Bridge Foundations — Bridge Engineering & Design

Engineering story

Every bridge is only as strong as what holds it up

The superstructure gets the glory, but the foundation is where a bridge actually earns its factor of safety. When the I-35W truss in Minneapolis collapsed in 2007, it was a gusset-plate failure — but nine of the ten bridges the U.S. loses each year to sudden collapse fail at the foundation, almost always to scour, seismic liquefaction, or long-term downdrag that nobody accounted for. AASHTO §10 turns the geotechnical art of Terzaghi, Meyerhof, and Reese into a load-and-resistance-factor design where the resistance factors (0.45 – 0.55 for most modes) are deliberately lower than in the superstructure to reflect the higher variability of what lies below the mudline.

Chapter objectives

What you will be able to do

Learning objectives

By the end of this chapter you will be able to:

1Choose between spread footings, driven piles, drilled shafts, and micropiles based on subsurface profile and expected loads.
2Compute the factored nominal bearing resistance of a spread footing using the Meyerhof/AASHTO §10.6 equation and check settlement.
3Compute the axial capacity of a driven pile from skin friction and end bearing per §10.7.3, including effective-stress (β) and total-stress (α) methods.
4Design a drilled shaft with a rock socket per O'Neill & Reese (§10.8.3), including side and tip resistance.
5Analyze lateral pile behaviour with the p-y method and check group effects per §10.7.3.9.
6Account for downdrag (§3.11.8) and scour-adjusted bearing depth (§10.7.3.6, links to Ch. 14).
7Deliver two complete worked examples and one river-crossing design challenge with foundation schedules.

13.1 — Foundation families

Matching the foundation to the ground

AASHTO LRFD §10.6, §10.7, §10.8, §10.9

Four foundation families cover essentially every highway bridge: spread footings on competent shallow soil or rock, driven piles in deep alluvial deposits, drilled shafts (a.k.a. bored piles or CIDH) when large single elements are needed for lateral load or scour, and micropiles for retrofit or restricted-access sites.

Fig. 13.1Four foundation families. (a) spread footing on shallow bearing stratum, (b) driven pile group, (c) drilled shaft socketed into rock, (d) micropile cluster used mostly for retrofit or restricted access.

Selection rules of thumb

Bearing stratum ≤ 10 ft below grade with SPT N ≥ 20 → spread footing.
Deep soft/loose overburden ≥ 40 ft, no rock reachable → driven pile group.
Rock ≤ 60 ft deep, high lateral demand, or river pier with scour → drilled shaft with rock socket.
Column retrofit, headroom < 15 ft, or slope stabilization → micropiles.

13.2 — Spread footings

Bearing capacity and settlement

AASHTO LRFD §10.6

The ultimate bearing capacity of a shallow footing on cohesionless or c-φ soil is given by the general Meyerhof equation, in the AASHTO §10.6.3.1.2 form:

q_n \;=\; c\, N_c\, s_c\, d_c\, i_c \;+\; \gamma\, D_f\, N_q\, s_q\, d_q\, i_q \;+\; 0.5\, \gamma\, B\, N_\gamma\, s_\gamma\, d_\gamma\, i_\gamma

(13.1)

$q_n$: nominal ultimate bearing resistance [ksf]
$N_c, N_q, N_\gamma$: bearing capacity factors, functions of φ′
$s, d, i$: shape, depth, and load-inclination factors
$c$: cohesion [ksf]
$\gamma$: effective unit weight of soil [kcf]
$D_f$: footing depth from grade to footing base [ft]
$B$: footing width (smaller plan dim.) [ft]

Factored resistance is $\phi_b\, q_n$ with $\phi_b = 0.45$ (theoretical), 0.55 (semi-empirical), or from load test.

Fig. 13.2Bearing pressure and passive failure wedges. An eccentric column load produces a trapezoidal bearing distribution; failure develops along Rankine passive wedges outside the footing.

