4. Major soil deposits in India
1. Alluvial deposits
2. Black cotton soils
3. Lateritic soils
4. Desert soils
5. Marine deposits
5.
6. •Soil Particle size
Based on the size of the particle soil classified into 4 types:
1. Clay: >0.002mm
2. Silt: 0.002mm to 0.075mm
3. Sand: 0.075mm to 4.75mm
4. Gravel: >4.75mm
/08/10
7. Soil particle size classification
Grain size(mm)
Name of Organization Gravel Sand Silt Clay
MIT classification (Massachusetts Institute ofTechnology) >2 2 to 0.06 0.06 to 0.002 <0.002
US dept. of Agriculture (USDA) >2 2 to 0.05 0.05 to 0.002 <0.002
AASHTO classification 76.2 to 2 2 to 0.075 0.075 to 0.002 <0.002
Unified soil classification system 76.2 to 4.75 4.75 to 0.075 0.075 to 0.002 <0.002
8. Specific Gravity (G)
• Specific gravity (G) : Ratio of the unit weight of the given material to unit
weight of the water
• Sp. gravity = gs / gw
• Sp. gravity of sandy soil = 2.65 to 2.8
• Sp. gravity of Clay soil = 2.6 to 2.9
•Application: It is required to solve the geotechnical
problems
10. • For coarse grained –
particle size & relative density
• For fine grained—
atterberg’s limits & consistency
11. Mechanical analysis of soil
• Objective: To determination of the size range of particles present in
a soil
Two methods:
1. Sieve analysis: for particles larger than 0.075mm in diameter
2. Hydrometer analysis: for particles smaller than 0.075mm in
diameter
12. Sieve Analysis
• Principle: Finding grading of coarse grained soil through the set
of sieves
• Apparatus/Procedure: Sieves sizes
•For Gravel : 80mm,40mm, 20mm,10mm, 4.75mm
•For Sands : 2mm, 1mm, 600m, 425m, 212m, 150m, 75m
• Output result/Application: Percentage finer
13.
14. Hydrometer Analysis
• Principle: Stokes law: v = (rs-rw),D2/18n
V= particle velocity
• Apparatus/Procedure: Hydrometer placed into cylinder with a liquid
of sodium hexameta phosphate
Output result / Application:
rs = density of solids
rw = density of water
Gs = Sp. Gravity of solids
u= Viscosity of liquids
L = distance
T = time
t
L
wGS
D .
).1(
18
r
15. Particle size distribution curve
•From graph: Graph analysis
•Effective size: D10
D10 = Particle size such that 10% of the soil is finer than
this size
D60= Particle size such that 60% of the soil is finer than
this size
•Uniform coefficient (Cu): D60/D10
Cu for sands - 6 - called well graded sands
Cu for gravel – 4 – called well graded gravels
•Coefficient of curvature (Cc): (D30)2 / (D60 x D30)
•Cc for well graded soils lies between 1 to 3
16. • The larger the numerical value of cu the more is the range of
particles.
• Soil with a value of cu less than 2 are uniform soils.
• Soil with a value of cu of 6 or more are well graded
24. Uses of Particle size distribution curve
Usefull for coarse grained soils & fine grained soils
Others:
•1. Classifying the coarse grained soils
•2. Coeff. Permeability will depend on the particle size.
It can determine accurately
•3. useful to design the drainage filters
•4. it provides an index to the shear strength of the soil.
•5. Soil stabilization and pavement design
25. Particle shape
• 1. Bulky (length,width,thickness of same order of magnitude)
-cohesionless soil
• 2. Flaky
26.
27.
28. • R.D. of a soil gives idea of the densenss and indicates how it would
behave under loads.
• If the deposit is dense, It can take heavy loads with little
settlement.
