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The image appears to depict a simply supported beam subject to a uniformly distributed load \( w = 200 \, \text{lb/ft} \) along its entire length and a concentrated load \( P = 1,500 \, \text{lb} \) placed at a point. The beam has a total length of 22 feet, distributed as 8 feet, 8 feet (where the concentrated load is placed), and 6 feet.

To solve this, let's calculate the reactions at the supports due to the distributed and concentrated loads.

### Calculations:

1. **Total Load due to Distributed Load**:
   The total distributed load \( W \) is calculated by multiplying the load per foot by the total length of the beam.
   \[
   W = w \times \text{Total Length} = 200 \, \text{lb/ft} \times 22 \, \text{ft} = 4400 \, \text{lb}
   \]

2. **Resultant of Distributed Load Position**:
   The resultant of the distributed load acts at the centroid of the distribution, which, since it's uniformly distributed, is at the midpoint of the beam.
   \[
   \text{Centroid Position} = \frac{22 \, \text{ft}}{2} = 11 \, \text{ft} \, \text{from either support}
   \]

3. **Calculating Reactions at Supports**:
   The beam is supported at the ends, so it's statically determinate. Using the equilibrium conditions \( \sum F_y = 0 \) and \( \sum M = 0 \):
   
   - **Sum of Vertical Forces**:
     \[
     R_A + R_B = W + P = 4400 \, \text{lb} + 1500 \, \text{lb} = 5900 \, \text{lb}
     \]

   - **Sum of Moments about Point A (counterclockwise positive)**:
     Moments due to the distributed load and the point load need to be considered.
     \[
     4400 \, \text{lb} \times 11 \, \text{ft} + 1500 \, \text{lb} \times 16 \, \text{ft} - R_B \times 22 \, \text{ft} = 0
     \]
     Solving for \( R_B \):
     \[
     R_B = \frac{4400 \times 11 + 1500 \times 16}{22} = 2950 \, \text{lb}
     \]
     Then, \( R_A \) is:
     \[
     R_A = 5900 \, \text{lb} - 2950 \, \text{lb} = 2950 \, \text{lb}
     \]

### Summary:
The reaction at support A, \( R_A \), and at support B, \( R_B \), are both 2950 lb. Each support carries half the total load since the concentrated load is symmetrically placed relative to the geometry of the beam. If you have any questions or need further clarification, please let me know!

Would you like to:
1. Calculate the shear force diagram for this beam?
2. Determine the bending moment diagram?
3. Understand the effect of changing the position of the point load?
4. Explore the impact of altering the distributed load intensity?
5. Calculate the maximum stress in the beam due to bending?

**Tip:** Always verify the units used in calculations to ensure they are consistent throughout the problem.

The beam is simply supported with a distributed load of 200 lb/ft and a concentrated load of 1,500 lb. The beam is divided into three segments of 8 ft, 8 ft, and 6 ft. Determine the reactions at the supports.

Learn how to calculate the reactions at supports for a simply supported beam subjected to a 200 lb/ft distributed load and a 1,500 lb concentrated load. This step-by-step guide explains the process using static equilibrium principles. Suitable for high school and undergraduate engineering students.

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