= 1.5 m in a soil with [5] The base of a square footing with 2 m wide rests at a depth Df parameters y = 19 kN/m³, ysat = 20 kN/m³, c = 20 kPa, = 30°. Using the unit weight of water 9.8 kN/m³, determine the ultimate bearing capacity of the soil under the following conditions. Y'w = 1) The water table is located deep enough and does not affect the bearing capacity of the soil. 2) The water table rises up to 1 m below the ground surface 3) The water table stays at the base of the foundation Note that the ultimate bearing capacity of square foundation can be basically estimated by Eq. (4), using the bearing capacity factors given below, where is in radians: Square foundation: qu = 1.3cN+yDN₁ +0.4yBN, 3π EXP [(2 4-6) 414] exp tan 1-sin N = (N-1) cot N₁ = (N₁ -1) tan (1.46) where Na Eq. (4)

= 1.5 m in a soil with [5] The base of a square footing with 2 m wide rests at a depth Df parameters y = 19 kN/m³, ysat = 20 kN/m³, c = 20 kPa, = 30°. Using the unit weight of water 9.8 kN/m³, determine the ultimate bearing capacity of the soil under the following conditions. Y'w = 1) The water table is located deep enough and does not affect the bearing capacity of the soil. 2) The water table rises up to 1 m below the ground surface 3) The water table stays at the base of the foundation Note that the ultimate bearing capacity of square foundation can be basically estimated by Eq. (4), using the bearing capacity factors given below, where is in radians: Square foundation: qu = 1.3cN+yDN₁ +0.4yBN, 3π EXP [(2 4-6) 414] exp tan 1-sin N = (N-1) cot N₁ = (N₁ -1) tan (1.46) where Na Eq. (4)

Principles of Foundation Engineering (MindTap Course List)

9th Edition

ISBN:9781337705028

Author:Braja M. Das, Nagaratnam Sivakugan

Publisher:Braja M. Das, Nagaratnam Sivakugan

Chapter6: Shallow Foundations: Ultimate Bearing Capacity

Section: Chapter Questions

Problem 6.21P: Consider a continuous foundation of width B = 1.4 m on a sand deposit with c = 0, = 38, and = 17.5...

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A footing of size 2m×2m transferring a pressure of 200 kN/m², is placed at a depth of 1.5 m below the ground as shown in the figure (not drawn to the scale). The clay stratum is normally consolidated. The clay has specific gravity of 2.65 and compression index of 0.3. 1.5m 1m 1.5 m 200 kN/m² Silty sand Clay Ya =15kN/m³ Y sat = 18kN/m³ Y sat = 17 kN/m³ GWT $0.5 m Dense sand Considering 2:1 (vertical to horizontal) method of load distribution and Y₁ = 10kN/m³, the primary consolida- tion settlement (in mm, round off to two decimal places) of the clay stratum is
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