--- title: "CU test NC clay 200/350/80, φ and φ'" subject: "Soil Mechanics" chapter: 10 key: "semester_4/soil_mechanics:cu-nc-clay-200-350-80" source: https://www.bctengineer.com/q/semester_4/soil_mechanics/cu-nc-clay-200-350-80 site: BCT Engineering --- # CU test NC clay 200/350/80, φ and φ' Soil Mechanics · Chapter 10: Shear Strength of Soil - Asked in: 2076 Baisakh, 4 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/cu-nc-clay-200-350-80 - Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-10 ## Question 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. ## 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. ## Answer **Data:** $\sigma_3 = 200$, $\sigma_d = 350$, so $\sigma_1 = 550$ kN/m²; $u = 80$ kN/m²; $c = c' = 0$. #### (i) Mohr circles - Total stress: $\sigma_3 = 200$, $\sigma_1 = 550$; centre 375, radius 175. - Effective stress: $\sigma_3' = 200-80 = 120$, $\sigma_1' = 550-80 = 470$; centre 295, radius 175. The effective circle has the same size but is shifted left by $u = 80$. Draw both, with a line through the origin tangent to each: ```text tau | / / | / / . total circle | / / .-'''-. | / /.' .-'''-\. | // eff.' ' +-+----+--+-------+-- sigma 0 120 200 470 550 ``` #### (ii) Friction angles $$ \sin\phi = \frac{\sigma_1-\sigma_3}{\sigma_1+\sigma_3} = \frac{350}{750} = 0.467 \;\Rightarrow\; \phi = 27.8^\circ $$ $$ \sin\phi' = \frac{\sigma_1'-\sigma_3'}{\sigma_1'+\sigma_3'} = \frac{350}{590} = 0.593 \;\Rightarrow\; \phi' = 36.4^\circ $$ #### (iii) Direction of the failure plane The failure plane is inclined at $\theta = 45^\circ + \phi'/2$ to the major principal plane (the horizontal plane here, since the vertical stress is the major principal stress): $$ \theta = 45^\circ + \frac{36.4^\circ}{2} = 63.2^\circ \text{ to the horizontal} $$ It makes $26.8^\circ$ with the vertical axis of the specimen. (Using the total stress angle, $45^\circ + 27.8^\circ/2 = 58.9^\circ$.) **Answer: $\phi = 27.8^\circ$, $\phi' = 36.4^\circ$; failure plane at about $63.2^\circ$ to the horizontal.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are sigma_1 and the effective principal stresses? A. sigma_1 = 550; sigma_3' = 120 and sigma_1' = 470 B. sigma_1 = 350; sigma_3' = 280 and sigma_1' = 630 C. sigma_1 = 550; sigma_3' = 280 and sigma_1' = 630 D. sigma_1 = 350; sigma_3' = 120 and sigma_1' = 470 Answer: A. sigma_1 = 200 + 350; subtract u = 80 from each. (Key point: Principal stresses) 2. How does the effective circle compare with the total circle? A. Larger radius B. Shifted right by u C. Same centre but smaller radius D. Same radius (175) but shifted left by u = 80 Answer: D. Pore pressure reduces both principal stresses equally, so the diameter is unchanged. (Key point: Effective circle) 3. What are phi and phi'? A. phi = 36.4° and phi' = 27.8° B. both 27.8° C. phi = 27.8° and phi' = 36.4° D. both 36.4° Answer: C. sin phi = 350/750 = 0.467 and sin phi' = 350/590 = 0.593. (Key point: Friction angles) 4. What is the failure plane direction? A. 45° - phi'/2 = 26.8° to the horizontal B. 45° + phi'/2 = 63.2° to the horizontal C. 45° to the horizontal D. 63.2° to the vertical Answer: B. The failure plane makes 45° + phi'/2 with the major principal plane (horizontal). 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