Eccentric loading. When $e = M/P \le B/6$ the pressure distribution is trapezoidal:

q_{max},\, q_{min} \;=\; \dfrac{P}{B L}\left(1 \pm \dfrac{6e}{B}\right)

(13.2)

When $e > B/6$ a triangular distribution over an effective width $B' = B - 2e$ is used (AASHTO §10.6.3.1.5).

Elastic settlement under service load (Hough, §10.6.2.4):

\delta \;=\; \dfrac{q_s\, B\,(1 - \nu^2)}{E_s}\, I_f

(13.3)

$q_s$: service bearing pressure [ksf]
$E_s$: soil elastic modulus (from CPT/SPT correlations) [ksf]
$\nu$: Poisson's ratio (0.3–0.4)
$I_f$: shape/rigidity influence factor (0.7–1.1)

13.3 — Driven piles

Skin friction + end bearing

AASHTO LRFD §10.7.3

A driven pile transfers load through side (skin) friction along its embedded length and end (tip) bearing at its toe:

R_n \;=\; R_s + R_p \;=\; \sum_{i} f_{s,i}\, A_{s,i} \;+\; q_p\, A_p

(13.4)

$R_s$: total nominal skin friction [kip]
$R_p$: nominal end-bearing at tip [kip]
$f_s$: unit skin friction in layer i [ksf]
$A_s$: pile perimeter × layer thickness [ft²]
$q_p$: unit end-bearing pressure at tip [ksf]
$A_p$: gross tip area [ft²]

Fig. 13.3Pile axial capacity. Skin friction acts along the pile-soil interface at every depth; end-bearing is developed at the pile tip in the founding stratum.

Cohesive soils (α-method). Unit skin friction:

f_s \;=\; \alpha\, s_u, \qquad \alpha \;=\; 1.0 \;\text{for}\; s_u \le 0.5\;\text{ksf}; \;\; \alpha \;=\; 0.5\text{–}0.8 \;\text{for}\; s_u = 1\text{–}4\;\text{ksf}

(13.5)

End bearing in clay: $q_p = 9\, s_u$ .

Cohesionless soils (β-method).

f_s \;=\; \beta\, \sigma'_v \;\le\; f_{s,lim} \qquad \text{(typically }\beta = 0.25\text{–}0.5\text{)}

(13.6)

End bearing (Meyerhof correlated to SPT): $q_p = 0.4\, N_{60}\, L_e/D \;\le\; 4\, N_{60}$ (ksf, with L_e/D limited by 10).

Factored: $\phi_{stat}\, R_n$ with $\phi_{stat} = 0.45$ without load test, 0.65 with dynamic monitoring, 0.80 with static load test.

13.4 — Drilled shafts

O'Neill & Reese with a rock socket

AASHTO LRFD §10.8

A drilled shaft — CIDH pile in California parlance — is a large-diameter (usually 3–10 ft) reinforced concrete column cast in a drilled hole, often socketed into rock for the last few diameters. Compared to a pile group of equal capacity, a single shaft eliminates the cap, resists lateral load through its own bending stiffness, and can be socketed into rock to isolate the foundation from scour.

Fig. 13.4Drilled shaft with rock socket. Side resistance in overburden follows α or β methods; the rock socket provides both side shear (f_s function of q_u) and tip bearing.

Side resistance in rock (§10.8.3.5.4).

f_s \;=\; 0.65\, p_a \left(\dfrac{q_u}{p_a}\right)^{0.5} \;\le\; 7.8\, p_a \left(\dfrac{f'_c}{p_a}\right)^{0.5}

(13.7)

$q_u$: uniaxial compressive strength of intact rock [ksf]
$p_a$: atmospheric pressure (2.12 ksf)
$f'_c$: shaft concrete compressive strength [ksi]

Tip resistance in rock: $q_p = 2.5\, q_u$ for RQD ≥ 50 %, unweathered.

13.5 — Lateral behaviour

p-y curves and effective fixity

AASHTO LRFD §10.7.3.12

Ship impact, seismic inertia, wind on the superstructure, and eccentric footing loads all put lateral shear at the top of the foundation. AASHTO §10.7.3.12 accepts the p-y method — the pile is modelled as a beam on nonlinear soil springs whose force-displacement curve p(y) depends on soil type and depth.

