Posted in Geotechnical Engineering, Soil Mechanics

p-q Diagrams and Mohr-Coulomb Failure

Students and practicioners of soil mechanics alike are used to seeing triaxial test results that look like this (from DM 7.01):

DM7 Triaxial Test Mohr-Coulomb Diagrams

Ideally, the Mohr-Coulomb failure line should be straight, but with real soils it doesn’t have to be that way.  With the advent of finite element analysis we also have the failure function to consider, thus (from Warrington (2016)):

Dissertation Presentation_Page_28

All of these involve constructing (or using) a line which is tangent to a circle at failure.  This can be confusing to understand completely.  The biggest problem from a “newbie” standpoint is that the maximum shear defined by the circle of stress (its radius) and the failure shear stress defined by the intersection of the circle with the Mohr-Coulomb failure envelope are not the same.

nhi16072-1Is there a better graphical way to represent the interaction of stresses with the Mohr-Coulomb failure criterion?  The answer is “yes” and it involves the use of p-q diagrams.  These have been around for a long time and are used in such things as critical state soil mechanics and stress paths.  A broad explanation of these is found in our new publication, Geotechnical Site Characterization.  The purpose of this article is to present these as a purely mathematical transformation of the classic Mohr-Coulomb diagram.  This is especially important since their explanation is frequently lacking in textbooks.

The Basics

Consider the failure function, which is valid throughout the Mohr-Coulomb plot.  It can be stated as follows:

f=\sigma_{{1}}-\sigma_{{3}}-2\,c\cos(\phi)-\left (\sigma_{{1}}+\sigma_ {{3}}\right )\sin(\phi)

(The main difference between the two formulations is multiplication by 2; the failure function can either be diametral or radial relative to Mohr’s Circle.  With a purely elasto-plastic model, the results are the same.)

Now let us define the following terms:

p=1/2\,\sigma_{{1}}+1/2\,\sigma_{{3}}

q=1/2\,\sigma_{{1}}-1/2\,\sigma_{{3}}

We should also define the following:

\sin(\phi)=\tan(\delta)

The physical significance of the last one is discussed in this post.  In any case we can start with \phi and solve for \delta or vice versa.  Solving for \phi and substituting this and the equations for p and q into the failure functions yields

f=2\,q-2\,c\sqrt {1-\left (\tan(\delta)\right )^{2}}-2\,p\tan(\delta)

For the failure line, f = 0 .  Let us set the p axis as the abscissa (x-axis) and the q axis as the ordinate (y-axis.)  For the failure line, if we substitute for f and solve for q, we have

q = p\tan(\delta) + c\sqrt {1-\left (\tan(\delta)\right )^{2}}

This is a classic “slope-intercept” form like y = mx + b , where in this case q = mp + b , m = \tan(\delta) and b = c\sqrt {1-\left (\tan(\delta)\right )^{2}} .  A sample plot of this kind is shown below.

p-q-diagram-illlustration

Some Observations

  1. For the case of a purely cohesive soil, where \phi = \delta = 0 , the failure envelope is horizontal, just like with a conventional Mohr-Coulomb diagram.
  2. For the case of a purely cohesionless soil, where c = 0 , the y-intercept is in both cases through the origin.
  3. The two diagrams are thus very similar visually, it’s just that the p-q diagram eliminates the circles and tangents, reducing each case to a single point.

Examples of Use

Drained Triaxial Test in Clay

Consider the example of a drained triaxial test in clay with the following two data points:

  1. Confining Pressure = 70 kPa; Failure Pressure = 200 kPa.
  2. Confining Pressure = 160 kPa; Failure Pressure = 383.5 kPa.

Determine the friction angle and cohesion using the p-q diagram.

