Sm Chapter VII

Chapter VII
CONSOLIDATION and COMPACTION
Specific Objectives
To differentiate between consolidation and compaction.
To know the concept and process of consolidation.
To be able to calculate consolidation settlements.
To understand time rate of consolidation.
To understand process of compaction.
To be able to compute earth work quantities.

When a soil mass is subjected to a compressive force, like all other materials, its volume
decreases. The property due to which decrease in volume of soil occurs is known as
compressibility of soil.
Compression of soil occurs due to following reasons.
1. Compression of solid particles and water in the voids.
2. Compression and expulsion of air in the voids.
3. Expulsion of water in the voids.
Compression of solid particles and water in the voids is extremely small in the range of
stresses involved in soil engineering.
Air exists only in partially saturated soils and dry soils. The compression of air is rapid as
it is highly compressible. It is also expelled quickly. Also it is irrelevant to saturated soils.
When the soil is fully saturated, compression of soil occurs mainly due to the third cause,
i.e. expulsion of water from the pores of soil.
The compression of saturated soil under a steady pressure is known as consolidation. It is
entirely due to expulsion of water from the voids. As the consolidation occurs, the water
escapes, the solid particles shift from one position to other by rolling and sliding and
attain a closer packing.
It is to be emphasized here that decrease in volume of soil occurs not due to compression
of solids or water but due to shifting of position of the particles as the water escapes.
The compression of soils due to expulsion of air due to dynamic methods such as rolling
and tamping is known as compaction.

INITIAL, PRIMARY AND SECONDARY CONSOLIDATION
When a load is applied to a partially saturated soil a decrease in volume occurs due
expulsion and compression of air in the voids. A small decrease in volume also occurs
due to compression of solid particles. This reduction in volume of the soil just after the
application of load is known as initial consolidation.
After initial consolidation, further reduction in volume occurs due to expulsion of water
from voids. When a saturated soil is subjected to pressure, initially, all applied pressure is
taken by water (as it is highly incompressible). This sets a hydraulic gradient between
water subjected to pressure and the water in surrounding soil. Generally water moves
from area of high pressure to low pressure, and excess pore water pressure decreases.
This increases the effective stress on solids and cause rearrangement of particles and
decrease in volume of soil. This is known as primary consolidation.

HAWASSA UNIVERSITY FACULTY OF TECHNOLOGY CIVIL ENGINEERING DEPARTMENT

The reduction in volume continues at a very slow rate even after the excess pore water
developed is fully dissipated and primary consolidation is over. This reduction in volume
of soil is known as secondary consolidation. It may be attributed to plastic readjustment
of particles and the adsorbed water to the new stress system.
Among these, primary consolidation is the most important component.

TERZEGHI’S SPRING ANOLOGY
To explain primary consolidation Terzaghi gave the spring- piston analogy. Fig.No.1
shows a cylinder fitted with a tight fitting piston having a valve. The cylinder is filled
with water and contains spring of specified stiffness.
P P P

Pw = P Pw = P - ΔP Pw = 0

Ps= 0 Ps= ΔP Ps= P

t=0 t = t1 t = t2 = ∞
Fig.No.1. spring piston analogy for consolidation process
When a load is applied on piston with valve closed all the load is taken by water and
spring will not take any load, as the stiffness of spring is negligible compared to that of
water.
Therefore P = 0 + P and Pw = P Ps= 0
If the valve is now gradually opened water starts escaping from the cylinder. The spring
starts sharing some load and a decrease in its length occurs. When a portion of load (ΔP)
is transferred from the water to the spring,
P = ΔP + ( P - ΔP) and Pw =( P - ΔP) and Ps = ΔP
And finally the entire load is transferred on to spring and water will not carry any load.
P=P+0 and Pw = 0 and Ps = P
The behavior of saturated soils, when subjected to a steady static pressure, is similar to
that of the spring – piston analogy model. The solid particles, in saturated soils, behave
like spring, while the water in the voids behaves like water in the cylinder. The
permeability of the soil controls the flow of water and it can be likened to valve in the
piston. The pore water pressure in the soil is analogous to the pressure carried by water in
the cylinder. Finally the stress developed in the spring is analogous to the effective stress
developed in the soil.
When load applied Δσ = 0 + ue
After some time Δσ = Δσ′ + ( ue - Δσ′ )
At the end Δσ = Δσ′ + 0
As the effective stress increases, the volume of the soil decreases. The decrease in
volume is generally expressed as change in void ratio. Fig. shows decrease in void ratio
with time, as effective stress increase due to transfer of pressure to the solid particles. It

