Steel I-Girder and Plate-Girder Bridges — Bridge Engineering & Design

Engineering story

Why steel dominates the medium-span crossing

Between 120 and 300 feet — the range that covers most highway grade separations, river crossings, and interchange ramps — the composite steel plate girder is the workhorse of North American bridge construction. It is light enough to ship and erect in single pieces, deep enough to control deflection over long spans, and tunable: the designer chooses each flange and web plate independently and varies them along the span so the section grows only where the moment demands it. A well-designed plate girder can weigh 30% less than a rolled shape of equivalent capacity, and that steel weight drives the substructure cost.

This chapter is the AASHTO §6.10 workflow for I-shaped steel girders acting compositely with a concrete deck. We classify the section (compact / non-compact / slender), compute the plastic moment $M_p$ , check lateral-torsional buckling for the non-composite construction stage, size web shear with or without tension-field action, design the shear-stud interface that makes the section composite, and check Category-C fatigue at every web-to-flange weld. Every calculation follows the Formula → Substitute → Result pattern.

Chapter objectives

What you will be able to do

Learning objectives

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

1Choose between rolled W-shapes and welded plate girders based on span, depth, and economy.
2Compute short-term (n) and long-term (3n) composite section properties for the deck-plus-steel system.
3Classify a composite section as compact or non-compact per AASHTO §6.10.6 and §B6.
4Locate the plastic neutral axis and compute Mp using the seven AASHTO PNA cases.
5Check lateral-torsional buckling of the non-composite construction stage using Lp, Lr, and the Cb moment-gradient factor.
6Design the web for shear per §6.10.9: nominal shear Vn, panel aspect do/D, and tension-field contribution.
7Size shear-stud connectors at strength (§6.10.10.4) and fatigue (§6.10.10.2) limit states.
8Detect and detail transverse and bearing stiffeners per §6.10.11.
9Check web-to-flange weld fatigue as Category C using the AASHTO nominal fatigue resistance ΔF(TH).
10Estimate live-load and dead-load deflections and meet the §2.5.2.6.2 optional deflection limits.

8.1 — Steel superstructure systems

Rolled shapes, plate girders, and box girders

AASHTO LRFD §6.10.1

Three families cover most highway steel bridges. Rolled W-shapes (e.g. W40×324) come out of the mill with fixed proportions; they are the fastest and cheapest choice up to about 100 ft spans. Welded plate girders are fabricated from three plates (top flange, web, bottom flange) with continuous fillet welds; they cover 100–300 ft simple and continuous spans and allow flange transitions where the moment envelope requires. Steel box girders add a second web and a closed cell — they are torsionally stiff, ideal for curved alignments and long single-cell crossings.

Comparison of a rolled W-shape and a welded plate girder cross section — Figure 8.1Rolled W-shape (left) has fixed flange/web proportions set by the mill; a welded plate girder (right) lets the designer pick each plate independently and vary them along the span.

Attribute	Rolled W-shape	Welded plate girder	Steel box girder
Typical span range	40 – 100 ft	100 – 300 ft	150 – 500 ft
Depth-to-span	1/25 – 1/30	1/28 – 1/33	1/30 – 1/40
Fabrication cost	Lowest	Medium	Highest
Curved alignments	Poor	Fair (multiple straight chords)	Excellent
Fracture-critical	No (redundant multi-girder)	No (redundant)	Sometimes (single tub)

8.2 — Steels used in bridges

ASTM grades and toughness

AASHTO LRFD §6.4.1

AASHTO M270 is the umbrella specification for bridge structural steel. Four grades cover 99% of practice:

Grade	Fy (ksi)	Fu (ksi)	Notes
36	36	58	Rare in new work; retrofits.
50	50	65	Default for rolled shapes.
50W	50	70	Weathering; no paint if detailed correctly.
HPS 70W	70	85	High performance; ideal for negative-moment regions of continuous girders.

AASHTO §6.4.1 — Charpy V-notch toughness

Every primary steel component in a tension zone must meet a minimum CVN energy at the site's Zone 1/2/3 service temperature. Fracture-critical members require higher CVN values with tighter test frequency. Specify the zone and the fracture-critical flag on the shop drawings — do not leave it to the mill.

8.3 — Composite behavior

Making the deck work with the steel

AASHTO LRFD §6.10.1.1, §4.6.2.6

When shear studs anchor the deck to the top flange, the deck concrete carries compression in positive-moment regions and the neutral axis rises up into (or near the top of) the deck. That single geometric change roughly doubles the section modulus of the bare steel girder. AASHTO uses a modular ratio $n = E_s / E_c$ to transform the deck into an equivalent steel area. For short-term live loads use $n$ ; for long-term dead loads (creep) use $3n$ so the composite stiffness is reduced to reflect creep of the concrete.

Composite steel plate girder cross section with concrete deck and shear studs — Figure 8.2Composite section: the deck's effective width b_eff (§4.6.2.6) is transformed to steel using n for live load or 3n for sustained load, and the neutral axis of the composite section moves up toward the deck.

n \;=\; \dfrac{E_s}{E_c}, \qquad E_c \;=\; 33{,}000 \, w_c^{1.5} \, \sqrt{f'_c}

$E_s$: Steel modulus of elasticity [ksi]
$E_c$: Concrete modulus of elasticity [ksi]
$w_c$: Concrete unit weight [kcf]
$f'_c$: Concrete compressive strength [ksi]

8.4 — Section classification

Compact, non-compact, slender

AASHTO LRFD §6.10.6.2, §B6

A compact section can develop and sustain the full plastic moment $M_p$ before local or lateral instability. A non-compact section reaches only the yield moment $M_y$ . A slender web buckles before yield and requires reduced-strength or hybrid design (§6.10.1.10). For composite sections in positive flexure the deck restrains the compression flange, so classification hinges on the web slenderness $2 D_{cp}/t_w$ :

\dfrac{2 \, D_{cp}}{t_w} \;\le\; 3.76 \, \sqrt{\dfrac{E}{F_{yc}}} \qquad \Rightarrow \;\; \text{compact web}

(6.10.6.2.2-1)

$D_{cp}$: Depth of web in compression at the plastic moment [in]
$t_w$: Web thickness [in]
$F_{yc}$: Compression-flange yield stress [ksi]

If the ratio exceeds this limit but is below $5.70\sqrt{E/F_{yc}}$ , the web is non-compact; above $5.70\sqrt{E/F_{yc}}$ it is slender.

8.5 — Flexural resistance

Plastic moment Mp — the seven PNA cases

AASHTO LRFD §D6.1, §6.10.7

The plastic moment of a composite section is found by locating the plastic neutral axis (PNA) so that the compressive force above equals the tensile force below. AASHTO tabulates seven cases depending on whether the PNA lies in the slab, the top flange, or the web. In every case the deck carries a rectangular Whitney block of depth $a$ at $0.85 f'_c$ , and each steel component acts at $F_y$ throughout its depth.

Plastic stress distribution across a composite steel girder cross section — Figure 8.3Plastic stress distribution: concrete deck at 0.85 f'c above the PNA, steel at Fy in compression above and Fy in tension below. Summing internal forces to zero locates the PNA.

M_p \;=\; C_{deck} \, d_1 \;+\; C_{tf} \, d_2 \;+\; T_w \, d_3 \;+\; T_{bf} \, d_4

(6.10.7.1.2-2)

$C_{deck}$: Deck compression = 0.85 f'c b_eff a [kip]
$C_{tf}, T_w, T_{bf}$: Steel force resultants above/below PNA [kip]
$d_1 \ldots d_4$: Lever arms from each resultant to the PNA [in]

Schematic form — the exact expression depends on which PNA case controls; see AASHTO Table D6.1-1.

Ductility check §6.10.7.3

Even compact composite sections in positive flexure must satisfy

D_p \le 0.42\, D_t

, where

D_p

is the distance from the top of the deck to the PNA and

D_t

is the total composite depth. A deep PNA (high

D_p

) means the section is barely ductile and the plastic resistance must be discounted.

8.6 — Lateral-torsional buckling

The non-composite construction stage

AASHTO LRFD §6.10.8.2, §6.10.3.4

Before the deck cures the girder is a bare steel I-beam with the wet-concrete load on top. The compression flange is braced only by the cross-frames, so LTB can govern. AASHTO frames LTB with two length limits:

Lateral-torsional buckling of a steel I-beam between cross-frames — Figure 8.4LTB of the compression flange between cross-frames spaced Lb apart. Bracing at the tenth-points during deck placement is the cheapest way to keep Lb ≤ Lp so the plastic branch controls.

L_p \;=\; 1.0 \, r_t \, \sqrt{\dfrac{E}{F_{yc}}}

(6.10.8.2.3-4)

L_r \;=\; \pi \, r_t \, \sqrt{\dfrac{E}{F_{yr}}}, \qquad F_{yr} \;=\; 0.7 \, F_{yc}

(6.10.8.2.3-5)

If $L_b \le L_p$ → $F_{nc} = R_b R_h F_{yc}$ (plastic branch).
If $L_p < L_b \le L_r$ → linear inelastic reduction.
If $L_b > L_r$ → elastic LTB, $F_{cr} = \dfrac{C_b R_b \pi^2 E}{(L_b/r_t)^2}$ .

