b-2011_2012_eg_shear strength.pdf

ENVIRONMENTAL GEOTECHNICS

Shear Strength of Sands and ClaysShear Strength of Sands and Clays

Prof. Ing. Marco FavarettigUniversit di Padova Facolt di Ingegneria

Dipartimento di Ingegneria Idraulica, Marittima, Ambientale e Geotecnica (I.M.A.GE.)

Via Ognissanti, 39 35129 Italia

Tel: +39.049.827.7901 Fax: +39.049.827.7988

E-Mail: [email protected]@ p

1

SHEAR STRENGTH OF SANDS AND CLAYS

The shear strength of soils is a most important aspect of geotechnical

engineering.

The bearing capacity of shallow or deep foundations, slope stability,

retaining wall design are all affected by the shear strength of the soil in a

slope, behind a retaining wail, or supporting a foundation or pavement.

Structures and slopes must be stable and secure against total collapseStructures and slopes must be stable and secure against total collapse

when subjected to maximum applied loads.

Thus limiting equilibrium methods of analysis are conventionally used for

their design, and these methods require determination of the ultimate or

2limiting shear strength of the soil.

SHEARSTRENGTHOFSANDSANDCLAYS

We define the shear strength of a soil as the ultimate or maximum shear stress

the soil can withstand.

Sometimesthelimitingvalueofshearstresswasbasedonamaximumallowable

strainordeformation.

Thisallowabledeformationactuallyoftencontrolsthedesignofastructure

becausewiththelargesafetyfactorsweuse,theactualshearstressesinthesoil

producedbytheappliedloadsaremuchlessthanthestressescausingcollapseor

failure.

Shearstrengthcanbedeterminedinseveraldifferentways;thenwedescribed

someofthemorecommonlaboratoryandfieldtests.

3

I it th d h th h t t t t id

SHEAR STRENGTH OF SANDS AND CLAYS

In situ methods such as the vane shear test or penetrometers avoidsome of the problems of disturbance associated with the extraction of soilsamples from the groundsamples from the ground.

These methods only determine the shear strength indirectly throughcorrelations with laboratory results or back calculated from actual failurescorrelations with laboratory results or back-calculated from actual failures.

Laboratory tests, on the other hand, yield the shear strength directly.

In addition, valuable information about the stress-strain behaviour and thedevelopment of pore pressures during shear can often be obtained.

We shall illustrate the fundamental stress-deformation and shear strengthresponse of soils with the results of laboratory tests for typical soils.

4

ANGLE OF REPOSE OF SANDS

If we were to deposit a granular soil by pouring it from a single pointabove the ground, it would form a conical pile.

As more and more granular material was deposited on the pile, the slopefor a short period of time might appear to be steeper, but then the soil

ti l ld li d lid d th l t th l fparticles would slip and slide down the slope to the angle of repose.

This angle of the slope with respect to the horizontal plane would remainconstant at some minimum value.

Since this angle is the steepest stable slope for very loosely packedsand, the angle of repose represents the angle of internal friction of thegranular material at its loosest state.

Sand dunes are an example from nature of the angle of repose.

12


Fi h h b thFigure shows how both astationary dune (SD) aswell as a migrating dunewell as a migrating dune(MD) are formed. On theleeward side (LS), the( )slope of the dune willhave an angle (of repose)which varies from 30 to35.

If the slope on the leeward side becomes steeper than 30 to 35, then theslope is unstable and sand grains will roll down the slope until the angle ofrepose is reachedrepose is reached.

An unstable condition is shown on the slope at the far right-hand side ofFi t ll th l t th l f ill f

13

Figure; eventually a smooth slope at the angle of repose will form.


The angle of repose depends on thetype of materials and other factors,

d it t th l fand it represents the angle ofinternal friction or shearingresistance at its loosest stateresistance at its loosest state.Recall that the terms loose ord l l ti tdense are only relative terms,especially with respect to theirbehaviour in shearbehaviour in shear.

As we shall soon see, the stress-strain and volume change responsestrain and volume change responsedepends on the confining pressureas well as on the relative density.

14

as well as on the relative density.

BEHAVIOUR OF SATURATED SANDS DURING DRAINED SHEAR

DIRECT SHEAR TEST

Purpose:

This test is performed to determinep

the consolidated-drained shear

strength of a sandy to silty soil.strength of a sandy to silty soil.

Standard Reference:Standard Reference:

ASTM D 3080 - Standard Test

Method for Direct Shear Test ofMethod for Direct Shear Test of

Soils Under Consolidated Drained

C15

Conditions

BEHAVIOUR OF SATURATED SANDSDURING DRAINED SHEARDURING DRAINED SHEAR

To illustrate the behaviour of sandsduring shear lets start by taking twoduring shear, let s start by taking twosamples of sand: (1) loose at a veryhigh void ratio; (2) dense at a very lowhigh void ratio; (2) dense at a very lowvoid ratio.

We could perform direct shear tests butWe could perform direct shear tests, butto better measure the volume changeswe shall use the triaxial apparatus.pp

We shall run two tests underconsolidated drained (CD) conditions,consolidated drained (CD) conditions,which means we will allow water tofreely enter or leave the sample during

16shear.


If we have a saturated sample, we caneasily monitor the amount of water thateasily monitor the amount of water thatenters or leaves the sample and equatethis to the volume change and thus thethis to the volume change and thus thevoid ratio change in the sample.

Water leaving the sample during shearWater leaving the sample during shearindicates a volume decrease, and viceversa. In both out tests the confininggpressure, c equals 3, is held constantand the axial stress is increased untilfailure occurs.

17

Failure may be defined as:


Failure may be defined as:

1. maximum principal stress difference (1 3)max;2. maximum principal effective stress ratio (1/ 3)max;3. = (1 3)/2 at a prescribed strain( 1 3) p

We will often define failure as the maximum principal stressdifference, equal to the compressive strength of the specimen.

Typical stress-strain curves for loose and dense sand are shown inFigure. When the loose sand is sheared, the principal stress differencegradually increases to a maximum or ultimate value (1 -3).Concurrently, as the stress is increased the void ratio decreases from el(e-loose) down to ecl (ec-loose), which is very close to the critical void

18ratio ecrit.


Casagrande called the ultimate voidCasagrande called the ultimate voidratio at which continuous deformationoccurs with no change in principaloccurs with no change in principalstress difference critical void ratio.

When the dense specimen isWhen the dense specimen issheared, the principal stressdifference reaches a peak orpmaximum, after which it decreases toa value very close to (1 - 3)ult forthe loose sand.

The void ratio-stress curve showsthat the dense sand decreases involume slightly at first, then expands

19or dilates up to ecd (ec- dense).


N ti th t th id ti t f ilNotice that the void ratio at failureecd is very close to ecl.Theoretically they both should beTheoretically, they both should beequal to ecrit critical void ratio.

Similarly the values of ( ) forSimilarly, the values of (1 - 3) for both tests should be the same. The differences are usually attributed todifferences are usually attributed to difficulties in precise measurement of ultimate void ratios as well as non uniform stress distributions in the test specimens.

Evidence of this latter phenomenon is illustrated by the different ways in

20which the samples usually fail.


Th l l j t b l hilThe loose sample just bulges, whilethe dense sample often fails along adistinct plane orienteddistinct plane orientedapproximately (45 + /2) from thehorizontal ( is the effective angle( gof shearing resistance of the densesand).

