
Physics I course sequence · Interactive lesson
Stress–Strain Curves, Elasticity & Failure
A tensile test pulls one specimen until it breaks and hands you a single graph. Read it the way a designer does: how much stretch springs back, where stretch turns permanent, the largest load the part can ever carry, the energy it soaks up before it snaps, and how far below all that a working stress belongs.
UPI-12.10 · Lesson 10 of 10 in this unit. Position is not a completion score.Build the model
Hooke's law has an edge. The tensile test finds it.
The key idea
A stress–strain curve plots engineering stress F/A₀ against strain ΔL/L₀ as a specimen is stretched to fracture; its landmarks mark where the material stops springing back, where it stops getting stronger, and where it breaks.
Example. Structural steel with E = 200 GPa yields near 250 MPa, at a strain of only 250 MPa ÷ 200 GPa = 0.125 %, peaks near 400 MPa and breaks at about 20 % strain, so at least 99 % of the stretch it survives is permanent.
The complete physical model and assumptions
Young modulus describes a solid only while its atoms are being pulled slightly apart and would spring back if released; the stress–strain curve measures how far that promise holds. Pull a specimen slowly, divide the load by the original area and the extension by the original gauge length, and the curve begins as a straight line of slope E, bends at the proportional limit, and stops being recoverable at the elastic limit. Beyond yield, planes of atoms slip past one another. That strain stays when the load is removed: unloading runs back down a line of slope E and returns only σ/E. Slip also tangles and hardens the metal, so the stress needed to keep stretching it rises, until at the ultimate tensile strength the hardening can no longer outpace the shrinking cross-section. Deformation then gathers into a neck, the load falls, and the specimen fractures. A ductile material spends most of its curve in that plastic range, absorbing energy and giving visible warning; a brittle one fractures near the top of its elastic line with neither. The curve belongs to slow, steady, one-directional tension at one temperature. Load the same material fast, cold, repeatedly, around a notch or as a slender strut in compression, and it can fail far below every stress this graph reports, which is why working stresses are set well beneath it.
1Proportional limit, elastic limit, yield: three claims
The start of the curve makes three claims that first-year answers often run together. The proportional limit is where the graph stops being straight: below it, σ = Eε exactly. The elastic limit is the largest stress after which the specimen still returns to its original length, a statement about unloading rather than about shape. Rubber shows how far apart the two can sit: its curve bends almost from the start, yet it recovers from strains of several hundred per cent. Yield is the start of plastic flow, planes of atoms slipping permanently past one another. In most metals all three fall within a fraction of a per cent of strain, too close to separate cleanly in a test, which is why engineers quote a defined yield strength instead. Annealed low-carbon steel adds a twist: at yield its stress dips slightly and the bar stretches at nearly constant stress for a while before hardening begins.
2Unload the specimen to separate elastic from plastic strain
The only way to find out how much of a strain is permanent is to take the load off. Below the elastic limit, the specimen retraces its loading line back to zero. Beyond it, the stress falls along a new straight line with the same slope E, parallel to the original elastic line, because the atomic bonds unload just as they loaded. The recovered strain is σ/E and what remains is the permanent set, εₚ = ε − σ/E. Load a steel wire with E = 200 GPa to a strain of 3.00 %, where the stress is 380 MPa, and let go: it springs back 380 MPa ÷ 200 GPa = 0.19 % and keeps 2.81 %, about 94 % of what it had. Reload it and it climbs that same unloading line elastically, not yielding until it again reaches 380 MPa. That is cold working: the wire now behaves elastically up to 380 MPa, well above its original yield strength, and has spent part of its ductility to get there.
