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 13, Problem 13.15P
To determine
Find the factor of safety of the slope and locate the critical circle.
Locate the critical circle for the given condition.
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An excavation is to be made in purely cohesive soil deposit having c = 20 kN/m² and unit weight
= 16 kN/m³. The depth to which the vertical sides of the excavation will remain stable without side
supports will be:
The figure below shows a proposed site where an excavation will be made. The 12 ft layer of sand will be removed, so that the top of the 30 ft normally consolidated clay layer will be exposed (Assume that changes in GWT do not affect γt of clay). a. Assume that the water-table location remains the same during excavation. Compute the σ, σ′v, and u values at the middle of the clay layer before and after the excavation b. Assuming 1-D conditions, compute how much the clay layer will deform due to this excavation, in inches. Specify whether this is settlement or heave
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Qus
Chapter 13 Solutions
Fundamentals of Geotechnical Engineering (MindTap Course List)
Ch. 13 - Prob. 13.1PCh. 13 - Prob. 13.2PCh. 13 - Prob. 13.3PCh. 13 - Prob. 13.4PCh. 13 - Prob. 13.5PCh. 13 - Prob. 13.6PCh. 13 - Prob. 13.7PCh. 13 - Prob. 13.8PCh. 13 - Prob. 13.9PCh. 13 - Prob. 13.10P
Ch. 13 - Prob. 13.11PCh. 13 - Prob. 13.12PCh. 13 - Prob. 13.13PCh. 13 - Prob. 13.14PCh. 13 - Prob. 13.15PCh. 13 - Prob. 13.16PCh. 13 - Prob. 13.17PCh. 13 - Prob. 13.18PCh. 13 - Prob. 13.19PCh. 13 - Prob. 13.20PCh. 13 - Prob. 13.21PCh. 13 - Prob. 13.22PCh. 13 - Prob. 13.23PCh. 13 - Prob. 13.25PCh. 13 - Prob. 13.26PCh. 13 - Prob. 13.27CTPCh. 13 - Prob. 13.28CTPCh. 13 - Prob. 13.29CTP
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- Complete and correct solution pleasearrow_forwardAn excavation is made with a vertical face in a clay soil which has C_{u} = 50kN /(m ^2) gamma_{1} = 18kN / (m^3) and S_{n} = 0.261. The maximum depth of a stable excavation will be nearly?arrow_forwardThe cohesion (c), angle of internal friction (p) and unit weight (y) of a soil are 15 kPa, 20° and 17.5 kN/m³, respectively. The maximum depth of unsupported excavation in the soil (in m, round off to two decimal places) is Enter Your Answer... Checkarrow_forward
- Problem- Geotech- The cohesion (c), angle of internal friction () and unit weight (y) of a soil are 15 kPa, 20° and 17.5 kN/m3, respectively. The maximum depth of unsupported excavation in the soil (in m, round off to two decimal places) isarrow_forwardTable 5. Adjustment factor for trench production Type of Material Adjustment Factor Loose (sand, gravel, loam) 0.60-0.70 Average (common earth) 0.90-0.95 Firm (firm plastic soils) 0.95-1.00arrow_forward3. In a large field with a flat surface, an excavation will take place. According to the drilling works, no groundwater was encountered, and a homogeneous cohesionless soil has been found throughout the excavation depth. As a result of the experiments, the internal friction angle of the relevant soil was found to be 24°, and the dry and saturated unit weights were 17 kN/m³ and 19.2 kN/m², respectively. According to the information given, what is the inclination of the steepest slope that can be excavated in the area?arrow_forward
- A cut slope was excavated in sand. The slope is shown in following attachment . To find the properties of the sand, representative samples were taken for the soil and tested in atriaxial apparatus. A consolidated drained triaxial test was conducted. The result is given below:Chamber pressure = 103 kN/m2Deviator stress at failure = 234.6 kN/m2We know the unit weight of the sand is 16 kN/m3.Use the ordinary method of slices (use 7 slices) and estimate the factor of safety against sliding forshown slip surface and comment on the calculated factor of safety.Please note that:1. The figure is not in scale. You may resketch it in scale and calculate the factor of safety2. The slip surface is circular and passes through the toe of the embankmentarrow_forwardAn excavation is rapidly made on site as depicted in the Figure below. The soil wedge above the critical slip surface is divided into thirteen slices. Based on the ordinary method of slices, calculate the ratio (R/S) between the Resisting Force (R) and the Driving Force (S) for slice #3, shortly after the excavation is completed. (Acceptable tolerance = 4%). The geometry of slice #3 and the soil properties are provided in the tables below. The height of water at the middle of the slice is hw = 3.0 m. Note that the total weight of the slice is equal to the weight of the saturated soil below the water table plus the weight of the moist soil above the water table. Hint: the weight of slice #3 is 495 kNarrow_forwardKindly solve as fast as possible.arrow_forward
- Civil Engineering The cohesion (c), angle of internal friction () and unit weight (y) of a soil are 15 kPa, 20° and 17.5 kN/m³, respectively. The maximum depth of unsupported excavation in the soil (in m, round off to two decimal places) isarrow_forwardCENTER OF FAILURE CIRCLE soil 1 unit weight = 16 kN/m3 c = 20 kPa ' = 25 deg soil 2 unit weight = 18 kN/m3 c = 30 kPa = 35 deg Scale 1 cm = 2m Figure shows an embankment consisting of two soils with the given soil properties. There is no ground water table in the vicinity. (a) Calculate the factor of safety using the Ordinary Method of Slices (b) Assuming a factor of safety of 1.5, perform one iteration to calculate factor of safety using the Bishop?s Method of Slicesarrow_forwardPlease solve all the three a,b,c as soon as possible. A trench is to be excavated in a clay soil with γsat =20 kN/m3 and su=15 kPa.(a) Determine the theoretical maximum depth of thetrench without lateral support.(b) It was decided to use a bentonite slurry of unitweight γf to fill the trench as excavation proceeds.Show that the theoretical maximum depth of thetrench is Hcr = 4su / γsat - γf where γsat is the saturatedunit weight of the soil and su is the undrainedshear strength.(c) How much deeper can the trench be excavated byusing a bentonite slurry of γf= 11 kN/m3?arrow_forward
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