m06 consolidation

CVNG 2002 Soil Mechanics M06 Consolidation Page 1 of 67

\\Civilserver02\temp\course notes 2005\CVNG 2002 (SoilMechanics)\CE22B RDEAN 2005\M06 Consolidation.doc

THE UNIVERSITY OF THE WEST INDIES

ST AUGUSTINE, TRINIDAD AND TOBAGO, WEST INDIES

FACULTY OF ENGINEERING

DEPARTMENT OF CIVIL AND ENVIRONMENTAL ENGINEERING

CVNG 2002 Soil Mechanics

Module 6. Consolidation Lectures by Richard Dean, Semester 2, 2006

Dump truck unloading fill material as part of a soil consolidation contact, Wilson Bridge, Maryland

(www.wilsonbridge.com/photos )

Oedometers for laboratory testing, UWI



Contents What is consolidation? ..................................................................................................................................3 The Oedometer Test .....................................................................................................................................8 Calculating the Compression Curve ...........................................................................................................10 Typical results .............................................................................................................................................12 Using the results to calculate long-term settlements ..................................................................................13 Using the results to calculate swelling ........................................................................................................14 Using the results to interpret geological history ..........................................................................................15 Simplifying the calculations (1) Index formulation.......................................................................................17 Simplifying the calculations (2) Compressibility formulation .......................................................................20 Casagrande's Construction.........................................................................................................................22 Interpreting the data of time-dependency ...................................................................................................23 Applying the square-law to the oedometer .................................................................................................28 Using oedometer data to estimate settlement-time responses...................................................................30 Terzaghi's Consolidation Equation Derivation .........................................................................................32 Exploring Terzaghi's Consolidation Equation .............................................................................................35 Solving Terzaghi's equation ........................................................................................................................36 Interpreting time-dependent oedometer data..............................................................................................44 Using consolidation theory to manage long-term settlements .................................................................... 47 Secondary consolidation.............................................................................................................................49 Consolidation Lab .......................................................................................................................................50 Review Questions .......................................................................................................................................51 Previous Exam Questions...........................................................................................................................55 Appendix A. Revision of Permeability and Darcy's law...............................................................................59 References and further reading ..................................................................................................................67



What is consolidation? If a load is applied to a foundation that rests on clay or silt soil, there is usually an immediate

settlement, followed by settlements that increase as time passes. This increase in settlement is usually due to consolidation of the silt or clay. The rate of increase usually reduces as time passes. Eventually the settlement rate is so small that the settlement has stopped for all intents and purposes.

For example, suppose a house is built on sand overlying clay overlying bedrock.

The house will take some time to build, and there will be some settlements during that time. There will also normally be some continuing settlements after the house-building has finished, as shown below.

Example of a settlement-time graph for a heavily loaded foundation with large long-term settlements

Time, years

Settlement, mm

0

50

100

150

1 2 3 0

Primary consolidation

Sand Soft clay

Rock

House being built



The increase of settlements with time occurs because the stresses in the clay due to the weight of the house tend to force water out of the soil, so that the soil layer compacts. The sketch below shows the results of this in an exaggerated form. The immediate settlement is usually estimated using elastic theory. The eventual, long-term settlement is estimated using compression theory, and the final settlement depends on the overall compressibility of the clay. The settlement takes place over some time because it takes time for the water to flow out of the soil. This is affected by the permeability of the clay. Long-term settlements can be quite large. For example, for a foundation on soft to firm clay, long-term settlements can easily exceed one foot. The sketch below shows the values of ultimate settlements, in mm, for square and strip footings on soft to firm clays. Typical relation between footing width and ultimate settlement. Footings loaded to q=50kPa, 1.5m below ground level on Mexico City Clay of 40% water content. Figure from page 407 of Terzaghi et al (1996)

Footing width, metres0 2 4 6

0

100

200

300 Long-term settlement,

mm

Strip footings

Square footings

Water movements



The settlement depends on the size of the footing. Also, the rate of settlement can depend on size, but the distance involved is the drainage path length. In the sketch below, the thickness of the clay layer on the right is twice as great as on the left. Because of the size effect, the maximum settlement is a little under twice as much for the house on the right. The drainage path length on the right is twice as long on the right as on the left. So the pressure gradients pushing the water are half as much on the right, and the distance to move is twice, so it takes 4 times as long to achieve maximum settlements. This square-law of drainage path lengths is quite general the time needed to achieve maximum settlement depends on the square of the drainage path length.

Typical relationship between consolidation time and drainage path length for a silty clay

(Note: time depends on coefficient of consolidation, which can be very different for different clays)

2m 4m

1 day

10 days

3 months

3 years

30 years

1foot 3m 30m

Time to achieve 90% of long-term

settlement

Drainage path length



As a result of the square-law, some large structures will still be settling at the end of their design lives. For instance, the design life of a large offshore concrete oil platform may typically be 20 to 30 years. But the time required to achieve 90% its ultimate settlement might be as long as 100 years.

One of the effects of consolidation can be to cause negative skin friction to develop on piles. In

the sketch below, a building has been founded on piles that transmit the weight of the building to the underlying stronger soil. But the general area has also been covered over with sand. The weight of the sand causes time-dependent compression of the soft clay. As the clay settles, it drags downwards on the piles, causing the "negative" skin friction.

Structures founded on overconsolidated clays are generally found to behave somewhat differently. If we compare the long-term settlements of several structures on stiff clays, the relation between foundation load and long-term settlement is generally non-linear.

Typical relation between load and long-term settlement for some clays

Building supported by piles

Sand

Clay

Sand

Magnitude of footing load

Magnitude of long-term

settlement



The difference between the behaviours of soft and stiff clays can be explained by the theory of compression for silts and clays.

There is also the opposite effect. If an excavation is made for the basement of a building, then one of the effects of the excavation is to reduce the stresses on the underlying soil. As a consequence, the soil can suck water in, and expand. The expansion is called swelling, and the upwards movement of the base of the excavation is called heave. In an extreme case, the base of the excavation will move significantly upwards. A similar effect can occur if it rains, or if there is a water pipe burst in the excavation. Swelling also occurs in expansive clays, except that the effect is primarily due to wetting rather than unloading per se.

One of the good things is that consolidation is generally associated with an increase in strength of the soil. This increase in strength is due to the effect of compaction. Conversely, there is a small decrease in strength if swelling occurs.

In summary, consolidation is the time-dependent settlement of silts or clays. It occurs whenever loads are applied or removed, and is due to the fact that the changes of stress due to the loads cause water to be pushed out or sucked in to the soil. This pushing out or sucking in takes time, due to Darcy's law. In general, a designer will need to estimate both the long-term settlements (or heave) and the time-evolution of those settlements. Consolidation theory is used for this purpose, consisting of the theory of compression and the theory of permeability . The main laboratory test to use to get parameters for the theory is the oedometer test.

Original base of excavation

Base after some timeSaturated sand: a

source of water



The Oedometer Test The oedometer is a laboratory device which is designed to reproduce the conditions in a layer of soil that is subjected to uniform vertical loading. These are the simplest conditions that an engineer will need to analyze. For other conditions, the engineer usually makes estimates based on the simpler conditions, then adjusts the estimate to take account of the increased complexity.

Simple field loading scenario ion which the oedometer test is based. A thin layer of clay is located between sand and bedrock. A wide uniformly pressured foundation is placed on top of the sand. In the limit, the foundation is "infinitely wide". The clay experiences the weight of the foundation as an additional load q.

The clay layer is being loaded uniformly and vertically. The clay layer cannot expand laterally if it did, then clay that was a long way left or right would have to travel fast! We normally assume that the clay experiences zero lateral strain. This is called one-dimensional condition (the one dimension being the vertical one). Thus we have one-dimensional loading, when an increase in vertical load is applied, and one-dimensional unloading, when the vertical load is decreased.

