In-situ stresses – Seepage condition ground level sheet Example: For the sheet pile which T is buried in sandy soil ,shown in 2.0m the figure below, find the total and effective vertical stress at piling water level Points 0, 6, and 12?. Note: Assume reasonable values 4.0m for saturated and dry unit weight equipotential datum of the soil College Of 3.0m flow line

Fundamentals of Geotechnical Engineering (MindTap Course List)
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Author:Braja M. Das, Nagaratnam Sivakugan
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Chapter7: Seepage
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In-situ stresses - Seepage condition
ground level
sheet
piling.
Example: For the sheet pile which
is buried in sandy soil ,shown in 20m
the figure below, find the total +
water level
and effective vertical stress at
Points 0, 6, and 12?.
4.0m
Note: Assume reasonable values
for saturated and dry unit weight
equipotential
datum
12
of the soil
Col
3.0 m
10
flow line
3.0 m
Soil Mechanics II: Lecture 2
Transcribed Image Text:In-situ stresses - Seepage condition ground level sheet piling. Example: For the sheet pile which is buried in sandy soil ,shown in 20m the figure below, find the total + water level and effective vertical stress at Points 0, 6, and 12?. 4.0m Note: Assume reasonable values for saturated and dry unit weight equipotential datum 12 of the soil Col 3.0 m 10 flow line 3.0 m Soil Mechanics II: Lecture 2
Let Ksat = 20 kN/m3
Point a on
rava=16 kvlm3
O -0
shtro: hP=o
6v= 2x16 -32 KPa
0*9,81 い0
6v : 32-0 = 32 K Pa
へ、へ へ
へ、
へ
へ へっ
Point « 6
h= 4-0
こo33 m
12
4-Co133 *6)
ニ 2.02 m
he: -3
hp: 2:02 +3=5:02 m
Gv=
3* 20+2火16+い 20
= 172 KPe
u : 5.02 * 9.81
- 49, 24 kPa
bý - 172- 49.24 =122.76 Kpa
ニ
へ、
Point
12 リ
4-0
こ033
ニ
12
hに u -
L0.33 x 12)=0.ou m
he-o hP= 0.0um
bv = LO* 2o) + (4x 20) + (2x16) = 112 kpa
u =0.ou x 9.81
こ o392u kPa
6v 112-0.3924= 111.6 KPa
%3D
Transcribed Image Text:Let Ksat = 20 kN/m3 Point a on rava=16 kvlm3 O -0 shtro: hP=o 6v= 2x16 -32 KPa 0*9,81 い0 6v : 32-0 = 32 K Pa へ、へ へ へ、 へ へ へっ Point « 6 h= 4-0 こo33 m 12 4-Co133 *6) ニ 2.02 m he: -3 hp: 2:02 +3=5:02 m Gv= 3* 20+2火16+い 20 = 172 KPe u : 5.02 * 9.81 - 49, 24 kPa bý - 172- 49.24 =122.76 Kpa ニ へ、 Point 12 リ 4-0 こ033 ニ 12 hに u - L0.33 x 12)=0.ou m he-o hP= 0.0um bv = LO* 2o) + (4x 20) + (2x16) = 112 kpa u =0.ou x 9.81 こ o392u kPa 6v 112-0.3924= 111.6 KPa %3D
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