Skip to main content

Hollow shaft 10/7 cm: power transmitted

Short answer

J = (π/32)(100⁴-70⁴) = 7.46×10⁶ mm⁴, T_max = τJ/R = 80×7.46×10⁶/50 = 11.94 kNm. Mean torque = 11.94/1.3 = 9.18 kNm. Power = 2π×300×9.18/60 = 288.5 kW.

A hollow steel shaft having 10 cm outer diameter and 7 cm inside diameter is rotating at a speed of 300 rpm. If the permissible shear stress is 80 N/mm280\ \text{N/mm}^2 and the maximum torque is 1.3 times the mean torque. Determine the power transmitted by the shaft.

Data: D=100D = 100 mm, d=70d = 70 mm, N=300N = 300 rpm, τallow=80 N/mm2\tau_{allow} = 80\ \text{N/mm}^2, Tmax=1.3 TmeanT_{max} = 1.3\,T_{mean}.

Maximum torque the shaft can carry

J=π32(D4−d4)=π32(1004−704)=7.4603×106 mm4J = \frac{\pi}{32}(D^4 - d^4) = \frac{\pi}{32}(100^4 - 70^4) = 7.4603\times10^{6}\ \text{mm}^4 Tmax=τJR=80×7.4603×10650=1.1936×107 N⋅mm=11.936 kN⋅mT_{max} = \frac{\tau J}{R} = \frac{80\times7.4603\times10^{6}}{50} = 1.1936\times10^{7}\ \text{N·mm} = 11.936\ \text{kN·m}

Mean torque

Tmean=Tmax1.3=11.9361.3=9.182 kN⋅mT_{mean} = \frac{T_{max}}{1.3} = \frac{11.936}{1.3} = 9.182\ \text{kN·m}

Power transmitted

P=2πNTmean60=2π(300)(9181.9 N⋅m)60=288.5 kWP = \frac{2\pi N T_{mean}}{60} = \frac{2\pi(300)(9181.9\ \text{N·m})}{60} = 288.5\ \text{kW}

Answer: power transmitted ≈288.5 kW\approx 288.5\ \text{kW}.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Strength of Materials chapter 6 questions →