Study Companion: MM04 — Hooke's Law and Poisson's Ratio

A steel bar 20\,\mathrm{mm} square and 300\,\mathrm{mm} long is squeezed by a 50\,\mathrm{kN} load. By how much does it shorten, and does it get fatter or thinner? The first answer needs the one relation that links the stress of Lesson MM02 to the strain of Lesson MM02, and the second needs a number that most people have never heard of but that decides whether a pipe leaks, a rubber seal grips or a press fit holds. This lesson is about the straight part of the curve of Lesson MM03: what its slope means, what it predicts and what the material does sideways while it is stretched.

Sources: Gere and Goodno, Mechanics of Materials, 9th edition (SI), section 1.7 (the syllabus for this lesson); Beer, Johnston, DeWolf and Mazurek, 8th edition, section 2.4; Liu, Mechanics of Materials: Formulations and Solutions with Python, section 5.2 (with 5.1 and 5.3, which carry the law itself); Hearn, Mechanics of Materials 1, sections 1.5–1.6 and 1.10–1.11; Venkataraman, Essential Mechanics, section 3.2 (Poisson's ratio itself is in his 3.4.3); Muvdi and Elhouar, section 1.4.4. Translations: Muvdi's Poisson's ratio \mu is our \nu; Gere's lateral strain \varepsilon' is Beer's \varepsilon_y = \varepsilon_z and Hearn's "lateral strain". All working is in SI. 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 bar 20\,\mathrm{mm} square and 300\,\mathrm{mm} long carries a 50\,\mathrm{kN} compression. The plausible answer to "does it get fatter or thinner" is "thinner, since it is being squeezed along its length"; it gets fatter. How much shorter, and how much fatter? The answers are in Nugget 01 and Nugget 03.

Nugget 01: Linear Elasticity: Stress Is Proportional to Strain

Claim. For most engineering materials, over the first part of the stress-strain curve, the stress is proportional to the strain: \sigma = E\varepsilon. The constant E, the slope of the straight line, is the modulus of elasticity (Young's modulus).

The picture. The straight start of the curve of Lesson MM03, with its slope marked, for two materials.

At the same stress the aluminium is strained 200/70 = 2.86 times as much as the steel. E is the stiffness of the material: how much stress a unit of strain costs. The straight part ends at the proportional limit (Lesson MM03); beyond it the law does not apply. Because strain has no unit, E has the unit of stress.

Check with numbers. The puzzle's bar: A = 20 \times 20 = 400\,\mathrm{mm^2}, so \sigma = -50\,000/400 = -125\,\mathrm{MPa} (compression). With E = 200\,\mathrm{GPa} the strain is \varepsilon = -125/200\,000 = -6.25\times10^{-4}, and over 300\,\mathrm{mm} the length changes by \varepsilon L = -6.25\times10^{-4}\times300 = -0.1875\,\mathrm{mm}. The bar is 0.19\,\mathrm{mm} shorter. The stress is half the 250\,\mathrm{MPa} yield of A36 steel, so the law applies.

The idea. Gere (1.7) calls Eq. 1-12 "a very limited version" of Hooke's law, because it covers only one axis of stress; Nugget 06 and Part 6 give the general form. Robert Hooke (1635–1703) measured the stretch of long wires under weights and found stretch proportional to weight; Thomas Young (1773–1829) later introduced the modulus that carries his name. The law holds for stress up to the proportional limit; "linearly elastic" is the working region for design because nothing is permanently deformed there. Hearn notes that strains in this range seldom exceed 0.003, which also justifies treating deformations as small. E is the same in tension and compression (Gere, Hearn). Liu (5.1–5.3) writes the same law as \sigma_{xx} = E\varepsilon_{xx} and then as the general tensor law \sigma_{ij} = C_{ijkl}\varepsilon_{kl} with 81 constants, of which an isotropic material needs only two; E is one, Poisson's ratio will be the other.

The equation (material model: a linear approximation to the curve below the proportional limit):

Typical values of E (Gere, Appendix I): steel 190–210\,\mathrm{GPa}, aluminium alloys 70–79 (about 73), copper alloys 110–120, titanium 100–120, glass 48–83, concrete in compression 17–31, nylon 2.1–3.4, rubber 0.0007–0.004\,\mathrm{GPa}.

