Seismic Design of Bridges — Bridge Engineering & Design

Engineering story

From Loma Prieta to capacity design

When the 1989 Loma Prieta earthquake collapsed the Cypress Viaduct and a section of the Bay Bridge — and when Northridge (1994) and Kobe (1995) followed — the profession stopped trying to make bridges elastic under strong ground motion and started making them ductile. Modern AASHTO seismic design accepts that a pier column will yield and rotate at a plastic hinge; the engineer's job is to choose where that hinge forms, detail it for rotation without loss of strength, and capacity-protect everything else — the bent cap, the connections, the foundation — so failure cannot migrate to a brittle element.

Chapter objectives

What you will be able to do

Learning objectives

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

1Look up mapped short-period (S_S) and 1-s (S_1) accelerations and build the design response spectrum per AASHTO §3.4.
2Classify a bridge into Seismic Design Category A–D and pick the required analysis and detailing.
3Compute the fundamental period of a single-mode SDOF pier and read the elastic base shear from the spectrum.
4Apply the response modification factor R to obtain design forces and displacements (§3.10).
5Detail a column plastic hinge for confinement, longitudinal splices, and shear (AASHTO §5.10.11).
6Apply capacity design so that overstrength column moments protect the cap, joints, and foundation.
7Deliver two worked examples (spectrum-based base shear + confinement schedule) and a full-bridge seismic design challenge.

15.1 — Seismic hazard

Mapped ground motion and the design earthquake

AASHTO LRFD §3.10.2

AASHTO uses a 7 % probability of exceedance in 75 years design event (~1000-yr return period). USGS maps give the two ground motion parameters that anchor the design spectrum:

$S_S$ — mapped short-period (0.2 s) spectral acceleration on Site Class B rock.
$S_1$ — mapped 1.0-s spectral acceleration on Site Class B rock.
$\mathrm{PGA}$ — mapped peak ground acceleration.

Fig. 15.1Schematic USGS hazard map — contours of design PGA/S₁. Sites in the highest zone (west coast, New Madrid, Charleston) land in SDC D.

Site coefficients F_pga, F_a, F_v (function of Site Class A–F per §3.10.3) scale the mapped values:

S_{DS} = F_a\, S_S, \qquad S_{D1} = F_v\, S_1, \qquad A_S = F_{pga}\, \mathrm{PGA}

(15.1)

15.2 — Design response spectrum

The three-region curve

AASHTO LRFD §3.10.4

The AASHTO design acceleration spectrum has three regions built from S_DS, S_D1, and A_S:

S_a = \begin{cases} A_S + (S_{DS} - A_S)\, T/T_0 & T \le T_0 \\ S_{DS} & T_0 \le T \le T_S \\ S_{D1} / T & T_S \le T \le T_L \\ S_{D1}\, T_L / T^{2} & T > T_L \end{cases}

(15.2)

$T_0$: 0.2 · T_S
$T_S$: S_{D1}/S_{DS}
$T_L$: long-period transition (from USGS map)

Fig. 15.2AASHTO design response spectrum. A short-period ramp, a plateau at S_DS, a 1/T descending branch, and a 1/T² long-period tail beyond T_L.

Seismic Design Categories (§3.10.6)

SDC is based on

S_{D1}

: A (<0.15), B (0.15–0.30), C (0.30–0.50), D (>0.50). SDC A requires only minimum support-length checks; SDC D requires full capacity design, ductile detailing, and displacement-based checks.

15.3 — Single-mode analysis

SDOF idealisation of a pier bent

AASHTO LRFD §4.7.4.3

For regular bridges (SDC B / C), a single-mode analysis is enough. The superstructure mass sits atop the pier acting as a lateral spring:

Fig. 15.3Bridge pier → SDOF. The superstructure mass m = W/g is lumped at the top of a lateral spring k (pier stiffness) with damping c.

T \;=\; 2\pi\, \sqrt{\dfrac{m}{k}}, \qquad k \;=\; \dfrac{3\, E_c\, I_e}{H^3} \;\text{(cantilever pier)}

(15.3)

$m$: tributary superstructure mass = W/g [kip·s²/ft]
$k$: lateral stiffness of pier [kip/ft]
$I_e$: effective moment of inertia (0.5–0.7 I_g cracked)
$H$: clear column height [ft]

The elastic base shear from the spectrum is:

V_e \;=\; S_a(T)\, W

(15.4)

15.4 — Response modification

Ductility reduces elastic demand

AASHTO LRFD §3.10.7

AASHTO permits reducing elastic shear by a response modification factor R that reflects the ductility of the substructure element:

V_{design} \;=\; V_e / R

(15.5)

R = 5 for multi-column bents (well-detailed).
R = 3 for single-column piers.
R = 1.5 for connections between substructure and superstructure — connections are not allowed to yield.

