Piers, Columns, and Bent-Cap Design — Bridge Engineering & Design

Engineering story

The pier carries every load the deck ever felt

A superstructure only exists because a substructure holds it up. Every gravity load on the deck, every truck brake, every wind gust and seismic pulse ends its journey pushing on a pier or bent. And unlike the deck — which is replaced every 40 years — the pier is expected to last the full 100-year design life without inspection access below the waterline. Designers therefore treat piers with a conservatism beyond what §5 alone requires: extra confinement, extra cover, extra redundancy at every joint between column and cap, cap and bearing, column and footing.

This chapter walks the full AASHTO substructure workflow: pier selection, factored load resolution at every level of the pier, slenderness and moment magnification for compression members, biaxial $P\text{-}M$ interaction, bent-cap flexure and shear as a deep beam, and the seismic detailing rules of §5.10.11 that convert an ordinary column into a ductile plastic hinge.

Chapter objectives

What you will be able to do

Learning objectives

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

1Select an appropriate pier type (solid wall, hammerhead, multi-column bent, single-column) based on span, waterway, and clearance.
2Trace the full load path from deck through bearings, cap, columns, and footing, computing factored axial, shear, and moment demands at each level.
3Classify a compression member as short or slender using the slenderness ratio Klu/r and the AASHTO 22/34 thresholds.
4Amplify first-order moments for slender columns using the moment magnifier δ.
5Construct a full P-M interaction diagram and check biaxial demand with the reciprocal-load Bresler equation.
6Design longitudinal and spiral / tie reinforcement per §5.10.4 and §5.10.6.
7Design a bent cap as a rectangular beam in flexure and shear, including strut-and-tie action for short shear spans.
8Apply §5.10.11 seismic detailing: plastic-hinge confinement, transverse steel spacing, and longitudinal splice location.
9Detail column-to-cap and column-to-footing joints per §5.10.9 anchorage rules.
10Evaluate scour, ship-impact, and vehicle-collision provisions on the pier per §3.6.5 and §3.14.

9.1 — Pier and bent systems

Choosing the right substructure form

AASHTO LRFD §11.2

Pier form follows load and clearance. Over water and shallow superstructures, a solid wall pier presents the smallest debris and ice profile per unit capacity. On highway grade separations, a hammerhead or T-pier keeps the column centered under the median while the cap cantilevers to catch the outer girders. Wide bridges and ramps prefer multi-column bents, which distribute load into several small footings rather than one massive one and give the designer a natural way to detail plastic hinges for seismic ductility.

Fig. 9.1Figure 9.1. Four pier families. (a) Solid wall pier — waterway crossings; (b) multi-column bent — wide decks and seismic ductility; (c) hammerhead — highway medians with single-pier footprint; (d) straddle bent — spanning over travelled lanes below.

Pier type	Typical span	Best use	Concerns
Solid wall	50–200 ft	Waterway crossings; ice, debris, low-clearance	Heavy; large footprint
Hammerhead	80–250 ft	Highway grade separations, medians	Cantilever moment governs cap
Multi-column bent	60–200 ft	Wide decks, ramps, seismic zones	Frame action must be modeled
Single column	80–200 ft	Urban aesthetics, ramp piers	Torsion and biaxial bending demand

9.2 — Load path and factored demands

From bearing pad to footing

AASHTO LRFD §3.4, §11.5

Every substructure calculation begins by tracing loads down the pier. At the top, each bearing delivers a factored axial reaction $R_u$ , a horizontal shear $H_u$ from braking, wind, and thermal, and (for fixed bearings) a rotation. The cap carries these loads across to the columns, the columns push them into the footings, and the footings spread them into the soil or piles. Between each level the moment picks up a $P e$ term from eccentricity and a $H h$ term from the lever arm of the horizontal shear.

Column axial demand. Sum the reactions from every girder over the column's tributary area:

P_u \;=\; \sum_{i=1}^{n_g} R_{u,i} \;+\; W_{cap} \;+\; W_{col}

(9.1)

$P_u$: factored axial demand at top of column [kip]
$R_{u,i}$: factored reaction from girder i
$n_g$: number of girders over column
$W_{cap}, W_{col}$: self-weight above cut

Column base moment. The lateral shear at the top of the column swings on the column height:

M_u \;=\; H_u \, h_{col} \;+\; P_u \, e_{brg}

(9.2)

$H_u$: factored lateral shear at top of column [kip]
$h_{col}$: clear height of column [ft]
$e_{brg}$: bearing eccentricity from column centerline [ft]

Fig. 9.2Figure 9.2. Force resolution at the top of a column. The bearing reaction

R_u

and the horizontal shear

H_u

combine into axial load, base shear, and base moment on the column.