Worked number. \sigma = -125\,\mathrm{MPa}, \varepsilon = -6.25\times10^{-4}, \delta = -0.1875\,\mathrm{mm}. If the same load were applied to an aluminium bar of the same size (E = 70\,\mathrm{GPa}), the shortening would be 125/70\,000\times300 = 0.536\,\mathrm{mm}.

Real example (one you can see). Hang a weight from a steel ruler's end and from a plastic ruler's end: the plastic bends far more under the same load. That is the ratio of the moduli, about 200/3. (Bending is a Part 4 matter, but the stiffness ranking is the same.)

Plausible wrong answer. "\sigma = E\varepsilon always holds." It fails above the proportional limit; a steel bar at 400\,\mathrm{MPa} is not at \varepsilon = 400/200\,000.

Check. A bar of E = 200\,\mathrm{GPa} is stressed to 300\,\mathrm{MPa} and the steel yields at 250\,\mathrm{MPa}. Can you find its strain from \sigma/E? (No: 300 > 250, beyond the straight line.)

Narration spine. The straight start of two curves; the steeper one labelled steel; a horizontal line at one stress meets the curves at two strains; E is the ratio.

Bridge. Stress and strain are properties of the material; to answer the practical question, how far does this bar stretch, we need to put the geometry back.

Nugget 02: A Bar Is a Spring: Stiffness and Elongation From E

Claim. Putting \sigma = P/A and \varepsilon = \delta/L into \sigma = E\varepsilon gives \delta = PL/(AE). A bar behaves like a spring of stiffness k = AE/L: E is a property of the material, k a property of the member.

The picture. A bar and a spring side by side.

Check with numbers. For a stepped bar carrying P = 18\,\mathrm{kN}, with E = 200\,\mathrm{GPa}:

Total elongation = 0.1343 + 0.0288 + 0.2785 = 0.4416\,\mathrm{mm}: the elongations of the sections add because they are in series. (Hearn's Example 1.1, with 20\,\mathrm{kN} and his own three sections of 20\,\mathrm{mm} round, 30\,\mathrm{mm} square and 15\,\mathrm{mm} round, with E = 210\,\mathrm{GPa}, gives 75.8 + 10.6 + 215.6 = 302\,\mu\mathrm m by the same steps.)

The idea. Liu (5.2) sets the spring law f = ke beside the bar. The bar's force is f = A\sigma, its elongation e = L\varepsilon, so A\sigma = (AE/L)(L\varepsilon) and k = AE/L. A stiffer bar needs a larger E, a larger area or a shorter length. The step "substitute the definitions into Hooke's law" is the whole derivation; Hearn gives it as Eq. 1.2, E = PL/(A\,\delta L).

The equation (geometric relationship combined with the material model):

Worked number. A 20\,\mathrm{mm} steel rod 1\,\mathrm m long: A = 314.16\,\mathrm{mm^2}, k = 314.16\times10^{-6}\,\mathrm{m^2}\times200\times10^{9}\,\mathrm{Pa}/1\,\mathrm m = 62.83\,\mathrm{MN/m}. The same rod in aluminium: 21.99\,\mathrm{MN/m}. Under 10\,\mathrm{kN} the steel rod stretches 10\,000/62.83\times10^{6} = 0.159\,\mathrm{mm}.

Real example (one you can see). Two cables lift a load: the shorter, thicker one is the stiffer spring and takes more of the load if they share it. Lesson MM08 uses exactly this.

Plausible wrong answer. "A steel bar of any size is stiffer than an aluminium one." E is higher, but k = AE/L also depends on area and length: a thin steel wire is a softer spring than a thick aluminium bar.

Check. A bar has twice the length and half the area of another of the same material. By what factor does the stiffness k change? (Quartered.)

Narration spine. A bar morphs into a spring; P and \delta are marked; the two definitions are substituted into \sigma = E\varepsilon and k appears.

Bridge. The bar stretches, but it also changes in a direction we have not looked at.

Nugget 03: A Stretched Bar Gets Thinner: the Poisson Ratio

Claim. A bar pulled along its axis gets longer and thinner; a bar squeezed gets shorter and fatter. The ratio of lateral to axial strain, with a minus sign, is a material constant \nu, Poisson's ratio.

