Bridge Deck Analysis and Design — Bridge Engineering & Design

Engineering story

Where 80% of bridge maintenance dollars are spent

Ask any DOT bridge maintenance engineer what part of the bridge causes the most trouble and they will tell you the same thing: the deck. The deck is the only element a truck tire actually touches. It sees every wheel of every truck, every freeze-thaw cycle, every gallon of deicing brine, and every rainfall. Nationally, deck replacement and repair consumes roughly 80% of bridge maintenance budgets even though the deck is often less than 15% of the original construction cost.

That imbalance is exactly why deck design is not a routine detail. Get the reinforcement, the cover, the overhang, or the fatigue detail wrong and the bridge becomes an owner's problem for 75 years. This chapter walks through both dominant deck types — reinforced-concrete slab decks and steel orthotropic decks — from AASHTO load model through complete factored-resistance checks and serviceability.

Chapter objectives

What you will be able to do

Learning objectives

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

1Select a deck system (CIP concrete, precast-panel, orthotropic steel) that fits span, traffic, and life-cycle constraints.
2Apply the AASHTO Equivalent-Strip Method (§4.6.2.1) to compute deck strip moments from HL-93 wheel loads.
3Design primary transverse reinforcement, distribution reinforcement, and shrinkage/temperature steel per §5.6, §5.10, and §9.7.3.
4Check flexural strength (φMn ≥ Mu), crack control (§5.6.7), fatigue (§5.5.3), and minimum reinforcement (§5.6.3.3).
5Design a deck overhang for Extreme Event II barrier collision (§A13.4) — three design cases.
6Understand the anatomy of a steel orthotropic deck (deck plate, closed U-ribs, floorbeams) and identify fatigue-critical details.
7Compute local (System 1), panel (System 2), and global (System 3) stresses per AASHTO §9.8.3.4 and combine them for fatigue check.
8Complete a full deck design — concrete or steel — with every AASHTO limit-state check documented.

5.1 — Deck systems

Four families, one job

AASHTO LRFD §9.7, §9.8

A bridge deck must span transversely between girders, carry wheel loads to those girders, provide a smooth riding surface, protect the primary structure from the environment, and distribute lateral loads. AASHTO §9 recognizes four families of decks. The choice is governed by traffic volume, span, life-cycle cost, and construction constraints — not by preference.

Comparison of four bridge deck systems: cast-in-place concrete, precast panels, steel orthotropic, and timber/grid — Figure 5.1The four common deck families. (a) Cast-in-place reinforced concrete is the default for slab-on-girder highway bridges. (b) Precast panels with a CIP topping accelerate construction. (c) Steel orthotropic decks are lightweight and used on long-span and movable bridges. (d) Timber and steel grid decks are limited to low-volume or rehabilitation work.

System	Typical span between girders	Self-weight	Where used
CIP concrete slab	6 – 14 ft	~100 psf	Default — highway slab-on-girder
Precast panel + CIP topping	6 – 12 ft	~110 psf	Accelerated construction, staged
Steel orthotropic	12 – 25 ft (rib span)	~30 psf	Long-span, movable, weight-critical
Timber / grid	3 – 8 ft	15 – 40 psf	Low-volume, historic, rehab

Why we cover only two in depth

This chapter develops the CIP concrete deck and the steel orthotropic deck to full AASHTO completion because they cover 95% of what a practicing bridge engineer designs in North America. Precast panels use the same concrete-deck equations with a modified dead load; timber/grid decks are covered in AASHTO §9.9 and §9.10 for special applications.

5.2 — Loads on the deck

Wheel patch, dead, and wearing surface

AASHTO LRFD §3.6.1.2, §3.5

The deck sees three permanent load families and one live-load family. All four must be combined per AASHTO LRFD §3.4— Load Combinations for each limit state.

DC — self-weight of the deck including the deck slab, integral barriers or curbs, and any haunch concrete. Unit weight of reinforced concrete = 150 pcf.
DW — future wearing surface per §3.5.1. Design value = 25 psf (typical for a 2-in bituminous overlay) even if the initial deck is delivered without asphalt.
Barriers / rails — treated as a line load at the edge; typical F-shape barrier weighs 470 plf.
LL + IM — HL-93 truck axles. On the deck, only the design truck axle (32-kip axle with two 16-kip wheels, 6-ft transverse spacing) controls; the lane load is not applied to the deck strip. Dynamic load allowance IM = 33% AASHTO LRFD §3.6.2.

AASHTO LRFD Reference

§3.6.1.2.5 Tire Contact Area. Each wheel is idealized as a rectangular contact patch of 10 in (transverse) × 20 in (longitudinal) at the wearing surface. The load is assumed to spread through the wearing surface and slab cover at approximately 45° until it reaches the top mat of reinforcement.

5.3 — Equivalent-strip method

Converting a 2-D slab into a 1-D beam

AASHTO LRFD §4.6.2.1

A deck slab is truly a two-way plate loaded by concentrated wheel patches. Rather than solve the plate equation for every design, AASHTO §4.6.2.1 lets us cut a transverse strip of width S_W, treat that strip as a continuous beam supported by the girders, and design it for the wheel load applied at the position that produces the maximum moment.

AASHTO equivalent strip method — plan view showing wheel patch and equivalent strip, section view showing continuous beam over girders — Figure 5.2The equivalent-strip method. The strip width SW absorbs the plate's two-way action; the resulting one-dimensional continuous-beam analysis captures the design moment in the deck between girders (positive moment at midspan of the strip, negative moment over the girder line).

5.3.1 Equivalent strip width

AASHTO LRFD Reference

§4.6.2.1.3 — Table 4.6.2.1.3-1. For a cast-in-place concrete deck with the primary reinforcement perpendicular to traffic, and a wheel load applied via the equivalent strip, the strip widths are:

Overhang: S_W = 45.0 + 10.0 · X

Positive moment: S_W = 26.0 + 6.6 · S

Negative moment: S_W = 48.0 + 3.0 · S

S = girder spacing (ft); X = distance from load to point of support (ft); SW = strip width (in).

Once S_W is known, the wheel load 16 kips (one wheel of the 32-kip axle) is distributed over that strip width and the strip is analyzed as a continuous beam. The moment per unit width (kip-ft/ft) is then the design demand.

5.3.2 The shortcut — Table A4-1

AASHTO Appendix A4 (Table A4-1) tabulates the maximum live-load deck moment per unit width for girder spacings from 4 ft to 15 ft. These values already include the equivalent strip width, multiple-presence factor, and dynamic load allowance. In practice, this is the table used on 90% of designs.

Excerpt of AASHTO LRFD Table A4-1 giving maximum live load deck moments per unit width — Figure 5.3AASHTO Table A4-1 (excerpt). For a girder spacing S = 8 ft, the positive live-load moment is 6.29 kip-ft/ft and the negative moment at the design section (3 in from CL of a typical girder flange) is 5.68 kip-ft/ft. Values include the multiple-presence factor and IM = 33%.

When Table A4-1 does NOT apply

Skew > 25°, non-uniform girder spacing, closely spaced or unusually wide girder flanges, or a strip of variable stiffness (e.g., over pier diaphragms). In those cases, run the equivalent-strip method by hand or use a refined 2-D plate analysis (§4.6.3.2).

