Prerequisites
Before studying stress and strain, make sure you understand:
- Newton’s Laws — Force and equilibrium concepts
- Work and Energy — Elastic potential energy
What is Elasticity?
Elasticity is the property of a material to regain its original shape and size after the deforming force is removed.
Types of Materials:
| Type | Behavior | Examples |
|---|---|---|
| Perfectly Elastic | Returns to original shape completely | Quartz, steel (within limits) |
| Perfectly Plastic | Doesn’t return, permanent deformation | Clay, putty, wet mud |
| Partially Elastic | Partial recovery | Most real materials, rubber |
Stress
Stress is the internal restoring force per unit area that develops in a material when it’s deformed.
$$\boxed{\text{Stress} = \frac{\text{Force}}{\text{Area}} = \frac{F}{A}}$$SI Unit: N/m² or Pascal (Pa)
Practical Units:
- 1 MPa (megapascal) = 10⁶ Pa
- 1 GPa (gigapascal) = 10⁹ Pa
Dimensions: [ML⁻¹T⁻²] (same as pressure)
Stress is a measure of internal resistance to deformation. It’s not the applied force itself, but the force per unit area acting inside the material.
Think of it like population density: same number of people (force) in a smaller area creates higher density (stress).
Types of Stress
1. Tensile Stress (Longitudinal Stress)
When forces try to stretch the material (pulling apart).
Example: Rope supporting a weight, wire hanging a load
$$\boxed{\text{Tensile stress} = \frac{F}{A}}$$where F = stretching force, A = cross-sectional area
Elevator cables experience tensile stress. A cable supporting a 1000 kg elevator with cross-section 5 cm² experiences:
Stress = (1000 × 10) / (5 × 10⁻⁴) = 2 × 10⁷ Pa = 20 MPa
Engineers ensure this is well below the breaking stress of steel (~400 MPa) for safety!
2. Compressive Stress (Longitudinal Stress)
When forces try to compress the material (pushing together).
Example: Pillar supporting building, hydraulic press
$$\boxed{\text{Compressive stress} = \frac{F}{A}}$$- Tensile: Forces pull outward, material stretches
- Compressive: Forces push inward, material compresses
Both are longitudinal (along the length) but opposite in direction.
3. Shearing Stress (Tangential Stress)
When forces act parallel to the surface, causing layers to slide over each other.
Example: Cutting with scissors, rivets in joints
$$\boxed{\text{Shearing stress} = \frac{F_{\text{tangential}}}{A}}$$4. Hydraulic Stress (Bulk Stress)
When uniform pressure acts on all surfaces of a material (like underwater).
Example: Submarine hull under water, gas in container
$$\boxed{\text{Hydraulic stress} = -\Delta P = -(P_{\text{applied}} - P_{\text{normal}})}$$Negative sign indicates compression.
Strain
Strain is the measure of deformation (change in dimension) relative to the original dimension.
$$\boxed{\text{Strain} = \frac{\text{Change in dimension}}{\text{Original dimension}}}$$Important Properties:
- Strain is dimensionless (no units)
- It’s a pure ratio
- Usually expressed as fraction or percentage
Types of Strain
1. Longitudinal Strain (Tensile/Compressive Strain)
Change in length per unit original length.
$$\boxed{\text{Longitudinal strain} = \frac{\Delta L}{L} = \frac{L - L_0}{L_0}}$$where:
- L₀ = original length
- L = final length (after deformation)
- ΔL = change in length = L - L₀
A wire of length 2.0 m is stretched to 2.002 m.
Strain = (2.002 - 2.0) / 2.0 = 0.002 / 2.0 = 0.001 = 0.1%
2. Shearing Strain (Tangential Strain)
Measure of angular deformation when shearing stress is applied.
$$\boxed{\text{Shearing strain} = \theta = \frac{\Delta x}{L}}$$where:
- Δx = lateral displacement
- L = perpendicular distance (height)
- θ = angle of deformation (in radians, for small angles)
For small deformations: tan θ ≈ θ (in radians)
Shearing strain is the angle (in radians) through which a face originally perpendicular to the fixed face gets turned.
It’s dimensionless because θ = arc length / radius = L/L = dimensionless.
3. Volume Strain (Bulk Strain)
Change in volume per unit original volume.
$$\boxed{\text{Volume strain} = \frac{\Delta V}{V} = \frac{V - V_0}{V_0}}$$where:
- V₀ = original volume
- V = final volume
- ΔV = change in volume
A 1000 cm³ rubber ball is compressed to 998 cm³ underwater.