Note that it is at least theoreticallypossible to set up a sample at aninitial void ratio such that thevolume change at failure would bezero.

This void ratio would, of course, be

21the critical void ratio ecrit.

EFFECT OF VOID RATIO AND CONFINING PRESSURE ON VOLUME CHANGE

I d ibi th b h i f th t d i d t i i l t t l dIn describing the behaviour of the two drained triaxial tests on loose and dense sands, we have mentioned the following physical quantities:

principal stress difference

strain volume change

critical void ratio ecrit and, indirectly,

relative density relative density

We have purposely avoided defining the terms loose and dense b th l h b h i d i h d d t lbecause the volume change behaviour during shear depends not only on the initial void ratio and relative density but also on the confining pressurepressure.

Now we shall consider the effect of confining pressure on the stress-strain and volume change characteristics of sands in drained shear

22

strain and volume change characteristics of sands in drained shear.


W th ff t f (i d i d t t thWe can assess the effects of 3 (in a drained test 3 = 3 as the excess pore water pressure u = 0) by preparing several samples at the same void ratio and testing them at different confining pressuresvoid ratio and testing them at different confining pressures.

We would find that the shear strength increases with 3. A convenient way to plot the principal stress difference versus strain data is to normalize the principal stress ratio 1/3 versus strain. For a drained test 1/3 = 1/3At failure the ratio (1/3)maxAt failure the ratio ( 1/ 3)max

''1'

+=+=

2'45tan

'sen1'sen1 2

max'3

1 1

23

max


h i th ff ti l f i t l f i tiwhere is the effective angle of internal friction.The principal stress difference is related to principal stress ratio by:

= 1'

'1'

331 2 3331

At failure the relationship is:

'1'( )

= 1

max'3

1'f3f31 3

24

EFFECT OF VOID RATIO AND CONFININGPRESSURE ON VOLUME CHANGE

Lets look first at the behaviour of loose sand.

Typical drained triaxial test results are shown for a loose sand in Figure.

The principal stress ratio is plotted versus axial strain for different effective

lid ti consolidation pressures 3c.Note that none of the curves has a distinct peak, and they have a shape similar to the loose curve shown in slides n 10/12

25

shown in slides n.10/12.

EFFECT OF VOID RATIO AND CONFININGPRESSURE ON VOLUME CHANGE

The volume change data is also normalized by dividing the volume change V by the original volume V0 to bt i th l t i t iobtain the volumetric strain:

4

100xVV(%) strain volumetric0

=26

V0


T b tt i t h t iTo better appreciate what is going on in Figure (a) let us compute the principal stresscompute the principal stress difference (1 - 3) at a strain of 5% for 3c = 3.9 MPa and 3c3c = = 0.1 MPa. The principal stress ratiosThe principal stress ratiosfor these conditions are 2.0 and 3.5, respectively.

Utilizing Eq. 2, we obtain the following results: g

27

= 1'3

'1'

331


It i i t ti t l k t th h f thIt is interesting to look at the shapes of the volumetric strain versus axial strain curves in Figurein Figure.

As the strain increases, the volumetric strain decreases for the most partstrain decreases for the most part.

This is consistent with the behaviour of a loose sand.

At low confining pressures (0.1 MPa -0.2 MPa) the volumetric strain is positive or DILATION is taking place

Even an initially loose sand behaves like a dense sand; that is dilates if 3c is low enough.

28

3c g


Behaviour of dense sandBehaviour of dense sand.

The result of several drained triaxial t t d d t dtests on dense sand are presented.

Results seem to be similar in appearance to the previous Figure.

Definite peaks are in the (1/3) strain curves - typical of dense sands.

Large increases of volumetric strain (dilation) are observed.

At higher confining pressures, dense sand exhibits the behaviour of loose sand by showing a decrease in volume

29or compression with strain.

FACTORS AFFECTING THE SHEAR STRENGTH OF SANDS

Since sand is a frictional material we would expect those factors that increase the frictional resistance of sand to lead to increases in the angle of internal friction.

First, let us summarize the factors that influence : 1. Void ratio or relative density 2. Particle shape p3. Grain size distribution 4. Particle surface roughness 5. Water 6. Intermediate principal stress 7 P ti l i7. Particle size 8. Overconsolidation or prestress

30


Void ratio, related to the density of the sand, is perhaps the most important single parameter that affects the strength of sands.

Generally speaking, for drained tests either in the direct shear or triaxial test apparatus, the lower the void ratio (higher density or higher relative density) the higher the sheardensity), the higher the shear strength.

Th M h i l f th t i i l t tThe Mohr circles for the triaxial test data presented earlier are shown in Figure for various confining pressures

31

Figure for various confining pressures and four void ratios.


A th id ti d thAs the void ratio decreases, or the density increases, the angle of internal friction or angle of shearinginternal friction or angle of shearing resistance increases.Mohr failure envelopes in Figure areMohr failure envelopes in Figure are curved; that is, is not a constant if the range in confining pressures isthe range in confining pressures is large.

We usually speak of as if it were aWe usually speak of as if it were a constant, but we understand that the Mohr failure envelope really is p ycurved.

32


Th ff t f l tiThe effects of relative density or void ratio, grain shape grain sizeshape, grain size distribution, and particle size on are summarized in Table.

Values were determinedValues were determined by TX tests on saturated samples at moderate confining pressures.

Generally speaking, with y p g,all else constant, increases with increasing

33angularity.


If t d h th l ti d it th il th t i b tt d dIf two sands have the same relative density, the soil that is better graded (for example, an SW soil as opposed to an SP soil) has a larger . Two sands, at the same void ratio, may not necessarily have the same relative density.

Particle size, at constant void ratio, does not seem to influence significantly. Thus a fine sand and a coarse sand, at the same void ratio, will probably have about the same .Another parameter (not included in Table) is surface roughness, which is very difficult to measure. It will, however, have an effect on .Generally, the greater the surface roughness, the greater will be . It has also been found that wet soils show a 1 to 2 lower than if the sands were dry.

34

y


Th fi l f t lid ti t f d h bThe final factor, overconsolidation or prestress of sands, has been found to not significantly affect , but it strongly affects the compression modulus of granular materialsmodulus of granular materials.

All the factors are summarized in Table.

35


S l ti b t d d d it l ti d it d ilSome correlations between and dry density, relative density, and soil classification are shown in Figure.

36

SHEAR STRENGTH OF SATURATED COHESIVE SOILS

What happens when shear stresses are applied to saturated cohesive soils?

First, lets briefly review what happens when saturated sands are sheared.

You know that volume changes can take place in a drained test and thatYou know that volume changes can take place in a drained test, and that the direction of the volume changes, whether dilation or compression, depends on the relative density as well as the confining pressure. p y g p

If shear takes place undrained, then the volume change tendencies produce pore pressures in the sandproduce pore pressures in the sand.

Basically, the same things happen when clay soils are sheared.

37


I d i d h h th th l h dil ti iIn drained shear, whether the volume changes are dilation or compression depends not only on the density and the confining pressure but also on the stress history of the soilstress history of the soil.

Similarly, in undrained shear, the developed pore pressures depend greatly on whether the soil is normally consolidated orgreatly on whether the soil is normally consolidated or overconsolidated.

Typically, engineering loads are applied much faster than the water can escape from the pores of a clay soil, and consequently excess hydrostatic or pore pressures are producedor pore pressures are produced.

If the loading is such that failure does not occur, then the pore pressures di i t d l h d l b th lldissipate and volume changes develop by the process we call consolidation.

