--- title: "Primary settlement, sand over clay, 110 kPa" subject: "Soil Mechanics" chapter: 9 key: "semester_4/soil_mechanics:primary-settlement-sand-clay-110kpa" source: https://www.bctengineer.com/q/semester_4/soil_mechanics/primary-settlement-sand-clay-110kpa site: BCT Engineering --- # Primary settlement, sand over clay, 110 kPa Soil Mechanics · Chapter 9: Compressibility of Soil - Asked in: 2079 Asoj, 5 marks - Video: [Consolidation_Primary Consolidation Settlement](https://www.youtube.com/watch?v=vp_p_jLBGTM) — Dr. Maria Cecilia Marcos - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/primary-settlement-sand-clay-110kpa - Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-9 ## Question The soil profile of the ground shows a sand layer (3.5 m thick, void ratio = 0.98, specific gravity $G_s$ = 2.62) lying above the clay layer (3.5 m thick, void ratio = 0.62, specific gravity $G_s$ = 2.7, $w_L$ = 50%). Ground water table lies 1.5 m below the ground surface. Assume an impervious layer lies below the clay layer. If a uniformly distributed load of 110 kPa is applied on the ground surface of this soil, find the primary settlement of the clay layer. For compressibility index, use $C_c$ = 0.0099 (LL − 10). ## Short answer Determine effective stress at mid-clay (5.25 m depth) using submerged unit weights, giving 53.54 kPa. With Cc = 0.396, apply settlement formula Sc = (Cc H/(1+e0)) log10((σ0'+Δσ)/σ0'). Primary settlement ≈ 0.415 m (415 mm). ## Answer **Assumptions:** no water content is given, so the sand above the water table (top 1.5 m) is taken as dry and the sand below it as saturated. The clay is normally consolidated, $\gamma_w = 9.81\ \text{kN/m}^3$, and settlement is computed at the middle of the clay layer. **Step 1: Unit weights** from $\gamma_d = \dfrac{G_s\gamma_w}{1+e}$ and $\gamma_{sat} = \dfrac{(G_s+e)\gamma_w}{1+e}$ $$ \begin{aligned} \gamma_{d,sand} &= \frac{2.62(9.81)}{1.98} = 12.98\ \text{kN/m}^3 \\ \gamma_{sat,sand} &= \frac{(2.62+0.98)(9.81)}{1.98} = 17.84\ \text{kN/m}^3 \\ \gamma_{sat,clay} &= \frac{(2.70+0.62)(9.81)}{1.62} = 20.10\ \text{kN/m}^3 \end{aligned} $$ **Step 2: Compression index** $$ C_c = 0.0099(w_L-10) = 0.0099(50-10) = 0.396 $$ **Step 3: Effective stress at mid-clay** (depth $3.5+1.75 = 5.25$ m) $$ \begin{aligned} \sigma_0' &= 12.98(1.5) + (17.84-9.81)(2.0) + (20.10-9.81)(1.75) \\ &= 19.47 + 16.06 + 18.01 = 53.54\ \text{kPa} \end{aligned} $$ **Step 4: Settlement**, with $\Delta\sigma = 110$ kPa $$ \begin{aligned} S_c &= \frac{C_c H}{1+e_0}\log_{10}\frac{\sigma_0'+\Delta\sigma}{\sigma_0'} \\ &= \frac{0.396(3.5)}{1.62}\log_{10}\frac{53.54+110}{53.54} \\ &= 0.8556 \times 0.4850 = 0.415\ \text{m} \end{aligned} $$ **Answer: primary settlement of the clay layer $\approx 0.415\ \text{m} \approx 415\ \text{mm}$.** (If the sand above the water table were taken as saturated by capillarity, $\sigma_0' = 60.8$ kPa and the settlement would be about 384 mm.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which formula gives the saturated unit weight used for the sand and clay layers? A. gamma_sat = Gs gamma_w / (1 + e) B. gamma_sat = (Gs + e) gamma_w / e C. gamma_sat = (Gs + e) gamma_w / (1 + e) D. gamma_sat = (Gs - 1) gamma_w / (1 + e) Answer: C. Saturated soil has solids plus water-filled voids, so weight per unit volume is (Gs + e) gamma_w / (1 + e). (Key point: Saturated unit weight) 2. With C_c = 0.0099 (LL - 10) and w_L = 50%, what is C_c? A. 0.495 B. 0.396 C. 0.0099 D. 0.594 Answer: B. C_c = 0.0099 x (50 - 10) = 0.396. (Key point: Compression index) 3. How is the initial effective stress at the middle of the clay layer found? A. Total stress at 5.25 m using gamma_sat for all layers, ignoring water B. Stress at the top of the clay (3.5 m) only C. Sum of gamma x thickness for each layer, using gamma_sat - gamma_w below the water table, up to 5.25 m depth D. Stress using gamma_sat - gamma_w for every layer including the sand above the water table Answer: C. Above the water table the sand uses its saturated weight (as assumed), below it the submerged weight, and the clay is taken at its mid-depth 5.25 m. (Key point: Effective stress at mid-clay) 4. Which expression gives the primary settlement of the normally consolidated clay? A. S = [C_c H/(1+e0)] log10[delta sigma/sigma0'] B. S = [C_c H/e0] log10[(sigma0' + delta sigma)/sigma0'] C. S = [C_c H] (sigma0' + delta sigma)/sigma0' D. S = [C_c H/(1+e0)] log10[(sigma0' + delta sigma)/sigma0'] Answer: D. For normally consolidated clay S = Cc H/(1+e0) times log of final to initial effective stress. 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