We first start by computing p and q for each case as follows:

p_1 = 200/2+70/2 = 135\,kPa

p_2 = 383.5/2 + 160/2 = 271.75\,kPa

q_1 = 200/2-70/2 = 65\,kPa

q_2 = 383.5/2 - 160/2 = 111.75\,kPa

The slope is simply

m = \frac {q_2 - q_1}{p_2 - p_1} = \frac {111.75 - 65}{271.5 - 135} = 0.342 = \tan(\delta)

from which

\delta = 18.9^o

\phi = sin^{-1}(tan(\delta)) = sin^{-1}(0.342) = 20.03^o

b = q - mp = 65 - 0.342 \times 135 = 18.83 (using values from the first point, just as easy to use the second one.)

b = c\sqrt {1-\left (\tan(\delta)\right )^{2}} = c \sqrt {1-0.342^{2}} = 0.94 c

b = 18.83 = 0.94 c

c = 20.03\,kPa

Use of this method eliminates the need to solve two equations in two unknowns, and the repetition of the quantity tan(\delta) makes the calculations a little simpler.  When c = 0 , the calculations are even simpler, as p_1 = q_1 = 0 .

Stress Paths

As mentioned earlier, p-q diagrams are commonly used with stress paths.  An example of this from DM 7.01 is shown below.

DM7 Triaxial Test Stress Path Diagram

We note that p and q are defined here exactly as we have them above.  (That isn’t always the case; examples of other formulations of the p-q diagram are here.  We should note, however, that for this diagram \phi" = \delta )  With this we can track the stress state of a sample from the start (where the deviator stress is zero, at the start of the triaxial test) around to its various points of stress.

As an example, consider the stress path example from Verruijt, A., and van Bars, S. (2007). Soil Mechanics. VSSD, Delft, the Netherlands.  The basic data from Test 1 are below:

\sigma_3 Deviator Stress Pore Water Pressure
40 0 0
40 10 4
40 20 9
40 30 13
40 40 17
40 50 21
40 60 25

Using the p-q diagram and performing some calculations (which are shown in the spreadsheet Stress Paths Verruijt Example)  the stress paths can be plotted as follows:

It’s worth noting that the q axis is unaffected by the drainage condition because the pore water pressures cancel each other out.  Only the p-axis changes.

Conclusion

The p-q diagram is a method of simplifying the analysis of triaxial and other stress data which are commonly used in soil mechanics.  It can be used in a variety of applications and solve a range of problems.

 

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Posted in Geotechnical Engineering

An Updated Version of Our Mohr’s Circle Routine Available, with Documentation

In 2016 we posted about two- and three-dimensional Mohr’s Circle problems and their solution using a strictly linear algebra solution.  We’ve done two things recently to update that: we’ve resolved some of the limitations of the original method (especially with the eigenvectors/direction cosines) and we’ve put the routine online for your use.  Both can be accessed here:

  1. Mohr’s Circle and Linear Algebra (the documentation)
  2. Online Routine

Image above from Verruijt, A., and van Bars, S. (2007). Soil Mechanics. VSSD, Delft, the Netherlands.

Posted in Geotechnical Engineering, Soil Mechanics, STADYN, TAMWAVE

Relating Hyperbolic and Elasto-Plastic Soil Stress-Strain Models

Note: this post has an update to it with a more rigourous and complete treatment here.

It is routine in soil mechanics to attempt to use continuum mechanics/theory of elasticity methods to analyse the stresses and strains/deflections in soil.  We always do this with the caveat that soils are really not linear in their response to stress, be that stress axial, shear or a combination of the two.  In the course of the STADYN project, that fact became apparent when attempting to establish the soil modulus of elasticity.  It is easy to find “typical” values of the modulus of elasticity; applying them to a given situation is another matter altogether.  In this post we will examine this problem from a more theoretical/mathematical side, but one that should vividly illustrate the pitfalls of establishing values of the modulus of elasticity for soils.

Although the non-linear response of soils can be modelled in a number of ways, probably the most accepted method of doing so is to use a hyperbolic model of soil response.  This is illustrated (with an elasto-plastic response superimposed in red) below.