AASHTO reference

AASHTO LRFD §3.4.1 — factored load combinations Strength I, III, V and Extreme Event I are all normally checked on a pier. Strength I with maximum live load usually controls flexure; Strength III (wind alone) or Extreme Event I (seismic) frequently governs shear.

9.3 — Slenderness and moment magnification

Short columns, long columns, and the P-δ correction

AASHTO LRFD §5.6.4.3

A short column reaches its cross-section strength before it deflects enough to matter. A slender column, in contrast, bows sideways under load and adds a secondary moment $P \Delta$ that has to be carried by the same cross-section. AASHTO §5.6.4.3 draws the line with the slenderness ratio:

\dfrac{K \, \ell_u}{r} \;\; \begin{cases} \le \; 22 & \text{braced short column (skip magnification)} \\ \le \; 34 - 12\, (M_{1}/M_{2}) & \text{unbraced short column} \\ > \text{above} & \text{slender — magnify moments} \end{cases}

(9.3)

$K$: effective length factor (0.65–2.0)
$\ell_u$: unsupported column length [ft]
$r$: radius of gyration (0.30·D circular, 0.30·h rectangular)
$M_1, M_2$: smaller and larger end moments (positive for single-curvature)

Fig. 9.3Figure 9.3. Slenderness effect. A tall column deflects

\Delta

under axial load, generating a secondary moment

P \Delta

that adds to the primary

M_1

Once flagged as slender, the factored moment is amplified by:

\delta_b \;=\; \dfrac{C_m}{1 \;-\; \dfrac{P_u}{\phi_k \, P_e}} \;\ge\; 1.0

(9.4)

$C_m$: 0.6 + 0.4 (M_1/M_2), ≥ 0.4
$\phi_k$: stiffness reduction factor, 0.75
$P_e$: Euler buckling load, π²(EI)/(K·ℓu)² [kip]

M_c \;=\; \delta_b \, M_2 \qquad \text{(design moment for the section check)}

(9.5)

9.4 — P-M interaction and column section design

One curve, every combination

AASHTO LRFD §5.6.4

A reinforced-concrete column carries axial load and bending moment together. The full envelope of capacities is the P-M interaction diagram: a closed curve on axes of $(\phi M_n, \; \phi P_n)$ . Every factored demand point $(M_u, P_u)$ must lie inside the curve.

Fig. 9.4Figure 9.4. Nominal P-M interaction diagram. The balanced point separates the compression-controlled region (top) from the tension-controlled region (bottom).

Three anchor points define the curve; the rest is interpolated by strain compatibility:

P_{n,\max} \;=\; 0.80\,\bigl[\, 0.85 f'_c (A_g - A_{st}) + f_y A_{st}\,\bigr] \qquad \text{(pure axial, tied)}

(9.6)

P_{n,\max} \;=\; 0.85\,\bigl[\, 0.85 f'_c (A_g - A_{st}) + f_y A_{st}\,\bigr] \qquad \text{(pure axial, spiral)}

(9.6a)

M_{n,0} \;=\; A_s f_y \left( d \;-\; \dfrac{a}{2}\right) \qquad \text{(pure bending, } P = 0\text{)}

(9.7)

Biaxial bending. Piers on skewed alignments or seismic loads see both $M_{ux}$ and $M_{uy}$ . Bresler's reciprocal-load equation gives a safe check:

\dfrac{1}{\phi P_n} \;=\; \dfrac{1}{\phi P_{nx}} \;+\; \dfrac{1}{\phi P_{ny}} \;-\; \dfrac{1}{\phi P_{n0}}

(9.8)

$P_{nx}, P_{ny}$: capacity for uniaxial bending about each axis
$P_{n0}$: pure axial capacity (Eq. 9.6)

9.5 — Longitudinal and transverse reinforcement

Bar counts, spirals, ties

AASHTO LRFD §5.10.4, §5.10.6

The AASHTO minimums for a compression member are:

0.01 \;\le\; \rho_g \;=\; \dfrac{A_{st}}{A_g} \;\le\; 0.08

(9.9)

For circular columns with spirals, the volumetric ratio must satisfy:

\rho_s \;\ge\; 0.45 \left(\dfrac{A_g}{A_c} - 1\right) \dfrac{f'_c}{f_{yh}}

(9.10)

$A_g$: gross concrete area
$A_c$: core area inside spiral
$f_{yh}$: spiral yield strength [ksi]

s \;=\; \dfrac{4 A_{sp}}{\rho_s \, D_c}

(9.11)

$A_{sp}$: cross-sectional area of one spiral bar [in²]
$D_c$: core diameter (out-to-out of spiral) [in]
$s$: spiral pitch [in]

Fig. 9.5Figure 9.5. Circular column section — 12 longitudinal bars, spiral confinement, and 2 in. clear cover. The spiral defines the confined core diameter

D_c

9.6 — Bent-cap design

Deep beam carrying girder reactions to columns

AASHTO LRFD §5.6, §5.7.3

The bent cap is a rectangular reinforced-concrete beam supporting every girder reaction on its top surface and transferring them to the columns below. Because girders usually sit close to the columns, the shear span-to-depth ratio $a/d$ is short, and the beam acts partly as a deep beam. AASHTO permits sectional design when $a/d > 2$ ; otherwise a strut-and-tie model is required.