29. Problem1: The results of sieve analysis of a soil are
given below, total massofsample900gm Draw the
particle size distribution curve hence determine the
uniformity coefficient and the Cc
IS Sieve 20m
m
10m
m
4.75mm 2mm 1.0mm 0.6mm 4.25m 212m 150
m
75m pan
Mass of
soil
retaine
d (gm)
35 40 80 150 150 140 115 55 35 25 75
31. weight – volume relationships
•Weight- volume relationship:
•Three phase diagram(block diagram): Soil mass consists of 3
constituents
•1. Solids(soil)
•2. Water
•3. Air
•Actually it can’t be segregated, a 3-phase diagram is an artifice
used for easy understanding and convenience for calculations.
•When the soil absolutely dry (no water content) – 2-
phase diagram.
•When the soil fully saturated (no air)- 2 phase diagram.
/08/10
32. Three Phases in Soils
S : Solid Soil particle
W: Liquid Water (electrolytes)
A: Air Air
34. Three Volumetric
Ratios
• (1)Void ratio e (given in decimal, 0.65)
•(2) Porosity n (given in percent 100%)
•(3) Degree of Saturation S (given in percent 100%, 65%)
)V(solidsofVolume
)V(voidsofVolume
e
s
v
)V(samplesoilofvolumeTotal
)V(voidsofVolume
n
t
v
%100
)V(voidsofvolumeTotal
)V(watercontainsvoidsofvolumeTotal
S
v
w
e1
e
)e1(V
eV
n
s
s
35. Engineering Applications (e)
•Typical values •Engineering applications:
•Volume change tendency
•Strength
(Lambe and Whitman, 1979)
Simple cubic (SC), e = 0.91, Contract
Cubic-tetrahedral (CT), e = 0.65, Dilate
Link: the strength of
rock joint
)itan(strengthShear n
i
36. 1. Void ratio (e): Volume of voids to the volume of solids.
e = Vv/Vs
Expressed in decimal like 0.4 , 0.5..etc
For coarse grained soil smaller than that for fine grained soils.
Some soils it may have a value even greater than unity.
2. Porosity (n): Volume of voids to the total volume
n = Vv/V
Also called percentage of voids, expressed in percentage,
The porosity of soil cannot exceed 100%
Reason: Vv cannot be greater than V
Both porosity and void ratio are measures of the denseness or
looseness of soil.
As the soil becomes more and more dense, e, n values will decrease
37. Relationship for ‘e’ and ‘n’:
1/n = V/Vv
1/n = Vs/Vv+Vv/Vv
1/n = e+1
n = e/(1+e),
e = n/(n-1)
3. Degree of saturation (s): Ratio of the volume of water to the
volume of voids.
S = Vw/Vv
Expressed in percentage (%)
It is zero when the soil is absolutely dry
It is 100% when the soil is fully saturated
38. 4. Percentage air voids (na): Ratio of the volume of air to the
total volume.
na = Va/V …….expressed in percentage (%)
5. Air content (ac): Ratio of the volume of air to the volume of
voids.
ac = Va/Vv….expressed in percentage (%)
Both air content and percentage air voids are zero when the
soil is saturated (Va = 0).
39. 6. Water content (w): Ratio of the weight of water to the weight of solids in a
given soil sample.
w = Ww/Ws
Expressed in percentage (%).
7. Unit weight (g): Weight of soil per unit volume
g = W/V
From the drawing, g = W / V,
g = Ws+Ww/V
g = Ws(1+Ww/Ws)/V
g = Ws(1+w)/V
8. Dry unit weight (gd): gd = Ws/V (Weight of solids per unit volume)
Relationship between moisture content (w, gd and g),
gd = g/(1+w)
40. Weight Relationships
• (1)Water Content w (100%)
•
• For some organic soils w>100%, up to 500 %
• For quick clays, w>100%
• (2)Density of water (slightly varied with
temperatures)
•(3) Density of soil
•a. Dry density
•b.Total,Wet, or Moist density (0%<S<100%,
Unsaturated)
•c. Saturated density (S=100%,Va =0)
•d. Submerged density (Buoyant density)
%100
)(
)(
s
w
MsolidssoilofMass
MwaterofMass
w
)V(samplesoilofvolumeTotal
)M(solidssoilofMass
t
s
d r
)V(samplesoilofvolumeTotal
)MM(samplesoilofMass
t
ws
r
)V(samplesoilofvolumeTotal
)MM(watersolidssoilofMass
t
ws
sat
r
wsat
'
rrr
333
w m/Mg1m/kg1000cm/g1 r
41. Relationship among unit weight, void ratio, moisture content
and sp.gravity
e
wGs
VWsd
e
wGsw
e
wGswwGs
V
WwWs
VW
wGswWswWw
wGsWs
1
.
