Engineering and true stress–strain measures

Two denominators describe different quantities

Engineering stress divides force by original area; true stress uses current area. Engineering strain uses extension divided by original length. True axial strain accumulates relative length changes and, for uniform extension, equals ln(L/L₀).

Imagine following the same two material marks through a tensile test. Their original separation is L₀ and their current separation is L. The original cross-section of a uniform gauge is A₀; its current section perpendicular to the loading axis is A. Use the force F recorded at the same instant. A later unloaded section is not the area that carried an earlier force. Deformation and relative strain and Applied force and stress over an area introduce the reference measurements.

Write engineering stress σ_eng = F/A₀ and engineering strain e = (L − L₀)/L₀. Define the axial stretch ratio λ = L/L₀ = 1 + e. Ratios λ and e are dimensionless; λ = 1 means unchanged length, whereas e = 0 means unchanged length. Engineering strain 0.10 is 10%, not 0.10%.

For a current section in quasistatic axial equilibrium, the section-mean axial stress is σ̄ = F/A. It is the axial force divided by the actual area of that section. In an approximately homogeneous uniaxial gauge this is the usual true axial stress σ_true. More generally it is an average of an axial stress component, not every local stress at every point. Here a bar over σ marks that section average.

Taking the ratio of the two stress definitions gives σ̄/σ_eng = A₀/A for nonzero force, or directly σ̄ = σ_eng(A₀/A). This identity needs concurrent F and A and a correctly identified section; it does not require volume conservation or an elastic/plastic law. Under tension, if A is smaller than A₀ the same force gives a larger current-area value. Neither definition is a different experiment or automatically a strength endpoint.

Accumulate small changes relative to the length that exists

Engineering strain uses one fixed reference length. Another useful axial measure adds very small length changes divided by the length present when each change occurs: dε_log = dL/L. The differential dL is a small signed length change, so extension contributes positively and shortening negatively. For a positive material length L, accumulating from L₀ to L gives ε_log = ∫dL/L = ln(L/L₀) = ln λ = ln(1 + e). The symbol ln is the natural logarithm, the inverse of the exponential function exp.

Why a logarithm? Divide a finite extension into many small successive steps, evaluating each relative change at its current length. As the steps become finer, their sum approaches this integral. A single finite quotient (L − L₀)/L is not that sum. The first figure compares the accumulated measure with the fixed-reference engineering measure; it plots definitions, not two material constitutive laws.

For a uniform axial gauge, the same λ applies to each small material interval, so ln λ is the local axial logarithmic strain as well as the gauge value. No volume assumption was used to obtain it. The inverse conversion is λ = exp(ε_log), hence e = exp(ε_log) − 1. A positive length ratio requires λ > 0 and e > −1. You can always calculate ln(L/L₀) for a positive measured mark separation, but in a nonuniform gauge that number alone does not identify the local logarithmic strains.

For sequential collinear changes of the same material interval, stretch ratios multiply. From 50.0 to 60.0 to 75.0 mm, λ₁ = 1.20 and λ₂ = 1.25, so the complete λ = 1.20 × 1.25 = 1.50. The two stage engineering strains, each resetting its denominator to the stage’s start, are 0.20 and 0.25. Their sum 0.45 is not the complete engineering strain 0.50: the second stage extended an already longer interval.

The logarithms add instead: ln(1.20) ≈ 0.182322 and ln(1.25) ≈ 0.223144. Adding before rounding gives ln(1.50) ≈ 0.405465. In general e_total = (1 + e₁)(1 + e₂) − 1, while ε_log, total = ε_log, 1 + ε_log, 2 under these axial conditions. Engineering increments referred throughout to the same original L₀ do add; the failed sum used two different denominators.

At small extension the measures are close: e = 0.0100 gives ln(1.0100) ≈ 0.00995033. At e = 0.500 they are 0.500 and ln(1.500) ≈ 0.405465. Neither comparison assigns an elastic or plastic part. If a length returns to L₀, its net signed logarithmic strain is zero, but that alone does not prove a reversible path or zero dissipated work. Plastic deformation and yield behaviour develops the needed unloading evidence.

The familiar conversion has limits

Before necking, approximate volume conservation gives true stress = engineering stress × (1 + engineering strain). Necking makes the area and strain nonuniform, so this simple global conversion no longer supplies the local state. State whether volume conservation and uniformity are justified.

Derive the familiar shortcut for a uniform prismatic gauge. Its volume is V = AL; initially V₀ = A₀L₀. Prismatic means that the area is constant along the gauge at the state being described. If that same material approximately preserves volume, AL ≈ A₀L₀, so A ≈ A₀L₀/L = A₀/λ. Extension then decreases area; inverting this ratio would incorrectly make an extending constant-volume gauge wider.

Insert A₀/A ≈ λ into the stress identity: σ_true ≈ σ_eng λ = σ_eng(1 + e). The assumptions entered when area was inferred from length, not when stress was defined. Logarithmic strain remains ε_log = ln(1 + e) for this uniform gauge, independently of whether its volume is constant.

For the retained original case e = 0.10 and σ_eng = 200 MPa, λ = 1.10 and the constant-volume estimate is σ_true = 220 MPa. Both measures of strain must also be reported: engineering strain 0.10 and logarithmic strain ln(1.10) ≈ 0.0953102, about 9.531% when written as a percentage. Writing 0.10% inside ln would describe a different length change.

Make the geometry explicit with original fictional L₀ = 50.0 mm and A₀ = 10.0 mm², so V₀ = 500 mm³. At L = 55.0 mm, the volume model gives A = 500/55.0 ≈ 9.09091 mm². A simultaneous 2000 N gives 200 MPa engineering and 220 MPa current-area stress. The table extends this invented uniform account; no named material, measured curve, stability criterion or failure endpoint is supplied.

Near-volume-preserving plastic flow can be a useful approximation for a suitably dense metal over an appropriate regime. It is not a definition of plasticity and need not describe elastic volume change, pore growth, cavitation, damage or other material/state changes. Evidence must support its use. Do not extend the approximation to an unmeasured area merely because a curve is labelled “plastic”.

Measure the area when the volume assumption is in question

A separate fictional uniform specimen starts from the same 50.0 mm and 10.0 mm², but simultaneous observations at 2400 N now give L = 60.0 mm and A = 8.50 mm². The current gauge volume is 8.50 × 60.0 = 510 mm³, so J = V/V₀ = 510/500 = 1.020. The letter J here denotes a dimensionless volume ratio; it is not a fracture-energy quantity.

Directly from the observations, σ_eng = 2400/10.0 = 240 MPa and σ̄ = 2400/8.50 ≈ 282.353 MPa. Multiplying engineering stress by the length ratio alone would give 240 × 1.20 = 288 MPa, which is not this measured-area result. It silently replaced the observed volume change by J = 1.

The general uniform geometry account is J = (A/A₀)λ, hence A₀/A = λ/J. Therefore σ̄ = σ_eng λ/J. For this record, 240 × 1.20/1.020 ≈ 282.353 MPa, agreeing with force divided by observed area. The logarithmic axial strain is still ln(1.20) ≈ 0.182322. A change in volume affects the inferred area conversion, not the length identity.

These supplied numbers establish a geometric difference, not its cause. Before diagnosing dilation, distinguish actual specimen evolution from area calibration, imaging direction, gauge selection, thermal change and measurement uncertainty. Here uniformity and the measurements are stipulated; an actual experiment must establish them. The same specimen, state and instant belong with every quantity in the account.

The same gauge extension can conceal different local states

A neck localises deformation. The whole gauge can lengthen while neighbouring material intervals have different stretches and current areas. Taking a logarithm of the whole mark separation does not recover those local intervals, and using its λ to infer a neck area is no longer a uniform-geometry calculation. Tensile curves, strength and ductility develops the associated maximum-load and ductility observations.

Construct an original counterexample using two adjoining original 25.0 mm material intervals, each initially 10.0 mm² in area. In U, both stretch by 1.20: their current lengths are 30.0 and 30.0 mm. In N, their stretches are 1.10 and 1.30: current lengths are 27.5 and 32.5 mm. Both whole gauges reach 60.0 mm from 50.0 mm, hence have e = 0.200 and ln(L/L₀) ≈ 0.182322.

Independently stipulate volume conservation in each interval for this counterexample. Each originally contained 250 mm³. U therefore has area 250/30.0 ≈ 8.33333 mm² in either zone. N has areas 250/27.5 ≈ 9.09091 mm² and 250/32.5 ≈ 7.69231 mm². All four intervals preserve their supplied material volume, so even knowing the total volume does not select U rather than N.

With the same transmitted axial force 2400 N, every complete specimen has engineering stress 240 MPa. But the current-section means are 288 MPa in either U zone, and 264 and 312 MPa in N’s zones. The global shortcut 240 × 1.20 = 288 MPa returns neither N zone’s mean. The second figure uses interval-length inventory bars, not a solved neck shape or stress field.

The local logarithmic strains are ln(1.20) ≈ 0.182322 throughout U, but ln(1.10) ≈ 0.0953102 and ln(1.30) ≈ 0.262364 in N. Their simple mean is about 0.178837, not the whole-gauge logarithm 0.182322. The whole stretch is the original-length-weighted mean of local stretches; applying a nonlinear logarithm after that mean is different from averaging the logarithms.

This piecewise account proves that global force and extension, even with preserved total volume, do not uniquely identify local geometry or deformation. It prescribes approximately uniform zones and ignores transition fields; it is not an equilibrium/compatibility solution for a sharp physical shoulder, a damage mechanism or a material flow law. Local mark tracking and concurrent area/shape observations supply different missing information.

In an actual neck, stress can also vary across a section and include lateral normal and shear components. F/A_min can still give a section-mean axial component at a measured minimum section under the stated equilibrium assumptions. That number alone is not the equivalent uniaxial flow stress or the maximum local stress. Inferring those quantities requires a suitable multiaxial model and evidence; changing the label to “true” does not remove the constraint.

Keep definitions and evidence attached to the numbers

Report the original area and gauge, the material marks or local interval tracked, current geometry, concurrent force, loading/observation state, temperature, direction and rate as relevant. Distinguish measured values from values inferred under a volume or uniformity model. Specimens, dimensions and test conditions prepares that evidence record.

Engineering and logarithmic strain can both describe the same positive length ratio. Neither automatically means plastic strain, and neither is a strength or lifetime. Elastic response and elastic modulus and Plastic deformation and yield behaviour develop how recoverability and the loading history are determined. At finite deformation, choosing a strain measure is not a replacement for a constitutive or measurement model.

An unloaded after-fracture neck area and a force from an earlier loaded frame are not concurrent inputs. Likewise, machine travel is not automatically specimen-gauge extension. A large numerical “true” stress reconstructed from those mixed measurements is not a corrected material property. Keep the raw force, geometry and interval observations so another question can be checked without silently changing the model.

Worked examples

Try the reasoning

1. Can the same conversion be applied unchanged after necking?

   Show the answer and reasoning

   No.

   Global extension does not identify the local neck geometry. The U/N counterexample supplies the same total length, force and conserved volume with different zone areas and stretches.

2. Is true strain exactly equal to engineering strain for large extension?

   Show the answer and reasoning

   No.

   One is logarithmic and the other a fractional total change. At 50% engineering extension, ln(1.50) ≈ 0.405465, not 0.50. The two measures approach each other only for small changes.

3. A new uniform gauge starts at 40.0 mm and 8.00 mm². At 1600 N it is 50.0 mm long; constant volume is explicitly stipulated. Find its current area, engineering/current-area stresses and engineering/logarithmic strains.

   Show the answer and reasoning

   Current area 6.40 mm²; engineering stress 200 MPa and true axial stress 250 MPa; e = 0.250 and ε_log = ln(1.25) ≈ 0.223144.

   Initial volume is 320 mm³, so A = 320/50.0 = 6.40 mm². Divide the same 1600 N by original and current areas. The stretch is 1.25, giving σ_true/σ_eng = 1.25. These model-based values need the supplied uniformity and volume assumptions.

4. A 50.0 mm interval grows uniformly to 60.0 mm. A learner calls (60.0 − 50.0)/60.0 = 0.166667 its true strain because the denominator is current length. Correct the claim.

   Show the answer and reasoning

   Engineering strain is 0.200 and logarithmic strain is ln(60.0/50.0) ≈ 0.182322. The single finite final-length quotient 0.166667 is neither.

   The logarithmic measure adds infinitesimal signed changes divided by the current length throughout the extension. Dividing the entire finite change by only the final length does not perform that accumulation.

5. In an ideal uniform axial shortening account, L/L₀ = 0.800 and volume is preserved. If engineering stress is −100 MPa with compression negative, find A/A₀, current-area mean stress and both strains. No buckling/strength conclusion is requested.

