A soil bearing test indicates a bearing capacity of 2,500 psf. What is the maximum allowable bearing load for a continuous footing that is 2 feet wide and 100 feet long?
Correct Answer
B) 500,000 pounds
Maximum bearing load = bearing capacity × footing area. 2,500 psf × (2 ft × 100 ft) = 2,500 × 200 = 500,000 pounds.
Why This Is the Correct Answer
The maximum allowable bearing load is calculated by multiplying the soil bearing capacity by the total contact area of the footing. With a bearing capacity of 2,500 psf and a footing area of 200 square feet (2 ft × 100 ft), the calculation is straightforward: 2,500 psf × 200 sf = 500,000 pounds. This represents the total load that can be safely distributed across the entire footing without exceeding the soil's capacity.
Why the Other Options Are Wrong
Option A: 250,000 pounds
This answer incorrectly uses only half the footing area in the calculation, possibly confusing width (2 ft) with total area, resulting in 2,500 × 100 = 250,000 pounds instead of using the full 200 square feet.
Option C: 750,000 pounds
This answer appears to multiply the bearing capacity by 300 instead of 200, possibly adding the length and width incorrectly (100 + 2 = 102, then rounding to 300) rather than multiplying them for area.
Option D: 1,000,000 pounds
This answer uses 400 square feet instead of 200 square feet, possibly doubling the correct area calculation or confusing the footing dimensions in some way.
Memory Technique
Remember 'BCA' - Bearing capacity × Contact Area = Allowable load. Think 'Before Construction Approval' to remember the sequence of the calculation.
Reference Hint
Florida Building Code, Chapter 18 - Soils and Foundations, or structural engineering reference sections covering soil bearing capacity calculations
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