HW Ch11_14 16 18

HW Ch11_14 16 18 - PROBLEM 11-14 Statement: shaft. If the...

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T g 1627 ft lbf = T g 19524 in lbf = T g N g N p T p = 4. We could have calculated the torque on the gear shaft directly without finding the gear shaft speed, T g 1627 ft lbf = T g 19524 in lbf = T g P ϖ g = g 403.509 rpm = g N p N g p = 3. For the gear shaft 2. The gear shaft will rotate at a lower speed, which is determined by the gear ratio. (The speed will be decreased i proportion to the ratio and the torque will be increased by the reciprocal of the ratio). T p 656.5 ft lbf = T p 7878 in lbf = T p P p = 1. For the pinion shaft Solution: See Mathcad file P1114. Assumptions: There is no loss of power in the gear mesh (100% efficiency). P 125 hp = Transmitted power N g 57 = gear p 1000 rpm = Pinion speed N p 23 = pinion Tooth numbers: Given: rpm 2 π rad min 1 - = Units: Statement: If the gearset in Problem 11-3 transmits 125 HP at 1000 pinion rpm, find the torque on each shaft. PROBLEM 11-14
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Pitch-line velocity (fpm) V t p d ( 29 d p d ( 29 n 2 = Transmitted load (lbf) W t p d ( 29 H V t p d ( 29 = 3. Set the application factor, K a K a 1 = 4. Write the equation for the dynamic load factor, K v B 0.25 12 Q v - ( 29 0.6667 = B 0.52 = A 50 56 1 B - ( ) + = A 76.878 = K v p d ( 29 A A V t p d ( 29 min ft + B = 5. Tentatively choose the mounting factor,
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HW Ch11_14 16 18 - PROBLEM 11-14 Statement: shaft. If the...

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