Answer to Question #200262 in Mechanical Engineering for Muhammad Idrees

Question

Two parallel shafts are to be connected by spur gearing. The approximate distance between the shafts is 600 mm. If one shaft runs at 120 r.p.m. and the other at 360 r.p.m., find the number of teeth on each wheel if the module is 8 mm. Also determine the exact distance apart of the shafts.

Answer 1

$\frac{N_1}{N_2}=\frac{d_2}{d_1}=\frac{360}{120}=d_2= 3d_1$

and,$x=\frac{1}{2}(d_1+d_2)=600=\frac{1}{2}*(d_1+d_2)$

$\therefore d_1=300mm,d_2=900mm$

No of teeth on the first gear, $T_1=D_1/m$

$=300/8=375.5\simeq 38$

No of teeth on the first gear, $T_2=D_2/m$

$=900/8=112.5$

$\therefore$ speed ratio is 3

No of teeth on second gear should be $38*3=114$

So,

Exact pitch circle diameter

$d_1=T_1*m=38*8=304mm$

$d_2=T_2*m=114*8=912mm$

Exact pitch distance between shafts

$x^6=\frac{d_1'+d_2'}{2}=\frac{304+912}{2}=608mm$