Page 61 - Unit2.docx

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Figure 6-3 Worked Example 6-2

SOLUTION

Calculate single pile bearing capacity:

R s = α ⋅ C s ⋅ A s = 0.8 ⋅ 50 ⋅ 28⋅ π ⋅ (0.4) = 904kN ♣

2
R b = N c C b ⋅ A b = 9 ⋅ 50⋅ π ⋅ (0.2) = 56.6kN
⋅

∴ R ci = R si + R bi = 904 + 56.6 = 960

2
(W p +W cap ) - W s = (60× 25+(600-20× 5.0× 5.0× 2.0)) - (20× 28 ⋅ π ⋅ (0.2) ⋅ 25 = 469kN ♣♣

∴ total load capacity of 25 piles = R uc25 = (R ci = R si + R bi ) ⋅ 25 - {(W p +W cap ) - W s } = 960⋅ 25