Civil Engineering - Soil Mechanics and Foundation Engineering - Discussion

By interpolation wL = 61.67.
By flow index method wL= 50.

Please clarify the answer.

We can get it from flow index value, I.

n1 = 10; w1 = 70%.
n2 = 100; w2 = 20%.

Flow index (I) = (w1-w2)/log(n2/n1) = 50;

wL = Ilog (n2/25)+w2;

log(2) = 0.301.

wL = 50.10%.

OPTION B.

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%.

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%.

@Aakash karoo.

Your simple steps helped me get the right answer which is 65%, the graph part is actually much easier!

Thanks.

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 = 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%.

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.

Correct answer is 50.1.

Simply use equation ll = -if(logx)+c.

Two equation, two unknown, solve it you will get ll = 50.1.