https://file.aiquickdraw.com/mathbot/file/1727334803989323z6tw1.jpg

The provided image shows a beam structure subjected to both a uniformly distributed load (UDL) and a point load. Here's a breakdown of the beam configuration:

- The beam is pinned at point \( A \) and has a roller support at point \( C \), indicating that there are vertical reactions at both points \( A \) and \( C \).
- The beam is loaded with:
  - A uniformly distributed load (UDL) of \( 30 \, \text{kN/m} \) over a length of \( 3.2 \, \text{m} \) from \( A \) to \( C \).
  - A point load of \( 24 \, \text{kN} \) acting vertically downward at point \( B \), which is \( 0.8 \, \text{m} \) from \( C \).

### Step 1: Free Body Diagram (FBD)

The FBD of the beam would include:
- A reaction force at point \( A \), denoted \( R_A \), acting vertically upward since it's a pinned support.
- A reaction force at point \( C \), denoted \( R_C \), acting vertically upward because it's a roller support.
- The uniformly distributed load (UDL) of \( 30 \, \text{kN/m} \), acting over the span from \( A \) to \( C \), for a total length of \( 3.2 \, \text{m} \).
- A concentrated load at point \( B \) of \( 24 \, \text{kN} \), acting vertically downward.

### Step 2: Calculation of Reaction Forces

#### Uniformly Distributed Load (UDL)
The total load from the UDL is calculated as:
\[
w_{\text{total}} = 30 \, \text{kN/m} \times 3.2 \, \text{m} = 96 \, \text{kN}
\]
This load acts at the centroid of the distributed load, which is located at the midpoint of \( A \) and \( C \) (i.e., \( 1.6 \, \text{m} \) from point \( A \)).

#### Reaction Force Calculations Using Equilibrium Equations

To calculate the reactions \( R_A \) and \( R_C \), we use the following static equilibrium equations:

1. **Sum of vertical forces (ΣF\(_y\)) = 0:**
   \[
   R_A + R_C = 96 \, \text{kN} + 24 \, \text{kN} = 120 \, \text{kN}
   \]

2. **Sum of moments about point \( A \) (ΣM\(_A\)) = 0:**
   Taking moments about point \( A \):
   \[
   R_C \times 3.2 \, \text{m} = 96 \, \text{kN} \times 1.6 \, \text{m} + 24 \, \text{kN} \times 4.0 \, \text{m}
   \]
   \[
   R_C \times 3.2 = 153.6 \, \text{kN} \cdot \text{m} + 96 \, \text{kN} \cdot \text{m}
   \]
   \[
   R_C \times 3.2 = 249.6 \, \text{kN} \cdot \text{m}
   \]
   \[
   R_C = \frac{249.6}{3.2} = 78 \, \text{kN}
   \]

3. **Reaction at \( A \):**
   Using the sum of vertical forces:
   \[
   R_A = 120 \, \text{kN} - R_C = 120 \, \text{kN} - 78 \, \text{kN} = 42 \, \text{kN}
   \]

### Final Reactions:
- \( R_A = 42 \, \text{kN} \)
- \( R_C = 78 \, \text{kN} \)

### Summary:

- The **Free Body Diagram (FBD)** includes:
  - Reaction forces \( R_A = 42 \, \text{kN} \) at point \( A \) and \( R_C = 78 \, \text{kN} \) at point \( C \).
  - A uniformly distributed load of \( 30 \, \text{kN/m} \) between points \( A \) and \( C \).
  - A point load of \( 24 \, \text{kN} \) at point \( B \).
  
Do you need further details or have any questions? Here are some related queries:

1. How would you calculate the bending moment at different points on the beam?
2. What is the shear force at point \( C \)?
3. How do the reaction forces change if the point load at \( B \) increases?
4. What would happen if the support at \( C \) was also pinned instead of a roller?
5. How would you draw the shear force and bending moment diagrams for this beam?

**Tip:** When working with distributed loads, always calculate the equivalent point load and its location to simplify the analysis.

Learn how to analyze a beam subjected to a uniformly distributed load (30 kN/m) and a point load (24 kN). The beam is pinned at point A and has a roller support at point C. The solution includes the calculation of reaction forces at points A and C using static equilibrium equations.

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