PE ExamTechnical KnowledgeGeotechnical Engineering

Bearing Capacity & Pile Capacity Calculations on the PE Geotechnical Exam

Bearing capacity and pile capacity carry up to 30% of the PE Geotechnical exam. Three worked NCEES-style problems — Terzaghi shallow footing, alpha-method pile in clay, and pile-group block-failure check — plus the Meyerhof bearing-capacity factor table.

PEwise Team
May 18, 2026

Foundations are the highest-weight calculation topic on the PE Civil Geotechnical exam. Per the NCEES April 2024 specification, Topic 9 (Shallow Foundations, ASD or LRFD) carries 6–9 questions and Topic 10 (Deep Foundations, ASD or LRFD) carries 10–15. Combined, that's 16–24 questions out of 80 — up to 30% of the exam hinging on bearing-capacity and pile-capacity calculations. No other topic block carries that much weight.

The good news: most foundation problems on the exam are direct calculation questions. The Terzaghi general bearing-capacity equation and the alpha / beta methods for pile axial capacity are reproduced in the NCEES PE Civil Reference Handbook. The skill the exam tests isn't formula recall — it's getting the inputs right (effective stress vs. total stress, drained vs. undrained, gross vs. net), picking the right shape factor for square / strip / circular footings, and remembering which method applies to which soil type.

This post walks through the five foundation problem types NCEES tests, with three fully solved worked examples — a Terzaghi shallow-footing calculation, an alpha-method pile in clay, and a pile-group capacity check including block failure. Every formula traces to its section in the handbook (§3.10 for shallow foundations, §3.11 for deep foundations) and its FHWA reference where applicable.

Why foundations dominate the PE Geotechnical exam

The April 2024 NCEES PE Civil Geotechnical specification puts foundations at the top of the exam blueprint. Topic 9 and Topic 10 together take up 16–24 questions out of 80 — more than soil mechanics and lab testing (Topic 2, 8–12 questions), more than retaining structures (Topic 8, 10–15), more than earth structures (Topic 5, 9–14). If you can reliably solve the standard bearing-capacity and pile-capacity problems, you've put a substantial fraction of a passing score on the table before exam day.

The other reason foundations dominate: they connect to almost every other geotechnical concept. Bearing capacity needs effective stress (Topic 2A), unit weight (Topic 2A), and friction angle or undrained shear strength (Topic 2C). Pile capacity in clay needs adhesion factors that depend on soil classification (Topic 1F) and the SPT or CPT data from site characterization (Topic 1A–1E). A well-designed exam problem chains these together. Practicing foundations means practicing the whole geotech curriculum.

Core concepts you must master

Terzaghi general bearing-capacity equation

For a continuous (strip) footing on homogeneous soil, the ultimate bearing capacity is given by the Terzaghi general equation (handbook §3.10):

qult = c·Nc + q·Nq + 0.5·γ·B·Nγ

Where c = soil cohesion, q = effective overburden pressure at footing base = γ·Df, γ = soil unit weight (use effective weight if below the water table), B = footing width, and Nc, Nq, Nγ are bearing-capacity factors that depend only on the friction angle φ.

Shape factors apply for non-strip footings:

Square:   qult = 1.3·c·Nc + q·Nq + 0.4·γ·B·Nγ
Circular:   qult = 1.3·c·Nc + q·Nq + 0.3·γ·B·Nγ

Bearing-capacity factors (Meyerhof)

The PE Civil Reference Handbook reproduces a table of bearing-capacity factors. Meyerhof's expressions are the most common modern form:

Nq = eπ·tan φ·tan2(45° + φ/2)
Nc = (Nq − 1)·cot φ  (or 5.14 if φ = 0)
Nγ = (Nq − 1)·tan(1.4φ)

For undrained loading on saturated clay (φ = 0): Nc = 5.14, Nq = 1, Nγ = 0. The strip-footing equation reduces to qult = 5.14·su + q; the square-footing version uses 1.3·5.14 = 6.68 in place of 5.14.

Effective stress, groundwater, and the unit-weight trap

The unit-weight term in q = γ·Df and the third term 0.5·γ·B·Nγ use effective unit weight. If the water table is above the footing base, use buoyant unit weight (γbuoyant = γsatγw). If the water table is between the footing base and one footing-width below, the third term gets a partial correction. If the water table is more than one footing-width below the base, no correction is needed. Missing this is one of the most common errors on bearing-capacity problems.

