Determine the values of the stress and strain in portions AC and CB of the steel
bar shown (Figure below) when the temperature of the bar is -45 0
C, knowing that
a close fit exists at both of the rigid supports when the temperature is +24 0
C. Use
the values E = 200 GPa and
a = 11.7 * 10 –6
/ 0
C for steel

Let's determine the stress and strain in portions \(AC\) and \(CB\) of the steel bar when the temperature changes from \(+24^\circ C\) to \(-45^\circ C\). The bar is in close fit with the rigid supports at the initial temperature of \(+24^\circ C\).

### Given Data:
- Initial temperature \(T_{\text{initial}} = +24^\circ C\)
- Final temperature \(T_{\text{final}} = -45^\circ C\)
- Young's modulus \(E = 200 \, \text{GPa} = 200 \times 10^9 \, \text{Pa}\)
- Coefficient of thermal expansion \(\alpha = 11.7 \times 10^{-6} \, /^\circ C\)

### Step 1: Calculate the Temperature Change
The temperature change \(\Delta T\) is:
\[
\Delta T = T_{\text{final}} - T_{\text{initial}} = -45^\circ C - 24^\circ C = -69^\circ C
\]

### Step 2: Calculate the Free Thermal Strain
The thermal strain \(\epsilon_{\text{thermal}}\) without any constraints is given by:
\[
\epsilon_{\text{thermal}} = \alpha \Delta T
\]
Substituting the values:
\[
\epsilon_{\text{thermal}} = 11.7 \times 10^{-6} \times (-69) = -8.073 \times 10^{-4}
\]

### Step 3: Calculate the Stress due to Thermal Strain
Since the steel bar is constrained (fixed at both ends), it cannot freely expand or contract. Therefore, a thermal stress will develop to counteract the thermal strain. This thermal stress \(\sigma_{\text{thermal}}\) is given by:
\[
\sigma_{\text{thermal}} = E \cdot \epsilon_{\text{thermal}}
\]
Substituting the values:
\[
\sigma_{\text{thermal}} = 200 \times 10^9 \, \text{Pa} \times (-8.073 \times 10^{-4}) = -1.6146 \times 10^8 \, \text{Pa}
\]
Converting to MPa:
\[
\sigma_{\text{thermal}} = -161.46 \, \text{MPa}
\]

### Step 4: Determine the Total Strain
Since the bar is in equilibrium, the total strain \(\epsilon_{\text{total}}\) is zero because the thermal strain is fully resisted by the induced mechanical strain. Therefore, the strain in both sections \(AC\) and \(CB\) is:
\[
\epsilon_{AC} = \epsilon_{CB} = \epsilon_{\text{thermal}} = -8.073 \times 10^{-4}
\]

### Conclusion:
- **Stress in both portions \(AC\) and \(CB\):** \(-161.46 \, \text{MPa}\) (compressive stress)
- **Strain in both portions \(AC\) and \(CB\):** \(-8.073 \times 10^{-4}\)

This result implies that the bar is under compressive stress due to the reduction in temperature, with a corresponding compressive strain.

Would you like more details or have any questions?

### Related Questions:
1. How would the stress and strain change if the bar was not perfectly rigid?
2. What if the bar was made of a material with a different coefficient of thermal expansion?
3. How would the bar's behavior change if the supports allowed some movement?
4. What effect would increasing the cross-sectional area have on the stress?
5. How would the analysis differ if the bar was subject to an external force in addition to the thermal stress?

### Tip:
When calculating thermal stresses, it's important to consider whether the material is free to expand or is constrained, as this significantly impacts the resulting stress and strain.

Learn how to calculate stress and strain in portions AC and CB of a steel bar due to temperature change from +24°C to -45°C. Detailed solution using Young's modulus and coefficient of thermal expansion for steel.

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