Study Companion: MM03 — The Tension Test and Mechanical Properties

A tensile test machine pulls a steel bar slowly until it breaks. The bar is 11.28\,\mathrm{mm} across and the technician records a load of 40\,\mathrm{kN} at the strongest moment, then watches the load fall to 28\,\mathrm{kN} as the bar goes on stretching and finally parts. How strong is this steel: 40\,\mathrm{kN}, 28\,\mathrm{kN}, or neither? The answer needs stresses, not loads, and it needs a vocabulary for the dozen distinct things the curve of one test can tell us. This lesson builds that vocabulary: the test, the curve, the points on it, and the numbers a designer takes away.

Sources: Gere and Goodno, Mechanics of Materials, 9th edition (SI), sections 1.5–1.6 and Appendix I (the syllabus for this lesson); Beer, Johnston, DeWolf and Mazurek, 8th edition, section 2.1A–F; Hearn, Mechanics of Materials 1, sections 1.7–1.9, 1.20–1.21 and Example 1.5; Venkataraman, Essential Mechanics, section 3.5 (a survey of material classes; his stress-strain curve is in section 3.2); Muvdi and Elhouar, section 1.4.4. Hooke's law, Poisson's ratio and \delta = PL/AE are the next lesson; allowable stress is Lesson MM06; creep and fatigue are only named here and treated in Lesson MM12. Stress is quoted in MPa; \varepsilon is dimensionless (1\,\mu\varepsilon = 10^{-6}). Tension positive.

How to use this companion

Each nugget is one idea. It opens with a claim, then a picture shows why it holds, a number checks it, and an equation states it, labelled with the kind of statement it is. A worked example with units, a real use, the most common wrong answer and a check question follow. A bridge says why the next idea is needed, and the Narration spine summarises the video beat. The synthesis, computed views, traps and self-check close the lesson.

The puzzle

A steel specimen 11.28\,\mathrm{mm} in diameter carries a peak load of 40\,\mathrm{kN}, then 28\,\mathrm{kN} at fracture. The plausible answer is "its strength is the load it broke at, 28\,\mathrm{kN}", or "40\,\mathrm{kN} because it is the most it carried". Neither is a property of the material. The resolution is in Nugget 03.

Nugget 01: A Tension Test Turns a Specimen Into a Material

Claim. A load-extension record belongs to one specimen; dividing the load by the original area and the extension by the original gauge length gives a stress-strain curve that belongs to the material and is the same for any size of specimen.

The picture. A standard specimen in the grips of a test machine.

The specimen is made with enlarged ends so the grips hold it without crushing it, and a narrow middle section where the stress is uniform (Lesson MM02). A short length of that section, the gauge length L_0, is marked. The machine pulls at a steady, slow rate and records P; an extensometer records the change \delta in the gauge length. Compression tests use stocky specimens: cubes of about 50\,\mathrm{mm}, cylinders of about 25\,\mathrm{mm} diameter, and for concrete a 152\,\mathrm{mm} by 305\,\mathrm{mm} cylinder cured 28 days.

Check with numbers. The ASTM tension specimen is 12.8\,\mathrm{mm} in diameter with a 50.8\,\mathrm{mm} gauge length. Its area is A_0 = \pi(6.4\,\mathrm{mm})^2 = 128.7\,\mathrm{mm^2}. A load of 20\,\mathrm{kN} with an extension of 0.0385\,\mathrm{mm} gives a stress of 20\,000/128.7 = 155.4\,\mathrm{MPa} and a strain of 0.0385/50.8 = 7.58\times10^{-4}. A specimen twice as thick, loaded to 40\,\mathrm{kN}, would give the same point on the curve.

The idea. Beer (2.1) makes the argument with three rods. Load 2P on a rod of area 2A stretches it exactly as P on A does, so stress P/A is the size-free measure of load. Load P on a rod of length 2L stretches it 2\delta, so strain \delta/L is the size-free measure of stretch. The curve of \sigma against \varepsilon is therefore the same for every specimen of one material. Venkataraman says the same in a sentence.

The equation (definition: engineering stress and strain; they use the original area and gauge length):

Worked number. Beer's microstrain illustration: a 0.600\,\mathrm m bar that stretches 150\,\mu\mathrm m has \varepsilon = 150\times10^{-6}/0.600 = 250\times10^{-6} = 250\,\mu\varepsilon.

Real example (one you can see). Pull a plastic carrier bag handle until it stretches and whitens at one point: the narrow middle of the handle does the stretching, and that is why tests use a narrow section.

