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 1 of 13.
ASHISH WAGHMARE said:
1 decade ago
STEP1: just draw the rough graph on which, no. of blows on X-axis and moisture content on Y-axis.
80%.
M C 70%. .A(10,70%)
O O 60%.
I N 50%.
S T 40%.
T E 30%.
U N 20%. .B(100,20%)
R T 10%.
E(%)0% . . . . . . . . . . . .
0 10 20 30 40 50 60 70 80 90 100 110
NO. OF BLOWS
STEP2: Join the coordinates A and B.
STEP3: draw the vertical line from 25 no. of blows to the line AB and mark it.
STEP4: Draw horizontal line from the marked point on line AB To the pt. on Y-axis, that same will give us the liquid limit of soil sample(ie. 65%).
(NOTE: ALL STEPS ARE TAKEN FROM EXPERIMENT OF DETERMINATION OF LIQUID LIMIT BY USING LIQUID LIMIT DEVICE)
80%.
M C 70%. .A(10,70%)
O O 60%.
I N 50%.
S T 40%.
T E 30%.
U N 20%. .B(100,20%)
R T 10%.
E(%)0% . . . . . . . . . . . .
0 10 20 30 40 50 60 70 80 90 100 110
NO. OF BLOWS
STEP2: Join the coordinates A and B.
STEP3: draw the vertical line from 25 no. of blows to the line AB and mark it.
STEP4: Draw horizontal line from the marked point on line AB To the pt. on Y-axis, that same will give us the liquid limit of soil sample(ie. 65%).
(NOTE: ALL STEPS ARE TAKEN FROM EXPERIMENT OF DETERMINATION OF LIQUID LIMIT BY USING LIQUID LIMIT DEVICE)
(1)
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.
Aakash Karoo said:
1 decade ago
STEP 1: Just draw the rough graph on which, no. of blows on X-axis and moisture content on Y-axis.
STEP 2: Join the coordinates A (20, 70) and B (100, 20).
STEP 3 : Draw the vertical line from 25 no. of blows to the line AB and mark it.
STEP 4 : Draw horizontal line from the marked point on line AB to the point on Y-axis, that same will give us the liquid limit of soil sample (i.e. 65%).
OR,
Liquid Limit = Wl1 = W1*(N2/25)^0.121.
Wl2 = W2*(N1/25)^0.121.
Where,
W = Water content at N blows.
N = No. of blows.
Liquid limit = 70*(100/25)^0.121 = 82.78%.
Liquid Limit = 20*(10/25)^0.121 = 17.90%.
Required liquid limit = 82.78-17.90 = 64.87%.
= 65.00%.
STEP 2: Join the coordinates A (20, 70) and B (100, 20).
STEP 3 : Draw the vertical line from 25 no. of blows to the line AB and mark it.
STEP 4 : Draw horizontal line from the marked point on line AB to the point on Y-axis, that same will give us the liquid limit of soil sample (i.e. 65%).
OR,
Liquid Limit = Wl1 = W1*(N2/25)^0.121.
Wl2 = W2*(N1/25)^0.121.
Where,
W = Water content at N blows.
N = No. of blows.
Liquid limit = 70*(100/25)^0.121 = 82.78%.
Liquid Limit = 20*(10/25)^0.121 = 17.90%.
Required liquid limit = 82.78-17.90 = 64.87%.
= 65.00%.
(2)
Faruque Abdullah said:
6 years ago
tanβ = 0.121 is for the range of 20 to 30 blows.
As the blows number is out of this range this value should be adjusted for the individual case.
For 20 to 30 blows, tanβ = 20/(30^1.5) = 0.121.
Case-I:
tanβ = 10/(25^1.5)=0.08.
w(LL) = 70 x (10/25)^0.08 = 65.05 which is tends to 65%.
Case-II:
tanβ = 100/(25^1.5) = 0.8.
w(LL) = 20 x (100/25)^0.08 = 60.62 which is close to 65%.
As the blows number is out of this range this value should be adjusted for the individual case.
For 20 to 30 blows, tanβ = 20/(30^1.5) = 0.121.
