Mechanical Properties of Solids

All real materials deform to some extent when an external (deforming) force acts on them. A body that returns completely to its original shape and size the instant the deforming force is removed is called perfectly elastic (e.g. quartz, phosphor bronze, steel over a limited range). A body that stays deformed even after the force is removed is called perfectly plastic (e.g. putty, wet clay).

When a deforming force is applied to a body, internal forces develop within it that resist the deformation and tend to restore the original configuration. The restoring force per unit area, acting normal or tangential to a surface inside the body, is called stress.

• Tensile stress — the deforming force stretches the body along its length (e.g. a wire being pulled).
• Compressive stress — the deforming force compresses/shortens the body (e.g. a pillar under a load).
• Shearing (tangential) stress — the deforming force acts tangentially/parallel to a surface, changing the shape without necessarily changing the volume (e.g. cutting with scissors, twisting a rod).

Tensile and compressive stresses together are often called longitudinal stress; when the deforming force is applied normally and uniformly on the entire surface (as by a surrounding fluid), the effect is called hydraulic (volume) stress.

The fractional change produced in the dimensions of a body by stress is called strain. Strain has no unit and no dimensions since it is a ratio of two similar quantities.

• Longitudinal strain = change in length / original length
• Volume strain = change in volume / original volume
• Shearing strain = angle (in radians) through which a face originally perpendicular to the fixed face turns, ≈ Δx/L for small angles

Worked example: A metal wire of original length 2 m and cross-sectional area 1×10^-6 m² is pulled by a force of 100 N and stretches by 1 mm. Find the stress and strain.

Given: F = 100 N, A = 1×10^-6 m², L = 2 m, ΔL = 1×10^-3 m

Formula: stress = F/A, strain = ΔL/L

Substitution: stress = 100 / (1×10^-6) = 1×10⁸ N/m²; strain = (1×10^-3) / 2 = 5×10^-4

Result: stress = 1×10⁸ Pa, strain = 5×10^-4 (dimensionless).

English physicist Robert Hooke found that, for small deformations, the stress developed in a body is directly proportional to the strain produced, provided the elastic limit is not exceeded. This is Hooke's law: stress ∝ strain, so stress = k × strain, where k is a constant of proportionality called the modulus of elasticity of the material (its value depends on the material and the type of deformation).

A stress–strain curve is obtained by plotting the stress developed in a test specimen (usually a wire or rod) against the strain produced, as the load is gradually increased until the specimen breaks. For a typical ductile metal such as annealed copper or mild steel, the curve shows the following regions (labelled O, A, B, D, E as in the standard NCERT stress–strain graph):

• Proportional/elastic region (O to A): stress is directly proportional to strain (straight line); Hooke's law holds and the slope of this line gives the modulus of elasticity.
• Elastic limit (point B, close to A): the maximum stress up to which the material returns completely to its original dimensions when unloaded; beyond this point permanent (plastic) deformation sets in even after the stress is removed.
• Yield point (near B): a small increase in stress beyond the elastic limit produces a disproportionately large increase in strain; the material starts to flow plastically.
• Plastic region (B to D) and ultimate tensile strength (point D): the maximum stress the material can withstand before necking (local thinning) begins; beyond D, strain increases even as stress is reduced.
• Fracture point (E): the point at which the specimen finally breaks.

Materials such as steel have a long, nearly linear elastic region and are called ductile; materials such as glass or cast iron break soon after the elastic limit with very little plastic deformation and are called brittle. Elastomers such as rubber can sustain very large strains without a proportional (straight-line) region and without a well-defined yield point, even though their elastic limit is reached only at a large strain.

Worked example: The linear (elastic) part of the stress–strain graph of a metal wire passes through a point where the stress is 150 MPa and the strain is 7.5×10^-4. Find the modulus of elasticity of the material and identify it.

Given: stress = 150×10⁶ Pa, strain = 7.5×10^-4 (within the linear region)

Formula: modulus of elasticity = stress / strain (Hooke's law, slope of the linear part)

Substitution: modulus = (150×10⁶) / (7.5×10^-4)

Result: modulus = 2×10¹¹ Pa, which is the typical value of Young's modulus for steel.

