Torsion

Mechanics of Materials · FE Reference Handbook section

Mechanics of Materials

3 formulas

10 exam-style examples

~51 min

All Mechanics of Materials lectures

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Torsional shear stress and twist in a shaft

A solid 60 mm diameter shaft, 2.0 m long, transmits 3.5 kN·m. With G = 80 GPa, find the maximum shear stress and the angle of twist.

Given

$d = 60 mm$
T = 3.5 kN·m
$L = 2.0 m$
$G = 80 GPa$

Find

τ_max and φ

Start with the thinking

Polar moment for a solid circle is πd⁴/32.
Twist is in radians — convert if degrees are asked.

Step-by-step solution

Polar moment — J = πd⁴/32 = π(60)⁴/32 = 1.272×10⁶ mm⁴
Torsion formula
$\tau = Tr/J with r = 30 mm$
Substitute
$\tau = 3.5\times10^{6}(30)/1.272\times10^{6} = 82.6 MPa$
Twist
$\phi = TL/(GJ) = 3.5\times10^{6}(2,000)/[80,000(1.272\times10^{6})]$
Result
$\phi = 0.0688 rad = 3.94^{\circ}$

Answer:

\tau_max = 82.6 MPa, \phi = 3.94^{\circ}

Why the other options are there

τ = 41.3 MPa (radius taken as diameter)
φ = 3.94 rad (radians reported as degrees)

Reference: FE Reference Handbook — Mechanics of Materials — Torsion

Example 2

Torsional shear stress — solve for shear stress — Torsion

A mechanics of materials problem uses Torsional shear stress. Given torque (T) = 400.0 kip·in; outer radius (c) = 4.2000 in; polar moment of inertia (J) = 440.0 in⁴, determine the shear stress (tau) in ksi.

Given

torque (T) = 400.0 kip·in
$outer radius (c) = 4.2000 in$
$polar moment of inertia (J) = 440.0 in^{4}$

Find

shear stress (tau), in ksi

Start with the thinking

The governing relation printed in this handbook section is Torsional shear stress.
Everything except tau is given, so isolate tau symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Mechanics of Materials items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$\tau = T c / J$
Step 2 — Rearrange the relation so that tau stands alone on the left-hand side.
Step 3 — List the givens: torque (T) = 400.0 kip·in, outer radius (c) = 4.2000 in, polar moment of inertia (J) = 440.0 in⁴.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$\tau = 3.8182\ \text{ksi}$
Step 6 — Check: returning tau = 3.8182 ksi to
$\tau = T c / J$
reproduces the given quantities, and both sides carry the same units.

Answer:

\tau = 3.8182\ \text{ksi}

Why the other options are there

7.6364 — kept a factor of two that cancels in the correct rearrangement.
1.9091 — dropped that same factor in the other direction.
4.2000 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Torsion

Example 3

Torsional shear stress — solve for torque — Torsion (2)

A mechanics of materials problem uses Torsional shear stress. Given outer radius (c) = 5.8000 in; polar moment of inertia (J) = 145.0 in⁴; shear stress (tau) = 12.3400 ksi, determine the torque (T) in kip·in.

Given

$outer radius (c) = 5.8000 in$
$polar moment of inertia (J) = 145.0 in^{4}$
$shear stress (tau) = 12.3400 ksi$

Find

torque (T), in kip·in

Start with the thinking

The governing relation printed in this handbook section is Torsional shear stress.
Everything except T is given, so isolate T symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Mechanics of Materials items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$\tau = T c / J$
Step 2 — Rearrange the relation so that T stands alone on the left-hand side.
Step 3 — List the givens: outer radius (c) = 5.8000 in, polar moment of inertia (J) = 145.0 in⁴, shear stress (tau) = 12.3400 ksi.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$T = 308.5\ \text{kip·in}$
Step 6 — Check: returning T = 308.5 kip·in to
$\tau = T c / J$
reproduces the given quantities, and both sides carry the same units.

Answer:

T = 308.5\ \text{kip·in}

Why the other options are there

617.0 — kept a factor of two that cancels in the correct rearrangement.
154.3 — dropped that same factor in the other direction.
339.4 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Torsion

Example 4

Torsional shear stress — solve for outer radius — Torsion (3)

A mechanics of materials problem uses Torsional shear stress. Given torque (T) = 369.0 kip·in; polar moment of inertia (J) = 289.0 in⁴; shear stress (tau) = 6.2800 ksi, determine the outer radius (c) in in.

