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ISRO Scientist ME 2020 Paper

Option 1 : 0.25

ST 1: Thermodynamics

2167

20 Questions
60 Marks
32 Mins

**Concept:**

Euler’s critical buckling load \(\left( {{P}_{e}} \right)=\frac{{{\pi }^{2}}\cdot E\cdot {{I}_{min}}}{L_{e}^{2}}\)

Where, Imin = minimum area moment of Inertia, E = young’s modulus of elasticity, Le = Equivalent length = L ⋅ α

α for different loading conditions are different

__Calculation:__

**Given:**

Two columns,

both ends are fixed, L_{e} = L × 0.5

both ends are hinged L_{e} = L × 1

\({{P}_{e}}\propto \frac{1}{{{\left( {{L}_{e}} \right)}^{2}}}\)

\(\frac{{{\left( {{P}_{e}} \right)}_{Both~hinged}}}{{{\left( {{P}_{e}} \right)}_{Both~clamped}}~~}=\frac{1}{{{\left( L× 1 \right)}^{2}}}× \frac{{{\left( L× 0.5 \right)}^{2}}}{1}=0.25\)