## Introduction

Shear Modulus of elastic Moduli is the measure of mechanical properties present in solids. Other types of elastic moduli are Young’s modulus and bulk modulus. Let us help you understand the shear modulus elastic moduli. Shear modulus of elasticity is also known as modulus of rigidity. It is used to measure the rigidity of a body. Shear modulus is the ratio of shear stress to shear strain. This property is denoted using G or S or μ.

Any element that has a definite shape and shape is a rigid body. All the things and shapes that we see around us cannot be said to be completely rigid. When an external force is applied, these shapes can be bent, stretched or even compressed. On the application of an external force, these shapes tend to deform.

Stress is defined as the internal force that acts per unit area and is applied to regain the original shape.

A strain is defined as the ratio of change in the dimension produced and the body's original dimension.

The relationship so established between stress and strain is called the modulus of elasticity.

E=Stress/Strain

SI unit= Nm-2

Dimensions= [M L-1 T -2]

We can use this to explain how transformation is resisted from transverse deformations. However, this is practically possible just for small deformations.

Corresponding to the three types of strain namely longitudinal, volumetric and shear, there are mainly three types of modulus of elasticity. These are:

Young’s modulus

Bulk modulus

Shear modulus

### Young’s Modulus of Elasticity

Young’s modulus is defined as the material’s ability to withstand the compression or expansion in accordance with its length.

It is represented using E or Y.

It is used to define the relationship between the longitudinal stress vs the longitudinal strain of an object. Whenever a load is applied to an object, its deformation takes place. The object can regain its original shape by removing the pressure and if it is elastic.

Young’s modulus formula is given by:

E= σ/ε

σ= F/A

ε= ΔL/L

Therefore,

\[E= \frac{\sigma (\epsilon )}{\epsilon }=\frac{\frac{F}{A}}{\frac{\Delta L}{L}}=\frac{FL}{A\Delta L}\]

Here,

E= Young’s modulus in Pascal

σ= Stress in Pascal

ε= Strain/ Proportional deformation

F= Force exerted on the object placed under tension

A= Cross-sectional area of the object

ΔL= Change in the length of the object

L= Original length of the object

### The Formula For Modulus of Rigidity:

\[G=\frac{\tau _{xy}}{\gamma _{xy}}=\frac{Fl}{A\Delta x}\]

\[\tau _{xy}=\frac{F}{A}\] which is the shear stress.

F= Force acting on the object

A= Area on which the force acts

Δx= Transverse displacement

l= initial length

### Difference Between Modulus of Rigidity and Modulus of Elasticity

Modulus of rigidity helps in calculating the deformation of an object when the deforming force is applied at right angles to the surface.

While modulus of elasticity helps in calculating the deformations of an object when the deforming force is applied parallel to the surface.

### Modulus of Elasticity and Shear Modulus Relationship

Modulus of rigidity= G

Modulus of elasticity= E

The relationship between modulus of elasticity and shear modulus is given by:

E= 2G(1+μ)

The SI unit of the relation is Pascal (Pa).

### Examples For Shear Modulus of Rigidity

For some material, the shear modulus of rigidity is given as follows:

Wood= 6.2 x 108 Pa

Steel= 7.2 x 1010 Pa

This shows that steel is highly rigid in comparison to wood.

1. Under a Shearing Force, a Block is Placed on the Table. Calculate the Shear Modulus of the Material of the Block, Given the Following Data:

Dimensions: 60mm*60mm*20mm

Ans: Shearing force= 0.245 N

Displacement= 5 mm

Shear stress= F/A= 2450/50 N/m^{2}

Shear strain= Δx/l= 5/60= 1/12

Therefore, shear modulus= G= Shear stress/Shear strain = 2450 N/m^{2}

2. A Wax Piece in the Shape of a Box has an Area of 10 cm^{2} and a Height of 2 cm. A Shearing Force of 0.5 N is Applied to its Upper Surface. It Displaces 4mm of its Bottom Surface Relatively. Find the Following:

Shearing stress

Shearing strain

Shear modulus

Ans:

Area= 10 cm^{2}

Height= 2cm

Displacement of box= 4mm

Shearing stress of the box= Tangential force/Area= 500 Pa

Shearing strain= Displacement(change)/Initial length = x/L= 0.2

Shear modulus= Shear strain/shear stress = 500/0.2 = 2500 Pa