Fig. 13.5Lateral p-y model. The pile is a beam on non-linear soil springs. Deflection y at each depth mobilises a soil reaction p per unit length; software (LPile/COM624) solves the coupled beam equation.

E I\, \dfrac{d^4 y}{d z^4} \;+\; p(y, z) \;=\; 0

(13.8)

For preliminary sizing, an effective fixity depth $L_f \approx 5\, T$ where $T = (EI/n_h)^{1/5}$ can be used to convert the pile to an equivalent cantilever for hand checks.

13.6 — Group effects

Efficiency and block failure

AASHTO LRFD §10.7.3.9

Piles in a group share the same soil wedge, so their combined capacity is less than $n \cdot Q_{single}$ . AASHTO defines a group efficiency factor that depends on spacing s and diameter D:

\eta \;=\; \dfrac{Q_{group}}{n\, Q_{single}}, \qquad \eta = 1.0 \;\text{for}\; s \ge 6D

(13.9)

Fig. 13.63 × 3 pile group in plan. Efficiency drops from ~1.0 at s = 6D to ~0.7 at s = 3D; in cohesive soils the group is also checked as a block failure with a footing perimeter equal to the group envelope.

13.7 — Downdrag

When the soil pulls down on the pile

AASHTO LRFD §3.11.8, §10.7.3.7

If a pile passes through a compressible layer that is still consolidating — because of new embankment fill, groundwater drawdown, or a nearby surcharge — the layer settles more than the pile and drags the pile downward through side shear. This downdrag (DD) is treated as a permanent load with load factor γ_DD = 1.4 and must be added to the structural axial demand.

Fig. 13.7Downdrag mechanism. New fill and soft clay consolidate around the pile. Above the neutral plane the soil moves down relative to the pile, adding load; below it the pile moves down relative to the soil, still resisting through skin friction.

13.8 — Worked example 1

Spread footing for a bent column

AASHTO LRFD §10.6

Problem statement

A 3-column pier bent (Ch. 9) rests on individual spread footings. Design the interior footing for the factored column reaction on a medium-dense sand founding stratum.

Given

Factored column load $P_u = 900\;\text{kip}, \; M_u = 300\;\text{kip-ft}$
Service load (for settlement) $P_s = 620\;\text{kip}$
SoilMedium-dense sand, φ′ = 34°, γ = 0.120 kcf, E_s = 400 ksf
Footing depth $D_f = 4\;\text{ft}$
Materialsf′_c = 4 ksi, f_y = 60 ksi
Target settlement≤ 1.0 in.

Required

Size a square footing that meets bearing (Strength I) and settlement (Service I), and detail the bottom mat reinforcement.

Step 1 — Bearing capacity factors (φ′ = 34°). From AASHTO Table 10.6.3.1.2a-1: $N_q = 29.4, \; N_\gamma = 41.1$ . Cohesion = 0.

Step 2 — Try B = 10 ft square. Shape/depth factors (Meyerhof):

Formula

s_q = 1 + \dfrac{B}{L}\tan\phi', \;\; s_\gamma = 1 - 0.4\dfrac{B}{L}

Substitute

s_q = 1 + (1)(0.675) = 1.675, \; s_\gamma = 0.6

Result

s_q = 1.68, \; s_\gamma = 0.60

Formula

d_q = 1 + 2\tan\phi'(1-\sin\phi')^2 \dfrac{D_f}{B}

Substitute

d_q = 1 + 2(0.675)(1-0.559)^2 (4/10)

Result

d_q = 1.11, \; d_\gamma = 1.0

Step 3 — Nominal ultimate bearing (Eq. 13.1, c = 0).

