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.