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Four identical hollow cylindrical column of steel support a big structure of mass $50000\;kg$. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming load distribution to be uniform , the compressional strain of each column is $ [y_{steel}=2.0 \times 10^{11} Nm^{-2}]$

\[\begin {array} {1 1} (a)\;7.2 \times 10^{-7} \\ (b)\;15.2 \times 10^{-7} \\ (c)\;8.4 \times 10^{-7} \\ (d)\;12.1 \times 10^{-7} \end {array}\]

1 Answer

$Mass= 50000\;kg$
Area of cross section of 4 cylindrical column
$A=4 \pi (0.6^2-0.3^2)$
$Y= 2.0 \times 10^{11} Nm^{-2}$
$Y=\large\frac{MgL}{A \Delta L}$
$\large\frac{\Delta L}{L}=\frac{Mg}{AY}$
$\qquad= \large\frac{50,000 \times 9.8}{4 \times 3.14 \times (0.6^2 -0.3)^2) \times 2 \times 10^{11}}$
$\qquad= 7.2 \times 10^{-7}$
Hence a is the correct answer.


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