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.
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.
Gilbert Khakhlari said:
1 decade ago
Read the question properly guys. You all did a common mistake.
Hint: Check the units. It's g/cm^3. It's the unit of density and not unit weight even though the question says so then the question is solved correctly.
BULK UNIT WEIGHT = 9.81* Density = 9.81*2 = 19.62 kN/m^3.
DRY UNIT WEIGHT = (19. 62/1+0. 1) = 17.836 kN/m^3.
So, e = (G*Yw/Yd)-1 = (2.7*9.81/17.836) = 0.485.
Therefore, S = w*G/e = (0.1*2.7/0.485) = 55.67%
So, Answer is Option (B).
Hint: Check the units. It's g/cm^3. It's the unit of density and not unit weight even though the question says so then the question is solved correctly.
BULK UNIT WEIGHT = 9.81* Density = 9.81*2 = 19.62 kN/m^3.
DRY UNIT WEIGHT = (19. 62/1+0. 1) = 17.836 kN/m^3.
So, e = (G*Yw/Yd)-1 = (2.7*9.81/17.836) = 0.485.
Therefore, S = w*G/e = (0.1*2.7/0.485) = 55.67%
So, Answer is Option (B).
(1)
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.
MD AYAT KARIM said:
1 decade ago
Given That,
Water Content w = 0.1
Unit Wt of Soil = 2.0 g/cc.
Sp. Gravity of Soil = 2.7.
Unit Wt of water = 1 g/cc.
Dry Unit Wt of Soil = Bulk unit wt/ (1+ w).
= 2/(1+0.1).
=1.818 g/cc.
Now,
Void Ratio (e)=(G * Unit Wt of Water/dry unit wt ) - 1.
= (2.7 * 1 / 1.818) - 1.
= 0.4851 .
Now,
Degree of Saturation (S) = G*w/e.
=2.7*0.1/0.4851.
=0.556.
=55.6% Ans.
But the answer showing is 69.6%.
Water Content w = 0.1
Unit Wt of Soil = 2.0 g/cc.
Sp. Gravity of Soil = 2.7.
Unit Wt of water = 1 g/cc.
Dry Unit Wt of Soil = Bulk unit wt/ (1+ w).
= 2/(1+0.1).
=1.818 g/cc.
Now,
Void Ratio (e)=(G * Unit Wt of Water/dry unit wt ) - 1.
= (2.7 * 1 / 1.818) - 1.
= 0.4851 .
Now,
Degree of Saturation (S) = G*w/e.
=2.7*0.1/0.4851.
=0.556.
=55.6% Ans.
But the answer showing is 69.6%.
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