Stress & Strain
Types, Formulas, Stress-Strain Diagram & Solved Problems — The Foundation of SOM
Last Updated: March 2026
📌 Key Takeaways
- Stress (σ, τ) = internal force per unit area. Normal stress: σ = F/A. Shear stress: τ = F/A.
- Strain (ε, γ) = deformation per unit length. Normal strain: ε = ΔL/L. Shear strain: γ = tan(φ) ≈ φ.
- Hooke’s law: σ = Eε (within elastic limit). E = Young’s modulus.
- The stress-strain curve for mild steel shows: proportional limit → elastic limit → yield → strain hardening → ultimate stress → necking → fracture.
- Tensile stress (pulling) is positive. Compressive stress (pushing) is negative.
- Factor of Safety = failure strength / working stress — always > 1 for safe design.
1. Stress — Definition & Types
When an external force is applied to a solid body, internal forces develop throughout the material to resist the applied load. Stress is the intensity of these internal forces — the force acting on a unit area of material.
Normal Stress
σ = F / A
Where: σ = normal stress (Pa or N/m²), F = force perpendicular to the cross-section (N), A = cross-sectional area (m²)
1 MPa = 10⁶ Pa = 1 N/mm²
| Type | Direction of Force | Sign Convention | Effect |
|---|---|---|---|
| Tensile stress | Pulling — away from the section | Positive (+) | Elongation |
| Compressive stress | Pushing — into the section | Negative (−) | Shortening |
Shear Stress
τ = F / A
Where F is the force acting parallel (tangential) to the cross-section.
Shear stress causes angular distortion, not elongation or shortening.
Bearing Stress
σb = F / (d × t)
Contact stress between a pin/bolt and the plate it passes through. d = pin diameter, t = plate thickness.
2. Strain — Definition & Types
Strain is the measure of deformation produced by stress. It is the ratio of change in dimension to the original dimension — and is therefore dimensionless (no units).
Normal (Linear) Strain
ε = ΔL / L₀
Where: ε = normal strain (dimensionless), ΔL = change in length (m), L₀ = original length (m)
Tensile strain: positive (elongation). Compressive strain: negative (shortening).
Shear Strain
γ = tan(φ) ≈ φ (for small angles)
Where φ = angular distortion in radians. Shear strain measures how much a rectangular element distorts into a parallelogram.
Volumetric Strain
εv = ΔV / V₀ = εx + εy + εz
Sum of linear strains in three perpendicular directions.
Lateral Strain & Poisson’s Ratio
ν = −(lateral strain) / (axial strain) = −εlateral / εaxial
Poisson’s ratio ν is typically 0.25–0.35 for metals. When you stretch a rubber band, it gets thinner — that thinning is the lateral strain.
Theoretical range: 0 ≤ ν ≤ 0.5. Cork: ν ≈ 0. Rubber: ν ≈ 0.5 (incompressible). Steel: ν ≈ 0.3.
3. Hooke’s Law
Within the elastic limit, stress is directly proportional to strain. This is Hooke’s law — the most fundamental relationship in strength of materials.
Hooke’s Law — Normal Stress
σ = E × ε
E = Young’s modulus (modulus of elasticity). Units: Pa, GPa.
E measures stiffness — higher E means less deformation for the same stress.
Hooke’s Law — Shear
τ = G × γ
G = shear modulus (modulus of rigidity). Units: Pa, GPa.
