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Elastic Behaviour of Solids
Deforming Force
A force which produces a change in configuration of an object on applying it is called a deforming force. The change in shape, size, or both of a body due to an external force is called deformation. Deformation can be: change in length, change in volume, or change in shape.
| Type of Body | Behaviour | Examples |
|---|
| Perfectly Elastic | Regains original configuration immediately and completely after removal of deforming force | Quartz fibre, steel |
| Partially Elastic (Inelastic) | Partially regains original shape | Most metals partially |
| Perfectly Plastic | Does not regain original configuration at all | Putty, paraffin, wax, clay |
| Rigid Body | Distance between any two points remains same (ideal; no real body is perfectly rigid) | Ideal concept only |
⚠️ Important Fact
Rubber is LESS elastic than steel. Elasticity measures how perfectly a body regains its original shape — not how much it stretches. Steel returns to its original shape far more perfectly than rubber → steel is more elastic.
Elasticity
- Property of a body by virtue of which it regains its original configuration after removal of deforming force
- Restoring force: F = −dU/dr (slope of potential energy curve)
- At mean position: F = 0; attractive and repulsive elastic forces balance
Definition
- Stress = internal restoring force per unit area developed inside a body
- When body is in equilibrium (altered shape): restoring force = deforming force
- SI unit = N/m² or Pascal (Pa); Dimension = [L⁻¹M¹T⁻²]
| Type of Stress | Nature of Force | Formula | Related Modulus |
|---|
| ① Longitudinal (Tensile / Compressive) | Along length of body | |F|/A | Young's Modulus Y |
| ② Volume (Hydrostatic) | Equal from all sides (pressure) | ΔP = |F|/A | Bulk Modulus B (or K) |
| ③ Tangential (Shearing) | Parallel to surface (tangential) | Ftangential/A | Rigidity Modulus η |
Tensile vs Compressive
- Tensile stress — rod pulled at two ends → length increases
- Compressive stress — rod pushed at two ends with equal and opposite forces → length decreases
- Both are longitudinal stress: = |Restoring Force| / Area of cross-section
Definition
- Strain = change in configuration per unit original configuration
- Dimensionless, unitless quantity
| Type of Strain | Formula | Description |
|---|
| Longitudinal strain | ΔL/L = ℓ/L | Change in length / original length |
| Volume strain | ΔV/V | Change in volume / original volume |
| Shear strain | θ = ΔL/L = tan θ ≈ θ (small angle) | Angular displacement of layers |
| Lateral strain | ΔD/D | Change in diameter / original diameter |
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Hooke's Law and Moduli of Elasticity
Hooke's Law
Within the proportional limit: Stress ∝ Strain
Modulus of Elasticity (E) = Stress / Strain = constant
Slope of stress-strain graph in proportional region = Modulus of Elasticity
① Young's Modulus (Y)
Definition
- Modulus of elasticity related to change in length of an object (wire, rod, beam)
- Also called elasticity of length
- Y = Longitudinal stress / Longitudinal strain
Y = (F/A) / (ΔL/L) = FL / (AΔL)
∴ F = YAΔl / L (restoring force)
SI unit: N/m² | Dimension: [L⁻¹M¹T⁻²] | Only for solids
Q: A wire of length 20 m, area 1.025×10⁻⁴ m² is subjected to load 2.5 kg (F = 2.5×9.8 N). Elongation = 1×10⁻⁴ m. Find Young's Modulus.
Y = FL/(AΔL) = (2.5×9.8×20) / (1.025×10⁻⁴ × 10⁻⁴) = 490 / (1.025×10⁻⁸) = 3.92×10¹⁰ N/m²
Ans: Y ≈ 3.92×10¹⁰ N/m² ✓
Q: A brass wire (L = 4.5 m, A = 3×10⁻⁵ m²) and copper wire (L = 5 m, A = 4×10⁻⁵ m²) stretched by same load produce same elongation. Find Ybrass/Ycopper.
Y = FL/(AΔL); same F and same ΔL: YB/YC = (LB/AB) / (LC/AC) = (4.5/3×10⁻⁵) / (5/4×10⁻⁵) = (4.5×4)/(3×5) = 18/15 = 1.2
Ans: YB/YC = 1.2 ✓
② Bulk Modulus (B or K)
Definition
- Modulus of elasticity related to change in volume
- Measures resistance offered by gases, liquids, or solids to change in volume
- Valid for: solids, liquids, and gases
B = −ΔP / (ΔV/V) = −V·ΔP / ΔV
Compressibility = 1/B (SI unit: m²/N or Pa⁻¹)
Negative sign because volume decreases when pressure increases. Bsolid > Bliquid > Bgas
Bulk Modulus in Different Processes (for Gases)
- Isothermal process (T = const, PV = const): B = P
- Adiabatic process (PVγ = const): B = γP
- Isobaric process (P = const, ΔP = 0): B = 0
- Isochoric process (V = const, ΔV = 0): B = ∞
Q: A metal cube (side 1 m) subjected to force. Volume changes by 1.5×10⁻⁵ m³. Bulk modulus = 6.6×10¹⁰ N/m². Find change in pressure.
