Question

four identical hollow cylindrical columns of steel support a big structure of mass 50,000 kg the inner and outer radii of each column is 30cm and 60cm assuming the load distribution to be uniform calculate the compressional strain of each column (Ys=2x10^11)

Answer 1

Let us assume that the weight of the structure is equally distributed among all the columns..

Stress = 50, 000 * 9.81 Newtons / π (0.60² - 0.30²) m²
Strain =  Stress / Ys =  578262.96 / 2  * 10¹¹
       =    2.891 * 10⁻⁶ meters  = 2.891  μ meter

Answer 2

$\sf \red{Given}$

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

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

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)

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

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

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

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

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

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