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5_ch 17 Mechanical Design budynas_SM_ch17

5_ch 17 Mechanical Design budynas_SM_ch17 - exp f φ − 1...

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424 Solutions Manual Instructor’s Solution Manual to Accompany Mechanical Engineering Design design by reducing the initial tension, which reduces F 1 and F 2 , thereby increasing belt life. This will bring f to 0.80 F 1 = ( F + F c ) exp( f θ ) F c exp( f θ ) 1 exp( f θ ) = exp(0 . 80 π ) = 12 . 345 Therefore F 1 = (456 . 1 + 77 . 4)(12 . 345) 77 . 4 12 . 345 1 = 573 . 7 lbf F 2 = F 1 F = 573 . 7 456 . 1 = 117 . 6 lbf F i = F 1 + F 2 2 F c = 573 . 7 + 117 . 6 2 77 . 4 = 268 . 3 lbf These are small reductions since f is close to f , but improvements nevertheless. dip = 3 C 2 w 2 F i = 3(192 / 12) 2 (0 . 393) 2(268 . 3) = 0 . 562 in 17-4 From the last equation given in the Problem Statement, exp( f φ ) = 1 1 − { 2 T / [ d ( a 0 a 2 ) b ] } 1 2 T d ( a 0 a 2 ) b exp( f φ ) = 1 2 T d ( a 0 a 2 ) b exp( f φ ) = exp( f φ ) 1 b = 1 a 0 a 2 2 T d
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Unformatted text preview: ) exp( f φ ) − 1 ² But 2 T / d = 33 000 H d / V Thus, b = 1 a − a 2 ³ 33 000 H d V ´± exp( f φ ) exp( f φ ) − 1 ² Q . E . D . 17-5 Refer to Ex. 17-1 on p. 870 for the values used below. (a) The maximum torque prior to slip is, T = 63 025 H nom K s n d n = 63 025(15)(1 . 25)(1 . 1) 1750 = 742 . 8 lbf · in Ans . The corresponding initial tension is, F i = T D ³ exp( f θ ) + 1 exp( f θ ) − 1 ´ = 742 . 8 6 ³ 11 . 17 + 1 11 . 17 − 1 ´ = 148 . 1 lbf Ans ....
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