Allowable vs. ultimate (and gross vs. net)

Ultimate bearing capacity qult is the load at which the soil shears. Allowable bearing capacity is reduced by a factor of safety (typically FS = 3 for shallow foundations under static loads): qall = qult / FS. The exam often distinguishes gross (includes the overburden replaced by the footing) from net (subtracts γ·Df). Read the question carefully — answers can differ by 10–15% depending on which one is asked.

Pile axial capacity: shaft + tip

The total ultimate axial capacity of a single pile is the sum of skin friction and end bearing (handbook §3.11):

Qult = Qs + Qp = fs·As + qp·Ap

Where fs = unit skin friction, As = side surface area along the pile shaft, qp = unit end-bearing resistance, Ap = pile tip cross-sectional area.

The unit values depend on the soil:

  • Cohesive soil (alpha-method): fs = α·su, where α is an adhesion factor that decreases with increasing su (typically 1.0 for soft clay, dropping to 0.3–0.5 for stiff clay per FHWA NHI-05-042). End bearing qp = 9·su.
  • Cohesionless soil (beta-method): fs = β·σv, where β = K·tan δ (lateral earth pressure coefficient times pile-soil friction angle, typically 0.25–0.45). End bearing qp = Nq·σv at the tip, capped at limiting values.
  • Lambda-method (mixed): fs = λ·(σv,avg + 2·su,avg), used for piles longer than ~50 ft passing through layered soils.

Pile group efficiency and block failure

For a pile group, the ultimate capacity is the lesser of (a) the sum of individual pile capacities reduced by a group efficiency factor η, and (b) the capacity of the entire pile group acting as a single block (block failure). For cohesive soils with center-to-center spacing ≥ 3D, FHWA NHI-16-009 generally takes η = 1.0 — but the block failure check is always required, especially for tight spacing or weak clays.

The 5 types of foundation problems on the PE Geotechnical exam

Type 1: Shallow square footing on granular soil

Given footing dimensions, embedment depth, soil unit weight, friction angle, and (often) a target service load. Use the square-footing form of Terzaghi with c = 0; compute qult, divide by FS = 3, and either report qall directly or check whether the footing carries the service load. Worked below.

Type 2: Shallow strip footing on cohesive soil (undrained)

Given footing width, embedment depth, total unit weight, and undrained shear strength su. Use the strip-footing form with φ = 0 (Terzaghi-Skempton): qult = 5.14·su + γ·Df. Note this uses total unit weight (not effective) because the analysis is undrained. The third term vanishes (Nγ = 0).

Type 3: Pile axial capacity in cohesionless soil (beta-method)

Given pile geometry, soil profile (sand layers with friction angle and unit weight), and water-table depth. Compute σv at the midpoint of each sand layer and at the tip; multiply by β for skin friction and by Nq for end bearing; sum over the pile length. Watch for limiting values per FHWA NHI-16-009.

Type 4: Pile axial capacity in cohesive soil (alpha-method)

Given pile geometry and undrained shear strength of the clay. Compute fs = α·su with α selected from the FHWA chart for the given su; multiply by shaft area. End bearing qp = 9·su; multiply by tip area. Sum. Worked below.

Type 5: Pile group capacity (block failure vs. sum of individual)

Given a pile group geometry (rows × columns, spacing) and the single-pile capacity. Compute the sum of individual pile capacities (with group efficiency factor); separately compute the block-failure capacity. Take the smaller. Worked below.

Worked example: Terzaghi bearing capacity (granular soil)

Worked example 1 — Terzaghi bearing capacity, square footing on sand. A 6-ft square footing is embedded 4 ft below the ground surface. The supporting sand has unit weight γ = 120 pcf, friction angle φ = 30°, cohesion c = 0. The water table is well below the footing (more than one footing-width below the base). Compute the ultimate and allowable bearing capacity using FS = 3.

Step 1 — Identify the equation. Square footing on cohesionless soil:

qult = 1.3·c·Nc + q·Nq + 0.4·γ·B·Nγ

Step 2 — Look up Meyerhof factors for φ = 30°: Nc = 30.14, Nq = 18.40, Nγ = 15.67.

Step 3 — Compute overburden pressure at the footing base.

q = γ·Df = 120 × 4 = 480 psf

Step 4 — Substitute. The cohesion term drops out (c = 0).

qult = 0 + 480·(18.40) + 0.4·120·6·(15.67)
    = 8,832 + 4,513
    = 13,345 psf ≈ 13.3 ksf

Step 5 — Apply factor of safety.

qall = qult / FS = 13,345 / 3 = 4,448 psf ≈ 4.45 ksf

Answer: qult ≈ 13.3 ksf, qall ≈ 4.45 ksf. For a 6 ft × 6 ft footing (A = 36 ft²), the allowable column load is roughly Pall = 4.45 × 36 = 160 kip.