3The 0.2 % offset turns a gradual knee into a number
Many alloys, aluminium and stainless steel among them, have no sharp yield point: the curve bends away from its straight line gradually, so any stress picked by eye is arbitrary. The offset method replaces judgement with a construction. Draw a line of slope E starting from ε = 0.002 on the strain axis; the stress where it crosses the curve is the 0.2 % offset yield strength σy. The number has a precise meaning: unload from that point, down a line of slope E, and the specimen keeps exactly 0.2 % of permanent strain. For an aluminium alloy with E = 70 GPa whose curve passes through 280 MPa at a total strain of 0.60 %, the offset line reaches 70 000 MPa × (0.0060 − 0.002) = 280 MPa at that strain, so σy = 280 MPa, made of 0.40 % elastic strain and 0.20 % plastic. Two tenths of a per cent is small enough to leave a part serviceable and large enough to measure reliably with an extensometer.
4Hardening, the peak load, and the neck
Past yield, slip piles up defects that obstruct further slip, so the stress needed to keep stretching the metal rises. The true stress σₜ = F/A tracks that honestly because it uses the current area; at constant volume A = A₀L₀/L, which gives σₜ = σ(1 + ε) until necking starts. A copper specimen at an engineering strain of 0.30 and 220 MPa carries a true stress of 286 MPa at a true strain of ln 1.30 = 0.262. The load F = σₜA is the product of a rising factor and a falling one. It peaks when dF = A dσₜ + σₜ dA = 0, and at constant volume dA/A = −dεₜ, so the peak comes where dσₜ/dεₜ = σₜ. Beyond that point a slightly thinner section carries a higher stress than its neighbours but can no longer harden fast enough to compensate, so it keeps thinning while the rest of the bar stops stretching. That local collapse is the neck, and the peak engineering stress is the ultimate tensile strength σᵤ.
5Ductile or brittle: count the strain and the energy
Ductility is measured after fracture, on the broken halves fitted back together: the percentage elongation (Lf − L₀)/L₀, or the percentage reduction of area at the neck. A material that fractures with less than about 5 % elongation is usually called brittle; glass, cast iron, ceramics and concrete in tension fail on or near their elastic line, often suddenly and with no neck to warn anyone. The energy absorbed per unit volume is the area under the curve, and it separates the two classes far more sharply than strength does. Steel that averages about 350 MPa over 20 % strain absorbs roughly 350 × 10⁶ Pa × 0.20 = 70 MJ m⁻³. Window glass with a fracture stress near 50 MPa and E = 70 GPa stores only σ²/(2E) = 1.8 × 10⁴ J m⁻³ before it breaks, about four thousand times less, although its stiffness matches aluminium's. Even the steel's elastic share, σy²/(2E) = 0.16 MJ m⁻³ for σy = 250 MPa, is under 1 % of its total: a ductile metal's toughness is almost all plastic work.
6From curve to design, and where the curve stops applying
A ductile part has failed its purpose once it yields, even if it has not broken, so its allowable stress is σy/n. A structural steel with σy = 250 MPa and n = 1.67 may carry 150 MPa, keeping the working point on the straight line where every strain springs back. Brittle materials give no plastic warning and scatter widely in strength, so designs divide σᵤ by a larger n. The factor also has to cover what the tensile test never saw. A small circular hole in a wide plate triples the local stress, so 150 MPa far from the hole becomes 450 MPa at its edge, past yield. Loads repeated millions of times grow fatigue cracks at stresses well below σy. Some steels turn brittle in the cold, which is how welded Liberty ships cracked in cold seas in the 1940s. A slender strut in compression buckles long before its material yields. The curve reports the material's limits; the design has to supply the rest.
Key equations
Read the units, assumptions and limits before substituting.
Springback and permanent set
εₚ = ε − σ/E
Unloading runs down a line of slope E, so only σ/E springs back and the rest of the strain stays.
ε and σ are the strain and stress (Pa) when unloading starts; E in Pa; εₚ is dimensionless0.2 % offset yield strength
σy is where the line σ = E(ε − 0.002) meets the curve
Turns a gradual knee into a reproducible number: the stress that leaves 0.2 % of permanent strain once unloaded.
