THREE-SPAN CONTINUOUS STRAIGHT COMPOSITE I GIRDER Load and Resistance Factor Design (Third Edition -- Customary U.S. Units)

EXAMPLE 1: THREE-SPAN CONTINUOUS STRAIGHT COMPOSITE I GIRDER Load and Resistance Factor Design (Third Edition -- Customary U.S. Units) by Michael A. Grubb, P.E. Bridge Sotware Development International, Ltd. Cranberry Township, PA and Robert E. Schmidt, E.I.T. SITE-Blauvelt Engineers Pittsburgh, PA

Disclaimer All data, speciications, suggested practices, and drawings presented herein, are based on the best available inormation and delineated in accordance with recognized proessional engineering principles and practices, and are published or general inormation only. Procedures and products, suggested or discussed, should not be used without irst securing competent advice respecting their suitability or any given application. Publication o the material herein is not to be construed as a warranty on the part o the National Steel Bridge Alliance or that o any person named herein that these data and suggested practices are suitable or any general or particular use, or o reedom rom inringement on any patent or patents. Further, any use o these data or suggested practices can only be made with the understanding that the National Steel Bridge Alliance makes no warranty o any kind respecting such use and the user assumes all liability arising thererom

LRFD rd Edition Table o Contents 1. INTRODUCTION... 1. OVERVIEW OF LRFD ARTICLE 6.10... 1. DESIGN PARAMETERS... 4. STEEL FRAMING... 4 4.1. Span Arrangement... 4 4.. Bridge Cross-Section... 5 4.. Cross-Frames... 6 4.4. Field Section Sizes... 7 5. PRELIMINARY GIRDER SIZES... 8 5.1. Girder Depth... 8 5.. Cross-section Proportions... 8 6. LOADS... 11 6.1. Dead Loads... 11 6.. Live Loads... 1 6..1. Design Vehicular Live Load (Article.6.1.)... 1 6... Loading or Optional Live-Load Delection Evaluation (Article.6.1..)... 14 6... Fatigue Load (Article.6.1.4)... 14 6.. Wind Loads... 15 6.4. Load Combinations... 17 7. STRUCTURAL ANALYSIS... 19 7.1. Multiple Presence Factors (Article.6.1.1.)... 19 7.. Live-Load Distribution Factors (Article 4.6..)... 0 7..1. Live-Load Lateral Distribution Factors - Positive Flexure... 7..1.1. Interior Girder - Strength Limit State... 7..1.. Exterior Girder - Strength Limit State... 4 7..1.. Distribution Factors or Fatigue Limit State... 6 7..1.4. Distribution Factor or Live-Load Delection... 7 7... Live-Load Lateral Distribution Factors - Negative Flexure... 7 7...1. Interior Girder - Strength Limit State... 8 7... Distribution Factors or Fatigue Limit State... 8 7.. Dynamic Load Allowance: IM (Article.6.)... 8 8. ANALYSIS RESULTS... 9 8.1. Moment and Shear Envelopes... 9 8.. Live Load Delection... 4 9. LIMIT STATES... 4 9.1. Service Limit State (Articles 1... and 6.5.)... 4 9.. Fatigue and Fracture Limit State (Articles 1... and 6.5.)... 4 9.. Strength Limit State (Articles 1...4 and 6.5.4)... 5 9.4. Extreme Event Limit State (Articles 1...5 and 6.5.5)... 5 10. SAMPLE CALCULATIONS... 5 10.1. Section Properties... 5 10.1.1. Section 1-1... 6 10.1.1.1. Eective Flange Width (Article 4.6..6): Section 1-1... 6 10.1.1.. Elastic Section Properties: Section 1-1... 7 10.1.1.. Plastic Moment: Section 1-1... 7 10.1.1.4. Yield Moment: Section 1-1... 8 10.1.. Section -... 9 10.1..1. Eective Flange Width (Article 4.6..6): Section -... 9 10.1... Minimum Negative Flexure Concrete Deck Reinorcement (Article 6.10.1.7)... 9 10.1... Elastic Section Properties: Section -... 40 May, 004 i

LRFD rd Edition 10.. Exterior Girder Check: Section 1-1... 4 10..1. Constructibility (Article 6.10.)... 4 10..1.1. Deck Placement Analysis... 4 10..1.. Deck Overhang Loads... 45 10..1.. Wind Loads... 49 10..1.4. Flexure (Article 6.10..)... 5 Top Flange... 5 Local Buckling Resistance (Article 6.10.8..)... 5 Lateral Torsional Buckling Resistance (Article 6.10.8..)... 5 Web Bend-Buckling Resistance (Article 6.10.1.9)... 54 Bottom Flange... 56 10..1.5. Shear (Article 6.10..)... 56 10..1.6. Concrete Deck (Article 6.10...4)... 57 10... Service Limit State (Article 6.10.4)... 58 10...1. Elastic Deormations (Article 6.10.4.1)... 58 10... Permanent Deormations (Article 6.10.4.)... 58 10... Concrete Deck (Article 6.10.1.7)... 60 10... Fatigue And Fracture Limit State (Article 6.10.5)... 60 10...1. Load Induced Fatigue (Article 6.6.1.)... 60 Top-Flange Connection-Plate Weld... 61 Bottom-Flange Connection-Plate Weld... 61 Stud Shear-Connector Weld... 6 10... Distortion Induced Fatigue (Article 6.6.1.)... 64 10... Fracture (Article 6.6.)... 64 10...4. Special Fatigue Requirement or Webs (Article 6.10.5.)... 64 10..4. Strength Limit State (Article 6.10.6)... 65 10..4.1. Flexure (Article 6.10.6.)... 65 Nominal Flexural Resistance (Article 6.10.7.1.)... 67 10..4.. Shear (6.10.6.)... 69 End Panel (Article 6.10.9..)... 70 Interior Panels (Article 6.10.9..)... 71 10.. Exterior Girder Check: Section -... 7 10..1. Strength Limit State (Article 6.10.6)... 7 10..1.1. Flexure (Article 6.10.6.)... 7 Bottom Flange... 74 Lateral Torsional Buckling Resistance (Article 6.10.8..)... 74 Elastic Section Properties: Flange transition... 77 Local Buckling Resistance (Article 6.10.8..)... 79 Stress Check... 80 Bottom Flange... 8 Top Flange... 84 10..1.. Shear (6.10.6.)... 85 10... Service Limit State (Article 6.10.4)... 85 10...1. Permanent Deormations (Article 6.10.4.)... 85 10... Fatigue And Fracture Limit State (Article 6.10.5)... 88 10...1. Load Induced Fatigue (Article 6.6.1.)... 88 10... Special Fatigue Requirement or Webs (Article 6.10.5.)... 89 10..4. Constructibility (Article 6.10.)... 90 10..4.1. Flexure (Article 6.10..)... 90 Web Bend-Buckling... 90 10..4.. Shear (Article 6.10..)... 9 10.4. Shear Connector Design (Article 6.10.10)... 9 10.4.1. Stud Proportions... 9 10.4.. Pitch (Article 6.10.10.1.)... 9 10.4.. Fatigue Limit State... 9 10.4.4. Strength Limit State (Article 6.10.10.4)... 95 ii May, 004

LRFD rd Edition 10.5. Exterior Girder: Field Section 1... 97 10.5.1. Transverse Intermediate Stiener Design (Article 6.10.11.1)... 97 10.5.1.1. Projecting Width (Article 6.10.11.1.)... 97 10.5.1.. Moment o Inertia (Article 6.10.11.1.)... 98 10.5.1.. Area (Article 6.10.11.1.4)... 98 10.6. Exterior Girder: Abutment 1... 99 10.6.1. Bearing Stiener Design (Article 6.10.11.)... 99 10.6.1.1. Projecting Width (Article 6.10.11..)... 100 10.6.1.. Bearing Resistance (Article 6.10.11..)... 100 10.6.1.. Axial Resistance (Article 6.10.11..4)... 101 10.6.1.4. Bearing Stiener-to-Web Welds... 10 10.7. Exterior Girder: Design Example Summary... 10 10.7.1. Positive-Moment Region, Span 1 (Section 1-1)... 10 10.7.1.1. Constructibility (Slender-web section)... 10 10.7.1.. Service Limit State... 10 10.7.1.. Fatigue and Fracture Limit State... 10 10.7.1.4. Strength Limit State (Compact Section)... 10 10.7.. Interior-Pier Section (Section -)... 104 10.7..1. Strength Limit State (Slender-web section)... 104 10.7... Service Limit State... 104 10.7... Fatigue and Fracture Limit State... 104 10.7..4. Constructibility (Slender-web section)... 104 Appendix A: Elastic Eective Length Factor or Lateral Torsional Buckling... A1 Appendix B: Moment Gradient Modiier, C b... B1 Appendix C: Lateral Torsional Buckling Resistance o Stepped Flanges...C1 May, 004 iii

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LRFD rd Edition 1. INTRODUCTION In 199, the American Association o State Highway and Transportation Oicials (AASHTO) adopted the Load and Resistance Factor Design (LRFD) speciications or bridge design. The First Edition o the design speciications was published by AASHTO in 1994. The publication o a Second Edition ollowed in 1998, along with the publication o the First Edition o a companion document the AASHTO LRFD Bridge Construction Speciications. The design speciications are available in either customary U.S. units or in SI (metric) units, whereas the construction speciications are currently only available in SI units. The LRFD speciications were approved by AASHTO or use as alternative speciications to the AASHTO Standard Speciications or Highway Bridges The LRFD speciications evolved in response to a high level o interest amongst the AASHTO Subcommittee on Bridges and Structures in developing updated AASHTO bridge speciications together with accompanying commentary. The goal was to develop more comprehensive speciications that would eliminate any gaps and inconsistencies in the Standard Speciications, incorporate the latest in bridge research, and achieve more uniorm margins o saety or reliability across a wide variety o structures. The decision was made to develop these new speciications in an LRFD-based ormat, which takes the variability o the behavior o structural elements into account through the application o statistical methods, but presents the results in a manner that is readily usable by bridge designers. A detailed discussion o the evolution o the LRFD design speciications and commentary is presented in NCHRP Research Results Digest 198 (available rom the Transportation Research Board) and elsewhere, and will not be repeated herein. The design o steel structures is covered in Section 6 o the AASHTO LRFD Bridge Design Speciications and only straight steel bridges are covered in the provisions as o this writing. The Third Edition o the design speciications, to be published in 004, will contain a complete set o new provisions or the design o straight steel I- and box-section lexural members within Articles 6.10 and 6.11, respectively. These new provisions have been structured to simpliy their logic, organization and application, while also improving their accuracy and generality. More importantly, the new provisions lay the necessary groundwork or a potential uture uniied design approach or both straight and horizontally curved girders within a single speciication, which will urther streamline and improve the overall eiciency o the design process or bridges that contain both straight and curved spans. The basic application o these new provisions to the design o straight steel I-section section lexural members is illustrated through the design example presented herein. The example illustrates the design o a typical three-span continuous straight steel I-girder bridge with spans o 140-0 175-0 140-0. Speciically, the example illustrates the design o selected critical sections rom an exterior girder at the strength, service and atigue limit states. Constructibility checks, stiener and shear connector designs are also presented.. OVERVIEW OF LRFD ARTICLE 6.10 The design o I-section lexural members is covered within Article 6.10 o the AASHTO LRFD Bridge Design Speciications. The new provisions o Article 6.10 contained in the Third Edition o the LRFD speciications have been organized to correspond more closely to the general low o the calculations necessary or the design o I-section lexural members. Each o the sub-articles are written such that they are largely sel-contained, thus minimizing the need or reerence to multiple sub-articles to address any o the essential design considerations. Many o the individual calculations and equations have been streamlined and selected resistance equations have been updated to improve their accuracy and generality. The Commentary to the provisions also covers a number o areas in which the previous speciications have been largely silent. The sub-articles within the Third Edition Article 6.10 are organized as ollows: 6.10.1 General 6.10. Cross-section Proportion Limits May, 004 Page 1 o 1

LRFD rd Edition 6.10. Constructibility 6.10.4 Service Limit State 6.10.5 Fatigue and Fracture Limit State 6.10.6 Strength Limit State 6.10.7 Flexural Resistance - Composite Sections in Positive Flexure 6.10.8 Flexural Resistance - Composite Sections in Negative Flexure and Noncomposite Sections 6.10.9 Shear Resistance 6.10.10 Shear Connectors 6.10.11 Stieners 6.10.1 Cover Plates In addition, our new appendices to Section 6 relevant to the design o lexural members are provided as ollows: Appendix A - Flexural Resistance - Composite Sections in Negative Flexure and Noncomposite Sections with Compact or Noncompact Webs Appendix B - Moment Redistribution rom Interior-Pier Sections in Continuous-Span Bridges Appendix C - Basic Steps or Steel Bridge Superstructures Appendix D - Fundamental Calculations or Flexural Members For composite I-sections in negative lexure and noncomposite I-sections, the provisions o Article 6.10.8 limit the nominal lexural resistance to a maximum o the moment at irst yield. As a result, the nominal lexural resistance or these sections is conveniently expressed in terms o the elastically computed lange stress. When these sections satisy speciic steel grade requirements and have webs that are classiied as either compact or noncompact, the optional provisions o Appendix A may be applied instead to determine the lexural resistance, which may exceed the moment at irst yield. Thereore, the lexural resistance is expressed in terms o moment in Appendix A. The provisions o Appendix A are a direct extension o and are ully consistent with the main provisions o Article 6.10.8. The previous Speciications deined sections as either compact or noncompact and did not explicitly distinguish between a noncompact web and a slender web. The proposed provisions make explicit use o these deinitions or composite I-sections in negative lexure and noncomposite I-sections because the noncompact web limit serves as a useul anchor point or a continuous representation o the maximum potential section resistance rom the nominal yield moment up to the plastic moment resistance. Because sections with compact or nearly compact webs are less commonly used, the provisions or sections with compact or noncompact webs have been placed in an appendix in order to simpliy and streamline the main provisions. The main provisions within the body o Article 6.10 may be used or these types o sections to obtain an accurate to somewhat conservative determination o the lexural resistance calculated using Appendix A. For girders that are proportioned with webs near the noncompact web slenderness limit, the provisions o Article 6.10 and Appendix A produce the same strength or all practical purposes, with the exception o cases with large unsupported lengths sometimes encountered during construction. In these cases, Appendix A gives a larger more accurate lexural resistance calculation. In the example to ollow, a slender-web section is utilized or both the composite section in regions o negative lexure and or the noncomposite section in regions o positive lexure beore the concrete deck has hardened. As a result, the main provisions o Article 6.10 must be applied or the strength limit state and constructibility checks or those sections and the optional Appendix A is not applicable. Minor yielding at interior piers o continuous spans results in redistribution o the moments. For straight continuous-span lexural members that satisy certain restrictions intended to ensure adequate ductility and robustness o the pier sections, the optional procedures o Appendix B may be used to calculate the redistribution moments at the service and/or strength limit states. These provisions replace the ormer ten-percent redistribution allowance as well as the ormer inelastic analysis procedures. They provide a simple calculated percentage redistribution rom interior-pier sections. This approach utilizes elastic Page o 1 May, 004

LRFD rd Edition moment envelopes and does not require the direct use o any inelastic analysis. As such, the new procedures are substantially simpler and more streamlined than the inelastic analysis procedures o the previous Speciications. Where appropriate, these provisions make it possible to use prismatic sections along the entire length o the bridge or between ield splices, which can improve overall atigue resistance and provide signiicant abrication economies. Although the necessary steps could be taken to allow moment redistribution in the example presented herein, the provisions o Appendix B are not applied. Flow charts or lexural design o I-sections according to the new provisions, along with a revised outline giving the basic steps or steel-bridge superstructure design, are provided in Appendix C. Fundamental section property calculations or lexural members have been relocated rom the speciication to a new Appendix D. The provisions o Article 6.10 and the optional Appendices A and B provide a uniied approach or consideration o combined major-axis bending and lange lateral bending rom any source in both straight and horizontally curved I-girders. As such, general design equations are provided that include the consideration o both major-axis bending and lange lateral bending. For straight girders, lange lateral bending is caused by wind and by torsion rom various origins. Sources o signiicant lange lateral bending due to torsion include eccentric slab overhang loads acting on cantilever orming brackets placed along exterior members, staggered cross-rames, and signiicant support skew. When the above eects are judged to be insigniicant or incidental, the lange lateral bending term,, is simply set equal to zero in the appropriate equations. The ormat o the equations then simply reduces to the more conventional and amiliar ormat or checking the nominal lexural resistance o I-sections in the absence o lange lateral bending. The example to ollow considers the eects o lange lateral bending caused by wind and by torsion due to the eects o eccentric slab overhang loads. Finally, it should be noted that the new Article 6.10 provisions do not incorporate all the necessary requirements or horizontally curved I-girder bridge design as o this writing. It is anticipated that these requirements will soon be incorporated under the work o NCHRP Project 1-5, and that some additional restrictions will be placed on the application o the provisions in Article 6.10 to curved bridge design. However, the new provisions lay the necessary groundwork or accomplishing that eort in an eicient manner, which was the primary impetus or the extensive revisions.. DESIGN PARAMETERS The ollowing data apply to this example design: Speciications: 004 AASHTO LRFD Bridge Design Speciications, Customary U.S. Units, Third Edition Structural Steel: AASHTO M 70 Grade HPS 70W (ASTM A 709 Grade HPS 70W) uncoated weathering steel with F y = 70 ksi (or the langes in regions o negative lexure) AASHTO M 70, Grade 50W (ASTM A 709, Grade 50W) uncoated weathering steel with = 50 ksi (or all other girder and cross-rame components) The example design utilizes uncoated weathering steel. Where site conditions are adequate or successul application, uncoated weathering steel is the most cost-eective material choice in terms o savings in both initial and uture repainting costs. In the years since its introduction into bridge construction by the Michigan DOT in the 1960's, uncoated weathering steel has become widely accepted as cost-eective, currently representing about 45 percent o the steel-bridge market. However, it has also requently been misused because o inexperience or ignorance about the properties o the material. To counter this and increase the conidence in its perormance, the FHWA issued a Technical Advisory (T5140.) in 1989 F y May, 004 Page o 1

LRFD rd Edition entitled Uncoated Weathering Steel in Structures. The guidelines contained in this document, developed in cooperation with the steel industry, are a valuable source o inormation on the proper environments or the use o weathering steel. The guidelines also suggest good detailing practice to help ensure successul application o the material. In regions o negative lexure, the example design utilizes a hybrid section consisting o ASTM A 709 Grade HPS 70W high-perormance steel (HPS) langes and an ASTM A 709 Grade 50W web. Grade HPS 70W was developed in the early 1990s under a successul cooperative research program between the Federal Highway Administration, the U.S. Navy, and the American Iron and Steel Institute. Grade HPS 70W possesses superior weldability and toughness compared to conventional steels o this strength range. Grade HPS 70W is currently produced by quenching and tempering (Q&T) or by thermo-mechanicalcontrolled-processing (TMCP). TMCP HPS is available in plate thicknesses up to inches and in maximum plate lengths rom approximately 600 to 1500 inches depending on weights. Q&T HPS is available in plate thicknesses up to 4 inches, but because o the urnaces that are used in the tempering process, is subject to a maximum plate-length limitation o 600 inches or less depending on weights. Thereore, when Q&T HPS is used, the maximum plate-length limitation should be considered when laying out lange and web transitions. Current inormation on maximum plate length availability can be obtained by contacting a steel producer. Guidelines or abrication using Grade HPS 70W steel are available in the AASHTO Guide Speciications or Highway Bridge Fabrication with HPS 70W Steel (Second Edition to be published in 004). HPS is inding increasing application in highway bridges across the U.S., with hybrid designs utilizing Grade HPS 70W langes in conjunction with a Grade HPS 50W web being the most popular application. Concrete: Slab Reinorcing Steel: c = 4.0 ksi AASHTO M 1, Grade 60 (ASTM A 615, Grade 60) with F y = 60 ksi Permanent steel deck orms are assumed between the girders; the orms are assumed to weigh 15.0 ps. The girders are assumed to be composite throughout. For the atigue design, the Average Daily Truck Traic (ADTT) in one direction, considering the expected growth in traic volume over the 75-year atigue design lie, is assumed to be,000 trucks/day. 4. STEEL FRAMING 4.1. Span Arrangement Proper layout o the steel raming is an important part o the design process. The example bridge has spans o 140-0 175-0 140-0, with the span lengths arranged to give similar positive dead load moments in the end and center spans. Such balanced span arrangements (i.e. end spans approximately 0.8 o the length o the center spans) in multiple continuous-span steel bridges result in the largest possible negative moments at the adjacent piers, along with smaller concomitant positive moments and girder delections. As a result, the optimum depth o the girder in all spans will be nearly the same resulting in a much more eicient design. Steel has the lexibility to be utilized or most any span arrangement. However, in some competitive situations, steel has been compelled to use a particular span arrangement that has been optimized or an alternate design. In a competitive situation, i the pier locations are lexible and i the spans have been optimized or the alternate design, the span arrangement or the steel design almost certainly will be dierent and must also be optimized. In situations where there are severe depth restrictions or where it is desirable to eliminate center piers (e.g. certain overpass-type structures), it may be desirable to provide short end spans. However, in cases where there are no such restrictions or needs, it will likely be more economical to extend the end spans to provide a balanced span ratio. This will avoid the costs associated with the possible need or tie-downs at the end bearings, ineicient girder depths and additional moment Page 4 o 1 May, 004

LRFD rd Edition in some spans. In curved structures, extension o the end spans may also permit the use o radial supports where skewed supports might otherwise have been necessary. It should be noted that the most eicient and cost-competitive steel bridge system can result only when the substructure or the steel design is evaluated and designed concurrently with the superstructure. Although the superstructure and substructure act in concert, each is oten analyzed or separate loads and isolated rom the other as much as possible both physically and analytically. Substructure costs represent a signiicant portion o the total bridge cost. The orm chosen or the substructure, oten based on past experience or the desire to be conservative, may unknowingly lead to an ineicient steel design. Substructure orm also has a marked eect on the overall aesthetic appeal o the structure. When the site dictates diicult span arrangements and pier designs, steel is oten the only material o choice. However, its eiciency oten suers when designed to conorm to oundations developed or other materials. For major projects, superstructure and substructure cost curves should be developed or a series o preliminary designs using dierent span arrangements. Since the concrete deck costs are constant and independent o span length, they need not be considered when developing these curves. The optimum span arrangement lies at the minimum o the sum o the superstructure and substructure costs. These curves should always be regenerated to incorporate changes in unit costs that may result rom an improved knowledge o speciic site conditions. While it is recognized that the locations o piers cannot be varied in many instances, or cases where pier locations are lexible, the use o poorly conceived span arrangements and/or substructure orm can have the greatest total cost impact on a steel-bridge design. 4.. Bridge Cross-Section The example bridge cross-section consists o our (4) girders spaced at 1-0 centers with -6 deck overhangs and an out-to-out deck width o 4-0. The 40-0 roadway width can accommodate up to three 1-oot-wide design traic lanes. The total thickness o the cast-in-place concrete deck is 9½ including a ½ -thick integral wearing surace. The concrete deck haunch is ½ deep measured rom the top o the web to the bottom o the deck. The width o the deck haunch is assumed to be 16.0 inches. Deck parapets are each assumed to weigh 50 pounds per linear oot. A uture wearing surace o 5.0 ps is also assumed in the design. A typical cross-section is shown in Figure 1: Figure 1: Typical Bridge Cross-Section The deck overhangs are approximately 9 percent o the girder spacing. Reducing the girder spacing below 1-0 would lead to an increase in the size o the deck overhangs, which may lead to larger May, 004 Page 5 o 1

