State the torsion formula for a circular shaft and name each symbol.

. torque, polar moment, shear stress at radius, shear modulus, twist in radians, length.

Why does the elementary torsion theory apply only to circular sections?

Plane sections remain plane and undistorted only for circular shafts; non-circular sections warp. Polar moment alone does not govern non-circular torsion.

Write for solid and hollow circular shafts.

Solid:. Hollow: where are outer and inner diameters.

Where is maximum shear stress in a solid circular shaft under torsion?

At the outer surface:. Stress is zero at the centre and varies linearly with.

Define torsional rigidity and torsional stiffness.

Torsional rigidity is. Torsional stiffness — torque per unit angle of twist.

How do you find angle of twist for a stepped shaft?

Segments in series:. Use the internal torque in each segment from free-body diagrams.

What is a composite shaft? Contrast series and parallel connections.

Series: same torque, twists add. Parallel (e.g. concentric shafts joined at ends): same, torques add with.

Write power transmitted by a rotating shaft.

where is rpm, in, in watts. Design often sizes shaft from required.

True/False: Maximum shear stress in a hollow shaft occurs at the inner surface.

False — maximum is at the outer surface; minimum (nonzero) shear is at the inner surface.

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Study notes

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Qwestrum Engineering360 · Mechanical Engineering · Strength of Materials (SOM)

Torsion of Circular Shafts

Q: Torsion of circular shafts

For elastic circular shafts, shear stress and twist follow. Maximum shear stress is at the outer radius:. Power transmitted (RK Bansal / VB Bhandari shaft design).

For elastic circular shafts, shear stress and twist follow

\dfrac{T}{J}=\dfrac{\tau}{r}=\dfrac{G\theta}{L}

. Maximum shear stress is at the outer radius:

\tau_{\max}=Tr/J

. Power transmitted

P=T\omega=2\pi NT/60

(RK Bansal / VB Bhandari shaft design).

Exam tip: keep units consistent — N with mm² gives MPa directly (1 MPa = 1 N/mm²).

Key formulas & points

Skim these first — then read the full notes below.

Torsion formula: $\dfrac{T}{J}=\dfrac{\tau}{r}=\dfrac{G\theta}{L}$ .
Polar moment: solid $J=\pi d^4/32$ ; hollow $J=\pi(D^4-d^4)/32$ .
Max shear: $\tau_{\max}=\dfrac{T R}{J}$ at outer fibre.
Angle of twist: $\theta=\dfrac{TL}{GJ}$ (radians).
Power: $P=\dfrac{2\pi NT}{60}$ with $T$ in N·m, $N$ in rpm → $P$ in W.
Shear strain: $\gamma=r\theta/L$ ; Hooke $\tau=G\gamma$ .
Design: size $d$ from $\tau_{\mathrm{allow}}$ and/or $\theta_{\mathrm{allow}}$ (Bhandari).

Topic details

Definition and physical meaning

Torsion is twisting of a member by torques (moments) about its longitudinal axis. In a circular shaft, shear stress varies linearly with radius under elastic torsion.

Governing relation (elastic circular shaft):

\frac{T}{J}=\frac{\tau}{r}=\frac{G\theta}{L}

Physical parameters

Symbol	Meaning	SI unit
$T$	Torque	$\mathrm{N\cdot m}$
$J$	Polar second moment of area	$\mathrm{m^4}$
$\tau$	Shear stress	$\mathrm{Pa}$
$r$ , $R$	Radius (general / outer)	$\mathrm{m}$
$G$	Shear modulus	$\mathrm{Pa}$
$\theta$	Angle of twist	$\mathrm{rad}$
$L$	Shaft length	$\mathrm{m}$
$N$	Rotational speed	$\mathrm{rpm}$
$P$	Power	$\mathrm{W}$

Polar moment of area
- Solid circular:

J=\frac{\pi}{32}d^4

- Hollow circular:

J=\frac{\pi}{32}(D^4-d^4)

Fig 6.1 — τ = Tr/J, θ = TL/(GJ). Solid: J = πd⁴/32. Hollow shaft: material at large r is most effective.

Schematic diagram for study — aligned with standard B.Tech / GATE syllabus.

Core assumptions (state these in exams)

1. Circular cross-section (solid or hollow concentric) — non-circular needs warping theory.
2. Shaft remains straight; axis of twist coincides with centroidal axis.
3. Plane sections remain plane and circular sections remain circular (no warping).
4. Radii remain straight — shear strain

\gamma=r\theta/L

.
5. Linear elastic, homogeneous, isotropic;

\tau=G\gamma

.
6. Constant torque along the segment considered (or analyse segment-wise).
7. Small twist angles for linear geometry.

These assumptions fail for open thin-walled non-circular sections (severe warping) and for plastic torsion (stress distribution no longer linear).

