Ge i-module3-rajesh sir

Geotechnical Engineering IGeotechnical Engineering - I
Compressibility
Dr. Rajesh K. N.
Assistant Professor in Civil EngineeringAssistant Professor in Civil Engineering
Govt. College of Engineering, Kannur
Dept. of CE, GCE Kannur Dr.RajeshKN

Module III
Compaction:
definition and objectives of compaction –definition and objectives of compaction –
Proctor test and modified Proctor test –
concept of Optimum Moisture Content and maximum dry density –
zero air voids line –zero air voids line
factors influencing compaction –
effect of compaction on soil properties –
field compaction methods - Proctor needle for field controlp
Consolidation:
definition - Compressibility –p y
coefficient of volume change and compression index –
Laboratory consolidation test –
e-log p curves - pre-consolidation pressure –
Terzaghi's theory of one dimensional consolidation –
Time rate of consolidation –
difference between consolidation and compaction

ConsolidationConsolidation
• Compression: Decrease in volume due to a stress• Compression: Decrease in volume due to a stress
• Compressibility: Property of a soil mass by virtue of which
compression happens in itcompression happens in it
• Consolidation: Decrease in volume and a consequent escape of pore
water
• Swelling (Opposite process of consolidation): Increase in water
content due to an increase in the volume of voids
• Principle of consolidation: Spring analogy (Terzaghi and Frohlich)

Consolidation may be due to:y
•External static loads from structures
•Self-weight of soil
•Lowering of groundwater tableg g

The total compression settlement of a clay strata consists of:
1. Immediate compression settlement
2. Primary consolidation settlement:. a y co so dat o sett e e t:
3. Secondary consolidation settlement
•Immediate compression settlement: Settlement occurs almost
simultaneously with the applied load – due to immediate compressionsimultaneously with the applied load due to immediate compression
of soil layer under undrained condition
• Soil mass assumed as elastic• Soil mass assumed as elastic

• Primary consolidation settlement:
• Due to primary consolidation
• Primary consolidation:
• If the rate of compression is controlled solely by the resistance
of the flow of water under the induced hydraulic gradients, the
process is called primary consolidation
• Reduction in volume due to squeezing out of water from
voids – till equilibrium (i.e., till all pore pressure is carried by
soil solids)
• Explained by means of one dimensional consolidation theory.

•Secondary consolidation settlement:
• Due to secondary consolidation, occurring at a very slow ratey g y
• Starts after primary consolidation stops, i.e., after excess pore
water pressure approaches zerop pp
• Secondary consolidation is assumed to proceed linearly with
logarithm of timeg

One dimensional consolidation
• Settlement of a structure many times happens due to the
compression of a soft clay layer between layers of sand or stiffer clay
• Adhesion between soft and stiff layers prevents lateral movement
of soft layers
• Theory based on this assumption: Terzaghi’s theory of one
dimensional consolidation
• In the laboratory, this condition is simulated by confined
compression or consolidation test

Consolidation of laterally confined soil
AA
Virgin compression curve
Recompression
sratioe
BC
E i
Recompression
Voids
D
Expansion
0 10
0
logCe e C
σ
σ
′
= −
′
E
log σ’ (log of normal pressure)
P id ti f ld d i
Pressure-voids ratio curve for remoulded specimen

loge e C
σ′
= −
e0 - initial voids ratio corresponding to 0σ′
loge e C
σ′
= −0 10
0
logCe e C
σ
=
′ e - final voids ratio corresponding to
CC – compression index
σ′
0 10
0
logCe e C
σ
=
′
CC=
( )10log
e
σ
Δ
′Δ
Slope of pressure-voids ratio curve
( )10log σΔ
Coefficient of compressibility 0
v
decrease in voids ratio e e e
a
i i
− −Δ
= = =
′ ′ ′Δ
ff f p y
0
v
increase in pressure σ σ σ′ ′ ′− Δ
Coefficient of volume change
change in volume
m =Coefficient of volume change vm
initial volume increase in pressure
=
×
( )1 1 1v v v Δ( )
( )
0
0
0
0 0 0
1 1 1
1 1 1
v v s v
v
v s s
v v v e e e a
m
e e ev v vσ σ σ
− − −Δ
= = = =
′ ′ ′Δ + Δ + Δ ++

