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, the shear modulus of rigidity measures the rigidity of a body. Conceptually, it is the ratio of shear stress to shear strain in a body.

## Shear Modulus Formula

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

### Notations Used In Shear Modulus Formula

- G is shear modulus in N.m
^{-2} - F is the force acting on the body
- l is the initial length
- âˆ†x is the change in length
- 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 7.9×10^{10} and for plywood is 6.2×10^{8}. This implies that steel is a lot more rigid than plywood, about 127 times more!

### Shear Modulus Unit And Dimension

SI unit | Pa |

Dimensional formula | M^{1}L^{-1}T^{-2} |

### Shear Modulus Is Related To Other Elastic Moduli Of A Material

2G (1+υ) = E =3K (1−2υ) |

### Notations Used In Elastic Moduli Of A Material

- E is the Young’s modulus
- G is the shear modulus
- K is the bulk modulus
- u is the 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

σ=FA

Where,

- F is the force applied
- σ is the stress applied
- 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:

Ïµ=tanθ=Δxl

Where,

- Ïµ is the strain caused due to the applied stress
- l is the original length
- âˆ†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:

ShearModulusG=ShearstressShearstrainG

=FlAΔx

### 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 the shear modulus is applied. One of the common examples might be to consider a pair of sneakers that are too stiff or tight, they run up and down on the back of the heel, which will eventually lead to a shearing injury on the skin as a blister. Other examples are painting, brushing teeth, the opening of a screw, bottle cap, friction between car tyre 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 the radius of 0.6 inches is to be drilled on the plate. The shear strength is FA=4×10^{4}lbinch^{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 inch^{2}

Given,FA=4×10^{4}lbinch^{2}

Thus, to drill the hole, the force needed= 4 x 10^{4} x 0.11304

Force = 4521.6lb

**Question 2:** Compute the Shear modulus, if the stress experienced by a body is 5×104Nm^{2} and strain is 4×10^{-2}.

**Solution: **Given,

Stress =5×10^{4}Nm^{2}

Strain = 4×10^{-2}

ShearModulusG=ShearstressShearstrainG=5×10^{4}NM 24×10^{-2}=1.25×10^{6}Nm^{2}

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