Posted in Soil Mechanics

## More Uses for p-q Diagrams

In our last post on p-q diagrams we discussed their basic concept and application.  In this post we’ll expand on that for two applications: using it to estimate the friction angle and cohesion for multiple triaxial tests, and using it to plot the failure function.

## Processing Triaxial Test Results

The process of determining internal friction angle and cohesion from successive triaxial tests (i.e., those where the confining stress is successively increased) is well known.  In the case of two tests, using the standard $\sigma - \tau$ diagram, the tangent line between the two circles is unique (well, there are two of them, but the slopes and intercepts have opposite signs) as shown below.

If we use the p-q diagram, as we saw earlier, the process is even simpler, as two points have a unique line between them.

But what happens with three tests?  Mathematically there is no guarantee of a unique line, and given the nature of geotechnical testing it is the extraordinary lab which could hit such as result.  It’s also possible that the failure envelope is non-linear, as shown below.

So is there a way to at least get a decent approximation without guesswork or graphics skills?  The answer is “yes” and it involves using p-q diagrams in conjunction with a spreadsheet.  The mathematical concept behind this is here and we have an example to show how it is done.  The problem is taken from Tchebotarioff’s (1951) classic soil mechanics test.  The results of three triaxial tests are as follows, the failure stresses are in tsf:

 Test $\sigma_3$ $\sigma_1$ p q 1 0.2 0.82 0.51 0.31 2 0.4 1.6 1 0.6 3 0.6 2.44 1.52 0.92

We’ve taken the liberty of computing the p and q values for each test.  Now we can plot these in our spreadsheet.

We’ve also taken the liberty to use the spreadsheet’s “trend line” feature to plot a linear “curve fit” for the points.  The slope of the equation $m=tan \delta = sin \phi = 0.6041$, which yields both $\delta = 31.1^\circ$ and $\phi = 37.2^\circ$.  For the intercept $b = c\sqrt {1-\left (\tan(\delta)\right )^{2}} = 0.0001$, which means we can solve for the cohesion, but in this case the quantity is so small it’s probably best to assume that the cohesion is zero.

The $R^2$ value for this problem is very high, so the correlation is good.  We can use this parameter to determine whether we have a good correlation or not.  We can also use least-squares trend line analysis for non-linear failure envelopes, although when we consider the “kink” caused by preconsolidation this may not be as meaningful as one would like.

## Plotting the Failure Function

As mentioned earlier, the Mohr-Coulomb failure function is define in this way:

A little math transforms this into

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

or

$f=2\,q-2\,c\sqrt {1-\left (\sin(\phi)\right )^{2}}-2\,p\sin(\phi)$

Since $\delta$ and $c$ are known, this suggests that we can plot the failure function three-dimensionally.  Consider the case where $\delta = \frac{\pi}{8}$ and $c = 5$.  The p-q diagram for the failure envelope $f = 0$ is shown below.

If we plot the failure function three-dimensionally, we obtain this result:

The failure envelope of the previous diagram is the contour line which stops at the q-axis at around q = 4.6.  Values below this line are negative and values above it are positive.  Positive values of $f$ indicate failure and an illegal stress state.  The failure function is used extensively in finite element analyses like this one.

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 Soil Mechanics

## Computing Pore Water Pressure and Effective Stress in Upward (and Downward) Flow in Soil

Water flow through soil–and the whole subject of permeability–is one of those topics that tends to mystify students in undergraduate soil mechanics courses.  This article will deal with one type of flow–flow that is purely vertical, downward or upward–and show how it is possible to compute the pore water pressure and effective stress in soils with vertical water flow.

## Hydrostatic Case

We’ll start with the hydrostatic case, classic in the determination of effective stresses in many soil strata.  The pore water pressure is computed by the equation usually written in this way:

$u=\gamma_w z$

where $u$ is the pore water pressure, $\gamma_w$ is the unit weight of the water, and $z$ is the distance from the phreatic surface/water table, where by definition $z = 0$.

Let us write this equation more generally, thus

$\Delta u=\gamma_w \Delta z$

where $\Delta u$ is the change in the pore water pressure from some elevation 1 in the soil to some other elevation 2 in the soil, and $\Delta z$ is the change in elevation from point 1 to point 2.  As a condition, since $z$ is positive in the downward direction, $\Delta z$ is likewise positive in the downward direction.

With soil layers and total stress, we routinely “pile on” the stresses from layer to layer, because the unit weight of the soil changes.  For hydrostatic water, we usually don’t because the unit weight of the water is considered a constant.

## Vertically Flowing Water

With flowing water, although the unit weight of the water is a constant, the effect it has on effective stress changes.  For this case we can expand the previous equation to read as follows (from Verruijt, A., and van Bars, S. (2007). Soil Mechanics. VSSD, Delft, the Netherlands.):

$\Delta u=\gamma_w \Delta z\left( i + 1 \right)$

Note that we have added the hydraulic gradient into the mix, defined in the figure to the right.

This drawing shows a classic case of vertical, downward flow.  The coefficient of permeability $k$ can be computed using methods described in Department of the Army (1986) — Laboratory Soils Testing for granular soils.  However, we can also use this test–or problems based on this test–to consider the effect of the flowing water on the effective stress, which in turn leads us to consider the topic of soil boiling when the flow is upward.  The best way to see how this works is to consider an example.

