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9_ch 07 Mechanical Design budynas_SM_ch07

# 9_ch 07 Mechanical Design budynas_SM_ch07 - can be lightly...

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186 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design k b = ± 1 0 . 3 ² 0 . 107 = 0 . 879 k c = k d = k e = k f = 1 S e = 0 . 869(0 . 879)(36) = 27 . 5 kpsi From Fig. A-15-8 with D / d = 1 . 25 and r / d = 0 . 03, K ts = 1 . 8 . From Fig. A-15-9 with D / d = 1 . 25 and r / d = 0 . 03, K t = 2 . 3 From Fig. 6-20 with r = 0 . 03 in, q = 0 . 65 . From Fig. 6-21 with r = 0 . 03 in, q s = 0 . 83 Eq. (6-31): K f = 1 + 0 . 65(2 . 3 1) = 1 . 85 K fs = 1 + 0 . 83(1 . 8 1) = 1 . 66 Using DE-elliptic, Eq. (7-11) with M m = T a = 0, 1 n = 16 π (1 3 ) ³ 4 ´ 1 . 85(360) 27 500 µ 2 + 3 ´ 1 . 66(192) 39 500 µ 2 1 / 2 n = 3 . 89 Perform a similar analysis at the proﬁle keyway under the gear. The main problem with the design is the undersized shaft overhang with excessive slope at the gear. The use of crowned-teeth in the gears will eliminate this problem. 7-7 (a) One possible shaft layout is shown. Both bearings and the gear will be located against shoulders. The gear and the motor will transmit the torque through keys. The bearings
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Unformatted text preview: can be lightly pressed onto the shaft. The left bearing will locate the shaft in the housing, while the right bearing will ﬂoat in the housing. (b) From summing moments around the shaft axis, the tangential transmitted load through the gear will be W t = T / ( d / 2) = 2500 / (4 / 2) = 1250 lbf The radial component of gear force is related by the pressure angle. W r = W t tan φ = 1250 tan 20 ◦ = 455 lbf W = [ W 2 r + W 2 t ] 1 / 2 = (455 2 + 1250 2 ) 1 / 2 = 1330 lbf Reactions R A and R B , and the load W are all in the same plane. From force and moment balance, R A = 1330(2 / 11) = 242 lbf R B = 1330(9 / 11) = 1088 lbf M max = R A (9) = (242)(9) = 2178 lbf · in...
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