SOIL MECHANICS-I CHAPTER-VII CONSOLIDATION AND COMPACTION M U JAGADEESHA 2008-09 69


must be noted that the curve shown is drawn for application of one pressure increment
Δσ1.

CONSOLIDATION TEST
The consolidation test is conducted in a laboratory to study compressibility of a soil. The
test is performed in a consolidometer or an oedometer shown in Fig.No.2. It consists of a
loading device and a cylindrical container called consolidation cell. The soil specimen is
placed in the cell between top and bottom porous stones. The inside surface of the ring
should smooth and polished to reduce friction. The ring imposes a condition of zero
lateral strain on the sample. Soil sample is usually 60 mm diameter and 20 mm. thickness

Dial Gauge
P
Porous stone

Water
Bath

Soil
Sample

Porous stone
Fig.No.2. Consolidometer
It has arrangement for the application of desired load increment, saturation of sample,
and measurement of change in thickness of the sample at every stage of consolidation
process.
An initial setting pressure of about 5.0 KN/m2 (2.5KN/m2 for soft clays) is applied. The
load is allowed to stand till there is no change in dial gauge reading or 24 hours
whichever is less. The final dial gauge reading is noted.
The first increment of load to give a pressure of 10KN/m2 is then applied to the
specimen. The dial gauge readings are taken after 0.25, 1, 2.25, 4, 6.25, 9, 12.25, 16,
20.25, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 289, 324, 400, 500, 600 and 1440
minutes (24hours) The primary consolidation is usually complete within 24hours
The second increment of load is then applied. Usual practice is to double the previous
load in each increment. (20, 40, 80, 160, 320, 640 KN/m2).
After the consolidation under final load increment is complete the load is reduced to ¼ of
the final load and allowed to stand for 24 hours the sample takes water and swells. The
reading of the dial gauge is taken when the swelling is complete. The load is again
decreased to ¼ value (40KN/m2) and allowed to swell. The load is further decreased to
10KN/m2 and swelling is noted. Finally it is reduced to initial setting load and kept for 24
hours and final dial gauge reading is noted.



Immediately after completion of unloading, the ring with the sample is taken out. The
excess surface water is dried using a blotting paper. The wt. of the ring and the sample is
taken. The sample is then dried in oven and its dry mass Ms and the water content is
determined.

Determination of void ratio at various load increment:
The results of a consolidation test are plotted in the form of a plot b/w void ratio and the
effective stress. It is therefore required to determine the void ratio at various load
increments.
Height of solids method:
Hs = Vs/A = (Ms/Gsγw) /A
Where
Hs =height of solids
Vs = volume of solids
Ms = dry mass of sample
Gs = sp. gr. of soil solids
A = c/s area of sample

e = Vv/Vs = (V – Vs )/Vs = [(AH) – (AHs)]/(AHs) = (H – Hs)/Hs
Where
H is the total height.
Thus the void ratio is determined from the total ht. of solids. The total height of the
sample is measured at least once during the test usually at start or at the end of the test. At
other stages of loading, the height H is worked out from the measured thickness and
difference in dial gauge readings.
Therefore H = H0 ±∑ ΔH where H0 = initial height ΔH = change in ht.
Change in void ratio method:
In this method, the final void ratio (el) corresponding to complete swelling condition after
load has been removed, is determined from its water content using equation
el = wGs -------------- (assuming soil is fully saturated)
The void ratio corresponding to intermediate loading stages is determined as explained
below.
e = Vv/Vs = (V – Vs )/V = (V/Vs) – 1
Where
V= total volume and
Vs = vol of solids
Therefore V = Vs( 1 + e ) or AH = Vs(1+e) ------------------------------(1)
Where A is the c/s area of specimen and H is the total ht.
By partial differentiation A dh = Vs de ----------------------------(2)
From (1) and (2) dH/H = de/(1+e)
Or Δe = [( 1 + e ) ΔH /H ]
As the void ratio el and total ht. ‘H’ of the sample are known at the end of the test, the
void ratio at any other stage can be determined from change in thickness ΔH measured by
the dial gauge.
Therefore Δe = [( 1 + el ) ΔH /Hl ]



