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

We will build up to this one line and understand every letter in it:

$Δ L = \frac{F L}{A E}$

But first we earn each letter.

Step 1 — The bar, the pull, and what "force" even is

WHAT. Start with the simplest thing: a solid bar (imagine an aluminium strut). We grab both ends and pull outward. The amount of "pull" is a force, which we write $F$ , measured in newtons (N). One newton is roughly the weight of a small apple sitting in your hand.

WHY start here. Everything downstream is a response to this pull. If we can't picture the pull cleanly, nothing else makes sense. The pull has a direction (along the bar) and a size (in N).

PICTURE. Look at the bar below. The two pink arrows are the force $F$ — same size, opposite directions, so the bar isn't flying off anywhere; it just sits there being stretched.

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

$how hard we pull, in newtons F ⟷ a bar of length original length, in metres L$

$F$ — the pull. Bigger $F$ = harder tug.
$L$ — how long the bar is before we touch it. Remember this word "before"; it matters later.

See Spacecraft Load Paths and Struts for where these forces come from during launch.

Step 2 — The imaginary cut: force inside the bar

WHAT. Take a knife and imagine slicing the bar straight across, perpendicular to the pull. Don't really cut it — just picture the cut face. Ask: what holds the two halves together?

WHY. We can't see stress from the outside. The trick is to look inside. If the bar is sitting still (equilibrium), the left half must be pulling the right half back with exactly $F$ , and vice versa. This is Newton's third law: the material's internal bonds carry that same force $F$ across the cut.

PICTURE. The dashed blue line is the imaginary cut. The blue arrows show the internal force — same size $F$ as the outside pull, now revealed crossing the cut face.

Step 3 — Crowding: dividing force by area gives stress

WHAT. That internal force $F$ is spread over the whole cut face. Call the area of that face $A$ (in square metres, m²). Now define stress:

$σ = \frac{F}{A}$

WHY divide by area — and why this operation, not another? We want a number that describes the material, not the particular chunky-or-skinny bar. Ask a sharper question: how crowded is the force among the bonds on the cut face? A wide face shares $F$ among many bonds (each carries a little); a narrow face crams $F$ into few bonds (each carries a lot). "Amount shared among an area" is exactly what division by area computes — that's why $F / A$ and not, say, $F \times A$ or $F - A$ .

PICTURE. Two faces, same $F$ . Left: fat face, arrows spread thin. Right: thin face, arrows packed tight. The pink density of arrows is the stress.

Step 4 — Stretching: dividing stretch by length gives strain

WHAT. Under the pull the bar grows from $L$ to $L + Δ L$ . The little extra bit $Δ L$ (the Greek $Δ$ just means "change in") is the stretch. Define strain:

$ε = \frac{Δ L}{L}$

WHY divide by original length? Same logic as stress: we want a material number, not a part number. Is 1 mm of stretch a lot? On a 1 mm chip it's enormous; on a 2 m strut it's nothing. "Stretch compared to how big it started" is exactly division by the original length $L$ . That's why $Δ L / L$ , and why we use the original $L$ (the "before" length from Step 1), not the new one.

PICTURE. Two bars stretched by the same $Δ L$ . The short one is badly deformed (big strain); the long one barely notices (tiny strain). The yellow bracket shows $Δ L$ against each bar's own length.

Why is the bar also getting thinner as it stretches? That's a sideways cousin of this same idea — a story for another page.

Step 5 — The experiment: stress and strain rise together

WHAT. Now measure. Hang more and more weight, record $σ$ (from Step 3) and $ε$ (from Step 4), and plot one against the other. For small pulls you get a straight line through the origin.

WHY plot them against each other? Because we suspect they're related, and a graph is the cleanest way to see the relationship. A straight line means: double the stress → double the strain. In symbols, $σ$ is proportional to $ε$ , written $σ \propto ε$ .

PICTURE. The straight sloped chalk line: strain on the horizontal axis, stress on the vertical. Every point is one experiment. The line's steepness is the whole story of Step 6.

This straight part is the domain of Hooke's Law. It does not last forever — see Step 8.

Step 6 — Naming the slope: Young's modulus $E$

WHAT. A straight line through the origin has an equation $y = (slope) \times x$ . Here $y = σ$ and $x = ε$ . Give the slope a name, $E$ :

$σ = E ε$

WHY invent $E$ ? The slope is a single number that captures the whole line — the material's stiffness. Steep line = stiff material (a lot of stress buys only a little strain). Shallow line = floppy material. That one number lets an engineer compare aluminium, steel, and composite at a glance.

PICTURE. Two lines on the same axes: steep blue (steel, big $E$ ) and shallow yellow (aluminium, small $E$ ). The rise-over-run triangle on each is $E = σ / ε$ .

Step 7 — Assembling the master formula $Δ L = F L / (A E)$

WHAT. Now we substitute Steps 3 and 4 into Step 6 and solve for the stretch $Δ L$ — the thing an engineer actually wants to predict.

WHY. We know $F$ , the geometry ( $A$ , $L$ ), and the material ( $E$ ). We want the deflection. So we chain the definitions together and rearrange.

PICTURE. A flow of the substitution: $σ = E ε$ with $σ$ and $ε$ expanded, then $Δ L$ pulled out alone.