/
1
.).1(
1
...
/
...
.
g
g
ggg
g
g
g
Considering the volume of soil solids is 1, If the volume of soil solids is 1,
then the volume of voids numerically equal to void ratio, e. The weights of soil
solids and water can be given as
Because the weight for the soil element under consideration is w.Gs.gw, the
volume occupied by water
Gsw
w
wGsw
w
Ww
Vw .
..
g
g
g
Hence from the definition of degree of saturation, S = Vw/Vv = w. Gs/ e
/08/10
42. • Various unit weight relationships: (Usefull for solveing the problem)
Moist unit weight (g) Dry unit weight(gd) Saturated unit weight(gsat)
wSnnwGs
wnwGs
S
Gsw
wGsw
e
weSGs
e
wGsw
ggg
gg
g
g
g
g
g
g
..)1(.
)1)(1(.
.
1
.).1(
1
)..(
1
.).1(
wnsatd
e
we
satd
nwGsd
e
wGs
d
w
d
ggg
g
gg
gg
g
g
g
g
.
1
.
)1(.
1
.
1
wnGsnsat
e
weGs
sat
gg
g
g
)].).1[(
1
).(
43. • Problem:1 The mass of a soil sample having a volume of 0.0057m3 is 10.5kg, the
moisture content (w) and the specific gravity of soil solids(Gs) were determined to
be 13% and 2.68, respectively, Determine
a. Moist density, r, b. Dry density, rd, c. Void ratio, e, d. Porosity, n, e. Degree of
saturation, S (%)
Solution: r = M/V, rd = r / (1+w), e = (Gs.gw/rd)-1, n = e/(1+e), S(%) =
(w.Gs/e)x100
Problem. 2, A soil has void ratio = 0.72, moisture content = 12% and Gs= 2.72.
Determine its
(a) Dry unit weight
(b) Moist unit weight, and the
(c) Amount of water to be added per m3 to make it saturated
Problem.3, The dry density of a sand with porosity of 0.387 is 1600 kg/m3. Find
the void ratio of the soil and the specific gravity of the soil solids.
44. Problem: A moist soil sample weighs 3.52N. After drying in an
oven, its weight is reduced to 2.9N. The specific gravity of solids
and the mass specific gravity are respectively 2.65 and 1.85.
Determine the water content, void ratio and the degree of
saturation. gw = 10kN/m3
Solution: From given data: Weight of water
Water content, w
g = Gm.gw
gd = g/(1+w)
gd = G.gw/(1+e)
n = e / (1+e)
S = w.G/e
45. Relative density
Application: In granular soils, the degree of compaction in the field can be
measured according to the relative density.
Relative density (Dr) = (emax – e)/(emax – emin)
Expressed in percentage (%)
emax = void ratio of the soil in the loosest state
emin = void ration of the soil in the densest state
e = insitu void ratio of the soil
The term relative density is commonly used to indicate the insitu denseness
or looseness of granular soil
Some values for granular soil
Relative density (%) Description of soil deposit
0 - 15 Very loose
15 - 50 Loose
50 -70 Medium
70 - 85 Dense
85 - 100 Very dense
/08/10
46. Problem:2- For a given sandy soil, emax = 0.82, emin = 0.42, Gs =
2.66. In the field the soil is compacted to a moist density of 1720 kg/m3
at a moisture content of 9%. Determine the relative density of
compaction.
