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 $A$ ($\Sigma M_A = 0$):