Volume strain = (998 - 1000) / 1000 = -0.002 = -0.2%
(Negative indicates compression)
Hooke’s Law
Hooke’s Law states that within the elastic limit, stress is directly proportional to strain.
$$\boxed{\text{Stress} \propto \text{Strain}}$$ $$\boxed{\text{Stress} = E \times \text{Strain}}$$where E is the modulus of elasticity (material constant).
Interactive Demo: Visualize Stress and Strain
See how materials deform under different types of stress.
Alternative forms:
For a spring: $\boxed{F = kx}$
where:
- F = applied force
- k = spring constant
- x = displacement from equilibrium
Hooke’s law is valid only:
- Within elastic limit (below yield point)
- For small deformations
- At constant temperature
- For isotropic materials (same properties in all directions)
Car suspension springs follow Hooke’s law. When you hit a bump:
- Spring compresses (applies force F = kx)
- Stores elastic energy
- Releases energy, smoothing the ride
Spring constant k determines how “stiff” the ride feels. Sports cars have higher k (stiffer), luxury cars have lower k (softer).
Stress-Strain Curve
When we gradually increase load on a material and plot stress vs strain, we get important information about material behavior.
Key Points on the Curve:
1. Proportional Limit (Point A)
- Hooke’s law is valid up to this point
- Stress ∝ Strain (linear region)
- Perfectly elastic behavior
2. Elastic Limit (Point B)
- Maximum stress for which material returns to original shape
- Beyond this: permanent deformation begins
- Usually very close to proportional limit
3. Yield Point (Point C)
- Material starts to flow or yield
- Large strain for small increase in stress
- Permanent deformation is significant
Upper Yield Point (C₁): Stress drops slightly after this Lower Yield Point (C₂): Stress at which yielding continues
4. Ultimate Tensile Strength (Point D)
- Maximum stress material can withstand
- After this, material begins to neck (cross-section reduces)
- Material weakens beyond this point
5. Breaking Point/Fracture Point (Point E)
- Material breaks or fractures
- Breaking stress < Ultimate stress (due to necking)
Ductile Materials (metals like copper, aluminum):
- Large plastic region (C to E)
- Considerable elongation before breaking
- Yield point clearly visible
- Example: Iron wire can stretch significantly
Brittle Materials (glass, ceramics):
- Very small or no plastic region
- Break suddenly with little deformation
- Fracture point close to elastic limit
- Example: Glass breaks suddenly without much bending
Elastic Hysteresis
When a material is loaded and unloaded, the stress-strain curves for loading and unloading are not identical. The material doesn’t return along the same path.
Characteristics:
- Forms a loop on stress-strain graph
- Area of loop = energy dissipated as heat
- Indicates internal friction in material
Elastic Potential Energy
When a material is deformed within elastic limit, work done is stored as elastic potential energy.
For a spring (or elastic material following Hooke’s law):
$$\boxed{U = \frac{1}{2}kx^2}$$For a wire or rod:
$$\boxed{U = \frac{1}{2} \times \text{Stress} \times \text{Strain} \times \text{Volume}}$$ $$\boxed{U = \frac{1}{2} \times \frac{F}{A} \times \frac{\Delta L}{L} \times (A \times L)}$$ $$\boxed{U = \frac{1}{2} F \cdot \Delta L}$$Energy density (energy per unit volume):
$$\boxed{u = \frac{U}{V} = \frac{1}{2} \times \text{Stress} \times \text{Strain}}$$A spring with k = 500 N/m is compressed by 10 cm. Find energy stored.
Solution:
$$U = \frac{1}{2}kx^2 = \frac{1}{2} \times 500 \times (0.1)^2$$ $$= \frac{1}{2} \times 500 \times 0.01 = 2.5 \text{ J}$$Common Mistakes to Avoid
Wrong: Stress and pressure are the same thing.
Correct:
- Pressure = external force per unit area
- Stress = internal restoring force per unit area
Pressure acts from outside, stress is the material’s internal response.
Wrong: Strain is measured in meters or cm.
Correct: Strain is dimensionless (ratio of lengths). Express as decimal or percentage.
Wrong: F = kx works for all deformations.
Correct: Hooke’s law is valid only within elastic limit. Beyond that, relationship becomes non-linear.
