875) 0.5369 = 9.78 kpsi ri: (σx)i = 6(0.75) 0.5369 = 8.38 kpsi Torsion: ro: (τxy)o = 8(0.875) 1.0738 = 6.52 kpsi ri: (τxy)i = 8(0.75) 1.0738 = 5.59 kpsi y x —2.997 kpsi 9.78 kpsi 22.58 kpsi 6.52 kpsi z 142 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design The inner radius is in plane stress: σx = 8.38 kpsi, σy = −22.59 kpsi, τxy = 5.59 kpsi σ ′ i = [8.382 − (8.38)(−22.59) + (−22.59)2 + 3(5.592)]1/2 = 29.4 kpsi ni = Sy σ ′ i = 60 29.4 = 2.04 Ans. Outer radius experiences a radial stress, σr σ ′ o = 1 √ 2 � (−19.60 + 2.997)2 + (−2.997 − 9.78)2 + (9.78 + 19.60)2 + 6(6.52)2�1/2 = 27.9 kpsi no = 60 27.9 = 2.15 Ans. 5-40 σp = 1 2 � 2 KI √ 2πr cos θ 2 � ± �� KI √ 2πr sin θ 2 cos θ 2 sin 3θ 2 �2 + � KI √ 2πr sin θ 2 cos θ 2 cos 3θ 2 �2�1/2 = KI √ 2πr � cos θ 2 ± � sin2 θ 2 cos2 θ 2 sin2 3θ 2 + sin2 θ 2 cos2 θ 2 cos2 3θ 2 �1/2� = KI √ 2πr � cos θ 2 ± cos θ 2 sin θ 2 � = KI √ 2πr cos θ 2 � 1 ± sin θ 2 � Plane stress: The third principal stress is zero and σ1 = KI √ 2πr cos θ 2 � 1 + sin θ 2 � , σ2 = KI √ 2πr cos θ 2 � 1 − sin θ 2 � , σ3 = 0 Ans. Plane strain: σ1 and σ2 equations still valid however, σ3 = ν(σx + σy) = 2ν KI √ 2πr cos θ 2 Ans. 5-41 For θ = 0 and plane strain, the principal stress equations of Prob. 5-40 give σ1 = σ2 = KI √ 2πr , σ3 = 2ν KI √ 2πr = 2νσ1 (a) DE: 1 √ 2 [(σ1 − σ1)2 + (σ1 − 2νσ1)2 + (2νσ1 − σ1)2]1/2 = Sy σ1 − 2νσ1 = Sy For ν = 1 3, � 1 − 2 �1 3 �� σ1 = Sy ⇒ σ1 = 3Sy Ans. Chapter 5 143 (b) MSS: σ1 − σ3 = Sy ⇒ σ1 − 2νσ1 = Sy ν = 1 3 ⇒ σ1 = 3Sy Ans. σ3 = 2 3σ1 Radius of largest circle R = 1 2 � σ1 − 2 3σ1 � = σ1 6 5-42 (a) Ignoring stress concentration F = Sy A = 160(4)(0.5) = 320 kips Ans. (b) From Fig. 6-36: h/b = 1, a/b = 0.625/4 = 0.1563, β = 1.3 Eq. (6-51) 70 = 1.3 F 4(0.5) � π(0.625) F = 76.9 kips Ans. 5-43 Given: a = 12.5 mm, KIc = 80 MPa · √m, Sy = 1200 MPa, Sut = 1350 MPa ro = 350 2 = 175 mm, ri = 350 − 50 2 = 150 mm a/(ro − ri) = 12.5 175 − 150 = 0.5 ri/ro = 150 175 = 0.857 Fig. 5-30: β .= 2.5 Eq. (5-37): KIc = βσ√πa 80 = 2.5σ � π(0.0125) σ = 161.5 MPa Eq. (3-50) at r = ro: σt = r2 i pi r2o − r2 i (2) 161.5 = 1502 pi(2) 1752 − 1502 pi = 29.2 MPa Ans. �1, �2 � �1 � 2 3 144 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 5-44 (a) First convert the data to radial dimensions to agree with the formulations of Fig. 3-33. Thus ro = 0.5625 ± 0.001in ri = 0.1875 ± 0.001in Ro = 0.375 ± 0.0002in Ri = 0.376 ± 0.0002in The