Strength of Materials

Chapter 4
Torsion of Shafts

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Torsion of Shafts

Introduction and Context

Shafts are machine elements that transmit torque and power between components such as motors, gears, pulleys, couplings, propellers, and wheels. When a shaft is subjected primarily to twisting moments about its longitudinal axis, the resulting state of stress is called torsion. Torsion analysis is central to the design of drive shafts, axles, transmission shafts, rotor assemblies, and power-transmission systems.

For circular shafts, the torsion problem has an especially useful closed-form solution. The relationship among applied torque, shear stress, geometry, and angle of twist is known as the torsion equation. This chapter develops that relation, applies it to solid, hollow, and stepped shafts, and introduces the behavior of non-circular and thin-walled sections.

Learning Objectives

After studying this chapter, you should be able to:

  1. Explain the physical behavior of a circular shaft under torsion.
  2. Derive and apply the elastic torsion equation.
  3. Determine maximum shear stress and angle of twist for solid and hollow shafts.
  4. Analyze stepped shafts subjected to torque.
  5. Calculate power transmitted by rotating shafts.
  6. Design shafts using strength and torsional-stiffness criteria.
  7. Explain the limitations of the circular-shaft theory.
  8. Describe torsion in non-circular and thin-walled sections.

1. Basic Concepts and Terminology

A torque, or twisting moment, is a moment applied about the longitudinal axis of a member. Its SI unit is newton-metre (N·m). A shaft may be subjected to a single torque, several concentrated torques, or a distributed torque caused by friction or electromagnetic action.

Important terms include:

  • Torque, T: the twisting moment applied to a shaft.
  • Shear stress, τ: tangential stress developed on a cross-section.
  • Angle of twist, θ: relative angular rotation between two cross-sections.
  • Torsional rigidity, GJ: resistance to elastic twisting.
  • Polar second moment of area, J: a geometric measure of resistance to torsional deformation.
  • Modulus of rigidity, G: material stiffness in shear.

A shaft can fail because its shear stress exceeds the allowable value, or because its angle of twist is excessive even though the stress remains acceptable. Thus, both strength and stiffness must be considered.

2. Physical Behavior of a Circular Shaft

Consider a circular shaft of length L subjected to torque T. Before loading, lines drawn along the shaft axis are straight and lines drawn around the circumference are perpendicular to the axis. Under elastic torsion:

  • cross-sections remain plane;
  • cross-sections remain circular;
  • radial lines remain straight;
  • the shaft experiences rotation of one section relative to another;
  • shear strain increases linearly with distance from the center;
  • shear stress also increases linearly with radius.

At the center, the radius is zero, so the idealized torsional shear stress is zero. The maximum stress occurs at the outer surface, where the radius is greatest.

These statements are valid for homogeneous, isotropic, circular shafts operating within the elastic range and subjected to pure torsion. They become inaccurate near abrupt geometric discontinuities, plastic yielding, severe warping, or non-circular cross-sections.

3. Derivation of the Elastic Torsion Equation

Let a shaft have length L and radius R. Suppose one end rotates through angle θ relative to the other. At a radius r, a longitudinal line moves through an arc of length rθ. Therefore, the shear strain is:

γ = rθ / L

where γ is shear strain, r is the distance from the center, θ is the angle of twist in radians, and L is the shaft length.

For a linearly elastic material, Hooke’s law in shear gives:

τ = Gγ

where τ is shear stress and G is the modulus of rigidity. Substitution gives:

τ = Grθ / L

Thus, τ/r is constant throughout the cross-section:

τ / r = Gθ / L

The elemental torque contributed by an area element dA at radius r is rτ dA. Integrating over the cross-section:

T = ∫A rτ dA

Using τ = (Gθ/L)r:

T = (Gθ / L) ∫A r² dA

The integral is the polar second moment of area:

J = ∫A r² dA

Therefore:

T / J = Gθ / L

Combining this with τ/r = Gθ/L gives the elastic torsion equation:

T / J = τ / r = Gθ / L

This equation applies to circular shafts under Saint-Venant torsion, provided the material is homogeneous and isotropic and the stresses remain within the elastic range.

Variable Definitions and Units

SymbolMeaningSI unit
TApplied torqueN·m
JPolar second moment of aream⁴ or mm⁴
τShear stressPa or N/mm²
rRadial distancem or mm
GModulus of rigidityPa or N/mm²
θTotal angle of twistradians
LShaft lengthm or mm

Dimensional verification, using N and mm, is straightforward: T/J has units N/mm³, while τ/r also has units (N/mm²)/mm = N/mm³. Furthermore, Gθ/L has units (N/mm²)/mm = N/mm³ because radians are dimensionless.