LRFD rd Edition loading on the exterior girders. The eect o a wider girder spacing would have to be evaluated with respect to any potential increase in the cost o the concrete deck. Wide girder spacings oer the advantages o ewer girders and pieces to abricate, inspect, ship and erect, and ewer bearings to purchase and set. 4.. Cross-Frames Cross-rames provide lateral stability to the top and bottom langes o the girder, distribute vertical dead and live loads applied to the structure, transer lateral wind loads rom the bottom o the girder to the deck and rom the deck to the bearings, reduce any lange lateral bending eects and transverse deck stresses and provide adequate distribution o load to ensure relatively equal girder delection during construction. Cost-eective design o steel-bridge superstructures requires careul attention to details, including the design o diaphragms and cross-rames. Although these members account or only a small percentage o the total structure weight, they usually account or a signiicant percentage o the total erected steel cost. Cross-rames in steel-girder bridges, along with the concrete deck, provide restoring orces that tend to make the steel girders delect equally. During erection and prior to curing o the deck, the cross-rames are the only members available to provide the restoring orces that prevent the girders rom delecting independently. The restoring orces will be very small i the stinesses o the adjacent girders at the cross-rame connection points are approximately equal and the applied loads to each girder are approximately the same. For the more general case where the girders delect by dierent amounts, the cross-rames and concrete deck will develop larger restoring orces, with the magnitude being dependent on the relative girder, cross-rame and deck stinesses. With ewer cross-rame lines, the orce in each cross-rame member will increase to some degree since the total restoring orce between two adjacent girders is the same regardless o the number o crossrames that are provided. Stresses in the concrete deck will also increase to a degree. For a tangent composite bridge with a regular raming plan, which is the case in this particular design example, the increases in these orces and stresses will typically be o less concern; particularly at the cross-rame spacings chosen or this example. However, the designer should be at least cognizant o these eects when ewer cross-rame lines are provided, especially or more irregular raming plans and when the bridge is non-composite. When reined methods o analysis are used and the cross-rames are included in the structural model to determine orce eects, the cross-rame members are to be designed or the calculated orce eects. When approximate methods o analysis are used (e.g., lateral distribution actors), cross-rame orce eects due to dead and live loads generally cannot be easily calculated. Thus, as a minimum, crossrames are designed to transer wind loads and to meet all applicable slenderness and minimum material thickness requirements. For the most part, such an approach has proven successul on tangent bridges without skewed supports or with small skews. For tangent bridges with moderate to highly skewed supports, where the eects o dierential delections between girders become more pronounced, and or all curved bridges, closer scrutiny o cross-rame orce eects is warranted. Since 1949, the AASHTO Standard Speciications or steel design have speciied a limit o 5'-0" on the longitudinal diaphragm or cross-rame spacing or I-girder bridges. While this limit has ensured satisactory perormance o these structures over the years, it is essentially an arbitrary limit that was based on the experience and knowledge that existed at that time. This arbitrary requirement has been removed in the LRFD speciications. Instead, the need or cross-rames at all stages o construction and the inal condition is to be established by rational analysis (Article 6.7.4.1). Article 6.7.4.1 urther states that the investigation should include, but not be limited to, consideration o the transer o lateral wind loads rom the bottom o the girder to the deck and rom the deck to the bearings, the stability o bottom langes or all loads when subject to compression, the stability o top langes in compression prior to curing o the deck and the distribution o vertical dead and live loads applied to the structure. Diaphragms or cross-rames required or conditions other than the inal condition may be speciied to be Page 6 o 1 May, 004

LRFD rd Edition temporary bracing. Based on the preceding considerations, the cross-rame spacings shown on the raming plan in Figure were chosen or this example. Although the AASHTO design speciications are generally member based, the overall behavior o the entire bridge system must also be considered, particularly during the various stages o construction. As will be demonstrated later on in the design example, the noncomposite bridge structure acts as a system to resist wind loads during construction. The example calculations will illustrate how a couple o panels o top lateral bracing, as shown in the interior bays adjacent to the interior piers in Figure, can be added, i necessary, to provide a stier load path or wind loads acting on the noncomposite structure during construction. The lateral bracing helps to reduce the lateral delections and lateral lange bending stresses due to the wind loads. A rational approach is presented to help the Engineer evaluate how many panels o lateral bracing might be necessary to reduce the lateral delections and stresses to a level deemed acceptable or the situation under consideration. Such a system o lateral bracing adjacent to supports can also help provide additional rigidity to an I-girder bridge system to help prevent signiicant relative horizontal movements o the girders that may occur during construction, particularly in longer spans (e.g. spans exceeding approximately 00 eet). Unlike building columns, which are restrained against the ground by gravity and cannot translate with respect to each other, bare steel bridge girders are generally ree to translate longitudinally with respect to adjacent girders. Lateral bracing provides a triangulation o the members to help prevent the rectangles ormed by the girders and cross-rames rom signiicantly changing shape and moving longitudinally with respect to each other. Bottom lateral bracing can serve similar unctions to those described above, but unlike top bracing, would be subject to signiicant liveload orces in the inished structure that would have to be considered should the bracing be let in place. 4.4. Field Section Sizes Field section lengths are generally dictated by shipping weight and length considerations. The Engineer should consult with abricators regarding any speciic restrictions that might inluence the ield-splice locations. For the example design, there is one ield splice assumed in each end span and two ield splices assumed in the center span resulting in ive (5) ield sections in each line o girders, or 0 ield sections or the bridge (Figure ). Figure : Framing Plan May, 004 Page 7 o 1

LRFD rd Edition 5. PRELIMINARY GIRDER SIZES 5.1. Girder Depth The proper girder depth is another extremely important consideration aecting the economy o steelgirder design. In the absence o any depth restrictions, Article.5..6. provides suggested minimum span-to-depth ratios. From Table.5..6.-1, the suggested minimum depth o the steel section in a composite I-section in a continuous span is given as 0.07L, where L is the span length in eet. Using the longest span o 175-0, the suggested minimum depth o the steel section is: 0.07(175.0) = 4.75 t = 56.7 in. Since there are no depth restrictions in this case, a deeper steel section is desired to provide greater stiness to the girders in their noncomposite condition during construction (it should be noted that the optimum web depth is usually also greater than the suggested minimum web depth). Thereore, the suggested minimum overall depth o the composite I-section in a continuous span, equal to 0.0L, rom Table.5..6.-1 will be used here or the steel section: 0.0(175.0) = 5.60 t = 67. in. A web depth o 69 inches is used. 5.. Cross-section Proportions Cross-section proportion limits or webs o I-sections are speciied in Article 6.10..1. In the span ranges given or this example, the need or longitudinal stieners on the web is not anticipated. For webs without longitudinal stieners, webs must be proportioned such that: Rearranging: D 150 Eq. (6.10..1.1-1) t w D 69 ( t w ) = = =0.46 in. min. 150 150 Because o concerns about the web bend-buckling resistance at the service limit state in regions o negative lexure and also the higher shears in these regions, try a web thickness o 0.565 inches in regions o negative lexure and a web thickness o 0.5 inches in regions o positive lexure. Note that the AASHTO/NSBA Steel Bridge Collaboration Guidelines or Design or Constructibility (hereater reerred to as the Guidelines - to be published in 004 and endorsed by the AASHTO T-14 Technical Committee or Structural Steel Design) recommend a minimum web thickness o 0.475 inches, with a minimum thickness o 0.5 inches preerred. Cross-section proportion limits or langes o I-sections are speciied in Article 6.10... The minimum width o langes is speciied as: Thereore: ( b ) min. The minimum thickness o langes is speciied as: Or: b D6 Eq. (6.10..-) = D 6=69 6=11.5 in. t 1.1t ( t ) ( ) min w =1.1t =1.1 0.565 =0.6 in. w Eq. (6.10..-) Page 8 o 1 May, 004

LRFD rd Edition However, the Guidelines recommend a minimum lange thickness o 0.75 inches. Thereore, use (t ) min = 0.75 inches. For the top lange in regions o positive lexure in composite girders, Article C6.10..4 provides the ollowing additional guideline or the minimum compression-lange width. This guideline is intended to provide more stable ield pieces that are easier to handle during erection without the need or special stiening trusses or alsework, and to help limit out-o-plane distortions o the compression lange and web during the deck-casting operation: L bc Eq. (C6.10..4-1) 85 where L is the length o the girder shipping piece in eet. From Figure, the length o the longest ield piece, which is assumed to also equal the length o the longest shipping piece in this case, is 100 eet. Thereore, or this particular shipping piece: L 100 ( b c ) = = =1.176 t=14.1 in. min 85 85 Based on the above minimum proportions, the trial girder shown in Figure is assumed or the exterior girder, which is assumed to control. Because the top lange o the exterior girders will be subject to lange lateral bending due to the eect o the eccentric deck overhang loads, and also due to wind loads during construction, top-lange sizes slightly larger than the minimum sizes are assumed in regions o positive lexure. The bottom lange plates in regions o positive lexure in this example are primarily sized based on the lange-stress limitation at the service limit state speciied in Article 6.10.4... However, in the end spans, the size o the larger bottom-lange plate in this region is controlled by the stress-range limitation on a cross-rame connection plate weld to the tension lange at the atigue and racture limit state, as will be demonstrated later. The bottom-lange sizes in regions o negative lexure are assumed controlled by either the lange local buckling or lateral torsional buckling resistance at the strength limit state. Top-lange sizes in these regions are assumed controlled by tension-lange yielding at the strength limit state. At this stage, the initial trial plate sizes in regions o negative lexure are primarily educated guesses based on experience. Because the girder is assumed to be composite throughout, the minimum one-percent longitudinal reinorcement required in Article 6.10.1.7 will be included in the section properties in regions o negative lexure. As a result, a top lange with an area slightly smaller than the area o the bottom lange can be used in these regions. Recall that the langes in regions o negative lexure are assumed to be Grade HPS 70W steel in this example. Because the most economical plate to buy rom a mill is between 7 and 96 inches wide, an attempt was made in the design to minimize the number o thicknesses o plate that must ordered or the langes. As recommended in the Guidelines, lange thicknesses should be selected in not less than 1/8-inch increments up to ½ inches in thickness and ¼-inch increments over ½ inches in thickness. Note that individual lange widths are kept constant within each ield piece, as recommended in the Guidelines. The Guidelines contain more detailed discussion on speciic issues pertinent to the sizing o girder langes as it relates to the ordering o plate and the abrication o the langes. Fabricators can also be consulted regarding these issues and all other abrication-related issues discussed herein. Flange transitions, or shop-welded splices, are located based on design considerations, plate length availability (as discussed earlier) and the economics o welding and inspecting a splice compared to the cost o extending a thicker plate. The design plans should consider allowing an option or the abricator to eliminate a shop splice by extending a thicker lange plate subject to the approval o the Engineer. Usually, a savings in weight o between 800 to 1000 pounds should be realized in order to justiy a lange butt splice. Again, the Guidelines contain more detailed discussion regarding this particular issue. At May, 004 Page 9 o 1

LRFD rd Edition Figure : Elevation o Exterior Girder Page 10 o 1 May, 004

LRFD rd Edition 6. LOADS lange splices, the cross-sectional area o the thinner plate should not be less than one-hal the crosssectional area o the thicker plate. Article 6.10.. contains two additional lange proportion limits as ollows: b t 1.0 Eq. (6.10..-1) Iyc 0.1 10 Eq. (6.10..-4) I where: I yc = moment o inertia o the compression lange o the steel section about the vertical axis in the plane o the web (in 4 ) I yt = moment o inertia o the tension lange o the steel section about the vertical axis in the plane o the web (in 4 ) These criteria are each checked or the most critical case (reer to Figure ): ( ) All other langes have a ratio o b /t less than 10.. yt b 18 = = 10.< 1.0 ok t 0.875 ( ) Iyc 1 yt 116 = = 0.51 I 1.75 18 ( ) 1 0.1< 0.51< 10 ok At all other sections, the ratio o I yc /I yt is greater than 0.51 and less than 10. 6.1. Dead Loads As speciied in Article.5.1, dead loads are permanent loads that include the weight o all components o the structure, appurtenances and utilities attached to the structure, earth cover, wearing suraces, uture overlays and planned widenings. In the LRFD speciication, the component dead load DC is assumed to consist o all the structure dead load except or any non-integral wearing suraces and any speciied utility loads. For composite steelgirder design, DC is assumed divided into two separate parts: 1) DC acting on the non-composite section (DC 1 ), and ) DC acting on the composite section (DC ). As speciied in Article 6.10.1.1.1a, DC 1 represents permanent component load that is applied beore the concrete deck has hardened or is made composite, and is assumed carried by the steel section alone. DC represents permanent component load that is applied ater the concrete deck has hardened or is made composite, and is assumed carried by the long-term composite section. For computing stresses rom moments, the stiness o the long-term composite section in regions o positive lexure is calculated by transorming the concrete deck using a modular ratio o n to account in an approximate way or the eect o concrete creep (Article 6.10.1.1.1b). In regions o negative lexure, the long-term composite section is assumed to consist o the steel section plus the longitudinal reinorcement within the eective width o the concrete deck (Article 6.10.1.1.1c). As discussed previously, cross-rames in steel-girder bridges, along with the concrete deck, provide restoring orces that tend to make the steel girders delect equally. Under the component dead load, DC 1, May, 004 Page 11 o 1

LRFD rd Edition applied prior to hardening o the deck or beore the deck is made composite, the cross-rames are the only members available to provide the restoring orces that prevent the girders rom delecting independently. Thereore, aside rom delections resulting rom elastic shortening o the cross-rames, which are generally negligible, it is reasonable to assume or typical deck overhangs and or bridges with approximately equal girder stinesses at points o connection o the cross-rames (e.g. straight bridges with approximately equal-size girders and bearing lines skewed not more than approximately 10 rom normal) that all girders in the cross-section will resist the DC 1 loads equally. This assumption has been borne out analytically and in the ield. Other assumptions may potentially lead to problems in the ield, particularly when the DC 1 delections are large. Thereore, in this example, the total DC 1 load will be assumed equally distributed to each girder in the cross-section. Note that Article 4.6...1 permits the permanent load o the deck to be distributed uniormly among the girders when certain speciied conditions are met. In the ollowing, the unit weight o concrete is taken equal to 0.150 kc (Table.5.1-1), the concrete deck haunch width is taken equal to 16.0 inches, and the deck haunch thickness is conservatively taken equal to.75 inches (reer also to Figure 1): Component dead load (DC 1 ): Concrete deck = 9.5 ( 4.0 )( 0.150 ) =5.106 kips/t (includes IWS) 1 Concrete deck overhang tapers = 1 1+10 16-9.5.5- ( 0.150 ) =0.14 kips/t 1 1 Concrete deck haunches = 16(.75) 4 ( 0.150 ) =0.18 kips/t 144 Stay-in-place orms = 16 1.0- (0.015)=0.480 kips/t 1 Cross-rames and details = 0.10 kips/t DC 1 load per girder = 6.01 kips/t 4 girders = 1.508 kips/t plus girder sel-weight DW in the LRFD speciication consists o the dead load o any non-integral wearing suraces and any utilities. DW is also assumed carried by the long-term composite section. DC and DW are separated because dierent permanent-load load actors γ p (Table.4.1-) are applied to each load. In this example, the wearing surace load, DW, is assumed applied over the 40-0 roadway width and equally distributed to each girder, which has been the customary practice or many years and is also permitted in Article 4.6...1 or bridges satisying speciied conditions. Over time, there has been a signiicant increase in the use o large concrete barriers that are oten placed at the outer edges o the concrete deck. When reined methods o analysis are employed, these concrete barrier loads (the DC loads in this case) should be applied at their actual locations at the outer edges o the deck, which results in the exterior girders carrying a larger percentage o these loads. Thus, in this example, the weight o each concrete barrier will be distributed equally to an exterior girder and the adjacent interior girder. The PennDOT DM-4 Design Manual ollows such a practice (others have assigned 60 percent o the barrier weight to the exterior girder and 40 percent to the adjacent interior girder, while others continue to distribute the barrier weight equally to each girder). In this particular case, with only our girders in the cross-section, this is equivalent to equal distribution o the total barrier weight to all the girders, but this would not be the case when there are more girders in the cross-section. Thereore, the DW and DC loads on a single exterior girder are computed as ollows or this particular example: Wearing surace load (DW) = [0.05 x 40.0]/4 girders = 0.50 kips/t Component dead load -- Barrier load (DC ) = 0.50/ = 0.60 kips/t Page 1 o 1 May, 004

LRFD rd Edition 6.. Live Loads In the LRFD speciications, live loads are assumed to consist o gravity loads (vehicular live loads, rail transit loads and pedestrian loads), the dynamic load allowance, centriugal orces, braking orces and vehicular collision orces. Live loads o interest in this example are the basic design vehicular live load, a speciied loading or optional live-load delection evaluation, and a atigue load, with the appropriate dynamic load allowance included. Live loads are considered to be transient loads that are assumed applied to the short-term composite section. For computing stresses rom moments, the short-term composite section in regions o positive lexure is calculated by transorming the concrete deck using a modular ratio o n (Article 6.10.1.1.1b). In regions o negative lexure, the short-term composite section is assumed to consist o the steel section plus the longitudinal reinorcement within the eective width o the concrete deck (Article 6.10.1.1.1c), except as permitted otherwise at the atigue and service limit states (see Articles 6.6.1..1 and 6.10.4..1) and when computing longitudinal lexural stresses in the concrete deck (see Article 6.10.1.1.1d). 6..1. Design Vehicular Live Load (Article.6.1.) The basic design vehicular live load in the LRFD speciications is designated as HL-9 and consists o a combination o the ollowing placed within each design lane: a design truck or design tandem. a design lane load. This represents a deviation rom the traditional AASHTO approach in which the design truck or tandem is applied independently rom the lane load. In the AASHTO Standard Speciications, the lane load is treated as a separate loading and one or two single concentrated loads are superimposed onto the lane loading to produce extreme orce eects. The design truck (Article.6.1..) is equivalent to the current AASHTO HS0 truck with the spacing between the kip rear-axle loads varied between 14 and 0 t to produce extreme orce eects (Figure.6.1..-1). The 8 kip ront axle is located at a constant distance o 14 t rom the closest rear axle. The transverse spacing o the wheels is 6 t. The truck is assumed to occupy a design lane 1 t in width with only one truck to be placed within each design lane (except as discussed below). The truck is to be positioned transversely within a lane to produce extreme orce eects; however, the truck is to be positioned no closer than t rom the edge o the design lane. For the design o the deck overhang, the truck is to be positioned no closer than 1 t rom the ace o the curb or railing (Article.6.1.. 1). The design tandem (Article.6.1..) consists o a pair o 5 kip axles spaced 4 t apart with a transverse spacing o wheels equal to 6 t. The design lane load (Article.6.1..4) consists o a 0.64 kips/t uniormly distributed load occupying a 10 t lane width positioned to produce extreme orce eects. The uniorm load may be continuous or discontinuous as necessary to produce the maximum orce eect. For continuous spans, live-load moments in regions o positive lexure and in regions o negative lexure outside the points o permanent-load contralexure are computed using only the HL-9 loading. For computing live-load moments in regions o negative lexure between the points o permanent-load contralexure, a special negative-moment loading is also considered. For this special negative-moment loading, a second design truck is added in combination with the design lane load (Article.6.1..1). The minimum headway between the lead axle o the second truck and the rear axle o the irst truck is speciied to be 50 t (a larger headway may be used to obtain the maximum eect). The distance between the two kip rear axles o each o the design trucks is to be kept at a constant distance o 14 t. In addition, all design loads (truck and lane) are to be reduced to 90 percent o their speciied values. The live-load negative moments between points o permanent-load contralexure are then taken as the larger o the moments caused by the HL-9 loading or this special negative-moment loading. The speciication May, 004 Page 1 o 1

LRFD rd Edition is currently silent regarding spans without points o permanent-load contralexure. It is presumed that the special negative-moment loading should be considered over the entire span in such cases. Live-load shears in regions o positive and negative lexure are to be computed using the HL-9 loading only. However, interior-pier reactions are to be calculated based on the larger o the shears caused by the HL-9 loading or the special negative-moment loading. In all cases, axles that do not contribute to the extreme orce eects under consideration are to be neglected. For strength limit state and live-load delection checks, a percent dynamic load allowance (or impact actor) is applied only to the design truck or tandem portion o the HL-9 design live load or to the truck portion o the special negative-moment loading (Article.6.). The dynamic load allowance is not to be applied to the lane portion o the loadings. As a result, the dynamic load allowance implicitly remains a unction o the span length, although the span length is not explicitly used to compute the allowance. The live-load models discussed above are not intended to represent a particular truck, but rather they are intended to simulate the moments and shears produced by groups o vehicles routinely permitted on highways o various states under "grandather" exclusions to weight laws. The moment and shear eects rom these notional live-load models were compared to selected weigh-in-motion data, the results o truck weight studies, the Ontario Highway Bridge Design Code live-load model, and statistical projections o 75-year vehicles, and were ound to be representative when scaled by appropriate load actors. The current HS0 and HS5 vehicles by themselves were not considered to be accurate representations o the exclusion loads over a wide range o spans that were studied. 6... Loading or Optional Live-Load Delection Evaluation (Article.6.1..) The vehicular live load or checking the optional live-load delection criterion speciied in Article.5..6. is taken as the larger o: the design truck alone. 5 percent o the design truck along with the design lane load. These loadings are used to produce apparent live-load delections similar to those produced by AASHTO HS0 design live loadings. It is assumed in the live-load delection check that all design lanes are loaded and that all supporting components are assumed to delect equally (Article.5..6.). The appropriate multiple presence actors speciied in Article.6.1.1. (discussed later) are to be applied. For composite design, Article.5..6. also permits the stiness o the design cross-section used or the determination o the delection to include the entire width o the roadway and the structurally continuous portions o any railings, sidewalks and barriers. The bending stiness o an individual girder may be taken as the stiness, determined as described above, divided by the number o girders. Live-load delection is checked using the live-load portion o the SERVICE I load combination (Table.4.1-1), including the appropriate dynamic load allowance. 6... Fatigue Load (Article.6.1.4) The vehicular live load or checking atigue in steel structures in the LRFD speciications consists o a single design truck (without the lane load) with a constant rear-axle spacing o 0 t (Article.6.1.4.1). The atigue load is used to represent the variety o trucks o dierent types and weights in actual traic. The constant rear-axle spacing approximates that or the 4- and 5-axle semi-trailers that do most o the atigue damage to bridges. The AASHTO atigue-design procedures given in the Standard Speciications do not accurately relect actual atigue conditions in bridges; these procedures combine an artiicially high atigue stress range with an artiicially low number o stress cycles to achieve a reasonable design. The speciied atigue load in the LRFD speciications produces a lower calculated stress range than produced by the loadings in the Page 14 o 1 May, 004

