Torsional Strain

Mechanics of Materials · FE Reference Handbook section

Mechanics of Materials

10 formulas

10 exam-style examples

~60 min

All Mechanics of Materials lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Torsional Strain within Mechanics of Materials. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what torsional strain describes physically and when it applies.
State every one of the 10 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: psi vs ksi and kip vs lb decide the answer choice.

Lecture

Why this section exists. Torsional Strain is the part of Mechanics of Materials that lets you connect an axially loaded, bent or twisted member to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as a stress or deformation at one point of one member. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. psi vs ksi and kip vs lb decide the answer choice. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Concrete cylinder under axial load in a compression testing machine. — Photo 1. Where this shows up in practice: torsional strain.
Wikimedia Commons, public domain

Mechanics of Materials — Torsional Strain: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes an axially loaded, bent or twisted member. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 10 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

czz	Quantity produced by "czz = limit r ^Dz/Dz h = r ^dz/dz h" — read its definition and unit from the handbook line directly above the equation.
τφz	Quantity produced by "τφz = Gγφz = Gr _dφ/dz i" — read its definition and unit from the handbook line directly above the equation.
φ	Quantity produced by "φ = total angle (radians) of twist" — read its definition and unit from the handbook line directly above the equation.
T	Quantity produced by "T = torque" — read its definition and unit from the handbook line directly above the equation.
L	Quantity produced by "L = length of shaft" — read its definition and unit from the handbook line directly above the equation.
x	Quantity produced by "x = 2A t" — read its definition and unit from the handbook line directly above the equation.
t	Quantity produced by "t = thickness of shaft wall" — read its definition and unit from the handbook line directly above the equation.
Am	Quantity produced by "Am = area of a solid shaft of radius equal to the mean radius of the hollow shaft" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

Dz " 0
The shear strain varies in direct proportion to the radius, from zero strain at the center to the greatest strain at the outside of the
shaft. dφ/dz is the twist per unit length or the rate of twist.
L TL
where
T/φ gives the twisting moment per radian of twist. This is called the torsional stiffness and is often denoted by the symbol k or c.
For Hollow, Thin-Walled Shafts
where

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

Elongation of a steel hanger rod

A 25 mm diameter steel rod, 6.0 m long, carries 120 kN in tension. With E = 200 GPa, find the stress, the strain and the elongation.

Given

d = 25 mm
L = 6.0 m
P = 120 kN
E = 200 GPa

Find

σ, ε and δ

Start with the thinking

Area first — every axial answer depends on it.
Keep units consistent: N and mm give MPa directly.

Figure for Elongation of a steel hanger rod

Step-by-step solution

Area — A = πd²/4 = π(25)²/4 = 491 mm²
Stress
$σ = P / A = 120, 000/491 = 244 M P a$
Strain
$ε = σ / E = 244/200, 000 = 0.00122$
Elongation — δ = PL/(AE) = εL = 0.00122(6,000)
Result
$δ = 7.33 mm$

Answer: σ = 244 MPa, ε = 0.00122, δ = 7.33 mm

Why the other options are there

δ = 0.733 mm (metres and millimetres mixed)
σ = 61 MPa (diameter used as area)

Reference: FE Reference Handbook — Mechanics of Materials — Uniaxial loading

Example 2

Thermal stress in a restrained member

A steel member is fully restrained between rigid abutments and heated 40 °C. With α = 11.7 × 10⁻⁶ /°C and E = 200 GPa, what stress develops?

Given

ΔT = 40 °C
α = 11.7 × 10⁻⁶ /°C
E = 200 GPa
Full restraint

Find

Thermal stress σ

Start with the thinking

Full restraint means the free thermal strain is cancelled by an equal mechanical strain.
Length cancels out — the answer does not depend on the span.

