A sample of argillaceous soil with a volume of 1x10-3 m3 and a mass of 1.762 kg,

Question

A sample of argillaceous soil with a volume of 1x10-3 m3 and a mass of 1.762 kg, after oven dried, has a mass of 1.368 kg. The relative density of the grains is equal to 2.69. Determine, for the sample in its original condition:

a) dry unit weight

b) moisture content

c) voids

d) voids filled with air (expressed as% of voids = Sa = degree of air saturation)

Considering a sample of this soil completely saturated by water, which has not undergone fading in the process of saturation (ie, assuming that it has the same voids above), determine, also:

e) moisture content

f) overall specific weight.

(R: 13.41 kN/m3, 28.80%, 0.97, 19.83%, 36.06%, 18.21 kN/m3)

Explanation / Answer

V= 1*10^-3 m^3

W = 1.762 kg

Ws = 1.368 kg

Gs = 2.69

Yw= 1000

a) dry unit weight Yd = Ws/V = 1.368/1*10^-3 = 1368 kg/m^3

b) moisture content w = W/Ws = (W-Ws)/Ws = (1.762-1.368)/1.368 = 28.80%

c) voids e= Gs*Yw/Yd - 1 = 2.69*1000/1368 -1 = 0.97

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