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I I I I I I I I I I I
I
DEPARTMENT OF. MINE·RAlS AND ENERGY
BUREAU OF M~NERAlRESOUrRCESv GEOLOGY AND GEOPHYS~CS
BMR Record
1974/108 c.3
504949
Record 1974/108
TERZAGHI'S THEORY OF ONE DIMENSIONAL··
PRIMARy CONSOLIDATION OF SOILS AND ITS APPLlCATION
by
J.R. Kellett
~llUGrZ; >
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.
,I I I I I I I I I I I I I I I I I I I I
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^ 7
SETTLEMENT^ 10
THE CONSOLIDATION TEST^ 11
PRACTICAL APPLICATIONS OF THE THEORY OF CONSOLIDATION^13
Settlement Calculations^ 14
^
Example 1^ 14
^
2^ 15
11^3^ 16
Digression - Boussinesq Analysis^ 19
Discussion of basic assumptions^ 20
Time Calculations^ 21
CONCLUSION^ 23
REFERENCES^ 24
TABLE 1. Settlement analysis, Isabella Plains.
APPENDIX 1. Worked solutions of problems.
I I I I I I I I I I I I I I I I I I I I
FIGURES
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
Total head and hydraulic gradient.
Inflow and outflow through a small element of soil.
Consolidation model.
Relation between time factor and degree of consolidation.
Settlement of column of soil.
Consolidation apparatus.
Consolidation curve.
Void ratio-effective pressure curve.
Consolidation test on peaty clay.
Soil sequence at Isabella Plains.
Model for example 1.
Model for example 3.
Contours of equal vertical stress in example 3.
Boundary conditions at Isabella Plains.
Time-settlement curves for different boundary conditions.
Void ratio-effective pressure curve of an overconsolidated soil.
Increase in effective pressure at depth.
Compression-root time curve.
Compression-log time curve.
Site conditions for example 4. I
I I I I I I I I I I I I I I I I I I I I
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 - - dhdl
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 = vz + av
z.dz
3z
Cross-Sectional
area = 1
I Inflow = vz
Fig. 2. INFLOW AND OUTFLOW THROUGH A SMALL ELEMENT OF SOIL
e#.
Let vz = inflow velocity,
then total inflow = vz
(since cross-sectional area = 1), and
outflow depends upon the change in velocity, v z , of the water as
it flows through the soil.
The magnitude of the change in flow is given by Dv
z .dz
azTherefore, Outflow = [ Inflow ] + [(rate of change of velocity of
water in z direction) x (distance through which
it travels)]
That is, Outflow = vz + ay
z.dz
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
Vz + 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 = vzdxdy
Volume of water leaving the element in unit time = (vz + 3v
zdz)dxdy
3 zHence, rate of volume change = net decrease in volume of water.
i.e.av (v + 3vzdz)dxdy - vzdxdyz
at^3z
avz dxdydz
k3hFrom Darcy's Law, vz = ki = —az
Whence, rate of volume change 3V = a kah dxdydz3z 3z
3Vthat is, (assuming constant k)^= k3
2h dxdydz
at 3 2
Now,Now, volume of solids in the element, V s = dxdydz 1+e
^(i)
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-Vv
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.
Vi.e. Porosity, n = vV
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 Vs is constant.at
^
that is, 3V .(dxdydz).3e ^ (ii)Dt^1 + e at
Now, rate of reduction of voids = net rate of flow of water from
the element,
that is,
^
(dxdydz )3e^ka 2h dxdydz
^
1 + e /at^a 2
^
/ 1 \ae^k3 2h ^.
^
ie. L -i_re-5—t-^az2
But from (1), h = z + uYw
and Dh . 13u yw
whence, ah = 3uYw
(equating (i) and (ii))
Substituting into (iii), we have:
(
' 1^ e^3 2u ^ (iv)1+e )at = yw a z 2
Now, the drop in pore-water pressure (-du) = increase in effective
pressure (dp);
6.
and from the identity, MV = - de ^1 *, we havedp (lie)
de = Mvdu (1+e)^
(since dp = -du)
and substituting in to (iv),
1^Mv(1+e) all k 3 2u
1+e^at y^2w z
that^is, au ^k^3 2u^ (v)
^
at^Yw v^3 z2 :) ^
Equation (v) is Terzaghi's differential equation for one-dimensional
consolidation.
