Mechanical Properties of Solids

Imagine you pull a rubber band in your room — it gets longer and when you leave it, it comes back. Now press a lump of dough kept in kitchen — it stays pressed, does not come back. Why difference? Because every solid has some internal property to fight back when you deform it. That property is called elasticity.

Till now we treated bodies as rigid — fixed shape, fixed size. But real life, even a steel rod stretches a little when you hang weight on it. The cycle spoke becomes tight, the fan rod bends a little when wind pushes, the gym spring expands when you pull. This chapter is all about that small but very important deformation and how to calculate it.

We will learn what is stress, what is strain, why steel is more elastic than rubber even though rubber stretches more, why bridges use I-shaped beams, why rope for crane has to be thick, why mountains cannot be taller than about 10 km. All from one simple idea: stress is proportional to strain for small deformations.

Think of your gym resistance band. You pull it with both hands — your hands apply force, band resists inside. That internal resisting force per unit area is called stress. Formula: stress = magnitude of restoring force per area = F/A. Unit is N/m² which we call Pascal Pa, same as pressure. Dimensional formula is ML⁻¹T⁻².

But force can be applied in three different ways, so stress has three types. First, you pull a cylinder from both ends normal to its cross-section — this is tensile stress, length increases. If you push from both ends, it is compressive stress, length decreases. Both are called longitudinal stress. Second, you apply force parallel to surface, like pushing top of a book while bottom is fixed — the book slants. This is shearing or tangential stress. Third, you put a ball inside water deep — water presses from all sides equally — volume shrinks but shape stays same. This is hydraulic stress which equals pressure.

Now strain — how much deformation happened compared to original size. Since different stresses produce different deformations, we have three strains. Longitudinal strain = change in length ΔL divided by original length L. Shearing strain = relative sideways shift Δx divided by original length L = tanθ ≈ θ when θ small, angle in radians. Volume strain = change in volume ΔV divided by original volume V. Strain has no unit — it is ratio, dimensionless.

Daily example: When you stretch a rubber band, longitudinal strain is big, you can see it. When you push a stack of cards sideways, shearing strain you see as slant. When you press a plastic bottle full of air inside water, volume strain but shape same. All these are small for metals — you need microscope to see, but they are there.

For small pulls, if you double the force, the stretch also doubles. This linear relation was first noted by Robert Hooke: stress is proportional to strain. So stress = k × strain, where k is a constant called modulus of elasticity which depends on material, not on shape. This is Hooke's law, valid only for small deformations, empirical — most materials follow it up to a limit.

If you actually do experiment — take a wire, hang weights step by step, measure ΔL and plot stress vs strain — you get a typical curve for metal. Let's understand it point by point. O to A is straight line — here Hooke's law obeyed perfectly, if you remove load, wire comes back to original length. This is ideal elastic region. A to B — curve bends, still elastic, wire still comes back, but stress not proportional to strain. B is very important — yield point or elastic limit, stress is called yield strength σy. Up to B, body is elastic.

Beyond B, if you increase load further, strain increases very fast for small stress — B to D region. If you unload at point C between B and D, wire does NOT come back fully, some permanent strain remains — this is plastic deformation, like dough. D is ultimate tensile strength σu — maximum stress material can bear. After D, even if you reduce force, strain keeps increasing and at E wire breaks — fracture point. If D and E are close, material breaks soon after ultimate — brittle like glass. If D and E far, it stretches a lot before breaking — ductile like copper, you can make wires.

But there are special materials called elastomers — rubber, elastic tissue of aorta in heart. Their curve is very different — you can stretch rubber to several times its length and it still comes back, no plastic region, very large elastic region, but Hooke's law not obeyed over most of it. That is why aorta can expand and contract with blood pressure.

When you talk about longitudinal pull, the modulus we use is Young's modulus Y. Definition: Y = tensile stress / tensile strain = (F/A) / (ΔL/L) = F×L / A×ΔL. Rearranged, ΔL = F×L / A×Y. So for same F, L, A — larger Y means smaller ΔL, less stretch. Unit of Y is same as stress Pa.

Here comes the confusion that scares first-timers: Rubber stretches more, so it feels more elastic in daily language. But in physics, more elastic means it resists stretch more, needs more force for same strain. So steel with Y≈200 GPa is much more elastic than rubber with Y≈0.05 GPa. To get 0.1% strain in steel wire of 0.1 cm² you need 2000 N, same strain in aluminium needs 690 N, brass 900 N, copper 1100 N — so steel needs biggest force.

Table values you must remember roughly: Steel 200×10⁹ Pa, Iron 190, Copper 110, Aluminium 70, Bone 9.4, Glass ~65. All are large, so metals need large force for small change. That is why steel is used in heavy machines, bridges, buildings — it does not deform easily.

Numerical trick: Many questions give two wires copper and steel same load and area, total elongation given, find load. Use W/A = Yc ΔLc/Lc = Ys ΔLs/Ls. Ratio ΔLc/ΔLs = Ys/Yc × Lc/Ls. Then use sum given to find each.

Example: Steel rod radius 10 mm length 1 m force 100 kN, Y=2e11 Pa. Stress = F/A = 100kN / πr² = 100000 / 3.14e-4 = 3.18e8 Pa. Elongation ΔL = FL/AY = (F/A)L/Y = 3.18e8×1/2e11=1.59 mm. Strain = 1.59e-3 =0.16% very small.

