Budynas ch07

# Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)

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Chapter 7 7-1 (a) DE-Gerber, Eq. (7-10): A = 4[2 . 2(600)] 2 + 3[1 . 8(400)] 2 1 / 2 = 2920 lbf · in B = 4[2 . 2(500)] 2 + 3[1 . 8(300)] 2 1 / 2 = 2391 lbf · in d = 8(2)(2920) π (30 000) 1 + 1 + 2(2391)(30 000) 2920(100 000) 2 1 / 2 1 / 3 = 1 . 016 in Ans . (b) DE-elliptic, Eq. (7-12) can be shown to be d = 16 n π A 2 S 2 e + B 2 S 2 y 1 / 3 = 16(2) π 2920 30 000 2 + 2391 80 000 2 1 / 3 = 1 . 012 in Ans. (c) DE-Soderberg, Eq. (7-14) can be shown to be d = 16 n π A S e + B S y 1 / 3 = 16(2) π 2920 30 000 + 2391 80 000 1 / 3 = 1 . 090 in Ans. (d) DE-Goodman: Eq. (7-8) can be shown to be d = 16 n π A S e + B S ut 1 / 3 = 16(2) π 2920 30 000 + 2391 100 000 1 / 3 = 1 . 073 in Ans. Criterion d (in) Compared to DE-Gerber DE-Gerber 1.016 DE-elliptic 1.012 0.4% lower less conservative DE-Soderberg 1.090 7.3% higher more conservative DE-Goodman 1.073 5.6% higher more conservative

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Chapter 7 179 7-2 This problem has to be done by successive trials, since S e is a function of shaft size. The material is SAE 2340 for which S ut = 1226 MPa, S y = 1130 MPa, and H B 368 . Eq. (6-19): k a = 4 . 51(1226) 0 . 265 = 0 . 685 Trial #1 : Choose d r = 22 mm Eq. (6-20): k b = 22 7 . 62 0 . 107 = 0 . 893 Eq. (6-18): S e = 0 . 685(0 . 893)(0 . 5)(1226) = 375 MPa d r = d 2 r = 0 . 75 D 2 D / 20 = 0 . 65 D D = d r 0 . 65 = 22 0 . 65 = 33 . 8 mm r = D 20 = 33 . 8 20 = 1 . 69 mm Fig. A-15-14: d = d r + 2 r = 22 + 2(1 . 69) = 25 . 4 mm d d r = 25 . 4 22 = 1 . 15 r d r = 1 . 69 22 = 0 . 077 K t = 1 . 9 Fig. A-15-15: K ts = 1 . 5 Fig. 6-20: r = 1 . 69 mm, q = 0 . 90 Fig. 6-21: r = 1 . 69 mm, q s = 0 . 97 Eq. (6-32): K f = 1 + 0 . 90(1 . 9 1) = 1 . 81 K f s = 1 + 0 . 97(1 . 5 1) = 1 . 49 We select the DE-ASME Elliptic failure criteria. Eq. (7-12) with d as d r , and M m = T a = 0, d r = 16(2 . 5) π 4 1 . 81(70)(10 3 ) 375 2 + 3 1 . 49(45)(10 3 ) 1130 2 1 / 2 1 / 3 = 20 . 6 mm
180 Solutions Manual Instructor’s Solution Manual to Accompany Mechanical Engineering Design Trial #2 : Choose d r = 20 . 6 mm k b = 20 . 6 7 . 62 0 . 107 = 0 . 899 S e = 0 . 685(0 . 899)(0 . 5)(1226) = 377 . 5 MPa D = d r 0 . 65 = 20 . 6 0 . 65 = 31 . 7 mm r = D 20 = 31 . 7 20 = 1 . 59 mm Figs. A-15-14 and A-15-15: d = d r + 2 r = 20 . 6 + 2(1 . 59) = 23 . 8 mm d d r = 23 . 8 20 . 6 = 1 . 16 r d r = 1 . 59 20 . 6 = 0 . 077 We are at the limit of readability of the figures so K t = 1 . 9, K ts = 1 . 5 q = 0 . 9, q s = 0 . 97 K f = 1 . 81 K f s = 1 . 49 Using Eq. (7-12) produces d r = 20 . 5 mm . Further iteration produces no change. Decisions : d r = 20 . 5 mm D = 20 . 5 0 . 65 = 31 . 5 mm, d = 0 . 75(31 . 5) = 23 . 6 mm Use D = 32 mm, d = 24 mm, r = 1 . 6 mm Ans. 7-3 F cos 20°( d / 2) = T , F = 2 T / ( d cos 20°) = 2(3000) / (6 cos 20°) = 1064 lbf M C = 1064(4) = 4257 lbf · in For sharp fillet radii at the shoulders, from Table 7-1, K t = 2 . 7, and K ts = 2 . 2 . Examining Figs. 6-20 and 6-21, with S ut = 80 kpsi, conservatively estimate q = 0 . 8 and q s = 0 . 9 . These estimates can be checked once a specific fillet radius is determined. Eq. (6-32): K f = 1 + (0 . 8)(2 . 7 1) = 2 . 4 K f s = 1 + (0 . 9)(2 . 2 1) = 2 . 1 (a) Static analysis using fatigue stress concentration factors: From Eq. (7-15) with M = M m , T = T m , and M a = T a = 0, σ max = 32 K f M π d 3 2 + 3 16 K f s T π d 3 2 1 / 2

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Chapter 7 181 Eq. (7-16): n = S y σ max = S y 32 K f M π d 3 2 + 3 16 K f s T π d 3 2 1 / 2 Solving for d , d = 16 n π S y 4( K f M ) 2 + 3( K f s T ) 2 1 / 2 1 / 3 = 16(2 . 5) π (60 000) 4(2 . 4)(4257) 2 + 3(2 . 1)(3000) 2 1 / 2 1 / 3 = 1 . 700 in Ans . (b) k a = 2 . 70(80) 0 . 265 = 0 . 845 Assume d = 2 . 00 in to estimate the size factor, k b = 2 0 . 3 0 .
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