LRFD rd Edition Standard Speciications. This reduction in calculated stress range is oset by an increase in the number o cycles o loading to be considered in the LRFD speciications. The lower stress range and the increased number o cycles are believed to be more relective o actual conditions experienced by many bridges. The number o cycles to be considered is the number o cycles due to the trucks actually anticipated to cross the bridge in the most heavily traveled lane in one direction averaged over its design lie. This Average Daily Truck Traic (ADTT) can be estimated as a reasonable raction o the Average Daily Traic (including all vehicles), which research has shown to be limited to about 0,000 vehicles per lane per day under normal conditions. In the absence o site-speciic data, Table C.6.1.4.-1 in the Commentary to Article.6.1.4. may be used to estimate the raction o trucks in the traic. The requency o the atigue load is then taken as the single lane average daily truck traic, (ADTT) SL. In the absence o better inormation, (ADTT) SL can be computed by multiplying the ADTT by the raction o truck traic in a single lane p given in Table.6.1.4.-1. It is believed adequate to assume that only one atigue truck is on the bridge at a given time. In the FATIGUE load combination given in the LRFD speciications (Table.4.1-1), a load actor o 0.75 is applied to the atigue load to urther represent the eective stress range due to this truck population. The actored atigue load is nearly equivalent to the current AASHTO HS15 loading. The live-load stress range caused by passage o this atigue load is considered to be approximately one-hal that caused by the heaviest single truck expected to cross the bridge over a 75-year period. I the maximum stress range (twice the stress range due to the actored atigue load) that a welded steel detail experiences in its lietime is less than the constant amplitude atigue threshold or that detail, the detail is believed to not experience atigue crack growth. As a result, the detail is considered to have an ininite lie. Rather than speciying a separate load case, this is accounted or in the speciication by dividing the constant amplitude atigue threshold resistance or a particular detail (equivalent to the allowable stress range or over two million cycles given in the Standard Speciications) by one-hal. This resistance is then checked against the stress range caused by the actored atigue load (Article 6.6.1..5). This modiied constant amplitude atigue threshold resistance will govern the atigue resistance o most details, except possibly or lower category details and details on bridges subjected to very low truck-traic volumes. For these cases, the inite-lie atigue resistance o the detail is computed rom an equation deining the slope o the log S- log N curve or that detail (equivalent to the allowable stress ranges or two million cycles or less given in the Standard Speciications). It is important to remember that atigue is only to be considered i the maximum tensile stress due to twice the actored atigue load at a particular detail is greater than or equal to the unactored permanent-load compressive stress at that detail, as speciied in Article 6.6.1..1. Where the bridge is analyzed using approximate analysis methods, the speciied lateral live-load distribution actors or one traic lane loaded are to be used in the atigue check. Where the bridge is analyzed by any reined method, the single design truck is to be positioned transversely and longitudinally to maximize the stress range at the detail under consideration. A reduced dynamic load allowance o 15 percent is to be applied to the atigue load (Article.6.). 6.. Wind Loads The design horizontal wind pressure, P D, used to compute the wind load on the structure, WS, is determined as speciied in Article.8.1. It will be assumed that the example bridge superstructure is 5 eet above the low ground and that it is located in open country. In the absence o more precise data, the design horizontal wind pressure is to be determined as ollows: V VDZ P=P =P V 10,000 DZ D B B B Eq. (.8.1..1-1) May, 004 Page 15 o 1

LRFD rd Edition where: P B = base wind pressure 0.050 ks or beams (Table.8.1..1-1) V DZ = design wind velocity at design elevation, Z (mph) V B = base wind velocity at 0 t height 100 mph For bridges or parts o bridges more than 0 eet above low ground, V DZ is to be adjusted as ollows: V Z Z 0 V DZ =.5Vo ln V B o Eq. (.8.1.1-1) where: V o = riction velocity = 8.0 mph or open country (Table.8.1.1-1) V 0 = wind velocity at 0 eet above low ground = V B = 100 mph in the absence o better inormation Z = height o the structure measured rom low ground (> 0 eet) Z o = riction length o upstream etch = 0. eet or open country (Table.8.1.1-1) Thereore, 100 5 V DZ =.5( 8.0) ln =10.0 mph 100 0. ( 10.0) P D =0.050 = 10,000 0.05 ks P D is to be assumed uniormly distributed on the area exposed to the wind. The exposed area is to be the sum o the area o all components as seen in elevation taken perpendicular to the assumed wind direction. The direction o the wind is to be varied to determine the extreme orce eect in the structure or its components. For cases where the wind is not taken as normal to the structure, lateral and longitudinal components o the base wind pressure, P B, or various angles o wind direction (assuming V B = 100 mph) are given in Table.8.1..-1. The angles are assumed measured rom a perpendicular to the longitudinal axis. As speciied in Article.8.1..1, the total wind load, WS, on girder spans is not to be taken less than 0. kl. Assuming no superelevation or the example bridge and a barrier height o 4 inches above the concrete deck, the minimum exposed height o the composite superstructure is computed as: h exp. =(0.875+69.0+.5+9.5+4.0)/1=10.41 t The total wind load per unit length, w, or the case o wind applied normal to the structure is computed as: w=p h D exp. =0.05(10.41)=0.55 kips/t>0. kips/t Wind pressure on live load, WL, is speciied in Article.8.1.. Wind pressure on live load is to be represented by a moving orce o 0.1 kl acting normal to and 6 eet above the roadway, which results in an overturning orce on the vehicle similar to the eect o centriugal orce on vehicles traversing horizontally curved bridges. The horizontal line load is to be applied to the same tributary area as the design lane load or the orce eect under consideration. When wind on live load is not taken normal to the structure, the normal and parallel components o the orce applied to the live load may be taken rom Table.8.1.-1. Finally, or load cases where the direction o the wind is taken perpendicular to the bridge and there is no wind on live load considered, a vertical wind pressure o 0.00 ks applied to the entire width o the deck is to be applied in combination with the horizontal wind loads to investigate potential overturning o the bridge (Article.8.). This load case is not investigated in this example. ok Page 16 o 1 May, 004

LRFD rd Edition 6.4. Load Combinations Four limit states are deined in the LRFD speciications to satisy the basic design objectives o LRFD; that is, to achieve saety, serviceability, and constructibility. Each o these limit states is discussed in more detail later on. For each limit state, the ollowing basic equation (Article 1...1) must be satisied: Ση i γ i Q i φr n = R r Eq. (1...1-1) where: η i = load modiier related to ductility, redundancy and operational importance γ i = load actor, a statistically based multiplier applied to orce eects φ = resistance actor, a statistically based multiplier applied to nominal resistance Q i = orce eect R n = nominal resistance R r = actored resistance The load actors are speciied in Tables.4.1-1 and.4.1- o the speciications. For steel structures, the resistance actors are speciied in Article 6.5.4.. As evident rom the above equation, in the LRFD speciications, redundancy, ductility, and operational importance are considered more explicitly in the design. Ductility and redundancy relate directly to the strength o the bridge, while the operational importance relates directly to the consequences o the bridge being out o service. The grouping o these three eects on the load side o the above equation through the use o the load modiier η i represents an initial attempt at their codiication. Improved quantiication o these eects may be possible in the uture. For loads or which a maximum value o γ i is appropriate: η i=ηdηrηi 0.95 where: η D = ductility actor speciied in Article 1.. η R = redundancy actor speciied in Article 1..4 η I = operational importance actor speciied in Article 1..5 For loads or which a minimum value o γ i is appropriate: i D R I Eq. (1...1-) 1 η = 1.0 Eq. (1...1-) η η η For typical bridges or which additional ductility-enhancing measures have not been provided beyond those required by the speciications, and/or or which exceptional levels o redundancy are not provided, the three η actors have deault values o 1.0 speciied at the strength limit state. At all other limit states, all three η actors must be taken equal to 1.0. Thereore, or the example design, η i will be taken equal to 1.0 at all limit states. The load combinations are also presented in Table.4.1-1; there are ewer load combinations speciied in LRFD than in Load Factor Design (LFD). STRENGTH I is the load combination to be used or checking the strength o a member or component under normal use in the absence o wind. The basic STRENGTH I load combination is 1.5 times the permanent load o member components (e.g. the concrete deck and parapets), plus 1.5 times the load due to any non-integral wearing suraces and utilities, plus 1.75 times the design live load. When evaluating the strength o the structure during construction, the load actor or construction loads, or equipment and or dynamic eects (i.e. temporary dead and/or live loads that act on the structure during construction) is not to be taken less than 1.5 in the STRENGTH I load combination (Article.4.). Also, the load actor or any non-integral wearing surace and utility loads may be reduced rom 1.5 to 1.5 when evaluating the construction condition. May, 004 Page 17 o 1

LRFD rd Edition To check the strength o a member or component under special permit loadings in the absence o wind, the STRENGTH II load combination should be used. The STRENGTH II load combination is the same as the STRENGTH I load combination with the live-load load actor reduced to 1.5. The STRENGTH III load combination is to be used or checking strength o a member or component assuming the bridge is exposed to a wind velocity exceeding 55 miles per hour in the absence o live load. The basic STRENGTH III load combination is 1.5 times the permanent load o member components, plus 1.5 times the load due to any non-integral wearing suraces and utilities, plus 1.4 times the wind load on the structure. Note that the load actor or wind may be reduced to not less than 1.5 when checking the STRENGTH III load combination during construction (Article.4.). Also, or evaluating the construction condition, the load actor or temporary dead loads that act on the structure during construction is not to be taken less than 1.5 and the load actor or any non-integral wearing surace and utility loads may be reduced rom 1.5 to 1.5. In the STRENGTH IV load combination, all permanent-load eects (or both the construction and inal conditions) are actored by 1.5 and both live- and wind-load eects are not included. For the bridge in its inal condition, the STRENGTH IV load combination basically relates to very high dead-to-live load orce eect ratios. For longer-span bridges in their inal condition, the ratio o dead-to-live load orce eects is very high and could result in a set o resistance actors dierent rom those determined to be suitable or the sample o smaller-span bridges (with spans not exceeding 00 t) that were used in the calibration o the speciication. Rather than using two sets o resistance actors with the STRENGTH I load combination, it was decided that it would be more practical to include this separate load case. It has also been ound that this particular load combination can control during the investigation o various construction stages. Finally, the STRENGTH V load combination is to be used to check the strength o a member or component assuming the bridge is exposed to a wind velocity equal to 55 miles per hour under normal use. The basic STRENGTH V load combination is 1.5 times the permanent load o member components, plus 1.5 times the load due to any non-integral wearing suraces and utilities, plus 1.5 times the design live load (or any temporary live loads acting on the structure when evaluating the construction condition), plus 0.4 times the wind load on the structure, plus 1.0 times the wind on the live load. For evaluating the construction condition under the STRENGTH V load combination, the load actor or temporary dead loads that act on the structure during construction is not to be taken less than 1.5 and the load actor or any non-integral wearing surace and utility loads may be reduced rom 1.5 to 1.5. EXTREME EVENT I is the load combination including earthquake loading. EXTREME EVENT II is the load combination relating to vehicle and ship collisions and ice loads. SERVICE I relates to normal operational use o the bridge and would be used primarily or crack control in reinorced concrete structures. However, the live-load portion o the SERVICE I load combination is used or checking live-load delection in steel bridges. SERVICE II is used only or steel structures and corresponds to the Overload level in Standard Speciications. In the SERVICE II load combination, the permanent-load load actors are all reduced to 1.0 and the live-load load actor is reduced to 1.. I the SERVICE II load combination is to be applied to a permit-load situation, consideration should be given to reducing the live-load load actor urther. SERVICE III is used or crack control in prestressed concrete structures. Finally, there is the FATIGUE load combination, which has previously been discussed. In strength load combinations where one orce eect decreases another orce eect, the speciied minimum values o the load actors γ p in Table.4.1- are to be applied instead to the permanent-load orce eects. For example, when checking or uplit at end supports, the load actor applied to the permanent load o member components would be reduced rom 1.5 to 0.90. The load actor applied to the non-integral wearing surace loads (i considered in this check) and utility loads would be reduced rom 1.50 to 0.65. Page 18 o 1 May, 004

LRFD rd Edition In this particular example, the ollowing load combinations will be evaluated. Only the maximum permanent-load load actors γ p (rom Table.4.1-) are used in the ollowing load combinations since uplit is not a concern or this particular bridge geometry. STRENGTH I: STRENGTH III: STRENGTH IV: STRENGTH V: 1.5DC + 1.5DW + 1.75(LL+IM) 1.5DC + 1.5DW + 1.4WS 1.5(DC+DW) 1.5DC + 1.5DW + 1.5(LL+IM) + 0.4WS + 1.0WL Load actors are modiied as speciied as speciied in Article.4. when checking the strength o a member or component during construction. No permit vehicle is speciied in this example; thereore, load combination STRENGTH II is not checked. The eect o the thermal gradient is not included. Extreme event limit state checks are also not demonstrated in this example. SERVICE II: 1.0DC + 1.0DW + 1.(LL+IM) In the above, LL is the HL-9 vehicular live load or the special negative-moment loading, WS is the wind load on the structure, and WL is the wind on the live load. FATIGUE: 0.75(LL+IM) where LL is the atigue load speciied in Article.6.1.4.1. SERVICE I and SERVICE III are not directly applicable to steel girder structures. However, the liveload delection check will be perormed as speciied in Article.5..6. using the live-load portion o load combination SERVICE I, including the dynamic load allowance, as ollows: 1.00(LL+IM) where LL is the live loading or live-load delection evaluation speciied in Article.6.1... 7. STRUCTURAL ANALYSIS Structural analysis is covered in Section 4 o the LRFD speciications. Both approximate and reined methods o analysis are discussed in detail. Reined methods o analysis are given greater coverage in the LRFD speciications than they have been in the past recognizing the technological advancements that have been made to allow or easier and more eicient application o these methods. However, or this particular example, approximate methods o analysis (discussed below) are utilized to determine the lateral live-load distribution to the individual girders, and the girder moments and shears are determined rom a line-girder analysis. 7.1. Multiple Presence Factors (Article.6.1.1.) Multiple presence actors to account or the probability o coincident loadings are presented in Section o the LRFD speciications (Table.6.1.1.-1). The actors are dierent than the actors given in the Standard Speciications. The extreme live-load orce eect is to be determined by considering each possible combination o number o loaded lanes multiplied by the corresponding multiple presence actor. However, the speciied multiple presence actors are only to be applied when the lever rule (discussed below), the special requirement or exterior girders assuming rigid rotation o the cross-section (also discussed below), or reined analysis methods are employed. The actors are not to be applied when the tabularized equations or live-load distribution actors given in the speciication are used, as the multiple presence eect has already been actored into the derivation o the equations. As speciied in Article.6.1.1., multiple presence actors are also not to be applied to the atigue limit state check or which one design truck is used. Thereore, when using the tabularized equation or the May, 004 Page 19 o 1

LRFD rd Edition distribution actor or one-lane loaded in the atigue limit-state check, the 1. multiple presence actor or one-lane loaded must be divided out o the calculated actor. Or, when using the lever rule or the special analysis to compute the actor or one-lane loaded or the exterior girder or the atigue checks, the 1. multiple presence actor is not to be applied. The speciied 1. multiple presence actor or one-lane loaded results rom the act that the statistical calibration o the LRFD speciications was based on pairs o vehicles rather than a single vehicle. The actor o 1. accounts or the act that a single vehicle that is heavier than each one o a pair o vehicles (in two adjacent lanes) can still have the same probability o occurrence. The proper use o the multiple presence actors is demonstrated below in the calculation o the live-load distribution actors or the example bridge. 7.. Live-Load Distribution Factors (Article 4.6..) New and more reined equations or the lateral live-load distribution actors or I-girders, based on research done under NCHRP Project 1-6, are incorporated in the LRFD speciications. The actors vary according to the type o deck and girders, the number o design lanes loaded, and whether the girder is an interior or exterior girder. The actors are generally dependent on the span length, transverse girder spacing, and the stiness o the member. For example, the live-load distribution actor or the interior-girder bending moment or steel I-girder bridges with a concrete deck loaded by two or more design lanes is given as ollows (Table 4.6...b-1): where: g = 0.075 + 0.6 0. 0.1 S K g S S 9.5 L 1.0Lt g = live-load distribution actor or bending moment (in units o lanes) S = girder spacing (.5 t S 16 t) L = span length (0 t L 40 t) (see Table C4.6...1-1 or suggested values o L to use) t s = structural concrete deck thickness (4.5 in. t s 1 in.) K g = n(i+ae g ) n = modular ratio I = moment o inertia o the steel girder A = cross-sectional area o the steel girder e g = distance rom the centroid o the steel girder to the mid-point o the concrete deck A dierent equation is given to compute the distribution actor or one-lane loaded. Note that the results rom all the ormulas are given in terms o lanes rather than wheels. Since the stiness o the girders is usually not known in advance, the stiness term (K g /1.OLt s ) may be taken as unity or preliminary design. The above equation is to be used when designing in Customary U.S. units. The use o the approximate equations or I-girder bridges is limited to bridges where the deck is supported on our or more girders. The use o these equations is also subject to the limitations on girder spacing, span length, slab thickness, etc., as noted above. For cases outside these limits, engineering judgment should be employed in extending the application o the ormulas beyond the limits, or else other approaches such as reined analysis methods may be used. When the upper limitation on girder spacing is exceeded, Article 4.6...1 requires that the lever rule (discussed below) be used to compute the lateral distribution o load to the individual girders, unless otherwise speciied. The distribution actor or interior girders, as determined rom the above equation, will generally result in lower live-load bending moments than when the moments are computed using a actor o S/5.5, except possibly or very short spans. Page 0 o 1 May, 004

LRFD rd Edition For exterior girders when two or more design lanes are loaded, a correction actor is applied to the computed distribution actor or the interior girders to compute the raction o the wheel loads distributed to the exterior girders. The correction actor depends on the distance rom the centerline o the exterior girder to the edge o the curb (Table 4.6...d-1). To compute the distribution actor or an exterior girder when one lane is loaded, the lever rule is applied. The lever rule involves the use o statics to determine the wheel-load reaction at the exterior girder by summing moments about the adjacent interior girder assuming the concrete deck is hinged at the interior girder. For steel girders utilizing diaphragms or cross-rames, it is also speciied that the distribution o live load to the exterior girders is not to be less than that computed rom a special analysis assuming the entire bridge cross-section delects and rotates as a rigid body. This latter clause was instituted into the speciications primarily because the distribution-actor ormulas were developed without consideration o diaphragms or cross-rames and their eect on the distribution o load to the exterior girders o steel I-girder bridges. A ormula to determine the reaction at an exterior girder under one or more lanes o loading based on the above assumption is given in the Commentary to Article 4.6...d [Eq. (C4.6...d-1)]; the procedure is equivalent to the conventional procedure used to approximate loads on pile groups. When utilizing the lever rule and the special analysis, vehicles must be placed within their design lanes. As speciied in Article.6.1..1, the HL-9 live loading is assumed to occupy a load lane width o 10 t transversely within a 1-t-wide design lane. Figure.6.1..-1 shows that or the assumed transverse wheel spacing o 6 t, a distance o t remains rom the center o each wheel to each edge o the speciied load lane width (note that the 6 t transverse wheel spacing is also conservatively assumed to apply to the design lane load). The number o design traic lanes to be placed on the bridge is determined by taking the integer part o w/1.0, where w is the roadway width measured between curbs. As speciied in Article.6.1.1.1, roadway widths rom 0 to 4 t shall have two design lanes, each equal to one-hal the roadway width. In the computation o the exterior-girder distribution actor according to the above procedures, the live loads occupying their individual load lane widths are to be placed within their design lanes. The design lanes are then to be placed within the roadway width to maximize the wheel-load reaction at the exterior girder. As shown in Figure.6.1..-1, a wheel load can be no closer than t rom the edge o a barrier or curb. These same rules or positioning o the live loads on the bridge would apply when perorming reined analyses. Also, as speciied in Article.5..7.1, unless uture widening o the bridge is virtually inconceivable, the total load carrying capacity o an exterior girder (considering dead plus live load) is not to be less than the total load carrying capacity o an interior girder. However, it should be noted that the use o the reined distribution actors given in the LRFD Speciications, along with the assumption o equal distribution o the DC 1 loads to each girder and the suggested increase in the percentage o the barrier weight assigned to the exterior girders (as discussed above), will typically result in larger total actored moments in the exterior girders than the interior girders, unless the deck overhangs are very small. For this reason, it is recommended that deck overhangs be limited to approximately 5 percent (or less) o the transverse girder spacing, i possible, to ensure a reasonable balance o the total moments in the interior and exterior girders. Separate distribution actors are also now given or determining the bending moment and shear in individual I girders. The distribution actors or shear are speciied in Tables 4.6...a-1 and 4.6...b-1 or interior and exterior girders, respectively. Correction actors, given in Tables 4.6...e-1 and 4.6...c-1, may be applied to the individual distribution actors or bending moment and shear to account, in a limited way, or the eects o skewed supports. Dead-load eects are currently not adjusted or the eects o skew. May, 004 Page 1 o 1

LRFD rd Edition The computation o the live-load distribution actors or an interior and exterior girder rom the example bridge, utilizing the approximate methods discussed above, is now illustrated. 7..1. Live-Load Lateral Distribution Factors - Positive Flexure The ollowing preliminary cross-section (Figure 4) is assumed to determine the longitudinal stiness parameter K g that is utilized in the approximate ormulas to compute the live-load distribution actors or regions in positive lexure (reer also to Figure ): Figure 4: Preliminary Cross-Section - Positive Flexure Component A d Ad Ad I o I Top Flange 1" x 16" 16.00 5.00 560.0 19,600 1. 19,601 Web ½" x 69" 4.50 1,688 1,688 Bottom Flange 1 8" x 18" 4.75 5.19 871.0 0,649.90 0,65 75.5 11.0 6,94 4.1(11.0) = 1,84-11.0 I = 6,658 in. 4 NA d= s =-4.1 in. 75.5 d TOP OF STEEL =5.50+4.1=9.6 in. d BOT OF STEEL =5.88-4.1=1.75 in. 6,658 6,658 S TOP OF STEEL = =1,581 in. S BOT OF STEEL = =1,97 in. 9.6 1.75 Compute the modular ratio n (Article 6.10.1.1.1b): E n = Eq. (6.10.1.1.1b-1) E where E c is the modulus o elasticity o the concrete determined as speciied in Article 5.4..4. A unit weight o 0.145 kc will be used or the concrete in the calculation o the modular ratio (since 0.005 kc o the speciied unit weight o 0.150 kc is typically assumed to account or the weight o the reinorcement). c c E =,000w c Eq. (5.4..4-1) ( ) 1.5 E =,000 0.145 c 1.5 ' c 4.0=,644 ksi Page o 1 May, 004

LRFD rd Edition 9,000 n= =7.96,644 Note that or normal-density concrete, Article C6.10.1.1.1b permits n to be taken as 8 or 4.0-ksi concrete. Thereore, n = 8 will be used in all subsequent computations. 9.0 e g = +.5+9.6-1.0=46.6 in. ( ) ( g) ( ) 6 K g=n I+Ae =8 6,658+75.5 46.6 =1.81 x 10 in. 4 For preliminary design, the entire term containing K g in the approximate ormulas may be taken as 1.0. Although the K g term varies slightly along the span and between spans, the value at the maximum positive moment section in the end span is used in this example to compute the distribution actor to be used in all regions o positive lexure. Other options are to compute a separate K g in each span based on the average or a weighted average o the properties along each span in the positive-lexure region, or to compute K g based on the actual values o the section properties at each change o section resulting in a variable distribution actor along each span within the positive-lexure region. However, the distribution actor is typically not overly sensitive to the value o K g that is assumed. The girders satisy the limitations deining the range o applicability o the approximate ormulas; these limitations are speciied in the individual tables containing the ormulas. For example, the number o girders in the cross-section is greater than or equal to our, the transverse girder spacing is greater than or equal '-6" and less than or equal to 16'-0", and the span length is greater than or equal to 0'-0" and less than or equal to 40'-0". The limitations on K g (speciied or the shear distribution actor only) and on the slab thickness are also satisied. The computation o the distribution actors (in units o lanes) is illustrated below. 7..1.1. Interior Girder - Strength Limit State The live-load distribution actors or an interior girder or checking the strength limit state are determined using the approximate ormulas given in the indicated tables. Multiple presence actors (Article.6.1.1.) are not explicitly applied because these actors were included in the derivation o these ormulas. Separate actors are given to compute the bending moment and shear. For regions in positive lexure, Table C4.6...1-1 suggests using the length o the span under consideration or L. Bending Moment (Table 4.6...b-1): One lane loaded: Two or more lanes loaded: 0.06+ 14 L 1.0Lt 0.4 0. 0.1 S S Kg S 0.1 0.4 0. 6 1.0 1.0 1.81 x 10 0.06+ =0.58 lanes 14 140.0 1.0( 140.0)( 9.0 ) 0.075+ 9.5 L 1.0Lt 0.6 0. 0.1 S S Kg S 0.1 0.6 0. 6 1.0 1.0 1.81 x 10 0.075+ =0.807 lanes (governs) 9.5 140.0 1.0( 140.0)( 9.0 ) May, 004 Page o 1

