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 TAMWAVE

TAMWAVE: Cavity Expansion Theory and Soil Set-Up

One of the things that was attempted in the TAMWAVE project is the use of cavity expansion theory to estimate soil set-up in cohesive soils.  Doing this, however, brought some complications that need some explanation.

Cavity expansion theory is basically the study of what happens when one body expands inside of another.  When this takes places, additional radial stresses (most analyses center around a cylinder or sphere) are generated.  In the case of driven piles, these additional stresses add to the pile’s resistance to load.  It can be argued that cavity expansion is one of the key advantages of driven piles.  In the case of drilled piles such as drilled shafts or auger-cast piles, this does not take place, as the soil is removed either before or during the actual pile installation.  The result of this is that driven piles, for the same length and diameter as a corresponding drilled pile, a driven pile will have a greater resistance to load (ultimate capacity.)

Applying cavity expansion theory to piles has a long history and is detailed in documents such as Randolph, Carter and Wroth (1979) and Yu and Houlsby (1991).  Our particular interest is with clays because, in addition to the changes in the soil from cavity expansion, the pore water pressure increases.  This is the primary (but not the only) reason why the SRD (soil resistance to driving) in cohesive soils is significantly less than the ultimate capacity; this fact inevitably complicates drivability studies.

The increase in pore water pressure is a dynamic phenomenon; it experiences a sudden increase during driving and then gradually dissipates after installation.  How gradually the latter takes place depends on many factors such as the permeability of the soil.  Study of this phenomenon is well represented in the literature; however, for the TAMWAVE project it is not really of interest.  The primary interest here is the value of SRD after the immediate increase of pore water pressures during driving.

Both cavity expansion theory and practice show that excess pore water pressures can easily exceed the effective stresses.  In principle, considering that we have established the beta (effective-stress dependent) method for both cohesionless and cohesive soils, this can mean a complete loss of SRD.  Although dramatic drops in SRD are not unknown, for most piles this is unrealistic.  The reason for this is that, like the dissipation, the build-up of pore water pressures is a dynamic phenomenon, albeit in a much smaller time frame.  Dissipation, hindered as it is by the low permeability of cohesive soils, begins immediately.  Pore water pressures (along with any other stresses induced by cavity expansion) also vary with the distance from the pile.

The result of all this is that prediction of both the increase of pore water pressure and its effect on the SRD of the pile during driving is a complicated phenomenon that is not completely represented by closed-form cavity expansion based solutions.  For a project such as this, what we need is something that will give a reasonable representation of soil set-up for cohesive soils.  To accomplish this, we stick with computing the excess pore water pressure, but with a different methodology.  We assume the following:

1. The basic validity of our effective-stress based beta methods of shaft friction calculation.
2. All of the decrease from static ultimate capacity to SRD takes place due to pore water pressure increase.
3. The excess pore water pressure increases affect the effective stress used to compute the shaft friction.
4. Only those pile segments under the water table are subject to this analysis.

Assuming all that, for a soil set-up factor (from this source, loosely adapted) $S_r$, the excess pore water pressures that affect the effective stress (which in turn determine the shaft friction) are computed by the equation

$\Delta u = \sigma'_{vo}\left( 1 - \frac{1}{S_r} \right)$

This gives identical results to those when the $S_r$ are applied directly.

In practice this phenomenon is still subject to investigation.  Some of the research involves use of numerical methods (such as finite element methods) to simulate cavity expansion effects.  This is doubtless an advance over the closed-form solutions of the past, and a necessity given the complexity of the physics of the problem.  Empirical methods are also still being developed, such as are documented by Wang, Verma and Steward (2009).

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 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.