38

Oedometric Tests

wVvolumesolid Vvolume watere =

es

initial initial loadingloading

Consolidationstress, p

ehigher vertical stress which acted on the cohesive layer

Virgin consolidationline of clayy

reloadingreloading'c

ec =unloadingunloading

'c log

0o e1

eHH +=loadingloading

loglog

lid ti ti OCR''pp

Oedometric Test

overconsolidation ratio OCR = pp''vvexisting pressure

eNC normally consolidated

existing pressure, vNC normally consolidatedv = p OCR = 1

OC - overconsolidated

consolidation v < p OCR >1

stress, p

log

Soil NC OC

wVvolumesolid Vvolume watere =

OCee

sVvolumesolid

NCOC

NC

NCNC

OC

z

OC

41( ) z or sat = ( ) z'z' wsat ==


Th i diff i b h i b t d d l i i th tiThe primary difference in behaviour between sands and clays is in the timeit takes for these volume changes to occur.

The time aspect strictly depends on, or is a function of the difference inpermeability between sands and clays.

Since cohesive soils have a much lower permeability than sands and gravels, it takes much longer for the water to flow in or out of a cohesive soil mass.

What happens when the loading is such that a shear failure is imminent?

Since (by definition) the pore water cannot carry any shear stress, all the applied shear stress must be resisted by the soil structure.

42


P t th th h t th f th il d d l thPut another way, the shear strength of the soil depends only on the effective stresses and not on the pore water pressures.

This does not mean that the pore pressures induced in the soil are unimportant.

On the contrary, as the total stresses are changed because of some engineering loading, the pore water pressures also change, and until equilibrium of effective stresses occurs instability is possible.

These observations lead to two fundamentally different approaches to the solution of stability problems in geotechnical engineering:

(1) the total stress approach

(2) the effective stress approach.

43


I th t t l t h ll d i t t k l d i thIn the total stress approach we allow no drainage to take place during the shear test, and we make the assumption, admittedly a big one, that the pore water pressure and therefore the effective stresses in the testpore water pressure and therefore the effective stresses in the test specimen are identical to those in the field.

The method of stability analysis is called the total stress analysis and itThe method of stability analysis is called the total stress analysis, and it utilizes the total or the undrained shear strength f of the soil. The undrained strength can be determined by either laboratory or field tests.

If field tests such as the vane shear, Dutch cone penetrometer, or pressuremeter test are used, then they must be conducted rapidly enough

th t d i d diti il i itso that undrained conditions prevail in situ.

44


Th d h t l l t th t bilit f f d tiThe second approach to calculate the stability of foundations, embankments, slopes, etc., uses the shear strength in terms of effective stressesstresses.

In this approach, we have to measure or estimate the excess hydrostatic pressure both in the laboratory and in the fieldpressure, both in the laboratory and in the field.

Then, if we know or can estimate the initial and applied total stresses, we may calculate the effective stresses acting in the soil.

Since we believe that shear strength and stress-deformation behaviour of soils is really controlled or determined by the effective stresses, this second approach is philosophically more satisfying.

But, it does have its practical problems. For example, estimating or measuring pore pressures, especially in situ, is not easy to do.

45


Th th d f t bilit l i i ll d th ff ti t l i dThe method of stability analysis is called the effective stress analysis, and it utilizes drained shear strength or shear strength in terms of effective stresses The drained shear strength is ordinarily only determined bystresses. The drained shear strength is ordinarily only determined by laboratory tests.

You probably recall that there are limiting conditions of drainage in the testYou probably recall that there are limiting conditions of drainage in the test which model real field situations.

We mentioned that you could have three different conditions:

consolidated-drained (CD)

consolidated-undrained (CU)

unconsolidated-undrained (UU) unconsolidated-undrained (UU).

It is also convenient to describe the behaviour of cohesive soils at these limiting drainage conditions

46

limiting drainage conditions.


It i t diffi lt t t l t th t t diti i t ifi fi ld it tiIt is not difficult to translate these test conditions into specific field situations with similar drainage conditions.

We mentioned that the unconsolidated-drained test (UD) is not a meaningful test.

First, it models no real engineering design situation.

Second, the test cannot be interpreted because drainage would occur p gduring shear, and you could not separate the effects of the confining pressure and the shear stress.

As we did with sands, we shall discuss the shear behaviour of cohesive soils with reference to their behaviour during triaxial shear tests.

47


Y thi k f th l i th t i i l ll ti t i l ilYou can think of the sample in the triaxial cell as representing a typical soil element in the field under different drainage conditions and undergoing different stress pathsdifferent stress paths.

In this manner, we hope you will gain some insight into how cohesive soils behave in shear both in the laboratory and in the fieldbehave in shear, both in the laboratory and in the field.

Keep in mind that the following discussion is somewhat simplified, and that real soil behaviour is much more complicated.

48

CONSOLIDATED-DRAINED (CD) TEST BEHAVIOUR

Th d i t lid t th t t i d t t fThe procedure is to consolidate the test specimen under some state of stress appropriate to the field or design situation.

The consolidation stresses can either be hydrostatic (equal in all directions, sometimes called isotropic) or non-hydrostatic (different in different directions sometimes called anisotropic)different directions, sometimes called anisotropic).

Another way of looking at this second case is that a stress difference or (from the Mohr circles) a shear stress is applied to the soil.

When consolidation is over, the C part of the CD test is complete.

During the D part, the drainage valves remain open and the stress difference is applied very slowly so that essentially no excess pore water pressure develops during the test.

Professor A. Casagrande termed this test the S-test (for slow test).

49

g ( )

CD TEST BEHAVIOUR

50


I Fi t t l t l d ff ti t diti i i lIn Figure total, neutral, and effective stress conditions in an axial compression CD test at the end of consolidation, during application of axial load and at failure are shownload, and at failure are shown.

The subscripts v and h refer to vertical and horizontal, respectively while cmeans consolidationmeans consolidation.

For conventional axial compression tests, the initial consolidation stresses are hydrostatic.

Thus v = h = 3c cell pressure, which is usually held constant during the application of the axial stress . In the axial compression test = 1 3 adn at failure f = (1 3)f.

51


Th i l t b li d ith b i i th l d th i tThe axial stress can be applied either by increasing the load on the piston incrementally (stress controlled loading) or through a motor-jack system which deforms the sample at a constant rate (called a constant rate ofwhich deforms the sample at a constant rate (called a constant rate of strain test).

During the CD test pore water pressure is essentially zeroDuring the CD test pore water pressure is essentially zero.

This means that the total stresses in the drained test are always equal to the effective stresses.

Thus 3c = 3c = 3f = 3f and 1f = 1f = 3f + fIf isotropic consolidation stresses were applied to the specimen, then 1f=1f would be equal to a 1c + f.

52


T i l t t i d lTypical stress-strain curves and volume change versus strain curves for a remoldedor compacted clay are shown in Fig 24or compacted clay are shown in Fig. 24

Even though the two samples were tested at the same confining pressure OC specimenthe same confining pressure, OC specimen has a greater strength than the NC clay.

It has a higher modulus and failure [maximum = (1 3)f for TX test ] occurs at a much lower strain than for NC specimen

Figura 24

a much lower strain than for NC specimen.

Note analogy to drained behaviour of sands.

OC clay expands during shear while the NC clay compresses or consolidates

53during shear.