LRFD rd Edition Shear (Table 4.6...a-1): One lane loaded: Two or more lanes loaded: S 0.6+ 5.0 1.0 0.6+ =0.840 lanes 5.0 S S 0.+ - 1 5 1.0 1.0 0.+ - =1.08 lanes (governs) 1 5 7..1.. Exterior Girder - Strength Limit State The live-load distribution actors or an exterior girder or checking the strength limit state are determined as the governing actors calculated using a combination o the lever rule, approximate ormulas, and a special analysis assuming that the entire cross-section delects and rotates as a rigid body. Each method is illustrated below. As stated in Article.6.1.1., multiple presence actors are included at the strength limit state when the lever rule and the special analysis are used. Separate actors are again computed or bending moment and shear. Bending Moment: One lane loaded: Use the lever rule (Table 4.6...d-1) The lever rule involves the use o statics to determine the lateral distribution to the exterior girder by summing moments about the adjacent interior girder to ind the wheel-load reaction at the exterior girder assuming the concrete deck is hinged at the interior girder (Figure 5). A wheel cannot be closer than '-0" to the base o the curb (Article.6.1..1). For the speciied transverse wheel spacing o 6'-0", the wheelload distribution to the exterior girder is computed as: Figure 5: Exterior-Girder Distribution Factor - Lever Rule Page 4 o 1 May, 004

LRFD rd Edition 9.0 = 0.750 1.0 Multiple presence actor m = 1. (Table.6.1.1.-1) ( ) 1. 0.750 = 0.900 lanes Two or more lanes loaded: Modiy interior-girder actor by e (Table 4.6...d-1) The actor e is computed using the distance d, where d is the distance rom the exterior girder to the e edge o the curb or traic barrier (must be less than or equal to 5.5 t). d e is negative i the girder web is outboard o the curb or traic barrier (must be greater than or equal to -1.0 t). d e = 0.77 + e 9.1.0 e = 0.77 + = 0.990 9.1 0.990 0.807 = 0.799 lanes ( ) The multiple presence actor is not applied. Special Analysis (C4.6...d - Commentary): Assuming the entire cross-section rotates as a rigid body about the longitudinal centerline o the bridge, distribution actors or the exterior girder are also computed or one, two and three lanes loaded using the ollowing ormula: N R = N NL L Xext e + Nb b x e Eq. (C4.6...d-1) where: R = reaction on exterior beam in terms o lanes N L = number o loaded lanes under consideration e = eccentricity o a lane rom the center o gravity o the pattern o girders (t) X = horizontal distance rom the center o gravity o the pattern o girders to each girder (t) X ext = horizontal distance rom the center o gravity o the pattern o girders to the exterior girder (t) N b = number o beams or girders Figure 6: Exterior-Girder Distribution Factor - Special Analysis May, 004 Page 5 o 1

LRFD rd Edition Multiple presence actors (Table.6.1.1.-1): 1 lane: m 1 = 1. lanes: m = 1.0 lanes: m = 0.85 Reerring to Figure 6: Shear: One lane loaded: Two lanes loaded: Three lanes loaded: 1 ( 1.0 + 6.0)( 1.0 +.0) 1 R = + = 0.65 4 18.0 6.0 ( + ) m R = 1.(0.65) = 0.750 lanes ( 1.0 + 6.0)( 1.0 +.0 +.0) R = + = 0.950 4 18.0 6.0 ( + ) m R = 1.0(0.950) = 0.950 lanes (governs) ( ) ( ) (1.0 + 6.0) 1.0 +.0 +.0-9.0 R= + = 0.975 4 18.0 6.0 ( + ) m R = 0.85 0.975 = 0.89 lanes One lane loaded: Use the lever rule (Table 4.6...b-1) 0.900 lanes (See previous computation) Two or more lanes loaded: Modiy interior-girder actor by e (Table 4.6...b-1) d e e= 0.6+ 10.0 e= 0.6+ = 0.80 10 0.80 1.08 = 0.866 lanes ( ) Special Analysis (C4.6...d - Commentary): The actors computed or bending moment are also used or shear: One lane loaded: 0.750 lanes Two lanes loaded: 0.950 lanes (governs) Three lanes loaded: 0.89 lanes The resulting distribution actors used to check the strength limit state in regions o positive lexure are: Interior Girder Exterior Girder Bending Moment 0.807 lanes 0.950 lanes Shear 1.08 lanes 0.950 lanes 7..1.. Distribution Factors or Fatigue Limit State When checking atigue, the atigue load is placed in a single lane. Thereore, the distribution actors or one-lane loaded are used when computing the stress and shear ranges due to the atigue load, as speciied in Article.6.1.4.b. According to Article.6.1.1., multiple presence actors shall not be applied when checking the atigue limit state. Thereore, the ollowing values o the distribution actors or checking the atigue limit state in regions o positive lexure relect the preceding values or one-lane loaded divided by the speciied multiple presence actor o 1. or one-lane loaded (Table.6.1.1.-1): Page 6 o 1 May, 004

LRFD rd Edition Interior Girder Exterior Girder Bending Moment 0.440 lanes 0.750 lanes Shear 0.700 lanes 0.750 lanes 7..1.4. Distribution Factor or Live-Load Delection According to Article.5..6., when investigating the maximum absolute live-load delection, all design lanes should be loaded, and all supporting components should be assumed to delect equally. For multigirder bridges, this is equivalent to saying that the distribution actor or computing live-load delection is equal to the number o lanes divided by the number o girders. Also, the appropriate multiple presence actor rom Article.6.1.1. shall apply as stated in Article.5..6.. N DF m N L = b = 0.85 = 0.68 lanes 4 7... Live-Load Lateral Distribution Factors - Negative Flexure The ollowing preliminary cross-section (Figure 7) is assumed to determine the longitudinal stiness parameter K g that is utilized in the approximate ormulas to compute the live-load distribution actors or regions in negative lexure (reer also to Figure ): Figure 7: Preliminary Cross-Section - Negative Flexure Component A d Ad Ad I o I Top Flange " x 18" 6.00 5.50 1,78 45,69 1.00 45,81 Web 9 16" x 69" 8.81 15,99 15,99 Bottom Flange " x 0" 40.00 5.50 1,40 50,410 1. 50,4 114.8 14.0 111,0-1.4(14.0) = 176.1 14.0 I = 111,07in. 4 NA ds = = 1.4 in. 114.8 dtop OF STEEL = 6.50 + 1.4 = 7.74 in. dbot OF STEEL = 6.50 1.4 = 5.6 in. 111,07 111,07 STOP OF STEEL = =,94 in. SBOT OF STEEL = =,149 in. 7.74 5.6 May, 004 Page 7 o 1

LRFD rd Edition 9.0 eg = +.5 + 7.74.0 = 4.74 in. n= 8 g 6 g K = n(i + Ae ) = 8(111,07 + 114.8(4.74) ) =.65 x 10 in. Again, or preliminary design, the entire term containing K g in the approximate ormulas may be taken as 1.0. In the preceding calculation, K g is based on the section properties o the interior-pier section. K g may instead be computed based on the section properties at each change o section resulting in a variable distribution actor along the span within the negative-lexure region, or K g may be based on the average or weighted average o the properties along each span in the negative-lexure region. 7...1. Interior Girder - Strength Limit State For regions in negative lexure between points o contralexure, Table C4.6...1-1 suggests using the average length o the two adjacent spans or L. Bending Moment (Table 4.6...b-1): 4 One lane loaded: 0.1 0.4 0. 6 1.0 1.0.65 x 10 0.06 + = 0.54 lanes 14 157.5 1.0(157.5) ( 9.0) Two or more lanes loaded: 0.1 0.6 0. 6 1.0 1.0.65 x 10 0.075 + = 0.809 lanes (governs) 9.5 157.5 1.0(157.5) ( 9.0) All other distribution actors or regions in negative lexure or the interior girder and or the exterior girder are independent o the span length and the stiness o the girder; thereore, they are identical to the values calculated earlier or regions in positive lexure. The resulting distribution actors used to check strength limit state in regions o negative lexure are: Interior Girder Exterior Girder Bending Moment 0.809 lanes 0.950 lanes Shear 1.08 lanes 0.950 lanes 7... Distribution Factors or Fatigue Limit State The ollowing values o the distribution actors or checking the atigue limit state in regions o negative lexure relect values computed previously or one-lane loaded divided by the speciied multiple-presence actor o 1. or one-lane loaded (Table.6.1.1.-1): Interior Girder Exterior Girder Bending Moment 0.47 lanes 0.750 lanes Shear 0.700 lanes 0.750 lanes 7.. Dynamic Load Allowance: IM (Article.6.) The dynamic load allowance is an increment applied to the static wheel load to account or wheel-load impact rom moving vehicles. For the strength limit state and live-load delection checks: IM = % (Table.6..1-1) Factor = 1+ = 1. 100 Page 8 o 1 May, 004

LRFD rd Edition This actor is applied only to the design truck or tandem portion o the HL-9 design live load, or to the truck-train portion o the special negative-moment loading discussed previously. For the atigue limit state checks: IM = 15% (Table.6..1-1) 15 Factor = 1+ = 1.15 100 This actor is applied to the atigue load. 8. ANALYSIS RESULTS 8.1. Moment and Shear Envelopes The analysis results or the exterior girder (Figure ) are shown in the ollowing igures. As speciied in Article 6.10.1.5, the ollowing stiness properties were used in the analysis: 1) or loads applied to the noncomposite section, the stiness properties o the steel section alone, ) or permanent loads applied to the composite section, the stiness properties o the long-term composite section assuming the concrete deck to be eective over the entire span length, and ) or transient loads applied to the composite section, the stiness properties o the short-term composite section assuming the concrete deck to be eective over the entire span length. The entire cross-sectional area o the deck associated with the exterior girder was assumed eective in the analysis or loads applied to the composite section. Note that or a continuous span with a nonprismatic member, changes to individual section stinesses can have a signiicant eect on the analysis results. Thus, or such a span, whenever plate sizes or a particular section are revised, it is most always desirable to perorm a new analysis. In the irst series o plots (Figures 8 and 9), moment and shear envelopes due to the unactored dead and live loads are given. Live-load moments in regions o positive lexure and in regions o negative lexure outside points o permanent-load contralexure are due to the HL-9 loading (design tandem or design truck with the variable axle spacing combined with the design lane load; whichever governs). Live-load moments in regions o negative lexure between points o permanent-load contralexure are equal to the larger o the moments caused by the HL-9 loading or a special negative-moment loading (90 percent o the eect o the truck-train speciied in Article.6.1..1 combined with 90 percent o the eect o the design lane load). Live-load shears are due to the HL-9 loading only. However, it should be noted that interior-pier reactions are to be calculated based on the larger o the shears caused by the HL-9 loading or the special negative-moment loading. The indicated live-load moment and shear values include the appropriate lateral distribution actor and dynamic load allowance or the strength limit state, computed earlier. DC 1 is the component dead load acting on the noncomposite section and DC is the component dead load acting on the long-term composite section. DW is the wearing surace load. Note that the liveload shears in Figure 9 are controlled by the interior girder in this example (the distribution actor or shear or the interior girder at the strength limit state is 1.08 lanes versus 0.950 lanes or the exterior girder). The second series o plots (Figures 10 and 11) shows the moment and shear envelopes due to the unactored atigue load speciied in Article.6.1.4.1. The appropriate lateral distribution actor and reduced dynamic load allowance or the atigue limit state are included in the indicated values. May, 004 Page 9 o 1

LRFD rd Edition Figure 8: Dead- and Live-Load Moment Envelopes Page 0 o 1 May, 004

LRFD rd Edition Figure 9: Dead- and Live-Load Shear Envelopes May, 004 Page 1 o 1

LRFD rd Edition Figure 10: Fatigue-Load Moments Page o 1 May, 004

LRFD rd Edition Figure 11: Fatigue-Load Shears May, 004 Page o 1

LRFD rd Edition 8.. Live Load Delection As discussed previously, the optional live-load delection check consists o evaluating two separate liveload conditions to simulate current AASHTO HS0 loadings. Again, the two load conditions are (Article.6.1..): The design truck. The design lane load plus 5 percent o the design truck. The dynamic load allowance o percent is applied to the design truck in each case. A load actor o 1.0 is applied to the live load since the live-load portion o the SERVICE I load combination is to be used in the check. The lateral distribution actor or live-load delection, computed earlier, is also used. The actual n-composite moments o inertia along the entire length o the girder are used in the analysis. Because live-load delection is not anticipated to be a signiicant concern or the example bridge, the stiness o the barriers is not included in the composite stiness used in determining the live-load delections. However, the ull width o the concrete deck associated with the exterior girder (versus the eective lange width) is used in determining the composite stiness, as recommended in Article 4.6..6 or the calculation o live-load delections. The maximum live-load delections in the end span and center span due to the design truck plus the dynamic load allowance are: ( ) end span = 0.91 in. (governs) LL+ IM ( ) center span = 1. in. (governs) LL+ IM The maximum live-load delections in the end span and center span due to the design lane load plus 5 percent o the design truck plus the dynamic load allowance are: ( ) end span = 0.60 + 0.5(0.91) = 0.8 in. LL+ IM ( ) center span = 0.85 + 0.5(1.) = 1.16 in. LL+ IM 9. LIMIT STATES 9.1. Service Limit State (Articles 1... and 6.5.) To satisy the service limit state, restrictions on stress and deormation under regular service conditions are speciied to ensure satisactory perormance o the bridge over its service lie. As speciied in Article 6.10.4.1, optional live load delection criteria and span-to-depth ratios (Article.5..6) may be invoked to control deormations. Steel structures must also satisy the requirements o Article 6.10.4. under the SERVICE II load combination, which corresponds to the Overload level in current LFD procedures. The intent o the design checks speciied in Article 6.10.4. is to prevent objectionable permanent deormations, caused by localized yielding and potential web bend-buckling under expected severe traic loadings, which might impair rideability. The live-load portion o the SERVICE II load combination is intended to be the design live load speciied in Article.6.1.1 (discussed previously). For a permit load situation, a reduction in the speciied load actor or live load under the SERVICE II load combination should be considered or this limit-state check. 9.. Fatigue and Fracture Limit State (Articles 1... and 6.5.) To satisy the atigue and racture limit state, restrictions on stress range under regular service conditions are speciied to control crack growth under repetitive loads and to prevent racture during the design lie o the bridge (Article 6.6.1). Material toughness requirements are also addressed (Article 6.6.). Page 4 o 1 May, 004

LRFD rd Edition For checking atigue in steel structures, the atigue load and FATIGUE load combination (discussed previously) apply. Fatigue resistance o details is discussed in Article 6.6. A special atigue requirement or webs (Article 6.10.5.) is also speciied to control out-o-plane lexing o the web that might potentially lead to atigue cracking under repeated live loading. 9.. Strength Limit State (Articles 1...4 and 6.5.4) At the strength limit state, it must be ensured that adequate strength and stability is provided to resist the statistically signiicant load combinations the bridge is expected to experience over its design lie. Extensive structural damage may occur, but overall structural integrity is maintained. The applicable STRENGTH load combinations (discussed previously) are used to check the strength limit state. Although not speciied as a separate limit state, constructibility is one o the basic design objectives o LRFD. The bridge must be saely erected and have adequate strength and stability during all phases o construction. Speciic design provisions are given in Article 6.10. o the LRFD speciications to help ensure constructibility o steel I-girder bridges; in particular, when subject to the speciied deck-casting sequence. The constructibility checks are typically made on the steel section only under the actored non-composite dead loads using the appropriate strength load combinations. 9.4. Extreme Event Limit State (Articles 1...5 and 6.5.5) At the extreme event limit state, structural survival o the bridge must be ensured during a major earthquake or lood, or when struck by a vessel, vehicle, or ice low. Extreme event limit states are not covered in this example. 10. SAMPLE CALCULATIONS Sample calculations or two critical sections in an exterior girder rom the example bridge ollow. Section 1-1 (reer to Figure ) represents the section o maximum positive lexure in the end spans, and Section - represents the section at each interior pier. The calculations are intended to illustrate the application o some o the more signiicant provisions contained in Article 6.10. The sample calculations illustrate calculations to be made at the service, atigue and racture and strength limit states. Detailed constructibility checks are also illustrated. Sample stiener designs and the design o the stud shear connectors are included as well. The calculations make use o the moments and shears shown in Figures 8 through 11 and the section properties calculated below. In the calculation o the vertical bending stresses throughout the sample calculations, compressive stresses are always shown as negative values and tensile stresses are always shown as positive values. This convention is ollowed regardless o the expected sign o the calculation result, in which the sign o the major-axis bending moment is maintained. Note that a direct comparison should not be made between the unit weight o the example design contained herein and the unit weight o the design given in Example o the AISI/NSBA publication entitled Four LRFD Design Examples o Steel Highway Bridges. Although the cross-section and span lengths are the same, the assumed component dead loads are signiicantly dierent in the two designs and a hybrid section is also used in regions o negative lexure in the design contained herein. This example design is NOT intended to provide a direct comparison between a girder designed using the new Article 6.10 provisions and the Article 6.10 provisions contained in preceding editions o the LRFD Speciications. 10.1. Section Properties The calculation o the section properties or Sections 1-1 Figure 11: Section 1-1 May, 004 Page 5 o 1

LRFD rd Edition and - is illustrated below. In computing the composite section properties, the structural slab thickness, or total thickness minus the thickness o the integral wearing surace, is used. The modular ratio was computed earlier to be n =7.96 say n = 8. 10.1.1. Section 1-1 Section 1-1 is shown in Figure 1. For this section, the longitudinal reinorcement is conservatively neglected in computing the composite section properties. 10.1.1.1. Eective Flange Width (Article 4.6..6): Section 1-1 As speciied in Article 6.10.1.1.1e, the eective lange width is to be determined as speciied in Article 4.6..6. According to Article 4.6..6, or exterior girders, the eective lange width may be taken as onehal the eective width o the adjacent interior girder, plus the least o: 1) one-eighth o the eective span length, L, where L may be taken as the distance between points o permanent load contralexure or continuous spans, ) 6.0 times the average thickness o the slab, plus the greater o hal the web thickness or one-quarter o the width o the top lange o the girder, or ) the width o the overhang. For interior girders, the eective lange width may be taken as the least o: 1) one-quarter o the eective span length, L, where L may be taken as the distance between points o permanent load contralexure or continuous spans, ) 1.0 times the average thickness o the slab, plus the greater o the web thickness or one-hal the width o the top lange o the girder, or ) the average spacing o adjacent girders. For an interior girder rom the example bridge in regions o positive lexure, b e is the least o: or or L 100.0 x 1 = = 00.0 in. 4 4 b 16.0 ( ) t 1.0ts + = 1.0 9.0 + = 116.0 in. (governs) average spacing o girders = 144.0 in. Thereore, or an exterior girder, b e is the least o: or 116.0 L 100.0 x 1 + = 58.0 + = 08.0 in. 8 8 116.0 b 16.0 + + = + ( ) + = 4 4 or t 6.0ts 58.0 6.0 9.0 116.0 in. 116.0 + width o the overhang = 58.0 + 4.0 in. = 100.0 in. (governs) Page 6 o 1 May, 004

LRFD rd Edition 10.1.1.. Elastic Section Properties: Section 1-1 Steel Section Component A d Ad Ad I o I Top Flange 1" x 16" 16.00 5.00 560.0 19,600 1. 19,601 Web ½" x 69" 4.50 1,688 1,688 Bottom Flange 1 8" x 18" 4.75 5.19 871.0 0,649.90 0,65 75.5 11.0 6,94 4.1(11.0) = 1,84 11.0 I = 6,658 in. 4 NA ds = = 4.1 in. 75.5 dtop OF STEEL = 5.50 + 4.1 = 9.6 in. dbot OF STEEL = 5.88 4.1 = 1.75 in. 6,658 6,658 STOP OF STEEL = = 1,581 in. SBOT OF STEEL = = 1,97 in. 9.6 1.75 Composite Section; n=4 Component A d Ad Ad I o I Steel Section 75.5 11.0 6,94 Concrete Slab 9" x 100"/ 4 7.50 4.50 1,594 67,74 5.1 67,987 11.8 1,8 11,99 11.7(1,8) = 14,588 1, 8 I = 117,41 in. 4 NA dn = = 11.7 in. 11.8 dtop OF STEEL = 5.50 11.7 = 4.1 in. dbot OF STEEL = 5.88 + 11.7 = 47.5 in. 117,41 117,41 STOP OF STEEL = = 4,86 in. SBOT OF STEEL = =,48 in. 4.1 47.5 Composite Section; n=8 Component A d Ad Ad I o I Steel Section 75.5 11.0 6,94 Concrete Slab 9" x 100"/ 8 11.5 4.50 4,781 0,0 759.4 0,96 187.8 4,470 67,904. 80(4,470) = 106,86 4,470 I = 161,518 in. 4 NA dn = =.80 in. 187.8 dtop OF STEEL = 5.50.80 = 11.70 in. d BOT OF STEEL =5.88+.80=59.68 in. 161,518 STOP OF STEEL = = 1,805 in. 11.70 10.1.1.. Plastic Moment: Section 1-1 161,518 SBOT OF STEEL = =,706 in. 59.68 Determine the plastic-moment M p o the composite section using the equations provided in Appendix D to Section 6 o the speciication (Article D6.1). The longitudinal deck reinorcement is conservatively neglected. May, 004 Page 7 o 1

LRFD rd Edition P + P + P = A F = 75.5(50) =,76 kips t w c steel y P = 0.85 ' b t = 0.85(4.0)(100.0)(9.0) =,060 kips s c e s,060 kips <,76 kips PNA is in the top lange, use Case II in Table D6.1-1 t c Pw + Pt P s y= + 1 P c 1.0 50( 69.0)( 0.5) + 50( 1.75)( 18.0),060 y= 1 + 50( 1.0)( 16.0) = 0.44 in. rom the top o the top lange P M = y + t y + Pd + P d + Pd ( ) [ ] c p c s s w w t t t c Calculate the distances rom the PNA to the centroid o each element: 9.0 ds = +.5 + 0.44 1.0 = 7.44 in. 69.0 dw = 1.0 + 0.44 = 5.06 in. 1.75 dt = 1.0 + 69.0 + 0.44 = 70.5 in. p 50 1.0 16.0 ( )( ) ( ) ( ) 1.0 ( ) Mp = 0.44 + 1.0 0.44 + [(,060)(7.44) + 69.0(0.5)(50)(5.06) + 1.75(18.0)(50)(70.5) ] M = 170,8 kip-in. = 14,199 kip-t 10.1.1.4. Yield Moment: Section 1-1 Calculate the yield moment M y o the composite section using the equations provided in Appendix D to Section 6 o the speciication (Article D6..). Essentially, M is taken as the sum o the moments due to the actored loads at the strength limit state applied separately to the steel, long-term, and short-term composite sections to cause irst yield in either steel lange. Flange lateral bending is to be disregarded in the calculation. F M M M y D1 D AD y = + + Eq. (D6.-1) SNC SLT SST where MD1, MD and MAD are the moments applied to the steel, long-term and short-term composite sections, respectively, actored by η and the corresponding load actors. Solve or M AD (bottom lange governs by inspection): Page 8 o 1 May, 004

LRFD rd Edition 10.1.. Section - AD y D1 D AD y ( )( ) ( )( ) + ( )( ) 1.5,0 1 1.5 5 1 1.50 1 M AD 50 = 1.0 + + 1,97,48,706 M = 78,06 kip-in. = 6,517 kip-t M = M + M + M Section - is shown in Figure 1. ( ) ( ) ( ) My = 1.0 1.5,0 1.5 5 1.50 6,517 + + + M = 10,171 kip-t 10.1..1. Eective Flange Width (Article 4.6..6): Section - The eective lange width or Section - is calculated using the procedures discussed previously or Section 1-1. For an interior girder rom the example bridge in regions o negative lexure, b e is the least o: Eq. (D6.-) or or L 8.0 x 1 = = 46.0 in. 4 4 b 18.0 ( ) t 1.0ts + = 1.0 9.0 + = 117.0 in. (governs) average spacing o girders = 144.0 in. Thereore, or an exterior girder, b e is the least o: or 117.0 L 8.0 x 1 + = 58.5 + = 181.5 in. 8 8 117.0 b 18.0 + + = + ( ) + = 4 4 or t 6.0ts 58.5 6.0 9.0 117.0 in. Figure 1: Section - 117.0 + width o the overhang = 58.5 + 4.0 in. = 100.5 in. (governs) 10.1... Minimum Negative Flexure Concrete Deck Reinorcement (Article 6.10.1.7) To control concrete deck cracking in regions o negative lexure, Article 6.10.1.7 speciies that the total cross-sectional area o the longitudinal reinorcement must not be less than 1 percent o the total crosssectional area o the deck. This minimum longitudinal reinorcement must be provided wherever the longitudinal tensile stress in the concrete deck due to either the actored construction loads or Load Combination SERVICE II in Table.4.1-1 exceeds φ r, where r is the modulus o rupture o the concrete determined as speciied in Article 5.4..6 and φ is the appropriate resistance actor or concrete in tension speciied in Article 5.5.4..1. The reinorcement is to have a speciied minimum yield strength not less than 60 ksi and a size not exceeding No. 6 bars. The reinorcement should be placed in two layers May, 004 Page 9 o 1