When a solid rod or wire is stretched or compressed along its length, the ratio of longitudinal stress to longitudinal strain (within the elastic limit) is called Young's modulus, denoted Y. It applies only to solids, since only solids can sustain a definite length under a longitudinal (tensile or compressive) force.

A large value of Y means the material is very stiff — a large stress is needed to produce even a small strain (e.g. steel). A small value of Y means the material stretches easily under a small stress (e.g. rubber). Since strain is dimensionless, Y has the same SI unit as stress, the pascal (N/m²); typical values are of the order 10⁹–10¹¹ Pa for common engineering solids.

For a wire of original length L and uniform cross-sectional area A, if a stretching force F produces an extension ΔL, then:

Y = stress/strain = (F/A) / (ΔL/L) = FL / (A·ΔL)

Young's modulus governs the design of columns, beams, wires and cables — materials with a high Y (like steel) are chosen wherever minimal sag or stretch under load is required.

Worked example: A structural steel rod has a radius of 10 mm and a length of 1.0 m. A tensile force of 100 kN is applied along its length. If Young's modulus of steel is 2.0×10¹¹ Pa, find (a) the stress, (b) the strain, and (c) the elongation of the rod.

Given: r = 10 mm = 1.0×10^-2 m, L = 1.0 m, F = 100×10³ N, Y = 2.0×10¹¹ Pa

Formula: A = πr²; stress = F/A; strain = stress/Y; ΔL = strain × L

Substitution: A = π×(1.0×10^-2)² = 3.14×10^-4 m²

stress = (1.0×10⁵) / (3.14×10^-4) = 3.18×10⁸ Pa

strain = (3.18×10⁸) / (2.0×10¹¹) = 1.59×10^-3

ΔL = (1.59×10^-3) × 1.0 = 1.59×10^-3 m

Result: stress ≈ 3.18×10⁸ Pa, strain ≈ 1.59×10^-3, elongation ≈ 1.59×10^-3 m (about 1.6 mm).

When a tangential (shearing) force is applied to one face of a solid while the opposite face is held fixed, the body changes shape (its faces get sheared relative to each other) without appreciable change in volume. The ratio of shearing stress to the shearing strain produced, within the elastic limit, is called the shear modulus or modulus of rigidity, denoted G or η.

G = shearing stress / shearing strain

Shear modulus is defined only for solids, because fluids (liquids and gases) cannot sustain a static shearing stress — they simply flow instead of developing a restoring shear stress. For a given material, the shear modulus is generally smaller than its Young's modulus, since it is comparatively easier to change the shape of a solid than to change its length while its volume stays fixed.

Worked example: A square lead slab of side 50 cm and thickness 10 cm has its lower edge rigidly fixed. A shearing force of 9.0×10⁴ N is applied tangentially on a narrow face of area 0.5 m × 0.1 m. If the shear modulus of lead is 5.6×10⁹ Pa, find the horizontal displacement of the upper edge, given the height of the slab over which the shear acts is 0.5 m.

Given: F = 9.0×10⁴ N, area of application A = 0.5 m × 0.1 m = 0.05 m², G = 5.6×10⁹ Pa, height L = 0.5 m

Formula: shear stress = F/A; shear strain = stress/G; Δx = shear strain × L

Substitution: shear stress = (9.0×10⁴)/(0.05) = 1.8×10⁶ Pa

shear strain = (1.8×10⁶)/(5.6×10⁹) = 3.2×10^-4

Δx = (3.2×10^-4) × 0.5 = 1.6×10^-4 m

Result: the upper edge is displaced by about 1.6×10^-4 m, i.e. 0.16 mm.

When a body is subjected to a uniform pressure (force per unit area applied normally over its entire surface, as by a surrounding fluid), it undergoes a change in volume without any change in shape. The ratio of the hydraulic (volume) stress to the volume strain produced, within the elastic limit, is called the bulk modulus, denoted B.

