Introduction
A rigid body is an idealisation. In reality, every solid changes its shape or size, however slightly, when forces are applied to it. A steel wire lengthens when a load is hung from it, a rubber band stretches, and a spring compresses. This ability of solids to resist deformation and recover their original form when the applied force is removed is called elasticity. In this chapter we study the relation between force and deformation, stress and strain, elastic moduli, and the energy stored in deformed bodies.
8.1 Hooke's law and the force constant
Deforming force and restoring force
A deforming force is an external force that changes the length, volume or shape of a body. When a body is deformed, its atoms or molecules are displaced from their equilibrium positions, and the body develops internal restoring forces that oppose the deformation.
Statement of Hooke's law
Robert Hooke (1678) found experimentally that, for small deformations, the deformation produced is directly proportional to the deforming force.
Within the elastic limit, the extension (or compression) produced in a body is directly proportional to the deforming force applied.
For a spring or a wire, if a force \(F\) produces an extension \(x\):
Force constant
The force constant is the force required per unit extension:
SI unit: N m⁻¹. Dimensional formula: [M L⁰ T⁻²]. A large \(k\) means a stiff spring, which needs a large force to produce a small extension.
Force constant of a wire
A wire of natural length \(L\) and cross-sectional area \(A\), made of a material of Young modulus \(Y\) (section 8.3), obeys \(Y = \dfrac{F/A}{\Delta L/L}\). Rearranging gives \(F = \dfrac{YA}{L}\,\Delta L\), and comparing with \(F = kx\) (with \(x = \Delta L\)):
The force constant of a wire is directly proportional to its cross-sectional area and inversely proportional to its length. A thicker wire is stiffer; a longer wire is softer.
Combination of springs
Series. Each spring carries the same force \(F\), so the extensions add:
\[ \frac{F}{k_s} = \frac{F}{k_1} + \frac{F}{k_2} \]
\[ \frac{1}{k_s} = \frac{1}{k_1} + \frac{1}{k_2}, \quad k_s = \frac{k_1k_2}{k_1 + k_2} \]
Parallel. Both springs have the same extension \(x\), so the forces add: \(F = k_1x + k_2x\).
\[ k_p = k_1 + k_2 \]
Cutting a spring. The force constant of a spring is inversely proportional to its length. If a spring of force constant \(k\) is cut into \(n\) equal parts, each part has force constant \(nk\). (Check: the \(n\) parts joined in series give back \(\dfrac{nk}{n} = k\).) So each half has \(2k\), and each third has \(3k\).
Limits of Hooke's law
Hooke's law is not a universal law of nature but an approximation valid for small deformations. Beyond the proportional limit, the extension is no longer proportional to the force. If the material is loaded beyond the elastic limit, it does not return fully to its original shape when the force is removed.
The force constant \(k\) is not the same as the Young modulus \(Y\). \(Y\) is a property of the material only; \(k\) also depends on the length and cross-section of the particular specimen (\(k = YA/L\) for a wire).
- Spring balances measure weight because the extension of the spring is proportional to the load.
- Vehicle suspension springs and shock absorbers are chosen with a suitable \(k\) for the load they must carry.
- Mechanical watches, weighing scales and force gauges all rely on a linear elastic response.
A force of 10 N stretches a spring by 5 cm. Find (a) the force constant, (b) the force needed to stretch the same spring by 8 cm.
Show solution
(a) \(k = F/x = 10/0.05 = 200\ \text{N m}^{-1}\)
(b) \(F = kx = 200 \times 0.08 = 16\ \text{N}\)
Answer: \(k = 200\ \text{N m}^{-1}\); \(F = 16\ \text{N}\).
Two springs of force constants 200 N m⁻¹ and 300 N m⁻¹ are joined (a) in parallel, (b) in series. Find the equivalent force constant in each case and the extension produced by a force of 30 N.
Show solution
(a) Parallel: \(k_p = 200 + 300 = 500\ \text{N m}^{-1}\); \(x = 30/500 = 0.06\ \text{m} = 6\ \text{cm}\).