Sand

Clay

Rock

Wide foundation applying a uniform vertical load q

Small sample, taken for testing in the oedometer



In the oedometer, a sample is confined in a stiff ring typically a lubricated, highly polished stainless steel ring with a diameter-to-height ratio of about 3. There will be a porous stone at the top and/or base of the soil sample, so allowing water to flow into the sample or out of it. In some devices, drainage is allowed only at the top of the sample, and pore pressure is measured at the base. There is a load cell and a settlement gauge to measure vertical load and settlement. In some devices there is a lateral load cell to measure the lateral force applied by the soil to the steel ring.

Key features of the oedometer cell The typical test procedure is to apply a load increment rapidly, and to the measure the settlement at various times after the increment has been applied. This is called the incremental loading procedure. A graph of settlement versus the square-root of time is then constructed. This usually allows a straightforward estimate to be made of when sufficient time has elapsed to achieve the long-term settlement value for that increment.

Load, usually applied by a hanger and deadweight

Settlement gauge

Drainage line, maintaining constant pore pressure at top of sample

Drainage line, or pore pressure measurement line

Porous stone

Porous stone

Soil sample

Steel ring

Base plate

Loading piston



Typical result for settlement versus time for several load increments in an oedometer After the full amount of settlement has been reached or nearly reached, a further increase of load is applied, and the procedure is repeated. This incremental loading procedure is usually repeated five to eight times, with the load being doubled each time. Unloading can also be done. Usually this involves halving the load in each increment. It is also possible to apply load cycles, with the load being increased, then decreased, then increased again, etc.

Calculating the Compression Curve The data for the long-term settlements are usually plotted on a diagram of voids ratio versus

vertical effective stress. This is called the compression curve. If the test included some unloading and reloading sequences, these are called swelling and re-compression.

To calculate the voids ratio, we use the simple phase diagram, noting that air is ideally not

present, so that there is only soil and water in the oedometer. Suppose that the voids ratio at the beginning of the test was eo this is the ratio of the volume of voids in the sample to the volume of solids.

time

settlement

Load increment 1 Load increment 2 Load increment 3



Actual sample at start of test Phase diagram for the start of the test Then, if the initial height was Ho, we can say that the equivalent eight of solids was Ho/(1+eo). Suppose that the height was H at a later stage of the test, and the voids ratio was e. Then the equivalent height of solids is H/(1+e). So, if the soils are incompressible, Ho/(1+eo) = H/(1+e). Hence we have

e = (1+eo).oH

H 1 (1)

Suppose the settlement was s. then H = Ho s. Putting this into the equation gives:

e = (1+eo).

oHs1 1 (2)

The vertical effective stress on the soil is equal to the vertical total stress less the pore pressure. We shall assume that the pore pressure is zero at the end of each increment, when the long-term settlement for that increment has occurred. Then the vertical effective stress equals the vertical total stress, which is just the load on the sample divided by its cross-sectional area. Note that an area-correction is not needed, because of the one-dimensional condition.

The results are usually plotted on semi-log paper or double-log paper, and usually consist of several parts. The first part shows an increase of vertical effective stress without much change of voids ratio. This means that the sample is quite stiff at this point. At some point the slope of the graph changes, and the change of voids ratio per unit change of stress becomes much larger. The point at which this occurs is called the yield point or pre-consolidation point, the vertical effective stress at this point is called the pre-consolidation stress. Sometimes, it is difficult to estimate where this point is. A method called the Casagrande construction is often used to do this in a formal way.

Ho Water in proportion eo Solids in proportion 1



Typical results The figure below shows some typical results for a clay. The clay was from Lagunillas in Venezula.

In this case, the in-situ voids ratio was 1.92. The sample was set up in the oedometer at point P, at a vertical effective stress of about 25 kPa.

The sample was then loaded, and the initial response was quite stiff, with a relatively rapid increase in stress with relatively small reduction in voids ratio. However, the response became less stiff around point A, which is interpreted as the end of the elastic phase and the beginning of the elasto-plastic phase. On further compression, the response followed the line ABD. On unloading from D, the sample "swelled" slightly, with a small increase in voids ratio as the stress was reduced along curve DE. The response was again stiff, with relatively little change in voids ratio for a relatively large change in stress.

On re-loading, the response was stiff from E to F, but with a small hysteresis loop. There was another yield point at F, after which the sample responded in the less stiff, elasto-plastic manner. On unloading from G to H, the selling response was again quite stiff.

Oedometer test results for Lagunillas clay, adapted

from Lambe (1961) and Lambe and Whitman, p.385, 1979)

P

D E

F

G

H



Using the results to calculate long-term settlements We can use the results to calculate long-term settlements. In the particular case that was

investigated by labe 91961), the in-situ vertical effective stress was about 60 kPa. For interest, however, let us consider a site nearby, where the same clay exists at the same in-situ voids ratio, but at lower in-situ stress of say 10 kPa. This would be represented by the on the vertical axis where the arrow marked "e in situ" points.

Suppose a house is to be built on a 3-metre thick layer of the Langunillas clay. Suppose also that we have calculated that, in the long term, the vertical effective stress on the soil will increase to 50 kPa. The we can look along the measured curve to find the point corresponding to 50 kPa which is at a voids ratio of about 1.84. Hence, using the phase relations, we can calculate the settlement of the 3m layer:

o

o

eH+1 = 92.11

3+

m = 1.027m = esHo

+

1 =

1.841sm

+3

The first calculation gives the equivalent height of solids in the layer, which turns out to be 1.027m. The second calculation must also give the same value, since the soil particles themselves are incompressible. From this we can work out the settlement s, which turns out to be 3 2.84 x 1.027 m = 8.3cm.

What happens if we load the soil even more? If we apply a stress greater than the value at A, then there will be much more settlement, because of the lower stiffness along the curve ABDFG. If we tabulate the settlements versus the stress level.

Long-term bearing stress, kPa

Final voids ratio from oedometer test results

Implied long-term vertical settlement, cm

Notes

50 1.84 8.3 Just before A 80 1.74 18.6 Just after A 100 1.63 29.9 200 1.33 60.7 400 1.09 85.4 Extrapolated just past D

We can also plot the results.



0

10

20

30

40

50

60

70

80

90

0 100 200 300 400 500

Long-term bearing stress, kPa

Long-term component of

settlement, cm

Results for a 3-metre thick layer of Lagunillas clay with an in-situ voids ratio of 1.92

We can use these results to interpret the graph of settlement versus load on page 6. For small loads, the material response is stiff and elastic, so there is not much settlement. For larger loads, the yield point (such as A for Lagunillas clay) is exceeded, and there is an increase in settlement relative to load. At larger loads still, because of the curvature of the line ABFGH for Lagunillas clay, the increase of settlement with load decreases slightly.

Using the results to calculate swelling We can also use the results to calculate the swelling potential for the base of an excavation.

Suppose that, at a nearby site in a deeper layer of the same clay, the in-situ stress is about 166.4 kPa, and the clay at that location has reached the point B. This means that he in-situ voids ratio here will be about 1.4. Suppose that the excavation will be 10m deep, and that it will reduce the long term vertical effective stress on the Lagunillas clay layer by about 100 kPa.

To calculate the amount of swelling or heave, we can either do another oedometer test, or we can "interpolate" from the data that we have. If we interpolate, then we can probably expect that the unloading curve from point B will be similar in shape and slope to the unloading curves from D and from G. SO we might expect an unloading curve like BC.

At point C, the stress has reduced by 100 kPa, and the voids ratio is now 1.45 instead of 1.40. Suppose the thickness of the clay layer here is 4m. Then the phase relation calculation looks like this:



o

o

eH+1 = 40.11

4+

m = 1.667m = esHo

+

1 =

1.451sm

+4

On the left side, the calculation fro the in-situ height of solids now uses the in-situ voids ratio of 1.4 instead of 1.92. For a 4m deep layer, this gives an equivalent height of solids of 1.667m. On the right, we can use the same equation, but the "settlement" will actually come out as a negative value. In this case, s = - 8.4cm downwards, which is equivalent to a "heave" upwards of 8.4 cm.