Case-I:
tanβ = 10/(25^1.5)=0.08.
w(LL) = 70 x (10/25)^0.08 = 65.05 which is tends to 65%.
Case-II:
tanβ = 100/(25^1.5) = 0.8.
w(LL) = 20 x (100/25)^0.08 = 60.62 which is close to 65%.
DaKiht said:
1 year ago
@All.
Here, we don't use the Empirical formula i.e- WL=WP*(N/25)^x [where x = 0.1-0.3] the value of x is not given in the question.
The Main formula is preferred over the empirical.
Flow Index (If) = (w1-w2)/log(N2/N1) = (70-20)/log(100/10) = 50.
The same formula was used to find moisture content at 25 No. of blows is 50%.
If = (w1-w2)/log(N2/N1),
50 = (w1-20)/log(100/25),
w1 = 50%.
Here, we don't use the Empirical formula i.e- WL=WP*(N/25)^x [where x = 0.1-0.3] the value of x is not given in the question.
The Main formula is preferred over the empirical.
Flow Index (If) = (w1-w2)/log(N2/N1) = (70-20)/log(100/10) = 50.
The same formula was used to find moisture content at 25 No. of blows is 50%.
If = (w1-w2)/log(N2/N1),
50 = (w1-20)/log(100/25),
w1 = 50%.
(8)
Sneha Mathai said:
5 years ago
Take log(no of blows) only for finding flow index or for finding liquid limit using a flow index value.
If the graph is given and we want to find LL, then just do interpolate without taking the log of no of blows. Because we are plotting as such in semilog to get a straight line.
ie, (70-20)/(100-10) = (LL-20)/(100-25).
LL = 61.66.
If the graph is given and we want to find LL, then just do interpolate without taking the log of no of blows. Because we are plotting as such in semilog to get a straight line.
ie, (70-20)/(100-10) = (LL-20)/(100-25).
LL = 61.66.
(7)
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%.
Abdul Quader said:
6 years ago
N1=10, W1=70% & N2=100, W2=20%.
WL1 = W1*(N1/25)^0.121 = 70*(10/25)^0.121 = 62.65%,
WL2 = W2*(N2/25)^0.121 = 20*(100/25)^0.121 = 23.65,
Required Liquid Limit= WL1 - WL2.
= 62.65 - 23.65,
= 39%.
So, Answer (D) None of These.
WL1 = W1*(N1/25)^0.121 = 70*(10/25)^0.121 = 62.65%,
WL2 = W2*(N2/25)^0.121 = 20*(100/25)^0.121 = 23.65,
Required Liquid Limit= WL1 - WL2.
= 62.65 - 23.65,
= 39%.
So, Answer (D) None of These.
(1)
Tomin Nyodu said:
8 years ago
@Hemant Mahar.
Method 1
I = (w1-w2) /log(n2/n1)
I= (70-20)/log(100/10)
I= 50
Hence, the ans is option (b)
Method 2
W1=70(10/25)^0.121=62.65
W2=20(100/25)^0.121=23.65
W1-W2=38.99 (Wrong)
Why? This is an empherical Formula and the value of the exponent is between 0.068 to 1.210.
Method 1
I = (w1-w2) /log(n2/n1)
I= (70-20)/log(100/10)
I= 50
Hence, the ans is option (b)
Method 2
W1=70(10/25)^0.121=62.65
W2=20(100/25)^0.121=23.65
W1-W2=38.99 (Wrong)
Why? This is an empherical Formula and the value of the exponent is between 0.068 to 1.210.
Haseeb said:
4 years ago
The Correct Answer is 50%. Flow Index is 50. Flow Index = (w1-w2)/log(N2/N1) = (70-20)/log(100/10) = 50.
Now, using that Flow Index of 50 and the same formula to find moisture content at 25 No. of blows is 50%.
Flow Index = (w1-w2)/log(N2/N1),
50 = (w1-20)/log(100/25),
w1 = 50.1%.
Now, using that Flow Index of 50 and the same formula to find moisture content at 25 No. of blows is 50%.
Flow Index = (w1-w2)/log(N2/N1),
50 = (w1-20)/log(100/25),
w1 = 50.1%.
(19)
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