8.7 — Web shear resistance

With and without tension-field action

AASHTO LRFD §6.10.9

Shear in an I-girder is carried almost entirely by the web. Two mechanisms exist: pure elastic shear buckling (thin web, no stiffeners), and post-buckling tension-field action once the web is stiffened at spacing $d_o$ . AASHTO gives:

Steel web with transverse and bearing stiffeners and the tension-field pattern — Figure 8.5Transverse stiffeners at spacing do divide the web into panels that develop tension-field action after elastic buckling. A bearing stiffener over each support transfers the reaction into the web without local crippling.

V_n \;=\; V_p \left[\, C \;+\; \dfrac{0.87 \,(1 - C)}{\sqrt{1 + (d_o / D)^2}}\,\right]

(6.10.9.3.2-2)

V_p \;=\; 0.58 \, F_{yw} \, D \, t_w

(6.10.9.2-1)

The ratio $C$ is the elastic web-shear-buckling coefficient — a function of $D/t_w$ and $d_o/D$ . If $d_o \ge 3D$ the panel is “unstiffened” and only the elastic $C \, V_p$ term is used.

8.8 — Shear-stud connectors

Making composite action real

AASHTO LRFD §6.10.10

Composite action lives or dies with the shear studs. AASHTO requires two checks: fatigue (§6.10.10.2) — the studs handle the horizontal shear range from every truck cycle — and strength (§6.10.10.4) — the cumulative stud capacity between the point of zero moment and the point of maximum moment must equal or exceed the total horizontal thrust needed to develop $M_p$ .

Welded shear stud connectors on the top flange embedded in the concrete deck — Figure 8.67/8-in diameter × 6-in tall welded shear studs are the industry standard. Center-to-center pitch p is set by fatigue at the girder end and grows toward mid-span.

Q_r \;=\; \phi_{sc} \, Q_n, \qquad Q_n \;=\; 0.5 \, A_{sc} \, \sqrt{f'_c \, E_c} \;\le\; A_{sc} \, F_u

(6.10.10.4.3-1)

p \;\le\; \dfrac{n \, Z_r}{V_{sr}}

(6.10.10.2-1)

$Z_r$: AASHTO fatigue shear resistance per stud [kip]
$V_{sr}$: Horizontal shear range per unit length at section [kip/in]
$n$: Number of studs per transverse row [—]

8.9 — Fatigue and fracture

Category C is not optional

AASHTO LRFD §6.6.1.2

Every weld detail is assigned a fatigue category A–E based on its geometry and residual-stress state. The web-to-flange fillet weld on a plate girder is Category C (or C′ where transverse stiffeners are welded to the tension flange). The stress range under the AASHTO Fatigue-I truck ( $\gamma = 1.5$ ) must not exceed the nominal fatigue resistance $\Delta F_n$ .

AASHTO fatigue S-N curves for detail categories A through E — Figure 8.7AASHTO S-N curves. Each detail category has a constant-amplitude fatigue threshold ΔF(TH) below which infinite life is assumed.

\Delta F_n \;=\; \left(\dfrac{A}{N}\right)^{1/3} \;\ge\; \dfrac{1}{2}\, \Delta F_{TH}

(6.6.1.2.5-1)

Category	A (× 10⁸)	ΔF(TH) (ksi)	Typical detail
A	250	24	Base metal, rolled surface
B	120	16	Longitudinal groove weld
C	44	10	Web-to-flange fillet, transverse stiffener terminations
D	22	7	Fillet-welded cover plate ends
E	11	4.5	Long attachments, cover plates > 4·flange width

8.10 — Worked example

Two-span continuous composite plate girder, 2 × 140 ft

AASHTO LRFD §6.10 full check

Plan and elevation of a two-span continuous composite steel plate girder bridge — Figure 8.8Two-span continuous composite plate girder bridge: 2 × 140 ft, 44 ft roadway, 4 girders at 11 ft, 9-in deck. Design the positive-moment interior girder for Strength I.

Problem statement

Design the interior plate-girder positive-moment section at mid-span of the first span. Bridge has two equal continuous spans.

Given

Span layout2 × 140 ft continuous
Girder spacing S11.0 ft (4 girders, 44 ft roadway)
Deck thickness t_s9.0 in (0.5-in wearing surface deducted)
Steel gradeAASHTO M270 Grade 50W (Fy = 50 ksi)
$f'_c$ 4.0 ksi
Modular ratio n8 (short-term)
Cross-frame spacing L_b20 ft (construction stage)
Trial sectionTop fl 16×1.0, Web 66×0.5, Bot fl 20×1.5
Strength-I moment M_uM(DC1)=1,180 + M(DC2)=310 + M(LL+IM)=2,540 → Mu ≈ 6,470 kip-ft
Strength-I shear V_u≈ 620 kip at first interior support

Required

Check section classification, plastic moment, LTB during deck placement, web shear (with stiffeners), shear-stud pitch, and Category-C fatigue at mid-span.

Step 1

Section properties

Trial welded plate girder cross section with dimensioned top flange, web, and bottom flange — Figure 8.9Trial welded section: top flange 16×1.0, web 66×0.5, bottom flange 20×1.5. Total depth 68.5 in; span/depth ≈ 140·12/68.5 ≈ 24.5 — within AASHTO §2.5.2.6.3 for continuous composite.

Area of steel section

A_s \;=\; b_{ft}\, t_{ft} \;+\; D\, t_w \;+\; b_{bf}\, t_{bf}

Substitute

A_s \;=\; 16(1.0) + 66(0.5) + 20(1.5) \;=\; 16 + 33 + 30

Result

A_s \;=\; 79.0 \;\text{in}^2

Effective flange width (AASHTO §4.6.2.6.1)

b_{eff} \;=\; \min\!\left(\, S,\;\; 12 t_s + \max(t_w, b_{ft}/2),\;\; \dfrac{L}{4}\,\right)

Substitute

b_{eff} \;=\; \min\!\left(\, 11(12),\;\; 12(9) + \max(0.5, 8),\;\; \dfrac{140(12)}{4}\,\right) = \min(132,\, 116,\, 420)

Result

b_{eff} \;=\; 116 \;\text{in}

Short-term transformed deck area

A_{deck,n} \;=\; \dfrac{b_{eff} \, t_s}{n}

Substitute

A_{deck,n} \;=\; \dfrac{116 \,(9)}{8} \;=\; \dfrac{1{,}044}{8}

Result

A_{deck,n} \;=\; 130.5 \;\text{in}^2 \; (\text{steel equivalent})

Step 2

Section classification (§6.10.6.2.2)

Assume PNA falls in the web; check web slenderness using $D_{cp} \approx 30 \;\text{in}$ (from Step 3).

Formula

\dfrac{2 \, D_{cp}}{t_w} \;\le\; 3.76 \, \sqrt{\dfrac{E}{F_{yc}}}

Substitute

\dfrac{2 \,(30)}{0.5} \;=\; 120 \quad \text{vs.} \quad 3.76 \, \sqrt{\dfrac{29{,}000}{50}} \;=\; 3.76 \, \sqrt{580} \;=\; 90.6

Result

120 \;>\; 90.6 \;\;\Rightarrow\;\; \text{non-compact web — use §6.10.7.2 (yield moment approach)}

What can go wrong

The web is non-compact, so

M_n

is capped at

R_b R_h M_y

rather than reaching

M_p

. To reach the plastic branch we would need

t_w \ge 66/90.6 = 0.73

in (call it 3/4 in) — a common design iteration.

Step 3

Elastic composite section (short-term, n = 8)

Neutral-axis location from bottom (transformed section)

\bar{y} \;=\; \dfrac{\sum A_i \, y_i}{\sum A_i}

Substitute

\bar{y} \;=\; \dfrac{30(0.75) + 33(1.5 + 33) + 16(68 + 0.5) + 130.5(68.5 + 4.5)}{79 + 130.5}

Result

\bar{y} \;\approx\; 54.1 \;\text{in from bottom of bot flange}

Composite moment of inertia (short-term)

I_{ST} \;=\; \sum \bigl[\, I_o \;+\; A_i \,(y_i - \bar{y})^2 \,\bigr]

Substitute

I_{ST} \;=\; 30(54.1 - 0.75)^2 + 33(54.1 - 17.5)^2 + 16(54.1 - 68)^2 + I_{web} + I_{deck,n}

Result

I_{ST} \;\approx\; 231{,}000 \;\text{in}^4

Section modulus, bottom fiber

S_{ST,\text{bot}} \;=\; \dfrac{I_{ST}}{\bar{y}}

Substitute

S_{ST,\text{bot}} \;=\; \dfrac{231{,}000}{54.1}

Result

S_{ST,\text{bot}} \;\approx\; 4{,}270 \;\text{in}^3

Step 4

Yield-moment resistance (§6.10.7.2)

Yield moment

M_y \;=\; \dfrac{F_y \, S_{ST,\text{bot}}}{12}

Substitute

M_y \;=\; \dfrac{50 \,(4{,}270)}{12}

Result

M_y \;\approx\; 17{,}790 \;\text{kip-ft}

Nominal flexural resistance (composite non-compact)

M_n \;=\; R_b \, R_h \, M_y

Substitute

R_h = 1.0 \; (\text{homogeneous 50W}), \quad R_b \approx 1.0 \; (\text{below slender-web threshold})

Result

M_n \;\approx\; 17{,}790 \;\text{kip-ft}

Factored resistance

\phi_f \, M_n \;\ge\; M_u

Substitute

1.0 \,(17{,}790) \;=\; 17{,}790 \;\ge\; 6{,}470

Result

\dfrac{M_u}{\phi_f M_n} \;=\; \dfrac{6{,}470}{17{,}790} \;=\; 0.36 \;\;\Rightarrow\;\; \text{OK (deflection-controlled)}

