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Materials Science · Chapter 5 of 10 · Intermediate

Mechanical Properties

Pull a metal bar until it breaks and it tells you almost everything: how stiff it is, when it yields, how strong it gets, and how much it stretches first. One test, the whole story.

01

Readiness check

This chapter turns loads into properties. Tick only what you can do closed-notes.

  • Compute stress as force over area.
  • Compute strain as change in length over length.
  • Read the slope of a line on a graph.
  • Convert mm² to m² and N to MPa.
  • Recall that defects (dislocations) carry plastic flow.
0 or 1 weak itemsContinue with this chapter.
2 weak itemsReview stress basics in Mechanics of Materials.
3 or more weak itemsRevisit dislocations in Chapter 4 first.
02

The core idea

A tensile test maps stress against strain. The straight part gives stiffness, the knee gives yield strength, the peak gives tensile strength, and the stretch to fracture gives ductility.

σ = F/A₀, ε = ΔL/L₀σ = Eε (elastic)yield at 0.2% offset

Below the yield point the bar deforms elastically: stress is proportional to strain, the constant being the elastic modulus E, and the deformation fully recovers. Beyond yield it deforms plastically as dislocations glide, and the change is permanent. The curve keeps rising (work hardening) to the tensile strength, then a neck forms and the bar fails. Reading these four landmarks is the core skill of the chapter.

The skill works when: you separate elastic (recoverable, slope E) from plastic (permanent), and read each property from its landmark.
The skill breaks down when: stiffness is confused with strength, or yield is read as the very first deviation rather than the 0.2% offset.
The concept. The engineering stress-strain curve. The linear region's slope is the modulus E; the knee is the yield strength; the peak is the tensile strength; the strain at fracture is the ductility. Four numbers from one test.
03

The skills, taught in order

Mechanical behaviour is read from the stress-strain curve. Five skills cover stress and strain, the elastic and plastic regions, and the hardness and toughness that round out the picture.

5.1 Stress and strain

Engineering stress is σ = F/A₀ (force over the original area) and engineering strain is ε = ΔL/L₀. Normalising by the specimen size makes the result a material property, independent of the bar's dimensions. True stress and strain use the instantaneous area and matter near necking.

5.2 Elastic deformation

Below yield, σ = Eε (Hooke's law). The elastic modulus E is the stiffness, fixed by the bonding (Chapter 2) and barely changed by processing. Pulling one way also contracts the bar sideways, captured by Poisson's ratio ν (about 0.3 for metals); shear has its own modulus G.

5.3 Yielding and plastic deformation

The proportional limit ends the elastic region; beyond it, dislocations glide and deformation becomes permanent. Because the transition is gradual, yield strength σy is defined at the 0.2% offset: draw a line parallel to the elastic slope from a strain of 0.002 and read where it meets the curve.

5.4 Tensile strength and ductility

The curve rises with work hardening to a maximum, the tensile strength (UTS), where a neck begins; the bar then thins locally and fractures. Ductility is how much it stretched, the percent elongation %EL = (Lf − L₀)/L₀ × 100, or the percent reduction in area.

PropertySymbolRead from
Elastic modulusEslope of the elastic line
Yield strengthσy0.2% offset intersection
Tensile strengthUTSpeak of the curve
Ductility%ELstrain at fracture

5.5 Hardness, resilience, and toughness

Hardness measures resistance to local indentation and tracks roughly with tensile strength, so it is a fast, non-destructive proxy. Resilience is the elastic energy stored (area under the elastic region); toughness is the total energy absorbed to fracture (area under the whole curve), where strength and ductility combine.

MaterialE (GPa)σy (MPa)UTS (MPa)%EL
Low-carbon steel20725040025
Aluminium alloy7028034012
Copper (annealed)1107022045
Gray cast iron110(brittle)200<1

Engineering connection: these properties feed directly into stress analysis and design; the next chapter explains how to raise σy, and Chapter 7 what happens when parts fail.

