1. A soil profile is shown in Figure 1. Assuming the water table is located at the surface and the profile is undisturbed. Show the variation of total vertical and horizontal stresses with the depth to a total depth of 12m. Assume there is a surcharge of 6kN/m2 and Groundwater is 5m below the surface. 5m Sandy clay c=25KN/m², =10°, y=18KN/m³ 7.5m Silt c=10kN/m²?, =15°, y=20KN/m³ Sand =35°, y=19KN/m³ Figure 1

Q: An embankment consists of clay fill for which c=25 kPa and angle of internal friction is 260 (from…

A: Given Data: c=25 kPa Angle of internal friction=260 Weight of fill per unit volume=18.64 kN/m3…

Q: 'Problem 5. A dense silt layer has the following properties. void ratio = 0.30 eff. Diameter D10 =…

A: Height of capillary riseh=CD10×e

Q: Problem 5. A dense silt layer has the following properties. void ratio = 0.30 eff. Diameter D1o =…

A:

Q: A loose, natural sand deposit has a saturated unit weight of 19.3 kN/m3 and an angle of internal…

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Q: The soil stress state is shown in the figure, σx = 10 kN/m2, σy = 50 kN/m2, τxy = -10 kN/m2 : (1)…

A: Note- Since we only answer up to 3 sub-parts, we’ll answer the first 3. Please resubmit the question…

Q: A river 5 m deep consists of a sand bed with a saturated unit weight of 20 kN/m3, yw = 9.81 kN/m3.…

A: Given- Depth of sand bed = 5m Saturated unit weight γsat=20 KN/m3 Unit weight of water γw=9.81 KN/m3…

Q: A 4-meter clay layer (e = 0.92, Gs = 2.89, Cc = 0.315) is overlain with 4-meter-thick sand layer (e…

A: For the given details,

Q: A loose, natural sand deposit has a saturated unit weight of 19.3 kN/m3 and an angle of internal…

A:

Q: 1. A soil has the following profile and a lateral stress coefficient at rest is 0.45. 0-4 G, = 2.75…

A: Given  K0 = 0.45  Take Unit weight of water  = 10Kn/m3 Water table is at 5m below the ground…

Q: Compute the vertical normal stress, o, at points A, B, and C in Figure 9.11. 15.0 kN/m3: 0.9 m 2.0 m…

A: Density of soil 1, γ1 =15 KNm3Density of soil, γ2 =16.8 KNm3Density of soil, γ3 =17.2 KNm3

Q: Head (m) (cm 5.7 In a certain sand deposit, with Ysat = 20 kN/m3, the water table is at the ground…

A: unit wt of soil    γ=20 KN/m3depth of the soil at bottom  H=4.5 mand hydraulic gradient   i=0.5⇒(a)…

Q: Calculate the active earth pressure coefficient for the backfill soil using both the Rankine and…

A: Rankine's theory also called as plastic theory assumed that no adhesion or friction between the…

Q: The soil stress state is shown in the figure, σx = 10 kN/m2, σy = 50 kN/m2, τxy = -10 kN/m2 What is…

A: The angle of intersection is given by:

Q: 7. For the soil profile shown below, plot the variation of total stress, effective stress and pore…

A: Effective stress: The term effective stress is defined as the total stress minus pore water…

Q: Compute the intensities of active and passive earth pressure at depth of 6 m in dry cohesionless…

A: Given that, H = 6m φ = 30°, Unit weight of the sand, Y = 18kN / m³   Find   Active and passive Earth…

Q: The soil profile consists of a 16 m layer of sand, 12 m layer of clay and incompressible bedrock.…

A: gyebehejis

Q: Example (4.3): An elastic soil medium of (4m x 3m) rectangular area is shown in figure. If the area…

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Q: SITUATION. From the soil profile shown below, Ground surface Dry Sand Clay A e = 0.49 Gs=2.68 Ground…

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Q: Example: Vertical Stresses due to Self Weight of Soil Mass A soil profile is shown in Figure 9.3.…

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Q: 3 m 1 m A seepage flow condition is shown in the figure. The saturated unit weight of the soil Yar…

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Q: 35 At the construction site, 3 m thick clay layer is followed by 4 m thick gravel layer which rest…

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Q: Refer to the soil profile shown in Figure P2.12. Determine the total stress, pore water pressure,…

A: For dry sand part,γd=G×γw1+e=2.66×9.811+0.55=   16.84kN/m3                               (…

Q: Following are the results of two consolidated-drained triaxial tests on a clay:Test I: σ3 = 140…

A:

Q: Determine values of total stress, pore water pressure and effective stress at each soil layer then…

A: Calculating the total stress: σ=γd1×h1+γt1×h2+γt2×h3   =13.1×2+18×2.5+19.8×4.5   =160.3 KN/m2

Q: A soil with a unit weight of 18 kN/m3 is loaded on the ground surface by a uniformly distributed…

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Q: *An embankment is proposed to construct on 10m thick clay deposit having the coefficient of volume…

A: Given data in question  Height of fill Change in effective stress Coefficient of volume…

Q: A soil profi le consists of a clay layer underlain by a sand layer, as shown in Figure P7.17. If a…

A: Given: The water rise above the ground surface is 1 m. The value of K0 is 0.5. Consider the figure.