Consolidation test results
(a)Dial gauge reading- time plot.
At each load increment we can plot dial gauge reading v/s time plot as shown in
Fig.No.3. For sand change in thickness occurs very quickly, for clays it will take 24 hours
This plot is required for determining co efficient of consolidation.
e e
σ' = σ1 σ' = σ2

Dial gauge reading Dial gauge reading
Fig.No.3. Variation of void ratio with dial gauge reading or compression
(b)Final void ratio v/s eff stress plot.
At each load increment at 24 hours we assume that full consolidation has occurred and
have obtained corresponding final void ratios. We can plot effective stress v/s
corresponding final void ratio as shown in Fig.No.4. This plot is useful for determination
of magnitude of consolidation settlement in the field.

e

σz'

Fig.No.4. Variation of final void ratio with effective stress.
(c) Final void ratio – log σ′ plot.
Void ratio v/s σ′ plot shown in Fig.No.5. shows concavity upwards. The slopes of curve
at different points are different. The slope decreases with increase in effective stress.
It is more common to plot results on a semi log plot. The plot is practically a straight line
for normally consolidated soil (defined later) with in the range of pressure usually
encountered in practice.



(d)Unloading and reloading plot.
Fig.No.5. shows curve AB indicating decrease in ‘e’ with an increase in effective stress.
After the sample has reached equilibrium at the effective stress σ2′ as shown by point B,
the pressure is reduced and sample is allowed to take up water and swell. Curve BEC
obtained in unloading. This is known as expansion curve or swelling curve. It may be
noted that the soil cannot attain the void ratio existing before the start of the test, and
there is always some permanent set (plastic) or residual deformation.

e
A

C
F

E B

D

G
log σ'

Fig.No.5. Variation of final void ratio with log of effective stress
If the specimen is again loaded it will follow CFD. As the load approaches maximum
value of load previously applied corresponding to point B, there is reversal of curvature
of the curve and the plot DG continues as an extension of the first loading curve AB.

BASIC DEFINITIONS.
(1) Coefficient of compressibility (av): is the decrease in void ratio per unit increase in
effective stress. It is equal to slope of the e-σ′ curve at the point under
consideration. Coefficient of compressibility decreases with an increase in the
effective stress; as such it should be reported with the effective stress value at
which it is determined. It means soil becomes stiffer as the effective stress is
increased.
av = Δe / Δσ′ Dimension of av is L2/F i.e. m2/KN i.e. inverse of pressure
(2) Coefficient of volume change: is the ratio of volumetric strain per unit increase in
effective stress.
mv = (-ΔV / V0 )/Δσ′
Where
mv = coefficient of volume change,
V0 = initial volume,
ΔV = volume change and



Δσ′ = change in effective stress.
Volumetric strain can be expressed in terms of ‘e’ or the thickness of the
specimen as given below.
ΔV/V0 = Δe/(1 + e0) therefore mv = -[Δe/(1+e0)]/ Δσ′
As c/s area of sample in consolidometer remains constant, the change in volume is
proportional to change in ht.
ΔV/V0 = ΔH/H0 therefore mv = - [ΔH/H0]/ Δσ′
Or ΔH =- mvH0 Δσ′
Therefore mv = av/ (1+e0)
Like av the coefficient of consolidation mv also depends on effective stress.
(3) Compression Index: Slope of the linear portion of the void ratio v/s log σ′ plot is
the compression index.
Cc = - Δe / [ log10 (σ′/σ0′ )
Sometimes it is also given as Cc = - Δe / [ log10 {(σ0′ +Δσ′) /σ0′ }]
Where Δσ′ is the change in eff stress.
Empirical relationships: Cc =0.009 ( wl – 10 ) undisturbed soils
Cc =0.007 ( wl – 10 ) remoulded soils
Cc lies between 0.3 for highly plastic clays to 0.075 for low plastic clays. Also
Cc =0.54( e0 – 0.35)
Cc = 0.0054 ( 2.6 w0 –35 )
av = 0.435 Cc / σ′
(4)Expansion Index: The expansion index or swelling index (Ce) is the slope of the
e-logσ′ plot obtained during unloading.
Ce =Δe / [ log10 {(σ0′ +Δσ′) /σ0′ }]
(5) Recompression Index: The recompression index is the slope of the recompression
curve. It is the change in voids ratio per tenfold increase in effective stress.