04

Worked example 1: elastic stretch of a rod

A 10 mm diameter steel rod, 2 m long (E = 200 GPa, σy = 250 MPa, ν = 0.3), carries a 15 kN tensile load. Find the stress, confirm it is elastic, and find the elongation and the lateral contraction.

Figure 1. A rod in tension. Stress and strain follow from the load and geometry; since the stress is below yield, Hooke's law gives the elongation directly.
  1. ProblemFind the stress, elongation, and lateral contraction of the rod in Figure 1.
  2. Given / findd = 10 mm, L = 2 m, F = 15 kN, E = 200 GPa, σy = 250 MPa, ν = 0.3. Find σ, ΔL, Δd.
  3. AssumptionsUniaxial tension, uniform rod, behaviour elastic if σ < σy.
  4. ModelStress from F/A₀, check against yield, then Hooke's law for axial strain and Poisson's ratio for lateral.
  5. Equationsσ = F/A₀ ΔL = (σ/E)L Δd = −ν(σ/E)d
  6. SolveA₀ = π(0.005)² = 7.85×10⁻⁵ m². σ = 15 000/7.85×10⁻⁵ = 191 MPa, below the 250 MPa yield, so elastic. ε = 191/200 000 = 9.55×10⁻⁴, giving ΔL = 9.55×10⁻⁴ × 2 = 1.91 mm. Lateral: Δd = −0.3 × 9.55×10⁻⁴ × 10 = −0.0029 mm.
  7. CheckThe stress sits comfortably below yield, so assuming elastic behaviour was valid. The lateral contraction is tiny, as Poisson's ratio of 0.3 implies, and the elongation of about 2 mm over 2 m is a strain of 0.1%, typical elastic territory.
  8. ConclusionWithin the elastic range, elongation scales linearly with load and length and inversely with E and area. Cross the yield stress and this linear rule fails, which is why the yield check comes first.
Result. σ = 191 MPa (elastic); elongation 1.91 mm; lateral contraction 0.0029 mm.
05

Worked example 2: reading a tensile test

A tensile specimen (diameter 12.8 mm, gauge length 50 mm) gives: 0.05 mm elongation at 25 kN (still elastic), yield at the 0.2% offset at 50 kN, a peak load of 75 kN, and a final gauge length of 63 mm. Find E, the yield strength, the tensile strength, and the ductility.

Figure 2. The 0.2% offset construction: a line parallel to the elastic slope, shifted by 0.002 strain, locates the yield strength. The peak gives UTS and the fracture strain gives ductility.
  1. ProblemFind E, σy, UTS, and %EL for the specimen in Figure 2.
  2. Given / findd₀ = 12.8 mm, L₀ = 50 mm; elastic point 25 kN at 0.05 mm; yield load 50 kN; peak 75 kN; Lf = 63 mm.
  3. AssumptionsStandard tensile test, properties from engineering stress and strain.
  4. ModelCompute A₀, then E from the elastic point, σy and UTS from their loads, and %EL from the gauge lengths.
  5. EquationsA₀ = πd₀²/4 E = (F/A₀)/(ΔL/L₀) %EL = (Lf − L₀)/L₀ × 100
  6. SolveA₀ = π(12.8)²/4 = 128.7 mm². E = (25 000/128.7×10⁻⁶)/(0.05/50) = 194.3 MPa/0.001 = 194 GPa. σy = 50 000/128.7×10⁻⁶ = 389 MPa. UTS = 75 000/128.7×10⁻⁶ = 583 MPa. %EL = (63 − 50)/50 × 100 = 26%.
  7. CheckE near 194 GPa, σy 389 MPa, UTS 583 MPa, and 26% elongation are all in range for a medium-carbon steel. UTS exceeds σy (work hardening), and the generous ductility marks a tough, formable metal.
  8. ConclusionOne test yielded four design numbers. The 0.2% offset is the agreed convention for yield because real curves bend gradually rather than turning a sharp corner.
Result. E = 194 GPa, σy = 389 MPa, UTS = 583 MPa, ductility 26% elongation.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Stiffness equals strengthHigh-E material assumed strong"Is this resistance to stretching or to yielding?"E is stiffness; σy is strength. They are independent.
Yield at first deviationYield read too low"Did I use the 0.2% offset?"Yield strength is the 0.2% offset value, by convention.
Using current areaEngineering stress computed with necked area"Original or instantaneous area?"Engineering stress uses the original A₀; true stress uses the current area.
Strong means toughBrittle high-strength part assumed safe"How much energy to fracture?"Toughness needs both strength and ductility (area under the curve).
07

