Civil Engineering - Soil Mechanics and Foundation Engineering - Discussion
Discussion Forum : Soil Mechanics and Foundation Engineering - Section 3 (Q.No. 33)
33.
A sample of saturated soil has 30% water content and the specific gravity of soil grains is 2.6. The dry density of the soil mass in g/cm3, is
Discussion:
14 comments Page 1 of 2.
Pavan said:
6 years ago
We have Yd=GYw/(1+e).
We have to calculate the value of e from S.e=W.G.
S=1, W=0.3,G=2.6 all values put and find e as 0.78.
Now all putting in eq.Yd = GYw/(1+e).
Yd = 2.6/(1+0.78).
Yd = 1.46.
We have to calculate the value of e from S.e=W.G.
S=1, W=0.3,G=2.6 all values put and find e as 0.78.
Now all putting in eq.Yd = GYw/(1+e).
Yd = 2.6/(1+0.78).
Yd = 1.46.
(9)
Reetu jayswal said:
4 years ago
C is correct.
For saturated soil,
Yd=(G+e)Yw/(1+e),
Yd=(2.6+.78)1/(1+.78),
Yd= 1.91.
For saturated soil,
Yd=(G+e)Yw/(1+e),
Yd=(2.6+.78)1/(1+.78),
Yd= 1.91.
(4)
Vikash said:
1 decade ago
Yd = GYw/1+WG.
= 2.65x10/1+(.3x2.65).
= 14.76.
= 2.65x10/1+(.3x2.65).
= 14.76.
(1)
Seazzzz said:
10 years ago
Sr.E = W.G.
Saturated soil, Sr = 1.
Water content, w = 30% = 0.3.
Specific gravity, G = 2.6.
Using formula, e = 0.78.
Now, Yd = (G.Yw)/(1 + e).
Solving this, we get dry density, Yd = (2.6X1)/(1+0.78).
= 1.46 gm/cc.
Saturated soil, Sr = 1.
Water content, w = 30% = 0.3.
Specific gravity, G = 2.6.
Using formula, e = 0.78.
Now, Yd = (G.Yw)/(1 + e).
Solving this, we get dry density, Yd = (2.6X1)/(1+0.78).
= 1.46 gm/cc.
(1)
DH satgam said:
9 years ago
According to me, the answer will be 1.12g/cc.
(1)
Mubbasher said:
7 years ago
According to me, the answer is 14.44.
(1)
Jeet said:
6 years ago
A is correct.
(1)
Shilpa said:
1 decade ago
yd = y/1+w.
= 2.6/1+0.3.
= 2.
= 2.6/1+0.3.
= 2.
Ankit said:
1 decade ago
Yd = GYw/(1+WG).
= 2.65*1/(1+.3*2.65) (here we will use Yw = 1 gm/cc).
= 1.476 gm/cc.
= 2.65*1/(1+.3*2.65) (here we will use Yw = 1 gm/cc).
= 1.476 gm/cc.
Bhupal said:
10 years ago
yd = (yw(1-na)G)/(1+wG).
yd = y/(1+w).
yd = y/(1+w).
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