Plausible wrong answer. "A stronger machine reading means a stronger material." A bigger specimen reads a bigger load; only the load per original area compares materials.

Check. A specimen of 100\,\mathrm{mm^2} carries 25\,\mathrm{kN}. Its stress? (250\,\mathrm{MPa}.)

Narration spine. A specimen in grips is stretched; the load-extension plot appears; every point is relabelled by dividing by A_0 and L_0 and the curve keeps its shape but loses its units.

Bridge. With the axes in stress and strain, the curve for mild steel has several distinct regions, each with a name.

Nugget 02: The Mild-Steel Curve Has a Name for Every Bend

Claim. The stress-strain curve of low-carbon steel has a straight start, a flat plateau, a rise, a peak and a fall. Each landmark is a property: proportional limit, elastic limit, yield stress, ultimate stress, fracture stress.

The picture. The engineering curve of mild steel, not to scale in the first part so the features can be seen.

Reading it from the left: the straight line is the elastic region, whose slope is the stiffness of the material, the modulus E (next lesson). At the yield point the stress stops rising and the steel stretches a great deal at about constant stress: the plateau. After it, strain hardening makes the stress rise again, to the peak at the ultimate stress \sigma_U. Past the peak the engineering stress falls while a local waist, the neck, forms, and the bar breaks.

Check with numbers. Hearn's Example 1.5 is a low-carbon steel specimen 11.28\,\mathrm{mm} in diameter (area 100\,\mathrm{mm^2}), gauge length 56\,\mathrm{mm}, tested to fracture. From its table of 28 load-extension pairs (in the Python view): the load rises almost linearly to 32.1\,\mathrm{kN}, jumps to 33.3\,\mathrm{kN} at 112\,\mu\mathrm m, then drops to 31.2\,\mathrm{kN} at 224\,\mu\mathrm m and wanders around 31–32\,\mathrm{kN} through to 1120\,\mu\mathrm m: the plateau. The upper yield stress is 33\,300/100 = 333\,\mathrm{MPa} and the lower 31\,200/100 = 312\,\mathrm{MPa}. The peak is 40\,\mathrm{kN} at 7840\,\mu\mathrm m: \sigma_U = 400\,\mathrm{MPa}. Fracture is at 28\,\mathrm{kN}: nominal 280\,\mathrm{MPa}. Stretching 7840\,\mu\mathrm m in 56\,\mathrm{mm} at the peak is \varepsilon = 0.14, which is over a hundred times the elastic strain at yield.

The idea. The names differ slightly between books, and this book fixes them as follows.

• Proportional limit \sigma_{pl}: the highest stress for which stress is proportional to strain.
• Elastic limit: the highest stress after which the material returns to its original length on unloading. It is at or just above the proportional limit and hard to find experimentally, so for mild steel the two are treated as equal (Beer, Muvdi, Hearn).
• Yield: the stress at which strain increases with no noticeable rise in load. Gere draws one yield point and a perfectly plastic plateau; Beer, Hearn and Muvdi distinguish the transient upper yield point from the steadier lower yield point. This book uses the lower yield stress as \sigma_Y for steels with a plateau, because the upper one depends on how smoothly the test is run.
• Ultimate stress \sigma_U (tensile strength): the maximum engineering stress, the maximum load over A_0.
• Fracture (breaking) stress: the engineering stress at rupture.

Yield and ultimate are called "strengths" although they are stresses.

The equation (definition, each feature a stress read off the plotted curve):

Worked number. Hearn's steel: \sigma_Y = 312\,\mathrm{MPa} (lower), upper yield 333\,\mathrm{MPa}, \sigma_U = 400\,\mathrm{MPa} (the book reads 402 from the plotted peak), nominal fracture 280\,\mathrm{MPa}.

Real example (one you can see). Bend a paperclip once: it springs back (elastic). Bend it farther: it stays bent (past yield). Bend it back and forth: it gets stiffer, then snaps (strain hardening, then fracture).

Plausible wrong answer. "The proportional limit and the yield point are the same thing." For mild steel they are close; for aluminium and most other metals there is no plateau and the proportional limit is well below any yield value (Nugget 04).

Check. What is the nominal stress at the peak of a curve with P_{\max} = 36\,\mathrm{kN} on A_0 = 90\,\mathrm{mm^2}? (400\,\mathrm{MPa}.)

Narration spine. The curve is drawn point by point; each landmark lights up with its name and the load behind it; the plateau is stretched horizontally to show how much it is.

