What will be the stress at `-20^(@)`C, if a steel rod with a cross-sectional area of 150 `mm^(2)` is stretched between two fixed points? The tensile load at `20^(@)`C is 5000 N (Assume, `alpha=11.7xx10^(-6)//^(@)C` and Y = `200 xx 10^(11)`N//m^(2)`)

Options

A. `12.7 xx 10^(6)N m^(-2)`
B. `84.2 xx 10^(6)N m^(-2)`
C. `127 xx 10^(6)N m^(-2)`
D. `0.842xx 10^(6)N m^(-2)`

Answer

To solve the problem, we need to calculate the stress in the steel rod at -20°C given the parameters provided. The stress is defined as the force applied per unit area.

### Step-by-Step Solution:

1. **Understand the Given Data:**
- Cross-sectional area (A) = 150 mm² = \(150 \times 10^{-6}\) m²
- Tensile load (F) at 20°C = 5000 N
- Coefficient of linear expansion (\(\alpha\)) = \(11.7 \times 10^{-6} \, ^\circ C^{-1}\)
- Young's modulus (Y) = \(200 \times 10^{9} \, \text{N/m}^2\)
- Initial temperature (T1) = 20°C
- Final temperature (T2) = -20°C
- Change in temperature (\(\Delta T\)) = T2 - T1 = -20 - 20 = -40°C

2. **Calculate the Change in Length due to Temperature Change:**
- The change in length (\(\Delta L\)) of the rod due to temperature change can be calculated using the formula:
\[ \Delta L = \alpha L \Delta T \]
- However, we do not have the original length (L), but we can express the change in length in terms of L.

3. **Relate the Stresses at Different Temperatures:**
- The stress (\(\sigma\)) is defined as:
\[ \sigma = \frac{F}{A} \]
- At 20°C, the stress is:
\[ \sigma_1 = \frac{5000 \, \text{N}}{150 \times 10^{-6} \, \text{m}^2} \]

4. **Calculate the Stress at 20°C:**
\[ \sigma_1 = \frac{5000}{150 \times 10^{-6}} = \frac{5000}{0.000150} = 33.33 \times 10^{6} \, \text{N/m}^2 \]

5. **Calculate the Stress at -20°C:**
- The stress at -20°C can be calculated using the relationship derived from Young's modulus:
\[ \sigma_2 = \sigma_1 + 2 \alpha Y \Delta T \]
- Substitute the values:
\[ \sigma_2 = 33.33 \times 10^{6} + 2 \times (11.7 \times 10^{-6}) \times (200 \times 10^{9}) \times (-40) \]

6. **Calculate the Additional Stress Contribution:**
- Calculate \(2 \alpha Y \Delta T\):
\[ 2 \times (11.7 \times 10^{-6}) \times (200 \times 10^{9}) \times (-40) = -1.176 \times 10^{7} \, \text{N/m}^2 \]

7. **Final Calculation of Stress at -20°C:**
\[ \sigma_2 = 33.33 \times 10^{6} - 1.176 \times 10^{7} = 21.153 \times 10^{6} \, \text{N/m}^2 \]

8. **Convert to Standard Form:**
\[ \sigma_2 \approx 21.15 \times 10^{6} \, \text{N/m}^2 \text{ or } 21.15 \, \text{MPa} \]

### Final Answer:
The stress at -20°C is approximately \(21.15 \times 10^{6} \, \text{N/m}^2\) or \(21.15 \, \text{MPa}\).