Stress and strain — σ = F - A, ε = ΔL - L, Young's modulus E: Worked examples

Cell	What makes it distinct	Example
A. Tension (pull)	$F > 0$ , bar lengthens, $Δ L > 0$	Ex 1
B. Compression (push)	$F < 0$ , bar shortens, $Δ L < 0$ — sign flips	Ex 2
C. Solve for unknown geometry	given $Δ L$ , find $A$	Ex 3
D. Zero-input / degenerate	$A \to 0$ , $Δ L = 0$ , or $F = 0$ — what the formula "says"	Ex 4
E. Limiting behaviour	load creeps toward yield; where does Hooke's law die?	Ex 5
F. Two materials / ratios	same load, compare — cancel the shared quantity	Ex 6
G. Real-world word problem	messy units, MPa↔Pa, mm↔m, mass→force	Ex 7
H. Exam-style twist	change area & length together; predict the net effect	Ex 8

Cell A — Tension

Worked example Ex 1 — A strut being pulled (Cell A)

A steel tie ( $E = 200$ GPa) is $L = 1.5$ m long, cross-section $A = 2 \times 1 0^{- 4}$ m², pulled by $F = + 8 \times 1 0^{4}$ N. Find $σ$ , $ε$ , and $Δ L$ .

Forecast: steel is stiff ( $E$ huge), so guess the stretch is sub-millimetre. Sign of $Δ L$ ? Positive — we are pulling.

Stress: $σ = F / A = \frac{8 \times 1 0 ^{4}}{2 \times 1 0 ^{- 4}} = 4 \times 1 0^{8}$ Pa $= 400$ MPa. Why this step? Force per area is the material-level "crowding" — the first normalized quantity.
Strain: $ε = σ / E = \frac{4 \times 1 0 ^{8}}{2 \times 1 0 ^{11}} = 2 \times 1 0^{- 3}$ . Why this step? Rearranged Hooke's law $σ = E ε$ — strain is stress ÷ stiffness.
Stretch: $Δ L = ε L = 2 \times 1 0^{- 3} \times 1.5 = 3 \times 1 0^{- 3}$ m $= 3$ mm. Why this step? We undo the "divide by $L$ " that defined strain.

Verify: direct route $Δ L = F L / (A E) = \frac{8 \times 1 0 ^{4} \cdot 1.5}{2 \times 1 0 ^{- 4} \cdot 2 \times 1 0 ^{11}} = 3 \times 1 0^{- 3}$ m ✓. Positive as forecast. (3 mm — a touch bigger than my "sub-mm" guess, because the load is heavy.)

Figure — Stress and strain — σ = F - A, ε = ΔL - L, Young's modulus E

Cell B — Compression (the sign flip)

Worked example Ex 2 — A landing-leg under compression (Cell B)

The same steel member as Ex 1 ( $E = 200$ GPa, $L = 1.5$ m, $A = 2 \times 1 0^{- 4}$ m²) is now pushed with $F = - 8 \times 1 0^{4}$ N (compression). Find $Δ L$ .

Forecast: identical magnitude of load → identical magnitude of length change, but with a minus sign. Guess $- 3$ mm.

Signed stress: $σ = F / A = \frac{- 8 \times 1 0 ^{4}}{2 \times 1 0 ^{- 4}} = - 4 \times 1 0^{8}$ Pa. Negative = compressive. Why this step? Keeping the sign lets one formula cover both push and pull.
Signed strain: $ε = σ / E = - 2 \times 1 0^{- 3}$ . Why this step? Hooke's law is linear and odd — flip the input sign, the output flips.
Length change: $Δ L = ε L = - 3 \times 1 0^{- 3}$ m $= - 3$ mm (it shortens). Why this step? Same undo-step as before; the sign carries the physics.

Verify: magnitude equals Ex 1 (3 mm), sign is negative ✓.

[!mistake] Reporting compression as a stretch A compressed bar has $Δ L < 0$ . If you drop the sign you'll accidentally add length to a member that is actually shrinking — dangerous when checking clearances. Keep the sign until the very last sentence.

Cell C — Solve for the geometry

Worked example Ex 3 — Sizing a strut for a stretch budget (Cell C)

An aluminium strut ( $E = 70$ GPa) is $L = 2$ m long and must not stretch more than $1$ mm under $F = 1.4 \times 1 0^{4}$ N. What is the minimum cross-section $A$ ?

Forecast: less allowed stretch ⇒ we need more area (stiffer path). Expect a small-but-finite $A$ of order a cm².