Hyberbolic-Stress-Strain

The difficulties of relating the two curves is apparent.  The value E1 is referred to as the “tangent” or “small-strain” modulus of elasticity.  (In this diagram axial modulus is shown; similar curves can be constructed for shear modulus G as well.)  This is commonly used for geophysical methods and in seismic analyses.

As strain/deflection increases, the slope of the curve decreases continuously, and the tangent modulus of elasticity thus varies continuously with deflection.  For larger deflections we frequently resort to a “secant” modulus of elasticity, where we basically draw a line between the origin (usually) and whatever point of strain/deflection we are interested in.

Unfortunately, like its tangent counterpart, the secant modulus varies too.  The question now arises: what stress/strain point do we stop at to determine a secant modulus?  Probably a better question to ask is this: how do we construct an elasto-plastic curve that best fits the hyperbolic one?

One solution mentioned in the original study is that of Nath (1990), who used a hardening model instead of an elastic-purely plastic model.  The difference between the two is illustrated below.

Elasto-Plastic Response

Although this has some merit, the elastic-purely plastic model is well entrenched in the literature.  Moreover the asymptotic nature of the hyperbolic model makes such a correspondence “natural.”

Let us begin by making some changes in variables.  Referring to the first figure,

y=\frac{\sigma}{\sigma_1-\sigma_3}=\frac{\sigma}{\sigma_0}

and

x = \epsilon

Let us also define a few ratios, thus:

A_1 = \frac {E_1}{\sigma_0}

A_2 = \frac {E_2}{\sigma_0}

A = \frac {E_2}{E_1}

Substituting these into the hyperbolic equation shown above, and doing some algebra, yields

y=\frac{x A_1}{1+x A_1}

One way of making the two models “close” to each other is to use a least-squares (2-norm) difference, or at least minimising the 1-norm difference.  To do the latter with equally spaced data points is essentially to minimise the difference (or equate if possible) the integrals of the two, which also equates the strain energy.  This is the approach we will take here.

It is easier to equate the areas between the two curves and the \sigma_0 line than to the x-axis.  To do this we need first to rewrite the previous equation as

y'=1-\frac{x A_1}{1+x A_1}

Integrating this with respect to x from 0 to some value x_1 yields

A_{hyp} = \frac{ln\left( 1+x_1 A_1 \right)}{A_1}

Turning to the elastic-plastic model, the area between this “curve” and the maximum stress is simply the triangle area above the elastic region.  Noting that

E_2 = \frac {\sigma_0} {x_2} ,

employing the dimensionless variables defined above and doing some additional algebra yields the area between the elastic line and the maximum stress, which is

A_{ep} = \frac {1}{2 A A_1}

Equation the two areas, we have

ln\left( 1+x_1 A_1 \right) - \frac {1}{2 A} = 0

With this equation, we have good news and bad news.

The good news is that we can (or at least think we can) solve explicitly for A , the ratio between the elastic modulus needed by elasto-plastic theory and the small-deflections modulus from the hyperbolic model.  The bad news is that we need to know A_1 , which is the ratio of the small deflections modulus to the limiting stress.  This implies that the limiting stress will be a factor in our ultimate result.  Even worse is that x_1 is an input variable, which means that the result will depend upon how far we go with the deflection.

This last point makes sense if we consider the two integrals.  The integral for the elasto-plastic model is bounded; that for the hyperbolic model is not because the stress predicted by the hyperbolic model is asymptotic to the limiting stress, i.e., it never reaches it.  This is a key difference between the two models and illustrates the limitations of both.

Some additional simplification of the equation is possible, however, if we make the substitution

x_1 = n x_2

In this case we make the maximum strain/deflection a multiple of the elastic limit strain/deformation of the elasto-plastic model.  Since

x_2 = \frac {sigma_0}{E_2} = \frac {1}{A_2} = \frac {1}{A A_1}

we can substitute to yield

ln\left( 1+\frac{n}{A} \right) - \frac{1}{2A} = 0

At this point we have a useful expression which is only a function of n and A .  The explicit solution to this is difficult; the easier way to do this is numerically.  In this case we skipped making an explicit derivative and use regula falsi to solve for the roots for various cases of n .  Although this method is slow, the computational time is really trivial, even for many different values of n .  The larger value of n , the more deflection we are expecting in the system.

