Question

A ${20}^{\circ }$ full depth involute spur pinion of 4 mm module and 21 teeth is to transmit 15 kW at 960 rpm. Its face width is 25 mm.
The tangential force transmitted (in N) is

• A. $1776$
• B. $3552$
• C. $2611$
• D. $1305$

Answer 1

The correct option is B $3552$

Module : $m=4mm$
Number of teeth : $T=21$
Speed : $N=960rpm$
Pressure angle : $\varphi ={20}^{\circ }$
Face width: $b=25mm$
Full depth involute spur pinion
Transmission of power,
$P=15kW$ at $960rpm$
Power = Torque $\times$ Angular speed
$P=T\omega =F_t\times r\times \omega$
= Tangential force $\times$ Radius $\times \omega$
Radius : $R=\frac{D}{2}=\frac{mT}{2}=\frac{4\times 21}{2}=42mm$
$P+F_t\times r\times \omega$
$\Rightarrow 15\times 1000=F_t\times 42\times {10}^{-3}\frac{2\pi \times 960}{60}$
$\Rightarrow F_t=3552.56N$