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This is a statics problem involving a simply supported beam subjected to a concentrated load and a concentrated moment. Let's go through the steps to solve it.

### 1. **Determine the Reaction Forces at the Supports**
Given:
- Concentrated load \( P = 20 \, \text{kN} \) at a distance of \( 4 \, \text{m} \) from the left support.
- Concentrated moment \( M = 16 \, \text{kN} \cdot \text{m} \) applied counterclockwise at the left support.
- The beam is simply supported, so it has two reaction forces: \( R_A \) at the left support and \( R_B \) at the right support.

#### Equilibrium Equations:
- **Sum of vertical forces:**
  \[
  R_A + R_B - P = 0
  \]
  \[
  R_A + R_B = 20 \, \text{kN}
  \]

- **Sum of moments about the left support \( A \):**
  \[
  -M + P \cdot 4 - R_B \cdot 8 = 0
  \]
  \[
  -16 + 20 \cdot 4 - R_B \cdot 8 = 0
  \]
  \[
  64 - 16 = 8R_B
  \]
  \[
  R_B = \frac{48}{8} = 6 \, \text{kN}
  \]

- **Substitute \( R_B = 6 \, \text{kN} \) into the vertical force equation:**
  \[
  R_A + 6 = 20
  \]
  \[
  R_A = 20 - 6 = 14 \, \text{kN}
  \]

### 2. **Shear Force Diagram**
- **Shear at the left support \( A \):** \( V_A = R_A = 14 \, \text{kN} \).
- **Shear just before the point load:** \( V = 14 \, \text{kN} \).
- **Shear just after the point load:** \( V = 14 \, \text{kN} - 20 \, \text{kN} = -6 \, \text{kN} \).
- **Shear at the right support \( B \):** \( V_B = -6 \, \text{kN} \).

### 3. **Bending Moment Diagram**
- **Moment at \( A \):** \( M_A = -16 \, \text{kN} \cdot \text{m} \).
- **Moment at any point \( x \) between \( A \) and the point load:**
  \[
  M(x) = -16 + 14x
  \]
  For \( x = 4 \, \text{m} \) (just before the point load):
  \[
  M(4) = -16 + 14 \times 4 = 40 \, \text{kN} \cdot \text{m}
  \]
- **Moment at the point load:** Subtract the moment caused by the load:
  \[
  M(4^+) = 40 \, \text{kN} \cdot \text{m} - 20 \times 0 = 40 \, \text{kN} \cdot \text{m}
  \]
- **Moment between the point load and right support:** \( M(x) = -6(x - 8) \).

### **Final Answers:**
a) **Reaction Forces:**
   - \( R_A = 14 \, \text{kN} \)
   - \( R_B = 6 \, \text{kN} \)

b) **Shear Force Diagram:**
   - \( V_A = 14 \, \text{kN} \)
   - Shear drops to \( -6 \, \text{kN} \) just after the point load.

c) **Bending Moment Diagram:**
   - \( M_A = -16 \, \text{kN} \cdot \text{m} \)
   - \( M_B = 0 \, \text{kN} \cdot \text{m} \)

Let me know if you need more details or have any questions.

---

**5 Related Questions:**
1. How do you calculate the deflection of a beam under a similar loading condition?
2. What would happen to the reaction forces if the moment was applied at the right support instead?
3. How does a simply supported beam compare to a cantilever beam under the same load?
4. What is the significance of the bending moment being zero at certain points on the beam?
5. How would the shear force diagram change if an additional distributed load was applied?

**Tip:** Always check the equilibrium equations when solving for reaction forces to ensure that all forces and moments are accounted for correctly.

Learn how to solve a statics problem involving a simply supported beam subjected to a concentrated load and a concentrated moment. This detailed analysis covers equilibrium equations, shear force and bending moment diagrams, and reaction forces calculation. Suitable for advanced engineering students and professionals.

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