For sideways force, we use shear modulus G also called modulus of rigidity. Imagine a square slab fixed at bottom, you push top face with force F parallel. Top moves by Δx, height is L, shear strain = Δx/L = tanθ ≈ θ. Shear stress = F/A where A is area of face. Then G = shear stress / shear strain = (F/A) / (Δx/L) = FL / AΔx = F / Aθ. SI unit Pa.

Typical values: Steel 84 GPa, Iron 70, Copper 42, Brass 36, Aluminium 25, Glass 23, Lead 5.6, Wood 10. Notice G is generally about one third of Y, G≈Y/3 for most materials. That means shear is easier than stretch.

Example: Lead slab side 50 cm thickness 10 cm, shearing force 9e4 N on narrow face. Area parallel = 0.5×0.1=0.05 m², stress = 1.8e6 Pa, strain = stress/G =1.8e6/5.6e9=3.2e-4, displacement Δx= strain×L=3.2e-4×0.5=0.16 mm very small.

Now hydraulic case — when pressure from all sides. Volume strain = ΔV/V, hydraulic stress = pressure p. Bulk modulus B = -p / (ΔV/V). Negative sign because if p positive, ΔV negative — volume decreases. B is always positive. Its reciprocal is compressibility k=1/B = -(1/Δp)(ΔV/V) — fractional volume change per unit pressure increase.

Values: Solids B much larger than liquids, liquids larger than gases. Aluminium 72 GPa, Copper 140, Glass 37, Iron 100, Steel 160, Water 2.2 GPa, Air at STP 1.0×10⁻⁴ GPa. So solids least compressible, gases most — about million times more compressible than solids! That is because atoms in solids tightly coupled.

Example: Average depth of ocean 3000 m, pressure p=hρg=3000×1000×10=3e7 Pa, B water 2.2e9 Pa, ΔV/V = p/B =3e7/2.2e9=1.36×10⁻² =1.36% compression at bottom — noticeable.

When you stretch a wire lengthwise, it becomes thinner sideways — diameter contracts. This sideways strain is called lateral strain Δd/d. Longitudinal strain is ΔL/L. Poisson pointed out that within elastic limit, lateral strain is proportional to longitudinal strain. Ratio is Poisson's ratio ν = lateral strain / longitudinal strain magnitude? More precisely ν = -(Δd/d)/(ΔL/L) negative sign to make ν positive because Δd negative when ΔL positive. So ν = -(lateral)/(longitudinal). It is pure number no unit, depends only on material. For steels 0.28-0.30, aluminium alloys about 0.33, rubber ~0.5.

Now energy — when you stretch a wire, you do work against interatomic forces. This work is stored as elastic potential energy. Let's derive: Wire original length L area A Young Y. At some extension l, force needed F = Y A × l/L from Y formula. For further small extension dl, work dW = F dl = Y A l dl /L. Integrate from 0 to l: W = ∫ Y A l dl /L =½ Y A l² /L. This is stored energy U. We can write U =½ × (Y A l/L) × l =½ × F × ΔL =½ × (F/A)×(ΔL/L)×(A L) =½ × stress×strain×volume.

Energy density u = energy per volume = U/Volume =½ σ ε =½ Y ε² =½ (stress²/Y). Same for shear: u=½ G θ². Unit J/m³.

Example: Steel wire Y=200 GPa strain 0.001 (0.1%) volume 1e-6 m³. u=½×200e9×(1e-3)²=½×200e9×1e-6=100 J/m³. Total U= u×Vol=1e-4 J=0.1 mJ tiny — because strain small. That is why elastic energy in metal is small but in spring large.

Elasticity is not just theory — it decides how we build things. Two classic applications you must know for exams and for life.

First — crane rope thickness: Suppose we want crane to lift 10 tonnes =10000 kg weight W=Mg=10⁴×9.8≈1e5 N. Rope must not exceed elastic limit, otherwise permanent stretch and it can break. Yield strength σy for mild steel about 300×10⁶ N/m². So minimum area A ≥ W/σy =1e5/3e8=3.3×10⁻⁴ m² radius about 1 cm for circular cross-section. But we always keep safety factor about 10 times — so design for 10× load — radius about 3 cm. That is why crane ropes are thick and made of many thin braided wires — flexibility plus strength.

Second — beam bending, why I-shape: A beam supported at ends loaded at centre sags by amount δ = W l³ / (4 b d³ Y) where l span, b breadth width, d depth height, Y Young's modulus. To reduce sag, you can increase b or d or use larger Y material. But formula shows δ ∝1/b and ∝1/d³ — depth is cubic more effective! Doubling depth makes beam 8 times stiffer, doubling breadth only 2 times stiffer. So increasing depth is much better.

But if you make bar very deep and thin, it can bend sideways — buckling as shown in diagrams. To avoid buckling and save material, we use I-section — top and bottom flanges have most material far from neutral axis, middle web thin. This gives large d for bending stiffness, high second moment of area I for low weight, resists buckling, saves 30-40% steel. That is why bridges, buildings, railway tracks have I shape.

Other examples: Pillars with distributed ends support more load than rounded ends. Mountain height limited — at bottom stress = hρg, if this exceeds critical shear stress for rock ~30×10⁷ N/m², rock flows. With ρ=3×10³ kg/m³ g=10, h=30e7/(3e3×10)=10 km — more than Everest! That is why mountains on Earth cannot be much taller than 10 km. Pyramidal shape of hill is also for stability.