15.5 — Plastic hinges in columns

Where ductility is delivered

AASHTO LRFD §5.10.11 · §8.4 (Guide Spec)

Fig. 15.4Column plastic hinge at the base. Peak moment forms just above the footing; hinge length L_p ≈ 0.08 H + 0.15 f_ye d_bl absorbs the plastic rotation.

L_p \;=\; 0.08\, H \;+\; 0.15\, f_{ye}\, d_{bl} \;\ge\; 0.3\, f_{ye}\, d_{bl}

(15.6)

$H$: distance to inflection point [in.]
$f_{ye}$: expected yield strength of longitudinal steel [ksi]
$d_{bl}$: longitudinal bar diameter [in.]

Plastic rotation capacity:

\theta_p \;=\; (\phi_u - \phi_y)\, L_p

(15.7)

15.6 — Transverse confinement

Spiral / hoop reinforcement in the hinge zone

AASHTO LRFD §5.10.11.4.1d

Minimum volumetric spiral ratio:

\rho_s \;\ge\; 0.12\, \dfrac{f'_c}{f_{yh}} \;\text{or}\; 0.45\left(\dfrac{A_g}{A_c} - 1\right)\dfrac{f'_c}{f_{yh}}

(15.8)

$\rho_s$: vol. spiral steel / vol. concrete core
$A_g, A_c$: gross and core section areas [in²]
$f_{yh}$: spiral yield stress [ksi]

Fig. 15.5Circular column spiral confinement. The concrete inside the spiral (dashed circle) becomes triaxially confined, sustaining much higher strain than plain concrete.

Maximum spiral pitch s ≤ min(6·d_bl, 6 in., D/4) inside the hinge zone.

15.7 — Capacity design

Overstrength column protects everything else

AASHTO LRFD §4.11 (Guide Spec)

The plastic hinge is the fuse. To prevent brittle failure elsewhere, every other component is designed for the column overstrength moment:

M_o \;=\; \lambda_{mo}\, M_p \qquad (\lambda_{mo} = 1.2\text{–}1.4)

(15.9)

Fig. 15.6Capacity-design hierarchy. Plastic hinges (shaded) form at column tops and bases only. The bent cap, joints, and foundations are designed for column overstrength moments so they remain elastic.

The corresponding column shear is $V_o = 2M_o/H$ for a double-fixed column; the joint, cap, and foundation are designed for at least V_o.

15.8 — Seismic isolation

Lengthening the period at the interface

AASHTO LRFD AASHTO Guide Spec for Isolation §7

For high-seismic sites or retrofits, seismic isolators (lead-rubber, friction pendulum) placed between deck and substructure push the fundamental period out to 2–3 s, moving the structure into the descending 1/T branch of the spectrum and dissipating energy in a stable bilinear loop.

Fig. 15.7Base-isolation concept. An LRB bearing between girder and pier top has an effective stiffness k_eff and equivalent damping ξ_eq (10–30 %) that reduce base shear substantially.

15.9 — Worked example 1

Design base shear on a two-column bent

AASHTO LRFD §3.10, §4.7.4

Problem statement

A 60-ft-span highway bridge in California carries W = 900 kip of superstructure tributary weight onto a multi-column bent. Compute the design seismic base shear per column.

Given

Mapped valuesS_S = 1.4 g, S₁ = 0.55 g, PGA = 0.55 g
Site ClassD (F_a = 1.0, F_v = 1.5, F_pga = 1.0)
PierH = 22 ft; two 4-ft-Ø columns, f′_c = 4 ksi
Effective inertiaI_e = 0.5 I_g
R-factorR = 5 (multi-column bent)

Required

Build the design spectrum, compute pier period T, elastic base shear V_e, and design shear V per column.

Step 1 — Design spectral accelerations (Eq. 15.1).

