When a deforming force acts on a body in static equilibrium, internal restoring forces develop equal and opposite to the applied force. The restoring force per unit area is called stress. SI unit: N/m² or Pascal (Pa). Dimensional formula: [ML⁻¹T⁻²].
Three types of stress & strain: (1) Tensile/Compressive (Longitudinal) Stress — force normal to cross-section; produces longitudinal strain ΔL/L (change in length). (2) Shearing (Tangential) Stress — force parallel to cross-section; produces shearing strain Δx/L = tan θ ≈ θ (angle of deformation). (3) Hydraulic Stress — pressure applied uniformly from all sides; produces volume strain ΔV/V (no shape change).
Strain is a dimensionless ratio (change in dimension / original dimension). Elasticity = tendency to regain original shape. Plasticity = permanent deformation retained (e.g., putty, mud). Substances like rubber/aorta tissue that stretch a lot are called elastomers.
- Stress = restoring force/area (NOT applied force — they're equal in magnitude, opposite in direction).
- Steel more elastic than rubber — it resists deformation more (needs more force for same strain).
- Shearing stress possible ONLY in solids (liquids and gases cannot maintain shear).
- Bulk modulus applicable to solids, liquids AND gases.
- Elastomers (rubber, aorta tissue): large elastic region but don't obey Hooke's law.
A = πr² = π × (10×10⁻³)² = π × 10⁻⁴ = 3.14×10⁻⁴ m²
Stress = F/A = 100×10³ / 3.14×10⁻⁴ = 3.18×10⁸ N/m²
Shearing strain = Δx/L = 2mm / 100mm = 2/100 = 0.02
This equals tan θ ≈ θ (in radians) for small angles.
Volume strain = ΔV/V = 0.5 cm³ / 1000 cm³ = 0.5/1000 = 5×10⁻⁴
Dimensionless quantity (ratio).
Hooke's Law: For small deformations, stress ∝ strain. Stress = k × strain, where k is the modulus of elasticity. This is valid only in the linear (proportional) region of the stress-strain curve.
Stress-Strain Curve regions: O→A: Linear, Hooke's law obeyed, elastic behaviour. A→B: Non-linear but still elastic; B is the yield point (elastic limit), stress = yield strength σᵧ. B→D: Plastic deformation; permanent set occurs. D: Ultimate tensile strength σᵤ (maximum stress). E: Fracture point.
Materials with D & E close together → brittle (glass). Materials with D & E far apart → ductile (steel, copper). Elastic potential energy stored per unit volume: u = ½ × stress × strain = ½σε.
- Hooke's law: stress ∝ strain, valid only in the linear part (O to A).
- Yield point B = elastic limit: beyond this, permanent deformation occurs.
- Ductile materials (steel, copper): large plastic region before fracture — used in bridges, buildings.
- Brittle materials (glass, cast iron): fracture suddenly near ultimate strength — not preferred for load-bearing.
- Elastic PE per unit volume = ½ × stress × strain.
- Steel is MORE elastic than rubber — not less! Counterintuitive but true.
Y = stress/strain, valid only where Hooke's law holds → linear region O to A.
Y = slope of stress-strain graph in the proportional/elastic region. In the plastic region, stress/strain ratio changes continuously.
Strain = ΔL/L = 2×10⁻³/2 = 10⁻³
Volume = A×L = 10⁻⁶ × 2 = 2×10⁻⁶ m³
u = ½σε = ½ × 10⁸ × 10⁻³ = 5×10⁴ J/m³
Total PE = u × V = 5×10⁴ × 2×10⁻⁶ = 0.1 J
Ductility = ability to deform plastically before fracture.
Larger gap between ultimate strength (D) and fracture point (E) → more ductile.
Second material has much larger plastic region → more ductile. First is more brittle.
Young's modulus Y is the ratio of tensile (or compressive) stress to longitudinal strain. It measures a material's resistance to being stretched or compressed lengthwise. Higher Y → stiffer material (needs more force for same strain).
