--- title: "Compaction test: curve, MDD, OMC, RC" subject: "Soil Mechanics" chapter: 5 key: "semester_4/soil_mechanics:compaction-test-curve-mdd-omc-rc" source: https://www.bctengineer.com/q/semester_4/soil_mechanics/compaction-test-curve-mdd-omc-rc site: BCT Engineering --- # Compaction test: curve, MDD, OMC, RC Soil Mechanics · Chapter 5: Soil Compaction - Asked in: 2078 Poush, 3+2 marks - Video: [What is the optimum moisture content test|| What is an OMC test|| What is the standard Proctor test](https://www.youtube.com/watch?v=oIeBsY9vCSA) — Water Resource 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-mdd-omc-rc - 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). ## Short answer Dry densities were computed from the wet mass and water content. The compaction curve peaks at maximum dry density ≈ 1.90 Mg/m³ (18.6 kN/m³) and optimum water content ≈ 13.1%. The field dry density is 1.717 Mg/m³, giving relative compaction = 90.4%. ## Answer #### (i) Compaction curve Mould volume $= 1000$ cm³ = 1 litre, so bulk density $\rho = m/1000$ g/cm³ (Mg/m³), and $$ \rho_d = \frac{\rho}{1+w}, \qquad \gamma_d = \rho_d \times 9.81 $$ | Test | $w$ (%) | $\rho$ (Mg/m³) | $\rho_d$ (Mg/m³) | $\gamma_d$ (kN/m³) | $\rho_{d,zav}$ (Mg/m³) | |---|---|---|---|---|---| | 1 | 11.0 | 1.9205 | 1.730 | 16.97 | 2.082 | | 2 | 12.1 | 2.0515 | 1.830 | 17.95 | 2.035 | | 3 | 12.8 | 2.1385 | 1.896 | 18.60 | 2.007 | | 4 | 13.6 | 2.1470 | 1.890 | 18.54 | 1.975 | | 5 | 14.6 | 2.1200 | 1.850 | 18.15 | 1.937 | | 6 | 16.3 | 2.0815 | 1.790 | 17.56 | 1.875 | ZAV: $\rho_{d,zav} = \dfrac{G\rho_w}{1+wG}$ with $G = 2.7$. Plot $\rho_d$ (or $\gamma_d$) against $w$ and draw a smooth curve through the points. It rises to a peak between tests 3 and 4 and then falls, staying below the ZAV values. ```text ρd (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 ``` From the smoothed curve (a parabola through the three points around the peak gives its crest at about $w = 13.1\%$): - **Maximum dry density** $\approx 1.90$ Mg/m³, i.e. **maximum dry unit weight $\approx 18.6$ kN/m³** - **OMC $\approx 13.1\%$** #### (ii) Relative compaction Field dry density: $$ \rho_{d,field} = \frac{2.318}{1 + 0.35} = 1.717\ \text{Mg/m}^3\ (\gamma_d = 16.84\ \text{kN/m}^3) $$ $$ RC = \frac{\rho_{d,field}}{\rho_{d,max}} \times 100 = \frac{1.717}{1.90} \times 100 = 90.4\% $$ **Answer: MDD $\approx 1.90$ Mg/m³ (18.6 kN/m³), OMC $\approx 13.1\%$, relative compaction $\approx 90\%$** (below the usual 95% requirement). 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 value is probably a misprint, but the RC above follows the data as given. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Dry density in each test: A. Mass/(1+w) volume B. Bulk density/(1+w) C. Bulk density x (1+w) D. Bulk density x w Answer: B. rho_d = rho/(1+w). (Key point: Dry density) 2. MDD and OMC from the curve: A. About 1.90 Mg/m3 at about 13.2% B. About 1.73 at 11% C. About 2.15 at 13.6% D. About 1.79 at 16.3% Answer: A. Peak lies between tests 3 and 4. (Key point: MDD and OMC) 3. Field dry density with rho = 2.318 and w = 35%: A. 1.717 Mg/m3 B. 3.13 C. 1.90 D. 2.318 Answer: A. 2.318/1.35 = 1.717. (Key point: Field density) 4. RC = 1.717/1.90: A. Around 75% B. Around 95% C. Around 90%, below 95% D. Around 105% Answer: C. 90.4%, not satisfactory. 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