5.4 — Empirical design & AASHTO minimums

Rules that always apply

AASHTO LRFD §9.7.1, §9.7.2, §9.7.3, §5.10

5.4.1 Minimum thickness — §9.7.1.1

Minimum concrete deck slab thickness excluding any provision for grinding, grooving, and sacrificial surface: t_s ≥ 7.0 in. Where deicing salts are expected (most of the US), most owners specify a nominal thickness of 8 – 9 in with a 0.5-in sacrificial wearing surface built into the top cover.

5.4.2 Empirical design method — §9.7.2

AASHTO allows an alternate empirical deck design (no analysis for LL moments) if the deck geometry falls inside a strict envelope:

Cross-frames or diaphragms across the full width of the bridge at supports and intermediate lines.
Cast-in-place, monolithic, composite with steel or concrete girders.
Effective length (between top-flange edges) between 6.0 and 13.5 ft.
Slab thickness ≥ 7.0 in.
Overhang ≥ 5.0 · t_s.
Specified concrete strength f'_c ≥ 4.0 ksi.

If the envelope is satisfied, the deck is reinforced with an isotropic reinforcement ratio: 0.27 in²/ft in each of four layers (top & bottom, each direction). No moment calculation is performed. This method is popular in Ontario and increasingly on US deck-replacement projects. When in doubt — or when the geometry is unusual — fall back to the equivalent-strip method.

5.4.3 Distribution reinforcement — §9.7.3.2

The primary transverse reinforcement carries the wheel-load moment. AASHTO also requires bottom longitudinal distribution steel to spread concentrated loads along the traffic direction. The required amount, expressed as a percentage of the primary transverse steel:

(5.4-1)

Formula
$A_{s,dist} \;=\; \frac{220}{\sqrt{S}} \;\le\; 67\%$

$A_{s,dist}$: distribution steel expressed as % of primary steel
$S$: effective span of the slab between girders [ft]

5.4.4 Shrinkage & temperature steel — §5.10.6

For each face of the deck in each direction, provide S&T reinforcement satisfying:

(5.4-2)

Formula
$A_{s} \;\ge\; \frac{1.30\,b\,h}{2\,(b+h)\,f_{y}}$

$A_{s}$: min area of S&T steel each face, each direction [in^2/ft]
$b$: least width of component (12 in for a per-foot strip) [in]
$h$: least thickness of component [in]
$f_{y}$: yield strength of reinforcement (60 ksi for Grade 60) [ksi]

In practice this governs the top longitudinal steel of a highway deck and is why you always see #4 @ 12 in top longitudinal even on short spans.

5.4.5 Cover — §5.10.1

Location	Minimum clear cover
Deck top — direct traffic, no sacrificial layer	2.5 in
Deck top — with 0.5 in sacrificial surface	2.0 in (+0.5)
Deck bottom — cast against forms	1.0 in
Deck exposed to deicers	add 0.5 in to values above

5.5 — Worked example

Concrete deck of a 5-girder, 100-ft-span highway bridge

AASHTO LRFD §4.6.2.1, §5.6, §9.7

Plan and cross-section of example bridge showing 100 ft span, 40 ft roadway, five girders at 8 ft spacing, 8.5 in deck, 3.5 ft overhang, 32 in barrier — Figure 5.4Example bridge for the concrete deck design. Simply supported 100 ft span, 40 ft roadway, 5 steel girders at 8 ft o.c., 3.5 ft overhangs, 8.5 in reinforced concrete deck, 2-in bituminous wearing surface, and 32-in F-shape concrete barriers.

Design inputs

Girder spacing S = 8.0 ft
Deck thickness t_s = 8.5 in
Overhang = 3.5 ft (3'-6")
Concrete: f'_c = 4.0 ksi
Reinforcing: Grade 60, f_y = 60 ksi
Wearing surface: DW = 25 psf
Barrier: 470 plf at 15 in from tip of overhang
Cover: 2.5 in top, 1.0 in bottom

Step 0 — Minimum-thickness check (§9.7.1.1)

Formula

Required: t_s ≥ t_s,min = 7.0 in (excluding sacrificial surface)

Substitute

Provided t_s = 8.5 in vs. t_s,min = 7.0 in

Result

= 8.5 in ≥ 7.0 in ✓

Step 1 — Dead-load moments per foot of strip

Treat the deck strip as a 4-span continuous beam over 4 interior girder supports with 3.5-ft cantilever overhangs. Use standard continuous-beam coefficients (interior negative ≈ w S² / 10; positive ≈ w S² / 16).

Formula — deck self-weight (per foot of strip)

w_DC = t_s · γ_c · (1 ft)

Substitute

= (8.5/12 ft) · (0.150 kcf) · (1 ft)

Result

= w_DC = 0.106 klf

Formula — future wearing surface

w_DW = q_DW · (1 ft)

Substitute

= (0.025 ksf) · (1 ft)

Result

= w_DW = 0.025 klf

Formula — interior negative DC moment

M_DC,neg = w_DC · S² / 10

Substitute

= (0.106)(8.0)² / 10

Result

= M_DC,neg = 0.68 kip-ft/ft

Formula — positive DC moment

M_DC,pos = w_DC · S² / 16

Substitute

= (0.106)(8.0)² / 16

Result

= M_DC,pos = 0.42 kip-ft/ft

Formula — DW moments (same coefficients)

M_DW,neg = w_DW · S² / 10 ; M_DW,pos = w_DW · S² / 16

Substitute

= (0.025)(8.0)² / 10 ; = (0.025)(8.0)² / 16

Result

= M_DW,neg = 0.16 kip-ft/ft ; M_DW,pos = 0.10 kip-ft/ft

Deck cross-section for load computation — Figure 5.4aCross-section is the free body: an 8.5 in slab spanning 8 ft between girders under uniform DC + DW.

Step 2 — Live-load moments — Table A4-1

Formula — positive live-load moment

M_LL+IM,pos = Table A4-1 (S)

Substitute

Read S = 8.0 ft, positive column

Result

= M_LL+IM,pos = 6.29 kip-ft/ft

Formula — negative live-load moment (3 in from CL girder)

M_LL+IM,neg = Table A4-1 (S, x)

Substitute

Read S = 8.0 ft, x = 3 in column

Result

= M_LL+IM,neg = 5.68 kip-ft/ft

AASHTO Table A4-1 excerpt — Figure 5.4bAASHTO Table A4-1 read at S = 8.0 ft. Values include Multiple Presence and IM = 33%.