Using the results to interpret geological history We know from the results of Lagunillas clay that, if a soil layer is subjected to an increase of load,

to a point like D, and then unloaded to E and reloaded, it will have an initially elastic response from E to F, followed by "yielding" at F and a less stiff elasto-plastic response from F to G.

Suppose that, at a nearby site for the same clay, we find that the clay is overlain by sand, and that the in-situ soil state is represented by point P, with an in-situ stress of 25 kPa. For this soil state, the yield stress can be measured from the oedometer data, and is about 67.4 kPa. On this basis, we can imagine a geological history which has three parts to it (see the following figure):

(1) In the original deposition, material was deposited to a higher ground surface, such that the

vertical effective stress reached 67.4 kPa. This might have been deposition of the clay itself, as shown above or it might have been another material.

(2) After the deposition, there was a process of erosion, probably due to wind or water action. The erosion removed sufficient clay down to the top of the clay layer as it is at present. Thus the material was unloaded, and its state changed from the point A to some point like U.

(3) After the erosion, the sand layer was deposited. This increased the vertical effective stress on the

clay layer to the present value of 25 kPa. Now this history is a little speculative, because there are some other more complicated histories that can have the same effect. For instance, there might have been a more complicated sequence for the sand. We can see, however, that the maximum vertical effective stress that the soil has experienced previously is an important parameter. We call it the pre-consolidation stress. Thus the pre-consolidation stress for the Lagunillas clay was about 67.4 kPa.



Historical deposition ... followed by erosion then deposition of overlying sand

Possible sequence of Geological Events for the Lagunillas clay Our interpretation also means that the curve ABDFG is a line of maximum historical stresses. For example, point D was reached during the oedoemeter test, and this was the first time that this sample for Lagunillas clay had ever experienced a vertical effective stress of just under 400 kPa. We call the curve ABDFG the virgin compression curve. It is the curve of first-time loading of the clay. We call curves like D-E and G-H swelling curves. Curves like PA and EF are called re-compression curves. These curves normally form a "hysteresis loops", as shown for DEF. In fact, in many calculations we assume that the hysteresis loop is absent, and that the swelling and re-compression curves are straight lines on the semi-log plot. We will look at this simplification shortly.

It's important to know the relationship between the in-situ state of the soil and the virgin compression line. For example, if the in-situ state is a long way below or to the left if the virgin compression line, then a relatively large increase in stress will be needed before the pre-consolidation stress is reached. We call the ratio of the pre-consolidation stress to the in-situ stress the over-consolidation ratio (OCR):

OCR = //

v

v

of value present of value historical maximum

(3)

For example for the point P for the Lagunillas clay, the in-situ stress is 25 kPa, and the pre-consolidation stress is 67.4 kPa. Therefore, the OCR is 67.4 / 25 = 2.7. We use the following terminology:

Present ground surface



OCR value terminology 1 Normally consolidated

Between 1 and about 3 Lightly over-consolidated Between about 3 and 8 Over-consolidated

Greater than 8 Heavily over-consolidated

Simplifying the calculations (1) Index formulation The calculations of settlement and heave can sometimes be complicated. This is fine if an

engineer has only one settlement to calculate. However, often times there are many calculations to be done. For example, in a feasibility study, the engineer might want to investigate several different types of foundation solution, and several different dimensions and depths for each solution. Also, if this is a major construction development with many different buildings, there may be many calculations to be done.

The next figure shows the usual simplification that is made. Essentially, the virgin compression curve is considered to be a straight line in the semi-log plot. The swelling and re-compression curves are assumed to be the same as each other, and are assumed to be straight lines of smaller slope. Because the actual data shows curves, the straight line that best fits the data depends on the range of voids ratio or stress over which the straight line is drawn. Generally, the engineer will estimate the range that is relevant for a particular design, and draw the straight lines accordingly.

There are several ways to describe the straight lines depending on what type of calculation is to

be done. One way is to use so called "indexes". In this way, the equation for the virgin compression line is written as:

e + Cc.log10( /v ) = constant (4) where Cc is the compression index. It is dimensionless. It is the slope of the virgin compression curve in semi-log space, and we can define it as the (negative of) the change of voids ratio per log cycle::

Cc = ))((log /10 vd

de

(5)

For the Lagunillas clay, the change of voids ratio over the entire simplified graph is from 2.2 to 0.8, i.e a change of 1.4. The change of stress along the virgin compression curve is from about 30 kPa to about 780 kPa, so the compression index is about 1.4 / log10(780/30) = 0.99.



0.8

1.0

1.2

1.4

1.6

1.8

2.0

2.2

10 100 1000

Consolidation stress, kPa

Voi

d ra

tio e

Virgin compression line

Elastic lines (swelling lines)

P

A

B

D,F

G

H

C

Simplification for the data of Lagunillas Clay

Similarly, we can define the swelling index of a soil by writing the equation for an elastic swelling line as: e + Cs.log10( /v ) = constant (6)

where Cs is the swelling index. It is dimensionless. It is the slope of the elastic lines in semi-log space, and we can define it as the (negative of) the change of voids ratio per log cycle along one such line

Cs = ))((log /10 vd

de

(7)

For the Lagunillas clay, the change of voids ratio from point G to H was about 0.16. The stress reduced from about 400 kPa to about 50 kPa. The swelling index in this case was 0.16 / log10(400/50) = 0.18.

Values of the compression and swelling indices can be very different for different soils. Some examples are given in the following table, where the Atterberg limits are also shown. Some Authors suggest that the compression index is correlated with liquid limit or with plasticity index, but such correlations may be limited in range. The best policy is always to make the measurement directly.

E



LL, % PI, % Cc Cs (a) Cs(b) Reference Clay minerals (exchangeable ion) Attapulgite (Mg+2) 270 120 0.77 0.24 (1) Illite (Na+) 120 67 1.1 0.15 (1) Illite (Mg+2) 94 48 0.56 0.18 (1) Kaolinite (Na+) 53 21 0.26 (1) Kaolinite (Mg+2) 54 23 0.24 0.08 (1) Montmorillonite (Na+) 710 656 2.6 (1) Montmorillonite (Fe+3) 290 215 1.6 0.03 (1) Clay soils - general Expansive soil A 84 36 0.14 0.25 (2) Expansive soil B 87 45 0.21 0.05 0.15 (2) Cincinatti clay 30 18 0.17 0.02 0.03 (3) Montana clay 58 30 0.21 0.04 0.07 (4) Siburua clay 70 44 0.21 0.08 0.12 (3) St.Lawrence clay 55 33 0.84 0.04 0.08 (3) Undisturbed samples Boston blue clay 41 21 0.35 0.07 0.09 (3) Chicago clay 58 37 0.42 0.07 0.12 (3) Fore river clay 49 28 0.36 0.09 0.09 (3) Louisiana clay 74 48 0.33 0.05 0.08 (3) New Orleans clay 79 53 0.29 0.04 0.08 (3) Remoulded samples Boston blue clay 41 21 0.21 0.07 0.07 (3) Chicago clay 58 37 0.22 0.07 0.09 (3) Fore river clay 49 28 0.25 0.04 0.04 (3) Louisiana clay 74 48 0.29 0.05 0.08 (3) New Orleans clay 79 53 0.26 0.04 0.09 (3)

Compression and swelling indices of some clay minerals and some natural soils. Swell indices for stress ranges (a) 1000 to 100 kPa, and (b) 100 to 10 kPa. From Lambe and Whitman (pp322-323, 1979). Refs (1)

Cornell (1951), (2) Dawson, 1957, (3), Mitchell, 1956, (4) Lambe-Martin, 1957



Let us re-calculate some settlements for Lagunillas clay using the compression and swelling index approach. Suppose that the in-situ soil stress is 10 kPa, and the pre-consolidation stress is 67 kPa. Suppose the final long-term stress after the building has been constructed will be 200 kPa. The long-term settlement will be composed of two parts:

(1) the elastic settlement associated with elastic re-compression, from the in-situ stress of 10 kPa to the pre-consolidation stress of 67 kPa. For this calculation, we use the swelling index, which we previous calculated as 0.18. The change of voids ratio e is Cs.log10(67/10) = 0.15

(2) the elasto-plastic settlement associated with elasto-plastic compression along the virgin compression line, from 67 kPa to 200 kPa. For this, we use the compression index. The change of voids ratio is Cc.log10(200/67) = 0.47

The total change of voids ratio is therefore 0.15+0.47 = 0.62. Recall that a 3m layer of material at an initial voids ratio of 1.92 contains an equivalent height of solids of 3 / (1+1.92) = 1.027m. Hence the settlement is 0.62 x 1.027m = 64cm. This compares well with the value of 62cm which we got previously using the data directly.