Step 5

LTB during deck placement (§6.10.8.2)

Effective radius of gyration

r_t \;=\; \dfrac{b_{ft}}{\sqrt{12 \left(1 \;+\; \dfrac{D_c \, t_w}{3\, b_{ft}\, t_{ft}}\right)}}

Substitute

r_t \;=\; \dfrac{16}{\sqrt{12 \left(1 \;+\; \dfrac{34 \,(0.5)}{3 \,(16)(1.0)}\right)}} \;=\; \dfrac{16}{\sqrt{12 \,(1.354)}}

Result

r_t \;\approx\; 3.97 \;\text{in}

Plastic bracing length

L_p \;=\; 1.0 \, r_t \, \sqrt{\dfrac{E}{F_{yc}}}

Substitute

L_p \;=\; 1.0 \,(3.97) \, \sqrt{\dfrac{29{,}000}{50}} \;=\; 3.97 \,(24.08)

Result

L_p \;\approx\; 95.6 \;\text{in} \;=\; 7.97 \;\text{ft}

Inelastic bracing length

L_r \;=\; \pi \, r_t \, \sqrt{\dfrac{E}{F_{yr}}}, \quad F_{yr} \;=\; 0.7\, F_{yc}

Substitute

L_r \;=\; \pi \,(3.97) \, \sqrt{\dfrac{29{,}000}{35}} \;=\; 12.47 \,(28.78)

Result

L_r \;\approx\; 359 \;\text{in} \;=\; 29.9 \;\text{ft}

Cross-frame spacing $L_b = 20\;\text{ft}$ lies between $L_p = 7.97\;\text{ft}$ and $L_r = 29.9\;\text{ft}$ → inelastic LTB.

Inelastic LTB resistance

F_{nc} \;=\; C_b \left[\, 1 \;-\; \left(1 - \dfrac{F_{yr}}{F_{yc}}\right) \dfrac{L_b - L_p}{L_r - L_p} \,\right] R_b R_h F_{yc} \;\le\; R_b R_h F_{yc}

Substitute

F_{nc} \;=\; 1.0 \left[\, 1 \;-\; 0.3 \,\dfrac{20 - 7.97}{29.9 - 7.97} \,\right] (1)(1)(50) \;=\; \bigl[1 - 0.3\,(0.549)\bigr]\, 50 \;=\; 0.835 \,(50)

Result

F_{nc} \;\approx\; 41.7 \;\text{ksi} \;\;\Rightarrow\;\; M_{n,\text{constr}} \;\approx\; \dfrac{41.7 \,(2{,}650)}{12} \;\approx\; 9{,}200 \;\text{kip-ft}

Construction-stage demand (DC1 only)

M_{u,\text{constr}} \;=\; 1.25 \, M_{DC1}

Substitute

M_{u,\text{constr}} \;=\; 1.25 \,(1{,}180)

Result

M_{u,\text{constr}} \;=\; 1{,}475 \;\text{kip-ft} \;\ll\; 9{,}200 \;\text{kip-ft} \;\;\Rightarrow\;\; \text{OK}

Step 6

Web shear (§6.10.9) — stiffened panel

Provide transverse stiffeners at $d_o = 8\;\text{ft} = 96\;\text{in}$ near the pier.

Plastic shear

V_p \;=\; 0.58 \, F_{yw} \, D \, t_w

Substitute

V_p \;=\; 0.58 \,(50)(66)(0.5)

Result

V_p \;=\; 957 \;\text{kip}

Shear-buckling coefficient k

k \;=\; 5 \;+\; \dfrac{5}{(d_o / D)^2}

Substitute

\dfrac{d_o}{D} \;=\; \dfrac{96}{66} \;=\; 1.455 \;\;\Rightarrow\;\; k \;=\; 5 \;+\; \dfrac{5}{2.117}

Result

k \;=\; 7.36

Elastic buckling ratio C (elastic regime)

C \;=\; \dfrac{1.57}{(D/t_w)^2} \, \dfrac{E \, k}{F_{yw}}

Substitute

\dfrac{D}{t_w} \;=\; 132; \quad 1.40\sqrt{\dfrac{E k}{F_{yw}}} = 1.40\sqrt{\dfrac{29{,}000 \,(7.36)}{50}} = 91.6 < 132 \;\Rightarrow\; \text{elastic}; \quad C \;=\; \dfrac{1.57 \,(4{,}269)}{132^2}

Result

C \;=\; \dfrac{6{,}702}{17{,}424} \;=\; 0.385

Nominal shear with tension-field action

V_n \;=\; V_p \left[\, C \;+\; \dfrac{0.87 \,(1 - C)}{\sqrt{1 + (d_o/D)^2}} \,\right]

Substitute

V_n \;=\; 957 \left[\, 0.385 \;+\; \dfrac{0.87 \,(0.615)}{\sqrt{1 + 2.117}} \,\right] \;=\; 957 \left[\, 0.385 \;+\; \dfrac{0.535}{1.766} \,\right] \;=\; 957 \,(0.688)

Result

V_n \;=\; 658 \;\text{kip}

Factored shear check

\phi_v \, V_n \;\ge\; V_u

Substitute

1.0 \,(658) \;=\; 658 \;\ge\; 620

Result

\dfrac{V_u}{\phi_v V_n} \;=\; \dfrac{620}{658} \;=\; 0.94 \;\;\Rightarrow\;\; \text{OK (tighten } d_o \text{ to 72 in for margin)}

Step 7

Shear studs — fatigue and strength (§6.10.10)

Strength: total horizontal shear P

P \;=\; \min\!\left(\, 0.85 \, f'_c \, b_{eff} \, t_s,\;\; F_y \, A_s \,\right)

Substitute

P \;=\; \min\!\left(\, 0.85 \,(4)(116)(9),\;\; 50 \,(79) \,\right) \;=\; \min(3{,}549,\, 3{,}950)

Result

P \;=\; 3{,}549 \;\text{kip}

Nominal stud capacity (7/8-in Ø, Fu = 60 ksi)

Q_n \;=\; 0.5 \, A_{sc} \, \sqrt{f'_c \, E_c} \;\le\; A_{sc} \, F_u

Substitute

A_{sc} = \tfrac{\pi}{4}(0.875)^2 = 0.601\;\text{in}^2; \;\; E_c = 33{,}000 \,(0.150)^{1.5} \sqrt{4} = 3{,}834\;\text{ksi}; \;\; Q_n = 0.5 \,(0.601)\sqrt{4 \,(3{,}834)} = 0.5 \,(0.601)(123.8)

Result

Q_n \;=\; 37.2 \;\text{kip}; \;\; \text{cap } A_{sc} F_u = 36.1 \;\text{kip} \;\Rightarrow\; \text{use } 36.1 \;\text{kip}

Studs required between zero and max moment

n \;=\; \dfrac{P}{\phi_{sc} \, Q_n}

Substitute

n \;=\; \dfrac{3{,}549}{0.85 \,(36.1)}

Result

n \;\approx\; 116 \;\text{studs (3 rows across flange → 39 rows over 70 ft)}

Fatigue pitch check (Fatigue-I, Category C studs, N > 10⁶)

p \;\le\; \dfrac{n_{studs} \, Z_r}{V_{sr}}, \quad Z_r \;=\; \dfrac{5.5 \, d^2}{2}

Substitute

d = 0.875\;\text{in}; \;\; Z_r = \dfrac{5.5 \,(0.875)^2}{2} = 2.11\;\text{kip}; \;\; V_{sr} \approx 1.3\;\text{kip/in near support}; \;\; p \le \dfrac{3 \,(2.11)}{1.3}

Result

p \;\le\; 4.87 \;\text{in near supports; open to 18 in max at mid-span}

Step 8

Web-to-flange fatigue (Category C, §6.6.1.2)

Live-load stress range at bottom fiber

\Delta f \;=\; \dfrac{\gamma \, M_{LL+IM,\,\text{fatigue}} \, c}{I_{LT}}

Substitute

\gamma = 1.5 \text{ (Fatigue I)}; \;\; M_{fatigue} = 720\;\text{kip-ft}; \;\; \Delta f = \dfrac{1.5 \,(720)(12)(54.1)}{220{,}000}

Result

\Delta f \;\approx\; 3.18 \;\text{ksi}

Fatigue resistance — Category C, infinite life

\Delta F_n \;=\; \Delta F_{TH} \;=\; 10 \;\text{ksi}

Substitute

3.18 \;\text{ksi} \;\le\; 10 \;\text{ksi}

Result

\text{Ratio} = 0.32 \;\;\Rightarrow\;\; \text{OK — Category-C web/flange weld has infinite life}

Step 9

Deflection check (§2.5.2.6.2)

Optional live-load deflection limit

\Delta_{LL} \;\le\; \dfrac{L}{800}

Substitute

\Delta_{allow} \;=\; \dfrac{140 \,(12)}{800}

Result

\Delta_{allow} \;=\; 2.10 \;\text{in}

Computed deflection (composite short-term, n = 8)

