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Shear Modulus Formula

When a shear force is applied on a body that results in its lateral deformation, then the elastic coefficient is referred to as the shear modulus of rigidity. Therefore, the shear modulus of rigidity measures the rigidity of a body. Also, it is the ratio of shear stress to shear strain in a body. In this topic, we will discuss the shear modulus formula with some examples.

Concept of Shear Modulus

Shear modulus is used to explain how a material resists transverse deformations. But this is practical for small deformations only, after which they are able to return back to the original state. This is due to the large shearing forces lead to permanent deformations i.e. no longer elastic body.

The value of G for steel is \(7.9\times 10^10\) and for plywood is \(6.2\times 10^8\). Hence steel is a lot more rigid than plywood, about 127 times more!

Shear Modulus Formula

The Formula for Shear Modulus

It is given as: \(G=\frac{Fl}{A\Delta x}\)


G Shear modulus
l Initial Length
\(\Delta\) Change in  length
A Area
F Force

SI unit of G is Pascal i.e. Pa. Shear Modulus is related to other Elastic Moduli of the Material. This relationship is given as below:

\(E= 2G ( 1+\mu )\)


\(E = 3K ( 1 – 2 \mu )\)


E Young’s Modulus
G Shear Modulus
K Bulk Modulus
\(\mu\) Poisson’s ration

Derivation of the Shear Modulus Formula

1] Shear Stress

Internal restoring forces because of the elastic bodies to get back 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 are parallel to it and thus the stress that’s acting on the surface also plots a tangent. Here stress is termed as a shearing or tangential stress. This stress is expressed as Newton per square meter.

Shearing Stress = Force / Surface Area

\(\sigma =FA\)

F Force Applied
\(\sigma\) Stress applied
A Area of force applied

2] Shear Strain

The strain is the measure of the deformation experienced by a body in the direction of the force applied. Further, it is divided by the body’s initial dimensions. We can express it as:

\(\varepsilon =tan \theta = \Delta xl =tan \theta = \Delta xl\)

\(\varepsilon =tan \theta = \Delta xl\)  , is the strain caused due to the applied stress

\(\varepsilon\) Shear Strain
l Original Length
\(\Delta xl\) change in length of the material

Note that the quantity strain does not have any dimension, as it is indicative of a relative change in the shape of the body. Hence, we can express shear modulus as:

\(Shear Modulus G= F l A \Delta x\)

Solved Examples for Shear Modulus Formula

Q.1: The thickness of a metal plate is 0.3 inches. We drill a hole of the radius of 0.6 inches on the plate. If, the shear strength is \(FA=4 \times10^4\) lb square inch, determine the force we need to make the hole.

Solution: The shear stress is exerted over the surface of the cylindrical shape.

Therefore, the area of the cylindrical surface,

\(= 2 \pi r h =  2 \times 3.14 \times 0.06 \times 0.30\)

=  0.11304 square inch

Given, \(FA=4 \times10^4\) lb square inch

Thus, to drill the hole, the force needed \(= 4 \times10^4 \times 0.11304\)

Force = 4521.6 lb

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