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DEPARTMENT OF.
MINE·RAlS AND ENERGY
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BUREAU OF M~NERAlRESOUrRCESv
GEOLOGY AND GEOPHYS~CS
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504949
Record 1974/108
TERZAGHI'S THEORY OF ONE DIMENSIONAL··
PRIMARy CONSOLIDATION OF SOILS AND ITS APPLlCATION
by
J.R. Kellett
~llUGrZ;
I
BMR
Record
1974/108
c.3
>
The information contained in this report has been obtained by the Department of Minerals and Energy
as part of the· policy of the Australian Government to assist in the exploration and development of
mineral. resources. It may not I?e published in any form or used in a companiprospectus or statement
withoutthe permission in writing of the Director, Bureau of Mineral Resources, Geology and Geophysics.
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Record 1974/108
TERZAGHI'S THEORY OF ONE DIMENSIONAL
PRIMARY CONSOLIDATION OF SOILS AND ITS APPLICATION
by
J.R. Kellett
piSInrcl&
^
CONTENTS
PREFACE
INTRODUCTION^
1
DARCY'S LAW
GENERAL CONDITIONS OF FLOW ^
3
THEORY OF CONSOLIDATION ^
4
CALCULATION OF SETTLEMENT AND TIME ^
7
TIME^
SETTLEMENT^
THE CONSOLIDATION TEST^
7
10
11
PRACTICAL APPLICATIONS OF THE THEORY OF CONSOLIDATION ^13
Settlement Calculations^
14
Example 1^
14
2^
15
11^3^
16
Digression - Boussinesq Analysis ^
Discussion of basic assumptions^
Time Calculations^
19
20
21
CONCLUSION^
23
REFERENCES^
24
TABLE 1. Settlement analysis, Isabella Plains.
APPENDIX 1. Worked solutions of problems.
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FIGURES
1.
Total head and hydraulic gradient.
2.
Inflow and outflow through a small element of soil.
3.
Consolidation model.
4.
Relation between time factor and degree of consolidation.
5.
Settlement of column of soil.
6.
Consolidation apparatus.
7.
Consolidation curve.
8.
Void ratio-effective pressure curve.
9.
Consolidation test on peaty clay.
10.
Soil sequence at Isabella Plains.
11.
Model for example 1.
12.
Model for example 3.
13.
Contours of equal vertical stress in example 3.
14.
Boundary conditions at Isabella Plains.
15.
Time-settlement curves for different boundary conditions.
16.
Void ratio-effective pressure curve of an overconsolidated soil.
17.
Increase in effective pressure at depth.
18.
Compression-root time curve.
19.
Compression-log time curve.
20.
Site conditions for example 4.
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PREFACE
This presentation of Terzaghi's theory of one-dimensional
primary consolidation and its application has been prepared for use
within the Engineering Geology Subsection of the Bureau as an
instructional document for geological and technical staff.
INTRODUCTION
In this paper, a non rigorous mathematical proof of
Terzaghi's theory of one-dimensional primary consolidation is set
out in simple language without omitting any of the basic steps;
in addition some examples of its use are included. ^Further
information can be found in soil mechanics textbooks.
DARCY'S LAW
Darcy's Law states that in the case of steady state
laminar flow, the apparent velocity of a fluid through a porous
medium is directly proportional to the hydraulic gradient.
That is, ^v = .ki
where v = apparent velocity of flow
k = coefficient of permeability
i = hydraulic gradient
A-
- dh
dh
- —
dl
Fig. 1. TOTAL HEAD AND HYDRAULIC GRADIENT
Consider a point B in a mass of saturated porous soil
(Fig. 1).^Let the pore pressure at B = u.
A column of water in equilibrium with the pore pressure at B will
rise to a height above B = u (where yw = density of water).
Yw
Now, the total head = position head + pressure head; that is,
h= z +!
Yw
The rate of flow is governed by the HYDRAULIC GRADIENT which is
defined as:
i - - dh
dl
It follows then, that if q is the rate of flow through an area of
cross-section A,
then^v =
A
whence, q = Aki^(since v = ki)
GENERAL CONDITIONS OF FLOW
•^
Consider an element of soil of unit cross-sectional area
and height dz (Fig. 2) through which water is flowing in the z
direction.
Outflow = v
z
+ av .dz
z
3z
Cross-Sectional
area = 1
I
Inflow = v z
Fig. 2. INFLOW AND OUTFLOW THROUGH A SMALL ELEMENT OF SOIL
e#.
Let v
z
= inflow velocity,
then total inflow = v z (since cross-sectional area = 1), and
outflow depends upon the change in velocity, v z , of the water as
it flows through the soil.
Dv .dz
z
az
Therefore, Outflow = [ Inflow ] + [(rate of change of velocity of
The magnitude of the change in flow is given by
water in z direction) x (distance through which
it travels)]
That is, Outflow = v
z
+ ay .dz
z
3z
Consider an element of dimensions dx, dy and dz through
which water is flowing parallel to the z axis (Fig. 3).
THEORY OF CONSOLIDATION
V z + 3V z .dz
az
Fig, 3. CONSOLIDATION MODEL
Volume of water entering at any time, t = v. t. (dxdy)
So, volume of water entering the element in unit time = v dxdy
z
Volume of water leaving the element in unit time = (v
z
+ 3v dz)dxdy
z
3z
Hence, rate of volume change = net decrease in volume of water.
i.e.
av (v z +
3v dz)dxdy - v dxdy
z
z
at^3z
av z dxdydz
From Darcy's Law, v
k3h
—
z = ki = az
Whence, rate of volume change 3V =
a
kah dxdydz
3z 3z
2 h dxdydz
3V
that is, (assuming constant k)^= k3
3 2
^(i)
at
Now,
Now, volume of solids in the element, V
s
= dxdydz
1+e
and, volume of voids in the element, V v = dxdydz. e
where e is the void ratio*
*FOOTNOTE
The void ratio, e, is defined as the ratio of the volume
of voids to the volume of solids.
