Answer:
=>> 167.3 kpa.
=>> 60° from horizontal face.
Explanation:
So, we are given the following data or parameters or information which is going to assist us in solving this kind of question;
=>> "A soil element is subjected to a minor principle stress of 50 kPa on a plane rotated 20 ° counterclockwise from vertical. "
=>>"If the deviator stress is 120 kPa and the shear strength parameters are a friction angle of 30° and a cohesion of 5 kPa."
The orientation of this plane with respect to the major principle stress plane = 50 tan^2 (45 + 30/2) + 10 tan ( 45 + 30/2).
magnitude of the stresses on the failure plane = 167.3 kpa.
The orientation of this plane with respect to the major principle stress plane => x = 60 cos 60° = 30kpa.
y = 60 sin 60° = 30√3 = sheer stress.
the orientation of this plane with respect to the major principle stress plane.
Theta = 45 + 15 = 60°.