The term C = k is denoted as the "COEFFICIENT OFV -YwM v
CONSOLIDATION" ^(It should be noted that the coefficient of perme-
ability can be deduced from consolidation test results).
Equation (v) is usually written as:
Du C a 2u—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 . 1dp H
-2 -de . 1^(assuming constantdp (1+e)^cross-sectional area)
i.e.^IMv1= de . ^1 dp^(1+e)
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" (Tv )
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 Tv 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, yw
change in void ratio, De
change in pressure, Dp
8.
t(l+e)k Terzaghi deduced that T -
v^2y 3e
ITI )
and from the identities:
^(vi)
^ and Me. 1Cv. -^v = —
y M 3p 1+eWV
we have, Tv - 2̂(H) ,y 3e.N^3p 1+e
1
t ^.^k
(11-111^Y MWV
i.e.— 2
( .111)
tk(dividing top and bottomby (1+e))
from which 2
=vCv
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
3u^C 3 2udifferential equation^= v
at^3z 2
0 1.41-2P.O
9,
.20
40
00P
60
:' so
1 00
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 = p2
Then, the consolidating pressure = p 2-pi
and total settlement, s = h-h 2 .
Now, height of solids =I' = volume of solids (since cross-sectional
area = 1).
Initial height of voids = 11 = initial volume of voids.
Final height of voids =^= final volume of voids.
Also, h = +
and h2 = +12
If el = initial void ratio, then:
h-/ he1 = —^= — - 1
and if^
= final void ratio, then:
/2
h2 -/ h
2e =^=^=^-12 —/^/
whence, e1-e2 - h - h
2
i.e., Total settlement, s = h.[ 1 + e
1 ]
and, for a suitably small Ae,
S = h( Ae1+e
Ap 1+e). Ap. Ae . 1
E el - e23
i.e.^ s = h.A p. N'
11.
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
1/8 ton/ft
2 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 = -dev —
dp
That is, av is the gradient of the curve in Fig. 8.
Pressure (P)
8, -VOID:RATIO ..'''EFFECTIVE PRESSURE CURVE
-de Note. Recall the coefficient of compressibility, My - dp(1
ewhich is the compressibility( --d--)divided by the total volumedp
•
(1+e).
I I I I I I I I I I I 0
~ a::
I 0 0 ->
I I. I I I I I I
.84
.8 I
.78
.75
.72
.69
.66
.63
.60
.57
PRESSURE Vs. VOID RATIO
PROJECT: Isabella Plains Stormwater System - Site Investigation
.. ; BORE HOLCTP 59
DEPTH ... .1.10": 1.15m
CLASSIFICATION .. CL-CH CLAY: medium plastici~y, dqrk grey, some fine sand, some lenses of light grey,
non-plastic silt, some decaying fine rO<?fs and root fibres, moist (Me )PL) soft.
Inttiol Wet Density ~ 115 Ib/tt3.
Initial M.C ... 30.60%
Final M.C .... 23.73 % .
Range (psf) 200-500
I Mv. (ft7' I b) 1S.8 x 10-6
--Cv. (ft2;yeor) 2.5
I i I ! I I I
1 !
i i !
I i ,
! --t--:1 ! I
t-.
I •
I ,
I
I
I I
I ~--
{
Ht.= _.0.751
Sample Size
.' Dio.= ... 1.766
500-1000 1000- 2000 I 2000-4000
17.2 x10- 6 17.4 K 10-6 14.5 K 10-6
3.7 4.4 6.6 ,
rt , I-..." 1':--...,.. '
~ i ~! . !
I\.
'\ :
!
- I-- - -:-- r-- _ -!