The picture. A bar before and after being stretched, with both strains marked.

Why does it have to? The material is not hollow and its atoms are not free to ignore their neighbours: pulling them apart along the axis lets them come closer across it. The lateral strain is a consequence of the axial one, not a separate response to a stress on the sides. A tempting but wrong shortcut (Beer) is to say that because the side faces carry no stress, the lateral strain is zero. The stress on the sides is zero; the strain is not.

Check with numbers. Venkataraman's Example 3.8: an aluminium bar 2\,\mathrm m long, 50\,\mathrm{mm}\times60\,\mathrm{mm}, with \nu = 0.32, carries a 120\,\mathrm{kN} tension and lengthens 1.2\,\mathrm{mm}. Axial strain \varepsilon = 1.2/2000 = 6.0\times10^{-4}. Lateral strain \varepsilon' = -0.32\times6.0\times10^{-4} = -1.92\times10^{-4}, the same in both lateral directions. The 50\,\mathrm{mm} side becomes 50(1 - 1.92\times10^{-4}) = 49.9904\,\mathrm{mm} and the 60\,\mathrm{mm} side 59.98848\,\mathrm{mm}. (The stress 120\,000/3000 = 40\,\mathrm{MPa} and strain 6.0\times10^{-4} imply E = 66.7\,\mathrm{GPa}, near aluminium's.) Back to the puzzle: the compressed steel bar has \varepsilon' = -0.3\times(-6.25\times10^{-4}) = +1.875\times10^{-4}, so each 20\,\mathrm{mm} side grows by 20\times1.875\times10^{-4} = 3.75\times10^{-3}\,\mathrm{mm}.

The idea. Gere (Eqs. 1-13, 1-14) and Beer (Eqs. 2.17–2.19) define \nu with the minus sign so that it is positive for ordinary materials. Hearn and Muvdi quote the ratio without the sign; the signs of the two strains are opposite and must be remembered. This book keeps the signed form. The range is a matter of what a real material can do: \nu is typically 0.25–0.35 for metals (steel 0.27–0.30, aluminium 0.33), 0.1–0.2 for concrete, near 0 for cork and near 0.5 for rubber. The upper limit is 0.5: with \nu > 0.5 a bar in tension would lose volume (Nugget 05), and the bulk modulus would be negative; at \nu = 0.5 the material is incompressible, which is why rubber needs special care in finite-element models (Liu). Foams can have negative \nu (they get fatter when stretched): Gere gives the elastic range as -1.0 to +0.5. Poisson's own molecular theory gave \nu = 1/4; better atomic models give 1/3.

When one \nu is enough. The same lateral strain in every lateral direction needs the material to be homogeneous (the same at every point) and the same in all directions perpendicular to the axis. Isotropic means the same in all directions. Wood is homogeneous in the sense of one material but not isotropic: its E along the grain is many times that across it, and so is its \nu. The relation applies to a bar in uniaxial stress only. Beyond the elastic range the ratio changes (for steel it approaches 0.5 at yielding) and is called the contraction ratio.

The equation (definition of \nu; the lateral strain follows from it):

Worked number. The puzzle's bar gets fatter: each 20\,\mathrm{mm} side grows by 3.75\times10^{-3}\,\mathrm{mm}, to 20.00375\,\mathrm{mm}.

Real example (one you can see). Stretch a rubber band: it visibly thins. Squeeze a rubber eraser in a vice: it bulges sideways. In steel the same happens, too little to see.

Plausible wrong answer. "The bar's volume stays constant, so the lateral strain is -\varepsilon/2." The volume does not stay constant unless \nu = 0.5 (Nugget 05); for steel the volume rises with tension.

Check. A bar with \nu = 0.3 is stretched by \varepsilon = 0.001. What is its lateral strain? (-3.0\times10^{-4}.)

Narration spine. The bar is pulled; a dashed outline shows the original; both strains appear with opposite signs, linked by -\nu.

Bridge. Both E and \nu can be measured. Where do the tabulated values come from?

Nugget 04: Two Measurements From One Test Give E and Poisson's Ratio

Claim. One tension test, with a measurement of the lengthening and of the change in diameter, gives both constants: E = \sigma/\varepsilon and \nu = -\varepsilon'/\varepsilon.