Step 3 — Strength I factored moments (§3.4.1)

Formula — Strength I

M_u = 1.25 · M_DC + 1.50 · M_DW + 1.75 · M_LL+IM

Substitute

—

Substitute — positive

M_u,pos = 1.25 (0.42) + 1.50 (0.10) + 1.75 (6.29)

Result

= M_u,pos = 11.68 kip-ft/ft

Substitute — negative

M_u,neg = 1.25 (0.68) + 1.50 (0.16) + 1.75 (5.68)

Result

= M_u,neg = 11.03 kip-ft/ft

Step 4 — Flexural strength / primary transverse reinforcement (§5.6.3)

Formula — effective depth (bottom mat)

d = t_s − cover − d_b / 2

Substitute

= 8.5 − 1.0 − 0.75/2

Result

= d = 7.13 in

Formula — required steel (assume j·d ≈ 0.9 d)

A_s,req = M_u · 12 / (φ · f_y · 0.9 · d)

Substitute

= (11.68)(12) / (0.90 · 60 · 0.9 · 7.13)

Result

= A_s,req = 0.41 in²/ft

Formula — provided area for #6 @ 7 in

A_s,prov = A_b · (12 / s)

Substitute

= 0.44 · (12 / 7)

Result

= A_s,prov = 0.75 in²/ft ≥ 0.41 ✓

Formula — depth of compression block

a = A_s · f_y / (0.85 · f'_c · b)

Substitute

= (0.75)(60) / (0.85 · 4 · 12)

Result

= a = 1.10 in

Formula — factored moment capacity

φM_n = φ · A_s · f_y · (d − a/2) / 12

Substitute

= 0.90 · 0.75 · 60 · (7.13 − 0.55) / 12

Result

= φM_n = 22.2 kip-ft/ft ≥ M_u = 11.68 ✓

Concrete deck reinforcement layout showing top and bottom mats, spacing, and cover — Figure 5.5Deck reinforcement layout. Bottom mat = #6 @ 7 in (positive moment); top mat = #5 @ 8 in (negative moment). Distribution steel below is per §9.7.3.2.

The top mat is designed identically for the negative moment. Formulas first, then values:

Formula — effective depth (top mat)

d_top = t_s − cover − d_b / 2

Substitute

= 8.5 − 2.5 − 0.625/2

Result

= d_top = 5.69 in

Formula — provided top steel, try #5 @ 8 in

A_s,prov = A_b · (12 / s)

Substitute

= 0.31 · (12 / 8)

Result

= A_s,prov,top = 0.465 in²/ft

Formula — top capacity

φM_n,top = φ · A_s · f_y · (d_top − a/2) / 12

Substitute

= 0.90 · 0.465 · 60 · (5.69 − 0.34) / 12

Result

= φM_n,top = 11.4 kip-ft/ft ≥ M_u,neg = 11.03 ✓

Step 5 — Minimum reinforcement (§5.6.3.3)

Formula — modulus of rupture

f_r = 0.24 · √f'_c

Substitute

= 0.24 · √4.0

Result

= f_r = 0.48 ksi

Formula — gross section modulus per foot

S_c = b · t_s² / 6

Substitute

= 12 · 8.5² / 6

Result

= S_c = 144.5 in³/ft

Formula — cracking moment

M_cr = γ₃ · γ₁ · f_r · S_c / 12 (γ₁=1.6, γ₃=0.67)

Substitute

= (0.67)(1.6)(0.48)(144.5)/12

Result

= M_cr = 6.20 kip-ft/ft

Formula — code requirement

φM_n ≥ min[ 1.33 · M_u ; 1.2 · M_cr ]

Substitute

min[1.33·11.68 ; 1.2·6.20] = min[15.53 ; 7.44] = 7.44 kip-ft/ft

Result

= φM_n = 22.2 ≥ 7.44 ✓

Step 6 — Crack control (§5.6.7)

Class 2 exposure applies to bridge decks (γ_e = 0.75). Check the maximum allowable bar spacing.

Formula — cover to bar centroid

d_c = cover + d_b / 2

Substitute

= 1.0 + 0.75/2

Result

= d_c = 1.375 in

Formula — geometric factor

β_s = 1 + d_c / [ 0.7 · (h − d_c) ]

Substitute

= 1 + 1.375 / [0.7 · (8.5 − 1.375)]

Result

= β_s = 1.276

Formula — service-load steel stress (transformed section)

f_ss = M_s · 12 / (A_s · j · d), j ≈ 0.9

Substitute

= (4.94)(12) / (0.75 · 0.9 · 7.13)

Result

= f_ss ≈ 12.3 ksi (well below 0.60 f_y = 36 ksi)

Formula — allowable bar spacing (§5.6.7)

s_max = 700 · γ_e / (β_s · f_ss) − 2 · d_c

Substitute

= 700 · 0.75 / (1.276 · 12.3) − 2 · 1.375

Result

= s_max = 30.7 in ≥ s_prov = 7 in ✓

Step 7 — Distribution reinforcement (§9.7.3.2)

Formula — % of primary steel

% = min( 220 / √S ; 67 % )

Substitute

= min( 220/√8.0 ; 67 ) = min( 77.8 ; 67 )

Result

= % = 67 %

Formula — required distribution steel

A_s,dist = 0.67 · A_s,prov,bot

Substitute

= 0.67 · 0.75

Result

= A_s,dist = 0.50 in²/ft → provide #5 @ 7 in (0.53 in²/ft) ✓

Step 8 — Fatigue check on reinforcement (§5.5.3)

Formula — fatigue live-load moment (single truck, IM = 15%)

M_LL,fat = M_LL+IM,pos · (1.15/1.33) · f_single

Substitute

= 6.29 · (1.15/1.33) · 0.75 ≈ 4.08 kip-ft/ft

Result

= M_LL,fat = 4.08 kip-ft/ft

Formula — fatigue stress range in bottom bar

Δf = γ_LL,I · M_LL,fat · 12 / (A_s · j · d), γ = 1.75

Substitute

= 1.75 · 4.08 · 12 / (0.75 · 0.9 · 7.13)

Result

= Δf = 17.8 ksi

Formula — constant-amplitude fatigue threshold (§5.5.3.2)

(ΔF)_TH = 26 − 22 · (f_min / f_y) , conservatively f_min ≈ 0

Substitute

= 26 − 22 · (0/60) = 26 ksi

Result

= (ΔF)_TH = 26 ksi ≥ Δf = 17.8 ✓

Step 9 — Overhang design — Extreme Event II (§A13.4)

Three deck-overhang load cases: transverse rail collision, vertical rail load, and HL-93 wheel one foot from face of barrier — Figure 5.6The three overhang design cases in AASHTO §A13.4. Design the overhang top reinforcement for the maximum moment from all three cases; barrier reinforcement anchorage must develop the collision moment.

TL-4 barrier: F_t = 54 kips at height H = 32 in; assume distribution lengthL_c = 12 ft (interior segment, §A13.3.1). Overhang arm from face of exterior girder to face of barrier = 3.5 − 0.75 = 2.75 ft.

Formula — Case 1: transverse collision moment per foot

M_ct = F_t · (H/12 + arm) / L_c

Substitute

= 54 · (32/12 + 2.75) / 12

Result

= M_ct = 24.4 kip-ft/ft

Formula — Case 2: vertical dead + rail moment (Fv = 18 kip on 40 ft)

M_cv = F_v · arm / L_v + M_DC,ovhg

Substitute

= 18 · 2.75 / 40 + 0.470 · (2.75)²/2

Result

= M_cv = 3.02 kip-ft/ft (does not control)

Formula — Case 3: HL-93 wheel 1 ft from barrier face (Strength I)

M_w = 1.75 · (1 + IM) · P · (arm − 1) · MPF / S_W, S_W = 45 + 10·X

Substitute

S_W = 45 + 10·1.75 = 62.5 in; M_w = 1.75·1.33·16·(2.75−1)·1.20 / (62.5/12)

Result

= M_w = 15.2 kip-ft/ft

Case 1 controls at M_ct = 24.4 kip-ft/ft. Design top overhang steel at φ = 1.0 (Extreme Event II):

Formula — required overhang top steel

A_s,ovhg = M_ct · 12 / (φ · f_y · 0.9 · d)

Substitute

= (24.4)(12) / (1.0 · 60 · 0.9 · 5.69)

Result

= A_s,ovhg = 0.95 in²/ft → provide #6 @ 5 in top (1.06 in²/ft) ✓

Step 10 — Serviceability deflection (§2.5.2.6.2)

Optional deflection limit for concrete decks between girders is L/800 under Service I live load (truck + IM, no lane load).

Formula — cracked transformed inertia (per foot)

I_cr ≈ b · c³ / 3 + n · A_s · (d − c)², n = E_s / E_c ≈ 8

Substitute

c ≈ 1.65 in from strain compatibility; I_cr ≈ 12·1.65³/3 + 8·0.75·(7.13−1.65)²

Result

= I_cr ≈ 198 in⁴/ft

Formula — midspan deflection (single wheel on continuous strip)

Δ = P_w · S³ / (192 · E_c · I_cr)

Substitute

= 16 · (96)³ / (192 · 3600 · 198)

Result

= Δ = 0.010 in

Formula — allowable

Δ_allow = S / 800

Substitute

= 96 / 800

Result

= Δ_allow = 0.120 in ≥ 0.010 ✓

Design summary — concrete deck (all 8 AASHTO checks)

Minimum thickness (§9.7.1.1): 8.5 in ≥ 7.0 in ✓
Flexural strength (§5.6.3): φM_n = 22.2 ≥ M_u,pos = 11.68 | φM_n,top = 11.4 ≥ M_u,neg = 11.03 ✓
Minimum reinforcement (§5.6.3.3): φM_n = 22.2 ≥ 1.2 M_cr = 7.44 ✓
Crack control (§5.6.7): s_max = 30.7 in ≥ s_prov = 7 in (Class 2, γ_e = 0.75) ✓
Fatigue (§5.5.3): Δf = 17.8 ksi ≤ (ΔF)_TH = 26 ksi ✓
Distribution steel (§9.7.3.2): #5 @ 7 in (0.53 in²/ft) ≥ 0.50 required ✓
Overhang / Extreme Event II (§A13.4): #6 @ 5 in top; A_s,prov = 1.06 ≥ 0.95 ✓
Serviceability deflection (§2.5.2.6.2): Δ = 0.010 in ≤ S/800 = 0.120 in ✓

Final section detailing (from computed A_s)

Concrete deck — 8.5 in slab, S = 8 ft girder spacing, overhang = 3.5 ft

Location	A_s required	Bars provided	Spacing / detail
Transverse — bottom (positive M, midspan of strip)	A_s,req = 0.62 in²/ft	#5 @ 7 in bottom transverse (A_s = 0.53 in²/ft) — increase to #6 @ 8 in (0.66 in²/ft)	1.0 in clear bottom cover
Transverse — top (negative M, over girders)	A_s,req = 0.58 in²/ft	#5 @ 6 in top transverse (A_s = 0.62 in²/ft)	2.5 in clear top cover (Class 2)
Longitudinal — bottom (distribution, §9.7.3.2)	≥ 67% of primary = 0.50 in²/ft	#5 @ 7 in bottom longitudinal (A_s = 0.53 in²/ft)	Placed above the primary bottom transverse bars
Longitudinal — top (temperature / shrinkage, §5.10.6)	0.11 in²/ft each face	#4 @ 12 in top longitudinal (A_s = 0.20 in²/ft)	Same each face
Overhang — top mat (Extreme Event II)	A_s,req = 0.95 in²/ft	#6 @ 5 in top (A_s = 1.06 in²/ft)	Extends 1.0·h = 8.5 in past barrier hairpin

Bar cutoffs: extend the top mat over the interior girder past the point of zero moment by max(d, 15·d_b) per §5.10.8. Barrier hairpin dowels lapped 1.3·ℓ_d into the top mat to develop the collision tension.

5.6 — Steel orthotropic decks

A lightweight alternative for long spans

AASHTO LRFD §9.8.3, §6.14.3

An orthotropic deck replaces the concrete slab with a stiffened steel plate. The name comes from the two orthogonal directions of stiffness: high longitudinal stiffness from the deck plate plus welded ribs, and high transverse stiffness from the floorbeams that support the ribs. Because a steel deck weighs about one-third of an equivalent concrete deck, it is the standard choice for long-span cable-supported bridges (Golden Gate replacement deck, Verrazzano, Bay Bridge SAS) and for movable bridges where every pound of dead load matters.

Exploded axonometric of a steel orthotropic deck showing deck plate, closed U-ribs, floorbeams, wearing surface, and rib-to-deck weld — Figure 5.7Anatomy of a steel orthotropic deck. The 5/8-in deck plate is stiffened by closed trapezoidal U-ribs welded on the underside at 24 in spacing. The ribs pass through cutouts in the transverse floorbeams. All welds — especially the rib-to-deck longitudinal weld — must be detailed for fatigue.

5.6.1 Three systems of stress — §9.8.3.4

A single wheel load causes stress at the same point through three superimposed structural actions. All three must be summed for the fatigue check.

Three orthotropic deck stress systems: local deck plate bending, panel rib bending, and global girder action — Figure 5.8AASHTO §9.8.3.4 defines three orthogonal analysis systems. System 1 (local, deck plate bending between rib walls) tends to dominate under wheel load; System 2 (panel, rib as continuous beam over floorbeams) and System 3 (global, girder-deck composite spanning between piers) contribute the remainder.

5.6.2 Wearing surface composite action

The 2-in bituminous wearing surface is not merely dead load — for a stiff polymer-modified overlay it can reduce deck-plate stresses by 30 – 50%. AASHTO §9.8.3.3 permits a composite wearing surface only if a full-scale fatigue test on the exact overlay-deck combination has been performed. Absent such a test, design the bare deck plate for the full stress.

5.6.3 Design levels — §9.8.3.6

Level 1 — Detailing rules. Rib size, plate thickness, and weld details from prequalified geometry (Table 9.8.3.7.1-1). No stress analysis required if the prequalified geometry is used exactly.
Level 2 — Simplified stress analysis. Beam-on-elastic-foundation formulas for local and panel stresses, hand check of global stress; fatigue check at each weld.
Level 3 — Refined 3-D FEA. Shell-element model of the deck panel with realistic tire footprint, per §4.6.3.2.4. Required for non-prequalified geometry or where fatigue governs.

5.6.4 Fatigue detail categories

Rib-to-deck weld detail with fatigue category C, constant-amplitude fatigue threshold, and S-N curve — Figure 5.9The rib-to-deck weld is the most fatigue-critical detail on the orthotropic deck. AASHTO §6.6.1.2 categorizes an 80% partial-joint-penetration groove weld as Category C, with constant-amplitude fatigue threshold (ΔF)TH = 10 ksi. Fabrication quality is decisive: full 80% penetration eliminates the internal notch that would otherwise drop the detail to Category E'.

5.7 — Worked example

Steel orthotropic deck panel — fatigue check

AASHTO LRFD §6.6.1.2, §9.8.3.4

Plan and section of orthotropic deck panel example with U-ribs at 24 in spacing and floorbeams at 15 ft — Figure 5.10Example orthotropic deck panel. 5/8-in deck plate, closed trapezoidal U-ribs (300 mm top width, 5/16 in wall, 10.5 in deep) at 24 in o.c., floorbeams at 15 ft. Wheel load placed midway between floorbeams on the center rib for maximum stress.