Normally consolidated and over consolidated clays
From the plots of e-log σ, unloading and reloading of specimen we can say that soil
behaves in a different way when subjected to a load, which it has already experienced
once. That is why it compresses to different void ratio when it is subjected to an effective
stress σ′ for the first time and subjected to same pressure second time.
The maximum pressure a soil had been subjected to in its history is known as pre
consolidation pressure If the present pressure a soil is subjected to is the maximum
pressure in the history of soil it is known as normally consolidated soil. If the present
pressure is less than the maximum pressure the soil has experienced is known as over
consolidated soils.
The ratio of pre consolidation pressure to present pressure is known as over consolidation
ratio. For normally consolidated soils OCR = 1.
If the clay deposit has not reached equilibrium under the applied overburden loads it is
said to be under consolidated.



TERZAGHI’S THEORY OF CONSOLIDATION
Assumptions;
1. The soil is homogeneous and isotropic.
2. The soil is fully saturated.
3. The solid particles and water in the voids are incompressible. The consolidation
occurs due to expulsion of water from the voids.
4. The k, has the same value at all points, and it remains constant during the entire
period of consolidation.
5. Darcy’s law is valid throughout the consolidation process.
6. Soil is laterally confined and the consolidation takes place only in axial direction.
Drainage of water also occurs only in the vertical direction.
7. The time lag in consolidation is due entirely to the low permeability of the soil.
8. There is unique relationship between the void ratio and the effective stress and
this relationship remains constant during the load increment.

t =0
t = t1
t = t2

Applied load

t=0

soil

2d

Pervious layer

Fig.No.6. Derivation of time rate of consolidation equation

The partial differential equation relating excess pore water pressure, depth, time and
properties can be derived - by equating the change in pore water pressure with depth by
equating rate of change of volume to rate at which the water is squeezed out of the soil,
and change in pore water pressure with depth from seepage consideration.
∂ 2 u ∂u
cv 2 =
∂z ∂t
Where
cv = k/(γwmv) or
cv = k(1+e0)/(γw av)



The above equation is the basic differential equation of one-dimensional consolidation. It
gives the distribution of hydrostatic excess pressure ‘ue’ with depth z and time t.
The solution of above differential equation can be obtained by using Fourier series and
boundary conditions as below.
ue = 4ui /π [∑N=0∞ 1/(2N+1) {sin (2N+1)πz/2d}e-(2N+1) (2N+1) ππ T / 4
Where T= Tv= cvt/d2
A series of isochrones indicating the variation of ue with z can be plotted for different
values of Tv as shown in Fig.No7. The shape of the isochrones depends upon the initial
distribution of excess pore water pressure ui and the drainage condition at the boundaries
of clay layer.

ue= 0 ue =ui
z=0

Tv=∞ Tv = 0

z=d

z = 2d
t=∞ t=0
Fig.No.7. Dissipation of excess pore water pressure with time
The progress of consolidation at any point depends upon the pore water pressure ue at that
point. The degree of consolidation (Uz) at any point is equal to the ratio of the dissipated
excess pore water pressure to the initial excess pore water pressure i.e.
Uz = (ui – ue )/ ui = 1 – ue/ui
Substituting the value of ue/ ui from equation of consolidation we get
Uz = 1 - ∑N=0∞ (2/m) sin (Mz/d)e- M M T
where M= π/2 (2N+1)
The above equation gives the degree of consolidation at a point . in practical problems the
main interest is to know the average degree of consolidation of the whole layer. The
average degree of consolidation is defined as
U = (Ui – Ut ) / Ui
Where Ui is the initial average excess hydrostatic pressure over entire depth.
Ui = (1/2d )∫02d ui dz
and Ut is the average excess pore water pressure after time t over entire depth
Ut = (1/2d) ∫02d ue dz
Therefore U= 1 – [(1/2d) ∫0 ue dz]/[ (1/2d )∫02d ui dz]
2d

