- Strain is a measure of deformation representing the displacement between particles in the body to a reference length.
- It tells us how and what changes takes place when a body is subjected to strain.
- Mathematically:- Strain = ΔL/L , where ΔL=change in length L= original length
- It is dimensionless quantity because it is a ratio of two quantities.
- For example: – If we have a metal beam and we apply force from both sides the shape of the metal beam will get deformed.
- This change in length or the deformation is known as Strain.
Types of Strain: Longitudinal Strain
- Change in length to the original length of the body due to the longitudinal stress.
- If we apply longitudinal stress to a body either the body elongates or it compresses this change along the length of the body. This change in length is measured by Longitudinal Strain.
- Longitudinal Strain = ΔL/L
- Consider a rod whose initial length is L after elongation length becomes L’.
- So the change in length is ΔL= L’ – L
- So Strain= ΔL/L
- Strain occurs as a result of stress.
- Shearing strain is the measure of the relative displacement of the opposite faces of the body as a result of shearing stress.
- If we apply force parallel to the cross – sectional area because of which there was relative displacement between the opposite faces of the body.
- Shearing strain measures to what extent the two opposite faces got displaced relative to each other.
- Consider a cube whose initial length was L which is at some position and when it gets displaced by an angle θ.
- Let the small relative displacement be x.
- Shearing strain= x/L
- In terms of tan θ,
- Shearing strain = tan θ = x/L
- tan θ is equal to θ (as θ is very small)
- Therefore, x/L = θ
- Volume strain is defined as ratio of change in volume to the original volume as a result of the hydraulic stress.
- When the stress is applied by a fluid on a body there is change in the volume of body without changing the shape of the body.
- Volume strain = ΔV/V
- For example:-
- Consider a ball initially at volume V.
Problem:- Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 cm and compressional strain of each column.
Mass of the big structure, M = 50,000 kg
Inner radius of the column, r = 30 cm = 0.3 m
Outer radius of the column, R = 60 cm = 0.6 m
Young’s modulus of steel, Y = 2 × 1011 Pa
Total force exerted, F = Mg = 50000 × 9.8 N
Stress = Force exerted on a single column = 122500 N
Young’s modulus, Y =𝑆𝑡𝑟𝑒𝑠𝑠/𝑆𝑡𝑟𝑎𝑖𝑛
Area, A = π (R2 – r2) = π ((0.6)2 – (0.3)2)
Strain = 122500/ (π [0.6)2 – (0.3)2] x 2×1011)
= 7.22 × 10–7
Hence, the compressional strain of each column is 7.22 × 10–7