Thi i l t th b h i

Figura 24

This is analogous to the behaviour described earlier for sands: N.C. clays behave similarly to looseclays behave similarly to loose sands, whereas O.C. clays behave like dense sands.

The Mohr failure envelopes for CD tests of typical clay soils are showntests of typical clay soils are shown in Figures 24.

The envelope for a remolded clay as Figura 25The envelope for a remolded clay as well as a N.C. undisturbed clay is shown in Figure 25.

g

g

54


E th h l M h i lEven though only one Mohr circle (representing the stress conditions at failure) is shown the results of 3 or more CD tests onis shown, the results of 3 or more CD tests on identical specimens, at different consolidation pressures, would ordinarily be required to plot Figura 25p , y q pthe complete Mohr failure envelope.

If the consolidation stress range is large orIf the consolidation stress range is large or the specimens do not have exactly the same initial water content, density, and stress history, then the 3 failure circles will not exactly define a straight lime, and an average best-fit lime by eye is drawn. The slope of the line determines the Mohr-Coulomb strength parameter , f i t f ff ti t

55

of course, in terms of effective stresses.


Wh th f il l i t l t d tWhen the failure envelope is extrapolated to the shear axis, it will show a surprisingly small interceptsmall intercept.

It is usually assumed that the c parameter for N C clays is essentiallyparameter for N.C. clays is essentially zero for all practical purposes.

For O.C. clays the c parameter is greater than zero, as indicated by Figure 26.

The O.C. portion of the strength envelope (DEC) lies above the N.C. envelope (ABCF). Figura 26

This portion (DEC) of the Mohr failure envelope is called the preconsolidation

56hump.


Th l ti f thi b h i i hThe explanation for this behaviour is shown in e vs. curve. We begin consolidation of a sedimentary clay at a very high watera sedimentary clay at a very high water content and high void ratio. As we continue to increase the vertical stress we reach point pA on the virgin compression curve and conduct a CD TX test.

The strength of the sample consolidated to point A on the virgin curve would correspond to point A on the N.C. Mohr failure envelope.

Figura 26

If we consolidate and test another otherwise identical specimen which is loaded to point B, then we would obtain the strength, again N.C., at point B on the failure envelope in Figure 26

57

on the failure envelope in Figure 26


If we repeat the process to point C (p the preconsolidation stress), then rebound the specimen to point D then reload it to point Especimen to point D, then reload it to point E and shear, we would obtain the strength shown at point E in the lower figureshown at point E in the lower figure.

Note that the shear strength of specimen E is greater than specimen B even thoughis greater than specimen B, even though they are tested at exactly the same effective consolidation stresses.

The reason for the greater strength of E than B is suggested by the fact that E is at a

Figura 26

than B is suggested by the fact that E is at a lower water content, has a lower void ratio, and thus is denser than B.

58


If th i l d d t CIf another specimen were loaded to C, rebounded to D, reloaded back past E and C and on to F it would have the strength asC and on to F, it would have the strength as shown in the figure at point F.

Note that it is now back on the virginNote that it is now back on the virgin compression curve and the N.C. failure envelopeenvelope.

The effects of the rebounding and reconsolidation have been in effect erasedreconsolidation have been in effect erased by the increased loading to point F.

O th il h b l d d ll t th

Figura 26

Once the soil has been loaded well past the preconsolidation pressure p,, it no longer remembers its stress history

59

remembers its stress history.

TYPICAL VALUES OF DRAINED STRENGTH PARAMETERS

Average values of for undisturbed clays range from around 20 for N.C. highly plastic clays up to 30 or more for silty and sandy clays.

The value of for compacted clays is typically 25 or 30 and occasionally as high as 35.

The value of c for N.C. non-cemented clays is very small and can be neglected for practical work.

If the soil is O.C. then would be less, and the c intercept greater than for the N.C. part of the failure envelope.

For natural O.C. non-cemented clays with a preconsolidation stress of less than 500 to 1000 kPa, c will probably be less than 5 to 10 kPa at low , p ystresses.

60

TYPICAL VALUES OF DRAINED STRENGTH PARAMETERS

For compacted clays at low stresses, c will be much greater due to the prestress caused by compaction.

For stability analyses, the Mohr-Coulomb effective stress parameters and c are determined over the range of effective normal stresses likely to

fbe encountered in the field.

It has been observed that there is not much difference between determined on undisturbed or remolded samples at the same water content.

Apparently, the development of the maximum value of requires so much strain that the soil structure is broken down and almost remolded in the region of the failure plane.

61

USE OF CD STRENGTH IN ENGINEERING PRACTICE Figura 28

Where do we use the strengths determined from the CD test?the CD test?

The limiting drainage conditions modeled in the TX test refer to real field situations CD conditions aresituations. CD conditions are the most critical for the long-term steady seepage case y p gfor embankment dams and the long-term stability of excavations or slopes in both soft and stiff clays.

62Examples of CD analysis

USE OF CD STRENGTH IN ENGINEERING PRACTICE

You should be aware that practically speaking it is not easy to actuallyYou should be aware that, practically speaking, it is not easy to actually conduct a CD test on a clay in the laboratory. To ensure that no pore pressure is really induced in the specimen during shear for materials with pressure is really induced in the specimen during shear for materials with very low permeability, the rate of loading must be very slow. The time required to fail the specimen ranges from a day to several weeks.

Such a long time leads to practical problems in the laboratory such as leakage of valves, seals, and the membrane surrounding the g , , gsample.

Consequently, since it is possible to measure the induced pore pressuresConsequently, since it is possible to measure the induced pore pressures in a consolidated-undrained (CU) test and thereby calculate the effective stresses in the specimen, CU tests are more practical for obtaining the effective stress strength parameters.

CD triaxial tests are not very popular in most soils laboratories.

63

y p p

Test specimen is first consolidated (drainage valves open) under the

CONSOLIDATED-UNDRAINED (CU) TEST BEHAVIOUR

Test specimen is first consolidated (drainage valves open) under the desired consolidation stresses. These can either be hydrostatic or non-hydrostatic consolidation stresses. hydrostatic consolidation stresses.

After consolidation is complete, drainage valves are closed, and the specimen is loaded to failure in undrained shearspecimen is loaded to failure in undrained shear.

Pore water pressures developed during shear are measured and both total and effective stresses may be calculated during shear and at failuretotal and effective stresses may be calculated during shear and at failure.

This test can either be a total or an effective stress test.

Total, neutral, and effective stress conditions in the specimen during the several phases of the CU test are shown in Fig. 29.

64

CU TEST BEHAVIOUR

65Figura 29

CU TEST BEHAVIOUR

66Figura 29

The general case of unequal consolidation is shown but typically for


The general case of unequal consolidation is shown, but typically for routine testing the specimen is consolidated hydrostatically under a cell pressure which remains constant during shear. Thus:pressure which remains constant during shear. Thus:

'f3f3

'c3

'c1hcvccell =====

( )f3f3c3c1hcvccell

( )f31f Like the CD test, the axial test stress can be increased incrementally or at a constant rate of strain up to failure.a constant rate of strain up to failure.

67

Excess pore water pressure u developed during shear can either beCONSOLIDATED-UNDRAINED (CU) TEST BEHAVIOUR

Excess pore water pressure u developed during shear can either be positive (increase) or negative (decrease). This happens because the sample tries to either contract or expand during shear. sample tries to either contract or expand during shear.

We are not allowing any volume change (an undrained test) and therefore no water can flow in or out of the specimen during shearno water can flow in or out of the specimen during shear.

Because volume changes are prevented, the tendency towards volume change induces a pressure in the pore waterchange induces a pressure in the pore water.

If the specimen tends to contract or consolidate during shear, then th i d d t i itithe induced pore water pressure is positive.

It wants to contract and squeeze water out of the pores, but cannot; thus the induced pore water pressure is positive.

68

Positive pore pressures occur in N C clays


Positive pore pressures occur in N.C. clays.

If the specimen tends to expand or swell during shear, the induced pore t i tiwater pressure is negative.

It wants to expand and draw water into the pores, but cannot; thus the pore water pressure decreases and may even go negative (that is, below zero gage pressure).

Negative pore pressures occur in O.C. clays.

Thus, as noted in Fig.29, the direction of the induced pore water pressure u is important since it directly affects the magnitudes of the effective stresses.

Also you might note that in actual testing the initial pore water pressure typically is greater than zero.

69

In order to ensure full saturation a backpressure u is usually applied to


In order to ensure full saturation, a backpressure u0 is usually applied to the test specimen (Fig.29).

Wh b k i li d t l th ll tWhen a back pressure is applied to a sample, the cell pressure must also be increased by an amount equal to the back pressure so that the effective consolidation stresses will remain the samethe effective consolidation stresses will remain the same.

Since the effective stress in the specimen does not change, the strength of the specimen is not supposed to be changed by the use of backof the specimen is not supposed to be changed by the use of back pressure.

I ti thi t b tl t b t th d t f h iIn practice this may not be exactly true, but the advantage of having 100% saturation for accurate measurement of induced pore water pressures far outweighs any disadvantages of the use of back pressurepressures far outweighs any disadvantages of the use of back pressure.

Typical stress-strain, u, and 1/3 curves for CU tests are shown in Fig 30 for both N C and O C cla s

70

Fig.30, for both N.C. and O.C. clays.

u vertu

'3

'1 /

vert

71Figura 30 vert

Also shown for comparison is a


Also shown for comparison is a stress-strain curve for an O.C. clay at low effective consolidation at low effective consolidation stress.

Note the peak then the drop-off ofNote the peak, then the drop-off of stress as strain increases (work-softening material). g )

The pore pressure versus strain curves illustrate what happens tocurves illustrate what happens to the pore pressures during shear.

The N C specimen developsThe N.C. specimen develops positive pore pressure.

Figura 30

72

In O C specimen after a slight initial


In O.C. specimen after a slight initial increase, the pore pressure goes negative in this case, negative with negative in this case, negative with respect to the back pressure u0.

Another quantity that is useful forAnother quantity that is useful for analyzing test results is the principal (effective) stress ratio 1/3( ) 1 3Note how this ratio peaks early, just like the stress difference curve, for thelike the stress difference curve, for the O.C. clay.

Similar test specimens having similarSimilar test specimens having similar behaviour on an effective stress basis will have similarly shaped 1/3

Figura 30

73

y p 1 3curves.

They are simply a way of


They are simply a way of normalizing the stress behaviour with respect to the effective minor with respect to the effective minor principal stress during the test.

Sometimes too the maximum ofSometimes, too, the maximum of this ratio is used as a criterion of failure.

However we will continue to assume failure occurs at theassume failure occurs at the maximum principal stress difference (compressive strength).

What do the Mohr failure envelopes look like for CU tests?

Figura 30

74

p

Since we can get both the total and


Since we can get both the total and effective stress circles at failure for a CU test when we measure the CU test when we measure the induced pore water pressures, it is possible to define the Mohr failure envelopes in terms of both total and effective stresses from a series of

Figura 31

triaxial tests conducted over a range of stresses, as illustrated in Fig. 31 for a N C clay Only one set of Mohrfor a N.C. clay. Only one set of Mohr circles is shown.

Eff ti t i l i di l d t th l ft t d th i i f thEffective stress circle is displaced to the left, towards the origin, for the N.C. case, because the specimens develop positive pore pressure during shear and = - u

75

during shear and = - u.

Both circles have the same diameter


Both circles have the same diameter because of our definition of failure at maximum (1 3) = (1 3)maximum (1 3) ( 1 3)You should verify that this equation is true.

Once the two failure envelopes are drawn, the Mohr-Coulomb strength parameters are readily definable in terms of both total

Figura 31

are readily definable in terms of both total (c, or sometimes cT, T) and effective stresses (c, ).stresses (c , ). As with the CD test, the envelope for N.C. clay passes essentially through the origin, and thus for pratical purposes c can be taken to be zero, which g , p p p ,is also true for the total stress c parameter.

Note that T is less than and often it is about one-half of 76

Note that T is less than and often it is about one half of .

Things are different if the clay is O C

CONSOLIDATED-UNDRAINED (CU) TEST BEHAVIOUR Figura 30

Things are different if the clay is O.C..

Since an O.C. specimen tends to expand d i h th tduring shear, the pore water pressure decreases or even goes negative, as shown in Fig 30shown in Fig.30.

Because 3f = 3f (uf) or 1f = 3f (uf) the effective stresses are greater thanthe effective stresses are greater than the total stresses, and the effective stress circle at failure is shifted to the right of thecircle at failure is shifted to the right of the total stress circle (Fig. 11.32).

The shift of the effective stress circle atThe shift of the effective stress circle at failure to the right sometimes means that the is less than T. Figura 32

77

T

Complete Mohr failure envelopes are


Complete Mohr failure envelopes are determined by tests on several specimens consolidated over the specimens consolidated over the working stress range of the field problem. Figure 33 shows Mohr failure envelopes over a wide range of stresses spanning the

Figura 32

preconsolidation stress.

Thus some of the specimens are O.C. and others are N.C..

You should note that the break in the total stress envelope (point z) occurs roughly about twice the p for

78typical clays.

Figura 33

The two sets of Mohr circles at


The two sets of Mohr circles at failure (Fig.33) correspond to tests shown in Fig.30 for N.C. specimen shown in Fig.30 for N.C. specimen and specimen O.C. at low hc.You may have noticed that an angleYou may have noticed that an angle 1 was indicated on effective stress Mohr circles of Figs. 31, 32, and 33.

Figura 32

g

Do you recall the Mohr failure hypothesis wherein the point ofhypothesis wherein the point of tangency of the failure envelope with the Mohr circle at failure defined the angle of the failure plane in the specimen?

79Figura 33

CONSOLIDATED-UNDRAINED (CU) TEST BEHAVIOURFigura 32

Since we believe that the shear strength is controlled by the effective stresses in the specimen at failure, Mohr failure hypothesis i lid i t f ff tiis valid in terms of effective stresses only.

80Figura 33

Earlier we gave some typical values for c and determined by CD triaxialTYPICAL VALUES OF THE UNDRAINED STRENGTH PARAMETERS

Earlier, we gave some typical values for c and determined by CD triaxialtests.

Th f l i di t d i t i l f ff ti t t thThe range of values indicated is typical for effective stress strengths determined in CU tests with pore pressure measurements, with the following reservationfollowing reservation.

We have tacitly assumed that Mohr-Coulomb strength parameters in terms of effective stresses determined by CU tests with pore pressureterms of effective stresses determined by CU tests with pore pressure measurements would be the same as those determined by CD tests.

W d th b l d f th t d t i d b thWe used the same symbols, c and , for the parameters determined both ways. This assumption is not strictly correct. The problem is complicated by alternative definitions of failurealternative definitions of failure.

81

We have used the maximum principal stress

TYPICAL VALUES OF THE UNDRAINED STRENGTH PARAMETERS

We have used the maximum principal stress difference (1 3)max to define failure throughout this chapter, but often in the throughout this chapter, but often in the literature and sometimes in practice you will find failure defined in terms of the maximum principal effective stress ratio (1/3)maxwhich is the same as the maximum obliquity.

Bjerrum and Simons (1960) studied this problem in some detail, and their results are Figura 36summmarized in Fig.36.

Here, as defined at (1/3)max and (1 3)max are plotted versus d, the effective stress parameter determined in

82drained tests.

Note that from the maximum principalTYPICAL VALUES OF THE UNDRAINED STRENGTH PARAMETERS

Note that from the maximum principal effective stress ratio (the dots) is from 0 to 3greater than d. greater than d. Also note that at maximum principal stress difference (the squares) is less than both ddifference (the squares) is less than both dand at the maximum principal effective stress ratio.

In one case the difference is about 7.

Th i t i th t h ld b f l hFigura 36

The point is that you should be careful when studying published data or engineering test reports to determine exactly how the strengthreports to determine exactly how the strength tests were conducted, how failure was defined, and how any reported Mohr-Coulomb

83

, y pparameters were determined.

Note that from the maximum principal effective stress ratio (the dots) isTYPICAL VALUES OF THE UNDRAINED STRENGTH PARAMETERS

Note that from the maximum principal effective stress ratio (the dots) is from 0 to 3 greater than d. Al t th t t i i i l t diff (th ) iAlso note that at maximum principal stress difference (the squares) is less than both d and at the maximum principal effective stress ratio. In one case the difference is about 7.

The point is that you should be careful when studying published data or engineering test reports to determine exactly how the strength tests were conducted, how failure was defined, and how any reported Mohr-C l b t d t i dCoulomb parameters were determined.

For the Mohr-Coulomb strength parameters in terms of total stresses, the problem of definition of failure doesnt come up.

Failure is defined at the maximum compressive strength (1 - 3). 84

For N C clays seems to be about half of ; thus values of 10 to 15 orTYPICAL VALUES OF THE UNDRAINED STRENGTH PARAMETERS

For N.C. clays, seems to be about half of ; thus values of 10 to 15 or more are typical.

Th t t l t i l tThe total stress c is very close to zero.

For O.C. and compacted clays, may decrease and c will often be significant.

When the failure envelope straddles the preconsolidation stress, proper interpretation of the strength parameters in terms of total stresses is difficult.

This is especially true for undisturbed samples which may have some variation in water content and void ratio, even within the same geologic stratum.

85

Where do we use the CU strength in engineering practice?

USE OF CU STRENGTH IN ENGINEERING PRACTICE

Where do we use the CU strength in engineering practice?

As mentioned before, this test, with pore pressures measured, is l d t d t i th h t th t i t fcommonly used to determine the shear strength parameters in terms of

both total and effective stresses.

CU strengths are used for stability problems where the soils have first become fully consolidated and are at equilibrium with the existing stress systemsystem.

Then, for some reason, additional stresses are applied quickly, with no d i idrainage occurring.

Practical examples include rapid drawdown of embankment dams and the slopes of reservoirs and canals.

86

Also in terms of effective stresses CU test results are applied to the field


Also, in terms of effective stresses, CU test results are applied to the field situations mentioned in the earlier discussion of CD tests.

S f th ti l l ill t t d i Fi 37 J t ithSome of these practical examples are illustrated in Fig. 37. Just as with CD tests, there are some problems with CU tests on clay.

For proper measurement of the pore pressures induced during shear, special care must be taken to see that the sample is fully saturated, that no leaks occur during testing and that the rate of loading (or rate of strain)no leaks occur during testing, and that the rate of loading (or rate of strain) is sufficiently slow so that the pore pressures measured at the ends of the specimen are the same as those occurring in the vicinity of the failurespecimen are the same as those occurring in the vicinity of the failure plane.

Use of back pressure is common to assure 100% saturationUse of back pressure is common to assure 100% saturation.

The effects of the other two factors can be minimized by proper testing techniq es hich are described at length b Bishop and Henkel (1962)

87

techniques, which are described at length by Bishop and Henkel (1962).

Fi 37Figura 37

88

Another problem not often mentioned results from trying to determine the


Another problem, not often mentioned, results from trying to determine the long-term or effective stress strength parameters and the short-term or CU-total stress strength parameters from the same test series. CU total stress strength parameters from the same test series.

The rates of loading or strain required for correct effective stress strength determination may not be appropriate for the short-term or undraineddetermination may not be appropriate for the short-term or undrained loading situation.

The stress deformation and strength response of clay soils is rateThe stress-deformation and strength response of clay soils is rate-dependent; that is, usually the faster you load a clay, the stronger it becomes.becomes.

89

In the short term case the rate of loading in the field may be quite rapid


In the short-term case, the rate of loading in the field may be quite rapid, and therefore for correct modeling of the field situation, the rates of loading in the laboratory sample should be comparable.loading in the laboratory sample should be comparable.

Thus the two objective to do, though rarely done in practice, would be have two sets of tests one set tested CD modeling the long-term situationhave two sets of tests, one set tested CD modeling the long-term situation and the other CU set modeling the short-term undrained loading.

90

In this test the specimen is placed in the triaxial cell with the drainage

UNCONSOLIDATED-UNDRAINED (UU) TEST BEHAVIOUR

In this test, the specimen is placed in the triaxial cell with the drainage valves closed from the beginning.

Th h fi i i li d lid tiThus, even when a confining pressure is applied, no consolidation can occur if the sample is 100% saturated. Then, as with the CU test, the specimen is sheared undrainedspecimen is sheared undrained.

The sample is loaded to failure in about 10 to 20 min; usually pore water pressures are not measured in this testpressures are not measured in this test.

This test is a total stress test and it yields the strength in terms of total tstresses.

Total, neutral, and effective stress conditions in the specimen during the several phases of the UU test are shown in Fig. 38.

91

92Figura 38

The test illustrated in Fig 38 is quite conventional in that hydrostatic cell


The test illustrated in Fig.38 is quite conventional in that hydrostatic cell pressure is usually applied, and the specimen is failed by increasing the axial Ioad, usually at a constant rate of strain. axial Ioad, usually at a constant rate of strain.

The principal stress difference at failure is (1 3)max.Note that initially for undisturbed samples, the pore pressure is negative, and it is called the residual pore pressure -ur which results from stress release during samplingrelease during sampling.

Since the effective stresses initially must be greater than zero (otherwise th i ld i l di i t t ) d th t t l tthe specimen would simply disintegrate) and the total stresses are zero (atmospheric pressure = zero gage pressure), the pore pressure must be negativenegative.

93

When the cell pressure is applied with the drainage valves closed a


When the cell pressure is applied with the drainage valves closed, a positive pore pressure uc is induced in the specimen, which is exactly equal to the applied cell pressure c. equal to the applied cell pressure c. All the increase in hydrostatic stress is carried by the pore water because (1) the soil is 100% saturated (2) the compressibility of the water and(1) the soil is 100% saturated, (2) the compressibility of the water and individual soil grains is small compared to the compressibility of the soil structure, and (3) there is a unique relationship between the effective ( ) q phydrostatic stress and the void ratio.

