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The problem you're presenting is about finding the shear and moment for a beam with a uniformly distributed load (UDL). In the image, we have:

- A fixed-end beam subjected to a uniformly distributed load of \( w = 15 \, \text{kN/m} \).
- The span of the beam is \( L = 6 \, \text{m} \).

### Step 1: Reactions at the Supports
The beam appears to be fixed at one end (left end), meaning it has both a vertical reaction and a moment at that support.

For a cantilever beam with a uniformly distributed load, the reactions are:
- **Vertical reaction at A** (shear force) \( V_A \):
  \[
  V_A = w \times L = 15 \, \text{kN/m} \times 6 \, \text{m} = 90 \, \text{kN}
  \]
  
- **Moment at A** (fixed moment) \( M_A \):
  \[
  M_A = \frac{w \times L^2}{2} = \frac{15 \, \text{kN/m} \times (6 \, \text{m})^2}{2} = 270 \, \text{kNm}
  \]

### Step 2: Shear Force Distribution
The shear force varies linearly along the length of the beam. At the fixed end, the shear force is equal to the reaction \( V_A \), and at the free end, the shear force is zero.

- **Shear force at any point \( x \) along the beam**:
  \[
  V(x) = V_A - w \times x = 90 \, \text{kN} - 15 \times x
  \]
  This gives a linear decrease in shear force along the length of the beam.

### Step 3: Bending Moment Distribution
The moment at the free end is zero, and at the fixed end, the moment is maximum.

- **Bending moment at any point \( x \) along the beam**:
  \[
  M(x) = V_A \times x - \frac{w \times x^2}{2} = 90 \times x - \frac{15 \times x^2}{2}
  \]
  This is a quadratic equation that represents the moment along the beam.

### Results:
1. The **maximum shear force** occurs at the fixed end and is \( V_A = 90 \, \text{kN} \).
2. The **maximum bending moment** occurs at the fixed end and is \( M_A = 270 \, \text{kNm} \).

Would you like me to proceed with a detailed shear and moment diagram, or provide further clarifications on the calculations?

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**Relative Questions:**
1. What is the significance of the shear force at different points along the beam?
2. How do fixed and free supports affect the bending moment in beams?
3. Why does the bending moment decrease towards the free end of a cantilever beam?
4. How is the shear force related to the load distribution in a beam?
5. What are the implications of a larger span or load on the shear force and moment?

**Tip:** Always remember that for a cantilever beam under a uniformly distributed load, the shear and moment equations follow predictable patterns, with shear being linear and moment being quadratic.

Solve the shear and moment for a beam with a uniformly distributed load (w = 15 kN/m) and a span of 6 meters. The beam is fixed at one end.

Learn how to solve for shear force and bending moment in a cantilever beam with a uniformly distributed load of 15 kN/m and a span of 6 meters. This problem covers key structural engineering concepts, including shear force equations, bending moment equations, and reactions at supports. Suitable for undergraduate engineering students.

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