MEC-25

Shaft twist θ = TL/GJ

θ = T L /(G J) with J = π d⁴/32 for a circular shaft.

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ShaftsCoulomb–Navier

Governing equation

\theta=\dfrac{T L}{G J},\quad J=\dfrac{\pi d^4}{32}

where

T: Torque (N·m)
L: Length (m)
d: Diameter (mm)
G: Shear modulus (GPa)
\theta: Twist angle (°)
\tau_{max}: Max shear (MPa)

Lecture brief

Historical brief

Machine design grew from Coulomb torsion and Hertz contact (1881) through Soderberg fatigue and the heat-engine cycle. The lab writes shaft, bearing, contact and thermodynamic limits in SI. This sheet (MEC-25 — Shaft twist θ = TL/GJ) is the form associated with Coulomb–Navier. Working symbols: $T$ , $L$ , $d$ , $G$ $\rightarrow$ $\theta$ , $\tau_{max}$ . Integrating γ = r θ/L with τ = G γ and τ = T r / J recovers the torsion formula.

Purpose

Purpose: compute

\theta

\tau_{max}

from

T

L

d

G

in Mechanical via

\theta=\dfrac{T L}{G J},\quad J=\dfrac{\pi d^4}{32}

θ = T L /(G J) with J = π d⁴/32 for a circular shaft. Use it when a real mechanical question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given

T = 250.000\,\mathrm{N·m}

L = 0.800\,\mathrm{m}

d = 30.000\,\mathrm{mm}

G = 80.000\,\mathrm{GPa}

, the governing relation

\theta=\dfrac{T L}{G J},\quad J=\dfrac{\pi d^4}{32}

yields

\theta = 1.801\,\mathrm{^{\circ}}

\tau_{max} = 47.16\,\mathrm{MPa}

. A shaft, a torque arrow, a twist angle. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Torque T250 N·mLength L0.8 mDiameter d30 mmShear modulus G80 GPa

Outputs

Twist angle \theta1.801 °
Max shear \tau_{max}47.16 MPa

Reading speed

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Narration of this film

A shaft, a torque arrow, a twist angle.

Integrating γ = r θ/L with τ = G γ and τ = T r / J recovers the torsion formula.

Reading speed

Watch on YouTube