If laterally restrained
1H
m
−Δ
= H m H σ′∴Δ = − ΔIf laterally restrained,
0
vm
H σ′Δ 0vH m H σ∴Δ Δ
CONSOLIDATION SETTLEMENT
0vH m Hρ σ′= Δ = Δ
0
0 0 10
0 0 0
log
1 1
Ce e C
H H
e e
σ
ρ
σ
′−
= =
′+ +0 0 01 1e e σ+ +
Empirical formulae for compression index:
0.007( 10)c lC w= −Skempton’s formula:
(for remoulded clays) wl - liquid limit in %wl liquid limit in %
0 009( 10)C w= −Terzaghi and Peck formula:
0.009( 10)c lC w=g
(for normally consolidated clays)
wl - liquid limit in %

Problem X: A soil has a compression index of 0.28. At a stress of 120 kN/m2,
the void ratio was 1 02 Calculate:the void ratio was 1.02. Calculate:
a) void ratio if the stress on the soil is increased to 180 kN/m2
b) total settlement of the stratum of 6m thickness
a) Void ratio if the stress on the soil is increased to 180 kN/m2
0 10logCe e C
σ′
= −
′ 10
180
1.02 0.28log
120
= − 0.97=0 10
0
gC
σ′ 10g
120
b) Total settlement of the stratum of 6m thickness
logCC
H
σ
ρ
′
b) Total settlement of the stratum of 6m thickness
0.28 180
6 log 0 1460 10
0 0
log
1
C
H
e
ρ
σ
=
′+
106 log
1 1.02 120
=
+
0.146 m=

Terzaghi’s Theory of One-dimensional Consolidationg y
rate of expulsion of excess pore waterRate of change of rate of expulsion of excess pore water
from a unit volume of soil during the
same time interval
Rate of change of
excess hydrostatic
pressure
α
( )01
v
v w v w
k ek
c
m aγ γ
+
= = Coefficient of consolidation
2v w v wγ γ 2
cm s
Coefficient of consolidation combined effect of permeability
and compressibility on the rate of volume change

Laboratory consolidation testy
• Consolidometer (Oedometer) - Devised by Terzaghi
• Fixed ring type and floating ring type consolidometers• Fixed ring type and floating ring type consolidometers

• In a fixed ring type consolidometer, the brass ring is fixed and the
t t i f t d dtop porous stone is free to move downwards
• In floating ring type, the brass ring is floating and both the top and
bottom porous stones are free to move towards the middle
Measurements:
• Specimen allowed to consolidate under vertical pressure increments
10, 20, 50, 100, 200, 400, 800, 1000 kN/m2
• Each pressure increment kept until compression virtually ceases
(say, 24 hrs.)
• Vertical compression measured at various time intervals for each
pressure increment: 0.25 hrs, 1 hr, 2.25 hrs …, 24 hrs.
• Final compression for each pressure increment also recorded

Pressure-voids ratio curves
• Obtained from the observations of laboratory consolidation test
• Voids ratio of the sample under consolidation test is determined at the endVoids ratio of the sample under consolidation test is determined at the end
of each load increment
h f l d h d
• If hs is the thickness of the solid matter of the sample and h is the
thickness of the sample at the end of a load increment
1. Height of solids method
sh h
e
−
=Voids ratio at the end of the load increment
thickness of the sample at the end of a load increment,
s
e
h
Voids ratio at the end of the load increment
s dV W
h = =hs can be determined as: sh
A GA
hs can be determined as:
Volume of solids, Dry weight of soil,s dV W
Specific gravity, Cross sectional area of sampleG A

2. Change of voids ratio method
• Assuming full saturation at the end of the test, final voids ratio can be
found as:
2. Change of voids ratio method
found as:
.f fe w G= Final voids ratio, Final water content= =f fe w
• Change of voids ratio under each pressure increment can be found as
follows:
( )
( )
( )
( ) 1 1
vf v vf v s f
f fvf s vf s s
v v v v v e eH A H e
H AH e ev v v v v
− − −Δ Δ Δ
= = = = =
+ ++ +( ) ( ) f fvf s vf s s
( )1 fe H
e
+ Δ
∴Δ =e
H
∴Δ =
• Knowing Δe, working backwards from the known value of ef , the voids
f
ratio corresponding to each pressure can be found

• On a semi-log scale, pressure voids ratio curve (e-logp curve) is
drawn
• e-logp curve should pass through A, the point corresponding to
initial voids ratio, if the test conditions are exactly as in the field
• This rarely happens, as the samples will atleast slightly differ from
the site conditions