Formula

S_{DS} = F_a S_S, \; S_{D1} = F_v S_1

Substitute

S_{DS} = 1.0(1.4), \; S_{D1} = 1.5(0.55)

Result

S_{DS} = 1.40 g, \; S_{D1} = 0.825 g

Formula

T_S = S_{D1}/S_{DS}

Substitute

T_S = 0.825/1.40

Result

T_S = 0.59 s

Step 2 — Pier stiffness and period. Two 4-ft columns act in parallel; I_g = π(4)⁴/64 = 12.57 ft⁴; I_e = 6.28 ft⁴. E_c = 3,600 ksi = 519,000 ksf.

Formula

k = 2 \cdot \dfrac{3 E_c I_e}{H^3}

Substitute

k = 2 \cdot \dfrac{3(519{,}000)(6.28)}{22^3}

Result

k = 1{,}835\;\text{kip/ft}

Formula

T = 2\pi\sqrt{W/(g k)}

Substitute

T = 2\pi\sqrt{900/(32.2 \times 1{,}835)}

Result

T = 0.87\;\text{s} \approx 0.9\;\text{s}

Step 3 — Spectral acceleration. T > T_S, use 1/T branch:

Formula

S_a = S_{D1}/T

Substitute

S_a = 0.825/0.9

Result

S_a = 0.92 g \;\text{(before R)}

Wait: 0.825/0.9 = 0.917 g. That is above S_DS = 1.40 g? No, 0.917 < 1.40, still on the descending branch (correct).

Step 4 — Elastic and design base shear.

Formula

V_e = S_a W

Substitute

V_e = 0.917 \times 900

Result

V_e = 825\;\text{kip}

Formula

V_{design} = V_e / R

Substitute

V_{design} = 825/5

Result

V_{design} = 165\;\text{kip (per bent)}

Per column (two columns): V = 82.5 kip. Column moment at base for double-fixed action: M = V·H/2 = 82.5 × 11 = 908 kip-ft. Compare to plastic moment capacity M_p from the P-M interaction of the column section.

Fig. 15.8Figure 15.8. Design spectrum with the pier period T = 0.9 s plotted — S_a(0.9 s) ≈ 0.44 g on the descending 1/T branch.

Final section detailing (from computed A_s)

Two-column bent — seismic demand (SDC D)

Location	A_s required	Bars provided	Spacing / detail
Design spectrum	AASHTO Eq. 15.2, Site D	S_DS = 1.40 g, S_D1 = 0.825 g, T_S = 0.59 s	checked against USGS 2023 maps
Pier period T	Eq. 15.3 with cracked I_e	0.87 s (descending branch)	re-check after column resizing
Elastic base shear	Eq. 15.4	V_e = 825 kip / bent	govern in longitudinal dir.
R-factor	multi-column bent, ductile	R = 5	R = 1.5 at connections
Design column shear	V_design = V_e/R	165 kip / bent = 82.5 kip / column	detail M_o for capacity

The design shear is checked against the column shear capacity computed at the plastic-moment overstrength — cap and foundation are then designed for V_o > V_design.

15.10 — Worked example 2

Spiral confinement in the plastic-hinge zone

AASHTO LRFD §5.10.11.4.1

Problem statement

For the 4-ft-Ø columns of the previous example, detail the transverse spiral in the plastic-hinge zone. Longitudinal reinforcement: 20 – #10 bars, f_y = 60 ksi, f_yh = 60 ksi, f′_c = 4 ksi. Clear cover 2.5 in.

Given

D (gross diameter)48 in.
D_core (to c.l. of spiral)48 − 2(2.5) − 0.5 = 43 in.
A_gπ/4(48)² = 1,810 in²
A_cπ/4(43)² = 1,452 in²
Longitudinal bar diameter d_bl#10 → 1.27 in.

Required

Compute required spiral ratio and pitch for #4 spiral; specify the plastic-hinge length L_p and detail the transition to the regular zone.

Step 1 — Governing spiral ratio (Eq. 15.8).

Formula

\rho_s = 0.12\, f'_c/f_{yh}

Substitute

\rho_s = 0.12(4)/60

Result

\rho_s = 0.0080

Formula

\rho_s = 0.45(A_g/A_c - 1) f'_c/f_{yh}

Substitute

\rho_s = 0.45(1810/1452 - 1)(4/60)

Result

\rho_s = 0.0074

Governing: ρ_s = 0.0080 (larger).

Step 2 — Spiral pitch for #4 (A_sp = 0.20 in²).