The results of this survey are shown in the graph below.

hyperb

The lowest values we obtained results for were about n = \frac{x_1}{x_2} = 0.75 .  When n = \frac{x_1}{x_2} = 1 , it is the case when the anticipated deflection is approximately equal to the “yield point.”  For this case the ratio between the elasto-plastic modulus and the small-strain hyperbolic modulus is approximately 0.4.  As one would expect, as n increases the elasto-plastic system becomes “softer” and the ratio A = \frac {E_2}{E_1} likewise decreases.  However, as the deflection increases this ratio’s increase is not as great.

To use an illustration, consider pile toe resistance in a typical wave equation analysis.  Consider a pile where the quake (x_2 ) is 0.1″.  Most “traditional” wave equation programs estimate the permanent set per blow to be the maximum movement of the pile toe less the quake.  In the case of 120 BPF–a typical refusal–the set is 0.1″, which when added to the quake yields a total deflection of 0.2″ of a value of n = 2.  This implies a value of A = 0.2139950 .  On the other hand, for 60 BPF, the permanent set is 0.2″, the total movement is 0.3″, and n = 3, which implies a value of A = 0.1713409.  Cutting the blow count in half again to 30 BPF yields n = 5 or A = 0.1383195.  Thus, during driving, not only does the plastic deformation increase, the effective stiffness of the toe likewise decreases as well.

Based on all this, we can draw the following conclusions:

  1. The ratio between the equivalent elasto-plastic modulus and the small-strain modulus decreases with increasing deflection, as we would expect.
  2. As deflections increase, the effect on the the equivalent modulus decreases.
  3. Any attempt to estimate the shear or elastic modulus of soils must take into consideration the amount of plastic deformation anticipated during loading.  Use of “typical” values must be tempered by the actual application in question; such values cannot be accepted blindly.
  4. The equivalence here is with hyperbolic soil models.  Although the hyperbolic soil model is probably the most accurate model currently in use, it is not universal with all soils.  Some soils exhibit a more definite “yield” point than others; this should be taken into consideration.
Posted in Geotechnical Engineering, STADYN

Lateral Earth Pressure Coefficients for Beta Methods in Sands

In our last post we considered some basic concepts behind beta methods for determining beta coefficients for estimating shaft friction for piles in sands.  The idea is that the unit friction along the surface of the pile can be determined at any point by the relationship

f_s = \beta \sigma'_{vo}

where f_s is the unit shaft friction, \sigma'_{vo} is the vertical effective stress, and \beta is the ratio of the two, which can be further broken down as follows:

\beta = K tan \phi

where K is the lateral earth pressure coefficient and \phi is the internal friction angle of the soil.  Our last post showed that, when compared with empirically determined values of \beta , values of K determined from more conventional retaining wall considerations are not adequate to describe the interaction between the shaft of the pile and the soil.

Needless to say, there has been a good deal of research to refine our understanding of this relationship.  Also, needless to say, there is more than one way to express this relationship.  The formulation we will use here is that of Randolph, Dolwin and Beck (1994) and Randolph (2003), and was recently featured in Han, Salgado, Prezzi and Zaheer (2016).  The basic form of the lateral earth pressure equation is as follows:

K = K_{min} + (K_{max} - K_{min}) e^{-\mu \frac {L-z}{d}}

Let’s start on the right end of the equation; the exponential term is a way of representing the fact that the maximum shaft friction (with effective stress taken into account) is just above the pile toe and decays above that point to the surface of the soil.  This was first proposed by Edward Heerema (whose company was instrumental in the development of large steam and hydraulic impact hammers) in the early 1980’s.  (For another paper of his relating to the topic, click here.)

