9.16 The soil profile and the loading condition are given in the following figure. Total new footing stress including the foundation at z = 2 m is 400 kN/m2. The soil profile has a rather thick clay layer (15 m) so that the layer should be divided into several sublayers to estimate the consolidation settlement adequately. Divide it into three sublay- ers and compute the primary consolidation settlement at the mid- point of each layer, and then make a summation for the total final settlement. Assume that the clay is normally consolidated and use Newmark's rectangular footing solution for the stress increment computation. New footing load 4 m x4 m square footing Sand Ydry = 19.0 kN/m3 -W.T. Sand ywet = 19.4 kN/m³ 5 10 Clay Ywet = 18.5 kN/m³ Clay, H = 15 m en = 0.78, LL = 38, PL = 14 15 20 Gravel z (m)
9.16 The soil profile and the loading condition are given in the following figure. Total new footing stress including the foundation at z = 2 m is 400 kN/m2. The soil profile has a rather thick clay layer (15 m) so that the layer should be divided into several sublayers to estimate the consolidation settlement adequately. Divide it into three sublay- ers and compute the primary consolidation settlement at the mid- point of each layer, and then make a summation for the total final settlement. Assume that the clay is normally consolidated and use Newmark's rectangular footing solution for the stress increment computation. New footing load 4 m x4 m square footing Sand Ydry = 19.0 kN/m3 -W.T. Sand ywet = 19.4 kN/m³ 5 10 Clay Ywet = 18.5 kN/m³ Clay, H = 15 m en = 0.78, LL = 38, PL = 14 15 20 Gravel z (m)
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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CEN 333 Geotechnical
Transcribed Image Text:9.16 The soil profile and the loading condition are given in the following
figure. Total new footing stress including the foundation at z = 2 m
is 400 kN/m². The soil profile has a rather thick clay layer (15 m)
so that the layer should be divided into several sublayers to estimate
the consolidation settlement adequately. Divide it into three sublay-
ers and compute the primary consolidation settlement at the mid-
point of each layer, and then make a summation for the total final
settlement. Assume that the clay is normally consolidated and use
Newmark's rectangular footing solution for the stress increment
computation.
New footing load
4 m ×4 m
square footing
2
Sand
Ydry
= 19.0 kN/m³
3
---W.T.
Sand ywet = 19.4 kN/m³
5
10
Clay ywet = 18.5 kN/m³
Clay, H = 15 m
eo = 0.78, LL = 38, PL = 14
15
20
Gravel
z (m)
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