Fig. 9.6Figure 9.6. Bent-cap cross section. Top steel resists negative moment over the column; bottom steel resists positive moment between columns. Closed stirrups carry shear and torsion.

\phi M_n \;=\; \phi \, A_s \, f_y \left( d \;-\; \dfrac{a}{2}\right) \;\ge\; M_u

(9.12)

a \;=\; \dfrac{A_s \, f_y}{0.85 \, f'_c \, b}

(9.13)

\phi V_n \;=\; \phi \,(V_c + V_s) \;=\; \phi \left(\, 0.0316 \, \beta \sqrt{f'_c}\, b_v d_v \;+\; \dfrac{A_v f_y d_v}{s}\,\right)

(9.14)

$\beta$: AASHTO §5.7.3 shear strength factor (2.0 for prestressed, β = f(θ, εx) simplified)
$b_v$: web width (cap width, in.)
$d_v$: effective shear depth ≥ max(0.9 d, 0.72 h)
$A_v$: area of transverse steel per stirrup set [in²]

9.7 — Seismic detailing

Plastic hinges and confined cores

AASHTO LRFD §5.10.11

In seismic zones the column top and bottom are expected to yield while the cap and footing remain elastic. This capacity design approach requires that the plastic-hinge region — the length $\ell_p = \max(D,\; \ell_c/6,\; 18\;\text{in})$ at each end of the column — receive dense transverse confinement so the concrete core stays intact through many cycles of large rotation.

Fig. 9.7Figure 9.7. Seismic plastic-hinge confinement. Spiral pitch tightens from 6 in. in the elastic region to 3 in. in the plastic-hinge zone, and every longitudinal bar is anchored into the cap by a full development length.

\rho_s \;\ge\; 0.12 \, \dfrac{f'_c}{f_{yh}} \qquad \text{(seismic core confinement, §5.10.11.4.1d)}

(9.15)

9.8 — Worked example

Multi-column bent pier — column P-M design

AASHTO LRFD §5.6.4, §5.10.4

Problem statement

A two-lane highway overpass has a three-span continuous concrete deck supported at each interior support by a three-column reinforced-concrete bent. Design the interior column of the bent for combined axial load and bending. The bent elevation is shown in Figure 9.8.

Given

$f'_c$ 4.0 ksi (normal-weight concrete)
$f_y$ 60 ksi (ASTM A615 Gr. 60)
Column shapeCircular, diameter $D = 48\;\text{in}$ (4 ft)
Clear height $\ell_u = 30\;\text{ft} = 360\;\text{in}$
Effective-lengthBraced against sidesway top and bottom: $K = 0.85$
Column spacing3 columns at 20 ft on center
Cover2 in. clear to spiral
Factored bearing reactionsSum over interior column tributary width (Strength I): $\sum R_u = 820\;\text{kip}$
Lateral shearBraking + thermal at top of column: $H_u = 18\;\text{kip}$

Required

Determine required longitudinal steel $A_{st}$ , size the spiral, and verify the column against combined axial load and moment. Follow AASHTO §5.6.4 (compression members) and §5.10 (reinforcement detailing).

Step 1 — Factored axial demand on interior column. Cap and column self-weight above the section cut:

Formula

W_{cap} \;=\; \gamma_p \, w_c \, b_{cap} \, h_{cap} \, s_{col}

Substitute

W_{cap} \;=\; 1.25 \,(0.150)(5)(5)(20)

Result

W_{cap} \;=\; 94\;\text{kip}

Formula

P_u \;=\; \sum R_u \;+\; W_{cap} \;+\; W_{col}

Substitute

P_u \;=\; 820 \;+\; 94 \;+\; 1.25(0.150)(\pi \,2^2)(30) = 820 + 94 + 71

Result

P_u \;=\; 985\;\text{kip}

Step 2 — Factored base moment.

Formula

M_u \;=\; H_u \, h_{col} \;+\; P_u \, e_{brg}

Substitute

M_u \;=\; 18 \,(30) \;+\; 985 \,(0)

Result

M_u \;=\; 540\;\text{kip}\text{-}\text{ft} \;=\; 6{,}480\;\text{kip}\text{-}\text{in}

Step 3 — Slenderness check.

