Answer:
civil
Explanation:
mark be branilist
Answer:
Civil
Explanation:
civil engineering – the application of planning, designing, constructing, maintaining, and operating infrastructure while protecting the public and environmental health, as well as improving existing infrastructure that may
A sample of soil has a volume of 0.45 ft^3 and a weight of 53.3 lb. After being dried inan oven, it has a weight of 45.1 lb. It has a specific gravity of solids of 2.70. Compute its moisture content and degree of saturation before it was placed in the oven.
Answer:
a) the moisture content before it was placed in the oven is 18.18%
b) degree of saturation for soil is 72.19%
Explanation:
Given the data in the question;
Moisture Content = [(Weight of soil before dry - dry weight) / dry weight] × 100
so we substitute
Moisture content = [(53.3 - 45.1) / 45.1 ] × 100
= (8.2/45.1) × 100
= 18.18%
Therefore the moisture content before it was placed in the oven is 18.18%
Dry Unit Weight = dry weight / volume
Dry Unit Weight = 45.1 lb / 0.45 ft³
Dry Unit Weight = 100.22 lb/ft³
we know that;
dry unit weight = (Specific gravity × unit weight of water) / (1 + e)
we also know that; unit weight of water is 62.43 lbf/ft³
so we substitute
e = (2.70×62.43 / 100.22) - 1
e = 1.68 - 1
e = 0.68
so void ratio e = 0.68
Now we determine the degree of saturation using the equation;
degree of saturation = (Moisture content × specific gravity) / void ratio
we substitute
degree of saturation = ( 18.18% × 2.7) / 0.68
= 0.49086 / 0.68
= 0.7219 ≈ 72.19%
Therefore degree of saturation for soil is 72.19%