V = 1 m³; B = ΔP·V/ΔV → ΔP = B·ΔV/V = 6.6×10¹⁰ × 1.5×10⁻⁵/1 = 9.9×10⁵ N/m²
Ans: ΔP = 9.9×10⁵ N/m² ✓
③ Modulus of Rigidity (η) — Shear Modulus
Definition
- Modulus of elasticity related to change in shape of an object
- Also known as shear modulus of elasticity
- Valid only for solids (liquids and gases cannot sustain shear)
η = Shearing stress / Shearing strain = F / (A·θ)
where θ = shear strain = ΔL/L (angle in radians for small deformation)
Q: Metal cube side 40 cm, shearing force 2000 N, top surface displaced 0.5 cm. Find modulus of rigidity.
A = 0.40² = 0.16 m²; θ = ΔL/L = 0.005/0.40 = 0.0125; η = F/(Aθ) = 2000/(0.16×0.0125) = 2000/0.002 = 1.0×10⁶ N/m²
Ans: η = 1.0×10⁶ N/m² ✓
Definition
- Lateral strain = ΔD/D (ratio of change in diameter to original diameter)
- Longitudinal strain = Δℓ/L (ratio of change in length to original length)
- Poisson's Ratio σ = Lateral strain / Longitudinal strain = (ΔD/D) / (Δℓ/L)
- Dimensionless; no unit
- Theoretically: −1 ≤ σ ≤ 0.5 | Practically: 0 ≤ σ ≤ 0.5
- For incompressible material (V = const): σ = 0.5; B = ∞
Regions of the Stress-Strain Graph (for metals)
- Region OA (Proportional Limit): Straight line. Hooke's law valid; stress ∝ strain. Slope = Young's Modulus. Point A = Proportional Limit.
- Region OB (Elastic Limit / Yield Point): Beyond A, Hooke's law not valid, but body still regains shape. Point B = Elastic Limit or Yield Point. Corresponding stress = Yield Strength.
- Region BC (Plastic Flow / Permanent Set): Beyond B, load removed → body does NOT fully regain length. Permanent set occurs. Material undergoes plastic deformation.
- Region BD (Plastic Deformation to Fracture): Further increase in load → large increase in strain for small increase in stress (ductile materials) or immediate fracture (brittle). Point D = Fracture / Breaking Point.
- Breaking Stress = Minimum stress to break a wire = Fixed for a material; does NOT depend on area. Breaking Force = Breaking Stress × Area (depends on area).
| Material Type | Plastic Range | Examples | Uses |
|---|
| Ductile | Large plastic range beyond elastic limit | Copper, silver, aluminium, iron | Making springs, sheets, wires |
| Brittle | Very small plastic range | Glass, cast iron, ceramics | Used as they are; cannot be drawn |
| Elastomers | No plastic range | Rubber, aorta tissue | Large strain for small stress; used in shock absorbers |
| Malleable | Can be hammered into thin sheets | Gold, silver | Jewellery, foils |
Elastic After-Effect, Elastic Fatigue, Elastic Relaxation
- Elastic After-Effect: Temporary delay in regaining original configuration after removal of deforming force
- Elastic Relaxation Time: Time delay in restoring original configuration. For quartz, this time is negligible (near-perfect elastic)
- Elastic Fatigue: Property by which elastic body becomes less elastic under action of repeated alternation of deforming force (e.g. bridge metals fatigue over time)
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Strain Energy (Elastic Potential Energy in Wire)
Definition
Elastic potential energy gained by a wire during elongation by a stretching force is called strain energy. It equals the work done by the stretching force.
Work done = ½ × Load × Extension = ½ × F × ΔL
Strain Energy = ½ × F × ΔL = ½ × (YA/L) × (ΔL)² = ½ × Stress × Strain × Volume
Strain Energy per unit volume (u) = ½ × Stress × Strain
u = ½ × Y × (Strain)² = ½ × (Stress)² / Y
Area under stress-strain graph = energy per unit volume (energy density)
Derivation of Strain Energy
- Force at extension x: F = YAx/L
- Work for further extension dx: dW = F·dx = (YAx/L)dx
- Total work: W = ∫₀ˡ (YAx/L)dx = YA/L × [x²/2]₀ˡ = YAl²/(2L) = ½ × (YAl/L) × l = ½Fl
- Therefore: Strain Energy = ½ × F × ΔL (same as spring energy = ½kx²)
Q: A wire of length 2 m, area 1 mm² = 10⁻⁶ m², Y = 2×10¹¹ N/m² stretched by ΔL = 1 mm. Find strain energy.
