Tensile curves, strength and ductility

One curve contains several questions

An engineering stress–strain curve can show an initial elastic slope, yield behaviour, work hardening, a maximum and fracture. Ultimate tensile strength is the maximum engineering stress, not necessarily the engineering stress at final separation.

Imagine following a small metal coupon until it separates. Its force first rises, reaches a maximum, then falls while the gauge keeps extending. The largest force and the last force before separation are different events. A curve is a history of measured states; first decide which event a reported property names.

Use original gauge length L₀ = 50.0 mm and original cross-sectional area A₀ = 10.0 mm² for the invented record below. The specimen-gauge extension is u, measured between the same material marks. Engineering strain is ε = u/L₀ and engineering stress is σ_eng = F/A₀. Here N/mm² equals MPa. A strain of 0.10 is 10%, not 0.10%. Deformation and relative strain develops that measurement interval.

All main-record numbers are invented for one axial coupon, at fixed material state, orientation, temperature and prescribed loading rate. The full underlying record is stipulated to have no force above 3000 N. The table is an abridged set of landmarks; joining them in the figure guides reading rather than supplying a smooth constitutive law or an exact instability calculation. Gauge movement is measured on the specimen, with relevant zero and measurement corrections already applied.

At E₁, F = 500 N and u = 0.0250 mm give 50.0 MPa and ε = 0.000500 = 0.0500%. At E₂, 1000 N and 0.0500 mm give 100 MPa and 0.100%. The reversible initial interval is stipulated: its slope is (100 − 50)/(0.00100 − 0.000500) = 100000 MPa = 100 GPa. Elastic response and elastic modulus explains choosing and measuring that slope.

Point P is supplied as the 0.2% proof endpoint: 200 MPa at total strain 0.00400 = 0.400%. It is a chosen plastic-deformation comparison, not the maximum. Plastic deformation and yield behaviour constructs such an offset point and explains why it need not be the first microscopic plastic event. Here the endpoint is given, so concentrate on reading the later curve.

H and M are later rising-load states. M gives UTS = F_max/A₀ = 3000/10.0 = 300 MPa. Its gauge strain is 5.00/50.0 = 0.100 = 10.0%. Here that is the total elongation at maximum force, including recoverable and plastic contributions, rather than the later after-fracture elongation. At F the last resolved load-bearing frame gives 2200/10.0 = 220 MPa and 11.00/50.0 = 22.0%. Those loaded-frame values are not the same as unloaded measurements made on the broken pieces. After complete separation the tensile force becomes zero; 220 MPa names the supplied last frame, not an enduring stress across a fully open fracture.

Geometry evolves during the test

After necking begins, deformation becomes localised. The maximum load reflects competition between strengthening and area reduction. Elongation at fracture depends on gauge length and measurement convention; compare like definitions rather than one unexplained ductility number.

Plastic flow in a metal can increase its resistance to further flow through evolving dislocation interactions, as developed in the yield lesson. Simultaneously, tensile extension reduces a carrying section. Before localisation, thinning is approximately distributed along the gauge. A neck is a region where the section becomes especially small and subsequent extension becomes concentrated. The gauge strain averages that nonuniform extension; it is not the neck’s maximum local strain.

Force depends on both stress and carrying area. For a current cross-section perpendicular to the loading axis in quasi-static axial equilibrium, define its section-mean axial stress s̄ = F/A_current. Equivalently F = s̄A_current. The bar means an average across that section, not a new material constant. In an approximately uniform uniaxial region this is a useful current-area stress. In a neck, off-axis stresses and nonuniform fields can matter, so this average is not automatically the equivalent uniaxial flow stress of the material.

For M and N, independently supplied simultaneous section measurements are 9.10 and 6.50 mm². Spatial observations show approximately uniform deformation up to M and a local neck at N. At M, s̄ = 3000/9.10 ≈ 329.67 MPa. At N, s̄ = 2600/6.50 = 400 MPa. Engineering stress fell from 300 to 260 MPa, while this measured-section mean rose. A falling engineering branch therefore does not require the local stress measure to fall.

The area factor from M to N is 6.50/9.10 ≈ 0.7143; the mean-stress factor is 400/329.67 ≈ 1.2133. Their product is about 0.8667, the force factor 2600/3000. Area lost about 28.6%, while mean stress rose about 21.3%. Multiplication, rather than subtracting these two percentage changes, accounts for the roughly 13.3% force decrease.

This comparison is consistent with continued resistance to plastic flow while geometry reduces the supported force. It does not independently prove a local hardening law after necking: constraint, damage and spatial redistribution can also affect the measured mean. The force trace alone does not establish necking either; slip in the grips or other changes can produce a drop. Here the supplied spatial observations and concurrent area measurements provide the separate evidence.

