When a shear force is applied on a body which results in its lateral deformation, the elastic coefficient is called the shear modulus of rigidity. So, shear modulus of rigidity measures the rigidity of a body. Conceptually, it is the ratio of shear stress to shear strain in a body. It can be represented as:

\(Shear\;Modulus\;G= \frac{Shear\; stress}{Shear\; strain}\)

\(G= \frac{Fl}{A\Delta x}\)

Where, G is measured in N/m2, F is the force that acts on the body, l is the initial length, ∆x is the change in length and A is the area.

Shear modulus can be used to explain how a material resists transverse deformations but this is practical for small deformations only, following which they are able to return to original state. This is because large shearing forces lead to permanent deformations (no longer elastic body). The value of G for steel it is 7.9×1010 and for plywood is 6.2×108. This implies that steel is a lot more rigid than plywood, about 127 times more!

Shear modulus is related to other elastic moduli of a material and is given by the equation:

**2G (1+υ) = E =3K (1−2υ)**

Where E is Young’s Modulus, G is Shear Modulus, K is Bulk Modulus and υ is Poisson’s Ratio.

**Derivation:**

**Shear Stress:**

Internal restoring forces cause the elastic bodies to regain their initial shape. This restoring force that acts on per unit area of a deformed body is termed as stress. When the forces being applied on the surface is parallel to it and thus the stress that’s acting on the surface also plots a tangent. This kind of stress is termed as a shearing or tangential stress. Stress is expressed as N/m2.

Shearing Stress = Force / Surface Area

**\(\sigma =\frac{F}{A}\)**

Where F is the force applied, σ is the stress applied and A is the area of force applied.

**Shear Strain:**

The strain is the measure of deformation experienced by a body in the direction of the force applied, divided by the body’s initial dimensions. It can be expressed as:

\(\epsilon =tan\theta =\frac{\Delta x}{l}\)

Where ϵ is the strain caused due to the applied stress, l is the original length and ∆x is the change in length of the material.

It is to be noted here that the quantity strain does not have any dimension as it is indicative of a relative change in the shape of a body.

Hence, shear modulus can be expressed as:

\(Shear\;Modulus\;G= \frac{Shear\; stress}{Shear\; strain}\)

\(G= \frac{Fl}{A\Delta x}\)

**Shear Modulus in Real Life:**

The concept of shear modulus can be defined by the use of the word “friction”. You may observe a lot of examples where shear modulus is applied. One of the common examples might be to consider a pair of sneakers that are too stiff or tight. When you them, they run up and down on the back of the heel, which will eventually leads to a shearing injury on skin as a blister. Other examples are painting, brushing teeth, opening of screw, bottle cap, friction between car tyres and ground, between mouse feet and mouse pad.

**Solved Examples:**

**Question 1:** The thickness of an iron plate is 0.3 inches. A hole of radius 0.6 inches is to be drilled on the plate. The shear strength is \(\frac{F}{A}= \frac{4\times 10^{4}lb}{inch^{2}}\). Find the force needed to make the hole.

**Solution:** Here the shear stress is exerted over the cylindrical surface.

Therefore, the area of the cylindrical surface = 2πrh = 2 × 3.14 × 0.06 × 0.30 = 0.11304 inch2

Given,

\(\frac{F}{A}= \frac{4\times 10^{4}lb}{inch^{2}}\)

Thus, to drill the hole, the force needed= 4 x 104 x 0.11304

Force = 4521.6lb

**Question 2:** Compute the Shear modulus, if the stress experienced by a body is \(5\times\frac{10^{4}N}{m^{2}}\) and strain is \(4\times 10^{-2}\).

**Solution: **Given,

Stress =\(5\times\frac{10^{4}N}{m^{2}}\)

Strain = \(4\times 10^{-2}\)

\(Shear\;Modulus\;G= \frac{Shear\; stress}{Shear\; strain}\)

\(G= \frac{5\times 10^{4}\frac{N}{M^{2}}}{4\times 10^{-2}}=1.25\times \frac{10^{6}N}{m^{2}}\)

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