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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

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)
Fundamentals of Geotechnical Engineering (MindTap Course List)
5th Edition
ISBN:9781305635180
Author:Braja M. Das, Nagaratnam Sivakugan
Publisher:Braja M. Das, Nagaratnam Sivakugan
Chapter7: Seepage
Section: Chapter Questions
Problem 7.7CTP
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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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