Question

In terms of bulk modulus (K) and modulus of rigidity (C), the Poisson’s ratio can be expressed as

1. \(\frac{3K\ -\ 4C}{6K\ +\ 4C}\)
2. \(\frac{3K\ +\ 4C}{6K\ -\ 4C}\)
3. \(\frac{3K\ -\ 2C}{6K\ +\ 2C}\)
4. \(\frac{3K\ +\ 2C}{6K\ -\ 2C}\)

Answer 1

Correct Answer - Option 3 : \(\frac{3K\ -\ 2C}{6K\ +\ 2C}\)

Concept:

Elastic Modulus (E): When the body is loaded within its elastic limit, the ratio of stress and strain is constant. This constant is known as Elastic modulus

\({\rm{E}} = \frac{{{\rm{Stress}}}}{{{\rm{Strain}}}} = \frac{{\rm{\sigma }}}{\epsilon}\)

S.I unit: N/m²

Rigidity modulus (C): When a body is loaded within its elastic limit, the ratio of shear stress and shear strain is constant, this constant is known as the shear modulus.

\({\rm{C}} = \frac{{{\rm{Shear\;stress\;}}}}{{{\rm{Shear\;strain}}}} = \frac{{\rm{\tau }}}{\phi }\)

S.I unit: N/m²

Bulk modulus (K): When a body is subjected to three mutually perpendicular like stresses of the same intensity then the ratio of direct stress and the volumetric strain of the body is known as bulk modulus

\({\rm{K}} = \frac{{{\rm{Direct\;stress}}}}{{{\rm{Volumetric\;strain}}}} = \frac{{\rm{\sigma }}}{{\frac{{{\rm{\delta V}}}}{{\rm{V}}}}}\)

S.I unit: N/m²

Explanation:

The relation between E, C, μ is

E = 2C (1 + μ)           ---(1)

Relation Between E, K, μ is

E = 3K (1 – 2μ)         ---(2)

Equating (1) & (2)

2G (1 + μ) = 3K(1 – 2μ)

2C + 2μC = 3K – 6μK

μ (2C + 6K) = 3K – 2C

\(\mu = \frac{{3K\; -\; 2C}}{{2C\; + \;6K}}\)

Poisson’s ratio: 

 When the body is loaded within its elastic limit, the ratio of lateral strain and linear strain is constant. This constant is known as Poisson's ratio.

\({\rm{\mu }} = \frac{{{\rm{lateral\;strain}}}}{{{\rm{linear\;strain}}}}\)

Other relations between elastic constants are,

1) E = 2C (1 + μ)

2) E = 3K (1 – 2μ)

3) \({\bf{E}} = \frac{{9{\bf{KC}}}}{{3{\bf{K}}\; + \;{\bf{C}}}}\)