Formula

q_n = \gamma D_f N_q s_q d_q + 0.5 \gamma B N_\gamma s_\gamma d_\gamma

Substitute

q_n = (0.120)(4)(29.4)(1.68)(1.11) + 0.5(0.120)(10)(41.1)(0.60)(1.0)

Result

q_n = 26.3 + 14.8 = 41.1\;\text{ksf}

Formula

\phi_b q_n

Substitute

0.45 \times 41.1

Result

= 18.5\;\text{ksf}

Step 4 — Applied Strength I pressure. Eccentricity $e = 300/900 = 0.33\;\text{ft} < B/6 = 1.67$ , so trapezoidal:

Formula

q_{max} = \dfrac{P_u}{B^2}\left(1 + \dfrac{6e}{B}\right)

Substitute

q_{max} = \dfrac{900}{100}(1 + 6\cdot0.33/10)

Result

q_{max} = 11.0\;\text{ksf} < 18.5\;\text{ksf} \;\;\checkmark

Step 5 — Service-load settlement (Eq. 13.3). Service pressure $q_s = 620/100 = 6.2\;\text{ksf}$ . I_f = 0.88, ν = 0.30.

Formula

\delta = \dfrac{q_s B(1-\nu^2)}{E_s}I_f

Substitute

\delta = \dfrac{6.2(120)(1-0.09)}{400}(0.88)

Result

\delta = 1.49\;\text{in.}

Settlement 1.49 in. > 1.0 in. target. Increase B to 12 ft: repeat → δ ≈ 0.9 in ✓; new q_max ≈ 7.4 ksf ≪ 18.5 ✓. Adopt 12 ft × 12 ft × 3.5 ft.

Step 6 — Flexural reinforcement of bottom mat. Critical section at column face. Cantilever length = (12 − 2)/2 = 5 ft. Uniform equivalent pressure ≈ 7.4 ksf. Moment per ft width:

Formula

M_u = q\, l^2/2

Substitute

M_u = 7.4 \times 5^2 / 2

Result

M_u = 92.5\;\text{kip-ft/ft}

Formula

A_s \approx \dfrac{M_u}{\phi f_y (0.9 d)}

Substitute

A_s = \dfrac{92.5 \times 12}{0.9 (60)(0.9 \times 30)}

Result

A_s = 0.76\;\text{in}^2/\text{ft} \Rightarrow \text{use #8 @ 8 in. (0.79 in}^2/\text{ft)}

Fig. 13.8Figure 13.8. Trial footing — 10 ft × 10 ft × 3.5 ft with a #8 @ 8 in. bottom mat each way.

Final section detailing (from computed A_s)

Bent-column spread footing (interior)

Location	A_s required	Bars provided	Spacing / detail
Plan dimensions	B ≥ 12 ft for settlement	12 ft × 12 ft square	concentric under column
Thickness	d ≥ 24 in. for one-way and punching shear	3.5 ft (42 in.)	3 in. clear cover bottom
Bottom mat (both ways)	A_s = 0.76 in²/ft	#8 @ 8 in. each way (0.79 in²/ft)	extend full width less 3 in. cover
Top mat	shrinkage/temperature #5 @ 12 in.	same each way	hooked ends into edges
Dowels into column	match column longitudinal	8 – #9 dowels, 40 db lap	development length 4 ft-6 in. into footing

Bearing pressure q_max ≈ 7.4 ksf ≪ φ_b q_n = 18.5 ksf, so strength is not close. Settlement was the driver — a common outcome for granular founding strata under bent columns.

13.9 — Worked example 2

Driven-pile group for a river pier

AASHTO LRFD §10.7

Problem statement

A river pier column carries a factored Strength I axial load and passes through 40 ft of medium-stiff clay overlying dense sand. Design a driven-pile group to support the load.

Given

Factored axial load $P_u = 2{,}400\;\text{kip}$
PilesHP 14×89 (A = 26.1 in², perimeter p = 4.28 ft)
Clay (0 – 40 ft)s_u = 1.5 ksf, α = 0.55
Dense sand (40 – 65 ft)β = 0.40, γ′ = 0.055 kcf (below GWT)
Spacings = 3D = 3.5 ft
Resistance factorφ_stat = 0.65 (dynamic monitoring)

Required

Design a rectangular pile group (n and layout), verify group efficiency, and check block failure in clay.

Step 1 — Single-pile skin friction in clay (0 – 40 ft).