Steel (Y ≈ 200 GPa) > Iron > Copper > Aluminium > Glass > Wood. To increase length of a thin steel wire of 0.1 cm² cross-section by 0.1%, a force of 2000 N is required — compared to just 690 N for aluminium. This is why steel is used in heavy-duty machines and structures.
| Material | Y (10⁹ N/m²) | Yield Strength (10⁶ N/m²) |
|---|---|---|
| Steel | 200 | 250 |
| Iron (wrought) | 190 | 170 |
| Copper | 110 | 200 |
| Aluminium | 70 | 95 |
| Glass | 65 | — |
| Bone | 9.4 | — |
| Wood | 13 | — |
- Y = FL/(AΔL) — memorise this form for numericals.
- Steel most elastic among metals (highest Y) — preferred for structural use.
- Y and G are for SOLIDS only; B is for solids, liquids, and gases.
- Poisson's ratio = lateral strain / longitudinal strain (dimensionless, 0.28–0.33 for metals).
- Elastic PE per unit volume = ½σε = ½Y(ε)² = σ²/(2Y).
A = π × (10⁻²)² = 3.14 × 10⁻⁴ m²
ΔL = FL/(AY) = (10⁵ × 1) / (3.14×10⁻⁴ × 2×10¹¹)
= 10⁵ / (6.28×10⁷) = 1.59×10⁻³ m = 1.59 mm
Same load F and same ΔL: ΔL = FL/(AY) → AY/L = constant
Ys/Yc = (Ls/Lc) × (Ac/As) = (4.7/3.5) × (4×10⁻⁵/3×10⁻⁵)
= 1.343 × 1.333 = 1.79 ≈ 1.8
A = π × (2×10⁻²)² = 1.257×10⁻³ m²
ΔL = FL/(YA) = (1078 × 0.5) / (9.4×10⁹ × 1.257×10⁻³)
= 539 / (11.82×10⁶) = 4.56×10⁻⁵ m ≈ 4.55×10⁻⁵ m (very small!)
Shear modulus G (also called modulus of rigidity) is the ratio of shearing stress to shearing strain. It measures a material's resistance to shape change without volume change. For most materials, G ≈ Y/3. It is relevant only for solids.
Examples: Turning a bolt, pushing a book horizontally, the effect of wind on tall buildings — all involve shearing stress. Shear modulus of steel = 84 GPa; lead = 5.6 GPa (easily sheared).
- G = shearing stress / shearing strain — applicable to SOLIDS only.
- G ≈ Y/3 for most materials (G always less than Y).
- Twisting a wire involves shear modulus, NOT Young's modulus.
- A material with high G resists shape change → rigid (e.g., tungsten G=150 GPa).
- Shear modulus of steel (84 GPa) >> lead (5.6 GPa) → steel is much harder to shear.
Area (narrow face) = 50cm × 10cm = 0.5 × 0.1 = 0.05 m²
Stress = F/A = 9×10⁴ / 0.05 = 1.8×10⁶ N/m²
Shearing strain = Stress/G = 1.8×10⁶ / 5.6×10⁹ = 3.21×10⁻⁴
Δx = strain × L = 3.21×10⁻⁴ × 0.5 = 1.6×10⁻⁴ m = 0.16 mm
F = mg = 100 × 9.8 = 980 N
A (face area) = (0.1)² = 0.01 m², L = 0.1 m
Stress = F/A = 980/0.01 = 9.8×10⁴ N/m²
Δx = (Stress × L)/G = (9.8×10⁴ × 0.1) / 25×10⁹ = 9.8×10³ / 25×10⁹ = 3.92×10⁻⁷ m
For most materials: G ≈ Y/3
G ≈ 200/3 ≈ 66.7 GPa ≈ 67 GPa
Shear modulus is always less than Young's modulus.
Bulk modulus B = −p/(ΔV/V). The negative sign ensures B is positive (pressure increase → volume decrease). It measures resistance to volume change. Applicable to solids, liquids, AND gases.
Compressibility k = 1/B = fractional change in volume per unit pressure increase. Solids: least compressible (B large). Gases: most compressible (B small). Gases are ~10⁶ times more compressible than solids! B(steel) = 160 GPa; B(water) = 2.2 GPa; B(air) = 10⁻⁴ GPa.
