Civil Engineering - Soil Mechanics and Foundation Engineering - Discussion
Discussion Forum : Soil Mechanics and Foundation Engineering - Section 3 (Q.No. 29)
29.
If the unit weight of sand particles is 2.696 g/cc. and porosity in loose state is 44%, the critical hydraulic gradient for quick sand condition, is
Discussion:
8 comments Page 1 of 1.
Abhay said:
10 years ago
G = Υs/Υw => G= 2.696/1.
Υs = Unit weight of sand particle.
Ic = (G-1)/(1+e).
e = n/(1-n) => e = 0.44/(1 = 0.44) = 0.78.
Ic = (2.696-1)/(1+0.78) = 0.9528.
Υs = Unit weight of sand particle.
Ic = (G-1)/(1+e).
e = n/(1-n) => e = 0.44/(1 = 0.44) = 0.78.
Ic = (2.696-1)/(1+0.78) = 0.9528.
(6)
Swapnil said:
1 decade ago
Ic = G-1/1+e.
e = n/1-n.
e = 0.44/1-0.44 = 0.78.
Ic = 2.696-1/1+0.78.
Ic = 0.9528.
e = n/1-n.
e = 0.44/1-0.44 = 0.78.
Ic = 2.696-1/1+0.78.
Ic = 0.9528.
(4)
Rosh said:
10 years ago
G is not given. Unit weight is not equal to G. Data insufficient can't solve.
(1)
Saurabh said:
2 years ago
G = Υs/Υw;
G= 2.696/1.
Ic = (G-1)(1-n),
Ic = (2.696-1)(1-0.44) = 0.9497.
G= 2.696/1.
Ic = (G-1)(1-n),
Ic = (2.696-1)(1-0.44) = 0.9497.
(2)
Nayan said:
8 years ago
Unit weight of water = 1. So directly specific gravity of soil is = 1.
(1)
Atchaya said:
8 years ago
We can also use this formula, Ic = (g-1)(1-n).
(1)
Suraj said:
1 decade ago
Ic = (Gs-1)/1+e.
Gs = 2.696.
e = n/1-n.
Gs = 2.696.
e = n/1-n.
Shaheer said:
1 year ago
You are correct, thanks @Swapnil.
(1)
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