Torsion of shafts
Torsion is twisting about the long axis. Shafts that drive pumps, curved bridge girders, edge beams that support slabs on one side and the spandrels of buildings all carry torque. For a solid or hollow circular shaft the theory is simple because plane sections stay plane and radii stay straight, so shear strain grows linearly from zero at the centre to a maximum at the surface.
The torsion equation T / J = τ / r = G θ / L links torque T, polar moment J, shear stress τ at radius r, shear modulus G and angle of twist θ over length L. A hollow shaft is more efficient than a solid one, since the material near the centre carries little stress. Power transmitted by a shaft is P = 2π N T / 60 with N in rpm.
Non-circular sections warp and need other methods. A thin-walled closed tube carries torque by a constant shear flow q = T / (2 A₀), with A₀ the area enclosed by the wall's centre line, which is why box girders are so torsionally stiff. Open sections such as I-beams and channels have very small St Venant (uniform) torsional stiffness; where warping is restrained, for example at a fixed end, warping resistance adds considerably to their torsional capacity. In reinforced concrete, torsion is resisted by closed stirrups plus extra longitudinal bars, combined with shear and bending in design.
T / J = τ / r = G θ / LSolid shaft: J = π D⁴ / 32, τ_max = 16 T / (π D³)Hollow: J = π (D⁴ − d⁴) / 32Power: P = 2π N T / 60; thin closed tube: q = T / (2A₀)
- T/J = τ/r = Gθ/L applies only to circular (solid or hollow) shafts in linear elastic torsion with uniform torque and section.
- Non-circular sections warp; use St Venant, thin-walled or warping torsion theory instead.
Problem. A solid steel shaft transmits 60 kW at 300 rpm. Allowable shear is 50 MPa. Find the minimum diameter, then the twist over 2 m with G = 80 GPa.
- T = 60 000 × 60 / (2π × 300) = 1 910 Nm = 1.91 × 10⁶ Nmm.
- D³ = 16 T / (π τ) = 16 × 1.91 × 10⁶ / (π × 50) = 194 600, so D = 58 mm; adopt 60 mm.
- J = π × 60⁴ / 32 = 1.272 × 10⁶ mm⁴.
- θ = T L / (G J) = 1.91 × 10⁶ × 2 000 / (80 000 × 1.272 × 10⁶) = 0.0375 rad ≈ 2.15°.
Answer. Use 60 mm diameter; twist about 2.2° over 2 m
Quick quiz
1Where is torsional shear stress largest in a circular shaft?
At the outer surface.
2Why are box girders good in torsion?
A closed cell carries torque by shear flow round a large enclosed area.
3What reinforcement resists torsion in RC beams?
Closed stirrups plus longitudinal bars.
- IS 456 (India): Plain and reinforced concrete: general rules for materials, durability, design and construction.
Clauses and documents to read
- IS 456:2000 cl. 41 (torsion)
Cited by number and clause only. We do not reproduce clause text, tables or figures (BIS Act 2016, s.10(5) and s.11). Check the current edition and amendments before use.
References and further reading
Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.
- J. M. Gere, B. J. Goodno, Mechanics of Materials, Cengage, latest edition.
- F. P. Beer, E. R. Johnston, J. T. DeWolf, Mechanics of Materials, McGraw Hill, latest edition.
- E. P. Popov, Engineering Mechanics of Solids, Prentice Hall, 2nd edition.
Free and official sources
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