Bridge. The curve falls after the peak, yet the steel does not weaken. What changes?

Nugget 03: Engineering Stress Falls After the Peak While True Stress Keeps Rising

Claim. After the ultimate point the engineering stress falls only because the load is divided by the original area. The actual area of the neck is smaller, so the true stress still rises until fracture. Strength is read at the peak of the engineering curve, not at fracture.

The picture. Two curves: engineering (falls) and true (keeps rising).

Once the neck begins, almost all the further stretch happens in a short length where the area keeps shrinking. The same bar carries a smaller load but over a much smaller area, so the stress in the neck is still rising.

Check with numbers. In Hearn's test at the peak, P = 40\,\mathrm{kN} over 100\,\mathrm{mm^2}: \sigma = 400\,\mathrm{MPa}, \varepsilon = 7840\,\mu\mathrm m/56\,\mathrm{mm} = 0.14. If the volume stays constant before necking, A = A_0/(1 + \varepsilon), so the true stress is 400 \times 1.14 = 456\,\mathrm{MPa} and the true strain \ln 1.14 = 0.131. At fracture the nominal stress is 28\,\mathrm{kN}/100\,\mathrm{mm^2} = 280\,\mathrm{MPa}, but the necked bar's diameter is 6.45\,\mathrm{mm}, so the actual stress is 28\,000/(\pi/4 \times 6.45^2) = 856.9\,\mathrm{MPa}, about three times the nominal. The puzzle's bar broke at 28\,\mathrm{kN}, but the steel's strength is the 400\,\mathrm{MPa} at the peak.

The idea. Gere and Beer distinguish nominal (engineering, conventional) stress and strain, which use the original area and gauge length, from true stress, which uses the actual area, and true (natural) strain, \varepsilon_t = \int_{L_0}^{L} dL/L = \ln(L/L_0). In tension the true stress is larger than the nominal and the true strain is smaller. The engineering curve is what design uses, because the original dimensions are known. Hearn adds that the actual fracture stress is "seldom of any practical value".

The equation (definition of true stress and strain; the constant-volume form is an approximation valid before necking):

Worked number. \sigma_t = 456\,\mathrm{MPa}, \varepsilon_t = 0.131 at the peak; at fracture \sigma_t = 856.9\,\mathrm{MPa}.

Real example (one you can see). Pull a strip of chewing gum: a thin waist forms and runs away while your hands pull less hard. The waist is thin, so its stress is high even as the force needed falls.

Plausible wrong answer. "The steel is weakest at fracture because the load is lowest." The load is low because the cross-section has shrunk; the material is at its highest stress.

Check. At what strain do the engineering and true strain differ by 1\,\% of the engineering value? (About \varepsilon = 0.02: \ln 1.02 = 0.0198.)

Narration spine. A necked bar next to the curve; the load arrow shrinks while the neck shrinks faster; the true-stress curve is drawn above the engineering one.

Bridge. Many metals have no yield plateau, so there is no yield point to read.

Nugget 04: Without a Yield Plateau, Draw a Line Offset by a Fixed Strain

Claim. For a material whose curve bends smoothly, yield is defined by convention: draw a line parallel to the initial slope, shifted along the strain axis by 0.002 (the 0.2\,\% offset); the stress where it meets the curve is the offset yield stress.

The picture. A rounded curve, its initial tangent, and the offset line.

The 0.2 % rule is arbitrary, a convention chosen because it corresponds to a small and agreed permanent set. It is not a physical property like a melting point, but every engineer who quotes a yield strength of aluminium or high-strength steel means this.

Check with numbers. Gere's Example 1-7 gives a copper-alloy wire's curve by a formula: \sigma = \frac{124\,000\,\varepsilon}{1 + 240\,\varepsilon}\ \mathrm{MPa}. At \varepsilon = 0.03 the stress is 124\,000(0.03)/8.2 = 453.7\,\mathrm{MPa}. The initial slope, the derivative at 0, is E = 124\,000\,\mathrm{MPa} = 124\,\mathrm{GPa}. The offset line is \sigma = E(\varepsilon - 0.002). Set them equal: \varepsilon/(1+240\varepsilon) = \varepsilon - 0.002, which reduces to 240\varepsilon^2 - 0.48\varepsilon - 0.002 = 0, with positive root \varepsilon_Y = 0.004056. Then \sigma_Y = 124\,000(0.004056 - 0.002) = 254.8\,\mathrm{MPa}, which the Python and MATLAB views confirm.