If V'= total volume
and^Vv= total volume of voids;
then^e = Vv by definition.
V-V
v
Void ratio should not be confused with porosity, which is
defined as the ratio of the volume of voids to the total volume of
soil aggregate.
i.e. Porosity, n = Vv
V
The relationship between void ratio and porosity is:
e =
1-n
^
^
5.
If the original volume of the element is V, then the time rate
of volume change in terms of the change in the void ratio is:
av^a^(Vv)
at = at
=
a
(dxdydz. e )
at^. 1+e)
. a (V e)
at
= V 3e assuming V s is constant.
at
that is, 3V .(dxdydz).3e
Dt^1 + e at
^
(ii)
Now, rate of reduction of voids = net rate of flow of water from
the element,
that is,
(dxdydz )3e^ka 2 h dxdydz
(equating (i) and (ii))
1 + e /at^a 2
2
/ 1 \ae^k3 h ^
ie.
.
- r_ e- —t-^az2
Li 5
But from (1), h = z + u
Yw
and Dh . 1
3u y
w
whence, ah = 3u
Yw
Substituting into (iii), we have:
(
2
' 1^
e^3 u
1+e )at = yw a z 2
^
Now, the drop in pore-water pressure (-du) = increase in effective
pressure (dp);
(iv)
^
6.
and from the identity, MV = - de ^1 *, we have
dp (lie)
^
de = M du (1+e)
(since dp = -du)
v
and substituting in to (iv),
1^M (1+e) all k 3 2 u
v
1+e^at y^2
w z
2
^is,
a u ^k^3 u^
that
at^Yw v^3 z 2 :) ^
(v)
Equation (v) is Terzaghi's differential equation for one-dimensional
consolidation.
The term C = k is denoted as the "COEFFICIENT OF
V YwM v
CONSOLIDATION" ^(It should be noted that the coefficient of permeability can be deduced from consolidation test results).
Equation (v) is usually written as:
Du C a 2 u
—v--at 3 2
*FOOTNOTE
The COEFFICIENT OF VOLUME COMPRESSIBILITY, M v , is
defined as the compression of the soil, per unit of original
thickness, due to a unit increase of pressure.
If the thickness of the soil is H, then the rate of
change of thickness with respect to pressure, (as a proportion
of the original thickness) is -dH . 1
dp H
-2
i.e.^IMv1= de . ^1
dp^(1+e)
-de . 1^(assuming constant
dp (1+e)^cross-sectional area)
7.
This partial differential equation can be solved by a Fourier
series which relates the drop in pore-water pressure to the
original pressure at time of loading.^But it is more useful
to express the solution in terms of:
(a) the average degree of consolidation (U), and
(b) the "time factor" (T v )
Figure 4 shows the graphical relationship between T v and U.
CALCULATION OF SETTLEMENT AND TIME
TIME
The percentage of primary consolidation, U, at
any time t, is a function of a dimensionless ratio, which
Terzaghi called the "time factor", T.
U = f (Tv )
Now T
v
depends on all those factors which influence the rate of
seepage from the soil.^These are:
void ratio, e
permeability, k
thickness of the compressible stratum, H
number of drainage faces of the stratum, N
density of water, y w
change in void ratio, De
change in pressure, Dp
8.
t(l+e)k
Terzaghi deduced that T v^2
y 3e
ITI )
^(vi)
and from the identities:
^ and Me. 1
C . -^v = —
y M
v
3p 1+e
WV
we have, T
v
tk
- ^
2
(H) ,y 3e. 1
N^3p 1+e
t
(dividing top and bottom
by (1+e))
^.^k
YWV
M
(11-111^
i.e.
—
2
( .111 )
from which
2
=
v
C
v
The functional relationship between T and U is shown in
Figure 4. The curves represent different boundary conditions
which will be explained later; they are derived from the
2
3u^C 3 u
differential equation ^= v
at^3z 2
9,
.20
40
0
0
P
60
:' so
1 00
P.O
0
1-2
1.4
Fig. 4. RELATION BETWEEN THE AVERAGE DEGREE OF CONSOLIDATION (U) AND
THE TIME FACTOR
Crld_
2
Fig. 5. SETTLEMENT OF COLUMN OF SOIL
10.
SETTLEMENT
Consider a column of soil (Fig. 5) of unit cross-sectional
area, under pressure p l , and let the final pressure = p 2
Then, the consolidating pressure = p 2 -p i
and total settlement, s = h-h 2 .
Now, height of solids =I' = volume of solids (since cross-sectional
area = 1).
Initial height of voids = 1 1 = initial volume of voids.
Final height of voids =^= final volume of voids.
Also, h = +
and h 2 = +1
2
If e
l
= initial void ratio, then:
e
and if
h-/ h
= —^= — - 1
1
^
= final void ratio, then:
/
h -/ h
2
2
2
e2=^=^=^-1
—
/^/
- h - h
whence, e 1 -e 2
2
E e l - e 23
i.e., Total settlement, s = h.
[ 1 + e ]
1
and, for a suitably small Ae,
S = h( Ae
1+e
. Ap. Ae . 1
Ap 1+e)
i.e.^
s = h.A p. N'
1 1.
THE CONSOLIDATION TEST
The standard consolidation test is carried out on
silts and clays in the oedometer apparatus shown in Fig. 6.