F = YAΔl/L = 2×10¹¹×10⁻⁶×10⁻³/2 = 100 N; Strain energy = ½×F×ΔL = ½×100×10⁻³ = 0.05 J
Ans: U = 0.05 J ✓
Q: Breaking stress of steel = 8×10⁸ N/m², density = 8000 kg/m³. Find max length of wire before breaking under own weight.
Stress = ρgL → 8×10⁸ = 8000×10×L → L = 10⁴ m = 10 km
Ans: Lmax = 10 km ✓
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Properties of Common Materials
| Material | Y (×10¹⁰ N/m²) | B (×10¹⁰ N/m²) | η (×10¹⁰ N/m²) | Poisson's σ |
|---|
| Steel | 21.0 | 16.0 | 8.3 | 0.28 |
| Copper | 11.0 | 14.0 | 4.04 | 0.37 |
| Brass | 9.0 | 6.0 | 3.5 | 0.37 |
| Aluminium | 7.0 | 7.5 | 2.5 | 0.36 |
| Gold | 8.1 | 18.0 | 2.9 | 0.42 |
| Silver | 7.6 | 10.0 | 2.7 | 0.38 |
| Glass (crown) | 6.0 | 6.0 | 2.5 | 0.2 |
| Lead | 1.5 | 4.1 | 0.6 | — |
Methods of Reducing Friction
- Use of lubricants, oil and grease in different parts of a machine
- Use of ball bearings converts kinetic friction into rolling friction (rolling friction is much less)
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Fluid Mechanics — Pressure and Pascal's Law
Pressure in Fluids
- Pressure = Force / Area; scalar; SI unit = Pa (N/m²)
- Pressure at depth h: P = P₀ + ρgh (P₀ = atmospheric pressure = 1.013×10⁵ Pa)
- Pressure depends only on depth, fluid density, and g — NOT on shape or volume of container
- Pressure is same at all points at same horizontal level in a connected fluid
Pascal's Law: Pressure applied to enclosed fluid is transmitted equally and undiminished in all directions
Hydraulic press: F₁/A₁ = F₂/A₂ → F₂ = F₁(A₂/A₁) (mechanical advantage = A₂/A₁)
Archimedes' Principle: Upthrust = weight of fluid displaced = ρfluid·Vsubmerged·g
| Condition | Result |
|---|
| ρbody < ρliquid | Floats; fraction submerged = ρbody/ρliquid |
| ρbody = ρliquid | Neutral buoyancy; stays anywhere in liquid |
| ρbody > ρliquid | Sinks; apparent weight = actual weight − buoyancy |
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Fluid Dynamics — Bernoulli's Theorem
Equation of Continuity: A₁v₁ = A₂v₂ (mass conservation; incompressible fluid)
Bernoulli's Theorem: P + ½ρv² + ρgh = constant (along a streamline)
Torricelli's Theorem: Speed of efflux from hole at depth h = √(2gh)
Applications
- Venturimeter: measures flow rate using pressure difference at throat and wide section
- Pitot tube: measures speed of aircraft using pressure difference
- Atomizer / Spray pump: fast air over nozzle → low pressure → liquid rises
- Aerofoil / Aircraft wing: faster flow above wing → lower pressure → upward lift
- Magnus Effect: spinning ball curves due to pressure difference on two sides
Definition
- Viscosity = internal friction between layers of fluid in relative motion
- Coefficient of viscosity η; SI unit = Pa·s = N·s/m²; Dimension [ML⁻¹T⁻¹]
- Temperature effect: for liquids, η decreases with temperature; for gases, η increases with temperature
Newton's Viscosity Law: F = ηA(dv/dy) (viscous force = η × area × velocity gradient)
Stokes' Law: F = 6πηrv (drag on sphere of radius r at velocity v)
Terminal velocity: vT = 2r²(ρ−σ)g / (9η)
Poiseuille's Law: Q = πr⁴ΔP / (8ηL) (volume flow rate through tube)
Re = ρvD/η (Reynolds number); streamline if Re < 2000; turbulent if Re > 3000
⚡ Terminal Velocity — Key Facts
- At terminal velocity: Weight = Buoyancy + Drag → mg = ρfluidVg + 6πηrvT
- vT ∝ r² (larger radius → much faster terminal velocity)
- vT ∝ (ρ−σ) → same speed in vacuum regardless of density
- vT ∝ 1/η → lower viscosity fluid → faster terminal velocity
Definition
- Surface tension T = Force per unit length = F/L
