--- title: "CU test on NC clay, three samples" subject: "Soil Mechanics" chapter: 10 key: "semester_4/soil_mechanics:cu-test-nc-clay-envelopes" source: https://www.bctengineer.com/q/semester_4/soil_mechanics/cu-test-nc-clay-envelopes site: BCT Engineering --- # CU test on NC clay, three samples Soil Mechanics · Chapter 10: Shear Strength of Soil - Asked in: 2078 Poush, 5 marks - Video: [CU (CIU) Triaxial Test: Data Analysis Part 1](https://www.youtube.com/watch?v=dKy6z3GmBuI) — GeoTechSoil - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/cu-test-nc-clay-envelopes - Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-10 ## Question The following results were obtained from a consolidated-undrained test on normally consolidated clay. Plot the strength envelope in terms of total stress and effective stress and determine the strength parameters. | Sample No. | Cell pressure (kN/m²) | Deviator Stress (kN/m²) | Pore Water Pressure (kN/m²) | |---|---|---|---| | 1 | 200 | 244 | 55 | | 2 | 300 | 314 | 107 | | 3 | 400 | 384 | 159 | ## Short answer Total stress: using σ1 = Nσ3 + 2c√N, N=1.70 gives φcu = 15°, ccu = 40 kN/m². Effective stress: N′=2.458 gives φ′ = 25°, c′ = 10 kN/m². For ideal normally consolidated clay the cohesion intercepts would be zero. ## Answer **Step 1: Principal stresses.** $\sigma_1 = \sigma_3 + \sigma_d$, effective: $\sigma' = \sigma - u$. | Sample | $\sigma_3$ | $\sigma_1$ | $u$ | $\sigma_3'$ | $\sigma_1'$ | |---|---|---|---|---|---| | 1 | 200 | 444 | 55 | 145 | 389 | | 2 | 300 | 614 | 107 | 193 | 507 | | 3 | 400 | 784 | 159 | 241 | 625 | **Step 2: Plot.** Draw three total stress circles (diameters 200 to 444, 300 to 614, 400 to 784) and three effective stress circles (145 to 389, 193 to 507, 241 to 625). Draw a common tangent for each set. ```text tau | _.-'''-._ total-stress envelope | _.-' _.-''-. `-. (phi_cu, c_cu) | / .-' effective \ envelope is steeper +-+--+----+-----+----+---+---- sigma 145 200 241 389 444 ... ``` **Step 3: Parameters.** Both sets of points fit straight lines of the form $\sigma_1 = N\sigma_3 + 2c\sqrt{N}$. *Total stress:* slope $N = \dfrac{614-444}{300-200} = 1.70$, intercept $= 104$ kPa. $$ \phi_{cu} = \sin^{-1}\frac{N-1}{N+1} = \sin^{-1}\frac{0.70}{2.70} = 15.0^\circ, \qquad c_{cu} = \frac{104}{2\sqrt{1.70}} = 39.9\ \text{kPa} $$ *Effective stress:* slope $N' = \dfrac{507-389}{193-145} = 2.458$, intercept $= 32.5$ kPa. $$ \phi' = \sin^{-1}\frac{1.458}{3.458} = 24.9^\circ, \qquad c' = \frac{32.5}{2\sqrt{2.458}} = 10.4\ \text{kPa} $$ **Answer: total stress $c_{cu} \approx 40$ kN/m², $\phi_{cu} \approx 15^\circ$; effective stress $c' \approx 10$ kN/m², $\phi' \approx 25^\circ$.** (For an ideal normally consolidated clay the intercepts would be close to zero; the small cohesion shown here comes from the given data.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How are effective stresses computed from the data? A. sigma' = sigma - u for both sigma_3 and sigma_1 B. sigma' = sigma + u C. sigma' = sigma / u D. sigma' = sigma_d - u Answer: A. Sample 1: sigma_3' = 200 - 55 = 145 and sigma_1' = 444 - 55 = 389. (Key point: Effective stress) 2. What relation do the failure points satisfy? A. sigma_1 = sigma_3 + 2c B. sigma_1 = N sigma_3 - 2c N C. sigma_1 = c + sigma_3 tan phi D. sigma_1 = N sigma_3 + 2c sqrt(N), N = tan^2(45° + phi/2) Answer: D. The slope N and intercept 2c sqrt(N) of the line give phi and c. (Key point: Line form) 3. What is phi_cu from slope N = 1.70? A. phi = 24.9° B. phi = 45° C. sin phi = (N-1)/(N+1) = 0.70/2.70, so phi_cu = 15.0° D. phi = 36.4° Answer: C. phi = sin^-1[(1.70 - 1)/(1.70 + 1)] = 15.0°. (Key point: Total stress phi) 4. How do effective parameters compare with total ones? A. phi' is smaller than phi_cu B. phi' (about 25°) is larger than phi_cu (15°), with an envelope steeper than the total one C. They are equal D. phi' is zero Answer: B. Pore pressure shifts circles left; the effective envelope is steeper, giving phi' = 24.9° and c' about 10 kPa. 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