The maximum shear stress occurs at r = R:

τmax = TR / J

The angle of twist is:

θ = TL / GJ

If torque, diameter, or material varies along the shaft, the constant-property form cannot be used for the complete shaft without modification.

4. Polar Second Moment of Area

4.1 Solid Circular Shaft

For a solid shaft of diameter d:

J = πd⁴ / 32

The maximum shear stress becomes:

τmax = 16T / πd³

The angle of twist becomes:

θ = 32TL / πGd⁴

The strong dependence on diameter is important. Strength capacity varies with d³, whereas torsional stiffness varies with d⁴. A modest increase in diameter can therefore produce a large increase in stiffness.

4.2 Hollow Circular Shaft

For a hollow shaft with outside diameter d_o and inside diameter d_i:

J = π(do⁴ − di⁴) / 32

The maximum stress at the outside surface is:

τmax = 16T do / [π(do⁴ − di⁴)]

The angle of twist is:

θ = 32TL / [πG(do⁴ − di⁴)]

A hollow shaft removes material near the center, where its contribution to J is small. Consequently, hollow shafts can provide favorable strength-to-mass and stiffness-to-mass ratios. However, local buckling, manufacturing, corrosion, connection details, and stress concentrations must also be considered.

5. Torque, Power, and Rotational Speed

The mechanical power transmitted by a rotating shaft is the product of torque and angular velocity:

P = Tω

where P is power in watts and ω is angular velocity in radians per second. If rotational speed is n revolutions per minute:

ω = 2πn / 60

Therefore:

P = 2πnT / 60

When P is in kilowatts, n is in revolutions per minute, and T is in N·m:

P(kW) = Tn / 9550

The constant 9550 follows from 2π/60 and conversion from watts to kilowatts. This relation assumes steady rotational speed and torque. In real systems, transmitted torque may fluctuate because of combustion, gear meshing, start-up, or cyclic loading.

Worked Example 1: Solid Shaft in Torsion

A solid steel shaft is 50 mm in diameter and 1.8 m long. It transmits 4 kW at 600 rpm. Take G = 80 GPa. Determine the torque, maximum shear stress, and angle of twist.

Solution. First calculate torque:

T = 9550P/n = 9550(4)/600 = 63.67 N·m

Use N and mm: T = 63,670 N·mm, d = 50 mm, and L = 1800 mm. The polar second moment is:

J = πd⁴/32 = π(50)⁴/32 = 613,592 mm⁴

Maximum shear stress:

τmax = 16T/(πd³)
= 16(63,670)/[π(50)³]
= 2.59 N/mm²

Since 80 GPa = 80,000 N/mm², the angle of twist is:

θ = TL/GJ
= (63,670)(1800)/(80,000 × 613,592)
= 0.00234 rad

In degrees, θ = 0.00234 × 180/π = 0.134°. The shaft is therefore lightly stressed, but a stiffness requirement could still govern its design in another application.

6. Design for Strength and Stiffness

For a solid circular shaft designed for an allowable shear stress τ_allow, the required diameter is obtained from:

d ≥ [16T/(πτallow)]^(1/3)

For a specified allowable angle of twist θ_allow:

d ≥ [32TL/(πGθallow)]^(1/4)

The selected diameter must satisfy both conditions, together with practical requirements such as standard sizes, keyways, shoulders, fatigue, corrosion allowance, and manufacturing tolerances.

A keyway removes material and creates a stress concentration, so the simple circular-shaft formula generally overestimates the actual local strength. Design standards or stress-concentration factors should be used where appropriate.

7. Stepped Shafts

A stepped shaft contains segments with different diameters, lengths, or materials. For a shaft carrying the same internal torque T through each segment, the total angle of twist is the sum of the twists of the segments:

θtotal = ∑(Ti Li / Gi Ji)

If the torque is constant through all segments:

θtotal = T ∑(Li / GiJi)

The maximum shear stress in segment i is:

τi = Ti Ri / Ji

For a stepped shaft, torque diagrams should be drawn before applying formulas. The internal torque in a segment is determined by the external torques acting on one side of a cut. The smallest-diameter segment often governs stress, but the longest or least stiff segment may govern angle of twist.

Worked Example 2: Stepped Shaft

A two-segment steel shaft carries a constant torque of 500 N·m. Segment 1 has length 0.8 m and diameter 40 mm; segment 2 has length 1.2 m and diameter 60 mm. Take G = 80 GPa. Determine the total angle of twist and the maximum stress in each segment.

Use T = 500,000 N·mm, L₁ = 800 mm, L₂ = 1200 mm, and G = 80,000 N/mm².

