Module 15 · Lesson 15.4
Strain energy in shear and torsion
The other two ways a member stores the work you do on it.
Why this matters
Strain energy in axial members and in bending covers most of what you need. But a shaft stores energy in torsion, a stubby beam stores a non-trivial amount in shear, and once those terms are written down the unit-load method extends to torsion and to shear deflection with no new ideas at all.
By the end of this lesson you should be able to
- Write the strain energy stored in shear and in torsion
- Use the shear form factor and say where it comes from
- Compare the four strain-energy terms and judge which matter
- Extend the unit-load method to a member carrying torsion
What you should already know
- Strain energy in axial members and bending (this module)
- The unit-load method (this module)
- Torsion: T/J = τ/r = Gθ/L (Module 11)
All four strain-energy expressions have the same shape: the action squared, over twice the relevant rigidity, integrated along the member.
| Action | Rigidity | Strain energy |
|---|---|---|
| Axial force N | AE | ∫ N²/2AE dx |
| Bending moment M | EI | ∫ M²/2EI dx |
| Torque T | GJ | ∫ T²/2GJ dx |
| Shear force V | GA | ∫ kV²/2GA dx |
The torsional term needs no comment: it is the exact analogue of the axial one, with GJ in place of AE.
The shear term carries an extra factor k, the shear form factor, and that needs explaining. The naive expression V²L/2GA assumes the shear stress is uniform at V/A across the section. It is not — it is parabolic in a rectangle, peaking at 1.5 times the average. Because energy goes as the square of the stress, the true energy exceeds the uniform-stress estimate, and k is the correction:
- Rectangle: k = 6/5 = 1.2
- Circle: k = 10/9 ≈ 1.11
- I-section: roughly the gross area divided by the web area, since the web carries nearly all the shear
What it calculates: energy stored by the two shear-type actions
- T
- torque (N·mm)
- V
- shear force (N)
- G
- shear modulus (N/mm²)
- J
- polar second moment of area (mm⁴)
- k
- shear form factor (dimensionless)
This assumes
- Linearly elastic material
- Circular section for the torsion term, or a closed thin-walled one
- k depends only on the shape of the cross-section
In plain terms: For constant T over a length, U = T²L/2GJ, and equating that to ½Tθ recovers the twist θ = TL/GJ — the same self-consistency check that ½Pδ gave for an axial member.
Predict first
For an ordinary floor beam of span-to-depth ratio 18, which strain-energy term dominates?
With these terms written down, the unit-load method extends directly. Alongside the bending and axial contributions, add:
δ = Σ ∫ (T t)/GJ dx for torsion, and Σ ∫ (k V v)/GA dx for shear
where T and V are the real actions and t and v those under the unit load, exactly as N and n were for a truss.
The practical rule for which terms to keep:
- Trusses: axial only. Bending is negligible if the joints really do behave as pins.
- Beams and frames: bending only, unless the members are unusually stubby.
- Grillages and curved members in plan: bending and torsion — the torsion term is essential, not optional.
- Shear walls and deep beams: shear becomes significant, and often dominant.
Including every term is never wrong, only laborious. Omitting one that matters is wrong, and quietly so.
Worked example
Comparing the shear and torsion energy terms
Given
- Case A: a length of beam 2.00 m long carrying constant shear V = 50.0 kN, section 200 × 300 mm (A = 60 000 mm²), k = 6/5
- Case B: a solid circular shaft of diameter 80 mm, 2.00 m long, carrying T = 5.00 kNm
- G = 79 000 N/mm² for both
Find
The strain energy stored in each case, and a self-consistency check on the torsional one.
Practice
A member 2.00 m long carries a constant shear force of 50.0 kN. Its area is 60 000 mm², G = 79 000 N/mm² and k = 6/5. How much strain energy does shear store, in N·mm?
Practice
A solid circular shaft of diameter 80.0 mm and length 2.00 m carries a torque of 5.00 kNm, with G = 79 000 N/mm². How much strain energy does it store, in N·mm?
Practice
What is the shear form factor k for a solid circular section?
Summary
- All four strain-energy terms share the form ∫(action)²/(2 × rigidity) dx
- Torsion: ∫T²/2GJ dx, the exact analogue of the axial term
- Shear: ∫kV²/2GA dx, with k correcting for the non-uniform stress distribution
- k = 6/5 for a rectangle, 10/9 for a circle, roughly A/Aweb for an I-section
- ½Tθ reproduces the torsional energy, just as ½Pδ does the axial
- Keep the terms that matter: axial for trusses, bending for beams, bending plus torsion for grillages
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