Question 9.7:
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.
Solution:
Mass of the big
structure, M = 50,000 kgInner 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 X 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) X 2 X 1011 ] = 7.22 X 10-7
Hence, the compressional strain of each column is 7.22 × 10–7.
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