Question

Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg, The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column. (The Young's modulus of steel is 2.0 × 10^11 N/m^2)

Answer 1

$\sf \red{Given}$

$\sf{Total \: mass \: to \: be \: supported \: = \: 50,000 \: kg}$

$\sf{Total \: weight \: to \: be \: supported \: = \: 50,000 \: kg.wt \: = \: 50,000 \times 9.8 \: N}$

Since this weight is to be supported by 4 columns, therefore compressional force on one column is

$\boxed{ \boxed{F = \frac{50000 \times 9.8}{4} \: N}}$

$\sf{inner \: radius \: of \: each \: column \: (r_{1}) \: = 30 \: cm \: = 0.3 \: m}$

$\sf{outer \: radius \: of \: each \: column \: (r_{2}) \: = 60 \: cm \: = 0.6 \: m}$

$\sf{Cross-sectional \: area \: of \: each \: column, \: A \: = \pi ( {r_{2}}^{2} - {r_{1}}^{2} )}$

$\sf{ \implies \:A = \frac{22}{7} \bigg[( {0.6})^{2} - ({0.3})^{2} \bigg] \: {m}^{2} }$

$\sf{ \implies \: A = \frac{22}{7} (0.6 - 0.3)(0.6 + 0.3)}$

$\sf{ \implies \: A = \frac{22}{7} \times 0.3 \times 0.9 \: {m}^{2} }$

$\boxed{ \boxed{\sf{ \implies \: A = 0.84\: {m}^{2} } }}$

$\sf \red{Now : }$

$\boxed{ \boxed{Y = \: \frac{ F/A}{compressional \: strain} }}$

(or)

$\boxed{ \boxed{compressional \: strain \: = \: \frac{F}{AY} }}$

$\implies \: \frac{50000 \times 9.8}{4 \times 0.84 \times 2.0 \times {10}^{11} }$

$\implies \: \frac{5 \times 98 \times {10}^{3}}{4 \times 84 \times 2 \times {10}^{9} }$

$\implies \: 0.72 \times {10}^{-6}$

$\approx 7.2 \times {10}^{-7}$

$\boxed{ \boxed{compressional \: strain \: = \: 7.2 \times {10}^{ - 7} }}$

Answer 2

Answer:

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Explanation:

Have a good night....