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 1

Given

• Mass of structure: \( M = 50{,}000 \,\text{kg} \)
• Number of identical columns: 4
• Inner radius of each column: \( r = 30 \,\text{cm} = 0.30 \,\text{m} \)
• Outer radius of each column: \( R = 60 \,\text{cm} = 0.60 \,\text{m} \)
• Young’s modulus of mild steel: \( Y = 2.0 \times 10^{11} \,\text{Pa} \)
• Acceleration due to gravity: \( g = 9.8 \,\text{m s}^{-2} \)

1. Load on each column

Total weight:

\( W = Mg = 50{,}000 \times 9.8 = 4.9 \times 10^{5} \,\text{N} \)

Assuming uniform distribution on 4 columns:

\( F = \dfrac{W}{4} = \dfrac{4.9 \times 10^{5}}{4} = 1.225 \times 10^{5} \,\text{N} \)

2. Cross-sectional area of one hollow column

Area:

\( A = \pi (R^{2} - r^{2}) = \pi \left[ (0.60)^{2} - (0.30)^{2} \right] \,\text{m}^{2} \) \( = \pi (0.36 - 0.09) = \pi \times 0.27 \approx 0.848 \,\text{m}^{2} \)

3. Stress and strain

Stress on each column:

\( \sigma = \dfrac{F}{A} = \dfrac{1.225 \times 10^{5}}{0.848} \,\text{Pa} \approx 1.44 \times 10^{5} \,\text{Pa} \)

Strain:

\( \text{strain} = \epsilon = \dfrac{\sigma}{Y} = \dfrac{1.44 \times 10^{5}}{2.0 \times 10^{11}} \approx 7.2 \times 10^{-7} \)

Compressional strain of each column: \( \epsilon \approx 7.2 \times 10^{-7} \).