0.002 is an agreed permanent strain; σy and E in Pa, strains dimensionlessTrue stress and true strain
σₜ = F/A = σ(1 + ε) · εₜ = ∫ dL/L from L₀ to L = ln(1 + ε)
Shows the metal still strengthening while the engineering curve flattens: at ε = 0.30 the true stress is 30 % higher.
A is the current area in m²; σ(1 + ε) assumes constant volume, AL = A₀L₀, so use it only before neckingOnset of necking (Considère)
dF = 0 ⇔ dσₜ/dεₜ = σₜ
The load peaks, at the ultimate tensile strength, once the hardening rate has fallen to the true stress itself.
F = σₜA with dA/A = −dεₜ at constant volume; both sides in PaEnergy absorbed per unit volume
toughness = ∫ σ dε from 0 to εf · resilience = σy²/(2E)
Area under the curve: resilience is the elastic triangle you get back, toughness is all the energy a cubic metre absorbs before it breaks.
J m⁻³, the same unit as Pa; εf is the strain at fractureAllowable (working) stress
σallow = σy/n for ductile materials, σᵤ/n for brittle ones
Keeps service loads on the straight, recoverable part of the curve, with a margin for overloads, flaws and scatter in strength.
n is the dimensionless factor of safety, n > 1; σᵤ is the ultimate tensile strength in PaInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Drag the unloading strain past 0.4 %: below it the line runs back to the origin, above it the permanent set grows from nothing to almost the whole strain. Then pull the fracture strain under 0.4 %: stiffness is unchanged, but toughness falls from 35 to a few tenths of a MJ m⁻³ and the allowable stress drops too.
Live results
- STRESS AT UNLOADING σ
- 307 MPa
- PERMANENT SET ε − σ/E
- 4.56 %
- TOUGHNESS, AREA TO FRACTURE
- 35.23 MJ m⁻³
- ALLOWABLE STRESS
- 140 MPa
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Strain when unloaded ε: 5.0 %; Fracture strain (ductility): 12.0 %; Factor of safety n: 2.0
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
Reading the falling end of the curve as the metal getting weaker. Past the peak, the engineering stress still divides the load by the original area A₀, but the neck is losing area faster than the load is falling, so the true stress in the neck keeps rising right up to fracture. Nothing softened; one section got thinner. The same misreading suggests a part fails at its fracture stress. It fails at the ultimate tensile strength: that is the largest load the bar can ever carry, and a load held at that value keeps the neck thinning until the bar breaks, even though the engineering stress recorded at fracture is lower.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A round aluminium alloy test piece has a diameter of 12.0 mm and gauge marks 50.0 mm apart. The 0.2 % offset line meets its curve at a load of 31.7 kN, and the largest load recorded is 35.1 kN. Fitted back together after fracture, the halves measure 56.0 mm between the marks. Find the yield strength, the ultimate tensile strength and the elongation at fracture, and classify the alloy.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Engineering stresses divide every load by the original area: A₀ = πd²/4 = π(0.0120 m)²/4 = 1.131 × 10⁻⁴ m².
- Yield strength: σy = 31.7 × 10³ N ÷ 1.131 × 10⁻⁴ m² = 2.80 × 10⁸ Pa = 280 MPa.
- Ultimate tensile strength, from the largest load: σᵤ = 35.1 × 10³ N ÷ 1.131 × 10⁻⁴ m² = 3.10 × 10⁸ Pa = 310 MPa. Necking began at that load.
- Elongation at fracture: (56.0 − 50.0) mm ÷ 50.0 mm = 0.120 = 12 %. The halves are measured with no load on them, so the elastic strain has already sprung back and this 12 % is all permanent.
- Twelve per cent is well above the 5 % or so that marks a brittle material, so the alloy is ductile: it stretches visibly and necks before it breaks.
Answer. σy = 280 MPa, σᵤ = 310 MPa, elongation at fracture 12 %: a ductile alloy.
