Question

stem. Give group symbol and group name. 5.7 Classify the following soils using the Unified soil classification system. Give the group symbols and the group names Percent passing Sieve size E A C No, 4 94 98 100 100 100 No. 10 63 86 100 100 100 No. 20 21 50 98 100 100 No. 40 10 28 93 99 94 No. 60 7 95 18 88 82 No. 100 14 90 66 8.5 ON 2 29 d or o Copyrigt 2ca e gnt day eoed mmo 153 Problems I Percent passing A Sieve size 45 77 86 No. 200 10 42 26 65 001 mm 17 60 0.002 mm 55 36 63 Liquid limit Plasticity index 22 28 25 NP NP

Answer 1

Unified Soil Classification System:

		%finer
Sieve Designation	Sieve size (mm)	Soil A	Soil B	Soil C	Soil D	Soil E
No.4	4.75	94	98	100	100	100
No.10	2	63	86	100	100	100
No.20	0.84	21	50	98	100	100
No.40	0.425	10	28	93	99	94
No.60	0.25	7	18	88	95	82
No.100	0.15	5	14	83	90	66
No.200	0.075	3	10	77	86	45
0.01mm	0.01	0	0	65	42	26
0.002mm	0.002	0	0	60	17	21

-Soil A Soil B Soil C -Soil D Soil E 0.01 0.001 0.1 10 Sieve size (mm) Jau%

[Vertical scale: 1 unit = 20%]

Soil A:

% finer for sieve No.4 (4.75 mm) = 94%

% finer for sieve No.200 (0.075 mm) = 03%

So it is clear that more than 50% of the materials are coarser than sieve No.200. Hence it is coarse grained soil with 97% coarse fraction and 03% fine fraction.

As more than 50% of the coarse fraction is finer than No.4 sieve, hence it is sand.

Also, as the % finer for No.200 sieve is (3%) less than 5%, hence it is clean sand.

From the graph,

D₆₀ = 1.9, D₃₀ = 1 and D₁₀ = 0.4

Coefficient of uniformity (C_u) = D₆₀ / D₁₀ = 1.9/0.4 = 4.75 < 6

Coefficient of curvature (C_c) = (D₃₀)²/ (D₆₀ * D₁₀)= (1)²/ (1.9 * 0.4) = 1.32 (1<1.32<3)

For well graded sand coefficient of uniformity is more than 6. In comparison our obtained value is 4.75.

Hence the given soil A is poorly graded sand whose group symbol is (SP)

Soil B:

% finer for sieve No.4 (4.75 mm) = 98%

% finer for sieve No.200 (0.075 mm) = 10%

So it is clear that more than 50% of the materials are coarser than sieve No.200. Hence it is coarse grained soil with 90% coarse fraction and 10% fine fraction.

As more than 50% of the coarse fraction is finer than No.4 sieve hence it is sand.

Also, as the % finer for No.200 sieve is (10%) more than 5%, hence it is sand with fines (more than 12% passes No.200 sieve).

From the graph,

D₆₀ = 1.1, D₃₀ = 0.45 and D₁₀ = 0.075

Coefficient of uniformity (C_u) = D₆₀ / D₁₀ = 1.1/0.075 = 14.66 > 6

Coefficient of curvature (C_c) = (D₃₀)²/ (D₆₀ * D₁₀)= (0.45)²/ (1.1 * 0.075) = 2.45 (1<2.45<3)

For well graded sand coefficient of uniformity is more than 6. In comparison our obtained value is 14.66 which is more than the required value.

Also, for well graded sand coefficient of curvature is between 1 and 3. In comparison our obtained value is 2.45 which satisfies the required condition.

Hence the given soil B is well graded sand whose group symbol is (SW)

Soil C:

% finer for sieve No.4 (4.75 mm) = 100%

% finer for sieve No.200 (0.075 mm) = 77%

So it is clear that more than 50% of the materials are finer than sieve No.200. Hence it is fine grained soil with 23% coarse fraction and 77% fine fraction.

Liquid limit = 63, Plastic limit = 25

Therefore, Plasticity index = 63 - 25 = 38

And, LL = 63% which is more than 50%, the soil is heavily compressible

In order to use plasticity chart

Plasticity index of A-line = 0.73 * (LL - 20) = 0.73 (63-25) = 0.73 * 43 = 31.4 < 38

Therefore Plasticity index of soil is more than the plasticity index of A-line. Hence the soil is clayey soil.

Also liquid limit is more than 50%. Hence it is Heavy compressible clay whose group symbol is (CH)

From the graph it is clear that D₁₀ of the soil can not be determined. But by looking at the graph it is clear that all the different size particles are present in suitable proportion which is a good indication of well graded soil.

Soil D:

% finer for sieve No.4 (4.75 mm) = 100%

% finer for sieve No.200 (0.075 mm) = 86%

So it is clear that more than 50% of the materials are finer than sieve No.200. Hence it is fine grained soil with 14% coarse fraction and 86% fine fraction.

Liquid limit = 55, Plastic limit = 28

Therefore, Plasticity index = 55 - 28 = 27

And, LL = 55% which is more than 50%, the soil is heavily compressible

In order to use plasticity chart

Plasticity index of A-line = 0.73 * (LL - 20) = 0.73 * (55 - 20) = 25.55 < 27

Therefore Plasticity index of soil is more than the plasticity index of A-line. Hence the soil is clayey soil.

Also liquid limit is more than 50%. Hence it is Heavy compressible clay whose group symbol is (CH)

From the graph it is clear that D₁₀ of the soil can not be determined. But by looking at the graph it is clear that some of the particle sizes are not present in suitable proportion which means it is not a well graded soil.

Soil E:

% finer for sieve No.4 (4.75 mm) = 100%

% finer for sieve No.200 (0.075 mm) = 45%

So it is clear that less than 50% of the materials are finer than sieve No.200. Hence it is coarse grained soil with 55% coarse fraction and 45% fine fraction.

As more than 50% of the coarse fraction is finer than No.4 sieve hence it is sand.

Also, as the % finer for No.200 sieve is (45%) more than 5%, hence it is sand with fines (more than 12% passes No.200 sieve).

Liquid limit = 36, Plastic limit = 22

Therefore, Plasticity index = 36 - 22 = 14

In order to use plasticity chart

Plasticity index of A-line = 0.73 * (LL - 20) = 0.73 * (36 - 20) = 11.68 < 14

Therefore Plasticity index of soil is more than the plasticity index of A-line. Hence the soil is clayey soil.

Therefore the given soil is clayey sand whose group symbol is (SC)

From the graph it is clear that D₁₀ of the soil can not be determined. But by looking at the graph it is clear that some of the particle sizes are not present in suitable proportion which means it is not a well graded soil.