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 cm and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.
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Answer 1

Answer:

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 = 2 × 1011 Pa

Total force exerted, F = Mg = 50000 × 9.8 N

Stress = Force exerted on a single column = 50000 × 9.8 / 4 = 122500 N

Young’s modulus, Y = Stress / Strain

Strain = (F/A) / Y

Where,

Area, A = π (R2 – r2) = π ((0.6)2 – (0.3)2)

Strain = 122500 / [ π ((0.6)2 – (0.3)2) × 2 × 1011 ] = 7.22 × 10-7

Hence, the compressional strain of each column is 7.22 × 10–7.

Answer 2

Explanation:

total mass support by cylindrical column(m) = 50000 kg

so, total weight support by cylindrical column= mg = 50000×9.8 N

                    = 490000 N

loading for each cylindrical support = 490000/4 = 122500 N

inner radius of each column(r1) = 30cm = 0.3 m

outer radius of each column (r2) = 60 cm = 0.6m

so, CSA of each column = π(r₂² -r₁²)

                       = 3.14 (0.6²-0.3²)

                       = 3.14×0.27 m²

young's modulus (Y) = 2×10¹¹ N/m²

we know,

young's modulus = stress/strain

so,

so, the compressional strain of each column = F/AY

                = 122500/(3.14×0.27)2×10¹¹

                =7.22 ×⁻⁷