Question

$\text{A steel spur pinion has a module (m) of 1.25 mm, 20 teeth and}{20}^{\circ }\text{pressure angle. The pinion rotates at 1200 rpm and transmits power to a 60 teeth gear. The face width (F) is 50 mm, Lewis form factor Y= 0.322 and a dynamic factor}{K}_v=\mathrm{1.26.}\text{The bending stress}\left(\sigma \right)\text{induced in a tooth can be calculated by using the Lewis formula given below.}$
$\text{If the maximum bending stress experienced by the pinion is 400MPa. The power transmitted is}$
$\text{(round off to one decimal place),}$
$\text{Lewis formula:}\sigma =\frac{K_v{W}^t}{FmY},where{W}^t\text{is the tangential load acting on the pinion.}$

Answer 1

As per given date : $m=1.25mm$
$Z_p=20\text{teeth};\varphi ={20}^{\circ }$
$N_P=1200rpm;Z_G=60\text{teeth}$
$F_t\times c_{v}s=bmy[{\sigma }_b{]}_{max}$
$c_v=1.26$
$F_t\times 1.26\times 1=50\times 1.25\times 0.322\times 400$$F_t=6388.88N$
Power Transmitted = $F_t\times v=\frac{F_t\times \pi D_{p}N}{60}$
$\frac{6388.88\times \pi \times 1.25\times 20\times 1200}{60}=10kW$