For segment 1:

J₁ = π(40)⁴/32 = 251,327 mm⁴
θ₁ = TL₁/GJ₁
= (500,000)(800)/(80,000 × 251,327)
= 0.01989 rad
τ₁ = 16T/(πd₁³) = 39.79 N/mm²

For segment 2:

J₂ = π(60)⁴/32 = 1,272,345 mm⁴
θ₂ = (500,000)(1200)/(80,000 × 1,272,345)
= 0.00589 rad
τ₂ = 16T/(πd₂³) = 11.79 N/mm²

Thus:

θtotal = θ₁ + θ₂ = 0.02578 rad = 1.48°

The smaller segment governs the stress, and it also contributes approximately 77% of the total twist.

8. Shafts with Distributed or Variable Torque

For a shaft whose torque T(x), polar moment J(x), or shear modulus G(x) varies continuously along its length, the twist is obtained by integration:

θ = ∫₀ᴸ T(x) / [G(x)J(x)] dx

This is the general elastic relation for a circular shaft. It is particularly useful for tapered shafts, shafts carrying distributed torque, and composite shafts. The internal torque must first be determined as a function of position x.

For a torque diagram, choose a sign convention, cut the shaft at a general location, and apply rotational equilibrium. The area under a plot of T(x)/GJ(x) against x gives the angle of twist.

9. Compatibility and Composite Shafts

When two materials are bonded so that they twist together, their angle of twist per unit length is the same. For materials A and B:

θA/L = θB/L

Since τ = Grθ/L, the shear stresses at the same radius are proportional to their shear moduli:

τA / τB = GA / GB

The total torque is the sum of the torques carried by the materials:

Ttotal = TA + TB

Composite-shaft analysis therefore requires both equilibrium and compatibility. It is important not to assume that torque is shared equally unless the geometry and material properties justify that assumption.

10. Non-Circular Sections

The simple equation T/J = τ/r = Gθ/L is not generally valid for non-circular sections. In a non-circular bar, cross-sections warp out of their original planes. The stress distribution is not simply proportional to distance from a central axis, and a single radius r cannot describe the boundary stress.

For rectangular sections, torsional rigidity is commonly written as:

T = GJt θ/L

where J_t is the torsion constant, not necessarily the polar second moment of area. For a rectangle with sides a and b, where a ≥ b, an approximate Saint-Venant torsion constant is:

Jt ≈ ab³[1/3 − 0.21(b/a)(1 − b⁴/(12a⁴))]

Here a and b are the rectangle dimensions. This approximation is most useful for solid rectangles and should not be confused with the circular-section value J = πd⁴/32.

For open thin-walled sections, warping can be substantial. If warping is restrained, additional normal stresses may occur. Such members require more advanced thin-walled or warping-torsion theory.

11. Thin-Walled Closed Sections

Thin-walled closed sections are efficient in torsion because the wall can develop a circulating shear flow. For a single-cell closed section subjected to torque T, the Bredt–Batho relation is:

q = T/(2Am)

where q is shear flow, with units N/m, and A_m is the area enclosed by the median line of the wall.

If wall thickness t is approximately uniform, the shear stress is:

τ = q/t = T/(2Am t)

For variable thickness, q remains approximately constant around a single-cell section under the basic assumptions, while τ varies as 1/t.

The angle of twist per unit length is:

θ/L = 1/(4Am²G) ∮(ds/t)

where ds is an element of the median-line perimeter. These relations assume a thin wall, a closed section, elastic material, and Saint-Venant torsion without significant distortion or restrained warping.

Open thin-walled sections are much less torsionally stiff than closed sections of comparable material area. Their response is often governed by warping, and their torsion constant is approximately the sum of contributions from individual wall strips rather than the enclosed-area relation above.

12. Stress Concentrations and Practical Design

The elementary torsion equation gives nominal stress away from discontinuities. Actual shafts commonly contain shoulders, grooves, splines, keyways, holes, threads, and abrupt changes in diameter. These features disturb the stress field and cause local stress concentration.

The local elastic stress may be estimated using a torsional stress-concentration factor K_t:

τlocal,max = Kt τnominal

The value of K_t depends on geometry, such as fillet radius-to-diameter ratio and groove dimensions. For fatigue design, a fatigue concentration factor may be more appropriate than the theoretical elastic factor.

Good practice includes:

  • using generous fillet radii at shoulders;
  • avoiding abrupt changes in diameter;
  • checking keyways and splines separately;
  • considering fatigue under fluctuating torque;
  • checking both maximum stress and angular deflection;
  • accounting for corrosion, wear, temperature, and manufacturing defects;
  • using consistent units throughout the calculation;
  • verifying the design against applicable standards and material data.