\Delta_{LL} \;=\; \beta_{cont} \, \dfrac{5 \, w \, L^4}{384 \, E \, I_{ST}}

Substitute

\beta_{cont} \approx 0.55; \;\; w_{LL+IM,\,g} \approx 0.85\;\text{kip/ft}; \;\; \Delta_{LL} = 0.55 \, \dfrac{5 \,(0.85/12)\,(140 \cdot 12)^4}{384 \,(29{,}000)(231{,}000)}

Result

\Delta_{LL} \;\approx\; 1.32 \;\text{in} \;\le\; 2.10 \;\text{in} \;\;\Rightarrow\;\; \text{OK}

Final section detailing (from computed A_s)

Interior plate girder — positive-moment section (mid-span, 1st span)

Location	A_s required	Bars provided	Spacing / detail
Top flange	≥ 15.2 in² (from Mu)	PL 16 × 1.0 in = 16.0 in²	Grade 50W; full length
Web	≥ 33 in² for shear + LTB	PL 66 × 0.50 in = 33.0 in²	Trans. stiffeners at 8 ft near pier, 12 ft in mid-span
Bottom flange	≥ 27 in² (tension, Mu)	PL 20 × 1.5 in = 30.0 in²	Full length; splice at 0.72·L per moment envelope
Shear studs	116 studs / half-span	3 rows of 7/8-in Ø × 6-in studs	5-in pitch at support, 12-in at mid-span
Bearing stiffeners	R_u = 620 kip	2 × PL 7 × 3/4 in each face, full-depth	Both girder ends and at pier

All fillet welds ≥ 5/16 in per §6.13.3.4. CVN toughness per Zone-2, non-fracture-critical. Camber = dead-load deflection + 3/8-in profile grade correction; shown on shop drawings.

8.10b — Design Example 3

Composite steel–concrete bridge (simple-span, L = 40 ft)

AASHTO LRFD Strength I, Service II, Fatigue II — full workflow

This worked example — reproduced from Simplified LRFD Bridge Design (Design Example 3) — walks the complete AASHTO LRFD workflow for a simple-span composite steel–concrete bridge with rolled W24×76 girders. Every check is presented as Equation → Substitute → Result so the formula is stated first before any numbers are entered.

Figure 8.11Fig. 2.21 — Composite steel–concrete bridge example. Six W24×76 girders spaced at S = 8 ft support a 44-ft roadway with an 8-in deck slab and 2-in haunch. Overhang de = 2 ft to inside face of barrier.

Problem statement

Design the superstructure of a simple-span composite steel–concrete bridge for Strength I, Service II, and Fatigue II Limit States.

Given

Span L40 ft (simple span)
Beam spacing S8 ft (6 girders, 44 ft roadway)
Slab thickness t_s8 in
Haunch2 in
Barrier weight BW0.5 kip/ft (each)
Future wearing surface w_FWS25 lbf/ft²
Stay-in-place forms7 lbf/ft²
Concrete f'_c4 kip/in² (0.85·f'_c uses 4.5 ksi per source table)
Steel yield F_y60 kip/in²
Concrete unit weight w_c150 lbf/ft³
ADTT (one direction)2,500
Design fatigue life75 yr
Overhang de2 ft
Rolled sectionW24×76 → d = 23.9 in, bf = 9 in, tf = 0.68 in, tw = 0.44 in, D = 22.54 in, A = 22.4 in², Ix = 2,100 in⁴, w = 76 lbf/ft

Required

Compute effective flange widths, dead / live load moments and shears with distribution factors, then check Strength I flexure and shear, Service II flange stresses, and Fatigue II (moment + special shear provisions of §6.10.5.3) for both interior and exterior girders.

Step 1

Effective flange width (§4.6.2.6)

Interior beam — b_e = beam spacing

b_{e,\text{int}} \;=\; S

Substitute

b_{e,\text{int}} \;=\; (8\;\text{ft})\,(12\;\text{in/ft})

Result

b_{e,\text{int}} \;=\; 96\;\text{in}

Exterior beam — half spacing plus overhang

b_{e,\text{ext}} \;=\; \dfrac{S}{2} \;+\; \text{overhang}

Substitute

b_{e,\text{ext}} \;=\; \dfrac{96\;\text{in}}{2} \;+\; 39\;\text{in}

Result

b_{e,\text{ext}} \;=\; 87\;\text{in}

Figure 8.12Fig. 2.23 — Composite section for the interior girder: 96-in effective flange × 8-in slab acting with the W24×76 beam through a 2-in haunch.

Step 2

Non-composite dead load DC₁ (per interior girder)

Each dead-load component is a distributed weight $w = \gamma \cdot A_{trib}$ applied to the bare steel girder before the deck cures.

Slab weight

DC_{slab} \;=\; b_{e,\text{int}} \cdot t_s \cdot w_c

Substitute

DC_{slab} \;=\; \tfrac{96}{12}\,(\tfrac{8}{12})\,(0.15)

Result

DC_{slab} \;=\; 0.80\;\text{kip/ft}

Haunch weight (2 in × 9 in)

DC_{haunch} \;=\; b_f \cdot h_{haunch} \cdot w_c

Substitute

DC_{haunch} \;=\; \tfrac{9}{12}\,(\tfrac{2}{12})\,(0.15)

Result

DC_{haunch} \;=\; 0.02\;\text{kip/ft}

Steel self-weight (+ 5% for diaphragms & stiffeners)

DC_{steel} \;=\; 1.05 \, w_{beam}

Substitute

DC_{steel} \;=\; 1.05\,(0.076)

Result

DC_{steel} \;=\; 0.08\;\text{kip/ft}

Stay-in-place metal forms — spread over 6 girders

DC_{forms} \;=\; \dfrac{w_{form} \cdot w_{roadway}}{n_{girders}}

Substitute

DC_{forms} \;=\; \dfrac{0.007\,(44)}{6}

Result

DC_{forms} \;=\; 0.051\;\text{kip/ft}

Total non-composite DC₁

DC_1 \;=\; DC_{slab} + DC_{haunch} + DC_{steel} + DC_{forms}

Substitute

DC_1 \;=\; 0.80 + 0.02 + 0.08 + 0.051

Result

DC_1 \;=\; 0.951\;\text{kip/ft}

Non-composite shear at support

V_{DC_1} \;=\; \dfrac{w \, L}{2}

Substitute

V_{DC_1} \;=\; \dfrac{0.951\,(40)}{2}

Result

V_{DC_1} \;=\; 19.02\;\text{kip}

Non-composite mid-span moment

M_{DC_1} \;=\; \dfrac{w \, L^2}{8}

Substitute

M_{DC_1} \;=\; \dfrac{0.951\,(40)^2}{8}

Result

M_{DC_1} \;=\; 190.2\;\text{kip-ft}

Step 3

Composite dead load DC₂ and future wearing surface DW

Barrier load (2 barriers, 6 girders)

DC_2 \;=\; \dfrac{BW \cdot n_{barriers}}{n_{girders}}

Substitute

DC_2 \;=\; \dfrac{0.5\,(2)}{6}

Result

DC_2 \;=\; 0.167\;\text{kip/ft}

Shear from DC₂

V_{DC_2} \;=\; \dfrac{w \, L}{2}

Substitute

V_{DC_2} \;=\; \dfrac{0.167\,(40)}{2}

Result

V_{DC_2} \;=\; 3.34\;\text{kip}

Moment from DC₂

M_{DC_2} \;=\; \dfrac{w \, L^2}{8}

Substitute

M_{DC_2} \;=\; \dfrac{0.167\,(40)^2}{8}

Result

M_{DC_2} \;=\; 33.4\;\text{kip-ft}

Future wearing surface (spread over 6 girders)

DW \;=\; \dfrac{w_{FWS} \cdot w_{roadway}}{n_{girders}}

Substitute

DW \;=\; \dfrac{0.025\,(44)}{6}

Result

DW \;=\; 0.183\;\text{kip/ft}

DW shear and moment

V_{DW} = \tfrac{wL}{2}, \quad M_{DW} = \tfrac{wL^2}{8}

Substitute

V_{DW} = \tfrac{0.183\,(40)}{2}, \quad M_{DW} = \tfrac{0.183\,(40)^2}{8}

Result

V_{DW} = 3.66\;\text{kip}, \quad M_{DW} = 36.6\;\text{kip-ft}

Engineering note

Total dead-load effects per interior girder:

V_{DC} = 22.4\;\text{kip}

M_{DC} = 223.6\;\text{kip-ft}

V_{DW} = 3.66\;\text{kip}

M_{DW} = 36.6\;\text{kip-ft}

. Assume exterior girders carry the same DL (conservative).

Step 4

Longitudinal stiffness parameter Kg (§4.6.2.2.1)

Modulus of elasticity of concrete

E_c \;=\; 33{,}000\, w_c^{1.5}\, \sqrt{f'_c}

Substitute

E_c \;=\; 33{,}000\,(0.15)^{1.5}\,\sqrt{4}

Result

E_c \;=\; 3{,}834\;\text{ksi}

Modular ratio n = E_s / E_c

n \;=\; \dfrac{E_s}{E_c}

Substitute

n \;=\; \dfrac{29{,}000}{3{,}834} \;=\; 7.56 \;\Rightarrow\; \text{use } n = 8

Result

n \;=\; 8

Longitudinal stiffness parameter

K_g \;=\; n\,(I \;+\; A\, e_g^{\,2})

Substitute

K_g \;=\; 8\,\bigl[2{,}100 \;+\; 22.4\,(17.95)^2\bigr]

Result

K_g \;=\; 74{,}539\;\text{in}^4

Step 5

Live-load distribution factors (§4.6.2.2)

Cross-section type (a): concrete slab on steel beams.

