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Mechanics of Materials · Chapter 2 of 10 · Beginner

Axial Loading

How much does a loaded bar stretch? The answer, δ = PL/AE, packs the load, the geometry, and the material into one tidy formula, and it opens the door to indeterminate and thermal problems.

Readiness check

This chapter turns stress and strain into deformation. Tick only what you can do closed-notes.

Use σ = P/A and ε = δ/L.
Apply Hooke's law σ = Eε.
Find internal forces in a multi-part member.
Solve two simultaneous equations.
Recall thermal expansion δ = αΔT·L.

0 or 1 weak itemsContinue with this chapter.

2 weak itemsReview stress and strain in Chapter 1.

3 or more weak itemsRevisit Hooke's law and internal forces first.

The core idea

An axially loaded bar stretches by δ = PL/AE; when supports over-constrain it, compatibility of deformation supplies the extra equation, and temperature change adds its own strain.

δ = PL/AEδ = Σ P_iL_i/(A_iE)δ_T = αΔT·L

Combining ε = δ/L with σ = Eε and σ = P/A gives the elongation directly: more load or length stretches it, more area or stiffness resists. For a member made of segments, the elongations simply add. When the supports leave the forces undetermined by statics alone (statically indeterminate), a compatibility condition on the deformation closes the system. Temperature change causes free expansion αΔT·L, and if that expansion is restrained it becomes stress.

The skill works when: you sum PL/AE over segments, and add a compatibility or thermal condition when statics is not enough.

The skill breaks down when: an indeterminate member is solved by statics alone, or restrained thermal expansion is ignored.

The concept. A bar fixed at one end stretches by δ = PL/AE under load P. The elongation grows with load and length, and shrinks with cross-section and stiffness.

The skills, taught in order

Axial loading is one formula extended to segments, redundant supports, and temperature. Five skills cover them.

2.1 Axial deformation

The elongation of a uniform bar is δ = PL/AE. The group AE is the axial rigidity; PL is the load times length. The formula follows directly from stacking σ = P/A, ε = σ/E, and δ = εL.

Increase	Effect on δ
Load P	increases (proportional)
Length L	increases (proportional)
Area A	decreases
Modulus E	decreases

2.2 Multi-segment and varying load

When the bar has segments of different load, area, or material, the total elongation is the sum δ = Σ P_iL_i/(A_iE_i). If the load or area varies continuously, integrate δ = ∫P(x)/(A(x)E) dx. Each segment's internal force comes from a free body.

2.3 Statically indeterminate members

If the reactions cannot be found by equilibrium alone (more supports than equations), the member is statically indeterminate. Add a compatibility equation: the deformations must fit the geometry (for instance, the total length change between two fixed walls is zero). Equilibrium plus compatibility then solves it.

2.4 Thermal stress and strain

A free bar expands by δ_T = αΔT·L with no stress. If it is restrained, the prevented expansion produces stress. For a bar held rigidly at both ends, the stress is σ = EαΔT, independent of length and area, often surprisingly large.

2.5 Poisson's ratio and multiaxial stress

Stretching a bar axially contracts it sideways: ε_lateral = −νε_axial, with ν ≈ 0.3 for metals. Under several stresses at once, the generalized Hooke's law superposes the strains. Geometric features (holes, fillets) raise the local stress by a concentration factor K_t.

Engineering connection: tie-rods, bolts, truss members, and bimetallic and restrained assemblies are all axial-loading problems; thermal stress alone explains expansion joints in bridges and rails.

Worked example 1: elongation of a stepped bar

A steel bar (E = 200 GPa) has two segments. The first carries an internal force of 80 kN over 1.2 m with a 600 mm² area; the second carries 30 kN over 0.9 m with a 300 mm² area. Find the total elongation.

Figure 1. Each segment elongates by its own PL/AE; the total stretch is their sum. The thinner, lighter-loaded segment still contributes because its area is smaller.

ProblemFind the total elongation of the stepped bar in Figure 1.
Given / findE = 200 GPa; segment 1: N = 80 kN, L = 1.2 m, A = 600 mm²; segment 2: N = 30 kN, L = 0.9 m, A = 300 mm². Find δ.
AssumptionsUniform stress in each segment, linear-elastic, internal forces as given.
ModelCompute each segment's PL/AE and add them.
Equationsδ = Σ N_iL_i/(A_iE)
Solveδ₁ = (80 000 × 1200)/(600 × 200 000) = 0.80 mm. δ₂ = (30 000 × 900)/(300 × 200 000) = 0.45 mm. Total δ = 1.25 mm.
CheckUnits: (N·mm)/(mm²·MPa) = mm, correct. The first segment dominates despite its larger area because it carries far more load over a longer length.
ConclusionElongations of segments simply add, each computed with its own internal force, area, and length. This summation handles any stepped or multi-material axial member.

Result. Total elongation δ = 1.25 mm (0.80 + 0.45).

Worked example 2: thermal stress in a restrained bar

A steel bar is fixed rigidly between two walls and heated by 60 °C. With E = 200 GPa and α = 12×10⁻⁶ /°C, find the thermal stress.

Figure 2. The walls prevent the heated bar from expanding, so the would-be expansion is converted into compressive stress, independent of the bar's length and area.

