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.
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.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6abdb772-2674-4ccf-b97c-7a87bba16f16%2F8a91886e-3bff-42ec-9c6f-26f15ce14e16%2Fu8pi46_processed.png&w=3840&q=75)
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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