Dote _ 16.8.73
Specific GrOllity _ .2.57
4000-8000 8000-16000
8.9 K 10-6 4.9 K 10-6
6.2 6.3
:
1 I I I I I !
I i i I I I I I I ,
~
1\ ~ ,
1\ 1\
i\ I
\ --\ - -- I-
.54 100 2 3 4 6 7 89
i 2 3456789
250 500 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
Depth(Feet)
0
.^•Water table5
30
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.
Peaty clays with sand lenses
70
Massive blue-grey sandy clay
(g2)10 5
• °. .° e°° °Q°° ^0
.0°0 0^°^06° 0^
if 00 •^000 0000o00• 00 •
0000 0 0 0:^Po°^0^:t^0: 0 0 Oa • • 0
0^°o0.000,,O*1°1; °:*g.' °
0 e 0* * e* 0 0 0l
0 0:: 00 ...70:0: 10°0:0e a 0 "0 • Q e O• .00 - 0 e^0
aqu i fer grading to o °.to^.0. 0 0 0 o 0 .0
c,_0 0 ^: ,','sandy aqui^ :00 0° 0 1:i:0 0: 00 7,:o^gra.vel and cabbies':00:o 0 0 000000 0 "„ 0000
^
00:00 : 00e: 0000 „ 0 00 0
::::: ei 000 :: 0°:^°c o:0000 : 0 0°000 0 „^, 0%0: 0
18-0^co.;;.^oe. o eoo^° e t': 6 0." 7°70:
(to. 00 0:0 °40. ° 00 0.0:0 000:000• 00 .:0:0000
A < -1^>^< Weathered tuff
(9 1 )
12 0
Fig. 10. SOIL SEQUENCE AT ISABELLA PLAINS
Present positionof water table
Density (moist) :102 pcf
00 0 0 0 0 0 0 0 0 0 0 0 0 0 0I 2 0 -
Proposed newwater table
Density (submerged):115pcf
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), (g2 ) and (g3) are grouped
together as layer (g). • Our amended model is shown in Fig. 11.
Aquifer (assumed incompressible)
Fig. 11. MODEL FOR EXAMPLE 1 (not to scale)
I I
15.
I Calculation of effective stress at base of clay layer:
I Before Lowering Water Table After Lowering Water Table
~il 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.
I Water pressure
= (1.5 x 125)+(9 x 115) 1222.5 psf
= (-10.5 x 62.4) = -655.2 psf
Soil below W.T. = (3 x 115) = 345 psf
Water pressure = (-3 x 62.4) =-187.2 psf
I Effective stress 717.3 psf Effective stress 1069.8 psf
I I I I I I I I I I I I I
Whence, the increase in effective stress due, to ,
lowering of water table = 352.5 psf.
From Fig. 9, the coefficient of volume compressibility
-6 2 within these stress ranges is MY = 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 Mv
= [(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.
••..•.
eriiityAmoist):=:100.p.c
•^•...• .^•.• • •'."•.•.•• .•••
••^ .
ensity (submerged) 125 pet
Density (submerged) :115 pcf
.^17.
W.T.
t^t^tAssumed incompressible
Density (moist) = 102 pcf
W T.
Density (sumerged)= 115 pet
0 0 OOOOOOO 0 0 0 0 0
t^t^t
Depth(Feet)
0-.
1-5 - It )
3-0 -
(q )
9-0 -
12-0 -
INITIAL CONDITIONS^ FINAL CONDITIONS
Fig. 12. MODEL FOR EXAMPLE 3.
The best procedure is to divide the clay layer into 9 x 1 foot-
thick 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.
I I I Table 1 SETTLEMENT ANALYSIS
I 1 Z 3 4 , 6 7 8 '9 10
La)'u h Depth to Initial preuure Pressure after Pre.,ura . t.o.d 6p MY h 6p My (feet) mid-plane lowcrina vater reductlotl p\'II •• ure ·PZ+P,+P4-Pl (xlO-6 (xlO-6 ft)
I (feet) table due to at mld- ft Z/ll1) excavatlotl plene depth
PI Pz ,., '. I lOOx1.5 100x1.5
125x1.5 102xO.5 115xO.5 100xl.5 "62.4x2
I 1 1 ,., (270.2) (351.0) -300 3600 3380.8 14.' 49021.6
lOOxl.' -62.4x1.5 100x1.'
