Question

Choose the option showing the correct relation between Poisson’s ratio (σ), Bulk modulus (B) and modulus of rigidity (G).

• A. \(\sigma=\frac{3B-2G}{2G+6B}\)
• B. \(\sigma=\frac{6B+2G}{3B-2G}\)
• C. \(\sigma=\frac{9BG}{3B+G}\)
• D. \(B=\frac{3\sigma-3G}{6\sigma+2G}\)

Answer 1

The Correct Option is A

 To find the correct relation between Poisson's ratio \(\sigma\), Bulk modulus \(B\), and modulus of rigidity \(G\), we will determine the expressions and relationships between these mechanical properties.

The relationships between different elastic constants in material science are key to solving this problem. Let's start by understanding the basics:

• Poisson’s Ratio (\(\sigma\)): It is the ratio of transverse strain to axial strain under axial loading. It is defined as: \(\sigma = -\frac{\text{lateral strain}}{\text{longitudinal strain}}\)
• Bulk Modulus (B): It is a measure of a material's resistance to uniform compression and is defined as: \(B = \frac{\text{normal stress}}{\text{volumetric strain}}\)
• Modulus of Rigidity (G): Also known as the shear modulus, it measures the material's response to shear stress and is defined as: \(G = \frac{\text{shear stress}}{\text{shear strain}}\)

For isotropic materials, the relationships between these moduli and Poisson's ratio are given by:

• \(E = 2G(1+\sigma)\) where \(E\) is the modulus of elasticity.
• \(E = 3B(1-2\sigma)\)

By eliminating \(E\), we can relate \(B\), \(G\), and \(\sigma\):

Given:	\(E = 2G(1+\sigma)\)
	\(E = 3B(1-2\sigma)\)

Setting the equations equal gives:

\(2G(1+\sigma) = 3B(1-2\sigma)\)

Expanding both sides and solving for \(\sigma\):

\(2G + 2G\sigma = 3B - 6B\sigma\)
\(2G\sigma + 6B\sigma = 3B - 2G\)
\(\sigma(2G + 6B) = 3B - 2G\)
\(\sigma = \frac{3B - 2G}{2G + 6B}\)

Therefore, the correct relation is:

The correct option showing the correct relation is:

\(\sigma=\frac{3B-2G}{2G+6B}\)

Hence, the answer is correct.