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 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.
Answer:
Mass of the bid structure (M) = 50000 Kg
Inner radius of the column (r) = 30 cm = 0.3 m
Inner radius of the column (R) = 60 cm = 0.6 m
Youngs modulus of steel (Y) = 2 x 1011 Pa
Force (F) = Mg = 50000 x 9.8 N
Stress = Force on column = 50000 x 9.8/4 = 122500 N
Youngs 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] x 2 x 1011 = 7.2 x 10-7 Pa
Hence, the compressional strain of each column is 7.2 x 10-7
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