Fig. 6. CONSOLIDATION APPARATUS
An undisturbed circular slice of soil 3 /4 inch thick and 3 inches
in diameter is placed in a cell between two porous stones and
connected to a water reservoir so that it is always saturated.
Loads are applied in increments to simulate pressures ranging from
1
2
/8 ton/ft to 16 ton/ft 2 and a dial gauge measures compression of
the sample.
The resulting curve for each pressure increment is of
the shape shown in Fig. 7.
Fig. 7. CONSOLIDATION CURVE
12.
From the known dimensions of the sample . , its void ratio at the
end of each loading stage can be found. The rate of change of
.void ratio with respect to pressure is known as the COMPRESSIBILITY:
a = -de
v —
dp
That is, av is the gradient of the curve in Fig. 8.
Pressure (P)
8,
-
VOID:RATIO ..'''EFFECTIVE PRESSURE CURVE
Note. Recall the coefficient of compressibility, My
-de
- dp(1
e
which is the compressibility( -d
- --)divided
by the total volume
dp
(1+e).
•
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.. ; BORE HOLCTP 59
DEPTH ... .1.10": 1.15m
PRESSURE Vs. VOID RATIO
PROJECT: Isabella Plains Stormwater System - Site Investigation
of
CLASSIFICATION .. CL-CH CLAY: medium plastici~y, dqrk grey, some fine sand, some lenses
light grey,
non-plastic silt, some decaying fine rO<?fs and root fibres, moist (Me )PL) soft.
Inttiol Wet Density
~ 115
3
Ib/tt .
Initial M.C ... 30.60%
Final M.C .... 23.73 %
I
Sample Size
Range (psf)
200-500
500-1000
Mv. (ft7' Ib)
1S.8 x 10- 6
17.2 x10- 6
--
!
I
i
i
1
.84
I
.78 t-.
~
a::
.75
->
10-6
8.9 K 10-6
6.6
,
, I-..."
1':--...,..
I
~
K
10-6
6.3
6.2
1
i
.~!!
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I\.
'\
:
.69
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!
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~
1\ ,
~
.66
1\
!
.63
-
I
.60
I
1\
i\
I
I--
I
.57
4.9
'
I
.72
14.5 K 10- 6
4.4
I,
0
0
I
•
0
K
!
:1 rt
!
17.4
4000-8000 8000-16000
:
--t--
!
Specific GrOllity _.2.57
!
,
i
.8 I
Dio.= ... 1.766
II
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Dote _ 16.8.73
1000- 2000 I 2000-4000
3.7
2.5
Cv. (ft2;yeor)
I
{
.'
.
Ht.= _.0.751
-
-:--
r-- _
-
I
~--
!
\
- -- --\
I-
i
.54
100
3
2
4
250
2
6 7 89
500
3456789
1000
10000
16000
32000
PRESSURE (pst)
Fig. 9
CONSOLIDATION TEST ON PEATY CLAY
(After Coffey and Hollingsworth Pty Ltd)
Record 1974/108
M(G) 432
^
13.
PRACTICAL APPLICATIONS OF THE THEORY OF CONSOLIDATION
Fig. 9 shows the consolidation test results of a peaty
clay sample from Isabella Plains with pressure plotted on a log
scale. These clays occur in swampy areas and are probably the most
compressible soils in the Plains. Therefore, the following
examples of settlement and time rates are the maxima that can be
expected from primary consolidation.
The geological conditions are shown in Fig. 10.
Depth
(Feet)
0
5
.^•Water table
30
Peaty clays with sand lenses
(9 1 )
70
Massive blue-grey sandy clay
10 5
12 0
(g2)
0
if
00 •^000 0000o00• 00 •
^.0°0 0^
°^
06° 0^
Po°^0^:t^
0: 0 0 Oa • • 0
0e 0
°
o0.000,,O*1°1; ° : *g.' °
*
*
l
e*
0
0
0
::
0
0^
e a 0 "0 • e O• .00 - 0 e^0
0 ...70:0: 10°0:0
0 0 to^.
0. 0 0 0 o 0Q.0
c,_0 0
^: ,','sandy aqui^
aqu i fer grading to o ° . :00 0° 0 1:i
• °.
.°
e°° °Q°°
0000
0
0 0:^
:0
0: 00 7,:o^gra.vel
and cabbies':00:o 0 0 0 00 000 0 "„ 0 000
00 :00 000
: 0 0e: 0
„ 0 00 0 c: ei 000 :: 0°:^° o:000 0 : 0 00
° 0 0 0 „^, 0 %0 : 0
18-0^
40.
:::: ° 00 0.0:0 0 00:000• 00
0:0 °^
(to. 00 co.;;.
oe. o eoo^° e t':
A < -1^>^<
6
.: 0000
0:
0." 7°70:
Weathered tuff
Fig. 10. SOIL SEQUENCE AT ISABELLA PLAINS
14.
SETTLEMENT CALCULATIONS
Example 1
Calculate settlement that will be caused by lowering
the water table to 9 feet.
Solution. We only have consolidation coefficients for
layer (g1 ) - so, for simplicity, (g 1 ), (g 2 ) and (g 3 ) are grouped
together as layer (g). • Our amended model is shown in Fig. 11.
Present position
of water table
Density (moist) :102 pcf
Proposed new
water table
Density (submerged):115pcf
I2000
0
0
0
0
0
0
0
0
0
0
0
0
0 0
Aquifer (assumed incompressible)
Fig. 11. MODEL FOR EXAMPLE 1 (not to scale)
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~il
15.
Calculation of effective stress at base of clay layer:
Before Lowering Water Table
After Lowering Water Table
above W.T. = (1.5 x 100) - 150 psf
Soil above W.T. = (3 x 100)+(6 x 102) = 912 psf
Soil below W. T. = (1.5 x 125)+(9 x 115)
1222.5 psf
Soil below W.T.