LRFD rd Edition uniormly distributed across the deck width, and two-thirds should be placed in the top layer. The individual bars must be spaced at intervals not exceeding 1 in. Article 6.10.1.1.1c states that or calculating stresses in composite sections subjected to negative lexure at the strength limit state, the composite section or both short-term and long-term moments is to consist o the steel section and the longitudinal reinorcement within the eective width o the concrete deck. Reerring to the cross-section shown in Figure 1: 9.0 1.0 18 Adeck = ( 4.0) + + 0.5.5 - =.17 t = 4,776 in. 1 1 1 0.01(4,776) = 47.76 in. 47.76 1.11in. t 0.096 in. in. 4.0 = = 0.096(100.5) = 9.0 in. For the purposes o this example, the longitudinal reinorcement in the two layers is assumed to be combined into a single layer placed at the centroid o the two layers (with each layer also including the assumed transverse deck reinorcement). From separate calculations, the centroid o the two layers is computed to be 4.6 in. rom the bottom o the concrete deck. Also in this example, the area o the longitudinal reinorcement is conservatively taken equal to the minimum required area o longitudinal reinorcement, although a larger area may be provided in the actual deck design. Although not required by speciication, or stress calculations involving the application o long-term loads to the composite section in regions o negative lexure in this example, the area o the longitudinal reinorcement is conservatively adjusted or the eects o concrete creep by dividing the area by (i.e. 9.0/ =.10 in. ). The concrete is assumed to transer the orce rom the longitudinal deck steel to the rest o the cross-section and concrete creep acts to reduce that orce over time. Finally, or members with shear connectors provided throughout their entire length that also satisy the provisions o Article 6.10.1.7, Articles 6.6.1..1 and 6.10.4..1 permit the concrete deck to also be considered eective or negative lexure when computing stress ranges and lexural stresses acting on the composite section at the atigue and service limit states, respectively. Thereore, section properties or the short-term and long-term composite section, including the concrete deck but neglecting the longitudinal reinorcement, are also determined or later use in the calculations or Section - at these limit states. 10.1... Elastic Section Properties: Section - Steel Section Component A d Ad Ad I o I Top Flange " x 18" 6.00 5.50 1,78 45,69 1.00 45,81 Web 9 16" x 69" 8.81 15,99 15,99 Bottom Flange " x 0" 40.00 5.50 1,40 50,410 1. 50,4 114.8 14.0 111,0 1.4(14.0) = 176.1 14.0 I = 111,07 in. 4 NA ds = = 1.4 in. 114.8 dtop OF STEEL = 6.50 + 1.4 = 7.74 in. dbot OF STEEL = 6.50 1.4 = 5.6 in. 111,07 111,07 STOP OF STEEL = =,94 in. SBOT OF STEEL = =,149 in. 7.74 5.6 Page 40 o 1 May, 004

LRFD rd Edition Steel Section + Long. Reinorcement/ Component A d Ad Ad I o I Steel Section 114.8 14.0 111,0 Long. Reinorcement/.10 4.6 1. 5,64 5,64 117.9-9.80 116,87 0.08(9.80) = 0.78 9.80 I = 116,86 in. 4 NA drein/ = = 0.08 in. 117.9 dtop OF STEEL = 6.50 + 0.08 = 6.58 in. dbot OF STEEL = 6.50 0.08 = 6.4 in. 116,86 116,86 STOP OF STEEL = =,194 in. SBOT OF STEEL = =,08 in. 6.58 6.4 Steel Section + Long. Reinorcement Component A d Ad Ad I o I Steel Section 114.8 14.0 111,0 Long. Reinorcement 9.0 4.6 96.5 16,901 16,901 14.1 54.5 18,104.05(54.5) = 51.7 54.5 I = 17,58 in. 4 NA drein = =.05 in. 14.1 dtop OF STEEL = 6.50.05 = 4.45 in. dbot OF STEEL = 6.50 +.05 = 8.55 in. 17,58 17,58 STOP OF STEEL = =,70 in. SBOT OF STEEL = =,10 in. 4.45 8.55 Composite Section; n=4 Component A d Ad Ad I o I Steel Section 114.8 14.0 111,0 Concrete Slab 9" x 100.5"/ 4 7.69 4.50 1,60 68,078 54.4 68, 15.5 1,460 179,55 1, 460 dn = = 9. 57 in. 15.5 9.57(1,460) = 1,97 I NA = 165,56 in. 4 dtop OF STEEL = 6.50 9.57 = 6.9 in. dbot OF STEEL = 6.50 + 9.57 = 46.07 in. 165,56 165,56 STOP OF STEEL = = 6,148 in. SBOT OF STEEL = =,594 in. 6.9 46.07 Composite Section; n=8 Component A d Ad Ad I o I Steel Section 114.8 14.0 111,0 Concrete Slab 9" x 100.5"/ 8 11.1 4.50 4,807 04,87 76. 05,050 7.9 4,665 16,5 0.47(4,665) = 95,49 4,665 I = 0,760 in. 4 NA dn = = 0. 47 in. 7.9 dtop OF STEEL = 6.50 0.47 = 16.0 in. dbot OF STEEL = 6.50 + 0.47 = 56.97 in. May, 004 Page 41 o 1

LRFD rd Edition 0,760 STOP OF STEEL = = 1,77 in. 16.0 10.. Exterior Girder Check: Section 1-1 10..1. Constructibility (Article 6.10.) 0,760 SBOT OF STEEL = =,875 in. 56.97 Article 6.10..1 states that in addition to providing adequate strength, nominal yielding or reliance on post-buckling resistance is not to be permitted or main load-carrying members during critical stages o construction, except or yielding o the web in hybrid sections. This is accomplished by satisying the requirements o Article 6.10.. (Flexure) and 6.10.. (Shear) under the applicable Strength load combinations speciied in Table.4.1-1, with all loads actored as speciied in Article.4.. For the calculation o delections during construction, all load actors are to be taken equal to 1.0. As speciied in Article 6.10..4, sections in positive lexure that are composite in the inal condition, but noncomposite during construction, are to be investigated during the various stages o the deck placement. The eects o orces rom deck overhang brackets acting on ascia girders are also to be considered. Wind-load eects on the noncomposite structure prior to casting o the deck are also an important consideration during construction, and are considered herein. Potential uplit at bearings should be investigated at each critical construction stage. 10..1.1. Deck Placement Analysis During the deck placement, parts o the girders become composite in sequential stages. Temporary moments induced in the girders during the deck placement can be signiicantly higher than the inal noncomposite dead load moments ater the sequential placement is complete. A separate analysis was conducted using the BSDI, Ltd. Line Girder System (LGS) to determine the maximum moments in the exterior girders o the example bridge caused by the ollowing assumed deck-placement sequence (Figure 14). Note the sequence assumes that the concrete is cast in the two end spans at approximately the same time. A check is not made or uplit should the cast in one end span be completed beore the cast in the other end span has started. Article 6.10..4 requires that changes in the stiness during the various stages o the deck placement be considered. Thereore, in the analysis, all preceding deck casts are assumed composite or the casts that ollow. Should the deck not be cast in separate stages, but instead be cast rom one end o the bridge to the other, the end span must still be checked or the critical instantaneous unbalanced case where wet concrete exists over the entire end span, with no concrete cast yet on the remaining spans. Figure 1: Deck-Placement Sequence Page 4 o 1 May, 004

LRFD rd Edition Unactored dead-load moments in Span 1 rom the abutment to the end o Cast 1, including the moments resulting rom the preceding deck-placement sequence, are summarized in Table 1. In addition to the moments due to each o the individual casts, Table 1 gives the moments due to the steel weight, the moments due to the weight o the SIP orms, the sum o the moments due to the three casts plus the weight o the SIP orms, the maximum accumulated positive moments during the sequential deck casts (not including the steel weight), the sum o the moments due to the dead loads DC and DW applied to the inal composite structure, and the moments due to the weight o the concrete deck, haunches and SIP orms assuming that the concrete is placed all at once on the noncomposite girders. The assumed weight o the SIP orms includes the weight o the concrete in the orm lutes. Although the orms are initially empty, the weight o the deck reinorcement is essentially equivalent to the weight o the concrete in the orm lutes. The slight dierences in the moments on the last line o Table 1 and the sum o the moments due to the three casts plus the weight o the SIP orms are due to the changes in the girder stiness with each cast. The principle o superposition does not apply directly in the deck-placement analyses since the girder stiness changes at each step o the analysis. However, note the signiicant dierences between the moments on the last line o Table 1 and the maximum accumulated positive moments during the sequential deck casts. In regions o positive lexure, the noncomposite girder should be checked or the eect o this larger maximum accumulated deck-placement moment. This moment at Section 1-1 is shown in bold in Table 1, along with the moment due to the steel weight. The sum o these moments is computed as: M = 5 +,57 =,889 kip-t Table 1: Moments rom Deck-Placement Analysis Span -> 1 Unactored Dead-Load Moments (kip-t) Length (t) 0.0 1.0 4.0 4.0 48.0 56.0 7.0 84.0 96.0 100.0 Steel Weight 0 14 50 41 5 5 96 06 74 1 SIP Forms (SIP) 0 6 110 147 151 150 14 84 7 4 Cast 1 0 870 1544 189 06 87 86 198 1484 175 0-168 -6-589 -67-786 -1010-1179 -147-140 0 14 8 50 57 67 86 101 115 10 Sum o Casts + SIP 0 779 146 1797 1841 1818 1486 989 79-4 Max. +M 0 9 1654 6 457 57 410 067 1511 179 DC + DW 0 75 477 64 661 657 551 86 148 5 Deck, haunches + SIP 0 786 160 18 1870 1850 158 108 5 5 The unactored vertical dead-load delections in Span 1 rom the abutment to the end o Cast 1, including the delections resulting rom the preceding deck-placement sequence, are summarized in Table. Negative values are downward delections and positive values are upward delections. Table : Vertical Delections rom Deck-Placement Analysis Span ->1 Unactored Vertical Dead-Load Delections (In.) Length (t) 0.0 1.0 4.0 4.0 48.0 56.0 7.0 84.0 96.0 100.0 Steel Weight 0 -.17 -. -.47 -.50 -.51 -.47 -.9 -.9 -.5 SIP Forms (SIP) 0 -.07 -.14 -.0 -.1 -.1 -.0 -.16 -.1 -.10 Cast 1 0-1. -.50 -.78-4.04-4.7-4.0 -.95 -. -.08 0.7.5.86.96 1.08 1.5 1. 1. 1.1 0 -.01 -.0 -.04 -.04 -.05 -.05 -.05 -.04 -.0 Sum o Casts + SIP 0-1.14 -.14 -.16 -.4 -.46 -.0 -.84 -.17-1.91 DC + DW 0 -.17 -. -.46 -.48 -.49 -.45 -.8 -.8 -.4 Total 0-1.48 -.78-4.09-4. -4.46-4. -.61 -.74 -.40 Deck, haunches + SIP 0 -.9-1.71 -.47 -.59 -.64 -.4 -.0-1.47-1.7 May, 004 Page 4 o 1

LRFD rd Edition Since the deck casts are relatively short-term loadings, the actual moments and delections that occur during construction are more likely to correspond to those computed using a modular ratio o n or determining the stiness o the sections that are assumed composite. Thereore, the n-composite stiness is used or all preceding casts in computing the moments and delections shown or Casts and in Tables 1 and. The moments and delections on the inal composite structure due to the sum o the DC and DW loads shown in Tables 1 and are computed using the n-composite stiness to account or the long-term eects o concrete creep. The entire cross-sectional area o the deck associated with the exterior girder was assumed eective in the analysis in determining the stiness o the composite sections. Note the dierences in the calculated delections on the last line o Table (assuming the deck is cast all at once on the noncomposite structure) and the sum o the accumulated delections during the sequential deck casts. In many cases, the delections shown on the last line can be used to estimate the girder cambers, as required in Article 6.10..5 to account or the dead-load delections. When the dierences in these delections are not signiicant, the delections due to the accumulated deck casts will eventually converge toward the delections shown on the last line as concrete creep occurs. However, i the dierences in the delections are deemed signiicant, the Engineer may need to evaluate which set o delections should be used, or else estimate delections somewhere in-between to compute the girder cambers and avoid potential errors in the inal girder elevations. It is interesting to note that a reined D analysis o the example bridge yielded a maximum delection in Span 1 (at Section 1-1) due to the weight o the concrete deck, haunches and SIP orms (assuming that the concrete is placed all at once on the noncomposite girders) o.61 inches in the exterior girders and.65 inches in the interior girders. From Table 1, the comparable maximum delection rom the line-girder analysis is.64 inches, which indicates the assumption o equal distribution o the DC 1 loads to all the girders is the proper assumption in this case. The unactored vertical dead-load reactions resulting rom the deck-placement analysis are given in Table. Negative reactions represent upward reactions that resist the maximum downward orce at the support under consideration. Conversely, positive reactions represent downward reactions that resist the maximum uplit orce at the support. Table : Unactored Vertical Dead-Load Reactions rom Deck-Placement Analysis (kips) Abut. 1 Pier 1 Pier Abut. Steel Weight -1. -5. -5. -1. sum -1. -5. -5. -1. SIP Forms (SIP) -6. -1. -1. -6. sum -19. 74. -74. -19. Cast 1-80. -55. -55. -80. sum -99. -19. -19. -99. Cast 14. -75. -75. 14. sum -85. -04. -04. -85. Cast -1. -110. -110. -1. sum -86. -14. -14. -86. Sum o Casts + SIP -7. -61. -61. -7. DC + DW -6. -90. -90. -6. Total -11. -404. -404. -11. Deck, haunches + SIP -74. -61. -61. -74. Shown in Table (under sum ) are the accumulated reactions or the steel weight plus the individual deck casts, which should be used to check or uplit under the deck placement. A net positive reaction indicates that the girder may lit-o at the support. Lit-o does not occur in this particular example; lito is most common when end spans o continuous units are skewed or relatively short. I the girder is Page 44 o 1 May, 004

LRFD rd Edition permitted to lit-o its bearing seat, the staging analysis is incorrect unless a hold-down o the girder is provided at the location o a positive reaction. Options to consider when uplit occurs include: 1) rearranging the concrete casts, ) speciying a temporary load over that support, ) speciying a tie-down bearing, or 4) perorming another staging analysis with zero bearing stiness at the support experiencing lit-o. Note that the sum o the reactions rom the analysis o the staged deck casts may dier somewhat rom the reactions assuming the deck is cast all at once on the noncomposite structure (as given on the last line o Table ); however, in most cases, the reactions should not dier greatly. Calculate the maximum lexural stresses in the langes o the steel section due to the actored loads resulting rom the deck-placement sequence. As speciied in Article 6.10.1.6, or design checks where the lexural resistance is based on lateral torsional buckling, bu is to be determined as the largest value o the compressive stress throughout the unbraced length in the lange under consideration, calculated without consideration o lange lateral bending. For design checks where the lexural resistance is based on yielding, lange local buckling or web bend buckling, bu may be determined as the stress at the section under consideration. From Figure, cross-rames adjacent to Section 1-1 are located 48 t and 7 t rom the let abutment. From inspection o Table 1, since the girder is prismatic between the two cross-rames, the largest stress within the unbraced length occurs right at Section 1-1. As discussed previously, the η actor is taken equal to 1.0 in this example. Thereore, For STRENGTH I: For STRENGTH IV: Top lange: Bot. lange: Top lange: Bot. lange: 10..1.. Deck Overhang Loads 1.0(1.5)(,889)(1) bu = =-7.41 ksi 1,581 bu 1.0(1.5)(,889)(1) = =1.96 ksi 1,97 1.0(1.5)(,889)(1) bu = =-.89 ksi 1,581 bu 1.0(1.5)(,889)(1) = =6.6 ksi 1,97 Assume the deck overhang bracket coniguration shown in Figure 15 with the brackets extending to the bottom lange, which is preerred. Alternatively, the brackets may bear on the girder web i means are provided to ensure that the web is not damaged and that the associated deormations permit proper placement o the concrete deck. Although the brackets are typically spaced at to 4 eet along the exterior girder, all bracket loads except or the inishing machine load are assumed applied uniormly. Calculate the vertical loads acting on the overhang brackets. Because in this case the bracket is assumed to extend near the edge o the deck overhang, assume that hal the deck overhang weight is placed on the exterior girder and hal the weight is placed on the overhang brackets. Conservatively include one-hal the deck haunch weight in the total overhang weight. Thereore: Figure 14: Deck Overhang Bracket May, 004 Page 45 o 1

LRFD rd Edition Deck Overhang Weight: 9.5 1.0 16.75 16 P=0.5*150 (.5 ) + +0.5.5- + =55 lbs/t 1 1 1 1 1 The other hal o the overhang weight can be assumed to act at the edge o the top lange (at a distance o 8.0 inches rom the shear center o the girder in this case). The eective deck weight acting on the other side o the girder can be assumed applied at the other edge o the top lange. The net torque can be resolved into lange lateral moments that generally act in the opposite direction to the lateral moments caused by the overhang loads. This eect is conservatively neglected in this example. Construction loads, or dead loads and temporary loads that act on the overhang only during construction, are assumed as ollows: Overhang deck orms: Screed rail: Railing: Walkway: Finishing machine: P = 40 lbs/t P = 85 lbs/t P = 5 lbs/t P = 15 lbs/t P = 000 lbs The inishing machine load is estimated as one-hal o the total inishing machine truss weight, plus some additional load to account or the weight o the engine, drum and operator assumed to be located on one side o the truss. Note that the above loads are estimated loads used here or illustration purposes only. It is recommended that the Engineer consider talking to local Contractors to obtain more accurate values or these construction loads. The lateral orce on the top lange due to the vertical load on the overhang brackets is computed as: where: F= Ptanα 1.5 t α = tan = 1. 5.75 t Note that the calculated lateral orce and the design calculations that ollow are dependent on the assumed angle o the deck overhang brackets. Thus, a sketch similar to Figure 15 with the assumed angle should be shown on the contract plans. Should the Contractor deviate signiicantly rom this assumed angle, an additional investigation by the Contractor may be necessary. In the absence o a more reined analysis, the equations given in Article C6.10..4 may be used to estimate the maximum lange lateral bending moments in the langes due to the lateral bracket orces. Assuming the langes are continuous with the adjacent unbraced lengths and that the adjacent unbraced lengths are approximately equal, the lateral bending moment due to a statically equivalent uniormly distributed lateral bracket orce may be estimated as: FL M = b Eq. (C6.10..4-) 1 The lateral bending moment due to a statically equivalent concentrated lateral bracket orce assumed placed at the middle o the unbraced length may be estimated as: PL M = b Eq. (C6.10..4-) 8 As speciied in Article 6.10.1.6, or design checks where the lexural resistance is based on lateral torsional buckling, the stress,, is to be determined as the largest value o the stress due to lateral bending throughout the unbraced length in the lange under consideration. For design checks where the lexural resistance is based on yielding or lange local buckling, may be determined as the stress at the Page 46 o 1 May, 004

LRFD rd Edition section under consideration. For simplicity in this example, the largest value o within the unbraced length will conservatively be used in all design checks. is to be taken as positive in sign in all resistance equations. The unbraced length, Lb, containing Section 1-1 is equal to 4.0 eet (Figure ). According to Article 6.10.1.6, lateral bending stresses determined rom a irst-order analysis may be used in discretely braced compression langes or which-: CR Eq. (6.10.1.6-) b b Lb 1.Lp bm F yc L p is the limiting unbraced length speciied in Article 6.10.8.. determined as: L=1.0r p t E F yc Eq. (6.10.8..-4) where r t is the eective radius o gyration or lateral torsional buckling speciied in Article 6.10.8.. determined as: r t = b c 1 Dt c w 1 1+ b c t c Eq. (6.10.8..-9) For the steel section, the depth o the web in compression in the elastic range, D c, at Section 1-1 is 8.6 inches. Thereore, 16 rt = =.90 in. 18.6(0.5) 1 1+ 16(1) 1.0(.90) 9,000 Lp = = 7.8 t 1 50 C b is the moment gradient modiier speciied in Article 6.10.8... Separate calculations show that mid / > 1 in the unbraced length under consideration. Thereore, C b must be taken equal to 1.0. According to Article 6.10.1.10., the web load-shedding actor, R b, is to be taken equal to 1.0 when checking constructibility since web bend buckling is prevented during construction by a separate limit state check. Finally, bm is the largest value o the compressive stress due to the actored loads throughout the unbraced length in the lange under consideration, calculated without consideration o lange lateral bending. In this case, use bm = bu = -.89 ksi, as computed earlier or the STRENGTH IV load combination (which controls in this particular computation). Thereore: 1.0(1.0) 1.( 7.8) 11.59 t Lb 4.0 t.89 50 = < = Because the preceding equation is not satisied, Article 6.10.1.6 requires that second-order elastic compression-lange lateral bending stresses be determined. The second-order compression-lange lateral bending stresses may be determined by ampliying irst-order values (i.e. 1 ) as ollows: 0.85 = Eq. (6.10.1.6-4) 1 F cr 1 1 bm May, 004 Page 47 o 1

LRFD rd Edition or: = (AF) 1 where AF is the ampliication actor and F cr is the elastic lateral torsional buckling stress or the lange under consideration speciied in Article 6.10.8.. determined as: 1 CRπ E = Lb r b b cr F 1.0(1.0) π (9,000) cr F = = 5.49 ksi 4(1).90 t Eq. (6.10.8..-8) Note that the calculated value o F cr or use in Eq. 6.10.1.6-4 is not limited to R b R h F yc as speciied in Article 6.10.8... The ampliication actor is then determined as ollows: For STRENGTH I: For STRENGTH IV: 0.85 AF = = 1.78> 1.0 ok 7.41 1 5.49 0.85 AF = =.8> 1.0 ok.89 1 5.49 AF is taken equal to 1.0 or tension langes. The above equation or the ampliication actor conservatively assumes an elastic eective length actor or lateral torsional buckling equal to 1.0. Article C6.10.8.. provides reerences to a relatively simple method that can be used in certain situations to potentially calculate a lower elastic eective length actor or the unbraced length under consideration. Appendix A (to this design example) illustrates the application o this method to this particular unbraced length. Should the unbraced length under consideration end up being the critical unbraced length or which K is less than 1.0, the lower value o K can then subsequently be used to appropriately modiy F cr in the ampliication actor ormula and also L b when determining the lateral torsional buckling resistance. Note that irst- or second-order lange lateral bending stresses, as applicable, are limited to a maximum value o 0.6F y according to Eq. 6.10.1.6-1. In the STRENGTH I load combination, a load actor o 1.5 is applied to all construction loads (Article.4.). For STRENGTH I: Dead loads: P = 1.0[ 1.5(55) + 1.5(40 + 85 + 5 + 15) ] = 71. lbs / t F = F = P tanα = 71.tan(1. ) = 444.6 lbs / t ( ) FL 0.4446 4 l b Ml = = = 1.4 kip-t 1 1 M 1.4(1) Top lange: l = = = 6.00 ksi S 1(16) 6 Page 48 o 1 May, 004