The idea. Gere (Fig. 1-35), Beer (Fig. 2.8) and Muvdi offer the same construction. Muvdi allows offsets from 0.002 to 0.0035. Hearn's convention is the proof stress, the stress that leaves a permanent set of 0.1\,\% after being held for 15 seconds and removed; the logic is the same, the number differs. This book uses 0.2\,\%.

The equation (definition by construction):

Worked number. \sigma_Y = 254.8\,\mathrm{MPa} at \varepsilon_Y = 0.004056. For an aluminium-like curve \sigma = \frac{70\,000\,\varepsilon}{1 + 300\,\varepsilon} the same steps give \varepsilon_Y = 0.003768 and \sigma_Y = 123.8\,\mathrm{MPa}.

Real example (one you can see). Bend a thin aluminium strip, a drinks can ring-pull: it has no sudden give; it simply feels softer and softer. There is no yield point to feel, so engineers agree on one.

Plausible wrong answer. "The offset yield stress is where the material first deforms plastically." Plastic strain begins earlier; 0.2\,\% is a convenient measurable amount, with 0.002 of permanent strain left on unloading.

Check. Why is the offset line parallel to the initial slope? (Because unloading from any point follows the elastic slope, the line is the unloading path that leaves 0.002 of permanent strain; Nugget 06.)

Narration spine. A smooth curve, a dashed tangent; a copy of the tangent slides right by 0.002 and stops where it meets the curve.

Bridge. Yield alone does not say whether a material gives warning before it breaks.

Nugget 05: Ductile or Brittle: How Much a Material Stretches Before It Breaks

Claim. A ductile material stretches a lot (and often necks) before it fractures; a brittle one breaks at small strain. Ductility is measured by the percent elongation and the percent reduction in area, and brittle materials are usually much stronger in compression than in tension.

The picture. Fracture shapes and curves.

A ductile bar fails with a cup-and-cone surface inclined at about 45^\circ: Beer reads this as shear doing the work, since the maximum shear stress in a bar in tension acts at 45^\circ to the axis (Lesson MM10), and yielding is slipping along such surfaces. A brittle bar breaks on a flat plane perpendicular to the load: normal stress separates it.

Check with numbers. Hearn's specimen had a gauge length L_0 = 56\,\mathrm{mm} that grew by 14.56\,\mathrm{mm} at fracture, so the percent elongation is 14.56/56 \times 100 = 26.0\,\%; the diameter fell from 11.28\,\mathrm{mm} to 6.45\,\mathrm{mm}, so the percent reduction in area is (11.28^2 - 6.45^2)/11.28^2 \times 100 = 67.3\,\%. Elongation depends on the gauge length because most of the stretch happens in the neck, so a shorter gauge length gives a larger percentage: always quote it. Reduction in area is independent of gauge length (Hearn calls it the more useful measure).

The idea. Beer explains why brittle materials are stronger in compression: they contain flaws, and tension opens flaws while compression closes them. Gere adds that a ductile metal's compression curve follows the tension curve to yield, after which the specimen bulges into a barrel (friction at the platens) and the curve climbs steeply. Venkataraman's survey classifies materials as metals (mostly ductile and tough), ceramics (brittle, stiff, hard), polymers (plastics and elastomers) and composites (a matrix with reinforcement). Beer's Fig. 2.11 shows the point that matters most for design: pure iron and different steels have very different yield, ultimate and ductility but the same E, so a stronger steel gives more load capacity, not more stiffness. Gere's property tables (Appendix I) give typical values only, never for specific design:

The equation (definition, two measures of ductility):

Worked number. 26.0\,\% elongation and 67.3\,\% reduction in area: a ductile steel. A36 structural steel is rated about 30\,\% elongation; steels with \sigma_Y \le 345\,\mathrm{MPa} must meet at least 21\,\% in 51\,\mathrm{mm}.

Real example (one you can see). Drop a ceramic mug: it shatters into flat-faced pieces. Drop a steel mug: it dents. Chalk snaps flat; a thin copper wire necks before parting.

Plausible wrong answer. "A higher-strength steel is also stiffer." Not so: E is nearly the same for all steels; only the yield and ultimate values change.

Check. A cast iron has tension strength 200\,\mathrm{MPa} and compression strength 800\,\mathrm{MPa}. Is it better used as a tie or as a column? (A column, if buckling is not the limit: it is four times stronger in compression.)

