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To determine the reactions at points \( A \) and \( B \) in this beam structure, we must resolve the forces and moments acting on the beam. The beam is subjected to various forces and distributed loads. Let’s analyze the given information and solve step by step.

### Step 1: Known Loads and Dimensions
- A **point load** of \( 50 \, \text{kN} \) at 3 m from the left support \( A \).
- A **uniformly distributed load (UDL)** of \( 20 \, \text{kN/m} \) over a span of 6 m starting from the left support.
- A **point load** of \( 80 \, \text{kN} \) located 2 m from the roller support \( B \).
- A **linearly varying load** starting at \( 0 \, \text{kN/m} \) and increasing to \( 30 \, \text{kN/m} \) over 6 m from \( B \) to the right.

### Step 2: Equivalent Loads for Distributed Loads
1. **For the \( 20 \, \text{kN/m} \) UDL** over a span of 6 m:
   - Equivalent point load: \( 20 \times 6 = 120 \, \text{kN} \)
   - This load acts at the centroid, i.e., at \( 3 \, \text{m} \) from the center of the UDL (or 6 m from \( A \)).

2. **For the triangular load starting at 0 and increasing to \( 30 \, \text{kN/m} \):**
   - Equivalent point load: The total load for a triangular distribution is \( \frac{1}{2} \times \text{base} \times \text{height} \), which gives \( \frac{1}{2} \times 6 \times 30 = 90 \, \text{kN} \).
   - This load acts at \( \frac{1}{3} \) the base from the larger end, so it acts \( 2 \, \text{m} \) from the right end.

### Step 3: Reaction Forces
We'll apply equilibrium conditions (sum of forces and sum of moments) to find reactions at \( A \) and \( B \).

#### Equilibrium Equations
- **Sum of vertical forces:**
  \[
  R_A + R_B = 120 \, \text{kN} + 50 \, \text{kN} + 80 \, \text{kN} + 90 \, \text{kN}
  \]
  \[
  R_A + R_B = 340 \, \text{kN} \quad \text{(Equation 1)}
  \]

- **Sum of moments about point \( A \):**
  Taking moments about \( A \) (clockwise positive):
  \[
  \text{Moment at } A = 0
  \]
  \[
  - R_B \times 12 + (120 \, \text{kN} \times 6) + (50 \, \text{kN} \times 3) + (80 \, \text{kN} \times 10) + (90 \, \text{kN} \times 16) = 0
  \]
  \[
  -12R_B + 720 + 150 + 800 + 1440 = 0
  \]
  \[
  -12R_B + 3110 = 0
  \]
  \[
  R_B = \frac{3110}{12} = 259.17 \, \text{kN} \quad \text{(Equation 2)}
  \]

- **Using Equation 1:**
  \[
  R_A + 259.17 = 340
  \]
  \[
  R_A = 340 - 259.17 = 80.83 \, \text{kN}
  \]

### Step 4: Results
- Reaction at \( A \), \( R_A = 80.83 \, \text{kN} \)
- Reaction at \( B \), \( R_B = 259.17 \, \text{kN} \)

Let me know if you need further details or explanations! Here are some related questions to consider:

1. How do we determine the position of the resultant force for non-uniform loads?
2. What is the difference between point loads and distributed loads in beam analysis?
3. How do the reactions change if the beam is cantilevered at one end?
4. What are shear force and bending moment diagrams, and how are they derived?
5. How do triangular loads affect the shear and moment at different points in a beam?

**Tip:** Always ensure that the sum of moments and forces is zero when solving for reactions in static equilibrium problems.

Explore how to calculate the reactions at supports A and B for a beam subjected to point loads, uniformly distributed loads, and triangular loads. This analysis uses basic principles of static equilibrium conditions in beam theory, suitable for university-level engineering studies.

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