Engineering and true stress–strain measures develops the distinction between reference and current geometry. Do not take the whole-gauge extension after necking and silently use a uniform-deformation conversion to reconstruct a local true curve. The current area at F has not been supplied, so no section-mean stress at that loaded frame can be calculated here.

Ductility needs an endpoint and a measurement convention

Ductility describes an ability to sustain plastic deformation before fracture under stated conditions. Two common tensile measures ask different geometric questions. Elongation after fracture compares the final unloaded separation L_f of the original gauge marks with L₀: elongation (%) = 100(L_f − L₀)/L₀. In this example the broken pieces are properly fitted together and aligned for that reading; an open fracture gap is not added as material extension.

Reduction in area compares the original section with the minimum section measured after fracture, A_f: reduction in area (%) = 100(A₀ − A_f)/A₀. It concentrates on the fractured region. It is not a percent change in length and is not the current loaded area at an earlier frame. For a noncircular or irregular section, measure the relevant area rather than inserting an arbitrary single diameter.

After the main-record test, suppose the fitted pieces give L_f = 60.85 mm and A_f = 4.00 mm² at the stated observation conditions, with no further recovery in this example. Elongation is 100(60.85 − 50.0)/50.0 = 21.7%; reduction in area is 100(10.0 − 4.00)/10.0 = 60.0%. Neither 60.85 mm nor 4.00 mm² is a loaded-frame input.

The last loaded-frame extension was 11.00 mm, or 22.0%. The after-fracture retained extension is 10.85 mm, or 21.7%. The 0.150 mm endpoint difference is supplied by these separate observations. Do not estimate its value by applying σ_eng/E uniformly to a necked gauge: the local stress and geometry are no longer uniform. A fracture-extensometer reading and an after-fracture fitted-pieces reading need their own labels.

Two specimens can rank differently by elongation and reduction in area because one averages extension over a gauge while the other examines the fractured section. Neither establishes resistance to an existing crack. Strength, toughness and fracture resistance develops that separate question; a long tensile curve is not by itself a valid crack-resistance test.

Retain gauge marks, measurement method, fracture location, original geometry, actual loading conditions and uncertainty with the result. A fracture outside the intended gauge or a gauge that loses its measurement points changes what can be inferred. Specimens, dimensions and test conditions explains that evidence record. For time-dependent materials also specify recovery time; immediate residual extension need not be final permanent extension.

The same localised deformation can give different gauge percentages

Use a separate invented, reassembled specimen to isolate the gauge effect. Let x be original material position in mm, from 0 to 100. The retained local axial engineering strain is prescribed as 0.10 outside x = 45–55 mm and 0.30 inside that 10 mm region. These are residual values after unloading at the stated observation time. No fracture gap is included, and no further recovery occurs in this model.

An original segment of length ℓ at local strain e retains extension eℓ. Add those segment extensions across a gauge, then divide by its original total length. The 100 mm interval contains 90 mm at 10% and 10 mm at 30%: retained extension is 90 × 0.10 + 10 × 0.30 = 12 mm. Its reported elongation is 12%.

The centred 50 mm gauge from x = 25 to 75 contains 40 mm at 10% and the same 10 mm at 30%. It retains 7 mm, giving 14%. The centred 25 mm gauge from x = 37.5 to 62.5 contains 15 mm at 10% and 10 mm at 30%. It retains 4.5 mm, giving 18%. Nothing about the physical specimen changed; the shorter denominator gives the localised region more weight.

The second figure plots cumulative retained displacement u_r(x), measured relative to x = 0. Its rise between any two material positions is the retained extension of the interval between them. Its steeper central segment means more retained extension per original millimetre. The three brackets identify actual intervals on that same position scale, rather than three differently ductile materials.

This prescribed field explains the mechanism of gauge dependence. It is not a fitted conversion rule for other alloys, thicknesses, fracture locations or gauge shapes. If deformation were uniform at 10% everywhere, every contained gauge would report 10%, although their absolute extensions would differ. Compare like endpoint and gauge conventions; retain local measurements when localisation itself is the question.

Compare the property, the history and the evidence

A larger maximum force is not automatically a larger UTS if original areas differ. A larger fracture elongation is not automatically a larger reduction in area if deformation localises differently. State which feature and denominator you used before ranking curves. Keep material processing state, orientation, temperature, loading rate and specimen geometry comparable when the aim is to isolate a material difference.

A complete recorded maximum is different from the largest point in an incomplete or coarsely sampled trace. Without evidence that the peak was captured, the largest sampled F/A₀ is only the largest observed value. Likewise, a curve drawn to its last available point does not establish fracture if the test stopped earlier.

The first figure deliberately shows two strain scales. The full curve makes the maximum and subsequent loaded-frame endpoint visible. Its separate low-strain panel reveals the initial slope and supplied proof endpoint that would be crowded near the origin on the full scale. Read axis units before comparing apparent steepness; plotting on a different scale does not change E.

