Determine the values of the stress in portions AC and CB of the steel bar shown when the temperature of the bar is -50°F, knowing that a close fit exists at both of the rigid supports when the temperature is +75°F. Use the values E = 29 x 106 psi and a = 6.5 x 106/°F for steel.

To solve this problem, we need to determine the stresses in portions \(AC\) and \(CB\) of the steel bar due to the temperature change. Here's a step-by-step approach to solving it:

### Step 1: Understanding the scenario

- **Given data**:
  - The steel bar is initially at a temperature of \( +75^\circ F \), where it fits snugly between the rigid supports.
  - The temperature decreases to \( -50^\circ F \), which will cause the bar to contract.
  - The elastic modulus \( E = 29 \times 10^6 \) psi (modulus of elasticity for steel).
  - The coefficient of thermal expansion \( \alpha = 6.5 \times 10^{-6} \) per °F.

- Since the supports are rigid, the bar is constrained and cannot contract freely. Therefore, thermal stresses will develop as the bar tries to contract.

### Step 2: Formula for thermal strain and stress

The thermal strain \( \varepsilon_{\text{thermal}} \) due to a temperature change \( \Delta T \) is given by:

\[
\varepsilon_{\text{thermal}} = \alpha \Delta T
\]

Where:
- \( \alpha \) is the coefficient of thermal expansion,
- \( \Delta T = T_{\text{final}} - T_{\text{initial}} \) is the change in temperature.

The thermal stress \( \sigma_{\text{thermal}} \) is related to the thermal strain by Hooke's Law:

\[
\sigma_{\text{thermal}} = E \varepsilon_{\text{thermal}} = E \alpha \Delta T
\]

### Step 3: Determine the temperature change

\[
\Delta T = -50^\circ F - 75^\circ F = -125^\circ F
\]

### Step 4: Calculate the thermal strain

Substitute the values of \( \alpha \) and \( \Delta T \) into the equation for thermal strain:

\[
\varepsilon_{\text{thermal}} = (6.5 \times 10^{-6} /^\circ F) \times (-125^\circ F) = -8.125 \times 10^{-4}
\]

### Step 5: Calculate the thermal stress

Now, using the value of \( E = 29 \times 10^6 \) psi, calculate the thermal stress:

\[
\sigma_{\text{thermal}} = (29 \times 10^6 \, \text{psi}) \times (-8.125 \times 10^{-4}) = -23,562.5 \, \text{psi}
\]

Thus, the thermal stress in both portions \(AC\) and \(CB\) is \( \sigma_{\text{thermal}} = -23,562.5 \, \text{psi} \).

### Step 6: Interpretation of the stress

The negative sign indicates that the stress is compressive, as expected due to the contraction caused by the temperature decrease.

### Final Answer:
- The stress in portions \(AC\) and \(CB\) is **\(-23,562.5 \, \text{psi}\)**, which is compressive.

---

Would you like further details on any part of the process, or do you have any questions?

### 5 Relative Questions:
1. How does thermal stress differ from mechanical stress in materials?
2. How would the stress change if the coefficient of thermal expansion was larger?
3. What happens if the supports are not rigid but can move slightly?
4. How would this problem change if the bar was made of a different material, like aluminum?
5. What are the long-term effects of cyclic thermal stresses on materials?

### Tip:
Thermal stresses can cause significant strain in materials, especially if they are constrained. Always account for temperature variations when designing structures in environments with extreme temperatures.

Learn how to calculate the thermal stress in portions AC and CB of a steel bar due to a temperature drop from +75°F to -50°F. This problem covers key thermodynamic concepts like thermal strain, Hooke's Law, and the impact of thermal expansion coefficients in constrained materials.

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