The shear strength of a normally consolidated clay can be given by the equation f =...

The shear strength of a normally consolidated clay can be given by the equation f =  tan 30. A consolidated-undrained triaxial test was conducted on the clay. Following are the results of the test:
Chamber confining pressure = 100 kN/m2
Deviator stress at failure = 80 kN/m2
Determine
a) Consolidated Undrained friction angle (CU), (6 points)
b) Using formula, calculate shear and normal stresses on the failure plane inside the soil sample, (6 points)
c) Pore water pressure developed in the clay specimen at failure. (8 points)

Expert Solution

Ally Wells answered 7 months ago

QUESTION 5 (25 marks) A consolidated drained triaxial test was conducted on a normally consolidated clay....

QUESTION 5 A consolidated drained triaxial test was conducted on a normally consolidated clay. Following are the results of the test: • Cell or Chamber pressure = 130 kN/m2 • Deviator stress at failure = 230 kN/m2 a) Determine the consolidated drained friction angle b) What is the pore water pressure developed in the clay specimen at failure? c) What is the angle θ that the failure plane makes with minor principal stress? d) Determine the normal stress and shear...

Q.3) USE ONLY GRAPHICAL METHOD: A consolidated-drained triaxial test was conducted on a normally consolidated clay...

Q.3) USE ONLY GRAPHICAL METHOD: A consolidated-drained triaxial test was conducted on a normally consolidated clay with a chamber pressure, σ3 = 172 kN/m2 . The deviator stress at failure, (Δσd)f = 227 kN/m2 . Determine (12 points): a. The angle of friction, φ' b. The angle θ that the failure plane makes with the major principal plane c. The normal stress, σf', and the shear stress, τf, on the failure plane

A layer of soft, normally consolidated clay is 15 m thick and lies under a newly...

A layer of soft, normally consolidated clay is 15 m thick and lies under a newly constructed building. The overburden pressure of the sand overlying the clay layer is 500 kN/m2 and the new construction increases it by 180 kN/m2. If the compression index is 0.25 and the recompression index is 0.015, compute the settlement of the building. The water content of the clay is 30%, the specific gravity of the soil grains is 2.7, and the clay is fully...

A Normally Consolidated 10 Ft. clay layer is surcharged, which causes a decrease in thickness. The...

A Normally Consolidated 10 Ft. clay layer is surcharged, which causes a decrease in thickness. The coefficient of consolidation is 0.16 ft^2/day and the time factor is 1.2 for U=50%. The clay layer is confined between 2 layers of dense sand. The time (days) required for 50% consolidation is most nearly? This is a question from a practice FE exam. The reference book and the solution to this leaves out some information. The equation is Tv=Cv(t)/Hdr^2) The solution has Hdr=5...

A 20 m thick saturated clay layer is normally consolidated. At mid-depth in this layer, the...

A 20 m thick saturated clay layer is normally consolidated. At mid-depth in this layer, the void ratio (e) is 1.110, the coefficient of compressibility (av) is 5.74 x103 cm2=N. Recall that w = 9.81 kN/m3. a) Assuming that a 122 cm lowering of the ground water table (GWT) occurs in the soil overlying the clay layer, what settlement will occur in the soil overlying the clay layer because of the ultimate primary consolidation of the clay layer? Express your...

Q2. Assume a rigid strip footing is constructed on a normally consolidated clay layer under a...

Q2. Assume a rigid strip footing is constructed on a normally consolidated clay layer under a uniformly distributed vertical load. The depth of footing is zero and the water table is a few metres below the surface ground. The average shear strength parameters of the clay layer are: At the beginning of loading: fu = 0, cu = 50 kPa Long time after loading:, f’ = 25° and c’ = 0 kPa Will the bearing capacity of this footing change,...

Assume the measurements of nominal shear strength (in kN) are normally distributed. To estimate the mean...

Assume the measurements of nominal shear strength (in kN) are normally distributed. To estimate the mean of this distribution, a sample of 26 measurements was collected with sample mean 678.27 and the sample standard deviation 82.89. Find a 90% confidence interval. Please show work. Thanks.

At the end of the CD triaxial pressure experiment on a normal consolidated clay, the sample...

At the end of the CD triaxial pressure experiment on a normal consolidated clay, the sample was defeated along a plane of 570. When another sample with the same properties was tested in the CU experiment, the pore pressure in the defeat was measured as 85 kPa. According to the cell pressures (63) 200 kPa in both experiments; a) At the end of the A = CD experiment, find the effective internal friction angle and the difference in prime stress....

low carbon steel having a tensile strength of 350 MPa and shear strength of 250 MPa...

low carbon steel having a tensile strength of 350 MPa and shear strength of 250 MPa is cut in a turning operation with a cutting speed of 200 m/min. The feed is 0.3 mm/rev and the depth of cut is 4.8 mm. The rake angle of the tool is 12 degrees in the direction of chip flow. The resulting chip ratio is 0.8. Using the orthogonal model as an approximation of turning determine... a.The shear plane angle. b. Shear force...

5.18 The shear strength of an adhesive is thought to be affected by the application pressure...

5.18 The shear strength of an adhesive is thought to be affected by the application pressure and temperature. A factorial experiment is performed in which both factors are assumed to be fixed. Analyze the data and draw conclusions. Perform a test for nonadditivity on minitab. Temperature (°F) Pressure (lb/in2) 250 260 270 120 9.60 11.28 9.00 130 9.69 10.10 9.57 140 8.43 11.01 9.03 150 9.98 10.44 9.80

Question