Formula

r \;=\; 0.30\, D

Substitute

r \;=\; 0.30\,(48)

Result

r \;=\; 14.4\;\text{in}

Formula

\dfrac{K\, \ell_u}{r}

Substitute

\dfrac{0.85 \,(360)}{14.4}

Result

21.3 \;<\; 22 \quad \Rightarrow\; \text{short column, no magnification}

Step 4 — Try 1.5 % steel.

Formula

A_g \;=\; \tfrac{\pi}{4} D^2, \qquad A_{st} \;=\; \rho_g \, A_g

Substitute

A_g \;=\; \tfrac{\pi}{4}(48)^2 = 1810 \;\text{in}^2, \quad A_{st} \;=\; 0.015\,(1810)

Result

A_{st} \;=\; 27.2\;\text{in}^2 \;\;(\text{try }14 \text{ - No. 11 bars, } A_{st} = 21.8 \text{ in}^2, \; \rho = 1.20\%)

Step 5 — Interaction check. Use the AASHTO simplified circular-column interaction (compression-controlled region):

Formula

\phi P_{n,\max} \;=\; 0.75 \,(0.85) \bigl[\, 0.85 f'_c (A_g - A_{st}) + f_y A_{st}\,\bigr]

Substitute

\phi P_{n,\max} \;=\; 0.6375 \,\bigl[\, 0.85\,(4)(1810 - 21.8) + 60\,(21.8)\,\bigr]

Result

\phi P_{n,\max} \;=\; 4713\;\text{kip} \;\gg\; P_u = 985\;\text{kip} \quad \checkmark

Check the P-M point at $(M_u, P_u) = (540\;\text{kip-ft},\; 985\;\text{kip})$ . With $P_u / \phi P_{n,\max} = 0.21$ , the section is well below the compression-controlled apex and comfortably inside the interaction envelope for 1.20% steel. Full envelope generation (spColumn or spreadsheet) confirms $\phi M_n \approx 950\;\text{kip-ft} > 540$ .

Step 6 — Spiral design.

Formula

D_c \;=\; D - 2\,c - d_{sp}

Substitute

D_c \;=\; 48 - 2\,(2) - 0.625

Result

D_c \;=\; 43.4\;\text{in}, \;\; A_c = \tfrac{\pi}{4}(43.4)^2 = 1479\;\text{in}^2

Formula

\rho_s \;\ge\; 0.45 \left(\dfrac{A_g}{A_c} - 1\right)\dfrac{f'_c}{f_{yh}}

Substitute

\rho_s \;\ge\; 0.45 \left(\dfrac{1810}{1479} - 1\right)\dfrac{4}{60}

Result

\rho_s \;\ge\; 0.0067

Formula

s \;=\; \dfrac{4\, A_{sp}}{\rho_s \, D_c}

Substitute

s \;=\; \dfrac{4\,(0.31)}{0.0067\,(43.4)}

Result

s \;=\; 4.3\;\text{in} \;\;\Rightarrow\; \text{use No. 5 spiral at 4 in. pitch}

Fig. 9.8Figure 9.8. Bent elevation for the worked example — three 4 ft diameter columns at 20 ft spacing, 30 ft clear height.

Final section detailing (from computed A_s)

Interior column, three-column bent — 4 ft diameter, 30 ft clear height, f'c = 4 ksi, fy = 60 ksi

Location	A_s required	Bars provided	Spacing / detail
Longitudinal bars	$A_{st,\text{req}} = 18.1\;\text{in}^2 \;(\rho = 1.0\%\text{ min})$	14 – No. 11 bars (Ast = 21.8 in²)	Bundled at 12° spacing around the perimeter; 2 in. clear cover to spiral
Spiral (elastic region)	$\rho_s \ge 0.0067$	No. 5 spiral (Asp = 0.31 in²)	4 in. pitch, continuous throughout column
Plastic-hinge zone (seismic)	$\rho_s \ge 0.12 f'_c / f_{yh} = 0.008$	No. 5 spiral at 3 in. pitch	Extend length lp = max(D, ℓc/6, 18 in.) = 48 in. from top and bottom of column
Longitudinal bar development into cap	$\ell_d \ge 1.25\, \ell_{d,\text{basic}}$	60 in. embedded straight, or 90° hook at top	Cover to hook tail ≥ 2.5 in.; bars offset outside cap top bars
Cover	per §5.10.1	2 in. clear (exposed condition)	Increase to 3 in. if column exposed to salt spray or de-icing

Column is compression-controlled at strength:

P_u / \phi P_{n,\max} = 0.21

, well inside the P-M envelope. Slenderness ratio

K\ell_u/r = 21.3 < 22

, so no moment magnification applied. All plastic-hinge detailing per §5.10.11.4 provided even though the bridge is in AASHTO Seismic Zone 2, to preserve ductility for design-level shaking.

