Civil Engineering - Soil Mechanics and Foundation Engineering - Discussion

Discussion Forum : Soil Mechanics and Foundation Engineering - Section 1 (Q.No. 20)
20.
A partially saturated sample of soil has a unit weight of 2.0 g/cm3 and specific gravity of soil particles is 2.6. If the moisture content in the soil is 20%, the degree of saturation is
20%
77%
92%
none of these.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
19 comments Page 1 of 2.

Nehal said:   2 years ago
Ws = Gs * 1 = 2.6
Wc = Ww/Ws.
0.2 = Ww/2.6.
Ww = 0.52.
Vw = 0.52.
Density = Wt/Vt.
2 = 2.6 + 0.52/Vt.
Vt = 1.56.

e = Vv/vs = 1.56-1/1 = 0.56,
S = Vw/Vv = 0.52/1.56-1 = 0.92.

Guri said:   2 years ago
92.85% is an accurate value.

Hassan Bilal said:   2 years ago
According to me, 77 is the correct answer.

Shravani said:   3 years ago
Not getting this, anyone explain me clearly.

UJJVAL said:   4 years ago
W=.2, bulk density = 2g/cc.
Dry density = 2/1 +.2 = 1.667g/cc.
e = [(2.6 * 1)/1.67-1] = .56.
S = (2.6 * 0.1)/.56 = .9265.
OR
S = 92.6%.

Jitendra gkp said:   4 years ago
b=G*w(1+w)/(1+e).
e=.56.
Sr=wG/e=92%.

Anil said:   5 years ago
77% is correct I think.

Kiruthika said:   5 years ago
@Himanshu Gupta.

The question is to find the degree saturation;

I am getting the answer as S=87%,
By using for:eS=wG,
e=.62.

Himanshu gupta said:   5 years ago
@Souvik.

Partially saturated means the soil have a degree of saturation less than 100%. If it is completely saturated then there is no need to find a degree of saturation.

Sub said:   5 years ago
Unit weight of soil = 2.0g/cm3.
Dry unit weight = unit weight/(1+w).
Here w = 0.2.
Hence dry unit weight = 2/(1+0.2).
= 1.66.
Void ratio = ((specific gravity * unit weight of water )/dry unit weight)-1.
= 0.566.
Degree of saturation = w * specific gravity /void ratio.
= 0.9182.
= 91.82%.


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