=
Water pressure
Water pressure
= (-3 x 62.4)
=-187.2 psf
Effective stress
1069.8 psf
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= (-10.5 x 62.4) =
-655.2 psf
Effective stress
717.3 psf
= 345 psf
(3 x 115)
Whence, the increase in effective stress due,, to
lowering of water table = 352.5 psf.
From Fig. 9, the coefficient of volume compressibility
within these stress ranges is
MY
-6
2
= 17.3 x 10
f+ /lb~
Therefore, from the settlement equation (viii), s =
h.~p.
M,
v
we have, s = (9 x 352.5 x 17.3 x 10- 6 ) ft.
= 0.055 ft.
= 0.66 inches
Example 2
At the site where this sample was obtained on Isabella
Plains, it is also proposed to construct the major roads above flood
level.
So now we consider the
settlem~nt
due to emplacement of
4 feet of compacted sandy fill (moist density 130 pcf) overlain
by 1 foot of compacted gravel (moist density 140 pet), in addition
to that due to drainage.
16.
Solution. The pressure due to the fill is:
(130 x 4) + (1 x 140) psf
= 660 psf
and assuming that the weight of the fill stresses the clay evenly
over its whole thickness*,
then final settlement = (increase in overburden pressure + pressure
of fill) x h x M
v
= [(352.5 + 660) x9 x 17.4 x
io
6 i , ft.
= 0.16 ft.
= 1.9 inches
Example 3
We now consider the settlement due to:
(a) dewatering to 9 feet,
(b) excavation of top 3'feet of soil,
and^(c) installation of a 2 ft x 2 ft pier of load 16,000 lbs.
Solution. The initial and final conditions are set out in
Fig. 12.
*This assumption is not strictly correct. ^The effective stress
due to the fill will vary throughout the clAI profile — see
Boussinesq analysis in Example 3.
.^17.
Depth
(Feet)
0-.
••..•.
eriiityAmoist):=:100.p.c
1-5 -
It
)
W.T.
•^•...• .^•
.• • •'."•.•.•• .•••
••^
.
ensity (submerged) 125 pet
3-0 -
Density (moist) = 102 pcf
(q )
Density (submerged) :115 pcf
W T.
9-0 -
Density (sumerged)= 115 pet
12-0 -
0
0
OOOOOOO
t^t^t
0
0
0
t^t^t
Assumed incompressible
INITIAL CONDITIONS^
FINAL CONDITIONS
Fig. 12. MODEL FOR EXAMPLE 3.
The best procedure is to divide the clay layer into 9 x 1 footthick strips and sum the average settlements at their mid-planes.
The data is shown in Table 1. ^Stresses due to the pier are
calculated by Boussinesq analysis.
0
0
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Table 1 SETTLEMENT ANALYSIS
1
La)'u
Z
3
4
h
Depth to
mid-plane
(feet)
Initial preuure
(feet)
,
Pressure after
lowcrina vater
table
Pz
PI
1
1
2
1
3
1
4
,
6
7
8
9
1
1
1
1
1
1
,.,
4.5
,.,
..,
7.'
8.5
9.'
10.5
11.'
lOOx1.5
125x1.5
115xO.5
"62.4x2
100x1.5
102xO.5
100xl.5
(270.2)
(351.0)
lOOxl.' -62.4x1.5
125xl.5 -62.4x1.5
,115x1.S
100x1.'
10Od.S
102x1. S
(322.8)
(453.0)
100x1.5 -62.4x1.'
125xl.S -62.4x2.5
115x2.S
100xl.'
100x1. ,
102x2.5
(375.4)
(555.0)
100x1.5 -62.4xl.5
125x1.5 -62.4x3.5
ll'x: .,
100x1.5
100x1.'
102x3.5
(428.0)
(657 .0)
100x1.5 -62.4x1.'
125xl.5 -62.4x4.5
11',,4.5
10Ox1.'
100x1.5
102x4.5
(480.6)
(759.0)
100xl.' -62.4xl.5
125x1.5 -62.4x5.5
115,,5.5
100x1. 5
100,,1.'
102,,4.5
(533.2)
(861.0)
100x1.5 -62.4x1.5
125x1.5 -62.4,,6;5'
ll'x6.5
100,,1. 5
100,,1.5
102,,6
115>=0.5
-62,1, xO. 5
(585.8)
(938.3)
100x1. 5 -62.41.:9
125x1. 5_
115,,7.5
100x1.5
100x1. 5
102x6
115,,1.5
-62.4xl.5
(638.4)
(990.9)
100,,1.5 -62.4xlO
l25x1.5
115,,8 • .5
100x1.5
100,,1.5
102,,6
115x2.5.
-62.4x1.5
(691. 0)
(1043.')
~hApMy • 126,228.64 x 10-6 feet.
• 1.5 inches.
6
7
8
Pre.,ura
. t.o.d
6p
reductlotl
p\'II •• ure ·PZ+P,+P4-Pl
due to
at mldexcavatlotl plene depth
'9
10
MY
(xlO- 6
ft Z/ll1)
h 6p My
(xlO- 6 ft)
,.,
'.
-300
3600
3380.8
14.'
49021.6
-300
ZOOO
1830.2 ;
'17.3
3184'.48
-300
1000
879.6
17.2
15129.12
·300
600
529.0
17.2
9098.8
-300
3GO
338.4
11.2
5820.48
-300
Z40
261.8
17.2
4606.16
.;.300
200
252.'
17.2
4343.0
-300
145
191.'
17.2
3397.0
-300
120
172.'
17.Z
2967.0
:1
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footing .
