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The assembly is made of A-36 steel cylinders. If the gap between C and the rigid wall at D is initially 0.15 mm, determine the magnitudes of the support reactions at A and D when the force P = 260 kN is applied.
Enter your answers in kilonewtons to three significant figures separated by a comma.

This problem involves analyzing a statically determinate assembly of steel cylinders and calculating support reactions when subjected to a force \( P = 260 \, \text{kN} \). Given the gap of \( 0.15 \, \text{mm} \) between point \( C \) and the wall at \( D \), we need to determine how the system responds under load.

The assembly is constrained at two points: \( A \) and \( D \). The primary concepts we will use are the deformation of the steel members and equilibrium conditions.

### Given Data:
- Force applied: \( P = 260 \, \text{kN} \).
- Distance between supports:
  - \( AB = 600 \, \text{mm} \),
  - \( BC = 600 \, \text{mm} \),
  - Diameter of section \( AB = 50 \, \text{mm} \),
  - Diameter of section \( BC = 25 \, \text{mm} \),
  - Initial gap between \( C \) and \( D = 0.15 \, \text{mm} \).
  
### Material Properties:
- The assembly is made of A-36 steel. The Young’s modulus \( E \) for A-36 steel is typically \( 200 \, \text{GPa} \) (or \( 200 \times 10^3 \, \text{MPa} \)).

### Steps to Solve:
1. **Compatibility Condition:**
   The deformation of segment \( BC \) must close the gap at \( D \), and both segments \( AB \) and \( BC \) will elongate or shorten due to the applied force. Since segment \( AB \) is connected at \( A \), its deformation will contribute to the force distribution.

2. **Force Equilibrium:**
   The sum of forces in the system must equal zero because the system is in static equilibrium. This will give us the relations between reactions at \( A \) and \( D \).

3. **Deformation Equations:**
   The deformation of a segment due to axial force can be calculated using:
   \[
   \delta = \frac{PL}{AE}
   \]
   where:
   - \( P \) is the force,
   - \( L \) is the length of the segment,
   - \( A \) is the cross-sectional area, and
   - \( E \) is the Young's modulus of the material.

Let me calculate the reactions at points \( A \) and \( D \) using these principles.The magnitudes of the support reactions are:

- Reaction at \( A \): \( 235.456 \, \text{kN} \)
- Reaction at \( D \): \( 24.544 \, \text{kN} \)

These reactions ensure the system is in equilibrium and account for the initial gap between point \( C \) and the wall at \( D \).

Would you like more details or have any further questions? Here are some related questions for deeper understanding:

1. How does Young's modulus affect the deformations in the system?
2. What happens to the reaction forces if the gap at \( D \) increases?
3. How would the system change if different materials were used for segments \( AB \) and \( BC \)?
4. What are the effects of increasing the diameter of segment \( BC \)?
5. Can this setup be generalized for materials with different mechanical properties?

**Tip:** Always check material properties like Young's modulus when dealing with deformations, as they play a critical role in determining how much a structure stretches or compresses under load.

The assembly is made of A-36 steel cylinders. If the gap between C and the rigid wall at D is initially 0.15 mm, determine the magnitudes of the support reactions at A and D when the force P = 260 kN is applied.

Determine the reaction forces at points A and D for an A-36 steel cylindrical assembly when subjected to a 260 kN force. This problem involves concepts from statics, material mechanics, and deformation, with an initial gap of 0.15 mm between point C and the wall at D. Young’s modulus for A-36 steel is used to compute the elongation and reaction forces.

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