Interior — moment, one design lane loaded

DFM_{si} \;=\; 0.06 \;+\; \left(\dfrac{S}{14}\right)^{0.4} \left(\dfrac{S}{L}\right)^{0.3} \left(\dfrac{K_g}{12\,L\,t_s^{3}}\right)^{0.1}

Substitute

DFM_{si} \;=\; 0.06 + \left(\tfrac{8}{14}\right)^{0.4}\left(\tfrac{8}{40}\right)^{0.3}\left(\tfrac{74{,}539}{12(40)(8)^3}\right)^{0.1}

Result

DFM_{si} \;=\; 0.498\;\text{lanes}

Interior — moment, two or more design lanes

DFM_{mi} \;=\; 0.075 \;+\; \left(\dfrac{S}{9.5}\right)^{0.6}\left(\dfrac{S}{L}\right)^{0.2}\left(\dfrac{K_g}{12\,L\,t_s^{3}}\right)^{0.1}

Substitute

DFM_{mi} \;=\; 0.075 + \left(\tfrac{8}{9.5}\right)^{0.6}\left(\tfrac{8}{40}\right)^{0.2}\left(\tfrac{74{,}539}{12(40)(8)^3}\right)^{0.1}

Result

DFM_{mi} \;=\; 0.655\;\text{lanes [controls interior]}

Interior — shear, one lane

DFV_{si} \;=\; 0.36 \;+\; \dfrac{S}{25}

Substitute

DFV_{si} \;=\; 0.36 + \tfrac{8}{25}

Result

DFV_{si} \;=\; 0.68\;\text{lanes}

Interior — shear, two or more lanes

DFV_{mi} \;=\; 0.2 \;+\; \dfrac{S}{12} \;-\; \left(\dfrac{S}{35}\right)^{2}

Substitute

DFV_{mi} \;=\; 0.2 + \tfrac{8}{12} - \left(\tfrac{8}{35}\right)^2

Result

DFV_{mi} \;=\; 0.814\;\text{lanes [controls interior]}

Figure 8.13Fig. 2.25 — Lever rule for the exterior-beam distribution factor with one lane loaded. Taking ΣM about the interior-side hinge and dividing by S gives R = 0.625·P.

Exterior — moment, one lane (lever rule + m = 1.2)

DFM_{se} \;=\; R \cdot m

Substitute

DFM_{se} \;=\; 0.625\,(1.2)

Result

DFM_{se} \;=\; 0.75\;\text{lanes [controls exterior]}

Exterior — moment, two or more lanes (correction e)

e \;=\; 0.77 \;+\; \dfrac{d_e}{9.1}, \quad DFM_{me} \;=\; e \cdot DFM_{mi}

Substitute

e \;=\; 0.77 + \tfrac{2}{9.1} = 0.9898; \quad DFM_{me} = 0.9898\,(0.655)

Result

DFM_{me} \;=\; 0.648\;\text{lanes}

Exterior — shear (lever rule 1-lane; e-factor 2-lanes)

DFV_{se} = R\,m; \quad e = 0.6 + \dfrac{d_e}{10}; \quad DFV_{me} = e\,DFV_{mi}

Substitute

DFV_{se} = 0.75; \quad e = 0.6 + \tfrac{2}{10} = 0.8; \quad DFV_{me} = 0.8\,(0.814)

Result

DFV_{se} = 0.75\;[\text{controls}], \quad DFV_{me} = 0.651

Step 6

Unfactored live-load moments (HL-93)

Figure 8.14Fig. 2.26 — HS-20 design truck positioned for maximum moment at midspan of a 40-ft simple span.

Truck reaction (ΣM about B = 0)

R_A \;=\; \dfrac{32\,(34) + 32\,(20) + 8\,(6)}{L}

Substitute

R_A \;=\; \dfrac{1{,}088 + 640 + 48}{40}

Result

R_A \;=\; 44.4\;\text{kip}

Design truck midspan moment

M_{tr} \;=\; R_A\,(L/2) \;-\; 32\,(14)

Substitute

M_{tr} \;=\; 44.4\,(20) - 32\,(14)

Result

M_{tr} \;=\; 440\;\text{kip-ft}

Figure 8.15Figs. 2.27 & 2.28 — Design tandem (2 × 25 kip @ 4 ft) and design lane load (0.64 kip/ft) positioned for maximum midspan moment.

Design tandem midspan moment

M_{tandem} \;=\; R_A \,(L/2), \quad R_A = \tfrac{25}{2} + 25\,\tfrac{16}{40}

Substitute

R_A = 22.5\;\text{kip}; \quad M_{tandem} = 22.5\,(20)

Result

M_{tandem} \;=\; 450\;\text{kip-ft [controls]}

Design lane moment

M_{ln} \;=\; \dfrac{w\,L^{2}}{8}

Substitute

M_{ln} \;=\; \dfrac{0.64\,(40)^2}{8}

Result

M_{ln} \;=\; 128\;\text{kip-ft}

Total unfactored live-load moment per lane (IM = 33%)

M_{LL+IM} \;=\; M_{tandem}\,(1 + IM) \;+\; M_{ln}

Substitute

M_{LL+IM} \;=\; 450\,(1.33) + 128

Result

M_{LL+IM} \;=\; 726.5\;\text{kip-ft per lane}

Step 7

Unfactored live-load shear at the support

Figure 8.16Figs. 2.29–2.31 — Truck, tandem, and lane loads placed for maximum shear at the support of the 40-ft span.

Design truck shear reaction

R_A \;=\; \dfrac{32\,(40) + 32\,(26) + 8\,(12)}{L}

Substitute

R_A \;=\; \dfrac{1{,}280 + 832 + 96}{40}

Result

V_{tr} \;=\; 55.2\;\text{kip [controls]}

Design tandem shear

V_{tandem} \;=\; \dfrac{25\,(40) + 25\,(36)}{L}

Substitute

V_{tandem} \;=\; \dfrac{1{,}000 + 900}{40}

Result

V_{tandem} \;=\; 47.5\;\text{kip}

Design lane shear

V_{ln} \;=\; \dfrac{w\,L}{2}

Substitute

V_{ln} \;=\; \dfrac{0.64\,(40)}{2}

Result

V_{ln} \;=\; 12.8\;\text{kip}

Total unfactored live-load shear per lane

V_{LL+IM} \;=\; V_{tr}\,(1 + IM) \;+\; V_{ln}

Substitute

V_{LL+IM} \;=\; 55.2\,(1.33) + 12.8

Result

V_{LL+IM} \;=\; 86.22\;\text{kip per lane}

Table 8.4 — Live-load effect summary (per-girder = per-lane × DF).
Location	Lanes	DFM	M_LL+IM (kip-ft/girder)	DFV	V_LL+IM (kip/girder)
Interior	2+	0.655	475.9	0.814	70.2
Exterior	1	0.75	544.9	0.75	64.7

Step 8

Strength I factored effects (Tbl. 3.4.1-1)

Strength I combination

U \;=\; 1.25\,DC \;+\; 1.50\,DW \;+\; 1.75\,(LL + IM)

Interior girder — factored moment

M_u \;=\; 1.25\,M_{DC} + 1.50\,M_{DW} + 1.75\,M_{LL+IM}

Substitute

M_u \;=\; 1.25\,(223.6) + 1.50\,(36.6) + 1.75\,(475.9)

Result

M_u \;=\; 1{,}167\;\text{kip-ft}

Interior girder — factored shear

V_u \;=\; 1.25\,V_{DC} + 1.50\,V_{DW} + 1.75\,V_{LL+IM}

Substitute

V_u \;=\; 1.25\,(22.4) + 1.50\,(3.66) + 1.75\,(70.2)

Result

V_u \;=\; 156.3\;\text{kip}

Exterior girder — factored moment

M_u \;=\; 1.25\,M_{DC} + 1.50\,M_{DW} + 1.75\,M_{LL+IM}

Substitute

M_u \;=\; 1.25\,(223.6) + 1.50\,(36.6) + 1.75\,(544.9)

Result

M_u \;=\; 1{,}288\;\text{kip-ft}

Exterior girder — factored shear

V_u \;=\; 1.25\,V_{DC} + 1.50\,V_{DW} + 1.75\,V_{LL+IM}

Substitute

V_u \;=\; 1.25\,(22.4) + 1.50\,(3.66) + 1.75\,(64.7)

Result

V_u \;=\; 146.7\;\text{kip}

Step 9

Plastic moment capacity Mₚ — interior girder (App. D6.1)

Figure 8.17Fig. 2.32 — Composite steel–concrete section used for the PNA search. Reinforcement layers at Crt = 3 in (top) and Crb = 5 in (bottom) from the top of deck.

Compute the plastic forces before searching for the PNA:

Slab compressive force

P_s \;=\; 0.85\, f'_c \, b_s \, t_s

Substitute

P_s \;=\; 0.85\,(4.5)(96)(8)

Result

P_s \;=\; 2{,}937.6\;\text{kip}

Compression-flange force

P_c \;=\; F_{yc}\, b_c\, t_c

Substitute

P_c \;=\; 60\,(9)(0.68)

Result

P_c \;=\; 367.2\;\text{kip}

Web force

P_w \;=\; F_{yw}\, D\, t_w

Substitute

P_w \;=\; 60\,(22.54)(0.44)

Result

P_w \;=\; 595\;\text{kip}

Tension-flange force

P_t \;=\; F_{yt}\, b_t\, t_t

Substitute

P_t \;=\; 60\,(9)(0.68)

Result

P_t \;=\; 367.2\;\text{kip}

Engineering note

Rebar contribution neglected (conservative):

P_{rb} = P_{rt} = 0

Screen the seven AASHTO cases:

Case I (PNA in web)

P_t + P_w \;\ge\; P_c + P_s ?