For constant initial excess pore water pressure over entire depth fig
U = 1 – (1/2dui) ∫02d ue dz
It may be noted that the average degree of consolidation U is equal to the area of the
hatched portion of the rectangle to the total area.



Substituting the value of ue
U = 1 – (1/2dui) ∫02d∑N=0∞ (2ui/M) sin (Mz/d)e- MM T dz
Where M = (π/2 )(2N+1)
Thus U = 1 - ∑N=0∞ (2/M2) e – MM T
Or U = f(Tv)
Therefore average degree of consolidation U depends upon non dimensional time factor
Tv.
U 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Tv 0.0 0.008 0.031 0.071 0.126 0.196 0.287 0.403 0.567 0.848 ∞

The values of the average degree of consolidation can be computed by using the
following equations also.
U = √ [ 4Tv /π ] 100 for Tv ≤0.217 (U≤ 52.6 %)
U = [1 –10 – (0.085 + T ) / 0.933 ]100 for Tv >0.217 (U> 52.6 %)

CONSOLIDATION SETTLEMENT PREDICTIONS
Normally consolidated soils:
de = - Cc d(logσ′)
Δe = - Cc log (σ′/ σ0′)
εz = -Δe / (1+e0)
Therefore εz = [Cc/(1+ e0)] log (σ′/ σ0′)
Integrating the above equation over the depth of the soil gives the consolidation
settlement at the ground surface.
(sc)ult = ∫ εz dz
= ∫ [Cc/(1+ e0)] log (σ′/ σ0′)
Usually the above integral is evaluated by the geo tech engineers as the
following summation.
(sc)ult = ∑ [Cc/(1+ e0)] log (σ′/ σ0′)
Where the Cc is the compression index for the virgin curve.

Over consolidated soils:
If entire consolidation occurs on recompression curve.
(sc)ult = ∑ [Cr /(1+ e0)] log (σ′/ σ0′)
If entire consolidation occurs on virgin curve
(sc)ult = ∑ [Cc/(1+ e0)] log (σ′/ σ0′)
If consolidation spans from recompression curve to virgin curve
(sc)ult = ∑ [Cr /(1+ e0)] log (σc′/ σ0′) + Cc/(1+ e0)] log (σ′/ σc′)

Adjustment to laboratory consolidation data
Consolidation tests are very sensitive to sample disturbance. Very high quality samples
produce distinct consolidation curves as shown in fig. Less than ideal sampling and
handling techniques, drying during storage and other effects can alter the sample and
make test results more difficult to Interpret. It is difficult to obtain σc′ , pre consolidation



pressure from poor quality samples, because the transition between the recompression
and virgin curves becomes much more rounded.
Casagrande and Schmertman developed methods of adjusting lab consolidation test
results in an attempt to compensate for nominal sample disturbance effects. These
methods are developed for soft clays and difficult to implement for stiffer soils.
Casagrande’s procedure determines pre consolidation pressure, σc′ , from the lab data as
shown in Fig.No.8. The following is the procedure to correct the pre consolidation
pressure.
1. Locate the point of min. radius on the consolidation curve. (point A)
2. Draw a horizontal line from Point A.
3. Draw a line tangent to the laboratory curve at point A.
4. Bisect the angle formed by the lines from steps 2 & 3.
5. Extend the portion of the virgin curve upward until it intersects the line formed in
step 4. This identifies point B, which is the pre consolidation stress σc′ .

e
point B

Hor. through A

Angle Bisector

Point A

Tangent to curve at A

log σ'
σc'

