Mechanical Properties of Solids — NEET Notes

Elasticity, plasticity, and the limit of rigidity

A "rigid body" was a convenient simplification of earlier mechanics chapters — useful for studying rotation, but not exactly true. Every solid yields under sufficient force. Stretch a helical spring gently and it returns to its original length when released; that recovery is the defining feature of elasticity. Press a lump of putty and it stays deformed; that permanence defines plasticity. Real solids sit between these idealisations. Steel is highly elastic up to a point and then plastic; rubber is elastic over enormous strains but doesn't follow Hooke's law over most of its range; mud and wet clay are nearly perfect plastics. NCERT frames the entire chapter around answering one practical question: when we design a bridge, a building, a thighbone replacement, or a chairlift cable, how much will the material give, and where will it break?

Elastic behaviour is a molecular phenomenon. Inside a solid, atoms sit at equilibrium spacings, held by electrostatic forces. A small displacement from equilibrium develops a restoring force that pulls them back — exactly the situation Hooke's law describes. Push too hard and the atoms slide past one another into a new arrangement that does not return when the load is removed. That sliding is the microscopic picture of plastic deformation.

Stress and strain — the two state variables

The chapter is built on two quantities that describe what is happening inside a deformed solid. Stress is the restoring force developed per unit area when a body is held in static equilibrium under a deforming load. Numerically, if a force F acts normal to a cross-section of area A, the magnitude of stress is

stress = F / A   [ N/m² = pascal ]

Restoring force per unit cross-sectional area

The SI unit of stress is the newton per square metre, also called the pascal (Pa). Its dimensional formula is [ML⁻¹T⁻²] — identical to that of pressure. Numerical values run from kilopascals for everyday loads to gigapascals (10⁹ Pa) inside heavy structural members. NEET 2020 asked this dimensional formula as a single-line question; you should know it without thinking.

Strain is the response — the fractional change in dimension. For a wire of original length L stretched by ΔL, the longitudinal strain is simply ΔL/L. Because it is a ratio of two like quantities, strain has no units and no dimensions. This is a frequent NEET trap: students assume strain has dimensions of length and trip on dimensional-analysis questions. Strain is a pure number — typically a small one. A 1 m steel rod stretched by 1 mm under load corresponds to a strain of just 10⁻³, or 0.1%.

Three types of stress and strain

A solid can be deformed in three geometrically distinct ways, and NCERT distinguishes them with care. Each has its own pair of stress and strain.

Longitudinal stress arises when two equal and opposite forces act normal to opposite faces of a body along the same line, stretching it (tensile stress) or squeezing it (compressive stress). The body changes length without changing shape. The associated strain is longitudinal strain = ΔL/L. A copper wire suspended from the ceiling with a weight at the bottom is the classic example.

Shearing stress arises when two equal and opposite forces act parallel to opposite faces of a body — like sliding the cover of a thick book sideways while the back stays fixed. The body changes shape (becomes a parallelepiped) but not volume. The associated strain is shearing strain = Δx/L = tan θ ≈ θ, where θ is the small angular displacement. Shearing stress exists only in solids: liquids and gases cannot sustain shape and so cannot develop shear.

Hydraulic (volumetric) stress arises when a body is submerged in a fluid that presses uniformly on every surface. The internal restoring force equals the hydraulic pressure. The body changes volume without changing shape. The associated strain is volume strain = ΔV/V. This is the only kind of deformation that liquids and gases can also experience.

Hooke's law — the linear elastic regime

The central observation of the chapter, first published by Robert Hooke in 1676 as the anagram ceiiinosssttuv (unscrambled: ut tensio, sic vis — "as the extension, so the force"), is breathtakingly simple. For small deformations, stress is directly proportional to strain.

stress = Y × strain

Hooke's law — valid only in the linear region

The constant of proportionality is the modulus of elasticity, a number characteristic of the material. The same law takes three specific forms — one for each kind of stress — and gives three moduli: Young's modulus Y for longitudinal deformation, shear modulus G for shape change, bulk modulus K for volume change. Hooke's law is an empirical statement, not a derivation from deeper principles. It holds beautifully for metals at small strains, fails for rubber and biological tissue over almost their entire elastic range, and even for steel ceases to apply once you cross the proportional limit.

It is important to be clear about what Hooke's law is not. It does not say a material is elastic; it says that within the elastic regime, stress and strain are linear. A material like rubber is highly elastic — it returns to shape after enormous strain — yet violates Hooke's law over most of that range. Steel obeys Hooke's law up to a point and then deviates well before it actually breaks.

