Let's analyze the image and solve for the reactions at the supports.

Step 1: Identify the elements in the diagram.

• Beam: A simply supported beam with point load and uniformly distributed load (UDL).
• Supports:
  • Support A: A pinned support (which can take both vertical and horizontal reactions).
  • Support B: A roller support (which can take only vertical reactions).

Step 2: Loadings on the beam.

• Uniformly Distributed Load (UDL): 30 kN/m, acting from A to C over a span of 3.2 m.
• Point Load: 24 kN, applied vertically at point B.

Step 3: Free Body Diagram (FBD)

In the FBD, we can represent the loads and reactions as follows:

• Vertical reaction at A: $R_A$
• Vertical reaction at B: $R_B$
• Horizontal reaction at A: $H_A$

We can ignore the horizontal reaction $H_A$, as there are no horizontal forces acting on the system.

Step 4: Equilibrium Equations

We will use the static equilibrium conditions:

1. Sum of Forces in the Vertical Direction (ΣF_y = 0): $R_A + R_B = \text{Total load on the beam}$ The total load is the sum of the UDL and the point load.

   UDL load: $30 \, \text{kN/m} \times 3.2 \, \text{m} = 96 \, \text{kN}$

   Point load: $24 \, \text{kN}$

   Therefore, the total vertical load is: $R_A + R_B = 96 + 24 = 120 \, \text{kN}$ So, $R_A + R_B = 120 \, \text{kN}$

2. Sum of Moments about Point A (ΣM_A = 0): The moment created by each force around point A must sum to zero.

   Moment due to UDL: The UDL creates a resultant force of 96 kN acting at the center of its distribution (i.e., 1.6 m from A). Thus, the moment is: $\text{Moment of UDL} = 96 \, \text{kN} \times 1.6 \, \text{m} = 153.6 \, \text{kN·m}$

   Moment due to point load at B: The point load of 24 kN acts at a distance of 4.0 m from A. The moment is: $\text{Moment of point load} = 24 \, \text{kN} \times 4.0 \, \text{m} = 96 \, \text{kN·m}$

   Moment due to reaction at B: $R_B$ creates a clockwise moment about A, and its distance from A is 4.0 m: $\text{Moment of } R_B = R_B \times 4.0 \, \text{m}$

   Using $ΣM_A = 0$: $R_B \times 4.0 = 153.6 + 96$ $R_B \times 4.0 = 249.6$ $R_B = \frac{249.6}{4.0} = 62.4 \, \text{kN}$

3. Sum of Forces in the Vertical Direction (Revisited): Now that we have $R_B$, we can solve for $R_A$: $R_A + 62.4 = 120$ $R_A = 120 - 62.4 = 57.6 \, \text{kN}$

Step 5: Final Reactions

• Reaction at A ($R_A$) = 57.6 kN (upward)
• Reaction at B ($R_B$) = 62.4 kN (upward)

Step 6: Free Body Diagram (FBD)

• UDL of 30 kN/m over 3.2 m acting from A to C.
• Point load of 24 kN acting at B.
• Reaction forces $R_A = 57.6 \, \text{kN}$ and $R_B = 62.4 \, \text{kN}$.

Would you like the detailed drawing of the FBD?

Here are 5 related questions to expand on this topic:

1. How would the reaction forces change if the UDL was increased to 40 kN/m?
2. What would be the reactions if the point load was moved 0.5 m to the left?
3. How does the distance between the supports affect the reaction forces?
4. Can you calculate the bending moment at the midpoint between A and B?
5. What would happen if the roller support at B was replaced with a pinned support?

Tip: Always double-check your moment calculations by verifying the dimensions and force magnitudes, especially when dealing with distributed loads.