,
r,
ground surface (feet)
~------------------------------~O
-----I
Q= 16000lbs
(°1
q~
4000psf
----------------~~2
l
I I 21\ j
2ft
4ft
."
T---------~------=r---------,----------''r-----------l3
-~rHY-~~--------------------------~_4
.,,
----f----\-
I
~----+---------46
I
1----- ---- -. -
7 .
8
----19
------"~ 1 6 0 . . - = - - - - - - - - - - + - -
I
------------10 '
.I
----------'----'------'---~____/_----'-------____c..--'-----------III
~8:--·-
i
i
.-
---+------+-----------l12
i
i
i
Fig. 13 CONTOURS OF EQUAL VERTICAL STRESS
IN EXAMPLE 3
'1
I
Depth below .
X 2' square
Record 1974/108
M(G)
433
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18.
To find our Cv value, we note that the pressure range in which
we are interested is 200 psf to 4,000 psf.
Hence, the weighted
"
average C value from Fig. 9 is:
v
C,v
= (200 x 2.5)
4000
= 4.99
+
(500 x
4000
3.7)
+
1000 x 4,.4 \,+ (2000 x
' 4000
)
4000
C'
2
ft /year
whence, substituting into (ix), we have
t =
=
34.02
4.99
6.8 years.
Now consider the effect of double drainage (i.e. the soil is now
tr'eated as an open layer).
From Curve 1 (Fig. 4), T = 0.6
v
,
2
and t ... (0.6)
x(t)
4.99 '
=
2.4 years.
Which time ,estimate is correct?
We must now go back to our logs
and determine the true boundary conditions (Fig. 14).
Consolidating pressure
rO
Deplh (fll
~30
Drain
I
Clays
I
o
I
_ _ _ _E_q'-u_i_1i_b_r_iu_m
__
W, T.
~. 9.0
r
I
10,5
!
~
,I
'
12,0
I
L
180
'
19.
DIGRESSION — BOUSSINESQ ANALYSIS
The load pressures in column 7 of Table I were calculated
from the Boussinesq equation, which gives the increase in vertical
stress at any depth due to a point load placed on the surface of a
homogeneous, isotropic, elastic material of infinite thickness.
The equation is:
3
Aa =
z
2 • ^
2 Tr [ 1 1.(1)2]
5/
2
in which Au = increase in vertical stress
Q = point load
= depth below load
= horizontal distance from the point of application
The stresses at depth in example 3 are obtained by integrating the
Boussinesq equation over the (2 x 2) feet square area.
Contours of equal vertical stress in the clay are shown
in Fig. 13.
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.1.
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20.
.
Discussion of basic assumptions
In assessing the reliability of the settlement result
in Example 3, an important variable is P4' the load pressure.
Bearing in mind that this parameter has been calculated by the
Boussinesq theory, we must now consider whether the use of this,
the simplest of all the point load-stress relationships, is justified.
The thoery depends on the following soil properties:
(i)
HOMOGENEITY:
the soil is not homogeneous as can
be seen from our simplified log.
..
Even the individual strata are complex
soils with lenses and wedges.
(U)
ISOTROPY:
the Isabella Plains soils are generally
stratified.
This tends to spread the
load further horizontally thus reducing
the stress concentration immediately
below the loaded area.
(iii)
ELASTICITY:
plasto-elastic is a more apt description
of the soil's behaviour as indicated by
the decompression curve on the
consolidation test sheet (Fig. 9) which
shows only a partial expansion of the
soil with release of pressure •
(iv)
INFINITE THICKNESS:
we have assumed a rigid boundary at 12 feet
below ground surface.
Therefore, the
stress concentration near the boundary is
increased.
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21.
It is evident that we should use a more sophisticated
technique to accurately estimate stress at depth, but these methods
are beyond our scope.
Nevertheless, the Boussinesq analysis provides
us with an approximate magnitude of settlement and we would normally
report expected settlement of 1 to 2 inches in Example 3.
TIME CALCULATIONS
Recall the consolidation-time relation,
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ (vii)
where,
t
= time taken for a certain percentage, U, of primary
consolidation to occur.
Tv
H
= time factor (real number)
= thickness of compressible stratum
v
= coefficient of consolidation
N
= number of drainage faces (since we are considering
C
vertical drainage, N can only be 1 or 2).·
In Fig. 4, the curves are different boundary solutions of
the differential equation:
22.
Curve No. 1 (Fig. 4) represents consolidation of an open* layer of
soil under a consolidation stress that is uniform throughout the
profile. Curve No. 2 represents consolidation of a half-closed layer
of soil whose thickness is greater than the width of its loaded area
(Example 3).'
Now assume that the maximum expected settlement is
inches and foundation design is such that a 1 /2 inch tolerance is
permitted.
Then the degree of consolidation, U = 80%, and if we
assume that the 9 feet of compressible soil underlying the pier is
half-closed, we find from Curve 2 that T= 0.42,
and substituting into (vii),
t = 0.42 x(1)
Cv
=(3
4.02
C
v
)
Years
(ix)
*If the soil is free to drain through both it's upper and lower
surfaces, it is said to be an "open layer".^If water can escape
through only one surface, the layer is said to be "half-closed".
23.
From geological evidence; we can infer that double drainage will
occur within the clay when subjected to a consolidating pressure
after lowering of the water table. Hence, our most realistic model
is a 7.5 feet-thick open layer, for which:
7 51
t = (0.6) x ( 2
4.99
.
i.e.^t = 1.7 years
CONCLUSION
This example clearly illustrates the importance of determining the
correct field boundary conditions. ^For this reason, it is the
engineering geologist or hydrologist who should define the geological
conditions and deduce time rates of settlement rather than the
engineer. The latter will normally accept the worst solution unless
he has a full understanding of the geological conditions. The
comparison between single and double drainage is shown graphically in
Fig. 15.
