---
title: "Elastic-plastic torsion of a circular shaft"
subject: "Mechanics of Solids"
chapter: 12
key: "semester_5/mechanics_of_solids:plastic-torsion-circular-shaft"
source: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/plastic-torsion-circular-shaft
site: BCT Engineering
---

# Elastic-plastic torsion of a circular shaft

Mechanics of Solids · Chapter 12: Introduction to Plastic Range Stress and Strain

- Asked in: year not given, 8 marks
- Video: [Plastic Torsion](https://www.youtube.com/watch?v=_6s-XZdmkIU) — Salah Alnomani
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/plastic-torsion-circular-shaft
- Whole chapter: https://www.bctengineer.com/md/semester_5/mechanics_of_solids/chapter-12

## Question

A solid circular shaft of diameter 50 mm is made of an elastic-perfectly plastic material with shear yield stress $\tau_y = 150$ MPa and $G = 80$ GPa. Find (a) the torque at first yield, (b) the fully plastic torque, (c) the torque and the twist per metre when the elastic core has a radius of 15 mm, and (d) the residual shear stress at the surface and at the centre when the shaft is unloaded after being fully plastic.

## Short answer

Torque at first yield Ty = 3.68 kN m. Fully plastic torque Tp = 4.91 kN m. With core radius 15 mm, torque T = 4.64 kN m, twist per metre = 0.125 rad/m. After full plastic unloading, residual shear stress: -50 MPa at surface, +150 MPa at centre.

## Answer

#### Data

$R = 25$ mm, $\tau_y = 150$ MPa, $G = 80\times10^3$ MPa, $J = \pi R^4/2 = 613592$ mm$^4$.

#### (a) Torque at first yield

$$
T_y = \frac{\tau_y J}{R} = \frac{\pi R^3\tau_y}{2} = \frac{150\times613592}{25} = 3.682\times10^6\ \text{N mm} = 3.68\ \text{kN m}
$$

#### (b) Fully plastic torque

All fibres carry $\tau_y$:

$$
T_p = \int_0^R \tau_y\,r\,(2\pi r\,dr) = \frac{2\pi}{3}\,\tau_yR^3 = \frac{2\pi}{3}\times150\times15625 = 4.909\ \text{kN m}
$$

$$
\frac{T_p}{T_y} = \frac43 = 1.333
$$

so the shape factor of a solid circular shaft in torsion is 1.333.

#### (c) Elastic-plastic stage, core radius $r_c = 15$ mm

The stress is $\tau_y r/r_c$ for $r < r_c$ and $\tau_y$ for $r_c<r<R$:

$$
T = \underbrace{\frac{\pi\tau_y r_c^3}{2}}_{\text{core}} + \underbrace{\frac{2\pi\tau_y}{3}\left(R^3 - r_c^3\right)}_{\text{plastic ring}} = \frac{\pi\tau_y}{6}\left(4R^3 - r_c^3\right)
$$

$$
T = \frac{\pi\times150}{6}\left(4\times15625 - 3375\right) = \frac{\pi\times150}{6}\times59125 = 4.644\ \text{kN m}
$$

Twist per unit length is governed by the core yielding at $r_c$:

$$
\theta = \frac{\tau_y}{G\,r_c} = \frac{150}{80\times10^3\times15} = 1.25\times10^{-4}\ \text{rad/mm} = 0.125\ \text{rad/m}
$$

(At first yield $\theta_y = \tau_y/(GR) = 0.075$ rad/m, so the twist is $1.67$ times the yield twist.)

#### (d) Residual stresses after unloading from $T_p$

Unloading is elastic, so the recovery stress at radius $r$ is $\tau' = T_p r/J$.

$$
\tau'(R) = \frac{4.909\times10^6\times25}{613592} = 200.0\ \text{MPa}
$$

Residual stress $= \tau_y - \tau'$:

| Radius | Plastic stress | Recovery | Residual |
|---|---|---|---|
| $r = 25$ mm (surface) | 150 MPa | 200 MPa | -50 MPa |
| $r = 0$ (centre) | 150 MPa | 0 | +150 MPa |

The residual stress reverses sign between the centre and the surface (zero at $r = 18.75$ mm, where $200r/25 = 150$). The torque from the residual stresses is zero, as the shaft is unloaded.

**Answer:** (a) $T_y = 3.68$ kN m; (b) $T_p = 4.91$ kN m; (c) $T = 4.64$ kN m, $\theta = 0.125$ rad/m; (d) residual shear stress $-50$ MPa at the surface and $+150$ MPa at the centre.

## Check yourself

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

1. What is the ratio of fully plastic torque to first-yield torque for a solid circular shaft?
   A. 1.0
   B. 1.5
   C. 1.333
   D. 2.0
   Answer: C. Tp/Ty = (2 pi/3 tau_y R^3)/(pi/2 tau_y R^3) = 4/3. (Key point: Fully plastic torque ratio)

2. What is the torque at first yield for R = 25 mm and tau_y = 150 MPa?
   A. 4.91 kN m
   B. 3.68 kN m
   C. 4.64 kN m
   D. 1.84 kN m
   Answer: B. T_y = pi R^3 tau_y/2 = 3.68 x 10^6 N mm. (Key point: Torque at first yield)

3. How is the twist per metre found when the elastic core radius is 15 mm?
   A. theta = tau_y/(G R) = 0.075 rad/m
   B. theta = T L / (G J) using the fully plastic torque
   C. theta = tau_y/(G r_c) = 0.125 rad/m
   D. theta = tau_y r_c / G = 28 rad/m
   Answer: C. The core edge just yields at tau_y, so theta = tau_y/(G r_c) = 1.25e-4 rad/mm = 0.125 rad/m. (Key point: Twist with elastic core)

4. How are residual stresses found after unloading from the fully plastic state?
   A. Plastic stress minus an elastic recovery stress T_p r/J
   B. Plastic stress plus the yield stress
   C. Recovery stress is zero because the shaft unloads plastically
   D. Add the elastic stress T_p r/J to the plastic stress
   Answer: A. Unloading is elastic, so subtract tau' = Tp r/J (200 MPa at the surface) from tau_y. (Key point: Residual stress method)

5. What are the residual shear stresses at the surface and centre?
   A. +50 MPa at the surface and 0 at the centre
   B. 0 at the surface and +150 MPa at the centre
   C. -50 MPa at the surface and +150 MPa at the centre
   D. -150 MPa at the surface and -50 MPa at the centre
   Answer: C. Surface: 150 - 200 = -50 MPa; centre: 150 - 0 = +150 MPa. (Key point: Residual stress values)

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