Strain

• 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
• Mathematically
  • 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

• 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.
• Mathematically:-
  • 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

• 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.

Answer:

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

Where,

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