Posted in Geotechnical Engineering

## Engineering Geology in the Civil Engineering Curriculum

For this post I’m featuring an article by J. David Rogers, Professor and Karl F. Hasselmann Chair in Geological Engineering, Department of Geosciences and Geological and Petroleum Engineering, Missouri University of Science and Technology, which he entitled  Disappearing Practice Opportunities: Why Are Owners And Engineers Taking Increased Risks? What Can Be Done To Counter This Threat?  It’s been around a while but bears repeating, especially for one thing: why we need to restore engineering geology to the civil engineering curriculum.

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):

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)):

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.

Is 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.

### 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.

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.

Posted in Geotechnical Engineering

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

In 2016 we posted (and updated two years later) 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:

## 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.

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.

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.

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.

## 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.

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.