---
title: "Plastic torque of solid shaft is 4/3"
subject: "Strength of Materials"
chapter: 6
key: "semester_3/strength_of_materials:plastic-torque-solid-shaft-4-3"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/plastic-torque-solid-shaft-4-3
site: BCT Engineering
---

# Plastic torque of solid shaft is 4/3

Strength of Materials · Chapter 6: Torsion

- Asked in: 2066 Jestha (old course), 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/plastic-torque-solid-shaft-4-3
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-6

## Question

Prove that plastic torque of a solid circular shaft made of elasto-plastic material is 4/3 times the maximum elastic torque of the shaft.

## Short answer

Maximum elastic torque T_e = πτ_y R³/2. Fully plastic torque T_p = 2πτ_y R³/3. Hence T_p = (4/3) T_e.

## Answer

**Elasto-plastic material:** stress rises linearly with strain up to the yield stress in shear $\tau_y$ and then stays constant at $\tau_y$ (perfectly plastic).

#### Maximum elastic torque
The shaft is on the point of yielding when the surface stress reaches $\tau_y$. For a solid shaft of radius $R$ with $J = \dfrac{\pi R^4}{2}$:

$$
T_e = \frac{\tau_y J}{R} = \frac{\tau_y}{R}\cdot\frac{\pi R^4}{2} = \frac{\pi\tau_y R^3}{2}
$$

#### Fully plastic torque
When the whole section yields, the stress is $\tau_y$ at every radius. Take a ring of radius $r$, thickness $dr$, area $2\pi r\,dr$:

$$
dT = \tau_y\,(2\pi r\,dr)\,r = 2\pi\tau_y r^2\,dr
$$

$$
T_p = \int_0^R 2\pi\tau_y r^2\,dr = 2\pi\tau_y\frac{R^3}{3} = \frac{2\pi\tau_y R^3}{3}
$$

#### Ratio
$$
\frac{T_p}{T_e} = \frac{2\pi\tau_y R^3/3}{\pi\tau_y R^3/2} = \frac{2}{3}\times 2 = \frac{4}{3}
$$

Hence the plastic torque is $\dfrac{4}{3}$ times the maximum elastic torque.

## Check yourself

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

1. What does an elasto-plastic material do beyond the shear yield stress?
   A. Stress keeps rising linearly
   B. Stress drops to zero
   C. Stress stays constant at tau_y
   D. Stress doubles
   Answer: C. Perfectly plastic means stress stays at tau_y while strain increases. (Key point: Elasto-plastic behaviour)

2. What is the maximum elastic torque of a solid shaft of radius R?
   A. 2 pi tau_y R^3/3
   B. pi tau_y R^3/2
   C. pi tau_y R^4/2
   D. pi tau_y R^3/4
   Answer: B. T_e = tau_y J/R with J = pi R^4/2 gives pi tau_y R^3/2. (Key point: Maximum elastic torque)

3. How is the fully plastic torque obtained?
   A. Average of elastic and zero
   B. Multiply T_e by 2
   C. Use J = pi R^4/2 with tau_y at centre
   D. Integrate tau_y * 2 pi r dr * r from 0 to R
   Answer: D. The ring torque dT = 2 pi tau_y r^2 dr integrates to 2 pi tau_y R^3/3. (Key point: Plastic torque derivation)

4. What is T_p/T_e?
   A. 2
   B. 3/2
   C. 5/4
   D. 4/3
   Answer: D. (2/3)/(1/2) = 4/3. (Key point: Ratio)

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