(b) Series: \(k_s = \dfrac{200 \times 300}{200 + 300} = 120\ \text{N m}^{-1}\); \(x = 30/120 = 0.25\ \text{m} = 25\ \text{cm}\).
Answer: \(k_p = 500\ \text{N m}^{-1}\), \(x = 6\) cm; \(k_s = 120\ \text{N m}^{-1}\), \(x = 25\) cm.
A spring of force constant 600 N m⁻¹ is cut into three equal parts. Two of the parts are connected in parallel and a mass of 6 kg is hung from the combination. Find the extension.
Show solution
Each part: \(k' = nk = 3 \times 600 = 1800\ \text{N m}^{-1}\).
Two parts in parallel: \(k_p = 1800 + 1800 = 3600\ \text{N m}^{-1}\).
Load: \(F = mg = 6 \times 9.8 = 58.8\ \text{N}\).
\(x = F/k_p = 58.8/3600 = 0.0163\ \text{m}\)
Answer: \(k' = 1800\ \text{N m}^{-1}\); \(x \approx 1.63\ \text{cm}\).
8.2 Stress, strain, elasticity and plasticity
Elasticity and plasticity
Elasticity is the property of a material by which it tends to regain its original size and shape when the deforming force is removed. A body that completely regains its original form is perfectly elastic.
Plasticity is the property by which a material does not regain its original size and shape when the deforming force is removed, but keeps the deformation wholly or partly. Such a body is called plastic.
Stress
Stress is the restoring force developed per unit area of cross-section of a deformed body. At equilibrium it equals the applied force divided by the area.
SI unit: N m⁻² or pascal (Pa). Dimensional formula: [M L⁻¹ T⁻²]. Although stress is force per unit area, it is a tensor quantity, not a vector.
- Longitudinal (normal) stress: the force acts perpendicular to the area. It is tensile if it tends to lengthen the body and compressive if it tends to shorten it.
- Volume (bulk) stress: a normal force acts uniformly over the whole surface, as when a body is immersed in a fluid. The volume stress equals the increase in pressure \(\Delta P\).
- Shear (tangential) stress: the force acts tangentially to the surface: shear stress \(= F_t/A\).
Strain
Strain is the fractional deformation produced in a body under stress. It is a ratio of two like quantities, so it has no unit and no dimensions.
- Longitudinal strain \(= \dfrac{\Delta L}{L}\)
- Volume strain \(= \dfrac{\Delta V}{V}\)
- Shear strain \(= \theta \approx \tan\theta = \dfrac{\Delta x}{h}\), where \(\Delta x\) is the relative displacement of two faces a distance \(h\) apart.
Stress–strain curve
When a wire is loaded gradually and the stress is plotted against the resulting strain, a characteristic curve is obtained.
- O to A (proportional region): stress is proportional to strain and Hooke's law holds. The slope is the Young modulus. A is the proportional limit.
- A to B: the wire is still elastic (returns to its original length when unloaded), but stress is no longer proportional to strain. B is the elastic limit: the maximum stress up to which the material regains its original form.
- B to C: permanent (plastic) deformation sets in. At the yield point C, the strain increases rapidly with little or no increase in stress.
- C to D: strain hardening: more stress is needed for further strain. D is the ultimate tensile strength, the maximum stress the material can withstand.
- D to E: the wire narrows locally (necking), the stress falls, and the wire breaks at E. The stress at which it breaks is the breaking stress.
For many materials the proportional limit and the elastic limit are very close and are often not distinguished at school level. Strictly, they are different points.
Ductile and brittle materials; related terms
- Ductile materials (copper, mild steel, aluminium) have a large plastic region between the elastic limit and the breaking point, and can be drawn into wires.
- Brittle materials (glass, cast iron, ceramics) break soon after the elastic limit with very little plastic deformation.
- Elastic hysteresis: when a material is loaded and then unloaded, the loading and unloading curves may not coincide (very pronounced in rubber). The enclosed area is the energy lost as heat.