Formula

R_{s,clay} = \alpha\, s_u\, p\, L_c

Substitute

R_{s,clay} = 0.55 (1.5)(4.28)(40)

Result

R_{s,clay} = 141\;\text{kip}

Step 2 — Skin friction in sand (40 – 60 ft). Average σ′_v in that layer ≈ 3.4 ksf (accounting for γ′ below GWT).

Formula

R_{s,sand} = \beta\, \bar\sigma'_v\, p\, L_s

Substitute

R_{s,sand} = 0.40 (3.4)(4.28)(20)

Result

R_{s,sand} = 116\;\text{kip}

Step 3 — End bearing at tip (60 ft in dense sand, N_60 ≈ 45).

Formula

q_p = \min\Big(0.4\, N_{60}\, L_e/D,\; 4\, N_{60}\Big)

Substitute

q_p = \min(0.4\cdot45\cdot10,\; 4\cdot45) = \min(180,\;180)

Result

q_p = 180\;\text{ksf}

Formula

R_p = q_p A_p

Substitute

R_p = 180 \times (26.1/144)

Result

R_p = 32.6\;\text{kip}

Step 4 — Nominal single-pile capacity.

Formula

R_n = R_{s,clay} + R_{s,sand} + R_p

Substitute

R_n = 141 + 116 + 33

Result

R_n = 290\;\text{kip}, \; \phi_{stat} R_n = 0.65(290) = 189\;\text{kip}

Step 5 — Required number of piles.

Formula

n = \dfrac{P_u}{\phi R_n}

Substitute

n = 2400 / 189

Result

n = 12.7 \Rightarrow \text{try 4 × 4 group (16 piles)}

Step 6 — Group efficiency. s/D = 3.0 in clay ⇒ η ≈ 0.70 (AASHTO Fig. 10.7.3.9-1). Group capacity:

Formula

Q_{group} = \eta\, n\, \phi R_n

Substitute

Q_{group} = 0.70(16)(189)

Result

Q_{group} = 2{,}117\;\text{kip} < 2{,}400 \;\; \text{not enough}

Try 5 × 4 = 20 piles: Q_group = 0.70(20)(189) = 2,646 kip > 2,400 ✓.

Step 7 — Block failure in clay. Group envelope 14 ft × 10.5 ft × 60 ft. Block skin friction + block end bearing = large; check dominates only for very close spacing in soft clay — verified adequate here.

Fig. 13.9Figure 13.9. Pile group elevation — HP 14×89 piles driven 60 ft through clay into 20 ft of dense sand.

Final section detailing (from computed A_s)

Driven pile group — river pier

Location	A_s required	Bars provided	Spacing / detail
Pile section	HP 14×89, F_y = 50 ksi	same, ASTM A572 Gr 50	5 rows × 4 columns = 20 piles
Layout	s ≥ 3D per §10.7.1.5	3.5 ft o.c. (s = 3D)	cap 14 × 10.5 × 4.5 ft
Embedment	L ≥ 60 ft to reach dense sand	60 ft driven length	tip elev. verified by wave-equation
Pile cap reinforcement	strut-and-tie, both ways	#11 @ 12 in. bottom mat, #8 @ 12 in. top	3 in. clear cover
Pile-to-cap connection	shear stud + 6 in. embed	12 – 3/4 in. Ø × 6 in. studs per pile	welded to top flange

Group efficiency (η = 0.70) drove the count from 13 required to 20 provided. Consider PDA monitoring during driving to raise φ_stat from 0.45 to 0.65 as assumed here.

13.10 — Guided practice

Rock-socketed drilled shaft

A 6-ft-diameter drilled shaft is socketed 15 ft into shale with $q_u = 500\;\text{ksf}$ . Concrete f′_c = 4 ksi. Ignore side friction in the overburden. Compute the factored socket capacity with $\phi = 0.55$ .

Expected result

p_a = 2.12 ksf; f_s = 0.65(2.12)√(500/2.12) = 21.2 ksf. Cap: 7.8(2.12)√(4/2.12·144) = 21.6 ksf, so f_s = 21.2 ksf governs. R_s = 21.2·π·6·15 = 5,990 kip. Tip: q_p = 2.5(500) = 1,250 ksf, R_p = 1,250·π/4·36 = 35,300 kip. φR_n = 0.55(5,990 + 35,300) = 22,700 kip — governed by shaft structural capacity, not soil.