In any case the variables in the exponential term are as follows:

  • \mu = rate of exponential decay, typically 0.05
  • L = embedded length of pile into the soil
  • z = distance from soil surface to a given point along the pile shaft.  At the pile toe, L = z and L-z = 0 , and the exponential term becomes unity.
  • d = “diameter” of the pile, more commonly designated as B in American textbooks.

K_{min} is the minimum lateral earth pressure coefficient.  It, according to Randolph, Dolwin and Beck (1994) “can be linked to the active earth pressure coefficient.” Randolph (2003) states that its value lies in the range 0.2-0.4. We stated in our previous post that

K_a = \frac {1 - sin \phi} {1 + sin \phi}

How do these two relate?  Although in the last post we produced extensive parametric studies on these, a simpler representation is to compare the active earth pressure coefficient with Jaky’s at-rest coefficient, which is done below.

Beta Image 4

The at-rest coefficient from Jaky is in blue and the active coefficient from Rankine is in red.  The range of 0.2 < K_a < 0.4 approximately translates into 25^\circ < \phi < 45^\circ , which is a wide range for granular soils but reasonable.

That leaves us K_{max}.  Randolph, Dolwin and Beck (1994) state that

K_{max} = S_t N_q

N_q , of course, is the bearing capacity factor at the toe.  It may seem odd to include a toe bearing capacity factor in a shaft equation, but keep in mind that cavity expansion during pile installation begins (literally) with an advancing toe.  Typically 8 < N_q < 40 depending upon whether the sand is loose (low end) or dense (high end.)  S_t “is the ratio of the radial effective stress acting in the vicinity of the pile tip at shaft failure to the end-bearing capacity.”  Values for S_t vary somewhat but generally centre around 0.02.  This in turn implies that 0.16 < K_{max} < 0.8 .  Inspection of the complete equation for K shows that, if L = z and the exponential term is at its maximum, K_{min} cancels out and the range of K_{max} is a range for K .

Comparing this result to the graph above, for larger values of \phi these values of K are greater than those given by Jaky’s Equation, which is what we were looking for to start with.  To compute \beta , we obviously will need to multiply this by tan \phi (or tan \delta ).  For, say, \delta = 35^\circ , this leads to \beta_{max} = 0.8 \times tan 35^\circ = 0.56 .  By way of comparison, using Jaky’s Equation for K, \beta = (1 - sin 35^\circ) tan 35^\circ = 0.30 .

From this we have “broken out” of Burland’s (1973) limitation on \beta , which was useful for him (and will be useful to us) for some soils but creates problems with higher values of \phi   Although some empirical methods indicate higher values for \beta , if we consider variations in S_t and other factors, this differential can be minimised, and in any case this is not a rigourous excercise but a qualitative one.

One thing we should further note–and this is important as we move forward–is that there is more than one way to compute K_{max} .  Randolph (2003) states that, when CPT data is available, it can be computed as follows for open-ended piles:

K_{max} = 0.01 \frac {q_c}{\sigma'_{vo}}

where q_c is the cone tip resistance.  Randolph (2003) recommends the coefficient be increased to 0.015 for closed-ended piles.  Making generalisations from this formulation is more difficult than the other, but the possibility of using this in conjunction with field data is attractive indeed.

At this point we have a reasonable method of computing \beta coefficients.  However, we still have the issue of clay soils to deal with, and this will be done in a subsequent post.

References

In addition to those previously given, we add the following:

  • Han, F., Prezzi, M., Salgado, R. and Zaheer, M., (2016), “Axial Resistance of Closed-Ended Steel-Pipe Piles Driven in Multilayered Soil“, Journal of Geotechnical and Geoenvironmental Engineering, DOI: 10.1061/(ASCE)GT.1943-5606.0001589.
  • Randolph, M., Dolwin, J., Beck, R. 1994, ‘Design of driven piles in sand’, GEOTECHNIQUE, 44, 3, pp. 427-448.
  • Randolph, M. 2003, ‘Science and empiricism in pile foundation design’, GEOTECHNIQUE, 53, 10, pp. 847-875.

