Module 11 · Lesson 11.3
Torque diagrams and plastic torsion
The fourth internal action, drawn like the others — and what a shaft does after it yields.
Why 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)
A torque diagram is built exactly like a shear force diagram, and is simpler because torque is a single scalar along the member.
Work from one end. The internal torque at any section equals the algebraic sum of all applied torques to one side of it. Applied torques are point actions, so the diagram is a series of horizontal steps, jumping by the applied torque at each point where one acts.
The conventions worth fixing before you start:
- Adopt a sign convention and keep it: a torque is positive if its vector points away from the cut face by the right-hand rule.
- The diagram closes at the far end. If it does not, either a torque has been missed or the reaction torque is wrong.
- The largest absolute value governs the design, and it need not be at a support. Where a shaft drives several machines, the peak is often in a middle span.
With the design torque known, everything from the torsion formula follows: the surface shear stress τ = Tr/J and the twist θ = TL/GJ, computed span by span because both T and J may change along the shaft.
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?
Now the plastic side. Twist a ductile circular shaft and follow the shear-stress distribution through the same four stages as bending.
Fully elastic. τ varies linearly with radius, zero at the centre and greatest at the surface.
First yield at the surface. τ reaches τy at r = R. The torque here is the yield torque:
Ty = τy J/R = π τy R³/2
Partially plastic. The outer annulus is at τy and cannot take more, but it can go on straining. The yielded zone spreads inwards, leaving a shrinking elastic core.
Fully plastic. The elastic core vanishes and the whole section is at τy. Integrating the torque of a constant stress over the circle:
Tp = ∫₀^R τy · r · 2πr dr = 2π τy R³/3
The ratio is the torsional shape factor:
Tp/Ty = (2/3)/(1/2) = 4/3
A third as much again, exactly, for any solid circular shaft of any size and any material. Compare that with 1.5 for a rectangle in bending — a solid circle has rather less understressed material near its axis to recruit, but it still has a useful reserve.
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
- A torque diagram is a series of steps, jumping at each applied torque
- It must close at the far end — if it does not, something has been missed
- The design torque is the largest absolute value, not necessarily at a support
- First yield: Ty = π τy R³/2; fully plastic: Tp = 2π τy R³/3
- The torsional shape factor of a solid circle is exactly 4/3
- A thin tube has almost no plastic reserve, because none of its material is understressed
This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint