Materials Science · Elasticity, plasticity and tensile strength

Elastic response and elastic modulus

Find a modulus from an appropriate stress–strain interval, distinguish specimen extension from machine movement, and explain why recoverable deformation need not have one constant slope.

Useful background: Units, loads, stress and strain; Atoms, bonding, structure and defects; Measuring and checking material claims. Each worked case explains the quantities it uses.

Back to the chapter

After reading, you can

Calculate a linear elastic modulus from stress–strain increments or force and specimen-gauge extension, with consistent units.
Select a supported data interval and distinguish an additive zero offset from a changed slope.
Explain and calculate a bounded series-compliance correction without treating machine travel as specimen strain.
Distinguish linearity, recoverability and the stated slope of a nonlinear response.

A reversible small deformation

In a uniaxial linear-elastic region, stress is proportional to strain: σ = Eε. E is Young’s modulus, the slope of that region. Elastic means recoverable; linear means proportional. A material can be elastically nonlinear, so the two words are not interchangeable.

Imagine pulling a small coupon, measuring the separation of two marks, and releasing it. In an ideal elastic excursion the deformation is recovered when the force is removed. A loading graph alone shows how stress and strain changed; recovery requires unloading evidence or an independently supported elastic model. A straight segment does not by itself establish how the body will unload.

For many solids, a sufficiently small reversible excursion is approximately linear. The same extra stress then produces the same extra strain within that range. Elastic deformation stores recoverable mechanical energy; permanent rearrangement and time-dependent relaxation require additional descriptions. Do not extend a small-strain straight line to every load or duration.

Deformation and relative strain develops the gauge-length and recovery-time distinctions. Here the linear calculations assume a uniform uniaxial gauge region, small strain and matching direction, temperature and loading rate. A residual measured immediately after unloading in a time-dependent material is not automatically permanent.

Turn the measured force and extension into a slope

Use original area A₀ and original gauge length L₀. Engineering stress is σ = F/A₀, and engineering strain is ε = u/L₀, where u is the change in separation of the specimen gauge marks. Applied force and stress over an area explains why the original cross-section and region matter. The gauge interval is not automatically the full distance between grips.

Across a selected linear interval, E = Δσ/Δε. The symbol Δ means the difference between the upper and lower recorded values, not the upper value alone. Substituting the definitions gives E = (L₀/A₀)(ΔF/Δu). With force in N, lengths in mm and area in mm², the result is N/mm² = MPa. Divide MPa by 1000 to express GPa. Strain is a ratio: 0.001 is 0.1%, not 0.001%.

The specimen’s axial stiffness is instead k_s = ΔF/Δu = EA₀/L₀, in N/mm for these units. Doubling the gauge length of the same uniform material and area halves k_s; it does not halve E. Stiffness and strength answer different questions develops this distinction between a material response and a body’s force–displacement response.

Measure the appropriate slope

Use strain from the specimen region and account for machine compliance. Specify orientation, temperature and rate where relevant. A slope fitted across yield or significant viscoelastic evolution does not automatically give the same modulus.

Consider the invented record in the table: A₀ = 10.0 mm² and L₀ = 50.0 mm. The gauge-extension indication contains a supplied constant positive zero offset of 0.0100 mm. Thus the corrected extension is u = u_ind − 0.0100 mm. This offset is an instrument datum in the example, not measured permanent strain. It is distinct from force-dependent machine movement.

The first two force increments are both 200 N, and the corresponding indication increments are both 0.0100 mm. Each gives Δσ = 20.0 MPa and Δε = 0.000200, hence E = 100000 MPa = 100 GPa. Combining the 100–500 N interval gives the same slope. A constant offset cancels in a difference, so this slope can be found without falsely forcing the indicated values through the origin.

At the first point, simply dividing 10.0 MPa by u_ind/L₀ = 0.000300 would give about 33.3 GPa. That quotient uses the wrong origin. After the supplied zero correction, the first point has strain 0.000100 and the same 100 GPa ratio as the other points in the linear region. Finding an arbitrary intercept is not, by itself, a calibration of a real instrument.

The 500–700 N interval is different: Δu_ind = 0.0250 mm, giving an incremental slope of 40.0 GPa. It should not be pooled silently into the initial 100 GPa fit. The loading record alone does not say whether this change comes from nonlinear elasticity, plasticity, damage or a measurement problem. The example stipulates elastic reversibility only over the first three points; unloading and measurement checks are needed to interpret a later departure.

In a real record, examine multiple points and deviations from a fitted line, the selected strain interval, loading/unloading history and uncertainty. Two endpoints determine a chord, but do not prove that every point between them is linear. The first figure keeps the zeroed original model separate from this offset-containing record.

Invented specimen-gauge indications, not crosshead travel. Original area 10.0 mm², gauge length 50.0 mm; supplied additive extension offset +0.0100 mm. The first three points are stipulated to belong to a reversible linear response.
Force F (N)	Gauge indication u_ind (mm)	Engineering stress σ (MPa)
100	0.0150	10.0
300	0.0250	30.0
500	0.0350	50.0
700	0.0600	70.0

Two distinct original teaching panels. Panel A shows a correctly zeroed linear example: engineering strain zero to 0.001 horizontally, engineering stress zero to 100 megapascals vertically. A solid line goes through the origin and a filled endpoint at strain 0.001 and stress 100 megapascals, giving 100 gigapascals. Panel B is a separate invented specimen-gauge record with original area 10 square millimetres and gauge length 50 millimetres. The horizontal axis is the indicated gauge extension divided by 50 millimetres, not corrected strain: an additive positive extension offset of 0.0100 millimetre is supplied. Filled blue circles occur at indicated ratios 0.000300, 0.000500 and 0.000700 with stresses 10, 30 and 50 megapascals; their solid connecting segment describes the selected initial linear region. The later observation is an open orange diamond at ratio 0.001200 and stress 70 megapascals and is not joined by an assumed fitted curve. The first region gives 100 gigapascals from increments; the later interval gives 40 gigapascals. Shapes distinguish these observations without colour. The shifted origin is not evidence of permanent strain, and the later loading point alone does not establish plasticity. Both panels are invented data/models, not measurements of a named material.

Machine travel can include deformation outside the gauge

The force passes through the load train as well as the specimen. The machine frame, grips and specimen portions outside the selected gauge can move elastically. Seating or slip can add further movement. A crosshead displacement is therefore not automatically u, even if its display has length units. An extensometer or suitable optical measurement follows defined specimen points; its own calibration, attachment and alignment still matter.

For a bounded calculation, replace all elastic movement outside the gauge by one separately characterised, constant series compliance C_f. Compliance is displacement change per force change. The gauge and this additional element carry the same force increment, and their displacement increments add: Δu_m = Δu + C_fΔF, where Δu_m is the machine movement. Substituting Δu = L₀ΔF/(EA₀) gives Δu_m/ΔF = L₀/(EA₀) + C_f.

Use the same 100–500 N interval: ΔF = 400 N and true gauge increment Δu = 0.0200 mm. If C_f = 0.0000200 mm/N, the extra movement is 0.00800 mm, so Δu_m = 0.0280 mm. Treating all of that as gauge extension gives E_app = (50.0/10.0)(400/0.0280) ≈ 71429 MPa = 71.4 GPa. The material did not soften; the displacement included another compliant element.

Under this specified series model, subtract the independently known increment C_fΔF: Δu = 0.0280 − 0.00800 = 0.0200 mm, recovering 100 GPa. This is not a universal correction for any crosshead trace. C_f must describe the same configuration and range; variable seating, slip, bending, specimen deformation outside the gauge or nonlinear fixtures may invalidate a single constant. The second figure shows the displacement account rather than assigning an actual machine a stiffness.

Small extension increments also make small absolute errors consequential. If the true 0.0200 mm gauge increment were instead reported as 0.0220 mm while the force and dimensions stayed exact, the computed modulus would be 90.9 GPa, about 9.09% low. This is one sensitivity calculation, not an uncertainty budget. Calibration, uncertainty and repeatability explains how correction, repeat scatter and other uncertainty contributions differ.

Original conceptual series-displacement account for the initial 400 newton force increment. Two compliance elements are linked in a single series path with opposing external tensile arrows: outside-gauge compliance C_f and specimen gauge compliance C_s equals L₀ divided by the product of E and A₀. This is not a physical drawing of a testing machine, its frame or its electrical wiring. Both elements carry the same force increment. The gauge compliance is 0.0000500 millimetre per newton and the independently supplied additional compliance is 0.0000200 millimetre per newton. A solid blue block represents 0.0200 millimetre of gauge movement; an adjoining orange dashed-outline block represents 0.00800 millimetre outside the gauge. Their widths have the correct ratio of 5 to 2. They sum to machine movement 0.0280 millimetre and total compliance 0.0000700 millimetre per newton. Solid versus dashed-outline blocks distinguish the parts without colour. Using all the machine movement as gauge extension gives about 71.4 gigapascals, whereas subtracting the valid independently supplied extra increment gives 100 gigapascals. These are ideal teaching inputs, not measured performance of equipment or a universal correction for slip, seating, bending or variable fixtures.

A recoverable curve can have more than one useful slope

A nonlinear elastic response can retrace a curved loading path on unloading without retaining strain. As a separate invented example, stipulate an instantaneous reversible uniaxial law σ = 1000ε + 10000ε² MPa for 0 ≤ ε ≤ 0.0200, with ε inserted as a ratio. It is a constructed normal stress–strain relation, not a fit for a named polymer or a claim beyond that interval. Loading and unloading follow the same law here; delay and hysteresis are excluded by the model.

At 1% strain the model gives 11.0 MPa; at 2% it gives 24.0 MPa. The secant from the correctly zeroed origin to the first point is 11.0/0.0100 = 1100 MPa. The chord between 1% and 2% is (24.0 − 11.0)/(0.0200 − 0.0100) = 1300 MPa. They answer different questions and need not be equal. A tangent modulus is the local slope near a specified strain, rather than a chord over a finite interval.

For a truly straight, correctly zeroed linear elastic law, secant, local tangent and chords within the region agree. For a curve, report which slope and interval are meant instead of claiming one constant E describes every point. A departure from a straight line does not alone prove plastic flow; Plastic deformation and yield behaviour develops unloading evidence and operational proof-stress definitions.

Separate invented nonlinear elastic law. Engineering strain is inserted as a ratio, and the same states are recovered on ideal unloading.
Strain ε	Percent strain	Stress σ (MPa)
0	0%	0
0.0100	1.00%	11.0
0.0200	2.00%	24.0

Keep the region and measurement with the result

A modulus result should identify the specimen and material state, loading direction, area and gauge definition, test temperature and timescale, strain measurement, selected interval and slope method. A good-looking straight fit over a chosen window does not automatically establish recoverability, uniform uniaxial loading or an accurate strain scale. Check those conditions with appropriate observations rather than replacing an unexpected result by a handbook value.

A large E means a larger stress increment is needed for a given elastic strain increment in the stated comparison. It does not determine yield stress, tensile strength, ductility or safe service load. Rate, direction and temperature effects extends how the conditions can change the response; the next yield lesson treats permanent deformation rather than another name for stiffness.

Worked examples

Work through the case

An invented linear-elastic response reaches 100 MPa at strain 0.001. Find E.

E = σ/ε.
100/0.001 = 100,000 MPa.
100,000 MPa = 100 GPa.

E = 100 GPa within the stated regime.

Use the data region without using the wrong origin

For the invented 10.0 mm², 50.0 mm gauge record, calculate the 100–500 N slope from the indications. Check the two adjacent intervals. What does the later 700 N point change?

ΔF = 500 − 100 = 400 N. Δu_ind = 0.0350 − 0.0150 = 0.0200 mm. The constant 0.0100 mm indication offset cancels.
Δσ = 400/10.0 = 40.0 MPa and Δε = 0.0200/50.0 = 0.000400. Therefore E = 40.0/0.000400 = 100000 MPa = 100 GPa.
The 100–300 N and 300–500 N intervals each give 20.0 MPa/0.000200 = 100 GPa. The specimen stiffness over this region is 400/0.0200 = 20000 N/mm.
The 500–700 N increment is 200 N over 0.0250 mm, hence (50.0/10.0)(200/0.0250) = 40000 MPa = 40.0 GPa. It is a changed incremental response, not a reason to rename the initial slope or assign a particular irreversible mechanism from this loading record alone.

The supported initial slope is 100 GPa over the stipulated reversible first-three-point region. The later increment gives 40.0 GPa and needs separate interpretation.

Account for where the movement occurred

For the same initial 400 N increment, machine movement is 0.0280 mm and an independently supplied constant outside-gauge compliance is 0.0000200 mm/N. Compare the uncorrected and corrected modulus estimates.

The outside-gauge movement is C_fΔF = 0.0000200 × 400 = 0.00800 mm. In this series model, subtract it from the total: Δu = 0.0280 − 0.00800 = 0.0200 mm.
Using machine movement alone gives E_app = (50.0/10.0)(400/0.0280) = 71428.6 MPa ≈ 71.4 GPa.
Using the gauge increment gives E = (50.0/10.0)(400/0.0200) = 100000 MPa = 100 GPa. The uncorrected result is 100(1 − 71.4286/100) ≈ 28.6% low relative to this stipulated model.
The total compliance is 0.0280/400 = 0.0000700 mm/N: gauge compliance 0.0000500 plus outside-gauge compliance 0.0000200. Their sum, not their average, is appropriate because the increments add at the same force.

71.4 GPa if all machine travel is misassigned to the gauge; 100 GPa after the valid supplied series correction. This does not establish a correction for another fixture.

Transfer the calculation to a different gauge and force unit

A different uniform coupon has original area 20.0 mm² and gauge length 80.0 mm. In a separately established linear elastic interval, force rises from 0.400 to 1.200 kN and a specimen-gauge indication rises from 0.0120 to 0.0280 mm. Assume any indication offset is constant and no compliance correction is needed for that gauge measurement. Find E and k_s.

Convert the force increment, not just its label: ΔF = 0.800 kN = 800 N. The gauge increment is 0.0160 mm.
Stress increment is 800/20.0 = 40.0 MPa. Strain increment is 0.0160/80.0 = 0.000200, or 0.0200%.
E = 40.0/0.000200 = 200000 MPa = 200 GPa. Equivalently, (80.0/20.0)(800/0.0160) gives 200000 N/mm².
k_s = 800/0.0160 = 50000 N/mm. Checking EA₀/L₀ = 200000 × 20.0/80.0 gives the same stiffness. No yield or maximum-strength endpoint was supplied.

E = 200 GPa in the stated interval, while specimen stiffness is 50000 N/mm. The different quantities have different units.

Try the reasoning

Is modulus measured in percent strain?
Show the answer and reasoning
No.
It has stress units because strain is dimensionless. For example, use 0.001 rather than 0.1 when substituting a strain labelled 0.1% into σ = Eε.
Does a large E establish a high yield stress?
Show the answer and reasoning
No.
Yield and elastic stiffness are different properties. The elastic interval defines a slope; it does not supply the separate stress at which irreversible flow starts.
A separate linear elastic interval rises from 35.0 to 105.0 MPa while a correctly measured strain axis rises from 0.0500% to 0.1500%. Find E. What goes wrong if the percent numbers are used without conversion?
Show the answer and reasoning
Δε = 0.00100, so E = 70.0/0.00100 = 70000 MPa = 70.0 GPa. Using 0.100 as the strain increment would make the result 100 times too small.
The plotted increment is 0.100 percentage point, equal to 0.100/100 = 0.00100 as a ratio. Stress units do not cancel the missing factor of 100.
Use the initial 100 GPa material and 10.0 mm² section, but double the uniform gauge length from 50.0 to 100.0 mm. Predict the extension increment at ΔF = 400 N and the new specimen stiffness. Did the material modulus change?
Show the answer and reasoning
The extension increment doubles to 0.0400 mm and specimen stiffness halves to 10000 N/mm. E remains 100 GPa under the matched material and conditions.
Δu = ΔFL₀/(EA₀). The longer gauge contains twice the extension at the same uniform strain; its larger displacement is not evidence of a softer material.
A report uses ΔF = 400 N and C_f = 0.0000200 mm/N, but the total machine movement is only 0.00600 mm. What does the series subtraction produce, and should it be reported as a modulus for this stable positive-compliance model?
Show the answer and reasoning
It produces 0.00600 − 0.00800 = −0.00200 mm. That is inconsistent with the assumed tensile gauge response; the report cannot extract a valid E from these inputs under this model.
Check units, sign, measurement interval, calibration and whether the stated C_f applies to the configuration. Do not replace the negative increment by its absolute value or quietly call it a material property.
At Δσ = 40.0 MPa and L₀ = 50.0 mm, compare an exact gauge increment of 0.0200 mm with an indication increment of 0.0220 mm, treating all other inputs as exact for this sensitivity exercise.
Show the answer and reasoning
The first gives 100 GPa; the second gives about 90.9 GPa, an underestimate of about 9.09%.
Δε changes from 0.000400 to 0.000440, and E divides by that increment. A 10% extension overestimate therefore does not produce a 10% modulus increase. The one-error exercise is not a full uncertainty statement.
For the separate stipulated reversible law, calculate the secant to 1% strain and chord from 1% to 2%. Does their difference prove permanent deformation?
Show the answer and reasoning
The secant is 1100 MPa = 1.10 GPa; the chord is 1300 MPa = 1.30 GPa. No: the model is explicitly nonlinear elastic and retraces its path.
The two slopes use different intervals on a curve. Recovery is supplied by the loading–unloading law, not by whether either finite slope equals the initial one. Actual material recovery and time dependence would need observation.

What to carry forward

Use a supported reversible linear interval and stress–strain increments; do not force an offset indication through the origin.
E has stress units, while specimen stiffness EA₀/L₀ has force-per-length units.
Machine movement can include deformation outside the chosen gauge. A compliance correction needs its own supported model.
Elasticity concerns recovery. A curved reversible response needs a specified secant, chord or local tangent rather than one universal constant slope.

References

Teaching resources and research supporting this lesson’s explanations and examples.