Derivation summary

Kinematics. A generator on the surface twists through angle

\theta

over length

L

. Arc displacement at radius

r

r\theta

, so

\gamma=\frac{r\theta}{L}

Constitutive.

\tau=G\gamma=G r\theta/L

— linear in

r

Equilibrium. Torque equals moment of shear stresses:

T=\int_A \tau\, r\, dA=\frac{G\theta}{L}\int_A r^2\, dA=\frac{G\theta}{L}J

Hence

\frac{T}{J}=\frac{G\theta}{L}=\frac{\tau}{r}

Maxima:

\tau_{\max}

and

\gamma_{\max}

occur at

r = R

(outer surface). For hollow shafts, stress at inner radius is smaller:

\tau_i=\tau_o(r_i/r_o)

Power transmission and design

Rotating shaft transmitting power:

P=T\omega=T\cdot\frac{2\pi N}{60}\qquad\Rightarrow\qquad T=\frac{60 P}{2\pi N}

Strength design (allowable shear):

\tau_{\max}=\frac{T R}{J}\le\tau_{\mathrm{allow}}\Rightarrow\text{size }d\text{ or }D,d

Stiffness design (allowable twist):

\theta=\frac{TL}{GJ}\le\theta_{\mathrm{allow}}

Often both criteria are checked; the larger required diameter governs (VB Bhandari).

Strain energy in torsion:

U=\frac{T^2 L}{2GJ}=\frac{1}{2}T\theta

Stepped and composite shafts

Series (stepped): same

T

if no intermediate torque gears; twists add:

\theta_{\mathrm{total}}=\sum_i\frac{T_i L_i}{G_i J_i}

Parallel (composite concentric shafts connected to same end plates): same

\theta

, torques add:

T=T_1+T_2,\qquad\theta=\frac{T_1 L}{G_1 J_1}=\frac{T_2 L}{G_2 J_2}

\frac{T_1}{T_2}=\frac{G_1 J_1}{G_2 J_2}

Statically indeterminate: fixed ends with intermediate torque — equilibrium +

\theta

compatibility.

Step-by-step problem approach

1. Convert power/speed to torque if needed:

T=60P/(2\pi N)

.
2. Compute

J

(solid or hollow); watch

d^4

D^4-d^4

.
3.

\tau_{\max}=TR/J

; compare with allowable.
4.

\theta=TL/(GJ)

in radians; convert to degrees if asked (

\times 180/\pi

).
5. For stepped shafts, split into segments; sum

\theta_i

.
6. Units:

T

in N·mm with

J

in mm⁴ and

\tau

in N/mm² is consistent; or all SI (N·m, m⁴, Pa).
7. State assumptions for circular elastic torsion.

Common mistakes in exams

• Using

J=\pi d^4/64

(that is

I

, not polar

J=\pi d^4/32

).
• Forgetting hollow formula uses

D^4-d^4

, not

(D-d)^4

.
• Mixing degrees and radians in

\theta=TL/(GJ)

.
• Using diameter instead of radius in

\tau=Tr/J

.
• Applying circular torsion formulas to rectangular shafts.
• Power formula with inconsistent units (kW vs W, rpm).

Calculator

Torsion shear (solid shaft)

\tau = \frac{16T}{\pi d^{3}}

Torque TN·mmDiameter dmm

Result

79.5775N/mm² (MPa)

τ = 16T/(πd³) = 16×1.000e+6 / (π×40³) = 79.58 N/mm²

Worked examples

Try the problem first — open the solution when you are ready to check.

Shear stress and twist in a solid shaft

Problem

A solid steel shaft has

d=80\,\mathrm{mm}

L=1.2\,\mathrm{m}

G=80\,\mathrm{GPa}

, and transmits

T=2.5\,\mathrm{kN\cdot m}

. Find

\tau_{\max}

and

\theta

in degrees.

Solution

Formulas (Indian textbook notation)

$J=\frac{\pi}{32}(0.08)^4=4.021\times 10^{-6}\,\mathrm{m^4}$

Formulas (Indian textbook notation)

$\tau_{\max}=\frac{T R}{J}=\frac{2500\times 0.04}{4.021\times 10^{-6}}=24.87\times 10^6\,\mathrm{Pa}=24.87\,\mathrm{MPa}$

Formulas (Indian textbook notation)

$\theta=\frac{TL}{GJ}=\frac{2500\times 1.2}{(80\times 10^9)(4.021\times 10^{-6})}=9.33\times 10^{-3}\,\mathrm{rad}=0.535^\circ$

Power and shaft diameter

Problem

Design a solid shaft to transmit

P=50\,\mathrm{kW}

N=300\,\mathrm{rpm}

\tau_{\mathrm{allow}}=60\,\mathrm{MPa}

Hollow vs solid — same material and τ_max

Problem

Compare polar moduli: solid shaft

d=100\,\mathrm{mm}

vs hollow

D=100\,\mathrm{mm}

d=60\,\mathrm{mm}

. Ratio of torque capacity at same

\tau_{\max}

Practice questions

Most-asked interview and GATE questions for this topic — expand any item for a model answer.

