To solve this problem, we need to analyze the beam using static equilibrium equations and draw the shear force and bending moment diagrams based on the given loading conditions.

1. Identify the forces and their locations:

• Concentrated load (P) = 1,000 lb, located at 2 ft from the left support.
• Uniform distributed load (w) = 150 lb/ft, spanning the 6 ft length from the concentrated load to the right support.

2. Support types:

• The beam appears to have a cantilever support on the right side, meaning it is fixed and resists vertical forces, horizontal forces, and moments.

3. Calculate the reactions:

Since the right side is fixed, we will calculate:

• Vertical reaction at the fixed support ($R_y$),
• Moment reaction at the fixed support ($M_A$).

Using static equilibrium equations: $\Sigma F_y = 0 \quad \text{(Sum of vertical forces)}$ $\Sigma M_A = 0 \quad \text{(Sum of moments about the fixed end)}$

Step 1: Vertical forces

The total vertical forces consist of:

1. The concentrated load of $1,000 \, \text{lb}$,
2. The uniformly distributed load over 6 ft ($w = 150 \, \text{lb/ft}$): $\text{Total distributed load} = 150 \, \text{lb/ft} \times 6 \, \text{ft} = 900 \, \text{lb}.$ The sum of vertical forces is: $R_y - 1,000 \, \text{lb} - 900 \, \text{lb} = 0$ $R_y = 1,900 \, \text{lb}.$

Step 2: Moments about the fixed end

Taking moments about the fixed support (assuming counterclockwise is positive):

• The moment due to the concentrated load $1,000 \, \text{lb}$ located 2 ft from the fixed support: $M_{\text{concentrated}} = 1,000 \, \text{lb} \times 2 \, \text{ft} = 2,000 \, \text{lb-ft}.$
• The moment due to the distributed load is calculated by taking the load's resultant (acting at the center of the distribution) at a distance of 3 ft from the fixed support: $M_{\text{distributed}} = 900 \, \text{lb} \times \frac{6}{2} \, \text{ft} = 900 \times 3 = 2,700 \, \text{lb-ft}.$ The moment equilibrium equation is: $M_A - 2,000 \, \text{lb-ft} - 2,700 \, \text{lb-ft} = 0$ $M_A = 4,700 \, \text{lb-ft}.$

4. Shear Force (V) Diagram:

• From 0 to 2 ft, there is only the concentrated load.
• At $x = 2 \, \text{ft}$, there is a drop due to the $1,000 \, \text{lb}$ load.
• From 2 to 8 ft, there is a uniform load causing a linearly decreasing shear.

5. Bending Moment (M) Diagram:

• The moment starts at $M_A = 4,700 \, \text{lb-ft}$.
• Between 0 and 2 ft, the moment decreases due to the concentrated load.
• Between 2 and 8 ft, the moment decreases further due to the distributed load, following a parabolic curve.

Summary of Results:

• Reaction at the fixed end: $R_y = 1,900 \, \text{lb}$
• Moment at the fixed end: $M_A = 4,700 \, \text{lb-ft}$

Would you like a more detailed breakdown of the shear force and bending moment diagrams or explanations on how to plot these?

5 Relative Questions:

1. How does the location of the concentrated load affect the bending moment?
2. What is the significance of a cantilever beam in structural analysis?
3. How do you determine the maximum shear force in a beam with combined loading?
4. Why does the bending moment diagram have a parabolic shape under a uniform load?
5. What would happen to the reactions if the beam were simply supported instead of cantilevered?

Tip:

In cantilever beams, the maximum bending moment always occurs at the fixed support. This is a critical point for designing safe structures.