Results of a standard compaction test (mold volume 943.9 cm3 ) are:

Trial No. 1 Mass of soil + mold 5619g
Mass of mold 3735g
Water content test
Mass of sample + container 288.26g
Mass of dry sample + container 265.39g
Mass of container 104.31g

Trial No. 2 Dry density 1795 kg/m^3 w=14.9%
Trial No. 3 Dry density 1828 kg/m^3 w=15.6%
Trial No. 4 Dry density 1832 kg/m^3 w=16.5%
Trial No. 5 Dry density 1796 kg/m^3 w=17.6%

a. Complete the calculations for trial No. 1.
b. Draw the moisture-density curve using Excel or other spreadsheet programs. Include ZAV curves at w=16%, 17%, and 18%. (The relative density of the soil is 2.65.) Find the maximum dry density and the optimum water content.

Question

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oeerivona oeerivona · Answer 1 · 2020-03-02T18:52:24+01:00

Answer:

The answers to the question are

Dry density for trial number 1 = 1747.82 kg/m³

Water Content for trial number 1 = 14.1979 %

Explanation:

To solve the question we have

Moist Unit Weight (γm) = [tex]\frac{Mass. of wet .soil .in. mold}{Volume}[/tex]

= [tex]\frac{5619-3735}{943.9}[/tex] = 19573.82 N/m³

Water Content ω = [tex]\frac{(Mass. of . wet . soil)-(Mass .of.container+dry.soil)}{(Mass.of.can+drysoil)-(Mass.of.can)}[/tex]

= [tex]\frac{288.26-265.39}{265.39-104.31}[/tex] × 100 = 14.1979 %

Dry unit weight, [tex]\gamma_d[/tex] = [tex]\frac{\gamma_m}{1+\omega}[/tex] = [tex]\frac{19.573}{1+0.141979}[/tex] = 17140.26 N/m³

Bulk density = mass / density = [tex]\frac{5619-3735}{943.9}[/tex]=1995.97 kg/m³

Dry density = [tex]\frac{1995.97}{1.1419}[/tex] = 1747.82 kg/m³

Please find the Moisture Density Curve