A circular foundation 2m in diameter is shown in the figure below. A normally consolidated clay layer 5m thick is located below the foundation. Determine the consolidation settlement of the clay. method 2:1 (1) As one layer of clay of 5m thick: - Divide the clay layer into (5) sub-layers each of 1m thick: (2) Calculation of increase of stress below the center of each sub-layer Aoo(): (3) Weighted average pressure increase (Simpson's rule): Circular foundation diameter B = 2m G.S. 1.0m Sand q = 150 kN/m y = 17 kNhm² 0.5m W.T. Sand 19 kNiw! 0.5m Normally consolidated clay Y sat. = 18.5 kN/m² C_ =0.16, e, =0.85 5.0m %3D Solution :
A circular foundation 2m in diameter is shown in the figure below. A normally consolidated clay layer 5m thick is located below the foundation. Determine the consolidation settlement of the clay. method 2:1 (1) As one layer of clay of 5m thick: - Divide the clay layer into (5) sub-layers each of 1m thick: (2) Calculation of increase of stress below the center of each sub-layer Aoo(): (3) Weighted average pressure increase (Simpson's rule): Circular foundation diameter B = 2m G.S. 1.0m Sand q = 150 kN/m y = 17 kNhm² 0.5m W.T. Sand 19 kNiw! 0.5m Normally consolidated clay Y sat. = 18.5 kN/m² C_ =0.16, e, =0.85 5.0m %3D Solution :
Principles of Foundation Engineering (MindTap Course List)
8th Edition
ISBN:9781305081550
Author:Braja M. Das
Publisher:Braja M. Das
Chapter2: Geotechnical Properties Of Soil
Section: Chapter Questions
Problem 2.14P
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"please resolve the question but using method 2:1"
note / the value load (q) is KN/m2 is stree not load
method 2:1 q/(B+z) why the q * area
b is circular
"note in the picture solve the question but please solve in method 2:1"
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