https://file.aiquickdraw.com/mathbot/file/1730024124224fbl8f22x.jpg

To solve this problem, we'll follow these steps:

1. **Calculate the reactions at the supports.**
2. **Calculate the shear force at key points along the beam.**
3. **Calculate the bending moment at key points along the beam.**
4. **Draw the shear force and bending moment diagrams.**

Given the beam configuration:
- The beam is simply supported at \(A\) and \(D\).
- Point loads: \(50 \, \text{kN}\) at \(B\) (2m from \(A\)), \(30 \, \text{kN}\) at \(C\) (2m from \(B\)), and \(60 \, \text{kN}\) at \(D\) (3m from \(C\)).

### Step 1: Calculate the reactions at supports

Using the equilibrium conditions:
- \(\sum F_y = 0\) (Sum of vertical forces)
- \(\sum M_A = 0\) (Sum of moments about point \(A\))

First, calculate the total moments about point \(A\):

\[ R_D \cdot 7 - 50 \cdot 2 - 30 \cdot 4 - 60 \cdot 7 = 0 \]

\[ 7R_D = 100 + 120 + 420 \]

\[ 7R_D = 640 \]

\[ R_D = \frac{640}{7} \approx 91.43 \, \text{kN} \]

Using \(\sum F_y = 0\):

\[ R_A + R_D = 50 + 30 + 60 \]

\[ R_A + 91.43 = 140 \]

\[ R_A = 140 - 91.43 = 48.57 \, \text{kN} \]

So, the reactions are:
- \( R_A = 48.57 \, \text{kN} \)
- \( R_D = 91.43 \, \text{kN} \)

### Step 2: Calculate the shear force at key points

1. **Shear force just to the left of \(B\)**: \( V = R_A = 48.57 \, \text{kN} \)
2. **Shear force just to the right of \(B\)**: \( V = 48.57 - 50 = -1.43 \, \text{kN} \)
3. **Shear force just to the left of \(C\)**: \( V = -1.43 \, \text{kN} \)
4. **Shear force just to the right of \(C\)**: \( V = -1.43 - 30 = -31.43 \, \text{kN} \)
5. **Shear force just to the left of \(D\)**: \( V = -31.43 \, \text{kN} \)
6. **Shear force just to the right of \(D\)**: \( V = -31.43 + 91.43 - 60 = 0 \, \text{kN} \)

### Step 3: Calculate the bending moment at key points

1. **Moment at \(A\)**: \( M_A = 0 \)
2. **Moment at \(B\)**:
\[ M_B = R_A \cdot 2 = 48.57 \cdot 2 = 97.14 \, \text{kN}\cdot\text{m} \]
3. **Moment at \(C\)**:
\[ M_C = R_A \cdot 4 - 50 \cdot 2 = 48.57 \cdot 4 - 100 = 194.28 - 100 = 94.28 \, \text{kN}\cdot\text{m} \]
4. **Moment at \(D\)**:
\[ M_D = 0 \]

### Step 4: Draw the shear force and bending moment diagrams

**Shear Force Diagram (SFD):**
1. Start at \(A\) with \(48.57 \, \text{kN}\)
2. Drop to \(-1.43 \, \text{kN}\) after \(B\)
3. Remain at \(-1.43 \, \text{kN}\) until \(C\)
4. Drop to \(-31.43 \, \text{kN}\) after \(C\)
5. Increase to \(60 \, \text{kN}\) after \(D\)

**Bending Moment Diagram (BMD):**
1. Start at \(A\) with \(0 \, \text{kN}\cdot\text{m}\)
2. Increase linearly to \(97.14 \, \text{kN}\cdot\text{m}\) at \(B\)
3. Drop to \(94.28 \, \text{kN}\cdot\text{m}\) at \(C\)
4. Return to \(0 \, \text{kN}\cdot\text{m}\) at \(D\)

Would you like a detailed graphical representation or have any specific questions?

### Related Questions:
1. How are the shear force and bending moment affected if the positions of the loads are changed?
2. How would the shear force and bending moment diagrams change if there was a uniformly distributed load instead of point loads?
3. What is the significance of the points where the shear force changes sign?
4. How do you calculate the maximum bending moment in a beam with multiple point loads?
5. What are the effects of different support conditions on the shear force and bending moment diagrams?

**Tip:** Always ensure that the sum of moments about any point equals zero when dealing with equilibrium in beam problems.

Calculate shear force and bending moment values at all salient points and draw shear force and bending moment diagrams for the beam with loads shown in Figure 1.

This guide provides a detailed solution for calculating shear force and bending moment at key points in a simply supported beam with point loads. Steps include calculating support reactions, shear force, and bending moment, followed by drawing shear force and bending moment diagrams. Suitable for engineering students.

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