A consolidated drained triaxial test was conducted on a granular soil. At failure (σ1^')/(σ3^' ) = 4.0. The effective minor principal stress at failure was 100 kN/m2. Compute the principal stress difference at failure in kPa.  a. 100 b. 200 c. 300 d. 400 e. 500

Q: During an oedometer test on a specimen of saturated clay, the thickness of the specimen decreased…

A: Oedometer test: This is a geotechnical investigation used for measuring properties of consolidation…

Q: At failure, the major and minor principal stresses on a triaxial test are 200 kPa and 110 kPa…

A: Given data Major principal stress : 200 kpa Minor principal stress :110 kpa Pore water pressure at…

Q: A triaxial test was performed on a saturated soil and failed under a normal stress of 150 kPa. If…

A:

Q: - A sand specimen is isotropically consolidated in a triaxial test apparatus under a cell pressure…

A:

Q: Q1/[B]: For the clayey sand shown in the figure, the saturated unite weight is 20.69kN/m³ and the…

A:

Q: Example 3 The equation of effective stress failure envelope for normally consolidated clayey soil is…

A:

Q: A triaxial test was performed on a dry. cohesionless sand. The sample fails when the confining…

A:

Q: A consolidated un-drained triaxial test was performed on a specimen of saturated clay with a kg…

A: Given : σ3 = 2.0 kg/cm2 σ1 - σ3 = 2.8 kg/cm2 u = 1.8 kg/cm2 θ = 57 o This is a problem of Soil where…

Q: A consolidated tri-axial test was conducted on a normally consolidated clay. The results were as…

A:

Q: A consolidated tri-axial test was conducted on a normally consolidated clay. The results were as…

A: To find out the shear stress on failure plane.  Confining stress = 326 kPa Deviator stress = 414 kPa

Q: The results of two consolidated-drained test triaxial tests on a clay are given below: Specimen No.…

A:

Q: Question attached

A: Given Data:- Depth = 6 m  Depth of water table = 2 m Depth considered (U) = 4 m  γd=14.5…

Q: need the ans

A:

Q: A sample of dry sand was placed in a direct shear testing apparatus and was tested until the sample…

A: Given that: Shear stress at failure, τf=70 kN/m2Normal stress at failure, σf=96 kN/m2

Q: 7. A specimen of soil was collected from a depth of 10 m in a deposit of clay. The ground water…

A:

Q: The table shows the results of a consolidated undrained (CU) test on a overconsolidated clay soil…

A: For sample 1 σ1 =15+20-10σ1 =25 kpaσ3 =15-10 σ3 =5 kpaRelation of major and minor stress is given…

Q: 3-Following are the results of consolidated-undrained triaxial tests on undisturbed soils: Excess…

A:

Q: ted tri-axial test was conducted on a normally consolidated clay. The results were as follows:…

A:

Q: Three consolidated-undrained tri-axial tests were performed on samples retrieved from a 4 meter…

A: For the test 1 -  Value of σ'3σ'3 = σ3 - ∆udf⇒σ'3 = 80 - -12⇒σ'3 = 92 KNm2   Value of σ'1σ'1 = σ'3 +…

Q: # 3. Following data are given for a direct shear test conducted on dry sand: Specimen dimensions: 63…

A:

Q: Tri-axial Test SIT Q. A consolidated drained tri-axial test was conducted on a normally consolidated…

A:

Q: The equation of the effective  stress failure envelope for normally  consolidated clay soil is  30…

A:

Q: A cylindrical sample of cohesive soil tested in the unconfined compression test. At failure, the…

A: Unconfined compressive stress =80kPa The failure plane makes an angle = 50o

Q: The results of two drained triaxial tests on a saturated clay are given next: Specimen I: Chamber…

A:

Q: 12.9 The relationship between the relative density D, and the angle of friction o' of a sand can be…

A:

Q: An unconfined compression (UC) test was conducted on a 50-mm-diameter and 100-mm-high cylindrical…

A: "Since you have posted a question with multiple sub-parts, we will solve first three sub-parts for…

Q: consolidated drained triaxial test was conducted on a granular soil. At failure (σ1^')/(σ3^' ) =…

A:

Q: In a tri –axial test of a silty soil, the sample failed at normal stress of 475 kPa and a shear…

A:

Q: A CU triaxial test was carried out on a silty clay that was isotopically consolidated using a cell…

A: Given:- Cell pressure = 125 kPa Axial load (kPA) Axial strain ε1 % Δu kPa 0 0 0 5.5…

Q: A consolidated-undrained tri-axial test was conducted on a normally consolidated clay sample and the…

A: Step: 1 Friction angle: - It is a shear strength parameter and describes the shear resistance of…

Q: Compute the angle of resistance in degrees.

A:

Q: A SOmple oJ OJ ary SO WAS tri axial test with a confining pressure of 250 kN/m², The angle of…

A:

Q: A dilatometer test (DMT) was conducted in a clay deposit. The water table was located at a depth of…

A:

Q: A compression test was conducted on a soil sample under drained conditions. The deviator stress at…

A: Given: Deviator stress at failure = 55 lb/in2 Angle of internal friction= 27°

Methods to Determine Vertical Stress

The load applied to the soil is transferred to the underground. The presence of water at soil pores and solid grains above the soil are the primary components that are responsible for the effective load transfer. Due to the load acting, the internal component of the soil or the underground formations undergoes deformations or a tectonic shift. The amount of deformation depends upon the amount of load acting on the soil surface, which indeed, depends on the weight of everything present above the soil. Due to this, the underground soil element develops internal stresses which try to resist the tectonic changes.