The picture. The same bar, now with two gauges.

Check with numbers. Beer's Concept Application 2.7: a rod 500\,\mathrm{mm} long and 16\,\mathrm{mm} in diameter carries 12\,\mathrm{kN}; its length increases by 300\,\mu\mathrm m and its diameter decreases by 2.4\,\mu\mathrm m. A = \pi(8\,\mathrm{mm})^2 = 201.06\,\mathrm{mm^2}, \sigma = 12\,000/201.06 = 59.7\,\mathrm{MPa}. \varepsilon = 300\,\mu\mathrm m/500\,\mathrm{mm} = 600\times10^{-6} and \varepsilon' = -2.4/16\,000 = -150\times10^{-6}. So E = 59.7/600\times10^{-6} = 99.5\,\mathrm{GPa} and \nu = 150/600 = 0.25. Gere's brass specimen (Problem 1.7-8): 10\,\mathrm{mm} diameter, 50\,\mathrm{mm} gauge, 20\,\mathrm{kN}, extends 0.122\,\mathrm{mm}, diameter falls 0.0083\,\mathrm{mm}: \sigma = 254.65\,\mathrm{MPa}, \varepsilon = 2.44\times10^{-3}, E = 104.4\,\mathrm{GPa}, \varepsilon' = -8.3\times10^{-4}, \nu = 0.34.

The idea. Muvdi's Example 1.5 extracts E and \mu (our \nu) from a steel rod at its proportional limit in the same way, giving E \approx 200\,\mathrm{GPa} and 0.286. The reasoning is the definitions read backwards: stress from the load and the original area, strains from the measured changes of length and diameter. A tensile test therefore yields the whole of the linear-elastic character of an isotropic material.

The equation (definitions read backwards):

Worked number. E = 99.5\,\mathrm{GPa}, \nu = 0.25 (Beer's rod); E = 104.4\,\mathrm{GPa}, \nu = 0.34 (Gere's brass).

Real example (one you can see). A strain-gauge rosette on a machine member reads two strains at right angles; the ratio of the readings under a simple pull is \nu. Lesson MM38 uses rosettes.

Plausible wrong answer. "Poisson's ratio can be found from the load and the length change alone." It needs the diameter change as well.

Check. A rod stretches 0.15\,\% and its diameter falls 0.045\,\%. Find \nu. (0.045/0.15 = 0.30.)

Narration spine. A rod in a rig; a dial reads the length, a calliper the diameter; the two ratios are written under the picture and become E and \nu.

Bridge. Once \nu is known, the size change of the whole bar follows: its diameter, its area, its volume.

Nugget 05: Lateral Strain Changes Diameter, Area and Volume

Claim. Axial strain \varepsilon and lateral strain \varepsilon' = -\nu\varepsilon together set every dimension of the loaded bar. The relative volume change is (1 - 2\nu)\varepsilon to first order, which is positive in tension for \nu < 0.5.

The picture. A plastic liner squashed to fill the bore of an iron pipe.

Check with numbers. Gere's Example 1-8 as a clean fit problem. A plastic pipe liner of outer diameter 110.0\,\mathrm{mm} and wall 0.1\,\mathrm{mm} (E_p = 2.1\,\mathrm{GPa}, \nu_p = 0.4) must be shortened until its Poisson swelling just fills a cast-iron bore of 110.2\,\mathrm{mm}, with the iron pipe 0.25\,\mathrm m long. The lateral strain needed is \varepsilon' = (110.2 - 110.0)/110.0 = 1.818\times10^{-3}. The axial strain that produces it is \varepsilon = -\varepsilon'/\nu = -4.545\times10^{-3}. The liner must therefore be L_p long before squeezing, with L_p(1 + \varepsilon) = 0.25\,\mathrm m: L_p = 0.25/(1 - 4.545\times10^{-3}) = 0.25114\,\mathrm m. The stress is \sigma = E_p\varepsilon = 2100\,\mathrm{MPa}\times(-4.545\times10^{-3}) = -9.55\,\mathrm{MPa} and the liner area is A_p = \frac{\pi}{4}(110.0^2 - 109.8^2) = 34.526\,\mathrm{mm^2}, so the force is -9.55\times34.526 = -330\,\mathrm N (compression). The wall thickens by 1.818\times10^{-3}\times0.1\,\mathrm{mm} = 1.8\times10^{-4}\,\mathrm{mm}. The liner's volume falls from 251.14\times34.526 = 8671\,\mathrm{mm^3} to 250\times34.652 = 8663\,\mathrm{mm^3}, a ratio of 0.99908 — it is a compressed plastic, not a volume-conserving one.