Simplifying the calculations (2) Compressibility formulation Another approach that is often used to simplify the settlement calculations is the compressibility approach. It is based on three concepts that are related to each other: (1) The coefficient of volume change, mv, is the most important. It is defined as the volume strain

divided by the change of vertical effective stress. In one-dimensional compression, the volume strain is the same as the vertical strain v, which equals the (negative of) the change of voids ratio divided by 1 plus the initial voids ratio:

mv = /v

v

= /).1( voe

e+

(8)

The symbol denotes a large change, giving a "secant" value. We can also use tangent values, in which case mv is the tangent of the graph of volume strain versus vertical effective stress.

mv is useful because e/(1+eo) is related to the settlement via the phase relation In fact, the settlement s is just the initial depth of the layer, Ho, multiplied by e/(1+eo). Hence the settlement due to a change of vertical stress /v can be calculated simply, as s = mv. /v .Ho. Its a sixty second calculation, or even less!



(2) The coefficient of compressibility, av, is defined as the change of voids ratio divided by the change of vertical effective stress. It is related to the coefficient of volume change:

av = /v

e = (1+eo).av (9)

(3) The constrained modulus, D, is defined as the change of vertical effective stress divided by the volume strain. It is just the inverse of mv

D = )1/(

/

o

vee +

= vm

1 (10)

These quantities can be used either with the oedometer data directly, or with the simplificatoons. Let us do some calculations using the data of Lagunillas clay.

(A) Small changes of stress: Suppose the in-situ soil state is at point U, with a voids ratio of 1.92 and a vertical effective stress of 10 kPa. Suppose the foundation loads will increase the vertical effective stress to 50 kPa. From the data, the voids ratio will reduce to about 1.84. On this basis: The coefficient of compressibility is the change of voids ratio, 0.08, divided by the change of

vertical stress, 40 kpa. Hence av = 0.002 kPa1. The coefficient of volume change is this divided by (1+1.92), so mv = 0.00069 kPa1. The constrained modulus is the inverse of this, which comes to 1460 kPa.

(B) Large change of stress. Suppose the foundation loads increase the vertical effective stress to 200 kPa. From the data, the void ratio will reduce to about 1.34, due to the fact that the pre-consolidation stress of 67 kPa has been exceeded. Therefore: The coefficient of compressibility is the change of voids ratio, (1.92 1.34 = 0.58), divided by

the change of vertical stress, (200-10=190 kPa. Hence av = 0.58/190 = 0.0031 kPa1. The coefficient of volume change is this divided by (1+1.92), so mv = 0.0010 kPa1. The constrained modulus is the inverse of this, which comes to 956 kPa.

This shows that the coefficients and the constrained modulus varies considerably with stress level. Is therefore always wise to note the stress range over which the values apply.



Casagrande's Construction The pre-consolidation pressure is an important parameter, but it is sometimes rather difficult to

estimate from oedometer data. There are two reasons for this. Firstly, the re-compression curve is estimated from only a few data points, which are obtained at the end of the first few load increments. One of the consequences is that there are gaps between data points. Another reason is that the response of the soil is not always simple to interpret. Professor Arthur Cagagrande, a famous Austrian engineer who was professor at Harvard University, proposed a way of estimating pre-consolidation pressure from oedometer data. His method is illustrated and described below.

Casagrande's construction The method consists of 4 steps:

(1) Draw the curve of voids ratio versus the logarithm of effective stress. Assuming that the virgin compression line BC is a straight line, extrapolate it backwards

(2) Determine the point D of maximum curvature on the re-compression part (AB) of the curve (3) Draw the tangent to the curve at D. Bisect the angle between this tangent and the horizontal.

Draw the bisector (DE) (4) Find the point of intersection of the bisector DE and the extrapolated virgin compression curve

BC. The stress at this point is the pre-consolidation stress. The construction provides a definite procedure for calculating pre-consolidation pressure. However, it is not without drawbacks. You are asked to use it in your lab report for the Consolidation lab (page 50 below). Based on this experience, you are asked to give your opinion of it.

Effective consolidation pressure (log scale)

Voids ratio

A

B

C

D

E

Pre-consolidation pressure



Interpreting the data of time-dependency Let us now look at the time-dependent aspect of consolidation. Recall that, in the field, if a load is

applied to a body of soil, one of the effects is normally that there are changes of pore pressures within the soil body. For example, if a vertical force is applied to a pad footing, the change of load causes the total stresses in the soil to change. Some or all of this change is taken up by the pore water, so the pore water pressure changes in response to the load.

Normally, the change of stress is different in different parts of the foundation, so the changes of pore pressure are also different. This means that a field of excess pore pressures is set up within the soil body. The sketch below shows the magnitudes of excess pore pressures induced in a later of clay when the foundation is subjected to a load F that is applied "rapidly", where "rapid" means fast in comparison to the time needed for drainage to occur for the clay layer.

Magnitude of excess pore pressure effects in clay due to rapid application of footing load F Because the excess pore pressures are different, Darcy's law implies that there will be a flow of water from high pressure regions to low pressure regions. The flow of water through the clay layer takes time, and gradually tails off because there is only a certain amount of water that can flow. Consequently, the

CLAY

Vertical load F

SAND

High

Interrmediate Interrmediate

Low Low

Low



foundations settles at a relatively high rate at first, but the rate of settlement reduces and is eventually zero for practical purposes.

For sands, we normally consider that no excess pore pressures occur. In fact, excess pore pressures can arise in sand, but they dissipate rapidly due to the high permeability of sand. However, this is a relative matter, not an absolute one, and the effect depends also on whether the sand is silty or not. Exceptions also occur when the rate of loading is very rapid or cyclic, such as in an earthquake or due to machinery vibrations or impact loading. Wave loading of coastal structures can also induce significant excess pore pressure effects.

We can use the oedometer to investigate the magnitude of the pore pressures generated when a

load is applied. To do this, we close the lower drainage line, and place a pore pressure transducer either in the line or immediately beneath the porous stone. We can then apply a load to the sample quickly, and measure the pore pressure response.

When we do this, we find a remarkable result. The change in pore pressure measured at the base of the sample is initially exactly equal to the change in stress due to the application of the load. Then, as time

LoadDrainage line, maintaining constant pore pressure at top of sample

Porous stone

Porous stone

Soil sample

Electrical transducer to measure pore pressure Electrical wires to/from transducer

C B A



passes, the pore pressure remains roughly constant for a while, then begins to reduce at an increasing rate, then slows down and eventually sops. While this is happening, the sample compresses.

Typical response for excess pore pressures at the base of a half-closed oedometer We can explain the first part of the time record by using the compression data, together with Terzaghi's Principle of Effective Stress. When the load is first applied, there is no time for the pore fluid to move out of the soil, so the void ratio of the soil is the same immediately after the load is applied as it was immediately before the load is applied. But this means, from the compression curve, that the vertical effective stress must be the same just after compared to just before. Now Terzaghi's Principle of Effective Stress implies that the total vertical stress ( v ) equals the sum of the effective vertical stress ( /v ) and the pore pressure (u):

v = /v + u (11) By applying the load quickly, we have increased the total stress, but the effective stress remains as it is instantaneously. It follows that the change of pore pressure must equal the change of vertical effective stress. This is exactly what is observed when we make those measurements in the oedometer.

