Module 10 · Lesson 10.1

How much, and how fast

Stress history sets the magnitude; the drainage path sets the clock.

Why this matters

Immediate settlement in Module 9 happened as the load was applied. In clay, most of the movement has not happened yet. Water has to be squeezed out before the soil skeleton can compress, and that can take years — long enough that the building is finished, occupied, and being blamed on the contractor before the settlement is complete.

By the end of this lesson you should be able to

Compute primary consolidation settlement from Cc, Cr and the stress history
Explain the effect of the preconsolidation stress
Predict the rate of settlement from the time factor

Two independent questions, two independent sets of parameters, and confusing them is a common source of nonsense.

How much comes from compressibility — the compression index C_c on the virgin line, the recompression index C_r below the preconsolidation stress σ′_c, and where the loading sits relative to σ′_c.

How fast comes from the coefficient of consolidation c_v and the drainage path. A soil can be very compressible and very quick, or barely compressible and very slow. Nothing about the magnitude tells you the rate.

Primary consolidation settlement, crossing the preconsolidation stress

s_{c} = \frac{H}{1 + e _{0}} [C_{r} lo g_{10} \frac{σ _{c}^{'}}{σ _{0}^{'}} + C_{c} lo g_{10} \frac{σ _{f}^{'}}{σ _{c}^{'}}]

What it calculates: Final settlement of a clay layer under a stress increase

H: Layer thickness (m)
e₀: Initial void ratio (—)
C_r: Recompression index — the flat part of the curve (—)
C_c: Compression index — the steep virgin line (—)
$σ_{0}^{'}$: Initial vertical effective stress (kPa)
$σ_{c}^{'}$: Preconsolidation stress — the highest the soil has seen (kPa)
$σ_{f}^{'}$: Final vertical effective stress (kPa)

This assumes

One-dimensional compression, with no lateral strain
Primary consolidation only — creep is additional and not included here

In plain terms: The soil is stiff until it is reloaded past its own history, then it becomes several times softer. The bracket splits the loading at σ′_c precisely because the two parts of the curve have different slopes.

Worked example

The same layer, two stress histories

Given

Clay layer, H = 4.0 m, e₀ = 0.90
C_c = 0.35, C_r = 0.07 — the virgin line is five times steeper
σ′₀ = 80 kPa, and the foundation adds 100 kPa, so σ′_f = 180 kPa
Case A: overconsolidated, σ′_c = 150 kPa. Case B: normally consolidated, σ′_c = σ′₀

Find

How much difference the stress history makes

Rate. Terzaghi's solution is expressed through a dimensionless time factor

T_v = c_v · t / H_dr²

where H_dr is the longest drainage path — the full layer thickness if water can only escape one way, half of it if there is a permeable layer above and below.

The average degree of consolidation U depends on T_v alone, so the same curve serves every problem. Reaching 50% requires T_v = 0.196, and 90% requires T_v = 0.848.

For c_v = 2 m²/year and a 4 m layer draining both ways, H_dr = 2 m and 90% consolidation takes 1.70 years. Take away the lower drainage layer and H_dr becomes 4 m — the same 90% now takes 6.78 years, four times as long, because time scales with the square of the drainage path.

Practice

A 4 m clay layer drains both ways, with c_v = 2 m²/year. What time factor is needed for 50% consolidation?

Check yourself

A clay layer is found to have a continuous sand seam at its base, so it drains both ways rather than only upwards. What happens to the time to reach 90% consolidation?

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This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint