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To solve for the reactions at points A and B in the given beam structure, we need to apply static equilibrium conditions. These conditions are:

1. The sum of all vertical forces must be zero:
   \[
   \Sigma F_y = 0
   \]

2. The sum of moments about any point must be zero:
   \[
   \Sigma M = 0
   \]

### Step-by-Step Breakdown

#### 1. Identify the forces acting on the beam:
- A vertical reaction at \( A \) (denoted as \( R_A \)).
- A vertical reaction at \( B \) (denoted as \( R_B \)).
- A 50 kN point load acting at a distance of 3 m from point \( A \).
- A 20 kN/m uniform distributed load (UDL) of 3 m acting from \( A \) to the 3 m mark.
- An 80 kN point load acting 5 m from point \( A \) (or 2 m from the end of the UDL).
- A 30 kN/m triangular load starting at point \( B \) and acting over the final 6 m of the beam.

#### 2. Calculate the resultant forces of the distributed loads:
- For the 20 kN/m UDL (between \( A \) and 3 m):
  \[
  \text{Resultant force} = 20 \, \text{kN/m} \times 3 \, \text{m} = 60 \, \text{kN}
  \]
  The resultant of this UDL acts at the centroid, which is 1.5 m from point \( A \).

- For the triangular load of 30 kN/m (acting over 6 m):
  \[
  \text{Resultant force} = \frac{1}{2} \times 30 \, \text{kN/m} \times 6 \, \text{m} = 90 \, \text{kN}
  \]
  The location of the resultant force of a triangular load is \( \frac{1}{3} \) of the base length from the larger side, i.e., \( 2 \, \text{m} \) from point \( B \).

#### 3. Apply the equilibrium equations:

**Sum of vertical forces (\( \Sigma F_y = 0 \)):**
\[
R_A + R_B = 50 + 60 + 80 + 90 = 280 \, \text{kN}
\]

**Sum of moments about point \( A \) (\( \Sigma M_A = 0 \)):**
\[
- (50 \, \text{kN}) \times 3 \, \text{m}
- (60 \, \text{kN}) \times 1.5 \, \text{m}
- (80 \, \text{kN}) \times 5 \, \text{m}
- (90 \, \text{kN}) \times (6 \, \text{m} + 2 \, \text{m})
+ R_B \times 10 \, \text{m} = 0
\]

Let me compute the moments and solve for \( R_B \).The reaction at point \( B \) is \( R_B = 136 \, \text{kN} \).

Now, using the equilibrium equation for vertical forces:

\[
R_A + R_B = 280 \, \text{kN}
\]

Substituting \( R_B = 136 \, \text{kN} \):

\[
R_A + 136 = 280
\]
\[
R_A = 280 - 136 = 144 \, \text{kN}
\]

Thus, the reactions at the supports are:
- \( R_A = 144 \, \text{kN} \)
- \( R_B = 136 \, \text{kN} \)

Would you like a detailed breakdown of any specific step or further questions?

Here are 5 related questions to consider:
1. How are the centroids of different types of loads (UDL, triangular) calculated?
2. How would the reactions change if the beam was cantilevered at one end instead of being supported at both ends?
3. What are shear force and bending moment diagrams, and how are they constructed for this beam?
4. How would the solution change if there was an additional horizontal load acting on the beam?
5. How do different boundary conditions (e.g., fixed, roller, hinge) affect the reactions in a beam?

**Tip:** Always calculate the moments about a point where one unknown reaction is present to simplify the solution process.

Solve for the reactions on point A and B in the beam structure.

Learn how to solve for the reactions at points A and B in a beam structure using static equilibrium equations. This problem involves calculating the resultant forces of distributed loads and applying the sum of forces and moments to find reactions. Suitable for college-level engineering mechanics students.

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