Strain and material response

Pull on a steel wire and it stretches — barely, invisibly, but it stretches. Strain is how engineers put a number on that invisible stretch, and once you have that number, a single material property connects it back to the stress causing it: Hooke's law. Every bolt, beam and bridge cable is designed inside this relationship, because it lets an engineer predict deformation before the structure is ever loaded for real. This article continues straight on from stress in the "Properties of Areas and Axial Loading" unit. You will define normal strain and shear strain, apply Hooke's law in both tension ( $\sigma = E\varepsilon$ ) and shear ( $\tau = G\gamma$ ), read a stress–strain diagram, and — new here — connect strain back to the axial deformation formula $\delta = PL/AE$ that ties force, geometry and material together in one line. Everything stays linear-elastic and isotropic; plastic behavior, strain hardening and necking are out of scope at this stage.

The core insight

Normal strain $\varepsilon = \delta/L$ is a dimensionless stretch ratio, and within the elastic range it is tied to stress by Hooke's law ( $\sigma = E\varepsilon$ , $\tau = G\gamma$ ), which combines with the axial-load formula $\sigma = P/A$ to give the deformation of a loaded bar, $\delta = PL/AE$ .

The core idea

Normal strain. For a line element of original length $L$ that changes length by $\delta$ under load,

\varepsilon = \frac{\delta}{L}

where $\delta$ is elongation (positive) or contraction (negative), and $L$ is the original, unloaded length. Because it is a length divided by a length, strain is dimensionless — it carries no unit, though it is often reported as a percentage or in microstrain ( $\mu\varepsilon = 10^{-6}$ ).

Hooke's law in tension/compression. Within the elastic range, stress is directly proportional to strain:

\sigma = E\varepsilon

where $E$ is Young's modulus (the modulus of elasticity), a material constant with units of stress — typically quoted in GPa for metals ( $E_{steel} \approx 200$ GPa, $E_{aluminium} \approx 70$ GPa). Geometrically, $E$ is the slope of the initial straight-line portion of the stress–strain diagram: a stiffer material has a steeper line and needs more stress to produce the same strain.

Hooke's law in shear. Shear strain $\gamma$ is the change in angle, in radians, between two originally perpendicular line segments in the material. It obeys the analogous relation

\tau = G\gamma

where $G$ is the shear modulus (modulus of rigidity), the shear-stiffness counterpart of $E$ .

Reading the stress–strain diagram. Load a tension specimen from zero and the curve first rises as a straight line through the origin — the linear-elastic region, slope $E$ , up to the proportional limit. Just beyond is the elastic limit, the highest stress from which the material still fully recovers on unloading (for most engineering materials this sits so close to the proportional limit that the two are often treated as coincident at this level). For ductile metals like mild steel a yield point follows, where strain grows with almost no rise in stress — the onset of permanent (plastic) deformation, whose detailed mechanics (strain hardening, necking) lie beyond this course. The curve then climbs to the ultimate stress $\sigma_{ult}$ , the maximum stress the specimen carries, before falling to the fracture point where it breaks. Every formula in this article — $\sigma = E\varepsilon$ , $\delta = PL/AE$ — assumes loading stays within the initial straight line, where unloading retraces the same path back to zero strain.

Figure

A stress strain diagram for a ductile metal showing the straight line elastic region up to the proportional and elastic limit then a yield plateau then a rising curve to the ultimate stress peak then a falling curve to the fracture point

Derivation AAxial deformation from stress, area and Hooke's law

The goal is to combine three separately learned facts — the definition of strain $\varepsilon = \delta/L$ , the definition of stress under axial load $\sigma = P/A$ , and Hooke's law $\sigma = E\varepsilon$ — into a single formula that gives the elongation $\delta$ of a bar directly from the applied force $P$ , its length $L$ , its cross-sectional area $A$ , and its material's Young's modulus $E$ , with no intermediate lookup of strain or stress needed.

Step 1. Start from the definition of normal strain for the bar as a whole, treating it as one line element of length $L$ : $\varepsilon = \delta/L$ . This is a purely geometric statement — it says nothing yet about force or material.

Step 2. Bring in Hooke's law for the linear-elastic range, $\sigma = E\varepsilon$ , and solve it for strain: $\varepsilon = \sigma/E$ . This is the bridge between the mechanical quantity (stress) and the geometric quantity (strain), valid only up to the proportional limit.

Step 3. Substitute the axial stress definition $\sigma = P/A$ (force divided by cross-sectional area) into the result of Step 2: $\varepsilon = \dfrac{P}{AE}$ .