Practice ladder

Level 1 · Direct skill

A 20 mm² bar carries 8 kN elastically with E = 70 GPa over a 250 mm length. Find the stress and elongation.

Show answer

σ = 8000/20×10⁻⁶ = 400 MPa; ε = 400/70 000 = 5.7×10⁻³; ΔL = 5.7×10⁻³ × 250 = 1.43 mm. The aluminium-like modulus gives a larger strain than steel would for the same stress.

Level 2 · Mixed concept

Two steels have the same E but yield strengths of 250 and 600 MPa. How do their elastic stretch and their dent resistance compare?

Show answer

For the same stress below yield, both stretch identically (same E), since stiffness is set by bonding, not strength. But the 600 MPa steel resists permanent deformation and indentation far better, so it is harder. Strength and stiffness are decoupled.

Level 3 · Independent problem

For the Worked Example 2 specimen, estimate the modulus of resilience (elastic energy per volume, Ur = σy²/2E).

Show answer

Ur = (389×10⁶)²/(2 × 194×10⁹) = 1.51×10¹⁷/3.88×10¹¹ = 3.9×10⁵ J/m³ = 0.39 MJ/m³. This is the elastic energy a spring made of this steel could store and return.

Level 4 · Transfer to real engineering

Find a published stress-strain curve (or a datasheet) for a real alloy. Read off E, σy, UTS, and %EL, and judge whether the material is strong, stiff, ductile, or tough.

What good work looks like

All four properties read with correct landmarks, units checked, and a verdict that distinguishes stiffness from strength and notes ductility's role in toughness.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I read yield at the 0.2% offset, not the first deviation."
"Give me five curves; I will label E, σy, UTS, and fracture on each."
"Find the yield strength." Doing the offset construction yourself is the skill.
"Is this material strong?" Distinguishing strength, stiffness, and toughness is the point.

Portfolio task

Take one real tensile dataset through the four landmarks, compute E, σy, UTS, and %EL, and write a one-line verdict on the material's character.

Must include: engineering stress and strain, the 0.2% offset for yield, and a strength-versus-stiffness distinction.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Define engineering stress and strain.

σ = F/A₀ and ε = ΔL/L₀, using the original area and length.

2. What does the elastic slope give, and what sets it?

The elastic modulus E (stiffness), set by atomic bonding and little changed by processing.

3. How is yield strength defined?

At the 0.2% offset: a line parallel to the elastic slope from ε = 0.002.

4. Where are UTS and ductility on the curve?

UTS is the peak stress; ductility is the strain at fracture (%EL).

5. What is toughness, and what does it require?

The energy absorbed to fracture (area under the curve); it needs both strength and ductility.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-read a stress-strain curve from a blank page.
+3 daysCompute E, σy, UTS, %EL from one dataset.
+7 daysSee how strengthening raises σy in Chapter 6.
+30 daysCarry these properties into failure analysis, Chapter 7.
10

Textbook mapping

ItemMapping
Primary sourceCallister and Rethwisch, Materials Science and Engineering: An Introduction, Chapter 6 (Mechanical Properties of Metals)
Cross-referenceAskeland, Ch. 6 · Shackelford, Ch. 6 · Mechanics of Materials
Core topics5.1 Stress and strain · 5.2 Elastic deformation · 5.3 Yielding · 5.4 Tensile strength and ductility · 5.5 Hardness and toughness
Engineering connectionThe property inputs to all stress analysis and design.
Read nextChapter 6: Strengthening Mechanisms.