I 125xl.5 -62.4x1.5 10Od.S ,115x1.S 102x1. S
2 1 4.5 (322.8) (453.0) -300 ZOOO 1830.2 ; '17.3 3184'.48
I 100x1.5 -62.4x1.' 100xl.' 125xl.S -62.4x2.5 100x1. , 115x2.S 102x2.5
3 1 ,., (375.4) (555.0) -300 1000 879.6 17.2 15129.12
I 100x1.5 -62.4xl.5 100x1.5 125x1.5 -62.4x3.5 100x1.' ll'x: ., 102x3.5
I 4 1 .. , (428.0) (657 .0) ·300 600 529.0 17.2 9098.8
100x1.5 -62.4x1.' 10Ox1.' 125xl.5 -62.4x4.5 100x1.5
I 11',,4.5 102x4.5
, 1 7.' (480.6) (759.0) -300 3GO 338.4 11.2 5820.48
100xl.' -62.4xl.5 100x1. 5
I 125x1.5 -62.4x5.5 100,,1.' 115,,5.5 102,,4.5
6 1 8.5 (533.2) (861.0) -300 Z40 261.8 17.2 4606.16
I 100x1.5 -62.4x1.5 100,,1. 5 125x1.5 -62.4,,6;5' 100,,1.5 ll'x6.5 102,,6
115>=0.5
I -62,1, xO. 5
7 1 9.' (585.8) (938.3) .;.300 200 252.' 17.2 4343.0
100x1. 5 -62.41.:9 100x1.5
I 125x1. 5_ 100x1. 5 115,,7.5 102x6
115,,1.5 -62.4xl.5
I 8 1 10.5 (638.4) (990.9) -300 145 191.' 17.2 3397.0
100,,1.5 -62.4xlO 100x1.5 l25x1.5 100,,1.5
I 115,,8 • .5 102,,6
115x2.5. -62.4x1.5
9 1 11.' (691. 0) (1043.') -300 120 172.' 17.Z 2967.0
I ~hApMy • 126,228.64 x 10-6 feet.
I • 1.5 inches.
:1 I I I. I I I I I I I I I I I I I I '1 I
Depth below . i X 2' square ground surface (feet)
~------------------------------~O
r -----I , ,
(°1
footing .
Q= 16000lbs
q~ 4000psf ----------------~~2
l II 21\ j 2ft 4ft ." T---------~------=r---------,----------''r-----------l3
-~rHY-~~--------------------------~_4
, ----f----\., I
~----+---------46
I 1----- ---- -. - 7 .
8
----19
I ------"~ 160..-=----------+--------------10 '
. I ----------'----'------'---~____/_----'-------____c .. --'-----------III
~8:--·- .. -i i
---+------+-----------l12
i i
Fig. 13 CONTOURS OF EQUAL VERTICAL STRESS IN EXAMPLE 3
i
Record 1974/108 M(G) 433
I I I I I I I I I I I I I I I I I I I
18.
To find our Cv value, we note that the pressure range in which
we are interested is 200 psf to 4,000 psf.
average C value from Fig. 9 is: v
Hence, the weighted
"
C, = (200 x 2.5) + (500 x 3.7) + C' 1000 x 4,.4 \,+ (2000 x v 4000 4000 ' 4000 ) 4000
2 = 4.99 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
and t ...
, 2
(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
Drain
Clays o
Deplh (fll rO ~30 I
I I
____ E_q'-u_i_1 i_b_r_iu_m __ W, T. ~. 9.0
I r 10,5
! ' ~ 12,0
, I ' 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:
3Aa =5/ 2z
2 • ^2 Tr [ 1 1.(1)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.
I I I I I I I I I I I I I I I I . 1. I I I
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.
(U) ISOTROPY:
(iii) ELASTICITY:
(iv) INFINITE THICKNESS:
.. Even the individual strata are complex
soils with lenses and wedges.
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.
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 •
we have assumed a rigid boundary at 12 feet
below ground surface. Therefore, the
stress concentration near the boundary is
increased.