13.11 — Mini design challenge

River-crossing drilled-shaft foundation with scour

Fig. 13.10Design challenge geometry. River pier at design flood — evaluate scour envelope, size a drilled shaft socketed into shale, and check lateral capacity for the barge impact case.

Deliver:

Predicted scour depth y_s from AASHTO §2.6.4 (link to Ch. 14) — pier + contraction scour.
Drilled-shaft diameter and socket depth that ignores side friction above the scour envelope.
Axial capacity check (Strength I) and lateral p-y check for a 1,000-kip barge impact at pier cap (Extreme Event II).
Shaft reinforcement (longitudinal + spiral) for combined axial + flexure.
Foundation schedule and a one-page design memo.

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13.12 — Chapter summary

What you leave with

Four foundation families and the selection triggers that pick between them.
Meyerhof bearing capacity (Eq. 13.1) and Hough elastic settlement (Eq. 13.3) for spread footings.
α- and β-methods for driven pile skin friction, plus SPT-based end bearing.
O'Neill–Reese rock socket for drilled shafts (Eq. 13.7).
p-y lateral analysis and group efficiency for pile groups.
Downdrag (γ_DD = 1.4) as a permanent load in every settling profile.

AASHTO LRFD §10 — Foundations

Section 2

Fully Worked Examples

Complete AASHTO LRFD solutions with knowns, assumptions, step calculations, verification, and design commentary. Difficulty rises from basic to consulting-grade.

Worked Example 1

Concept: bearing pressure under a spread footing

Basic

Problem

Compute

q_{avg}

(ksf).

Step-by-Step

A = B \cdot L = 80\ \text{ft}^{2}

q = \frac{Q}{A} = \frac{400}{80} = 5.0\ \text{ksf}

Result

q_{avg} = 5.0\ \text{ksf}

Design Verification

Order-of-magnitude sanity: typical medium-dense sand allowable is 3–6 ksf. Marginal — refine with Meyerhof.

Discussion

Uniform contact pressure is a design idealization. Real distribution beneath rigid footings on sand is parabolic (max at center) and on clay is saddle-shaped (max at edges).

Worked Example 2

Simple hand calc: ultimate bearing capacity (Meyerhof, cohesionless)

Basic

Problem

Compute

q_{ult}

(ksf) and factored

q_R

Step-by-Step

N_q
$N_q = e^{\pi \tan\phi} \cdot \tan^{2}\!\left(45 + \tfrac{\phi}{2}\right) = e^{2.12}\cdot \tan^{2}(62^\circ) = 8.3 \cdot 3.55 = 29.5$
N_γ
$N_\gamma = (N_q - 1)\cdot \tan(1.4\phi) = 28.5 \cdot \tan(47.6^\circ) = 31.3$

s_q
$s_q = 1 + \tfrac{B}{L}\tan\phi = 1 + 0.8 \cdot 0.675 = 1.54$
s_γ
$s_\gamma = 1 - 0.4\tfrac{B}{L} = 0.68$

Design Verification

Applied q_avg = 5.0 ksf ≪ q_R = 14.4 ksf → Bearing capacity OK, settlement will govern.

Discussion

In cohesionless soils, settlement (not bearing) almost always controls spread-footing size.

Worked Example 3

AASHTO settlement of spread footing on granular soil (Hough)

Intermediate

Problem

Compute

\rho

(in).

Step-by-Step

Consider 2 layers each of thickness $H_i = 0.5\cdot B = 4\ \text{ft}$ .

Formula
$\rho_i = \frac{H_i}{C'}\log_{10}\!\left(\frac{\sigma_o + \Delta\sigma}{\sigma_o}\right)$
Layer 1
$\sigma_o = 0.480 + 0.240 = 0.72\ \text{ksf};\ \Delta\sigma \approx 4.0\ \text{ksf}$
Result
$\rho_1 = \frac{4\cdot 12}{60}\log_{10}\!\left(\frac{4.72}{0.72}\right) = 0.80\cdot 0.817 = 0.65\ \text{in}$

Design Verification

1-in service settlement is at AASHTO tolerable limit for spread footings on multi-span bridges (§10.5.2.1). Consider deepening or widening if angular distortion > 0.004.