Another way of thinking is the so-called spring analogy. In the picture below, the soil particles are represented by springs inside a piston that is filled with water. The piston has a small hole in it. When a load is applied quickly to the piston, there is no time to the water to flow out through the hole. Therefore, the springs do not compress immediately. Instead, all of the applied load is taken up by an increase of the pressure in the water. As time passes, the extra water pressure forces water to flow out through the hole.

Time

Excess pore

pressure

Time that load is applied



So, the piston moves downwards, and the springs take up some of the pressure. This means that part of the load is now being taken by the springs, so the part taken by the water reduces, so the water pressure reduces. This cause the flow of water to slow down. Eventually, all of the applied load is transferred into the springs. All of the long-term compression is achieved, and the water pressure returns to zero.

Spring analogy for consolidation

Let us return to the oedometer on page 24. To investigate the experimental results further, we can put additional pore pressure transducers at various levels in the walls of the oedometer, such as at levels A, B, and C. At each of these positions, there is also a distance from the position to the top of the sample, where water can flow out. So we should also expect that the change of pore pressure, when the load is applied, will initially equal the change of stress. This is exactly what is observed experimentally.

Typical responses for excess pore pressures at various levels in a half-closed oedometer; A is closest to the base of the oedometer, C is closest to the top drainage boundary

This kind of experimental data shows that the dissipation of t he excess pore pressures starts earliest at the position closest to the drainage boundary. This helps to explain why there is a delay for the pore pressure at the base of the oedometer.

Time

Excess pore

pressure base of oedometer AB

C



Another way of looking at the data is to plot the values of the excess pore pressures versus position, at a given time. Typical results are shown below.

Typical excess pore pressure isochrones in an oedometer with the base drainage

line closed off, and development with time. Time T3 > Time T2 > Time T1 > 0

We can interpret these results as follows. At time T1, a short time after the load is applied, there has been sufficient time for some of the pore water to flow out of the soil, though the top drainage boundary. Because of this, the void ratio of the soil there has reduced slightly. The soil has compacted slightly, and the state of the soil has moved along its re-compression curve. As a result, the effective stress near the drainage layer has increased slightly, and the pore pressure has reduced slightly.

The effect is largest closest to the drainage boundary, and least at positions like A or at the base of the half-closed oedometer. Because of this, there is a gradient of excess pore pressure, and the gradient is greatest nearest the drainage boundary, and least at the base of the sample. Now Darcy's law states that the rate of flow of water through soil is proportional to the gradient of the excess pore pressure, in other words, the change of excess pore pressure with reference to position. This is why the pore pressure reduces fastest near the drainage boundary, and why there is hardly any change at first near the base of the sample. The excess pore pressure gradient is least nearest the base.

As time passes, more water flows out of the soil, and eventually the rate of compression of the base of the sample begins to increase. At time T3, the rate of flow of water upwards away from the base of the sample is relatively large.

Base

Excess pore pressure

Top Time T0, immediately after load application

T1

T2 T3

A

B

C



Not all the water flows out of the soil. As further time passes, the excess pore pressures get gradually smaller, and eventually all excess pore pressures have dissipated. At this point, all of the soil in the sample has moved along the recompression line, and has reached the place on the line where the vertical effective stress equals the equilibrium value, with no excess pore pressure.

Applying the square-law to the oedometer Now that we understand what is happening in the oedometer, we can start to apply our

knowledge to field situations. Before that, it is useful to consider one further aspect. In most normal oedometer tests, the drainage line at the base of the oedometer is open. What effect does this have?

One of the effects is that settlements in the oedometer will take 4 times less time. The reason can be understood by considering the drainage path lengths in the fully-open and half-closed situations. In the fully open situation, a molecule of water in the middle of the sample has to flow only half the sample height to get out of the sample. In the half-closed situation, the furthest that a molecule has to travel is twice as much, from the base of the sample to its top. Therefore, according to the square-law for drainage path lengths, settlements should take four times as long when the base drainage line is closed compared to when it is open. Fully open: drainage top and bottom half-closed: only one drainage boundary

Pathways for a molecule of water in the fully open and half closed situations Another was of thinking about this is shown below. For the half-closed case, we consider second oedometer sample, immediately below the first, open at its bottom end. By symmetry, a molecule at the centre of the double-height sample has a 50% chance of exiting upwards and a 50% chance of exiting downwards, so the effect is that there is no flow across the centerline! This makes the situation similar to the fully-open case, except that the height of the combined real-plus-imaginary sample is twice as much. Hence the square-law again predicts that the settlement rates will be four times slower for the half-closed sample as for the fully-one one of the same height.



half-closed fully open

Half-closed situation of depth D, and fully open situation of depth 2D

Using these ideas, let us work out what lies behind the square law. Suppose that the drainage path length is D, and an excess pore pressure uxs has been set up in the sample. Now the excess pore pressure gradients will be different at different parts of the sample, but we can characterize the gradients as uxs/D. For example, in the following situations, the same excess pore pressures are occurring in the two samples, but distances are twice as great in the sample on the right. The pore pressure gradient at level X-X on the left is related to the characteristic gradient uxs/D. The gradient at the corresponding level X-X on the right is half as much, and the characteristic gradient uxs/D is also half as much.

"Characterizing" the pore pressure gradients using uxs/D. The drainage path length D on the right is twice as much as on the left, but the excess pore pressures at corresponding levels are the same. So the

excess pore pressure gradients on the right are half as much as on the left.

D

2D

D

uxs

X X

D X X

uxs



Using Darcy's law, we can characterize the velocity of the flow of fluid by k times uxs/D, where k is the permeability of the soil. Therefore, velocities are half as much if the drainage path length is twice as long. But, if the drainage path length is twice as long, there is twice as far for a molecule of water to travel. Hence the time will be four times as much. Similarly, for a drainage path length that is N times longer, the velocities will be N times smaller, and the distance to be traveled will be N times larger. So the time will be N2 times longer.

Using oedometer data to estimate settlement-time responses Consider the situation of a footing that is to be placed on the surface of ground, where the soil

profile consists of sand over clay over sand. Most of the long term settlement will be in the clay. How can we estimate its value. And how can we estimate how long it will take?

Let us start by estimating the change of stress that will be applied to the clay layer. Of course this

depends on position in the clay layer, but we can take a representative value by calculating the change of stress underneath the centerline of the footing, at the level of the centre of the clay. If we use the 1:2 load-spreading method, we can say that the load spreads out at a gradient of 1:2, so that at 4m depth, the load of 3200 kN is being supported on an area of 8m x 8m, giving a stress increase of 3200/64 = 50 kPa.

Calculations for settlements and times for a footing on sand over clay over sand

SAND

CLAY

SAND TO GREAT DEPTH

0m

-2 m

-6 m

4mx4m footing, load =320 tons



Now, we take a sample of the clay from the centre of the layer. We set the sample up in the oedometer, and we apply the same stress increase. Our first task is to calculate the initial stress. Suppose the bulk unit weights for the sand and clay are 17 and 16 kN/m3 respectively, and the water table is at the soil surface. The in-situ total vertical stress 42m below ground level is then 17x2m + 16x2m = 66 kPa. If the unit weight of water is 10 kN/m3, the in-situ vertical effective stress is 66 40 = 26 kPa.

So, we set the sample up in the oedometer at a vertical effective stress of 26 kPa. Now the

process of sampling may have disturbed the sample a little, and will certainly have altered the pore pressures within it, We will need to allow for some time to elapse while the pore pressures in the oedometer return to equilibrium values. Then, starting at 26 kPa, we increase the stress on the sample by 50 kPa, to 76 kPa. This "models" the process that will occur in the field when the building is constructed.