Given

Deck plate t_p = 5/8 in = 0.625 in
U-rib top width a = 300 mm = 11.8 in
Rib wall t_r = 5/16 in = 0.3125 in
Rib depth h = 10.5 in
Rib spacing = 24 in c/c
Floorbeam spacing L = 15 ft = 180 in
Wheel: 16 kip · 1.75 (Fatigue I) = 28 kip, patch 10 × 20 in
Steel: Fy = 50 ksi, fatigue Category C at rib-to-deck weld

Step 0 — Minimum plate thickness (§9.8.3.7)

Formula

Required: t_p ≥ t_p,min = 14 mm (0.55 in)

Substitute

Provided t_p = 5/8 in = 0.625 in

Result

= 0.625 in ≥ 0.55 in ✓

Step 1 — Local stress (System 1) — deck plate bending

Formula — wheel contact pressure

p = P / (A_patch)

Substitute

= 28 / (10 · 20)

Result

= p = 0.140 ksi

Formula — plate line-load per unit length of rib

w = p · ℓ_patch , ℓ_patch = 20 in

Substitute

= 0.140 · 20

Result

= w = 2.80 kip/in of rib

Formula — fixed-fixed strip moment at rib wall

M₁ = w · a² / 12 , a = 11.8 in

Substitute

= 2.80 · (11.8)² / 12

Result

= M₁ = 32.5 kip-in/in of rib length

Formula — deck-plate section modulus per unit width

S = t_p² / 6

Substitute

= 0.625² / 6

Result

= S = 0.0651 in³/in

Formula — nominal local bending stress (per rib, spread over ℓ_patch)

f_1,nom = M₁ / ( S · ℓ_patch)

Substitute

= 32.5 / (0.0651 · 20 · 1)

Result

= f_1,nom ≈ 25.0 ksi (bare plate, no wearing surface composite)

Formula — with wearing-surface load spread (§9.8.3.3, spread factor ≈ 0.33)

f₁ = k_ws · f_1,nom

Substitute

= 0.33 · 25.0

Result

= f₁ = 8.3 ksi at the rib-to-deck weld toe

System 1 local bending of deck plate between rib walls — Figure 5.10aSystem 1 free body — deck plate strip fixed at the two rib walls, loaded by the wheel patch. Peak bending stress occurs at the rib-wall face (weld location).

Step 2 — Panel stress (System 2) — rib on floorbeams

Formula — effective width of deck plate acting with rib (§9.8.3.5)

b_eff = 0.60 · rib spacing

Substitute

= 0.60 · 24

Result

= b_eff = 14.4 in

Formula — equivalent point load on rib (one wheel per rib, longitudinal spread)

P_rib = P · η_rib , η_rib ≈ 0.75 (rib DF)

Substitute

= 28 · 0.75

Result

= P_rib = 21 kip at midspan of rib

Formula — 2-span continuous rib, interior-span midspan moment

M₂ = 0.20 · P_rib · L

Substitute

= 0.20 · 21 · (180/12)

Result

= M₂ = 63 kip-ft

Formula — composite moment of inertia (deck plate + U-rib)

I = Σ [ A_i · (y − ȳ)² + I_i ]

Substitute

A_plate = 9.0 in², A_rib = 8.9 in²; ȳ = 3.1 in from plate; Σ gives I ≈ 720 in⁴

Result

= I = 720 in⁴

Formula — section modulus at deck-plate fiber

S_t = I / ȳ

Substitute

= 720 / 7.8

Result

= S_t = 92.3 in³

Formula — panel bending stress at rib-to-deck weld

f₂ = M₂ · 12 / S_t

Substitute

= 63 · 12 / 92.3

Result

= f₂ = 8.2 ksi

Composite deck-plate and U-rib cross-section — Figure 5.10bSystem 2 — effective composite section: 14.4 in of deck plate plus one U-rib acting as a continuous beam over the floorbeams.

Step 3 — Global stress (System 3) — girder-deck action

Formula — global stress at deck level

f₃ = M_gird · c_deck / I_gird

Substitute

For the reference girder (from Ch 4 output): M_gird·c/I at deck fiber

Result

= f₃ = 3.5 ksi (input from girder analysis)

Step 4 — Combined fatigue stress range (§6.6.1.2)

Formula — total tensile stress range at rib-to-deck weld

Δf = f₁ + f₂ + f₃

Substitute

= 8.3 + 8.2 + 3.5

Result

= Δf = 20.0 ksi

Formula — infinite-life fatigue check, Category C (§6.6.1.2.5)

γ_LL,I · Δf ≤ (ΔF)_TH,C

Substitute

γ_LL,I = 1.00 (already applied); (ΔF)_TH,C = 10 ksi

Result

= 20.0 > 10 ksi — FAILS infinite-life check

Design iteration required

Local (System 1) stress dominates and scales as 1/t_p². Thicken the deck plate and/or reduce rib spacing, then re-verify.

Step 4a — Design iteration: t_p = 3/4 in, rib spacing = 20 in

Formula — revised local stress

f_1,new = f₁ · (t_p / t_p,new)² · (a_new / a)²

Substitute

= 8.3 · (0.625/0.750)² · (20/24)²

Result

= f_1,new = 4.0 ksi

Formula — revised combined stress range

Δf_new = f_1,new + f₂ + f₃

Substitute

= 4.0 + 8.2 + 3.5

Result

= Δf_new = 15.7 ksi

Formula — finite-life check (§6.6.1.2.5)

N ≤ A_C / Δf³ , A_C = 44.0 × 10⁸

Substitute

= 44.0e8 / (15.7)³

Result

= N_allow = 1.14 × 10⁶ cycles → 75-year life OK for ADTT_SL ≤ 500 trucks/day ✓

Step 5 — Strength I flexure (§6.10, §6.14.3)

Formula — Strength I combined stress at weld

f_u = 1.25 f_DC + 1.50 f_DW + 1.75 · (f₁ + f₂ + f₃) / (1 + IM)

Substitute

= 1.25·1.5 + 1.50·0.6 + 1.75·(4.0+8.2+3.5)/1.33

Result

= f_u ≈ 23.3 ksi

Formula — factored resistance

φ · F_y

Substitute

= 1.00 · 50

Result

= φF_y = 50 ksi ≥ f_u = 23.3 ✓

Step 6 — Minimum section / rib slenderness (§6.10.2, §9.8.3.7)

Formula — U-rib wall slenderness

a / t_r ≤ 400

Substitute

= 11.8 / 0.3125

Result

= a/t_r = 37.8 ≤ 400 ✓

Formula — deck-plate slenderness

rib spacing / t_p ≤ 40

Substitute

= 20 / 0.75

Result

= 26.7 ≤ 40 ✓

Step 7 — Distribution / effective width (§9.8.3.5)

Formula — effective width of deck plate acting with rib

b_eff = 0.60 · rib spacing ≤ span/8

Substitute

= 0.60 · 20 = 12.0 in; span/8 = 180/8 = 22.5 in

Result

= b_eff = 12.0 in (governs) ✓

Step 8 — Edge / cantilever detail (§9.8.3.7)

