Vertical drains: the smear zone doubles the time

A clay layer 10 m thick, drained at one face, takes years to consolidate because the water has 10 m to travel. Install vertical drains at 1.2 m centres and the water has 0.6 m to travel — horizontally, which is also the direction in which the clay is usually more permeable. The improvement is dramatic and it is the reason prefabricated vertical drains are installed by the million.

The complication is that installing a drain requires pushing a mandrel through the clay, which remoulds an annulus of soil around it and drops its permeability. The water arriving from the far field must cross that annulus to reach the drain, and the annulus is the least permeable thing in the problem.

Every number in this article is recomputed from its inputs by a script that runs on each build of this site.

Hansbo’s solutionPermalink to “Hansbo’s solution”#

For radial consolidation to a central drain in a cylindrical unit cell, the average degree of consolidation is

$$U_h = 1 - \exp\!\left(-\frac{8 T_h}{\mu}\right), \qquad T_h = \frac{c_h\,t}{d_e^{2}}$$

where $d_e$ is the diameter of the unit cell and $\mu$ is a dimensionless factor that collects the whole geometry. Ignoring well resistance,

$$\mu = \ln\!\left(\frac{n}{s}\right) + \frac{k_h}{k_s}\ln s - \frac{3}{4}, \qquad n = \frac{d_e}{d_w}, \qquad s = \frac{d_s}{d_w}$$

with $d_w$ the equivalent drain diameter, $d_s$ the smear zone diameter and $k_h/k_s$ the ratio of undisturbed to smeared permeability. Setting $s = 1$ removes the smear zone and recovers the classical result, $\mu_0 = \ln n - 3/4$.

Note the structure: $\mu$ sits in the denominator of the exponent, so time to any given degree of consolidation is directly proportional to $\mu$. Anything that doubles $\mu$ doubles the time.

The unit cellPermalink to “The unit cell”#

Drains at 1.2 m spacing on a triangular grid give an equivalent cell diameter

$$d_e = 1.05 \times 1.2 = 1.260\ \text{m}$$

A standard band drain, 100 mm by 4 mm in section, has an equivalent circular diameter taken from its perimeter:

$$d_w = \frac{2(a + b)}{\pi} = \frac{2(0.100 + 0.004)}{\pi} = 0.0662\ \text{m}$$

so $n = 1.260/0.0662 = 19.03$.

What the smear zone doesPermalink to “What the smear zone does”#

Take the smear zone to extend to three drain diameters, $s = 3$, with the permeability inside it reduced threefold, $k_h/k_s = 3$. Both are middle-of-the-range values; the literature supports $s$ between about 2 and 6, and $k_h/k_s$ between about 2 and 10.

$$\mu_0 = \ln(19.03) - 0.75 = 2.196$$ $$\mu = \ln(6.3436) + 3\ln(3) - 0.75 = 1.8474 + 3.2958 - 0.75 = 4.393$$

The factor is 2.001 times larger. The smear term $\frac{k_h}{k_s}\ln s = 3.2958$ is on its own larger than the entire undisturbed factor, which is the point: a thin annulus of damaged clay across which all the water must pass dominates a flow path twenty times its thickness.

Time to 90%Permalink to “Time to 90%”#

Solving for $T_h$ at $U_h = 0.9$ gives $T_h = -\mu\ln(0.1)/8$, and with $c_h = 3.0$ m²/yr and $d_e^2 = 1.5876$ m²:

Four months against eight. On a project where a surcharge must stay in place until 90% consolidation is reached before construction can proceed, that is the difference between one season and two.

The error is also in the optimistic direction, which is the worst kind. A design that ignores smear predicts the ground will be ready sooner than it will be, and the discrepancy is discovered by settlement monitoring some months into a programme that has already been committed.

What to do about itPermalink to “What to do about it”#

Do not calibrate $c_h$ to fit the observed rate. This is the common response and it hides the problem: back-figuring $c_h$ from a monitored embankment with the smear term omitted returns a value roughly half the true one, which is then carried onto the next project where the spacing is different and the error changes size.

Do measure what you can. $s$ and $k_h/k_s$ are properties of the mandrel and the installation, not of the clay alone, so they transfer between sites with the same contractor and equipment better than they transfer between soils. A trial area with piezometers between drains resolves both, because the shape of the pore pressure decay is sensitive to them in a way the average settlement is not.

Do check whether well resistance matters too. The term omitted above is $\pi z(2l - z)k_h/q_w$, in which $l$ is the drainage length — the full drain length where it drains at one end, half the layer thickness where it drains at both. It becomes significant for long drains with low discharge capacity, over roughly 20 m, or where the drain is folded or crimped by large settlement. It adds to $\mu$ in the same way and therefore has the same proportional effect on time.

None of this is exotic. It is the same discipline as everywhere else on a site: the time factor is only as good as the drainage path assumed, and the parameters that enter a settlement prediction are stated assumptions with sources rather than constants. What makes drains a particularly sharp example is that the dominant term — $3.2958$ out of $4.393$ — comes from the installation method, which is the one thing the soil report has nothing to say about.