For the field situation, there is drainage top and bottom, so it is a fully-open situation. Let us

suppose that the oedometer sample was 25mm deep at the time when the stress has equilibrated to 26 kPa, and that the bottom drainage line was open so that the sample was also in the fully-open state. Suppose the graph of settlement versus time was as follows.

We can deduce many things from these results. Firstly, the maximum, long-terms settlement in

the test was about 1.4mm. This was for a sample of initial thickness 25mm. The actual clay layer has a

Time after the application of this load increment, hours

settlement, mm

0.5

1.0

1.5

0 0 1 2 3 4 5



thickness of 4 metres, which is 160 times deeper than the sample. Therefore the long-term settlement for the real clay layer will be about 160 x 1.4mm = 0.224 metres. This is quiet a significant amount. For instance, if the footing supports a house and the floor of the house was at street level immediately after the house is built, then the floor will be 0.224 metres below street level in the long term. We will look later at ways of reducing such large long-term settlements.

We can also make an estimate of the time that will elapse for various degrees of settlement. For example, suppose we want to know how long it will take for half the settlement to take place. In the oedometer, half the long-term settlement had occurred after about 2.2 hours. For the field situation, the drainage conditions are the same, because water can easily flow in the sand layers top and bottom. Hence we simply need to use the square law for drainage path lengths. The path length in the oedometer was half the sample height (because there was drainage top and bottom in the oedometer). Hence the length was 12.5mm. The path length in the field was half the layer depth, i.e. 2m. The ratio of drainage path lengths is 160. Hence the ratio of times is 1602 = 25600. Hence it will take 25600 x 2.2 hours to reach 50% consolidation in the field. This is equivalent to 6.4 years.

Terzaghi's Consolidation Equation Derivation

Considering a thin element of soil To derive Terzaghi's consolidation equation, we consider a thin element of soil, between vertical

coordinates z and z+z, where z is a small distance. We shall suppose that the element is fully saturated, and the particles are incompressible. This means that the only way that the element can experience compaction is if water is forced out of the soil. For simplicity we will assume that the properties of the soil element do not depend on the position coordinate z. For instance, the permeability at coordinate z equals the permeability at coordinate z+z.

z

z + z

excess pore pressure gradient in eqn 12

excess pore pressure gradient in eqn 13

Total stress v Element of soil



We will also assume, initially, that the total stress is constant. This is true in the oedometer, for example, in the period from immediately after a load increment has been applied, to the time immediately before the next load increment. Later on, we shall consider the time during which the load is changed.

Calculating the flow of water into and out of the soil Suppose the excess pore fluid pressure in the soil is given by a function u of z (and of time t).

Then we know from Darcy's law that the rate of flow of pore fluid is proportional to the rate of change of this excess pore pressure with position z. Consider the top surface of the element. The rate of fluid flow into the element is given by:

vin = z at evaluated

xs

w zuk

. (12)

where k is the permeability of the soil, and w is the unit weight of water. (Appendix A contains a reminder of how this equation comes about). The derivative is evaluated at the depth z. Now consider the bottom of the element. The rate of fluid flow out of the element is:

vout = zz at evaluated

xs

w zuk

+

(13)

The derivative is evaluated at position z+z. We can evaluate the derivative using Taylor's theorem, which states that the value of a general quantity q, near a point at which it is known to be qo, is just qo plus the derivative of q times the distance. In our case, the value of vout will be the value of vin plus the derivative of the right hand side of equation 12, multiplied by z. So:

vout = vin

zu

zk xsw

. x z (14)

The net rate of volume flow into the element equals vin less vout, which is just the last term in the above equation. But we can also calculate this by a different method which will involve a small approximation.

Calculating the rate of volume change in terms of voids ratio Let t be time, measured from the start of this load increment. Suppose the voids ratio of the soil at

time t is e, and the voids ratio at time t+t is e+e. The phase diagram at time t is shown below

z Water, in proportion e Solids, in proportion 1



Based on this diagram, the equivalent height of solids in the element is z/(1+e). Now if the voids ratio changes by de over times t, then the overall height of the soil element will become this times (1+e+de). Hence the element will expand by an amount e x z/(1+e). If this expansion occurs in time t, then the rate of volume decrease is the negative, so:

vin vout = te

e

+ .11 x z (15)

Now we can equate (vin vout) calculated in this way with (vin vout) calculated from the flow of water into the element. This gives:

te

e

+ .11 =

zu

zk xsw

. (16)

Relating voids ratio, excess pore pressure, and vertical effective stress As mentioned earlier, we shall assume that the total stress is constant. This is true, for example,

in the period from immediately after a load increment has been applied, to the time immediately before the next load increment. Now we consider two things

(1) In effect, we have assumed that the sum of the effective stress and the pore pressure is constant. This means that, if the equilibrium level of the pore pressure is fixed, then any change of excess pore pressure (uxs) must be balanced by an equal and opposite change of vertical effective stress ( /v ). Hence we can say that uxs in the above equation is equal to /v .

(2) Second, we use the compression curve. This curve was inferred from the experimental data in terms of effective stresses instead of total stresses, so this curve still applies, even though the settlement is continuing. We already defined the coefficient of volume change mv. Based on our definition, we can replace e / (1+e) in the above equation by mv. /v .

Making these replacements, and then taking the mathematical limit in which the size z of the element tends to zero, and in which the interval t of time tends t zero, we get:

t

u xs

= cv. 22

zuxs

(17)

with: cv = vw m

k. (18)

This is Terzaghi's equation for one-dimensional consolidation, for constant total stress. The quantity cv is called the coefficient of consolidation. It depends on the compressibility and on the permeability.



Exploring Terzaghi's Consolidation Equation Terzaghi's discovery of the consolidation equation was quite remarkable. Several concepts were

needed to derive it:

(1) The particulate nature of soils, so that compression requires change of void volume (2) Flow of water through the voids of soils, permeability, and Darcy's law (3) The phase diagram - relation between volume change and voids ratio (4) The compression curve relation between voids ratio and effective stress, and mv (5) The principle of effective stress, that total stress equals effective stress plus pore pressure

There are several interesting features of the equation itself. One feature is that it is approximate,

partly because soil properties vary with the amount of compression. This is particularly the case with the coefficient of volume change mv, and to some extent true for permeability as well. This means that the coefficient of consolidation changes as consolidation takes place. In practice we get round this problem by making adjustments for it.

The dimensions of the coefficient of consolidation can be calculated by replacing the quantities on

the right side of equation 18 with their dimensions:

dim(cv) = )dim().dim(

)dim(

vw mk

= )../(1]).((/

2122 TLMTMLTL =

TL2 (19)

Typical units are m2/year or cm2/sec. Typical values of the coefficient of consolidation depend on the Atterberg limits, and also on the moisture content of the soil and on the effective stresses. Lambe and Whitman (1979, page.412) quote the US Navy (1962) who give the following ranges:

Liquid limit, % Range for cv in cm2/sec Range for cv in m2/year 30 % 0.001 to 0.04 3 to 0.120 60 % 0.0003 to 0.004 1 to 12 100 % 0.0001 to 0.0004 0.3 to 1.2

These values are not hard and fast rules and other values are possible. In fact, cv probably depends considerably on the plastic limit. Measurement of cv is one of the most important reasons why the oedometer test is so useful.



Terzaghi's equation is linear in excess pore pressure. In other words, if we multiply the pore pressures on left and right by a factor of 2, say, then the equation would be the same. The equation is quadratic in distance. This is because z2 appears on the right hand side, not z. This quadratic variation will help to explain the square-law for drainage path length that was mentioned earlier.

