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15_ch 17 Mechanical Design budynas_SM_ch17

# 15_ch 17 Mechanical Design budynas_SM_ch17 - 2 ²± 4 5489...

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434 Solutions Manual Instructor’s Solution Manual to Accompany Mechanical Engineering Design Table 17-13: Angle θ = θ d 180° π = (2 . 957 rad) 180° π = 169 . 42° The footnote regression equation gives K 1 without interpolation: K 1 = 0 . 143 543 + 0 . 007 468(169 . 42°) 0 . 000 015 052(169 . 42°) 2 = 0 . 9767 The design power is H d = H nom K s n d = 3(1 . 3)(1) = 3 . 9 hp From Table 17-14 for B90, K 2 = 1. From Table 17-12 take a marginal entry of H tab = 4, although extrapolation would give a slightly lower H tab . Eq. (17-17): H a = K 1 K 2 H tab = 0 . 9767(1)(4) = 3 . 91 hp The allowable F a is given by F a = 63 025 H a n ( d / 2) = 63 025(3 . 91) 3100(6 . 2 / 2) = 25 . 6 lbf The allowable torque T a is T a = F a d 2 = 25 . 6(6 . 2) 2 = 79 . 4 lbf · in From Table 17-16, K c = 0 . 965. Thus, Eq. (17-21) gives, F c = 0 . 965 5031 . 8 1000 2 = 24 . 4 lbf At incipient slip, Eq. (17-9) provides: F i = T d exp( f θ ) + 1 exp( f θ ) 1 = 79
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Unformatted text preview: 2 ²± 4 . 5489 + 1 4 . 5489 − 1 ² = 20 . 0 lbf Eq. (17-10): F 1 = F c + F i ³ 2 exp( f θ ) exp( f θ ) + 1 ´ = 24 . 4 + 20 ³ 2(4 . 5489) 4 . 5489 + 1 ´ = 57 . 2 lbf Thus, F 2 = F 1 − ± F a = 57 . 2 − 25 . 6 = 31 . 6 lbf Eq. (17-26): n f s = H a N b H d = (3 . 91)(1) 3 . 9 = 1 . 003 Ans . If we had extrapolated for H tab , the factor of safety would have been slightly less than one. Life Use Table 17-16 to ﬁnd equivalent tensions T 1 and T 2 . T 1 = F 1 + ( F b ) 1 = F 1 + K b d = 57 . 2 + 576 6 . 2 = 150 . 1 lbf T 2 = F 1 + ( F b ) 2 = F 1 + K b D = 57 . 2 + 576 12 = 105 . 2 lbf...
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