Given

torque (T) = 369.0 kip·in
$polar moment of inertia (J) = 289.0 in^{4}$
$shear stress (tau) = 6.2800 ksi$

Find

outer radius (c), in in

Start with the thinking

The governing relation printed in this handbook section is Torsional shear stress.
Everything except c is given, so isolate c symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Mechanics of Materials items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$\tau = T c / J$
Step 2 — Rearrange the relation so that c stands alone on the left-hand side.
Step 3 — List the givens: torque (T) = 369.0 kip·in, polar moment of inertia (J) = 289.0 in⁴, shear stress (tau) = 6.2800 ksi.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$c = 4.9185\ \text{in}$
Step 6 — Check: returning c = 4.9185 in to
$\tau = T c / J$
reproduces the given quantities, and both sides carry the same units.

Answer:

c = 4.9185\ \text{in}

Why the other options are there

9.8370 — kept a factor of two that cancels in the correct rearrangement.
2.4592 — dropped that same factor in the other direction.
5.4103 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Torsion

Example 5

Torsional shear stress — solve for polar moment of inertia — Torsion (4)

A mechanics of materials problem uses Torsional shear stress. Given torque (T) = 45.0000 kip·in; outer radius (c) = 1.9000 in; shear stress (tau) = 17.6500 ksi, determine the polar moment of inertia (J) in in⁴.

Given

torque (T) = 45.0000 kip·in
$outer radius (c) = 1.9000 in$
$shear stress (tau) = 17.6500 ksi$

Find

polar moment of inertia (J), in in⁴

Start with the thinking

The governing relation printed in this handbook section is Torsional shear stress.
Everything except J is given, so isolate J symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Mechanics of Materials items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$\tau = T c / J$
Step 2 — Rearrange the relation so that J stands alone on the left-hand side.
Step 3 — List the givens: torque (T) = 45.0000 kip·in, outer radius (c) = 1.9000 in, shear stress (tau) = 17.6500 ksi.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$J = 4.8442\ \text{in⁴}$
Step 6 — Check: returning J = 4.8442 in⁴ to
$\tau = T c / J$
reproduces the given quantities, and both sides carry the same units.

Answer:

J = 4.8442\ \text{in⁴}

Why the other options are there

9.6884 — kept a factor of two that cancels in the correct rearrangement.
2.4221 — dropped that same factor in the other direction.
5.3286 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Torsion

Example 6

Torsional shear stress — solve for shear stress (case 2) — Torsion (5)

A mechanics of materials problem uses Torsional shear stress. Given torque (T) = 434.0 kip·in; outer radius (c) = 5.0000 in; polar moment of inertia (J) = 469.0 in⁴, determine the shear stress (tau) in ksi.

Given

torque (T) = 434.0 kip·in
$outer radius (c) = 5.0000 in$
$polar moment of inertia (J) = 469.0 in^{4}$

Find

shear stress (tau), in ksi

Start with the thinking

The governing relation printed in this handbook section is Torsional shear stress.
Everything except tau is given, so isolate tau symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Mechanics of Materials items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$\tau = T c / J$
Step 2 — Rearrange the relation so that tau stands alone on the left-hand side.
Step 3 — List the givens: torque (T) = 434.0 kip·in, outer radius (c) = 5.0000 in, polar moment of inertia (J) = 469.0 in⁴.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$\tau = 4.6269\ \text{ksi}$
Step 6 — Check: returning tau = 4.6269 ksi to
$\tau = T c / J$
reproduces the given quantities, and both sides carry the same units.

Answer:

\tau = 4.6269\ \text{ksi}

Why the other options are there

9.2537 — kept a factor of two that cancels in the correct rearrangement.
2.3134 — dropped that same factor in the other direction.
5.0896 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Torsion

Example 7

Torsional shear stress — solve for torque (case 2) — Torsion (6)

A mechanics of materials problem uses Torsional shear stress. Given outer radius (c) = 5.2000 in; polar moment of inertia (J) = 38.0000 in⁴; shear stress (tau) = 8.4300 ksi, determine the torque (T) in kip·in.

Given

$outer radius (c) = 5.2000 in$
$polar moment of inertia (J) = 38.0000 in^{4}$
$shear stress (tau) = 8.4300 ksi$

Find

torque (T), in kip·in

Start with the thinking

The governing relation printed in this handbook section is Torsional shear stress.
Everything except T is given, so isolate T symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Mechanics of Materials items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$\tau = T c / J$
Step 2 — Rearrange the relation so that T stands alone on the left-hand side.
Step 3 — List the givens: outer radius (c) = 5.2000 in, polar moment of inertia (J) = 38.0000 in⁴, shear stress (tau) = 8.4300 ksi.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$T = 61.6038\ \text{kip·in}$
Step 6 — Check: returning T = 61.6038 kip·in to
$\tau = T c / J$
reproduces the given quantities, and both sides carry the same units.

