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AAI JE (Technical) Official Paper 2018

Option 3 : 31.25 mm^{3}

CT 1: Current Affairs (Government Policies and Schemes)

54676

10 Questions
10 Marks
10 Mins

__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 mm^{2}, 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 mm^{3}