Worked example: Pile axial capacity in cohesive soil (alpha-method)

Worked example 2 — Alpha-method pile in clay. A closed-end steel pipe pile, diameter D = 12 in (1.0 ft), is driven 50 ft into uniform medium clay with undrained shear strength su = 1500 psf. From the FHWA adhesion chart, α = 0.5 for su = 1500 psf. Compute the ultimate axial capacity and the allowable capacity at FS = 3.

Step 1 — Unit skin friction (alpha-method).

fs = α·su = 0.5 × 1,500 = 750 psf

Step 2 — Pile shaft surface area.

As = π·D·L = π × 1.0 × 50 = 157.1 ft2

Step 3 — Total skin friction.

Qs = fs · As = 750 × 157.1 = 117,810 lb ≈ 118 kip

Step 4 — End-bearing unit resistance and tip area. For piles in clay: qp = 9·su.

qp = 9 × 1,500 = 13,500 psf
Ap = π·D2/4 = π/4 = 0.785 ft2

Step 5 — End-bearing capacity.

Qp = qp·Ap = 13,500 × 0.785 = 10,598 lb ≈ 10.6 kip

Step 6 — Total and allowable.

Qult = Qs + Qp = 118 + 10.6 = 128.6 kip
Qall = Qult / FS = 128.6 / 3 ≈ 43 kip

Answer: Qult ≈ 129 kip, Qall ≈ 43 kip. Note that skin friction (118 kip) accounts for over 90% of the capacity — typical for piles in uniform clay, where the alpha-method skin term dominates.

Worked example: Pile group capacity (block failure check)

Worked example 3 — 3 × 3 pile group, block failure check. Continuing from Worked Example 2: a 3 × 3 group of the same piles (D = 1 ft, L = 50 ft, Qult,single = 129 kip) is arranged with center-to-center spacing s = 3 ft (= 3D). The clay has su = 1,500 psf throughout. Compute the group ultimate capacity.

Step 1 — Sum of individual pile capacities. For cohesive soil with s ≥ 3D, FHWA NHI-16-009 takes group efficiency η = 1.0:

ΣQsingle = 9 × 129 = 1,161 kip

Step 2 — Block dimensions. The group acts as a block of plan dimensions:

Bg = Lg = (rows − 1)·s + D = 2·(3) + 1 = 7 ft
Block depth = pile length = 50 ft

Step 3 — Block ultimate capacity. For block failure in clay, use full su on the block perimeter (not the alpha-reduced value used for individual piles), and Nc = 9 for end bearing:

Block skin: Qs,b = su·perimeter·L = 1,500·(4·7)·50 = 2,100,000 lb = 2,100 kip
Block tip: Qp,b = 9·su·Ag = 9·1,500·49 = 661,500 lb ≈ 662 kip
Block total: Qblock = 2,100 + 662 = 2,762 kip

Step 4 — Group capacity is the smaller of the two.

Qgroup = min(ΣQsingle, Qblock) = min(1,161, 2,762) = 1,161 kip

Step 5 — Allowable group capacity.

Qgroup,all = 1,161 / 3 ≈ 387 kip

Answer: Qgroup,ult ≈ 1,161 kip, Qgroup,all ≈ 387 kip. Sum-of-individual governs over block failure here. For tighter spacing (s < 3D), softer clays, or longer pile lengths, block failure can govern instead — always check both.

Common errors that cost points

Reporting ultimate when the question asks for allowable

Bearing-capacity and pile-capacity questions usually ask for the allowable value with a stated factor of safety (typically FS = 3 for shallow foundations, FS = 2–3 for deep foundations). Computing qult or Qult and forgetting to divide is a one-step arithmetic error that lands on a wrong-but-close-looking distractor. Read the question stem for "allowable," "service load," "design load," "FS," or "factor of safety."

Forgetting groundwater corrections to effective stress

If the water table is above the footing base, the overburden pressure q uses buoyant unit weight (γ′ = γsatγw) below the water table and total unit weight above it. The 0.5·γ·B·Nγ term needs the same treatment if the water table is within one footing-width below the base. Skipping the correction overestimates capacity by 30–50% on saturated sands.

Using alpha-method on sand or beta-method on clay

Alpha-method (fs = α·su) applies only to cohesive soils. Beta-method (fs = β·σv) applies only to cohesionless soils. Mixing them up is a single-step error that lands on the wrong answer family. If the problem gives you su, it's clay → alpha-method. If it gives you φ and γ with no cohesion, it's sand → beta-method.