Solving Terzaghi's equation

Plan of solution Terzaghi's Consolidation equation is not easy to solve. Some solutions were developed by Taylor (1948). Suppose we can find two solutions u1 and u2, both of which satisfy the equation:

t

u 1 = cv. 21

2

zu

(20)

t

u 2 = cv. 22

2

zu

(21)

Then consider the linear combination u* = 1.u1+2.u2, where 1 and 2 are constants:

t *u

=

t uu( 1

+ ). 221 (substituting for u*) (22)

= 1t

u 1 + 2.

tu 2 (since 1, 2 independent of t) (23)

= 1.cv. 212

zu

+ 2.cv. 22

2

zu

(since each part satisfies the eqn) (24)

= cv. 2 22112 )..(

zuu

+ (since 1, 2 independent of z) (25)

= cv. 22 *zu

(substituting for u*) (26)

Hence u* also satisfies the Terzaghi equation. Extending this further, Taylor realized that it might be possible to find solutions in which the excess pore pressure was composed, mathematically, of the sum of a number of terms. This idea led to the Fourier series solutions in which the actual solution uxs is taken to be the sum of a finite or infinite number of components, the nth of which is un:

uxs = =n

nu (27)

As well as satisfying the Terzaghi equation itself, the solutions must also satisfy certain boundary



conditions. For example, the solution for a soil layer that has sand above and below it is different from the solution for a clay layer on impermeable bedrock. In addition to these geometrical constraints, we must consider the initial conditions, which represent the excess pore pressures at the start of an analysis.

Fourier series solutions Taylor's (1948) solutions were based on the idea of separation of variables. Let us suppose un is one of the terms in the Fourier series. Taylor assumed that un could be expressed as the product of two functions. One would be some function Pn(t) which would be a function of time alone. The other would be a function Qn(z) of position alone. Thus Taylor assumed: un = Pn(t).Qn(z) (28) Putting uxs=un in Terzaghi's equation, the left hand side of the equation evaluates to Qn.dPn/dt, because Qn is independent of z. The right side evaluates to cv.Pn.d2Qn/dz2, because Pn is independent of z. Dividing the results by PnQn then gives:

t d

dPP

n

n

1 = n

vQc . 2

2

dzQd n (29)

Since the left side is a function of time alone, and the right side is a function of z alone, it follows that both sides must be constant. We normally represent the constant as 2nM .cv/D2, where Mn is a number to be determined, and D is the drainage path length. Then we can write:

t d

dPP

n

n

1 = n

vQc . 2

2

dzQd n = vn c

DM

2

2 (30)

We now solve the two sides separately. For the left side, the equation gives:

n

nPdP = vn c

DM

2

2.dt (31)

This can be integrated directly to give:

Pn = Pn,o.exp( 2nM . 2.

Dtc v ) (32)

where Pn,o is some initial value which we will consider soon. For the right side, the equation gives:

22

dzQd n = 2

2

DMn .Qn (33)



This is a second-order differential equation whose solution will in general involve a particular integral and a complementary function. The complementary function is:

Qn = Qn,o1.cos

DzMn + Qn,o2.sin

DzMn (34)

Where Qn,o1 and Qn,o2 are some constants. Putting the two parts of the solution together, and putting the results into the summation, gives:

uxs = +==

+

n

n

vnnonnon

Dtc

MDzMu

DzMu )

..(..... 2

22,1, expcossin (35)

where un,o1 = Pn,o.Qn,o1 and un,o2 = Pn,o.Qn,o2. This turns out to be a general solution for constant total stress. We can simplify the result by defining dimensionless variables z/D for position, and cv.t/D2 for time. The latter is called the time factor Tv:

Tv = 2.

Dtc v (36)

The time factor is one of the most important numbers in soil mechanics. It is dimensionless, because cv has dimensions L2/T, time t has dimensions T, and D2 has dimensions L2. Notice that the D-squared appears on the bottom line. This is the reason for the square-law for drainage path lengths, because for a given time factor we have:

t = v

vcT .D2 (37)

Thus the drainage time for a particular time factor is proportional to the square of the drainage path length (assuming cv is constant). We will find later that the time factor is related to the degree of consolidation.

The time factor appears in many different geotechnical calculations. For many calculations, there are standard solutions to the equations, expressed in terms of the time factor. We develop one of these solutions below. The solutions mean that the engineer just has to be able to calculate time factors. To do this, the coefficient of consolidation needs to be known, together with the drainage path length and the time t of interest. Once Tv is known, the engineer will look up the relevant standard solution in a text book, and use the time factor to select the answer to the question being asked. In some case, the engineer has to back-calculate Tv, and then deduce the time t. The procedures will be illustrated later, but first we will develop one of the standard solutions.



Solution for the fully-open, two-way drainage conditions In this situation, a molecule of water in the center of a soil layer can move either upwards towards

the upper drainage boundary, or downwards towards the lower drainage boundary. It is convenient to set up the z-coordinate so that one of the drainage boundaries corresponds to z=0.

Suppose a load is applied rapidly to the soil at time t=0. We already know that the excess pore pressures will immediately rise to the value of the applied stress. The applied load is taken by an increase in pore pressure, and this increase causes water to start to flow out of the soil. In fact, at the drainage boundaries, the excess pore pressure returns to zero immediately. Hence we have a mathematical condition that uxs=0 at z=0 and z=2D for all times t>0.

In the general solution, we can make this happen by just using the sine terms, and by arranging that the values of Mn are such that, for the lower boundary, sin(Mn.2D/D) is always zero. This means 2Mn is a multiply of . We could also arrange it by including cosine terms, but these terms turn out to be not needed for the case we are considering. Thus we have:

Mn = 2.n (38)

These results mean that the series solution has now reduced to:

SAND

SAND

CLAY

v

z

Excess pore pressure

At t=0

immediately after t=0



uxs = ( )+==

n

nvnnon TMD

zMu1

21, .... expsin (39)

The series now starts at n=1 because n=0 gives sin(0) which is 0 for the zeroeth term, and because the negative values of n give sine values that are the negative of the positive values. Consequently, we can include all of the possibilities by just considering positive values of n.

To find the coefficients un,o1, we use the method of Fourier analysis, applied at time t=0. According to this method, the integral of the product of uxs with any sine function can be calculated in two ways (a) by using the know conditions at t=0, or (b) from the right side of the equation. For example, suppose we multiply both sides by sin(z/D). Since uxs = v everywhere at t=0, the left side evaluates to:

D

zxs dzD

zu0

..sin. =

0

.sin..

dDv = Dv .2 (40)

Where the substitution z/D = has been made. The right side evaluates to a sum of integrals, but we find that only one of the integrals is non-zero, the one with n=1. Making the same substitution gives:

D

z

dzDzRHS

0

..sin. =

0

21,1 .sin.

dDu o =

2.1,1 Du o (41)

Equating the two results gives:

U1,o1 = 4 v (42)

Doing a similar calculation for all of the terms in the series, we find that all of the even terms vanish, and that the odd terms are proportional to 4/(n). We can then tidy up by putting n = (2m+1), where m goes from 0 to infinity. The final solution is:

uxs = v . ( ) ( )=

0

2 ...2

mvTMMZM

expsin (43)

with: M = )12(2

+ m (44)

Z = Dz (45)

Plotting the results in terms of consolidation ratio Uz Results for excess pore pressures are usually plotted in terms of the consolidation ratio Uz, defined as 1 (uxs /v). The consolidation ratio goes from zero at time t=0, to 1 at time t=infinity.



Each curve is an isochrone of excess pore pressure (from the greek, iso=same, chronos=time). The curve for Tv=0.05 is shortly after the load is applied, where "shortly" depends on the value of cv and D2. For instance, if cv=0.5 m2/year, and D=10 metres, Tv=0.05 corresponds to t = 0.05 x 102 / 0.5 = 10 years!

Values of cv for silts can be several orders of magnitude larger than for clays. A typical value for a silt could be 1000 times large, say 500 m2/year, in which case Tv=0.05 would correspond to 0.05 x 102/500 years = 3 days. Values of cv can also be measured for sands, and can be 1000 times greater than for silts. Thus Tv=0.05 for a 10m thick sand layer would correspond to only 3/1000 days, or about a minute and a half. We can normally ignore time effects in sands. Exceptions arise for very silty sands or clayey sands, or in situations of rapid loading, such as earthquake loading, wave loading, machine vibrations, impact loading (eg if a ship hits a quay wall), or blast loading from explosions.

