Civil Engineering - Soil Mechanics and Foundation Engineering - Discussion
Discussion Forum : Soil Mechanics and Foundation Engineering - Section 1 (Q.No. 1)
1.
In a liquid limit test, the moisture content at 10 blows was 70% and that at 100 blows was 20%. The liquid limit of the soil, is
Discussion:
123 comments Page 6 of 13.
Masoom said:
10 years ago
50.1 is the correct answer, if solved by semi-log graph method. And this method is proper method to solve.
Manjusha Shinde said:
4 years ago
LL = w*(N/25)^0.121.
LL = 70*(100/25)^0.121 = 82.78%.
LL = 20*(10/25)^0.121 = 17.90%.
LL = 83 - 18 = 65%.
LL = 70*(100/25)^0.121 = 82.78%.
LL = 20*(10/25)^0.121 = 17.90%.
LL = 83 - 18 = 65%.
(47)
Ganesh Jadhav said:
9 years ago
I think the one point method is not valid rather not necessary because two points are given for example.
Siva said:
8 years ago
Why should we take the liquid limit value at 25 blows only? If there any relation, please explain it.
Md. Suruzzaman said:
6 years ago
The water content at which soil changes from liquid limit to plastic limit is called liquid limit.
Harshdeep said:
8 years ago
Simply take x= 0.068.
and WL = Wn (N÷25)^x .
WL = 70 (10÷25 )^0.068 = 68.77 approx 65.
and WL = Wn (N÷25)^x .
WL = 70 (10÷25 )^0.068 = 68.77 approx 65.
RAHUL R K said:
9 years ago
Why all of them are taking the difference between liquid limit for 10 and 100 Nos of blows?
Navya said:
6 years ago
Interpolation method is correct method for liquid limit at 25 blows. So the answer is 50%.
Mea said:
8 years ago
65% is the right answer. Just draw the flow curve graph roughly. You will get it easily.
Pratima Yadav said:
1 decade ago
Liquid Limit= W * (N/25)^0.121.
where,
W= water content at N blows.
N= no.of blows.
where,
W= water content at N blows.
N= no.of blows.
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