Formula

\rho_s = 4 A_{sp}/(D_{core} s)

Substitute

0.0080 = 4(0.20)/(43 s)

Result

s = 4(0.20)/(43 \times 0.0080) = 2.33\;\text{in.}

Round down to 2 in. pitch would be conservative — but check upper bound: min(6·d_bl, 6 in., D/4) = min(7.6, 6, 12) = 6 in. So 3 in. pitch satisfies ρ_s only if computed ratio ≥ required.

Formula

\rho_s^{prov} = 4(0.20)/(43 \times 3)

Substitute

\rho_s^{prov} = 0.80/129

Result

\rho_s^{prov} = 0.0062 < 0.0080 \;\text{NG}

Use #4 spiral @ 2.25 in. pitch → ρ_s = 0.80/(43·2.25) = 0.0083 > 0.0080 ✓.

Step 3 — Plastic-hinge length (Eq. 15.6), H = 22 ft = 264 in.

Formula

L_p = 0.08 H + 0.15 f_{ye} d_{bl}

Substitute

L_p = 0.08(264) + 0.15(66)(1.27)

Result

L_p = 21.1 + 12.6 = 33.7\;\text{in.} \approx 3\;\text{ft}

Step 4 — Provide dense spiral over 2·L_p = 6 ft at each column end. Transition to regular pitch (#4 @ 4 in., ρ_s = 0.0047 ≥ min §5.10.6 requirement) above the hinge.

Fig. 15.9Figure 15.9. Column elevation showing dense #4 @ 3 in. spiral in the plastic-hinge zone at the base and #4 @ 4 in. above.

Final section detailing (from computed A_s)

Column confinement — plastic-hinge zone

Location	A_s required	Bars provided	Spacing / detail
Longitudinal steel	ρ_l ≥ 0.008, ≤ 0.04	20 – #10 bars, ρ_l = 0.014	clear cover 2.5 in.
Hinge spiral	ρ_s ≥ 0.0080	#4 @ 2.25 in. pitch, ρ_s = 0.0083	over 2·L_p = 6 ft from base
Above hinge	ρ_s ≥ 0.0045	#4 @ 4 in. pitch, ρ_s = 0.0047	full remaining column height
Splices	no lap splices in hinge zone	mechanical couplers only in L_p	staggered 4 ft
Cap-column joint	designed for M_o = 1.2 M_p	additional #6 stirrups + headed bars	AASHTO §5.10.11.4.4

The pitch (2.25 in.) is tight but constructible. If concrete placement quality becomes a concern, increase column diameter to 4.5 ft to relax pitch to 3 in.

15.11 — Guided practice

R-factor sensitivity check

Repeat Example 1 with a single-column pier (R = 3). Everything else the same. What is the design column shear? What does this tell you about substructure selection in high-seismic zones?

Expected result

V_e stays 825 kip; V_design = 825/3 = 275 kip — 67 % more than the multi-column case. Single-column piers require larger sections and heavier confinement; in SDC D the extra column of a bent is often the economical choice.

15.12 — Mini design challenge

Four-span highway crossing in SDC D

Fig. 15.10Design challenge elevation. Four continuous 160-ft spans on three multi-column bents; abutments seat on isolation bearings.

Deliver:

Design response spectrum from mapped values (S_S = 1.5 g, S₁ = 0.60 g, Site D). Classify SDC.
Single-mode analysis in longitudinal and transverse directions; period, spectral acceleration, and elastic base shear.
R-factor reduction, column P-M interaction check at each bent.
Plastic-hinge length and spiral confinement schedule (both hinge and regular zones).
Capacity-design demands (M_o, V_o) on bent cap, cap-column joint, and foundation.
Isolation-bearing sizing at abutments and check of resulting deck displacements/support length.
Foundation schedule and a one-page seismic design memo.

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15.13 — Chapter summary

What you leave with

USGS-mapped ground motion → design spectrum (Eq. 15.2) → SDC assignment.
SDOF period from cracked pier stiffness (Eq. 15.3) and elastic base shear V_e = S_a W.
Response modification factor R reduces elastic demand to design forces (multi-column R = 5, single R = 3, connections R = 1.5).
Plastic hinges at column ends — length L_p (Eq. 15.6), rotation θ_p, and spiral confinement ρ_s (Eq. 15.8).
Capacity design: cap, joints, and foundation sized for column overstrength moment M_o = 1.2–1.4 M_p.
Base isolation for retrofit or high-seismic new construction — lengthen T, add damping.

AASHTO LRFD §3.10 — Seismic · §5.10.11 · Guide Spec for LRFD Seismic Bridge Design

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: seismic hazard from AASHTO map (PGA, Ss, S1)

Basic

Problem

Report

PGA, S_s, S_1

Step-by-Step

PGA \approx 0.72\,g,\quad S_s \approx 1.80\,g,\quad S_1 \approx 0.70\,g

Result

High-seismic (SDC D)

Design Verification

Any site with S_D1 > 0.50 is SDC D; expect ductile detailing and capacity design.

Discussion

Hazard is a site property, not a design choice. Everything downstream flows from these three numbers.

Worked Example 2

Simple hand calc: design response spectrum

Basic

Problem

Compute

A_s, S_{DS}, S_{D1}

and plot the design spectrum breakpoints.

Step-by-Step

A_s
$A_s = F_{pga}\cdot PGA = 0.72$
S_DS
$S_{DS} = F_a\cdot S_s = 1.80$
S_D1
$S_{D1} = F_v\cdot S_1 = 1.05$

T_s = \frac{S_{D1}}{S_{DS}} = 0.58\ \text{s};\quad S_a = \frac{S_{D1}}{T}\ \text{for}\ T > T_s

Result

SDS=1.80, SD1=1.05

Design Verification

Long-period ordinate at T = 1.0 s is exactly S_D1 = 1.05 g — makes intuitive sense.

Discussion

Site Class D is a common default; if a real Vs30 shear-wave survey is done, F_a and F_v drop by 10–25%.

Worked Example 3

Intermediate: single-mode elastic-force method (2-span bridge)

Intermediate

Problem

Compute period

T

and

V_{long}

Step-by-Step

Formula
$T = 2\pi\sqrt{\frac{W}{k\,g}} = 2\pi\sqrt{\frac{1{,}800}{1{,}200\cdot 386}}$
Result
$T = 2\pi\sqrt{0.00388} = 0.39\ \text{s}$

0.39\ \text{s} < T_s = 0.58\ \text{s} \Rightarrow S_a = S_{DS} = 1.80\,g

Design Verification

Order of magnitude: pier base shear ≈ 1.8× deck weight — typical for SDC D short-period bridges.

Discussion

Long-period bridges (T > 1 s) fall on the descending spectrum branch and attract much less force; period-shifting via isolation exploits this.

Worked Example 4

Complete design: RC column longitudinal reinforcement (SDC D)

Intermediate

Problem

Longitudinal reinforcement layout.

Step-by-Step

M_u = \frac{M_e}{R} = \frac{71{,}300}{3} = 23{,}760\ \text{k-ft}

A_{st} = 0.02\cdot\frac{\pi\cdot 60^{2}}{4} = 56.5\ \text{in}^{2} \rightarrow 40\ \#11\ \text{bars evenly}

Design Verification

ρ_l = 2% is within Guide Spec allowable 1–4%.

Discussion

SDC D forces capacity-based design of shear, hoops, joint. Longitudinal ρ is only step 1.

Worked Example 5

Multiple limit states: capacity-protected shear and transverse reinforcement

Advanced

Problem

Hoop size, spacing in plastic hinge region.

Step-by-Step

V_u = V_{po} = 1{,}418\ \text{kip}

Ductility-dependent. Assume μ_D = 4 → α′ = 0. V_c = 0 in the plastic hinge zone.

Design Verification

Outside the plastic hinge, spacing may relax to 6 in.

Discussion

Modern SDC D detailing produces hoop-dominated columns — the goal is confinement, not just shear strength.

Worked Example 6

Design optimization: base isolation to reduce column demand

Advanced

Problem

Compute new

V_e

and reduction in column reinforcement.

Step-by-Step

S_a = \frac{S_{D1}}{T\cdot B} = \frac{1.05}{2\cdot 1.5} = 0.35\,g

V_e = 0.35\cdot 1{,}800 = 630\ \text{kip} \rightarrow 5\times\ \text{reduction}

Design Verification

Isolator displacement $d = \frac{S_{D1}\,T}{4\pi^{2}B}$ → ~ 8 in. Design isolator with 12-in displacement capacity for safety.

Discussion

Isolation is the single biggest lever in seismic design — but adds first cost, expansion joint complications, and inspection.