Q: 1. At the bottom of the Lean Clay layer described below, find the total and effective horizontal…

A: Data given: First layer: Sand with silt (SP-SM) γt = 19 kN/m3 h1 = 9 m Second layer, Lean clay γt=…

Q: Q2/ Determine the variation of total and effective stresses and pore water pressure with depth for…

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Q: A river 5m deep consists of a sand bed with saturated unit weight of 20 kN/m³, Yw = 9.81 kN/m³. The…

A:

Q: An embankment consists of clay fill for which c=25 kPa and angle of internal friction is 260 9from…

A:

Q: If the internal friction angle of a backfill soil is equal to 28°. Calculate the Rankine active…

A: Given: Internal friction angle = 28°

Q: An embankment is being built of a soil whose unit weight is equal to 18 kN/m³. The pore pressure…

A: given values- Pore pressure parameters-A=0.6B=0.8change in height 6 m to 9 m

Q: A soil profile is shown in Figure Q3(b)(i). The groundwater level is located on the ground surface.…

A: We are going to solve only three parts . for fourth part you have to re post this question. given…

Q: A soil profile is shown in Figure 1. Assuming the water table is located at the surface, and the…

A: Surcharge, q = 6 kN/m3 Total vertical stress calculations: At A, σA=qσA=6 kN/m2 At B,…

Q: Layer No. Thickness, H Soil Parameters e = 0.52, Gs = 2.62 e = 0.56, Gs = 2.65 w = 20%, e = 0.51 1…

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Q: A thin layer of silt exists at a depth of 18 m below the surface of the ground. The soil above this…

A: Given: The slit is at the depth of 18 m. The dry unit weight of soil is 15.1 kN/m3. The average…

Q: SITUATION 6: A deposit of clay underlain a 4.5 m thick sand Degree of saturation of sand is 37% and…

A: Given data Thickness of sand = 4.5m  Void ratio of sand (e) = 0.52 Degree of saturation of sand…

Q: The soil in a deposit underlying an area where a level ground surface exists has a unit weight of 22…

A: Given data: Unit weight of soil = 22 kN/m3. Depth = 5m Lateral pressure = 0.5 vertical pressure.

Q: Details of soil retained behind a smooth retaining wall are given in (Figure Q1). Using Rankine's…

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Q: 7.2 Compute the total vertical stress o, pore water pressure u, and then the effective vertical…

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Q: A soil profile is shown in Figure 1. Assuming the water table is located at the surface and the…

A: Solution: When the water table is at surface: Points Total stress Effective stress Pure water…

Q: 2 A granular soil is subjected to a minor principal stress of 200 kN/m2. If the angle of internal…

A: In question number 2- given soil is a granular soil so cohesion c will be zero. The angle with major…

Q: 7.17 A soil profile consists of a clay layer underlain by a sand layer, as shown in Figure P7.iz If…

A: Given :- a soil profile as shown

Q: Compute the immediate settlement beneath the center of a 5 m x 5 m flexible footine resting at 0.5 m…

A: Assume stress-strain modulus,E= 300 qu  where qu is undrained strength 150 kN/m2  E=300×150=45×103…

Q: Question-2: (a) A soil profile is shown in Figure 2.1. Illustrate the variation of total vertical…

A: step 1-

Q: e soil profile in the figure below. Given H= 4m and H2=3m; lculate the total stress, pore water…

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Q: An embankment consists of clay fill for which c=25 kPa and angle of internal friction is 260 9from…

A:

Q: Q1) Data obtained from a drained triaxle test are cell pressure o3 = 50 kN/ m? and deviator stress o…

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Q: onsider the soil profile shown in Figure 1.4: How high should the groundwater table rise/drop so…

A: calculate effective stress at c    so calculate dry unit weight of dry sand…

Calculation of Settlement

The vertical displacement of the structural foundation due to the loading from the superstructure is known as settlement. It is due to the reduction of air voids present in the soil. The amount of vertical displacement depends on the bearing capacity of the soil on which the foundation is laid. The effect of the settlement depends on the magnitude of settlement, uniformity, and the structural design of the structure. The maximum allowable settlement for residential and office buildings should be less than 18 mm, for industrial buildings, it is 38 mm and for commercial buildings, it is 25 mm.