The strain energy due to volumetric strain __________.

SOLUTION

Strain energy due to volumetric strain (ϵ_v)

\({\rm{U}} = \frac{1}{2}\left( {{\rm{Average\;stress}}} \right) \times \left( {{\rm{Voltmetric\;strain\;}}\left( {{_{\rm{o}}}} \right)} \right) \times \left( {{\rm{Volume}}} \right)\)

\({\epsilon_v} = \left( {\frac{{{\sigma _x} + {\sigma _y} + {\sigma _z}}}{E}} \right)\left( {1 - 2\mu } \right)\)

\({\rm{\mu }} = \frac{1}{2}\left( {\frac{{{{\rm{\sigma }}_{\rm{x}}} + {{\rm{\sigma }}_{\rm{y}}} + {{\rm{\sigma }}_{\rm{z}}}}}{3}} \right)\left( {\frac{{{{\rm{\sigma }}_{\rm{x}}} + {{\rm{\sigma }}_{\rm{y}}} + {{\rm{\sigma }}_{\rm{z}}}}}{{\rm{E}}}} \right)\left( {1 - 2{\rm{\mu }}} \right) \times {\rm{V}}\)

\({\rm{\mu }} = \frac{1}{2}\left( {\frac{{{{\left( {{{\rm{\sigma }}_{\rm{x}}} + {{\rm{\sigma }}_{\rm{y}}} + {{\rm{\sigma }}_{\rm{z}}}} \right)}^2}}}{{3{\rm{E}}}}} \right)\left[ {\left( {1 - 2{\rm{\mu }}} \right) \times {\rm{V}}} \right]\)

⇒ μ ∝ V

⇒ The strain energy due to volumetric strain is directly proportional to the volume.

μ ∝ (σ_x + σ_y + σ_z)²

⇒ The strain energy due to volumetric strain is directly proportional to the square of exerted pressure.

Now we know that E = 3K (1 - 2μ) or \(\left( {\frac{{1 - 2{\rm{\mu }}}}{{3{\rm{E}}}}} \right) = \frac{1}{{\rm{K}}}\)

\(\therefore {\rm{\mu }} = \frac{{{{\left( {{{\rm{\sigma }}_{\rm{x}}} + {{\rm{\sigma }}_{\rm{y}}} + {{\rm{\sigma }}_{\rm{z}}}} \right)}^2}}}{{2{\rm{K}}}} \times {\rm{V}}\)

\(\Rightarrow {\rm{\mu }} \propto \frac{1}{{\rm{K}}}\)