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 = 3 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

$\sigma_t\mid_{z=3.5} = 0.5 m \times 9.8 \frac{kN}{m^3} + 3\times 19\frac{kN}{m^3} = 61.9 kPa$

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

$u\mid_{z=3.5} = \left( 2.5 + 0.5 + 3 \right)\times 9.8 \frac{kN}{m^3} = 58.8 kPa$

The effective stress at this point is 61.9 – 58.8 = 3.1 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

$\Delta u\mid_{z=3.5} = 58.8 - 4.9 = 53.9 kPa = 9.8 \times 3 \left( 1+i \right)$

Solving for the hydraulic gradient yields $i = 0.833$.

Now we substitute this result back into the equation, changing the distance $\Delta z = 1 m$.  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

$\Delta u\mid_{z=2} = 9.8 \times 1.5 \left( 1 + 0.833 \right) = 26.95 kPa$

Adding the pore water pressure at the soil’s upper surface yields u = 4.9 + 26.95 = 31.85 kPa.  The total stress at this point is

$\sigma_t\mid_{z=2} = 0.5 m \times 9.8 \frac{kN}{m^3} + 1.5\times 19\frac{kN}{m^3} = 33.4 kPa$

The effective stress is simply 33.4 – 31.85  = 1.55 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.

We can show this by using a simpler method, by linearly interpolating between the properties of the top of the soil and the bottom.  Since the point of interest is in the middle, we can use a simple average of the properties of the top and bottom.  The averages are as follows:

Total Stress: $\frac{4.9 + 61.9}{2} = 33.4 kPa$

Pore Water Pressure: $\frac{4.9+58.8}{2} = 31.85 kPa$

Effective Stress: $\frac{0 + 3.1}{2} = 1.55 kPa$

which are the same answers without some of the computational effort.

• The hydraulic gradient is very high; in fact, the critical hydraulic gradient for this soil is 0.833, leaving us with a factor of safety of 1.13.  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.

## Author:Don Warrington

This site uses Akismet to reduce spam. Learn how your comment data is processed.