Narration spine. A ductile bar necks and cones; a brittle bar snaps flat; the two curves are drawn side by side, and the area to the right of the brittle curve's end is shaded as the warning the ductile material gives.

Bridge. Loading a ductile bar past yield and then unloading it shows what the curve means for the shape afterwards.

Nugget 06: Unloading Follows the Elastic Slope, So Plastic Strain Stays

Claim. If a bar is loaded past yield and then unloaded, the stress-strain path back is a straight line parallel to the initial elastic line. The strain that remains at zero stress is the permanent set; the part that returns is the elastic recovery \sigma/E.

The picture. Unloading from a point beyond yield.

Loading past yield breaks atomic bonds and re-forms them; unloading releases only the elastic part, which is why the line back has the same slope as the line up. Reload from C and the path is again linear, to B, then follows the original curve: the elastic limit has risen to the stress reached, the linear range is longer, and the remaining ductility is shorter. This is strain hardening (also called work hardening). Beer adds the Bauschinger effect: loading in the opposite direction after tension yielding gives a rounded curve with no sharp yield point, and the total stress change from tension yield to compression yield is 2\sigma_Y even after hardening; reversed loading into the plastic range is seldom allowed.

Check with numbers. The copper wire of Nugget 04 (\sigma = \frac{124\,000\varepsilon}{1 + 240\varepsilon}, d = 3.5\,\mathrm{mm}) carries a tension of 3.20\,\mathrm{kN}. Area 9.621\,\mathrm{mm^2}, so \sigma = 3200/9.621 = 332.6\,\mathrm{MPa}, above the 254.8\,\mathrm{MPa} yield. Solve the curve for strain: \varepsilon = \sigma/(124\,000 - 240\sigma) = 332.6/44\,176 = 7.53\times10^{-3}. Elastic recovery on unloading: \sigma/E = 332.6/124\,000 = 2.68\times10^{-3}. Permanent set: 7.53\times10^{-3} - 2.68\times10^{-3} = 4.85\times10^{-3}. On a wire length of 1.282\,\mathrm m the wire stays 4.85\times10^{-3} \times 1282\,\mathrm{mm} = 6.2\,\mathrm{mm} longer.

The idea. Gere (Figs. 1-39, 1-40) names the points as in the picture; the same-slope fact is what the earlier offset construction uses. Gere's worked example prints the elastic recovery as 3.895\times10^{-4} and a permanent set of 9.18\,\mathrm{mm}; with the book's own inputs the elastic strain is 333/124\,000 = 2.685\times10^{-3} and the set about 6.2\,\mathrm{mm}, so the printed number is a slip.

The equation (material model: elastic unloading line of slope E):

Worked number. \varepsilon_{\text{res}} = 4.85\times10^{-3}, set = 6.2\,\mathrm{mm} in 1.282\,\mathrm m. For \sigma = 300\,\mathrm{MPa} on a 1.5\,\mathrm m wire of the same material: strain 5.77\times10^{-3}, recovery 2.42\times10^{-3}, set 3.35\times10^{-3}\times1500\,\mathrm{mm} = 5.0\,\mathrm{mm}.

Real example (one you can see). Stretch a paper clip's wire by hand and let go: some of it springs back, some does not. The part that does not is the permanent set, and the clip is now harder to bend at that spot.

Plausible wrong answer. "After yielding, unloading returns the bar to zero strain along the same path." It returns along a straight line parallel to the elastic line and leaves a permanent strain.

Check. A bar is loaded to 300\,\mathrm{MPa} (E = 200\,\mathrm{GPa}, yield 250\,\mathrm{MPa}) and the strain at that point is 0.0030. What is the permanent strain? (0.0030 - 300/200\,000 = 0.0015.)

Narration spine. The load climbs the curve; a hand lets go and the path drops down a straight line; the gap on the strain axis opens as the permanent set; the line is replayed upward to show the raised elastic limit.

Bridge. The area under the curve is the energy absorbed; and a stress held for a long time does something a quick test does not show.

Nugget 07: Energy and Time: Resilience, Toughness, Creep and Fatigue

Claim. The area under the stress-strain curve is energy per unit volume. Up to the proportional limit it is the modulus of resilience; up to fracture it is the modulus of toughness. Over time, a steady load can produce creep, and a repeated load can break a bar well below \sigma_U (fatigue).

The picture. The two areas under a curve.

Energy. Stretching a bar by d\delta with force F does work F\,d\delta per bar; per unit volume it is \sigma\,d\varepsilon, so the energy density is the area under the curve. In the elastic range \sigma = E\varepsilon and the area of the triangle is \frac{1}{2}\sigma\varepsilon.