Consolidation ratio as a function of normalized depth and time factor. Diagram from Lambe and Whitman (1979, p.408). In this case, the time factor is represented as T

rather than Tv. The drainage path length is H rather than D.



Solutions for other drainage conditions and initial conditions Solutions for one-way drainage can be adapted from the solutions for two-way drainage, as long

as the initial excess pore pressure distribution is a uniform increase of pore pressure. For the situations below, the one on the left would use the upper half of the diagram on the previous page, from Z=0 to Z=1. The situation of the right would use the lower half, from Z=1 to Z=2.

In some situations, it is possible for the initial excess pore pressure distribution to have a

triangular shape. The sketch below shows solutions for the excess pore pressure isochrones for this case, for the case of two-way drainage.

Calculating the degree of consolidation ratio Uv The consolidation ratio Uz at a particular time factor Tv depends on the depth in the clay. To

calculate the settlement associated with the whole layer, we need to take the average of Uz at a given time factor Tv. This average is called the degree of consolidation, denoted as Uv. To calculate Uv, we

Development of consolidation ratio with time factor for the case of an initially triangular distribution of

excess pore pressures, with drainage top and bottom. Figure from Lambe and Whitman (1979)



integrate Uz with respect to depth z, and divide the result by 2D. For the case of an initially constant profile of excess pore pressure versus depth, the solution is :

Uv = 1 ( )=

0

22 ..

2

mvTM

Mexp (46)

The same solution also applies fro triangular distributions if there is two-way drainage. Uv is normally expressed as a %, so the above value is multiplied by 100. Notice that Uv depends only on Tv. Therefore we can plot a unique curve of Uv versus Tv. Curves of Uv versus Tv for several drainage conditions and initial conditions are shown below. The use of these standard solutions is very straightforward. Suppose that the long term-settlement has been estimated at 0.4 metres, based on the long-term compression curve from the oedometer. Suppose

Relations between degree of consolidation and time factor, for various drainage conditions

and initial conditions. Diagrams from pp.250-251 of Craig (2005).



there is two-way drainage, so that Curve 1 applies. Suppose that cv = 2 m2/year, and D=5 metres. How long will it take to achieve 50% consolidation? To find out, we carry out the following steps:

(1) Find the value of Tv on the relevant curve corresponding to the required degree of consolidation,. For the present case, curve 1 applies. For Uv=50%, the Tv value is about 0.2 (the exact value is 0.196, but 0.2 is accurate enough for most purposes)

(2) Back-calculate time from the value of Tv and the other parameters. In this case, t = 0.196 x 52 / 2 = 2.5 years.

We can also do the calculations the other way round. For example, to calculate the settlement after 1 year, we first calculate Tv = 2 x 1 / 52 = 0.08. Looking at Curve 1, this corresponds to a degree of consolidation of Uv = 0.31. Hence the settlement after 1 year will be 0.31 x 0.4 = 0.12 metres.

It is obviously important to know the coefficient of consolidation. It can be measured directly from the time-varying stages of an oedometer test, as follows.

Interpreting time-dependent oedometer data Two methods have been devised for interpreting oedometer test data and calculating the

coefficient of consolidation. The log-time method was developed by Arthur Casagrande. The root-time method was developed by Taylor (1948). Both methods involve plotting the time-dependent data from a load increment, and deducing the coefficient of consolidation from the result.

The log-time method is illustrated in the following figure. The bold curve represents data, plotted as the settlement gauge reading versus the logarithm of time. The method involves two estimates:

(1) 0% consolidation point. In theory, the initial part of the curve should be approximately parabolic. Tow points on the initial part of the curve are selected, such as A and B, for which the values of time are in the ratio 4:1. The vertical distance between them is measured. An equal distance above the first point is then set off. This gives a gauge reading a0. This is taken as the point of 0% consolidation. If this is the first load increment, then the difference between a0 and the measured start of the test may be due to initial compression of some air in the sample, for instance.

(2) 100% consolidation point. The middle of the curve can normally be approximated as a straight line. Also, the end of the curve, at large time, can normally be approximated by a different straight lies. The two straight lines are drawn and their intersection is taken to be the point of 100% primary consolidation (a100).



Since the two points represent the change from 0% to 100% consolidation, it is possible to estimate the coefficient of volume change, mv. This parameter was defined on page 20 above, and is such that the settlement s is equal to mv times the change of effective stress. For instance, the settlement in the above example is about 2.2mm (a0=4.8mm, a100=2.6mm). If the height of the sample at the end of the increment was 20mm, then the initial height was 22.2mm, and the vertical strain was 2.2/22.2 = 10%. If the change of stress was 100 kPa, then the coefficient of volume change was 0.1/100 = 0.001 kPa1.

The point corresponding to 50% consolidation is located at a settlement that is half-way between a0 and a100. According to Terzaghi's theory, the value of Tv corresponding to Uv=50% is 0.196. Consequently, we can estimate the coefficient of consolidation as:

cv = 50

2.196.0t

D (47)

where D is the drainage path length, and t50 is the time at 50% consolidation. For example, for the data shown, t50 is about 11 minutes. If the sample height was 25mm and there was drainage top and bottom,

The log-time method. From page 253 of Craig (2005), t is time after the load is applied



then drainage path length was D=12.5mm, so the coefficient of consolidation is 0.196 x 12.52 / 11 = 2.78 mm2/minute, equivalent to about 1.5 m2/year.

The second method of interpreting the time-dependent data is Taylor's root-time method. In this method, the dial gauge reading is plotted against the square-root of time after load application.

Two constructions are then made:

(1) 0% consolidation point. A straight line is drawn through the main part of the curve, and is extrapolated back to zero time. The dial gauge reading at this point is interpreted as the reading (a0) for 0% consolidation

(2) 90% consolidation point. A second line is drawn through the zero point, with a slope equal to 1.15 times the slope of the previous line. The point at which this second line intersects the experimental curve is interpreted as the point of 90% consolidation.

The root-time method. From page 255 of Craig (2005)



The difference between the two points corresponds to 90% of the estimated long-term primary consolidation. Hence the long terms settlement can be estimated as 10/9 times the difference between the dial gauge reading at the 0% and 90% points. From this, the coefficient of volume change can be calculated in a manner similar to the log-time method.

According to Terzaghi's theory, the value of Tv corresponding to 90% consolidation is 0.848.

Hence the coefficient of consolidation can be estimated as: cv =

50

2.848.0t

D (48)

where t90 is the time at 90% consolidation. For the example shown t90 is about 7.22 = 51.84 minutes. If the drainage path length was 12.5mm, then cv was about 0.848 x 12.52 / 51.84 = 2.56 mm2/min, equivalent to about 1.3 m2/year.

The last part of the interpretation of a load increment is to work out the permeability. Suppose that cv = 2.56 mm2/minute, and the coefficient of volume change came out as 0.001 kPa1. From the equation for cv on page 34, we can solve for the permeability k:

k = w.mv.cv = 10 kN/m3 x 0.001 kPa-1 x 2.56 mm2/minute (49)

We obviously have to be careful about units! One solution is to notice that the unit weight of water can actually be expressed as 10 kPa/m. Also, a coefficient of consolidation of 2.56mm2/minute is equivalent to 2.56/60 mm2/second or 2.56/60 x 106 metres2/second. Hence in his case:

k = 10 kPa/m x 0.001 kPa-1 x 2.56/60 x106 m2/sec (50) = 4.3 x 1010 m/sec

This is equivalent to 4.3 x 108 cm/sec, and is in the range of typical values for clay (see Appendix A).

Using consolidation theory to manage long-term settlements We have seen that the long-term settlements due to consolidation can sometimes be quite large.

As engineers, we need to develop ways to solve this problem.

One method is by preloading. Suppose that a footing will be subjected to a design load F, and the long-term settlement would be S. In preloading, we apply additional load to the soil, before or during construction. By doing this, we accelerate the settlements. We can sometime arrange that the


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