Let's break down the solution step by step for the highlighted beam loaded by a uniform floor load. --- ## **Step 1: Understand the Problem** - **Uniform load intensity**: \( 100 \, \text{psf} \) (pounds per square foot) - **Highlighted beam length**: \( 36 \, \text{ft} \) - **Tributary width for the beam**: \( 12 \, \text{ft} \) (distance between beams) - **Goal**: Find the total **uniform load** (plf) on the highlighted beam and the **total load**. --- ## **Step 2: Find the Tributary Area** - The **tributary width** for the highlighted beam is the distance to the next beam, which is \( 12 \, \text{ft} \). - The **length** of the beam is \( 36 \, \text{ft} \). So, **tributary area** = \( \text{Beam length} \times \text{Tributary width} = 36 \, \text{ft} \times 12 \, \text{ft} = 432 \, \text{ft}^2 \) --- ## **Step 3: Calculate the Total Load on the Beam** - **Uniform load on floor** = \( 100 \, \text{psf} \) - **Total load on beam** = \( \text{Uniform load} \times \text{Tributary area} = 100 \, \text{psf} \times 432 \, \text{ft}^2 = 43,200 \, \text{lb} \) --- ## **Step 4: Calculate the Uniform Load on the Beam (plf)** The beam "sees" a uniform line load (plf): \[ \text{Uniform load on beam (plf)} = \text{Floor load (psf)} \times \text{Tributary width (ft)} \] \[ = 100 \, \text{psf} \times 12 \, \text{ft} = 1,200 \, \text{plf} \] --- ## **Step 5: Free Body Diagram** - The beam is loaded with a **uniform line load** of \( 1,200 \, \text{plf} \) over its entire length (\( 36 \, \text{ft} \)). --- ## **Step 6: Check Given Answers** The values **135 lb; 2025 lb** seem to be for different tributary widths or partial spans (perhaps per bay or half the beam). --- ## **Final Answers** ### **Uniform Line Load on Beam** \[ \boxed{1,200 \, \text{plf}} \] ### **Total Load on Beam** \[ \boxed{43,200 \, \text{lb}} \] --- ### **Summary Table** | Quantity | Value | |----------------------------|---------------------| | Floor load | \( 100 \, \text{psf} \) | | Tributary width (beam) | \( 12 \, \text{ft} \) | | Beam length | \( 36 \, \text{ft} \) | | Uniform load on beam (plf) | \( 1,200 \, \text{plf} \) | | Total load on beam | \( 43,200 \, \text{lb} \) | --- ### **Free Body Diagram Description** - A simply supported beam, \( 36 \, \text{ft} \) long, with a uniform distributed load of \( 1,200 \, \text{plf} \) across its length. --- If you need the **reactions at supports** or a sketch of the free body diagram, let me know!