- Elastic fatigue: the loss of strength of a material after repeated cycles of stress. Bridges, aircraft parts and machine components can fail by fatigue.
- Factor of safety: the ratio of the breaking stress to the maximum working stress allowed in a design. Structures are built with a factor of safety greater than 1.
- A material that stretches more is not more elastic. Steel is more elastic than rubber in the physics sense: for the same stress, steel shows a much smaller strain and regains its shape more completely.
- Stress is not the same as pressure, although both are measured in N m⁻². Pressure is a normal force per unit area exerted by a fluid; stress is an internal restoring force per unit area within a solid.
- Strain is dimensionless. Never attach a unit to it.
- Crane ropes are made by twisting many thin wires together. Thin wires are more flexible, and a thin wire of a given metal has a higher breaking stress than a thick rod.
- Bridge beams and building girders have an I-shaped cross-section to resist bending with minimum material.
- Engineers apply a factor of safety when designing lifts, cables and pressure vessels.
A steel wire of cross-sectional area 2 mm² and length 2 m is stretched by a force of 100 N and its length increases by 0.5 mm. Find the tensile stress, the longitudinal strain and the Young modulus.
Show solution
\(A = 2 \times 10^{-6}\ \text{m}^2\), \(\Delta L = 5 \times 10^{-4}\ \text{m}\).
Stress \(= F/A = 100/(2 \times 10^{-6}) = 5 \times 10^{7}\ \text{N m}^{-2}\)
Strain \(= \Delta L/L = 5 \times 10^{-4}/2 = 2.5 \times 10^{-4}\)
\(Y = \text{stress}/\text{strain} = 5 \times 10^{7}/2.5 \times 10^{-4} = 2 \times 10^{11}\ \text{Pa}\)
Answer: stress \(= 5 \times 10^7\) Pa; strain \(= 2.5 \times 10^{-4}\); \(Y = 2 \times 10^{11}\) Pa.
8.3 Elastic moduli: Young, bulk and shear
Modulus of elasticity
Within the limit of proportionality, Hooke's law can be restated: stress is proportional to strain. The constant of proportionality is the modulus of elasticity.
Young modulus (Y)
The Young modulus is the ratio of longitudinal (tensile or compressive) stress to longitudinal strain.
Physical meaning: \(Y\) measures the stiffness of a material against stretching or compression. A larger \(Y\) means a larger stress is needed for a given strain. \(Y\) is defined for solids only.
Extension of a rod under its own weight
Assumptions: a uniform rod of length \(L\), cross-section \(A\), density \(\rho\) and Young modulus \(Y\) hangs vertically from a rigid support; strains are small.
Take a small element of length \(dy\) at a distance \(y\) from the lower (free) end. The tension there supports the weight of the rod below it: \(T(y) = \rho A y g\). The extension of the element is
Integrating over the whole rod:
This is half the extension that would result if the whole weight \(Mg\) hung from the free end, because the tension falls uniformly from \(Mg\) at the top to zero at the bottom.
Bulk modulus (K)
When a uniform pressure change \(\Delta P\) acts over the whole surface of a body and its volume changes by \(\Delta V\), the bulk modulus is the ratio of volume stress to volume strain.
Physical meaning: \(K\) measures resistance to uniform compression. It is defined for solids, liquids and gases. Solids and liquids have large \(K\) (nearly incompressible); gases have small \(K\).
Shear modulus (η), or modulus of rigidity
When a tangential force \(F\) acts on the top face (area \(A\)) of a body whose base is fixed, the body is sheared through an angle \(\theta\). The shear modulus is the ratio of shear stress to shear strain.
Physical meaning: \(\eta\) measures resistance to a change of shape at constant volume. It applies only to solids, since liquids and gases cannot sustain a shear stress.