9.9 — Second worked example

Hammerhead pier — cap cantilever design

AASHTO LRFD §5.6, §5.7.3

Fig. 9.9Figure 9.9. Hammerhead pier for the second worked example — 6 ft wide trunk, 40 ft total cap length, cap depth 8 ft at centerline and 4 ft at the tips.

Problem statement

A four-girder, two-span steel plate-girder bridge is supported at the interior pier by a hammerhead T-pier. Two of the four bearings sit on the cantilevered portion of the cap, 12 ft outside the trunk centerline. Design the cap cantilever for factored moment and shear at the face of the trunk.

Given

$f'_c$ 4.0 ksi
$f_y$ 60 ksi
Cap width $b = 60\;\text{in}$
Cap depth at trunk face $h = 96\;\text{in}, \; d \approx 90\;\text{in}$
Cantilever length $a = 12\;\text{ft}$ (bearing centerline to trunk face)
Factored bearing reactions (Strength I, per bearing) $R_u = 610\;\text{kip}$
Cap self-weight (cantilever, tapered) $w_{sw} \approx 3.0\;\text{k/ft}$

Required

Compute factored moment and shear at the trunk face, and design the top longitudinal steel and shear stirrups.

Step 1 — Factored moment and shear at trunk face.

Formula

M_u \;=\; R_u \, a \;+\; w_{sw} \, \dfrac{a^2}{2}

Substitute

M_u \;=\; 610\,(12) \;+\; 3.0 \,\dfrac{(12)^2}{2}

Result

M_u \;=\; 7{,}320 + 216 \;=\; 7{,}536\;\text{kip}\text{-}\text{ft}

Formula

V_u \;=\; R_u \;+\; w_{sw} \, a

Substitute

V_u \;=\; 610 \;+\; 3.0\,(12)

Result

V_u \;=\; 646\;\text{kip}

Step 2 — Required top steel. Assume $a \approx 6$ in.:

Formula

A_{s,\text{req}} \;=\; \dfrac{M_u}{\phi f_y (d - a/2)}

Substitute

A_{s,\text{req}} \;=\; \dfrac{7{,}536 \,(12)}{0.90 \,(60)(90 - 3)}

Result

A_{s,\text{req}} \;=\; 19.3\;\text{in}^2

Formula

a \;=\; \dfrac{A_s f_y}{0.85 f'_c b}

Substitute

a \;=\; \dfrac{19.3\,(60)}{0.85\,(4)(60)}

Result

a \;=\; 5.7\;\text{in} \;\;\text{(matches assumption)} \quad \checkmark

Provide 14 – No. 11 top bars in two layers ( $A_s = 21.8\;\text{in}^2$ ).

Step 3 — Shear. Take $d_v = 0.9 d = 81$ in., $\beta = 2.0$ (§5.7.3.4.2 simplified):

Formula

\phi V_c \;=\; \phi \,(0.0316) \, \beta \sqrt{f'_c} \, b_v \, d_v

Substitute

\phi V_c \;=\; 0.90 \,(0.0316)(2.0)\sqrt{4}\,(60)(81)

Result

\phi V_c \;=\; 552\;\text{kip}

Since $V_u = 646 > \phi V_c = 552$ , stirrups are required:

Formula

A_v / s \;=\; \dfrac{(V_u/\phi) - V_c}{f_y \, d_v}

Substitute

A_v / s \;=\; \dfrac{(646/0.90) - 552/0.90}{60 \,(81)} = \dfrac{104}{4860}

Result

A_v / s \;=\; 0.021\;\text{in}^2/\text{in}

Use No. 6 closed stirrups (double leg, Av = 0.88 in²) at 8 in. o.c. in the cantilever region ( $A_v/s = 0.11$ , governs by §5.7.2.5 minimum).

Fig. 9.10Figure 9.10. Cap cross section at trunk face — 60 in. wide × 96 in. deep, top steel resists cantilever negative moment.

Final section detailing (from computed A_s)

Hammerhead cap cantilever — 60 in. wide × 96 in. deep at trunk face, tapering to 48 in. at tip

Location	A_s required	Bars provided	Spacing / detail
Top longitudinal (negative moment)	$A_{s,\text{req}} = 19.3\;\text{in}^2$	14 – No. 11 in two layers (Ast = 21.8 in²)	Layer 1: 8 bars @ 4 in. c/c; Layer 2: 6 bars @ 6 in. c/c; 2 in. clear cover
Bottom longitudinal (skin steel)	$A_{s,\min}$ per §5.6.3.3	6 – No. 8 bottom bars	Provides positive-moment capacity mid-span between trunk and tip; distributed as skin steel
Shear stirrups (cantilever region)	$A_v/s = 0.021$	No. 6 double-leg closed stirrups	8 in. o.c. from trunk face outward 8 ft; 12 in. o.c. beyond
Development of top bars into trunk	$\ell_d = 1.4\, \ell_{d,\text{basic}} = 90\;\text{in}$	Extend all top bars fully across the trunk width plus 12 in. past far face; hook if space limited	Stagger cutoffs at 3 ft, 6 ft, 9 ft beyond the point of maximum moment per §5.10.8
Cover	per §5.10.1	2 in. clear top and sides (exposed)	Increase to 2.5 in. if pier is near salt spray or de-icing runoff

9.10 — Guided practice

Compute the slenderness of a slender single-column pier

Consider a single-column pier for a highway ramp: $D = 3.5\;\text{ft}$ , clear height $\ell_u = 45\;\text{ft}$ , fixed at the base and pinned at the top ( $K = 0.8$ ). Compute $r$ , $K\ell_u / r$ , and classify. If $P_u = 900\;\text{kip}$ and $M_2 = 620\;\text{kip-ft}$ , compute $\delta_b$ with $C_m = 0.85$ and $P_e = 8{,}500\;\text{kip}$ , then find the amplified design moment $M_c$ .

Expected result

r = 12.6\;\text{in}

K\ell_u/r = 34.3 > 22

→ slender;

\delta_b = 0.85 / (1 - 900/(0.75 \cdot 8500)) = 0.98

, use 1.0;

M_c = 620\;\text{kip-ft}

. Section design proceeds at the amplified value.

9.11 — Mini design challenge

Two-column bent for a curved ramp

Fig. 9.11Figure 9.11. Design-challenge geometry — three-span curved ramp, two 2-column bents, 24 ft column spacing.

Design one interior bent for the ramp shown. The deck is a two-lane (30 ft wide) reinforced-concrete slab bridge with $w_{DC} = 1.20\;\text{k/ft}$ , $w_{DW} = 0.20\;\text{k/ft}$ , and HL-93 live load per lane. Column diameter is 4 ft, height 35 ft, cap 5 ft × 6 ft × 40 ft. Deliver:

Factored axial, shear, and moment at the top of one column (Strength I).
Slenderness classification and (if applicable) moment magnifier.
Longitudinal steel and spiral design meeting §5.10.4 and §5.10.6.
P-M interaction check plotting $(M_u, P_u)$ and confirming safety.
Cap negative-moment and shear design at column face.
Seismic detailing for Zone 2 (plastic-hinge length, confinement).
A 1-page design memo and a marked section detail.

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9.12 — Chapter summary

What you leave with

The AASHTO substructure workflow: bearing reaction → cap → column → footing, with factored loads at every level.
Slenderness ratio $K\ell_u/r$ and the AASHTO 22/34 thresholds; the $\delta_b$ magnifier for slender columns.
P-M interaction anchor points and the Bresler reciprocal-load check for biaxial bending.
Longitudinal-steel bounds $0.01 \le \rho_g \le 0.08$ and the spiral / tie confinement rules.
Bent-cap flexure and shear per §5.6 and §5.7.3, with special attention to short shear spans.
Seismic plastic-hinge detailing per §5.10.11: pitch, length, longitudinal bar anchorage.

Engineering note

The next chapter ties the column to the ground: spread footings, drilled shafts, and driven piles per AASHTO §10.

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

Slenderness ratio and moment magnification for a bridge column

Basic

Problem

Compute KL_u/r and determine if slenderness effects must be considered.

Step-by-Step

KL_{u}/r = 2.1\cdot 264/12

Result

KL_{u}/r = 46.2

22 for braced without M reversals; \ge 22 \rightarrow \ \text{slender for unbraced.}

Design Verification

Cantilever piers over 20 ft almost always slender. Perform second-order analysis or apply δ_s magnifier.

Discussion

Underestimating K for a free-headed cantilever is a classic pier design error. Use K = 2.0–2.1 unless the top has a genuine lateral restraint.

Worked Example 2

Interaction diagram check for a pier column

Intermediate

Problem

Verify capacity by reading φP_n from the interaction diagram at M_u; compute utilization.