Volume of a stretched bar. Take \nu = 0.3 and \varepsilon = 0.001. The exact ratio of new to old volume is (1+\varepsilon)(1+\varepsilon')^2 = 1.001\times(1 - 0.0003)^2 = 1.000399, a change of 3.9949\times10^{-4}; the first-order estimate (1 - 2\nu)\varepsilon = 0.4\times0.001 = 4.0\times10^{-4} is within 0.13\,\% of it.

The idea. Gere's example is the most inventive use of Poisson action: what looks like a sideways effect fixes the axial dimension. For the volume, expand (1+\varepsilon)(1 - \nu\varepsilon)^2 and keep first-order terms: 1 + \varepsilon - 2\nu\varepsilon. A tension must not shrink the volume, so 1 - 2\nu \ge 0 and \nu \le 0.5, which is the bound quoted in Nugget 03. Dimensional changes under normal loads are tiny, but they matter in statically indeterminate structures (Lesson MM08), in fits like this one, and in strain-gauge work.

The equation (geometric relationship, exact and first-order):

Worked number. Liner free length 251.14\,\mathrm{mm}, force -330\,\mathrm N; volume ratio 0.99908 (first-order: (1 - 0.8)(-4.545\times10^{-3}) = -9.09\times10^{-4}, ratio 0.99909, agreeing).

Real example (one you can see). A tight rubber stopper pressed into a bottle neck bulges outward to seal it: axial squeeze produces the lateral push that grips the glass.

Plausible wrong answer. "The volume of a loaded bar does not change." It rises under tension and falls under compression unless \nu = 0.5.

Check. For \nu = 0.5 what is \Delta V/V to first order? (Zero: incompressible.)

Narration spine. A liner is pressed; its wall bulges outward until it touches the bore; the numbers \varepsilon', \varepsilon, L_p roll down the right of the picture.

Bridge. All of this was for a stress on one axis only. Often two stresses act together.

Nugget 06: Two Stresses at Once: Each Adds Its Own Strain and Its Own Contraction

Claim. If a plate has normal stresses \sigma_x and \sigma_y in two perpendicular directions, the strain along x is the sum of the strain \sigma_x/E it would have alone and the Poisson contraction -\nu\sigma_y/E caused by the other stress; likewise for y.

The picture. A small element with stresses on two pairs of faces.

Each stress acts independently on a linearly elastic body and the effects are added. This is the principle of superposition; its two conditions are a linear stress-strain relation and small deformations. A compressive stress enters each formula with its own minus sign.

Check with numbers. \sigma_x = 120\,\mathrm{MPa}, \sigma_y = -40\,\mathrm{MPa}, E = 200\,\mathrm{GPa}, \nu = 0.3. The stress \sigma_x alone gives \varepsilon_x = 120/200\,000 = 6.0\times10^{-4} and \varepsilon_y = -0.3 \times 6.0\times10^{-4} = -1.8\times10^{-4}. The stress \sigma_y alone gives \varepsilon_y = -40/200\,000 = -2.0\times10^{-4} and \varepsilon_x = -0.3\times(-2.0\times10^{-4}) = +0.6\times10^{-4}. Sum: \varepsilon_x = 6.6\times10^{-4} and \varepsilon_y = -3.8\times10^{-4}. The compressive \sigma_y increases the x strain, by its Poisson swelling.

The idea. Hearn (1.11, Eqs. 1.5–1.6) builds the result case by case and Beer (2.5) by superposition; both give the same two lines. In matrix form the result is the compliance relation \{\varepsilon\} = \frac{1}{E}\begin{bmatrix}1 & -\nu\\ -\nu & 1\end{bmatrix}\{\sigma\}, which the MATLAB view computes. Liu's tensor law (5.3) is the full three-dimensional version of the same thing; Lessons MM35–MM37 develop the plane and three-dimensional forms. Hearn's footnote warns that the one-axis \sigma = E\varepsilon is not safe when two stresses act together.