Worked Example 7

Construction-stage: staged deck placement seismic risk

Advanced

Problem

Compute construction-stage

V_e

vs completed-bridge

V_e

Step-by-Step

T = 2\pi\sqrt{\frac{900}{1200\cdot 386}} = 0.28\ \text{s}

S_a = S_{DS} = 1.80\ \text{(still plateau)}

Design Verification

Guide Spec §4.7: check construction stages when duration > 1 yr; reduce design event to 500-yr return.

Discussion

Long-duration construction on high-seismic sites must be staged with braces or temporary shear connections.

Worked Example 8

Consulting: seismic retrofit of pre-1971 non-ductile column

Consulting

Problem

Recommend and size retrofit.

Step-by-Step

(1) Steel shell jacket 3/8 in; (2) FRP wrap 4 layers of CFRP; (3) Concrete jacket +6 in with #6 spiral.

Formula
$\rho_{s,eq} = \frac{2\,n\,t\,E_f\,\varepsilon_f}{D\,f_y}$
Result
$\rho_{s,eq} \approx 1.4\% \ge 1.2\%\ \checkmark$

Design Verification

CFRP retrofit validated by extensive PEER database of column tests (Priestley/Seible 1996).

Discussion

Cost per column: FRP $8k, steel jacket $18k, concrete jacket $25k. FRP dominates 1990s→ retrofit programs (Caltrans Phase II).

Worked Example 9

Case-study reproduction: Cypress Viaduct collapse (Loma Prieta 1989)

Consulting

Problem

Compute the outrigger joint principal tensile stress at recorded PGA.

Step-by-Step

v_j = \frac{V_j}{A_j} = \frac{950}{3{,}600} = 0.264\ \text{ksi}

f_a = \frac{P_j}{A_j} = \frac{1{,}200}{3{,}600} = 0.333\ \text{ksi}

Design Verification

Reproduces PEER back-analysis of Priestley et al. Direct driver of Guide Spec §8.13 joint shear provisions in 1995.

Discussion

Joint shear is invisible in code books before 1990; every pre-1990 outrigger bent in high-seismic zone should be re-checked.

Worked Example 10

Comprehensive: seismic design of a 3-span RC bridge from hazard to detailing

Consulting

Problem

Complete design report outline & key values.

Step-by-Step

Multi-mode spectral; T_1,long = 1.1 s, T_1,trans = 0.85 s.

V_{e,long} = \frac{0.9}{1.1}\cdot 4{,}200 = 3{,}436\ \text{kip};\quad M_e \approx 68{,}700\ \text{k-ft/pier}

Design Verification

Push-over analysis shows plastic hinge ductility μ_D = 4.2 < μ_D,cap = 6 per Guide Spec.

Discussion

Modern SDC D bridges are 20–35% more expensive than SDC A equivalents; isolation typically pays back through smaller foundations.

Section 4

Independent Practice

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

Practice 1

Spectral acceleration lookup

Guide Spec §3.4

S_s

S_s = 0.8 g

F_a

F_a = 1.3

Step 1Compute S_DS.

Answer (g)

Randomized inputs, symbolic grading (±2%).

Practice 2

Design period of a single-column pier

Guide Spec §5.4.2

Trib weight W

W = 1200 kip

Pier stiffness k

k = 2000 kip/in

Step 1Compute period T.

Answer (s)

Randomized inputs, symbolic grading (±2%).

Practice 3

Elastic base shear

Guide Spec §5

S_a

S_a = 0.4 g

W = 4000 kip

Step 1Compute elastic base shear.

Answer (kip)

Randomized inputs, symbolic grading (±2%).

Practice 4

Reduced design force with R-factor

Guide Spec §4.7

V_e

V_e = 1100 kip

R = 3

Step 1Compute design shear.

Answer (kip)

Randomized inputs, symbolic grading (±2%).

Practice 5

Column overstrength moment

Guide Spec §4.11

M_n

M_n = 32500 k·ft

Overstrength λ

lambda = 1.25

Step 1Compute overstrength moment.

Answer (k·ft)

Randomized inputs, symbolic grading (±2%).

Practice 6

Capacity-protected shear demand

Guide Spec §8.6

M_po

M_po = 10000 k·ft

Column height H

H = 25 ft

Step 1Compute V_po = M_po/H.

Answer (kip)

Randomized inputs, symbolic grading (±2%).

Practice 7

Confinement steel ratio requirement

Guide Spec §8.8.7

f′_c

f_c = 4.5 ksi

f_yh

f_yh = 90 ksi

Step 1Compute required ρ_s.

Answer

Randomized inputs, symbolic grading (±2%).

