Dry sand y = 16.5 kN/m³ 1.5 m Groundwater table Sand 1.5 m Ysat = 19 kN/m3 N60 :1.5 m Clay Ysat = 16.8 kN/m3 1.5 m * A•8 1.5 m 1.5 m 10 Sand Figure P3.2 Си 0.29N72 Ра (3.8) where pa = atmospheric pressure ( 100 kN/m?; - 2000 lb/in²). The overconsolidation ratio, OCR, of a natural clay deposit can also be correlated with the standard penetration number. On the basis of the regression analysis of 110 data points, Mayne and Kemper (1988) obtained the relationship N60 OCR = 0.193 0.689 (3.9) where o, = effective vertical stress in MN/m?.
Dry sand y = 16.5 kN/m³ 1.5 m Groundwater table Sand 1.5 m Ysat = 19 kN/m3 N60 :1.5 m Clay Ysat = 16.8 kN/m3 1.5 m * A•8 1.5 m 1.5 m 10 Sand Figure P3.2 Си 0.29N72 Ра (3.8) where pa = atmospheric pressure ( 100 kN/m?; - 2000 lb/in²). The overconsolidation ratio, OCR, of a natural clay deposit can also be correlated with the standard penetration number. On the basis of the regression analysis of 110 data points, Mayne and Kemper (1988) obtained the relationship N60 OCR = 0.193 0.689 (3.9) where o, = effective vertical stress in MN/m?.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question
A soil profile is shown in Figure P3.2 along with the standard penetration numbers in the clay layer. Use Eqs. (3.8) and (3.9) to determine the variation of cu and OCR with depth. What is the average value of cu and OCR?
Transcribed Image Text:Dry sand
y = 16.5 kN/m³
1.5 m
Groundwater
table
Sand
1.5 m
Ysat
= 19 kN/m3
N60
:1.5 m
Clay
Ysat = 16.8 kN/m3
1.5 m
* A•8
1.5 m
1.5 m
10
Sand
Figure P3.2
Transcribed Image Text:Си
0.29N72
Ра
(3.8)
where pa = atmospheric pressure ( 100 kN/m?; - 2000 lb/in²).
The overconsolidation ratio, OCR, of a natural clay deposit can also be correlated
with the standard penetration number. On the basis of the regression analysis of 110 data
points, Mayne and Kemper (1988) obtained the relationship
N60
OCR = 0.193
0.689
(3.9)
where o, = effective vertical stress in MN/m?.
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