Question

A solid steel shaft has to transmit 100 kW at 160 RPM. Taking allowable shear stress at 70 Mpa, find the suitable diameter of the shaft. The maximum torque transmitted in each revolution exceeds the mean by 20%.

Answer 1

Answer:

The diameter of the shaft is 80.5 mm.

Explanation:

Torsion equation is applied for the diameter of the solid shaft.

Step1

Given:

Power of the shaft is 100 kw.

Revolution per minute is 160 RPM.

Allowable shear stress is 70 Mpa.

Maximum torque is 20% more than the mean torque.  

Step2

Mean torque is calculated as follows:

$P=T\omega$

$P=T(\frac{2\pi N}{60})$

$100\times 1000=T(\frac{2\pi 160}{60})$

T=5968.31 N-m

Step3

Maximum torque is calculated as follows:

$T_{max}=(1+\frac{20}{100})T$

$T_{max}=1.2T$

$T_{max}=1.2\times 5968.31$

T_{max}=7161.97 N-m

Step4

Apply torsional equation for diameter of shaft as follows:

$\tau _{max}=\frac{T_{max}}{\frac{\pi d^{3}}{16}}$

$70\times 10^{6}=\frac{7161.97}{\frac{\pi d^{3}}{16}}$

$d^{3}=5.211\times 10^{-4}$

d=0.0805 m

or,

d=80.5 mm

Thus, the diameter of the shaft is 80.5 mm.