It is also important to determine whether high permeability
layers in the soil profile are continuous. ^For instance, in the
above example, if one assumes a continuous sand layer at, say, 7 feet
from the surface, then the time calculation becomes:
2
f = (0.6) x2
(4 )
4.99
= 0.5 years
In the initial investigation, sand layers were detected
throughout the peaty clay, but additional augering revealed that
these layers were in fact lenses which would have negligible effect
on pore-water drainage.
20
40
0.
03
c,
-
80
25
3
01^05
4
5
LU
100
TIME (Years)
Fig. 15 TIME- SETTLEMENT CURVES FOR DIFFERENT BOUNDARY CONDITIONS
Record 1974/108
.M (G) 434
24.
REFERENCES
CAPPER, P.L. & CASSIE, W.F., 1969 - THE MECHANICS OF ENGINEERING
SOILS.^Spon, London.
SCOTT, C.R., 1969 - AN INTRODUCTION TO SOIL MECHANICS AND FOUNDATIONS.
Maclaren & Sons, London.
TAYLOR, D.W., 1948 - FUNDAMENTALS OF SOIL MECHANICS. Wiley, New York.
TERZAGHI, K. & PECK, R.B., 1967 - SOIL MECHANICS IN ENGINEERING
PRACTICE, 2nd edn. Wiley, New York.
'APPENDIX 1
Worked solutions of consolidation
problems posed by Professor E.H. Davis during
his lectures on soil mechanics, MR, 1972.
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Example
1.
No~
The following results were obtained from a consolidation
test carried out on a sample of clay.
Void ratio (e) is.re1ate~
to effective pressure (Pe) in kips*/sq.
ft~
P
3
4
e
0.705
0.698
1
2
1-
0.688
Calculate the
3
6
12
0.673
0.645
0.600
c~mpreasion
48
0.550
0.500
index of the soil and the
preconso1idation pressure.
Solution
The graph of voids ratio va effective pressure is shown
in Fig. 16.
The 'compression index', Cc, is the gradient of the
e-1og
10
p curve.
Analytically,
Cc =
__
-~d~e__~_
d(10g10 P )
For a "normally consolidated" soil,
ther~
is a linear relationship
between e and log10 P and hence Cc is constant.
However, in our
example; normal consolidation does not occur until we have reached
8,000 psf pressure.
(1. e. we can only calculate the gradient of
the curve between points A andB in Fig. 16).
*kip
= 1000
1bs.
I= NM MO NM I= 11111 MI I= NMI MN MI MI MI MI 11=1 OM MI
0.75
0.70
0.65
0.60
0.55
0.50
100
^
1000
^
Po^10 000^24 000^48 000^100 000
Pe ( p s f ) (Log scale)
FIG.I 6-VOID RATIO-EFFECTIVE PRESSURE CURVE OF AN OVERCONSOLIDATED SOIL
.
Record 1974/108^
M(G) 436
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(ii)
Taking two arbitrary points between A and B, we have:
..
el
= 0.550
PI
= 24,000
e2
= 0.500
P2
= 48,000
Cc
=
(0.550 - 0.500l
(logio 24,000
10gi0 48,000)
=
~0.050~1
(loglO /2)
=-
0.050
0.3010
=
0.166
So we would normally report Cc = 0.17;
or, more precisely Cc =
e
375 ~p ~ 1,500
1,500 <p < 8,000
8,000 ~p ~48,000
o03
variable,> 0
0.17
which gives us a far better picture of the behaviour 6f the curve.
The preconsolidation pressure, Po' is determined geometrically from
the following Gonstruction after Cassagrande:'
1)
Select point of maximum curvature (C).
2)
Draw the tangent to C (CC") and a horizontal line through C (CC').
3)
Bisect C'C" (CC I " ) .
4)Proj ect the normally consolidated part (AB) back to D.
5)
The intersection of CC"' and BAD gives the preconsolidation
pressure (p ) - in this example p
o
0
= 5,450
psf.
Summary
Compression Index, Cc
= 0.17
Preconsolidation pressure, p
o
=
5,450 psf.
Example No.'2.
Using the soil data from Example 1, calculate the
maximum differential settlement of a flexible rectangular foundation
10 ft. x 20 ft. located on the upper surface of a stratum of the clay
15 ft. thick.^The stress on the foundation is 10 kips/sq. ft.
The clay overlies an incompressible stratum. For the purpose of
calculation, divide the clay into three layers, each 5 ft. thick.
Take saturated density of clay throughout as 120 lb/cu. ft. ^(Water,
table at surface).
SOLUTION
10 000psf
20'^
Depth below
ground surface
(Feet)
0
0 9560 psf
—
/MID - PLANE 1
2-5
MID-PLANE 1
2390.p0
5-0
06400psf
/MID-PLANE 2
7.5
MID-PLANE 2
0 3720 psf
10•0 --
25
1600 psf
MID-PLANE 3
15.0--
Incompressible
Fig. 17. INCREASE IN EFFECTIVE PRESSURE AT DEPTH
(iv
)
The increase in effective stress due to the weight of the foundation
at the mid-planes of the three strips is shown in Fig. 17. These
values were obtained by-integrating the Boussinesq equation.
Settlement Calculations
(a) Mid-Plane No. 1.
(i) Initial pressure, P o^= 120 x 2.5
-62.4 x 2.5
144 psf
Final pressure, P 1 (centre) = 9560
144
9704 psf
From Fig. (16), e 0 = 0.717
e
1
= 0.615
whence, S (centre) = h e 0 -e l
1 + e
o
= 5 (0.717 - 0.615)
1 + 0.717
= 0.297 feet.
