A gear train is to be designed to transmit power from shaft 1 to shaft 2 with th

Question

A gear train is to be designed to transmit power from shaft 1 to shaft 2 with the following condition: (i) Shaft 1 (driven by an electric motor) rotates at 100 revs min%u20131. (ii) Shaft 2 is to rotate at 400 revs min%u20131 in the opposite direction to shaft 1 against a load of 200 Nm. (iii) The centre of shaft 1 is 300 mm from shaft 2. (iv) The minimum number of teeth on any gear is 15, all gears must have a multiple of 5 teeth and have a module of 2 mm. (v) The maximum number of gears permissible is 4 gears and the diameter of the maximum gear must be minimised. (vi) The centres of the gears should lie on a line which is as close to straight as possible. (vii) All shafts have a frictional resistance of 5 Nm. Carry out the following: (a) Design a gear train which satisfies the above criteria. Use a sketch to illustrate your design, label the gears A, B, C, etc. from the driver to the driven gear, and state the number of teeth on each gear. (b) Determine the input power required at shaft 1. (c) Specify the efficiency of the gear train as a percentage. (d) Determine an equation for the efficiency of the gear train in terms of the load (torque) on shaft 2 (all other factors remaining constant).

Explanation / Answer

I have worked out a gear train with 4 gears (2 idlers) so the 1st and 4th gear turn in opposite directions. This gives me the correct final speed 400rpm but not the correct center distance 300mm. N = no of teeth N2/N1 = 15/60 = 0.25 100rpm/0.25 = 400rpm at gear 2 N2/N3 = 15/30 = 0.5 400rpm x 0.5 = 200rpm at gear 3 N3/N4 = 30/15 = 2 200rpm x 2 = 400rpm at gear 4 Also I have considered 2 gears but the statement in the question diameter of the maximum gear must be minimised. makes me think this is not correct, but it gives me the correct center distance! Center = (PCD + PCD)/2 Pitch Circle Diameter PCD = N x M N=no of teeth M = module ie 2 PCD1= 240 x 2 = 480 PCD2= 60 x 2 = 120 center = (480+120)/2 = 300mm

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