PE ExamTechnical KnowledgeGeotechnical Engineering

Consolidation & Settlement Calculations on the PE Geotechnical Exam

Primary consolidation, time-rate analysis, and the OCR distinction that trips up most PE candidates — two worked NCEES-style problems plus a Tv vs. U% reference table.

PEwise Team
May 21, 2026

Settlement calculations are where geotech engineers commonly lose points on the PE Civil Geotechnical exam — not because the math is hard, but because the OCR distinction trips them up. The right formula for a normally consolidated (NC) clay differs from the right formula for an overconsolidated (OC) clay. The OC formula itself splits into two sub-cases depending on whether the final stress stays inside or crosses out of the recompression range. Pick the wrong branch and the answer is off by a factor of 2 or more.

The good news: the e–log p curve, the Cc and Cr definitions, the time-factor Tv tabulation, and the Boussinesq stress-distribution influence factors are all reproduced in the NCEES PE Civil Reference Handbook (§3.9). The skill the exam tests isn't formula recall — it's checking the OCR, picking the right branch, reading the right chart, and converting between feet and inches without arithmetic slips.

This post walks through the four consolidation problem types NCEES tests, with two fully solved worked examples (NC primary consolidation, and an OC two-step calculation that crosses the preconsolidation pressure) plus a Tv vs. U% reference table. Every formula traces to its section in the handbook and to FHWA NHI-06-088 / NHI-06-089 (Soils and Foundations) where applicable.

Why consolidation matters on the PE Geotechnical exam

Per the April 2024 NCEES PE Civil Geotechnical specification, settlement appears under Topic 9 (Shallow Foundations, ASD or LRFD), sub-topic 9B (settlement, including induced stress distribution). Topic 9 carries 6–9 questions overall. Beyond the direct settlement questions, consolidation also feeds Topic 8 (Retaining Structures — backfill settlement) and Topic 5 (Earth Structures — embankment time-rate). When the exam writers want to integrate two topics into one question, settlement is one of the most common bridges.

The challenge is that consolidation problems test multiple concepts in series: identify the OCR, compute the initial effective stress at the midpoint of the compressible layer, compute the stress increase using Boussinesq or 2:1 distribution, pick the right primary-consolidation formula based on the stress trajectory, then convert to inches and check against tolerance. A single sign error or wrong-branch mistake propagates through the rest of the calculation. Practicing consolidation means practicing the full chain.

Core concepts you must master

Elastic, primary, and secondary settlement

Total settlement of a foundation has three components:

  • Elastic (immediate) settlement — the instantaneous deformation as load is applied. Computed from elastic theory using Young's modulus E; usually small compared to consolidation in saturated clays. Important for shallow footings on sand (where it can dominate).
  • Primary consolidation — the time-dependent expulsion of water from the clay pores under the new load. This is governed by Terzaghi's 1-D consolidation theory and is what most "settlement" PE-exam problems are about.
  • Secondary compression (creep) — continued slow settlement after primary consolidation completes, driven by viscous reorientation of the clay structure. Computed using the secondary compression index Cα.

The e–log p curve, Cc, and Cr

A laboratory consolidation test plots void ratio e against the log of effective stress σ′. The curve has two key slopes:

Cc = compression index = −Δe / Δlog(σ′)  (virgin loading branch)
Cr = recompression index = −Δe / Δlog(σ′)  (reload / recompression branch)

The recompression branch is much flatter than the virgin branch — typically Cr ≈ 0.1·Cc to 0.2·Cc. That ratio is why OCR matters so much: a clay loaded entirely within its recompression range settles 5–10× less than the same clay loaded along its virgin curve.

OCR and the preconsolidation pressure

The preconsolidation pressure σc is the maximum effective stress the soil has ever experienced. It marks the transition between recompression and virgin compression on the e–log p curve. The overconsolidation ratio:

OCR = σc / σ0

OCR = 1 → normally consolidated (NC); OCR > 1 → overconsolidated (OC). σ0 is the current in-situ effective stress at the midpoint of the compressible layer.

The three primary-consolidation formulas

The right formula depends on where the loading trajectory sits relative to the preconsolidation pressure. There are three cases (handbook §3.9):

Case 1: Normally consolidated (OCR = 1, so σ0 = σc). Final stress is on the virgin curve.

