---
title: "Hollow shaft 10/7 cm: power transmitted"
subject: "Strength of Materials"
chapter: 6
key: "semester_3/strength_of_materials:hollow-shaft-power-300rpm-13-mean"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/hollow-shaft-power-300rpm-13-mean
site: BCT Engineering
---

# Hollow shaft 10/7 cm: power transmitted

Strength of Materials · Chapter 6: Torsion

- Asked in: 2071 Chaitra, 8 marks; 2068 Chaitra, 6 marks (asked 2 times)
- Video: [Problem on Torsion/Power Transmitted by hollow shaft/Diameter of Shaft/ Strength of Materials](https://www.youtube.com/watch?v=LT8wDQWD684) — Prof.Dr.Pravin Patil
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/hollow-shaft-power-300rpm-13-mean
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-6

## Question

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 $80\ \text{N/mm}^2$ and the maximum torque is 1.3 times the mean torque. Determine the power transmitted by the shaft.

## 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.

## Answer

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

#### Maximum torque the shaft can carry
$$
J = \frac{\pi}{32}(D^4 - d^4) = \frac{\pi}{32}(100^4 - 70^4) = 7.4603\times10^{6}\ \text{mm}^4
$$

$$
T_{max} = \frac{\tau J}{R} = \frac{80\times7.4603\times10^{6}}{50} = 1.1936\times10^{7}\ \text{N·mm} = 11.936\ \text{kN·m}
$$

#### Mean torque
$$
T_{mean} = \frac{T_{max}}{1.3} = \frac{11.936}{1.3} = 9.182\ \text{kN·m}
$$

#### Power transmitted
$$
P = \frac{2\pi N T_{mean}}{60} = \frac{2\pi(300)(9181.9\ \text{N·m})}{60} = 288.5\ \text{kW}
$$

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

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which torque does tau = 80 N/mm^2 limit?
   A. The mean torque
   B. The maximum torque, T_max = tau J/R
   C. Zero torque
   D. Torque at 1.3 times speed
   Answer: B. Permissible stress applies to the peak torque. (Key point: Torque capacity)

2. Which J is used for the hollow shaft?
   A. (pi/32)(D - d)^4
   B. (pi/64)(D^4 - d^4)
   C. (pi/32)(D^4 - d^4)
   D. (pi/32) D^4
   Answer: C. Subtract the hole's polar moment. (Key point: Polar moment)

3. How is mean torque obtained?
   A. 1.3 x T_max
   B. T_max
   C. T_max/2
   D. T_max/1.3 = 9.18 kN.m
   Answer: D. Max is 1.3 times mean. (Key point: Mean torque)

4. Which formula gives power?
   A. P = 2 pi N T_mean/60 = 288.5 kW
   B. P = 2 pi N T_max/60 = 375 kW
   C. P = N T
   D. P = T/N
   Answer: A. Power uses the mean torque. (Key point: Power)

## Related in Strength of Materials

- [Hollow shaft 1670 kW at 4500 rpm: size](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ahollow-shaft-1670kw-4500rpm) — Strength of Materials, chapter 6
- [Torsional rigidity; hollow shaft vs solid shaft strength](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Atorsional-rigidity-hollow-vs-solid-300rpm) — Strength of Materials, chapter 6
- [Steel shaft 200 hp: torque, shear, twist](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Asteel-shaft-200hp-150rpm) — Strength of Materials, chapter 6
- [Hollow shaft 30 kW at 240 rpm: diameters](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ahollow-shaft-30kw-240rpm) — Strength of Materials, chapter 6
- [Solid shaft 300 kW, 120 rpm with twist limit](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Asolid-shaft-300kw-120rpm-twist-limit) — Strength of Materials, chapter 6
- [Solid shaft 300 kW; hollow shaft saving](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Asolid-shaft-300kw-percent-saving) — Strength of Materials, chapter 6

## Related in other subjects

- [Hollow versus solid shaft for power](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Ahollow-versus-solid-shaft-power) — Strength of Materials, chapter 4
- [Shaft torque measurement using strain gauges](https://www.bctengineer.com/question.md?key=semester_4%2Finstrumentation_and_measurement%3Ashaft-torque-strain-gauge-numerical) — Instrumentation and Measurement, chapter 7
- [Torque and power loss in a shaft bearing](https://www.bctengineer.com/question.md?key=semester_4%2Ffluid_mechanics_mechanical%3Ashaft-in-bearing-torque-power) — Fluid Mechanics, chapter 1
- [Hollow shaft against solid shaft](https://www.bctengineer.com/question.md?key=semester_6%2Fmachine_design_i%3Ahollow-versus-solid-shaft) — Machine Design I, chapter 5