Check with numbers. For A36 structural steel, \sigma_{pl} \approx \sigma_Y = 250\,\mathrm{MPa} and E = 200\,\mathrm{GPa}: u_R = \sigma_Y^2/(2E) = (250\,\mathrm{MPa})^2/(2\times200\,000\,\mathrm{MPa}) = 0.156\,\mathrm{MJ/m^3}. A rod 12\,\mathrm{mm} in diameter and 1\,\mathrm m long has a volume of 1.131\times10^{-4}\,\mathrm{m^3}, so it can store 0.156\times10^{6}\,\mathrm{J/m^3}\times1.131\times10^{-4}\,\mathrm{m^3} = 17.7\,\mathrm J before yield. Muvdi's example (a steel rod at its proportional limit of about 702\,\mathrm{MPa} with E = 200.7\,\mathrm{GPa}) gives 1.23\,\mathrm{MJ/m^3}.

The idea. Muvdi defines both moduli. The modulus of toughness is approximately the average stress times the fracture strain, \sigma_{\text{avg}}\varepsilon_f. Hearn uses "toughness" in a second sense, resistance to crack growth, where a crack is a stress concentration and a ductile material blunts it by local yielding; the two meanings are related but not the same, and this book keeps them apart. Creep is the extra strain that grows with time under constant load; Hearn gives the creep curve (initial strain, primary stage with falling rate, secondary stage at constant rate, tertiary stage accelerating to fracture; design aims for the secondary stage), and Gere notes it matters at high temperatures, and that concrete decks sag slowly under constant load. Relaxation is the mirror: stress in a wire held at constant strain drops with time. Fatigue: a part loaded repeatedly can fail at stresses well below \sigma_U; for low-carbon steel the endurance limit is about half the ultimate stress (Beer). Both are only named here; fatigue and creep are treated in Lesson MM12.

The equation (definition of the two moduli; the first from the physical law of work):

Worked number. u_R = 0.156\,\mathrm{MJ/m^3} (A36) and 17.7\,\mathrm J in the 12\,\mathrm{mm} by 1\,\mathrm m rod. Toughness of the same steel, roughly \sigma_{\text{avg}} = 325\,\mathrm{MPa} over \varepsilon_f = 0.25: about 81\,\mathrm{MJ/m^3}, five hundred times the resilience: why a ductile steel survives an impact that a brittle one does not.

Real example (one you can see). A paper clip bent back and forth ten or twenty times snaps at the bend although a single pull of the same size does not.

Plausible wrong answer. "A stronger material is always tougher." Tough means absorbing energy; a high-strength but brittle ceramic has a large peak stress and a tiny area.

Check. Which has the larger toughness: a steel at \sigma_{\text{avg}} = 300\,\mathrm{MPa}, \varepsilon_f = 0.2 or a cast iron at \sigma_{\text{avg}} = 150\,\mathrm{MPa}, \varepsilon_f = 0.005? (The steel: 60 against 0.75\,\mathrm{MJ/m^3}.)

Narration spine. The curve fills with colour under the elastic part (resilience) and then under the whole curve (toughness); a clock runs and the strain creeps upward at constant stress.

Synthesis

What no single nugget shows. The tension test reduces an object to a material by removing size, and the single curve it gives holds the whole vocabulary of mechanical properties: where the material is elastic, where it yields, how strong it is, how much it stretches, how much energy it absorbs. The curve also teaches the limits of the convention: the engineering curve is false after the peak but is the right curve for design.

• Central idea. Divide by original area and gauge length to get a curve that belongs to the material, not the specimen.
• Key relationship. \sigma_Y, \sigma_U, percent elongation and percent reduction in area, and the area under the curve, are the numbers a designer carries away; the slope is E.
• Main assumption. Engineering values with the original dimensions; slow, uniaxial loading at room temperature, with typical values as guides and never for specific design.
• What to explore next. The straight part of the curve, Hooke's law and Poisson's ratio (Lesson MM04), and the use of \sigma_Y and \sigma_U in a factor of safety (Lesson MM06).

Computed views

Python (numpy). MM03_tension_test.py reduces Hearn's 28-pair table to E, yield, ultimate, reduction in area and elongation, finds offset yield stresses and the permanent set of the copper wire.