Solution: Calculate the ‘e’ value
r = (1+w).Gs.rw/(1+e)
Dr = (emax – e) / (emax – emin)
Problem:3 The laboratory test results of a sand are as follows, emax = 0.91,
emin = 0.48, and Gs = 2.67. What would be the dry and moist unit weights of
this sand when compacted at a moisture content of 10% to relative density of
65%? Ans: 16.07kN/m3, 17.68kN/m3
47. Problem: A fully saturated sample of soil has a volume of 25cc. And a weight 0f
40gm. After drying in the oven, its weight is 28gm. With the help of Phase
diagram, calculate the, 1.Void ratio, 2. Water content 3. Porosity, 4. degree of
saturation 5. Saturated unit weight. (Ans: e= 0.923, w = 42.9%, n = 48%, s
=100%, gsat = 1.6 g/cc
Problem: when a given soil sample of sand was tested in the laboratory, the
void ratio in the loosest and densest possible states were 0.95 and 0.4
respectively . Calculate the 1. Relative density 2. Degree of saturation (Ans:
e=, s = %, Dr = %)
Problem. Tests on fill reveal that one cubic metre of soil on the fill weighs 1624
kgs and after being dried weighs 1.40 tonnes. If the specific gravity of solida
is 2.65, determine w, e, n, s of the soil mass from the first principles.
48. Consistency of soil – Atterberg Limits
•Liquid limit (LL): It is the moisture where the soil sample starts
behaves like a liquid.
•Plastic limit (PL): It is the moisture where the soil sample starts
behaves like a plastic
•Shrinkage Limit (SL): The moisture content at which the
transaction from solid to semi solid state takes place.
49.
50.
51. • Plasticity of a soil is its ability to undergo deformation without
cracking or fracturing.
• - due to the presence of clay minerals.
• Plastic soil can be moulded into various shapes when it is wet.
52.
53.
54. Liquid Limit test
Objective: determination of liquid limit from clay silt
soil
Apparatus: Liquid limit test device, grooving tools
Output: Flow index, IF = (w1 – w2) / log(N2/N1)
w1 = moisture content of soil, in percent, corresponding
to N1 blows
w2 = Moisture content corresponding to N2 blows
57. Flow curve for liquid limit determination of
clayey siltGraph: X – Number of blows, N
y – Moisture content,w (%)
58.
59. • Flow index is the rate at which a soil mass loses its shear strength
with an increase in water content.
• -- soil with greater value of flow index has a steeper slope, hence
lower shear strength as compared to flatter.
60. Plastic Limit Test
Plastic limit: The moisture content in percent at which
the soil crumbles, when rolled into threads of 3.2mm
in diameter.
Test: Determination of plastic limit from the laboratory
61.
62. Plasticity Index
Plasticity Index: The difference between the Liquid
limit and plastic limit.
PI = LL – PL
When either LL or PL cannot be determined, the soil is non
plastic.
When the plastic limit is greater than Liquid limit, the plasticity
index reported as zero(not negative).
63.
64.
65.
66.
67. • CI indicates the consistency(firmness) of a soil.
• It shows the nearness of water content of the soil to its plastic limit.
• A soil with CI of zero is at L.L. it is extremely soft and has negligible shear
strength.
68.
69. • Amount of water in a soil mass depends upon the type of clay
mineral present.
• Activity is a measure of water holding capacity of clayey soil.
• The change in the volume of a clayey soil during swelling or
shrinkage depend upon the activity.
70.
71.
72. Uses Of Consistency Limits
• L.L. and P.L. depend upon the type and amount of clay in a soil.
However the P.I, depends on the amount of clay.
• The P.I.of a soil is a measure of amount of clay in a soil.
82. Classification based on Plasticity chart
Silts and clays Liquid limits is 50% or less
ML = Inorganic silts with low plasticity
CL = Inorganic clays of low to medium plasticity
OL = Organic silts of low plasticity
Aline , PI = 0.73(LL – 20)
Silts and clays Liquid limits is greater than 50%
MH - Inorganic silts of high plasticity
CH – Inorganic clays of high plasticity
OH – Organic clays of medium plasticity
Pt - Peat or muck or highly organic soils
83. Problem: The weight of a moist soil sample is 20kg and its volume is 0.011m3.
After drying in an oven, the weight of sample reduces to 16.5kg. Determine
the w, gd, g, e, n, s.