13. Failure Modes and Limitations

A shaft under torsion may fail through:

  1. Excessive elastic twist: the shaft remains intact but causes misalignment, vibration, or poor control.
  2. Yielding: the shear stress exceeds the material’s elastic limit.
  3. Ductile torsional fracture: commonly associated with a fracture surface approximately normal to the shaft axis.
  4. Brittle fracture: often occurs on an approximately helical surface because principal tensile stresses are inclined to the axis.
  5. Fatigue failure: develops under repeated or reversing torque, often beginning at a surface discontinuity.
  6. Buckling or local instability: important for thin-walled or hollow members.

The elementary theory does not account fully for plastic torsion, residual stresses, anisotropic composites, large deformation, dynamic effects, or three-dimensional stress concentrations. Numerical methods such as finite-element analysis may be needed for complex geometry, but the elementary equations remain valuable for preliminary design and checking computational results.

14. Applied Case: Automotive Drive Shaft

Background. Consider an illustrative design study for a rear-wheel-drive vehicle. A drive shaft transmits 90 kW at 3000 rpm. The shaft is 1.5 m long, made of steel with G = 80 GPa, and is to be designed with an allowable nominal shear stress of 60 MPa and an allowable twist of 1.0°.

Problem. Determine the minimum solid-shaft diameter based on strength and stiffness, then identify the governing criterion.

Analysis: Strength criterion. The torque is:

T = 9550P/n = 9550(90)/3000 = 286.5 N·m

Using N and mm, T = 286,500 N·mm and τ_allow = 60 N/mm²:

dstrength = [16T/(πτallow)]^(1/3)
= [16(286,500)/(π × 60)]^(1/3)
= 42.5 mm

Analysis: Stiffness criterion. The allowable twist is θ_allow = 1.0° = 0.01745 rad. With L = 1500 mm and G = 80,000 N/mm²:

dstiffness = [32TL/(πGθallow)]^(1/4)
= [32(286,500)(1500)/(π × 80,000 × 0.01745)]^(1/4)
= 59.0 mm

Approach and solution. The stiffness requirement governs because 59.0 mm is greater than 42.5 mm. A practical standard diameter would be selected above this value, followed by checks for critical speed, fatigue, universal-joint effects, balance, stress concentrations, and manufacturing constraints. A hollow shaft might reduce mass while satisfying both torsional requirements, but its wall thickness would require additional stability and connection checks.

Outcome. The case demonstrates that a shaft can be safe against nominal yielding yet unacceptable because of excessive twist. It also illustrates why power, speed, material stiffness, length, and geometry must be considered together.

15. Technology and Professional Practice

Modern shaft design combines classical mechanics with computer-aided design, finite-element analysis, material databases, and condition monitoring. A finite-element model can reveal local stresses near keyways, fillets, splines, and joints, while strain gauges or torque transducers can validate the design experimentally. Digital monitoring systems may detect torsional vibration, overload, or progressive fatigue damage.

Nevertheless, software does not remove the need for engineering judgment. Boundary conditions, material models, mesh quality, contact assumptions, and fatigue data strongly influence numerical results. A sound workflow begins with free-body diagrams and hand calculations, continues with refined analysis where justified, and ends with testing or field validation.

Chapter Summary

Torsion is the twisting response of a member subjected to torque. For a homogeneous, isotropic, circular shaft in elastic Saint-Venant torsion, the fundamental relation is T/J = τ/r = Gθ/L. It leads directly to maximum shear stress, angle of twist, and torsional rigidity calculations.

For solid circular shafts, J = πd⁴/32; for hollow circular shafts, J = π(do⁴ − di⁴)/32. Strength depends on the cube of diameter, whereas torsional stiffness depends on the fourth power. Power transmission is related to torque and speed by P = Tω = 2πnT/60.

Stepped shafts are analyzed by summing the twist of individual segments, while variable shafts require integration. Non-circular sections warp and require a torsion constant rather than the circular polar second moment. Thin-walled closed sections carry shear flow described by the Bredt–Batho relation. Real designs must also consider stress concentrations, fatigue, vibration, instability, manufacturing, and applicable standards.

Key Terms

Torque: A moment that twists a member about its longitudinal axis.

Torsion: The state of stress and deformation produced by twisting.

Shear stress: Tangential stress acting on a cross-section.

Shear strain: Angular distortion produced by shear stress.

Modulus of rigidity: Material constant G relating shear stress to shear strain.

Angle of twist: Relative rotation between two cross-sections.

