Stress, strain, and Hooke’s law

Start with two rods

Two supplied rods have the same uniform material and the same original cross-sectional area A₀ = 0.0001 m². One has original length L₀ = 2 m; the other has L₀ = 1 m. Pull each slowly along its length with the same tensile load. Which rod extends farther, and does that make its material different?

Extension means ΔL = L − L₀, where L is the length under load. In the linear model introduced below, a tensile force of 1200 N gives extensions of 0.2 mm and 0.1 mm. The longer rod extends twice as far. We will account for the difference by keeping the material property separate from the object’s dimensions.

Units, standards, and physical constants and Algebra, functions, graphs, and logarithms are published preparation for unit conversions, ratios and slopes. Newton’s laws and free-body diagrams is an optional return path for the force balance. The definitions needed for this tensile comparison are introduced here.

Divide by reference quantities

Let F ≥ 0 be the transmitted tensile-force magnitude. Engineering tensile stress is σ = F/A₀: force per original area. Its SI unit is the pascal, 1 Pa = 1 N/m². For the supplied area, F = 1200 N gives σ = 12000000 Pa = 12 MPa. One megapascal is 10⁶ Pa.

Engineering axial strain is ε = ΔL/L₀: extension divided by original length. Both lengths must use the same unit, so strain is dimensionless. An extension of 0.2 mm = 0.0002 m in a 2 m rod gives ε = 0.0001. The 1 m rod extended by 0.1 mm has the same strain.

Small strain numbers are easier to display after multiplying them. The upper graph and table show the number 10⁶ε, sometimes called the strain in microstrain. A displayed value of 100 means ε = 100 × 10⁻⁶ = 0.0001; it does not mean an extension of 100 micrometres. Strain remains dimensionless.

State the linear elastic law

For this main comparison, stipulate the uniaxial law σ = Eε with constant Young’s modulus E = 120 GPa = 120000000000 Pa. The load window is 0 ≤ F ≤ 2000 N. We also stipulate recovery to the original length when the load is removed. These inputs describe an invented teaching material, not measured material data.

The modulus has stress units because ε is dimensionless. At ε = 0.0001, the law gives σ = (120 GPa)(0.0001) = 12 MPa. On a stress-versus-strain graph, E is the slope with respect to ε. Multiplying the horizontal number by 10⁶ changes the displayed numerical slope, while the material modulus remains 120 GPa.

Linear response means proportional stress and strain in the stated window. Elastic recovery concerns what happens on unloading. Those statements are different: a recoverable response can be nonlinear. A separate supplied nonlinear example below makes the distinction explicit.

Compare material and rod stiffness

Substitute σ = F/A₀ and ε = ΔL/L₀ into σ = Eε. The result is F = (EA₀/L₀)ΔL, or ΔL = FL₀/(EA₀). Define the whole-rod stiffness k = EA₀/L₀, giving F = kΔL. Modulus E has units Pa; stiffness k has units N/m.

Here EA₀ = 12000000 N. The 2 m rod has k = 6000000 N/m; the 1 m rod has k = 12000000 N/m. Shortening the rod doubles its stiffness without changing E. At a fixed force, the shorter rod has half the extension but the same engineering stress and strain.

The upper panel has one material line. The lower panel has two whole-rod lines: solid blue with circles for 2 m, orange dashes with squares for 1 m. On this force-versus-extension plot, the steeper line means larger k. The exact table supplies the four marked positive-load states and the zero-load state.

These are relations between static load states, not histories in time or magnified drawings of the rods. The panels use different horizontal and vertical quantities. Their plotted angles cannot be compared to decide whether E equals k; the units already show that those are different quantities.

Distinguish original and current area

Engineering stress keeps the original area A₀ in its denominator. A true axial stress instead uses the current cross-sectional area A at the state being considered: σ_true = F/A. The choice of denominator matters even when the force is unchanged.

For a separate supplied geometric state, let F = 990 N, A₀ = 0.0001 m² and A = 0.000099 m². Engineering stress is 9.9 MPa, while true stress is 10 MPa. Their ratio is A₀/A = 100/99. Both areas are given independently; the state is not an extra point on the main model’s plots.

Axial extension alone does not supply the current area. A model for transverse deformation, or additional area data, would be needed. The main engineering law uses its reference area; it does not silently calculate a true stress or a large-deformation constitutive law.

Keep the tensile conditions

Use prismatic rods, meaning constant original cross-section along their length, and a slowly established equilibrium. Away from the grips, assume uniform axial loading and strain, negligible body-force and temperature effects, and the stated linear recoverable material response. Those conditions are what allow the simple ΔL formula.

Each rod has two applied end forces of equal magnitude F pointing in opposite directions. Their net vector sum is zero in equilibrium, while the rod is under tension. F in σ = F/A₀ is the transmitted magnitude; it is neither zero because the vectors balance nor 2F because two ends are loaded.

At 2000 N the largest main-model strain is 1/6000, about 0.000167. A small calculated strain does not establish the elastic limit, proportional limit or safe load of a real specimen. The example supplies its constitutive window and contains no measured yield or fracture threshold. Compression, buckling and nonuniform grip regions require other models.

