A residential foundation is being constructed on volcanic ash soil with a bearing capacity of 1,500 psf. If a continuous footing must support a load of 4,200 pounds per linear foot, what is the minimum required footing width?
Correct Answer
D) 2.8 feet
Required width = Load ÷ Bearing capacity = 4,200 lbs/ft ÷ 1,500 psf = 2.8 feet minimum. Round up to nearest practical dimension for construction.
Why This Is the Correct Answer
2.8 feet is correct. The required footing width is calculated by dividing the load per linear foot by the allowable soil bearing capacity: Width = Load ÷ Bearing Capacity = 4,200 lbs/ft ÷ 1,500 lbs/ft² = 2.8 feet. This is the minimum width needed so that the soil pressure under the footing does not exceed the allowable bearing capacity of 1,500 psf.
Why the Other Options Are Wrong
Option A: 4.0 feet
4.0 feet is more than 40% wider than required. This may result from a calculation error such as using the wrong bearing capacity value or confusing units. At 4.0 feet, the actual soil pressure would be 4,200 ÷ 4.0 = 1,050 psf, well below the allowable — the footing is over-designed.
Option B: 3.6 feet
3.6 feet is incorrect. This might come from rounding up excessively or misapplying a safety factor to the width directly. The calculated width of 2.8 feet is the minimum — rounding to 3.6 feet without cause produces an over-designed and more expensive footing than necessary.
Option C: 3.2 feet
3.2 feet is also incorrect. It represents an intermediate over-estimate that does not correspond to the correct formula result. Using 3.2 feet would give a soil pressure of 4,200 ÷ 3.2 = 1,312.5 psf, which is within the allowable but not the minimum required.
Memory Technique
Think of a footing as a distributor: it spreads load over area. Width = Load per foot ÷ Soil capacity per square foot. The units cancel perfectly: (lbs/ft) ÷ (lbs/ft²) = ft. The answer is always in feet — a good sanity check.
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