These are conditional tensile observations, not a complete account of a component’s response to a notch, cycling, corrosion or long service. Preserve the force, gauge and geometry records so that a new question can be checked against actual evidence rather than answered by the largest number on one curve.

Worked examples

Try the reasoning

1. Can a tensile curve fall while local material continues to harden?

   Show the answer and reasoning

   Yes.

   Necking reduces the section carrying load. For example, M → N has engineering stress 300 → 260 MPa but measured-section mean about 329.67 → 400 MPa. Concurrent geometry and local-state limits matter; the falling branch alone does not prove the mechanism.

2. Does high elongation establish high crack resistance?

   Show the answer and reasoning

   No.

   Ductility and fracture toughness use different assessments. Elongation and reduction in area describe this tensile specimen’s deformation; resistance to a pre-existing crack needs its own geometry, loading and valid measurement.

3. Coupon A has original area 16.0 mm² and a captured maximum force 4.80 kN. Coupon B has 20.0 mm² and a captured maximum 5.00 kN. Which has the larger UTS? Treat each stated maximum as complete for its test.

   Show the answer and reasoning

   A has UTS 300 MPa; B has 250 MPa. A has the larger tensile strength despite its smaller maximum force.

   Convert kN to N: 4800/16.0 = 300 N/mm², while 5000/20.0 = 250 N/mm². The ratio is 1.20, not the force ratio 4.80/5.00. Original area is the same reference throughout each engineering curve, but differs between the two coupons.

4. Two separate coupons each begin with L₀ = 50.0 mm and A₀ = 10.0 mm², tested with matched conventions. P gives fitted-pieces L_f = 60.0 mm and A_f = 6.00 mm². Q gives 57.5 mm and 4.00 mm². Compare their after-fracture elongations and reductions in area. Is one unconditionally more ductile?

   Show the answer and reasoning

   P: 20.0% elongation and 40.0% reduction in area. Q: 15.0% and 60.0%. The rankings differ; identify the ductility measure rather than claiming one unconditional ordering.

   P retains 10.0/50.0 = 0.20 of its original length and loses 4.00/10.0 = 0.40 of its area. Q retains 7.50/50.0 = 0.15 and loses 6.00/10.0 = 0.60. These metrics weight the deformation distribution differently. Neither supplies crack resistance or a forming-limit guarantee.

5. In the main record, a reader labels N as strain 0.16% and calls its 260 MPa the UTS because it is the last point they copied. Correct both conclusions using the complete supplied record.

   Show the answer and reasoning

   N has ε = 8.00/50.0 = 0.160 = 16.0%, not 0.16%. UTS is 300 MPa at M; 260 MPa is a later lower engineering stress.

   Multiplying the strain ratio by 100 gives its percent value. The complete underlying record is stated to reach its maximum at 3000 N, and engineering stress uses the unchanged 10.0 mm² reference. Neither N nor the final 220 MPa frame replaces that captured maximum.

6. On the separate residual field, compare original gauges x = 0–20 mm and x = 0–100 mm. Find retained extension and elongation for each. Why is the shorter gauge not the one with the larger percentage in this case?

   Show the answer and reasoning

   0–20 mm retains 2.00 mm, giving 10.0%; 0–100 mm retains 12.0 mm, giving 12.0%. The shorter interval excludes the high-strain region.

   The 20 mm gauge lies entirely in the region at strain 0.10. The full gauge includes the 10 mm region at 0.30. A smaller denominator alone does not guarantee a larger result: which deformation is included also changes. The centred nested examples all included the neck; this unfamiliar interval does not.

7. A separate logger stores only 1200, 1900 and 1600 N before the test stops. Original area is 8.00 mm². A report labels 237.5 MPa as the material’s UTS and the last point as fracture. What is established, and what additional evidence is needed?

   Show the answer and reasoning

   237.5 MPa is the largest observed engineering stress. The supplied readings do not establish a captured test maximum or fracture.

   1900/8.00 = 237.5 MPa is valid for the largest recorded force. A higher force may have occurred between sparse readings or outside the saved interval. Verify peak capture and the actual test endpoint; the final 1600 N reading alone does not show separation. Report specimen state and conditions before transferring any resulting property.

What to carry forward

• Name the curve feature behind each property.
• Track changes in specimen geometry.
• Engineering UTS uses the largest captured force and original area; it need not be the engineering stress at fracture.
• A current-area section mean and a local uniaxial flow stress are different interpretations once a neck develops.
• After-fracture elongation and reduction in area use unloaded geometric measurements; do not mix their state with an earlier force.
• A localised region changes its weight when gauge marks move. Keep interval, endpoint, state and actual conditions with a ductility percentage.