Numericals on Modulus of Rigidity
Numerical 1. The area of the upper face of a rectangular block is 0.5 m x 0.5 m and the lower face is fixed. The height of the block is 1 cm. a shearing force applied to the top face produces a displacement of 0.015 mm. Find the strain, stress and the shearing force. Modulus of rigidity = η = 4.5 × 1010 N/m².
Solution:
Given,
- Area under shear = A = 0.5 m x 0.5 m = 0.25 m²
- Height of the block = h = 1 cm = 1 × 10-2 m
- Displacement of top face = x = 0.015 mm = 0.015 × 10-3 m = 1.5 × 10-5 m
- Modulus of rigidity = η = 4.5 × 1010 N/m²
- Shear strain = tan θ = x/h = (1.5 × 10-5) / (1 × 10-2) = 1.5 × 10-3
Modulus of rigidity = η = Shear stress / Shear strain
∴ Shear stress = η × Shear strain = 4.5 × 1010 × 1.5 × 10-3
∴ Shear stress = 6.75 × 107 N/m².
Shear stress = F/A
∴ F = Shear stress × Area
∴ F = 6.75 × 107 × 0.25
∴ F = 1.69 × 107 N
Ans:
- Shear strain = 1.5 × 10-3
- Shear stress = 6.75 × 107 N/m²
- Shearing force = 1.69 × 107 N
Numerical 2. A metallic cube of side 5 cm, has its lower surface fixed rigidly. When a tangential force of 104 kg. wt. is applied to the upper surface, it is displaced through 0.03 mm. Calculate (1) the shearing stress (2) the shearing strain and (3) the modulus of rigidity of the metal.
Solution:
Given,
- Area under shear = A = 5 cm x 5 cm = 25 cm² = 25 × 10-4 m²
- Height of the block = h = 5 cm = 5 × 10-2 m
- Displacement of top face = x = 0.03 mm = 0.03 × 10-3 m = 3 × 10-5 m
- Shearing Force = 104 kg-wt = 104 × 9.8 N
Shear stress = F/A
∴ Shear stress = (104 × 9.8)/( 25 × 10-4)
∴ Shear stress = 3.92 × 107 N
Shear strain = tanθ = x/h = (3 × 10-5 ) / (5 × 10-2 ) = 6 × 10-4
Modulus of rigidity = η = Shear stress / Shear strain
η = (3.92 × 107) / (6 × 10-4) = 6.53 × 1010 N/m²
Ans:
- Shear stress = 3.92 × 107 N
- Shear strain = 6 × 10-4
- Modulus of rigidity = 6.53 × 1010 N/m²
Numerical 3. A metal plate has an area of face 1m x 1m and thickness of 1 cm. One face of a larger area is fixed and a tangential force is applied to the opposite face. The displacement of the edge produced thereby is 0.005 cm. Find the shearing stress, strain and magnitude of the tangential force applied. Modulus of rigidity of metal is ϒ = 8.4 × 1010 N/m²
Solution:
Given,
- Area under shear = A = 1 m x 1 cm = 1 m²
- Thickness of plate = h = 1 cm = 1 × 10-2 m
- Displacement of top face = x = 0.005 cm = 0.005 × 10-2 m = 5 × 10-5 m
- Modulus of rigidity = η = 8.4 × 1010 N/m²
Shear strain = tanθ = x/h = (5 × 10-5) / (1 × 10-2) = 5 × 10-3
Modulus of rigidity = η = Shear stress / Shear strain
∴ Shear stress = η × Shear strain = 8.4 × 1010 × 5 × 10-3
∴ Shear stress = 4.2 × 108 N/m².
Shear stress = F/A
∴ F = Shear stress × Area
∴ F = 4.2 × 108 ×1
∴ F = 4.2 × 108 N
Ans:
- Shear Strain = 5 × 10-3
- Shear Stress = 4.2 × 108 N/m²
- Shearing Force = 4.2 × 108 N
Modulus of Rigidity
Modulus of rigidity also known as shear modulus, is used to measure the rigidity of a given body. It is the ratio of shear stress to shear strain and is denoted by G or sometimes by S or μ. The modulus of rigidity of a material is directly proportional to its elastic modulus which depends on the material’s nature and properties.
In this article, we will talk about what Modulus of rigidity is, its formula, examples, and applications.
Table of Content
- What is the Modulus of Rigidity (Shear Modulus)?
- Modulus of Rigidity Formula
- Characteristics of Modulus of Rigidity
- Examples of Modulus of Rigidity
- Relation between Modulus of Elasticity and Modulus of Rigidity
- Modulus of Rigidity vs Modulus of Elasticity