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Step-by-Step Solution
Step 1: Identify the Given Data
• Total mass of the structure, $m = 50 \times 10^3\,\mathrm{kg}$
• Gravitational acceleration, $g = 9.8\,\mathrm{m/s^2}$
• Young’s modulus of the material, $Y = 2.0 \times 10^{11}\,\mathrm{Pa}$
• Number of supporting columns, $N = 4$
• Outer radius of each column, $R = 100\,\mathrm{cm} = 1.0\,\mathrm{m}$
• Inner radius of each column, $r = 50\,\mathrm{cm} = 0.5\,\mathrm{m}$
Step 2: Calculate the Force on Each Column
The total weight of the structure is:
$$ W = mg = \bigl(50 \times 10^3\,\mathrm{kg}\bigr)\,\bigl(9.8\,\mathrm{m/s^2}\bigr) = 4.9 \times 10^5\,\mathrm{N} $$
Since there are four identical columns sharing this load equally, the force on each column is:
$$ F_{\text{each}} = \frac{W}{N} = \frac{4.9 \times 10^5\,\mathrm{N}}{4} = 1.225 \times 10^5\,\mathrm{N} $$
Step 3: Compute the Cross-Sectional Area of Each Hollow Cylinder
For a hollow cylinder with outer radius $R$ and inner radius $r$, the cross-sectional area $A$ is:
$$ A = \pi \bigl(R^2 - r^2\bigr) $$
Substituting the given radii:
$$ A = \pi \bigl((1.0\,\mathrm{m})^2 - (0.5\,\mathrm{m})^2\bigr) = \pi \bigl(1 - 0.25\bigr) = 0.75\,\pi\,\mathrm{m^2} $$
Step 4: Calculate the Stress on Each Column
Stress ($\sigma$) is given by:
$$ \sigma = \frac{F_{\text{each}}}{A} $$
Therefore,
$$ \sigma = \frac{1.225 \times 10^5\,\mathrm{N}}{0.75\,\pi\,\mathrm{m^2}} $$
Step 5: Calculate the Compression Strain
Strain ($\epsilon$) is given by the ratio of stress to Young's modulus:
$$ \epsilon = \frac{\sigma}{Y} = \frac{F_{\text{each}}}{A \, Y} $$
Substitute the values:
$$
\epsilon
= \frac{1.225 \times 10^5\,\mathrm{N}}{\bigl(0.75\,\pi\,\mathrm{m^2}\bigr)\,\bigl(2.0 \times 10^{11}\,\mathrm{Pa}\bigr)}
$$
After calculation, this simplifies to approximately:
$$
\epsilon \approx 2.60 \times 10^{-7}
$$
Final Answer
The compression strain of each column is
$$2.60 \times 10^{-7}.$$
