Determine the magnitude of the internal bending moment at point C in the beam.

To determine the magnitude of the internal bending moment at point \( C \) in the beam, we will follow these steps:

### Recap of the Situation:
- The beam is subjected to a uniformly distributed load (UDL) \( w = 38 \, \text{kN/m} \) over a length of \( 2 \, \text{m} \).
- The beam is simply supported at points \( A \) and \( B \), with point \( C \) at the midpoint between \( A \) and \( B \).
- The distances between points \( A \), \( C \), and \( B \) are each \( 2 \, \text{m} \).

### Steps to Calculate the Bending Moment at Point \( C \):

1. **Calculate the reaction forces at the supports:**
   Since the beam and loading are symmetric, the reactions at \( A \) and \( B \) are each \( 38 \, \text{kN} \).

2. **Calculate the bending moment at point \( C \):**
   To find the bending moment at point \( C \), we can consider the left section of the beam from \( A \) to \( C \).

   - The reaction force at \( A \) is \( 38 \, \text{kN} \).
   - The distance from \( A \) to \( C \) is \( 2 \, \text{m} \).
   - The UDL over this section creates a uniformly distributed load with a total force of \( 76 \, \text{kN} \) (acting over \( 2 \, \text{m} \)).

   The bending moment at point \( C \) due to the reaction force at \( A \) and the UDL can be calculated as follows:
   
   **Moment due to the reaction force at \( A \):**
   \[
   M_C = R_A \times \text{distance from } A \text{ to } C = 38 \, \text{kN} \times 2 \, \text{m} = 76 \, \text{kN} \cdot \text{m}
   \]
   
   **Moment due to the UDL between \( A \) and \( C \):**
   The UDL creates a load of \( 76 \, \text{kN} \), which acts at the midpoint of the loaded section, i.e., \( 1 \, \text{m} \) from \( A \).
   \[
   M_C = - \left( w \times \text{length} \times \frac{\text{distance from } A \text{ to the UDL center}}{2} \right)
   \]
   \[
   M_C = - \left( 38 \, \text{kN/m} \times 2 \, \text{m} \times 1 \, \text{m} \right) = - 76 \, \text{kN} \cdot \text{m}
   \]

3. **Total Bending Moment at Point \( C \):**
   The total bending moment at point \( C \) is the sum of the moments due to the reaction force and the UDL:
   \[
   M_C = 76 \, \text{kN} \cdot \text{m} - 76 \, \text{kN} \cdot \text{m} = 0 \, \text{kN} \cdot \text{m}
   \]

### Conclusion:
The magnitude of the internal bending moment at point \( C \) in the beam is \( 0 \, \text{kN} \cdot \text{m} \).

Would you like to explore more on this topic or have any other questions?

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Here are 5 related questions to consider:
1. What would be the bending moment at the midpoint between \( C \) and \( B \)?
2. How would the bending moment change if the UDL were extended over the entire length of the beam?
3. What is the significance of a zero bending moment at a particular point in a beam?
4. How would the bending moment distribution look for a beam with different lengths between supports?
5. Can the shear force at a point in a beam be zero while the bending moment is non-zero? Explain.

**Tip:** The bending moment is often highest where the shear force changes sign. Analyzing the shear force distribution can provide insight into potential points of maximum bending moment.

Learn how to calculate the internal bending moment at point C in a beam subjected to a uniformly distributed load (UDL). Includes step-by-step calculations for static equilibrium and bending moment formulas. Suitable for university-level students and engineers.

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