(ii) Initial pressure, P o = 144 psf
Final pressure, P 1 (corner) = 2390
144
2534 psf
from Fig. (16), e o = 0.717
el
= 0.677
whence, S (corner) = 5 (0.717 - 0.677)
1 + 0.717
= 0.116 feet.
(v)
(b) Mid-Plane No. 2.
(i) P
o
= 120 x 7.5
-62.4 x 7.5
432 psf
P
1
(centre) = 6400^ e = 0.704
o
432
e = 0.636
6832 psf
1
S (centre) = 5 (0.704 - 0.636)
1 +0.704
= 0.200 feet.
(ii) P
1
(corner) = 1600^ e = 0.682
1
432
2032 psf
S (corner) = 5 (0.704 - 0.682)
1 + 0.704
= 0.064 feet.
(c) Mid-Plane No. 3.
(i) P
o
= 120 x12.5^
-62.4x 12.5
e
e
720 psf
P 1 (centre) = 3720
720
4440
S (centre) = 5 (0.698 - 0.660)
1 + 0.698
= 0.112 feet.
o
1
= 0.698
= 0.660
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(v~)
(i1)
PI (corner)
= 930
e 1 = 0.686
720
1650 psf
• S (corner)
= 5 (0.698 - 0.686)
1 + 0.698
= 0.035
feet.
• Total settlement (centre)
= 0.297
0.200
0.112
0.609
Total settlement (corner)
= 0.116
0.064
0.035
0.215
.·.Maximum differential settlement
feet
=
feet
(0.609 - 0.215) feet
=·0.394 feet
= 4.7 inches
(vii)
Example No. 3.
The following dial gauge readings were obtained in
an oedometer test on a sample 0.50 inches thick. The deflections
are in units of 10
Time:
-4
inches.
0
7.5s
15s
1800
1728
1714
4m
8m
16m
Reading:
1581
1555
1540
Time:
240m
480m
Reading:
1517
1514
Reading:
Time:
30s
lm
2m
1692
1660
1620
30m
60m
120m
1532
1528
1522
Find (a) the initial compression (b) the value of C
v
for this particular loading using (i) the root-time plot
(ii) the log-time plot.
On the basis of the root-time plot, how long would it
take a stratum of the same material, 20 feet thick, to reach
90 percent primary consolidation? .Assume that the stratum is free
to drain from the upper surface only.
Solution
The root-time plot is shown in Fig. 18 and the log time
-
•
plot is shown in Fig. 19.
(i) Root-time Plot
^•
(a) From Fig. 18, Initial compression = Si-So
= (1800 - 1760) x 10 -4 inches
= 40 x 10-4 inches
0^0 o^0^o^
o^to^o
o^to to^
rto
0^tr)^
CD^1,- r-^
to^to
rk-^to^o
co
.,,^..
..
o^
0
co
(say3u! 0 _01 x) NOISS3eldIAJOD
r•ro
-. - - - - - - - - - - - - - - - - .'- - '.
"
,.
5 1 =1800
So =1764
1750
<f)
Q)
.s:;
'U
.,
I
c
1700
0
X
S50=
1650
z
t50
-0
1652
= 1·17 min.
(/')
(/')
.11.1
a:
0-
1600
~
0
u
1550
5
90
=1562-4
t 90 =5,4 min.
~
. - --;-- - - - - - - 1 - -
-
10
_ _ --L_ _ _ _ _ _ .. ______... _. _ _ _
100,
~ __ ._._
. .. ..l._. . .
1000
t ( m.i n.) .Iqg scole
FIG 19 - CO M PRE S S I ON - LOG
TIM E
CUR V E
",.,'
,'.'
.,'"
Rpcord 1974/108
, :.
M(G) 435
^
The line SoS' is obtained by drawing a line with absciassae 1.15 times
'those of the siraight portion of the test curve (i.e. that part of
the curve between A and - B).
This construction works because the empirical relationship
between U and T is given by the following continuous function:
. v
:0^U $0.6
a log lo (1-U)tio^: 0.6< U.5 1 (a, b constants, <o)
Now, if we take square roots and rearrange to make U the dependent
variable, we have:
U=
2
/717'
'yrv
^(x)
(U
'Ea)
1 - 10 1 (( Vi7) 2 - b) ^ (xi)(U >
For U = 0.9, say, we have IC = 0.921 from (xi); however, if equation
(x) is extrapolated beyond it's co-domain, we have:
= 0.8 for U = 0.9 when a = 0.933 and b = 0.0851 as
determined experimentally by Taylor (1948).^Clearly then, if the
abscissae of the linear relation are multiplied by 0. 921
0.8
-
1.15, then
the intersection of the straight line so obtained and the line t = 0
(i.e. the compression axis) gives us the point So, corresponding to
U = 0; similarly the intersection of the straight line and the
laboratory curve gives the point S
90 corresponding to U = 0.9.
(b) From Fig. 4, T v = 0.85 for U = 0.9 under conditions of
double drainage (as in the consolidation test).
2
From the relationship C = T ^
V^V
.^V
)
(T2
(0.5Y
We have, C v = 0.85L 2
2
(2.2)
2
= 0.011 in /min.
(c) The second part of the question requires a time
estimate for U
90 under single drainage conditions:
Substituting into t 90 =
we have,
t
2^
min.
90 = (0.92) (20 x 12)
0.011
= 4817455 min.
= 9.2 years.
(ii) Log-Time Plot
(a) From Fig. 19, Initial compression = Si-So
= (1800-1764) x 10 -4 inches
= 36 x 10 -4 inches
The point So is derived from the fact that the theoretical
U-log10Tv curve is initially parabolic (i.e. the curve is of the
form U = -a(log 10Tv ) 2 ).
