Thermal Deformations

Mechanics of Materials · FE Reference Handbook section

Mechanics of Materials

6 formulas

10 exam-style examples

~57 min

All Mechanics of Materials lectures

Learning objectives

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

This chapter section covers Thermal Deformations 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 thermal deformations describes physically and when it applies.
State every one of the 6 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. Thermal Deformations 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: thermal deformations.
Wikimedia Commons, public domain

Mechanics of Materials — Thermal Deformations: 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 6 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

δt	Quantity produced by "δt = αL(T – To)" — read its definition and unit from the handbook line directly above the equation.
α	Quantity produced by "α = temperature coefficient of expansion" — read its definition and unit from the handbook line directly above the equation.
L	Quantity produced by "L = length of member" — read its definition and unit from the handbook line directly above the equation.
T	Quantity produced by "T = final temperature" — read its definition and unit from the handbook line directly above the equation.
To	Quantity produced by "To = initial temperature" — 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.

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

Thermal expansion and restrained stress — Thermal Deformations

A 71 ft steel member is heated 61°F (α = 6.5 × 10⁻⁶ /°F). Find the free elongation and the stress if expansion is fully restrained.

Given

L = 71 ft
ΔT = 61°F
α = 6.5e−6 /°F
E = 29,000 ksi

Find

δ_free and σ_restrained

Start with the thinking

Free expansion produces strain but no stress.
Full restraint converts all that strain into stress.

Step-by-step solution

Free elongation — δ = αΔT·L
Substituting
$δ = 6.5 e - 6 (61) (71 \times 12) = 0.3378 in$
Restrained stress — σ = EαΔT
Substituting
$σ = 29, 000 (6.5 e - 6) (61) = 11.50 k s i (co m p r ess i o n)$

Answer: δ ≈ 0.338 in.; σ ≈ 11.5 ksi compression

Why the other options are there

0.16 ksi (length wrongly divided in)
0.0282 in. (feet/inches conversion missed)

Reference: FE Reference Handbook — Mechanics of Materials → Thermal Deformations

Example 4

Thermal expansion and restrained stress — Thermal Deformations (2)

A 50 ft steel member is heated 75°F (α = 6.5 × 10⁻⁶ /°F). Find the free elongation and the stress if expansion is fully restrained.

Given

L = 50 ft
ΔT = 75°F
α = 6.5e−6 /°F
E = 29,000 ksi

Find

δ_free and σ_restrained

Start with the thinking

Free expansion produces strain but no stress.
Full restraint converts all that strain into stress.

Step-by-step solution

Free elongation — δ = αΔT·L
Substituting
$δ = 6.5 e - 6 (75) (50 \times 12) = 0.2925 in$
Restrained stress — σ = EαΔT
Substituting
$σ = 29, 000 (6.5 e - 6) (75) = 14.14 k s i (co m p r ess i o n)$

Answer: δ ≈ 0.293 in.; σ ≈ 14.1 ksi compression

Why the other options are there

0.28 ksi (length wrongly divided in)
0.0244 in. (feet/inches conversion missed)

Reference: FE Reference Handbook — Mechanics of Materials → Thermal Deformations

Example 5

Thermal expansion and restrained stress — Thermal Deformations (3)

A 93 ft steel member is heated 39°F (α = 6.5 × 10⁻⁶ /°F). Find the free elongation and the stress if expansion is fully restrained.

Given

L = 93 ft
ΔT = 39°F
α = 6.5e−6 /°F
E = 29,000 ksi

Find

δ_free and σ_restrained

Start with the thinking

Free expansion produces strain but no stress.
Full restraint converts all that strain into stress.

Step-by-step solution

Free elongation — δ = αΔT·L
Substituting
$δ = 6.5 e - 6 (39) (93 \times 12) = 0.2829 in$
Restrained stress — σ = EαΔT
Substituting
$σ = 29, 000 (6.5 e - 6) (39) = 7.35 k s i (co m p r ess i o n)$

Answer: δ ≈ 0.283 in.; σ ≈ 7.4 ksi compression

Why the other options are there

0.08 ksi (length wrongly divided in)
0.0236 in. (feet/inches conversion missed)

Reference: FE Reference Handbook — Mechanics of Materials → Thermal Deformations

Example 6

Thermal expansion and restrained stress — Thermal Deformations (4)

A 79 ft steel member is heated 66°F (α = 6.5 × 10⁻⁶ /°F). Find the free elongation and the stress if expansion is fully restrained.

Given

L = 79 ft
ΔT = 66°F
α = 6.5e−6 /°F
E = 29,000 ksi

Find

δ_free and σ_restrained

Start with the thinking

Free expansion produces strain but no stress.
Full restraint converts all that strain into stress.