Check a recoverable nonlinear response

Consider a separate memory-free material rule on 0 ≤ ε ≤ ε* with ε* = 1/6000: σ = E(ε + ε²/ε*), using E = 120 GPa as an initial-slope parameter. Supply the same rule for loading and unloading. It increases strictly with ε and has σ = 0 only at ε = 0, so unloading to zero stress returns to zero strain in this stipulated model.

At ε = ε*/2 = 1/12000, the stress is 15 MPa. At ε = ε* = 1/6000, it is 40 MPa. The ratios σ/ε are 180 GPa and 240 GPa, respectively. They are different, so one constant proportionality between stress and strain cannot describe the whole interval.

This example is recoverable by its supplied unloading rule and nonlinear by its changing ratio. The symbol E here is the slope at zero strain, not a constant ratio σ/ε at positive strain. It is a separate mathematical counterexample, not the main rod material, a plotted yield curve or a claim about a named real substance.

Use geometry in an inverse calculation

A known force–extension slope k alone does not identify a material modulus: geometry also matters. Within the stated uniform linear model, E = kL₀/A₀. Two different materials and geometries can give the same k. To infer E, supply both the original length and original area, together with a justified linear-response range.

For a real experiment, dimensions, force, extension, loading history and measurement uncertainty would need to be checked. A straight segment or a return to the starting length addresses a particular observed range; it does not establish every possible loading condition. The exact exercises here are calculations in supplied models, not specimen certification.

Worked examples

Try it yourself

1. For the main 2 m rod at F = 1200 N, find σ, ε, ΔL and the displayed number 10⁶ε. What does that displayed number represent?

   Show the answer and reasoning

   σ = 12 MPa, ε = 0.0001, ΔL = 0.0002 m = 0.2 mm and 10⁶ε = 100. The last value is a multiplied dimensionless strain number, not a length.

   Divide by the original area for stress, then by E for strain. Multiply strain by the original length for extension; only the display number is multiplied by 10⁶.

2. Keep the main material, area and F = 1200 N, but compare L₀ = 2 m and L₀ = 1 m. Which of E, k, σ, ε and ΔL changes?

   Show the answer and reasoning

   E stays 120 GPa; σ stays 12 MPa and ε stays 0.0001. Shortening from 2 m to 1 m changes k from 6000000 to 12000000 N/m and ΔL from 0.2 to 0.1 mm.

   The definitions of σ and the supplied material law contain no L₀. Whole-rod stiffness is inversely proportional to L₀, while extension at fixed force is proportional to L₀.

3. Keep E = 120 GPa, L₀ = 2 m and F = 1200 N, but double A₀ from 0.0001 to 0.0002 m². Find the new stress, strain, extension and stiffness.

   Show the answer and reasoning

   σ = 6 MPa, ε = 0.00005, ΔL = 0.1 mm and k = 12000000 N/m. E remains 120 GPa.

   At fixed force the doubled reference area halves engineering stress and hence strain. The extension also halves, while k = EA₀/L₀ doubles.

4. For the separate state F = 990 N, A₀ = 0.0001 m² and A = 0.000099 m², give engineering and true stress. Can the current area be inferred from an axial strain alone?

   Show the answer and reasoning

   The stresses are 9.9 MPa and 10 MPa, respectively. An axial strain alone does not specify the transverse deformation or current area. Both areas here are independent inputs.

   Use the denominator named by each definition. Inferring a current area would require additional data or a constitutive relation for transverse deformation.

5. Use the separate σ = E(ε + ε²/ε*) rule at ε*/2 and ε*. Give both stresses and both σ/ε ratios. Does the supplied recovery make the relation proportional?

   Show the answer and reasoning

   The stresses are 15 MPa and 40 MPa; σ/ε is 180 GPa and 240 GPa. The same-path zero-stress recovery is supplied, but the unequal ratios rule out a constant proportionality on this interval.

   For ε > 0, σ/ε = E(1 + ε/ε*). It changes with strain. Recovery follows from the supplied unloading rule, not from a constant stress-to-strain ratio.

6. A separate uniform rod is stipulated to have exact linear tensile stiffness k = 9000000 N/m, L₀ = 3 m and A₀ = 0.00015 m². Infer E within that model. Would k alone identify E?

   Show the answer and reasoning

   E = kL₀/A₀ = 180000000000 Pa = 180 GPa. Stiffness alone would not identify E because the ratio A₀/L₀ also matters.

   Rearrange k = EA₀/L₀ and retain the geometry. The result uses an exact supplied slope and a linear uniform-rod assumption; it is not a measured modulus or a safe-load claim.

What to carry forward

• Engineering stress and strain use original area and length. Strain is dimensionless, even when its displayed number is multiplied.
• Material modulus E and whole-rod stiffness k have different units. In the stated linear model, k = EA₀/L₀.
• Recovery after unloading does not by itself imply proportional stress and strain or certify an untested load range.