- #1

Moranovich

- 10

- 0

**1.**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–1.

(ii) Shaft 2 is to rotate at 400 revs min–1 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.

**3.**Attempt at solution:

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

Please help I have been at this for so long...

Many thanks.