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Chapter 6 of 9

Torsion and Shear Centre

In the OPSC AEE Civil syllabus under Strength of Materials · 2 parts

Torsion of Circular Shafts · Unsymmetrical Bending & Shear Centre

CIVILGYAN TEST SERIES · civilgyantests.com — free study notes for civil engineering exams

📑 Contents (22 sections)

Part 1 of 2

Torsion of Circular Shafts

Last reviewed 16 Sept 2026 · 14 min read

What torsion is

A member is in torsion when it carries a couple whose axis lies along the member's own axis. The couple is called the torque or twisting moment $T$ . Shafts that carry power from a motor to a pump, drive shafts, the spindle of a valve and a helical spring wire are all loaded mainly in torsion.

Under a torque every cross-section rotates about the axis by a small angle relative to its neighbour. The total rotation of one end relative to the other is the angle of twist $θ$ . The twisting produces shear stresses on the cross-section; there is no normal stress on a transverse section in pure torsion.

DefinitionKey terms

Torque, $T$ — twisting moment about the longitudinal axis (N·mm or kN·m).
Angle of twist, $θ$ — relative rotation of two sections a length $L$ apart (radians).
Shear strain, $γ$ — change in the right angle of an element on the surface.
Polar moment of inertia, $J$ — the section property that resists twisting (mm⁴).
Torsional rigidity, $G J$ — torque needed to produce unit twist per unit length.

Assumptions of the simple torsion theory

The classical theory (Coulomb) for circular shafts rests on these assumptions. Examiners like to ask which one is not an assumption.

The material is homogeneous and isotropic.
Stresses are within the elastic limit, so Hooke's law in shear holds: $τ = G γ$ .
The shaft is straight and of uniform circular (solid or hollow) cross-section.
Plane transverse sections remain plane after twisting (no warping). This is true only for circular sections.
Radii remain straight — a radial line on a section stays a straight line.
The twist along the shaft is uniform, and the angle of twist is small.
The torque acts in a plane perpendicular to the axis.

Common MistakeNon-circular sections

Assumption 4 fails for square, rectangular and open sections: their cross-sections warp. That is why the torsion equation below must not be used for a rectangular bar.

Derivation of the torsion equation

Consider a shaft of length $L$ and radius $R$ , twisted through $θ$ . A line AB drawn on the surface parallel to the axis moves to AB′.

Arc BB′ at the surface $= R θ$ .
The same arc, seen from the fixed end A, is $L γ$ (for small angles).

R θ = L γ \Rightarrow γ = \frac{R θ}{L}

At any radius $r$ the strain is $γ_{r} = r θ / L$ . Using Hooke's law $τ_{r} = G γ_{r}$ :

τ_{r} = \frac{Gθ}{L} r \Rightarrow \frac{τ _{r}}{r} = \frac{Gθ}{L}

So shear stress varies linearly with radius — zero at the centre, maximum at the outer surface.

The resisting torque of an elementary ring of area $d A$ at radius $r$ is $d T = τ_{r} d A r$ . Summing over the section,

T = \int \frac{Gθ}{L} r^{2} d A = \frac{Gθ}{L} \int r^{2} d A = \frac{Gθ}{L} J

Combining the two results gives the torsion equation:

FormulaTorsion equation

\frac{T}{J} = \frac{τ}{r} = \frac{G θ}{L}

$T$ = torque (N·mm), $J$ = polar moment of inertia (mm⁴), $τ$ = shear stress at radius $r$ (N/mm²), $G$ = modulus of rigidity (N/mm²), $θ$ = angle of twist (rad), $L$ = length (mm).

FigureShear stress distribution in pure torsion

Stress grows in proportion to radius. In a hollow shaft the least-stressed core material is simply absent, which is why hollow shafts use material more efficiently.

Polar moment of inertia and polar modulus

By the perpendicular axis theorem, $J = I_{xx} + I_{y y}$ . For a circle $I_{xx} = I_{y y} = π d^{4} /64$ , so $J = π d^{4} /32$ .

The polar section modulus (torsional section modulus) is $Z_{p} = J / R$ , so that $T = τ_{ma x} Z_{p}$ — the torque a section can carry at a given permissible stress.

