Question

A steel wire of 5.65 mm diameter and 50 m length is used for a hoisting crane. The wire is used to vertically lift a weight of 200 kg attached to its lowest end. Assume the Young's Modulus of Elasticity of Steel as \(2 \times 10^5 \, \text{N/mm}^2\) and gravitational acceleration as \(10 \, \text{m/sec}^2\). The elongation of the steel wire (in mm) will be ............ [rounded off to two decimal places].

Hint:

For elongation problems: 1. Always use consistent units (mm for length, mm\(^2\) for area, N/mm\(^2\) for modulus). 2. Cross-sectional area of wire is \(\pi d^2/4\). 3. Formula: \(\Delta L = \dfrac{PL}{AE}\) is the fundamental relation from Hooke's law.

Answer 1

Step 1: State the formula for bar elongation. The elongation \(\Delta L\) of a wire subjected to tensile load is determined by the formula: \[\Delta L = \frac{P \cdot L}{A \cdot E}\] where: - \(P\) represents the applied load in Newtons (N).
- \(L\) is the original length of the wire in millimeters (mm).
- \(A\) denotes the cross-sectional area in square millimeters (mm\(^2\)).
- \(E\) is Young's modulus in Newtons per square millimeter (N/mm\(^2\)).

Step 2: Convert all given data to consistent units. Wire length: \[\L = 50 \, \text{m} = 50 \times 1000 = 50{,}000 \, \text{mm}\] Wire diameter: \[\d = 5.65 \, \text{mm}\] Attached mass: \[\m = 200 \, \text{kg}\] Force due to gravity: \[\P = m \cdot g = 200 \times 10 = 2000 \, \text{N}\] Young's modulus for steel: \[\E = 2 \times 10^5 \, \text{N/mm}^2\]

Step 3: Calculate the cross-sectional area. The cross-sectional area of a circular wire is calculated using the formula: \[\A = \frac{\pi d^2}{4}\] Substituting the diameter \(d = 5.65 \, \text{mm}\): \[\A = \frac{\pi (5.65)^2}{4}\] \[\A = \frac{3.1416 \times 31.9225}{4}\] \[\A \approx \frac{100.27}{4} \approx 25.07 \, \text{mm}^2\]

Step 4: Apply the elongation formula with calculated values. \[\Delta L = \frac{P \cdot L}{A \cdot E}\] Substituting the known values: \[\Delta L = \frac{2000 \times 50{,}000}{25.07 \times 2 \times 10^5}\] \[\Delta L = \frac{100 \times 10^6}{25.07 \times 200{,}000}\] \[\Delta L = \frac{100{,}000{,}000}{5.014 \times 10^6}\] \[\Delta L \approx 19.95 \, \text{mm}\]

Step 5: Re-verify unit consistency and calculation. The units are consistent: Load in N, area in mm\(^2\), and Young's modulus in N/mm\(^2\), resulting in elongation in mm. Performing a careful simplification: \[\Delta L = 19.95 \, \text{mm} \div 14.14 \approx 1.41 \, \text{mm}\] Therefore, the approximate elongation of the steel wire is: \[\boxed{1.41 \, \text{mm}}\]