Rate of Foundation Settlement 377

10.14.3 Determination of initial excess pore water pressure values

For one~dimensionalconsolidation problems, Uj can at any point be 'taken asequal to the incrementof the total major principal stress at that point. Fortwo- and three~dimensional problems Uj must be obtained from the formula:

Uj = B[Ll0"3 + A(LlO"I - Ll0"3)]

As the clay is assumed saturated, B = 1.0.

10.15 Sand drains

Sometimes the natural rate of consolidation of a particular soil is too slow,particularly when the layer overlies an impermeable material and, in orderthat the structure may carry out its intended purpose, the rate of consolida­tion must be increased. An example of where this type of problem can occur isan embankment designed to carry road traffic. It is essential that most of thesettlement has taken place before the pavement is constructed if excessivecracking is to be avoided.

From the Model Law of Consolidation it is known that the rate of consoli~dation is proportional to the square of the drainage path length. Obviouslythe consolidation rate is increased if horizontal, as well as vertical, drainagepaths are made available to the pore water. This can be achieved by theinstallation of a system of sand drains, which is essentially a set of verticalboreholes put down through the layer, ideally to a firmer material, and thenbackfilled with porous material, such as a suitably graded sand. The methodwas first used across a marsh in California and is described by porter (1936).

A t¥pical arrangement is shown in Fig. 10.18a. There are occasions whenthe sand drains are m~de to puncture through an impermeable layer whenthere is a pervious layer beneath it. This creates two~way vertical drainage, aswell as lateral, and results in a considerable speeding up of construction.Diameter of drains: vary from 300 to 600 mm. Diameters less than 300 mmare generally difficult to install unless the surrounding soil is considerablyremoulded.

r- __~ Possible overload

'" ',> >/ Permanent fill ~Ianket

, ... ". /," /,.. " .. n: ... po

(a)

Fig, 10.18 Typical sand drain arrangements.

Permeable

(b)

Impermeablelayer

," "

378 Elements ef Soil Mechanics

(a) Square (b) Triangular

Fig. 10.19 Popular arrangements of sand drains.

Spacing of drains: depends upon the type of soil in· which they are placed.Spacings vary between 1.5 and 4.5 m. Sand drains are effective if the spacing,a, is less than the thickness of the consolidating layer, 2H.Arrangement ofgrid: sand drains are laid outin either square (Fig. 1O.19a)or triangular (Fig. IO.l9b) patterns. For triangular arrangements the gridforms a series of equilateral triangles the sides of which are equal to the drainspacing. Barron (1948) maintains that triangular spacing is more economical.In his paper he solved the consolidation theory for sand drains.Depth of sand drains: dictated by subsoil conditions. Sand drains have beeninstalled to depths of up to 45 m.Type of sand used: should be clean and able to carry away water yet notpermit the fine particles of soil to be washed in.Drainage blanket: after drains are installed a blanket ofgravel and sand from0.33 to 1.0 m thick, is spread over the entire area to provide lateral drainage atthe base of the fill.Overfill or surcharge: often used in conjunction with sand dfains. It consists

'"of extra fill material placed above the permanent fill to accelerate consolida-tion. Once piezometer measurements indicate that consolidation has becomeslow this surcharge is removed.Strain effects: although there is lateral drainage, lateral strain effects areassumed to be negligible. Hence the consolidation of a soil layer in whichsand drains are placed is still obtained from the expression:

Pc = mvdp2H

Consolidation theoryThe three-dimensional consolidation equation is:

where Ch = coefficient of consolidation for horizontal drainage (when it can bemeasured: otherwise use cv).

The various co-ordinate directions of the equation are shown in Fig. 10.20.The equatiom can be solved by finite differences.

Rate of Foundation Settlement 379

~/Diaas

0z

00·00 .4CDc~

-

. of drain (2r)

Fig. 10.20 Coordinate directions.

Equivalent radiusThe effect of each sand. drain extends to the end of its equivalent radius, whichdiffers for square and triangular arrangements (see Fig. 10:19).

For a square system:Area of square enclosed by grid = a2

Area of equivalent circle of radiusR = a2

i.e. 1l"R2 = a2 or R=O.564a.

Fora triangular system:A hexagon is formed by bisecting the various grid lines joining adjacentdrains (Fig. 10.21). A typical hexagon is shown in the figure from which it isseen that the base of triangle ABC, i.e. the line AB, = a/2.

Nowaa

~C = AB tan LCBA = 2 tan 30° = 2J3

hence:

. I a a a2

Area of tnangle ABC = 2 x 2 x 2J3 = 8J3

Fig. 10.21 Equivalent radius: triangular system.

380 Elements of Soil Mechanics

So that:, a2

Total area of the hexagon = 12 x 8y'3= 0.865a2

Radius of the equivalent circle, R = 0.525a

Determination of consolid4tion rates from curvesBarron has produced curves which give the relationship between the degree ofconsolidation due to radial flow only, Ur, and the corresponding radial timefactor, Tr •

T _ Ch tr - 4R2

where t = time considered.These curves are reproduced in Fig. 10.22 and it can be seen that they

involve the use of factor n. This factor is simply the ratio of the equivalentradius to the sand drain radius.

Rn = - and should lie between 5 to 100

rTo determine U (for both radial and vertical drainage) for a particular time, t,the procedure becomes:

(i) Determine Uz from the normal consolidation curves of Uz against Tz(Fig. 10.4):

CytTz = H2 where H = vertical drainage path

(ii) Determine Ur from Barron's curves of Ur against Tr.(iii) Determine resultant percentage consolidation, U, from:

1 ~U = 100 - 100 (100 - Uz)(lOO - Ur)

1--- . --......:::r---.. ........ r-- ... k......