- Also = Surface energy per unit area = U/A
- Scalar; SI unit = N/m; Dimension [MT⁻²]
- Arises due to cohesive forces between liquid molecules in surface layer
- With temperature↑: T decreases (becomes 0 at critical temperature)
- Impurities: soap/detergent/organic solutes decrease T; inorganic salts generally increase T
Excess pressure inside soap bubble: ΔP = 4T/r (two free surfaces)
Excess pressure inside water droplet / air bubble in liquid: ΔP = 2T/r
Capillary rise: h = 2T cosθ / (ρgr)
Work done forming soap bubble of radius r: W = 8πr²T
| Shape | Free Surfaces | Excess Pressure |
|---|
| Soap bubble in air | 2 | 4T/r |
| Water droplet in air | 1 | 2T/r |
| Air bubble in liquid | 1 | 2T/r |
| Soap bubble on glass plate | 1 (only curved surface) | 2T/r |
Contact Angle and Capillarity
- θ < 90° → liquid wets surface → concave meniscus → capillary rise (e.g. water in glass)
- θ > 90° → liquid does not wet → convex meniscus → capillary depression (e.g. mercury in glass)
- θ = 0° → complete wetting; maximum rise
- Capillary rise h ∝ 1/r → finer capillary = more rise
- Jurin's law: h·r = 2T cosθ/(ρg) = constant for a liquid-surface pair
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Thermal Properties — Expansion
Thermal Expansion
- On heating: solids, liquids, gases expand (exception: water 0°C–4°C contracts on heating)
- Linear expansion: ΔL = LαΔT; LT = L₀(1 + αΔT)
- Superficial (areal) expansion: ΔA = A(2α)ΔT = AβΔT; β = 2α
- Volumetric expansion: ΔV = V(3α)ΔT = VγΔT; γ = 3α
- Relation: α : β : γ = 1 : 2 : 3
- α depends on material; for steel ≈ 1.1×10⁻⁵ /°C; for copper ≈ 1.7×10⁻⁵ /°C
Thermal stress (rod fixed at both ends): σ = YαΔT
Thermal force: F = YAαΔT
Anomalous expansion of water: density maximum at 4°C; water expands on cooling below 4°C
Specific Heat and Calorimetry
- Q = msΔT (s = specific heat capacity; unit J/kg·K)
- Latent heat: Q = mL (L = latent heat; unit J/kg) — phase change without temperature change
- Lfusion (ice) = 336 kJ/kg = 80 cal/g
- Lvaporization (water) = 2260 kJ/kg = 540 cal/g
- Principle of calorimetry: Heat gained = Heat lost (thermally isolated system)
- Specific heat of water = 4200 J/kg·K; of ice = 2100 J/kg·K; of steam = 2010 J/kg·K
Modes of Heat Transfer
- Conduction: Q/t = KA(T₁−T₂)/L (K = thermal conductivity; unit W/m·K)
- Convection: transfer by actual movement of heated fluid (natural or forced)
- Radiation: transfer by electromagnetic waves; no medium required
Newton's Law of Cooling and Stefan's Law
Newton's Law of Cooling: dT/dt = −b(T − T₀) — rate of cooling proportional to excess temperature above surroundings. Valid for small temperature differences.
Stefan-Boltzmann Law: E = σT⁴ (energy radiated per unit area per unit time). σ = 5.67×10⁻⁸ W/m²·K⁴
Wien's Displacement Law: λmax·T = b = 2.898×10⁻³ m·K (peak wavelength of blackbody radiation shifts with temperature)
Q: A steel rod (L = 1 m, α = 1.1×10⁻⁵/°C, Y = 2×10¹¹ N/m², A = 1 cm²) is fixed at both ends. Temperature raised by 40°C. Find thermal force developed.
F = YAαΔT = 2×10¹¹ × 10⁻⁴ × 1.1×10⁻⁵ × 40 = 2×10¹¹ × 10⁻⁴ × 4.4×10⁻⁴ = 8800 N
Ans: F = 8800 N = 8.8 kN ✓
Q: A soap bubble of radius 2 cm has surface tension T = 0.03 N/m. Find excess pressure inside bubble.
ΔP = 4T/r = 4×0.03/0.02 = 6 Pa
Ans: ΔP = 6 Pa ✓
Q: A steel ball (r = 2 mm, ρ = 8000 kg/m³) falls through oil (η = 0.5 Pa·s, σ = 850 kg/m³). Find terminal velocity.
vT = 2r²(ρ−σ)g/(9η) = 2×(2×10⁻³)²×(8000−850)×10/(9×0.5) = 2×4×10⁻⁶×7150×10/4.5 = 0.127 m/s
Ans: vT ≈ 0.127 m/s ✓