The weight of a sample of saturated soil before drying is 2.8 lb and after drying is 2.1 lb. If the specific gravity of the soil particles is 2.6, determine: (a) moisture content, (b) void ratio, (c) porosity, (d) bulk density, and (e) dry density.
What will be an ideal response?
(a) Solve for moisture content, w, using Equation 17.6.
Calculate the weight of the water in the soil sample.
W = Ww + Ws
Ww = 2.8 – 2.1
Ww = 0.7 lbs
w = (Ww / Ws) (100)
w = (0.7 / 2.1) (100)
w = 33.33 %
(b) Solve for void ratio, e, using Equation 17.3. First, the volume of water and the
volume of solid must be determined.
?w = Ww / Vw
Vw = Ww / ?w
Vw = 0.7 / 62.4
Vw = 0.0112 ft3
Gs ?w = Ws / Vs
Vs = Ws / (Gs ?w)
Vs = 2.1 / (2.6)(62.4)
Vs = 0.0129 ft3
e = Vv / Vs = (Vw + Va) / Vs
Va = 0 (since the sample is saturated, no air voids are assumed to
be present).
e = 0.0112 / 0.0129
e = 0.867
(c) Solve for porosity, n, using Equation 17.4.
n = e / (1 + e)
n = (0.867)/(1 + 0.867)
n = 0.464
(d) Calculate the bulk density, ?, using Equation 17.9
? = W / V
? = 2.8 / (0.0112 + 0.0129)
? = 2.8 / 0.0241
? = 116.2 lb/ft3
(e) Solve for the dry density, ?d , using Equation 17.11
?d = Ws / V
?d = 2.1 / (0.0112 + 0.0129)
?d = 87.1 lb/ft3
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