Step 4. Equate the two expressions for $\varepsilon$ from Step 1 and Step 3, since both describe the same strain, and solve for $\delta$ :

\delta = \frac{PL}{AE}

where: $\delta$ is total elongation (m), $P$ is the axial force (N), $L$ is original length (m), $A$ is cross-sectional area (m²), and $E$ is Young's modulus (Pa). Read the formula physically: elongation grows with a larger force or a longer bar, and shrinks with a larger cross-section or a stiffer material — exactly the four levers an engineer has when sizing a tension member.

Worked example

A steel rod of circular cross-section, diameter $d = 10$ mm, has an original gauge length $L_0 = 50$ mm and Young's modulus $E = 200$ GPa. Under an axial tensile load it develops the same 300 MPa stress as before (from Worked example 1 of the base article, where strain came out to $\varepsilon = 1.5 \times 10^{-3}$ ). Find the axial force $P$ producing this stress, and confirm the elongation using $\delta = PL/AE$ .

Step 1. Cross-sectional area from the diameter: $A = \dfrac{\pi d^2}{4} = \dfrac{\pi (10\times10^{-3})^2}{4} = 7.854\times10^{-5}\ \text{m}^2$ .

Step 2. Force from the axial stress definition, $\sigma = P/A$ , rearranged: $P = \sigma A = (300\times10^{6})(7.854\times10^{-5}) = 2.356\times10^{4}\ \text{N} \approx 23.6\ \text{kN}$ .

Step 3. Elongation using the derived formula, with $L = 50\times10^{-3}$ m: $\delta = \dfrac{PL}{AE} = \dfrac{(2.356\times10^{4})(50\times10^{-3})}{(7.854\times10^{-5})(200\times10^{9})} = 7.5\times10^{-5}\ \text{m} = 0.075\ \text{mm}$ . This matches the 0.075 mm elongation given directly in the base article's example, confirming the two routes — strain-first or force-first — agree. A 10 mm diameter steel rod needs about 23.6 kN of axial force to stretch its 50 mm gauge length by 0.075 mm.

Common traps

Reporting strain with a unit (mm, %) without recognizing it is a dimensionless ratio $\delta/L$ — both quantities are lengths, so the units cancel.
Applying $\sigma = E\varepsilon$ or $\delta = PL/AE$ beyond the proportional limit — both formulas assume linear-elastic behavior only; they say nothing about the yield plateau or the region beyond it.
Mixing up the elastic limit (full recovery on unloading) with the yield point (onset of permanent set) — the two are numerically close for most metals but are conceptually different points on the curve.
Forgetting that shear strain $\gamma$ is measured in radians (an angle), unlike normal strain, which is a length-to-length ratio — both are dimensionless, but they measure different kinds of distortion.
Skipping the diameter-to-area conversion ( $A = \pi d^2/4$ ) and plugging diameter directly into $\sigma = P/A$ , which silently uses the wrong area.

Ask the tutor: if the force-first and strain-first routes to elongation feel disconnected, ask the AI tutor to walk through both starting from the same numbers side by side.

Why does the base article's example give stress directly from strain, but this one goes through force and area instead?

Both routes describe the same physical situation, just entered from different known quantities. If you are handed elongation and length, compute strain first, then get stress from Hooke's law — that is the base article's path. If you are handed force and cross-sectional area instead, compute stress directly from $\sigma = P/A$ , then get strain (and elongation) from Hooke's law rearranged. The formula $\delta = PL/AE$ derived here is simply the force-first path pre-combined into one equation, so you do not need to compute stress and strain as separate steps every time.

Does delta equals PL over AE work for compression as well as tension?

Yes, with a sign convention: treat compressive force P as negative (or track magnitude and note the bar shortens), and the same formula gives the shortening $\delta$ , provided the compressive stress also stays within the linear-elastic range and the member is short/stocky enough that it does not buckle. Buckling of slender members under compression is a separate topic (columns) not covered by this simple axial formula.

If two rods have the same P, L and E but different diameters, why does the fatter one stretch less?

Because area appears in the denominator of $\delta = PL/AE$ : a fatter rod has more cross-sectional area, so the same force is spread over more material, producing lower stress ( $\sigma = P/A$ ) and therefore lower strain ( $\varepsilon = \sigma/E$ ) and less total elongation. This is the same reasoning engineers use to size a tension member — increase the diameter to keep both stress and deflection within limits for a given load.

Strain and material response

The core idea

Worked example

Common traps

Before this

Where this leads