Example 2: A soil Profile is shown in the figure A zone of capillary rise of 2.5m is on the sand layer overlaying clay. In this zone, the average degree of saturation is 60%
Example 2: A soil Profile is shown in the figure A zone of capillary rise of 2.5m is on the sand layer overlaying clay. In this zone, the average degree of saturation is 60%
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question
Solve letter C only
![Example 2:
A soil Profile is shown in the figure A zone of capillary rise of 2.5m is on the sand layer
overlaying clay. In this zone, the average degree of saturation is 60%
4m
25m
4m
2 5m
DRY SAND
B
3.5m CLAY
D
3.5m CLAY
D
DRY SAND
e=0.75
zone capillary rise
e=0.50
Gs=2.66
ground surface
G₁=2.71
ground water table
e=0.95
Gs=2.72
ground surface
e=0.50
Gs=2.66
e=0.75
zone capillary rise
G₁=2.71
ground water table
e=0.95
Gs=2.72
a.) Compute for the effective stress right below B.
b.) Compute the effective stress of C
c.) Compute the effective stress of D (Exercise Number 2)
Dry unit weight of sand
=
Yd=
Gs (Yw)
1+e
2.66(9.81)
Yd=
1 +0.50
Ya 17.40KN/m^3
b.) Compute the effective stress of C
(Gs+Se)yw
1+e
Y =
(2.71 +0.60(0.75))9.81
1 +0.75
y = 17.71KN/m^3
Total Stress:
Total Stress of Point B
PE=PT - Pw
Pe = 69.60 – (-14.72)
PE=84.32KPa
PT 17.40(4) + 17.71(2.5) =
PT = 17.40(4) =
Pore Pressure at B
Pw = -S(h)(Yw)
Pw = -0.60(2.5)(9.81)
Pw=-14.72Kpa
113.875KN
m3³
69.60KN
m³
PE
= PE](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F723ce8c7-e69d-4dc2-ad89-8bab285851e2%2Fed0db42b-4f48-4141-9476-de6098729a31%2F5g9wuj8r_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Example 2:
A soil Profile is shown in the figure A zone of capillary rise of 2.5m is on the sand layer
overlaying clay. In this zone, the average degree of saturation is 60%
4m
25m
4m
2 5m
DRY SAND
B
3.5m CLAY
D
3.5m CLAY
D
DRY SAND
e=0.75
zone capillary rise
e=0.50
Gs=2.66
ground surface
G₁=2.71
ground water table
e=0.95
Gs=2.72
ground surface
e=0.50
Gs=2.66
e=0.75
zone capillary rise
G₁=2.71
ground water table
e=0.95
Gs=2.72
a.) Compute for the effective stress right below B.
b.) Compute the effective stress of C
c.) Compute the effective stress of D (Exercise Number 2)
Dry unit weight of sand
=
Yd=
Gs (Yw)
1+e
2.66(9.81)
Yd=
1 +0.50
Ya 17.40KN/m^3
b.) Compute the effective stress of C
(Gs+Se)yw
1+e
Y =
(2.71 +0.60(0.75))9.81
1 +0.75
y = 17.71KN/m^3
Total Stress:
Total Stress of Point B
PE=PT - Pw
Pe = 69.60 – (-14.72)
PE=84.32KPa
PT 17.40(4) + 17.71(2.5) =
PT = 17.40(4) =
Pore Pressure at B
Pw = -S(h)(Yw)
Pw = -0.60(2.5)(9.81)
Pw=-14.72Kpa
113.875KN
m3³
69.60KN
m³
PE
= PE
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