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CU test NC clay 200/350/80, φ and φ'

Short answer

Total principal stresses are 200 and 550 kPa; effective are 120 and 470 kPa after subtracting pore pressure 80 kPa. For c=0, sin phi = 350/750 so phi = 27.8 degrees; sin phi' = 350/590 so phi' = 36.4 degrees. The failure plane is at 63.2 degrees to the horizontal.

A consolidated undrained triaxial test was performed on a normally consolidated saturated clay. During the consolidation stage, a cell pressure of 200 kN/m² was applied and drainage was allowed. In the shearing stage, a deviatoric stress of 350 kN/m² was applied in the vertical direction and a pore water pressure of 80 kN/m² was measured. Answer the following: i) Draw Mohr's circle for total and effective stresses. ii) Find the value of internal friction angle in total and effective stress conditions. Take the value of cohesion equal to zero for normally consolidated soil. iii) Determine the direction of failure plane that might occur within the specimen.

Data: σ3=200\sigma_3 = 200, σd=350\sigma_d = 350, so σ1=550\sigma_1 = 550 kN/m²; u=80u = 80 kN/m²; c=c′=0c = c' = 0.

(i) Mohr circles

  • Total stress: σ3=200\sigma_3 = 200, σ1=550\sigma_1 = 550; centre 375, radius 175.
  • Effective stress: σ3′=200−80=120\sigma_3' = 200-80 = 120, σ1′=550−80=470\sigma_1' = 550-80 = 470; centre 295, radius 175.

The effective circle has the same size but is shifted left by u=80u = 80. Draw both, with a line through the origin tangent to each:

 tau
  |       /  /
  |     /  /   . total circle
  |   /  / .-'''-.
  |  / /.'  .-'''-\.
  | //  eff.'      '
  +-+----+--+-------+-- sigma
   0  120 200  470  550

(ii) Friction angles

sin⁡ϕ=σ1−σ3σ1+σ3=350750=0.467  ⇒  ϕ=27.8∘\sin\phi = \frac{\sigma_1-\sigma_3}{\sigma_1+\sigma_3} = \frac{350}{750} = 0.467 \;\Rightarrow\; \phi = 27.8^\circ sin⁡ϕ′=σ1′−σ3′σ1′+σ3′=350590=0.593  ⇒  ϕ′=36.4∘\sin\phi' = \frac{\sigma_1'-\sigma_3'}{\sigma_1'+\sigma_3'} = \frac{350}{590} = 0.593 \;\Rightarrow\; \phi' = 36.4^\circ

(iii) Direction of the failure plane

The failure plane is inclined at θ=45∘+ϕ′/2\theta = 45^\circ + \phi'/2 to the major principal plane (the horizontal plane here, since the vertical stress is the major principal stress):

θ=45∘+36.4∘2=63.2∘ to the horizontal\theta = 45^\circ + \frac{36.4^\circ}{2} = 63.2^\circ \text{ to the horizontal}

It makes 26.8∘26.8^\circ with the vertical axis of the specimen. (Using the total stress angle, 45∘+27.8∘/2=58.9∘45^\circ + 27.8^\circ/2 = 58.9^\circ.)

Answer: ϕ=27.8∘\phi = 27.8^\circ, ϕ′=36.4∘\phi' = 36.4^\circ; failure plane at about 63.2∘63.2^\circ to the horizontal.

✓ Answer checked by Claude Opus 5.5 (high effort)

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