Discussion

Hough is conservative — refined constitutive analyses often show 50–70% of Hough predictions.

Worked Example 4

Complete design: reinforced concrete spread footing

Intermediate

Problem

Design flexural and punching-shear reinforcement.

Step-by-Step

Eccentricity
$e = \frac{M}{Q} = \frac{250}{560} = 0.446\ \text{ft}$
Result
$\tfrac{B}{6} = 1.67\ \text{ft} \ge e \Rightarrow \text{within kern, no uplift}$

Cantilever arm
$a = \frac{10 - 2}{2} = 4\ \text{ft}$
Moment per ft
$M_u/\text{ft} = \frac{7.5\cdot 4^{2}}{2} = 60\ \text{k-ft/ft}$
Result
$M_u = 60 \cdot 8 = 480\ \text{k-ft}$

Design Verification

Min steel §5.6.7 — provided A_s well exceeds 0.11·b·d/f_y·(4 ksi factor).

Discussion

Footing design almost never governed by punching for a stiff column-to-footing ratio; flexure usually rules.

Worked Example 5

Multi-limit-state: overturning, sliding, and eccentricity on an abutment footing

Advanced

Problem

All three checks.

Step-by-Step

x̄
$\bar{x} = \frac{M_r - M_{ot}}{V} = \frac{8{,}900 - 3{,}100}{780} = 7.44\ \text{ft from toe}$
e
$e = \frac{B}{2} - \bar{x} = 11 - 7.44 = 3.56\ \text{ft}$
Result
$e = 3.56\ \text{ft} \le \tfrac{B}{4} = 5.5\ \text{ft}\ \checkmark$

\frac{e}{B} = \frac{3.56}{22} = 0.16 \le 0.25\ \checkmark

Design Verification

All three checks pass with margin. If EH increases 25% (Strength IV), rerun.

Discussion

AASHTO retired classical FS_OT ≥ 2.0 in favor of eccentricity-based checks per §11.6 for footings on soil.

Worked Example 6

Design optimization: shallow vs deep foundation trade study

Advanced

Problem

Recommend a foundation type.

Step-by-Step

q_R
$q_R = \phi\,q_{ult} = 0.5\cdot 2.57 = 1.29\ \text{ksf}$
Area
$A = \frac{3{,}200}{1.29} = 2{,}480\ \text{ft}^2 \rightarrow 50\times 50\ \text{ft}$
Result
$\text{Impractical; also} > 4\ \text{in settlement} \Rightarrow \text{NG}$

42-in shafts, 45 ft to dense sand tip. Axial per shaft ≈ 700 kip (400 side + 300 tip); 12 shafts × 700 = 8,400 kip; more than enough.

Design Verification

Soft clay always drives foundations deep. Settlement, not strength, is the deciding factor.

Discussion

Rule of thumb: if S_u < 1 ksf near surface, go deep.

Worked Example 7

Construction-stage: temporary casing removal on a drilled shaft

Advanced

Problem

Minimum concrete head above slurry to guarantee positive pressure.

Step-by-Step

\gamma_c H_c = \gamma_{slurry} H_s + FS\,(\gamma_{slurry} H_s),\quad FS = 1.2

Result

Δh_c ≥ 5 ft during withdrawal

Tremie pipe kept 5–10 ft embedded; slow withdrawal (1 ft/min) with continuous supply.

Design Verification

Matches FHWA IF-99-025 standard practice.

Discussion

Shaft defects almost always trace to casing extraction, not to bar placement. Field control is the whole ballgame.

Worked Example 8

Consulting: scour analysis and re-founding of an existing pier

Consulting

Problem

Is the pier still stable? Options?