Approximate values
| Material | Y (Pa) | K (Pa) | η (Pa) |
|---|---|---|---|
| Steel | 2.0 × 10¹¹ | 1.6 × 10¹¹ | 8 × 10¹⁰ |
| Copper | 1.1 × 10¹¹ | – | 4 × 10¹⁰ |
| Aluminium | 7 × 10¹⁰ | – | 2.5 × 10¹⁰ |
| Water | – | 2.2 × 10⁹ | 0 |
| Rubber | 10⁶ to 10⁷ | – | – |
These are typical values; exact figures vary slightly between reference books. In examination problems, use the values given in the question.
- The Young, bulk and shear moduli are constants of the material, unlike the force constant \(k\), which depends on the specimen.
- Only the bulk modulus is defined for liquids and gases. The Young and shear moduli apply to solids only.
- The larger the elastic modulus, the stiffer (less deformable) the material, not the more stretchable.
- The Young modulus governs the choice of material for cables, columns and beams.
- The bulk modulus of a fluid determines the speed of sound in it (\(v = \sqrt{K/\rho}\)); the compressibility of water matters in the design of submarines and deep-sea equipment.
- The shear modulus governs the twisting of shafts and the design of rubber mounts and rivets.
The pressure on a sample of water is increased by 2 × 10⁷ Pa. If the bulk modulus of water is 2.2 × 10⁹ Pa, find the fractional decrease in its volume.
Show solution
\(\left|\dfrac{\Delta V}{V}\right| = \dfrac{\Delta P}{K} = \dfrac{2 \times 10^{7}}{2.2 \times 10^{9}} = 9.09 \times 10^{-3}\)
Answer: \(\Delta V/V \approx 9.1 \times 10^{-3}\), a decrease of about 0.91%.
The top face of a cube of side 10 cm, of shear modulus 2 × 10¹⁰ Pa, is acted on by a tangential force of 5 × 10³ N while the base is fixed. Find the angle of shear and the displacement of the top face.
Show solution
\(A = 0.1 \times 0.1 = 0.01\ \text{m}^2\), \(h = 0.1\ \text{m}\).
Shear stress \(= F/A = 5 \times 10^{3}/0.01 = 5 \times 10^{5}\ \text{Pa}\)
\(\theta = \text{stress}/\eta = 5 \times 10^{5}/2 \times 10^{10} = 2.5 \times 10^{-5}\ \text{rad}\)
\(\Delta x = \theta h = 2.5 \times 10^{-5} \times 0.1 = 2.5 \times 10^{-6}\ \text{m}\)
Answer: \(\theta = 2.5 \times 10^{-5}\) rad; \(\Delta x = 2.5\ \mu\text{m}\).
A steel wire of length 2 m and diameter 1 mm supports a mass of 10 kg. Calculate (a) the stress, (b) the strain, (c) the extension. (\(Y = 2 \times 10^{11}\) Pa, \(g = 9.8\) m s⁻²)
Show solution
\(F = mg = 98\ \text{N}\); \(A = \pi d^2/4 = \pi (10^{-3})^2/4 = 7.854 \times 10^{-7}\ \text{m}^2\).
(a) Stress \(= 98/7.854 \times 10^{-7} = 1.248 \times 10^{8}\ \text{Pa}\)
(b) Strain \(= 1.248 \times 10^{8}/2 \times 10^{11} = 6.24 \times 10^{-4}\)
(c) \(\Delta L = 6.24 \times 10^{-4} \times 2 = 1.248 \times 10^{-3}\ \text{m}\)
Answer: stress \(\approx 1.25 \times 10^8\) Pa; strain \(\approx 6.24 \times 10^{-4}\); extension \(\approx 1.25\) mm.
Two wires A and B of the same material carry the same load. Their lengths are in the ratio 1 : 2 and their diameters in the ratio 2 : 1. Find the ratio of their extensions.
Show solution
\(\Delta L = \dfrac{FL}{AY}\). With \(F\) and \(Y\) the same and \(A \propto d^2\), \(\Delta L \propto L/d^2\).
\(\dfrac{\Delta L_A}{\Delta L_B} = \dfrac{L_A}{L_B}\left(\dfrac{d_B}{d_A}\right)^2 = \dfrac12 \times \left(\dfrac12\right)^2 = \dfrac18\)
Answer: \(\Delta L_A : \Delta L_B = 1 : 8\).
