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 5 of 13.
Mantu kumar patel said:
1 decade ago
We can solve it by one point method formula of liquid limit determination.
W1=Wn (N/25) ^n where, value of n is 0.12.
Liquid Limit = W*(N/25)^0.121.
Where,
W = water content at N blows.
N = no. of blows.
Liquid limit = 70*(10/25)^0121 = 89.55%.
Liquid Limit = 20*(100/25)^0.121 = 23.65%.
Required liquid limit = 88-23 = 65%.
W1=Wn (N/25) ^n where, value of n is 0.12.
Liquid Limit = W*(N/25)^0.121.
Where,
W = water content at N blows.
N = no. of blows.
Liquid limit = 70*(10/25)^0121 = 89.55%.
Liquid Limit = 20*(100/25)^0.121 = 23.65%.
Required liquid limit = 88-23 = 65%.
Moha cool said:
1 decade ago
@Aakash karoo.
Your simple steps helped me get the right answer which is 65%, the graph part is actually much easier!
Thanks.
Your simple steps helped me get the right answer which is 65%, the graph part is actually much easier!
Thanks.
Masud said:
10 years ago
W1 = 70, N1 = 10, W2 = 20, N2 = 100, W3 = ?, N3 = 25.
W1-W2/N1-N2 = W2-W3/N2-N3 => 70-20/10-100 = 20-W3/100-25- => 50/-90 = 20-W3/75 => W3 = 61.66%.
W1-W2/N1-N2 = W2-W3/N2-N3 => 70-20/10-100 = 20-W3/100-25- => 50/-90 = 20-W3/75 => W3 = 61.66%.
Masud said:
10 years ago
W1 = 70, N1 = 10, W2 = 20, N2 = 100, W3 = ?, N3 = 25.
W1-W2/N1-N2 = W2-W3/N2-N3 => 70-20/10-100 = 20-W3/100-25- => 50/-90 = 20-W3/75 => W3 = 61.66%.
W1-W2/N1-N2 = W2-W3/N2-N3 => 70-20/10-100 = 20-W3/100-25- => 50/-90 = 20-W3/75 => W3 = 61.66%.
Ishita Bhatnagar said:
10 years ago
First the graph between water content and the corresponding number of blows is NOT a straight line! when the points are plotted on a SEMI LOG graph, that is, water content versus log (N), where N = no. of blows required to close the 2 mm groove at that water content, this water content versus log (N) plot comes out to be a STRAIGHT LINE!
Hence the answer according to flow index will be the same as got from the semi log graph using interpolation. Because remember interpolation is valid only as long as the two quanties have a linear relationship!. Hence answer should be 50.1%.
As far as the one point method is concerned, this method is used only when only one measurement is given to you.
Hence the answer according to flow index will be the same as got from the semi log graph using interpolation. Because remember interpolation is valid only as long as the two quanties have a linear relationship!. Hence answer should be 50.1%.
As far as the one point method is concerned, this method is used only when only one measurement is given to you.
MAYANK said:
10 years ago
Correct answer is 50.1.
Simply use equation ll = -if(logx)+c.
Two equation, two unknown, solve it you will get ll = 50.1.
Simply use equation ll = -if(logx)+c.
Two equation, two unknown, solve it you will get ll = 50.1.
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.
Sravya iiit basara said:
9 years ago
No, the answer is wrong.
Ammar said:
9 years ago
Wrong answer.
By linear interpolation, for blows 25 find L L.
L. L = 61%.
By linear interpolation, for blows 25 find L L.
L. L = 61%.
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?
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