Section	Polar moment of inertia $J$	Polar modulus $Z_{p} = J / R$	Maximum shear stress
Solid, diameter $d$	$\frac{π d ^{4}}{32}$	$\frac{π d ^{3}}{16}$	$τ_{ma x} = \frac{16 T}{π d ^{3}}$
Hollow, outer $D$ , inner $d$	$\frac{π ( D ^{4} - d ^{4} )}{32}$	$\frac{π ( D ^{4} - d ^{4} )}{16 D}$	$τ_{ma x} = \frac{16 T D}{π ( D ^{4} - d ^{4} )}$
Thin tube, mean radius $r$ , thickness $t$	$\approx 2 π r^{3} t$	$\approx 2 π r^{2} t$	$τ \approx \frac{T}{2 π r ^{2} t}$

Remember

For a hollow shaft the stress at the inner surface is $τ_{inn er} = τ_{ma x} \frac{d}{D}$ — the stress is still linear in radius.

Angle of twist, torsional rigidity and stiffness

From the torsion equation,

θ = \frac{T L}{G J}

Torsional rigidity $= G J$ — a property of material and section (units N·mm²).
Torsional stiffness $k_{t} = \frac{T}{θ} = \frac{G J}{L}$ — torque per radian of twist for a given length.
Torsional flexibility $= \frac{L}{G J}$ — the reciprocal of stiffness.

Design of a shaft usually checks two things: strength ( $τ_{ma x}$ within the permissible shear stress) and stiffness (twist within a limit such as a stated number of degrees per metre). The larger diameter from the two checks is adopted.

Common MistakeUnits of twist

$θ$ in the formula is in radians. Convert a limit like "1° per metre" to $π /180$ rad per 1000 mm before substituting. Using degrees directly is the commonest error in twist problems.

Power transmitted by a shaft

A torque $T$ rotating at angular speed $ω$ delivers power $P = T ω$ . With speed $N$ in revolutions per minute, $ω = 2 π N /60$ rad/s.

FormulaPower and torque

P = \frac{2 π N T}{60}

$P$ in watts when $T$ is in N·m; $P$ in kW when $T$ is in kN·m. Equivalently $T = \frac{60 P}{2 π N}$ .

If the torque fluctuates, the shaft must be designed for the maximum torque, not the mean: $T_{ma x} = (ratio) \times T_{m e an}$ as given in the question.

Solid versus hollow shafts

Let a hollow shaft have outer diameter $D$ and inner diameter $d = k D$ ( $0 < k < 1$ ). Compare it with a solid shaft of diameter $d_{s}$ , same material and length.

Equal strength (same torque, same $τ_{ma x}$ ) requires equal polar moduli:

\frac{π D ^{3} ( 1 - k ^{4} )}{16} = \frac{π d _{s}^{3}}{16} \Rightarrow \frac{D}{d _{s}} = (\frac{1}{1 - k ^{4}})^{1/3}

The weight ratio equals the area ratio:

\frac{W _{h o l l o w}}{W _{so l i d}} = \frac{D ^{2} ( 1 - k ^{2} )}{d _{s}^{2}} = \frac{1 - k ^{2}}{( 1 - k ^{4} ) ^{2/3}}

Equal weight (same area) instead: the hollow shaft carries more torque in the ratio

\frac{T _{h o l l o w}}{T _{so l i d}} = \frac{1 + k ^{2}}{1 - k ^{2}}

Exam TipStandard exam results

For the same weight, a hollow shaft is stronger and stiffer than a solid shaft.
For the same strength, a hollow shaft is lighter.
At the same outer diameter, a hollow shaft carries a smaller torque than the solid one — it has less material. Read the comparison condition carefully.

Shafts in series and in parallel

Series (stepped or composite along the length)

The same torque passes through every segment; the twists add.

θ_{t o t a l} = θ_{1} + θ_{2} = \frac{T L _{1}}{G _{1} J _{1}} + \frac{T L _{2}}{G _{2} J _{2}}

The equivalent stiffness follows the "springs in series" rule: $\frac{1}{k} = \frac{1}{k _{1}} + \frac{1}{k _{2}}$ .