   Show the answer and reasoning

   A/A₀ = 1.25, current-area mean −80.0 MPa, e = −0.200 and ε_log = ln(0.800) ≈ −0.223144.

   Area increases by the reciprocal stretch, so the same negative force is divided by a larger current area. Use σ̄ = σ_eng λ = −80.0 MPa under this specific account; its compression magnitude is 80.0 MPa rather than 100 MPa because the current area is larger. Keep the sign distinct from the magnitude. A specimen’s stability/contact response would need separate evidence.

6. For a separate uniform interval, F = 1800 N, A₀ = 9.00 mm², L/L₀ = 1.20 and simultaneous measured A = 7.80 mm². Does the constant-volume conversion reproduce F/A? Give a consistent account.

   Show the answer and reasoning

   σ_eng = 200 MPa; F/A ≈ 230.769 MPa. J = (7.80/9.00) × 1.20 = 1.040, so 200 × 1.20/1.040 ≈ 230.769 MPa. The J = 1 shortcut would give 240 MPa.

   The supplied current area is not 9.00/1.20 = 7.50 mm². Distinguish the observed uniform geometric account from an assumed volume. The logarithmic axial strain remains ln(1.20) ≈ 0.182322; these geometric numbers alone do not identify why J differs from one.

7. Only N’s second original 25.0 mm interval is tracked, reaching 32.5 mm while its model area is 7.69231 mm² and the force 2400 N. Calculate its engineering/logarithmic strains and section mean. Why should the whole-gauge 20% extension not replace it?

   Show the answer and reasoning

   For that material interval, e = 0.300, ε_log = ln(1.30) ≈ 0.262364 and section mean 312 MPa. The whole-gauge e = 0.200 averages a different interval.

   Its own stretch is 32.5/25.0 = 1.30; its exact model area is 250/32.5 mm². Dividing 2400 N by it gives 312 MPa. This local geometric account does not prove a uniaxial material flow stress inside a real neck.

8. A material interval goes from 50.0 to 60.0 mm and later returns to 50.0 mm. What is the net signed logarithmic strain, and does that prove zero plastic history or zero dissipated work?

   Show the answer and reasoning

   Net signed logarithmic strain is ln(60/50) + ln(50/60) = 0. Neither history nor dissipated work follows from the final length alone.

   The stretches 1.20 and 5/6 multiply to one. A work calculation needs the force–displacement path, and a mechanism/recovery interpretation needs loading-state and time evidence. Net strain is not accumulated absolute strain or accumulated plastic strain.

9. An earlier loaded frame carried 2400 N, but only an unloaded after-fracture minimum area 5.00 mm² is measured later. Is 2400/5.00 = 480 MPa an established loaded true flow stress?

   Show the answer and reasoning

   No. 480 MPa is arithmetic on nonconcurrent records and is not established as that earlier section mean or a uniaxial flow stress.

   The area that carried the earlier force was not supplied. Even a simultaneous neck-area mean requires its own multiaxial interpretation. Preserve the time/state of each measurement instead of repairing a curve with an unrelated final area.

10. A new uniform constant-volume state is reported as ε_log = ln(1.40) and σ_true = 280 MPa. Its original L₀ = 30.0 mm and A₀ = 12.0 mm². Reconstruct engineering strain/stress, current length/area and force.

    Show the answer and reasoning

    λ = 1.40, e = 0.400, σ_eng = 200 MPa, L = 42.0 mm, A ≈ 8.57143 mm² and F = 2400 N.

    Exponentiating the supplied logarithm recovers 1.40. Constant volume gives A = 12.0/1.40, while engineering stress is 280/1.40 = 200 MPa. Force is 200 × 12.0 = 2400 N, which also equals 280 × (12.0/1.40). The matching force checks that the two stress measures describe the same supplied state.

What to carry forward

• Keep the chosen stress and strain definition.
• Uniformity and volume assumptions bound the conversion.
• Current-area stress needs the current section and concurrent force; volume conservation is only one way of estimating an otherwise unknown area.
• Logarithmic axial strain comes from relative length accumulation and requires no volume assumption.
• Sequential stretch ratios multiply; collinear logarithmic increments add for the same material interval.
• Whole-gauge extension cannot identify local stretches or a neck area, even when total volume is conserved.
• A section-mean axial stress in a neck is not automatically an equivalent uniaxial flow stress.