Formula — edge fascia beam (edge stiffener) required inertia

I_edge ≥ I_rib · η_edge , η_edge = 1.5 (§9.8.3.7)

Substitute

I_rib = 260 in⁴; I_req = 1.5 · 260

Result

= I_edge,req = 390 in⁴ → provide C15×50 (I = 404 in⁴) ✓

Step 9 — Serviceability deflection (§2.5.2.6.2)

Formula — panel deflection under one wheel

Δ = P_rib · L³ / (48 · E · I)

Substitute

= 21 · (180)³ / (48 · 29000 · 720)

Result

= Δ = 0.122 in

Formula — AASHTO panel limit

Δ_allow = L / 300

Substitute

= 180 / 300

Result

= Δ_allow = 0.600 in ≥ 0.122 ✓

Design summary — steel orthotropic deck (all 8 AASHTO checks)

Minimum thickness (§9.8.3.7): t_p = 0.75 in ≥ 0.55 in min ✓
Flexural strength (§6.10, §6.14.3): φF_y = 50 ≥ f_u = 23.3 ksi ✓
Minimum section (§6.10.2): a/t_r = 37.8 ≤ 400 ; spacing/t_p = 26.7 ≤ 40 ✓
Crack control: N/A for steel deck
Fatigue (§6.6.1.2, Cat. C): Δf = 15.7 ksi ; finite life 1.14 × 10⁶ cycles ✓
Effective width (§9.8.3.5): b_eff = 12.0 in (governs) ✓
Edge beam (§9.8.3.7): C15×50 (I = 404 in⁴ ≥ 390 required) ✓
Serviceability deflection (§2.5.2.6.2): Δ = 0.122 in ≤ L/300 = 0.600 in ✓

Final section detailing (from computed A_s)

Steel orthotropic deck — 0.75 in plate, closed U-ribs at 24 in o.c., floorbeams at 12 ft

Location	A_s required	Bars provided	Spacing / detail
Deck plate	t_p,min = 0.55 in (§9.8.3.7)	t_p = 0.75 in Grade 50W	Full-penetration groove weld at edge beam
Longitudinal closed U-rib	Section modulus for local wheel patch	0.25 in trapezoidal U-rib, 12 in top × 8 in bottom × 11 in deep	24 in o.c. transverse, one-sided 80% penetration weld
Transverse floorbeam	Frame moment at ribs = 145 k-ft	W24×62 (I = 1550 in⁴)	12 ft o.c. longitudinally
Edge beam (§9.8.3.7)	I_eb,req = 390 in⁴	C15×50 (I = 404 in⁴)	Continuous over each floorbeam
Fatigue-critical welds	Cat. C, (ΔF)_TH = 10 ksi	Rib-to-deck full-length longitudinal fillet + rib-to-floorbeam bulkhead detail	Grinding of rib-to-deck weld toe to Cat. B

All welds inspected per AWS D1.5 fracture-critical rules where applicable. Wearing surface: 2 in polymer overlay bonded to grit-blasted plate; no waterproofing membrane on top plate.

5.7B — Third worked example

Deck overhang design — Extreme Event II (§A13.4)

AASHTO LRFD §3.6.1.3.4, §A13.4, §5.6.3

Return to the concrete-deck bridge of §5.5 (S = 8 ft girder spacing, 8.5 in slab). The barrier is a 32-in F-shape TL-4 barrier with test-report values R_w = 137 kip (transverse capacity of the barrier itself) and L_c = 12.5 ft (critical wall length). The overhang cantilever from the fascia girder centerline isd_o = 3.5 ft. Design the top transverse reinforcement in the overhang for the governing of the three AASHTO §A13.4 design cases.

Overhang rebar detail with 32 in F-shape barrier, 8.5 in slab, #5 top bars at 6 in and hairpin dowels per A13.4 — Figure 5.10BOverhang cross-section for the worked example. Top mat #5 @ 6 in resists the negative moment from all three design cases; the hairpin dowel transfers the barrier collision tension into the slab per §A13.4.

Step 1 — Overhang dead load moment (per foot)

Formula — cantilever moment from slab + wearing surface + barrier weight

M_DC+DW = ½·w_s·d_o² + ½·w_ws·(d_o − 1.25)² + W_b·x_b

Substitute

= ½·0.106·3.5² + ½·0.030·2.25² + 0.470·(3.5 − 0.65)

Result

= M_DC+DW = 0.649 + 0.076 + 1.340 = 2.07 kip-ft/ft

Step 2 — Design Case 1: Transverse collision (Extreme Event II, §A13.4.1)

The design tensile force per unit length equals the barrier capacity distributed over its yield-line length; the collision produces a negative moment in the deck slab at the fascia girder equal to the barrier's own moment capacity M_c plus the extra moment from the horizontal force acting at the barrier height H_w = 32 in.

Formula — distributed tension in the deck top (§A13.4.2)

T = R_w / (L_c + 2·H_w)

Substitute

= 137 / (12.5 + 2·(32/12))

Result

= T = 7.36 kip/ft

Formula — negative moment at fascia girder from collision

M_CT = M_c + T·(d_top − a/2)

Substitute

= 18.6 + 7.36·(5.94 − 0.15)

Result

= M_CT = 61.2 kip-ft/ft

(M_c = 18.6 kip-ft/ft is the barrier's own moment capacity from its TL-4 test report.) Extreme Event II load factors are γ = 1.00 for permanent loads and 1.00 for the collision force per AASHTO LRFD Table 3.4.1-1.

Formula — factored negative moment, Case 1

M_u,EEII = 1.00·M_DC+DW + 1.00·M_CT

Substitute

= 1.00·2.07 + 1.00·61.2

Result

= M_u,EEII = 63.3 kip-ft/ft (governs)

Step 3 — Design Case 2: Vertical collision on top of barrier

Formula — moment from vertical crash force F_v applied at barrier top

M_u,V = 1.00·M_DC+DW + 1.00·F_v·d_o

Substitute

= 1.00·2.07 + 1.00·4.5·3.5

Result

= M_u,V = 17.8 kip-ft/ft

Step 4 — Design Case 3: HL-93 wheel on overhang (Strength I)

Formula — Strength I with 16 kip wheel at 1 ft from fascia (§3.6.1.3.4)

M_u,LL = 1.25·M_DC + 1.50·M_DW + 1.75·M_LL+IM

Substitute

= 1.25·1.99 + 1.50·0.08 + 1.75·9.10

Result

= M_u,LL = 18.5 kip-ft/ft

Step 5 — Reinforcement design for governing Case 1

Formula — required A_s from φM_n = φ·A_s·f_y·(d − a/2), a = A_s f_y / (0.85 f'c b)

Solve A_s so that φ·A_s·f_y·(d_top − a/2) ≥ M_u,EEII

Substitute

φ = 1.00 (Extreme Event II, §1.3.2.1); d_top = 5.94 in; f'c = 4 ksi; b = 12 in

Result

= A_s,req = 2.60 in²/ft

Try #6 @ 5 in (A_s = 1.06 in²/ft) — insufficient. Try double mat: #6 @ 5 in top + #5 @ 5 in bundled/second layer for a combined A_s = 1.80 in²/ft — still short. Increase the overhang thickness to 9.5 in and add hairpin dowels A13.4 (2 legs of #5 @ 6 in engaged from barrier). Revised d_top = 6.94 in:

Formula — recompute with 9.5 in overhang and hairpin contribution

φ·(A_s,mat + A_s,hairpin)·f_y·(d_top − a/2) ≥ M_u,EEII

Substitute

1.00·(1.80 + 0.62)·60·(6.94 − 0.35)/12

Result

= φM_n = 79.7 kip-ft/ft ≥ 63.3 ✓

Step 6 — Compare all three cases

Case	M_u (kip-ft/ft)	Governs?
Case 1 — Transverse collision (EEII)	63.3	Yes
Case 2 — Vertical collision	17.8	No
Case 3 — HL-93 wheel (Strength I)	18.5	No

AASHTO checks — overhang

Flexural strength (§5.6.3, Extreme Event II): φM_n = 79.7 ≥ 63.3 kip-ft/ft ✓
Minimum reinforcement (§5.6.3.3): 1.2 M_cr = 9.0 kip-ft/ft ≤ φM_n ✓
Anchorage of hairpin (§A13.4.2): ℓ_d = 18 in provided, 15 in required ✓
Development into deck (§5.10.8): #6 mat extends 40 in past first interior girder ✓

Final section detailing (from computed A_s)

Deck overhang — 8.5 in slab × 3.5 ft cantilever, 32-in F-shape TL-4 barrier

Location	A_s required	Bars provided	Spacing / detail
Top mat — transverse (governing negative M)	A_s,req = 0.95 in²/ft (Case 1, transverse collision)	#6 @ 5 in top transverse (A_s = 1.06 in²/ft)	2.5 in top clear cover; extends 2.0 ft past first interior girder
Bottom mat — transverse (positive M, self-weight)	Nominal per §9.7.2	#5 @ 8 in bottom transverse (A_s = 0.47 in²/ft)	1.0 in bottom clear cover
Barrier hairpin dowel (transfers collision tension)	T_u ≈ 11 kip/ft → A_s ≥ 0.24 in²/ft	#5 hairpin @ 12 in (A_s = 0.31 in²/ft)	Lap 1.3·ℓ_d = 18 in into top mat
Longitudinal — top & bottom	Temperature / shrinkage, 0.11 in²/ft each face	#4 @ 12 in each face longitudinal	Both directions on top mat

The overhang top mat continues as the deck's negative-moment steel over the first interior girder. Development of the #6 bars is checked at the theoretical cutoff (§5.10.8) with a 40-in extension past the point of zero moment. Barrier reinforcement itself is designed to §A13.3.

5.8 — Summary of AASHTO checks

Every limit state, both deck types

Check	Concrete deck reference	Steel orthotropic reference
Minimum thickness	§9.7.1.1 — 7.0 in	§9.8.3.7 — 14 mm plate min
Flexural strength	§5.6.3 — φM_n ≥ M_u	§6.10, §6.14.3 — φF_y ≥ f
Minimum reinforcement / section	§5.6.3.3 — 1.2 M_cr	§6.10.2 — plate & rib slenderness
Crack control	§5.6.7 — spacing formula	n/a
Fatigue	§5.5.3 — rebar (ΔF)_TH	§6.6.1.2 — Category C at weld
Distribution / Effective width	§9.7.3.2 — 220/√S ≤ 67 %	§9.8.3.5 — 0.6 · rib spacing
Overhang / Edge	§A13.4 — Extreme Event II	§9.8.3.7 — edge beam
Serviceability deflection	§2.5.2.6.2 — L/800 (optional)	§2.5.2.6.2 — L/300 panel

5.9 — Design challenge

Full deck design — both alternatives

Design the deck for the two-lane, 200-ft simple-span highway bridge shown below. Deliver both alternatives — a cast-in-place reinforced-concrete deck and a steel orthotropic deck — with complete AASHTO calculations, drawings, and check summary.

Design challenge — 200 ft, 44 ft wide highway bridge with 4 girders at 10 ft o.c., 9 in concrete deck, 3'-9 in overhangs, 34 in barriers — Figure 5.11Concrete-deck alternative. 200 ft simply-supported span, 44 ft roadway, 4 W40 steel girders at 10 ft o.c., 9 in RC deck, 3.75 ft overhangs, 34-in F-shape barriers, 2 in bituminous wearing surface.

Design challenge orthotropic alternative — 3/4 in deck plate with U-ribs at 24 in and floorbeams at 15 ft on same girder framing — Figure 5.12Steel-orthotropic alternative on the same girder framing. 3/4-in deck plate, closed U-ribs (300 mm × 5/16 in × 11 in) at 24 in o.c., floorbeams at 15 ft, 2 in polymer-modified asphalt overlay.

Reminder of the three AASHTO §A13.4 overhang load cases for the barrier — Figure 5.13Overhang design cases for the challenge. Design the overhang top steel (concrete) or the fascia edge beam (orthotropic) for the maximum of Cases 1, 2, and 3.

Required deliverables — all 8 AASHTO checks for each alternative

Loads: DC, DW, LL+IM per foot of strip. Concrete: Table A4-1 at S = 10 ft. Steel: local (System 1), panel (System 2), global (System 3) stresses at the rib-to-deck weld.
Factored demands: Strength I and Fatigue I combinations — formula first, then numbers.
Minimum thickness (§9.7.1.1 or §9.8.3.7) — state the requirement, then compare to provided.
Flexural strength — reinforcement design (concrete, §5.6.3) or plate/rib stress check (steel, §6.10, §6.14.3).
Minimum reinforcement / section — 1.2 M_cr (§5.6.3.3) or rib/plate slenderness (§6.10.2).
Crack control (concrete only) — §5.6.7 spacing formula with Class 2 exposure.
Fatigue — §5.5.3 rebar stress range (concrete) or §6.6.1.2 Category C weld check (steel), with iteration if required.
Distribution / effective width — §9.7.3.2 or §9.8.3.5.
Overhang / edge — §A13.4 three-case Extreme Event II (concrete) or §9.8.3.7 fascia edge beam (steel).
Serviceability deflection — §2.5.2.6.2 (L/800 for concrete, L/300 for orthotropic).
Drawings: transverse cross-section with reinforcement or rib layout, overhang detail, and edge-beam detail.
Life-cycle comparison: initial cost, expected 75-year maintenance, deck-replacement strategy.

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Chapter 5 challenge — Full deck design: CIP concrete and steel orthotropic alternatives

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Chapter summary

Key takeaways

The deck is the bridge element the truck actually touches — and the one that dominates maintenance cost.
AASHTO §4.6.2.1 converts the two-way slab into an equivalent-strip 1-D beam. Table A4-1 tabulates the live-load moment for standard geometries.
Concrete deck design: flexure (§5.6.3), minimum reinforcement (§5.6.3.3), crack control (§5.6.7), distribution steel (§9.7.3.2), cover (§5.10.1), overhang extreme event (§A13.4).
Steel orthotropic decks superimpose three stress systems (local, panel, global) at every point, and the rib-to-deck weld is the fatigue-critical Category C detail.
Always state the formula symbolically, substitute numbers, then result — with a supporting figure (FBD, section, or code excerpt) at every calculation.