Parallel (concentric composite shaft, or a shaft fixed at both ends)

The same twist occurs in every part; the torques add.

T = T_{1} + T_{2}, θ_{1} = θ_{2} \Rightarrow \frac{T _{1} L _{1}}{G _{1} J _{1}} = \frac{T _{2} L _{2}}{G _{2} J _{2}}

For a uniform shaft fixed at both ends with a torque $T$ applied at a distance $a$ from end A and $b$ from end B ( $a + b = L$ ), compatibility of twist gives

T_{A} = \frac{T b}{L}, T_{B} = \frac{T a}{L}

The nearer support carries the larger share of the torque — the same pattern as the reactions of a beam.

Part 2 of 2

Unsymmetrical Bending & Shear Centre

Last reviewed 16 Sept 2026 · 6 min read

Symmetrical and unsymmetrical bending

The simple bending formula $σ = M y / I$ assumes the load acts in a plane of symmetry, so the beam bends in that same plane. Bending is unsymmetrical when:

the section has no axis of symmetry (unequal angle, Z-section), or
the section is symmetric but the plane of loading is inclined to the axes of symmetry — a purlin on a sloping roof, a rectangular beam loaded diagonally.

Then the neutral axis is not perpendicular to the plane of loading, the beam deflects out of the load plane, and stresses must be found about the principal axes.

Product of inertia and principal axes

For axes $x$ and $y$ through the centroid:

I_{xx} = \int y^{2} d A, I_{y y} = \int x^{2} d A, I_{x y} = \int x y d A

$I_{x y}$ may be positive, negative or zero.
If either axis is an axis of symmetry, $I_{x y} = 0$ .
Principal axes are the pair of perpendicular centroidal axes about which $I_{x y} = 0$ ; the moments of inertia about them are the maximum and minimum.

FormulaPrincipal axes and principal moments of inertia

tan 2 α = \frac{2 I _{x y}}{I _{y y} - I _{xx}}

I_{u, v} = \frac{I _{xx} + I _{y y}}{2} \pm (\frac{I _{xx} - I _{y y}}{2})^{2} + I_{x y}^{2}

$I_{xx} + I_{y y} = I_{u} + I_{v}$ (invariant, equal to the polar moment $J$ ).

These equations have the same form as the stress transformation equations, so Mohr's circle can also be drawn for moments of inertia.

Bending stress for a moment inclined to the principal axes

Let $U$ and $V$ be the principal axes, with $I_{u}$ and $I_{v}$ . A moment $M$ acts in a plane making angle $θ$ with the $V$ -axis. Resolve it:

M_{u} = M cos θ (about U), M_{v} = M sin θ (about V)

FormulaStress at a point (u, v)

σ = \frac{M _{u} v}{I _{u}} + \frac{M _{v} u}{I _{v}} = M (\frac{v cos θ}{I _{u}} + \frac{u sin θ}{I _{v}})

with signs chosen by inspection (tension or compression) for each component. The maximum stress occurs at the point farthest from the neutral axis — for a rectangle, at a corner.

Neutral axis

Setting $σ = 0$ gives a straight line through the centroid inclined at angle $ϕ$ to the $U$ -axis:

tan ϕ = \frac{I _{u}}{I _{v}} tan θ

Since $I_{u} \neq = I_{v}$ in general, $ϕ \neq = θ$ : the neutral axis is not perpendicular to the plane of loading. It swings towards the axis of minimum moment of inertia. The beam deflects perpendicular to the neutral axis.

General formula (non-principal axes)

If stresses are wanted directly in $x$ – $y$ axes with $I_{x y} \neq = 0$ :

σ = \frac{M _{x} I _{y y} - M _{y} I _{x y}}{I _{xx} I _{y y} - I _{x y}^{2}} y + \frac{M _{y} I _{xx} - M _{x} I _{x y}}{I _{xx} I _{y y} - I _{x y}^{2}} x

(signs depend on the convention for $M_{x}$ and $M_{y}$ ; with $I_{x y} = 0$ it reduces to the principal-axis form).

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