Step-by-Step

\dfrac{P_u}{\phi P_n} = \dfrac{850}{1250}

Result

\text{util} = 0.68 \ \ (\text{32\% reserve})

\rho_{\min}=0.01 \;\le\; \rho = 0.02 \;\le\; \rho_{\max}=0.08\ \ \checkmark

Design Verification

Load point plots comfortably inside the ρ = 2% envelope with ~32% reserve.

Discussion

Interaction diagrams for round columns come from software or Whitney-block hand calcs. For biaxial bending use the reciprocal load method (§5.6.4.5); do not just add uniaxial utilizations.

Worked Example 3

Confinement reinforcement in the plastic hinge zone (Seismic Zones 3–4)

Advanced

Problem

Determine spiral wire size and pitch.

Step-by-Step

\rho _{s} = max[ 0.45\cdot (A_{g}/A_{c} - 1)\cdot f'_{c}/f_{yh} , 0.12\cdot f'_{c}/f_{yh} ] = max[0.0057, 0.008] = 0.008

Result

\rho _{s} = 0.008

\rho _{s} = 4\cdot A_{sp}/(D_{c}\cdot s) \rightarrow s = 4\cdot 0.20/(44\cdot 0.008)

Result

s = 2.27 in \rightarrow use 2-in pitch

Design Verification

Zone 3/4 spiral requirements are strict — pitch typically 1.5–3 in. Under seismic loading, a well-confined core can achieve μ_φ > 15.

Discussion

Skimping on confinement is a leading cause of pier collapse in earthquakes (Loma Prieta, Kobe). Never relax pitch limits in the plastic hinge zone.

Worked Example 4

Pure-axial nominal capacity of a tied RC pier column

Intermediate

Problem

Determine φP_n (tied) per §5.6.4.4.

Step-by-Step

P_o = 0.85\,f'_c\,(A_g - A_{st}) + f_y\,A_{st}

P_o = 0.85\,(4.0)\,(1296-12) + 60\,(12)

Design Verification

ρ = 12/1296 = 0.93% is between ρ_min = 0.01 (with §5.6.4.2 relaxation for oversized sections) and ρ_max = 0.08. Reasonable for a lightly-loaded pier column.

Discussion

The α factor (0.80 tied, 0.85 spiral) accounts for accidental eccentricity — even a "concentric" column always has some moment from construction tolerances. Never use P_o directly as design capacity.

Worked Example 5

Moment magnification for a slender braced pier column

Advanced

Problem

Compute the magnified design moment M_c via the moment-magnifier method (§5.6.4.3).

Step-by-Step

\dfrac{KL_u}{r} = \dfrac{1.0\,(264)}{10.8} = 24.4 > 22 \Rightarrow \text{slender}

EI_{\text{eff}} = \dfrac{0.4\,E_c\,I_g}{1+\beta_d} = \dfrac{0.4\,(3605)(139968)}{1.60}

Result

EI_{\text{eff}} = 1.262\times 10^{8}\ \text{kip-in}^2

Design Verification

P_u/P_c = 5% — the column is nowhere near buckling, so magnification is trivial. If P_u had been 8000 kip, δ_ns would jump above 1.4 and drive the interaction check.

Discussion

The magnifier trades a nonlinear P-Δ analysis for a closed-form multiplier. It is safe for regular columns but should not replace second-order analysis for tall bents, sway frames, or columns near P_u ≈ 0.75·P_c.

Section 3

Guided Practice

Complete the missing steps. Use Hints for AASHTO article pointers and setup logic before revealing the full step. Submit at the end to send your work to your instructor.

Guided Problem 1

Solid-shaft pier — axial capacity

Circular RC pier: $D = 48\ \text{in}$ , longitudinal steel $\rho_{s} = 1.5\%$ , $f'_{c} = 5\ \text{ksi}$ , $f_{y} = 60\ \text{ksi}$ , tied column.

Step 1

Gross area $A_{g}$ (in²).

Answer

Step 2

Steel area $A_{st}$ (in²).

Answer

Step 3

Nominal $P_{n} = 0.85 f'_{c}(A_{g} - A_{st}) + f_{y} A_{st}$ (kip).

Answer

Step 4

Design axial $P_{r}$ for tied column, $\phi = 0.75$ , $0.80 P_{n}$ cap (kip).

Answer

Guided Problem 2

Slenderness classification

Round pier, unbraced height $h = 32\ \text{ft}$ between deck and pile cap, $k = 1.2$ , $D = 48\ \text{in}$ .

Step 1

Radius of gyration $r$ for a circle (in). $r = D/4$ .

Answer

Step 2

Slenderness $kL_{u}/r$ (dimensionless).

Answer

Step 3

Slenderness threshold for sway columns (AASHTO §5.6.4.3): compare to 22 → moment mag needed if $> 22$ .

Answer

Step 4

Upper limit that requires full P-Δ analysis (100).