Substitute

962.2 \;\text{vs.}\; 3{,}305

Result

962.2 < 3{,}305 \;\Rightarrow\; \text{no good}

Case II (PNA in top flange)

P_t + P_w + P_c \;\ge\; P_s ?

Substitute

1{,}329.4 \;\text{vs.}\; 2{,}937.6

Result

1{,}329.4 < 2{,}937.6 \;\Rightarrow\; \text{no good}

Case III (PNA below bottom rebar in deck)

P_t + P_w + P_c \;\ge\; (C_{rb}/t_s)\,P_s ?

Substitute

1{,}329.4 \;\text{vs.}\; (5/8)(2{,}937.6) = 1{,}836

Result

1{,}329.4 < 1{,}836 \;\Rightarrow\; \text{no good}

Case V (PNA between reinforcement layers) — controls

P_t + P_w + P_c \;\ge\; (C_{rt}/t_s)\,P_s ?

Substitute

1{,}329.4 \;\text{vs.}\; (3/8)(2{,}937.6) = 1{,}102

Result

1{,}329.4 > 1{,}102 \;\Rightarrow\; \text{OK — Case V governs}

PNA depth from top of deck

\bar{Y} \;=\; t_s \left(\dfrac{P_{rb} + P_c + P_w + P_t - P_{rt}}{P_s}\right)

Substitute

\bar{Y} \;=\; 8\,\dfrac{0 + 367.2 + 595 + 367.2 - 0}{2{,}937.6}

Result

\bar{Y} \;=\; 3.62\;\text{in from top of deck}

Lever arms from PNA to steel components

d_c, d_w, d_t \;=\; (\text{mid-height distances})

Substitute

d_c = 8 + 2 + \tfrac{0.68}{2} - 3.62 = 6.72; \;\; d_w = 8 + 2 + \tfrac{23.9}{2} - 3.62 = 18.33; \;\; d_t = 8 + 2 + 23.9 - \tfrac{0.68}{2} - 3.62 = 29.94

Result

d_c = 6.72\;\text{in}, \; d_w = 18.33\;\text{in}, \; d_t = 29.94\;\text{in}

Figure 8.18Fig. 2.33 — Interior girder Case V PNA location and lever arms used in the Mp integral.

Plastic moment (Case V, App. Tbl. D6.1-1)

M_p \;=\; \dfrac{\bar{Y}^{\,2}\, P_s}{2\, t_s} \;+\; \bigl(P_c d_c + P_w d_w + P_t d_t\bigr)

Substitute

M_p \;=\; \dfrac{(3.62)^2\,(2{,}937.6)}{2\,(8)} + \bigl[367.2\,(6.72) + 595\,(18.33) + 367.2\,(29.94)\bigr]

Result

M_p \;=\; 27{,}432\;\text{kip-in} \;=\; 2{,}286\;\text{kip-ft (interior)}

Step 10

Plastic moment capacity Mₚ — exterior girder

Repeat with the exterior effective width $b_s = 87\;\text{in}$ .

Slab force (exterior)

P_s \;=\; 0.85\, f'_c\, b_s\, t_s

Substitute

P_s \;=\; 0.85\,(4.5)(87)(8)

Result

P_s \;=\; 2{,}662.2\;\text{kip}

Case V check (governs again)

P_t + P_w + P_c \;\ge\; (C_{rt}/t_s)\, P_s ?

Substitute

1{,}329.4 \;\text{vs.}\; (3/8)(2{,}662.2) = 998.3

Result

1{,}329.4 > 998.3 \;\Rightarrow\; \text{OK}

PNA depth (exterior)

\bar{Y} \;=\; t_s\, \dfrac{P_c + P_w + P_t}{P_s}

Substitute

\bar{Y} \;=\; 8\,\dfrac{367.2 + 595 + 367.2}{2{,}662.2}

Result

\bar{Y} \;=\; 4.00\;\text{in}

Exterior lever arms

d_c, d_w, d_t

Substitute

d_c = 6.34, \; d_w = 17.95, \; d_t = 29.56\;\text{in}

Figure 8.19Fig. 2.35 — Exterior girder Case V PNA (Ȳ = 4.0 in) and lever arms.

Plastic moment (exterior)

M_p \;=\; \dfrac{\bar{Y}^{\,2}\, P_s}{2\, t_s} \;+\; (P_c d_c + P_w d_w + P_t d_t)

Substitute

M_p \;=\; \dfrac{(4.0)^2\,(2{,}662.2)}{2\,(8)} + \bigl[367.2\,(6.34) + 595\,(17.95) + 367.2\,(29.56)\bigr]

Result

M_p \;=\; 26{,}660\;\text{kip-in} \;=\; 2{,}222\;\text{kip-ft (exterior)}

Step 11

Plastic shear resistance Vₚ (§6.10.9.2)

Unstiffened web plastic shear (both girders)

V_p \;=\; 0.58\, F_{yw}\, D\, t_w

Substitute

V_p \;=\; 0.58\,(60)(22.54)(0.44)

Result

V_p \;=\; 345.2\;\text{kip}

Step 12

Strength I flexure check (§6.10.7.1)

Web-compactness check

\dfrac{2\, D_{cp}}{t_w} \;\le\; 3.76\,\sqrt{\dfrac{E}{F_{yc}}}

Substitute

D_{cp} = 0\;\text{(PNA in slab)} \;\Rightarrow\; 0 \;\le\; 3.76\sqrt{29{,}000/60} = 82.7

Result

\text{Section is compact composite [OK]}

Composite depth Dt and 0.1 Dt threshold

D_t \;=\; d \;+\; h_{haunch} \;+\; t_s, \quad 0.1 D_t

Substitute

D_t \;=\; 23.9 + 2 + 8 = 33.9; \quad 0.1 D_t = 3.39

Result

D_t = 33.9\;\text{in}, \;\; 0.1 D_t = 3.39\;\text{in}

Nominal flexural resistance (Eq. 6.10.7.1.2-2, since D_p > 0.1 D_t)

M_n \;=\; M_p \left(1.07 \;-\; 0.7\,\dfrac{D_p}{D_t}\right)

Substitute

\text{Interior: } M_n = 2{,}286\,(1.07 - 0.7\,\tfrac{3.62}{33.9})

Result

M_{n,\text{int}} \;=\; 2{,}275\;\text{kip-ft}

Exterior nominal flexural resistance

M_n \;=\; M_p \left(1.07 \;-\; 0.7\,\dfrac{D_p}{D_t}\right)

Substitute

M_n = 2{,}222\,(1.07 - 0.7\,\tfrac{4.0}{33.9})

Result

M_{n,\text{ext}} \;=\; 2{,}235\;\text{kip-ft}

Strength I flexure demand vs. capacity (φ_f = 1.0)

M_u \;+\; \tfrac{1}{3}\, f_\ell\, S_{xt} \;\le\; \phi_f\, M_n

Substitute

\text{Int: } 1{,}167 \le 1.0\,(2{,}275); \quad \text{Ext: } 1{,}288 \le 1.0\,(2{,}235)

Result

\text{Both OK}\;(f_\ell \approx 0)

Step 13

Strength I shear check (§6.10.9)

Shear-buckling coefficient (unstiffened → k = 5)

k \;=\; 5 \;+\; \dfrac{5}{(d_o/D)^{2}} \;\to\; 5.0

Result

k \;=\; 5.0

Shear-yielding limit check

\dfrac{D}{t_w} \;\le\; 1.12\,\sqrt{\dfrac{E\,k}{F_{yw}}}

Substitute

\dfrac{22.54}{0.44} = 51.2 \;\le\; 1.12\,\sqrt{\tfrac{29{,}000\,(5)}{60}} = 55.1

Result

C \;=\; 1.0

Nominal shear resistance

V_n \;=\; C\, V_p

Substitute

V_n \;=\; 1.0\,(345.2)

Result

V_n \;=\; 345.2\;\text{kip}

Strength I shear check (both girders)

V_u \;\le\; \phi_v\, V_n

Substitute

\text{Int: } 156.3 \le 345.2; \quad \text{Ext: } 146.7 \le 345.2

Result

\text{OK — governs by a wide margin}

Step 14

Service II — transformed interior section (n = 8)

Figure 8.20Figs. 2.36–2.38 — Interior girder before and after transformation of the concrete deck (b/n = 96/8 = 12 in equivalent steel width).