87. Engineering classification of soils
AASHTO Classification system:
Soil classified into Seven major groups:
Soil classified under groups A-1, A -2, A -3 are
granular material of which 35% or less of the particles
pass through the no.200(0.075mm) sieve.
Soils of which more than 35% pass through the no.200
sieve are classified under groups
A -4, A – 5, A -6 & A -7
/08/10
88. AASHTO classification…..
1. Grain size
a. Gravel : Particle passing through 75mm sieve
and retained on the No.10(2mm) US sieve
b. Sand: Particle passing through the no.10(2mm)
and retained on the No.200 (0.075mm) US sieve.
c. Silt and Clay: Particle passing the no.200 US
sieve
90. Group Index (GI):
Significance: To evaluate the quality of soil as a
highway material.
GI = (F200 – 35)[0.2+0.005(LL – 40)]+0.01(F200 – 15)(PI -10)
Where, F200 = Percentage passing through the No. 200 sieve
LL = liquid limit
PI = Plastic limit
1st term - (F200 – 35)[0.2+0.005(LL – 40)] – Partial group index
determined from liquid limit
2nd term - 0.01(F200 – 15)(PI -10) – Partial group index
determined from plastic limit
91. Group Index…….
Note: In general the quality performance of soil as a
subgrade material is inversely proportional to the GI
1. If in the equation yields a negative value for GI, it is
taken as 0.
2. The group index calculated is rounded off to the nearest
whole number
3. There is no upper limit for the group index
4. The group index of soils belonging to groups A–1-a, A-
1-b, A-2-4, A-2-5 and A-3 is always zero
92. Problem 1: The result of the particle size analysis of a soils as
follows:
a. % passing through no.10 sieve = 100
b. % passing through no.40 sieve = 80
c. % passing through no.200 sieve = 58
liquid limit and plasticity index of the minus no.40 fraction of the
soil are 30 and 10 respectively. Calculate the GI
Solution:
GI = (F200 – 35)[0.2+0.005(LL – 40)]+0.01(F200 – 15)(PI -10)
93. Problem:2, Ninty five percent of a soil passes through the no.200
sieve and has a liquid limit of 60 and plasticity index of 40.
Calculate th GI.
Solution:
GI = (F200 – 35)[0.2+0.005(LL – 40)]+0.01(F200 – 15)(PI -10)
94. Unified Soil Classification System
Main points:
1. The classification is based on material passing a 75m
sieve.
2. Coarse fraction = percent retained above no.200 sieve =
100 – F200 = R200
3. Fine fraction = percent passing no. 200 sieve = F200
4. Gravel fraction = percent retained above no.4 sieve = R4
/08/10
95. Classification
•As per USCS soil divided into groups
1. Coarse grained: Gravel or sand in nature with less
than 50% passing through the no.200 sieve.
- It represents clear symbols – G (Gravel), S
(sand)
2. Fine grained soils: Soils with 50% or more passing
through the no.200 sieve.
- The group symbols starts with ‘M’(inorganic silts)
- ‘C’for inorganic clays, ‘o’organic clays or silts
- ‘Pt’used for peat, muck and other highly organic
soils
96. Other symbols used for the classification:
W – well graded
P - poorly graded
L – Low plasticity (LL is less than 50%)
H – High plasticity (LL more than 50%)
99. AASHTO / USCS classification
1. Both systems are divide the soil into two major groups,
that is coarse and fine grained soils as separated by sieve
no . 200 (75m)
2. a. According to the AASTHO system a soil is considered
fine grained when more than 35% passes through the
no.200 sieve
b. USCS more than 50% passes through the no.200
sieve.
3. a. USCS Gravelly and sandy soils clearly separated,
b. AASHTO it is A-2
4. Clear symbols used in USCS (GW, SM, CH) , but in A
in AASHTO classification
100. 5. a. AASHTO system, the no.10 sieve is used to separate gravel
from sand.
b. USCS no. 4 sieve is used, upper limits appropriate concrete
and highway technology using no. 10(2mm) sieve.
6. a. USCS clearly organic soil classification (OL, OH, Pt) are
given,
b. in AASHTO there is no classification for organic soils