1
State the torsion formula for a circular shaft and name each symbol.
Model answer
$\dfrac{T}{J}=\dfrac{\tau}{r}=\dfrac{G\theta}{L}$ . $T$ torque, $J$ polar moment, $\tau$ shear stress at radius $r$ , $G$ shear modulus, $\theta$ twist in radians, $L$ length.
2
Why does the elementary torsion theory apply only to circular sections?
Model answer
Plane sections remain plane and undistorted only for circular shafts; non-circular sections warp. Polar moment alone does not govern non-circular torsion.
3
Write $J$ for solid and hollow circular shafts.
Model answer
Solid: $J=\pi d^4/32$ . Hollow: $J=\pi(D^4-d^4)/32$ where $D,d$ are outer and inner diameters.
4
Where is maximum shear stress in a solid circular shaft under torsion?
Model answer
At the outer surface $r = R$ : $\tau_{\max}=TR/J=16T/(\pi d^3)$ . Stress is zero at the centre and varies linearly with $r$ .
5
Define torsional rigidity and torsional stiffness.
Model answer
Torsional rigidity is $GJ$ . Torsional stiffness $k_t=T/\theta=GJ/L$ — torque per unit angle of twist.
6
How do you find angle of twist for a stepped shaft?
Model answer
Segments in series: $\theta=\sum T_i L_i/(G_i J_i)$ . Use the internal torque in each segment from free-body diagrams.
7
What is a composite shaft? Contrast series and parallel connections.
Model answer
Series: same torque, twists add. Parallel (e.g. concentric shafts joined at ends): same $\theta$ , torques add $T = T_{1}+T_{2}$ with $\theta=T_1 L/(G_1 J_1)=T_2 L/(G_2 J_2)$ .
8
Write power transmitted by a rotating shaft.
Model answer
$P=T\omega=2\pi NT/60$ where $N$ is rpm, $T$ in $\mathrm{N\cdot m}$ , $P$ in watts. Design often sizes shaft from required $T=P/\omega$ .
9
What are assumptions of pure torsion theory for circular shafts?
Model answer
Circular cross-section, material homogeneous isotropic and linearly elastic, plane sections remain plane, radii remain straight, no warping, twist uniform along length for constant $T$ .
10
Compare strength of hollow vs solid shaft of same material and same weight (same length).
Model answer
For same mass, hollow shaft has larger outer radius and larger $J$ , so higher torque capacity and stiffness — material is farther from the axis.
11
What is polar modulus? How is it used?
Model answer
Polar modulus $Z_{p} = \frac{J}{R}$ . Then $\tau_{\max}=T/Z_p$ . For solid circular, $Z_p=\pi d^3/16$ .
12
Explain shear strain variation in a twisted circular shaft.
Model answer
Shear strain $\gamma=r\theta/L$ increases linearly with $r$ . At centre $\gamma=0$ ; maximum at outer fibre.
13
How is strain energy in pure torsion expressed?
Model answer
$U=T^2 L/(2GJ)=GJ\theta^2/(2L)=\tfrac{1}{2}T\theta$ for linear elastic shafts.
14
What is the difference between open and closed thin-walled tubes in torsion?
Model answer
Closed thin tubes carry Bredt shear flow efficiently ( $T=2q\bar{A}$ ). Open thin sections (slit tube) are torsionally weak; warping and low $J_{\mathrm{eff}}$ dominate.
15
How do you design a shaft for both strength and stiffness?
Model answer
Strength: $\tau_{\max}\le\tau_{\mathrm{allow}}$ fixes minimum $d$ . Stiffness: $\theta\le\theta_{\mathrm{allow}}$ gives another $d$ . Choose the larger diameter.

Exams & GATE

1
Textbook: RK Bansal (torsion); VB Bhandari for shaft design under torque and bending.
2
Assumptions (circular, plane sections, elastic) are mandatory in answers.
3
GATE favourites: hollow vs solid for same material/weight, stepped shafts ( $\theta=\sum TL_i/(G J_i)$ ), and power–torque conversion.

📖 Standard books (India)

Strength of Materials — RK Bansal
Read: Ch. 14–15
SOM — beams, torsion, columns, and deflection

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📚Key formulas & points

Topic details

Definition and physical meaning

Core assumptions (state these in exams)

Derivation summary

Power transmission and design

Stepped and composite shafts

Step-by-step problem approach

Common mistakes in exams

Torsion shear (solid shaft)

Worked examples

Shear stress and twist in a solid shaft

Power and shaft diameter

Hollow vs solid — same material and τ_max

Practice questions

📝Exams & GATE

📖 Standard books (India)

Explore related topics

See real mechanical engineering careers

Key formulas & points

Exams & GATE