Elastic Strain Energy

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 Elastic Strain Energy 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 elastic strain energy 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. Elastic Strain Energy 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: elastic strain energy.
Wikimedia Commons, public domain

Mechanics of Materials — Elastic Strain Energy: 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

U	Quantity produced by "U = W = Pδ/2" — read its definition and unit from the handbook line directly above the equation.
u	Quantity produced by "u = U/AL = σ2/2E (for tension)" — 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.

If the strain remains within the elastic limit, the work done during deflection (extension) of a member will be transformed into
potential energy and can be recovered.
If the final load is P and the corresponding elongation of a tension member is δ, then the total energy U stored is equal to the
work W done during loading.
The strain energy per unit volume is
Material Properties
Table 1 - Typical Material Proper�es
(Use these values if the speciﬁc alloy and temper are not listed on Table 2 below)
Modulus of Coefficient of Thermal
Modulus of Rigidity, G Density, ρ
[Mpsi (GPa)] [lb/in3 (Mg/m3)]
[Mpsi (GPa)] [10−6/ºF (10−6/ºC)]
Steel 29.0 (200.0) 11.5 (80.0) 0.30 6.5 (11.7) 0.282 (7.8)
Aluminum 10.0 (69.0) 3.8 (26.0) 0.33 13.1 (23.6) 0.098 (2.7)
Cast Iron 14.5 (100.0) 6.0 (41.4) 0.21 6.7 (12.1) 0.246−0.282 (6.8−7.8)
Wood (Fir) 1.6 (11.0) 0.6 (4.1) 0.33 1.7 (3.0) −
Brass 14.8−18.1 (102−125) 5.8 (40) 0.33 10.4 (18.7) 0.303−0.313 (8.4−8.7)
Copper 17 (117) 6.5 (45) 0.36 9.3 (16.6) 0.322 (8.9)
Bronze 13.9−17.4 (96−120) 6.5 (45) 0.34 10.0 (18.0) 0.278−0.314 (7.7−8.7)
Magnesium 6.5 (45) 2.4 (16.5) 0.35 14 (25) 0.061 (1.7)
Glass 10.2 (70) − 0.22 5.0 (9.0) 0.090 (2.5)
Polystyrene 0.3 (2) − 0.34 38.9 (70.0) 0.038 (1.05)
Polyvinyl Chloride (PVC) <0.6 (<4) − − 28.0 (50.4) 0.047 (1.3)
Alumina Fiber 58 (400) − − − 0.141 (3.9)

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 — Elastic Strain Energy

A steel rod of area 5.50 in² and length 170 in. carries 29 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

P = 29 kip
A = 5.50 in²
L = 170 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
$σ = 29/5.50 = 5.27 k s i$
Elongation
$δ = P L / (A E)$
Substituting
$δ = 29 (170) / (5.50 \times 29, 000)$
Evaluate
$δ = 0.0309 in$
Strain check
$ε = δ / L = 0.000182 = σ / E = 0.000182 ✓$

Answer: σ ≈ 5.27 ksi; δ ≈ 0.031 in.

Why the other options are there

5,273 ksi (psi/ksi confusion)
896.4 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Elastic Strain Energy

Example 4

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

A steel rod of area 2.00 in² and length 286 in. carries 100 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

P = 100 kip
A = 2.00 in²
L = 286 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
$σ = 100/2.00 = 50.00 k s i$
Elongation
$δ = P L / (A E)$
Substituting
$δ = 100 (286) / (2.00 \times 29, 000)$
Evaluate
$δ = 0.4931 in$
Strain check
$ε = δ / L = 0.001724 = σ / E = 0.001724 ✓$

Answer: σ ≈ 50.00 ksi; δ ≈ 0.493 in.