B = -p / (ΔV/V)

Here p is the increase in pressure and ΔV/V is the resulting fractional change in volume; the negative sign is included because an increase in pressure (positive p) always causes a decrease in volume (negative ΔV), making B itself a positive quantity. Unlike Young's modulus and shear modulus, bulk modulus is defined for all three states of matter — solids, liquids and gases — since all of them can be compressed. Solids have the largest B (they are the least compressible), liquids have intermediate values, and gases have very small B (they are highly compressible; for gases, B also depends on the process, e.g. isothermal or adiabatic).

The reciprocal of the bulk modulus is called compressibility, k = 1/B — it measures the fractional decrease in volume per unit increase in pressure.

When a wire or rod is stretched along its length (longitudinal strain), it simultaneously contracts slightly in the perpendicular directions (lateral strain). Within the elastic limit, the ratio of lateral strain to longitudinal strain is a constant for a given material, called Poisson's ratio, σ. It is a pure number (dimensionless) and has no unit. For most engineering materials σ lies roughly between 0.2 and 0.4; theoretically, for an isotropic elastic solid, σ must lie between -1 and 0.5.

Worked example: At a certain depth in the ocean, the pressure is greater than at the surface by 1.0×10⁷ Pa. If the bulk modulus of water is 2.2×10⁹ Pa, find the fractional decrease in volume of a sample of water taken to that depth, and its compressibility.

Given: p = 1.0×10⁷ Pa, B = 2.2×10⁹ Pa

Formula: ΔV/V = p/B ; compressibility k = 1/B

Substitution: ΔV/V = (1.0×10⁷) / (2.2×10⁹) = 4.5×10^-3

k = 1 / (2.2×10⁹) = 4.5×10^-10 Pa^-1

Result: the fractional decrease in volume is about 4.5×10^-3 (0.45%), showing that even under high pressure water is only slightly compressible; compressibility of water = 4.5×10^-10 Pa^-1.

When a wire is stretched by a gradually increasing force, work is done against the internal restoring forces, and this work is stored in the wire as elastic potential energy. Since the stretching force increases from 0 to F as the extension grows from 0 to ΔL, the average force is F/2, so the work done (and hence the energy stored) is:

U = (1/2) F·ΔL

This can also be expressed in terms of stress, strain and volume of the wire, which is useful for comparing different materials:

U = (1/2) × stress × strain × volume, so the elastic energy stored per unit volume (energy density) is u = (1/2) × stress × strain = stress²/(2Y) = (1/2)Y×strain².

Worked example: For the steel rod of the earlier example (F = 1.0×10⁵ N, elongation ΔL = 1.59×10^-3 m), find the elastic potential energy stored in it.

Given: F = 1.0×10⁵ N, ΔL = 1.59×10^-3 m

Formula: U = (1/2) F·ΔL

Substitution: U = (1/2) × (1.0×10⁵) × (1.59×10^-3)

Result: U ≈ 79.6 J is stored as elastic potential energy in the stretched rod.

The elastic properties of materials guide many practical engineering decisions:

• Beams and girders: a beam of length l, breadth b and depth (thickness) d, supported at its ends and loaded at the centre with weight W, sags (depresses) by δ = Wl³/(4bd³Y). Since δ ∝ 1/d³, increasing the depth is far more effective than increasing the breadth at reducing sag. This is why girders and rails are given an I-shaped (or H-shaped) cross-section — most of the material is concentrated in the top and bottom flanges (away from the neutral axis, maximising effective depth), while the thin connecting web keeps the total weight and material cost low.
• Ropes and cables (e.g. cranes, lifts): the thickness of a metal rope is chosen using its elastic limit together with a safety factor for the maximum load it must bear, not merely its breaking stress, so that the rope always operates well within the elastic (fully recoverable) region.
• Maximum height of mountains: the material at the base of a mountain must withstand the compressive stress due to the weight of rock above it without exceeding its elastic limit. For rock (granite) with an elastic limit of about 3×10⁸ Pa and density about 3×10³ kg/m³, equating the compressive stress ρgh to the elastic limit gives a maximum possible height of about 10 km — consistent with the tallest mountains on Earth being close to this limit.