Sc = (Cc·H0 / (1 + e0)) · log[(σ0 + Δσ) / σ0]

Case 2: Overconsolidated, final stress stays in recompression range (i.e., σ0 + Δσσc). All settlement is along the flatter recompression curve.

Sc = (Cr·H0 / (1 + e0)) · log[(σ0 + Δσ) / σ0]

Case 3: Overconsolidated, final stress crosses into virgin range (i.e., σ0 + Δσ > σc). Two-step: recompression up to σc, then virgin compression past it.

Sc = (Cr·H0 / (1 + e0)) · log(σc / σ0) + (Cc·H0 / (1 + e0)) · log[(σ0 + Δσ) / σc]

Time-rate of consolidation (Terzaghi 1-D)

Primary consolidation takes time. The time factor Tv relates elapsed time to the degree of consolidation U:

Tv = Cv·t / Hd2

where Cv is the coefficient of consolidation (from lab) and Hd is the drainage path length. The drainage path is the longest distance water has to travel to escape: for double-drainage (drainage layers above and below the clay) Hd = H0/2; for single-drainage (impervious layer on one side) Hd = H0. Squaring the drainage path means switching from double to single drainage quadruples the time to a given U%.

Settlement at time t is S(t) = U(t) · Sc,∞, where Sc,∞ is the ultimate primary consolidation settlement.

Boussinesq stress distribution

The stress increase Δσ at depth from a surface load isn't constant — it spreads out laterally as it travels downward. The Boussinesq theory gives the exact solution for elastic stress distribution from a point or distributed surface load. The handbook tabulates Boussinesq influence factors I for circular, rectangular, and strip loads, so you can read Δσ/q0 directly off a chart for a given depth-to-width ratio. The simpler "2:1 method" approximates the same physics by spreading the load at a 2 vertical : 1 horizontal slope:

Δσz = q0·B·L / [(B + z)(L + z)]

Use the 2:1 method if the question allows it (faster); use the handbook's Boussinesq chart if precision matters or if the question explicitly calls for it.

The 4 types of consolidation problems on the PE Geotechnical exam

Type 1: Primary consolidation settlement (normally consolidated)

Given clay-layer thickness, e0, Cc, current σ0, and applied stress increase Δσ. OCR = 1 (or implied by "normally consolidated" in the question). Apply the NC formula directly. Worked below.

Type 2: Primary consolidation settlement (overconsolidated)

Given the same parameters plus σc (or OCR) and Cr. First check whether σ0 + Δσσc (Case 2: pure recompression) or > σc (Case 3: two-step). Apply the right formula. Worked below for Case 3.

Type 3: Time to reach percent consolidation (Tvt)

Given Cv, drainage geometry (single or double drainage), layer thickness, and a target U%. Look up Tv at that U%, then solve t = Tv·Hd2/Cv. Watch the drainage-path doubling trap.

Type 4: Secondary compression

Given Cα, layer thickness Hp (at the end of primary), ep (void ratio at end of primary), and a time interval (t1, t2). Apply Ss = (Cα·Hp / (1 + ep))·log(t2/t1). Less common on the exam than Types 1–3 but still tested.

Worked example: NC primary consolidation

Worked example 1 — NC primary consolidation. A 12-ft thick layer of normally consolidated clay (e0 = 0.85, Cc = 0.30) underlies a proposed warehouse. The current effective stress at the midpoint of the clay layer is σ0 = 1,500 psf. The warehouse will impose a uniform stress increase of Δσ = 1,000 psf at the clay-layer midpoint. Compute the ultimate primary consolidation settlement.

Step 1 — Identify the formula. Normally consolidated clay → Case 1:

Sc = (Cc·H0 / (1 + e0)) · log[(σ0 + Δσ) / σ0]

Step 2 — Plug in.

Cc·H0 / (1 + e0) = 0.30 × 12 / (1 + 0.85) = 3.60 / 1.85 = 1.946 ft
log[(σ0 + Δσ) / σ0] = log(2,500 / 1,500) = log(1.667) = 0.2218

Step 3 — Multiply.

Sc = 1.946 × 0.2218 = 0.432 ft5.18 in

Answer: Ultimate primary consolidation settlement Sc5.2 inches. Settlement of this magnitude exceeds typical building tolerances (usually 1–2 in for slab-on-grade), so a real warehouse on this site would need preloading, surcharging with wick drains, or a deep-foundation alternative.