Practice 8

Isolated period target

Guide Spec Isolation §7

Isolated weight

W = 3400 kip

Isolator stiffness

k_iso = 47 kip/in

Step 1Compute isolated period.

Answer (s)

Randomized inputs, symbolic grading (±2%).

Practice 9

Displacement of isolated bridge

Guide Spec Isolation

S_D1

S_D1 = 1.05 g

T_iso

T = 2.2 s

Damping factor B

B = 1.8

Step 1Compute isolator displacement.

Answer (in)

Randomized inputs, symbolic grading (±2%).

Practice 10

Plastic hinge length (Priestley)

Guide Spec §4.11

H = 360 in

d_b

d_b = 1.7000000000000002 in

f_y

f_y = 60 ksi

Step 1Compute plastic hinge length L_p.

Answer (in)

Randomized inputs, symbolic grading (±2%).

Practice 11

Foundation seismic axial demand

Guide Spec §6

Dead-load axial

P_dl = 3100 kip

V_po

V_po = 1000 kip

Column-to-footing lever h

h = 16 ft

Foundation moment arm s

s = 11 ft

Step 1Compute worst-case shaft axial.

Answer (kip)

Randomized inputs, symbolic grading (±2%).

Practice 12

Displacement capacity check

Guide Spec §4.8

Displacement capacity Δ_c

d_c = 6 in

Displacement demand Δ_d

d_d = 11.5 in

Step 1Compute Δ_c/Δ_d (must be ≥ 1.5).

Answer

Randomized inputs, symbolic grading (±2%).

Section 5

Design Challenges

Multi-day projects mirroring real consulting scope. Submit a report package for review.

Project 1

Design a 3-span RC bridge for SDC D from scratch

Guide Spec §3, §4, §5, §6, §8

Scope

Green-field bridge on CA SR-58, three 140-ft continuous concrete box spans, $S_{DS} = 1.5\,g$ , $S_{D1} = 0.85\,g$ . Two single-column bents, drilled-shaft foundations. Deliver a full seismic design report to Caltrans SDC-2019 standards.

Deliverables

Modal response spectrum analysis (min 3 modes each direction) with tabulated periods and mass participation.
Column longitudinal & transverse reinforcement design incl. plastic-hinge detailing.
Capacity-protected design of joints, bent cap, and foundations to M_po.
Displacement demand vs capacity check (Δ_c/Δ_d ≥ 1.5).
Foundation drilled-shaft layout and axial-lateral capacity check.
Bearing/isolator selection with movement demands.
Bid quantities: concrete CY, rebar T, shaft feet.

Constraints

SDC D per AASHTO Guide Spec.
Ductility μ_D ≤ 6.
Foundation shaft group efficiency ≥ 0.85.
Column ρ_l between 1% and 4%.

Grading Rubric

Correct seismic hazard characterization10%
Modal analysis quality15%
Column design & capacity-based sizing25%
Capacity-protected member consistency20%
Foundation design15%
Quantities & deliverables completeness15%

Submit your design challenge

Design a 3-span RC bridge for SDC D from scratch

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Project 2

Seismic retrofit of a 1970 non-ductile viaduct

FHWA Seismic Retrofitting Manual

Guide Spec §8

Caltrans SDC-2019

Scope

Existing 1970 double-deck RC viaduct on I-880 approach: 15 continuous spans, non-ductile RC columns, outrigger bents, no joint shear reinforcement. Owner has $18M budget. Recommend and design a retrofit that brings the structure to SDC D life-safety performance.

Deliverables

Vulnerability screening: identify all deficiencies vs current Guide Spec.
Retrofit alternatives matrix (column FRP, steel jacket, concrete overlay, isolation, replacement).
Preferred alternative full design: column wrap thickness, joint retrofit, cap beam post-tensioning if needed.
Foundation adequacy check post-retrofit.
Construction staging plan preserving 1 lane in each direction.
Bid quantities and cost estimate versus budget.

Constraints

Budget cap $18M.
One lane must remain open each direction during construction.
Retrofit must not reduce vertical clearance below 15 ft.
Design must be re-inspectable per NBIS.

Grading Rubric

Vulnerability screening thoroughness20%
Alternatives evaluation & selection logic20%
Design correctness of preferred retrofit25%
Construction staging feasibility15%
Cost-benefit vs budget20%

Submit your design challenge

Seismic retrofit of a 1970 non-ductile viaduct

You must be signed in to upload a submission.