A foundation is being constructed on soil with an allowable bearing pressure of 2,000 psf. What is the minimum footing width required for a wall load of 4,500 pounds per linear foot?
Correct Answer
B) 27 inches
Required width = Load ÷ Bearing pressure = 4,500 plf ÷ 2,000 psf = 2.25 feet = 27 inches minimum.
Why This Is the Correct Answer
Required footing width = Wall load ÷ Allowable soil bearing pressure = 4,500 plf ÷ 2,000 psf = 2.25 feet = 27 inches. This is the fundamental footing sizing formula: the footing must be wide enough so the pressure exerted on the soil does not exceed the allowable bearing capacity.
Why the Other Options Are Wrong
Option A: 18 inches
18 inches (1.5 feet) would only support 1.5 ft × 2,000 psf = 3,000 plf — insufficient for a 4,500 plf load. This footing would overstress the soil.
Option C: 30 inches
30 inches (2.5 feet) would support 2.5 ft × 2,000 psf = 5,000 plf — adequate but larger than the minimum required. This is a safe design but not the minimum.
Option D: 24 inches
24 inches (2.0 feet) would support 2.0 ft × 2,000 psf = 4,000 plf — still 500 plf short of the 4,500 plf load. This footing is undersized and would overstress the soil.
Memory Technique
Footing formula: Width = Load / Bearing. Think of the soil as a 'budget' per square foot — you divide your total load by what each square foot can handle. 4,500 ÷ 2,000 = 2.25 ft = 27 inches.
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