Why the other options are there

50,000 ksi (psi/ksi confusion)
14,300 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Elastic Strain Energy

Example 5

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

A steel rod of area 8.00 in² and length 182 in. carries 53 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

P = 53 kip
A = 8.00 in²
L = 182 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
$σ = 53/8.00 = 6.63 k s i$
Elongation
$δ = P L / (A E)$
Substituting
$δ = 53 (182) / (8.00 \times 29, 000)$
Evaluate
$δ = 0.0416 in$
Strain check
$ε = δ / L = 0.000228 = σ / E = 0.000228 ✓$

Answer: σ ≈ 6.63 ksi; δ ≈ 0.042 in.

Why the other options are there

6,625 ksi (psi/ksi confusion)
1,206 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Elastic Strain Energy

Example 6

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

A steel rod of area 3.50 in² and length 67 in. carries 67 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

P = 67 kip
A = 3.50 in²
L = 67 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
$σ = 67/3.50 = 19.14 k s i$
Elongation
$δ = P L / (A E)$
Substituting
$δ = 67 (67) / (3.50 \times 29, 000)$
Evaluate
$δ = 0.0442 in$
Strain check
$ε = δ / L = 0.000660 = σ / E = 0.000660 ✓$

Answer: σ ≈ 19.14 ksi; δ ≈ 0.044 in.

Why the other options are there

19,143 ksi (psi/ksi confusion)
1,283 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Elastic Strain Energy

Example 7

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

A steel rod of area 7.00 in² and length 80 in. carries 104 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

P = 104 kip
A = 7.00 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
$σ = 104/7.00 = 14.86 k s i$
Elongation
$δ = P L / (A E)$
Substituting
$δ = 104 (80) / (7.00 \times 29, 000)$
Evaluate
$δ = 0.0410 in$
Strain check
$ε = δ / L = 0.000512 = σ / E = 0.000512 ✓$

Answer: σ ≈ 14.86 ksi; δ ≈ 0.041 in.

Why the other options are there

14,857 ksi (psi/ksi confusion)
1,189 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Elastic Strain Energy

Example 8

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

A steel rod of area 7.50 in² and length 235 in. carries 104 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

P = 104 kip
A = 7.50 in²
L = 235 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
$σ = 104/7.50 = 13.87 k s i$
Elongation
$δ = P L / (A E)$
Substituting
$δ = 104 (235) / (7.50 \times 29, 000)$
Evaluate
$δ = 0.1124 in$
Strain check
$ε = δ / L = 0.000478 = σ / E = 0.000478 ✓$

Answer: σ ≈ 13.87 ksi; δ ≈ 0.112 in.

Why the other options are there

13,867 ksi (psi/ksi confusion)
3,259 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Elastic Strain Energy

Example 9

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

A steel rod of area 5.25 in² and length 67 in. carries 97 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

P = 97 kip
A = 5.25 in²
L = 67 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
$σ = 97/5.25 = 18.48 k s i$
Elongation
$δ = P L / (A E)$
Substituting
$δ = 97 (67) / (5.25 \times 29, 000)$
Evaluate
$δ = 0.0427 in$
Strain check
$ε = δ / L = 0.000637 = σ / E = 0.000637 ✓$

Answer: σ ≈ 18.48 ksi; δ ≈ 0.043 in.

Why the other options are there

18,476 ksi (psi/ksi confusion)
1,238 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Elastic Strain Energy

Example 10

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

A steel rod of area 7.25 in² and length 297 in. carries 103 kip of tension. Find the stress and elongation (E = 29,000 ksi).

Given

P = 103 kip
A = 7.25 in²
L = 297 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
$σ = 103/7.25 = 14.21 k s i$
Elongation
$δ = P L / (A E)$
Substituting
$δ = 103 (297) / (7.25 \times 29, 000)$
Evaluate
$δ = 0.1455 in$
Strain check
$ε = δ / L = 0.000490 = σ / E = 0.000490 ✓$

Answer: σ ≈ 14.21 ksi; δ ≈ 0.145 in.

Why the other options are there

14,207 ksi (psi/ksi confusion)
4,219 in. (E omitted)

Reference: FE Reference Handbook — Mechanics of Materials → Elastic Strain Energy

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

Elastic Strain Energy 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.