Question

The stresses at failure on the failure plane in a cohesionless soil mass were:
Shear stress = 4 kN/m2; normal stress = 10 kN/m2. Determine the resultant stress on the
failure plane, the angle of internal friction of the soil and the angle of inclination of the failure plane to the major principal plane.

Answer 1

Answer:

The correct answer for the given question is 21°48',55°54'.

Explanation:

The condition stated in the question is the same as that stated in point F

tanФ = $\frac{FN}{NO}$ = $\frac{FN}{FT}$ = $\frac{4}{10}$

Ф = 21°48'

Now for ΔDFC, external angel = sum of two internal angle

90 + ∅ = Ф + 90

∅ = Ф = 21°48'

i.e. friction angle of soil = 21°48'

The angle of failure plane inclination to the major primary plane is given as: 45 + ∅/2

= 45 + (21°48')/2 = 55°54'

Thus, the angle of internal friction (∅) and cohesion are two significant physical properties of soil that determine the angle of breakage, shearing strength, safety factor, and slope material stability.