A soil strata can be described as 6 meters of sand underlain by 4 meters of clay with the water table located 7 meters from the bottom of the said strata. The sand has a dry and saturated unit weight of 14 and 18 resp. On the other hand, pertinent clay properties are saturated unit weight of 19, void ratio of 0.8 and liquid limit of 40. Compute the kN primary consolidation settlement considering the following conditions if a uniform load of 90 kPa is acting on the soil. If necessary, assume a logical value of Cs. A. The clay is normally consolidated. B. The pre-consolidation pressure is 190 kN kN C. The pre-consolidation pressure is 170 m2

A soil strata can be described as 6 meters of sand underlain by 4 meters of clay with the water table located 7 meters from the bottom of the said strata. The sand has a dry and saturated unit weight of 14 and 18 resp. On the other hand, pertinent clay properties are saturated unit weight of 19, void ratio of 0.8 and liquid limit of 40. Compute the kN primary consolidation settlement considering the following conditions if a uniform load of 90 kPa is acting on the soil. If necessary, assume a logical value of Cs. A. The clay is normally consolidated. B. The pre-consolidation pressure is 190 kN kN C. The pre-consolidation pressure is 170 m2

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Q2-A) A normally consolidated clay layer is 5 m thick (one-way drainage). From the application of a given pressure, the total anticipated primary consolidation settlement will be 160 mm. a. What is the average degree of consolidation for the clay layer when the settlement is 50 mm? b. If the average value of c, for the pressure range is 0.003 cm²/sec, how long will it take for 50% settlement to occur? c. How long will it take for 50% consolidation to occur if the clay layer is drained at both top and bottom?
1. A soil profile is shown in the figure. A uniformly distributed load of 75kPa is applied at the ground surface. The clay is normally consolidated. Compute the secondary settlement 8 years after the completion of primary consolidation. Time for completion of primary settlement is 2 years. Assume secondary compression index = 0.025
In a reclamation project, the middle of a clay layer is under an initial effective vertical stress of 200 kPa and initial void The clay is ratio (eo) of 0.90. overconsolidated with OCR of 1.4, compression index of 0.34 and expansion index of 0.07. With an applied load of 299 kPa, what will be the final void ratio (e1) for the clay?
A 3 m thick saturated soil stratum has a compression index of 0.25 and coefficient of permeability 3.2 x 10-3 mm/sec. If the voids ratio is 1.5 at vertical stress of 0.10 N/mm2, compute the voids ratio when the vertical stress is increased to 0.25 N/mm2. Also calculate settlement due to above stress increase and time required for 50% consolidation. Assume single drainage.
13. Situation 13 - The soil is acted upon by a uniformly distributed load ΔP = 60 kPa at the ground surface. a. Compute the settlement of the clay layer, in mm, caused by primary consolidation if the preconsolidation pressure of clay is 230 kPa. Use Cs = Cc/5. Round off to the nearest whole number. b. Compute the settlement of the clay layer, in mm, caused by primary consolidation if the clay is normally consolidated. Round off to the nearest whole number. c. Compute the settlement of the clay layer, in mm, caused by primary consolidation if the preconsolidation pressure of clay is 200 kPa. Use Cs = Cc/5. Round off to the nearest whole number.
For a normally consolidated clay layer in the field, the following values are given: Thickness of clay layer = 8.5 ft ● Void ratio = 0.8 ● ● ● ● Compression index (CC) = 0.28 Average effective pressure on the clay layer is 2650 lb/ft² Additional Overburden Pressure is 970 lb/ft² Secondary compression index (Ca) is 0.02 Determine the total consolidation settlement considering primary and secondary settlement. Select the correct response: 2.6425" 2.5631" 2.8866" 2.3856"