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Concept 02 / 07Mechanical Properties of Solids

Hooke's Law and the Stress-Strain Curve

Within the proportional limit stress is proportional to strain, and the full stress-strain curve maps the elastic limit, yield point, ultimate tensile strength and fracture that separate ductile, brittle and elastomeric materials.

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Figure 2
Robert Hooke stated the law of elasticity in 1676 as an anagram, ut tensio sic vis: as the extension, so the force. In modern language: within the proportional limit, stress is directly proportional to strain. The constant of proportionality is called a modulus of elasticity, stress=(modulus)×strain\text{stress} = (\text{modulus})\times\text{strain} and because strain has no units, every modulus has the units of stress, namely pascal. Hooke law is not a fundamental law of nature; it is the first term of a Taylor expansion, and knowing that tells you exactly when it fails. Why the law is linear at small strain, and why it must fail later. Let U(r)U(r) be the interatomic potential energy, with a minimum at r=r0r = r_0. Expand about the minimum: U(r)=U(r0)+12U′′(r0)(r−r0)2+16U′′′(r0)(r−r0)3+…U(r) = U(r_0) + \dfrac{1}{2}U''(r_0)(r-r_0)^2 + \dfrac{1}{6}U'''(r_0)(r-r_0)^3 + \dots, where the linear term vanishes because U′(r0)=0U'(r_0) = 0 at a minimum. The interatomic force is F=−dUdr=−U′′(r0)(r−r0)−…F = -\dfrac{dU}{dr} = -U''(r_0)(r-r_0) - \dots, which is linear in the displacement provided the displacement is small enough that the cubic and higher terms are negligible. So Hooke law is the harmonic approximation to the bond, and k=U′′(r0)k = U''(r_0). Stretch the bond far enough and the anharmonic terms take over: the curve bends, the material softens, and eventually bonds begin to break or planes of atoms begin to slide past one another (dislocation glide), which is plastic flow. Incidentally, the same anharmonicity is what causes thermal expansion, so elasticity and thermal expansion are two faces of one shape. The stress-strain curve. The first graph above shows the standard curve for a ductile metal such as mild steel, obtained by pulling a specimen slowly and plotting load divided by original area against extension divided by original length. Its landmarks, in order: O to A is straight. Hooke law holds exactly, and the slope of this line is the Young modulus. A is the proportional limit. A to B is slightly curved but still fully recoverable: remove the load anywhere before B and the specimen returns to its original length along the same path. B is the elastic limit. For many metals A and B are so close that they are treated as one point. C is the yield point, with yield stress σy\sigma_y. Past C the specimen deforms plastically: unload it now and it follows a line parallel to OA back to a nonzero strain, the permanent set. Many mild steels show a small drop and a flat plateau here (upper and lower yield points) as dislocations break free of their pinning atmospheres. C to D is the strain hardening region. Plastic flow multiplies and tangles dislocations, which makes further flow harder, so more stress is needed for more strain. D is the ultimate tensile strength (UTS), the largest engineering stress the specimen withstands. D to E: a neck forms, the cross-section at the neck shrinks rapidly, and the load that the specimen can carry falls even though the true stress in the neck is still climbing. E is the fracture point. The falling portion is therefore an artefact of dividing by the original area; a true-stress curve rises monotonically to fracture. Ductile, brittle and elastomeric materials. The second graph above compares the three. A ductile material (copper, mild steel, gold) has a long plastic region, so its fracture strain is large and it can be drawn into wires or beaten into sheets. A brittle material (glass, cast iron, ceramics, chalk) fractures at or barely past the elastic limit, with almost no plastic region. An elastomer (rubber, elastic tissue) is the odd one out: it has no straight portion at all, so it never obeys Hooke law, yet it is highly elastic because it returns to its original length after strains of several hundred percent. Elastic does not mean Hookean; these are two independent ideas. Elastic hysteresis. For rubber the unloading curve lies below the loading curve, so the two enclose a loop. The area under the loading curve is the work done per unit volume on the material; the area under the unloading curve is the work recovered; the enclosed area is therefore the energy dissipated as heat per unit volume per cycle. This is why car tyres are built from rubber compounds with small hysteresis loops (little heating) while shock absorbing pads use rubber with large loops (lots of dissipation). Energy names worth knowing. Resilience is the elastic energy stored per unit volume up to the elastic limit, and the modulus of resilience is σe22Y\dfrac{\sigma_e^2}{2Y}. Toughness is the total area under the stress-strain curve up to fracture, the work per unit volume needed to break the material. Strength is the stress at failure. Stiffness is the slope YY. These four are independent: glass is strong and stiff but not tough; copper is tough but neither very strong nor very stiff. Other real-material effects. Elastic after-effect: some materials do not return instantly, but creep back over time (glass is bad, quartz and silver are almost perfect). Elastic fatigue: repeated cyclic loading