Sand Drains 4fv6e

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 as equal to the incrementof the total major principal stress at that point. For two- 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 order that 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 is an embankment designed to carry road traffic. It is essential that most of the settlement has taken place before the pavement is constructed if excessive cracking 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. Obviously the consolidation rate is increased if horizontal, as well as vertical, drainage paths are made available to the pore water. This can be achieved by the installation of a system of sand drains, which is essentially a set of vertical boreholes put down through the layer, ideally to a firmer material, and then backfilled with porous material, such as a suitably graded sand. The method was 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 when the sand drains are m~de to puncture through an impermeable layer when there is a pervious layer beneath it. This creates two~way vertical drainage, as well as lateral, and results in a considerable speeding up of construction. Diameter of drains: vary from 300 to 600 mm. Diameters less than 300 mm are generally difficult to install unless the surrounding soil is considerably remoulded.

r- __ ~ Possible overload

'" /

',> Permanent fill

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

...

Impermeable layer

>

~Ianket

po

Permeable (a)

Fig, 10.18 Typical sand drain arrangements.

(b)

,"

"

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 grid forms a series of equilateral triangles the sides of which are equal to the drain spacing. 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 been installed to depths of up to 45 m. Type of sand used: should be clean and able to carry away water yet not permit the fine particles of soil to be washed in. Drainage blanket: after drains are installed a blanket of gravel and sand from 0.33 to 1.0 m thick, is spread over the entire area to provide lateral drainage at the 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 become slow this surcharge is removed. Strain effects: although there is lateral drainage, lateral strain effects are assumed to be negligible. Hence the consolidation of a soil layer in which sand drains are placed is still obtained from the expression: Pc =

mv dp2H

Consolidation theory

The three-dimensional consolidation equation is:

where Ch = coefficient of consolidation for horizontal drainage (when it can be measured: 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

~

as 0

00· 00

CD

c

379

. of drain (2r) z /Dia

.4

~

Fig. 10.20 Coordinate directions.

Equivalent radius The effect of each sand. drain extends to the end of its equivalent radius, which differs 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 ing adjacent drains (Fig. 10.21). A typical hexagon is shown in the figure from which it is seen that the base of triangle ABC, i.e. the line AB, = a/2. Now aa ~C = AB tan LCBA = 2 tan 30° = 2J3

hence: .

Area of tnangle ABC =

Fig. 10.21

I a a 2 x 2 x 2J3 =

Equivalent radius: triangular system.

a2 8J3

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 curves Barron has produced curves which give the relationship between the degree of consolidation due to radial flow only, Ur, and the corresponding radial time factor, T r • Ch t

T _

4R2

r -

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 equivalent radius to the sand drain radius.

R

n= -

and should lie between 5 to 100

r

To determine U (for both radial and vertical drainage) for a particular time, t, the procedure becomes: (i) Determine U z from the normal consolidation curves of U z against Tz (Fig. 10.4): Cyt

Tz = H2

where H = vertical drainage path

(ii) Determine U r from Barron's curves of U r against T r. (iii) Determine resultant percentage consolidation, U, from: 1

~

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

o 10 1---

;g

20

~

30

c:

40

~

50

g

60

§

70

=>~

o

:2

.

--......:::

r---.. ........ r-- ... ...... ........

~

k

~""

" r-. r-.

'"~ "

~'voOO

'\.

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

()

80

t'-.....

90 100 0.004

"-

0.01

0.04

0.10

Time factor, T,

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

\

" 0.40

r-.. 1\ ~

t--. 1.0

Rate of Foundation SetUement

331·

Smear effects The curves in Fig. 10.22 are for idealised drains, perfectly installed, clean and working correctly. Wells are often installed by driving cased holes and then backfilling as the casing is withdrawn, a procedure that causes distortion and remoulding in the adjacent soil. In varved clays (clays with sandwich type layers of silt and sand within them) the finer and more impervious layers are dragged down and smear over the more pervious layers to create a zone of reducedpenneability around the perimeter of the drain. This smeared zone reduces the· rate of consolidation, and in situ measurements to .check on the estimated settlement rate are necessary on all but the smallest of jobs. Effectiveness of sand drains Sand drains are particularly suitable for soft clays but have little effect on soils with small primary but large secondary effects, such as peat. See Lake (1963). EXAMPLE 10.7

A soft clay layer, mv = 2.5 x 10-4 m2/kN; Cv = 0.187m2/month, is 9.2m thick and overlies impervious shale. An embankment, to be constructed in six months, will subject the centre of the layer to a pressure increase of 100 kN/m 2. It is expected that a roadway will be placed on top of the embankment one year after the start of construction and maximum allowable settlement after this is to be 25 mm. Determine a suitable sand drain system to achieve the requirements. Solution

2.5

Pc = mvdp2H = 10000 x 100 x 9.2 x 1000 = 230mm ,J

therefore, minimum settlement that must have occurred by the time the roadway is constructed == 230 - 25 = 205 mm. i.e. 205 U = 230 = 90 per cent Assume that settlement commences at half the construction time for the embankment. Then time to reach U = 90 per cent = 12 - ~ = 9 months. T = cvt = 0.187 x 9 = 0020 z . H2 9.22 . From Fig. 10.4 U z = 16 per cent. Try 450 mm (0.45 m) diameter drains in a triangular pattern. Select n = 10. Then R/r

= 10

and

R

hence 2.25 a = 0.525 = 4.3m

= 2.25m

382

Elements of Soil Mechanics

Select a grid spacing of 3 m. R = 0.525 x 3 = 1.575m _ 1.575 _ 7 n - 0.225Cyt 0.187 x 9 T r = 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 = 0 302 ' r 4 X 1.182 From graph, U r = 90 per cent. Total consolidation percentage = 100 -

1~0 (100

16)(100 - 90)

-1

= 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 of installation costs must be considered and several schemes would have to"'be closely examined before a final arrangement could be decided upon.

Exercises EXERCISE 10.1

A 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 cent consolidation? Answer

80min

EXERCISE 10.2

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

(i) 200 years, (ii) 800 years

Rate of Foundation Settlement

383

EXERCISE 10.3

In a consolidation test the following readings were obtained for a pressure increment: Sample thickness (mm)

Time (min)

o

16.97 16.84 16.76 16.61 16.46 16.31 16.15 16.08 16.03 15.98 15.95

(i)

(ii)

I

4 1

4 9 16

25 36 49 64 81

Determine the coefficient of consolidation of the sample. 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 your value from the formula TsoH2 tso=-Cv

Answer

Cy

= 1.28 mm2 /min, tso = 10.2 min

EXERCISE 10.4

A sample in a consolidation test had· a· mean thickness of 18.1 mm during a pressure increment of 150 to 290 kN/m2 . The sample achieved 50 per cent consolidation in 12.5 min. If the initial and final void ratios for the increment were 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 10.5

A 2m thick layer of clay, drained at its upper surfac~ only, is subjected to a triangular distribution of initial excess pore water pressure varying from 1000 kN/m 2 at the upper surface to 0.0 at the base. The Cy value of the clay is 1.8 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 Answer

U = 15 per cent

sev~n

increments, r = 0.494.