Question

A point within an embankment slope of sand has a total vertical normal stress (y) of 100 kPa, a total horizontal normal stress (x) of 60 kPa, a horizontal shear stress (xy) of 20 kPa, and a pore water pressure of uo. The sand has a peak effective friction angle of 30 degrees. (a) Determine the value of uo that would cause the sand to fail in drained shearing. Assume the total normal stresses and shear stress remain constant. (b) For the above value of uo, determine the major and minor principle effective stresses. (c) For the above value of uo, determine the orientation of the major principle stress (in degrees from vertical, clockwise or counterclockwise)

Answer 1

a) We have

Vertical normal stress = = 100 kPa

Horizontal normal stress = = 60 kPa

Shear stress = = 20 kPa

Major principal stress is given as

= (100+60)/2 + [{(100-60)/2}² + 20²)^1/2

= 108.28 kPa

Minor principal stress is given as

= (100+60)/2 - [{(100-60)/2}² + 20²)^1/2

  = 51.71 kPa

Effective major pricipal stress is

Effective minor principal stress is

Mohr coloumb failure criterion of Sand of is given as

(108.28 - u) = (51.71 - u)xtan²(45 + 30/2)

108.28 - u = (51.71 - u)x3

3u - u = 3x51.71 - 108.28

=> u = 23. 43 kPa

b) Therefore

= 108.28 - 23.43 = 84.45 kPa

= 51.71 - 23.43 = 28.28 kPa

c) Orientation of major principal stress with horizontal is given by

= 2x20/(100 - 60)

=>    = 22.5 degrees

With vertical