The equation (material model, plane stress, superposition):

Worked number. \varepsilon_x = 6.6\times10^{-4}, \varepsilon_y = -3.8\times10^{-4}. With \sigma_y removed, the one-axis answer is \varepsilon_x = 6.0\times10^{-4}: neglecting the other stress would be 10\,\% wrong in \varepsilon_x and would miss the y strain's size entirely.

Real example (one you can see). A thin-walled balloon or pipe under pressure is stressed along two perpendicular directions at once; its change of diameter reflects both. Lesson MM39 returns to it.

Plausible wrong answer. "The strain in x depends only on \sigma_x." It also depends on \sigma_y through \nu.

Check. A plate has \sigma_x = \sigma_y = 100\,\mathrm{MPa} (tension), E = 200\,\mathrm{GPa}, \nu = 0.3. Find \varepsilon_x. ((100 - 30)/200\,000 = 3.5\times10^{-4}.)

Narration spine. An element gets stress on its x faces, then separately on its y faces; the two sets of strains are drawn and added.

Synthesis

What no single nugget shows. Hooke's law is not a theorem but a fit to the first part of the curve; it takes two constants, E and \nu, to describe how an isotropic elastic solid responds to any stress state, and the whole of linear strength-of-materials is that pair of constants plus equilibrium plus geometry. E measures the response along the stress, \nu the response across it.

• Central idea. In the elastic range, strain is proportional to stress, with E along the axis and -\nu E across it.
• Key relationship. \sigma = E\varepsilon, \delta = PL/(AE) and \varepsilon' = -\nu\varepsilon; with two stresses, each adds its own strain and its own Poisson contraction.
• Main assumption. Stress below the proportional limit, a homogeneous isotropic material and small deformations.
• What to explore next. The same law for shear (Lesson MM05, with G), and why E, G and \nu are related (Lesson MM16).

Computed views

Python (numpy). MM04_hooke_poisson.py computes the strains and elongations above, the two constants from the two tests, the volume change and the liner.

20 mm square, 50 kN compression: -125 MPa, -6.250e-04
  shortening over 300 mm: -0.1875 mm
stepped bar stresses (MPa): [ 89.52  28.8  159.15]
stepped bar elongations (mm): [0.1343 0.0288 0.2785] total 0.4416
stiffness k = AE/L, 20 mm steel rod 1 m: 62.83 MN/m
  same rod in aluminium: 21.99 MN/m
aluminium bar: 40 MPa, E = 66.7 GPa
  lateral strain -1.920e-04, h' = 49.9904 mm
  w' = 59.98848 mm
test rod: sigma 59.7 MPa, E 99.5 GPa, nu 0.25
brass test: E 104.4 GPa, nu 0.34
volume change at strain 0.001, nu 0.3: 3.9949e-04 exact, 4.0000e-04
liner: lateral 1.818e-03, axial -4.545e-03, free length 251.14 mm
  stress -9.55 MPa, force -330 N
plane stress 120/-40 MPa: eps_x 6.600e-04, eps_y -3.800e-04

MATLAB (Octave-compatible). MM04_hooke_poisson.m tabulates the strains of eight materials under the same 10\,\mathrm{MPa} and computes the plane-stress strains as a matrix product.

material            E (GPa)     nu     axial (ue)   lateral (ue)
steel                 200.0   0.30           50.0          -15.0
aluminium 6061-T6      70.0   0.33          142.9          -47.1
brass                 100.0   0.34          100.0          -34.0
copper                117.0   0.34           85.5          -29.1
titanium              110.0   0.33           90.9          -30.0
cast iron             100.0   0.25          100.0          -25.0
concrete               25.0   0.15          400.0          -60.0
nylon                   2.8   0.40         3571.4        -1428.6
plane stress 120/-40 MPa: eps_x 6.600e-04, eps_y -3.800e-04
stiffness of a 20 mm rod, 1 m: steel 62.83 MN/m, aluminium 21.99 MN/m

The moduli and ratios are typical mid-range values from Gere's Appendix I, not design data.