Skipping the block-failure check on tight pile groups

For pile groups with center-to-center spacing < 3D, or in soft clay, block failure can govern over the sum of individual pile capacities. Reporting only ΣQsingle·η without checking the block capacity is a category-of-answer error. Always compute both; report the smaller.

Confusing gross and net allowable bearing capacity

Gross allowable: qall,gross = qult/FS (applies to total pressure under the footing, including overburden). Net allowable: qall,net = (qultγ·Df)/FS (subtracts the overburden the footing replaces). Both are correct — but only one is what the question asks for. The difference is typically 5–15%.

How to study foundations for the PE Geotechnical exam

Phase 1 — Equation fluency (Week 1)

Read handbook §3.10 (shallow foundations) and §3.11 (deep foundations) end-to-end. Practice writing the Terzaghi general bearing-capacity equation with shape factors for strip / square / circular until it's automatic. Practice the alpha-method and beta-method pile equations. Drill until you can pick the right equation for a given problem in under 30 seconds.

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

Work twenty problems across the five problem types: four shallow on granular, four shallow on cohesive (including φ = 0 cases), four pile in sand (beta-method), four pile in clay (alpha-method), four pile groups including at least two with block failure governing. Time yourself: four to six minutes per problem on the exam. PEwise's Module 14 (Shallow Foundations: Bearing Capacity) walks the Terzaghi equation with shape factors, groundwater corrections, and Meyerhof / Hansen modifications problem by problem.

Phase 3 — Integration with site characterization (Week 4)

Solve five integration problems where you start from raw site-investigation data — SPT N-values, CPT cone resistance, or laboratory triaxial / unconfined compression results — and back out φ or su, then compute foundation capacity. That chain (site characterization → soil parameter → foundation capacity) is the realistic exam pattern, especially for Topic-9 / Topic-10 questions that pull from Topic 1A–1E. PEwise's Module 18 (Deep Foundations — Drilled Shafts, Driven Piles, Micropiles) covers the alpha, beta, and lambda methods with worked NCEES-style problems for both straight and chained question formats.

Quick reference: bearing-capacity factors and pile capacity formulas

Meyerhof bearing-capacity factors

Friction angle φ Nc Nq Nγ
0° (saturated clay, undrained)5.141.000.00
20°14.836.402.87
25°20.7210.666.77
30°30.1418.4015.67
35°46.1333.3037.15
40°75.3164.2093.69

Computed from Meyerhof closed-form expressions. The handbook reproduces the same table for exam-day reference.

Pile axial capacity formulas

Method Use for Skin friction fs End bearing qp
Alpha (total stress)Cohesive soilα·susu
Beta (effective stress)Cohesionless soilβ·σvNq·σv,tip
Lambda (mixed)Long piles in layered soilsλ·(σv,avg + 2·su,avg)As applicable per soil

Sourced from FHWA NHI-16-009 (Driven Pile Foundations Vol I) and FHWA NHI-18-024 (Drilled Shafts).

Foundations Are Where the Exam Is Won

PEwise's PE Geotechnical course walks through bearing capacity, alpha and beta methods, and pile group design with the same step-by-step rigor as the worked examples above — on real NCEES-style problems with detailed solutions.

Connecting this to your overall PE Geotechnical exam strategy

Foundations sit downstream of soil classification (Topic 1F) and soil mechanics / lab testing (Topic 2A–2C), and upstream of settlement, slope stability, and retaining-structure problems. Get the foundation calculations automatic, and the integrated questions become a sequence of clean steps rather than a puzzle. 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 slope-stability extension that pulls bearing capacity, lateral earth pressure, and shear strength into a single problem, see our slope-stability problem-types post.

Final thoughts

Foundations reward engineers who treat the calculation as a clean recipe: identify the equation form (strip / square / circular for shallow; alpha / beta / lambda for piles), look up the right factor for the right friction angle or undrained shear strength, watch the groundwater correction, divide by the right factor of safety, and read whether the question wants gross or net, ultimate or allowable. The candidates who pass run that recipe in four to six minutes per question. The candidates who don't burn time second-guessing which equation to use. Drill the recipe until you don't have to think about it.

Master Foundations with PEwise

PEwise's Modules 14 and 18 of the PE Geotechnical course (30+ lessons) cover bearing capacity for shallow foundations and axial-plus-lateral capacity for deep foundations with worked NCEES-style problems — including the alpha-method for clays and the beta-method for sands. Course author Mahdi Bahrampouri, Ph.D., Geotechnical Earthquake Engineer, built the curriculum directly against FHWA NHI-16-009 (Driven Piles), FHWA NHI-18-024 (Drilled Shafts), and the NCEES PE Civil Reference Handbook.