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A drilled pier foundation extends 25 feet into clay soil with an allowable bearing capacity of 4,000 psf. If the pier diameter is 24 inches, what is the maximum allowable load on the pier considering end bearing only?

Correct Answer

A) 12,566 pounds

Pier area = π × (1 ft)² = 3.14159 sq ft. Maximum load = 3.14159 × 4,000 psf = 12,566 pounds. Note that 24-inch diameter = 2-foot diameter = 1-foot radius.

Answer Options
A
12,566 pounds
B
50,265 pounds
C
100,531 pounds
D
25,133 pounds

Why This Is the Correct Answer

The end bearing load equals the bearing area times the allowable bearing pressure. Pier diameter = 24 inches = 2 feet, so radius = 1 foot. Area = π × r² = π × (1)² = 3.14159 sq ft. Maximum load = 3.14159 sq ft × 4,000 psf = 12,566 pounds. Note that the 25-foot depth is a distractor — end bearing capacity depends only on the tip area and bearing pressure, not depth when using allowable bearing capacity directly.

Why the Other Options Are Wrong

Option B: 50,265 pounds

50,265 lb results from using a radius of 2 feet (treating the 24-inch diameter as the radius instead of converting correctly): π × (2)² × 4,000 = 50,265. This is a unit conversion error — always halve the diameter to get the radius before applying the area formula.

Option C: 100,531 pounds

100,531 lb results from using a diameter of 4 feet (doubling the correct value) or some other dimensional error. This figure is approximately 8× the correct answer and would represent an unrealistically large pier capacity for a 24-inch diameter pier.

Option D: 25,133 pounds

25,133 lb results from using a diameter of 2 feet as if it were a radius (area = π × 2² = 12.566 sq ft), then multiplying by 4,000 / 2 — or from another combination of unit errors. The critical step is correctly converting 24 inches to 2 feet for diameter, then halving to get 1-foot radius.

Memory Technique

Pier end bearing = 'Pizza formula': Area = π × r² (like calculating pizza area from the radius, not diameter). For a 24-inch pier: diameter = 2 ft → radius = 1 ft → area = π × 1² = 3.14 sq ft. Then: Load = Area × Bearing pressure. The 25-foot depth? Red herring for end bearing only.

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