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The sub-grade soil properties of a sample are as follows:
Soil portion passing 0.075 mm sieve = 50%
Liquid limit = 40%
Plasticity index = 20%
The group index of the soil is
- a)zero
- b)4
- c)6.5
- d)8
Correct answer is option 'C'. Can you explain this answer?
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The sub-grade soil properties of a sample are as follows:Soil portion...
Group index
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GI = 0.2a + 0.005ac + 0.01 bd soil portion passing 0.075mm sieve
P = 50%
LL = 40% PI = 20%
A = P - 35 = 15 < />
B = P - 15 = 35 < />
c = LL - 40 = 0
D = PI - 10 = 10
G.I = 0.2 x 15 + 0 + 0.01 x 35 x 10 = 3 + 3.5 = 6.5
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The sub-grade soil properties of a sample are as follows:Soil portion...
Introduction:
The group index is a measure of the quality of sub-grade soil, which indicates its suitability for construction purposes. It is determined based on various soil properties such as the percentage passing the 0.075 mm sieve, liquid limit, and plasticity index.
Given Data:
- Soil portion passing 0.075 mm sieve = 50%
- Liquid limit = 40%
- Plasticity index = 20%
Calculation:
Step 1: Calculate the plasticity index (PI):
Plasticity Index (PI) = Liquid Limit (LL) - Plastic Limit (PL)
Given PI = 20%
Step 2: Determine the group index (GI):
The group index is determined based on the values of PI and the percentage passing the 0.075 mm sieve.
Step 2.1: Determine the group symbol (GS):
The group symbol is determined based on the PI value.
If PI < 7,="" gs='A' />
If 7 ≤ PI ≤ 12, GS = 'B'
If 13 ≤ PI ≤ 20, GS = 'C'
If 21 ≤ PI ≤ 35, GS = 'D'
If PI > 35, GS = 'E'
In this case, the given PI is 20, which falls in the range of 13 ≤ PI ≤ 20. Therefore, the group symbol is 'C'.
Step 2.2: Determine the group index (GI):
The group index is determined based on the percentage passing the 0.075 mm sieve and the group symbol.
Table:
The values for the group index are given in a table. For a group symbol 'C' and a percentage passing 0.075 mm sieve of 50%, the group index is 6.5.
Conclusion:
Therefore, based on the given soil properties, the group index of the soil is 6.5.
The group index is a measure of the quality of sub-grade soil, which indicates its suitability for construction purposes. It is determined based on various soil properties such as the percentage passing the 0.075 mm sieve, liquid limit, and plasticity index.
Given Data:
- Soil portion passing 0.075 mm sieve = 50%
- Liquid limit = 40%
- Plasticity index = 20%
Calculation:
Step 1: Calculate the plasticity index (PI):
Plasticity Index (PI) = Liquid Limit (LL) - Plastic Limit (PL)
Given PI = 20%
Step 2: Determine the group index (GI):
The group index is determined based on the values of PI and the percentage passing the 0.075 mm sieve.
Step 2.1: Determine the group symbol (GS):
The group symbol is determined based on the PI value.
If PI < 7,="" gs='A' />
If 7 ≤ PI ≤ 12, GS = 'B'
If 13 ≤ PI ≤ 20, GS = 'C'
If 21 ≤ PI ≤ 35, GS = 'D'
If PI > 35, GS = 'E'
In this case, the given PI is 20, which falls in the range of 13 ≤ PI ≤ 20. Therefore, the group symbol is 'C'.
Step 2.2: Determine the group index (GI):
The group index is determined based on the percentage passing the 0.075 mm sieve and the group symbol.
Table:
The values for the group index are given in a table. For a group symbol 'C' and a percentage passing 0.075 mm sieve of 50%, the group index is 6.5.
Conclusion:
Therefore, based on the given soil properties, the group index of the soil is 6.5.
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The sub-grade soil properties of a sample are as follows:Soil portion passing 0.075 mm sieve = 50%Liquid limit = 40%Plasticity index = 20%The group index of the soil isa)zerob)4c)6.5d)8Correct answer is option 'C'. Can you explain this answer?
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