Engineering Strain

Mechanics of Materials · FE Reference Handbook section

Mechanics of Materials

4 formulas

10 exam-style examples

~53 min

All Mechanics of Materials lectures

Learning objectives

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

This chapter section covers Engineering 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 engineering strain describes physically and when it applies.
State every one of the 4 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. Engineering 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: engineering strain.
Wikimedia Commons, public domain

Mechanics of Materials — Engineering 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 4 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

ε	Quantity produced by "ε = ∆L/Lo" — read its definition and unit from the handbook line directly above the equation.
∆L	Quantity produced by "∆L = change in length (units) of member" — read its definition and unit from the handbook line directly above the equation.
Lo	Quantity produced by "Lo = original length (units) of member" — 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

Axial stress and elongation of a steel rod — Engineering Strain

A steel rod of area 4.50 in² and length 274 in. carries 47 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

P = 47 kip
A = 4.50 in²
L = 274 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
$σ = 47/4.50 = 10.44 k s i$
Elongation
$δ = P L / (A E)$
Substituting
$δ = 47 (274) / (4.50 \times 29, 000)$
Evaluate
$δ = 0.0987 in$
Strain check
$ε = δ / L = 0.000360 = σ / E = 0.000360 ✓$

Answer: σ ≈ 10.44 ksi; δ ≈ 0.099 in.

Why the other options are there

10,444 ksi (psi/ksi confusion)
2,862 in. (E omitted)

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

Example 4

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

A steel rod of area 2.50 in² and length 237 in. carries 105 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

P = 105 kip
A = 2.50 in²
L = 237 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
$σ = 105/2.50 = 42.00 k s i$
Elongation
$δ = P L / (A E)$
Substituting
$δ = 105 (237) / (2.50 \times 29, 000)$
Evaluate
$δ = 0.3432 in$
Strain check
$ε = δ / L = 0.001448 = σ / E = 0.001448 ✓$

Answer: σ ≈ 42.00 ksi; δ ≈ 0.343 in.

Why the other options are there

42,000 ksi (psi/ksi confusion)
9,954 in. (E omitted)

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

Example 5

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

A steel rod of area 2.25 in² and length 256 in. carries 93 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

P = 93 kip
A = 2.25 in²
L = 256 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
$σ = 93/2.25 = 41.33 k s i$
Elongation
$δ = P L / (A E)$
Substituting
$δ = 93 (256) / (2.25 \times 29, 000)$
Evaluate
$δ = 0.3649 in$
Strain check
$ε = δ / L = 0.001425 = σ / E = 0.001425 ✓$

Answer: σ ≈ 41.33 ksi; δ ≈ 0.365 in.

Why the other options are there

41,333 ksi (psi/ksi confusion)
10,581 in. (E omitted)

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

Example 6

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

A steel rod of area 3.50 in² and length 134 in. carries 62 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

P = 62 kip
A = 3.50 in²
L = 134 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
$σ = 62/3.50 = 17.71 k s i$
Elongation
$δ = P L / (A E)$
Substituting
$δ = 62 (134) / (3.50 \times 29, 000)$
Evaluate
$δ = 0.0819 in$
Strain check
$ε = δ / L = 0.000611 = σ / E = 0.000611 ✓$

Answer: σ ≈ 17.71 ksi; δ ≈ 0.082 in.

Why the other options are there

17,714 ksi (psi/ksi confusion)
2,374 in. (E omitted)

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

Example 7

Axial stress and elongation of a steel rod — Engineering Strain (5)

A steel rod of area 5.25 in² and length 154 in. carries 55 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

P = 55 kip
A = 5.25 in²
L = 154 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
$σ = 55/5.25 = 10.48 k s i$
Elongation
$δ = P L / (A E)$
Substituting
$δ = 55 (154) / (5.25 \times 29, 000)$
Evaluate
$δ = 0.0556 in$
Strain check
$ε = δ / L = 0.000361 = σ / E = 0.000361 ✓$

Answer: σ ≈ 10.48 ksi; δ ≈ 0.056 in.

Why the other options are there

10,476 ksi (psi/ksi confusion)
1,613 in. (E omitted)

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

Example 8

Axial stress and elongation of a steel rod — Engineering Strain (6)

A steel rod of area 4.50 in² and length 143 in. carries 47 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

P = 47 kip
A = 4.50 in²
L = 143 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
$σ = 47/4.50 = 10.44 k s i$
Elongation
$δ = P L / (A E)$
Substituting
$δ = 47 (143) / (4.50 \times 29, 000)$
Evaluate
$δ = 0.0515 in$
Strain check
$ε = δ / L = 0.000360 = σ / E = 0.000360 ✓$

Answer: σ ≈ 10.44 ksi; δ ≈ 0.052 in.

Why the other options are there

10,444 ksi (psi/ksi confusion)
1,494 in. (E omitted)

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

Example 9

Axial stress and elongation of a steel rod — Engineering Strain (7)

A steel rod of area 6.75 in² and length 77 in. carries 47 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

P = 47 kip
A = 6.75 in²
L = 77 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
$σ = 47/6.75 = 6.96 k s i$
Elongation
$δ = P L / (A E)$
Substituting
$δ = 47 (77) / (6.75 \times 29, 000)$
Evaluate
$δ = 0.0185 in$
Strain check
$ε = δ / L = 0.000240 = σ / E = 0.000240 ✓$

Answer: σ ≈ 6.96 ksi; δ ≈ 0.018 in.

Why the other options are there

6,963 ksi (psi/ksi confusion)
536.1 in. (E omitted)

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

Example 10

Axial stress and elongation of a steel rod — Engineering Strain (8)

A steel rod of area 2.00 in² and length 197 in. carries 116 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

P = 116 kip
A = 2.00 in²
L = 197 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
$σ = 116/2.00 = 58.00 k s i$
Elongation
$δ = P L / (A E)$
Substituting
$δ = 116 (197) / (2.00 \times 29, 000)$
Evaluate
$δ = 0.3940 in$
Strain check
$ε = δ / L = 0.002000 = σ / E = 0.002000 ✓$

Answer: σ ≈ 58.00 ksi; δ ≈ 0.394 in.

Why the other options are there

58,000 ksi (psi/ksi confusion)
11,426 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Engineering 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

Engineering Strain contains 4 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.