SOLIDWORKS Simulation exercise (our own; lateral contraction of a bar). Read the Poisson contraction off the displacement plot.

• Geometry. Steel bar 20\,\mathrm{mm}\times20\,\mathrm{mm}\times300\,\mathrm{mm}; model one eighth of it: a 10\,\mathrm{mm}\times10\,\mathrm{mm} section 150\,\mathrm{mm} long, cut by the three planes of symmetry.
• Material. E = 200\,\mathrm{GPa}, \nu = 0.3.
• Restraints. Symmetry (roller) restraints on the three cut faces: the faces through the axis prevent motion normal to them, the mid-length face prevents axial motion.
• Load. 30\,\mathrm{kN}/4 = 7.5\,\mathrm{kN} tensile on the free end face (one quarter of the section).
• Mesh. Standard solid mesh, element size 3\,\mathrm{mm}.
• Hand numbers the result must match. \sigma = 30\,000/400 = 75\,\mathrm{MPa}, \varepsilon = 3.75\times10^{-4}. Axial displacement of the end face = \varepsilon\times150 = 0.05625\,\mathrm{mm}. Lateral displacement of the outer side face = -\nu\varepsilon\times10 = -1.125\times10^{-3}\,\mathrm{mm}.
• What to look for. Plot the displacement in the axial direction (UX or UZ) and in a lateral direction. The end face should read 0.0563\,\mathrm{mm} and the outer side faces should have moved inward by 0.0011\,\mathrm{mm} (a number too small to see without scaling the plot). Change \nu to 0.0: the lateral displacement should vanish.

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

Traps

• Using \sigma = E\varepsilon above the proportional limit. The law is the straight part of the curve only. (Nugget 01.)
• Confusing E with stiffness of the member. E is the material; k = AE/L is the member. (Nugget 02.)
• Mixing units in \delta = PL/AE. Use N, mm and MPa together, or N, m and Pa, never a blend. (Nugget 02.)
• Saying the lateral strain is zero because the side stresses are zero. The side stress is zero; the lateral strain is -\nu\varepsilon. (Nugget 03.)
• Dropping the minus sign in \nu. The two strains have opposite signs; \nu \ge 0 for ordinary materials. (Nugget 03.)
• Applying one \nu to wood or a composite. It needs a homogeneous isotropic material. (Nugget 03.)
• Assuming the volume is constant. Only for \nu = 0.5. (Nugget 05.)
• Finding \nu without the diameter change. (Nugget 04.)
• Using the one-axis law when two stresses act. (Nugget 06.)

Self-check

1. A steel bar (E = 200\,\mathrm{GPa}, \nu = 0.3) is 40\,\mathrm{mm} in diameter. Under compression its diameter may increase by at most 0.012\,\mathrm{mm}. Find the largest compressive load.
2. An aluminium tube (E = 73\,\mathrm{GPa}, \nu = 0.33) carries a compressive stress of 57\,\mathrm{MPa}. Find the axial and lateral strains.
3. A steel bar (E = 200\,\mathrm{GPa}, \nu = 0.3) 2\,\mathrm m long and 12\,\mathrm{mm} in diameter lengthens 1.5\,\mathrm{mm}. Find the load and the change in diameter.
4. A steel bolt (E = 200\,\mathrm{GPa}, \nu = 0.3) of diameter 60\,\mathrm{mm} shrinks in diameter by 13\,\mu\mathrm m under a tensile load. Find the load.
5. A steel rod 12\,\mathrm{mm} in diameter and 2\,\mathrm m long carries 5\,\mathrm{kN}. Find its stiffness and its elongation.
6. A bar has \varepsilon = 2.0\times10^{-3}. Find the relative volume change to first order for \nu = 0.3, 0.4 and 0.5.
7. A plate (E = 70\,\mathrm{GPa}, \nu = 0.33) has tensile stresses 90\,\mathrm{MPa} in x and 60\,\mathrm{MPa} in y. Find \varepsilon_x and \varepsilon_y.
8. Explain, with a sketch, why a bar with \nu > 0.5 would lose volume in tension.
9. Why is a wooden pole homogeneous but not isotropic? Which of the formulas of this lesson fail for it?
10. A nylon bar (E = 2.8\,\mathrm{GPa}, \nu = 0.4) of diameter 50.0\,\mathrm{mm} must be squeezed axially until it fills a bore of 50.025\,\mathrm{mm}. Find the axial strain, the stress and the force.
11. A steel with yield 250\,\mathrm{MPa} is stressed to 300\,\mathrm{MPa}. Can its strain be found from \sigma/E?