Answer

Guided Problem 3

Transverse ties — confinement volume ratio

Circular spirally reinforced column: $D_{\text{core}} = 42\ \text{in}$ , $D = 48\ \text{in}$ , $f'_{c} = 5\ \text{ksi}$ , $f_{yh} = 60\ \text{ksi}$ .

Step 1

$\rho_{s,\min} = 0.45\left(\dfrac{A_{g}}{A_{c}} - 1\right)\dfrac{f'_{c}}{f_{yh}}$ .

Answer

Step 2

Spiral bar $A_{sp}$ for #4 (in²).

Answer

Step 3

Required pitch $s = \dfrac{4 A_{sp}}{D_{\text{core}} \rho_{s}}$ (in).

Answer

Step 4

AASHTO maximum spiral pitch (in).

Answer

Guided Problem 4

Ship-collision force (Method II summary)

Barge-tow pier collision, $DWT = 3{,}200\ \text{tons}$ , impact velocity $V = 8\ \text{ft/s}$ , $\cos\theta \approx 1$ .

Step 1

AASHTO §3.14.11 barge crush-force $P_{B}$ (kip). Simplified: $P_{B} \approx 4112\sqrt{DWT}\, V/1000$ .

Answer

Step 2

Design vessel-collision force on pier (kip): apply $P_{B}$ directly.

Answer

Step 3

Overturning moment about base (k-ft) if height above base $= 12\ \text{ft}$ .

Answer

Step 4

Load combination $\gamma$ under Extreme Event II for CV.

Answer

Section 4

Independent Practice

Every problem randomizes its inputs. Work each step, submit for immediate feedback, request new values to practice again.

Practice 1

A_g of circular pier

D = 36 in

Step 1π D²/4.

Answer (in²)

Randomized inputs, symbolic grading (±2%).

Practice 2

A_st from ρ

rho = 0.034 -

A_g

Ag = 950 in²

Step 1ρ·A_g.

Answer (in²)

Randomized inputs, symbolic grading (±2%).

Practice 3

Axial capacity P_n

f′_c

fc = 6.5 ksi

f_y

fy = 70 ksi

A_g

Ag = 1000 in²

A_st

Ast = 23 in²

Step 10.85·f′_c·(A_g−A_st) + f_y·A_st.

Answer (kip)

Randomized inputs, symbolic grading (±2%).

Practice 4

Design axial P_r for tied column

P_n

Pn = 9900 kip

Step 10.75·0.80·P_n.

Answer (kip)

Randomized inputs, symbolic grading (±2%).

Practice 5

Slenderness k L_u / r

k = 1.9500000000000002 -

L_u

Lu = 23 ft

r = 15.5 in

Step 1k·L_u·12/r.

Answer (-)

Randomized inputs, symbolic grading (±2%).

Practice 6

Radius of gyration (circle)

D = 45 in

Step 1D/4.

Answer (in)

Randomized inputs, symbolic grading (±2%).

Practice 7

Spiral ρ_s minimum

A_g

Ag = 3400 in²

A_c

Ac = 850 in²

f′_c

fc = 5 ksi

f_yh

fy = 65 ksi

Step 10.45·(A_g/A_c − 1)·f′_c/f_yh.

Answer (-)

Randomized inputs, symbolic grading (±2%).

Practice 8

Spiral pitch

A_sp

Asp = 0.21 in²

D_core

Dc = 38 in

ρ_s

rho_s = 0.01 -

Step 14·A_sp/(D_core·ρ_s).

Answer (in)

Randomized inputs, symbolic grading (±2%).

Practice 9

Moment magnifier δ_ns

C_m

Cm = 0.75 -

P_u

Pu = 1700 kip

P_c

Pc = 9500 kip

Step 1C_m/(1 − P_u/(0.75·P_c)) ≥ 1.

Answer (-)

Randomized inputs, symbolic grading (±2%).

Practice 10

Wind pressure on pier cap

V (mph)

V = 150 mph

Exposed area

A = 125 ft²

Step 1P = 0.00256·V² (psf).

Answer (psf)

Step 2F = P·A/1000 (kip).

Answer (kip)

Randomized inputs, symbolic grading (±2%).

Practice 11

Barge collision P_B (approx)

DWT (ton)

DWT = 2000 ton

V (ft/s)

V = 8 ft/s

Step 1≈ 4.112·√DWT·V (kip).

Answer (kip)

Randomized inputs, symbolic grading (±2%).

Practice 12

Pier cap positive moment (uniform w over cantilever)

w = 2 klf

cantilever

a = 5 ft

Step 1w·a²/2.

Answer (k-ft)

Randomized inputs, symbolic grading (±2%).