Question

Four identical hollow cylindrical columns of mild steel support a big structure of mass $50 \times 10^3\,\text{kg}$. The inner and outer radii of each column are 50 cm and 100 cm respectively. Assuming uniform local load distribution, calculate the compression strain of each column. [use $Y = 2.0 \times 10^{11}\,\text{Pa}$, $g = 9.8\,\text{m/s}^2$]

• A. $2.60 \times 10^{-7}$
• B. $3.60 \times 10^{-8}$
• C. $1.87 \times 10^{-3}$
• D. $7.07 \times 10^{-4}$

Hint:

In multi-support problems, always divide the total weight by the number of supports first. Also, ensure all units (like cm to m) are converted to SI before starting calculations.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The compression strain is defined as the ratio of stress to Young's modulus. Stress is the force acting per unit area. In this case, the total weight of the structure is shared equally by four columns.
Step 2: Key Formula or Approach:
1. Force per column: \( F = \frac{Mg}{4} \)
2. Area of cross-section of hollow cylinder: \( A = \pi (r_{outer}^2 - r_{inner}^2) \)
3. Strain \(\epsilon = \frac{\text{Stress}}{Y} = \frac{F}{AY}\)
Step 3: Detailed Explanation: Given: - Total mass \(M = 50 \times 10^3\,\text{kg}\). - Number of columns = 4. - Radii: \(r_1 = 0.5\,\text{m}\), \(r_2 = 1.0\,\text{m}\). Force on one column: \[ F = \frac{50 \times 10^3 \times 9.8}{4} = 122500\,\text{N} \] Area of cross-section: \[ A = \pi (1.0^2 - 0.5^2) = \pi (1 - 0.25) = 0.75\pi\,\text{m}^2 \] Calculating strain: \[ \epsilon = \frac{122500}{0.75\pi \times 2.0 \times 10^{11}} \] \[ \epsilon = \frac{122500}{1.5\pi \times 10^{11}} \approx \frac{122500}{4.71 \times 10^{11}} \] \[ \epsilon \approx 2.6 \times 10^{-7} \]
Step 4: Final Answer:
The compression strain of each column is \(2.60 \times 10^{-7}\).