Question:medium

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 $Y = 2.0 \times 10^{11}\,\text{Pa}$, $g = 9.8\,\text{m/s}^2$]

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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.
Updated On: Feb 16, 2026
  • $2.60 \times 10^{-7}$
  • $3.60 \times 10^{-8}$
  • $1.87 \times 10^{-3}$
  • $7.07 \times 10^{-4}$
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The Correct Option is A

Solution and Explanation

To solve this problem, we need to calculate the compression strain for each hollow cylindrical column supporting the structure. We are given the following parameters:

  • Mass of the structure, m = 50 \times 10^3\,\text{kg}
  • Outer radius of the column, R_o = 100\,\text{cm} = 1\,\text{m}
  • Inner radius of the column, R_i = 50\,\text{cm} = 0.5\,\text{m}
  • Young's modulus, Y = 2.0 \times 10^{11}\,\text{Pa}
  • Acceleration due to gravity, g = 9.8\,\text{m/s}^2

First, calculate the gravitational force (weight) acting on the structure:

F = mg = 50 \times 10^3 \times 9.8 = 4.9 \times 10^5 \,\text{N}

This force is distributed equally among the four columns. Thus, the force on one column is:

F_{\text{column}} = \frac{4.9 \times 10^5}{4} = 1.225 \times 10^5 \,\text{N}

Next, we calculate the cross-sectional area (A) of one cylinder:

A = \pi (R_o^2 - R_i^2) = \pi (1^2 - 0.5^2) = \pi (1 - 0.25) = \pi \times 0.75

A = 0.75 \pi\,\text{m}^2

Now, we calculate the stress in the column:

\text{Stress} = \frac{F_{\text{column}}}{A} = \frac{1.225 \times 10^5}{0.75 \pi} \approx 5.198 \times 10^4 \,\text{Pa}

Finally, we calculate the strain using Young's Modulus:

\text{Strain} = \frac{\text{Stress}}{Y} = \frac{5.198 \times 10^4}{2.0 \times 10^{11}} \approx 2.60 \times 10^{-7}

Therefore, the compression strain of each column is 2.60 \times 10^{-7}, which matches the correct answer provided.

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