Step-by-step solution

Free elongation — δ = αΔT·L
Substituting
$δ = 6.5 e - 6 (66) (79 \times 12) = 0.4067 in$
Restrained stress — σ = EαΔT
Substituting
$σ = 29, 000 (6.5 e - 6) (66) = 12.44 k s i (co m p r ess i o n)$

Answer: δ ≈ 0.407 in.; σ ≈ 12.4 ksi compression

Why the other options are there

0.16 ksi (length wrongly divided in)
0.0339 in. (feet/inches conversion missed)

Reference: FE Reference Handbook — Mechanics of Materials → Thermal Deformations

Example 7

Thermal expansion and restrained stress — Thermal Deformations (5)

A 102 ft steel member is heated 108°F (α = 6.5 × 10⁻⁶ /°F). Find the free elongation and the stress if expansion is fully restrained.

Given

L = 102 ft
ΔT = 108°F
α = 6.5e−6 /°F
E = 29,000 ksi

Find

δ_free and σ_restrained

Start with the thinking

Free expansion produces strain but no stress.
Full restraint converts all that strain into stress.

Step-by-step solution

Free elongation — δ = αΔT·L
Substituting
$δ = 6.5 e - 6 (108) (102 \times 12) = 0.8592 in$
Restrained stress — σ = EαΔT
Substituting
$σ = 29, 000 (6.5 e - 6) (108) = 20.36 k s i (co m p r ess i o n)$

Answer: δ ≈ 0.859 in.; σ ≈ 20.4 ksi compression

Why the other options are there

0.20 ksi (length wrongly divided in)
0.0716 in. (feet/inches conversion missed)

Reference: FE Reference Handbook — Mechanics of Materials → Thermal Deformations

Example 8

Thermal expansion and restrained stress — Thermal Deformations (6)

A 43 ft steel member is heated 42°F (α = 6.5 × 10⁻⁶ /°F). Find the free elongation and the stress if expansion is fully restrained.

Given

L = 43 ft
ΔT = 42°F
α = 6.5e−6 /°F
E = 29,000 ksi

Find

δ_free and σ_restrained

Start with the thinking

Free expansion produces strain but no stress.
Full restraint converts all that strain into stress.

Step-by-step solution

Free elongation — δ = αΔT·L
Substituting
$δ = 6.5 e - 6 (42) (43 \times 12) = 0.1409 in$
Restrained stress — σ = EαΔT
Substituting
$σ = 29, 000 (6.5 e - 6) (42) = 7.92 k s i (co m p r ess i o n)$

Answer: δ ≈ 0.141 in.; σ ≈ 7.9 ksi compression

Why the other options are there

0.18 ksi (length wrongly divided in)
0.0117 in. (feet/inches conversion missed)

Reference: FE Reference Handbook — Mechanics of Materials → Thermal Deformations

Example 9

Thermal expansion and restrained stress — Thermal Deformations (7)

A 74 ft steel member is heated 89°F (α = 6.5 × 10⁻⁶ /°F). Find the free elongation and the stress if expansion is fully restrained.

Given

L = 74 ft
ΔT = 89°F
α = 6.5e−6 /°F
E = 29,000 ksi

Find

δ_free and σ_restrained

Start with the thinking

Free expansion produces strain but no stress.
Full restraint converts all that strain into stress.

Step-by-step solution

Free elongation — δ = αΔT·L
Substituting
$δ = 6.5 e - 6 (89) (74 \times 12) = 0.5137 in$
Restrained stress — σ = EαΔT
Substituting
$σ = 29, 000 (6.5 e - 6) (89) = 16.78 k s i (co m p r ess i o n)$

Answer: δ ≈ 0.514 in.; σ ≈ 16.8 ksi compression

Why the other options are there

0.23 ksi (length wrongly divided in)
0.0428 in. (feet/inches conversion missed)

Reference: FE Reference Handbook — Mechanics of Materials → Thermal Deformations

Example 10

Thermal expansion and restrained stress — Thermal Deformations (8)

A 72 ft steel member is heated 97°F (α = 6.5 × 10⁻⁶ /°F). Find the free elongation and the stress if expansion is fully restrained.

Given

L = 72 ft
ΔT = 97°F
α = 6.5e−6 /°F
E = 29,000 ksi

Find

δ_free and σ_restrained

Start with the thinking

Free expansion produces strain but no stress.
Full restraint converts all that strain into stress.

Step-by-step solution

Free elongation — δ = αΔT·L
Substituting
$δ = 6.5 e - 6 (97) (72 \times 12) = 0.5448 in$
Restrained stress — σ = EαΔT
Substituting
$σ = 29, 000 (6.5 e - 6) (97) = 18.28 k s i (co m p r ess i o n)$

Answer: δ ≈ 0.545 in.; σ ≈ 18.3 ksi compression

Why the other options are there

0.25 ksi (length wrongly divided in)
0.0454 in. (feet/inches conversion missed)

Reference: FE Reference Handbook — Mechanics of Materials → Thermal Deformations

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

Thermal Deformations contains 6 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.