A vertical steel rod 100 m long hangs from a rigid support. Find its extension due to its own weight. (\(\rho = 7800\) kg m⁻³, \(Y = 2 \times 10^{11}\) Pa, \(g = 9.8\) m s⁻²)
Show solution
\(\Delta L = \dfrac{\rho g L^2}{2Y} = \dfrac{7800 \times 9.8 \times 100^2}{2 \times 2 \times 10^{11}} = \dfrac{7.644 \times 10^{8}}{4 \times 10^{11}} = 1.91 \times 10^{-3}\ \text{m}\)
Answer: \(\Delta L \approx 1.9\) mm.
8.4 Poisson's ratio
Lateral strain and Poisson's ratio
When a wire or bar is stretched along its length it becomes thinner; when compressed it becomes thicker. The strain produced perpendicular to the applied force is the lateral strain.
If a bar of diameter \(d\) and length \(L\) is stretched so that its length increases by \(\Delta L\) and its diameter decreases by \(\Delta d\):
- Longitudinal strain \(= \Delta L/L\)
- Lateral strain \(= -\Delta d/d\) (negative because the diameter decreases as the length increases)
Poisson's ratio \(\sigma\) is the ratio of lateral strain to longitudinal strain, taken with a sign that makes it normally positive.
Physical meaning: \(\sigma\) tells how much a material narrows for a given lengthening. A material with \(\sigma = 0.3\), stretched by 1%, contracts sideways by 0.3%.
Change in volume under tension
Assumptions: a cylindrical wire of an isotropic material; small strains. The volume is \(V = \pi r^2 L\). Taking logarithms and differentiating:
A body stretched by tension cannot decrease in volume, so \(\Delta V/V \ge 0\), which requires \(1 - 2\sigma \ge 0\), i.e. \(\sigma \le 0.5\). The value \(\sigma = 0.5\) corresponds to a perfectly incompressible material.
Relations among Y, K, η and σ
(a) Y, K and σ. Assumption: a cube of an isotropic material is under a uniform pressure \(P\) on all six faces.
Along one edge, the compressive stress \(P\) on that edge produces a compressive strain \(P/Y\). The stresses \(P\) along the other two directions each produce a lateral (extensional) strain \(\sigma P/Y\) along this edge. The net linear strain is
For small strains the volume strain is three times the linear strain (\(V = L^3\), so \(dV/V = 3\,dL/L\)):
Since \(K = -P/(\Delta V/V)\):
Because \(K\) is positive, this again requires \(\sigma < 0.5\).
(b) Y, η and σ. A pure shear stress \(\tau\) is equivalent to a tensile stress \(+\tau\) along one diagonal of the element together with a compressive stress \(-\tau\) along the other. This leads to
Since \(\eta > 0\) and \(Y > 0\), this requires \(\sigma > -1\). Theoretically, therefore, \(-1 < \sigma < 0.5\); in practice \(\sigma\) for real materials lies between 0 and 0.5.
(c) Eliminating σ. From the two relations, \(1 - 2\sigma = Y/3K\) and \(2(1 + \sigma) = Y/\eta\). Adding gives \(3 = Y/3K + Y/\eta\), so
Poisson's ratio is not the ratio of the change in diameter to the change in length. It is the ratio of the fractional changes, \(\Delta d/d\) divided by \(\Delta L/L\).
- Rubber gaskets and seals rely on \(\sigma \approx 0.5\): squeezing them in one direction makes them bulge sideways and fill gaps.
- Cork stoppers (\(\sigma \approx 0\)) can be pushed into a bottle neck without bulging sideways.
- Engineers use \(\sigma\) to compute the change in cross-section of cables and columns under load.
A wire 2 m long and 2 mm in diameter is stretched so that its length increases by 1 mm. If Poisson's ratio is 0.30, find the decrease in diameter.
