Principles of Geotechnical Engineering (MindTap Course List)
9th Edition
ISBN: 9781305970939
Author: Braja M. Das, Khaled Sobhan
Publisher: Cengage Learning
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Chapter 9, Problem 9.13P
To determine
Find and plot the total stress, pore water pressure, and effective stress for various depth.
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Subject: Geotechnical engineering- Soil Mechanics
The figure below shows the zone of capillary rise within a clay layer above the groundwater table. For the
following variables, calculate and plot o, u, and ơ' with depth.
G = 2.69; e = 0.47
H2
G¸ = 2.73; e = 0.68
H3
G¸ = 2.7; e = 0.89
3
Dry sand
O Clay; zone of capillary rise
Clay
Rock
Figure 9.29
Degree of saturation in
capillary rise zone, S (%)
H,
H2
3.05 m
2.43 m
4.88 m
40
Chapter 9 Solutions
Principles of Geotechnical Engineering (MindTap Course List)
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- 1. (30 pts) The soil profile shown below consists of 10 meters of sandy silt overlying gravel. The pore water pressure at the top surface of the silty sand is zero and can be assumed to remain zero. a) Calculate the level to which water would rise in a piezometer tube inserted into the top of the gravel if the silty sand is just stable? Use submerged unit weights and seepage forces to arrive at your answer (do not calculate total stresses and pore water pressures). Express your answer as an elevation, e.g. "Elev. 130". (Note: Elevations are in meters) (10 pts) b) Using the piezometric elevation calculated in part (a), calculate the pore water pressure at the bottom of the silty sand if the silty sand is just stable (10 pts). c) Calculate the total stress at the base of the silty sand and show that it is equal to the pore water pressure calculated in part (b) (10 pts) Elev. 120 m. Elev. 110 m. Sandy Silt (saturated) Void ratio, e = 0.68 G = 2.65 Gravelarrow_forward(Situation 3) Determine the equivalent hydraulic conductivity of the soil strata given in the figure considering direction of flow of water parallel and perpendicular to the layers. k = 2.3 x 102 cm/s k, = 8 x 102 cm/s k, = 5.7 x 10 cm/s, k, = 25.5 x 104 cm/s 10 m 2 m! = 9.2 x 10 cm/s k, = 27 x 107 cm/s 10 m %3!arrow_forwardA soil profile is shown in Figure 6.24. Calculate the values of o, u, and o at points A, B, C, and D. Plot the variation of ơ, u, and ơ with depth. We are given the values in the table. Layer No. Thickness (m) Unit weight (kN/m³) H = 2 H2 = 3 = 7 I Ydry = 15 Ysat = 17.8 II %3D III %3D Ysat = 18.6arrow_forward
- 1. Calculate the variation of σ, u, and σ’ with depth. Given: H1 = 12 ft, H2 = 8 ft, H3 = 5 ft, S = 50% in capillary rise zone. 2. Plot the variations of σ, u, and σ’ with depth.arrow_forwardRead the question carefully and give me the right solution according to the questionarrow_forwardP 2: A soil profile is shown in Figure below. Note the zone of capillary in the sand layer overlying clay. In the zone the average degree of saturation and the moisture content are 60% and 17.6kN/m³, respectively. Calculate and plot the variation of o, u and o' with depth at points A, B, C and D. Sand 3m Y = 16.5kN/m³ Zone of capillary rise y = 17.6kN/m² S,= 60% 1m Sand Ysat = 18.9kN/m³ Sáturated clay 3m Rock ANSWERS: o' At A =0, at B before capillary =49.5kPa, at B after capillary = 55.4kPa, at C = 67.1kPa , at D=94.4kPaarrow_forward
- Q2) Figure 2 shows the zone of capillary rise within a clay layer above the groundwater table. For the following variables, calculate and plot o, u, and ở with depth. Note: H1=4m, H2=2.5m, H3=4.5m, Degree of saturation in capillary rise zone, S (%)=60% G = 2.69; e = 0.47 G = 2.73; e = 0.68 G = 2.7; e = 0.89 Dry sand O Clay; zone of capillary rise Clay Rock Figure 2arrow_forwardA canal is cut into a soil with a stratigraphy as shown. Assume that flow takes place horizontally and vertically through the sides of the canal and vertically below the canal. The values of k in each layer is given. a. What is the equivalent permeability in the horizontal direction through the sides of the canal in cm/day? b. What is the equivalent permeability in the horizontal direction through the sides of the canal in cm/day? c. Determine the equivalent permeability in the vertical directions below the bottom of the canal, in cm/day. 1.0 m k = 2.3 x 10* cm/s 1.5 m T. k = 5.2 x 10 cm/s 3.0 m 2.0 m k = 2 x 10 cm/s 1.2 m k- 0.3 x 10 cm/s 3.0 m k = 0.8 x 10° cmn/sarrow_forwardEXAMPLE 9.8 A soil profile is shown in Figure 9.24. Given: H₁ Plot the variation of o, u, and o' with depth. H₁ H₂ B Zone of capillary rise G₁ = 2.66; e = 0.55 Degree of saturation = S = 50% Sand Saturated clay = 2 m, H₂ = 1.8 m, H3 = 3.2 m. Rock G₁ = 2.66 e = 0.55 Sand Groundwater table w 42% (moisture content) G₁ = 2.71 Clayarrow_forward
- Please solve with a clear explaintion, thank youarrow_forwardThe figure below shows the zone of capillary rise within a clay layer above the groundwater table. For the following variables, calculate and plot o, u, and o' with depth. G,-2.69; e0.47 H2 G,- 2.73; e= 0.68 H3 G, = 2.7: e= 0.89 Dry sand Clay; zone of capillary rise Clay Rock Figure 9.29 Degree of saturation in capillary rise zone, S (%) H, H2 3.05 m 2.43 m 4.88 m 40arrow_forwardProblem 5 From the figure shown, the void ratio of the sand is 0.55 with a specific gravity of 2.68. The cross- sectional area of the tank is 0.5 m² and hydraulic conductivity of sand = 0.1 cm/s. 1. What is the rate of upward seepage in m3/hour? 2. Determine the critical hydraulic gradient for zero effective stress. 3. Determine the value of h to cause boiling. 4. Determine the value of x. h=1.2 m H₁=1 m z =0.8 H₂=2 m In flow Soil C B H₂O valve open Karrow_forward
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