Transformed centroid from beam bottom

\bar{y} \;=\; \dfrac{\sum A_i\, \bar{Y}_i}{\sum A_i}

Substitute

\bar{y} \;=\; \dfrac{3{,}191.5}{120.4}

Result

\bar{y} \;=\; 26.5\;\text{in from beam bottom}

Transformed moment of inertia

I \;=\; \sum \bigl(\tfrac{b_i h_i^{3}}{12} \;+\; A_i\, d_i^{\,2}\bigr)

Substitute

I \;=\; 4{,}188.4 + 2{,}520.0 + 53.1 + 6.5 + 1{,}756.2

Result

I \;=\; 8{,}524.2\;\text{in}^4

Section moduli — top and bottom

S_{top} = \dfrac{I}{c_{top}}, \quad S_{bot} = \dfrac{I}{\bar{y}}

Substitute

S_{top} = \tfrac{8{,}524.2}{2.6}, \quad S_{bot} = \tfrac{8{,}524.2}{26.5}

Result

S_{top} = 3{,}279\;\text{in}^3, \;\; S_{bot} = 321.7\;\text{in}^3

Service II combination

f_f \;=\; 1.0\, f_{DC} \;+\; 1.0\, f_{DW} \;+\; 1.30\, f_{LL+IM}

Substitute

\text{Bottom: } f_f = 1.0\,(8.34) + 1.0\,(1.36) + 1.30\,(17.75)

Result

f_{f,\text{bot}} \;=\; 32.78\;\text{ksi}

Service II bottom-flange limit

f_f \;+\; \tfrac{f_\ell}{2} \;\le\; 0.95\, R_h\, F_{yf}

Substitute

32.78 + 0 \;\le\; 0.95\,(1.0)(60) = 57

Result

32.78 < 57\;\text{ksi [OK]}

Top flange (compression) similarly yields $f_f = 3.21\;\text{ksi} < 57\;\text{ksi}$ — OK.

Step 15

Service II — transformed exterior section

Figure 8.21Figs. 2.39–2.41 — Exterior girder transformed section (b/n = 87/8 = 10.875 in).

Exterior centroid and inertia

\bar{y}, \; I

Substitute

\bar{y} = \tfrac{2{,}922.15}{111.4} = 26.23\;\text{in}; \;\; I = 8{,}230.0\;\text{in}^4

Result

\bar{y} = 26.23\;\text{in}, \;\; I = 8{,}230.0\;\text{in}^4

Exterior section moduli

S_{top}, \; S_{bot}

Substitute

S_{top} = \tfrac{8{,}230}{2.33} = 3{,}532, \;\; S_{bot} = \tfrac{8{,}230}{26.23} = 313.8

Result

S_{top} = 3{,}532\;\text{in}^3, \;\; S_{bot} = 313.8\;\text{in}^3

Bottom-flange Service II stress

f_f \;=\; 1.0\, f_{DC} + 1.0\, f_{DW} + 1.30\, f_{LL+IM}

Substitute

f_f \;=\; 8.55 + 1.40 + 1.30\,(20.84)

Result

f_{f,\text{bot}} \;=\; 37.04\;\text{ksi} < 57\;\text{ksi [OK]}

Step 16

Fatigue II — load-induced fatigue (§6.6.1.2)

Figure 8.22Fig. 2.42 — Single-lane fatigue truck (constant 30-ft axle spacing) placed for maximum moment at midspan.

Fatigue combination and check

\gamma\,(\Delta f) \;\le\; (\Delta F)_n, \quad \gamma = 0.75\;\text{(Fatigue II)}

Fatigue moment reaction

R_A \;=\; \dfrac{32\,(20) + 8\,(6)}{L}

Substitute

R_A \;=\; \dfrac{640 + 48}{40}

Result

R_A \;=\; 17.2\;\text{kip}, \;\; M_c = 17.2\,(20) = 344\;\text{kip-ft}

Fatigue distribution factor (remove m = 1.2)

DFM_{fat} \;=\; \dfrac{DFM_{s}}{1.2}

Substitute

\text{Int: } 0.498/1.2 = 0.415; \;\; \text{Ext: } 0.75/1.2 = 0.625

Distributed fatigue moment (IM = 15%)

M_{fat} \;=\; M_c\, DFM_{fat}\,(1 + IM)

Substitute

\text{Int: } 344\,(0.415)(1.15) = 164.2; \;\; \text{Ext: } 344\,(0.625)(1.15) = 247.3

Result

M_{fat,\text{int}} = 164.2, \;\; M_{fat,\text{ext}} = 247.3\;\text{kip-ft}

Live-load stress range (use S_bot; max stress)

\Delta f \;=\; \dfrac{M_{fat}\,(12)}{S_{bot}}

Substitute

\text{Int: } \tfrac{164.2\,(12)}{321.7}; \;\; \text{Ext: } \tfrac{247.3\,(12)}{313.8}

Result

\Delta f_{int} = 6.12\;\text{ksi}, \;\; \Delta f_{ext} = 9.46\;\text{ksi}

Single-lane ADTT (p = 0.80 for 3 lanes)

ADTT_{SL} \;=\; p\,\cdot\,ADTT

Substitute

ADTT_{SL} \;=\; 0.80\,(2500)

Result

ADTT_{SL} \;=\; 2000\;\text{trucks/day}

Fatigue cycles at 75 yr, n = 2 cycles/truck

N \;=\; 365\,\cdot\,\text{years}\,\cdot\, n\,\cdot\, ADTT_{SL}

Substitute

N \;=\; 365\,(75)(2)(2000)

Result

N \;=\; 1.095\times 10^{8}\;\text{cycles}

Nominal fatigue resistance — Detail Category A

(\Delta F)_n \;=\; \left(\dfrac{A}{N}\right)^{1/3}, \;\; A = 250\times 10^{8}\;\text{ksi}^{3}

Substitute

(\Delta F)_n \;=\; \left(\dfrac{250\times 10^{8}}{1.095\times 10^{8}}\right)^{1/3}

Result

(\Delta F)_n \;=\; 6.11\;\text{ksi}

Fatigue II factored stress check

\gamma\,(\Delta f) \;\le\; (\Delta F)_n

Substitute

\text{Int: } 0.75\,(6.12) = 4.59; \;\; \text{Ext: } 0.75\,(9.46) = 7.09

Result

\text{Int: } 4.59 < 6.11\;[\text{OK}], \;\; \text{Ext: } 7.09 > 6.11\;[\text{NG}]

What can go wrong

Exterior girder fails the Fatigue II stress range. Distortion-induced fatigue of the web must be investigated per §6.10.5.3, and/or the exterior flange should be thickened (or an unsymmetric section adopted).

Step 17

Special fatigue requirement for webs (§6.10.5.3)

Figure 8.23Fig. 2.43 — Single-lane fatigue truck placed for maximum shear at the support.

Fatigue shear reaction (one truck)

R_A \;=\; 32 \;+\; \dfrac{10}{40}\,(32)

Substitute

R_A \;=\; 32 + 8

Result

V \;=\; 40\;\text{kip}

Fatigue shear DF (remove m = 1.2)

DFV_{fat} \;=\; \dfrac{DFV_{s}}{1.2}

Substitute

\text{Int: } 0.68/1.2 = 0.567; \;\; \text{Ext: } 0.75/1.2 = 0.625

Factored fatigue shear (special rule — doubled load)

V_f \;=\; 2\,(DFV)(\gamma)(V)(1 + IM)

Substitute

\text{Int: } 2\,(0.567)(0.75)(40)(1.15) = 39.3; \;\; \text{Ext: } 2\,(0.625)(0.75)(40)(1.15) = 43.1

Result

V_{f,\text{ext}} = 43.1\;\text{kip [controls]}

Permanent-load shear

V_{perm} \;=\; V_{DC} + V_{DW}

Substitute

V_{perm} \;=\; 22.44 + 3.66

Result

V_{perm} \;=\; 26.1\;\text{kip}

Interior-panel web-buckling check

V_u \;=\; V_{perm} + V_f \;\le\; V_{cr} \;=\; C\, V_p

Substitute

V_u \;=\; 26.1 + 43.1 = 69.2 \;\le\; 1.0\,(345.2)

Result

69.2 \;\ll\; 345.2\;\text{kip [OK]}

Design Example 3 — Summary

Results at a glance

Check	Interior	Exterior
Strength I — Mu vs φMn (kip-ft)	1,167 / 2,275 ✓	1,288 / 2,235 ✓
Strength I — Vu vs φVn (kip)	156.3 / 345.2 ✓	146.7 / 345.2 ✓
Service II — bottom flange (ksi)	32.8 / 57 ✓	37.0 / 57 ✓
Fatigue II — Δf vs (ΔF)ₙ (ksi)	4.59 / 6.11 ✓	7.09 / 6.11 ✗
§6.10.5.3 web-panel Vu vs Vcr (kip)	69.2 / 345.2 ✓	69.2 / 345.2 ✓

The W24×76 rolled section satisfies every Strength I and Service II check for both interior and exterior girders, but the exterior girder narrowly exceeds the Fatigue II stress range at the bottom flange. In practice the fix is to add a bottom cover plate, upsize the flange, or reduce the exterior distribution factor by widening the interior stringers — a common outcome that illustrates why fatigue often controls the final selection of rolled composite sections on short spans.

8.11 — Mini design challenge

Curved 3-span continuous steel plate-girder ramp

Three-span continuous curved steel plate girder ramp — Figure 8.10Design challenge: 3-span continuous horizontally curved steel plate girder ramp, 130 – 170 – 130 ft with 700-ft centerline radius. 4 girders at 10.5 ft, 34-ft roadway.

Problem statement

Design the interior plate girder of a 3-span continuous curved ramp. Use AASHTO §6.10 + §6.10.1.2 for curvature effects.

