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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