To find coefficient of consolidation
Square root of time Logarithm of timeq
fitting method
g
fitting method
2
T d
Casagrande Taylor
v
v
T d
c
t
=
vT Time factor→
2
U⎧ ⎛ ⎞
, 60%
4 100
v
U
U
T
U
π⎧ ⎛ ⎞
<⎪ ⎜ ⎟
⎪ ⎝ ⎠= ⎨
⎛ ⎞⎪
100.9332log 1 0.0851, 60%
100
U
U
⎛ ⎞⎪− − − >⎜ ⎟⎪ ⎝ ⎠⎩
100t
f
degree of compressionU
ρ
ρ
→ = ×
,
2
i
average drainage path
H H
for double dd rainage
− Δ
→ =

Square root of time fitting method
2
vT d
From a plot of compression dial reading Vs square root of time
0 848 f 90% lid tiTv
vc
t
=
2
d
0.848 for 90% consolidationvT =
90
90
0.848v
d
c
t
∴ =

Logarithm of time fitting method
2
T d
From a plot of compression dial reading Vs log of time
v
v
T d
c
t
=
2
50
50
0.197v
d
c
t
∴ =

Problem 1:

Problem 2: Two clay layers A and B are respectively 4m and 5m thick. The
i k f l A h 50 % lid i i 6 h C l l htime taken for layer A to reach 50 % consolidation is 6 months. Calculate the
time taken by the layer B to reach the same degree of consolidation. The
coefficient of consolidation of layer B is half that of layer A. Both layers have
d bl d idouble drainage.
50 6monthst =Layer A
50 m2
2
H
d = =
2
50
50
0.197v
d
c
t
=
2
2
0.197
6
vc = ×
50
2
m/months0.13133=
6
50 ?t =
Layer B
50 m
5
2.5
2
d = =50 ?t
m/months
0.13133
2
vc =
50
2
2 2
2 0.13133 2.5
0.197 0.197c = × ⇒ = ×2
50
0.197 0.197
6 2
vc
t
⇒
50 18.75monthst∴ =
50 18.75monthst∴

Problem 3: A 2.5 cm thick soil sample of clay was taken from field for
predicting the time of settlement for a proposed building which exerts ap g p p g
uniform pressure 100 kN/m2 over the clay stratum. The sample was loaded to
100 kN/m2 and proper drainage was allowed from top and bottom. It was
seen that 50 % of the total settlement occurred in 3 minutes. Find the time
required for 50 % of the total settlement of the building if it is to stand on 6m
thick layer of clay which extends from ground surface and is underlain by
sand.
2
50
0 197
d50
50
0.197vc
t
∴ =
2
50
50 0.197
d
t =50
vc

Problem 4: A clay layer 3.6 m thick is sandwiched between layers of sand.
Calculate the time the clay layer will take to reach 50% consolidation TheCalculate the time the clay layer will take to reach 50% consolidation. The
coefficient of consolidation is 4x10-4 cm2/s.
2
50d
50 m
3.6
1.8
2
d = =
50
50
0.197v
d
c
t
=
4
cm/s4 10vc −
= ×
2
50 ?t =
2
50 4
0.197 180
4 10
t −
×
=
×
s15957000= days185=

Based on consolidation history, a soil can be:y
• Pre-consolidated (over-consolidated) soil:
– In the past, it has been subjected to a pressure in excess of its
present overburden pressure
• Normally consolidated soil:Normally consolidated soil:
– It has never been subjected to an effective pressure greater than
its present overburden pressure,
– It is completely consolidated by the existing overburden
• Under-consolidated soil:
– not fully consolidated under the present overburden pressure– not fully consolidated under the present overburden pressure

Pre-consolidation pressure: The temporary overburden pressure to
which a soil has been subjected and under which it got consolidated
Determination of pre-consolidation pressure
which a soil has been subjected, and under which it got consolidated
Undisturbed sample of clay is
Casagrande
Undisturbed sample of clay is
consolidated in lab pressure-
voids ratio curve plotted (semilog)
Point A of maximum curvature
(min radius) selected
AB – horizontal
AC - tangent
AD - bisector
Straight portion of virgin curve
extended to meet AD at P
pσ′ is the pre-consolidation pressure

Secondary consolidation
• After excess pore pressure is dissipated, change in voids ratio continues at
a reduced rate – secondary consolidation
• Highly viscous water forced out, plastic readjustment, progressive fracture
of some particles
• Terzaghi’s theory not applicable• Terzaghi s theory not applicable
• It is assumed that secondary consolidation is linearly increasing with log
of time
2
10log
t
e
t
αΔ = −
1t
t2 Total elapsed time since load was applied
t1 Period required for primary consolidation to complete
– taken as time for 90% consolidation
α Coefficient representing rate of consolidation

