Module 11 · Lesson 11.1
Deriving the torsion formula
From the geometry of twisting to T/J = τ/r = Gθ/L.
Try it
Torsion: the core does almost nothing — so take it out
Shear stress rises straight-line with radius: zero at the centre, maximum at the surface. The material near the axis is barely working. Bore it out and watch how little strength you lose for how much weight.
- Outer diameter
- 100 mm
- Bore diameter
- 0 mm
- Polar J
- 9.82 ×10⁶ mm⁴
- Surface shear stress
- 25.5 N/mm²
- Stress at the bore
- 0.0 N/mm²
- Twist per metre
- 0.360 °/m
- Weight vs solid same OD
- 100 %
- Stress rise vs solid same OD
- 0.0 %
Things worth trying
- Start with the solid shaft. The stress profile on the right is a straight triangle: zero at the centre, maximum at the surface. That is τ = Tr/J — linear in radius, because the shaft twists like a stack of rigid discs and the outer fibres slide furthest.
- Look at what that means: the material at the centre carries almost no stress. It is weight without work. The torque is ∫τr dA, and since τ ∝ r the integrand goes with r³ — heavily weighted to the surface.
- Now bore out the centre. Drag the bore to 50%. You have removed a QUARTER of the material, and the surface stress rose by under 7%. The greyed triangle on the right is what you took out — the lightly stressed core.
- Why so little penalty? J goes with the FOURTH power of diameter, so J = π(D⁴−d⁴)/32 barely notices a bore at half the diameter: (0.5)⁴ is only 1/16. Weight goes with the square, so it falls much faster than strength.
- Bore it right out to 85% — a thin tube. Even now the surface stress is only about 40% higher than solid, for a fraction of the weight. There is a limit not shown here: a very thin wall buckles.
- Read the 'same weight' rows once you have a bore. They spend the saved weight the smart way: on a bigger solid shaft would be wasteful, but a hollow shaft of the same weight has a larger outer diameter, and torque capacity goes with the CUBE of diameter.
- That comparison is the whole reason drive shafts, axles, scaffold tubes and bicycle frames are hollow. Not to save material for its own sake — to put every gram where the stress actually is.
- Notice the profile band for a hollow shaft: it never touches zero. The stress runs from the bore value up to the surface, a narrow high band. The wall works nearly uniformly, which is exactly what efficient use of material looks like.
Why this matters
Torsion turns up in curved beams, in edge beams supporting cantilevers, in crane runway girders and in every drive shaft. The circular case is one of the few problems in solid mechanics with an exact, simple solution — and the derivation is a clean example of geometry, compatibility and equilibrium working together.
By the end of this lesson you should be able to
- Describe how a circular shaft deforms under torque
- Derive shear strain from the geometry of twist
- Combine with Hooke's law and equilibrium to get the torsion formula
What you should already know
- Shear stress and shear strain (Module 7)
- Shear modulus G and Hooke's law in shear (Module 7)
- Second moment of area (Module 9)
From first principles
Torsion of a circular shaft
We want to show: that T/J = τ/r = Gθ/L for a circular shaft in torsion.
Take a circular shaft and draw a straight line along its surface. Twist one end. The line becomes a shallow helix, but two things stay true, and they are the whole reason this problem has an exact solution. First, plane cross-sections stay plane — the shaft does not warp. Second, radii stay straight: a radius drawn on the end face simply rotates, it does not bend. That only holds because a circle looks the same from every direction about its centre. If radii stay straight, then a point twice as far from the centre must be dragged twice as far round. So the shear strain, and therefore the shear stress, grows linearly with radius: zero at the centre, greatest at the surface.
Worked example
Stress and twist in a solid shaft
Given
- Solid circular shaft, diameter 80 mm
- Applied torque 6.0 kN·m
- Length 1.5 m, G = 79 000 N/mm²
Find
The maximum shear stress and the angle of twist.
Assumptions
- Elastic behaviour
- Pure torsion
- Prismatic circular shaft
Worked example
Why a hollow shaft is more efficient
Given
- Compare a solid 80 mm shaft with a hollow shaft, 80 mm outside and 50 mm inside
- Same material and same length
Find
How much torsional stiffness is lost, and how much weight is saved.
Assumptions
- Same G and L for both
Practice
A solid circular shaft of diameter 60 mm carries a torque of 2.5 kN·m. What is the maximum shear stress?
Practice
That same 60 mm shaft is 2.0 m long with G = 79 000 N/mm². What is the angle of twist in degrees?
Check yourself
Where is the shear stress zero in a solid circular shaft under torque?
Practice
A solid circular shaft of diameter 80 mm carries a torque of 5.0 kNm. What is the maximum shear stress, in N/mm²?
Summary
- Radii stay straight, so γ = rθ/L grows linearly with radius
- Hooke's law gives τ = Grθ/L, and equilibrium introduces J = ∫r² dA
- T/J = τ/r = Gθ/L, so τmax = Tr/J and θ = TL/GJ
- GJ is torsional rigidity; hollow shafts keep most of the stiffness for far less weight
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