--- title: "USCS classification from PSD curves" subject: "Soil Mechanics" chapter: 3 key: "semester_4/soil_mechanics:uscs-classify-psd-curves-a-b" source: https://www.bctengineer.com/q/semester_4/soil_mechanics/uscs-classify-psd-curves-a-b site: BCT Engineering --- # USCS classification from PSD curves Soil Mechanics · Chapter 3: Soil Identifications and Classification - Asked in: 2075 Bhadra, 5 marks - Video: [Chapter 2 Example 1 - Particle size distribution curve](https://www.youtube.com/watch?v=Oy9J1GCNYFo) — uSeeGeo - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/soil_mechanics/uscs-classify-psd-curves-a-b - Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-3 ## Question Particle size distribution curves for two types of soil, Soil A and Soil B, are shown in the figure [Figure: percentage passing versus particle size (mm), log scale; labelled points on the curves: 99% and 55% for soil A, 63% and 13% for soil B, with the 0.075 mm and 4.75 mm sizes marked on the axis]. Water contents measured at the boundaries between the liquid state-plastic state and plastic state-semi solid for soil A are 45% and 15% respectively. Similarly, for Soil B, they are 25% and 10% respectively. Classify these soils based on the Unified Soil Classification System. Draw the plasticity chart if required. ## Short answer Soil A: CL, sandy lean clay; Soil B: SC, clayey sand with gravel. A has 55% fines (fine-grained) and plots above A-line with wL 45; B has 13% fines with sand dominating and Ip=15 above A-line. ## Answer #### Reading the curves For each soil, the larger percentage is taken as the percent passing the 4.75 mm sieve and the smaller as the percent passing the 0.075 mm sieve. | Soil | Passing 4.75 mm | Passing 0.075 mm | $w_L$ | $w_P$ | $I_p$ | |---|---|---|---|---|---| | A | 99% | 55% | 45 | 15 | 30 | | B | 63% | 13% | 25 | 10 | 15 | #### Soil A Fines $= 55\% > 50\%$, so it is **fine-grained**. $$ I_{p,A\text{-line}} = 0.73(45 - 20) = 18.25\% $$ $I_p = 30 > 18.25$: above the A-line (clay); $w_L = 45 < 50$ (low plasticity). **Soil A: CL, sandy lean clay.** #### Soil B Fines $= 13\%$, which is more than 12% but less than 50%, so it is **coarse-grained with fines**. - Gravel $= 100 - 63 = 37\%$ - Sand $= 63 - 13 = 50\%$ Sand is more than half of the coarse fraction ($50/87 = 57\%$), so it is **S**. $$ I_{p,A\text{-line}} = 0.73(25 - 20) = 3.65\% $$ $I_p = 15 > 7$ and above the A-line: clayey fines. **Soil B: SC, clayey sand with gravel.** #### Plasticity chart ```text Ip (%) 35 | | / A-line 30 | A | / 25 | | / 20 | /| 15 | B / | 10 | / | 5 | / | 0 | / | +------------------------------ wL (%) 0 20 25 50 ``` Point A (45, 30) and point B (25, 15) both lie above the A-line. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Soil A has 55% passing 75 micron; so it is A. fine-grained B. coarse-grained C. dual symbol D. organic Answer: A. > 50%. (Key point: Soil A type) 2. Soil A (wL 45, Ip 30) is A. CL B. CH C. ML D. MH Answer: A. Above A-line (18.25), wL < 50. (Key point: Soil A) 3. Soil B: 63% passes 4.75 and 13% passes 75 micron; sand is A. 37%, so G B. 50%, 57% of coarse fraction, so S C. 13%, so S D. 63%, so G Answer: B. 50/87. (Key point: Soil B sand) 4. Soil B fines 13% > 12%, Ip=15 > 7 above A-line gives A. SM B. SW C. SC D. GC Answer: C. Clayey sand. 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