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resuelva fuerzas y momentos con diagrama cortante y momento

To solve the problem of determining the forces and moments on the beam with the diagram provided, we need to go through several steps. Let's break them down:

### **1. Identifying the Loads and Reactions:**
- **Point A** is a pinned support (reaction forces \( R_A \) in the vertical direction and \( H_A \) in the horizontal direction).
- **Point E** is a roller support (reaction force \( R_E \) in the vertical direction).
- There are two types of loads on the beam:
  - A **point load** of \( 8 \, \text{kN} \) at point B.
  - A **uniformly distributed load (UDL)** of \( 2 \, \text{kN/m} \) acting from point C to point D (over a length of 2 meters).

### **2. Calculation of Support Reactions:**
We will begin by calculating the support reactions using equilibrium equations.

- **Equilibrium of forces in the vertical direction**:
  \[
  \sum F_y = 0 \quad \Rightarrow \quad R_A + R_E = 8 + (2 \times 2)
  \]
  The total load from the UDL is \( 2 \, \text{kN/m} \times 2 \, \text{m} = 4 \, \text{kN} \), so the equation becomes:
  \[
  R_A + R_E = 12 \, \text{kN} \tag{1}
  \]

- **Equilibrium of moments about point A**:
  \[
  \sum M_A = 0 \quad \Rightarrow \quad -8 \times 1 - 4 \times (1 + 2/2) + R_E \times 5 = 0
  \]
  The moment from the UDL is taken as if it acts at the midpoint of the 2-meter span, i.e., 2 meters from A.
  \[
  -8 - 4 \times 2 + 5 R_E = 0 \quad \Rightarrow \quad -16 + 5 R_E = 0
  \]
  Solving for \( R_E \):
  \[
  R_E = 3.2 \, \text{kN}
  \]

- **Substitute \( R_E \) into Equation (1)**:
  \[
  R_A + 3.2 = 12 \quad \Rightarrow \quad R_A = 8.8 \, \text{kN}
  \]

### **3. Shear Force Diagram (SFD):**
We will now construct the shear force diagram by calculating shear at key points along the beam:

- **At A**: Shear starts at \( +8.8 \, \text{kN} \) (upward reaction).
- **At B**: Shear drops by \( 8 \, \text{kN} \) due to the point load, so:
  \[
  V_B = 8.8 - 8 = 0.8 \, \text{kN}
  \]
- **At C**: Shear decreases due to the UDL. The UDL contributes \( -2 \, \text{kN/m} \) over a 2-meter span, so:
  \[
  V_C = 0.8 - (2 \times 2) = -3.2 \, \text{kN}
  \]
- **At D**: No change in shear since the UDL ends.
- **At E**: The shear force rises back to 0 due to the reaction at E:
  \[
  V_E = -3.2 + 3.2 = 0 \, \text{kN}
  \]

### **4. Bending Moment Diagram (BMD):**
Next, we calculate moments at the key points:

- **At A**: \( M_A = 0 \) (pinned support).
- **At B**: The moment is due to the reaction at A and the point load at B:
  \[
  M_B = 8.8 \times 1 = 8.8 \, \text{kN} \cdot \text{m}
  \]
- **At C**: The moment here is affected by both the reaction at A and the UDL:
  \[
  M_C = 8.8 \times 2 - 8 \times 1 = 9.6 \, \text{kN} \cdot \text{m}
  \]
- **At D**: The moment at D decreases due to the UDL acting between C and D:
  \[
  M_D = 9.6 - (2 \times 2 \times 2/2) = 5.6 \, \text{kN} \cdot \text{m}
  \]
- **At E**: \( M_E = 0 \) (roller support).

### **Summary:**
- \( R_A = 8.8 \, \text{kN} \)
- \( R_E = 3.2 \, \text{kN} \)
- The maximum shear is \( 8.8 \, \text{kN} \) at A.
- The maximum moment is \( 9.6 \, \text{kN} \cdot \text{m} \) near C.

Would you like a detailed explanation of any step? Here are five related questions:

1. How do we determine the location of maximum bending moment in a beam?
2. What is the significance of the UDL in shear force and moment calculations?
3. Why is the moment zero at roller supports?
4. Can we predict failure points based on the moment diagram?
5. What changes if the UDL acts over a longer span of the beam?

**Tip:** When constructing a bending moment diagram, areas with uniform loads create parabolic curves, while point loads result in linear segments.

Learn how to resolve forces and moments in a statically determinate beam, with detailed steps for constructing shear force and bending moment diagrams. The problem involves calculating reactions at supports and analyzing the effects of a point load and uniformly distributed load.

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