- B = −p/(ΔV/V): negative sign because pressure increase → volume decrease.
- Incompressibility of solids: due to tight coupling between neighbouring atoms.
- Compressibility = 1/B. Lower B → higher compressibility (gases most compressible).
- Liquids: B between solids and gases — molecules bound but less tightly than solids.
- Fractional compression ΔV/V = p/B — used to find volume change at ocean depths.
p = hρg = 3000 × 1000 × 10 = 3×10⁷ N/m²
ΔV/V = p/B = 3×10⁷ / 2.2×10⁹ = 1.36×10⁻² = 1.36%
p = 10 × 1.013×10⁵ = 1.013×10⁶ N/m²
ΔV/V = p/B = 1.013×10⁶ / 37×10⁹ = 2.74×10⁻⁵
Very small fractional change — glass is quite incompressible.
p = 100 × 1.013×10⁵ = 1.013×10⁷ Pa
ΔV/V = 0.5/100 = 5×10⁻³
B = p/(ΔV/V) = 1.013×10⁷ / 5×10⁻³ = 2.026×10⁹ Pa ≈ 2.0 GPa
(Matches tabulated value of 2.2 GPa for water — close!)
Crane rope design: For a 10-tonne crane, minimum cross-section A ≥ Mg/σᵧ = 3.3×10⁻⁴ m² (radius ≈ 1 cm). Safety factor of 10 used → rope of ~3 cm radius. Ropes made of thin braided wires for flexibility and strength.
I-shaped beams: A beam of length l, breadth b, depth d, loaded at centre sags by δ = Wl³/(4bd³Y). Increasing depth d is far more effective than breadth b (δ ∝ d⁻³ vs b⁻¹). But deep bars may buckle → I-shape compromise: large load-bearing surface + sufficient depth + reduced weight.
Maximum mountain height: At bottom of mountain height h, shear stress ≈ hρg. For rocks, elastic limit ≈ 3×10⁷ N/m². Setting hρg = 3×10⁷ gives h ≈ 10 km — which explains why no mountain on Earth exceeds this! (Mt. Everest ≈ 8.85 km).
- I-beam cross-section: reduces weight while maintaining bending resistance — used in bridges and buildings.
- Crane ropes: braided thin wires (not single thick wire) for flexibility + strength.
- Sag δ ∝ 1/d³: doubling depth reduces sag by 8 times — much better than doubling breadth (2 times).
- Maximum mountain height ~10 km explained by elastic limit of rocks (shear stress).
- Pillars with distributed (flared) ends support more load than rounded-end pillars.
A = πr² = π × (1.5×10⁻²)² = π × 2.25×10⁻⁴ = 7.069×10⁻⁴ m²
Max load F = stress × A = 10⁸ × 7.069×10⁻⁴ ≈ 70,686 N ≈ 7.07 × 10⁴ N
δ = Wl³/(4bd³Y) → δ ∝ 1/d³
New δ/Old δ = (d/2d)³ = (1/2)³ = 1/8
New sag = 1/8 cm = 0.125 cm
Doubling depth reduces sag by 8 times — very effective!
Load per column = (50000 × 9.8)/4 = 122,500 N
A = π(R²−r²) = π[(0.6)²−(0.3)²] = π[0.36−0.09] = π×0.27 = 0.848 m²
Stress = F/A = 122500/0.848 = 1.445×10⁵ N/m²
Strain = Stress/Y = 1.445×10⁵ / 2×10¹¹ ≈ 7.22×10⁻⁷
Restoring force/area = F/A (Pa) | Tensile, Shear, Hydraulic
ΔL/L (longitudinal) | Δx/L = θ (shear) | ΔV/V (volume)
Y = FL/(AΔL) | Tensile/Compressive | Solids only
G = F/(Aθ) ≈ Y/3 | Shearing stress only | Solids only
B = −p/(ΔV/V) | Hydraulic | Solid, Liquid & Gas
I-beam: δ ∝ d⁻³ | Crane: A≥Mg/σᵧ | Mountain max ≈ 10 km
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