Show solution
Longitudinal strain \(= 1/2000 = 5 \times 10^{-4}\)
Lateral strain \(= \sigma \times 5 \times 10^{-4} = 1.5 \times 10^{-4}\)
\(\Delta d = 1.5 \times 10^{-4} \times 2\ \text{mm} = 3 \times 10^{-4}\ \text{mm} = 3 \times 10^{-7}\ \text{m}\)
Answer: the diameter decreases by \(3 \times 10^{-7}\) m.
For the wire in Example 10, find the fractional change in its volume.
Show solution
\(\dfrac{\Delta V}{V} = (1 - 2\sigma)\dfrac{\Delta L}{L} = 0.40 \times 5 \times 10^{-4} = 2 \times 10^{-4}\)
Answer: \(\Delta V/V = 2 \times 10^{-4}\) (the volume increases by 0.02%).
For a metal, \(Y = 2.0 \times 10^{11}\) Pa and \(K = 1.67 \times 10^{11}\) Pa. Find Poisson's ratio and the shear modulus.
Show solution
\(1 - 2\sigma = \dfrac{Y}{3K} = \dfrac{2.0 \times 10^{11}}{3 \times 1.67 \times 10^{11}} = 0.399\), so \(\sigma = 0.30\).
\(\eta = \dfrac{Y}{2(1 + \sigma)} = \dfrac{2.0 \times 10^{11}}{2.60} = 7.7 \times 10^{10}\ \text{Pa}\)
Check: \(9/Y = 4.5 \times 10^{-11}\); \(3/\eta + 1/K = 3.9 \times 10^{-11} + 0.6 \times 10^{-11} = 4.5 \times 10^{-11}\ \text{Pa}^{-1}\).
Answer: \(\sigma \approx 0.30\); \(\eta \approx 7.7 \times 10^{10}\) Pa.
8.5 Elastic potential energy
To deform an elastic body, work must be done against the internal restoring forces. If the body is deformed slowly and within its elastic limit, this work is stored as elastic potential energy.
Energy stored in a stretched spring
Assumptions: the spring obeys Hooke's law throughout; the stretching is slow, so kinetic energy is negligible.
At extension \(x\) the applied force is \(F = kx\). The work done for a further small extension \(dx\) is \(dW = kx\,dx\), so
Because the force increases linearly from 0 to \(F\), the work equals the average force \(F/2\) times the extension. It is the area under the force–extension graph.
Energy stored in a stretched wire
For a wire of length \(L\), cross-section \(A\) and Young modulus \(Y\), the force for an extension \(l\) is \(F = (YA/L)\,l\). Stretching it from 0 to \(\Delta L\):
Energy per unit volume
Dividing by the volume \(AL\):
- When a load \(Mg\) is hung on a wire and let go, gravity does work \(Mg\,\Delta L\), but the wire stores only \(\tfrac12 Mg\,\Delta L\). The other half is dissipated as heat and sound as the load oscillates and settles.
- Elastic potential energy is proportional to the square of the extension. Doubling the extension of a spring quadruples the stored energy.
- For a given load, a thicker wire stores less energy than a thin wire of the same length and material, since \(U = F^2L/(2AY) \propto 1/A\).
- Bows, catapults, trampolines and pole-vault poles store elastic energy and release it as kinetic energy.
- Wind-up toys and mechanical watches store energy in coiled springs.
- Elastic energy stored in tendons makes running more efficient in animals.
- Shock absorbers convert elastic energy into heat through hysteresis and damping.
A spring of force constant 200 N m⁻¹ is stretched by 5 cm. Find the elastic potential energy stored.
