Consolidation and Differential Settlement

NCEES FE Civil spec: Geotechnical Engineering, I. Consolidation and differential settlement. FE Reference Handbook: Civil Engineering chapter, Geotechnical section (consolidation settlement, time factor table).

What the FE Civil Spec Covers

Section I covers how much a saturated clay layer settles under a new load (primary consolidation), how long it takes, and differential settlement between parts of a structure. Typical questions use the compression index to get settlement, compute OCR, or use the time factor to find the time to a given percent consolidation.

Key Concepts

Consolidation is the slow squeezing of water out of a saturated clay as excess pore pressure dissipates and load transfers to the soil skeleton.
Preconsolidation stress $σ_{p}^{'}$ is the largest effective stress the clay has felt. $O C R = σ_{p}^{'} / σ_{0}^{'}$ .
Compression index $C_{c}$ is the slope of the virgin line on an e vs log stress plot; the recompression index $C_{r}$ is much smaller (often about one fifth to one tenth of $C_{c}$ ).
Differential settlement causes damage more than uniform settlement. It is often expressed as angular distortion, the settlement difference divided by the distance between points.

Three Parts of Settlement

Immediate (elastic): happens as load is applied; dominant in sands.
Primary consolidation: time-dependent drainage of excess pore pressure in saturated clay.
Secondary compression: creep after primary ends; important in organic soils and peat.

To get the stress increase at the middle of a clay layer under a footing, the 2:1 method spreads the load at 2 vertical to 1 horizontal on each side. For a B by L footing carrying P, at depth z below the base:

Δ σ = \frac{P}{( B + z ) ( L + z )}

A wide fill applies its full pressure at every depth, which is why fills over soft clay cause large settlements.

Core Formulas

Normally consolidated clay:

S_{c} = \frac{C _{c} H}{1 + e _{0}} lo g_{10} \frac{σ _{0}^{'} + Δ σ}{σ _{0}^{'}}

Overconsolidated clay, final stress beyond the preconsolidation stress:

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

If the final stress stays below $σ_{p}^{'}$ , use only the $C_{r}$ term with the final stress in place of $σ_{p}^{'}$ . A common correlation is $C_{c} \approx 0.009 (LL - 10)$ .

Time rate of consolidation:

t = \frac{T _{v} H _{d r}^{2}}{c _{v}}, T_{50} = 0.197, T_{90} = 0.848, S (t) = U_{a v g} S_{c}

Angular distortion between two points a distance L apart:

β = \frac{Δ S}{L}

Where It Is in the FE Reference Handbook

The Civil Engineering chapter, Geotechnical section, has the consolidation settlement equations, the compression and recompression index definitions with correlations, and a table of time factor versus degree of consolidation with sketches of one-way and two-way drainage.

Worked Example: Settlement and Time

A 3 m thick normally consolidated clay layer lies between two sand layers. At mid-depth, $σ_{0}^{'} = 80$ kPa. A fill adds 60 kPa. The clay has $C_{c} = 0.36$ , $e_{0} = 1.10$ and $c_{v} = 0.90 m^{2} / yr$ . Find the primary settlement, the time to 50% and 90% consolidation, and the settlement at 50%.

Step 1: ultimate settlement.

S_{c} = \frac{0.36 ( 3 )}{1 + 1.10} lo g_{10} \frac{80 + 60}{80} = 0.5143 \times 0.2430 = 0.125 m = 125 mm

Step 2: drainage path. Sand above and below means two-way drainage, so $H_{d r} = 3/2 = 1.5$ m.

Step 3: times.

t_{50} = \frac{0.197 ( 1.5 ) ^{2}}{0.90} = 0.49 yr \approx 5.9 months

t_{90} = \frac{0.848 ( 1.5 ) ^{2}}{0.90} = 2.12 yr

Step 4: settlement at 50%. $S = 0.50 (125) = 62.5$ mm.

Differential check: if two columns 6 m apart settle 25 mm and 40 mm, the angular distortion is $15/6000 = 1/400$ . A commonly cited safe limit for buildings that must not crack is about 1/500, so this would be a concern.

Common Mistakes

Using the full layer thickness as the drainage path for a doubly drained layer, which overstates time by a factor of 4.
Using natural log instead of base-10 log in the settlement equation.
Applying $C_{c}$ to an overconsolidated clay over the whole stress range.
Computing initial effective stress with total unit weight below the water table.
Mixing units in $c_{v}$ (cm²/s vs m²/yr) when solving for time.

Exam Tips

Settlement problems are usually three quick steps: initial effective stress at the middle of the clay, final stress after the load, then the log formula. Compute OCR first so you pick the right equation. Time problems scale with the square of the drainage path, so halving the drainage path cuts the time to one quarter; this is why wick drains work. If you need a time factor other than 50% or 90%, look it up in the handbook table rather than interpolating by eye, and remember that settlement at any time equals the average degree of consolidation times the ultimate settlement.

Practice Problems

Try these from the free PolyCalc FE Civil problem bank. Each one has a full worked solution.

A normally consolidated clay layer has compression index C_c = 0.30, initial thickness H = 4 m, and initial void ratio e₀ = 0.8. Its representative vertical effective stress increases from 100 to 200 kPa. What is the primary consolidation settlement using the one-dimensional log₁₀ formula? See the worked solution or practice the Soils & Foundations deck
A normally consolidated clay has Cc = 0.30, H = 4 m, e0 = 0.90, initial effective stress 100 kPa and a stress increase of 80 kPa. Primary consolidation settlement is most nearly: See the worked solution or practice the Soil Mechanics deck
A 4-m clay layer is double drained and has cv = 2.0 × 10⁻⁷ m²/s. With Tv = 0.197 at 50% consolidation, t50 is most nearly: See the worked solution or practice the Soil Mechanics deck
A 3-m-thick clay layer drains only from its top and has cv = 1.0 × 10⁻⁷ m²/s. With Tv = 0.848 at 90% consolidation, t90 is most nearly: See the worked solution or practice the Soil Mechanics deck
A clay has a preconsolidation stress of 240 kPa and a current effective overburden stress of 120 kPa. Its overconsolidation ratio (OCR) is: See the worked solution or practice the Soil Mechanics deck

Related Calculators and Examples

Frequently Asked Questions

How do I find the drainage path length for consolidation?

If the clay drains at both top and bottom (sand above and below), the drainage path is half the layer thickness. If it drains only one way, for example clay over rock, the drainage path is the full thickness.

What are the time factors for 50% and 90% consolidation?

For a uniform initial excess pore pressure, the time factor is 0.197 at 50% average consolidation and 0.848 at 90%.

What is the difference between normally consolidated and overconsolidated clay?

A normally consolidated clay has never carried more effective stress than it does now (OCR = 1). An overconsolidated clay was loaded more heavily in the past (OCR greater than 1), so it recompresses along the much flatter recompression line until the preconsolidation stress is reached.

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