Given

Spans130 – 170 – 130 ft continuous
Centerline radius R700 ft
Roadway width34 ft (2 lanes, 4 girders at 10.5 ft)
Deck t_s9.0 in
SteelHPS 70W for negative-moment regions; 50W elsewhere
Load combosStrength I, Service II, Fatigue I
Cross-framesSpaced ≤ 20 ft (curvature limit)
DeliverablesTrial section table (positive & negative), Mp check, LTB check, curvature flange-lateral bending fl, shear + stud design, fatigue at pier, camber diagram

Required

Submit a design memo with full Formula → Substitute → Result calculations, a section-detailing table, and a plan showing stiffener and cross-frame layout.

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Summary

What to remember

Composite steel plate girders are the workhorse of 120–300 ft spans. Choose depth first (≈ L/30), then tune each plate.
Classify the web using $2 D_{cp}/t_w$ . Non-compact webs cap $M_n$ at $R_b R_h M_y$ .
LTB governs during deck placement — brace the compression flange with cross-frames spaced so $L_b \approx L_p$ .
Web shear with transverse stiffeners uses tension-field action (§6.10.9) and can double the elastic buckling capacity.
Shear studs require both fatigue (pitch) and strength (total count) checks.
The web-to-flange fillet weld is Category C; verify $\Delta f \le \Delta F_{TH}$ at Fatigue I.
Meet the optional L/800 live-load deflection limit; camber for the full dead-load deflection plus profile grade.

AASHTO LRFD §6.10 — I-Section Flexural Members; §6.6.1.2 — Fatigue

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: composite vs non-composite section modulus

Basic

Problem

Compute S_x,steel and S_x,composite (bottom fiber).

Step-by-Step

A_deck,tr = 8·96/8 = 96 in² (n = 8 for long-term).

Compute ȳ from bottom flange: ~ 27 in.

Design Verification

Composite action nearly doubles the resisting modulus at the tension fiber — this is why steel-composite is efficient for medium spans.

Discussion

Short-term n = 8 for LL+IM; long-term n = 24 (3n) for DC2/DW. Always compute both.

Worked Example 2

Hand calc: web slenderness classification

Basic

Problem

Is the section compact, non-compact, or slender-web?

Step-by-Step

2·D_cp/t_w = 2·36/0.5 = 144

3.76·√(E/F_yc) = 3.76·√(29,000/50) = 90.6

Design Verification

Slender-web governs almost every plate girder over 60-in web depth — that's why plate girders live in §6.10.1.9.

Discussion

Adding a longitudinal stiffener at D/5 from compression flange can rescue slender webs and boost R_b close to 1.0.

Worked Example 3

Intermediate: distribution factor and design moment

Intermediate

Problem

Compute g_moment,interior (2+ lanes loaded).

Step-by-Step

[K_g/(12·L·t_s³)]^0.1 = [470,000/(12·130·8³)]^0.1 = [235]^0.1 = 1.68... wait recompute: 470,000/(12·130·512) = 0.588 → ^0.1 = 0.949.

g = 0.075 + (9.5/9.5)^0.6·(9.5/130)^0.2·0.949 = 0.075 + 1·0.598·0.949 = 0.643

Result

g = 0.643

Design Verification

One-lane loaded formula gives 0.51 → controls the fatigue-only case. Multi-lane 0.643 controls Strength.

Discussion

DF drops if K_g rises — deeper girders with more composite deck stiffness attract more load; that's the aim.

Worked Example 4

Complete member design: welded plate girder at strength

Intermediate

Problem

Verify φM_n ≥ M_u.

Step-by-Step

a_wc = 2·D_c·t_w/(b_fc·t_fc) with D_c ≈ 30 in (composite). ≈ 1.25. R_b = 1 − [a_wc/(1200+300·a_wc)]·(2·D_c/t_w − λ_rw) ≈ 0.96.

M_yc = F_y·S_bot,comp = 50·940/12 = 3,917 k·ft. M_p from plastic analysis of composite section ≈ 12,600 k·ft.

Design Verification

Ductility check (§6.10.7.3): D_p/D_t ≤ 0.42 — verify with plastic-analysis geometry.

Discussion

Plate girder design is iterative: pick flanges, check §6.10.7 or §6.10.8, adjust and re-check R_b.

Worked Example 5

Multi-limit-state: negative moment region of continuous girder

Advanced

Problem

Check flange local buckling (FLB) and lateral-torsional buckling (LTB) of the compression flange.

Step-by-Step

λ_f = b_fc/(2·t_fc) = 5.5. λ_pf = 0.38·√(29,000/50) = 9.15 → compact. F_nc,FLB = R_b·R_h·F_yc.

r_t = b_fc/√(12·(1 + D·t_w/(3·b_fc·t_fc))) = 22/√(12·(1.13)) = 5.98 in. L_p = 1.0·r_t·√(E/F_yc) = 5.98·24.1 = 144 in = 12 ft. L_b = 15 ft > L_p.

Design Verification

φF_nc·S_bot ≥ M_u/12 → satisfied with margin if R_b ≈ 0.96.

Discussion

Cross-frame spacing is the fastest LTB fix — dropping L_b to 10 ft eliminates the reduction.

Worked Example 6

Design optimization: HPS 70W vs A709 Gr 50

Advanced

Problem

By what % does HPS 70W increase φM_n and does the cost premium pencil?

Step-by-Step

F_nc,HPS/F_nc,Gr50 ≈ 70/50 = 1.40.

Reduce flange from 2.0 to 2.0·0.71 = 1.42 in to keep M_n. New self-weight ~ −20% on negative-moment section (~250 lb/lf saved).

Design Verification

HPS is often specified only in negative-moment fields; hybrid girders (HPS flanges, Gr 50 web) are the common optimum.

Discussion

Never assume "higher-strength steel = cheaper" — deflection and fatigue may still govern the Gr 50 solution.

Worked Example 7

Construction-stage: deck-pour sequence and unbraced-length control

Advanced

Problem

Confirm top-flange compression during pour 3 (positive-moment field of main span) does not exceed 0.6·F_y.

Step-by-Step

f_c = M/S = 3,200·12/2,100 = 18.3 ksi ≤ 0.6·50 = 30 ✓

Set screed elevations to match cambered profile + expected pour 3 deflection (~ 3.5 in).

Design Verification

Cross-frames every 20 ft keep L_b well below L_p during pour — LTB not a construction concern here.

Discussion

Deck-pour sequence is chosen to keep positive-moment fields poured LAST so freshly cured concrete rides tension zones, not the compression ones.

Worked Example 8

Consulting: fatigue check at a Category C detail (transverse stiffener)

Consulting

Problem

Compare (ΔF)_eff to Δ(F)_TH.

Step-by-Step

Δf = 1.75·0.42·1.15·950·12/940 = 9.2 ksi (approx)

9.2 ksi < 12 ksi → infinite life ✓

Result

Passes Fatigue I

Design Verification

If detail were Category E (transverse stiffener welded to tension flange), Δ(F)_TH = 4.5 ksi → would fail; redesign detail (§6.6.1.2.4 mandates 4·t_w web gap).

Discussion

Category management is the whole game in steel-girder fatigue. Detail-neutral to Cat C is the target for tension flanges.

Worked Example 9

Case-study reproduction: I-35W gusset-plate underdesign

Consulting

Problem

Compute the gusset plate compressive resistance vs actual demand at collapse.

Step-by-Step

P_r = φ·F_cr·A_w = 0.9·50·(0.5·20) = 450 kip

Combined member forces ≈ 620 kip (dead load + construction stockpile).

Design Verification

Reproduces NTSB and FHWA Gusset Plate Guide 2009 forensic conclusion.

Discussion

Every non-load-path-redundant truss connection must be re-checked independently of the members. Design errors can lie latent for 40 years.

Worked Example 10

Comprehensive: single-span 145-ft welded plate girder full design

Consulting

Problem

Produce cross-section and plate schedule.

Step-by-Step

D = L/25 = 145·12/25 = 70 in → use 72-in web × 0.5625 in.

Bottom PL 20 × 1.75 mid; 20 × 1.0 ends. Top PL 16 × 1.25 mid; 16 × 0.75 ends.

Design Verification

Take-off: 65,000 lb per girder ≈ 4·65 = 260 kip fabricated steel; typical MDOT bid at $1.75/lb → $455k. Consistent with recent MD contract awards.

Discussion

Variable-depth flange plates ("shop splices") reduce steel weight by 12–18% versus prismatic. Every splice costs ~ $2,000 fabricated — solve the optimum splice count.

Worked Example 11

Horizontally curved girder — refined analysis check

Consulting

Problem

Trigger refined analysis? Compute L_a/R.

Step-by-Step

L_a/R = 18/800 = 0.0225 → below 0.03 threshold for V-load; refined analysis strongly recommended when 200-ft arc is critical.

ℓ_b lateral moment from curvature must be added to major-axis bending per §6.10.1.6 (fℓ term).

Design Verification

Grillage or 3D FE required for final. V-load is prelim sizing only.

Discussion

Curved steel girders always require §6.10.1.6 lateral flange bending — a scalar addition to the strong-axis stress check.

Worked Example 12

Capstone: bid-quality drawings and pay-item takeoff for a 3-span steel bridge

Consulting

Problem

Complete bid takeoff.

Step-by-Step

4·420·350 = 588,000 lb.

76·250 = 19,000 lb.

Design Verification

Steel unit weight 145 lb/lf/girder average matches recent MDOT projects for L = 420 ft continuous.

Discussion

Bid takeoff is what separates a design report from a biddable contract. Every quantity must reconcile to a drawing.