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
1.47
1.82
1.91
none of these.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
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.
(9)

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.
(1)

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.

Abhay said:   10 years ago
For saturated soil:

Υd = GΥw/(1+wG).

= 2.6*1/(1+0.3*2.6).

= 1.46067.

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.
(4)

Vikash said:   1 decade ago
Yd = GYw/1+WG.

= 2.65x10/1+(.3x2.65).

= 14.76.
(1)

DH satgam said:   9 years ago
According to me, the answer will be 1.12g/cc.
(1)

Bhupal said:   10 years ago
yd = (yw(1-na)G)/(1+wG).
yd = y/(1+w).

Shilpa said:   1 decade ago
yd = y/1+w.

= 2.6/1+0.3.

= 2.

Mubbasher said:   7 years ago
According to me, the answer is 14.44.
(1)


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