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.
Muhammad Ali said:
4 years ago
Why not Yd= GYw/1+w? Explain please.
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)
Jeet said:
6 years ago
A is correct.
(1)
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)
Mubbasher said:
7 years ago
According to me, the answer is 14.44.
(1)
DH satgam said:
9 years ago
According to me, the answer will be 1.12g/cc.
(1)
Sudarsan sahoo said:
9 years ago
Yeah. The correct answer is A.
Saunak said:
9 years ago
Answer is 1.461 so A.
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.
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)
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