E from the first ten points: 202.9 GPa
upper yield: 333 MPa
lower yield: 312 MPa
ultimate: 400 MPa
reduction in area: 67.3 %
elongation: 26.0 %
nominal fracture stress: 280 MPa
actual fracture stress: 856.9 MPa
strain at ultimate: 0.140, true stress 456 MPa
true strain at ultimate: 0.1310
offset yield (E 124 GPa, k 240): 254.8 MPa at 0.00406
offset yield (E 70 GPa, k 300): 123.8 MPa at 0.00377
wire 3.5 mm, 3.20 kN: stress 332.6 MPa, strain 0.00753
  recovery 0.00268, residual 0.00485, set 6.21 mm
resilience A36: 0.156 MJ/m3

The fit through the first ten data points gives E = 202.9\,\mathrm{GPa}; the book reads the slope off a plot and gets 203.6\,\mathrm{GPa}.

MATLAB (Octave-compatible). MM03_tension_test.m fits the elastic slope with polyfit and finds the offset yield point with fzero.

E from the first ten points: 202.9 GPa
offset yield (E 124 GPa, k 240): 254.8 MPa at 0.00406
offset yield (E 70 GPa, k 300): 123.8 MPa at 0.00377
sigma 300 MPa: strain 0.00577, residual 0.00335, set on 1.5 m 5.02 mm
resilience A36: 0.156 MJ/m3

SOLIDWORKS Simulation exercise (our own; a tension coupon with a tabulated curve). Compare the load at which the coupon yields with the hand value.

• Geometry. Flat coupon 10\,\mathrm{mm} wide, 4\,\mathrm{mm} thick, 60\,\mathrm{mm} long in the parallel section, with generous radius transitions to wider ends.
• Material. A custom elastic-plastic material: E = 200\,\mathrm{GPa}, yield 250\,\mathrm{MPa}, with a tabulated stress against plastic-strain curve rising from 250 to 400\,\mathrm{MPa} at plastic strain 0.14 (use the von Mises yield criterion).
• Study. Nonlinear static study, end displacement applied in steps up to 6\,\mathrm{mm} on one end face; the other end fixed.
• Mesh. Standard solid mesh with 1\,\mathrm{mm} elements in the parallel section.
• Hand number the result must match. Yield load = \sigma_Y A_0 = 250\times40 = 10.0\,\mathrm{kN}; load at the end of the tabulated curve (entered as engineering stress) = 400\times40 = 16.0\,\mathrm{kN}.
• What to look for. The reaction force against displacement should be linear to about 10\,\mathrm{kN} then flatten; in the stress plot the parallel section should be uniform at 250\,\mathrm{MPa} at the first yield, with higher values in the radius transitions.

(This exercise was written for the lesson and has not been run in SOLIDWORKS here.)

Traps

• Calling the fracture load the strength. Strength is the peak, P_{\max}/A_0; fracture is at a lower load but a higher true stress. (Nugget 03.)
• Dividing by the current area in an engineering curve. Engineering stress uses A_0 always. (Nugget 01.)
• Reading yield off a curve with no plateau without the offset rule. (Nugget 04.)
• Confusing the proportional, elastic and yield limits. They coincide only for mild steel. (Nugget 02.)
• Quoting elongation without the gauge length. The neck makes the percentage depend on it. (Nugget 05.)
• Expecting a stronger steel to be stiffer. E is the same for all steels. (Nugget 05.)
• Unloading along the original curve. Unloading is along a line of slope E, which leaves a permanent set. (Nugget 06.)
• Confusing the two toughnesses. Area under the curve (energy per volume) is not the same as crack resistance. (Nugget 07.)
• Treating a property table as design data. Appendix I values are typical, never for a specific design. (Nugget 05.)

