Question

A compound gear train with gears P, Q, R, S, T and U has number of teeth 10, 20, 02 40, 50, 80 and 100 respectively. Show the arrangement of gears for a speed reduction of 50% and determine the center distance between driving and driven shafts if the diametrical pitch of the gears is 0.5

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Answer 1

Let T denote number of teeth.

Given:

Tp = 10

Tq = 20

Tr = 40

Ts = 50

Tt = 80

Tu = 100

Now,

Diametral pitch = 0.5

For Gear P:

-> 0.5 = Tp / Dp

-> 0.5 = 10 / Dp

-> Dp = 20 inch -> Rp = 10 inch

For Gear Q:

-> 0.5 = Tq / Dq

-> 0.5 = 20 / Dq

-> Dq = 40 inch -> Rq = 20 inch

For Gear R:

-> 0.5 = Tr / Dr

-> 0.5 = 40 / Dr

-> Dr = 80 inch -> Rr = 40 inch

For Gear S:

-> 0.5 = Ts / Ds

-> 0.5 = 50 / Ds

-> Ds = 100 inch -> Rs = 50 inch

For Gear T:

-> 0.5 = Tt / Dt

-> 0.5 = 80 / Dt

-> Dt = 160 inch -> Rt = 80 inch

For Gear U:

-> 0.5 = Tu / Du

-> 0.5 = 100 / Du

-> Du = 200 inch -> Ru = 100 inch

For 50% Reduction, there can be multiple arrangements possible with the given set of Gears. One among those is as mentioned below:

-> Gear P mounted on Shaft 1.

-> Gears Q and T mounted on shaft 2.

-> Gears R and S mounted on shaft 3.

-> Gear U mounted on shaft 4.

-> P meshes with Q.

-> T meshes with R.

-> S meshes with U.

Now,

Speed Reduction = 0.5 = Nu / Np = (Nu / Ns) * (Nr / Nt) * (Nq / Np)

= (Ts / Tu) * (Tt / Tr) * (Tp / Tq)

= (50 / 100) * (80 / 40) * (10 / 20)

= 0.5

Centre Distance between driving and driven shafts ( i.e. shaft 1 and shaft 4 )

= Rp + Rq + Rt + Rr + Rs + Ru

= 10 + 20 + 80 + 40 + 50 + 100

= 300 inches

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