--- title: "Time-settlement curve, 3 m clay" subject: "Soil Mechanics" chapter: 9 key: "semester_4/soil_mechanics:time-settlement-curve-3m-clay" source: https://www.bctengineer.com/q/semester_4/soil_mechanics/time-settlement-curve-3m-clay site: BCT Engineering --- # Time-settlement curve, 3 m clay Soil Mechanics · Chapter 9: Compressibility of Soil - Asked in: 2078 Poush - Video: [The Mechanics of Soil Consolidation and Time Settlement Curves](https://www.youtube.com/watch?v=9JOLJhn4DY4) — Geotech Guide - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/time-settlement-curve-3m-clay - Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-9 ## Question A structure built on a 3 m thick single drained clay layer settled 5 cm in 60 days after it was built. If this settlement corresponds to 20 percent average consolidation of the clay layer, plot the time settlement curve of the structure for a period of 3 years from the time it was built. ## Short answer Final settlement is 25 cm. From t=60 days at U=20%, cv = 47.1 cm²/day. At 3 years (1095 days) the time factor is 0.573, giving U≈80% and settlement ≈20 cm. The curve rises steeply at first and asymptotically approaches 25 cm. ## Answer **Given:** single-drained layer, $H_{dr} = 3$ m $= 300$ cm. Settlement 5 cm at $t = 60$ days corresponds to $U = 20\%$. **Step 1: Final settlement** $$ S_{final} = \frac{5}{0.20} = 25\ \text{cm} $$ **Step 2: Coefficient of consolidation.** For $U=20\%$, $T_v = \frac{\pi}{4}(0.20)^2 = 0.0314$. $$ c_v = \frac{T_v H_{dr}^2}{t} = \frac{0.0314 \times 300^2}{60} = 47.1\ \text{cm}^2/\text{day} $$ **Step 3: Time for various $U$.** $t = T_v H_{dr}^2 / c_v = T_v \times 1910.0$ days. Use $T_v = \frac{\pi}{4}U^2$ for $U<60\%$ and $T_v = 1.781 - 0.933\log_{10}(100-U\%)$ for $U>60\%$. Settlement $= U \times 25$ cm. | $U$ (%) | $T_v$ | Time (days) | Settlement (cm) | |---|---|---|---| | 10 | 0.008 | 15 | 2.5 | | 20 | 0.031 | 60 | 5.0 | | 30 | 0.071 | 135 | 7.5 | | 40 | 0.126 | 240 | 10.0 | | 50 | 0.196 | 375 | 12.5 | | 60 | 0.286 | 547 | 15.0 | | 70 | 0.403 | 769 | 17.5 | | 80 | 0.567 | 1083 | 20.0 | | 90 | 0.848 | 1620 | 22.5 | **Step 4: Settlement at 3 years** ($t = 1095$ days): $$ T_v = \frac{c_v t}{H_{dr}^2} = \frac{47.1 \times 1095}{300^2} = 0.573 \;\Rightarrow\; U \approx 80\%, \quad S \approx 20\ \text{cm} $$ **Plot:** draw time (days, 0 to 1095) on the horizontal axis and settlement (cm) downward on the vertical axis; join the points above with a smooth curve. The curve is steep at first and flattens towards 25 cm. ```text t (days) 0 200 400 600 800 1000 1095 S (cm) 0 | * 5 | * 10 | * 15 | * 20 | * * <- 3 yr 25 - - - - - - - - - - - - - - (final) ``` **Answer: the structure settles about 20 cm in 3 years; the curve approaches the final value of 25 cm.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the final settlement found from the observed data? A. S_final = 5 cm x 0.20 = 1 cm B. S_final = 5 cm / 0.20 = 25 cm C. S_final = 5 + 20 = 25 cm D. S_final = 5 / 60 cm Answer: B. 5 cm is 20% of the final settlement, so the final is 25 cm. (Key point: Final settlement) 2. How is c_v found from the 60-day observation? A. Tv = 0.20 and cv = Tv H^2/t B. Tv = (pi/4)(0.20)^2 = 0.0314 and cv = Tv H_dr^2 / t = 47.1 cm^2/day C. Tv = 0.848 for U = 20% and cv = Tv H_dr^2/t D. Tv = (pi/4)(0.20)^2 and cv = t/(Tv H_dr^2) Answer: B. U = 20% is below 60%, so Tv = (pi/4)U^2 = 0.0314, then cv = 0.0314 x 300^2/60. (Key point: Finding c_v) 3. Why is H_dr taken as the full 3 m? A. The layer is double drained B. Clay always drains over its full thickness C. The layer is single drained D. Because the structure is 3 m wide Answer: C. Only one face drains in a single-drained layer, so the drainage path equals the layer thickness. (Key point: Drainage path) 4. What does the time-settlement curve look like? A. A straight line reaching 25 cm at 3 years B. Flat at first then steepening C. Rising steeply and exceeding 25 cm D. Steep at first, flattening towards the 25 cm final value Answer: D. Settlement rate falls as excess pore pressure dissipates, so the curve flattens; at 3 years U is about 80%, about 20 cm. 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