## Upward Flow Example

Consider the permeameter setup below.  We will concentrate on the constant head permeameter on the left.  The soil sample is in grey, with a length L and an area A.

There is a distance H1 from the top of the soil sample to the surface of the water above it.  There is an additional distance H2 from that water surface to the water surface of the constant head tank.

Now consider an example with the following parameters:

• H1 = 0.5 m
• H2 = 2.5 m
• L = 2.5 m
• $\gamma_{sat} = 19 \frac{kN}{m^3}$

Compute the effective stress at a point halfway between the upper and lower surfaces of the soil sample.

First, we compute the total stress at the top of the soil, thus

$\sigma_t\mid_{z=0.5} = 0.5 m \times 9.8 \frac{kN}{m^3} = 4.9 kPa$

Because the total stress at this point is due to free water, the pore water pressure $u\mid_{z=0.5} = 4.9 kPa$, and thus $\sigma'_{vo} = 0$.

On the lower surface of the soil sample, the total stress is

$\sigma_t\mid_{z=3} = 0.5 m \times 9.8 \frac{kN}{m^3} + 2.5\times 19\frac{kN}{m^3} = 52.4 kPa$

The pore water pressure, however, is due to the free water that begins in the constant head tank and ends at the bottom surface of the soil, thus

$u\mid_{z=3} = \left( 2.5 + 0.5 + 2 \right)\times 9.8 \frac{kN}{m^3} = 49 kPa$

The effective stress at this point is 52.4 – 49 = 3.4 kPa.

So how do we compute the effective stress at the midpoint in the soil sample?  Let us revisit the equation

$\Delta u=\gamma_w \Delta z\left( i + 1 \right)$

And determine the pore water pressure at the midpoint.  We first want to compute the hydraulic gradient of the entire specimen, substituting yields

$\Delta u\mid_{z=3} = 49 - 4.9 = 44.1 kPa = 9.8 \times 2.5 \left( 1+i \right)$

Solving for the hydraulic gradient yields $i = 0.8$.

Now we substitute this result back into the equation, changing the distance $\Delta z = 1.25 m$.  Keeping in mind that positive z is downwards, we start from the top of the soil sample.  The change in pore water pressure from the surface is

$\Delta u\mid_{z=1.75} = 9.8 \times 1.25 \left( 1 + 0.8 \right) = 22.25 kPa$

Adding the pore water pressure at the soil’s upper surface yields u = 4.9 + 22.25 = 26.95 kPa.  The total stress at this point is

$\sigma_t\mid_{z=1.75} = 0.5 m \times 9.8 \frac{kN}{m^3} + 1.25\times 19\frac{kN}{m^3} = 28.65 kPa$

The effective stress is simply 28.65 – 26.95 = 1.7 kPa.  Since this is the middle of the layer, we would expect this stress to be the average of the effective stress at the top of the soil and the bottom, which in fact is the case.  But we can use this technique to compute the pore water pressure at any point in the soil.

• The hydraulic gradient is very high; in fact, the critical hydraulic gradient for this soil is 0.94, leaving us with a factor of safety of 1.17.  This is reflected in the very low effective stresses that result.  Had the critical hydraulic gradient been exceeded, the effective stresses would have been negative.  Many “textbook” problems of this nature actually exceed any sensible range of hydraulic gradients because they don’t compute it as a part of the solution.  The soil in this case is about to “boil” (or at least put significant upward pressure on the filter material.)
• Many students wonder why the formula for the hydraulic gradient $i=\frac{\Delta h}{\Delta l}$ cannot be applied directly.  The reason is simple: even with moving water, the direct hydrostatic effect due to gravity does not go away, and has to be considered.  Thus we have the term $\left( i + 1 \right)$ rather than just $i$.
• Had the flow been downward, the hydraulic gradient would have been negative, and the effective stresses would have increased relative to hydrostatic stresses rather than decreased.
• As long as the flow is vertical, this equation can be used with flow net type problems as well.
• The critical hydraulic gradient equation can be derived using this equation.  As mentioned above, the critical hydraulic gradient is reached when the effective stresses in the soil are zero.  Assuming that we’re starting at the upper surface where the effective stress is zero, at the lower surface of the soil sample (or soil element in a flow net) the effective stress is zero when the total stress and pore water pressure is zero, or

$\gamma_{sat} \Delta z = \gamma_w \Delta z\left( i + 1 \right)$

Solving for $i_{crit}$ yields

$i_{crit} = \frac{\gamma_{sat}}{\gamma_w} - 1$

which is in fact the case.

• We can also solve the problem to determine the hydraulic head at a point in the soil.  We start by modifying our equation as follows:

$\Delta h=\Delta z\left( i + 1 \right)$

For this problem, at the centre of the layer, we would start by solving for the hydraulic gradient, or

$5 - 0.5=2.5\left( i + 1 \right)$

where the left hand side represents the total change in hydraulic head from the upper to the lower surface of the soil.  As before $i = 0.8$.

Now we use the equation directly to solve for the hydraulic head at the centre of the layer, thus

$\Delta h=1.25\left( 0.8 + 1 \right) = 2.25 m$

This must be added to the hydraulic head already at the surface, or 2.25 + 0.5 = 2.75 m.  By changing the value of $\Delta z$ we can compute this change at any point and add it to the head at the upper surface.

## 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.