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Mechanics of Materials · Chapter 3 of 10 · Intermediate

Torsion

Twist a shaft and the stress is pure shear, zero at the center and greatest at the surface. The torsion formula and the angle of twist size every drive shaft and axle.

Readiness check

This chapter applies shear stress and strain to twisted shafts. Tick only what you can do closed-notes.

Recall shear stress τ and shear modulus G.
Compute areas and fourth powers of a diameter.
Relate angular speed ω to rev/min.
Recall power P = Tω.
Work in N·mm and MPa.

0 or 1 weak itemsContinue with this chapter.

2 weak itemsReview shear stress in Chapter 1.

3 or more weak itemsRevisit shear and the shear modulus first.

The core idea

A torque on a circular shaft produces shear stress that grows linearly from the axis to the surface, τ = Tρ/J, and twists the shaft by φ = TL/JG.

τ = Tρ/J, τ_max = Tr/JJ = πd⁴/32 (solid)φ = TL/JG

Under torsion, plane cross-sections stay plane and rotate rigidly, so the shear strain (and stress) is proportional to the distance from the axis. The polar moment of inertia J measures how the area is spread from the center; because stress is highest at the rim, material near the surface does the most work, which is why hollow shafts are efficient. The angle of twist φ = TL/JG is the torsional analogue of δ = PL/AE.

The skill works when: you use τ = Tr/J at the surface and φ = TL/JG, with J for the actual cross-section.

The skill breaks down when: the bending second moment is used instead of the polar J, or torsion formulas are applied to non-circular sections.

The concept. Shear stress in a twisted circular shaft rises linearly from zero at the axis to a maximum at the surface. The polar moment J sets how that torque maps to stress.

The skills, taught in order

Torsion mirrors axial loading with shear instead of normal stress. Five skills cover the model, the torsion formula, the polar moment, the twist, and power.

3.1 Torsion of circular shafts

For a circular shaft, cross-sections rotate as rigid disks and remain plane, so the shear strain varies linearly with radius, γ = ρθ/L. With Hooke's law in shear (τ = Gγ), the shear stress is also linear in radius. This clean result holds only for circular sections.

3.2 The torsion formula

The shear stress at radius ρ is τ = Tρ/J, greatest at the surface: τ_max = Tr/J. The quantity J/r = πd³/16 for a solid shaft is the torsional section modulus, the torsion analogue of S in bending.

3.3 The polar moment of inertia

J measures how area is distributed about the axis.

Section	Polar moment J
Solid circle	πd⁴/32 = πr⁴/2
Hollow circle	π(d_o⁴ − d_i⁴)/32

Because the rim carries the most stress, removing the lightly stressed core gives a hollow shaft almost the same strength at much less weight.

3.4 Angle of twist

The total twist over a length is φ = TL/JG (radians), with JG the torsional rigidity. For segments, φ = Σ T_iL_i/(J_iG_i). Twist often governs precision drives even when stress is acceptable.

3.5 Power transmission

A rotating shaft transmits power P = Tω, where ω = 2πN/60 for N in rev/min. Given the power and speed, the torque follows, and the shaft is sized from τ_max = Tr/J against an allowable shear stress.

Engineering connection: drive shafts, axles, couplings, and torsion bars are all sized here; the hollow-shaft efficiency drives aerospace and automotive design.

Worked example 1: stress and twist in a shaft

A solid steel shaft 40 mm in diameter and 1.5 m long (G = 77 GPa) carries a torque of 600 N·m. Find the maximum shear stress and the angle of twist.

Figure 1. A solid shaft under torque. The polar moment sets the surface shear stress, and the torsional rigidity JG sets the twist over the length.

ProblemFind τ_max and φ for the shaft in Figure 1.
Given / findd = 40 mm, L = 1.5 m, T = 600 N·m = 600 000 N·mm, G = 77 GPa. Find τ_max and φ.
AssumptionsCircular solid shaft, linear-elastic, plane sections remain plane.
ModelCompute J, then τ_max = Tr/J and φ = TL/JG.
EquationsJ = πd⁴/32 τ_max = Tr/J φ = TL/JG
SolveJ = π(40)⁴/32 = 251 000 mm⁴. τ_max = 600 000 × 20/251 000 = 47.7 MPa. φ = (600 000 × 1500)/(251 000 × 77 000) = 0.0465 rad = 2.66°.
CheckBoth are modest for steel, so the shaft is safe and reasonably stiff. Note J grows as d⁴, so a small diameter increase sharply lowers both stress and twist.
ConclusionOne polar moment delivers both the strength check (τ) and the stiffness check (φ). Many shafts are governed by the twist limit, not the stress.

Result. τ_max = 47.7 MPa; angle of twist φ = 2.66° over 1.5 m.

Worked example 2: sizing a power-transmission shaft

A motor delivers 15 kW at 1500 rev/min through a solid shaft. For an allowable shear stress of 40 MPa, find the torque and the minimum shaft diameter.

Figure 2. Power and speed fix the torque; the allowable shear stress then sizes the shaft through the torsional section modulus πd³/16.

