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.
Answers
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:
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.