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The problem you have provided shows a rigid bar \( AB \) attached to two vertical rods, one made of steel and the other made of aluminum. The bar is initially horizontal before a load \( P = 70 \, \text{kN} \) is applied vertically downward at the center of the bar. We are tasked with determining the vertical movement of the bar.

Let's summarize the key information for both materials:

- **Steel rod:**
  - Length \( L_s = 4 \, \text{m} \)
  - Cross-sectional area \( A_s = 300 \, \text{mm}^2 \)
  - Modulus of elasticity \( E_s = 200 \, \text{GPa} \)

- **Aluminum rod:**
  - Length \( L_a = 5 \, \text{m} \)
  - Cross-sectional area \( A_a = 500 \, \text{mm}^2 \)
  - Modulus of elasticity \( E_a = 70 \, \text{GPa} \)

### Approach:
1. **Assumption of compatibility**: The rigid bar stays horizontal after the load is applied, so the vertical displacement of both rods at points A and B must be the same.
  
2. **Force distribution**: The load \( P \) will be distributed between the steel and aluminum rods. Let's denote the force in the steel rod as \( F_s \) and in the aluminum rod as \( F_a \). Therefore:
   \[
   F_s + F_a = P = 70 \, \text{kN}
   \]

3. **Deformation relationship**: The deformation \( \delta \) in each rod is related to the force, the material properties, and the geometry by the formula:
   \[
   \delta = \frac{F L}{A E}
   \]
   where \( F \) is the force in the rod, \( L \) is the length, \( A \) is the cross-sectional area, and \( E \) is the modulus of elasticity.

Since the bar remains horizontal, the deformations in both rods must be equal, i.e., \( \delta_s = \delta_a \). We can set up an equation using the deformation formulas for both rods and solve for the forces.

I'll calculate the forces and the vertical displacement of the bar next.The forces in the steel and aluminum rods are as follows:

- Force in the steel rod: \( F_s = 47.73 \, \text{kN} \)
- Force in the aluminum rod: \( F_a = 22.27 \, \text{kN} \)

The vertical displacement of the bar (which is the same for both rods) is:
\[
\delta = 0.00318 \, \text{m} = 3.18 \, \text{mm}
\]

Thus, the vertical movement of the rigid bar is \( 3.18 \, \text{mm} \) downward.

Do you need more details or have any questions?

Here are five related questions to expand on this:
1. How would the displacement change if the cross-sectional area of the steel rod was increased?
2. What happens to the distribution of forces if the modulus of elasticity of aluminum changes?
3. How would the system behave if the lengths of both rods were the same?
4. How do different materials' properties affect the overall stiffness of the system?
5. Can we analyze this system if the bar is not initially horizontal?

**Tip:** Remember, the material's modulus of elasticity directly influences how much it stretches under a load. A higher modulus means less deformation for the same applied force.

The rigid bar AB, attached to two vertical rods as shown in the figure, is horizontal before the load is applied. If the load P = 70 kN, determine its vertical movement.

Learn how to calculate the vertical movement of a rigid bar attached to steel and aluminum rods under a 70 kN load. This problem involves key concepts in statics and material deformation, including stress-strain relationships and force balance.

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