Step-by-step solution

Free thermal strain — ε_T = αΔT = 11.7×10⁻⁶(40) = 4.68×10⁻⁴
Restraint condition
$ε_{m} ec h = - ε_{T}$
Stress
$σ = E ε = 200, 000 (4.68 \times 1 0^{- 4})$
Result
$σ = 93.6 M P a co m p r ess i o n$

Answer: σ = 93.6 MPa compression

Why the other options are there

0 MPa (free expansion assumed)
93.6 MPa tension (sign of restraint reversed)

Reference: FE Reference Handbook — Mechanics of Materials — Thermal deformations

Example 3

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
$τ = T r / J w i t h r = 30 mm$
Substitute
$τ = 3.5 \times 1 0^{6} (30) /1.272 \times 1 0^{6} = 82.6 M P a$
Twist
$ϕ = T L / (G J) = 3.5 \times 1 0^{6} (2, 000) / [80, 000 (1.272 \times 1 0^{6})]$
Result
$ϕ = 0.0688 r a d = 3.9 4^{\circ}$

Answer: τ_max = 82.6 MPa, φ = 3.94°

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 4

Axial stress and elongation of a steel rod — Torsional Strain

A steel rod of area 1.75 in² and length 222 in. carries 88 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

P = 88 kip
A = 1.75 in²
L = 222 in.
E = 29,000 ksi

Find

σ and δ

Start with the thinking

Stress is load over area; deformation adds length and stiffness.
Keep kips and inches so ksi comes out directly.

Step-by-step solution

Stress
$σ = P / A$
Substituting
$σ = 88/1.75 = 50.29 k s i$
Elongation
$δ = P L / (A E)$
Substituting
$δ = 88 (222) / (1.75 \times 29, 000)$
Evaluate
$δ = 0.3849 in$
Strain check
$ε = δ / L = 0.001734 = σ / E = 0.001734 ✓$

Answer: σ ≈ 50.29 ksi; δ ≈ 0.385 in.

Why the other options are there

50,286 ksi (psi/ksi confusion)
11,163 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Torsional Strain

Example 5

Torsional shear stress in a solid shaft — Torsional Strain

A solid circular shaft 5.25 in. in diameter transmits 174 kip·ft of torque. Find the maximum shear stress.

Given

d = 5.25 in.
T = 174 kip·ft

Find

τ_max

Start with the thinking

Torsion formula uses the polar moment of inertia.
Maximum shear occurs at the outer radius.

Step-by-step solution

Polar moment — J = πd⁴/32 = π(5.25)⁴/32 = 74.583 in⁴
Torque in kip·in — T = 174 × 12 = 2088 kip·in
Shear
$τ = T r / J$
Substituting
$τ = 2088 (2.63) /74.583 = 73.49 k s i$

Answer: τ_max ≈ 73.49 ksi

Why the other options are there

36.74 ksi (radius and diameter confused)
147.0 ksi (I used instead of J)

Reference: FE Reference Handbook — Mechanics of Materials → Torsional Strain

Example 6

Axial stress and elongation of a steel rod — Torsional Strain (2)

A steel rod of area 5.25 in² and length 125 in. carries 43 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

P = 43 kip
A = 5.25 in²
L = 125 in.
E = 29,000 ksi

Find

σ and δ

Start with the thinking

Stress is load over area; deformation adds length and stiffness.
Keep kips and inches so ksi comes out directly.

Step-by-step solution

Stress
$σ = P / A$
Substituting
$σ = 43/5.25 = 8.19 k s i$
Elongation
$δ = P L / (A E)$
Substituting
$δ = 43 (125) / (5.25 \times 29, 000)$
Evaluate
$δ = 0.0353 in$
Strain check
$ε = δ / L = 0.000282 = σ / E = 0.000282 ✓$

Answer: σ ≈ 8.19 ksi; δ ≈ 0.035 in.

Why the other options are there

8,190 ksi (psi/ksi confusion)
1,024 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Torsional Strain

Example 7

Torsional shear stress in a solid shaft — Torsional Strain (2)

A solid circular shaft 4.00 in. in diameter transmits 76 kip·ft of torque. Find the maximum shear stress.