Step-by-Step

D_{f,eff} = 6 - 12 = -6\ \text{ft} \Rightarrow \text{footing undermined}

(1) Micropile underpinning 4×80 ft; (2) Sheet-pile scour countermeasure ring; (3) Rip-rap Class III per HEC-23. Countermeasure cheapest but only Level 1; underpinning fixes long-term.

Design Verification

Post-underpinning FS ≥ 2.0; scour no longer engages foundation reactions.

Discussion

Scour is the #1 cause of bridge failure in the US. Every consulting inspection must verify scour depth vs original foundation depth.

Worked Example 9

Case-study reproduction: Schoharie Creek Bridge failure back-analysis

Consulting

Problem

Depth of scour to cause overturning about the downstream edge?

Step-by-Step

M_{stab} = V\cdot\frac{B}{2} = 3{,}400\cdot 9 = 30{,}600\ \text{k-ft}

H·(h_arm) + moment from asymmetric bearing pressure as scour advances. When 12+ ft of scour occurs on upstream side, the effective footprint reduces to ≈ 10 × 18 = 180 ft² all on the downstream half — no bearing capacity margin left.

Design Verification

Matches NTSB back-analysis. Failure occurred at ~ Q100 flow with foundation designed to Q50 without scour analysis.

Discussion

Post-Schoharie, AASHTO §2.6.4.4.2 mandated scour analysis for all new bridges and required Plans of Action for scour-critical existing bridges.

Worked Example 10

Comprehensive: full foundation design for a 3-column interior pier

Consulting

Problem

Deliver complete foundation package.

Step-by-Step

Spread footing feasible (dense sand near surface). Trial 42 × 18 ft × 5 ft footing.

e_T
$e_T = \frac{8{,}200}{4{,}500} = 1.82\ \text{ft} \le \tfrac{B}{4} = 4.5\ \text{ft}\ \checkmark$
e_L
$e_L = \frac{1{,}600}{4{,}500} = 0.36\ \text{ft} \le \tfrac{L}{4} = 10.5\ \text{ft}\ \checkmark$

Design Verification

Bearing 71% utilized, settlement 70% utilized — balanced design. If ADT increases require an additional 1,000 kip, deepen to 6 ft footing.

Discussion

Complete foundation design integrates geotechnical, structural, and construction perspectives. Every quantity must round-trip to a drawing and a schedule.

Worked Example 11

Skin-friction capacity of a single steel H-pile

Intermediate

Problem

Compute

Q_s

and

\phi Q_s

Step-by-Step

\bar{\sigma}'_v = \tfrac{1}{2}\,\gamma'\,L = \tfrac{1}{2}(0.0576)(55) = 1.584\ \text{ksf}

f_s = K\,\bar{\sigma}'_v\,\tan\delta = 0.8\,(1.584)\,\tan 23.1^\circ = 0.8\,(1.584)(0.427) = 0.541\ \text{ksf}

Design Verification

For medium sand, f_s of 0.3–0.7 ksf averaged over the shaft is typical — the 0.54 ksf value is in-range.

Discussion

Buoyant unit weight cuts side resistance nearly in half compared to dry conditions. Always confirm water-table elevation and use consistent stress units.

Worked Example 12

Pile-group efficiency and settlement of a 3×3 driven pile group

Advanced

Problem

Compute the group efficiency η, verify capacity, and estimate the group settlement ρ (Meyerhof method).

Step-by-Step

\theta = \arctan(D/s) = \arctan(12/36) = 18.43^\circ

\eta = 1 - \dfrac{\theta}{90}\cdot \dfrac{(m-1)k + (k-1)m}{m\,k} = 1 - \dfrac{18.43}{90}\cdot \dfrac{2\,(3)+2\,(3)}{9}

Result

\eta = 1 - 0.205\,(1.333) = 0.727

Design Verification

Efficiency ≈ 0.73 is normal for s = 3D in sand; tightening spacing to 2D drops η toward 0.55 and drives group capacity to control over single-pile capacity.

Discussion

Widen spacing to 4D whenever cap geometry permits — the efficiency gain almost always outweighs the modest cap-size penalty, and settlement drops with the square root of B_g.