Analytical Chem Soil HW Solutions 22

Analytical Chem Soil HW Solutions 22 - At the soil surface...

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Unformatted text preview: At the soil surface At the water table At the base of the pile z, m 0 5 25 σv, kPa 0 95 475 u, kPa 0 0 196.2 σ'v, kPa σ'h, kPa τ, kPa 0 95 278.8 0 47.5 139.4 0 23.17 67.99 The surface area of the upper 5m of the pile is (π × 0.3)m × 5m = 4.71m2, and the average shear stress over this area is 23.17kPa ÷ 2 = 11.59kPa. The surface area of the lower 20m of the pile is (π × 0.3)m × 9m = 18.85m2, and the average shear stress over this area is (23.17kPa + 67.99kPa) ÷ 2 = 45.58kPa. Thus the overall frictional resistance is (4.71m2 × 11.59kPa) + (18.85m2 × 45.58kPa) = 914kN Q2.6 The depth of the friction uplift pile described in main text Example 2.4 is increased to 20m, where the undrained shear strength of the clay is 40 kPa. Calculate the short- and longterm uplift resistance of the 20m pile. Q2.6 Solution The total shear resistance of the clay/pile interface is given by T = average shear stress × surface area of pile (a) In the short term, the average shear stress is the average undrained shear strength on the interface, so that T = [(0 + 40kPa) ÷ 2] × [(π × 0.5m) × 20m] = 628 kN (b) In the long term, the ultimate shear stress on the interface is given by τult = σ'h.tanδ where σ'h = 0.5 × σ'v is the horizontal effective stress and δ is the angle of friction between the clay and the pile At a depth z, σv (kPa) = {γ (kN/m3) × z (m)} = {18 (kN/m3) × z (m)} u (kPa) = {γw (kN/m3) × z (m)} = {9.81 (kN/m3) × z (m)}, and σ'v = σv - u As in (a), T = average shear stress × surface area of pile The shear stress τ on the soil/pile interface is now 22 ...
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This note was uploaded on 12/28/2011 for the course CHM 4302 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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