Civil Engineering - Soil Mechanics and Foundation Engineering - Discussion
Discussion Forum : Soil Mechanics and Foundation Engineering - Section 1 (Q.No. 19)
19.
A compacted soil sample using 10% moisture content has a weight of 200 g and mass unit weight of 2.0 g/cm3. If the specific gravity of soil particles and water are 2.7 and 1.0, the degree of saturation of the soil is
Discussion:
27 comments Page 3 of 3.
DH SARGAM said:
9 years ago
W=.1, bulk density = 2g/cc.
Dry density = 2/1 +.1 = 1.8g/cc.
e = [(2.7 * .1)/1.8-1] = .485.
S = (2.7 * 0.1)/.485 = .556.
OR
S = 55.6%.
Dry density = 2/1 +.1 = 1.8g/cc.
e = [(2.7 * .1)/1.8-1] = .485.
S = (2.7 * 0.1)/.485 = .556.
OR
S = 55.6%.
R@ghv said:
9 years ago
S = W/ (Yw * (1+W)/Y ) - 1/G.
Use it by W = 0.1 ; Yw = 1.0 ; Y = 2 ; G = 2.7.
Then answer is 55.6 (b).
Use it by W = 0.1 ; Yw = 1.0 ; Y = 2 ; G = 2.7.
Then answer is 55.6 (b).
Prashant said:
10 years ago
Y = Yw*(G+e*Sr)/(1+e).
Putting values we get: e(2-Sr) = 0.7.
Then put e = W*G/Sr.
Get Sr = 55.6%.
Putting values we get: e(2-Sr) = 0.7.
Then put e = W*G/Sr.
Get Sr = 55.6%.
Paris said:
10 years ago
Option B is right refer 1st solution.
Sayandip Basak said:
1 decade ago
We know,
Y = (G+S*e)*Yw/(1+e).
Also, G*w = S*e => e = G*w/S.
Therefore,
Y = (G+G*w)*Yw/(1+G*w/S).
Substitute the values and find S.
You will get answer B as the right option.
Y = (G+S*e)*Yw/(1+e).
Also, G*w = S*e => e = G*w/S.
Therefore,
Y = (G+G*w)*Yw/(1+G*w/S).
Substitute the values and find S.
You will get answer B as the right option.
Azaz said:
1 decade ago
I got 77%.
Yd = G.Yw/ 1+ e.
I got e = 0.35.
Sr = W.G / e.
So, Sr = 0.777.
Yd = G.Yw/ 1+ e.
I got e = 0.35.
Sr = W.G / e.
So, Sr = 0.777.
Shukla said:
1 decade ago
Yep @Karim you are right me also getting same answer.
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