Compaction

Compaction
• Rapid reduction in voids by expulsion of air deliberately produced by
mechanical means in the field or in the laboratory
• Objectives:
– increase in shear strength
– reduction in permeability
– reduction in settlement under loading
• In the field, compaction can be done by varying water content, amount of
compaction and type of compaction
• Proctor (1933) showed that there is a definite relation between water
content and degree of the attained dry density
• For a particular compactive effort, soil attains a maximum dry density at a
water content known as ‘optimum moisture content’(OMC)
p ( )

Factors affecting compactiong p
• Water content
• Amount and type of compaction
• Type of soil: Well-graded coarse-grained soils have lower OMC
than fine-grained soils having larger specific surface
• Admixtures: can be incorporated in the soil mass to improve its
compaction properties

Measurement of Compaction
1
d
w
γ
γ =
+
In terms of dry density
1 w+
• Standard Proctor Test (AASHTO Test) R. R. Proctor, 1933( )
• Modified Proctor Test
,
Procedure for Proctor Tests
• Sample with a known water content is filled in a mould by applying ap y pp y g
specific compactive effort
• Compaction(dry density) is measured
• For the same sample, the above steps are repeated for varying water
contents
• From the curve drawn water content corresponding to maximum dry
• From the curve drawn, water content corresponding to maximum dry
density(OMC) is found

STANDARD PROCTOR TESTSTANDARD PROCTOR TEST
Compaction Apparatus
• Weight of hammer = 2.5 kg, free fall of hammer= 30 cm, no. of
l 3 f bl l 25
p pp
layers=3, no. of blows per layer = 25

MODIFIED PROCTOR TESTMODIFIED PROCTOR TEST
• Weight of hammer = 4.5 kg, free fall of hammer = 45cm, no. of
l 5 f bl l 25layers=5, no. of blows per layer = 25
• More compactive effortMore compactive effort

Effect of moisture content and compactive effort on
compaction (cohesive soils)compaction (cohesive soils)

• Optimum moisture content (OMC) is the water content at which soil attains• Optimum moisture content (OMC) is the water content at which soil attains
a maximum dry density for a particular compactive effort.
• Curve shifts to the left with increase in compactive effort

Comparison of moisture content and shear strength curves
•Highest shear strength attained at a moisture content lower than Optimum
g g p
moisture content (OMC) at which soil attains a maximum dry density

Zero air voids line
• Line showing relationship between moisture content and dry
density of a soil having zero air voids ( all the voids are filled with
water)
( )1 n Gγ−
)
• Otherwise called 100% saturation line
( )1
1
a w
d
n G
wG
γ
γ =
+
an % air voids
0
1
w
a d
G
n
wG
γ
γ= ⇒ =
+
Equation of zero air voids line
• Imaginary line
It i i ibl t l ll th t d i– It is impossible to expel all the entrapped air
– Hence, no compaction curve will cross the zero air voids line

• Air voids line can be drawn for any value of % air voids
w
d
G
wG
γ
γ =
⎛ ⎞
1
r
wG
S
⎛ ⎞
+ ⎜ ⎟
⎝ ⎠
• Sr = 1 corresponds to zero air voids (100 % saturation)

Compaction curve for sandp
• Films of water around particles can keep them away and
decrease density, initiallyy y
• More compactive efforts have less effect on cohesionless soils
than on cohesive soils

Problem 3: In a standard proctor test, the following values were obtained.
Volume of the mould was 940x10-6 m3 and specific gravity of the soil solidsg y
was 2.6.
Weight of wet sample (N) 16.5 17.25 17.75 17.9 17.75
%Water content % 19.1 20.5 22.3 22.5 24
i) Determine maximum dry density and OMC
ii) Determine percentage air voids, degree of saturation and air content at) p g , g
maximum dry density
iii) Draw zero air voids line and find theoretical max dry density at OMC
iv) Draw 10% air voids line)
Water content % 19.1 20.5 22.3 22.5 24
18.35 18.89 19.04 18.89Wet density 3
16 5 10 kN−
×Wet density
Wet weight
Volume
γ =
3
6
16.5 10
940 1
17.553
0 3
3
kN
m
kN m
γ −
=
=
×
×
15.23 15.45 15.54 15.2217.553
1 0.191
dγ =
+
Dry density
γ
14.738 3
kN m=
1
d
w
γ
γ =
+

Plot moisture content Vs dry density curve
Max
sity
Max
dry densitydrydens
i t t t
OMC
moisture content
3
⎫3
max
( )
16
22.6
kN/mMax dr
%
y density
from cur
O C
ve
M
dγ ⎫=
⎬
= ⎭

