--- title: "Consolidation test, 90% time, k, Cc" subject: "Soil Mechanics" chapter: 9 key: "semester_4/soil_mechanics:consolidation-test-50-90-percent-time" source: https://www.bctengineer.com/q/semester_4/soil_mechanics/consolidation-test-50-90-percent-time site: BCT Engineering --- # Consolidation test, 90% time, k, Cc Soil Mechanics · Chapter 9: Compressibility of Soil - Asked in: 2079 Jestha, 6 marks - Video: [How to Calculate Coefficient of Consolidation (Casagrande & Taylor Methods) Root vs. Log Time Method](https://www.youtube.com/watch?v=PkhKfvYBj0Y) — Soil Mechanics and Engineering Geology - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/consolidation-test-50-90-percent-time - Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-9 ## Question In a one dimensional consolidation test the time required for 50% consolidation has been measured at 154 seconds (through the observation and measurement of pore water pressure). The settlement of the sample at the end of the test was 2.5 mm. $\sigma_0'$ = 60 kPa, $\sigma_1'$ = 120 kPa, $e_0$ = 0.65, $H_0$ = 20 mm. Take $T_v$ = 0.197 and 0.848 for 50% and 90% consolidation respectively. Determine: (i) the time required for 90% consolidation, (ii) the coefficient of permeability in m/s, (iii) the compression index. ## Short answer Time for 90% consolidation t90 = t50 × (T90/T50) = 154 × (0.848/0.197) = 663 s (≈11 min). Coefficient of permeability k = cv mv γw = 2.30×10⁻⁹ m/s. Compression index Cc = Δe / log10(σ1'/σ0') = 0.685. ## Answer **Assumptions:** the specimen is drained at top and bottom (two porous stones), so $H_{dr}$ is half the average specimen thickness during the test. Average thickness $= 20 - 2.5/2 = 18.75$ mm, so $H_{dr} = 9.375$ mm. $\gamma_w = 9.81$ kN/m³. #### (i) Time for 90% consolidation Since $t \propto T_v$ for the same specimen: $$ t_{90} = t_{50}\,\frac{T_{90}}{T_{50}} = 154 \times \frac{0.848}{0.197} = 663\ \text{s} \approx 11\ \text{min} $$ #### (ii) Coefficient of permeability $$ c_v = \frac{T_{50}H_{dr}^2}{t_{50}} = \frac{0.197 (9.375)^2}{154} = 0.1124\ \text{mm}^2/\text{s} = 1.124\times10^{-7}\ \text{m}^2/\text{s} $$ Volume compressibility, with strain $= 2.5/20 = 0.125$ and $\Delta\sigma' = 120-60 = 60$ kPa: $$ m_v = \frac{0.125}{60} = 2.083\times10^{-3}\ \text{m}^2/\text{kN} $$ $$ k = c_v m_v \gamma_w = (1.124\times10^{-7})(2.083\times10^{-3})(9.81) = 2.30\times10^{-9}\ \text{m/s} $$ #### (iii) Compression index Change in void ratio: $$ \Delta e = \frac{\Delta H}{H_0}(1+e_0) = 0.125 \times 1.65 = 0.206 $$ $$ C_c = \frac{\Delta e}{\log_{10}(\sigma_1'/\sigma_0')} = \frac{0.206}{\log_{10}(120/60)} = \frac{0.206}{0.301} = 0.685 $$ **Answer: (i) $t_{90} \approx 663$ s (11 min); (ii) $k \approx 2.3\times10^{-9}$ m/s; (iii) $C_c \approx 0.685$.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is t90 found from t50 for the same specimen? A. t90 = t50 x (0.9/0.5) = 277 s B. t90 = t50 x (T90/T50) = 154 x 0.848/0.197, about 663 s C. t90 = t50 x (T90/T50)^2 = about 2850 s D. t90 = t50 + (0.848 - 0.197) = 154.65 s Answer: B. Since t = Tv H^2/cv with H and cv fixed, t is proportional to Tv, giving about 663 s. (Key point: Time scaling with Tv) 2. The specimen is drained at top and bottom. What is H_dr used for c_v? A. The average thickness, 18.75 mm B. The initial thickness, 20 mm C. Half the average thickness, 9.375 mm D. Half the initial thickness, 10 mm Answer: C. With two porous stones the water travels half the thickness; the average thickness during the test is 20 - 2.5/2 = 18.75 mm, so H_dr = 9.375 mm. (Key point: Drainage path in the oedometer) 3. How is k obtained from the test data? A. k = c_v / (m_v gamma_w) B. k = c_v gamma_w / m_v C. k = c_v m_v gamma_w D. k = m_v gamma_w / c_v Answer: C. Because c_v = k/(m_v gamma_w), k = c_v m_v gamma_w, about 2.3e-9 m/s. (Key point: Permeability from consolidation) 4. How is C_c computed here? A. Delta e = delta H/H0 divided by log10(120-60) B. Delta e = (delta H/H0)(1+e0) divided by (120 - 60) C. Delta e = delta H/(1+e0) divided by log10(120/60) D. Delta e = (delta H/H0)(1+e0) divided by log10(120/60) Answer: D. C_c is the change in void ratio per log cycle of stress: 0.206/0.301 = 0.685. 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