Question

A bronze, spur pinion rotating at 500 rpm drives a cast iron spur gear at a transmission ratio of 4 : 1. The allowable static stresses for bronze pinion and cast iron gear are 85 MPa and 103 MPa respectively. The pinion has 16 standard ${20}^{\circ }$ full depth invloved teeth of module 8 mm. The face width of both the gear is 90 mm. What is the power that can be transmitted from gear? Take velocity factor $C_V=\frac{3}{3+V}$ (where V is speed in m/s)

• A. 62.47 kW
• B. 29.52 kW
• C. 146.312 kW
• D. 53.93 kW

Answer 1

The correct option is B 29.52 kW

$D_p=m{T}_p=8\times 16=128mm$

$T_G=4\times T_P=4\times 16=64$

$D_G=m{T}_G=8\times 64=512mm$

We-know that tooth form factor is :

$y=0.154-\frac{0.912}{T}$

For pinion $y_p=(0.154-\frac{0.912}{16})=0.097$

$y_p\times {\sigma }_P=0.097\times 85=8.245$

For gear, $y_G=0.514-\frac{0.912}{64}=0.13975$

$y_G\times {\sigma }_G=0.13975\times 103=14.394$

As $y_p\times {\sigma }_P$ is less than $y_G\times {\sigma }_G$, then design is to be done with respect to pinion,

$P_{eff}=\pi \times y_P\times {\sigma }_P\times m\times b=\pi \times 8.245\times 8\times 90$

$=18649.75N$

$P_{eff}=\frac{P_t}{C_v}$

$P_t=P_{eff}\times C_v$

Power = $P_t\times V$

where $V=\frac{\pi \times D_p\times N}{60}=\frac{\pi \times 0.128\times 500}{60}=3.351m/s$

So, Power = $18649.75\times \frac{3}{3+3.351}\times 3.351$

$=29520.69W=29.521kW$