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To solve this problem, let's determine the reactions at supports \( A \) and \( B \), as well as the internal shear force and bending moment at point \( C \). Here's a breakdown of how we can approach it:

### Step 1: Analyze the Loads
1. **Point Loads**:
   - There is a 200 lb load applied at both ends of the beam (4 ft from each support, A and B).

2. **Distributed Load**:
   - A triangular distributed load acts over the central section (between A and B), spanning 18 ft. 
   - This load varies from 0 lb/ft at points near supports to 50 lb/ft at the midpoint of the triangle.

   To find the equivalent point load of this triangular load:
   \[
   W_{\text{triangular}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 18 \, \text{ft} \times 50 \, \text{lb/ft} = 450 \, \text{lb}
   \]
   This equivalent point load acts at a distance of one-third the base from the larger end of the triangle, so:
   \[
   \text{Distance from A to load} = 9 \, \text{ft} + \frac{1}{3} \times 9 \, \text{ft} = 12 \, \text{ft} \, \text{from A}
   \]

### Step 2: Set Up Equilibrium Equations
Since the beam is in static equilibrium, we can use the following conditions:
1. Sum of vertical forces (\( \sum F_y = 0 \)).
2. Sum of moments around any point (\( \sum M = 0 \)).

#### Equilibrium Equations

1. **Sum of Moments about Point A**:
   \[
   \sum M_A = 0 = -200 \times 4 - 450 \times 12 + R_B \times 17 + 200 \times 17
   \]
   Solving for \( R_B \) will give the reaction at support \( B \).

2. **Sum of Vertical Forces**:
   \[
   \sum F_y = 0 = R_A + R_B - 200 - 450 - 200
   \]
   Solving for \( R_A \) after determining \( R_B \) will give the reaction at support \( A \).

### Step 3: Calculate Internal Loads at Point C
Once the reactions at \( A \) and \( B \) are known, we can calculate the shear force and bending moment at point \( C \) by considering the left or right section of the beam up to point \( C \).

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### Further Exploration Questions
1. How would the reactions change if the distributed load were rectangular rather than triangular?
2. What if the beam had a cantilever at one end instead of being simply supported?
3. How would the bending moment diagram for this beam look after determining the reactions?
4. What is the effect of the location of point \( C \) on the internal shear and moment?
5. How would adding an additional support in the middle affect the reactions?

**Tip**: For triangular distributed loads, the resultant force acts at a distance of one-third the base from the larger end.

Determine the reactions as well as the internal loads at point C of the beam shown. Support A is a pin and B is a roller.

Learn how to calculate the reactions and internal loads at point C of a simply supported beam with a triangular distributed load and point loads. This solution includes equilibrium equations, triangular load calculation, and internal shear force and bending moment analysis. Suitable for engineering students.

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