Fig.No.8. Casagrande’s procedure for determining pre-consolidation pressure
Sample disturbance also affects the slope of the curves. So the Schmertmann
procedure is an attempt to reconstruct the field consolidation curve.
1. Determine σc′ using Casagrande procedure.
2. Compute the initial vertical effective stress σ0′ , at the sample depth. This is
the vertical effective stress prior to placement of proposed load.
3. Draw a horizontal line at e = e0. Through σ′ = σ0′ draw a vertical line. Point of
intersection of these two lines locates point C.
4. Beginning at point C, draw a line parallel to the rebound curve. Continue to
the right until reaching vertical line through σc′ , pre consolidation pressure.
This forms point D. in some cases σc′ ≈ σ0′ , so this step may not be necessary.
5. Extend the virgin curve downward to e = 0.42 e0 thus locating point E. If no
void ratio data is included on the consolidation plot, locate E at εz = 0.42,
which is same as e = 0.42 e0.



6. Draw a line connecting points D & E. this is reconstructed virgin curve.

e
e0 C
D

Parallel to
recompression
curve

0.42e E

σ0' σc' log σ'

Fig.No.9. Schemertmann’s method of adjusting consolidation curve

Determination of coefficient of consolidation
Coefficient of consolidation cv can be determined by assessing each of the parameters,
which cv is a function.
cv = [2.3 σ′k / γw ] (1+ e) / Cc
But coefficient of consolidation is rarely computed so. Instead engineers usually measure
the rate of consolidation in a lab consolidation test and back- compute cv by performing
time settlement analysis in reverse.
In principle it should be a simple matter to obtain cv from laboratory time- settlement
data. The stress condition in the laboratory sample is such that U – Tv relationship is
exactly as shown by the solid line and eqns.
U = √ [ 4Tv /π ] 100 for Tv ≤0.217 (U≤ 52.6 %)
U = [1 –10 – (0.085 + T ) / 0.933 ]100 for Tv >0.217 (U> 52.6 %)
U% (Average degree of consolidation)
0
10 0.008
20 0.031
30 0.071
40 0.126
50 0.197
60 0.287
70 0.403
80 0.567
90 0.848
100
0 Tv Time factor 0.9

Fig.No.10. Variation of Degree of consolidation with time factor



Thus we might expect to simply select an appropriate point in lab time–settlement plot,
identify the corresponding values of U, t and Tv and use eqn. Tv = cv t / d2 to compute cv.
In practice this task is slightly more complicated because the time settlement behavior in
the lab is slightly different than that in the field.
Therefore special curve fitting technique has been developed to determine cv.

Square root time method
*Plot the soil compression against square root of time.
*The initial portion of curve will be fairly straight. Extrapolate it back to √t = 0. This
locates the pt. A.
*Beginning at point A, draw a line that has a slope of 1.15 times that of the initial portion
of the curve.
*Note the point where the line in step 3 crosses the lab curve. This is point B, which
represents U=90%. Read corresponding time √t90.
*Using equation for U=90%, Tv = 0.848, t = t90 and d= one half of the sample height.
compute cv using Tv = cvt / d2.

A U = 0%
Dial gauge reading

B
U = 90%

t
t 90

Fig.No.11. Determination of cv

Soil Compaction
Compaction improves the engineering properties of the fill in many ways, including.
Increased shear strength, which reduces the potential for slop stability problems,
such as land slides, and enhance the fill’s capacity for supporting load, such as
foundation.
Decreased compressibility, which reduces the potential for excessive settlement.



Decreased the hydraulic conductivity, which inhabits the flow of water through
the soil.
Decreased void ratio, which reduces the amount of water that can be held in the
soil.

Proctor compaction test

1. Obtain a bulk sample of the soil to be used in the compacted fill and prepare it by
breaking lumps.
2. Place some of the prepared soil into a standard 1/30ft3(9.44 x 10-4m3) cylindrical steel
mould until it is about 40% full.
3. Compact the soil by applying 25 blows from a special 5.5lb (2.49Kg) hammer that
from a height of 12in (305mm).
4. Place a second layer of soil into the mould until it is about 75% full and compact it
using 25blows.
5. Place third layer into mould and compact it in the same fashion. Thus we have
applied a total of 75 hammer blows.
6. Trim the sample so that its volume is exactly 1/30ft3. Then weight it.
W − Wm
γ = ms
Vm
Where Wms = Weight of the mould + soil
Wm = Weight of mould
Vm = Volume of mould
7. Perform moisture content test in a representative portion of the compacted sample.
Then compute γd.
8. Repeat step 2-7 three or four times each with the soil at a different water content.