Show solution
\(U = \tfrac12 kx^2 = \tfrac12 \times 200 \times (0.05)^2 = 0.25\ \text{J}\)
A steel wire of length 2 m and cross-sectional area 1 mm² is stretched by a load of 100 N. Find the energy stored and the energy per unit volume. (\(Y = 2 \times 10^{11}\) Pa)
Show solution
\(\Delta L = \dfrac{FL}{AY} = \dfrac{100 \times 2}{10^{-6} \times 2 \times 10^{11}} = 10^{-3}\ \text{m}\)
\(U = \tfrac12 F\,\Delta L = \tfrac12 \times 100 \times 10^{-3} = 0.05\ \text{J}\)
Volume \(= AL = 2 \times 10^{-6}\ \text{m}^3\); \(u = 0.05/2 \times 10^{-6} = 2.5 \times 10^{4}\ \text{J m}^{-3}\)
Check: stress \(= 10^8\) Pa, strain \(= 5 \times 10^{-4}\); \(\tfrac12 \times 10^8 \times 5 \times 10^{-4} = 2.5 \times 10^4\ \text{J m}^{-3}\).
Answer: \(U = 0.05\) J; \(u = 2.5 \times 10^4\) J m⁻³.
A metal wire of length 3 m and cross-sectional area 1.5 mm² is stretched by 3 mm. If \(Y = 2 \times 10^{11}\) Pa, find (a) the tension, (b) the energy stored, (c) the energy density.
Show solution
(a) \(F = \dfrac{YA\,\Delta L}{L} = \dfrac{2 \times 10^{11} \times 1.5 \times 10^{-6} \times 3 \times 10^{-3}}{3} = 300\ \text{N}\)
(b) \(U = \tfrac12 F\,\Delta L = \tfrac12 \times 300 \times 3 \times 10^{-3} = 0.45\ \text{J}\)
(c) Volume \(= 4.5 \times 10^{-6}\ \text{m}^3\); \(u = 0.45/4.5 \times 10^{-6} = 1 \times 10^{5}\ \text{J m}^{-3}\)
Check: strain \(= 10^{-3}\); \(\tfrac12 Y(\text{strain})^2 = \tfrac12 \times 2 \times 10^{11} \times 10^{-6} = 10^5\ \text{J m}^{-3}\).
Answer: \(F = 300\) N; \(U = 0.45\) J; \(u = 10^5\) J m⁻³.
Exam tips
- Formulas to memorise: \(F = kx\); \(k = YA/L\); \(k_s = k_1k_2/(k_1 + k_2)\), \(k_p = k_1 + k_2\); stress \(= F/A\); strain \(= \Delta L/L\); \(Y = FL/A\Delta L\); \(K = -\Delta P/(\Delta V/V)\); \(\eta = F/A\theta\); \(U = \tfrac12 kx^2\); \(u = \tfrac12\) stress × strain.
- Common board questions: state Hooke's law and define force constant; define stress and strain and distinguish the three types; draw and explain the stress–strain curve; derive relations among \(Y\), \(K\), \(\eta\) and \(\sigma\).
- Typical numericals: stress, strain and extension from load, length and diameter; ratio of extensions of two wires; energy stored in springs or wires.
- Convert mm² to m² (1 mm² = 10⁻⁶ m²) and lengths to metres before substituting. If only the diameter is given, use \(A = \pi d^2/4\).
- Do not attach units to strain or Poisson's ratio.
- In energy problems, remember the factor ½. Using \(F\,\Delta L\) instead is the most frequent error.
- In stress–strain diagrams, put stress on the vertical axis and strain on the horizontal axis, mark all the key points, and state the meaning of each region.
Key points
- Hooke's law: within the elastic limit, \(F = kx\); for a wire \(k = YA/L\).
- Springs in series: \(1/k_s = 1/k_1 + 1/k_2\); in parallel: \(k_p = k_1 + k_2\); each of \(n\) equal parts has \(nk\).
- Stress \(= F/A\) (Pa); strain is a dimensionless ratio.
- \(Y\), \(K\) and \(\eta\) are stress/strain for length, volume and shape changes; only \(K\) applies to fluids.
- Poisson's ratio \(\sigma = (\Delta d/d)/(\Delta L/L)\); \(Y = 3K(1-2\sigma) = 2\eta(1+\sigma)\); \(9/Y = 3/\eta + 1/K\).
- Elastic energy \(U = \tfrac12 F\,\Delta L\); energy density \(u = \tfrac12\) stress × strain.