Structural Analysis Fundamentals / Torsion
Torque diagrams and plastic torsion
The fourth internal action, drawn like the others — and what a shaft does after it yields.
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Read the lesson register and outstanding review notesWhy this matters
Shear force and bending moment diagrams are second nature by now. Torque deserves the same treatment: it is the fourth internal action, it varies along a member, and the place it peaks is not always where you would guess. And as with bending, a ductile shaft carries substantially more torque after its surface has yielded than elastic theory allows.
By the end of this lesson you should be able to
- Draw a torque diagram for a shaft with several applied torques
- Identify the design torque from the diagram
- Describe how the yielded zone spreads inwards in plastic torsion
- Calculate the fully plastic torque and the torsional shape factor
What you should already know
- The torsion formula T/J = τ/r = Gθ/L (this module)
- Shear force and bending moment diagrams (Module 3)
- Plastic bending and the yield plateau (Module 9)
For a straight shaft with concentrated torques, internal torque is piecewise constant. Sum torques on one side of each cut using a consistent right-hand sign convention and check equilibrium at the far end. Distributed torques produce a changing diagram between points.
For a uniform shaft, the greatest absolute torque gives the greatest elastic torsional stress. Where section, material or restraints change, check each part separately. Integrate T/(GJ) along the length to obtain twist; TL/(GJ) applies to a segment with constant T and GJ.
Predict first
A shaft is driven at one end by a 12 kNm motor torque, and drives machines taking 8 kNm and 4 kNm at two points along its length. Where is the internal torque greatest?
For a solid circular shaft made from an elastic-perfectly plastic material with shear yield stress τy, elastic shear stress increases linearly with radius. First yield occurs at R, giving Ty = πτyR³/2.
As twist increases, an outer yielded annulus spreads inwards. The fully plastic limiting distribution has τy throughout the section apart from the axis. Integrating gives Tp = 2πτyR³/3, hence Tp/Ty = 4/3. The elastic core approaches zero as twist increases; the fully plastic distribution is a limiting idealisation.
This ratio assumes sufficient ductility and no strain hardening, instability or fracture. It is not a universal reserve for every material.
Worked example
Torque diagram and plastic capacity
Given
- Solid circular shaft, diameter 80 mm, τy = 150 N/mm²
- Driven by a 12.0 kNm torque at end A; machines take 8.00 kNm at B and 4.00 kNm at C
Find
The torque diagram, and the elastic and plastic capacities of the section.
Practice
A shaft is driven by 12.0 kNm at one end and drives machines taking 8.00 kNm and then 4.00 kNm. What is the internal torque in the middle span, between the two machines, in kNm?
Practice
A solid circular shaft of radius 40.0 mm has τy = 150 N/mm². What is its fully plastic torque, in kNm?
Practice
What is the torsional shape factor Tp/Ty of a solid circular shaft?
Summary
- Concentrated applied torques produce steps in the torque diagram; distributed torque changes it continuously
- It must close at the far end — if it does not, something has been missed
- Locate peak torque and check each segment against its section, material and restraint conditions
- First yield: Ty = π τy R³/2; fully plastic: Tp = 2π τy R³/3
- For an elastic-perfectly plastic solid circle, the torsional shape factor is 4/3
- A thin tube has almost no plastic reserve, because none of its material is understressed