LRFD rd Edition M 1.4(1) Bot. lange: = = =.45 ksi S 1.75(18) 6 Finishing machine: P = 1.0[ 1.5(000) ] = 4,500 lbs F = P = P tanα = 4,500 tan(1. ) =,76 lbs ( ) PL b.76 4 M = = = 8.1 kip-t 8 8 M 8.1(1) Top lange: = = =.1 ksi S 1(16) 6 M 8.1(1) Bot. lange: = = = 1. ksi S 1.75(18) 6 Top lange: total = 6.00 +.1= 8.1 ksi * AF = (8.1)(1.78) = 14.79 ksi < 0.6Fy = 0 ksi ok l Bot. lange: total =.45 + 1.= 4.78 ksi * AF = (4.78)(1.0) = 4.78 ksi < 0.6Fy = 0 ksi ok For STRENGTH IV: Dead loads: P = 1.0[ 1.5(55 + 40 + 85 + 5 + 15) ] = 795 lbs/t Finishing machine: F = F = Ptanα = 795tan(1. ) = 48.4 lbs/t ( ) FL b 0.484 4 M = = =.0 kip-t 1 1 M.0(1) Top lange: = = = 6.5 ksi S 1(16) 6 M.0(1) Bot. lange: = = =.75 ksi S 1.75(18) 6 Not considered Top lange: total = 6.5 ksi * AF = 6.5(.8) = 14.87 ksi < 0.6Fy = 0 ksi ok Bot. lange: total =.75 ksi * AF =.75(1.0) =.75 ksi < 0.6Fy = 0 ksi ok To account or potential uncertainties in the estimation o the preceding deck overhang loads, it may be desirable to ensure that the actored lateral orces determined above exceed, as a minimum, a percentage o the total actored weight o the structure. A reasonable percentage would be between three to ive percent. 10..1.. Wind Loads Wind load acting on the noncomposite structure prior to casting o the concrete deck will be investigated. Conservatively using the smallest steel section, the total wind load per unit length, w, or the case o wind applied normal to the structure assuming no superelevation is computed as: D exp. [ ] w = P h = 0.05 (0.875 + 69.0 + 1.0) /1 = 0.1 kips/t > 0. kips/t ok Note that the ull design horizontal wind pressure, calculated earlier to be P D = 0.05 ks, is conservatively used here in this illustration. For the actual temporary construction condition however, consideration might be given to using a smaller design wind pressure depending on the speciic situation and anticipated maximum wind velocity at the bridge site. Determine the maximum lexural stress, bu, in the top and bottom langes due to the actored steel weight within the unbraced length containing Section 1-1. The largest moment due to the steel weight within the May, 004 Page 49 o 1

LRFD rd Edition unbraced length is equal to 5 kip-eet right at Section 1-1 (Table 1). Thereore, since the member is prismatic in-between these two cross-rames, the largest stress in both langes also occurs at Section 1-1. The STRENGTH III load case applies to the case o dead plus wind load with no live load on the structure. η is taken equal to 1.0 at the strength limit state in this example. Thereore, For STRENGTH III: Top lange: Bot. lange: 1.0(1.5)(5)(1) bu = =.4 ksi 1,581 1.0(1.5)(5)(1) bu = =.68 ksi 1,97 Since there is no deck to provide horizontal diaphragm action, assume the cross-rames act as struts in distributing the total wind orce on the structure to the langes on all girders in the cross-section. The orce is then assumed transmitted through lateral bending o the langes to the ends o the span or to the closest point(s) o lateral wind bracing. Determine the total actored wind orce on the structure assuming the wind is applied to the deepest steel section and normal to the structure (with no superelevation). For the STRENGTH III load combination, the load actor or wind during construction is not to be taken less than 1.5. ( ) 1.0 1.5 (0.05)(.0 + 69.0 +.0) W = = 0.40 kips/t 1 To illustrate the eect that a couple o panels o top lateral bracing can have in providing a stier load path or wind loads acting on the noncomposite structure during construction, assume the system o top lateral bracing shown in Figure ; that is, top lateral bracing in the interior bays on each side o each interior-pier section. Assume that Span 1 o the structure (acting as a system) resists the lateral wind orce as a propped cantilever, with an eective span length, L e, o 10.0 eet. That is, the top lateral bracing is assumed to provide an eective line o ixity at the cross-rame 0.0 eet rom the pier or resisting the lateral orce. Calculate the moment on the propped cantilever at Section 1-1: 9 9 M = 1 1 WL e (0.40)(10.0) 408.0 kip-t 18 = 18 = Calculate the moment on the propped cantilever at the assumed line o ixity (call it Section - -- 0.0 eet rom the pier): 1 1 M = WL e (0.40)(10.0) 75.4 kip-t 8 = 8 = Note that a reined D analysis o the example noncomposite structure subjected to the actored wind load yielded a total lateral moment in the top and bottom langes o all our girders o 405 kip-t at Section 1-1 and 659 kip-t at Section -. Proportion the total lateral moment to the top and bottom langes at Section 1-1 according to the relative lateral stiness o each lange. Assume that the total lange lateral moment is then divided equally to each girder. The single bay o top bracing along with the line o cross rames adjacent to that bay (acting as an eective line o ixity) permits all the girders to work together as a system to resist the lateral wind orce along the entire span. 1(16) 4 Section 1-1: Top lange I = = 41. in. 1 1.75(18) 4 Bottom lange I = = 668. in. 1 Page 50 o 1 May, 004

LRFD rd Edition Top lange Bottom lange 408.0(41.) M = = 4.48 kip t (41.+ 668.)4 408.0(668.) M = = 67.5 kip t (41.+ 668.)4 A similar computation can be made at Section - (however, this section is not checked or this condition in this example). According to Article 6.10.1.6, lateral bending stresses determined rom a irst-order analysis may be used in discretely braced compression langes or which: CR Eq. (6.10.1.6-) b b Lb 1.Lp bm F yc bm is the largest value o the compressive stress due to the actored loads throughout the unbraced length in the lange under consideration, calculated without consideration o lange lateral bending. In this case, use bm = bu = -.4 ksi. Earlier, it was determined that the moment gradient modiier, C b, and the web load-shedding actor, R b, are equal to 1.0. The limiting unbraced length, L p, was also determined earlier to be 7.8 eet. Thereore, 1.0(1.0) 1.( 7.8) 6.5 t Lb 4.0 t.4 50 = > = Thereore, lateral bending stresses determined rom a irst-order analysis may be used. First- or secondorder lange lateral bending stresses, as applicable, are limited to a maximum value o 0.6F y according to Eq. 6.10.1.6-1. 4.48(1) Section 1-1: Top lange: = = 9.70 ksi < 0.6F y = 0.0 ksi ok 1(16) 6 Bottom lange: 67.5(1) = = 10.91 ksi < 0.6F = 0.0 ksi ok 1.75(18) 6 Calculate the shear in the propped cantilever at Section -: 5 5 V = WL e (0.40)(10.0) 0. kips 8 = 8 = Resolve the shear into a compressive orce in the diagonal o the top bracing: (0.0) (1.0) + P = 0. = 58.76 kips 1.0 In addition, the member carries a orce due to the steel weight. Calculate the average stress in the top lange adjacent to the braced bay using the average moment due to the actored steel weight along the 0- oot unbraced length adjacent to the pier section (rom Table 4) assumed applied to the larger section within this unbraced length (i.e. Section -): Resolve this stress into the diagonal: 1.0(1.5)(1)( 1 + 777) / t avg. = =.78 ksi,94 y May, 004 Page 51 o 1

LRFD rd Edition 0.0 diag. =.78 =.8 ksi ( 0.0) + ( 1.0) Assuming an area o 8.0 in. or the diagonal yields a compressive orce o 19.04 kips resulting in a total estimated compressive orce o (-58.76) +(-19.04) = -77.80 kips. The diagonal must be designed to carry this orce. Note that the reined D analysis, mentioned previously, yielded a total compressive orce in the diagonal bracing member o approximately -67.0 kips. Estimate the maximum lateral delection o Span 1 o the structure (i.e. the propped cantilever) due to the actored wind load using the total o the lateral moments o inertia o the top and bottom langes o all our girders at Section 1-1. For simplicity, this section is assumed to be an average section or the span (a weighted average section would likely yield greater accuracy): 4 4 WLe 0.40(10.0) (1,78) = = = 6.7 in. max. 185EI 185(9,000)(41.+ 668.)4 Note that the reined D analysis yielded a maximum lateral delection o approximately 7.0 inches in Span 1. I the top bracing were not present, L e would increase to 140.0 eet and the estimated maximum lateral delection calculated rom the above equation would increase to 1. inches. Large lateral delections may potentially result in damage to the bearings. Thereore such an approach may be helpul to determine how many panels o top lateral bracing, i any, might be necessary to reduce the lateral delection to a level deemed acceptable or the particular situation under consideration. To analyze the center span or this condition, a similar approach can be taken using the actions o an assumed ixed-ixed beam rather than a propped cantilever. 10..1.4. Flexure (Article 6.10..) For critical stages o construction, Article 6.10...1 requires that discretely braced langes in compression satisy the ollowing requirements, except that: 1) or slender-web sections, Eq. 6.10...1-1 need not be checked when is equal to zero, and ) or sections with compact or noncompact webs, Eq. 6.10...1- need not be checked. + φ R F bu bu h yc Eq. (6.10...1-1) 1 + φf nc Eq. (6.10...1-) φ F rw Eq. (6.10...1-) bu c Article 6.10... requires that discretely braced langes in tension satisy: + φ R F bu h yt Eq. (6.10...-1) where: φ = resistance actor or lexure = 1.0 (Article 6.5.4.) R h = hybrid actor speciied in Article 6.10.1.10.1 (= 1.0 at homogeneous Section 1-1) F crw = nominal bend-buckling resistance or webs determined as speciied in Article 6.10.1.9 F nc = nominal lexural resistance o the compression lange determined as speciied in Article 6.10.8. (i.e. local or lateral torsional buckling resistance, whichever controls). For sections with compact or noncompact webs, the provisions o Article A6.. may optionally be used to determine the lateral torsional buckling resistance. First, determine i the noncomposite Section 1-1 is a compact or noncompact web section according to Eq. 6.10.6..-1 (or alternatively, see Table C6.10.1.10.-): Page 5 o 1 May, 004

LRFD rd Edition yc w D t w c 5.7 E F yc Dc (8.6) = = 154.5 t 0.5 E 9,000 5.7 = 5.7 = 17.< 154.5 F 50 Eq. (6.10.6..-1) Thereore, the noncomposite Section 1-1 is a slender-web section. As a result, or the top lange, Eq. 6.10...1-1 must be checked since is not zero, Eq. 6.10...1- must also be checked, and the optional provisions o Appendix A (to LRFD Section 6 -- Article A6..) cannot be used to determine the lateral torsional buckling resistance o the lange. Top Flange Local Buckling Resistance (Article 6.10.8..) Determine the slenderness ratio o the top lange: b c λ = Eq. (6.10.8..-) tc 16 λ = = 8.0 1 Determine the limiting slenderness ratio or a compact lange (alternatively, see Table C6.10.8..-1): Since λ < λ p, () E λ p = 0.8 Eq. (6.10.8..-4) F 9,000 λ p = 0.8 = 9. 50 F = R R F yc nc b h yc Eq. (6.10.8..-1) As speciied in Article 6.10...1, in computing F nc or constructibility, the web load-shedding actor R b is to be taken equal to 1.0 because the lange stress is always limited to the web bend-buckling stress according to Eq. 6.10...1-. Thereore, Fnc = 1.0(1.0)(50) = 50.0 ksi Lateral Torsional Buckling Resistance (Article 6.10.8..) The limiting unbraced length, L p, was computed earlier to be 7.8 eet. The eective radius o gyration or lateral torsional buckling, r t, or the noncomposite Section 1-1 was also computed earlier to be.90 inches. Determine the limiting unbraced length, L r : where: Fyr = 0.7Fyc Fyw E Lr = π rt Eq. (6.10.8..-5) F yr May, 004 Page 5 o 1

LRFD rd Edition Fyr = 0.7(50) = 5.0 ksi < 50 ksi F yr must also not be less than 0.5F yc = 0.5(50) = 5.0 ksi ok. ok Thereore: π(.90) 9,000 Lr = = 9.9 t 1 5.0 Since L p = 7.8 eet < L b = 4.0 eet < L r = 9.9 eet, F L L F C 1 1 R R F R R F yr b p nc = b b h yc b h yc RF h yc Lr L p Eq. (6.10.8..-) As discussed previously, since mid / > 1 in the unbraced length under consideration, the moment-gradient modiier, C b, must be taken equal to 1.0. Thereore, 5.0 4.0 7.8 Fnc = 1.0 1 1 ( 1.0 )(1.0)(50) = 8.75 ksi < 1.0(1.0)(50) = 50 ksi 1.0(50) 9.9 7.8 ok F nc is governed by the lateral torsional buckling resistance, which is less than the local buckling resistance o 50.0 ksi computed earlier. Thereore, F nc = 8.75 ksi. Web Bend-Buckling Resistance (Article 6.10.1.9) Determine the nominal elastic web bend-buckling resistance at Section 1-1 according to the provisions o Article 6.10.1.9.1 as ollows: but not to exceed the smaller o R h F yc and F yw /0.7, where: Thereore, F 0.9Ek = D t crw w ( D D) c Eq. (6.10.1.9.1-1) 9 k = Eq. (6.10.1.9.1-) 9 k= = 8.7 ( 8.6 69.0) 0.9(9,000)(8.7) Fcrw = = 9. ksi < min ( R hf yc,fyw 0.7) = R hfyc = 1.0( 50) = 50 ksi ok 69.0 0.5 Now that all the required inormation has been assembled, check the requirements o Article 6.10...1: For STRENGTH I: bu + φr F bu h yc h yc + = 7.41 ksi + 14.79 ksi = 4.0 ksi φ R F = 1.0(1.0)(50) = 50.0 ksi ( ) 4.0 ksi < 50.0 ksi ok Ratio = 0.844 bu Eq. (6.10...1-1) 1 + φf nc Eq. (6.10...1-) Page 54 o 1 May, 004

LRFD rd Edition 1 14.79 bu + = 7.41 ksi + ksi =.4 ksi φ F = 1.0(8.75) = 8.75 ksi nc.4 ksi < 8.75 ksi ok (Ratio = 0.85) crw φ F rw Eq. (6.10...1-) bu c φ F = 1.0(9.) = 9. ksi 7.41 ksi < 9. ksi ok (Ratio = 0.697) For STRENGTH III: bu + φr F bu h yc h yc + =.4 ksi + 9.70 ksi = 1.04 ksi φ R F = 1.0(1.0)(50) = 50.0 ksi ( ) 1.04 ksi < 50.0 ksi ok Ratio = 0.61 nc bu Eq. (6.10...1-1) 1 + φf nc Eq. (6.10...1-) 1 9.70 bu + =.4 ksi + ksi = 6.57 ksi φ F = 1.0(8.75) = 8.75 ksi 6.57 ksi < 8.75 ksi ok (Ratio = 0.170) crw φ F rw Eq. (6.10...1-) bu c φ F = 1.0(9.) = 9. ksi.4 ksi < 9. ksi ok (Ratio = 0.085) For STRENGTH IV: bu + φr F bu h yc h yc + =.89 ksi + 14.87 ksi = 47.76 ksi φ R F = 1.0(1.0)(50) = 50.0 ksi 47.76 ksi < 50.0 ksi ok (Ratio = 0.955) nc bu Eq. (6.10...1-1) 1 + φf nc Eq. (6.10...1-) 1 14.87 bu + =.89 ksi + ksi = 7.85 ksi φ F = 1.0(8.75) = 8.75 ksi 7.85 ksi < 8.75 ksi ok (Ratio = 0.977) crw φ F rw Eq. (6.10...1-) bu c φ F = 1.0(9.) = 9. ksi.89 ksi < 9. ksi ok (Ratio = 0.86) Options to consider should the web bend-buckling resistance be exceeded include: 1) providing a larger compression lange or a smaller tension lange to decrease D c, ) adjusting the deck-placement sequence to reduce the compressive stress in the web, ) providing a thicker web, and 4) adding a longitudinal web stiener should the preceding options not prove to be practical or cost-eective. May, 004 Page 55 o 1

LRFD rd Edition Bottom Flange For STRENGTH I: bu + φr F bu h yc h yt + = 1.96 ksi + 4.78 ksi = 6.74 ksi φ R F = 1.0(1.0)(50) = 50.0 ksi 6.74 ksi < 50.0 ksi ok (Ratio = 0.55) STRENGTH III: bu + φr F bu h yc h yt + =.68 ksi + 10.91 ksi = 1.59 ksi φ R F = 1.0(1.0)(50) = 50.0 ksi 1.59 ksi < 50.0 ksi ok (Ratio = 0.7) For STRENGTH IV: bu + φr F bu h yc h yt + = 6.6 ksi +.75 ksi = 0.11 ksi φ R F = 1.0(1.0)(50) = 50.0 ksi 0.11 ksi < 50.0 ksi ok (Ratio = 0.60) vv cr p Eq. (6.10...-1) Eq. (6.10...-1) Eq. (6.10...-1) Although the checks are illustrated here or completeness, the bottom lange will typically not control in this region. 10..1.5. Shear (Article 6.10..) For critical stages o construction, Article 6.10.. requires that interior panels o stiened webs satisy the ollowing requirement: V φ u Eq. (6.10..-1) where: φ v = resistance actor or shear = 1.0 (Article 6.5.4.) V u = shear in the web at the section under consideration due to the actored permanent loads and actored construction loads applied to the noncomposite section V cr = shear buckling resistance determined rom Eq. 6.10.9..-1 Only the interior panels o stiened webs are checked because the shear resistance o the end panel o stiened webs and the shear resistance o unstiened webs are already limited to the shear buckling resistance at the strength limit state. For this example, the critical panel in Field Section 1 will be checked. The critical panel or this check is the panel immediately to the let o the ourth intermediate cross-rame rom the abutment, which is located 96.0 eet rom the abutment. The transverse stiener in this panel is assumed to be located at the maximum permitted spacing o d o = D = (69.0) = 07.0 inches to the let o this cross-rame (see later shear calculations). Since shear is rarely increased signiicantly due to deck staging, the actored DC 1 shear at the cross-rame will be used in this check (the STRENGTH IV load combination governs by inspection): ( ) u DC 1 V = 1.0(1.5)( 79) = 119 kips at 96-0 rom the abutment The shear buckling resistance o the 07-inch-long panel is determined as: V = V = CV n cr Eq. (6.10.9..-1) C is the ratio o the shear buckling resistance to the shear yield strength determined rom Eq. 6.10.9..-4, 6.10.9..-5 or 6.10.9..-6, as applicable. First, compute the shear buckling coeicient, k Page 56 o 1 May, 004

LRFD rd Edition Since, yw 5 k= 5+ do D 5 k = 5 + = 5.56 07.0 69.0 Ek 9, 000(5.56) D 69.0 1.40 = 1.40 = 79.5 < = = 18.0 F 50 t 0.5 ( 18.0) V p is the plastic shear orce determined as ollows: Thereore, 10..1.6. Concrete Deck (Article 6.10...4) 1.57 Ek C = D Fyw t w 1.57 9, 000(5.56) C = = 0.66 50 V p yw w w Eq. (6.10.9..-7) Eq. (6.10.9..-6) = 0.58F Dt Eq. (6.10.9..-) Vp = 0.58(50)(69.0)(0.5) = 1,001 kips Vcr = 0.66(1,001) = 66 kips φ V = 1.0(66) = 66 kips v cr 119 kips < 66 kips ok (Ratio = 0.447) Article 6.10...4 requires that the longitudinal tensile stress in a composite concrete deck due to the actored loads not exceed φ r during critical stages o construction, unless longitudinal reinorcement is provided according to the provisions o Article 6.10.1.7. r is the modulus o rupture o the concrete determined as ollows or normal weight concrete (Article 5.4..6): = 0.4 = 0.4 4.0 = 0.480 ksi r ' c φ is the appropriate resistance actor or concrete in tension speciied in Article 5.5.4..1. For reinorced concrete in tension, φ is equal to 0.90. φ r = 0.90(0.480) = 0.4 ksi Check the tensile stress in the concrete deck at the end o Cast 1 in Span 1 (100.0 eet rom the abutment) caused by the negative moment due to Cast. From Table 1, the negative moment at the end o Cast 1 due to Cast is 1,40 kip-eet. The longitudinal concrete deck stress is to be determined as speciied in Article 6.10.1.1.1d; that is, using the short-term modular ratio n = 8. The STRENGTH IV load combination controls by inspection. 1.0(1.5)( 1,40)(.0)(1) deck = = 0.45 ksi > 0.4 ksi 161,518(8) May, 004 Page 57 o 1

LRFD rd Edition Thereore, the minimum one percent longitudinal reinorcement is required at this section. The reinorcement is to be No. 6 bars or smaller spaced at not more than 1 inches. Separate calculations similar to those shown above indicate that the minimum longitudinal reinorcement in Span 1 must extend rom the interior-pier section (Section -) to a section approximately 95.0 eet rom the abutment in order to satisy this requirement or the construction condition. Although not done in this example, a more accurate estimate o the concrete strength at the time Cast is made, and the resulting modular ratio, can be used in this check. Note that the total tensile orce in the concrete deck at the end o Cast 1 is (0.45)(100.0)(9.0) = 408 kips. This orce will be transerred rom the deck through the shear connectors to the top lange. Suicient shear connectors should be present at this location to resist this orce and prevent potential crushing o the concrete around the studs or racturing o the studs. To estimate the length over which this orce must be transmitted, assume a 45-degree angle rom the end o the cast to where the concrete deck is assumed eective. Thereore, the length in this particular case is estimated to be 50.0 inches (< 100.0-95.0 = 5.0 eet). Later calculations show that the pitch o the studs is 1.0 inches in this region and that there are three studs per row. The nominal shear resistance o an individual stud is computed to be 6.0 kips ' (or c equal to 4.0 ksi). The orce resisted by the 1 studs within the 50-inch length is 1(6.0) = 4 kips > 408 kips. I necessary, the tensile orce in the deck can be lowered by modiying the placement sequence. 10... Service Limit State (Article 6.10.4) Article 6.10.4 contains provisions related to the control o elastic and permanent deormations at the service limit state. 10...1. Elastic Deormations (Article 6.10.4.1) For control o elastic deormations, Article 6.10.4.1 reers back to Article.5..6, which contains optional live-load delection criteria and criteria or span-to-depth ratios. The suggested span-to-depth ratios were utilized earlier to establish a reasonable minimum web depth or the example girder design. The maximum computed live-load delections at the service limit state or the example girder were reported earlier to be 0.91 inches in the end spans and 1. inches in the center span. The suggested liveload delection limit or a vehicular load is Span/800 (Article.5..6.). Thereore, End Spans: 140.0(1) ALLOW = =.10 in. > 0.91 in. ok (Ratio = 0.4) 800 175.0(1) Center Span: ALLOW = =.6 in. > 1. in. 800 ok (Ratio = 0.468) 10... Permanent Deormations (Article 6.10.4.) Article 6.10.4. contains criteria intended to control objectionable permanent deormations due to expected severe traic loadings that would impair rideability. As speciied in Article 6.10.4..1, these checks are to be made under the SERVICE II load combination speciied in Table.4.1-1. According to Article 6.10.4.., langes must satisy the ollowing requirements: Top steel lange o composite sections: 0.95R F Eq. (6.10.4..-1) h y Bottom steel lange o composite sections: + 0.95R hfy Eq. (6.10.4..-) where is the lange stress at the section under consideration due to the SERVICE II loads calculated without consideration o lange lateral bending, and is the lange lateral bending stress due to the SERVICE II loads determined as speciied in Article 6.10.1.6. Note that a resistance actor is not shown in these equations because Article 1...1 speciies that the resistance actor be taken equal to 1.0 at the service limit state. Page 58 o 1 May, 004