 

Posted in Geotechnical Engineering, STADYN

A First Look at Estimating Beta Factors for Determining Pile Shaft Capacity of Driven Piles

In the last posting about STADYN, we put forth considerations for interface elements between the pile shaft and the soil.  Before we formally incorporate these into the model (or whether we will incorporate them or not) some consideration of how the interface actually works.  We will start those considerations by looking at methods by which the static capacity of driven piles is computed, and specifically the so-called “beta” methods which are used for cohesionless and sometimes cohesive soils.

Beta methods assume that the shaft resistance of the pile is a function of the effective stress of the soil along the pile shaft.  They assume that the horizontal stress that results from the vertical stress acts perpendicular to the surface of the pile.  The pile surface thus acts like a block on a surface with some kind of Coulombic friction acting against the downward settlement of the pile.  The beta coefficient is the ratio between the vertical effective stress and the horizontal friction on the pile, or

f_s = \beta \sigma'_o

\beta is in turn broken down into two components: the lateral earth pressure coefficient, which is the ratio between the horizontal and vertical stresses,

K = \frac {\sigma_h}{\sigma'_o}

and the coefficient of friction, or

\mu = tan \phi

We put these together to yield

\beta = K tan \phi

At this point let’s make two assumptions.  The first is that the lateral earth pressure coefficient is in fact the at-rest lateral earth pressure coefficient.  (For some discussion of this, you can view this slide presentation.)  The second is that the friction angle between the pile and the soil is in fact the same as the soil’s internal friction angle.  If we use Jaky’s formula for the at-rest condition, these assumptions yield

\beta = \left (1-\sin(\phi)\right )\tan\phi

The various components of this equation are plotted below.

Beta Image 1

The three lines are as follows:

  • K_o = 1 - sin \phi is in red.
  • tan \phi is in blue.
  • \beta is in green.

It’s interesting to note that, as K_o increases, tan \phi decreases, and so \beta is within a surprisingly narrow range of values.  This plot is similar to one shown in Burland (1973), which we will discuss later.

If this were the case in practice, estimating \beta would be a straightforward proposition.  We’ll take two examples to show that this is not the case.

Let’s start with the Dennis and Olson Method for cohesionless soils, which is described here.  To arrive at \beta they do the following:

  1. They add a depth factor, which we will not consider.  Depth factors and critical lengths are common in static methods, but they are not well documented in the field.
  2. They assume K_o = 0.8 if their values for friction angle are used.
  3. They vary the friction angle from 15-35 degrees depending upon the type of soil.

Leaving out the depth factor, for this method \beta ranges from 0.21 to 0.56.  This is a considerably wider variation than is indicated above.  Since the depth factor is frequently greater than unity, this range is even larger.

An easier way to see this is to consider the method of Fellenius.  His values for \beta are as follows:

  • 0.15-0.35 for clay
  • 0.25-0.50 for silt
  • 0.30-0.90 for sand
  • 0.35-0.80 for gravel

Again the range of values is greater than the figure above would indicate.  Why is this?

Although it’s tempting to use a straight empirical approach, let’s back up and consider the structure of the basic equation about and the assumptions behind it.  There are several ways we can alter these equations in an attempt to match field conditions better by considering these assumptions and seeing what changes might be made.

The Two Friction Angles Aren’t the Same

The first one is suggested by the notation in Dennis and Olson: the internal friction angle of the soil and that of the soil-pile interface are not the same.  Retaining wall theory (when it considers friction) routinely makes this assumption; in fact, the ratio \frac {\delta}{\phi} routinely appears in calculations.  Let us rewrite the equation for \beta as

\beta = (1 - sin \phi) tan \delta

and be defining the ratio

m = \frac {\delta}{\phi}

we have

\beta = (1 - sin \phi) tan (m \phi)

If we plot this in a three-dimensional way, we get the following result.

