Civil Engineering - Soil Mechanics and Foundation Engineering - Discussion
Discussion Forum : Soil Mechanics and Foundation Engineering - Section 3 (Q.No. 19)
19.
A soil sample has passing 0.075 mm sieve = 60% liquid limit = 65% and plastic limit = 40%. The group index of the soil, is
Discussion:
48 comments Page 4 of 5.
Roop Narayan (NIT SURAT) said:
9 years ago
GI = 0.2 a + 0.005 ac + 0.01 bd.
a = P - 35 = 60 - 35 = 25.
b = P - 15 = 60 - 15 = 45.
c = LL - 40 = 65 - 40 = 25.
d = PI - 10 = 40 - 10 = 30.
GI = (0.2 * 25) + (0.05 * 25 * 25) + (0.01 * 45 * 30) = 29.125.
Therefore the correct answer is (D)
a = P - 35 = 60 - 35 = 25.
b = P - 15 = 60 - 15 = 45.
c = LL - 40 = 65 - 40 = 25.
d = PI - 10 = 40 - 10 = 30.
GI = (0.2 * 25) + (0.05 * 25 * 25) + (0.01 * 45 * 30) = 29.125.
Therefore the correct answer is (D)
Roop Narayan (nit surat) said:
9 years ago
GI = 0.2a + 0.005ac + 0.01bd.
a = P - 35 = 60 - 35 = 25.
b = P - 15 = 60 - 15 = 45.
c = LL - 40 = 65 - 40 = 25.
d = PI - 10 = 40 - 10 = 30.
GI = (0.2 * 25) + (0.05 * 25 * 25) + (0.01 * 45 * 30) = 29.125 ans.
a = P - 35 = 60 - 35 = 25.
b = P - 15 = 60 - 15 = 45.
c = LL - 40 = 65 - 40 = 25.
d = PI - 10 = 40 - 10 = 30.
GI = (0.2 * 25) + (0.05 * 25 * 25) + (0.01 * 45 * 30) = 29.125 ans.
GiRi said:
9 years ago
GI = (F-35) (0.2 + 0.05(wL - 40)) + 0.01(F - 15)(Ip - 10).
Where,
Ip = wL - Wp.
We get answer is 12.
Where,
Ip = wL - Wp.
We get answer is 12.
SAMKIT said:
9 years ago
The correct method is;
A = p-35 shud not be greater than 40,
B = p-15 shud not be greater than 40,
C = LL-40 shud not be greater than 20,
D = Ip-10 shud not be greater than 20,
Put in GI = 0.2a + 0.01bd + 0.005ac.
We get GI = 0.2*25+0.01 * 40 * 15 + 0.005 * 25 * 25.
= 14.125.
A = p-35 shud not be greater than 40,
B = p-15 shud not be greater than 40,
C = LL-40 shud not be greater than 20,
D = Ip-10 shud not be greater than 20,
Put in GI = 0.2a + 0.01bd + 0.005ac.
We get GI = 0.2*25+0.01 * 40 * 15 + 0.005 * 25 * 25.
= 14.125.
Ravi Kant said:
9 years ago
a = 60-35 = 25%
b = 60-15 = 45%
c = 65-40 = 25%
d = PI-10 = (65-40)-10 = 15%.
GI = 0.2a + 0.005ac + 0.01bd = 0.2 * 25 + 0.005 * 25 * 25 + 0.01 * 45 * 15 = 14.875.
b = 60-15 = 45%
c = 65-40 = 25%
d = PI-10 = (65-40)-10 = 15%.
GI = 0.2a + 0.005ac + 0.01bd = 0.2 * 25 + 0.005 * 25 * 25 + 0.01 * 45 * 15 = 14.875.
CHAUDHARY said:
8 years ago
@Samkit
GI = .2 * 25 + .01 * 40 * 15 + .005 * 20 * 25 = 13.5.
GI = .2 * 25 + .01 * 40 * 15 + .005 * 20 * 25 = 13.5.
Deep said:
1 decade ago
F = fineness constant.
L.l = liquid limit.
P.l = plastic limit.
Group index = (F-35)[0.2+(0.005)(Ll-40)]+0.01(F-15)(Pl-10).
G I = (60-35)[.2+(0.005)(65-40)]+0.01(60-15)(40-10).
Group index = 21.625.
L.l = liquid limit.
P.l = plastic limit.
Group index = (F-35)[0.2+(0.005)(Ll-40)]+0.01(F-15)(Pl-10).
G I = (60-35)[.2+(0.005)(65-40)]+0.01(60-15)(40-10).
Group index = 21.625.
Akash D Sajjanar said:
8 years ago
GI= 0.2a+0.005ac+0.01bd.
a= % passing 0.075 mm sieve (min 35 max 75)= 60-35= 25 (max value is 40 if >40 take a value as 40).
b= % passing 0.075 mm sieve (min 15 max 55)= 60-15= 45 > 40 so take b=40.
c= liquid limit max 40 = 65-40 = 25 > 20 so take c=20.
d= plasticity index = 25-10 = 15 < 20 so d=15.
GI = (0.2*25)+(0.005*25*25)+(.01*40*15)=14.125.
So, answer is GI=14.125.
a= % passing 0.075 mm sieve (min 35 max 75)= 60-35= 25 (max value is 40 if >40 take a value as 40).
b= % passing 0.075 mm sieve (min 15 max 55)= 60-15= 45 > 40 so take b=40.
c= liquid limit max 40 = 65-40 = 25 > 20 so take c=20.
d= plasticity index = 25-10 = 15 < 20 so d=15.
GI = (0.2*25)+(0.005*25*25)+(.01*40*15)=14.125.
So, answer is GI=14.125.
Praful sl said:
8 years ago
Not 14.875, 14.125 is correct.
Ganesh sharma said:
8 years ago
Answer will be 13.5.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers