To solve for the vertical reaction at Support B, we need to analyze the beam and apply static equilibrium conditions.

Step 1: Analyze the load distribution

The load is a uniformly varying load, starting at 15 kN/m at $A$ and increasing to 30 kN/m at $B$. This creates a triangular distributed load over the beam.

Total Load ($W$) of the triangular distribution:

The area under the triangular load is: $W = \frac{1}{2} \times \text{base} \times \text{height}$ $W = \frac{1}{2} \times 3 \, \text{m} \times (15 + 30) \, \text{kN/m} = \frac{1}{2} \times 3 \times 45 = 67.5 \, \text{kN}$

Location of the Resultant Load:

For a triangular load, the resultant acts at $\frac{2}{3}$ of the length from the smaller end ($A$): $x_{\text{resultant}} = \frac{2}{3} \times 3 \, \text{m} = 2 \, \text{m} \, \text{from } A$

Step 2: Apply Equilibrium Conditions

Static equilibrium involves:

1. Summation of vertical forces ($\Sigma F_y = 0$): $R_A + R_B - W = 0$ $R_A + R_B = 67.5 \, \text{kN}$

2. Summation of moments about $A$ ($\Sigma M_A = 0$): Taking moments counterclockwise as positive: $R_B \cdot 3 - W \cdot x_{\text{resultant}} = 0$ $R_B \cdot 3 - 67.5 \cdot 2 = 0$ $R_B = \frac{135}{3} = 45 \, \text{kN}$

Step 3: Solve for $R_A$:

Substituting $R_B = 45 \, \text{kN}$ into $R_A + R_B = 67.5$: $R_A + 45 = 67.5$ $R_A = 22.5 \, \text{kN}$

Final Answer:

The vertical reaction at Support $B$ is: $\boxed{45 \, \text{kN (Upwards)}}$

This does not match the provided options, so the correct answer is None of the Above.

Let me know if you have questions or want further clarification! Here are related questions for exploration:

1. What is the reaction at Support $A$?
2. How would the reactions change if the load distribution was uniform?
3. How do you calculate the moment at the supports in such problems?
4. Can you explain how the location of the resultant force changes with load type?
5. What are common mistakes in analyzing triangular loads?

Tip: Always check for consistent units and carefully determine the location of resultant forces for distributed loads.