Question 4 of 13intermediate🔧 ApplyNumerical4 marks

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.

Correct Answer

The compressional strain of each column is approximately 7.2×107.2 \times 10⁻⁷.

Exercise: EXERCISES | Q: 8.7
For More Understanding

Explanation

This numerical problem requires calculating the compressional strain using the relation: Strain = Stress / Young's Modulus. The stress is obtained from the force per unit area on each column. The context provides Young's modulus of structural steel as 2.0×102.0 \times 10¹¹ N m⁻² from Example 8.1, which applies to mild steel columns. The cross-sectional area of a hollow cylinder is π(R² - r²). Since the load is uniformly distributed among 4 columns, each column bears one-fourth of the total weight.

Solution Steps

  1. Step 1: Calculate cross-sectional area of each hollow column Area = π(R² - r²) = π[(0.60)² - (0.30)²] = π(0.36 - 0.09) = π × 0.27 = 3.14×0.273.14 \times 0.27 = 0.848 m²

  2. Step 2: Calculate total weight (force) Weight = Mass × g = 50,000×9.8000 \times 9.8 = 4.9×104.9 \times 10⁵ N

  3. Step 3: Calculate force on each column Since load is uniformly distributed among 4 columns: Force per column = (4.9×104.9 \times 10⁵) / 4 = 1.225×101.225 \times 10⁵ N

  4. Step 4: Calculate stress on each column Stress = Force / Area = (1.225×101.225 \times 10⁵) / 0.848 = 1.44×101.44 \times 10⁵ N m⁻²

  5. Step 5: Calculate compressional strain Young's modulus of steel (Y) = 2.0×102.0 \times 10¹¹ N m⁻² Strain = Stress / Y = (1.44×101.44 \times 10⁵) / (2.0×102.0 \times 10¹¹) = 7.2×107.2 \times 10⁻⁷

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