Q. 74.0( 36 Votes )

# Four identical ho

Answer :

Given Data,

Mass of the big structure, M = 50,000 kg.

Inner radius of the column, r = 30 cm = 0.3 m

Outer radius of the column, R = 60 cm = 0.6 m

Young’s modulus of steel, Y_{s}, from the standard chart is 2×10^{11} Pa

Total force exerted on all the columns, F = Mg = 5000kg × 9.8 N

Amount of force exerted on each column, F_{1} = (5000Kg× 9.8N)/4

⇒ F_{1} = 122500 N

Young’s modulus, Y = Stress/ Strain

⇒ Y = (/Strain

⇒ Strain = …………………………… (1)

Where, A = area of each column = π (R^{2}-r^{2})

⇒ A = π ((0.6)^{2} - (0.3)^{2})

Substituting value of Area in equation (1), we get,

Strain = 122500 N/ π ((0.6)^{2}-(0.3)^{2}) × 2×10^{11} Pa

⇒ Strain = 7.22× 10^{-7}

Hence, the compressional strain on each column is 7.22 × 10^{-7}.

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