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.