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.
Vikash said:
1 decade ago
Yd = GYw/1+WG.
= 2.65x10/1+(.3x2.65).
= 14.76.
= 2.65x10/1+(.3x2.65).
= 14.76.
(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).
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)
Abhay said:
10 years ago
For saturated soil:
Υd = GΥw/(1+wG).
= 2.6*1/(1+0.3*2.6).
= 1.46067.
Υd = GΥw/(1+wG).
= 2.6*1/(1+0.3*2.6).
= 1.46067.
Saunak said:
9 years ago
Answer is 1.461 so A.
Sudarsan sahoo said:
9 years ago
Yeah. The correct answer is A.
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)
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