---
title: "Derive the torsional equation"
subject: "Strength of Materials"
chapter: 6
key: "semester_3/strength_of_materials:derive-torsional-equation"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/derive-torsional-equation
site: BCT Engineering
---

# Derive the torsional equation

Strength of Materials · Chapter 6: Torsion

- Asked in: 2075 Asoj, 8 marks; 2068 Baisakh (old course), 6 marks (asked 2 times)
- Video: [TORSION EQUATION : DERIVATION](https://www.youtube.com/watch?v=vtybG1Or620) — TECHNICAL CLASSES
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/derive-torsional-equation
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-6

## Question

Derive torsional equation. $\frac{T}{J} = \frac{\tau_s}{R} = \frac{G\varphi}{L}$

## Short answer

From geometry, shear strain γ = Rφ/L; by Hooke's law τ = Gγ, so τ/r = Gφ/L. Torque T = ∫ τ r dA = (τ_s/R)∫ r² dA = τ_s J/R. Combining yields T/J = τ_s/R = Gφ/L.

## Answer

**Torsional equation:** $\dfrac{T}{J} = \dfrac{\tau_s}{R} = \dfrac{G\varphi}{L}$, where $T$ is the torque, $J$ the polar moment of inertia, $\tau_s$ the shear stress at the surface (radius $R$), $G$ the modulus of rigidity, $\varphi$ the angle of twist and $L$ the length.

#### Assumptions
1. The shaft is straight, of uniform circular section (solid or hollow) and of homogeneous, isotropic material.
2. Plane sections perpendicular to the axis remain plane and circular after twisting; radii remain straight.
3. The twist is uniform along the length and the stress is within the elastic limit (Hooke's law).
4. Shear stress is proportional to shear strain; the shaft is loaded only by torque.

#### Derivation
```text
  fixed end                 free end
  |---------------------------|  A -> A' (twist)
  |   O ======================|=> torque T
  |      L          B         B'
  |
  section: OB = R, BB' = R*phi, AB' tilts by gamma
```

Consider a shaft of length $L$ fixed at one end and twisted by torque $T$ at the other. A line AB on the surface moves to AB'. Let $\varphi$ be the angle of twist and $\gamma$ the shear strain at radius $R$. From the geometry:

$$
BB' = R\varphi = L\gamma \Rightarrow \gamma = \frac{R\varphi}{L}
$$

For a point at radius $r$, the shear strain is $\gamma_r = \dfrac{r\varphi}{L}$.

From Hooke's law in shear, $\tau = G\gamma$:

$$
\tau_r = G\frac{r\varphi}{L} \quad\Rightarrow\quad \frac{\tau_r}{r} = \frac{G\varphi}{L} \quad(1)
$$

so the shear stress varies linearly with radius, and at the surface $\dfrac{\tau_s}{R} = \dfrac{G\varphi}{L}$.

Consider an elemental ring of radius $r$ and thickness $dr$, area $dA = 2\pi r\,dr$. The force on it is $\tau_r\,dA$ and its moment about the axis is $\tau_r\,dA\cdot r$. The torque resisted by the entire section is

$$
T = \int \tau_r\,r\,dA = \int \frac{\tau_s}{R}\,r^2\,dA = \frac{\tau_s}{R}\int r^2\,dA
$$

Since $J = \int r^2\,dA$ (polar moment of inertia):

$$
T = \frac{\tau_s}{R}J \quad\Rightarrow\quad \frac{T}{J} = \frac{\tau_s}{R} \quad(2)
$$

Combining (1) and (2):

$$
\boxed{\frac{T}{J} = \frac{\tau_s}{R} = \frac{G\varphi}{L}}
$$

**Polar moment of inertia:** solid shaft $J = \dfrac{\pi D^4}{32}$; hollow shaft $J = \dfrac{\pi}{32}(D^4-d^4)$. Then $\tau_{max} = \dfrac{16T}{\pi D^3}$ for a solid shaft and $\varphi = \dfrac{TL}{GJ}$, where $GJ$ is the torsional rigidity.

## Check yourself

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

1. Which assumption is made?
   A. Sections warp
   B. Plane sections remain plane and radii remain straight
   C. Material is anisotropic
   D. Stress exceeds elastic limit
   Answer: B. This gives strain proportional to radius. (Key point: Assumption)

2. What relation do geometry give?
   A. gamma = L phi / R
   B. gamma = phi / (R L)
   C. gamma = R phi / L
   D. gamma = R L / phi
   Answer: C. Arc BB' = R phi = L gamma. (Key point: Shear strain)

3. How does shear stress vary with radius?
   A. Parabolic
   B. Constant
   C. Inversely with r
   D. Linearly, tau_r/r = G phi/L
   Answer: D. tau = G gamma and gamma is proportional to r. (Key point: Stress distribution)

4. How is T/J = tau_s/R obtained?
   A. T = integral tau_r r dA = (tau_s/R) integral r^2 dA, and J = integral r^2 dA
   B. T = tau_s x area
   C. T = integral tau_r dA
   D. T = G phi J
   Answer: A. Moment of ring force about axis, summed over the section. (Key point: Torque integral)

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