Shear Modulus

To understand shear modulus, we must first explore the concepts of shear stress and shear strain. Both describe how an object deforms when forces act parallel to its surface. For simplicity, consider a solid block (as shown in the figure).

Shear stress (τ) is the force applied per unit area in a direction parallel to the surface of the block. The following formula gives the shear stress:

τ = F/A

Where:

F: Applied force 

A: Cross-sectional area over which the force is distributed

Shear strain (γ) measures the deformation caused by shear stress. It is defined as the ratio of the displacement of one layer of the block to its original length. If the top surface of a block is displaced by a distance Δx while the bottom remains fixed, the shear strain is given by:

γ = Δx/h = tan θ ~ θ (for small θ)

Where:

h: Height of the block

θ: Angle of shear

Shear modulus, denoted by the letter G, is represented as the ratio of shear stress to shear strain:

\[\text{Shear Modulus} = \frac{\text{Shear Stress}}{\text{Shear Strain}} \\

=> G = \frac{\tau}{\gamma} \\

=> G = \frac{\frac{F}{A}}{\frac{\Delta x}{h}}\\

=> G = \frac{F \cdot h}{A \cdot \Delta x} \]

A higher shear modulus means a material is more rigid because more force (stress) is needed to produce the same horizontal displacement (strain). This is why the shear modulus is sometimes referred to as the modulus of rigidity. 

To a first approximation, this type of deformation does not result in a change in volume; instead, atomic planes simply slide sideways over one another. Therefore, the area (A) (which determines the number of atomic bonds) is crucial in defining stress rather than just the applied force (F).

SI Unit and Dimension of Shear Modulus

The SI unit of shear modulus is the Pascal (Pa), which is equivalent to Newton per square meter (N/m²).

Since shear modulus is the ratio of stress (force per unit area) to strain (which is dimensionless), its dimensional formula is:

[G] = [Force] [Area]^-1

=> [G] = [MLT⁻²] [L²]^-1

=> [G] = [ML^-1T⁻²]

Where M represents mass, L represents length, and T represents time.

Material	Shear Modulus (in GPa)
Rubber	0.0006
Polyethylene	0.117
Plywood	0.62
Nylon	4.1
Lead	13.1
Concrete	21
Glass	26.2
Iron	52.5
Steel	79.3
Diamond	478

The shear modulus is closely related to Young’s modulus (E), bulk modulus (K), and Poisson’s ratio (ν). These constants describe different types of material deformation:

For isotropic (uniform) materials, the relationship between these constants is:

\[ G = \frac{E}{2(1 + \nu)} \\

K = \frac{E}{3(1 – 2\nu)} \]

Problem 1: A metal rod is subjected to a shear force of 500 N over a cross-sectional area of 0.01 m². The top layer of the rod shifts 2 mm while the rod’s height is 0.5 m. Calculate the shear modulus of the material.

Solution

Given:

Shear force: F = 500 N  

Cross-sectional area: A = 0.01 m²  

Lateral displacement: Δx = 2 mm = 0.002 m  

Height of the rod: h = 0.5 m

The shear modulus (G) is given by:

\[ G = \frac{F \cdot h}{A \cdot h} \\

=> G = \frac{500 \text{ N} \cdot 0.5 \text{ m}}{0.01 \text{ m}^2 \cdot 0.5 \text{ m}} = 12.5 \times 10^6 \text{ Pa} = 12.5 \text{ MPa} \]

The shear modulus of the material is 12.5 MPa.

Problem 2: A 10 cm tall rubber block is fixed at the bottom and subjected to a shear force on the top surface, causing a 5 mm lateral displacement. If the rubber has a shear modulus of 0.8 MPa, calculate the shear stress acting on the block.  

Solution

Given:  

Height of the rubber block: h = 10 cm = 0.1 m  

Lateral displacement: Δx = 5 mm = 0.005 m  

Shear modulus: G = 0.8 MPa = 8 x 10⁵ Pa

The shear modulus is given by:

\[ G = \frac{\tau}{\gamma} \\

=> \tau = G \times \frac{x}{h}\\

=> \tau = 8 x 10⁵ \text{ Pa} \times \frac{0.005 \text{ m}}{0.1 \text{ m} = 40000 \text{ Pa} = 40 \text{ kPa} \]

The shear stress acting on the block is 40 kPa.