https://file.aiquickdraw.com/mathbot/file/1727341675464hlfiu9iy.jpg

To solve this problem, we need to create the shear and moment diagrams for the beam shown. Here’s a step-by-step breakdown:

### 1. **Identifying Forces and Loads**
- **Distributed Load:** There is a uniform distributed load of \(20 \, \text{kN/m}\) over a span of \(3 \, \text{m}\) from \(A\) to \(B\).
  - Total load from this section: \( 20 \times 3 = 60 \, \text{kN}\) acting at the midpoint of the distributed load (i.e., at \(1.5 \, \text{m}\) from \(A\)).
- **Point Load at \(C\):** A concentrated load of \(40 \, \text{kN}\) is applied at point \(C\) (at a distance of \(5 \, \text{m}\) from \(A\)).
- **Support Reactions:**
  - **At \(A\):** Assume a vertical reaction \(R_A\).
  - **At \(D\):** Since point \(D\) is a fixed support, it provides both a vertical reaction \(R_D\) and a moment \(M_D\).

### 2. **Static Equilibrium Equations**
To solve for the reactions at the supports, we use the conditions for static equilibrium:
- Sum of vertical forces \( \sum F_y = 0 \)
- Sum of moments about point \(A\), \( \sum M_A = 0 \)

#### a) **Sum of Moments about \(A\):**
Let’s take moments around point \(A\), where counterclockwise moments are positive.

\[
M_A = (R_D \times 7) - (60 \times 1.5) - (40 \times 5) = 0
\]
\[
M_A = 7R_D - 90 - 200 = 0
\]
\[
7R_D = 290 \implies R_D = \frac{290}{7} = 41.43 \, \text{kN}
\]

#### b) **Sum of Vertical Forces:**
Now apply the vertical equilibrium equation:

\[
R_A + R_D - 60 - 40 = 0
\]
\[
R_A + 41.43 = 100
\]
\[
R_A = 100 - 41.43 = 58.57 \, \text{kN}
\]

### 3. **Shear Force Diagram (SFD)**
- **From \(A\) to \(B\) (0 to 3 m):**
  - Start with \(V_A = +58.57 \, \text{kN}\).
  - Subtract the distributed load of \(20 \, \text{kN/m}\), which gives a linear drop in shear.
  - Shear at \(B\) (after 3 m): \( 58.57 - 20 \times 3 = -1.43 \, \text{kN}\).

- **From \(B\) to \(C\) (3 to 5 m):**
  - Constant shear because no load is applied in this section.
  - Shear remains at \( -1.43 \, \text{kN}\).

- **At \(C\) (5 m):**
  - Apply the point load of \(40 \, \text{kN}\): \( -1.43 - 40 = -41.43 \, \text{kN}\).

- **From \(C\) to \(D\) (5 to 7 m):**
  - Again, no load in this section, so shear remains constant at \( -41.43 \, \text{kN}\).

### 4. **Bending Moment Diagram (BMD)**
- **At \(A\) (0 m):** Moment is zero because it's a pin support.
- **From \(A\) to \(B\) (0 to 3 m):**
  - Moment is quadratic (due to the distributed load). The moment at \(B\) is:
    \[
    M_B = 58.57 \times 3 - \frac{1}{2} \times 20 \times 3^2 = 175.71 - 90 = 85.71 \, \text{kNm}
    \]

- **From \(B\) to \(C\) (3 to 5 m):**
  - Moment at \(C\):
    \[
    M_C = 85.71 + (-1.43) \times 2 = 85.71 - 2.86 = 82.85 \, \text{kNm}
    \]
  - At \(C\), subtract the effect of the point load:
    \[
    M_C = 82.85 - 40 \times 2 = 82.85 - 80 = 2.85 \, \text{kNm}
    \]

- **At \(D\) (7 m):**
  - Moment at \(D\) is zero since it’s a fixed support.

With these calculations, you can now sketch the shear and moment diagrams.

Let me know if you'd like a detailed sketch of the diagrams or any further clarifications!

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### Related Questions:
1. How do distributed loads affect the shape of moment diagrams?
2. What is the significance of zero-shear points in shear force diagrams?
3. How do you determine the maximum bending moment in a beam?
4. What is the difference between a pinned and a fixed support?
5. How do point loads alter both shear and moment diagrams?

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### Tip:
When drawing moment diagrams, remember that the slope of the moment diagram at any point corresponds to the value of the shear force at that point.

Draw the shear and moment diagram of the figure shown.

This tutorial demonstrates how to draw the shear and moment diagrams for a beam subjected to a uniform distributed load and a point load. Learn step-by-step calculations, including the determination of support reactions and using equilibrium conditions. Suitable for undergraduate civil engineering students.

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