The stress-strain curve — five regions, one diagram

The stress-strain curve is the single most informative diagram in the chapter. Take a wire, stretch it gradually with increasing load, plot stress (F/A) on the y-axis against strain (ΔL/L) on the x-axis, and the line records the full life story of the specimen — from elastic obedience to plastic flow to rupture. NCERT's Figure 8.2 (for a typical metal) shows five distinct phases.

The shape of the curve is a signature of the material. Ductile metals — copper, mild steel, gold — have a long plastic region between yield and fracture; you can draw them into wires or beat them into sheets. Brittle materials — glass, cast iron, ceramics — break almost at the elastic limit, with D and E nearly coincident. Elastomers such as rubber and the aortic tissue (NCERT Figure 8.3) show a different curve entirely: an enormous elastic region with little linearity and no defined plastic phase.

The three elastic moduli

Within the proportional region, each kind of deformation has its own modulus of elasticity — the slope of the stress-strain line. The unit of every modulus is the pascal (Pa), the same as stress, because strain is dimensionless. There are three to know.

Young's modulus in practice

Young's modulus is the most heavily tested of the three. Its formula in expanded form is

Y = (F × L) / (A × ΔL)

Force times original length, divided by area times extension

where F is the deforming force, L the original length, A the cross-sectional area, and ΔL the extension. Plug this into a typical setup — a wire of length 1 m and area 0.1 cm² stretched by 0.1% — and you find F = 2000 N is needed if the wire is steel. For aluminium the answer is 690 N, for brass 900 N, for copper 1100 N. The lesson: steel is more elastic than copper, brass, or aluminium, because a larger force is needed to produce the same strain. This is exactly the reason steel dominates structural design.

A common student confusion: "rubber stretches more than steel, so rubber must be more elastic." NCERT explicitly corrects this in its "Points to Ponder." Elasticity in physics is the resistance to deformation, not the willingness to deform. The material that stretches less under a given load is more elastic. Rubber's Young's modulus is roughly 10⁶ Pa — five orders of magnitude smaller than steel's. It stretches dramatically because Y is small, which means it is much less elastic.

Shear modulus — why coil springs are governed by G

Shear modulus G (also called the modulus of rigidity) relates a tangential stress to the shearing strain it produces. Numerically G = (F·L)/(A·Δx) = (F/A)/θ, where θ is the small angular displacement. Shear deformation involves no change in volume — only a change in shape. The lower face stays still, the upper face slides sideways, and a square becomes a parallelogram. For metals, the rule of thumb is G ≈ Y/3.

Why does a coil spring obey shear modulus rather than Young's? When you stretch the spring vertically, individual coil segments do not stretch lengthwise — they twist. The wire experiences a torsion (a kind of shear) all along its length. So the spring constant of a coil spring depends on G, not Y. NEET 2022 turned this into an assertion-reason trap: the stretching of a spring is determined by shear modulus (true), and a copper coil has more tensile strength than a steel coil (false — steel is stronger). Watch for it.

Bulk modulus — the only one that applies to fluids

When a body is squeezed equally from all sides — submerged deep in a fluid, for instance — the relevant modulus is bulk modulus K (NCERT also writes it B). It is defined with an explicit negative sign:

K = −p × V / ΔV

Bulk modulus — volume drops as pressure rises

The negative sign accommodates the fact that a positive pressure produces a negative volume change; without it, K would come out negative, which is unphysical for stable matter. Bulk modulus is the only modulus that applies to all three states of matter — solid, liquid, and gas — because every state can experience uniform pressure and respond with a volume change. The reciprocal of K is called the compressibility: 1/K = −(1/Δp)(ΔV/V).

The numbers tell a story. Solids have bulk moduli around 10¹¹ Pa — they barely compress. Liquids have K around 10⁹ Pa, two orders of magnitude smaller. Gases at STP have K around 10⁵ Pa, six orders of magnitude below solids. Gases are about a million times more compressible than solids. NCERT's example: at the bottom of the Indian Ocean (depth 3000 m), the fractional compression of water is just ΔV/V = 1.36% — small even there, because water's bulk modulus is 2.2 × 10⁹ Pa.

NEET 2017 used bulk modulus in a slightly disguised form: a spherical object under uniform pressure p has its volume change as ΔV/V = p/K, and since V ∝ r³, the fractional change in radius is Δr/r = p/(3K). The trick is to remember that volume scales as the cube of length, so the relative volume change is three times the relative radius change.

Poisson's ratio — the lateral squeeze

Stretch a rubber band — it gets longer along the pull and thinner across. The shrinkage perpendicular to the applied force is a lateral strain, and Simon Poisson observed that within the elastic limit it is proportional to the longitudinal strain. The ratio of the two is Poisson's ratio, denoted σ.

