Four identical hollow cylindrical columns of mild steel support a big structure of mass 50 $\times$ 103 kg. The inner and outer radii of each column are 50 cm and 100 cm respectively. Assuming uniform local distribution, calculate the compression strain of each column. [Use Y = 2.0 $\times$ 1011 Pa, g = 9.8 m/s2]
Solution
Force on each column = ${{mg} \over 4}$<br><br>Strain = ${{mg} \over {4AY}}$<br><br>$$ = {{50 \times {{10}^3} \times 9.8} \over {4 \times \pi (1 - 0.25) \times 2 \times {{10}^{11}}}}$$<br><br>= 2.6 $\times$ 10<sup>$-$7</sup>
About this question
Subject: Physics · Chapter: Properties of Solids and Liquids · Topic: Elasticity
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