Given

d = 4.00 in.
T = 76 kip·ft

Find

τ_max

Start with the thinking

Torsion formula uses the polar moment of inertia.
Maximum shear occurs at the outer radius.

Step-by-step solution

Polar moment — J = πd⁴/32 = π(4.00)⁴/32 = 25.133 in⁴
Torque in kip·in — T = 76 × 12 = 912 kip·in
Shear
$τ = T r / J$
Substituting
$τ = 912 (2.00) /25.133 = 72.57 k s i$

Answer: τ_max ≈ 72.57 ksi

Why the other options are there

36.29 ksi (radius and diameter confused)
145.1 ksi (I used instead of J)

Reference: FE Reference Handbook — Mechanics of Materials → Torsional Strain

Example 8

Axial stress and elongation of a steel rod — Torsional Strain (3)

A steel rod of area 3.25 in² and length 133 in. carries 50 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

P = 50 kip
A = 3.25 in²
L = 133 in.
E = 29,000 ksi

Find

σ and δ

Start with the thinking

Stress is load over area; deformation adds length and stiffness.
Keep kips and inches so ksi comes out directly.

Step-by-step solution

Stress
$σ = P / A$
Substituting
$σ = 50/3.25 = 15.38 k s i$
Elongation
$δ = P L / (A E)$
Substituting
$δ = 50 (133) / (3.25 \times 29, 000)$
Evaluate
$δ = 0.0706 in$
Strain check
$ε = δ / L = 0.000531 = σ / E = 0.000531 ✓$

Answer: σ ≈ 15.38 ksi; δ ≈ 0.071 in.

Why the other options are there

15,385 ksi (psi/ksi confusion)
2,046 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Torsional Strain

Example 9

Torsional shear stress in a solid shaft — Torsional Strain (3)

A solid circular shaft 5.50 in. in diameter transmits 141 kip·ft of torque. Find the maximum shear stress.

Given

d = 5.50 in.
T = 141 kip·ft

Find

τ_max

Start with the thinking

Torsion formula uses the polar moment of inertia.
Maximum shear occurs at the outer radius.

Step-by-step solution

Polar moment — J = πd⁴/32 = π(5.50)⁴/32 = 89.836 in⁴
Torque in kip·in — T = 141 × 12 = 1692 kip·in
Shear
$τ = T r / J$
Substituting
$τ = 1692 (2.75) /89.836 = 51.79 k s i$

Answer: τ_max ≈ 51.79 ksi

Why the other options are there

25.90 ksi (radius and diameter confused)
103.6 ksi (I used instead of J)

Reference: FE Reference Handbook — Mechanics of Materials → Torsional Strain

Example 10

Axial stress and elongation of a steel rod — Torsional Strain (4)

A steel rod of area 7.50 in² and length 80 in. carries 47 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

P = 47 kip
A = 7.50 in²
L = 80 in.
E = 29,000 ksi

Find

σ and δ

Start with the thinking

Stress is load over area; deformation adds length and stiffness.
Keep kips and inches so ksi comes out directly.

Step-by-step solution

Stress
$σ = P / A$
Substituting
$σ = 47/7.50 = 6.27 k s i$
Elongation
$δ = P L / (A E)$
Substituting
$δ = 47 (80) / (7.50 \times 29, 000)$
Evaluate
$δ = 0.0173 in$
Strain check
$ε = δ / L = 0.000216 = σ / E = 0.000216 ✓$

Answer: σ ≈ 6.27 ksi; δ ≈ 0.017 in.

Why the other options are there

6,267 ksi (psi/ksi confusion)
501.3 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Torsional Strain

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given an axially loaded, bent or twisted member, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Torsional Strain contains 10 relations; you must be able to find this page in under 15 seconds.
Exam style: a stress or deformation at one point of one member.
Unit rule: psi vs ksi and kip vs lb decide the answer choice.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

psi vs ksi and kip vs lb decide the answer choice
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.