A floor has a load of 150psf distributed to columns spaced at 24ft in one direction and 40ft in the other. Find the floor load on an EXTERIOR CORNER column.

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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### Understanding Floor Load Distribution

In structural engineering, it's crucial to calculate the floor load distributed to various columns in a building to ensure stability and safety. The given problem addresses a specific scenario:

"A floor has a load of 150 psf distributed to columns spaced at 24 ft in one direction and 40 ft in the other. Find the floor load on an EXTERIOR CORNER column."

#### Problem Breakdown:

1. **Floor Load (psf)**: 
   - The floor has a load of 150 pounds per square foot (psf). This unit measures the pressure applied over a unit area.
   
2. **Column Spacing (ft)**:
   - Columns are spaced 24 feet apart in one direction and 40 feet apart in the other. This gives us the area each column supports.
   
3. **Exterior Corner Column**:
   - Specifically, we need to find the load on the exterior corner column, which typically supports less area compared to interior columns.

#### Calculation Steps:

1. **Determine the Area Supported by the Column**:
   - The total area supported by each column can be calculated using:
     \[ \text{Area} = \text{Spacing}_1 \times \text{Spacing}_2 \]
   - For the given problem:
     \[ \text{Area} = 24 \, \text{ft} \times 40 \, \text{ft} = 960 \, \text{sq ft} \]

2. **Calculate the Total Load**:
   - Convert the area to the total load by multiplying with the uniform load:
     \[ \text{Total Load} = \text{Load per Unit Area} \times \text{Area} \]
   - Given:
     \[ \text{Total Load} = 150 \, \text{psf} \times 960 \, \text{sq ft} = 144,000 \, \text{pounds} \]

3. **Adjust for an Exterior Corner Column**:
   - An exterior corner column typically supports a quarter of the area an interior column does because it’s at the two edges of the building.
   - Therefore, we divide the total load by 4:
     \[ \text{Load on Exterior Corner Column} = \frac{144,000 \, \text{pounds}}{4} = 36,000 \, \text{p
Transcribed Image Text:### Understanding Floor Load Distribution In structural engineering, it's crucial to calculate the floor load distributed to various columns in a building to ensure stability and safety. The given problem addresses a specific scenario: "A floor has a load of 150 psf distributed to columns spaced at 24 ft in one direction and 40 ft in the other. Find the floor load on an EXTERIOR CORNER column." #### Problem Breakdown: 1. **Floor Load (psf)**: - The floor has a load of 150 pounds per square foot (psf). This unit measures the pressure applied over a unit area. 2. **Column Spacing (ft)**: - Columns are spaced 24 feet apart in one direction and 40 feet apart in the other. This gives us the area each column supports. 3. **Exterior Corner Column**: - Specifically, we need to find the load on the exterior corner column, which typically supports less area compared to interior columns. #### Calculation Steps: 1. **Determine the Area Supported by the Column**: - The total area supported by each column can be calculated using: \[ \text{Area} = \text{Spacing}_1 \times \text{Spacing}_2 \] - For the given problem: \[ \text{Area} = 24 \, \text{ft} \times 40 \, \text{ft} = 960 \, \text{sq ft} \] 2. **Calculate the Total Load**: - Convert the area to the total load by multiplying with the uniform load: \[ \text{Total Load} = \text{Load per Unit Area} \times \text{Area} \] - Given: \[ \text{Total Load} = 150 \, \text{psf} \times 960 \, \text{sq ft} = 144,000 \, \text{pounds} \] 3. **Adjust for an Exterior Corner Column**: - An exterior corner column typically supports a quarter of the area an interior column does because it’s at the two edges of the building. - Therefore, we divide the total load by 4: \[ \text{Load on Exterior Corner Column} = \frac{144,000 \, \text{pounds}}{4} = 36,000 \, \text{p
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