S=80% S=100%

γd max
γd

Zero air void

wo w
Fig.No.12. Compaction curve



A plot of Water Content versus dry density is shown in the Fig. No.12. above. So there is
a certain moisture content that produces greater γd.
Gγ
γd = s w
1+ e

Gsγ w
γd =
1 + WGs / s
The mechanics behind the shape of this curve is complex. For a dry soil, we achieve
compaction by first adding water to raise its moisture content to near optimum. This
water provides lubrication, softens clay bonds, and reduces surface tension forces with in
the soil. However, if the soil is too wet little air voids are left and thus it becomes very
difficult or impossible to compact.

Relative Compaction
Usually specifications for earth compaction works are given as fraction of the dry density
achieved in laboratory standard proctor test. The dry density to be achieved in the field is
stated as 90,95,98% etc. of the maximum dry density achieved in lab test. this
compaction achieved in field which is a fraction of laboratory compaction maximum dry
density is expressed as relative compaction and given by
γd
CR = ------× 100
γd,max
Modified Proctor test

Standard compaction Modified compaction
Hammer Weight 5.5lb(2.4Kg) 10.0lb(4.54Kg)
Drop height 12’’(305mm) 18’’(457mm)
No. of layer 3 5
No. of blows 25 25
Energy 12400 ft-lb/ft3(600KN- 56000 ft-lb/ft3(2700KN-
m/m3) m/m3)

Soils compacted on dry side of OMC shows a flocculated fabric while those compacted
on wet side of OMC shows oriented or dispersed fabric. Soils compacted on dry side also
shows high hydraulic conductivity and greater shear strength while soils compacted on
wet side shows low shear strength. Also compaction method produces a different fabric.
Pressure compaction produces a different fabric compared that produced by manipulation
compaction. at same dry density. Although fabric is not an important factor in general
projects it may be critical in special projects.



Earthwork Quantity Computations
In many engineering works we have to bring soil from a barrow pit (bank) and compact
the soil to required density in some engineering construction. In the barrow pit soil will
be at an in-situ density with natural moisture content and it has to be compacted to a
different density at different moisture content. In this problem we have to compute how
much of barrow pit soil produces how much compacted fill. that can be computed using
the shrinkage factor. However the student should not confuse it with the shrinkage
associated with natural soil.
ΔV (γ d ) f
Shrinkage factor = =( − 1)100
Vf (γ d ) c

Where ΔV= Change in volume during grading.
Vf = Volume of fill.
(γd )f = average dry unit weight of fill.
(γd)c = average dry unit weight of cut.
The following Fig.No.13 indicates the different stages involved in excavation and
compaction of soil in to a new fill, and also the change in state (densities and volume) of
soil in various stages. A problem of numerical nature solved at the end illustrates how we
can make use of shrinkage factor in earth work computations.

Excavate Compact
V=1m 3
V= 1.25m3 V=0.8m3

Bank Loose Compacted

Fig.No.13. soil quantity from barrow pit to emankment

PROBLEMS
A footing has a size of 3.0m by 1.5m and it causes a pressure increment of 200
KN/m2 at its base. Determine the consolidation settlement at middle of the clay
layer. Assume 2:1 pressure distribution and consider the variation of pressure
across the depth of the clay layer. γ=10KN/m3.
Soln:
Initial preesure at the center of clay layer
σ0′ = 2.5 (16) + ( 0.5 ( 18 –10) = 51.5 KN/m2
The overburden pressure at the center of the clay stratum
= 18×2 + 10.33×1.5 = 51.5 KN/m2
The pressure increase at the top, middle and bottom of the clay layer

200 x3x1.5
(Δσ ' ) t = = 51.4 Kn / m 2
(3 + 2)(1.5 + 2)



200 x3 x1.5
(Δσ ' ) m = = 27.7 Kn / m 2
(3 + 3.5)(1.5 + 3.5)