Deformation Formula
From σ = F/A and ε = ΔL/L and σ = Eε:
ΔL = FL / (AE)
This is the most-used deformation formula in SOM. It gives the elongation (or shortening) of a bar under axial load.
| Material | E (GPa) | G (GPa) | ν | σyield (MPa) |
|---|---|---|---|---|
| Mild steel | 200 | 80 | 0.30 | 250 |
| Aluminium | 70 | 26 | 0.33 | 270 |
| Copper | 120 | 44 | 0.34 | 70 |
| Cast iron | 100–170 | 40–65 | 0.26 | — (brittle) |
| Rubber | 0.01–0.1 | — | ~0.50 | — |
| Concrete | 20–40 | — | 0.15 | — (brittle) |
4. Stress-Strain Curve for Mild Steel
The stress-strain diagram is obtained from a standard tensile test. For mild steel (the most important material in engineering), the curve has distinct regions:
| Region / Point | What Happens | Key Property |
|---|---|---|
| O to A: Proportional limit | Stress ∝ strain (perfectly linear). Hooke’s law is valid. | Slope = E (Young’s modulus) |
| A to B: Elastic limit | Slight deviation from linearity, but deformation is still fully recoverable. | Remove load → returns to original shape |
| B to C: Upper yield point | Material begins to yield — permanent deformation starts. Stress drops suddenly. | Upper yield stress (σuy) |
| C to D: Lower yield / Yield plateau | Material deforms at approximately constant stress. Strain increases with no stress increase. | Lower yield stress (σly) — used for design |
| D to E: Strain hardening | Material strengthens as crystal structure rearranges. Stress rises again. | Material becomes harder and stronger |
| E: Ultimate Tensile Strength (UTS) | Maximum stress the material can withstand. Necking begins. | σu = UTS (engineering stress) |
| E to F: Necking & Fracture | Cross-section reduces locally (neck forms). Engineering stress drops. Specimen fractures at F. | Fracture stress (apparent drop due to reduced area) |
Important: The curve described above uses engineering stress (σ = F/A₀, based on original area). True stress (σ = F/Aactual) continues to increase right up to fracture because the actual cross-sectional area is shrinking during necking.
5. Ductile vs Brittle Materials
| Feature | Ductile Material | Brittle Material |
|---|---|---|
| Behaviour before failure | Large plastic deformation — warns before breaking | Little or no plastic deformation — breaks suddenly |
| Stress-strain curve | Long plastic region, distinct yield point, necking | Nearly linear to fracture, no yield point |
| Failure mode | Shear failure (45° fracture surface in tension) | Normal stress failure (flat fracture surface) |
| % Elongation at fracture | > 5% (often 20–40%) | < 5% |
| Design criterion | Yield strength (σy) | Ultimate strength (σu) |
| Examples | Mild steel, aluminium, copper, gold | Cast iron, glass, concrete, ceramics |
6. Factor of Safety
Factor of Safety (FOS)
For ductile materials: FOS = σyield / σworking
For brittle materials: FOS = σultimate / σworking
FOS must always be greater than 1. Higher FOS = safer but heavier/costlier design.
| Application | Typical FOS |
|---|---|
| Aircraft structures | 1.5–2.0 |
| Machine components (static) | 2–3 |
| Machine components (fatigue) | 3–5 |
| Building structures | 3–5 |
| Pressure vessels | 3–4 |
| Brittle materials, unknown loads | 5–10 |
7. Thermal Stress
When a material is heated or cooled and is prevented from expanding or contracting freely, internal stresses develop. These are called thermal stresses.
Free Thermal Expansion
ΔL = αLΔT
Where: α = coefficient of linear thermal expansion (1/°C or 1/K), L = original length, ΔT = temperature change
Thermal Stress (Fully Constrained Bar)
σ = EαΔT
Heating a constrained bar → compressive stress. Cooling a constrained bar → tensile stress.
Thermal stress problems are common in GATE, especially for composite bars (two materials joined together with different α values) and bars between rigid walls.
8. Worked Numerical Examples
Example 1: Axial Stress and Deformation
Problem: A steel rod (E = 200 GPa) has diameter 20 mm and length 1.5 m. It carries a tensile load of 50 kN. Find the stress and elongation.
