Percent Elongation

Mechanics of Materials · FE Reference Handbook section

Mechanics of Materials

1 formulas

10 exam-style examples

~47 min

All Mechanics of Materials lectures

Learning objectives

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

This chapter section covers Percent Elongation 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 percent elongation describes physically and when it applies.
State every one of the 1 relation 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. Percent Elongation 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: percent elongation.
Wikimedia Commons, public domain

Mechanics of Materials — Percent Elongation: 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 1 relation 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.

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

Axial stress and elongation of a steel rod — Percent Elongation

A steel rod of area 2.50 in² and length 148 in. carries 47 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

P = 47 kip
A = 2.50 in²
L = 148 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
$σ = 47/2.50 = 18.80 k s i$
Elongation
$δ = P L / (A E)$
Substituting
$δ = 47 (148) / (2.50 \times 29, 000)$
Evaluate
$δ = 0.0959 in$
Strain check
$ε = δ / L = 0.000648 = σ / E = 0.000648 ✓$

Answer: σ ≈ 18.80 ksi; δ ≈ 0.096 in.

Why the other options are there

18,800 ksi (psi/ksi confusion)
2,782 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Percent Elongation

Example 2

Percent reduction in area and elongation from a tension test — Percent Elongation

A steel tensile coupon starts with a 0.420 in diameter and necks to 0.332 in at fracture. The 2 in gauge length becomes 2.440 in. Compute the percent reduction in area and the percent elongation.

Given

d₀ = 0.420 in
d_f = 0.332 in
L₀ = 2 in, L_f = 2.440 in

Find

%RA and %elongation

Start with the thinking

Percent reduction in area uses areas, not diameters — the ratio is squared.
Both measures describe ductility; RA is the more sensitive of the two.

Step-by-step solution

Areas
$A_{0} = (π /4) (0.420)^{2} = 0.13854 i n^{2}, A_{f} = (π /4) (0.332)^{2} = 0.08647 i n^{2}$
Formula
Substituting
Formula
Substituting

Answer: %RA = 37.6%, %elongation = 22.0%

Why the other options are there

21.0% (diameters instead of areas)
62.4% (remaining area, not the reduction)

Reference: FE Reference Handbook — Mechanics of Materials → Percent Elongation

Example 3

Axial stress and elongation of a steel rod — Percent Elongation (2)

A steel rod of area 2.25 in² and length 169 in. carries 55 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

P = 55 kip
A = 2.25 in²
L = 169 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
$σ = 55/2.25 = 24.44 k s i$
Elongation
$δ = P L / (A E)$
Substituting
$δ = 55 (169) / (2.25 \times 29, 000)$
Evaluate
$δ = 0.1425 in$
Strain check
$ε = δ / L = 0.000843 = σ / E = 0.000843 ✓$

Answer: σ ≈ 24.44 ksi; δ ≈ 0.142 in.

Why the other options are there

24,444 ksi (psi/ksi confusion)
4,131 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Percent Elongation

Example 4

Percent reduction in area and elongation from a tension test — Percent Elongation (2)

A steel tensile coupon starts with a 0.575 in diameter and necks to 0.391 in at fracture. The 2 in gauge length becomes 2.440 in. Compute the percent reduction in area and the percent elongation.

Given

d₀ = 0.575 in
d_f = 0.391 in
L₀ = 2 in, L_f = 2.440 in

Find

%RA and %elongation

Start with the thinking

Percent reduction in area uses areas, not diameters — the ratio is squared.
Both measures describe ductility; RA is the more sensitive of the two.

Step-by-step solution

Areas
$A_{0} = (π /4) (0.575)^{2} = 0.25967 i n^{2}, A_{f} = (π /4) (0.391)^{2} = 0.12007 i n^{2}$
Formula
Substituting
Formula
Substituting

Answer: %RA = 53.8%, %elongation = 22.0%

Why the other options are there

32.0% (diameters instead of areas)
46.2% (remaining area, not the reduction)

Reference: FE Reference Handbook — Mechanics of Materials → Percent Elongation

Example 5

Axial stress and elongation of a steel rod — Percent Elongation (3)

A steel rod of area 3.75 in² and length 196 in. carries 119 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

P = 119 kip
A = 3.75 in²
L = 196 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
$σ = 119/3.75 = 31.73 k s i$
Elongation
$δ = P L / (A E)$
Substituting
$δ = 119 (196) / (3.75 \times 29, 000)$
Evaluate
$δ = 0.2145 in$
Strain check
$ε = δ / L = 0.001094 = σ / E = 0.001094 ✓$

Answer: σ ≈ 31.73 ksi; δ ≈ 0.214 in.