200 x3 x1.5
(Δσ ' ) b = = 17.3Kn / m 2
(3 + 5)(1.5 + 5)

Average pressure can be found by Simpson rule

Δσ =
1
6
[
Δσ t + 4(Δσ m ) + Δσ b
' '
]
1
= [51.4 + 4 * 27.7 + 17.3] = 29.9 KPa
6

0.3 ⎛ 51.5 + 29.9 ⎞
∴S f = x3 log10 ⎜ ⎟ = 0.09941m = 99.41mm
1 + 0.8 ⎝ 51.5 ⎠

A stratum of clay is 2m thick and has an initial overburden pressure of 50KN/m2 at
its middle. Determine the final settlement due to an increase in pressure of 40KN/m2
at the middle of the clay layer. The clay is over consolidated with a pre consolidation
pressure of 75KN/M2. The value of the coefficient of recompression index are 0.05 &
0.25 respectively. Take initial void ratio as 1.4.

Cr ⎛ σ c' ⎞ C ⎛ σ + Δσ ' ⎞
Sf = ⎜
H o log⎜ ' ⎟ ⎟ + c H 0 log⎜ c ' ⎟
1 + e0 σ 0 ⎠ 1 + e0 ⎜ σ ' ⎟
⎝ ⎝ c ⎠

0.05 x 2 ⎛ 75 ⎞ 0.25 x 2 ⎛ 50 + 40 ⎞
Sf = log⎜ ⎟ + log⎜ ⎟
1 + 1.4 ⎝ 50 ⎠ 1 + 1.4 ⎝ 75 ⎠

= 23.84mm

A clay layer 4m thick has final settlement of 6.0m. The layer has double drainage. If
the coefficient of consolidation is 0.02cm2/minute, determine the time required for
different percentage of consolidation from 10% up to 90.1%, and hence plot time
settlement curve.

τ v = C v t / d 2 =0.02 x t /200 or t=2x106 τv
if time is taken in years
2 x10 6
t= = 3.805τ v
60 x 24 x365

The calculations are given in the tabular form below.



V% 10 20 30 40 50 60 70 80 90
Τ 0.008 0.031 0.071 0.126 0.196 0.287 0.403 0.567 0.848
T 0.030 0.118 0.27 0.479 0.746 0.092 1.533 2.157 3.227
S 0.6 1.2 1.8 2.4 3.0 3.6 4.2 4.8 5.4

Twelve undisturbed soil sample were obtained from boring in a proposed cut area.
These samples had an average γd of 108lb/ft3 and an average water content of 9.1%.
A proctor compaction test performed on a representative bulk sample produced
(γd)max = 124lb/ft3 and water content of 12.8%. A proposed grading plan calls for
1200yd3 of fill, and the specifications call for a relative compaction pf at least 90%.
A. Compute the shrinkage factor.
B. Compute the required quantity of import or export soils based on unit of the
cut.
C. Compute the weight of import or export in tons using the moisture content of
the cut.
D. Compute the quantity of water in gallons to bring the fill soils to the
optimum moisture content.
Solution

A) Using average relative compaction of 92%

(γd)f=(γd)max xCR = 124lb/ft3 (0.92)=114lb/ft3

ΔV (γ d ) f
=( − 1)100 = [114/108 – 1]100=6%
Vf (γ d ) c

ΔV
B) ΔV = V f = 6% (11500) yd3
Vf
Cut required to produce 11500yd3 of fill = 11500+690=12190yd3
Required import = 12190-12000=190yd3

C) Ws = V γd = (190 yd3)(3ft/ayd)3(1080lb/ft3)15/2000lb = 277T
W = Ws (1+w)=(277T)(1+0.091)=302T
D) (Ws)fill = (γd)fill Vfill = 114lb/ft3 x 11500yd3 (3ft/ayd)3=35.4x106lb.
Wt. of water to be added
= ΔV = ΔWxWs (0.128 − 0.091)(35.4 x10 6 lb) = 1.31x10 6 lb
Water a unit weight of 8.34lb/gal
There fore, Vw =Ww / γw=1.31x106/8.34=157000gal.


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