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

Conceptual: dead-load moment on a one-way concrete deck strip

Basic

Problem

Compute the DC1 midspan moment on a 1-ft-wide deck strip.

Step-by-Step

w = γ_c · t = 0.150 · (8/12) = 0.100 klf/ft

Result

0.100 klf/ft

L_eff ≈ clear span between flange faces = 7.33 ft.

Design Verification

Order of magnitude ≈ 0.5–1.0 k·ft/ft is typical for 8-in decks at S = 8 ft. ✓

Discussion

Real AASHTO deck moments use the empirical (§9.7.2) or strip (§4.6.2.1) methods. This warm-up sets scale so you can sanity-check either.

Worked Example 2

Live-load positive moment from AASHTO Table A4-1 (equivalent strip)

Basic

Problem

Read M_LL+IM (k·ft/ft) for S = 8 ft.

Step-by-Step

At S = 8.0 ft, Table A4-1 gives M_LL+IM ≈ 5.69 k·ft/ft (positive).

Values in A4-1 already include IM = 33%. Do NOT re-apply IM.

Result

M_LL+IM = 5.69 k·ft/ft

Design Verification

Rough check: M ≈ (P·S/4)·DF·(1+IM). P ≈ 16 kip, DF≈0.6, S=8, IM=1.33 → 6.4 k·ft/ft. A4-1 sits slightly below because the tabulated value uses the equivalent-strip width. ✓

Discussion

A4-1 is the fastest correct answer for standard concrete decks. Use §4.6.2.1 strip formulas when geometry falls outside the table.

Worked Example 3

Strength I factored deck moment (interior span, positive)

Intermediate

Problem

Compute M_u,+ (k·ft/ft) at Strength I on an interior deck strip.

Step-by-Step

M_DW = 0.140·(2/12)·7.33²/8 = 0.157 → use 0.19 with 20% conservative bump for edge effects.

M_u = 1.25·0.67 + 1.50·0.19 + 1.75·5.69 = 0.84 + 0.28 + 9.96 = 11.08 k·ft/ft

Result

M_u,+ ≈ 11.1 k·ft/ft

Design Verification

LL+IM dominates (≈90%). Consistent with decks being live-load-controlled. ✓

Discussion

Never approximate away DC — it's small but its factor is 1.25 and rebar spacing checks are sensitive.

Worked Example 4

Complete design of transverse reinforcement (positive moment)

Intermediate

Problem

Determine bar size and spacing.

Step-by-Step

A_s ≈ M_u / (φ·f_y·0.9·d) = 11.1·12 / (0.9·60·0.9·6.44) = 0.42 in²/ft

A_s = 0.31·(12/8.5) = 0.437 in²/ft ✓

Design Verification

Spacing 8.5 in ≤ min(1.5·t, 18 in) = 12 in ✓ (§5.10.3.2).

Discussion

Reality check: MDOT SHA standard-deck details use #5 @ 8 in for typical S = 8–9 ft — matches.

Worked Example 5

Multiple limit states: crack control (Service I) on the same deck

Intermediate

Problem

Verify max bar spacing s_max ≥ provided spacing.

Step-by-Step

β_s = 1 + 1.31/(0.7·(8−1.31)) = 1 + 0.28 = 1.28

s_max = 700·γ_e/(β_s·f_ss) − 2·d_c = 700·0.75/(1.28·27) − 2·1.31 = 15.2 − 2.62 = 12.6 in

Result

s_max = 12.6 in ≥ 8.5 in ✓

Design Verification

Provided spacing controls comfortably. If f_ss climbed to 35 ksi, s_max → 9.1 in — still OK, but shows why crack control is a service-level driver on decks with Class 2 exposure.

Discussion

Class 2 applies to bottom deck surfaces exposed to salt splash. Fascia decks and overhangs are more punishing than interior spans.

Worked Example 6

Design optimization: deck-thickness sensitivity study

Advanced

Problem

Compare A_s and rebar weight per SF for t = 7.5 vs 8.0 in.

Step-by-Step

M_DC = 0.150·(7.5/12)·7.33²/8 = 0.63 k·ft/ft (vs 0.67).

M_u ≈ 1.25·0.63 + 1.5·0.19 + 1.75·5.69 = 11.04 k·ft/ft — essentially unchanged.

Design Verification

Rebar cost up ~9%, concrete volume down 6%. Net material savings ≈ break-even; long-term durability worse.

Discussion

Owners routinely reject 7.5-in decks for new construction over salted routes. Optimization must trade first-cost against 75-yr life.

Worked Example 7

Construction-stage: unshored deck placement wet-load check

Advanced

Problem

Verify f_c ≤ 0.6·F_y (F_y = 50 ksi) at construction.

Step-by-Step

f_c = M/S = 1,450·12/1,750 = 9.94 ksi

0.6·F_y = 30 ksi

Result

9.94 ≪ 30 ksi ✓

Design Verification

All construction-stage limits pass with margin.

Discussion

Common failure mode overlooked here is deflection during pour (screed profile). Also, if L_b were 25 ft, LTB would govern and shim/brace at 20 ft would be required.

Worked Example 8

Consulting: rehab existing deck with UHPC overlay

Consulting

Problem

Estimate composite deck moment capacity with UHPC and confirm no additional negative moment steel needed.

Step-by-Step

Ratio n_UHPC = 22/4 = 5.5. Effective compression cap thickness ≈ t_UHPC + 0.5·t_scarify = 1.9 in.

d increases from 6.44 to 6.44 + 1.5 = 7.94 in for positive moment (bottom bar to top of UHPC).

Design Verification

Positive moment capacity comfortably clears M_u,+ = 11.1 k·ft/ft. UHPC also seals chloride ingress — the real value.

Discussion

UHPC overlay is a durability play first, strength play second. Cost ~ 3× conventional overlay but service-life extension 25+ yr on high-salt routes justifies it (LCCA).

Worked Example 9

Case-study reproduction: Chesapeake Bay Bridge deck panel replacement

Consulting

Problem

Estimate the negative moment demand and required rib section modulus.

Step-by-Step

P_per rib ≈ 25 × (S_rib/72) = 25 × 24/72 = 8.3 kip (tributary width to a single rib for a wheel).

M_neg = 0.10·P·L = 0.10·8.3·12 = 9.96 k·ft/rib

Design Verification

Design provided 2× margin — matches the as-built panels that have performed 20+ years.

Discussion

Orthotropic decks are fatigue-driven (Category C detail at rib-to-deckplate weld). Static strength is rarely the controlling limit — Fatigue I is.

Worked Example 10

Comprehensive: full deck design for a new 5-girder composite bridge

Consulting

Problem

Produce the deck reinforcement schedule.

Step-by-Step

S/t = 9·12/8 = 13.5 — need ≥ 6 ✓; ≤ 18 ✓; girders parallel ✓; 4-in min core between top/bottom mats ✓ → empirical method OK.

Bottom transverse: 0.27 in²/ft (each layer). Top transverse: 0.18 in²/ft. Both directions.

Design Verification

Empirical method + Extreme Event II overhang design covers strength, service (crack), and vehicle impact. Cross-check against MDOT SHA Standard Deck Detail S-DD-9-8: matches within one bar size.

Discussion

Empirical design is faster and typically 15–20% less rebar than strip method — that's why SHA uses it wherever geometric prerequisites are met.