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 2 of 2.
Meghana said:
8 years ago
@Souvik.
If we use that formula, we get e = 0.6.
By using Gw=Sr.e. We get Sr = 86.6%.
If we use that formula, we get e = 0.6.
By using Gw=Sr.e. We get Sr = 86.6%.
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?
Anonymous said:
9 years ago
Look at the units.Unit weights are given in g/cm^3 and unit weight of water is 1 g/cm^3 in CGS system.
Pratyush said:
9 years ago
Yes @Priyanka.
I'm getting the same answer can anyone justify why Yw =1 not 9.81?
I'm getting the same answer can anyone justify why Yw =1 not 9.81?
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)
Prashant said:
10 years ago
Y = Yw*(G+e*Sr)/(1+e).
Put e*Sr = W*G.
Y = Yw*(G+G*W)/(1+e).
Put values: e = 0.56.
Sr = W*G/e.
Sr = 92%
Put e*Sr = W*G.
Y = Yw*(G+G*W)/(1+e).
Put values: e = 0.56.
Sr = W*G/e.
Sr = 92%
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%.
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%.
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%.
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