Fully worked solutions

1. Lateral strain \varepsilon' = 0.012/40 = 3.0\times10^{-4}. Axial strain \varepsilon = -\varepsilon'/\nu = -1.0\times10^{-3} (compression). \sigma = E\varepsilon = -200\,\mathrm{MPa}. Area = \pi/4\times40^2 = 1256.6\,\mathrm{mm^2}, so P = 200\times1256.6 = 251\,300\,\mathrm N = 251\,\mathrm{kN}.

2. \varepsilon = -57/73\,000 = -7.81\times10^{-4}. \varepsilon' = -\nu\varepsilon = +0.33\times7.81\times10^{-4} = +2.58\times10^{-4}.

3. \varepsilon = 1.5/2000 = 7.5\times10^{-4}; \sigma = 200\,000\times7.5\times10^{-4} = 150\,\mathrm{MPa}; A = \pi/4\times144 = 113.1\,\mathrm{mm^2}; P = 150\times113.1 = 16.96\,\mathrm{kN}. \Delta d = -\nu\varepsilon d = -0.3\times7.5\times10^{-4}\times12 = -2.7\times10^{-3}\,\mathrm{mm}.

4. \varepsilon' = -13\times10^{-3}/60 = -2.167\times10^{-4}; \varepsilon = 2.167\times10^{-4}/0.3 = 7.22\times10^{-4}; \sigma = 200\,000\times7.22\times10^{-4} = 144.4\,\mathrm{MPa}; A = \pi/4\times3600 = 2827\,\mathrm{mm^2}; P = 144.4\times2827 = 408\,\mathrm{kN}.

5. A = 113.1\,\mathrm{mm^2}; k = AE/L = 113.1\times10^{-6}\times200\times10^{9}/2 = 11.31\times10^{6}\,\mathrm{N/m} = 11.31\,\mathrm{MN/m}. \delta = P/k = 5000/11.31\times10^{6} = 4.42\times10^{-4}\,\mathrm m = 0.442\,\mathrm{mm}.

6. \Delta V/V = (1 - 2\nu)\varepsilon: \nu = 0.3: 0.4\times2.0\times10^{-3} = 8.0\times10^{-4}; \nu = 0.4: 0.2\times2.0\times10^{-3} = 4.0\times10^{-4}; \nu = 0.5: 0.

7. \varepsilon_x = (90 - 0.33\times60)/70\,000 = 70.2/70\,000 = 1.003\times10^{-3}; \varepsilon_y = (60 - 0.33\times90)/70\,000 = 30.3/70\,000 = 4.33\times10^{-4}.

8. \Delta V/V \approx (1 - 2\nu)\varepsilon. For \nu > 0.5 the factor is negative, so a positive \varepsilon (tension) gives a negative \Delta V: the cross-section would shrink more than the length grew. Sketch: a bar stretched by \varepsilon with sides contracting by 0.6\varepsilon each.

9. A single wood species is the same material everywhere along the pole (homogeneous), but its properties along the grain are very different from those across it (anisotropic, specifically orthotropic). \sigma = E\varepsilon with one E fails (there are different Es), and \varepsilon' = -\nu\varepsilon fails because the two lateral directions are not equivalent.

10. Lateral strain needed = 0.025/50.0 = 5.0\times10^{-4}; axial strain = -5.0\times10^{-4}/0.4 = -1.25\times10^{-3}; \sigma = 2800\times(-1.25\times10^{-3}) = -3.5\,\mathrm{MPa}; A = \pi/4\times2500 = 1963.5\,\mathrm{mm^2}; P = -3.5\times1963.5 = -6.87\,\mathrm{kN} (compression).

11. No. 300\,\mathrm{MPa} is above the 250\,\mathrm{MPa} yield (and above the proportional limit), so the stress is outside the straight part of the curve and \varepsilon \ne \sigma/E; the strain must be read from the actual curve.