The exact relationship between modulus of rigidity C, modulus of elasticity E and Poisson’s ratio ν is expressed as

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ISRO VSSC Technical Assistant Mechanical 14 July 2021 Official Paper
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  1. E = (1 + v)
  2. C = E(2 + v)
  3. C = \(\rm {E \over 2(1+v)}\)
  4. C = \(\rm {E \over 1+2v}\)

Answer (Detailed Solution Below)

Option 3 : C = \(\rm {E \over 2(1+v)}\)
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Detailed Solution

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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 or Young's Modulus.

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

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.

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

Bulk modulus (K)

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

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

The relationship between E, K, G, and μ is:

\(E = 2C(1 + μ) \)

\(C ={ E\over 2(1 + μ) }\)

\(E = 3K (1 – 2μ) \)

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

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

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