Introduction to Shear Modulus Formula
Shear modulus tells how effectively a body will resist the forces applied to change its shape. When a shear force is applied on a body which results in its lateral deformation, the elastic coefficient is called the shear modulus of elasticity. It measures the rigidity of a body. Conceptually, it is the ratio of shear stress to shear strain in a body.
Let, on application of a force F tangentially on the top surface of a box fixed at the bottom, the top surface gets displaced by x and a plane perpendicular to the force gets turned by an angle θ as shown.
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Then, the shear stress: \[\sigma =\frac{F}{A}\]
Shear strain: θ =\[\frac{x}{l}\] (As θ is very very small, tan θ=θ)
L is the perpendicular distance (on a plane perpendicular to the force) to the layer that gets displaced by an extent x, from the fixed layer.
Then, shear modulus:
G=shear stress/shear strain=F/A/x/L=FL/Ax
Unit of shear modulus is Nm–2 or pascals (Pa).
Example 1
A thin square plate of dimensions 80 cm × 80 cm × 0.5 cm is fixed vertical on one of its smaller surfaces. It is subjected to a shearing force of 2.8 ×104 N at the top. The top face of the cube is displaced through 0.16 mm with respect to the bottom surface. Find the shear modulus of elasticity of the material of the plate.
Solution:
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F = 2.8 ×104 N, x = 0.16 mm = 0.16 ×10–3 m, L = 0.8 m, A = 0.8 × 0.5×10–2 m2, G = ?
\[G=\frac{FL}{Ax}=\frac{2.8\times 10^4\times 0.8}{0.8\times 0.5\times 10^{-2}\times 0.6\times 10^{-3}}=3.5\times 10^{10}Pa\]
What is the effect of temperature and pressure on the shear modulus values?
A material's response to an applied force changes with temperature and pressure. In metals, the shear modulus decreases with increasing temperature. So, the rigidity decreases as pressure increases.
Mechanical Threshold Stress (MTS) plastic flow stress model, the Nadal and LePoac (NP) shear modulus model, and the Steinberg-Cochran-Guinan (SCG) shear modulus model are the three models used to predict how pressure and temperature affect shear modulus. It is observed that in metals, there is an area of temperature and pressure where there is a linear change in the shear modulus. Modelling the behaviour outside of this range though is trickier.
Practice question
On the application of a shear force of 8 kN on a cube of edge 4 cm and made of material of shear modulus 2 ×109 Pa, the upper face of a cube gets displaced by:
(a) 0.1 mm (b) 0.1 cm (c) 0.4 mm (d) 0.2 cm
Ans (a)
FAQs on Shear Modulus Formula
1. What is shear stress in everyday life?
Shear stress is induced in an object when two equal and opposite forces act on it tangentially to the area. It is the ratio of shear resistance to the shear area. It causes a change in the shape of the body. Some of the many examples of shear stress are cutting vegetables, writing on a blackboard, painting, applying creams/soaps etc, chewing food, walking or running. Other examples include when water flows river beds experience shear stress.
2. What are isotropic and anisotropic materials?
Materials are isotropic based on their response to shear stress. This means that the deformation as a result of the force is the same regardless of the orientation. Examples of isotropic materials are metals, alloys and glass.
Glass, metals and alloys are examples of isotropic materials.
Anisotropic materials respond differently to stress or strain depending on the orientation. Anisotropic materials are thus a lot more susceptible to shear along one axis than they would be on another.
For example, the response of a block of wood to a force applied parallel to the wood grain would be very different from its response to a force applied perpendicular to the grain.
The same is true of how a diamond responds to an applied force. How easily the crystal shears depends on the force’s orientation.
3. What are the shear modulus values of materials?
Here’s a list of shear modulus values of 15 materials at room temperature.
While soft, flexible materials tend to have low shear modulus values, transition metals and alloys have high values. Diamond has an extremely high shear modulus. Alkaline and basic metals have intermediate values.
Material:Shear Modulus (GPa)
Diamond (C): 478.0
Iron (Fe): 52.5
Steel: 79.3
Aluminum (Al): 25.5
Lead (Pb): 13.1
Magnesium (Mg): 16.5
Brass: 40
Titanium (Ti):41.1
Copper (Cu): 44.7
Glass: 26.2
Rubber: 0.0006
Polyethene: 0.117
Plywood: 0.62
Nylon:4.1
Concrete: 21
4. What is Hooke's Law?
Hooke’s law was discovered by the English scientist Robert Hooke in 1660. It is the law of elasticity. This principle of physics states that, in the smaller deformations of an object, the size of the deformation is directly proportional to the deforming force. So it’s the relation in which the stress σ is directly proportional to the strain ε (σ=E ε). The law also holds true when it comes to shear stress and strain in the linear elastic range (τ=Gγ).
A deforming force can be applied to a solid in different ways: by compressing, stretching, squeezing, twisting or bending.
In this case, the object returns to its original shape and size once the load or force is removed. Hooke’s law explains the elastic behaviour of solids with the fact that small displacements of the object’s ions, atoms or molecules from their normal positions is proportional to the load that causes that displacement.
5. What is Young’s Modulus?
The modulus is named after the British scientist, Thomas Young, who described its calculation in 1807. But the basic concept behind Young's modulus was described by a Swiss scientist and engineer Leonhard Euler in 1727. And in 1782, Giordano Riccati, an Italian scientist performed experiments that led to the modern calculations of the modulus.
It is a measure of a solid material’s resistance to elastic deformation under force or load. It relates the stress to the strain (calculated as force per unit area) or the proportion of deformation along an axis or line.
The principle states that a material goes through elastic deformation when it is compressed or extended, and returns to its original shape when the load is taken away.
SI unit for Young's modulus is Pa
The equation for Young's modulus is:
σ(ε) F/A FL0
Young’s Modulus = ---- = ------ = -----
ε ΔL/L0 A ΔL
(σ is the uniaxial stress, ε is strain, F is the force of compression/extension, A is the cross-sectional surface area of the applied force), Δ L is the change in length and L0 is the original length).