stochastic nature of the dimensions affects the δ = |Ri| − |Ro| relation in Eq. (3-57) but not the others. Set R = (1/2)(Ri + Ro) = 0.3755. From Eq. (3-57) p = Eδ R �� r2 o − R2� � R2 − r2 i � 2R2 � r2o − r2 i � � Substituting and solving with E = 30 Mpsi gives p = 18.70(106) δ Since δ = Ri − Ro ¯δ = ¯Ri − ¯Ro = 0.376 − 0.375 = 0.001 in and ˆσδ = ��0.0002 4 �2 + �0.0002 4 �2�1/2 = 0.000 070 7 in Then Cδ = ˆσδ ¯δ = 0.000 070 7 0.001 = 0.0707 The tangential inner-cylinder stress at the shrink-fit surface is given by σit = −p ¯R2 + ¯r2 i ¯R2 − ¯r2 i = −18.70(106) δ �0.37552 + 0.18752 0.37552 − 0.18752 � = −31.1(106) δ ¯σit = −31.1(106) ¯δ = −31.1(106)(0.001) = −31.1(103) psi Also ˆσσit = |Cδ ¯σit| = 0.0707(−31.1)103 = 2899 psi σit = N(−31 100, 2899) psi Ans. Chapter 5 145 (b) The tangential stress for the outer cylinder at the shrink-fit surface is given by σot = p � ¯r2 o + ¯R2 ¯r2o − ¯R2 � = 18.70(106) δ �0.56252 + 0.37552 0.56252 − 0.37552 � = 48.76(106) δ psi ¯σot = 48.76(106)(0.001) = 48.76(103) psi ˆσσot = Cδ ¯σot = 0.0707(48.76)(103) = 34.45 psi � σot = N(48 760, 3445) psi Ans. 5-45 From Prob. 5-44, 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(106) δ ¯p = 18.70(106)(0.001) = 18.7(103) psi ˆσp = Cδ ¯p = 0.0707(18.70)(103) = 1322 psi and so p = N(18.7, 1.32) kpsi and σor = −N(18.7, 1.32) kpsi These represent the principal stresses. The von Mises stress is next assessed. ¯σA = 48.8 kpsi, ¯σB = −18.7 kpsi k = ¯σB/¯σA = −18.7/48.8 = −0.383 ¯σ ′ = ¯σA(1 − k + k2)1/2 = 48.8[1 − (−0.383) + (−0.383)2]1/2 = 60.4 kpsi ˆσσ ′ = Cp ¯σ ′ = 0.0707(60.4) = 4.27 kpsi Using the interference equation z = − ¯S − ¯σ ′ � ˆσ 2 S + ˆσ 2 σ ′ �1/2 = − 95.5 − 60.4 [(6.59)2 + (4.27)2]1/2 = −4.5 pf = α = 0.000 003 40, or about 3 chances in a million. Ans. 146 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 5-46 σt = pd 2t = 6000N(1, 0.083 33)(0.75) 2(0.125) = 18N(1, 0.083 33) kpsi σl = pd 4t = 6000N(1, 0.083 33)(0.75) 4(0.125) = 9N(1, 0.083 33) kpsi σr = −p = −6000N(1, 0.083 33) kpsi These three stresses are principal stresses whose variability is due to the loading. From Eq. (5-12), we find the von Mises stress to be σ ′ = �(18 − 9)2 + [9 − (−6)]2 + (−6 − 18)2 2 �1/2 = 21.0 kpsi ˆσσ ′ = Cp ¯σ ′ = 0.083 33(21.0) = 1.75 kpsi z = − ¯S − ¯σ ′ � ˆσ 2 S + ˆσ 2 σ ′ �1/2 = 50 − 21.0 (4.12 + 1.752)1/2 = −6.5 The reliability is very high R = 1 − �(6.5) = 1 − 4.02(10−11) .= 1 Ans.