Answer:
a) the wet density of the CL sample is 0.0453 lb/in³
b) the water content in the sample is 65.37%
c) the dry density of the CL sample is 0.0274 lb/in³
Explanation:
Given that;
diameter d = 2.83 in
length L = 6 in
weight m = 1.71 lbs
A piece of clay sample had wet-weight of 140.9 grams and dry-weight of 85.2 grams
a) wet density of the CL sample
wet density can be expressed as p = M /v
V is volume of sample which is; π/4×d²×L
so p = M / π/4×d²×L
we substitute
p = 1.71 / (π/4 × (2.83)²× 6
p = 1.71 / 37.741
p = 0.0453 lbs/in³
so the wet density of the CL sample is 0.0453 lb/in³
b)
water content of sample is taken as;
w = (wet_weight - dry_weight) / dry_weight
we substitute
w = (140.9 - 85.2) / 85.2
w = 55.7 / 85.2
w = 0.6537 = 65.37%
therefore the water content in the sample is 65.37%
c)
dry density of the CL sample
to determine the dry density, we say;
Sd = p / ( 1 + w )
we substitute
Sd = 0.0453 / ( 1 + 0.6537)
Sd = 0.0453 / 1.6537
Sd = 0.0274 lb/in³
therefore the dry density of the CL sample is 0.0274 lb/in³
