Foundations & Retaining Walls · Study · PE Civil: Structural · FE → PE Prep
Foundations & Retaining Walls
8% of exam
Bearing capacity and settlement of spread footings, combined footings and mats, pile and drilled-shaft capacity, and retaining-wall stability against sliding, overturning, and bearing.
5 concepts
E. Shallow foundations
Shallow Foundation Bearing Capacity & Settlement
The general bearing-capacity equation with bearing-capacity, shape, depth and groundwater factors; net vs. gross allowable pressure with a factor of safety; and elastic plus consolidation settlement.
Every spread footing has to satisfy two completely separate limit states, and the exam loves to see whether you keep them apart. The first is a STRENGTH check: will the soil shear and the footing punch through (bearing capacity)? The second is a SERVICEABILITY check: how much will the footing settle, and is that settlement tolerable? A footing can pass one and fail the other — a wide footing on dense sand has enormous bearing capacity but may still settle too much, while a narrow footing on soft clay may have ample factor of safety against settlement-controlled service load yet not enough against a shear failure. The NCEES PE Civil Reference Handbook gives you the bearing-capacity machinery in §3.4 Bearing Capacity, the elastic-settlement machinery in §3.5 Foundation Settlement (specifically the elastic method §3.5.2), and the consolidation machinery — Cc
,
Cr
,
e0
,
pc
, and the NC/OC split — in §3.2 Consolidation; on exam day those sections, plus their factor tables, are where you live for this topic.
The general bearing-capacity equation
Ultimate bearing pressure is the sum of three contributions — cohesion, surcharge (the soil above the footing base), and the self-weight of the failure wedge below the base. Each is scaled by a dimensionless bearing-capacity factor that depends only on the friction angle ϕ, read from the handbook's Bearing Capacity Factors table (Nc, Nq, Nγ). For a strip footing the bare three-term form applies; for square or rectangular footings each term carries a shape factor sc, sq, sγ. Here q=γaDf is the total surcharge at the base, γ is the unit weight of the soil beneath the base, and Bf is the footing width.
qult=cNcsc+qNqsq+21γBfNγsγ
Reading the factors — and the groundwater correction
The factors grow steeply with ϕ: at ϕ=0 (saturated clay, undrained) Nc=5.14, Nq=1, Nγ=0, so for a clay the equation collapses to qult=5.14cusc+q. At ϕ=30∘ the factors jump to Nc=30.1, Nq=18.4, Nγ=22.4. The AASHTO shape factors the handbook tabulates are sc=1+(Bf/Lf)(Nq/Nc), sγ=1−0.4(Bf/Lf), and sq=1+(Bf/Lf)tanϕ. If the water table rises into the failure zone you must also apply the groundwater correction factors Cwq and Cwγ (and use buoyant γ′ for any term below the table), because submergence roughly halves the effective unit weight and therefore the Nγ and surcharge contributions.
The factor of safety is applied to the NET bearing capacity, not the gross. The net ultimate qnet=qult−q removes the surcharge that was already there before the footing existed (the soil you excavated is replaced by the footing). Divide qnet by the factor of safety — usually FS=3 for bearing — to get the net allowable bearing pressure, the value you compare against the net applied service pressure (P/A minus the overburden that the column load replaces). The cardinal rule is to keep the two sides of the check on the SAME basis: a net allowable goes against a net applied, and a gross allowable goes against a gross applied. Mixing the bases is the classic error, and the two ways of mixing them fail in opposite directions. Comparing a GROSS allowable (qult/FS, the larger capacity) against a NET applied pressure (the smaller demand) under-designs — you require a bigger number to beat a smaller one, crediting yourself capacity you do not actually have. The reverse mismatch — a NET allowable (the smaller capacity) against a GROSS applied pressure (the larger demand) — is merely over-conservative: you require the smaller number to beat the larger one, which oversizes the footing rather than endangering it. Separately, applying FS=3 to qult instead of qnet is conservative but wrong.
qnet=qult−γaDf,qall,net=FSqnet
Elastic (immediate) settlement
On sands and stiff overconsolidated clays the bulk of settlement happens essentially as the load is applied — it is elastic and immediate. The handbook's elastic method (§3.5.2) gives the surface settlement from the stress increase Δp at the base, the footing width Bf, the soil's Young's modulus Em and Poisson's ratio ν, and a shape-and-rigidity factor Cd tabulated by footing shape and the point of interest (center, corner, average). Picking Cd for the wrong point — corner versus center — is a classic slip; the center of a flexible footing settles roughly twice the corner.
δv=CdΔpBfEm1−ν2
Consolidation settlement of clay
Under a saturated clay, water must squeeze out before the soil compresses, so settlement is time-dependent and can take years. For a normally consolidated clay the primary consolidation settlement uses the compression index Cc, the initial void ratio e0, the layer thickness H0, the initial effective overburden po, and the final stress pf=po+Δp. If the clay is overconsolidated and the final stress stays below the preconsolidation pressure pc, use the much smaller recompression index Cr; if Δp pushes pf past pc, split the calculation into a recompression part (from po to pc using Cr) and a virgin part (from pc to pf using Cc). Always compute Δp at mid-depth of the clay layer, not at the footing base.