Answer:

T = 61.6038\ \text{kip·in}

Why the other options are there

123.2 — kept a factor of two that cancels in the correct rearrangement.
30.8019 — dropped that same factor in the other direction.
67.7642 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Torsion

Example 8

Torsional shear stress — solve for outer radius (case 2) — Torsion (7)

A mechanics of materials problem uses Torsional shear stress. Given torque (T) = 446.0 kip·in; polar moment of inertia (J) = 40.0000 in⁴; shear stress (tau) = 17.4400 ksi, determine the outer radius (c) in in.

Given

torque (T) = 446.0 kip·in
$polar moment of inertia (J) = 40.0000 in^{4}$
$shear stress (tau) = 17.4400 ksi$

Find

outer radius (c), in in

Start with the thinking

The governing relation printed in this handbook section is Torsional shear stress.
Everything except c is given, so isolate c symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Mechanics of Materials items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$\tau = T c / J$
Step 2 — Rearrange the relation so that c stands alone on the left-hand side.
Step 3 — List the givens: torque (T) = 446.0 kip·in, polar moment of inertia (J) = 40.0000 in⁴, shear stress (tau) = 17.4400 ksi.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$c = 1.5641\ \text{in}$
Step 6 — Check: returning c = 1.5641 in to
$\tau = T c / J$
reproduces the given quantities, and both sides carry the same units.

Answer:

c = 1.5641\ \text{in}

Why the other options are there

3.1283 — kept a factor of two that cancels in the correct rearrangement.
0.7821 — dropped that same factor in the other direction.
1.7205 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Torsion

Example 9

Torsional shear stress — solve for polar moment of inertia (case 2) — Torsion (8)

A mechanics of materials problem uses Torsional shear stress. Given torque (T) = 15.0000 kip·in; outer radius (c) = 2.6000 in; shear stress (tau) = 5.7800 ksi, determine the polar moment of inertia (J) in in⁴.

Given

torque (T) = 15.0000 kip·in
$outer radius (c) = 2.6000 in$
$shear stress (tau) = 5.7800 ksi$

Find

polar moment of inertia (J), in in⁴

Start with the thinking

The governing relation printed in this handbook section is Torsional shear stress.
Everything except J is given, so isolate J symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Mechanics of Materials items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$\tau = T c / J$
Step 2 — Rearrange the relation so that J stands alone on the left-hand side.
Step 3 — List the givens: torque (T) = 15.0000 kip·in, outer radius (c) = 2.6000 in, shear stress (tau) = 5.7800 ksi.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$J = 6.7474\ \text{in⁴}$
Step 6 — Check: returning J = 6.7474 in⁴ to
$\tau = T c / J$
reproduces the given quantities, and both sides carry the same units.

Answer:

J = 6.7474\ \text{in⁴}

Why the other options are there

13.4948 — kept a factor of two that cancels in the correct rearrangement.
3.3737 — dropped that same factor in the other direction.
7.4221 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Torsion

Example 10

Torsional shear stress — solve for shear stress (case 3) — Torsion (9)

A mechanics of materials problem uses Torsional shear stress. Given torque (T) = 447.0 kip·in; outer radius (c) = 1.2000 in; polar moment of inertia (J) = 234.0 in⁴, determine the shear stress (tau) in ksi.

Given

torque (T) = 447.0 kip·in
$outer radius (c) = 1.2000 in$
$polar moment of inertia (J) = 234.0 in^{4}$

Find

shear stress (tau), in ksi

Start with the thinking

The governing relation printed in this handbook section is Torsional shear stress.
Everything except tau is given, so isolate tau symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Mechanics of Materials items reward recognising the unknown before touching a calculator.

Step-by-step solution

Step 1 — State the governing relation:
$\tau = T c / J$
Step 2 — Rearrange the relation so that tau stands alone on the left-hand side.
Step 3 — List the givens: torque (T) = 447.0 kip·in, outer radius (c) = 1.2000 in, polar moment of inertia (J) = 234.0 in⁴.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$\tau = 2.2923\ \text{ksi}$
Step 6 — Check: returning tau = 2.2923 ksi to
$\tau = T c / J$
reproduces the given quantities, and both sides carry the same units.

Answer:

\tau = 2.2923\ \text{ksi}

Why the other options are there

4.5846 — kept a factor of two that cancels in the correct rearrangement.
1.1462 — dropped that same factor in the other direction.
2.5215 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Mechanics of Materials → Torsion

Torsion

Core formulas for this FE topic

τ=TrJ\tau = \frac{Tr}{J}τ=JTr​Verified

x=Trx = Trx=Tr

whereJ=polarmomentofinertiawhere J = polar moment of inertiawhereJ=polarmomentofinertia

Worked exam-style examples