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The problem is asking to determine the stress and strain in portions \(AC\) and \(CB\) of a steel bar when the temperature of the bar is reduced to \(-25^\circ C\), given that the initial temperature is \(+25^\circ C\). The problem also provides the following dataThe problem is asking to determine the stress and strain in portions \(AC\) and \(CB\) of a steel bar when the temperature of the bar is reduced to \(-25^\circ C\), given that the initial temperature is \(+25^\circ C\). The problem also provides the following 

- Force \(F = 200 \, \text{kN}\)
- Modulus of elasticity \(E = 200 \, \text{GPa}\)
- Coefficient of thermal expansion \(\alpha = 11.7 \times 10^{-6} \, /^\circ C\)

### Step 1: Calculate the thermal strain
The thermal strain \(\epsilon_{\text{thermal}}\) is given by:
\[
\epsilon_{\text{thermal}} = \alpha \Delta T
\]
where:
- \(\alpha = 11.7 \times 10^{-6} \, /^\circ C\)
- \(\Delta T = T_{\text{final}} - T_{\text{initial}} = -25^\circ C - (+25^\circ C) = -50^\circ C\)

\[
\epsilon_{\text{thermal}} = 11.7 \times 10^{-6} \times (-50) = -5.85 \times 10^{-4}
\]

### Step 2: Calculate the stress due to the applied force
The stress \(\sigma\) due to the applied force is given by:
\[
\sigma = \frac{F}{A}
\]
where:
- \(F = 200 \, \text{kN} = 200 \times 10^3 \, \text{N}\)
- Area \(A = 400 \, \text{mm}^2\) for the larger section and \(A = 300 \, \text{mm}^2\) for the smaller section

For the larger section (\(AC\)):
\[
\sigma_{AC} = \frac{200 \times 10^3 \, \text{N}}{400 \times 10^{-6} \, \text{m}^2} = 500 \, \text{MPa}
\]

For the smaller section (\(CB\)):
\[
\sigma_{CB} = \frac{200 \times 10^3 \, \text{N}}{300 \times 10^{-6} \, \text{m}^2} = 666.67 \, \text{MPa}
\]

### Step 3: Determine the total strain in both sections
The total strain \(\epsilon\) in each section is given by the sum of the mechanical strain due to the stress and the thermal strain:
\[
\epsilon = \frac{\sigma}{E} + \epsilon_{\text{thermal}}
\]
where \(E = 200 \, \text{GPa} = 200 \times 10^9 \, \text{Pa}\).

For \(AC\):
\[
\epsilon_{AC} = \frac{500 \times 10^6 \, \text{Pa}}{200 \times 10^9 \, \text{Pa}} + (-5.85 \times 10^{-4}) = 2.5 \times 10^{-3} - 5.85 \times 10^{-4} = 1.915 \times 10^{-3}
\]

For \(CB\):
\[
\epsilon_{CB} = \frac{666.67 \times 10^6 \, \text{Pa}}{200 \times 10^9 \, \text{Pa}} + (-5.85 \times 10^{-4}) = 3.333 \times 10^{-3} - 5.85 \times 10^{-4} = 2.748 \times 10^{-3}
\]

### Step 4: Calculate the stress considering the thermal effects
The stress considering thermal effects is:
\[
\sigma = E(\epsilon)
\]

For \(AC\):
\[
\sigma_{AC} = 200 \times 10^9 \times 1.915 \times 10^{-3} = 383 \, \text{MPa}
\]

For \(CB\):
\[
\sigma_{CB} = 200 \times 10^9 \times 2.748 \times 10^{-3} = 549.6 \, \text{MPa}
\]

### Conclusion:
- The stress in section \(AC\) is 383 MPa.
- The stress in section \(CB\) is 549.6 MPa.
- The corresponding strains are \(1.915 \times 10^{-3}\) for \(AC\) and \(2.748 \times 10^{-3}\) for \(CB\).

Would you like more details or have any questions?

### Related Questions:
1. How would the results change if the temperature drop was different?
2. What would happen to the stress if the material was aluminum instead of steel?
3. How do different cross-sectional areas affect the stress distribution?
4. What is the impact of thermal expansion on stress in composite materials?
5. How would the analysis differ if the bar was fixed at only one end?

### Tip:
Always double-check the units when calculating stress and strain, especially when converting between different unit systems.

Learn how to calculate stress and strain in sections AC and CB of a steel bar when the temperature changes from +25°C to -25°C. Includes detailed calculations for thermal strain, stress due to applied force, and final stress considering thermal effects. Suitable for advanced engineering and material science studies.

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