Start from $Δ L = \frac{F L}{A E}$ and solve for $A$ : $A = \frac{F L}{E Δ L}$ . Why this step? The unknown is now geometry, so isolate $A$ .
Plug in SI units ( $Δ L = 1 \times 1 0^{- 3}$ m): $A = \frac{1.4 \times 1 0 ^{4} \cdot 2}{70 \times 1 0 ^{9} \cdot 1 \times 1 0 ^{- 3}} = \frac{2.8 \times 1 0 ^{4}}{7 \times 1 0 ^{7}} = 4 \times 1 0^{- 4} m^{2} .$ Why this step? Any mismatch (mm vs m) would corrupt the answer — convert first.
That's $4 \times 1 0^{- 4}$ m² $= 4$ cm² — a modest strut.

Verify: put $A$ back: $Δ L = \frac{1.4 \times 1 0 ^{4} \cdot 2}{4 \times 1 0 ^{- 4} \cdot 7 \times 1 0 ^{10}} = 1 \times 1 0^{- 3}$ m $= 1$ mm ✓. Bigger area than a pure-strength design would need — that's the stiffness constraint biting, exactly as Spacecraft Load Paths and Struts warns.

Cell D — Zero and degenerate inputs

Worked example Ex 4 — What the formulas say at the edges (Cell D)

Read each formula literally at its boundary.

Forecast: guess which one "blows up" and which one just goes quietly to zero.

$F = 0$ : $σ = 0/ A = 0$ , so $ε = 0$ , so $Δ L = 0$ . No load, no stretch — a healthy sanity check. Why this step? A correct model must give "nothing happens" when nothing pulls.
$Δ L = 0$ with $F \neq = 0$ : possible only if $E \to \infty$ (perfectly rigid) or $A \to \infty$ . Real materials always stretch a little; a truly zero stretch is an idealization. Why this step? Distinguishes an ideal rigid body from real elastic behaviour.
$A \to 0$ (a hair-thin wire): $σ = F / A \to \infty$ . Stress diverges — the force is crowded into no material, so it always breaks. This is why thin members fail first. Why this step? Shows the formula predicting the real-world failure of thin sections — stress concentration.
$L \to 0$ (a very short block): $ε = Δ L / L$ — for a fixed $σ$ , strain is fixed, so $Δ L = ε L \to 0$ . Short things barely move even at high strain. Why this step? Separates strain (intensive) from displacement (extensive).

Verify (numeric): take $F = 8 \times 1 0^{4}$ N through $A = 1 0^{- 8}$ m²: $σ = 8 \times 1 0^{12}$ Pa $= 8$ TPa — thousands of times any metal's strength, i.e. instant fracture ✓. And $F = 0$ gives $Δ L = 0$ exactly ✓.

Cell E — Limiting behaviour toward yield

Worked example Ex 5 — How close are we to yield? (Cell E)

Aluminium 6061-T6 has yield stress $σ_{y} = 270$ MPa and $E = 70$ GPa. A strut carries $σ = 240$ MPa. Is Hooke's law still valid, and what is the Factor of Safety against yield?

Forecast: $240$ is close to $270$ — expect a factor barely above 1, i.e. uncomfortably marginal.

Check elasticity: is $σ < σ_{y}$ ? $240 < 270$ ✓, so we are still in the linear region and $σ = E ε$ holds. Why this step? Using Hooke's law past yield gives a wrong (too-small) strain.
Elastic strain: $ε = σ / E = \frac{2.4 \times 1 0 ^{8}}{7 \times 1 0 ^{10}} = 3.43 \times 1 0^{- 3}$ . Why this step? Valid because step 1 confirmed we're below yield.
Factor of safety: $FoS = σ_{y} / σ = 270/240 = 1.125$ . Why this step? FoS is the ratio of the failure ceiling to the actual load.

Verify: $ε = 3.43 \times 1 0^{- 3}$ back-check: $E ε = 7 \times 1 0^{10} \times 3.43 \times 1 0^{- 3} = 2.4 \times 1 0^{8}$ Pa $= 240$ MPa ✓. FoS $= 1.125$ — thin margin, forecast confirmed; an engineer would beef this up.

Cell F — Compare two materials (ratios)

Cell G — Real-world word problem (messy units)

Worked example Ex 7 — A hanging equipment box (Cell G)

A $340$ kg instrument hangs from a single titanium rod ( $E = 110$ GPa) of length $L = 800$ mm and diameter $d = 6$ mm. Take $g = 9.81$ m/s². Find the stress in MPa and the stretch in mm.

