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To calculate the shear force and bending moment for the beam in the given problem, we'll follow a standard approach for solving statically determinate beams subjected to a uniformly distributed load (UDL).

### Step-by-Step Approach:

1. **Beam Overview:**
   - The beam is 3 meters long and is subjected to a uniformly distributed load (UDL) of \( 5 \, \text{kN/m} \).
   - The left support at \( A \) is a pin (fixed in both directions), and the right support at \( B \) is a roller (only vertical support).
   - The UDL is spread over the entire span of the beam (3 m).

2. **Reactions at Supports:**
   First, we calculate the reactions at the supports using equilibrium equations.
   
   - Total load due to UDL: 
     \[
     w_{\text{total}} = 5 \, \text{kN/m} \times 3 \, \text{m} = 15 \, \text{kN}
     \]
     This load is uniformly distributed over the entire span and acts at the center (1.5 m from both ends).

   - Equilibrium of forces in the vertical direction (∑Fy = 0):
     \[
     R_A + R_B = 15 \, \text{kN}
     \]
     where \( R_A \) and \( R_B \) are the reaction forces at supports \( A \) and \( B \), respectively.

   - Taking moments about point \( A \) (∑M_A = 0):
     \[
     R_B \times 3 \, \text{m} = 15 \, \text{kN} \times 1.5 \, \text{m}
     \]
     Solving for \( R_B \):
     \[
     R_B = \frac{15 \times 1.5}{3} = 7.5 \, \text{kN}
     \]

   - Substituting \( R_B \) back into the force equilibrium equation:
     \[
     R_A + 7.5 = 15 \quad \Rightarrow \quad R_A = 7.5 \, \text{kN}
     \]

3. **Shear Force Calculation:**
   To plot the Shear Force Diagram (SFD), we analyze the beam at sections along its length:
   
   - At \( A \) (x = 0): \( V_A = R_A = 7.5 \, \text{kN} \)
   - Just before \( B \) (x = 3): The shear force drops linearly due to the UDL. At point \( B \), the shear force is zero.
   
   The shear force distribution is linear under a UDL.

4. **Bending Moment Calculation:**
   Now, for the Bending Moment Diagram (BMD), we calculate moments at key points:
   
   - At \( A \) and \( B \), the bending moment is zero (since these are supports).
   - At the center (x = 1.5 m):
     \[
     M = R_A \times 1.5 - \left( \frac{5 \times 1.5^2}{2} \right)
     \]
     \[
     M = 7.5 \times 1.5 - \frac{5 \times 2.25}{2} = 11.25 - 5.625 = 5.625 \, \text{kN} \cdot \text{m}
     \]
     This is the maximum bending moment at the midpoint.

### Final Diagrams:
- **Shear Force Diagram (SFD):** 
  The SFD will be a straight line decreasing from \( +7.5 \, \text{kN} \) at \( A \) to \( -7.5 \, \text{kN} \) at \( B \).
  
- **Bending Moment Diagram (BMD):**
  The BMD will be a parabolic curve, starting at 0 at both \( A \) and \( B \), and reaching a maximum of \( 5.625 \, \text{kN} \cdot \text{m} \) at the center.

Let me know if you'd like further clarifications or if you'd like me to draw the diagrams for this problem.

### Related Questions:
1. How would the reaction forces change if the load intensity of the UDL increased to \( 10 \, \text{kN/m} \)?
2. What would happen to the maximum bending moment if the span of the beam increased?
3. How do point loads differ in their effect on the shear force and bending moment diagrams compared to UDLs?
4. How does the bending moment vary along a beam that is simply supported at both ends with a triangular load?
5. How do you calculate the maximum deflection of a simply supported beam with a UDL?

### Tip:
Always ensure that the total applied load and reactions are in equilibrium, as this is fundamental for analyzing any beam structure.

Calculate the shear force and bending moment for the beam subjected to an uniformly distributed load (5 kN/m) over a span of 3 m, then draw the shear force diagram (SFD) and bending moment diagram (BMD).

Learn how to calculate shear force and bending moment for a beam subjected to a uniformly distributed load (5 kN/m) over a 3-meter span. This example demonstrates the application of equilibrium equations and shows how to plot shear force and bending moment diagrams (SFD and BMD).

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