Consider three layered soil strata as shown in the figure below. The total head loss for the given soil strata is: T H Flow 2m k₁ =2× 102cm/5 0.8 m 3m k=3x10cm/s L=1m 2 m k₁ = 4 × 102cm/s
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- Q.11 Consider the three layered soil strata shown in the figure below. The thickness and coefficient of vertical permeability of each layer is mentioned in the figure. Flow Ah = 0.50 m -3 H, = 3 m K, = 3 x 10° cm/s %3D -2 H2 = 2 m K, = 6.5 x 10 cm/s H = 9 m H3 = 4 m. L = 2 m -4 Kg = 7 x 10 cm/s The total head loss in three layers isHomeworks 2. The following results were obtained from an oedometer test on a specimen of saturated clay: Pressure (kN/m?) 27 54 107 214 429 214 107 54 Void ratio 1.243 1.217 1.144 1.068 0.994 1.001 1.012 1.024 A layer of this clay 8m thick lies below a 4m depth of sand, the water table being at the surface. The saturated unit weight for both soils is 19kN/m3. A 4m depth of fill of unit weight 21 kN/m3 is placed on the sand over an extensive area. Determine the final settlement due to consolidation of the clay. -1 e -eo 1+e, o'y-o'vo my S = i=lSolve the following problems using the figure. 1. Calculate the ultimate settlement of the clay layer. 2. Calculate the excess pore pressure (ue) and the effective stress at the midpoint of the clay layer after 3 years. 10' Fill, y = 118b/³ Sand, y 109 lb/ft³ Sand, Ysat 114 lb/ft³ eo = 1.1 Cc = 0.25 C₁ = 7 ft²/yr C₁ = 0.06 clay 14' N.C Ysat = 122 lb/ft³ sand 8. For the previous problem, report the predicted settlement after 365 days. 3. Create a plot (or plots) in a spreadsheet showing the hydrostatic pressure in the clay with depth and the excess pore pressure (ue) with depth after 3 years and the combined pore pressure after 3 years. (This is similar to an example worked in class) 4. Calculate the average percentage of consolidation (U%) for the clay layer after 3 years. 5. How long will it take for 80% of the ultimate settlement of the clay layer to take place? 6. Assume thin layers of drainable sand are present every two feet within the clay so it is actually five clay layers that…
- 2) A soil profile consisting of three layers is shown in the Figure. A) Calculate the values of σ, u and σ' at points A,B,C and D if Layer 1: H1=5m, e=0.7, Gs=2.69 Layer 2: H2=8m, e=0.55, Gs=2.7 Layer 3: H3=3m, w=38%, e=1.2 B) What is the change in effective stress at point C if: If the water table drops by 2m? If the water table rises to the surface up to point A? Water level rises 3 m above point A due to flooding?A 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 = 18.6A constant-head test was conducted on a sample of soil 15 cm long and 60 cm^2 in cross-sectional area. The quantity of water collected was 50 cm^3 in 20 seconds under a head difference of 24 cm. if the porosity of the sand is 55 % calculate the seepage velocity in cm/s. a. 1.0 x 10^-1 b. 4.3 x 10^-3 c. 3.47 x 10^-2 d. 2.6 x 10^-6
- A reservoir-rock sample has a porosity of 0. 18 measured at 14.7 psis. The rock compressibility is0.77 X 10 5psi *.What would the rock porosity be at 2,000 and at 3,000 psis? Plot the relationship be tween ¢ and p in a pressure range of 14.7 to 5,000 psis3. A civil engineer is interested in assessing the variation of shear strength of clay layers in a site of interest for different depths. After completing CU triaxial tests, the sample mean for the clay shear strength 40 kPa and the standard deviation was 15.5 kPa are obtained for the depth of 3 m from ground surface. For the following conditions: a. Calculate 99% CI for the population mean μ. Write down your interpretation.b. Explain why the size of CI changed between parts a - c.4. For a variable head permeability test, the following are given: 2. A borehole at a site reveals the soil profile shown. Assume Gs = 2.70 for all soil types. Length of soil specimen 200 mm Area of soil specimen a. What is the unit weight of the soil in layer 1 in kN/m³? 1000 mm? Area of standpipe b. What is the effective stress at a depth of 2m below the ground surface, in kPa? 40 mm? Head difference at time t = 0 c. What is the effective stress at a depth of 20.6m below the ground surface, in kPa? 500 mm Head difference at time t = 3 min 300 mm Elevation (m) Continuous supply Very fine wet sand with silt w = 5%, S = 40% Layer 1 2.0 Fine sand saturated by capillary action Layer 2 3.0 500 mm - Fine sand, w = 12% Layer 3 5.4 300 mm -Area of stand pipe. = 40 mm2 L=200 mm icimple -Porous plate Soft blue clay, w = 28% Layer 4 Area of soil sámple = 1000 mm? a. Compute the hydraulic conductivity of the soil in cm/sec. b. Compute the seepage velocity if the porosity of soil is 0.25. 20.6 c.…
- A layered soil is shown in Figure 7.33. Given: • H1 = 1.5 m k1 = 9 × 104 cm/sec• H2 = 2.5 m k2 = 7.8 × 10−3 cm/secQ2. A soil profile consisting of three layers is shown in Figure 1 below. Unit weight of water (yw) = 9.81 kN/m³ Layer no. 1 2 3 ↑ H₁ H₂ Dry sand Sand Clay Rock Figure 1: Soil Profile Thickness H₁ = 1.5 m H₂ = 2.53 m H3 = 2.64 m Layer 1 Groundwater table Layer 2 Layer 3 Soil parameters Yd = 16.5 kN/m³, Ysat = 17.42 kN/m³ Ya = 16.5 kN/m³, Ysat = 18.55 kN/m³ Yd = 17.50 kN/m³, Ysat = 19.23 kN/m³ a) Calculate the values of o, u and o' at points A, B, C and D. b) Calculate the effective stress at C when water table drops by 2 m. c) Calculate the effective stress at C when water table rises by 1 m above layer 1 due to flooding.As indicated in the following figure, an incompressible silty sand fill 16.4 ft thick was placed on top of a 49.2 ft thick layer of soft and compressible silty clay. Underlying the clay is a layer of incompressible sandy gravel. The soil properties of the silty clay layer are given in the following figure. The ultimate settlement, Sult = 5.64 ft, from the primary consolidation of silty clay layer due to the weight of the silty sand fill is calculated. O ft Fill Material - Silty Sand (SM), y = 124.86 pcf, 16.4 ft Silty Clay (CL) 94.89 pcf Y = Normally Consolidated 1.1 e, = Cc = 0.36 0.06 Cv = 9.235 ft/year 65.6 ft Dense Sandy Gravel (GP), Incompressible layer 82.0 ft A) Find the time rate of settlement by showing the Time [year] vs. Settlement [ft] curve. Show sample calculation when percentage of consolidation, Ugyg = 50%