Why the other options are there

31,733 ksi (psi/ksi confusion)
6,220 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Percent Elongation

Example 6

Percent reduction in area and elongation from a tension test — Percent Elongation (3)

A steel tensile coupon starts with a 0.515 in diameter and necks to 0.340 in at fracture. The 2 in gauge length becomes 2.440 in. Compute the percent reduction in area and the percent elongation.

Given

d₀ = 0.515 in
d_f = 0.340 in
L₀ = 2 in, L_f = 2.440 in

Find

%RA and %elongation

Start with the thinking

Percent reduction in area uses areas, not diameters — the ratio is squared.
Both measures describe ductility; RA is the more sensitive of the two.

Step-by-step solution

Areas
$A_{0} = (π /4) (0.515)^{2} = 0.20831 i n^{2}, A_{f} = (π /4) (0.340)^{2} = 0.09074 i n^{2}$
Formula
Substituting
Formula
Substituting

Answer: %RA = 56.4%, %elongation = 22.0%

Why the other options are there

34.0% (diameters instead of areas)
43.6% (remaining area, not the reduction)

Reference: FE Reference Handbook — Mechanics of Materials → Percent Elongation

Example 7

Axial stress and elongation of a steel rod — Percent Elongation (4)

A steel rod of area 7.50 in² and length 139 in. carries 109 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

P = 109 kip
A = 7.50 in²
L = 139 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
$σ = 109/7.50 = 14.53 k s i$
Elongation
$δ = P L / (A E)$
Substituting
$δ = 109 (139) / (7.50 \times 29, 000)$
Evaluate
$δ = 0.0697 in$
Strain check
$ε = δ / L = 0.000501 = σ / E = 0.000501 ✓$

Answer: σ ≈ 14.53 ksi; δ ≈ 0.070 in.

Why the other options are there

14,533 ksi (psi/ksi confusion)
2,020 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Percent Elongation

Example 8

Percent reduction in area and elongation from a tension test — Percent Elongation (4)

A steel tensile coupon starts with a 0.475 in diameter and necks to 0.361 in at fracture. The 2 in gauge length becomes 2.620 in. Compute the percent reduction in area and the percent elongation.

Given

d₀ = 0.475 in
d_f = 0.361 in
L₀ = 2 in, L_f = 2.620 in

Find

%RA and %elongation

Start with the thinking

Percent reduction in area uses areas, not diameters — the ratio is squared.
Both measures describe ductility; RA is the more sensitive of the two.

Step-by-step solution

Areas
$A_{0} = (π /4) (0.475)^{2} = 0.17721 i n^{2}, A_{f} = (π /4) (0.361)^{2} = 0.10235 i n^{2}$
Formula
Substituting
Formula
Substituting

Answer: %RA = 42.2%, %elongation = 31.0%

Why the other options are there

24.0% (diameters instead of areas)
57.8% (remaining area, not the reduction)

Reference: FE Reference Handbook — Mechanics of Materials → Percent Elongation

Example 9

Axial stress and elongation of a steel rod — Percent Elongation (5)

A steel rod of area 5.50 in² and length 123 in. carries 89 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

P = 89 kip
A = 5.50 in²
L = 123 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
$σ = 89/5.50 = 16.18 k s i$
Elongation
$δ = P L / (A E)$
Substituting
$δ = 89 (123) / (5.50 \times 29, 000)$
Evaluate
$δ = 0.0686 in$
Strain check
$ε = δ / L = 0.000558 = σ / E = 0.000558 ✓$

Answer: σ ≈ 16.18 ksi; δ ≈ 0.069 in.

Why the other options are there

16,182 ksi (psi/ksi confusion)
1,990 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Percent Elongation

Example 10

Percent reduction in area and elongation from a tension test — Percent Elongation (5)

A steel tensile coupon starts with a 0.550 in diameter and necks to 0.369 in at fracture. The 2 in gauge length becomes 2.440 in. Compute the percent reduction in area and the percent elongation.

Given

d₀ = 0.550 in
d_f = 0.369 in
L₀ = 2 in, L_f = 2.440 in

Find

%RA and %elongation

Start with the thinking

Percent reduction in area uses areas, not diameters — the ratio is squared.
Both measures describe ductility; RA is the more sensitive of the two.

Step-by-step solution

Areas
$A_{0} = (π /4) (0.550)^{2} = 0.23758 i n^{2}, A_{f} = (π /4) (0.369)^{2} = 0.10665 i n^{2}$
Formula
Substituting
Formula
Substituting

Answer: %RA = 55.1%, %elongation = 22.0%

Why the other options are there

33.0% (diameters instead of areas)
44.9% (remaining area, not the reduction)

Reference: FE Reference Handbook — Mechanics of Materials → Percent Elongation

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

Percent Elongation contains 1 relation; 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.