σ = lateral strain / longitudinal strain = (Δd/d) / (ΔL/L)

A pure number — no units, no dimensions

For an originally cylindrical wire of diameter d that contracts by Δd while elongating by ΔL on a length L, Poisson's ratio is σ = (Δd/d)/(ΔL/L). Like strain itself, Poisson's ratio is dimensionless. Its theoretical range is 0 to 0.5: 0 corresponds to a material with no lateral contraction (some types of cork come close), and 0.5 is the upper limit for an incompressible material like rubber. For ordinary structural materials the values cluster narrowly: steel sits between 0.28 and 0.30, aluminium alloys around 0.33, brass around 0.34.

Elastic potential energy in a stretched wire

Work done against inter-atomic forces during stretching is stored in the wire as elastic potential energy. For a wire of original length L, cross-section A, stretched by an extension l under force F = YA(l/L), the work done in increasing the length from 0 to l is found by integration:

W = ½ × Y × (l/L)² × A × L = ½ × stress × strain × volume

Elastic energy stored in a stretched wire

Dividing by the volume gives the elastic energy density:

u = (1/2) × σ × ε = (1/2) × Y × ε² = σ²/(2Y)

This is the elastic analogue of the spring-energy formula ½kx². It appears in problems on energy storage, wire vibrations, and the speed of elastic waves through solids — useful cross-links to the Oscillations and Waves chapters.

Applications — bones, beams, ropes, mountains

Elastic theory is what stops bridges from snapping and helicopters from disintegrating. NCERT closes the chapter with four applied examples that have generated NEET questions and that are worth grasping at the level of intuition.

1. Designing crane ropes. A crane lifting 10 tonnes (10⁴ kg) requires a steel cable whose cross-section can carry the load without exceeding the yield strength (σ_y ≈ 300 × 10⁶ Pa for mild steel). Setting A ≥ Mg/σ_y gives A ≈ 3.3 × 10⁻⁴ m², or a radius of about 1 cm. Engineers apply a safety factor of ten and use about 3 cm — but a single 3-cm wire would be rigid and impractical, so ropes are stranded from many thin wires braided together. This balances strength with flexibility, and exposes more surface for inspection.

2. Beams — why depth matters more than breadth. A beam of length l, breadth b, depth d, loaded at the centre with weight W sags by

δ = W l³ / (4 b d³ Y)

Sag of a centrally loaded beam — depth enters as cube

Notice that δ ∝ 1/d³ but only 1/b. Doubling the depth reduces sag by a factor of 8; doubling the breadth reduces it only by a factor of 2. So for a given amount of material, increasing depth wins. But a thin, deep beam is prone to buckling — sideways collapse under a misaligned load. The compromise is the familiar I-section used in railway tracks, girders, and bridge stringers: deep enough to resist bending, with flanges wide enough to resist buckling, and a thin web in between to save mass. The I-shape is one of the most economical structural cross-sections ever invented.

3. Bones, scaffolds, and the human pyramid. Bone has a Young's modulus of about 9.4 × 10⁹ Pa — about a tenth of steel. NCERT's worked example asks how much each thighbone (femur) of a circus performer at the base of a human pyramid compresses under 1078 N of load. The answer is about 4.5 × 10⁻⁵ m, or 0.0091% strain — biologically negligible. The same elastic theory underlies the design of orthopaedic implants and tissue scaffolds: the scaffold must match the modulus of the surrounding bone to prevent stress shielding, where one component carries all the load and the other atrophies.

4. Mountains and the limit of height. Why is Mount Everest about 9 km high and not 30? At the base of a mountain of height h, the weight per unit area is hρg. This is not pure compression — the sides of the mountain are free, so there is a shearing component approximately equal to hρg itself. If the shearing stress exceeds the rock's critical shear strength (about 30 × 10⁷ Pa for typical rock at ρ = 3 × 10³ kg/m³), the rock flows. Setting hρg equal to this limit gives h ≈ 10 km — neatly above Everest. The Earth's tallest mountain is set by its rocks' shear modulus and yield strength.

5. Bridge pylons and the longitudinal stress in a hanging wire. Suspend a wire from the ceiling with weight W at the bottom. The ceiling exerts a force W upward on the wire's top end, equal and opposite to the weight. But the tension at any cross-section of the wire is just W, not 2W. So the longitudinal stress is W/A, not 2W/A. NEET 2023 asked precisely this — and option (3) W/A is correct. This is the kind of "obvious" trap that catches students who confuse internal tension with external forces.