Self-check

1. A standard tension specimen (12.8\,\mathrm{mm} diameter, 50.8\,\mathrm{mm} gauge length) extends 0.0385\,\mathrm{mm} under 20\,\mathrm{kN}. Find the stress, strain and the slope \sigma/\varepsilon.
2. A 10\,\mathrm{mm} specimen with a 50\,\mathrm{mm} gauge length yields at 32\,\mathrm{kN}, carries a maximum of 41\,\mathrm{kN}, and breaks with a minimum diameter of 6\,\mathrm{mm} and a gauge length of 63\,\mathrm{mm}. Find \sigma_Y, \sigma_U, the percent reduction in area and the percent elongation.
3. A material follows \sigma = \frac{70\,000\varepsilon}{1 + 300\varepsilon}\ \mathrm{MPa}. Find the 0.2\,\% offset yield stress.
4. A 1.5\,\mathrm m wire of the copper alloy of Nugget 04 is loaded to 300\,\mathrm{MPa} and released. Find the permanent set.
5. At maximum load the engineering stress is 400\,\mathrm{MPa} and the engineering strain is 0.14. Find the true stress and strain (constant volume assumed).
6. Cast iron has a tensile strength near 200\,\mathrm{MPa} and a compressive strength near 800\,\mathrm{MPa}, with elongation under 1\,\%. Classify it and say whether a short stout support or a hanging tie suits it.
7. Find the modulus of resilience of A36 steel and the energy a 12\,\mathrm{mm} by 1\,\mathrm m rod absorbs before yield.
8. From the data pairs (7.40\,\mathrm{kN}, 18.2\,\mu\mathrm m) and (19.74\,\mathrm{kN}, 52.5\,\mu\mathrm m) on a 100\,\mathrm{mm^2} specimen of gauge length 56\,\mathrm{mm}, estimate E.
9. Explain why a ductile bar breaks along a cone at about 45^\circ and a brittle bar breaks flat.
10. Describe the difference between creep and relaxation, and name which one a bolt in a hot flange suffers.
11. What load can a 10\,\mathrm{mm} round rod of aluminium 6061-T6 carry before yield, using the table value?

Fully worked solutions

1. A_0 = \pi(6.4)^2 = 128.68\,\mathrm{mm^2}; \sigma = 20\,000/128.68 = 155.4\,\mathrm{MPa}; \varepsilon = 0.0385/50.8 = 7.58\times10^{-4}; slope = 155.4/7.58\times10^{-4} = 205\,\mathrm{GPa}.

2. A_0 = 78.54\,\mathrm{mm^2}. \sigma_Y = 32\,000/78.54 = 407.4\,\mathrm{MPa}; \sigma_U = 41\,000/78.54 = 522.0\,\mathrm{MPa}. Reduction in area = (100 - 36)/100 = 64.0\,\% (area scales as d^2). Elongation = (63 - 50)/50 = 26.0\,\%.

3. Offset line \sigma = 70\,000(\varepsilon - 0.002) meets the curve where \varepsilon/(1+300\varepsilon) = \varepsilon - 0.002. This gives 300\varepsilon^2 - 0.6\varepsilon - 0.002 = 0, so \varepsilon = 0.003768 and \sigma_Y = 70\,000(0.003768 - 0.002) = 123.8\,\mathrm{MPa}.

4. \varepsilon = 300/(124\,000 - 72\,000) = 5.769\times10^{-3}; recovery 300/124\,000 = 2.419\times10^{-3}; residual 3.350\times10^{-3}; set = 3.350\times10^{-3}\times1500 = 5.02\,\mathrm{mm}.

5. \sigma_t = 400 \times 1.14 = 456\,\mathrm{MPa}; \varepsilon_t = \ln 1.14 = 0.1310.

6. Ductility is far below the 5\,\% rule of thumb, and it is much stronger in compression than in tension: brittle. Use it in compression (a stout support, with buckling checked), never as a hanging tie.

7. u_R = 250^2/(2\times200\,000) = 0.15625\,\mathrm{MPa} = 0.156\,\mathrm{MJ/m^3}. Volume = \pi/4\times0.012^2\times1 = 1.131\times10^{-4}\,\mathrm{m^3}; energy = 0.15625\times10^{6}\times1.131\times10^{-4} = 17.7\,\mathrm J.

8. Slope = (19.74 - 7.40)\times10^{3}\,\mathrm N/[(52.5 - 18.2)\times10^{-6}\,\mathrm m] = 3.598\times10^{8}\,\mathrm{N/m}. E = \text{slope}\times L_0/A_0 = 3.598\times10^{8}\times0.056/(100\times10^{-6}) = 2.015\times10^{11}\,\mathrm{Pa} = 201.5\,\mathrm{GPa}.

9. The maximum shear stress in a bar in tension is on planes at 45^\circ to the axis; a ductile metal yields by slip, a shear process, so it fails on a cone near 45^\circ. A brittle material has little capacity to slip, so a flaw opens under the normal stress and the crack runs across the bar.

10. Creep: strain grows with time at constant stress. Relaxation: stress falls with time at constant strain. A tightened bolt is held at nearly fixed length, so in a hot flange its tension relaxes.

11. \sigma_Y = 270\,\mathrm{MPa}, A = \pi/4\times10^2 = 78.54\,\mathrm{mm^2}, P = 270\times78.54 = 21.2\,\mathrm{kN}.