https://file.aiquickdraw.com/mathbot/file/1727338408111fd5893ro.jpg

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?

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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.

Learn how to find the free body diagram (FBD) and calculate reactions for a simply supported beam with a uniform distributed load of 30 kN/m over 3.2 m and a point load of 24 kN. We use statics principles and equilibrium equations to solve for the reactions at the supports. Suitable for engineering students.

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