--- title: "Compaction test, RC and saturation" subject: "Soil Mechanics" chapter: 5 key: "semester_4/soil_mechanics:compaction-test-curve-saturation-degree" source: https://www.bctengineer.com/q/semester_4/soil_mechanics/compaction-test-curve-saturation-degree site: BCT Engineering --- # Compaction test, RC and saturation Soil Mechanics · Chapter 5: Soil Compaction - Asked in: 2073 Magh, 6 marks - Video: [Maximum Dry Unit Weight and Optimum Moisture Content |Compaction Test Calculations| Compaction Curve](https://www.youtube.com/watch?v=culCiOfbScs) — Master 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/compaction-test-curve-saturation-degree - Whole chapter: https://www.bctengineer.com/md/semester_4/soil_mechanics/chapter-5 ## Question The following results were obtained from a standard compaction test. | Test No. | 1 | 2 | 3 | 4 | 5 | 6 | |---|---|---|---|---|---|---| | Water content (%) | 11.0 | 12.1 | 12.8 | 13.6 | 14.6 | 16.3 | | Mass of compacted soil (gm) | 1920.5 | 2051.5 | 2138.5 | 2147.0 | 2120.0 | 2081.5 | The specific gravity of solids is 2.7 and the volume of the compaction mould is 1000 cm³. A field compacted soil sample showed water content of 35% and unit weight of 2.318 Mg/m³. i) Draw the compaction curve and determine the maximum dry unit weight and OMC. ii) Find the relative compaction (RC). iii) Find the degree of saturation at the maximum dry unit weight. ## Short answer Compute each dry density from bulk density and water content, plot gamma_d versus w; the peak gives gamma_dmax about 1.90 Mg/m3 at OMC about 13.1%. Field dry density is 2.318/(1.35)=1.717 Mg/m3, so RC is 90.4%. Degree of saturation at maximum dry density is about 84% (e=0.421). ## Answer The dry density of each test is found from the bulk density in the 1000 cm³ mould, $\gamma_d = \dfrac{\gamma_{bulk}}{1+w}$ (here $\gamma$ is used for density in Mg/m³). #### Computation of dry density $\gamma_{bulk} = \dfrac{\text{mass (g)}}{1000\ \text{cm}^3}$ (in Mg/m³ = g/cm³) | Test | w (%) | Mass (g) | $\gamma_{bulk}$ (Mg/m³) | $\gamma_d$ (Mg/m³) | |---|---|---|---|---| | 1 | 11.0 | 1920.5 | 1.9205 | 1.730 | | 2 | 12.1 | 2051.5 | 2.0515 | 1.830 | | 3 | 12.8 | 2138.5 | 2.1385 | 1.896 | | 4 | 13.6 | 2147.0 | 2.1470 | 1.890 | | 5 | 14.6 | 2120.0 | 2.1200 | 1.850 | | 6 | 16.3 | 2081.5 | 2.0815 | 1.790 | #### i) Compaction curve, $\gamma_{d,max}$ and OMC Plot $\gamma_d$ (y-axis) against $w$ (x-axis) and draw a smooth curve through the six points. The dry density rises to a peak between 12.8% and 13.6% and then falls. A parabola through the three points around the peak gives the crest at about $w = 13.1\%$ with $\gamma_d \approx 1.90$ Mg/m³. ```text γd (Mg/m³) 1.90 | * * <- peak ≈ 1.90 at w ≈ 13.1% 1.86 | * 1.82 | * 1.78 | * 1.74 | * +--+---+---+---+---+---+--- w (%) 11 12 13 14 15 16 ``` - Maximum dry unit weight, $\gamma_{d,max} \approx 1.90$ Mg/m³ ($\approx 18.6$ kN/m³) - Optimum moisture content, OMC $\approx 13.1\%$ #### ii) Relative compaction Field dry unit weight: $$ \gamma_{d,field} = \frac{2.318}{1+0.35} = 1.717\ \text{Mg/m}^3 $$ $$ RC = \frac{\gamma_{d,field}}{\gamma_{d,max}} \times 100 = \frac{1.717}{1.90} \times 100 = 90.4\% $$ #### iii) Degree of saturation at $\gamma_{d,max}$ $$ e = \frac{G\gamma_w}{\gamma_d} - 1 = \frac{2.7 \times 1}{1.90} - 1 = 0.421 $$ $$ S = \frac{wG}{e} = \frac{0.131 \times 2.7}{0.421} = 0.84 $$ **Answer: $\gamma_{d,max} \approx 1.90$ Mg/m³, OMC $\approx 13.1\%$; RC $\approx 90.4\%$; $S \approx 84\%$ at maximum dry density.** Note: the field data as printed are not physically consistent. At $w = 35\%$ the ZAV dry density is only $2.7/(1 + 0.35 \times 2.7) = 1.39$ Mg/m³, so a dry density of 1.717 Mg/m³ would need $S \approx 165\%$. The values are probably misprinted; the RC above simply follows the data as given and shows the fill is below the usual 95% requirement. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Test 3: mass 2138.5 g, w = 12.8%, dry density: A. 2.1385 B. 1.896 Mg/m3 C. 1.50 D. 2.41 Answer: B. 2.1385/1.128 = 1.896. (Key point: Dry density) 2. OMC from the curve: A. About 35% B. About 16% C. About 13% D. About 11% Answer: C. Peak is around 13%. (Key point: OMC) 3. e = G gw/gd - 1 with gd = 1.90 Mg/m3, G = 2.7: A. 0.1 B. 0.421 C. 1.42 D. 0.9 Answer: B. 2.7/1.9 - 1 = 0.421. (Key point: Void ratio) 4. S at MDD: A. 100% B. 83% C. 55% D. 95% Answer: B. 0.13*2.7/0.421 = 0.834. (Key point: Saturation) ## Close variants set in other exams - [Compaction test: curve, MDD, OMC, RC](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Acompaction-test-curve-mdd-omc-rc) — Soil Mechanics, chapter 5, asked 2× ## Related in Soil Mechanics - [Compacted soil 17 kN/m³, OWC 16%](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Amax-dry-weight-17-void-ratio) — Soil Mechanics, chapter 5, asked 2× - [Compacted soil 18 kN/m³, saturation](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Amax-dry-weight-18-saturation) — Soil Mechanics, chapter 5 - [Road compaction specification, sand cone](https://www.bctengineer.com/question.md?key=semester_4%2Fsoil_mechanics%3Aroad-compaction-sand-cone-check) — Soil Mechanics, chapter 5 ## Related in other subjects - [Standard error; population 12,19,13,16](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Astandard-error-population-12-19-13) — Probability and Statistics, chapter 4 - [K-map Σ(1,2,3,8,9,10,11,13,14)+d, POS](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Akmap-pos-form-12389) — Digital Logic, chapter 3