Physics, asked by PragyaTbia, 1 year ago

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.

Answers

Answered by abhi178
22
Given,
mass of the big structure , M = 50,000 kg
Total force applied on all the columns = Mg
= 50,000 × 9.8 N
so, amount of force exerted on each column , F = 50,000 × 9.8/4 N = 122500 N

inner radius of the column , r = 30cm
outer radius of the column , R = 60cm
so, area of each column, A = π(R² - r²)
= 22/7 × (60² - 30²)
= 22/7 × (3600 - 900)
= 22/7 × 2700 cm²
= 22/7 × 0.27 m²
= 0.848 m²

from data , Young's modulus of steel is Y = 2 × 10¹¹ Pa

now use formula, strain = F/AY
= 122500/(0.848 × 2 × 10¹¹)
= 7.22 × 10^-7 m
Answered by ITZBFF
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}  \: m}} \\

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