ProblemFind the transmitted torque and the minimum diameter for the shaft in Figure 2.
Given / findP = 15 kW, N = 1500 rev/min, τ_allow = 40 MPa, solid shaft. Find T and d_min.
AssumptionsSteady power, pure torsion, allowable stress at the surface.
ModelTorque from power and angular speed, then diameter from τ = T/(πd³/16) set to the allowable.
Equationsω = 2πN/60, T = P/ω τ_max = 16T/(πd³) d = (16T/πτ_allow)^1/3
Solveω = 2π(1500)/60 = 157.1 rad/s, so T = 15 000/157.1 = 95.5 N·m. Then d³ = 16(95 500)/(π × 40) = 12 160 mm³, so d = 23.0 mm, use 24 mm.
CheckTorque rises as power over speed, so a slower shaft at the same power needs a larger diameter. At 24 mm the actual stress is below 40 MPa, the safe side.
ConclusionPower transmission is a torque problem in disguise: convert to torque first, then size for shear. High-speed shafts can be slender; low-speed, high-torque shafts must be stout.

Result. Torque 95.5 N·m; minimum diameter 23.0 mm, rounded to 24 mm.

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Using I instead of J	Torsion stress wrong by a factor of 2	"Is this the polar moment J?"	Torsion uses J = πd⁴/32, not the bending I = πd⁴/64.
Max stress at the center	Stress placed where it is zero	"Where is ρ largest?"	Shear stress is maximum at the surface, zero on the axis.
Torsion formula on a square bar	Wrong stress for non-circular shaft	"Is the section circular?"	τ = Tr/J holds only for circular sections.
Forgetting to convert power to torque	Stress computed from power directly	"Did I find T = P/ω first?"	Convert power and speed to torque before sizing.

Practice ladder

Level 1 · Direct skill

A solid shaft of 30 mm diameter carries 250 N·m. Find the maximum shear stress.

Show answer

J = π(30)⁴/32 = 79 520 mm⁴; τ = Tr/J = 250 000 × 15/79 520 = 47.2 MPa. Or τ = 16T/πd³ directly.

Level 2 · Mixed concept

A hollow shaft has outer 40 mm, inner 30 mm. Compare its polar moment with a solid 40 mm shaft, and its weight.

Show answer

J_hollow = π(40⁴ − 30⁴)/32 = π(2.56×10⁶ − 8.1×10⁵)/32 = 171 800 mm⁴ versus J_solid = 251 000 mm⁴, so 68% of the strength. But its area (and weight) is (40² − 30²)/40² = 44% of the solid, a big weight saving for a modest strength loss.

Level 3 · Independent problem

A shaft transmits 30 kW at 600 rev/min. Find the torque and the diameter for τ_allow = 50 MPa.

Show answer

ω = 2π(600)/60 = 62.8 rad/s; T = 30 000/62.8 = 477 N·m. d³ = 16(477 000)/(π × 50) = 48 600 mm³, d = 36.5 mm, use 38 mm. The lower speed gives a much larger torque than the 1500-rev/min example.

Level 4 · Transfer to real engineering

Find a real shaft (a car drive shaft, a screwdriver, a stirrer). Estimate the torque it carries and check its diameter against an allowable shear stress, and comment on whether twist might matter.

What good work looks like

Torque from power and speed (or applied force and radius), τ = 16T/πd³ checked against an allowable, and a note on whether the twist angle is a concern.

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I used the polar moment J, not the bending I."

"Give me five shafts; I will say whether stress or twist is likely to govern."

"Compute the shear stress." Building J and applying τ = Tr/J is the skill.

"What diameter?" Converting power to torque and sizing is the point.

Portfolio task

Analyse one shaft: from a power and speed, find the torque, the surface shear stress, and the angle of twist, then size it to an allowable stress.

Must include: J for the section, τ = Tr/J, φ = TL/JG, and a power-to-torque conversion.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Write the torsion formula.

τ = Tρ/J, maximum τ = Tr/J at the surface.

2. Give J for a solid and a hollow circular shaft.

Solid: πd⁴/32. Hollow: π(d_o⁴ − d_i⁴)/32.

3. Write the angle of twist.

φ = TL/JG (radians), with JG the torsional rigidity.

4. How do you find torque from transmitted power?

T = P/ω, with ω = 2πN/60.

5. Why are hollow shafts efficient?

Stress is highest at the rim; the lightly stressed core can be removed, saving weight for little strength loss.

TodayFinish this quiz and Levels 1 and 2 of the ladder.

+1 dayRe-derive the shaft stress and twist from a blank page.

+3 daysOne analysis and one power-sizing problem.

+7 daysMove to bending stress, Chapter 4.

+30 daysCombine torsion with bending in Chapter 8.

Textbook mapping

Item	Mapping
Primary source	Beer, Johnston, DeWolf and Mazurek, Mechanics of Materials (6th ed), Chapter 3 (Torsion)
Cross-reference	Hibbeler, Ch. 5 · Gere and Goodno, Ch. 3
Core topics	3.1 Torsion model · 3.2 Torsion formula · 3.3 Polar moment · 3.4 Angle of twist · 3.5 Power transmission
Engineering connection	Drive shafts, axles, couplings, and torsion bars.
Read next	Chapter 4: Pure Bending.