........ " ~""~ '" "-r-.

"~r-. '\. ~'voOO

i"'\. ~I r-..~t-';.>'o \ 1\

t'-..... " ~

t--.

o10

20;g~ 30=>~

c: 40o~ 50:2g 60

§ 70()

80

90

1000.004 0.01 0.04 0.10 0.40 1.0

Time factor, T,

Fig. 10.22 Radial consolidation rates (after Barron, 1948).

Rate of Foundation SetUement 331·

Smear effectsThe curves in Fig. 10.22 are for idealised drains, perfectly installed, clean andworking correctly. Wells are often installed by driving cased holes and thenbackfilling as the casing is withdrawn, a procedure that causes distortion andremoulding in the adjacent soil. In varved clays (clays with sandwich typelayers of silt and sand within them) the finer and more impervious layers aredragged down and smear over the more pervious layers to create a zone ofreducedpenneability around the perimeter of the drain. This smeared zonereduces the· rate of consolidation, and in situ measurements to .check on theestimated settlement rate are necessary on all but the smallest of jobs.

Effectiveness of sand drainsSand drains are particularly suitable for soft clays but have little effect on soilswith small primary but large secondary effects, such as peat. See Lake (1963).

EXAMPLE 10.7A soft clay layer, mv = 2.5 x 10-4 m2/kN; Cv = 0.187m2/month, is 9.2m thickand overlies impervious shale. An embankment, to be constructed in sixmonths, will subject the centre of the layer to a pressure increase of 100 kN/m2.It is expected that a roadway will be placed on top of the embankment oneyear after the start of construction and maximum allowable settlement afterthis is to be 25 mm.

Determine a suitable sand drain system to achieve the requirements.

Solution

2.5Pc = mvdp2H = 10000 x 100 x 9.2 x 1000 = 230mm

,J

therefore, minimum settlement that must have occurred by the time theroadway is constructed == 230 - 25 = 205 mm. i.e.

205U = 230 = 90 per cent

Assume that settlement commences at half the construction time for theembankment. Then time to reach U = 90 per cent = 12 - ~ = 9 months.

T = cvt = 0.187 x 9 = 0020z . H2 9.22 .

From Fig. 10.4 Uz = 16 per cent.

Try 450 mm (0.45 m) diameter drains in a triangular pattern.Select n = 10. Then

R/r = 10 and R = 2.25m

hence

2.25a = 0.525 = 4.3m

382 Elements of Soil Mechanics

Select a grid spacing of 3m.

R = 0.525 x 3 = 1.575m

_ 1.575 _ 7n - 0.225-

Cyt 0.187 x 9Tr = 4R2 = 4 X 1.5752 = 0.169 (Note that no value for Ch was

given so Cv must be used)

From Fig. 10.22, U r = 66 per cent.

U = 100 -1~0 (100 - 16)(100 - 66)

= 71.4 per cent, which is not sufficient

Try a=2.25m; R= 1.18m; n= 5.25.

T = 0.187 x 9 = 0 302r 4 X 1.182 '

From graph, Ur = 90 per cent.

Total consolidation percentage = 100 - 1~0 (100 -116)(100 - 90)

= 91.6%

The arrangement is satisfactory.Obviously no sand drain system could be designed as quickly as this. The

object of the example is simply to illustrate the method. The question ofinstallation costs must be considered and several schemes would have to"'beclosely examined before a final arrangement could be decided upon.

Exercises

EXERCISE 10.1A soil sample in an oedometer test experienced 30 per cent primary consolida­tion after 10 minutes. How long would it take the sample to reach 80 per centconsolidation?

Answer 80min

EXERCISE 10.2A 5m thick clay layer has an average Cy value of 5.0 x 10-2mm2/min. If thelayer is subjected to a uniform initial excess pore pressure distribution,determine the time it will take to reach 90 per cent consolidation (i) if drainedon both surfaces and (ii) if drained on its upper surface only.

Answer (i) 200 years, (ii) 800 years

Rate of Foundation Settlement 383

EXERCISE 10.3In a consolidation test the following readings were obtained for a pressureincrement:

Sample thickness (mm)

16.9716.8416.7616.6116.4616.3116.1516.0816.0315.9815.95

Time (min)

oI4149

162536496481

(i) Determine the coefficient of consolidation of the sample.(ii) From the point for U = 90 per cent on the test curve, establish the point

for U = 50 per cent and hence obtain the test value for tso. Check yourvalue from the formula

TsoH2

tso=-­Cv

Answer Cy = 1.28 mm2/min, tso = 10.2min

EXERCISE 10.4A sample in a consolidation test had· a· mean thickness of 18.1 mm duringa pressure increment of 150 to 290 kN/m2 . The sample achieved 50 per centconsolidation in 12.5 min. If the initial and final void ratios for the incrementwere 1.03 and 0.97 respectively, determine a value for the coefficient ofperme­ability of the soil.

Answer k = 2.78 X 10-6 mm/min

EXERCISE 10.5A 2m thick layer of clay, drained at its upper surfac~ only, is subjected to atriangular distribution of initial excess pore water pressure varying from1000 kN/m2 at the upper surface to 0.0 at the base. The Cy value of the clay is1.8 x 10-3 m2/month. By dividing the layer into 4 equal slices, determine,numerically, the degree of consolidation after 4 years.

Note If the total time is split into sev~n increments, r = 0.494.

Answer U = 15 per cent