So if two time intervals, t l , and
t2, are taken such that t 2 = 4t 1
then, by the parabola function, the corresponding compression S 2 =
Algebraically, in our example we seek a number S, such
that:
So = 1728 + S = 1692 + 2S
i.e. S = 36
Whence, So = 1764
Note also in Fig. 19 that the point S 100 is given by the intersection
of the two tangents of the linear parts of the curve. This point
corresponds to the U 100 primary consolidation limit.^For S<1540,
the sample is undergoing secondary compression which is due to
plastic deformation of the soil particles and is not related to the
escape of pore water.
(x)
(b) By the log fitting method, C v = 0.011 in2 /min (which is
identical to that obtained by the root—time procedure). ^In this
method, C
v is usually calculated from the t 50 value.^From Fig. 4,
for U = 0.5 we have T = 0.2 under conditions of double drainage.
v
From C= T
v^v N
t
we have,
C
50
^(0.2) (
05 2
-1.)
2
1n /min.
1.17
= 0.011 in2 /min.
(c) The working is identical to that of part (i).
That is, t^= 9.2 years.
90
Example No. 4
Land is reclaimed in an estuary by placing sand in the
shallow water off-shore. Taking the R.L. of mean water level as
100, the sand is placed to R.L. 105. ^The sand rests on the
estuarine silty clay at R.L. 95, the clay in turn resting on
permeable sandstone at R.L. 75. After the soil has been in place
for a year, an oil storage tank is built on top. The diameter
of this tank is large compared with the depth of sand and clay.
If the tank exerts a pressure of 1,500 lb/sq. ft. on the underlying soil calculate the final settlement of the tank due to
consolidation of the clay.
Assume - (a) that the water-table remains at the previous water
level R.L. 100.
(b) that the bulk density of the sand is 130 lb/cu. ft.
both above and below the water-table.
(c) that the void ratio of the clay at mid-depth
(R.L. 85) is 2.) before any sand is placed, that
consolidation tests on a specimen 0.75 in. thick
give a compression index of 0.5 and that in these
tests 50 percent consolidation is achieved in about
10 minutes.^The specific gravity (GS) of the •clay
particles is 2.70.
SOLUTION
Silty^clay
STAGE 1
STAGE 2 (After i year)
Fig. 20. SITE CONDITIONS FOR EXAMPLE 4
"Stage 1
:
Settlement of Clay After 1 Year Due to Sand Loading.
.
e
o
= 2.0 (assumed constant throughout the whole clay
sequence)
= 2.70
y = 130 pcf
oedometer
test
results
Cc = 0.5
t = 10 min.
50
U =50%
H = 0.75 inches
N = 2
We assume that the clay has been normally consolidated and hence
Cc is constant at 0.5.
From the relationship, Cc = - de
d(log lo p)
E ^Ae^(since Cc is constant)
A(log lo p)
we have, 0.5 = - (e l - 2.0)
log lo p i,p
/ o
Now, P o (at R.L. 85) = 62.4 x 5
= 312
+ (submerged unit weight of clay (y'))x10 = lOy'
= 312 +^by'
To find y', we note that, by definition:
Submerged density = bulk density - density of water
that is,
Y' = Y Y w
(G T e).yw - yw
(Gs
=
hence,
=
+ el
(
2.7
1)62.4
1 + 2.0
-
= 35.4 pcf
whence,
P
o^= 312 + 10.(35.4)
= 666 pcf
and
P
1 = 130 x 5^= 650
130 x 5
(62.4 x 5) = 338
35.4 x 10^= 354
-
1342 psf
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So, upon substitution into (xi), we have, e1 = 1.85.
To find ultimate settlement due to the emplacement of sand, we use
the relation
Substituting our parameters, we have:
S
= 20 (2.0 - 1.85)
1
= 1.0
+ 2.0
ft.
But we require the amount of settlement which will have occurred
after 1 year.
From the consolidation
H
we have: C :. T ( _N)\2
.v .........v_...;...;._
test~
t
= [0.2
X(0'0~25f]
0.0000 19
=
and for t
=1
year, TV
=t
10.28
2
ft /year
Cv
aY
= 1 x 10.28
e~)2
= 0.10
From Fig. 4, U
28
= 0.36.
Therefore, settlement after 1 year of sand loading
= 0.36
ft.
2
ft /year
(xv)
Stage 2 : Settlement Due to Tank
Data:^P
P
o
1
= 666 psf
= (1342 + 1500) = 2842 psf
Ae
From the relation Cc = -
^, we have upon
A(loglop)
substitution of our parameters:
0.5 = - (el ' - 2:0)
(284
62
66)
log i o
whence, el " = 1.69
We now derive the coefficient of compressibility, M y .
Recall that M
V
is defined as:
Mv^--= -Ae . 1
Ap 1+e
In our example, Mv = - (2.0 - 1.69) x 1
(666 - 2842)^3
2
= 0.00 00 482 ft /lb
whence, total settlement, S = Ap.H.My
= (2176 x 20 x 0.0000482) ft.
= 2.098 ft.
But, the clay has already undergone 0.36 ft. of settlement due
to sand loading for 1 year.
Therefore, final settlement, S' = 2.098 -0.36
= 1.738 ft.
COMMONWEALTH OF AUSTRALIA
DEPARTMENT OF NATIONAL DEVELOPMENT
BUREAU OF MINERAL RESOURCES GEOLOGY AND GEOPHYSICS
CNR. CONSTITUTION AVENUE AND ANZAC PARADE. CANBERRA
Postal Address: Box 378, P.O. Canberra City
Telephone: 49 9111^Telegrams: Buromin^Telex: 62109
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