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Let's work through the questions step by step.

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### First Diagram: Cantilever Beam (Question #8)
- **Length of beam**: \( 4 \, \text{m} \)
- **Uniformly Distributed Load (UDL)**: \( 3 \, \text{kN/m} \)

#### 1. Shear force reaction at support A:
For a cantilever beam with a UDL, the total load \( W \) is calculated as:

\[
W = w \times L
\]
where:
- \( w = 3 \, \text{kN/m} \) (distributed load),
- \( L = 4 \, \text{m} \) (length of the beam).

\[
W = 3 \times 4 = 12 \, \text{kN}
\]

This total load acts vertically downwards, and since it's a cantilever beam, the entire load will be resisted by a vertical shear force at point \( A \). So, the **shear force at support \( A \)** is:

\[
V_A = 12 \, \text{kN}
\]

#### 2. Moment at support A:
For a cantilever beam subjected to a uniformly distributed load, the moment at the fixed end (support \( A \)) is calculated by:

\[
M_A = \frac{w \times L^2}{2}
\]

Substituting the given values:

\[
M_A = \frac{3 \times 4^2}{2} = \frac{3 \times 16}{2} = 24 \, \text{kNm}
\]

Thus, the **moment at support \( A \)** is \( 24 \, \text{kNm} \).

---

### Second Diagram: Simply Supported Beam (Question #9)
- **Length of beam**: \( 5.5 \, \text{m} \)
- **Uniformly Distributed Load (UDL)**: \( 3.5 \, \text{kN/m} \)

#### 1. Reaction at each support (A and B):
For a simply supported beam with a UDL over the entire span, the total load \( W \) is:

\[
W = w \times L
\]
where:
- \( w = 3.5 \, \text{kN/m} \),
- \( L = 5.5 \, \text{m} \).

\[
W = 3.5 \times 5.5 = 19.25 \, \text{kN}
\]

Since the beam is simply supported, the load is distributed equally between the two supports \( A \) and \( B \). The reaction at each support is:

\[
R_A = R_B = \frac{W}{2} = \frac{19.25}{2} = 9.625 \, \text{kN}
\]

#### 2. Maximum moment:
For a simply supported beam with a UDL, the maximum bending moment occurs at the midpoint of the beam and is given by:

\[
M_{max} = \frac{w \times L^2}{8}
\]

Substituting the given values:

\[
M_{max} = \frac{3.5 \times 5.5^2}{8} = \frac{3.5 \times 30.25}{8} = \frac{105.875}{8} = 13.23 \, \text{kNm}
\]

Thus, the **maximum moment** is \( 13.23 \, \text{kNm} \).

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Let me know if you need further details on any step!

### Related Questions:
1. What is the difference between shear force and bending moment in beams?
2. How does the load distribution change for non-uniformly distributed loads?
3. How would you calculate the deflection of the cantilever beam under the UDL in Question #8?
4. What factors affect the bending moment in beams with multiple spans?
5. How do you calculate the reactions for beams with overhangs?

### Tip:
In beam problems, always check if the loading is uniformly distributed or point load based, as this significantly impacts the calculations for reactions and moments.

1. A cantilever beam AB, with a fixed-end support at left end A, has a total length of 4 meters. If the full length of the beam is loaded with a uniformly distributed load equal to 3 kN/m, what is the shear force reaction at the support?
2. What is the moment at support A?

1. A 3.5 kN/m uniformly distributed load is placed over a simple beam with a length of 5.50 meters. If the beam is supported on both ends, what is the reaction for each support?
2. What is the maximum moment for the beam?

Solve cantilever and simply supported beam problems involving uniformly distributed loads. This example covers shear force reactions, bending moments, and reactions at supports for engineering students.

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