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
Discussion:
19 comments Page 1 of 2.
IIIT Basar said:
1 decade 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%.
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%.
Priyanka said:
1 decade ago
Unit weight of soil = 2g/cm3.
Dry unit weight = Unit weight/(1+w).
W = 0.2 & G = 2.6 unit weight of water = 9.81.
Dry unit weight = 2/(1+0.2) = 1.66.
E = g*9.81/dry unit weight-1 = (2.6*9.81/1.66)-1 = 14.36.
S, degree of saturation = (g*w)/e = (2.6*0.2)/14.30 = 0.036.
S = 3.6%.
Dry unit weight = Unit weight/(1+w).
W = 0.2 & G = 2.6 unit weight of water = 9.81.
Dry unit weight = 2/(1+0.2) = 1.66.
E = g*9.81/dry unit weight-1 = (2.6*9.81/1.66)-1 = 14.36.
S, degree of saturation = (g*w)/e = (2.6*0.2)/14.30 = 0.036.
S = 3.6%.
Sub said:
8 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%.
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%.
(3)
Nawaz said:
9 years ago
Here very simple given data, Y = 2g/cm^3,
G = 2.6, w (water content) = 0.2. Sr = ?,
We know-Dry unit weight (Yd) = Y/1+w.
So, 2/1 + 0.2 = 1.66.
Again we know, 1 + e = GYw/Yd,
So, 1+e = 2.6*1/1.66 = 1.56 - 1 = 0.56.
So, Sr = WG/e. So, Sr = 0.2*2.6/0.56 = 0.92.
So, 92% answer.
G = 2.6, w (water content) = 0.2. Sr = ?,
We know-Dry unit weight (Yd) = Y/1+w.
So, 2/1 + 0.2 = 1.66.
Again we know, 1 + e = GYw/Yd,
So, 1+e = 2.6*1/1.66 = 1.56 - 1 = 0.56.
So, Sr = WG/e. So, Sr = 0.2*2.6/0.56 = 0.92.
So, 92% answer.
(2)
RAMU VEMULA said:
1 decade ago
Unit weight of soil = 2g/cm3.
Dry unit weight = unit weight/(1+w).
w = 0.2 & G = 2.6.
Dry unit weight = 2/(1+0.2) = 1.81.
e = G*1/dry unit weight-1 = 2.6*1/1.81-1 = 0.566.
Deg of saturation = (G*w)/e = (2.6*0.2)/0.566 = 0.9187.
Deg of saturation = 91.87%.
Dry unit weight = unit weight/(1+w).
w = 0.2 & G = 2.6.
Dry unit weight = 2/(1+0.2) = 1.81.
e = G*1/dry unit weight-1 = 2.6*1/1.81-1 = 0.566.
Deg of saturation = (G*w)/e = (2.6*0.2)/0.566 = 0.9187.
Deg of saturation = 91.87%.
Souvik said:
8 years ago
Why don't we consider the saturated unit of soil because in question it is mentioned that the soil sample is partially saturated?
Why don't we ignore the formula Ysat = (G+e) * Yw/(1+e)?
Can anyone clarify the answer?
Why don't we ignore the formula Ysat = (G+e) * Yw/(1+e)?
Can anyone clarify the answer?
Nehal said:
4 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.
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.
(2)
Himanshu gupta said:
8 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.
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.
UJJVAL said:
7 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%.
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%.
(3)
Kiruthika said:
7 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.
The question is to find the degree saturation;
I am getting the answer as S=87%,
By using for:eS=wG,
e=.62.
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