Skip to main content

Embankment 95% compaction, 950 cc mould

Short answer

From the lab mould, dry densities peak at 1.865 g/cc at 14.6% water content. Required dry density at 95% compaction is 1.77 g/cc; field dry density 1.78 g/cc gives RC 95.4%, so embankment is satisfactory, though the margin is small.

Evaluate the construction of the embankment if the required degree of compaction is 95% and the dry density of the embankment was found as 1.78 g/cc. The result of the compaction test performed in the laboratory for the same material using a 950 cc mould are as follows.

w %7.711.514.617.519.721.2
Mass of wet soil (kg)1.701.892.031.991.961.92

Degree of compaction (relative compaction) is RC=γd,field/γd,maxRC = \gamma_{d,field}/\gamma_{d,max}. First find γd,max\gamma_{d,max} from the lab test, then compare with the field value.

Lab dry density (mould volume = 950 cm³)

γbulk=M950 cm3,γd=γbulk1+w\gamma_{bulk} = \frac{M}{950\ \text{cm}^3}, \qquad \gamma_d = \frac{\gamma_{bulk}}{1+w}
w (%)Mass (kg)γbulk\gamma_{bulk} (g/cc)γd\gamma_d (g/cc)
7.71.701.7891.662
11.51.891.9891.784
14.62.032.1371.865
17.51.992.0951.783
19.71.962.0631.724
21.21.922.0211.668

Plotting γd\gamma_d against ww gives a peak at

  • γd,max≈1.865\gamma_{d,max} \approx 1.865 g/cc, OMC ≈14.6%\approx 14.6\%

Check of the embankment

Required dry density =0.95×1.865=1.77= 0.95 \times 1.865 = 1.77 g/cc.

RC=1.781.865×100=95.4%RC = \frac{1.78}{1.865} \times 100 = 95.4\%

Answer: RC = 95.4% (> 95%), field dry density 1.78 g/cc > required 1.77 g/cc. The embankment construction is satisfactory.

The margin is small, so the result is acceptable but only just; the moisture content of the field fill should also be checked to be near the OMC.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Soil Mechanics chapter 5 questions →