Forecast: small rod, few-kN load → tens of MPa, sub-millimetre stretch.

Force from mass: $F = m g = 340 \times 9.81 = 3335.4$ N. Why this step? Weight is the load; convert mass to newtons before any stress.
Area from diameter: $A = π (d /2)^{2} = π (3 \times 1 0^{- 3})^{2} = 2.827 \times 1 0^{- 5}$ m². Why this step? Stress needs area, not diameter — and $d$ must be in metres.
Stress: $σ = F / A = \frac{3335.4}{2.827 \times 1 0 ^{- 5}} = 1.180 \times 1 0^{8}$ Pa $= 118.0$ MPa. Why this step? Report in MPa because that's the engineering habit — but compute in Pa.
Stretch: $Δ L = \frac{σ L}{E} = \frac{1.180 \times 1 0 ^{8} \times 0.8}{1.1 \times 1 0 ^{11}} = 8.58 \times 1 0^{- 4}$ m $= 0.858$ mm. Why this step? $L$ in metres (0.8, not 800); mixing mm here is the classic trap.

Verify: direct route $Δ L = F L / (A E) = \frac{3335.4 \times 0.8}{2.827 \times 1 0 ^{- 5} \times 1.1 \times 1 0 ^{11}} = 8.58 \times 1 0^{- 4}$ m ✓. 118 MPa and 0.86 mm — both in the forecast band.

Cell H — Exam-style twist

Worked example Ex 8 — Change area AND length together (Cell H)

A strut is redesigned so its length doubles ( $L \to 2 L$ ) and its cross-section triples ( $A \to 3 A$ ), with the same force and material. What happens to (a) stress and (b) stretch $Δ L$ ?

Forecast: stress depends only on $A$ , so tripling $A$ should cut stress to a third. Stretch depends on both $L$ (up) and $A$ (down) — guess it changes by $2/3$ .

Stress: $σ_{new} = \frac{F}{3 A} = \frac{1}{3} σ_{old}$ — length does not appear, so it's irrelevant to stress. Why this step? Stress is force ÷ area; changing $L$ can't touch it. A common exam trap is thinking longer bars are "more stressed."
Stretch: $Δ L_{new} = \frac{F ( 2 L )}{( 3 A ) E} = \frac{2}{3} \cdot \frac{F L}{A E} = \frac{2}{3} Δ L_{old}$ . Why this step? Both factors enter $Δ L = F L / (A E)$ : longer ⇒ more stretch ( $\times 2$ ), fatter ⇒ less ( $\div 3$ ), net $\times \frac{2}{3}$ .

Verify: with $F = 8 \times 1 0^{4}$ , $A = 2 \times 1 0^{- 4}$ , $L = 1.5$ , $E = 2 \times 1 0^{11}$ (Ex 1 numbers) old $Δ L = 3$ mm. New: $A = 6 \times 1 0^{- 4}$ , $L = 3$ → $Δ L = \frac{8 \times 1 0 ^{4} \cdot 3}{6 \times 1 0 ^{- 4} \cdot 2 \times 1 0 ^{11}} = 2 \times 1 0^{- 3}$ m $= 2$ mm $= \frac{2}{3} (3 mm)$ ✓. Stress old $= 400$ MPa, new $= 133.3$ MPa $= \frac{1}{3}$ ✓.

Recall Which cell was which?

Tension (positive stretch) ::: Ex 1 (Cell A) Compression (negative stretch) ::: Ex 2 (Cell B) Solving for the unknown area ::: Ex 3 (Cell C) What the formula says when $A \to 0$ or $F = 0$ ::: Ex 4 (Cell D) Checking we're still below yield ::: Ex 5 (Cell E) Same stress, compare two materials by ratio ::: Ex 6 (Cell F) Messy real units (mass→force, mm→m) ::: Ex 7 (Cell G) Change area and length at once ::: Ex 8 (Cell H)

Worked examples — Stress and strain — σ = F - A, ε = ΔL - L, Young's modulus E

The scenario matrix

Cell A — Tension

Cell B — Compression (the sign flip)

Cell C — Solve for the geometry

Cell D — Zero and degenerate inputs

Cell E — Limiting behaviour toward yield

Cell F — Compare two materials (ratios)

Cell G — Real-world word problem (messy units)

Cell H — Exam-style twist