A residential foundation in Hilo requires concrete footings on volcanic soil with a bearing capacity of 2,000 psf. Calculate the minimum footing width for a wall carrying 4,000 pounds per linear foot.
Correct Answer
B) 24 inches
Width = Load ÷ Bearing capacity = 4,000 lbf/ft ÷ 2,000 psf = 2 ft = 24 inches minimum width required.
Why This Is the Correct Answer
Width = Load ÷ Bearing Capacity = 4,000 lbf/ft ÷ 2,000 psf = 2 ft = 24 inches. This is the fundamental footing sizing formula: the footing area must be large enough so the soil pressure (load divided by contact area) does not exceed the allowable bearing capacity.
Why the Other Options Are Wrong
Option A: 18 inches
18 inches (1.5 ft) is too narrow. At 1.5 ft width, soil pressure = 4,000 ÷ 1.5 = 2,667 psf, which exceeds the 2,000 psf allowable bearing capacity and would risk settlement or failure.
Option C: 36 inches
36 inches (3 ft) is larger than necessary. While it would safely carry the load (soil pressure = 1,333 psf), it wastes material and money. The minimum required width is 24 inches — using 36 inches is overdesign.
Option D: 30 inches
30 inches (2.5 ft) would also work structurally (soil pressure = 1,920 psf < 2,000 psf), but the question asks for the minimum footing width. The exact minimum is 24 inches; 30 inches is not the minimum.
Memory Technique
Think of it as 'spreading the weight': Width = Load / Bearing Capacity. A 2,000 psf soil can hold 2,000 lb per square foot. If the wall pushes 4,000 lb per foot of length, you need 2 square feet of footing per foot of length, so the footing must be 2 ft (24 in) wide.
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