Number 1 is obvious. Number 2 means that no volume change can occurNumber 1 is obvious. Number 2 means that no volume change can occur unless water is allowed to flow out of (or into) the sample, and we are preventing that from occurring. Number 3 means basically that no secondary compression (volume change at constant effective stress) takes place.

94

You may recall from the discussion of the assumptions of the Terzaghi


You may recall from the discussion of the assumptions of the Terzaghitheory of consolidation that the same assumption was required; that is, that the void ratio and effective stress were uniquely related. that the void ratio and effective stress were uniquely related.

Thus there can be no change in void ratio without a change in effective stressstress.

Since we prevent any change in water content, the void ratio and effective stress remain the samestress remain the same.

Stress conditions during axial loading and at failure are similar to those for th CU t t (Fi 29)the CU test (Fig.29).

They may appear to be complex, but if you study Fig.38 you will see that the UU case is as readily understandable as the CU case.

95

Typically stress strain curves for UU


Typically, stress-strain curves for UU tests are not particularly different from CU or CD stress-strain curves for the CU or CD stress strain curves for the same soils.

For undisturbed samples especiallyFor undisturbed samples, especially the initial portions of the curve (initial tangent modulus), are strongly g ) g ydependent on the quality of the undisturbed samples.

The maximum stress difference often occurs at very low strains, usually

Figura 39

less than 0.5%.

Some typical UU stress-strain curves

96

ypare shown in Fig.39.

The Mohr failure envelopes for UU tests are shown in Fig 40 for 100%


The Mohr failure envelopes for UU tests are shown in Fig.40 for 100% saturated clays.

All t t i f f ll t t d l bl t thAll test specimens for fully saturated clays are presumably at the same water content (and void ratio), and consequently they will have the same shear strength since there is no consolidation allowedshear strength since there is no consolidation allowed.

Figura 40

97

All Mohr circles at failure will have the same diameter and Mohr failure


All Mohr circles at failure will have the same diameter and Mohr failure envelope will be a horizontal straight lime.

If d t d t d it f i t Fi 38 t th t i th UU t tIf you dont understand it, refer again to Fig.38 to see that in the UU test effective consolidation stress is the same throughout the test.

If all the samples are at the same water content and density (void ratio), then they will have the same strength.

UU test gives the shear strength in terms of total stresses, and the slope T of the UU Mohr failure envelope is equal to zero. The intercept of this envelope on the -axis defines the total stress strength parameter c, or f = c, where f is undrained shear strength.

98

The intercept of this envelope on the axis defines the total stressUNCONSOLIDATED-UNDRAINED (UU) TEST BEHAVIOUR

The intercept of this envelope on the -axis defines the total stress strength parameter c, or f = c, where f is undrained shear strength.F ti ll t t d il i f UU t t ill d fi i iti llFor partially saturated soils, a series of UU tests will define an initially curved failure envelope (Fig. 11 .40b) until the clay becomes essentially 100% saturated due simply to the cell pressure alone100% saturated due simply to the cell pressure alone.

Even though the drainage valves are closed, the confining pressure will compress the air in the voids and decrease the void ratiocompress the air in the voids and decrease the void ratio.

As the cell pressure is increased, more and more compression occurs and t ll h ffi i t i li d ti ll 100%eventually, when sufficient pressure is applied, essentially 100%

saturation is achieved.

99

Then as with the case for initially 100% saturated clays the Mohr failure


Then, as with the case for initially 100% saturated clays, the Mohr failure envelope becomes horizontal, as shown on the right side of Fig. 11.40b.

A th f l ki t th i f ti ll t t d l iAnother way of looking at the compression of partially saturated clays is shown in Fig. 11.41.

Figura 41

100

As the cell pressure is increased incrementally the measured increment of


As the cell pressure is increased incrementally, the measured increment of pore pressure increases gradually until at some point for every increment of cell pressure added, an equal increment of pore water pressure is of cell pressure added, an equal increment of pore water pressure is observed.

At this point the soil is 100% saturated and the solid (experimental) curveAt this point, the soil is 100% saturated and the solid (experimental) curve becomes parallel to the 45 line shown in the figure.

In principle it is possible to measure the induced pore water pressures inIn principle, it is possible to measure the induced pore water pressures in a series of UU tests although it is not commonly done.

Si th ff ti t t f il i d d t f th t t l llSince the effective stresses at failure are independent of the total cell pressures applied to the several specimens of a test series, there is only one UU effective stress Mohr circle at failureone UU effective stress Mohr circle at failure.

This point is illustrated in Fig.42.

101

Note that no matter what the confining pressure there is only one effective


Note that no matter what the confining pressure, there is only one effective stress Mohr circle at failure.

Th i ff ti i i l t t f il ( ) i th f ll t t lThe minor effective principal stress at failure (hf) is the same for all total stress circles shown in the figure.

Since we have only one effective circle at failure, strictly speaking, we need to know both and c in advance in order to draw the Mohr failure envelope in terms of effective stresses for the UU testenvelope in terms of effective stresses for the UU test.

We could perhaps measure the angle of the failure plane in the failed UU i d i k th M h f il h th i b t th ti lspecimens and invoke the Mohr failure hypothesis but there are practical

problems with this approach.

Angle of inclination of the failure plane f shown in Fig. 42 is defined by the effective stress envelope.

102Otherwise (Fig. 10.9c and Eq. 10-10) theory would predict f to be 45.

Since the strength ultimately is controlled or governed by the effective


Since the strength ultimately is controlled or governed by the effective stresses, we believe that the physical conditions controlling the formation of a failure plane in the test specimen must in some fashion be controlled of a failure plane in the test specimen must in some fashion be controlled by the effective stresses acting in the specimen at failure.

Thus Eq 10-10 should be in terms of instead of TThus Eq. 10-10 should be in terms of instead of T.

Figura 42

103

Undrained strength of clays varies widely

TYPICAL VALUES OF UU STRENGTH

Undrained strength of clays varies widely.

Of course, T is zero, but the magnitude of f can vary from almost zero for t l ft di t t l MP f tiff il d ft kextremely soft sediments to several MPa for very stiff soils and soft rocks.

Often, the undrained shear strength at a site is normalized with respect to the vertical effective overburden stress v0, at each sampling point. Then the f/v0 ratios are analyzed and compared with other data. This point is covered in more detail later in this chapter.

104

We can theoretically at least conduct an unconfined compression test

UNCONFINED COMPRESSION TEST

We can, theoretically at least, conduct an unconfined compression test and obtain the UU-total stress strength.

Thi t t i i l f th UU t t ith th fi i llThis test is a special case of the UU test with the confining or cell pressure equal to zero (atmospheric pressure).

The stress conditions in the unconfined compression test specimen are similar to those of Fig.38 for the UU test, except that is equal to zero, as shown in Fig 44shown in Fig.44.

If you compare these two figures, you will see that the effective stress diti t f il id ti l f b th t tconditions at failure are identical for both tests.

And if the effective stress conditions are the same in both tests, then the strengths will be the same!

105

Figura 44

106

Practically speaking for the unconfined compression test to yield the


Practically speaking, for the unconfined compression test to yield the same strength as the UU test, several assumptions must be satisfied.

Th f llThese are as follows: 1. The specimen must be 100% saturated; otherwise compression of the air in the voids will occur and cause a decrease in void ratio and anair in the voids will occur and cause a decrease in void ratio and an increase in strength. 2. The specimen must not contain any fissures, silt seams, valves, or p yother defects; this means that the specimen must be intact, homogeneous clay. Rarely are O.C. clays intact, and often even N.C. clays have some fissures.

