Big picture
A steel cable stretches slightly under load, while a soft clay lump may retain a changed shape after the force is removed. Elasticity concerns recovery of the original shape and size after deforming forces are removed. It is different from everyday stretchiness: a material can resist stretching strongly and still be highly elastic.
Engineering design uses small deformations to estimate loads and safety. A bridge cable, a crane wire or a building column must remain within acceptable stress and deformation limits. This chapter connects forces acting on a specimen with material properties that are independent of its particular dimensions.
Learning goals
Distinguish stress, strain, elasticity and plasticity.
Interpret a stress–strain curve and Hooke's law.
Apply Young's, shear and bulk moduli with correct units.
Calculate extension and elastic energy.
Explain design choices and limitations of ideal elastic models.
Core ideas
1. Deformation and restoring forces
A deforming force can change length, volume or shape. Internal interatomic interactions provide restoring forces when atoms are displaced from equilibrium separation. For small departures, the response is approximately linear; larger departures can cause permanent rearrangement or fracture.
An elastic deformation is recovered after unloading, while plastic deformation persists. An ideally rigid body does not deform, but real materials always deform to some extent. The model of a rigid body is useful when deformation is negligible for the question. Steel's high resistance to elastic strain does not mean it never stretches; its extension may simply be small compared with that of a rubber cord under comparable stress.
2. Stress and strain
Stress is internal restoring force per unit area, equal in magnitude to applied force per area in the simple equilibrium situations used here. Longitudinal stress is F/A for tension or compression. Shear stress is tangential force per area. Hydraulic stress is pressure change applied normally over the body's surface. Stress has SI unit Pa and dimensions M L⁻¹ T⁻².
Longitudinal strain is ΔL/L. Volume strain is ΔV/V, often signed negative during compression. Shear strain is the small angular deformation, approximately lateral displacement divided by perpendicular height. Strain is dimensionless because it compares like quantities. A 2 m wire extending 1 mm has strain 0.001/2 = 5 × 10⁻⁴, not 0.5, after converting millimetres to metres.
Quantity | Definition | Unit |
|---|---|---|
Tensile stress | F/A | Pa |
Longitudinal strain | ΔL/L | Dimensionless |
Shear strain | x/h for small deformation | Dimensionless |
Volume strain | ΔV/V | Dimensionless |
3. Hooke's law and elastic limits
Within the proportional range, Hooke's law states that stress is proportional to strain. Their ratio is an elastic modulus. This is a material relation; F = kx is the corresponding form for a particular spring or uniform specimen with an effective force constant k.
The proportional limit marks the end of strict linearity. The elastic limit marks the greatest stress after which the specimen can still recover its original dimensions. These need not be exactly identical. A material can be elastic but nonlinear over part of its response, as rubber often is. Do not apply a constant Young's modulus to arbitrary large rubber stretching simply because it returns to its original form.
4. Stress–strain curve
For a typical ductile metal under tensile testing, the initial straight segment has slope Young's modulus. Beyond proportional behaviour, the specimen may still recover until the elastic limit. At yield, substantial strain develops with relatively little stress increase, and permanent deformation becomes important. Further loading can lead through strain hardening to maximum engineering stress, called ultimate tensile strength, then necking and fracture.
Ductile materials show substantial plastic deformation before fracture; brittle materials show relatively little. A graph should have strain on the horizontal axis and stress on the vertical axis. Its steep initial slope represents stiffness, not ultimate strength. Two materials can therefore have different rankings for stiffness, toughness and breaking stress. Toughness relates to energy absorbed per volume before fracture, represented by the full area under a relevant stress–strain curve.
5. Young's modulus and wire extension
Young's modulus, Y, equals longitudinal stress divided by longitudinal strain: Y = FL/(AΔL). Hence ΔL = FL/(AY). Longer wires extend more, thicker wires extend less, and larger Y means smaller strain for the same stress.
For F = 120 N, L = 2.0 m, A = 1.0 × 10⁻⁶ m² and Y = 2.0 × 10¹¹ Pa, ΔL = 120 × 2/(10⁻⁶ × 2 × 10¹¹) = 0.0012 m = 1.2 mm. Stress is 120/10⁻⁶ = 1.2 × 10⁸ Pa and strain is 0.0012/2 = 6 × 10⁻⁴, whose ratio reproduces Y.
