The following are the results of different tests on soil samples are provided in the following table. Soil sample Percent passing sieve Liquid Limit plasticity Index No 10 No 40 No 200 A 100 75 25 35 15 B 100 85 30 45 22 C 80 45 10 25 12 Classify the soil according to AASTHO classification and determine the Group Index in each case. You may use the information provide

 

Q#:The following are the results of different tests on soil samples are provided in the following table.                      

Soil sample	Percent passing sieve			Liquid Limit	plasticity Index
	No 10	No 40	No 200
A	100	75	25	35	15
B	100	85	30	45	22
C	80	45	10	25	12

Classify the soil according to AASTHO classification and determine the Group Index in each case. You may use the information provided.[12 marks]

Solution#3 (to be typed here by the student)

 ANSWER

 SOIL Sample  A

      Percentage passing through  No. 10  (2mm) seive = 100%

      Percntage passing  through No. 40 (425 μm) seive = 75%

      Percentage passing through No. 200 (75 μm) seive = 25%

 

 

              Liquid limit = wL = 35

             Plasticity index = I_p = 15

 

Percenrage passing through no. 200 seieve  < 35%

        So there for soil is granual

 

because Percentage passing through no.10 (2mm) seive > 50%

Percentage passing through no. 40 (425 μm) seive > 50%

Percentage passing through no.200(75 μm) >10%  < 35%

 

There fore soil group is A -2

 Because  wL < 40    and    I_p > 11

 

There fore soil subgroup is A-2-6

 Group index of the soil is given as

GI = 0.2a = 0.005ac +0.01 bd

a = percentage passing through  75 μm seive  -35                     a  ≤  40

b = percentage passing through 75 μm seive  -15                       b ≤  40

c = wL -40        c  ≤  20

d = I_p– 10        d  ≤  20

 

Therefore,

                        a= 25-35= -10

                        b =25-15 =10

                        c = 35-40 = -5

                        d = 15-10 = 5

GI= 0.2 × (-10)+0.005 ×(-10)(-5)+0.01 × (10)(5)

GI = -2+0.25+0.5

GI = -1.5

There is no negative group index of soil is zero

There for soil classified is  A-2-6

SOIL Sample  B

      Percentage passing through  No. 10  (2mm) seive = 100%

      Percntage passing  through No. 40   (425 μm) seive = 85%

      Percentage passing through No. 200  (75 μm) seive = 30%    (My reg no.)

 

              Liquid limit = wL = 45

             Plasticity index = I_p = 22

Percenrage passing through no. 200 seieve  >35%

        So there for soil is silt clay type

 

because wL  > 41   and I_p   >  11  

There fore soil group is A -7

(wL  - 30) = 45 -30 =15

 Because  and    I_p > (wL  - 30)

 

There fore soil subgroup is A-7-6

 Group index of the soil is given as

GI = 0.2a = 0.005ac +0.01 bd

a = percentage passing through  75 μm seive  -35                     a  ≤  40

b = percentage passing through 75 μm seive  -15                       b ≤  40

c = wL -40        c  ≤  20

d = I_p– 10        d  ≤  20

 

Therefore,

                        a= 30-35=-5

                        b =30-15 =15

                    

                        c = 45-40 = 5

                        d = 22-10 = 12

GI= 0.2 × (-5) +0.005 × (-5)  ×5 +0.01 × 15 ×12

GI = 0.675

The group index of soil is = 0.675

 

There for soil classified is  A-7-6(0.675)        (clay soil)

 

 

SOIL Sample  C

      Percentage passing through  No. 10  (2mm) seive = 80%

      Percntage passing  through No. 40 (425 μm) seive = 45%

      Percentage passing through No. 75 (75 μm) seive = 10%

 

 

              Liquid limit = wL = 25

             Plasticity index = I_p = 12

 

Percentage passing through no. 200 seieve  < 35%

        So there for soil is granual

 

because Percentage passing through no.10 (2mm) seive > 50%

Percentage passing through no. 40 (425 μm) seive < 50%

Percentage passing through no.200(75 μm) >10%  < 25%

 

 

 

There fore soil subgroup is A-2-6

 

Group index of the soil is given as

 

GI = 0.2a = 0.005ac +0.01 bd

a = percentage passing through  75 μm seive  -35                     a  ≤  40

b = percentage passing through 75 μm seive  -15                       b ≤  40

c = wL -40        c  ≤  20

d = I_p– 10        d  ≤  20

 

Therefore,

                        a= 10-35= - 25

                        b =10-15 = -5

                        c = 25-40 = 15

                        d = 12-10 = 2

 

GI= 0.2 × (-25)+0.005 ×(-25)(-15)+0.01 × (-5)(2)

GI = -5+1.875+0.01

GI = -3.135

There for group index of soil is zero

There for soil classified is  A-1-b(0)