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It looks like you're interested in solving Problems 4.2, 4.3, and 4.4 which involve distributed loads and point loads on beams. Let's analyze each one step by step to find the reactions at the supports and other necessary calculations. 

### Problem 4.2
For Problem 4.2, you have a beam with several point loads and distributed loads:
- Point loads: \(5 \text{ kN at B}, 10 \text{ kN at C}, 5 \text{ kN at D}\)
- Distributed loads: \(3 \text{ kN/m over AB}, 2 \text{ kN/m over BC}, 4 \text{ kN/m over DE}\)
- Lengths: \(2.0 \text{ m between each point}\)

### Problem 4.3
For Problem 4.3, the beam's loads are:
- Point loads: \(8 \text{ kN at B}, 12 \text{ kN at D}\)
- Distributed loads: \(3 \text{ kN/m over AB}, 6 \text{ kN/m over DE}\)
- Lengths: \(2.0 \text{ m from A to B}, \(3.0 \text{ m from B to C and C to D}\), \(3.0 \text{ m from D to E}\)

### Problem 4.4
Finally, for Problem 4.4, the details are:
- Point loads: \(12 \text{ kN at C}\)
- Distributed loads: \(2 \text{ kN/m over AB}, \(6 \text{ kN/m over BC}\)
- Lengths: \(4.0 \text{ m from A to B}, \(2.0 \text{ m from B to C}\)

I'll begin with Problem 4.2, calculating the reactions at the supports due to both the distributed loads and the point loads. Let's start by calculating the total load from distributed forces and their moments about one end, then add the moments from the point loads, and use these to find the reactions. We will proceed to Problems 4.3 and 4.4 in a similar manner. Let's start with Problem 4.2.### Solution to Problem 4.2

For the given beam:

- **Total Load on Beam**: 38.0 kN
- **Total Moment about point A**: 202.0 kN·m

**Reactions at the supports:**
- **Reaction at A (R_A)**: 4.33 kN
- **Reaction at E (R_E)**: 33.67 kN

These reactions are based on the principle of equilibrium where the sum of vertical forces and the sum of moments about any point (in this case, point A) must be zero. Let's proceed with solving Problem 4.3 using a similar approach.### Solution to Problem 4.3

For the given beam:

- **Total Load on Beam**: 44.0 kN
- **Total Moment about point A**: 253.0 kN·m

**Reactions at the supports:**
- **Reaction at A (R_A)**: 21.0 kN
- **Reaction at E (R_E)**: 23.0 kN

These results are computed ensuring the conditions for static equilibrium, where both the sum of vertical forces and the moments about any point (here, point A) must equal zero. Now, let's solve Problem 4.4 in the same structured approach.### Solution to Problem 4.4

For the given beam:

- **Total Load on Beam**: 32.0 kN
- **Total Moment about point A**: 196.0 kN·m

**Reactions at the supports:**
- **Reaction at A (R_A)**: -0.67 kN (This value being negative suggests an upward reaction, indicating tension at this support rather than compression)
- **Reaction at C (R_C)**: 32.67 kN

The calculated results ensure that the conditions of static equilibrium are met, with forces summing to zero and moments around any point (such as point A) also summing to zero.

If you need further details or clarification on any part of these calculations, please let me know! Here are a few related questions you might find interesting:
1. How do changes in the length of spans affect the reactions at the supports?
2. What would be the effect of changing the location of the point loads on the beam's reactions?
3. How does adding additional supports within these beam setups affect the load distribution?
4. Could you explore the deflection profiles of these beams under the given loads?
5. What would be the shear force and bending moment diagrams for any of these beams?

**Tip:** When solving similar problems, always ensure to start by calculating total forces and moments about a point to simplify finding reactions at the supports.

Detailed solutions for beam reaction calculations in Problems 4.2, 4.3, and 4.4. These involve point loads and distributed loads on statically determinate beams. Learn how to calculate reactions, moments, and load distribution, with solutions based on static equilibrium principles. Suitable for engineering students.

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