ProblemFind the thermal stress in the fully restrained bar of Figure 2.
Given / findΔT = 60 °C, E = 200 GPa, α = 12×10⁻⁶ /°C, both ends fixed. Find σ.
AssumptionsRigid walls (zero net length change), uniform temperature, linear-elastic.
ModelFree expansion αΔT·L is fully prevented, so the compressive mechanical strain equals the thermal strain; stress follows from Hooke's law.
Equationsε_T = αΔT σ = E·ε_T = EαΔT
Solveε_T = 12×10⁻⁶ × 60 = 7.2×10⁻⁴. σ = 200 000 × 7.2×10⁻⁴ = 144 MPa, compressive.
CheckThe stress does not depend on length or area, only on E, α, and ΔT, which is why even a modest 60 °C rise produces 144 MPa, comparable to working stresses. Real supports flex a little, relieving some of it.
ConclusionRestrained thermal expansion is a major source of stress. Engineers add expansion joints, slots, or flexible supports precisely to avoid this EαΔT, in rails, pipes, and bridges.

Result. Thermal stress σ = 144 MPa (compressive), independent of length and area.

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
One formula for a stepped bar	Wrong elongation	"Does each segment have its own P, A, L?"	Sum PL/AE segment by segment.
Statics alone for indeterminate	More unknowns than equations	"Are there redundant supports?"	Add a compatibility (deformation) equation.
Ignoring restrained thermal stress	Buckled rail or cracked pipe	"Is expansion prevented?"	Restrained expansion gives σ = EαΔT; allow for it.
Forgetting lateral contraction	Volume or fit miscomputed	"Did I include Poisson's ratio?"	ε_lateral = −νε_axial.

Practice ladder

Level 1 · Direct skill

A 2 m aluminium rod (E = 70 GPa, A = 400 mm²) carries 28 kN. Find its elongation.

Show answer

δ = PL/AE = (28 000 × 2000)/(400 × 70 000) = 56×10⁶/2.8×10⁷ = 2.0 mm. Aluminium's lower E stretches more than steel would.

Level 2 · Mixed concept

The Worked Example 2 bar is heated the same 60 °C but one wall can move 0.5 mm; the bar is 2 m long. Does the full 144 MPa still develop?

Show answer

Free expansion is αΔT·L = 7.2×10⁻⁴ × 2000 = 1.44 mm. Only 1.44 − 0.5 = 0.94 mm is prevented, so ε = 0.94/2000 = 4.7×10⁻⁴ and σ = E·ε = 94 MPa, less than 144. A little give relieves a lot of stress.

Level 3 · Independent problem

A rod is fixed at both ends; a 40 kN load is applied at a point dividing it into 1 m and 2 m lengths (same A, E). Find the wall reactions (statically indeterminate).

Show answer

Compatibility: the point's displacement from each side must match, giving R₁L₁ = R₂L₂ shares inversely with length. With L₁ = 1, L₂ = 2: R₁/R₂ = L₂/L₁ = 2, and R₁ + R₂ = 40 kN, so R₁ = 26.7 kN, R₂ = 13.3 kN. Equilibrium alone could not split the 40 kN.

Level 4 · Transfer to real engineering

Find a real axially loaded or thermally restrained part (a turnbuckle tie, a rail, a bolted flange). Estimate its elongation or thermal stress and comment on whether it needs an expansion allowance.

What good work looks like

The right formula (PL/AE or EαΔT) applied, segments or restraint handled correctly, and a judgement on deformation or expansion allowance.

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check whether this member is statically indeterminate before I solve it."

"Give me five axial members; I will say which need a compatibility equation."

"Compute the elongation." Summing PL/AE yourself is the skill.

"What is the thermal stress?" Recognising restraint and applying EαΔT is the point.

Portfolio task

Analyse one axial member: find its elongation by segments, and if it is restrained or heated, compute the indeterminate or thermal stress.

Must include: a PL/AE summation, a statement of determinacy, and a thermal or compatibility check where relevant.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Write the axial deformation formula.

δ = PL/AE; for segments, δ = Σ PL/AE.

2. How do you solve a statically indeterminate member?

Equilibrium plus a compatibility condition on the deformations.

3. Give the thermal stress for a fully restrained bar.

σ = EαΔT, independent of length and area.

4. State Poisson's ratio relation.

ε_lateral = −νε_axial, with ν ≈ 0.3 for metals.

5. What is the axial rigidity?

AE, the product that resists axial deformation.

TodayFinish this quiz and Levels 1 and 2 of the ladder.

+1 dayRe-derive the stepped-bar and thermal results from a blank page.

+3 daysOne indeterminate and one thermal problem.

+7 daysMove to torsion, Chapter 3.

+30 daysReuse compatibility for indeterminate beams later.

Textbook mapping

Item	Mapping
Primary source	Beer, Johnston, DeWolf and Mazurek, Mechanics of Materials (6th ed), Chapter 2 (Stress and Strain, Axial Loading)
Cross-reference	Hibbeler, Ch. 4 · Gere and Goodno, Ch. 2
Core topics	2.1 Axial deformation · 2.2 Multi-segment · 2.3 Indeterminate members · 2.4 Thermal stress · 2.5 Poisson's ratio
Engineering connection	Tie-rods, bolts, and restrained or heated assemblies.
Read next	Chapter 3: Torsion.