Start from Step 6 and expand both sides:

$σ \frac{F}{A} = E ε \frac{Δ L}{L}$

Multiply both sides by $L$ (undoing the "divide by $L$ " from Step 4) and divide by $E$ :

$Δ L = \frac{F harder pull \to more stretch L longer bar \to more stretch}{fatter \to less stretch A stiffer \to less stretch E}$

Read the formula like a sentence: stretch grows with pull $F$ and length $L$ (top), and shrinks with area $A$ and stiffness $E$ (bottom). Every arrow above matches your physical intuition — that's the sign of a formula you actually understand.

Step 8 — Edge cases: each letter to zero AND to infinity

WHAT. A formula is only trustworthy if it behaves sensibly at both extremes of every letter. For each of $F, A, E, L$ we now send it to $0$ and to $\infty$ , and check the picture.

WHY. Real struts see zero load, huge loads, near-rigid materials, hair-thin wires, and long booms. If the formula gave nonsense at any limit, we couldn't trust it in between. Reading $Δ L = F L / (A E)$ : a letter on top ( $F, L$ ) drives $Δ L$ up when it grows; a letter on the bottom ( $A, E$ ) drives $Δ L$ down when it grows.

PICTURE. Panels showing the low-stretch limits (top row) and the runaway-stretch limits (bottom row), with the bar drawn in each extreme.

The stretch goes to zero ( $Δ L \to 0$ ) when:

$F \to 0$ (no pull): no force, no stretch. ✓
$A \to \infty$ (infinitely fat bar): a mountain doesn't stretch when you tug it. ✓
$E \to \infty$ (perfectly rigid, "unobtainium"): infinite stiffness means no deflection — the idealisation engineers call a rigid body. ✓
$L \to 0$ (a wafer-thin slice): almost nothing to stretch, so the total stretch vanishes.

The stretch blows up ( $Δ L \to \infty$ ) when:

$F \to \infty$ (unlimited pull): top letter runs away — infinite force gives (formally) infinite stretch. In reality the material yields and then snaps long before this; the math warns you the deflection is unbounded.
$L \to \infty$ (endless boom): top letter runs away — the same fractional strain applied over an ever-longer bar accumulates an ever-larger total stretch.
$A \to 0$ (hair-thin wire, or a crack tip): bottom letter vanishes, so $Δ L \to \infty$ . Physically the force is crammed into almost no material — the stress $σ = F / A$ explodes, the wire necks and fails. This is the mathematics of a failure mode.
$E \to 0$ (a floppy gel with no stiffness): bottom letter vanishes, so $Δ L \to \infty$ . A material that offers no resistance stretches without limit under any pull.

The one-picture summary

WHAT this figure does. Everything above, compressed onto a single board so you can rebuild the whole derivation from one glance. Use it as your recall trigger: if you can narrate this picture, you own the page.

HOW to read it, left to right. Start at the pink arrows — that is the pull $F$ of Step 1. Follow the dashed blue cut into the bar — that is Step 2's internal force, which we turned into stress $σ = F / A$ in Step 3. The bar's stretch over its length is the strain $ε = Δ L / L$ of Step 4. On the right, the yellow line is the Step 5 experiment; its slope is $E$ (Step 6). And at the bottom sits the payoff, $Δ L = F L / (A E)$ (Step 7), with the pink "up: $F, L$ " and blue "down: $A, E$ " tags reminding you which way each letter pushes the stretch (Step 8).

Recall Feynman: the whole walkthrough in plain words

We grabbed a bar and pulled it with a force $F$ . To see what's happening inside, we imagined slicing it — the two halves pull each other back with that same $F$ . Now, is that force a big deal? Depends how much material shares it: spread over a fat face it's chill, crammed into a thin face it's fierce. So we divided force by area and called the crowding stress.

Then the bar stretched a little bit, $Δ L$ . Is that stretch a big deal? Depends how long the bar was to start with — a millimetre is huge for a chip, nothing for a girder. So we divided stretch by the original length and called that fraction strain.

We did the experiment: plot stress against strain and, for gentle pulls, you get a straight line. The steepness of that line is one number that tells you how stubborn the material is — that's Young's modulus $E$ . Steep line, stubborn stuff. And "gentle" matters: the line only stays straight while the strain is small, a few percent at most.

Finally we glued it together. Write "stress equals $E$ times strain," open up both sides into $F / A = E Δ L / L$ , and juggle it until the stretch sits alone: $Δ L = F L / (A E)$ . Read it like a story — pull harder or use a longer bar and it stretches more; make it fatter or stiffer and it stretches less. Push instead of pull and every sign flips: the bar shortens. And we checked the corners on both ends: no force, or infinite fatness, or infinite stiffness give zero stretch; while infinite force, an endless boom, a hair-thin wire, or a floppy zero-stiffness gel all send the stretch racing to infinity — exactly as common sense demands. That whole chain is one idea: strip away the part's size, and what's left is the material talking.

Recall

The master deflection formula and each letter's effect ::: $Δ L = F L / (A E)$ ; up with $F$ and $L$ , down with $A$ and $E$ . Why we divide force by area to get stress ::: To measure how crowded the internal force is among the bonds, which is a material property, not a part property. What the slope of the stress–strain line is called ::: Young's modulus $E$ — the material's stiffness. What happens to $Δ L$ under a compressive (negative) force ::: It becomes negative — the bar shortens; the formula flips sign automatically. Does strain depend on length $L$ ::: No — $ε = F / (A E)$ has no $L$ in it; only the total stretch $Δ L = ε L$ scales with length.