Question
Part 1: A steel railroad track has a length of 22 m when the temperature is 0◦C. What is the increase in the length of the rail on a hot day when the temperature is 32 ◦C? The linear expansion coefficient of steel is 11 × 10−6 (◦C)−1. Answer in units of m.
Part 2: Suppose the ends of the rail are rigidly clamped at 0◦C to prevent expansion. Calculate the thermal stress in the rail if its temperature is raised to 32 ◦C. Young’s modulus for steel is 20 × 1010 N/m^2. Answer in units of N/m^2.
Part 2: Suppose the ends of the rail are rigidly clamped at 0◦C to prevent expansion. Calculate the thermal stress in the rail if its temperature is raised to 32 ◦C. Young’s modulus for steel is 20 × 1010 N/m^2. Answer in units of N/m^2.
Answers
GPT-4o mini
To solve these problems, we’ll use the formula for linear thermal expansion and the concept of thermal stress in a constrained material.
### Part 1: Increase in Length Due to Thermal Expansion
The formula for linear thermal expansion is given by:
\[
\Delta L = L_0 \cdot \alpha \cdot \Delta T
\]
Where:
- \(\Delta L\) = change in length
- \(L_0\) = original length (22 m)
- \(\alpha\) = coefficient of linear expansion (11 × 10⁻⁶ (°C)⁻¹)
- \(\Delta T\) = change in temperature (32°C - 0°C = 32°C)
Now, we can plug in the values:
\[
\Delta L = 22 \, \text{m} \cdot (11 \times 10^{-6} \, \text{°C}^{-1}) \cdot 32 \, \text{°C}
\]
Calculating this step-by-step:
1. Calculate the product of \(\alpha\) and \(\Delta T\):
\[
11 \times 10^{-6} \, \text{°C}^{-1} \cdot 32 \, \text{°C} = 352 \times 10^{-6} = 0.000352
\]
2. Now multiply by \(L_0\):
\[
\Delta L = 22 \, \text{m} \cdot 0.000352 = 0.007744 \, \text{m}
\]
Thus, the increase in the length of the rail is:
\[
\Delta L \approx 0.007744 \, \text{m}
\]
---
### Part 2: Thermal Stress Calculation
When the ends of the rail are rigidly clamped and unable to expand, a thermal stress is generated due to the increase in temperature, which we can calculate using the formula:
\[
\sigma = E \cdot \alpha \cdot \Delta T
\]
Where:
- \(\sigma\) = thermal stress (in N/m²)
- \(E\) = Young's modulus for steel (20 × 10¹⁰ N/m²)
- \(\alpha\) = coefficient of linear expansion (11 × 10⁻⁶ (°C)⁻¹)
- \(\Delta T\) = change in temperature (32°C - 0°C = 32°C)
Now we can plug in the values:
\[
\sigma = (20 \times 10^{10} \, \text{N/m}^2) \cdot (11 \times 10^{-6} \, \text{°C}^{-1}) \cdot (32 \, \text{°C})
\]
Calculating this step-by-step:
1. Calculate the product of \(\alpha\) and \(\Delta T\):
\[
11 \times 10^{-6} \, \text{°C}^{-1} \cdot 32 \, \text{°C} = 352 \times 10^{-6} = 0.000352
\]
2. Multiply by \(E\):
\[
\sigma = (20 \times 10^{10} \, \text{N/m}^2) \cdot 0.000352
\]
3. Calculating the final result:
\[
\sigma = 20 \times 10^{10} \cdot 0.000352 = 7.04 \times 10^6 \, \text{N/m}^2
\]
Thus, the thermal stress in the rail is approximately:
\[
\sigma \approx 7.04 \times 10^6 \, \text{N/m}^2
\]
In summary:
- The increase in length of the rail is approximately **0.007744 m**.
- The thermal stress in the rail, when clamped, is approximately **7.04 × 10⁶ N/m²**.
### Part 1: Increase in Length Due to Thermal Expansion
The formula for linear thermal expansion is given by:
\[
\Delta L = L_0 \cdot \alpha \cdot \Delta T
\]
Where:
- \(\Delta L\) = change in length
- \(L_0\) = original length (22 m)
- \(\alpha\) = coefficient of linear expansion (11 × 10⁻⁶ (°C)⁻¹)
- \(\Delta T\) = change in temperature (32°C - 0°C = 32°C)
Now, we can plug in the values:
\[
\Delta L = 22 \, \text{m} \cdot (11 \times 10^{-6} \, \text{°C}^{-1}) \cdot 32 \, \text{°C}
\]
Calculating this step-by-step:
1. Calculate the product of \(\alpha\) and \(\Delta T\):
\[
11 \times 10^{-6} \, \text{°C}^{-1} \cdot 32 \, \text{°C} = 352 \times 10^{-6} = 0.000352
\]
2. Now multiply by \(L_0\):
\[
\Delta L = 22 \, \text{m} \cdot 0.000352 = 0.007744 \, \text{m}
\]
Thus, the increase in the length of the rail is:
\[
\Delta L \approx 0.007744 \, \text{m}
\]
---
### Part 2: Thermal Stress Calculation
When the ends of the rail are rigidly clamped and unable to expand, a thermal stress is generated due to the increase in temperature, which we can calculate using the formula:
\[
\sigma = E \cdot \alpha \cdot \Delta T
\]
Where:
- \(\sigma\) = thermal stress (in N/m²)
- \(E\) = Young's modulus for steel (20 × 10¹⁰ N/m²)
- \(\alpha\) = coefficient of linear expansion (11 × 10⁻⁶ (°C)⁻¹)
- \(\Delta T\) = change in temperature (32°C - 0°C = 32°C)
Now we can plug in the values:
\[
\sigma = (20 \times 10^{10} \, \text{N/m}^2) \cdot (11 \times 10^{-6} \, \text{°C}^{-1}) \cdot (32 \, \text{°C})
\]
Calculating this step-by-step:
1. Calculate the product of \(\alpha\) and \(\Delta T\):
\[
11 \times 10^{-6} \, \text{°C}^{-1} \cdot 32 \, \text{°C} = 352 \times 10^{-6} = 0.000352
\]
2. Multiply by \(E\):
\[
\sigma = (20 \times 10^{10} \, \text{N/m}^2) \cdot 0.000352
\]
3. Calculating the final result:
\[
\sigma = 20 \times 10^{10} \cdot 0.000352 = 7.04 \times 10^6 \, \text{N/m}^2
\]
Thus, the thermal stress in the rail is approximately:
\[
\sigma \approx 7.04 \times 10^6 \, \text{N/m}^2
\]
In summary:
- The increase in length of the rail is approximately **0.007744 m**.
- The thermal stress in the rail, when clamped, is approximately **7.04 × 10⁶ N/m²**.