Fundamentals of Geotechnical Engineering (MindTap Course List)
5th Edition
ISBN: 9781305635180
Author: Braja M. Das, Nagaratnam Sivakugan
Publisher: Cengage Learning
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Chapter 9, Problem 9.4P
(a)
To determine
Calculate the primary consolidation settlement if the clay is normally consolidated.
(b)
To determine
Calculate the primary consolidation settlement if the pre-consolidation pressure is
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Consider the soil profile shown in Figure 11.44 subjected to the uniformly distributed load, Ds, on the ground surface. Given: Ds = 26 kN/m2; H1 = 1.83 m; H2 = 3.66 m; and H3 = 5.5 m. Soil characteristics are as follows:
Sand: gdry =17.9 kN/m3; gsat =18.5 kN/m3
Clay: gsat =18.3 kN/m3 ; LL = 38; e = 0.73; Cs = 1/5 Cc
Estimate the primary consolidation settlement in the clay if
The clay is normally consolidated
The preconsolidation pressure s’c = 105 kN/m2
The soil profile shown in Figure 4.2 is subjected to the uniformly distributed load, Ao,
on the ground surface. Given Ao = 15 kN/m2; H1 = 2 m; H2 = 3.5 m; and H3 = 5 m. Soil
characteristics are as follows:
*Sand: ya = 16.5 kN/ms; ysat = 19 kN/m3
*Clay: ysat = 18 kN/m:; LL = 31 %; e = 0.65; Cs = (1/5) C.
Use: Ce= 0.009( LL – 10)
a. Estimate the primary consolidation settlement in the clay (in mm.) if: The
preconsolidation pressure, oc= 621 kN/m2
b. Estimate the value of the void ratio at the end of the primary
consolidation which will be used for the analysis of the secondary consolidation
Aar
O Sand
Clay
A clay layer of 3 m thickness is covered by a 5 m sandy stratum. The properties of the soils are
shown in figure below.
5m
3 m
2m
Sand
Silty Clay
Rock
¥..
G= 2.65
Yary = 18 kN/m³
e = 0.40
c = 40 kN/m²; + = 30°
Ysat = 20 kN/m
The shear strength of the soil at the middle layer of silty clay is
Chapter 9 Solutions
Fundamentals of Geotechnical Engineering (MindTap Course List)
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.Similar questions
- Q: For the soil profile shown in figure. Determine: The effective stress (o') at level (A). Soil 1 k = 0.2 cm/sec Y sat = 19 kN/m³ Soil 2 k = 0.4 cm/sec Y sat = 21 kN/m³ Datum D EL + 2.0 m EL 0.0 m EL - 1.0 m EL - 4.0 m EL - 10.0 marrow_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_forwardThe arrangement of soil layers are as shown below. Lower layer of sand is under artesian pressure. If effective stress at top of clay layer is half of the effective stress at top of lower layer of sand, then amount by which piezometric head in lower layer of sand rises above water table is: 3 m Sand y= 17 kN/m 4 m y = 20 kN/m² [1m 6m Clay y= 21 kN/m' 4 m Sand y = 20 kN/marrow_forward
- A soil formation is composed of 5 m thick clay and 5 m thick sand being the sand above the clay. The ground water table (GWT) is located at 2 m below the ground surface. 40 kPa 2 m Sand Ydry = 17.66 kN/m^3 VGWT Sand 3 m Y sat = 20.93 kN/m^3 LI = 64% PL = 20% w = 40% 5 m Clay eo = 0.60 G. = 2.60 1. Calculate the primary compression index. [ Select ] 2. Calculate the primary consolidation settlement of the normally consolidated clay layer if there is a surcharge of 40 kPa acting on the ground surface. [ Select] 3. Calculate the secondary settlement of the clay layer 5 years after the completion of the primary consolidation settlement. Time for completion of primary settlement is 2 years. Use C, = 0.02. [Select ]arrow_forwardFor the soil profile shown, evaluate the vertical and horizontal geostatic stresses points A and B. 2m 3m 3m B sand Gs=2.65 Gs=2.68 silt e=0.65 n=0.35 Ko=0.45 clay Gs=2.72 e=0.5 S=40% Ko=0.4arrow_forwardA soil profi le consists of a clay layer underlain by a sandlayer, as shown in Figure P7.17. If a tube is inserted intothe bottom sand layer and the water level rises to 1 mabove the ground surface, determine the vertical effective stresses and porewater pressures at A, B, and C. IfKo is 0.5, determine the lateral effective and lateral totalstresses at A, B, and C. What is the value of the porewater pressure at A to cause the vertical effective stressthere to be zero?arrow_forward
- The compression curve (void ratio, e for a certain vs. effective stress, clayey soil is a straight line in a semi-logarithmic plot and it passes (e = 1.2; = 50kPa) through the points (e = 1.2; 0) V and (e=0.6;=800kPa). The compression index (up to two decimal places) of the soil isarrow_forward40 Refer to the soil profile shown. Given H1 = 9.89 m., and H2 = 4.4 m. If the ground water table rises by 3.13 meters, determine the change in effective stress (numerical value only, in kPa) at the bottom of the clay layer. Properties of dry sand: Gs = 2.54, e = 0.69. Properties of clay: Gs = 2.77, e = 0.85. Round off to two decimal places.arrow_forwardProblem 2 A soil element is shown in the figure below. 26 kN/m2 8 kN/m2 B to + 17 kN/m2 8 kN/m2 45° Determine the following: (in kPa) a. Maximum Principal Stress b. Minimum Principal Stress c. Normal Stress on plane AB d. Shear Stress on plane ABarrow_forward
- 5-4 For the given soil profile and the piezometric surfaces measured at the top of the two sand and gravel layers, determine the pore pressure and effective stress at points A, B, C, D, and E. Piezometers 2 m Ground surface 2 m Sand y = 18 kN/m3 Water table y_ 11m Y = 20 kN/m3 A 4 m Clay Y = 18 kN/m3 加。B 1Im 5 m Sand and gravel y = 22 kN/m3 2 m •D 5 m Clay Y = 19 kN/m3 Sand and gravel y = 23 kN/m3 %3D Cpyright byarrow_forwardstion 17 | Dry Sand H Groundwater Table Clay Refer to the soil profile shown. Given H1 = 8.93 m., and H2 = 5.32 m. If the ground water table rises by 2.01 meters, determine the change in effective stress (numerical value only, in kPa) at the bottom of the clay layer. Properties of dry sand: Gs = 2.56, e = 0.65. Properties of clay: Gs = 2.72, e = 0.87. Round off to two decimal places.arrow_forwardDry Sand H, Groundwater Table Clay H2 Refer to the soil profile shown. Given H1 = 9.48 m., and H2 = 4.41 m. If the ground water table rises by 2.94 meters, determine the change in effective stress (numerical value only, in kPa) at the bottom of the clay layer. Properties of dry sand: Gs = 2.58, e = 0.67. Properties of clay: Gs = 2.74, e = 0.87. Round off to two decimal places.arrow_forward
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