Four 12 inch diameter piles (each of whose frictional capacity is 300 psf) support a column carrying 120 kips total load. How deep must each pile be to support the load?
A. 24 ft
B. 30 ft
C. 32 ft
D. 40ft
C. 32 ft is correct.
From the diameter of each pile (r/2 = 12/2 = 6 in = 0.5 ft) the circumference is (C = 2∏r = 2x3.142x0.5) 3.142. Dividing the required frictional area by the circumference of the circular piles equals the required depth (100sf / 3.142 = 31.83 ft or 32 ft)
A. 24 ft
B. 30 ft
C. 32 ft
D. 40ft
C. 32 ft is correct.
From the diameter of each pile (r/2 = 12/2 = 6 in = 0.5 ft) the circumference is (C = 2∏r = 2x3.142x0.5) 3.142. Dividing the required frictional area by the circumference of the circular piles equals the required depth (100sf / 3.142 = 31.83 ft or 32 ft)
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