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.

Would you like more detailed calculations for each step, including drawing the shear force and bending moment diagrams?