Solution
A = π(0.02)²/4 = 3.142 × 10⁻⁴ m²
σ = F/A = 50,000 / 3.142 × 10⁻⁴ = 159.2 MPa
ΔL = FL/(AE) = 50,000 × 1.5 / (3.142 × 10⁻⁴ × 200 × 10⁹)
= 75,000 / 62,840,000 = 1.193 × 10⁻³ m = 1.19 mm
Example 2: Factor of Safety
Problem: A component made of mild steel (σy = 250 MPa) operates under a working stress of 80 MPa. Find the factor of safety.
Solution
FOS = σy / σworking = 250 / 80 = 3.125
The design has a FOS of 3.125 — adequately safe for most static loading applications.
Example 3: Thermal Stress
Problem: A steel rail (E = 200 GPa, α = 12 × 10⁻⁶ /°C) is fixed between two rigid walls at 20°C. If the temperature rises to 50°C, find the thermal stress.
Solution
ΔT = 50 − 20 = 30°C
σ = EαΔT = 200 × 10⁹ × 12 × 10⁻⁶ × 30 = 72 × 10⁶ Pa = 72 MPa (compressive)
The rail is in compression because it wants to expand but cannot.
Example 4: Poisson’s Effect
Problem: A copper bar (E = 120 GPa, ν = 0.34) of diameter 30 mm carries a tensile load of 80 kN. Find the change in diameter.
Solution
A = π(0.03)²/4 = 7.069 × 10⁻⁴ m²
σ = 80,000 / 7.069 × 10⁻⁴ = 113.2 MPa
εaxial = σ/E = 113.2 × 10⁶ / 120 × 10⁹ = 9.43 × 10⁻⁴
εlateral = −ν × εaxial = −0.34 × 9.43 × 10⁻⁴ = −3.207 × 10⁻⁴
Δd = εlateral × d = −3.207 × 10⁻⁴ × 30 = −0.00962 mm
Diameter decreases by about 0.0096 mm — the bar gets thinner when pulled.
9. Common Mistakes Students Make
- Using diameter instead of area in σ = F/A: Stress is force divided by cross-sectional AREA (πD²/4 for circular sections), not by diameter. This is the most common numerical error in SOM.
- Mixing up engineering and true stress: Engineering stress uses the original area (A₀). True stress uses the current area. The stress-strain curve in textbooks is usually engineering stress-strain. True stress never decreases before fracture.
- Forgetting sign conventions: Tensile stress/strain are positive; compressive are negative. Forgetting this leads to wrong answers in problems involving combined loading or thermal stresses.
- Applying Hooke’s law beyond the elastic limit: σ = Eε is only valid within the proportional/elastic limit. Beyond yielding, the relationship is no longer linear and Hooke’s law gives incorrect results.
- Using wrong units: If stress is in MPa (N/mm²), make sure area is in mm² and force in N. Or use Pa with m² consistently. Mixed units are the fastest path to wrong answers.
10. Frequently Asked Questions
What is the difference between stress and strain?
Stress (σ = F/A) is the internal force per unit area that resists deformation — measured in Pa or MPa. Strain (ε = ΔL/L) is the resulting deformation per unit original length — it is dimensionless. Stress is the cause; strain is the effect. They are connected by Hooke’s law (σ = Eε) within the elastic region.
What are the key points on the stress-strain curve for mild steel?
The key points are: proportional limit (end of linear region), elastic limit (start of permanent deformation), upper and lower yield points (material flows plastically), ultimate tensile strength (maximum stress before necking), and fracture point (specimen breaks). Mild steel has a distinctive yield plateau that makes it uniquely suitable for structural design — it gives warning before failure.
What is the factor of safety?
Factor of safety (FOS) is the ratio of failure strength to working stress. For ductile materials: FOS = σyield/σworking. For brittle materials: FOS = σultimate/σworking. It provides a safety margin against uncertainties in loading, material properties, and manufacturing. Typical values range from 1.5 (aerospace) to 10 (critical applications with brittle materials).