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To determine the shear force and moment at point C in the beam, we first need to break down the given dataTo determine the shear force and moment at point C in the beam, we first need to break down the given 

### Given:
1. **Beam AB** with a total length of \( 8 \, \text{m} \) (2 m from \( A \) to \( C \), 6 m from \( C \) to \( B \)).
2. The beam is subjected to a **triangular distributed load** starting from zero at \( A \) and reaching \( 4 \, \text{kN/m} \) at \( B \).

### Approach:
We'll follow these steps to determine the shear force and moment at \( C \):
1. **Calculate reactions at supports \( A \) and \( B \)** using static equilibrium.
2. **Determine the shear force** at \( C \).
3. **Determine the bending moment** at \( C \).

#### 1. Reaction Forces at \( A \) and \( B \)
We can treat the triangular load as a distributed load. The total load can be found by calculating the area under the triangular load.

The **total load** (\( W \)) due to the triangular distributed load is:
\[
W = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \, \text{m} \times 4 \, \text{kN/m} = 16 \, \text{kN}
\]

This total load acts at the **centroid** of the triangle, which is located at a distance of:
\[
\frac{1}{3} \times 8 \, \text{m} = 2.67 \, \text{m} \text{ from point B}.
\]

Now, applying the static equilibrium equations:
- Sum of vertical forces: \( R_A + R_B = 16 \, \text{kN} \)
- Sum of moments about point \( A \):
\[
R_B \times 8 \, \text{m} = 16 \, \text{kN} \times (2.67 \, \text{m}).
\]
From this, we can solve for \( R_B \) and \( R_A \).

#### 2. Shear Force at \( C \)
Once we know the reaction forces, the shear force at point \( C \) can be calculated by considering the portion of the load acting between \( A \) and \( C \).

#### 3. Moment at \( C \)
The moment at \( C \) can be determined by calculating the moment of forces to the left of point \( C \).

Let me compute the values based on these steps.### Results:
1. **Reaction at \( A \) ( \( R_A \) ):** \( 10.67 \, \text{kN} \)
2. **Reaction at \( B \) ( \( R_B \) ):** \( 5.33 \, \text{kN} \)
3. **Shear Force at \( C \):** \( 9.67 \, \text{kN} \)

Next, let's calculate the **bending moment at \( C \)** by taking moments of forces to the left of point \( C \).### Bending Moment at \( C \):
The bending moment at point \( C \) is **20.67 kN·m**.

### Summary:
- **Reaction at \( A \):** \( 10.67 \, \text{kN} \)
- **Reaction at \( B \):** \( 5.33 \, \text{kN} \)
- **Shear Force at \( C \):** \( 9.67 \, \text{kN} \)
- **Bending Moment at \( C \):** \( 20.67 \, \text{kN·m} \)

Would you like further details on any part of the solution? Here are some related questions you might want to explore:

1. How is the centroid of a triangular load distribution calculated?
2. How does the load distribution affect the reactions at the supports?
3. What is the significance of the shear force diagram for beam analysis?
4. How do boundary conditions (like supports) influence bending moments?
5. Can we use the same method for a different load distribution, such as a uniform or parabolic load?

**Tip:** Always verify equilibrium in static problems by checking both the sum of forces and the sum of moments.

Determine the shear force and moment acting at a section passing through point C in the beam.

Learn how to determine the shear force and moment acting at a section passing through point C in a beam subjected to a triangular distributed load. This step-by-step solution involves calculating reactions at supports, using static equilibrium, and finding shear force and bending moment. Suitable for college-level engineering students.

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