For a circular wire, A = πr² = πd²/4. Doubling diameter quadruples area, reducing extension to one-quarter for the same length, material and load. The effective spring constant is k = AY/L. Halving length doubles k without changing the material's Young's modulus.
6. Shear and bulk moduli
The shear modulus, G, is shear stress divided by shear strain: G = (F/A)/(x/h) = Fh/(Ax). If F = 40 N acts over 0.010 m² on a block of height 0.050 m with G = 2.0 × 10⁶ Pa, x = Fh/(AG) = 40 × 0.050/(0.010 × 2.0 × 10⁶) = 1.0 × 10⁻⁴ m, or 0.10 mm.
The bulk modulus, K, describes resistance to uniform compression: K = −ΔP/(ΔV/V). The minus sign makes K positive because increased pressure reduces volume. Compressibility is 1/K. For K = 2.0 × 10⁹ Pa and pressure increase 4.0 × 10⁶ Pa, ΔV/V = −4.0 × 10⁶/(2.0 × 10⁹) = −0.002, a 0.20% decrease. A liquid can have a large bulk modulus while having no static shear rigidity; volume resistance and shape resistance are distinct.
7. Poisson effect and material comparisons
A wire stretched longitudinally generally contracts laterally. Poisson's ratio, ν, is minus lateral strain divided by longitudinal strain for ordinary small deformations. If axial strain is 0.0010 and lateral strain is −0.00030, ν = −(−0.00030)/0.0010 = 0.30. The sign convention prevents the common error of reporting a negative value for an ordinary material's lateral contraction.
For an isotropic linear elastic material, moduli are related by Y = 2G(1 + ν) and Y = 3K(1 − 2ν). These relations require their material assumptions; they are not universal for anisotropic substances. In elementary comparisons, steel is called more elastic than rubber because it has a larger relevant modulus, meaning less strain under the same stress. This statement should not be confused with which specimen can be stretched farther without breaking.
8. Elastic potential energy
When a linear elastic wire is loaded gradually from zero to final force F, the average force is F/2. Work stored is U = ½FΔL = ½k(ΔL)². Dividing by specimen volume AL gives energy density u = ½ × stress × strain = stress²/(2Y).
For the wire with F = 120 N and ΔL = 0.0012 m, U = 0.5 × 120 × 0.0012 = 0.072 J. Its volume is 1.0 × 10⁻⁶ × 2 = 2.0 × 10⁻⁶ m³, so u = 0.072/(2.0 × 10⁻⁶) = 36,000 J m⁻³. Using final force times extension without the half-factor would double the result. Gradual loading is important; sudden application of a load can create additional dynamic extension.
9. Experiments and applications
In a wire-extension experiment, measure original length and diameter carefully, add loads gradually, allow readings to stabilise and unload to check recovery. A reference wire can help compensate for support movement and temperature changes. Plot force against extension: a straight line through the appropriate origin supports the linear model. The slope gives k, from which Y = kL/A.
For safe design, working stress must remain below an appropriate allowable limit, with a factor of safety accounting for uncertainty and service conditions. Increasing cable area lowers stress, while increasing length changes extension but not stress for a fixed axial load. Beams also undergo bending and nonuniform stress, so the uniform wire equation cannot automatically solve every bridge problem. A material's high modulus alone is insufficient to guarantee safe construction; strength, fatigue, corrosion and geometry also matter.
A reliable approach to questions
Identify tensile, shear or volume deformation and choose the matching modulus.
Convert all lengths and areas to SI units, checking diameter versus radius.
Calculate stress and strain separately before using a ratio if uncertain.
State the linear-elastic assumption for energy or extension formulas.
Distinguish a material property from a specimen property and report units.
Connections that matter
Atomic bonding explains restoring forces and material stiffness.
Work and energy explain strain-energy storage during loading.
Fluid mechanics uses bulk modulus and distinguishes it from shear rigidity.
Common mistakes
Treating strain as a length ignores that it is a dimensionless ratio.
Calling high breaking stress high modulus confuses strength and stiffness.
Using diameter directly as radius makes area four times too large.
Omitting the half-factor in gradual-loading energy doubles stored work.
Revision checklist
I can distinguish all stress and strain types.
I can interpret the main parts of a stress–strain graph.
I can calculate extension, displacement and volume change.
I can calculate energy and energy density.
I can explain why modulus is independent of specimen dimensions.