Practice Problems
Level 1: JEE Main Basics
A wire of length 2 m and cross-sectional area 2 mm² is stretched by 1 mm when a force of 100 N is applied. Calculate stress and strain.
Solution:
Area = 2 mm² = 2 × 10⁻⁶ m²
Stress:
$$\text{Stress} = \frac{F}{A} = \frac{100}{2 \times 10^{-6}} = 5 \times 10^7 \text{ Pa} = \boxed{50 \text{ MPa}}$$Strain:
$$\text{Strain} = \frac{\Delta L}{L} = \frac{0.001}{2} = 5 \times 10^{-4} = \boxed{0.05\%}$$A spring of spring constant 200 N/m is compressed by 5 cm. Find the restoring force and elastic energy stored.
Solution:
Force:
$$F = kx = 200 \times 0.05 = \boxed{10 \text{ N}}$$Energy:
$$U = \frac{1}{2}kx^2 = \frac{1}{2} \times 200 \times (0.05)^2$$ $$= 100 \times 0.0025 = \boxed{0.25 \text{ J}}$$Level 2: JEE Main/Advanced
Two wires of same material and length but cross-sectional areas in ratio 1:2 are stretched by same force. Find the ratio of: (a) Stress in two wires (b) Strain in two wires
Solution:
Let areas be A and 2A.
(a) Stress ratio:
$$\frac{\sigma_1}{\sigma_2} = \frac{F/A}{F/2A} = \frac{2A}{A} = \boxed{2:1}$$(b) Strain ratio:
Since same material: same Young’s modulus Y
From Hooke’s law: Strain = Stress/Y
$$\frac{\epsilon_1}{\epsilon_2} = \frac{\sigma_1/Y}{\sigma_2/Y} = \frac{\sigma_1}{\sigma_2} = \boxed{2:1}$$Thinner wire has more stress AND more strain!
A steel wire of length 4 m and diameter 2 mm is extended by 2 mm when loaded. If Young’s modulus of steel is 2 × 10¹¹ Pa, find the load applied.
Solution:
Radius r = 1 mm = 10⁻³ m
Area A = πr² = π × (10⁻³)² = π × 10⁻⁶ m²
Strain = ΔL/L = 0.002/4 = 5 × 10⁻⁴
From Y = Stress/Strain:
Stress = Y × Strain = 2 × 10¹¹ × 5 × 10⁻⁴ = 10⁸ Pa
Force = Stress × Area = 10⁸ × π × 10⁻⁶
$$F = 100\pi \approx \boxed{314 \text{ N}}$$Level 3: JEE Advanced
A wire is stretched by applying a force F at one end. If the same wire is stretched by forces F at both ends (holding one end fixed), what is the ratio of elongation in two cases?
Solution:
Case 1: Force F at one end, other end fixed
- Tension throughout = F
- Elongation = ΔL₁
Case 2: Force F at both ends
- By Newton’s third law, tension throughout = F (same as case 1!)
- One end is fixed, other pulled with F
- Elongation = ΔL₂
Since tension is same in both cases:
$$\boxed{\frac{\Delta L_1}{\Delta L_2} = 1:1}$$Important: Pulling with F from both ends creates same tension as pulling with F from one end (with other end fixed against wall).
A wire of length L and cross-sectional area A is stretched by force F. Another wire of same material has length 2L and area A/2. What force will produce the same extension?
Solution:
For wire 1: Extension = FL/(AY)
For wire 2: Let force be F'
Extension = F’(2L) / [(A/2)Y] = 4F’L/(AY)
For same extension:
$$\frac{FL}{AY} = \frac{4F'L}{AY}$$ $$F = 4F'$$ $$\boxed{F' = \frac{F}{4}}$$One-fourth the force produces same extension!
Quick Summary
| Property | Definition | Formula | Units |
|---|---|---|---|
| Stress | Force per unit area | σ = F/A | Pa (N/m²) |
| Strain | Relative deformation | ε = ΔL/L | Dimensionless |
| Hooke’s Law | Stress ∝ Strain | σ = Eε | — |
| Elastic Energy | Stored energy | U = ½kx² | Joule (J) |
Related Topics
Within Properties of Matter
- Elastic Moduli — Young’s, bulk, shear modulus
- Fluid Pressure — Pressure in liquids
- Viscosity — Resistance to flow
- Surface Tension — Surface energy
Connected Chapters
- Newton’s Laws — Force and equilibrium
- Work-Energy-Power — Elastic potential energy
- Simple Harmonic Motion — Spring oscillations