Torsional rigidity: Product GJ representing resistance to elastic twist.

Polar second moment of area: Geometric quantity J equal to ∫r² dA for a cross-section.

Solid shaft: Circular shaft containing material throughout its cross-section.

Hollow shaft: Circular shaft with an internal void.

Stepped shaft: Shaft composed of segments with different dimensions or materials.

Torque diagram: Plot showing internal torque along a shaft.

Torsion constant: Geometric constant used in the torsion of non-circular sections.

Warping: Out-of-plane displacement of a cross-section during torsion.

Shear flow: Shear force per unit length in a thin-walled section.

Bredt–Batho relation: Relation connecting torque and shear flow in a thin-walled closed section.

Stress concentration: Local increase in stress caused by a geometric discontinuity.

Saint-Venant torsion: Torsion theory for members where end effects and restrained warping are negligible.

Torsional stiffness: Resistance to angular deformation, quantified for a uniform shaft by GJ/L.

Review Questions

Short Answer Questions

  1. State the assumptions underlying the elastic torsion equation for circular shafts.
  2. Define the polar second moment of area and state its SI dimensions.
  3. Why does shear stress reach its maximum at the outer surface of a circular shaft?
  4. State the relationship between transmitted power, torque, and angular velocity.
  5. Why can a hollow shaft be structurally efficient in torsion?

Long Answer Questions

  1. Derive the elastic torsion equation for a circular shaft from the shear-strain distribution.
  2. Derive the expressions for J, maximum shear stress, and angle of twist for a hollow circular shaft.
  3. Explain how a stepped shaft is analyzed for internal torque, stress, and total angle of twist.
  4. Compare the torsional behavior of circular, rectangular, and thin-walled closed sections.
  5. Discuss the limitations of elementary torsion theory in practical shaft design.

Analytical / Application Questions

  1. A solid steel shaft of diameter 45 mm and length 1.2 m transmits 25 kW at 900 rpm. Using G = 80 GPa, calculate the maximum shear stress and angle of twist.
  2. A hollow shaft has outside diameter 80 mm and inside diameter 50 mm. Determine the torque that produces a maximum shear stress of 50 MPa.
  3. A two-segment shaft carries constant torque. Explain how you would decide whether strength or stiffness governs, and identify the additional checks required before approving the design.

Multiple Choice Questions with Answers

Question 1

For a circular shaft in elastic torsion, which quantity varies linearly with radius?

A. Shear stress

B. Torque

C. Polar second moment of area

D. Modulus of rigidity

Correct Answer: A. Shear stress

Question 2

The polar second moment of area of a solid circular shaft of diameter d is:

A. πd²/4

B. πd³/32

C. πd⁴/32

D. πd⁴/64

Correct Answer: C. πd⁴/32

Question 3

A shaft transmits 10 kW at 1000 rpm. Its torque is approximately:

A. 9.55 N·m

B. 95.5 N·m

C. 955 N·m

D. 9550 N·m

Correct Answer: B. 95.5 N·m

Question 4

For a uniform circular shaft, the angle of twist is given by:

A. TL/GJ

B. TG/JL

C. GJL/T

D. TJ/GL

Correct Answer: A. TL/GJ

Question 5

If the diameter of a solid circular shaft is doubled while torque remains constant, its maximum shear stress becomes:

A. Twice as large

B. Half as large

C. One-eighth as large

D. One-quarter as large

Correct Answer: C. One-eighth as large

Question 6

In a stepped shaft carrying the same torque through each segment, the total angle of twist is:

A. The twist of the longest segment only

B. The sum of the twists of all segments

C. The average of the segment twists

D. Always zero

Correct Answer: B. The sum of the twists of all segments

Question 7

For a thin-walled single-cell closed section, the shear flow under torque T is approximately:

A. T/A_m

B. T/(2A_m)

C. 2TA_m

D. A_m/(2T)

Correct Answer: B. T/(2A_m)

Question 8

A non-circular section differs from a circular section in torsion because it generally:

A. Has no shear stress

B. Cannot transmit torque

C. Experiences cross-sectional warping

D. Has infinite torsional rigidity

Correct Answer: C. Experiences cross-sectional warping

Question 9

For a solid shaft, torsional stiffness varies with diameter approximately as:

A. d

B. d²

C. d³

D. d⁴

Correct Answer: D. d⁴

Question 10

A keyway in a shaft generally:

A. Eliminates stress concentration

B. Reduces local stress below nominal stress

C. Creates a geometric discontinuity requiring additional design consideration

D. Makes the shaft immune to fatigue

Correct Answer: C. Creates a geometric discontinuity requiring additional design consideration

Strength of Materials
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