Effective stress: a worked profile, uplift and settlement

Terzaghi’s principle is one line:

$$\sigma' = \sigma - u$$

Strength and stiffness respond to $\sigma'$, not to $\sigma$. Soil does not know how much material is stacked above it; it knows how hard its grains are being pressed together, and water in the pores carries part of the load without pressing anything.

The line is easy. What makes it the source of so many wrong answers is that the two terms are of completely different kinds. Total stress is arithmetic on unit weights — nobody gets it wrong twice. Pore pressure is a boundary condition, and a boundary condition has to be established by going and looking. Assume it is hydrostatic and you will be right on most sites and catastrophically wrong on the rest.

Every number in this article is recomputed from its inputs by a script that runs on each build of this site, and the stress-profile diagram is checked against the same arithmetic.

The profilePermalink to “The profile”#

The water table stands at 2 m. Take $\gamma_w = 9.81$ kN/m³.

Total vertical stress accumulates downwards:

$$\sigma_v(11) = 2 \times 18.0 + 4 \times 20.2 + 5 \times 17.6 = 36.0 + 80.8 + 88.0 = 204.8\ \text{kPa}$$

Under hydrostatic conditions the pore pressure at 11 m is the weight of a 9 m column of water:

$$u = 9 \times 9.81 = 88.29\ \text{kPa}$$

and therefore

$$\sigma'_v(11) = 204.8 - 88.29 = 116.51\ \text{kPa}$$

So far this is bookkeeping. The three questions that follow all use the same subtraction, and all three turn on the pore pressure term.

Question one: what if the water is not hydrostatic?Permalink to “Question one: what if the water is not hydrostatic?”#

Suppose the clay is underlain by a confined aquifer whose piezometric head stands 3 m above ground level — an artesian condition, and not an exotic one. Then at the base of the clay the pore pressure is not the weight of a 9 m column but of a 14 m column:

$$u = (11 + 3) \times 9.81 = 137.34\ \text{kPa}$$ $$\sigma'_v(11) = 204.8 - 137.34 = 67.46\ \text{kPa}$$

The total stress has not changed by a single kilopascal. The effective stress has fallen by 49.05 kPa, which is 42.1% of it. Every strength that depends on effective stress falls with it, and nothing about the borehole log or the unit weights would have told you.

This is the whole argument for a piezometer. A water strike depth recorded by a driller is the level at which water was first seen, which in a layered profile is a different quantity from the head in any particular stratum. The two coincide in a uniform, freely draining deposit and nowhere else.

Question two: can the excavation be opened?Permalink to “Question two: can the excavation be opened?”#

Excavate the sand away down to 6 m, leaving 5 m of clay over the artesian aquifer. The clay is now a lid holding down water at 137.34 kPa, and the only thing holding it is its own weight:

$$\sigma_v = 5 \times 17.6 = 88.0\ \text{kPa}$$ $$F = \frac{88.0}{137.34} = 0.64$$

The base fails by uplift before anyone reaches the design level. Note what does not appear in that calculation: no strength, no friction angle, no undrained shear strength. Basal heave of this kind is a weight-against-pressure problem, and a clay’s strength enters only through the shear on the sides of the block being lifted, which for a wide excavation is negligible.

For a factor of safety of 1.1 the required clay thickness is

$$H = \frac{1.1 \times 137.34}{17.6} = 8.6\ \text{m}$$

which is more clay than the profile contains. The excavation is not made safe by digging carefully; it is made safe by relieving the head, which means pumping from the aquifer before the dig, not after the base lifts.

Question three: what does dewatering cost?Permalink to “Question three: what does dewatering cost?”#

Say the head is relieved and the water table is drawn down to 6 m for the works. The sand between 2 and 6 m drains, and its unit weight falls from 20.2 to a moist 18.6 kN/m³. At 11 m:

$$\sigma_v = 2 \times 18.0 + 4 \times 18.6 + 5 \times 17.6 = 198.40\ \text{kPa}$$ $$u = 5 \times 9.81 = 49.05\ \text{kPa}, \qquad \sigma'_v = 149.35\ \text{kPa}$$

Total stress went down by 6.4 kPa, because saturated soil was replaced by moist soil. Effective stress went up by 32.84 kPa, because the pore pressure fell by far more. That divergence is the point of the principle: the two quantities need not even move in the same direction.

An increase in effective stress on a clay is a consolidation load, and it is applied everywhere the drawdown reaches, not only under the works. At mid-clay, 8.5 m, effective stress rises from 97.035 to 129.875 kPa. For a normally consolidated layer with $C_c = 0.28$ and $e_0 = 0.92$:

$$s = \frac{C_c H}{1 + e_0} \log_{10}\frac{\sigma'_1}{\sigma'_0} = 0.7292 \times 0.1266 = 0.0923\ \text{m}$$

92 mm of settlement, delivered slowly, over whatever happens to be standing above the clay within the drawdown cone. Dewatering a site is not a neutral act of housekeeping; it is the application of a load, and the load is calculated the same way any other load is.

The patternPermalink to “The pattern”#

Three questions, three answers, one subtraction. What differs between them is only the pore pressure:

The spread in the last column is 82 kPa on a profile whose geometry and unit weights never changed. No laboratory test resolves that; it is decided by where the water stands and by nothing else.

Which is why the useful discipline is to write $u$ down as a separate, sourced line in every calculation — measured in which instrument, on which date, in which stratum — rather than letting it appear implicitly as “depth below the water table”. The subtraction is trivial. Knowing what to subtract is the engineering.

Once the effective stresses are established, the strength that depends on them is read through the stress state it was measured in, which is what a Mohr circle is for; and the classification that decides which of $C_c$, permeability or drainage even matters starts from the grading curve and the Atterberg limits.