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The left support is 6 ft from the left. The right support is at the rightmost end. The beam is 18 ft long. There is a concentrated load of 1500 lb that is 8 ft from the left. There is a distributed load of w=200lb/ft spans over 8 ft starting from the leftmost end.

To solve this beam problem, we'll perform the following steps:

1. **Determine the support reactions** at both the roller (left side) and pinned support (right side).
2. **Create shear force (V)** and **bending moment (M)** diagrams for the beam.

### Step 1: Determine Support Reactions

We have:
- Distributed load (\(w\)) of 200 lb/ft over 8 ft (from 0 to 6 ft).
- A concentrated load of 1,500 lb located 8 ft from the left end.

**Beam dimensions**:
- Distance from the left to the roller support: 6 ft.
- Total beam length: 18 ft.

### Total load from the distributed load:
The distributed load spans 8 ft. So the total load from this is:
\[
F_{\text{dist}} = w \times L = 200 \, \text{lb/ft} \times 8 \, \text{ft} = 1600 \, \text{lb}
\]
This acts at the centroid of the distributed load, which is at the midpoint of the 8 ft span, i.e., 4 ft from the left end.

### Support Reaction Calculation:

We will take moments about the right support (pinned) to find the reaction at the roller support (left):

Let \( R_A \) be the reaction at the roller (left support), and \( R_B \) be the reaction at the pinned (right support).

Taking moments about point B (right end support):

\[
\Sigma M_B = 0 \quad \Rightarrow \quad R_A \times 18 - (1600 \times 4) - (1500 \times 8) = 0
\]
\[
R_A \times 18 = 6400 + 12000
\]
\[
R_A = \frac{18400}{18} \approx 1022.22 \, \text{lb}
\]

Now, use vertical force equilibrium to find \( R_B \):

\[
\Sigma F_y = 0 \quad \Rightarrow \quad R_A + R_B = 1600 + 1500 = 3100
\]
\[
R_B = 3100 - R_A = 3100 - 1022.22 = 2077.78 \, \text{lb}
\]

### Step 2: Shear Force and Bending Moment Diagrams

#### Shear Force Calculation:

1. **From 0 to 6 ft** (Distributed load only):
   The shear force at \( x \) (distance from left):
   \[
   V(x) = R_A - w \times x = 1022.22 - 200 \times x
   \]
   At \( x = 6 \, \text{ft} \), just before the concentrated load:
   \[
   V(6) = 1022.22 - 200 \times 6 = 1022.22 - 1200 = -177.78 \, \text{lb}
   \]

2. **From 6 to 8 ft** (still under the distributed load):
   The load remains the same, and the shear will continue decreasing linearly.

3. **At 8 ft** (just before the concentrated load):
   Add the 1500 lb concentrated load:
   \[
   V(8^-) = V(8) = -177.78
   \]
   Just after the load:
   \[
   V(8^+) = -177.78 - 1500 = -1677.78 \, \text{lb}
   \]

#### Bending Moment Calculation:

We can integrate the shear force diagram to find the bending moment at key points:
- **At \( x = 0 \)**, \( M(0) = 0 \).
- **At \( x = 6 \, \text{ft} \)**, calculate the area under the shear force curve to get the moment.
  
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Would you like more detailed calculations for each step, including drawing the shear force and bending moment diagrams?

Solve for the support reactions and construct the shear force (V) and bending moment (M) diagrams for a beam subjected to a 200 lb/ft distributed load and a 1,500 lb concentrated load. Learn how to calculate reactions and create diagrams step-by-step.

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