below the elastic limit gradually degrades a material, which is why bridges and aircraft parts are retired on a schedule. Creep: slow flow under sustained load at high temperature. Working stress: engineers design for a stress well below the yield stress, the ratio of ultimate stress to working stress being the factor of safety, typically 22 to 1010. Key Takeaways and Formulae: Hooke law: stress∝strain\text{stress} \propto \text{strain} within the proportional limit; the constant is a modulus with units of pascal. Curve landmarks in order: proportional limit, elastic limit, yield point, strain hardening region, ultimate tensile strength, necking, fracture. Unloading from a plastic point follows a line parallel to the original elastic line, leaving a permanent set. The Young modulus does not change; the yield stress increases (work hardening). The engineering stress falls after D only because the original area is used in the denominator. Energy per unit volume =∫σ dϵ== \int \sigma\, d\epsilon = area under the stress-strain curve. Up to the elastic limit this is the resilience; up to fracture it is the toughness. Elastic means it recovers; Hookean means stress is proportional to strain. Rubber is the first but not the second. Brain-Teaser 1: Material P is linear right up to fracture, with YP=200Y_P = 200 GPa and fracture stress 500500 MPa. Material Q has YQ=70Y_Q = 70 GPa, yields at 100100 MPa, then flows with the stress climbing slowly to 200200 MPa at a fracture strain of 0.250.25. Which material is stiffer, which is stronger, and which is tougher? Quantify. Solution: Stiffness is the slope, so P is stiffer by the factor 20070≈2.9\dfrac{200}{70} \approx 2.9. Strength is the failure stress, so P is stronger, 500500 MPa against 200200 MPa. Toughness is the area under the curve. For P the fracture strain is ϵP=500×106200×109=2.5×10−3\epsilon_P = \dfrac{500\times10^6}{200\times10^9} = 2.5\times10^{-3}, and since the curve is a straight line the area is 12(500×106)(2.5×10−3)=6.25×105\dfrac{1}{2}(500\times10^6)(2.5\times10^{-3}) = 6.25\times10^5 J per cubic metre. For Q, the elastic part contributes 12(100×106)(100×10670×109)≈7×104\dfrac{1}{2}(100\times10^6)\left(\dfrac{100\times10^6}{70\times10^9}\right) \approx 7\times10^4 J per cubic metre, which is negligible; the plastic part contributes roughly the mean stress times the plastic strain, (150×106)(0.25−0.0014)≈3.7×107(150\times10^6)(0.25 - 0.0014) \approx 3.7\times10^7 J per cubic metre. So Q is about 6060 times tougher than P even though it is weaker and floppier. The moral, which JEE tests as a one-line conceptual question: strong, stiff and tough are three different properties, and a material can be excellent at one and hopeless at another. Aircraft windows are not made of the stiffest available material for exactly this reason. Brain-Teaser 2: A steel wire and a rubber band of identical dimensions are pulled with the same small force; the rubber stretches enormously more. A student concludes that rubber is more elastic. Where is the error? Also, why does a rubber band become noticeably warm if you stretch and release it rapidly many times? Solution: In physics, elasticity is measured by how strongly a body resists deformation and returns, quantified by the modulus: the larger the modulus, the more elastic the material. Steel has Y≈2×1011Y \approx 2\times10^{11} Pa and rubber only 10610^6 to 10710^7 Pa, so steel is 10410^4 to 10510^5 times more elastic. The student has confused elasticity with deformability; a large deformation for a given stress means a small modulus, that is, less elastic. For the heating: rubber shows elastic hysteresis, so the loading and unloading paths in the stress-strain plane do not coincide. Per cycle, the work you do on the band exceeds the work it returns by the area of the hysteresis loop, and that difference appears as internal energy. Cycle it rapidly and the heat has no time to escape, so the temperature rises. You can feel this by pressing a freshly stretched band to your lip. Note also the striking converse experiment: a stretched rubber band held under constant load contracts when heated, because rubber elasticity is entropic rather than energetic in origin. Brain-Teaser 3: A copper wire is loaded just past its yield point and then completely unloaded. (a) What does the unloading path look like? (b) Has its Young modulus changed? (c) If it is loaded again, at what stress will it now start to yield, and what has been lost in the bargain? Solution: (a) The unloading path is a straight line parallel to the original elastic line OA, ending on the strain axis at a nonzero strain, the permanent set. (b) The Young modulus is unchanged. YY is set by the curvature of the interatomic potential, Y≈U′′(r0)r0Y \approx \dfrac{U''(r_0)}{r_0}, and plastic flow only rearranges dislocations; it does not alter the bond stiffness. This is why the unloading line is parallel to the loading line, a fact you can quote directly in an exam. (c) On reloading, the wire behaves elastically all the way up to the stress from which it was unloaded, which is above the original yield stress. The material has been work hardened or strain hardened, and this is precisely how cold-drawn wires and cold-rolled sheets are strengthened commercially. What is lost is ductility: the remaining plastic strain available before fracture has been used up, so the hardened wire is stronger but more brittle, and it will snap with less warning.

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