107

3 The soil must be very fine grained; the initial effective confining stress


3. The soil must be very fine grained; the initial effective confining stress as indicated in Fig.44 is the residual capillary stress which is a function of the residual pore pressure -ur; this usually means that only clay soils are the residual pore pressure ur; this usually means that only clay soils are suitable for testing in unconfined compression. 4. The specimen must be sheared rapidly to failure; it is a total stress test and the conditions must be undrained throughout the test. If the time to failure is too long, evaporation and surface drying will increase the confining pressure and too high a strength will result. Typical time to failure is 5 to 15 mm. Be sure to distinguish unconfined compressive strength ( ) and the undrained shear strength which is:strength (1 3)f and the undrained shear strength which is:

( )1 ( )f31f 2 =108


109


110


111

EFFECTIVE STRESS (Karl Terzaghi, 1936)

The stresses in any point of a section through a mass of soil can be computed from the total principal stresses 1, 2, 3 which act at this point.

If the voids of the soil are filled with water under a stress u, the total principal stresses consist of two parts.

One part, u, acts in the water and in the solid in every direction with equal yintensity. It is called the neutral stress (or the pore water pressure).

The balance [1 = 1 u], [2 = 2 u], [3 = 3 u] represents an excessThe balance [ 1 1 u], [ 2 2 u], [ 3 3 u] represents an excess over the neutral stress u and it has its seat exclusively in the solid phase of the soil.

112

EFFECTIVE STRESS (Karl Terzaghi, 1936)

This fraction of the total principal stresses will be called the effective principal stresses....

A change in the neutral stress u produces practically no volume change and has practically no influence on the stress conditions for failure....

Porous materials (such as sand, clay and concrete) react to a change of u as if they were incompressible and as if their internal friction were equal to yzero.

All the measurable effects of a change of stress, such as compression,All the measurable effects of a change of stress, such as compression, distortion and a change of shearing resistance are exclusively due to changes in the effective stresses 1, 2, 3 .Hence every investigation of the stability of a saturated body of soil requires the knowledge of both the total and the neutral stresses.

113

q g

THE PRINCIPLES OF EFFECTIVE STRESS

The principle of effective stress was stated by Bishop (1959) in terms of two simple hypotheses: Volume change and deformation in soils depends on difference between the total stress and the pressure set up in the fluid in the pore space, not on the total stress applied This leads to the expression = uon the total stress applied. This leads to the expression = u Shear strength depends on the effective stress, not on the total normal stress on the plane considered. This may be expressed by the equationstress on the plane considered. This may be expressed by the equation f = c + tan The principle of effective stress as expressed above has proved to be vitalThe principle of effective stress, as expressed above, has proved to be vital in the solution of practical problems in soil mechanics.

114

RESIDUAL SHEAR STRENGTH direct shear test

The residual strength of an OC

clay is usually performed on a

test specimen cut from an

undisturbed sample. The

specimen is first sheared to

failure at a constant rate of

strain to measure the peak

strength. The two halves of the g

shear box are then returned to

the starting point by using athe starting point by using a

faster rate of strain or hand

cranking The test is repeated115

cranking. The test is repeated

(2 cycle).

RESIDUAL SHEAR STRENGTH

Multiple cycles are made until the

maximum shear stress is

unchanged from the previous

cycle, thus indicating that thecycle, thus indicating that the

residual strength has been

reached An example of a 5 cyclereached. An example of a 5-cycle

test is shown on Figure.

A i i f th di t b dA minimum of three undisturbed

specimens are tested at three

different normal stresses to obtain

Mohr envelopes of peak and

116residual strengths.

An annulus of soil is sheared by torque The

RESIDUAL SHEAR STRENGTH RING SHEAR TEST

An annulus of soil is sheared by torque. The

relative motion, and development of a slip plane,

t th id h i ht f th i Aoccurs at the mid-height of the specimen. An

advantage of the test (compared to the shear

box) is that shear movement is continuous,

without reversals, to residual condition.

The concept of performing torsion tests on soils

in the laboratory goes back to the 1930s, but the

impetus to use the test can be attributed to the

work of Bishop (1971).

The only commercial unit currently available is

the Bromhead apparatus..

117

pp

B h d i h


Bromhead ring shear

The test specimen is a

thin, annular ring of soil

with the following

dimensions: outside

diameter (100 mm);

inside diameter (70

mm) and thickness (5

mm).Thus, the annular

width of the specimen p

is (15 mm).

118

Th i i t ithi l th b l t


The specimen is set up within an annular groove on the base plate.

A loading platen and torque arm assembly is centered over the top of the

base plate so that the annular porous stone in the loading platen is aligned

over the specimen. Water is added to the perspex bowl to prevent the soil

from drying out during the test. A normal stress is applied by loading the

soil through a loading yoke and lever load arm.

In the shear test, the base plate containing the soil specimen is steadily

rotated by a multispeed motor located below it. This rotation is resisted by

the soil specimen and the shear stress is determined by loads measured

on two proving rings. A failure plane develops just below the loading p g g p p j g

platen/soil contact. The gearbox has a choice of 25 constant speeds

ranging from 60 per minute to 0 024 per minute

119

ranging from 60 per minute to 0.024 per minute.

R id l St th M t i th Ri Sh T t


Residual Strength Measurements in the Ring Shear Test

The test specimen is usually prepared from remoulded soil, which is placed

in the annular groove of the specimen container. It is kneaded into place

using a wood dowel or similar implement, and the upper surface is made

smooth and level. The loading yoke and torque arm assembly is placed

over the specimen and it is consolidated under the selected normal load.

The specimen is very thin and in double drainage, dissipation of excess

pore pressures will occur relatively quickly in all soil types.

Bromhead (1986) recommends a multistage test procedure. After applying

the 1st normal load, the specimen is strained until the residual strength , p g

appears to have been reached, then the next higher normal load is applied.

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The ring shear test shares a problem with the shear box test As the


The ring shear test shares a problem with the shear box test. As the

shearing continues, small amounts of soil squeeze out or slough from the

t t i i t th di t fill d b ltest specimen into the surrounding water-filled bowl.

Therefore, in a multistage test, the specimen becomes progressively

thinner. In the case of the ring shear, this loss of soil during shear, plus

earlier consolidation, causes the upper porous stone to move into the

annular groove. According to Stark and Vettel (1992), the wall friction

increases the measured residual strength. They recommend that

settlement from both sources be limited to 0.75 mm (15% of the original

height).

This can be accomplished by adding soil and reconsolidating as needed.

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The reliable measurement of peak strength of undisturbed samples in the


The reliable measurement of peak strength of undisturbed samples in the

ring shear apparatus presents a difficulty because the stresses developed

th t t i f th i t th t diacross the test specimen, from the inner to the outer radius, are

nonuniform. (For residual strength, there is no problemthe entire shear

surface is at the same stress.) Bishop et al. (1971) investigated this issue

in some detail. However the assumption that the normal stress and shear

stress are uniformly distributed across the plane of relative rotary motion

provides a reasonable interpretation of both the peak and residual

strengths.

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The relevant formulas are:


The relevant formulas are:

( )22'n rrW= ( )33 rr2 M3 =( )12 rr ( )12 rr2 W = vertical load on specimen

M = torque

r1, r2 = inner, outer radius respectively

F1, F2 = loads on the two proving rings

L = distance along torque arm between proving ring contacts

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b-2011_2012_eg_shear strength.pdf

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