MediumExample 2
A steel tie rod 3.20 m long must carry a steady tension of 45.0 kN. The steel has E = 200 GPa, σy = 355 MPa and σᵤ = 510 MPa. Using a factor of safety of 2.0 against yielding, choose the smallest whole-millimetre diameter, then find the rod's working stress and extension, and the loads at which it would begin to yield and would reach its peak.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- A ductile rod has failed its job once it yields, so the limit is σallow = σy/n = 355 MPa ÷ 2.0 = 177.5 MPa.
- Required area: A ≥ F/σallow = 45.0 × 10³ N ÷ 177.5 × 10⁶ Pa = 2.535 × 10⁻⁴ m², so d ≥ √(4A/π) = 17.97 mm. Take d = 18 mm, giving A = 2.545 × 10⁻⁴ m².
- Working stress: σ = 45.0 × 10³ N ÷ 2.545 × 10⁻⁴ m² = 1.77 × 10⁸ Pa = 177 MPa, just inside the allowable value and on the straight part of the curve.
- Extension: ΔL = σL/E = (1.768 × 10⁸ Pa × 3.20 m) ÷ (200 × 10⁹ Pa) = 2.83 × 10⁻³ m = 2.83 mm, all of it recoverable when the load comes off.
- Yield begins at Fy = σy A = 355 × 10⁶ Pa × 2.545 × 10⁻⁴ m² = 90.3 kN, and the largest load the rod can ever carry is Fₘₐₓ = σᵤA = 510 × 10⁶ Pa × 2.545 × 10⁻⁴ m² = 130 kN.
- The margins are 90.3 ÷ 45.0 = 2.01 against yield and about 2.9 against the peak. The second is not spare capacity: any load between 90 kN and 130 kN would leave the rod permanently longer.
Answer. d = 18 mm; σ = 177 MPa and ΔL = 2.83 mm at 45.0 kN; yield begins at 90.3 kN and the peak load is 130 kN.
HardExample 3
Beyond yield, the true stress in a low-carbon steel follows σₜ = K εₜⁿ with K = 540 MPa and n = 0.22, and its volume stays constant as it stretches. Show that necking begins at εₜ = n, then find the engineering strain there, the true stress, the ultimate tensile strength, and the largest load a bar of 10.0 mm original diameter can carry.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Constant volume gives AL = A₀L₀, and with εₜ = ln(L/L₀) the area is A = A₀e(−εₜ). The load is therefore F = σₜA = A₀K εₜⁿ e(−εₜ).
- The load peaks where dF/dεₜ = A₀K e(−εₜ)(n εₜⁿ⁻¹ − εₜⁿ) = 0, which needs n εₜⁿ⁻¹ = εₜⁿ, so εₜ = n = 0.22. This is the Considère condition dσₜ/dεₜ = σₜ for this material.
- Engineering strain at the peak: ε = e0.22 − 1 = 0.246, so the bar stretches uniformly by 24.6 % before the neck forms, and its area has fallen to e(−0.22) = 0.803 of A₀.
- True stress at the peak: σₜ = 540 MPa × 0.220.22 = 540 MPa × 0.7167 = 387 MPa. The engineering stress divides the same load by A₀ instead: σᵤ = σₜ/(1 + ε) = 387 MPa ÷ 1.246 = 311 MPa.
- Largest load: A₀ = π(5.00 × 10⁻³ m)² = 7.854 × 10⁻⁵ m², so Fₘₐₓ = σᵤA₀ = 311 × 10⁶ Pa × 7.854 × 10⁻⁵ m² = 2.44 × 10⁴ N.
Answer. Necking starts at εₜ = 0.22, an engineering strain of 24.6 %; true stress 387 MPa, σᵤ = 311 MPa, Fₘₐₓ = 24.4 kN.
Quick check
Select an answer and explain it before checking.
Review & continue
A stress–strain curve plots engineering stress F/A₀ against strain ΔL/L₀ as a specimen is stretched to fracture; its landmarks mark where the material stops springing back, where it stops getting stronger, and where it breaks.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.
Optional supporting lessons
These chapter resources can help with earlier ideas; they are not additional required Physics I completion tasks.