Sure, let's solve this problem step-by-step.
### Given Data
1. Length of the rod (L): 4 meters
2. Diameter of the rod (d): 30 mm (which is 0.03 meters after converting to meters)
3. Tensile load (F): 100,000 Newtons (100KN)
4. Modulus of Elasticity (E): 200 GPa (which is [tex]\(200 \times 10^9\)[/tex] Pascals after converting to Pascals)
### Step-by-Step Solution
Step 1: Calculate the Cross-sectional Area (A) of the Rod
The cross-sectional area of a circular rod is given by the formula:
[tex]\[ A = \pi \left(\frac{d}{2}\right)^2 \][/tex]
Here,
[tex]\[ d = 0.03 \text{ meters} \][/tex]
So,
[tex]\[ A = \pi \left(\frac{0.03}{2}\right)^2 \][/tex]
[tex]\[ A = \pi \left(0.015\right)^2 \][/tex]
[tex]\[ A \approx 0.000706858\ \text{square meters} \][/tex]
Step 2: Calculate the Extension (ΔL) of the Rod
The extension can be calculated using the formula for axial deformation:
[tex]\[ \Delta L = \frac{F \cdot L}{A \cdot E} \][/tex]
Here,
- [tex]\( F = 100,000 \)[/tex] N
- [tex]\( L = 4 \)[/tex] meters
- [tex]\( A \approx 0.000706858 \)[/tex] square meters
- [tex]\( E = 200 \times 10^9 \)[/tex] Pascals
So,
[tex]\[ \Delta L = \frac{100,000 \times 4}{0.000706858 \times 200 \times 10^9} \][/tex]
[tex]\[ \Delta L \approx 0.002829421 \text { meters} \][/tex]
### Final Answers
- Cross-sectional area (A): 0.000706858 square meters
- Extension (ΔL): 0.002829421 meters
So, the extension of the mild steel rod is approximately 0.002829 meters or 2.829 millimeters.