I I I I I I I I I I I I I I I I I I I I
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 = time factor (real number)
H = thickness of compressible stratum
C = coefficient of consolidation v
N = number of drainage faces (since we are considering
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
=(34.02 ) Years (ix)Cv
*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 .51t = (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:
f = (0.6) x2(4 )
2
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
0.40
03
c-,
80LU
510025
01^05TIME (Years)
Fig. 15 TIME- SETTLEMENT CURVES FOR DIFFERENT BOUNDARY CONDITIONS
Record 1974/108
3 4
.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.
I I I I I I I I I I I I I I I I I I I I
Example No~ 1.
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 e
3 4
1 1-2
0.705 0.698 0.688
3 6 12 48
0.673 0.645 0.600 0.550 0.500
Calculate the c~mpreasion 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-1og10
p curve.
Analytically,
Cc = __ -~d~e __ ~_ d(10g10P)
For a "normally consolidated" soil, ther~ is a linear relationship
between e and log10P 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
I I I I I I I I I I I I I I I I I I I I
(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 = eo03 375 ~p ~ 1,500 variable,> 0 1,500 <p < 8,000 0.17 8,000 ~p ~48,000
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 = 5,450 psf. o 0
Summary
Compression Index, Cc = 0.17
Preconsolidation pressure, p = 5,450 psf. o
20' ^
0 9560 psf/MID - PLANE 12390.p0
06400psf
/MID-PLANE 2 1600 psf
0 3720 psf
Depth belowground surface
(Feet)
0 —
2-5
5-0
7.5
10•0 --
25
15.0--
MID-PLANE 1
MID-PLANE 2
MID-PLANE 3
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
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, P1 (centre) = 9560144
9704 psf
From Fig. (16), e0 = 0.717
e1
= 0.615
whence, S (centre) = h e 0-el
1 + eo
= 5 (0.717 - 0.615) 1 + 0.717
= 0.297 feet.
(ii) Initial pressure, Po = 144 psf
Final pressure, P 1 (corner) = 2390144
2534 psf
from Fig. (16), eo = 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) Po
= 120 x 7.5-62.4 x 7.5
432 psf
P1 (centre) = 6400^ e
o = 0.704
432
6832 psf e1 = 0.636
S (centre) = 5 (0.704 - 0.636) 1 +0.704
= 0.200 feet.
(ii) P1 (corner) = 1600^ e
1 = 0.682
432
2032 psf
S (corner) = 5 (0.704 - 0.682) 1 + 0.704
= 0.064 feet.
(c) Mid-Plane No. 3.
(i) Po = 120 x12.5^ eo = 0.698-62.4x 12.5
e1 = 0.660720 psf
P1 (centre) = 3720720
4440
S (centre) = 5 (0.698 - 0.660) 1 + 0.698
= 0.112 feet.
I I I I I I I I I I I I I I I I I I I I
(v~)
(i1) PI (corner) = 930 720
1650 psf
• S (corner) = 5 (0.698 - 0.686) 1 + 0.698
= 0.035 feet.
• Total settlement (centre)
Total settlement (corner)
.·.Maximum differential settlement
e1 = 0.686
= 0.297 0.200 0.112
0.609 feet
= 0.116 0.064 0.035
0.215 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 10-4 inches.
Time: 0 7.5s 15s 30s lm 2m
Reading: 1800 1728 1714 1692 1660 1620
Time: 4m 8m 16m 30m 60m 120m
Reading: 1581 1555 1540 1532 1528 1522
Time: 240m 480m
Reading: 1517 1514
Find (a) the initial compression (b) the value of Cv
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^
tr-o^
to to^
o^oto
0^
tr)^o
CD^1,- r-^
to^to
rk-^to^
co.,,^
..o^..0
co(say3u! 0_01 x
) NOISS3eldIAJOD
r•-ro
-. - - - - - - - - - - - - - - - - .'- - -
5 1 =1800
So =1764
1750
<f)
Q)
.s:; 'U 1700 c
., I
0
X 1650
z 0 -(/')
(/')
.11.1 a:
1600 0-~ 0 u
1550
Rpcord 1974/108
" ,.
