Soil Mechanics - Shear Stress in a Soil Sample

[rstewart8]

Homework Statement

Two consolidated and drained (CO) triaxial compression tests (tests A and 8)
were conducted on dense dry sand at the same void ratio. Test A had a cell
pressure of 150 kPa, while in test B the cell pressure was 600 kPa (u=OkPa).
These stresses were held constant throughout the test. At failure, they had
maximum principal stress differences of 600 and 2550 kPa, respectively. You
are asked to:

a) Plot the Mohr circles for both tests at initial conditions and at failure.
b) Determine shear strength of this soil.
c) Determine the shear stress on the failure plane at failure for both tests?
d) Determine the orientation of the failure plane in each specimen (use
equations or graphical solution).
e) Determine the orientation of the major principal plane at failure.
f) Determine the orientation of the plane of maximum shear stress at failure
g) If these soil samples were tested in direct simple shear, would the soil
exhibit compression or dilation?

The Attempt at a Solution

IMG_0971[/url[/B]
] by Ryan Stewart, on Flickr

<< Fixed up image by Mentors >>

B. Determine the Shear Strength of the Soil

τ = (Ơ1 - Ơ3)/2 Sin(2θ)
=(2550)/2 * Sin (66.4 *2)
=935.5 KPa

C. Determine the shear stress

Sample B

θ = 45 + ∅/2
=45 + 42.8/2
θ=66.4

Cos∅ = Ơn/r
Ơn = Cos∅ x r
= Cos (180-2θ) x 1875
= 1273.95 KPA

τ = c + Ơn tan (2θ)
= 0 + 1273.95 x tan (2 x 66.4)
= 1375.7 KPA

This is as far as I've gotten just wanted to make sure I was on the correct path before continuing.

Thanks for any help!

 

[KataKoniK]

A. Plot the Mohr circles

Initial conditions:
Test A: (150, 0), (600, 0)
Test B: (600, 0), (2550, 0)

At failure:
Test A: (600, 600), (2550, 0)
Test B: (2550, 2550), (2550, 0)

B. Determine the shear strength

The shear strength of the soil can be calculated using the Mohr-Coulomb equation:
τ = c + Ơn tan ∅
Where:
τ = shear strength
c = cohesion
Ơn = normal stress on the failure plane
∅ = angle of internal friction

For Test A:
Ơn = (150 + 600)/2 = 375 kPa
∅ = tan^-1(600/600) = 45°
c = 0 (given in the question)

Therefore, the shear strength for Test A is:
τ = 0 + 375 x tan(45°) = 375 kPa

For Test B:
Ơn = (600 + 2550)/2 = 1575 kPa
∅ = tan^-1(2550/600) = 76.87°
c = 0 (given in the question)

Therefore, the shear strength for Test B is:
τ = 0 + 1575 x tan(76.87°) = 2550 kPa

C. Determine the shear stress on the failure plane at failure for both tests

For Test A:
The shear stress on the failure plane is equal to the maximum principal stress difference, which is given in the question as 600 kPa.

For Test B:
The shear stress on the failure plane is equal to the maximum principal stress difference, which is given in the question as 2550 kPa.

D. Determine the orientation of the failure plane in each specimen

The orientation of the failure plane can be determined using the Mohr's circle. The failure plane is perpendicular to the line connecting the center of the circle to the point where the Mohr circle intersects the x-axis.

For Test A:
The line connecting the center of the circle (150, 0) to the point where the circle intersects the x-axis (600, 600) has a slope of 1. Therefore, the angle between the failure plane and the x-axis is: