Civil Engineering - Soil Mechanics and Foundation Engineering - Discussion

GI= { 0. 2 a + 0. 005 ac + 0. 01 bd } ------> eq. (1).

A = % of soil passing through 0. 075 sieve - 35. {limit 1 < a < 40}.
A = 60 - 35 = 25 (ok within limit).

B = % of soil passing through 0. 075 sieve - 15. {limit 1 < b < 40}.
B = 60 - 15 = 45 (It's over limit) so we will take = 40.

C = Liquid limit - 40. {limit 1 < c < 20}.
C = 65 - 40 = 25 (It's over limit) so we will take = 20.

D = plasticity index - 10. {limit 1 < d < 20}.
{Plasticity index = Liquid limit - Plastic limit}.
D = (65 - 40) - 10 = 15 (ok within limit).

Putting the value of a, b, c, d in eq. (1).

GI = 0. 2 a + 0. 005 ac + 0. 01 bd.
GI = (0. 2 x 25) + (0. 005 x 25 x 20) + (0. 01 x 40 x 15).
GI = 13. 5.
So, The answer will be 13. 5.
And None of these is the correct one.

Right @Pkota.

I too agree, the answer is 13.5.

Agree @Vivek

14.875 is the right answer.

GI = 0.2a + 0.005ac + 0.01bd.
a = F-35 = 60-35 = 25.

Which is less than 40 so a=25
b = F-15=60-15 = 45 which is greater than 40 so take b=40
c = LL-40=65-40 = 25 which must not be greater than 20 so take c=20
d = Ip-10=(LL-PL)-10 = (65-40)-10=15
GI = 0.2*25 + 0.005*25*20 + 0.01*40*15,
= 5 + 2.5 + 6 = 13.5.

GI = 0.2a + 0.005ac + 0.01bd.
GI = 0.2(F-35) + 0.005(F-35)(WL-40) + 0.01(F-15)(IP-10).

You are right, thanks @Gaurav Sharma.

G.I = 0.2 * a +0.005 * a * c+ 0.01 * b * d.
G.I = 0.2 * 25 + 0.005 * 25 * 15 + 0.01 * 40 * 10 = 10.875.

a = P-35 = 60 - 35 = 25 ( 0-40 ) Hence adopt this value.
b = P- 15 = 60 - 15 = 45 (0-40) Hence adopt 40 .
c = LL- 40 = 65 - 40 = 15 (0-20 ) Hence adopt this value
d = PI - 10 = 65 - 40 - 10 = 15 (0-10) Hence adopt 10.

GI = .2 * a +.005* a*c +.01*b*d.

a = P (% passing from 75-micron sieve ) - 40 (value not more than 40.. if more than 40 then take 40).

a= 60-35= 25 ok not more than 40.
b= P (% passing from 75 micron sieve ) - 15 (value not more than 40.. if more than 40 then take 40),
b= 60 - 15 =45 ( not ok ) (take 40).

c= wL (liquid limit ) -40 (value not more than 20.. if more than 20 then take 20),
c= 65- 40 = 25 (not ok ) ( take 20).

d= IP ( plasiticity index ) - 10,
d=( WL liquid limit( 65) - (40)WP plastic limit )- 10,
= 25-10 = 15 ok value less than 20.

So , G.I = .2*25 +.005*25 * 25 + .01*40 * 15,
and 14.125 is the correct answer.

A=60-35=25 (0 to 40),
B=60-15=45 (0 to 40) = 40,
C=65-40=25 (0 to 20) =20,
Ip=65-40=25.

D=25-10=15 (0 to 20),
G.I = 0.2*a + 0.01*b*d +0.005*a*c.
= 0.2 * 25 + 0.01 * 40 * 15 + 0.005 * 25 * 20.
= 13.5.

So, here, option D is correct.

A is the correct answer guys.

0.2a+0.01ad+0.005ac.

a = %particle passing through 75micron sieve and their given % should be lie between (35<a>75) means whose value ranges from (0-40) in this question given value is 60% which means (a=60-35=25).

b=%particle passing through 75-micron sieve and lies between (15<b>55) which means (0-40) but in this question which is 60 % so it is not full fill our condition so b=0.

c = liquid limit lies between (40<c>60) which lies on range (0-20) in this question which is 65% so,

c = 0
d = plasticity index lies between (10<d>30) on range (0-20) in this question which is P.I = 65-40=25.

So, its range is d = 25 - 10 = 15.
d = 15.

Summary
a=25
b=0
c=0
d=15

G.I = 0.2a + 0.01bd + 0.005ac.
= 0.2 * 25+0.01 * 0 * 15 + 0.005 * 25 * 0.
= 5.0 + 0 + 0
G.I = 5.0.
So, A is the correct answer.