To get percentage air voids at maximum dry density (at OMC)
( )1
1
a w
d
n G
wG
γ
γ
−
=
+
( )1 2.6 9.81
16
1 0 226 2 6
an− ×
=
+ ×1 wG+ 1 0.226 2.6+ ×
0.0041 0.41%an∴ = =
To get degree of saturation Sr at max dry density
. . %a
w
d
G
wG
γ
γ =
⎛ ⎞
To get degree of saturation Sr at max dry density
2.6 9.81
16
0 226 2 6
×
=
⎛ ⎞×
1
r
wG
S
⎛ ⎞
+ ⎜ ⎟
⎝ ⎠
0.226 2.6
1
rS
⎛ ⎞×
+ ⎜ ⎟
⎝ ⎠
0 989 98 9%S 0.989 98.9%rS∴ = =
Hence, air content at maximum dry density
, y y
1.0 0.989 0.0109 1.0988 %ca = − = =

To plot zero air voids line
1
w
d
G
G
γ
γ =
+
Equation of zero air voids line
1 wG+
q
Water content % 19.1 20.5 22.3 22.5 24
Dry density for zero air
voids
16.64 16.15 16.09 15.71
1
w
d
G
wG
γ
γ =
+
2.6 9.81
17.04
1 0.191 2.6
dγ
×
= =
+ ×
Max dry density
Theo. max dry density at OMC
zero air voids line
density
Max dry density
dryd
Dept. of CE, GCE Kannur Dr.RajeshKNmoisture content
OMC

To plot 10 % air voids line (90 % saturation line)
Water content % 19.1 20.5 22.3 22.5 24
Dry density for 10% air
voids 2 6 9 81×voids
16.02 15.51 15.46 15.06
1
w
d
G
wG
S
γ
γ =
⎛ ⎞
+ ⎜ ⎟
⎝ ⎠
2.6 9.81
0.191 2.6
1
0.9
16 44
dγ
×
=
×⎛ ⎞
+ ⎜ ⎟
⎝ ⎠
=rS⎝ ⎠ 16.44=
ensity
10% i id li
dryde
10% air voids line
Dept. of CE, GCE Kannur Dr.RajeshKNmoisture content

Equipments for field compactionq p p
• Smooth wheel roller
• Rubber tyred roller
• Sheepsfoot roller
• Vibratory rollerVibratory roller

Field control of compactionp
• To attain a desired density by compaction, periodic measurements of
moisture content and dry density will be required during the compactionmoisture content and dry density will be required during the compaction
process
• The methods for determining moisture content and dry density in the field,g y y
several methods are available
– Core-cutter method
– Sand replacement method
Nuclear method– Nuclear method
– Proctor needle method

Proctor needle method
• Used for rapid determination of moisture contents in-
situ.
• Equipment consists of a needle attached to a spring
loaded plunger, calibrated to read penetration
resistance in kg/cm2
• Needle can have a suitable bearing area for the soil
(larger bearing area for cohesive soils)

• In the lab, sample is compacted in the mould and penetration resistance
measured using Proctor needlemeasured using Proctor needle.
• Water content and dry density are found
• The above process is repeated on the same sample for varying moisture
contents.
• Thus laboratory calibration curve is drawn

•To get moisture content in the site, sample of wet soil is compacted into
the mould and penetration resistance read off from the Proctor needle
M i t t t di t th t ti i t bt i d•Moisture content corresponding to the penetration resistance obtained
from the laboratory calibration curve
•Method is fast It gives fairly accurate results for fine grained cohesive•Method is fast. It gives fairly accurate results for fine-grained cohesive
soils
•Presence of gravel in the soil makes the readings less reliablePresence of gravel in the soil makes the readings less reliable
•Not very accurate for cohesionless soils

Summaryy
Consolidation:
definition - Compressibility –
coefficient of volume change and compression index –
Laboratory consolidation test –
l l de-log p curves - pre-consolidation pressure –
Terzaghi's theory of one dimensional consolidation –
Time rate of consolidation –
diff b t lid ti d tidifference between consolidation and compaction
Compaction:
definition and objectives of compactiondefinition and objectives of compaction –
Proctor test and modified Proctor test –
concept of Optimum Moisture Content and maximum dry density –
zero air voids line –zero air voids line –
factors influencing compaction –
effect of compaction on soil properties –
field compaction methods - Proctor needle for field control
field compaction methods Proctor needle for field control

Ge i-module3-rajesh sir

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