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Soil Mechanics and Foundation Engineering - Engineering

Q1:

A soil sample of mass specific gravity 1.92, has a moisture content 30%. If the specific gravity of solids is 2.75, the degree of saturation, is

A 95.4%

B 95.5%

C 95.6%

D 95.7%

ANS:D - 95.7%

G=Gs=Ys/Yw = 2.75.
w= .30.
Gm=Yb/Ye = 1.92,

We know;
Yb = ((G+Se)Yw)÷(1+e).
= >Yb/Ye = (G+Se)÷(1+e).
=>Gm = (G+Se)÷(1+e).
=>1.92 = (2.75+0.30*(wG))÷{1+(wG/S)}.
=>1.92 = (2.75+0.30*(0.30*2.75))÷{1+(0.30*2.75/S).
=>S=95.69 = 95.7 %. Bulk mass or apparent specific gravity(Gm) = γ/γw = 1.92.

γ= (G+eSr)γw/(1+e) => γ/γw = (G+eSr)/(1+e).

We know, eSr = wG => eSr = 0.3*2.75 = 0.825.

Therefore,
1.92 = (2.75 + 0.825)/(1+e) => 1+e = 3.575/1.92.

Therefore, e = 1.862 - 1 = 0.862.

Therefore, eSr = wG => 0.3*2.75/0.862 =>Sr = 0.957 i.e 95.7%

Note:
γ - The moist unit weight of the material.
γw - Unit weight of water.
G - Specific gravity of a Solid.
e - Void ratio.
Sr - Degree of saturation.