S50= 1652
t50 = 1·17 min.
5 90 =1562-4
t 90 =5,4 min.
.---;-- ------1--
10
t ( m.i n.) .Iqg scole
FIG 19 - CO M PRE S S I ON - LOG TIM E CUR V E
'.
~ - __ --L ______ .. ______ ... _. ___ ~ __ ._._ . .. ..l._ ...
100, 1000
",.,' ,'.'
.,'"
, :.
M(G) 435
:0^U $0.6
a loglo (1-U)tio^: 0.6< U.5 1 (a, b constants, <o)
U=
variable, we have:
2 'yr-/717' v ^(x) (U 'Ea)
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
Now, if we take square roots and rearrange to make U the dependent
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
921abscissae of the linear relation are multiplied by 0.- 1.15, then0.8
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 S90 corresponding to U = 0.9.
(b) From Fig. 4, Tv = 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, Cv = 0.85L 2
(2.2) 2
= 0.011 in2/min.
(c) The second part of the question requires a time
estimate for U90 under single drainage conditions:
Substituting into t 90 =
we have,
t90 = (0.92) (20 x 12)
2^min.
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(log10Tv ) 2 ).
So if two time intervals, t l , and t2, are taken such that t2 = 4t1
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 U100 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, Cv = 0.011 in2/min (which is
identical to that obtained by the root—time procedure).^In this
method, Cv is usually calculated from the t50 value.^From Fig. 4,
for U = 0.5 we have Tv = 0.2 under conditions of double drainage.
From C= Tv^v N
t50
we have, C ^(0.2) (05-1.)2 1n
2/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 under-
lying 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.
Silty^clay
STAGE 2 (After i year)STAGE 1
SOLUTION
Fig. 20. SITE CONDITIONS FOR EXAMPLE 4
"Stage 1 : Settlement of Clay After 1 .Year Due to Sand Loading.
eo = 2.0 (assumed constant throughout the whole clay
sequence)
oedo-metertestresults
= 2.70
y = 130 pcf
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(loglop)
E ^Ae^(since Cc is constant)A(loglop)
we have, 0.5 = - (e l - 2.0)
loglopi,p
/ o
Now, Po (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
hence,
whence,
that is, Y' = Y Yw
(GT e).yw - yw
= (Gs
+ el
= (2.7 - 1)62.41 + 2.0
= 35.4 pcf
Po^= 312 + 10.(35.4)
= 666 pcf
P1 = 130 x 5^= 650
130 x 5 - (62.4 x 5) = 33835.4 x 10^= 354
1342 psf
and
I I I I I I I I I I I I I I I I I I I I
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 + 2.0
= 1.0 ft.
But we require the amount of settlement which will have occurred
after 1 year.
( _NH\2 From the consolidation test~ we have: C :. T ) .v ......... v_...;...;._ t
= [0.2 X(0'0~25f] 0.0000 19
2 ft /year
= 10.28 2 ft /year
and for t = 1 year, T = t C V v
aY = 1 x 10.28
e~)2
= 0.10 28
From Fig. 4, U = 0.36.
Therefore, settlement after 1 year of sand loading = 0.36 ft.
(xv)
Stage 2 : Settlement Due to Tank
Data:^Po
= 666 psf
P1 = (1342 + 1500) = 2842 psf
From the relation Cc = - Ae ^, we have uponA(loglop)
substitution of our parameters:
0.5 = - (el ' - 2:0)
logio(2846 266)
whence, el " = 1.69
We now derive the coefficient of compressibility, M y .
Recall that MV
is defined as:
M = -Ae . 1v^---Ap 1+e
In our example, Mv = - (2.0 - 1.69) x 1(666 - 2842)^3
= 0.00 00 482 ft 2/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 GEOPHYSICSCNR. CONSTITUTION AVENUE AND ANZAC PARADE. CANBERRA
Postal Address: Box 378, P.O. Canberra CityTelephone: 49 9111^Telegrams: Buromin^Telex: 62109
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