Question

exercise
power is 10kw, speed of rotation is 500r/min for the rotor of a motor with the diameterd=40mm, the permissible shearing stress is τ=100mpa. try to check its strength.

Explanation:

Step1: Calculate Torque from Power

Power \( P = 10\mathrm{kW} = 10\times10^{3}\mathrm{W} \), angular velocity \( \omega = \frac{500}{60}\times2\pi\mathrm{rad/s} \). Power formula \( P = T\omega \), so torque \( T=\frac{P}{\omega}=\frac{10\times10^{3}}{\frac{500}{60}\times2\pi}=\frac{10\times10^{3}\times60}{500\times2\pi}=\frac{600000}{1000\pi}\approx190.99\mathrm{N\cdot m} \).

Step2: Calculate Shearing Stress

Diameter \( d = 40\mathrm{mm}=0.04\mathrm{m} \), radius \( r=\frac{d}{2}=0.02\mathrm{m} \). Polar moment of inertia for solid shaft \( I_{p}=\frac{\pi d^{4}}{32} \), shear stress formula \( \tau=\frac{T r}{I_{p}}=\frac{T\times\frac{d}{2}}{\frac{\pi d^{4}}{32}}=\frac{16T}{\pi d^{3}} \). Substitute \( T = 190.99\mathrm{N\cdot m} \), \( d = 0.04\mathrm{m} \): \( \tau=\frac{16\times190.99}{\pi\times(0.04)^{3}}=\frac{3055.84}{\pi\times6.4\times10^{-5}}\approx\frac{3055.84}{2.0106\times10^{-4}}\approx15.2\times10^{6}\mathrm{Pa}=15.2\mathrm{MPa} \).

Step3: Compare with Permissible Stress

Permissible shearing stress \( [\tau]=100\mathrm{MPa} \), calculated \( \tau = 15.2\mathrm{MPa}<100\mathrm{MPa} \), so the strength is sufficient.

Answer:

The calculated shearing stress is approximately \( 15.2\mathrm{MPa} \), which is less than the permissible shearing stress \( 100\mathrm{MPa} \), so the motor rotor has sufficient strength.