Beta Image 2.png

\beta is the vertical axis; m is varied from 0.25 to 1.75.  The results show that, for a given \phi , if we increase m we will increase \beta , and this increase is much more pronounced at higher values of \phi .

Although it’s certainly possible to have very high values of \delta = m \phi , as a practical matter in most cases m < 1.  Nordlund’s Method, for example indicates that m > 1 only with tapered piles, where a tapered pile face induces some compression in the soil in addition to shear.  In any case is m < 1 this will tend to depress values of \beta .  We should also note that using a ratio m does not mean that it will be a constant for any given soil.  This is especially true if \phi = 0 , where a multiplier is meaningless and we should have recourse to an additive term as well.

Jaky’s Equation Doesn’t Apply, or At-Rest Earth Pressure Conditions Are Not Present

Another assumption that can be challenged is that Jaky’s Equation doesn’t apply, or we don’t have at-rest earth pressure conditions.  Although Jaky’s Equation has done well, it is certainly not the last word on the subject, especially for overconsolidated soils (which we will discuss below.)  To try to “cover our bases” on this, let’s consider a range of lateral earth pressure coefficients by assuming that Jaky’s Equation is valid for the at-rest condition and that we need to somehow vary between some kind of active state and passive state.  The simplest way to do this is to assume Rankine’s conditions with level backfill, which just happens to be identical to Mohr-Coulomb relationships between confining and driving stresses.  (OK, it’s not all luck here…)  Thus,

K_a = \frac {1-sin\phi}{1+sin\phi}

and

K_p = \frac {1+sin\phi}{1-sin\phi}

Let us also define an active-passive factor called actpas, where actpas = -1 for the active state, 0 for the at-rest state and 1 for the passive state.  We then plot this equation

\beta = K(\phi,actpas) tan \phi

below.  Since we only have K values for three values of actpas, we’ll use a little Lagrangian interpolation in an attempt to achieve a smooth transition between the states.

Beta Image 3

We note from this the following:

  1. The dip in \beta for the high values of \phi and  -1 < actpas < 0 (states tending towards the active) may be more a function of the interpolation than the physics.  OTOH, if we look at NAVFAC DM 7.02, Chapter 3, Figure 1, we see a dip between the at-rest and active states for dense sands, which is what we would expect at higher values of \phi .
  2. Values of \beta for the active case show little variation.  Given that driven piles are subject to cavity expansion during installation, one would expect some passivity in earth pressures.  Drilled shafts are another story; however, if we look, for example, at O’Neill and Reese, values for \beta can certainly range higher than one sees with the active states above.  Bored piles, however, are beyond the scope of this discussion.
  3. For low values of \phi , there is little variation between the three states.
  4. If we compare these values with, say, those of Fellenius or Dennis and Olson, we cannot say that the fully passive state applies for most reasonable values of \phi , undrained or drained.  (Values in Nordlund, however, indicate higher values of K for larger displacements, approaching full passivity for large displacement piles.  Another look at this issue is here.)

Conclusion

If we compare the results we obtain above with empirical methods for determining \beta , we see that none of the variations shown above really allows us to match the theory we’ve presented with the empirical methods we’ve described (and others as well.)  As a general rule, \delta < \phi or m < 1 , so it’s safe to conclude that our assumption that the K can be determined using Jaky’s Equation only results in values of \beta that are too low.

It’s tempting to simply fall back on an empirical value for \beta , but for finite element analysis a more refined approach seems appropriate.  In subsequent posts we’ll look at such an approach, along with the issue of applying \beta methods to cohesive soils as well as cohesionless ones.

References

In addition to those in the original study, the following reference is mentioned here:

  • Burland, J.B. (1973) “Shaft friction of piles in clay – A simple fundamental approach.” Ground Engineering 6(3):30-42, January.