Correct Answer - Option 3 : 31.25 mm
3
Concept:
The volumetric strain occurred in material, ev is given by
\({{\rm{e}}_{\rm{v}}} = {\rm{\;}}\frac{{{\rm{Change\;in\;Volume\;}}\left( {{\rm{\;\Delta V}}} \right)}}{{{\rm{Original\;Volume\;}}\left( {\rm{V}} \right)}}\) OR
\({e_v} = \frac{{\left( {{\sigma _x} + \;{\sigma _y} + \;{\sigma _z}} \right)\left( {1 - 2\mu } \right)}}{E}\)
Calculation:
Given:
μ = 0.25; E = 2 × 107 N/mm2; L = 2.5 m = 2500 mm; Area = 20 mm2, Load P = 500 kN.
\({\sigma _x} = \;\frac{{Force\;in\;x\;direction\;}}{{cross\;sectional\;area}}\)
\({\sigma _x} = \;\frac{{500 \times 10^3}}{{20}}=25 \times 10^3\;N/mm^2\)
Volumetric strain:
\({e_v} = \frac{{\left( {{\sigma _x} + \;{\sigma _y} + \;{\sigma _z}} \right)\left( {1 - 2\mu } \right)}}{E}\)
\({e_v} = \frac{{\left( {{(25 \times 10^3)}\; + \;{0}\; + \;{0}} \right)\left( {1 - 2\times 0.25 } \right)}}{2\times10^7}=6.25\times10^{-4}\)
\({{\rm{e}}_{\rm{v}}} = {\rm{\;}}\frac{{{\rm{Change\;in\;Volume\;}}\left( {{\rm{\;Δ V}}} \right)}}{{{\rm{Original\;Volume\;}}\left( {\rm{V}} \right)}}\)
\({\rm{Δ }}V = V \times {e_v} = \left( {Area\;\times\;length} \right) \times 6.25 \times {10^{ - 4}}\)
\(\)\({\rm{Δ }}V = V \times {e_v} = \left( {20\;\times\;2500} \right) \times 6.25 \times {10^{ - 4}}\)
ΔV = 31.25 mm3