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MechanicalMechanics of MaterialsmediumFE ExamCoursework

Torsional Shear Stress in a Solid Shaft

A solid circular steel shaft of diameter d=50 mmd = 50\ \text{mm} transmits a torque of T=1.2 kN⋅mT = 1.2\ \text{kN·m}.

Find the maximum torsional shear stress in the shaft.

Given
d50 mmShaft diameter
T1.2 kN*mApplied torque
Hint 1

Torsion uses the polar second moment of area J, not the bending I. They differ by a factor of two for a circular section.

Hint 2

J = πd⁴/32 for a solid circular shaft; maximum shear occurs at the outer radius r = d/2.

Hint 3

There is a combined form worth remembering: τ_max = 16T/(πd³).

Worked solution — try the problem first

The torsion formula gives shear stress at radius rr:

τ=TrJ\tau = \frac{Tr}{J}

which is maximum at the outer surface, r=d/2r = d/2.

Polar second moment of area for a solid circular section:

J=πd432=π(0.050)432=π×6.25×10−632=1.9635×10−532=6.136×10−7 m4J = \frac{\pi d^4}{32} = \frac{\pi (0.050)^4}{32} = \frac{\pi \times 6.25\times10^{-6}}{32} = \frac{1.9635\times10^{-5}}{32} = 6.136\times10^{-7}\ \text{m}^4

Maximum shear stress:

τmax=1200×0.0256.136×10−7=306.136×10−7=4.889×107 Pa\tau_{max} = \frac{1200 \times 0.025}{6.136\times10^{-7}} = \frac{30}{6.136\times10^{-7}} = 4.889\times10^{7}\ \text{Pa}

τmax=48.9 MPa\boxed{\tau_{max} = 48.9\ \text{MPa}}

Faster route. Combining the two steps gives the standard solid-shaft result, worth memorising:

τmax=16Tπd3=16×1200π(0.050)3=19,2003.927×10−4=4.889×107 Pa ✓\tau_{max} = \frac{16T}{\pi d^3} = \frac{16 \times 1200}{\pi (0.050)^3} = \frac{19{,}200}{3.927\times10^{-4}} = 4.889\times10^{7}\ \text{Pa} \ \checkmark

Concepts:TorsionPolar moment of areaShear stress

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