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CU test: σ3 200, σ1 550, u 80

Short answer

Effective stresses are sigma3' = 120 kPa and sigma1' = 470 kPa. With c' = 0, sin phi' = 350/590 = 0.593, so phi' = 36.4 degrees. The failure plane is inclined at 45 + phi'/2 = 63.2 degrees to the horizontal, or 26.8 degrees to the specimen axis.

A consolidated undrained triaxial test was performed on a normally consolidated saturated clay and the cell pressure, σ3\sigma_3 = 200 kN/m², axial stress, σ1\sigma_1 = 550 kN/m² and pore water pressure, uwu_w = 80 kN/m² were measured. Answer the following: i) Plot the Mohr circle of stresses in regard to total stress. ii) Plot the Mohr circle of stresses in regard to effective stress. iii) Assume the condition of normal consolidation and c' = 0. Then obtain the value of ϕ′\phi'. iv) If Mohr-Coulomb's failure criterion is assumed to be valid, then determine the direction of failure plane that might occur within the specimen.

Data: σ3=200\sigma_3 = 200, σ1=550\sigma_1 = 550, u=80u = 80 kN/m². Deviator stress =350= 350 kN/m². Effective stresses: σ3′=200−80=120\sigma_3' = 200-80 = 120, σ1′=550−80=470\sigma_1' = 550-80 = 470 kN/m².

i) Mohr circle, total stress

Centre =200+5502=375= \frac{200+550}{2} = 375, radius =550−2002=175= \frac{550-200}{2} = 175. The circle cuts the σ\sigma-axis at 200 and 550.

ii) Mohr circle, effective stress

Centre =120+4702=295= \frac{120+470}{2} = 295, radius =175= 175. The circle cuts the σ\sigma-axis at 120 and 470. It is the total stress circle shifted left by u=80u = 80.

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

iii) Effective friction angle

For c′=0c' = 0:

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

(For information, the total stress angle is sin⁡ϕ=350/750\sin\phi = 350/750, ϕ=27.8∘\phi = 27.8^\circ.)

iv) Direction of the failure plane

The failure plane is inclined to the major principal plane (horizontal) at

θ=45∘+ϕ′2=45∘+18.2∘=63.2∘\theta = 45^\circ + \frac{\phi'}{2} = 45^\circ + 18.2^\circ = 63.2^\circ

So the plane is at about 63.2∘63.2^\circ to the horizontal, which is 26.8∘26.8^\circ to the axis of the specimen.

Answer: ϕ′≈36.4∘\phi' \approx 36.4^\circ; failure plane at about 63∘63^\circ to the horizontal.

✓ Answer checked by Claude Opus 5.5 (high effort)

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