Worked example: OC consolidation crossing the preconsolidation pressure

Worked example 2 — OC two-step consolidation. A 10-ft thick overconsolidated clay layer has e0 = 0.95, Cc = 0.40, and Cr = 0.05. At the midpoint of the layer, current effective stress σ0 = 2,000 psf and preconsolidation pressure σc = 3,000 psf (so OCR = 1.5). A new fill imposes Δσ = 2,000 psf at the midpoint. Compute the ultimate primary consolidation settlement.

Step 1 — Decide which case applies.

Final stress = σ0 + Δσ = 2,000 + 2,000 = 4,000 psf
4,000 psf > σc = 3,000 psf → loading crosses into virgin range → Case 3 (two-step)

Step 2 — Apply the two-step formula.

Sc = (Cr·H0 / (1 + e0)) · log(σc/σ0) + (Cc·H0 / (1 + e0)) · log[(σ0 + Δσ)/σc]

Step 3 — Recompression term.

Cr·H0 / (1 + e0) = 0.05 × 10 / 1.95 = 0.5 / 1.95 = 0.2564 ft
log(3,000/2,000) = log(1.5) = 0.1761
Recompression contribution = 0.2564 × 0.1761 = 0.0451 ft

Step 4 — Virgin compression term.

Cc·H0 / (1 + e0) = 0.40 × 10 / 1.95 = 4.0 / 1.95 = 2.051 ft
log(4,000/3,000) = log(1.333) = 0.1249
Virgin compression contribution = 2.051 × 0.1249 = 0.2562 ft

Step 5 — Sum.

Sc = 0.0451 + 0.2562 = 0.301 ft3.62 in

Answer: Sc3.6 in. Note the split: only 15% of the settlement is recompression (0.045 ft); 85% is virgin compression (0.256 ft) once the loading crosses σc. Crucial check: if you had incorrectly treated this as NC, you'd compute Sc = 2.051 × log(4,000/2,000) = 2.051 × 0.301 = 0.617 ft ≈ 7.4 in — over 2× the correct value. The OCR check is the difference between a defensible design and a serious overestimate.

Common errors that cost points

Confusing OCR with stress history

OCR is a ratio at a single point in time (now): OCR = σc/σ0. It tells you whether the soil is currently in its recompression or virgin-compression state. It does not tell you the stress trajectory under your applied load. The stress trajectory determines which formula (Case 1, 2, or 3) applies. A clay can be heavily overconsolidated (OCR = 4) but if Δσ is large enough to push the final stress past σc, you still need the two-step Case 3 formula.

Using Cc instead of Cr in the recompression branch (or vice versa)

Because Cr ≈ 0.1·Cc typically, mixing them up produces an order-of-magnitude error on the recompression contribution. Cue: any increment of stress on the recompression curve uses Cr; any increment on the virgin curve uses Cc; the two-step formula uses both, with Cr for the [σ0σc] segment and Cc for the [σcσ0σ] segment.

Misreading the e–log p plot

On a semi-log plot, void ratio is on the y-axis and effective stress on a log-x axis. The slope Cc = −Δe/Δlog(σ′) uses the change in e per log-cycle change in σ′. If a chart shows e dropping from 1.0 at 1 ksf to 0.7 at 10 ksf, then Cc = (1.0 − 0.7)/log(10/1) = 0.3 / 1 = 0.3. Reading the linear x-distance instead of the log cycle is a common failure mode that produces wildly wrong Cc values.

Drainage-path doubling trap on time-rate problems

For double drainage, Hd = H0/2. For single drainage, Hd = H0. Since tHd2, switching from double to single drainage quadruples the time to a given U%. The exam loves this trap because it's a clean factor-of-4 wrong answer that sits next to the right one in the multiple-choice list. Read the soil profile carefully: a clay sandwiched between two sand layers drains both ways; a clay sitting on bedrock drains only upward.

Forgetting to convert units (or the Boussinesq depth correction)

Settlement in feet vs. inches differs by a factor of 12. Stress in psf vs. tsf differs by 2,000. If the question gives e0 dimensionless (as it should be) but σ0 in tsf, you need to keep tsf throughout. For Boussinesq, the surface stress q0 isn't the stress at the clay-layer midpoint — you have to apply the depth-correction influence factor or the 2:1 method to get Δσ at depth before plugging into the consolidation formula. Skipping the correction overestimates settlement by 30–100%.

