--- title: "Cohesionless sample, deviator at 200 kPa" subject: "Soil Mechanics" chapter: 10 key: "semester_4/soil_mechanics:cohesionless-confining-100-200-deviator" source: https://www.bctengineer.com/q/semester_4/soil_mechanics/cohesionless-confining-100-200-deviator site: BCT Engineering --- # Cohesionless sample, deviator at 200 kPa Soil Mechanics · Chapter 10: Shear Strength of Soil - Asked in: 2078 Baisakh, 4 marks - Video: [Numerical on Mohrs Circle (shear strength of soil)](https://www.youtube.com/watch?v=guSU0Ddq-l0) — civil engineering - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/cohesionless-confining-100-200-deviator - Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-10 ## Question At a confining pressure of 100 kPa and deviator stress of 200 kPa, a cohesionless soil sample failed in a triaxial test. Determine the deviator stress if a sample of the same soil failed under a confining pressure of 200 kPa. Also, draw Mohr circles of stress along with the Mohr-Coulomb failure envelope. ## Short answer From the first test, sin phi = 200/400 = 0.5, so phi = 30 degrees. For sigma3 = 200 kPa, sigma1 = 200 tan^2(60) = 600 kPa. Therefore the deviator stress is 600 - 200 = 400 kPa; the Mohr circle from 200 to 600 is tangent to the envelope through the origin. ## Answer For a cohesionless soil ($c=0$), the failure circle touches a line through the origin. **Test 1:** $\sigma_3 = 100$ kPa, $\sigma_d = 200$ kPa, so $\sigma_1 = 300$ kPa. $$ \sin\phi = \frac{\sigma_1-\sigma_3}{\sigma_1+\sigma_3} = \frac{200}{400} = 0.5 \;\Rightarrow\; \phi = 30^\circ $$ **Test 2:** $\sigma_3 = 200$ kPa. $$ \sigma_1 = \sigma_3\tan^2\left(45^\circ+\frac{30^\circ}{2}\right) = 200 \times 3 = 600\ \text{kPa} $$ $$ \sigma_d = 600 - 200 = 400\ \text{kPa} $$ (The deviator stress is proportional to $\sigma_3$ for cohesionless soil, so doubling $\sigma_3$ doubles $\sigma_d$.) **Mohr circles and envelope** ```text tau / tau = sigma tan30 | / | / ..-'''-. circle 2: 200 to 600 | / .-'.-'''-. \ | / .' / c1 \ | circle 1: 100 to 300 +-+--+----+-+-----+---- sigma 0 100 200 300 600 ``` **Answer: $\phi = 30^\circ$; deviator stress at failure for $\sigma_3 = 200$ kPa is 400 kPa.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does the envelope pass through the origin? A. phi = 0 B. Pore pressure is zero C. The soil is cohesionless, so c = 0 D. The soil is saturated clay Answer: C. No cohesion means no intercept. (Key point: Cohesion) 2. What is phi from test 1? A. 26.6° B. 30°, since sin phi = 200/400 = 0.5 C. 45° D. 19.5° Answer: B. sigma_1 = 300, sigma_3 = 100, sin phi = 200/400. (Key point: Phi) 3. What is sigma_1 for sigma_3 = 200 kPa? A. 600 kPa from 200 x tan^2 60° = 200 x 3 B. 400 kPa C. 500 kPa D. 300 kPa Answer: A. tan^2(45° + 15°) = 3. (Key point: Second test) 4. How does deviator stress vary with sigma_3 for c = 0? A. It is independent of sigma_3 B. It halves when sigma_3 doubles C. It increases by a fixed 100 kPa D. It is proportional to sigma_3, so doubling sigma_3 doubles sigma_d to 400 kPa Answer: D. sigma_d = sigma_3(N - 1) with N constant. 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