First decide which limit state the question is testing — a bearing question gives you ϕ, c, γ, Df, B and asks for qult or an allowable load; a settlement question gives you Cc, e0, stresses or E and ν. For bearing, read Nc, Nq, Nγ from the table at your ϕ (do not interpolate the wrong row), apply shape factors only for square/rectangular footings, and apply FS to qnet. Keep both sides of the bearing check on the same basis — net allowable against net applied — because a gross allowable against a net applied is the dangerous, under-designing mismatch. For settlement, keep elastic (sand, immediate) separate from consolidation (clay, time-dependent), evaluate Δp at the layer mid-depth, and watch whether the clay is NC or OC — using Cc on an overconsolidated clay overstates settlement by a factor of three or more. Units: keep pressures in ksf or psf consistently, and remember log10 (not ln) in the consolidation equation.
Key equations
General bearing capacity (strip)qult=cNc+qNq+21γBfNγ
Ultimate bearing pressure of a concentrically loaded strip footing. q=γaDf surcharge; Nc,Nq,Nγ
Bearing capacity with shape factorsqult=cNcsc+qNqsq+21γBfNγsγ
Problem. A 7ft×7ft square footing bears at Df=5ft in a cohesionless sand with ϕ=32∘, c=0, and total unit weight γ=120pcf. The water table is deep. Using Nq=23.2, Nγ=30.2 and a factor of safety of 3, find the net allowable bearing pressure and the allowable column service load.
Problem. A 10ft thick normally consolidated clay layer has Cc=0.30 and e0=0.90
Common pitfalls
•Applying the factor of safety to the GROSS ultimate qult instead of the net qnet=qult−γaDf. The surcharge was already there before the footing; subtract it first, then divide by FS.
•Comparing a GROSS allowable (qult/FS) against a NET applied pressure — this under-designs, because you credit yourself the larger capacity while charging only the smaller demand. Keep both sides on the same basis: net allowable vs net applied, or gross allowable vs gross applied.
•Forgetting the shape factors on a square or rectangular footing — or applying them to a strip footing where Bf/Lf→0 makes them unity. Strip = bare three-term form; square = each term times its s factor.
•Reading the wrong N row: Nc,Nq,Nγ
•Using ln instead of log10 in the consolidation equation. The base-10 log is required; ln overstates settlement by a factor of 2.303.
•Using the compression index Cc on an overconsolidated clay whose final stress stays below pc. Use the much smaller recompression index Cr
•Evaluating Δp at the footing base rather than at the clay-layer mid-depth, which overstates the stress and the settlement; stress spreads and decays with depth.
References
NCEES PE Civil Reference Handbook — §3.4 Bearing Capacity (general equation, factor and shape-factor tables)
NCEES PE Civil Reference Handbook — §3.5.2 Foundation Settlement (elastic method)
FHWA-NHI-06-089, Soils and Foundations, Vol. II — source of the bearing-capacity and settlement relations the handbook reproduces
Combined Footings, Mats & Eccentric Loads
Soil-pressure distribution under an eccentric or combined footing, the e ≤ B/6 kern keeping the base in compression, uniform vs. trapezoidal vs. reduced-area pressure, and mat foundations.
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F. Deep foundations
Deep Foundations: Pile & Drilled-Shaft Capacity
Axial capacity as skin friction plus end bearing, allowable capacity with a factor of safety, pile-group efficiency and block failure, and the downdrag (negative skin friction) penalty.
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G. Retaining walls
Retaining-Wall Stability
Factors of safety against sliding, overturning, and bearing for a gravity or cantilever wall, the middle-third resultant location, and the active earth-pressure driving force.
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Cantilever Retaining-Wall Design
Proportioning the stem, heel, and toe; designing the stem and base slab for earth-pressure moments and shears per ACI 318-14; and detailing the flexural reinforcement.
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from the handbook table at
ϕ
.
Square/rectangular footing form; multiply each term by its AASHTO shape factor.
Bf/Lf=1 for a square footing, →0 for a strip. Per handbook §3.4.2.1.
Apply FS (typically 3) to the NET ultimate. Compare against net applied service pressure — same basis on both sides.
) is much smaller than the surcharge term — expected for a shallow, wide footing; and
FS
acts on the net, leaving a healthy allowable.
. At the layer mid-depth the initial effective overburden is
po=2.0ksf
, and a new footing increases the stress at mid-depth by
Δp=1.5ksf
. Compute the primary consolidation settlement.
Solution. Final stress: pf=po+Δp=2.0+1.5=3.5ksf.
Layer thickness in inches: H0=10ft=120in.
Settlement: Sc=1+e0CcH0log10popf=1.900.30(120)log102.03.5.
log10(1.75)=0.2430, so Sc=0.1579(120)(0.2430)=4.60in.
**Answer: Sc=4.60in.** Sanity check: the strain is Sc/H0=4.60/120=3.8% — a plausible primary-consolidation strain for a soft NC clay under a 75% stress increase. Had the clay been overconsolidated with Cr≈0.05, the settlement would have been roughly Cr/Cc=1/6 of this, about 0.77in.