How to study consolidation for the PE Geotechnical exam

Phase 1 — Formula fluency (Week 1)

Read handbook §3.9 (Settlement) end-to-end. Practice writing the three primary-consolidation cases and the time-rate equation from a blank page until you can identify the correct case in under 30 seconds from the problem statement. The OCR-to-case decision is the single most important diagnostic step.

Phase 2 — Worked-problem drills (Weeks 2–3)

Work fifteen problems across the four types: four NC primary consolidation, four OC two-step (Case 3), three OC recompression-only (Case 2), three time-rate (Tvt), and one secondary compression. Time yourself: four to six minutes per problem on the exam. PEwise's Modules 15 and 16 (Soil Deformation and Settlement Due to Consolidation, Parts 1 and 2) walk the e–log p derivation, all three primary-consolidation cases, and time-rate analysis with worked NCEES-style problems and reference citations.

Phase 3 — Integration with bearing capacity and stress distribution (Week 4)

Solve five integration problems where you start from a footing geometry and applied load, compute the stress increase at the clay-layer midpoint using Boussinesq or 2:1, and then compute the ultimate primary consolidation settlement. That chain (footing → stress distribution → settlement) is the realistic Topic-9 pattern on the exam, and it brings together Topics 9A (bearing capacity) and 9B (settlement) into a single question. PEwise's Module 17 (Settlement Due to Compression) covers elastic settlement, secondary compression, and the integration with bearing-capacity calculations.

Quick reference: time factor and stress-distribution summary

Time factor Tv vs. degree of consolidation U

U (%) Tv Closed-form approximation
100.008Tv = (π/4)·U2 for U ≤ 60%
200.031"
300.071"
400.126"
500.197"
600.286"
700.403Tv = −0.933·log(1−U) − 0.085 for U > 60%
800.567"
900.848"
100Asymptotic

Source: Terzaghi 1-D consolidation theory, reproduced in NCEES PE Civil Reference Handbook §3.9 and FHWA NHI-06-088.

Stress distribution methods (for Δσ at depth)

Method When to use Form
2:1 method (approximate)Quick estimate; uniform load on rectangular footingΔσz = q0BL/[(B+z)(L+z)]
Boussinesq influence factor IPrecision required; circular / strip / rectangular loadsΔσz = q0·I(B/z, L/z)
Westergaard (rare)Layered media with thin sand seams in claySpecial chart in handbook

See the e–log p Curve in Action

PEwise's PE Geotechnical course walks through the e–log p curve, the OCR check, and the three primary-consolidation cases with the loading trajectory animated — once you can SEE which branch of the curve you're on, the formula choice becomes automatic.

Connecting this to your overall PE Geotechnical exam strategy

Settlement sits downstream of soil classification (Topic 1F), soil mechanics / lab testing (Topic 2A–2C), and bearing capacity (Topic 9A), and feeds back into retaining-wall and embankment design. Get the OCR-to-case decision automatic, get the time-rate drainage path right, and the integrated questions become a sequence of clean steps. For the broader Topic 1 / Topic 2 fundamentals plus the full 24-module curriculum, our geotechnical PE exam study guide walks the syllabus end-to-end. For the bearing-capacity and pile-capacity calculations that pair with settlement to size shallow and deep foundations, see our bearing capacity and pile capacity post. For slope-stability problems that draw on the same effective-stress and shear-strength fundamentals, see the slope-stability problem-types post.

Final thoughts

Consolidation problems reward engineers who treat the OCR check as the first 30 seconds of every settlement question: NC, OC-recompression-only, or OC-two-step? Once the case is fixed, the formula is fixed, and the calculation is mechanical. The candidates who pass internalize the three-case taxonomy and pick the right branch without thinking. The candidates who don't second-guess between Cc and Cr at every problem and burn time on the wrong setup. Drill the OCR check until it's automatic.

Master Consolidation Calculations with PEwise

PEwise's Modules 15, 16, and 17 of the PE Geotechnical course (26 lessons combined) cover consolidation theory, time-rate analysis, and settlement calculations for shallow foundations — including the OCR distinction that NCEES tests. Course author Mahdi Bahrampouri, Ph.D., Geotechnical Earthquake Engineer, built the curriculum directly against FHWA NHI-06-088 / NHI-06-089 (Soils and Foundations), UFC 3-220-10 (Soil Mechanics), and the NCEES PE Civil Reference Handbook.