LRFD rd Edition The sign o and is always taken as positive in Eq. 6.10.4..-. is not included in Eq. 6.10.4..-1 because the top lange o composite sections is continuously braced by the concrete deck at the service limit state; thus, lange lateral bending stresses are small and may be neglected. For straight-girder bridges, lateral bending in the bottom lange at the service limit state is only a consideration or bridges with staggered cross-rames in conjunction with skews exceeding 0. Wind-load and deck overhang eects are not considered at the service limit state. Thereore, the term will be taken equal to zero in Eq. 6.10.4..- in this example. With the exception o composite sections in positive lexure in which the web satisies the requirement o Article 6.10..1.1 (i.e. D/t w 150) such that longitudinal stieners are not required, web bend-buckling o all sections under the SERVICE II load combination is to be checked as ollows: c F cr w h y Eq. (6.10.4..-4) where c is the compression-lange stress at the section under consideration due to the SERVICE II loads calculated without consideration o lange lateral bending, and F crw is the nominal bend-buckling resistance or webs determined as speciied in Article 6.10.1.9. Because Section 1-1 is a composite section subject to positive lexure without longitudinal web stieners, Eq. 6.10.4..-4 need not be checked. An explanation as to why these particular sections are exempt rom the above web bendbuckling check is given in Article C6.10.1.9.1. Check the lange stresses due to the SERVICE II loads at Section 1-1. η is speciied to always equal 1.0 at the service limit state (Article 1..): 0.95R F = 0.95(1.0)(50) = 47.50 ksi h y Top lange: 0.95R F Eq. (6.10.4..-1) 1.0(,0) 1.0(5 + ) 1.(,510) = 1.0 + + 1 =.0 ksi 1,581 4,86 1,805 Bot. lange:.0 ksi < 47.50 ksi ok (Ratio = 0.469) + 0.95R hfy Eq. (6.10.4..-) 1.0(,0) 1.0(5 + ) 1.(,510) = 1.0 + + 1 = 6.80 ksi 1,97,48,706 6.80 ksi + 0 < 47.50 ksi ok (Ratio = 0.775) Under the load combinations speciied in Table.4.1-1 and in the absence o lange lateral bending, the above lange-stress criterion will oten govern the size o the bottom lange or compact composite sections in positive lexure; that is, assuming atigue limit state criteria do not control. In this particular example, atigue limit state criteria control the size o the bottom lange at Section 1-1, as will be demonstrated in the next section. Regardless, it may be prudent and expedient in such cases to initially size the bottom lange to satisy this stress criterion and then subsequently check the nominal lexural resistance at the atigue and strength limit states. Finally, it should be noted that or continuous span lexural members that satisy the requirements o Article B6. to ensure adequate robustness and ductility o the pier sections, a calculated percentage o the negative moment due to the SERVICE II loads at the pier section under consideration may be redistributed prior to making the preceding checks. The moments may be redistributed using the optional procedures o Appendix B (to LRFD Section 6 -- speciically, Articles B6. or B6.6). When the redistribution moments are calculated according to these procedures, Eqs. 6.10.4..-1 and 6.10.4..- May, 004 Page 59 o 1

LRFD rd Edition need not be checked within the regions extending rom the pier section under consideration to the nearest lange transition or point o permanent-load contralexure, whichever is closest, in each adjacent span. Eq. 6.10.4..-4 must still be considered within these regions using the elastic moments prior to redistribution. At all locations outside o these regions, Eqs. 6.10.4..-1, 6.10.4..- and 6.10.4..-4, as applicable, must be satisied ater redistribution. 10... Concrete Deck (Article 6.10.1.7) As discussed previously, Article 6.10.1.7 requires the minimum one-percent longitudinal reinorcement in the concrete deck wherever the longitudinal tensile stress in the deck due to the actored construction loads and due to the SERVICE II load combination (Table.4.1-1) exceeds φ r. Earlier calculations showed that the minimum longitudinal reinorcement in Span 1 must extend rom the interior-pier section (Section -) to a section approximately 95.0 eet rom the abutment in order to satisy this requirement or the construction condition. Check the tensile stress in the concrete deck due to the SERVICE II load combination at the section 95.0 eet rom the abutment in Span 1. The longitudinal concrete deck stress is to be determined as speciied in Article 6.10.1.1.1d; that is, using the short-term modular ratio n = 8. Note that only DC, DW and LL+IM are assumed to cause stress in the concrete deck. [ + + ] 1.0 1.0(87) 1.0(8) 1.( 1,701) (.0)(1) deck = = 0.440 ksi > 0.90r = 0.4 ksi 161,518(8) Thereore, check the tensile stress in the concrete deck due to the SERVICE II load combination at a section 94.0 eet rom the abutment in Span 1. [ + + ] 1.0 1.0(98) 1.0(94) 1.( 1,68) (.0)(1) deck = = 0.40 ksi < 0.4 ksi ok 161,518(8) Extend the minimum one-percent longitudinal reinorcement one oot urther to a section 94.0 eet rom the abutment in Span 1. The Engineer should urther ensure that the reinorcement is adequately developed at this point. 10... Fatigue And Fracture Limit State (Article 6.10.5) As speciied in Article 6.10.5.1, details on I-section lexural members must be investigated or atigue as speciied in Article 6.6.1. For checking load-induced atigue, the FATIGUE load combination speciied in Table.4.1-1 and the atigue live load speciied in Article.6.1.4 apply. As speciied in Article 6.10.5., racture toughness requirements in the contract documents must be in conormance with the provisions o Article 6.6.. Finally, a special atigue requirement or webs must be checked according to the provisions o Article 6.10.5.. 10...1. Load Induced Fatigue (Article 6.6.1.) Fatigue o the base metal at the connection-plate welds to the langes at the third intermediate cross-rame in Span 1, located 7.0 eet rom the abutment, will be checked or the FATIGUE load combination (Table.4.1-1). Separate calculations indicate that this is the critical connection-plate weld detail in Field Section 1. Fatigue o the base metal at the stud shear-connector weld to the top lange at the right end o Field Section 1 (located 100.0 eet rom the abutment) will also be checked. The stress range due to the atigue live load modiied by the corresponding dynamic load allowance o 15 percent will be used to make this check. The lateral distribution actors or the atigue limit state, computed earlier, are also used. From Article.6.1.4., the single-lane average daily truck traic (ADTT) SL is: (ADTT) = p x ADTT Eq. (.6.1.4.-1) SL where: ADTT = number o trucks per day in one direction averaged over the design Page 60 o 1 May, 004

LRFD rd Edition lie (assumed to be,000 or this example) p = raction o truck traic in a single lane (Table.6.1.4.-1) For a -lane bridge; p = 0.80 (ADTT) = 0.80(,000) = 1,600 trucks/day SL The provisions o Article 6.6.1. apply only to details subject to a net applied tensile stress. According to Article 6.6.1..1, in regions where the unactored permanent loads produce compression, atigue is to be considered only i this compressive stress is less than twice the maximum tensile stress resulting rom the FATIGUE load combination. Note that the live-load stress due to the passage o the atigue load is considered to be approximately one-hal that o the heaviest truck expected to cross the bridge in 75 years. In this example, the eect o the uture wearing surace is conservatively ignored when determining i a detail is subject to a net applied tensile stress. According to Article 6.6.1..1, or lexural members with shear connectors provided throughout their entire length and with concrete deck reinorcement satisying the provisions o Article 6.10.1.7, lexural stresses and stress ranges applied to the composite section at the atigue limit state may be computed assuming the concrete deck to be eective or both positive and negative lexure. Shear connectors are assumed provided along the entire length o the girder in this example. Earlier computations were made to ensure that the longitudinal concrete deck reinorcement satisies the provisions o Article 6.10.1.7. Thereore, the concrete deck will be assumed eective in computing all stresses and stress ranges applied to the composite section in the subsequent atigue calculations. Top-Flange Connection-Plate Weld Check atigue o the base metal at the connection-plate welds to the langes at the third intermediate cross-rame in Span 1, located 7.0 eet rom the abutment. The total unactored permanent-load compressive stress at the top-lange weld at this location (neglecting the uture wearing surace) is computed as: ( )( ) 1,84 1 8.6 DC 1 = = 1.49 ksi 6,658 ( )( ) 81 1.1 DC = = 0.665 ksi 117,41 14.16 ksi Twice the maximum tensile stress at the top-lange weld at this location due to the negative moment caused by the actored atigue load (actored by the 0.75 load actor speciied or the FATIGUE load combination) is: LL+ IM ( ) ( )( ) 0.75 496 1 10.70 = = 0.591 ksi 161,518 14.16 ksi>0.591 ksi Thereore, atigue o the base metal at the connection-plate weld to the top lange at this location need not be checked. Bottom-Flange Connection-Plate Weld By inspection, it is determined that the base metal at the connection-plate weld to the bottom lange at this location is subject to a net applied tensile stress. Thus, the stress range γ( ƒ) at the connection-plate weld due to the actored atigue load (actored by the speciied 0.75 load actor) is computed using the properties o the short-term composite section as: May, 004 Page 61 o 1

LRFD rd Edition ( ) γ ( )( )( ) 0.75 496 ( 1)( 58.1) 0.75 1,7 1 58.1 = + 161,518 161,518 = 5.96 ksi Determine the atigue detail category rom Table 6.6.1..-1. Under the condition o illet-welded connections with welds normal to the direction o stress, the atigue detail category or base metal at transverse stiener-to-lange welds is Category C. According to Eq. 6.6.1..-1, γ( ) must not exceed the nominal atigue resistance ( F) n. Both the resistance actor φ and design actor η are speciied to be 1.0 at the atigue limit state (Article C6.6.1..). From Eq. 6.6.1..5-1, the nominal atigue resistance is determined as: 1 ( F) = ( ) n A 1 N F Eq. (6.6.1..5-1) TH Table C6.6.1..5-1 shows the values o (ADTT) SL or each atigue detail category above which the atigue resistance is governed by one-hal the constant amplitude atigue threshold, ( F) TH (such that the detail will theoretically provide ininite atigue lie). The values in the table assume a 75-year design lie and one stress cycle n per truck passage. For continuous spans with span lengths greater than 40.0 eet, the number o stress cycles per truck passage n is equal to 1.0 at sections away rom the pier (Table 6.6.1..5-). Sections away rom the pier are deined as sections greater than a distance o one-tenth the span on each side o the interior support. For other values o n, Table C6.6.1..5-1 should be modiied by dividing the values in the table by n. Thereore, rom Table C6.6.1..5-1, the 75-year (ADTT) SL equivalent to ininite atigue lie or a Category C detail is 745 trucks per day < 1,600 trucks per day. Thereore, For a Category C detail, ( F) TH 1 ( F) ( F) = n T H = 1.0 ksi (Table 6.6.1..5-). Thereore: 1 F = 1.0 = 6.00 ksi n ( ) ( ) ( ) F ( ) n γ 5.96 ksi < 6.00 ksi ok (Ratio = 0.99) Eq. (6.6.1..-1) The above atigue limit-state check at the connection-plate weld to the bottom lange ends up governing the design o the bottom lange in this region (see the tabulation o perormance ratios in the Design Example Summary at the end o the example). An alternative is to bolt the connection plates to the bottom lange, only in this region o high stress range, to raise the nominal atigue resistance to that or a Category B detail. Bolting these particular connection plates to the tension lange will raise the nominal atigue resistance to 8.00 ksi and may allow the designer to use a smaller bottom-lange plate in this region. However, the designer is cautioned that a Category C' detail still exists at the termination o the connection-plate weld to the web just above the bottom lange. Also, the bolted connections must be detailed properly to ensure a positive attachment to the lange that oers rotational ixity to prevent distortion-induced atigue caused by out-o-plane deormations (Article 6.6.1.). In most instances, bolting the connection plates to the lange is more expensive than welding the connection plates to the lange; thus, it is prudent or the Engineer to consult a abricator to determine the most overall costeective solution. Page 6 o 1 May, 004

LRFD rd Edition The Engineer is also reminded that the nominal atigue resistance o uncoated weathering steel base metal detailed in accordance with the Federal Highway Administration Technical Advisory (T5140.) Uncoated Weathering Steel in Structures is determined or atigue detail Category B (Table 6.6.1..-1). However, it should be noted that atigue considerations related to Category B details rarely control. Stud Shear-Connector Weld Check atigue o the base metal at the stud shear-connector weld to the top lange at the right end o Field Section 1 (located 100.0 eet rom the abutment). The total unactored permanent-load compressive stress in the top lange at this location (neglecting the uture wearing surace) is computed as: ( ) 74 1 DC 1 = = 0.56 ksi 1,581 ( ) 7 1 DC = = 0.067 ksi 4,86 0.67 ksi Twice the maximum tensile stress at the top-lange weld at this location due to the negative moment caused by the actored atigue load (actored by the 0.75 load actor speciied or the FATIGUE load combination) is: LL+ IM ( ) ( ) 0.75 688 1 = = 0.897 ksi 1,805 0.67 ksi < 0.897 ksi Thereore, atigue o the base metal at the stud shear-connector weld to the top lange at this location must be checked. The stress range γ( ƒ) at the stud shear-connector weld due to the actored atigue load (actored by the speciied 0.75 load actor) is computed using the properties o the short-term composite section as: ( )( ) 0.75 688 ( 1) 0.75 91 1 γ ( ) = + = 1.04 ksi 1,805 1,805 Determine the atigue detail category rom Table 6.6.1..-1. Under the condition o longitudinally loaded illet-welded attachments, the atigue detail category or base metal adjacent to welded stud-type shear connectors is Category C. From Table C6.6.1..5-1, the 75-year (ADTT) SL equivalent to ininite atigue lie or a Category C detail or n equal to 1.0 is 1,90 trucks per day < 1,600 trucks per day. Thereore, For a Category C detail, ( F) TH 1 ( F) ( F) = n T H = 10.0 ksi (Table 6.6.1..5-). Thereore: 1 F = 10.0 = 5.00 ksi n ( ) ( ) ( ) F ( ) n γ 1.04 ksi < 5.00 ksi ok (Ratio = 0.08) Eq. (6.6.1..-1) May, 004 Page 6 o 1

LRFD rd Edition 10... Distortion Induced Fatigue (Article 6.6.1.) To prevent distortion induced atigue, all transverse connection-plate details will provide a positive connection to both the top and bottom langes. 10... Fracture (Article 6.6.) Material or main load-carrying components subject to tensile stress under the STRENGTH I load combination is assumed or this example to be ordered to meet the appropriate Charpy V-notch racture toughness requirements or nonracture-critical material (Table 6.6.-) speciied or Temperature Zone (Table 6.6.-1). 10...4. Special Fatigue Requirement or Webs (Article 6.10.5.) Interior panels o stiened webs must satisy the ollowing requirement: where: Vu V cr p Eq. (6.10.5.-1) V u = shear in the web at the section under consideration due to the unactored permanent loads plus the actored atigue load V cr = shear buckling resistance determined rom Eq. 6.10.9..-1 In this check, the actored atigue load is to be taken as twice that calculated using the FATIGUE load combination (Table.4.1-1), with the atigue live load taken as speciied in Article.6.1.4. Again, the atigue live load is modiied by the dynamic load allowance o 15 percent and the lateral distribution actors or the atigue limit state are used. The live load stress or this check is intended to represent the heaviest truck expected to cross the bridge over a 75-year design lie. Satisaction o Eq. 6.10.5.-1 is intended to control elastic lexing o the web so that the member is assumed able to sustain an ininite number o smaller loadings without atigue cracking due to this eect. Only the interior panels o stiened webs are checked because the shear resistance o the end panel o stiened webs and the shear resistance o unstiened webs are already limited to the shear buckling resistance at the strength limit state. For this example, the critical panel in Field Section 1 will be checked. The critical panel or this check is the second panel rom the abutment, which is located adjacent to the end panel. The transverse stiener spacing in the end panel is d o = 7.5 eet (see later shear calculations). The stiener spacing in the second panel is d o = 16.75 eet = 01.0 inches (up to the irst intermediate cross-rame in Span 1). The shear 7.5 eet rom the abutment to be used in this check is computed as ollows: Vu = 7.0 + 11+ 10 + (0.75)(47) = 165 kips The shear buckling resistance o the 01-inch-long panel is determined as: V = V = CV n cr at 7 - rom the abutment Eq. (6.10.9..-1) C is the ratio o the shear buckling resistance to the shear yield strength determined rom Eq. 6.10.9..-4, 6.10.9..-5 or 6.10.9..-6, as applicable. First, compute the shear buckling coeicient, k 5 k= 5+ do D 5 k = 5 + = 5.59 01.0 69.0 Eq. (6.10.9..-7) Page 64 o 1 May, 004

LRFD rd Edition Since, Ek 9, 000(5.59) D 69.0 1.40 = 1.40 = 79.7 < = = 18.0 F 50 t 0.5 yw ( 18.0) V p is the plastic shear orce determined as ollows: 1.57 Ek C = D Fyw t w 1.57 9, 000(5.59) C = = 0.67 50 V p yw w w Eq. (6.10.9..-6) = 0.58F Dt Eq. (6.10.9..-) Vp = 0.58(50)(69.0)(0.5) = 1,001 kips Thereore, V = 0.67(1,001) = 67 kips > V = 165.0 kips ok (Ratio = 0.618) cr 10..4. Strength Limit State (Article 6.10.6) 10..4.1. Flexure (Article 6.10.6.) u For composite sections in positive lexure, Article 6.10.6.. reers to the provisions o Article 6.10.7 to determine the nominal lexural resistance at the strength limit state. Determine i Section 1-1 qualiies as a compact section. According to Article 6.10.6.., composite sections in positive lexure qualiy as compact when: 1) the speciied minimum yield strengths o the langes do not exceed 70 ksi, ) the web satisies the requirement o Article 6.10..1.1 such that longitudinal stieners are not required (i.e. D/t w 150), and ) the section satisies the ollowing webslenderness limit: D t w cp.76 E F yc Eq. (6.10.6..-1) where D cp is the depth o the web in compression at the plastic moment determined as speciied in Article D6... Earlier computations indicated that the plastic neutral axis o the composite section is located in the top lange. Thereore, according to Article D6.., D cp is taken equal to zero or this case, and thereore, Eq. 6.10.6..-1 is considered to be automatically satisied. Section 1-1 qualiies as a compact section. Compact sections must satisy the ollowing ductility requirement speciied in Article 6.10.7. to protect the concrete deck rom premature crushing: D 0.4D Eq. (6.10.7.-1) p where D p is the distance rom the top o the concrete deck to the neutral axis o the composite section at the plastic moment, and D t is the total depth o the composite section. At Section 1-1: Dp = 9.0 +.5 1.0 + 0.44 = 11.94 in. Dt = 1.75 + 69.0 +.5 + 9.0 = 8.88 in. 0.4Dt = 0.4(8.88) = 4.81 in. > 11.94 in. ok (Ratio = 0.4) According to Article 6.10.7.1.1, at the strength limit state, compact composite sections in positive lexure must satisy the ollowing relationship: t May, 004 Page 65 o 1

LRFD rd Edition M + 1 S φ M u xt n Eq. (6.10.7.1.1-1) where: φ = resistance actor or lexure = 1.0 (Article 6.5.4.) = lateral bending stress in the tension lange determined as speciied in Article 6.10.1.6 M n = nominal lexural resistance o the section determined as speciied in Article 6.10.7.1. M u = bending moment about the major-axis o the cross-section determined as speciied in Article 6.10.1.6 S xt = elastic section modulus about the major-axis o the section to the tension lange taken as M yt /F yt M yt = yield moment with respect to the tension lange determined as speciied in Article D6. As speciied in Article 6.10.1.6, or design checks where the lexural resistance is based on yielding (which is the case here), M u may be taken as the moment due to the actored loads at the section under consideration. In this example, lateral bending in the bottom lange due to wind-load eects will be considered at the strength limit state. For composite sections in positive lexure, lateral bending does not need to be considered in the compression lange at the strength limit state because the lange is continuously supported by the concrete deck. In Eq. 6.10.7.1.1-1, is the lange lateral bending stress determined as speciied in Article 6.10.1.6. According to Article 6.10.1.6, or design checks where the lexural resistance is based on yielding, may be determined as the stress at the section under consideration. For simplicity in this example, however, the largest value o within the unbraced length will conservatively be used in all design checks. is to be taken as positive in sign. In I-girder bridges with composite concrete decks, wind load on the upper hal o the exterior girder, the deck, the barriers and the vehicles may be assumed transmitted directly to the deck, which acts as a lateral diaphragm to carry the load to the supports. Wind load on the lower hal o the exterior girder may be assumed applied laterally to the bottom lange, which transmits the load to the adjacent cross-rames or diaphragms by lexural action. The rame action o the cross-rames or diaphragms then transmits the orces to the deck, which in turn transmits them to the supports through diaphragm action. Article C4.6..7.1 provides the ollowing ormula or the actored wind orce per unit length applied to the bottom lange o composite or noncomposite exterior members with cast-in-place concrete or orthotropic steel decks: W ηγpd D = Eq. (C4.6..7.1-1) where P D is the design horizontal wind pressure speciied in Article.8.1 and d is the depth o the girder. Earlier, P D was computed to be 0.05 ks. For the wind-load path identiied above, Article C4.6..7.1 also provides the ollowing approximate equation or computing the maximum lange lateral bending moment due to the actored wind load within the unbraced length under consideration: WLb Mw = Eq. (C4.6..7.1-) 10 Assemble the actored actions needed to check Eq. 6.10.7.1.1-1 at Section 1-1. The unbraced length, L b, at Section 1-1 is 4.0 eet. In this example, η is taken equal to 1.0 at the strength limit state. The wind load acting on the live load (WL) is assumed transmitted directly to the deck and is thereore not considered in the STRENGTH V load combination in this example. For simplicity, the eect o the Page 66 o 1 May, 004

LRFD rd Edition overturning orce due to WL on the vehicle wheel loads is also not considered in this example. The ampliication actor, AF, or (Article 6.10.1.6) is taken equal to 1.0 or langes in tension. Note again that irst- or second-order lange lateral bending stresses, as applicable, are limited to a maximum value o 0.6F y according to Eq. 6.10.1.6-1. For STRENGTH I: Dead and live loads: [ ] M = 1.0 1.5(,0 + 5) + 1.5() + 1.75(,510) = 9,797 kip t Wind loads: Not considered = 0 For STRENGTH III: u Dead loads: [ ] Mu = 1.0 1.5(,0 + 5) + 1.5() =,654 kip t Wind loads: 1.0(1.4)(0.05)(1.75 + 69.0 + 1.0) /1 W = = 0.1 kips / t 0.1(4.0) Mw = = 1.7 kip t 10 M 1.7(1) w = = =.06 ksi * AF =.06(1.0) =.06 ksi < 0.6F y = 0.0 ksi ok S 1.75 18 6 For STRENGTH IV: ( ) Dead loads: [ ] M = 1.0 1.5(,0 + 5 + ) = 4,89 kip t u Wind loads: Not considered = 0 For STRENGTH V: Dead and live loads: [ ] M = 1.0 1.5(,0 + 5) + 1.5() + 1.5(,510) = 8,9 kip t u Wind loads: 1.0(0.4)(0.05)(1.75 + 69.0 + 1.0) /1 W = = 0.06 kips / t M.6(1) ( ) 0.06(4.0) Mw = =.6 kip t 10 w = = = 0.587 ksi * AF = 0.587(1.0) = 0.587 ksi < 0.6F y = 0.0 ksi ok S 1.75 18 6 From an examination o the above lange lateral bending stresses, it is apparent that or typical crossrame spacings, the majority o the wind orce on the lower hal o a composite structure is transmitted directly to the deck through the cross-rames and only a small portion o the orce is resisted through lateral bending o the bottom lange. Nominal Flexural Resistance (Article 6.10.7.1.) According to the provisions o Article 6.10.7.1., the nominal lexural resistance o compact composite sections in positive lexure is determined as ollows: I D p 0.1D t, then: M M Eq. (6.10.7.1.-1) Otherwise: n = p D p Mn = Mp 1.07 0.7 D t where M p is the plastic moment o the composite section determined as speciied in Article D6.1. Eq. (6.10.7.1.-) May, 004 Page 67 o 1

LRFD rd Edition However, in a continuous span, the nominal lexural resistance o the section is limited to the ollowing: M n = 1.RhMy Eq. (6.10.7.1.-) where M y is the yield moment o the composite section determined as speciied in Article D6., unless the speciic steps outlined in Article 6.10.7.1. are taken to ensure suicient ductility and robustness o adjacent pier sections such that the redistribution o moments caused by partial yielding within the positive lexural regions is inconsequential. Speciically, Articles B6. and B6.6. in Appendix B (to LRFD Section 6) are reerred to or obtaining the requirements that must be satisied to avoid the limitation given by Eq. 6.10.7.1.-. For Section 1-1, M p and M y were computed earlier to be 14,199 kip-t and 10,171 kip-t, respectively. Thereore, Or, 0.1D = 0.1(8.88) = 8.9 in. < D = 11.94 in. t 11.94 Mn = 14,199 1.07 0.7 = 1,761 kip t 8.88 Mn = 1.(1.0)(10,171) = 1, kip t \ Mn = 1, kip t p (governs) Calculate S xt. The yield moment, M y, was calculated with respect to the tension lange; thereore, M yt = M y : S xt M 10,171(1) = = =,441 in F 50 yt yt Now that all the required inormation has been assembled, check Eq. 6.10.7.1.1-1: For STRENGTH I: For STRENGTH III: n n M + 1 S φ M u xt n 1 Mu + Sxt = 9,797 kip t + 0 = 9,797 kip t φ M = 1.0(1,) = 1, kip t 9,797 kip t < 1, kip t ok (Ratio = 0.741) ( ) 1 1.06 (,441) Mu + Sxt =,654 kip t + =,794 kip t 1 φ M = 1.0(1,) = 1, kip t,794 kip t < 1, kip t ok (Ratio = 0.87) For STRENGTH IV: For STRENGTH V: 1 Mu + Sxt = 4,89 kip t + 0 = 4,89 kip t φ M = 1.0(1,) = 1, kip t n 4,89 kip t < 1, kip t ok (Ratio = 0.4) Eq. (6.10.7.1.1-1) Page 68 o 1 May, 004

LRFD rd Edition 10..4.. Shear (6.10.6.) n ( ) 1 1 0.587 (,441) Mu + Sxt = 8,9 kip t + = 8,4 kip t 1 φ M = 1.0(1,) = 1, kip t 8,4 kip t < 1, kip t ok (Ratio = 0.68) Article 6.10.6. reers to the provisions o Article 6.10.9 to determine the nominal lexural resistance at the strength limit state. At the strength limit state, webs must satisy the ollowing: V φ V u v n Eq. (6.10.9.1-1) where: φ v = resistance actor or shear = 1.0 (Article 6.5.4.) V n = nominal shear resistance determined as speciied in Articles 6.10.9. and 6.10.9. or unstiened and stiened webs, respectively V u = shear in the web at the section under consideration due to the actored loads. A low chart or determining the shear resistance o I-sections is shown in Figure C6.10.9.1-1. The total design shears due to the actored loads, V u, at each tenth point along the interior girder or the STRENGTH I load combination are plotted in Figure 16. The STRENGTH I load combination controls or shear by inspection, and the total actored shears in the interior girder are larger under the STRENGTH I load combination. The η actor is again taken equal to 1.0 in this example at the strength limit state. Live-load shears are taken as the shear envelope values. A sample calculation o V u at the abutment is given below: u [ ] V = 1.0 1.5(87 + 1) + 1.5(1) + 1.75(19) = 88 kips The required spacing o transverse stieners in Field Section 1 will now be determined. First, determine the nominal shear resistance o an unstiened web according to the provisions o Article 6.10.9.. According to Article 6.10.9., the nominal shear resistance o an unstiened web is limited to the shear buckling resistance, V cr, determined as: V V CV n = cr = p Eq. (6.10.9.-1) C is the ratio o the shear buckling resistance to the shear yield strength determined rom Eq. 6.10.9..-4, 6.10.9..-5 or 6.10.9..-6, as applicable, with the shear buckling coeicient, k, taken equal to 5.0. Since, Ek 9,000(5.00) D 69.0 1.40 =1.40 =75.4< = =18.0 F 50 t 0.5 yw ( 18.0) V p is the plastic shear orce determined as ollows: 1.57 Ek C = D Fyw t w 1.57 9,000(5.0) C= = 0 50.9 V p yw w w Eq. (6.10.9..-6) = 0.58F Dt Eq. (6.10.9.-) Vp = 0.58(50)(69.0)(0.5) = 1,001 kips May, 004 Page 69 o 1

LRFD rd Edition Figure 15: Design Shears Due to the Factored Loads - STRENGTH I Shears shown are or the interior girder and are in kips Thereore, Vn = Vcr = 0.9(1,001) = 9 kips φ V = 1.0(9) = 9 kips v n The maximum value o V u in Field Section 1 is 88 kips (Figure 16), which exceeds φ v V n = 9 kips. Thereore, transverse stieners are required in Field Section 1 and the provisions o Article 6.10.9. apply. End Panel (Article 6.10.9..) According to Article 6.10.9.., the nominal shear resistance o a web end panel is limited to the shear buckling resistance, V cr, determined as: V V CV n = cr = p Eq. (6.10.9..-1) C is the ratio o the shear buckling resistance to the shear yield strength rom Eq. 6.10.9..-4, 6.10.9..- 5 or 6.10.9..-6, as applicable. First, compute the shear buckling coeicient, k. According to Article 6.10.9.., the transverse stiener spacing or end panels is not to exceed 1.5D = 1.5(69.0) = 10.5 inches. Assume the spacing rom the abutment to the irst transverse stiener is d o = 7.5 eet = 87.0 inches. Since, yw 5 k= 5+ = 8.15 87.0 69.0 Ek 9,000(8.15) D 69.0 1.40 = 1.40 = 96.< = = 18.0 F 50 t 0.5 w Page 70 o 1 May, 004

LRFD rd Edition 1.57 9,000(8.15) C= = 0 50.90 ( 18.0) V p = 0.58F Dt Eq. (6.10.9..-) Vp = 0.58(50)(69.0)(0.5) = 1,001 kips Thereore, Vn = Vcr = 0.90(1,001) = 90 kips Interior Panels (Article 6.10.9..) φ vvn = 1.0(90) = 90 kips > Vu = 88 kips yw w ok (Ratio = 0.995) According to Article 6.10.9.1, the transverse stiener spacing or interior panels without a longitudinal stiener is not to exceed D = (69.0) = 07.0 inches. For the irst interior panel to the right o the end panel, assume a transverse stiener spacing o d o = 16.75 eet = 01.0 inches, which is the distance rom the irst transverse stiener to the irst intermediate cross-rame in Span 1 (assume that the cross-rame connection plate serves as a transverse stiener). At the irst transverse stiener located d o = 7.5 eet rom the abutment, V u is equal to 45 kips. For interior panels o both nonhybrid and hybrid members with the section along the entire panel proportioned such that: Dt w ( b t + b t ) c c t t.5 Eq. (6.10.9..-1) The nominal shear resistance is to be taken as the sum o the shear buckling resistance and the postbuckling resistance due to tension-ield action, or: 0.87(1 C) Vn = Vp C+ Eq. (6.10.9..-) d o 1+ D Otherwise, the nominal shear resistance is to be taken as the shear resistance determined rom Eq. 6.10.9..-8. Previous speciications did not permit web panels o hybrid members to develop postbuckling resistance due to tension-ield action. Also, note that previous provisions related to the eects o moment-shear interaction are no longer included in the speciications or reasons discussed in Article C6.10.9... For the interior web panel under consideration: (69.0)(0.5).17.5 16(1.0) 18(0.875) = < [ + ] 5 Thereore: k = 5 + = 5.59 01.0 69.0 Since, Ek 9, 000(5.59) D 69.0 1.40 = 1.40 = 79.7 < = = 18.0 F 50 t 0.5 yw 1.57 9, 000(5.59) C = = 0.67 50 ( 18.0) w May, 004 Page 71 o 1

LRFD rd Edition Thereore, V p = 0.58F Dt Eq. (6.10.9..-) Vp = 0.58(50)(69.0)(0.5) = 1,001 kips 0.87(1 0.67) Vn = 1,001 0.67 + = 475 kips 01.0 1+ 69.0 φ vvn = 1.0(475) = 475 kips > Vu = 45 kips yw w ok (Ratio = 0.76) V u at the irst intermediate cross-rame in Span 1 located 4.0 eet rom the abutment is equal to 50 kips, which is greater than φ v V n = 9 kips or an unstiened web. Thereore, assume a transverse stiener spacing o d o = D = 17.5 eet = 07.0 inches rom the cross rame to the next stiener. Since, Thereore, yw 5 k = 5 + = 5.56 07.0 69.0 Ek 9, 000(5.56) D 69.0 1.40 = 1.40 = 79.5 < = = 18.0 F 50 t 0.5 1.57 9, 000(5.56) C = = 0.66 50 ( 18.0) Vp = 1,001 kips 0.87(1 0.66) Vn = 1,001 0.66 + = 468 kips 07.0 1+ 69.0 φ vvn = 1.0(468) = 468 kips > Vu = 50 kips w ok (Ratio = 0.54) V u at this stiener is equal to 16 kips, which is less than φ v V n = 9 kips or an unstiened web. Thereore, no additional transverse stieners are required at the let end o Field Section 1. At the right end o Field Section 1, V u at the ourth intermediate cross rame located 96.0 eet rom the abutment is equal to 0 kips, which exceeds φ v V n = 9 kips or an unstiened web. Assume a transverse stiener spacing o d o = D = 17.5 eet = 07.0 inches to the let o this cross rame. For this panel: (69.0)(0.5) 1.69.5 16(1.0) 18(1.75) = < [ + ] Thereore, the nominal shear resistance may be taken as the sum o the shear buckling resistance and the postbuckling resistance due to tension-ield action. As determined above or this stiener spacing, φ vvn = 1.0(468) = 468 kips > Vu = 0 kips ok (Ratio = 0.684) V u at this stiener is equal to kips, which is less than φ v V n = 9 kips or an unstiened web. Thereore, no additional transverse stieners are required at the right end o Field Section 1. Page 7 o 1 May, 004

LRFD rd Edition 10.. Exterior Girder Check: Section - 10..1. Strength Limit State (Article 6.10.6) 10..1.1. Flexure (Article 6.10.6.) For composite sections in negative lexure at the strength limit state, Article 6.10.6.. irst asks the Engineer to determine i the web o the section satisies the ollowing noncompact slenderness limit: D t w c < 5.7 E F yc Eq. (6.10.6..-1) where Dc is the depth o the web in compression in the elastic range. For composite sections, Dc is to be determined as speciied in Article D6..1. According to Article D6..1 (Appendix D to LRFD Section 6), or composite sections in negative lexure at the strength limit state, Dc is to be computed or the section consisting o the steel girder plus the longitudinal reinorcement. Thereore, at Section -, Dc is equal to 6.55 inches. Recall that Fyc at Section - is 70 ksi. Thereore, 9,000 5.7 = 116.0 70 (6.55) 10.0 116.0 0.565 = > Thus, Section - is classiied as slender-web section and the provisions o Article 6.10.8 must be used to compute the nominal lexural resistance. Since the speciied minimum yield strengths o the langes do not exceed 70 ksi, the optional provisions o Appendix A (to LRFD Section 6) could have been used to compute the nominal lexural resistance had Eq. 6.10.6..-1 been satisied. In Appendix A, which is applicable to either noncompact web or compact web sections, the nominal lexural resistance is permitted to exceed the moment at irst yield. The provisions o Article 6.10.8 may be used instead or these types o sections, i desired, but at the expense o some economy; in particular, or compact web sections. The potential loss in economy increases with decreasing web slenderness. According to Article 6.10.8.1, or composite sections in negative lexure, the ollowing relationship must be satisied or the discretely braced compression lange at the strength limit state: 1 bu + φf nc Eq. (6.10.8.1.1-1) where: φ = resistance actor or lexure = 1.0 (Article 6.5.4.) F nc = nominal lexural resistance o the compression lange determined as speciied in Article 6.10.8. (i.e. local or lateral torsional buckling resistance, whichever controls). The terms bu and are the same as deined earlier. At the strength limit state, the top (tension) lange is considered to be continuously braced by the composite concrete deck. According to Article 6.10.8.1., continuously braced langes in tension must satisy the ollowing relationship at the strength limit state: φ R F Eq. (6.10.8.1.-1) bu h As discussed in Article C6.10.1.6, any lange lateral bending stresses need not be considered once the lange is continuously braced. Compute the maximum lange lexural stresses at Section - due to the actored loads under the STRENGTH I load combination, calculated without consideration o lange lateral bending. As discussed previously, the η actor is taken equal to 1.0 in this example. Thereore: y May, 004 Page 7 o 1

LRFD rd Edition For STRENGTH I: 1.5( 4,840) 1.5( 690) 1.5( 664) 1.75( 4,040) Top lange: = 1.0 + + + 1 = 54.57 ksi,94,194,194,70 1.5( 4,840) 1.5( 690) 1.5( 664) 1.75( 4,040) Bot. lange: = 1.0 + + + 1 = 55.64 ksi,149,08,08,10 Calculate the nominal lexural resistance, F nc, o the bottom (compression) lange taken as the smaller o the local buckling resistance and the lateral torsional buckling resistance according to Article 6.10.8..1. Bottom Flange Lateral Torsional Buckling Resistance (Article 6.10.8..) For illustration purposes, initially assume an unbraced length, L b, on either side o the interior pier (Section -) equal to 17.0 eet. In both unbraced lengths, there is a lange transition located 15.0 eet rom the pier section (Figure ). According to Article 6.10.8.., or unbraced lengths containing a transition to a smaller section at a distance less than or equal to 0 percent o the unbraced length rom the brace point with the smaller moment, the lateral torsional buckling resistance may be determined assuming the transition to the smaller section does not exist. Based on this assumption, determine the limiting unbraced length, L p : E Lp = 1.0rt Eq. (6.10.8..-4) F where r t is the eective radius o gyration or lateral torsional buckling determined as: r t = b c yc 1 Dt c 1 1+ b t w c c 0 rt = = 5. in. 1 6.55(0.565) 1 1 + 0() 1.0(5.) 9,000 Lp = = 9.04 t 1 70 Eq. (6.10.8..-9) It should be emphasized here that the most economical solution is not usually achieved by limiting the unbraced length to L p in order to reach the maximum lateral torsional buckling resistance (i.e. F max in Figure C6.10.8..1-1). This is especially the case when the moment gradient modiier, C b, (discussed below) is taken equal to 1.0. Determine the limiting unbraced length, L r : where: Fyr = 0.7Fyc Fyw E Lr = π rt Eq. (6.10.8..-5) F Fyr = 0.7(70) = 49.0 ksi < 50 ksi F yr must also not be less than 0.5F yc = 0.5(70) = 5.0 ksi ok. yr ok Page 74 o 1 May, 004

LRFD rd Edition Thereore: π(5.) 9,000 Lr = =.95 t 1 49.0 For this unbraced length, since mid / is less than 1.0 and is not equal to zero, calculate the moment gradient modiier, C b, according to Eq. 6.10.8..-7 as ollows: 1 1 Cb = 1.75 1.05 + 0.. Eq. (6.10.8..-7) is generally taken as the largest compressive stress without consideration o lateral bending due to the actored loads at either end o the unbraced length o the lange under consideration, calculated rom the critical moment envelope value. is always taken as positive. I the stress is zero or tensile in the lange under consideration at both ends o the unbraced length, is to be taken equal to zero (in this case, C b = 1.0 and Eq. 6.10.8..-7 does not apply). For the STRENGTH I load combination, which is assumed to control or this calculation in this example, is equal to the largest compressive stress in the bottom lange at Section - calculated previously to equal 55.64 ksi ( is taken as positive or this calculation). The value o 1 is given by Eq. 6.10.8..-10 as: 1= mid o Eq. (6.10.8..-10) where mid is the stress without consideration o lateral bending due to the actored loads at the middle o the unbraced length. o is the stress without consideration o lateral bending due to the actored loads at the brace point opposite to the one corresponding to. Both mid and o are to be calculated rom the moment envelope value that produces the largest compression at the respective points, or the smallest tension i that point is never in compression, and both are to be taken as positive in compression and negative in tension. Note that Article C6.10.8.. states that or all cases where mid is smaller in magnitude than the average o o and (which is the case here since the lange transition is ignored and the moment envelope is concave in shape), 1 and in Eq. 6.10.8..-7 are always equal to the stresses at the ends o the unbraced length in the lange under consideration. Thus, in such cases, mid does not need to be calculated and Eq. 6.10.8..-10 need not be used to compute 1 (i.e. 1 = o ). The revisions to the deinitions o 1 and in Eq. 6.10.8..-7 along with the introduction o mid were done to remove ambiguities and to address a number o potentially important cases where the prior C b calculations were signiicantly unconservative relative to more reined solutions. To illustrate, Appendix B (to this design example) shows the values o C b calculated rom Eq. 6.10.8..-7 or a number o dierent potential cases. For the unbraced length under consideration in this example, calculate 1 = o assuming the lange transition does not exist. Separate calculations show that the stress at the brace point on the let side o Section - controls or the STRENGTH I load combination. Thereore, STRENGTH I: 1.5(,90) 1.5( 4) 1.5( 1) 1.75(,615) Bot. lange: 1 = o = 1.0 + + + 1 = 1.4 ksi,149,08,08,10 Note that o is taken as positive in compression. 1.4 1.4 Cb = 1.75 1.05 + 0. = 1.5. 55.64 55.64 < ok Determine the hybrid actor, R h. According to the provisions o Article 6.10.1.10.1, the hybrid actor is to be taken as: May, 004 Page 75 o 1

LRFD rd Edition R h ( ) 1 + β ρ ρ = 1 + β Eq. (6.10.1.10.1-1) Dntw where: β = Eq. (6.10.1.10.1-) A and ρ equals the smaller o F yw / n and 1.0. D n is taken as the larger o the distances rom the elastic neutral axis o the cross-section to the inside ace o either lange. For sections where the neutral axis is at the mid-depth o the web, consult Article 6.10.1.10.1. At Section -, D n is equal to 6.55 inches (use D n or the steel section plus the longitudinal reinorcement). A n is equal to the sum o the lange area and the area o any cover plates on the side o the neutral axis corresponding to D n. For composite sections in negative lexure, the area o the longitudinal reinorcement may be included in calculating A n or the top lange (when applicable). At Section -, A n is equal to the area o the bottom lange, or 0() = 40.0 in. Thereore, (6.55)(0.565) β = = 1.08 40.0 For sections where yielding occurs irst in the lange, a cover plate or the longitudinal reinorcement on the side o the neutral axis corresponding to D n, n is taken as the largest o the speciied minimum yield strengths o each component included in the calculation o A n. Otherwise, n is to be taken as the largest o the elastic stresses in the lange, cover plate or longitudinal reinorcement on the side o the neutral axis corresponding to D n at irst yield on the opposite side o the neutral axis. Separate calculations show that yielding occurs irst in the bottom lange at Section -. Thereore, n = 70.0 ksi. Fyw 50.0 ρ = = = 0.714 70.0 n 1 + 1.08 (0.714) (0.714) Rh = = 0.984 1 + (1.08) Determine the web load-shedding actor, R b. According to the provisions o Article 6.10.1.10., since: D t w c n E = 10.0 >λ rw = 5.7 = 116.0 Eqs. (6.10.1.10.-), (6.10.1.10.-4) F a wc c Rb = 1 1. 100 00a λrw + wc t 0 w yc D Eq. (6.10.1.10.-) Dctw where: a wc = Eq. (6.10.1.10.-5) b t c c (6.55)(0.565) awc = = 1.08 0() 1.08 R b = 1 (10.0 116.0) = 0.990 100 + 00(1.08) Since L p = 9.04 eet < L b = 17.0 eet < L r =.95 eet, F L L F C 1 1 R R F R R F yr b p nc = b b h yc b h yc RF h yc Lr L p Eq. (6.10.8..-) Page 76 o 1 May, 004

LRFD rd Edition 49.0 17.0 9.04 Fnc = 1.5 1 1 ( 0.990 )(0.984)(70) = 77.8 ksi > 0.990(0.984)(70) = 68.19 ksi 0.984(70.0).95 9.04 \F nc = 68.19 ksi For values o C b greater than 1.0, Article D6.4.1 (Appendix D to LRFD Section 6) allows the maximum lateral torsional buckling resistance, F nc = F max = R b R h F yc, to be reached at larger unbraced lengths. However, since F max is already reached at L b = 17.0 eet in this case, it is not necessary to utilize these provisions. A lateral torsional buckling resistance o 68.19 ksi is not required or this particular unbraced length. Thereore, try a larger unbraced length o L b = 0.0 eet on either side o Section -. In this case, the lange transition is now located at a distance greater than 0 percent o the unbraced length rom the brace point with the smaller moment. Thereore, according to Article 6.10.8.., the lateral torsional buckling resistance is to be taken as the smallest resistance within the unbraced length under consideration. This resistance is to be compared to the largest value o the compressive stress due to the actored loads, bu, throughout the unbraced length calculated using the actual properties at each section. Note also that the moment gradient modiier, C b, should be taken equal to 1.0, and L b should not be modiied by an elastic eective length actor when this approximate procedure is used. Calculate the elastic section properties o the smaller section at the lange transition: Elastic Section Properties: Flange transition Steel Section Component A d Ad Ad I o I Top Flange 1" x 18" 18.00 5.00 60.0,050 1.50,05 Web 9 16" x 69" 8.81 15,99 15,99 Bottom Flange 1" x 0" 0.00 5.00 700.0 4,500 1.67 4,50 76.81 70.00 61,95 0.91(70.00) = 6.70 70.00 I = 61,889 in. 4 NA ds = = 0.91 in. 76.81 dtop OF STEEL = 5.50 + 0.91= 6.41 in. dbot OF STEEL = 5.50 0.91= 4.59 in. 61,889 61,889 STOP OF STEEL = = 1,700 in. SBOT OF STEEL = = 1,789 in. 6.41 4.59 Steel Section + Long. Reinorcement/ Component A d Ad Ad I o I Steel Section 76.81 70.00 61,95 Long. Reinorcement/.10 4.6 1. 5,64 5,64 79.91 6.0 67,587 0.78(6.0) = 48.5 6.0 I = 67,58 in. 4 NA drein/ = = 0.78 in. 79.91 dtop OF STEEL = 5.50 0.78 = 4.7 in. dbot OF STEEL = 5.50 + 0.78 = 6.8 in. 67,58 67,58 STOP OF STEEL = = 1,945 in. SBOT OF STEEL = = 1,86 in. 4.7 6.8 May, 004 Page 77 o 1

LRFD rd Edition Steel Section + Long. Reinorcement Component A d Ad Ad I o I Steel Section 76.81 70.00 61,95 Long. Reinorcement 9.0 4.6 96.5 16,901 16,901 86.11 6.5 78,854.79(6.5) = 1,7 6.5 I = 77,617 in. 4 NA drein = =.79 in. 86.11 dtop OF STEEL = 5.50.79 = 1.71 in. dbot OF STEEL = 5.50 +.79 = 9.9 in. 77,617 77,617 STOP OF STEEL = =,448 in. SBOT OF STEEL = = 1,975 in. 1.71 9.9 Composite Section; n=4 Component A d Ad Ad I o I Steel Section 76.81 70.00 61,95 Concrete Slab 9" x 100.5"/ 4 7.69 4.50 1,60 68,078 54.4 68, 114.5 1,5 10,85 1,5 dn = = 1.8 in. 114.5 1.8(1,5) = 0,498 I NA = 109,787 in. 4 dtop OF STEEL = 5.50 1.8 =.1 in. dbot OF STEEL = 5.50 + 1.8 = 48.88 in. 109,787 109,787 STOP OF STEEL = = 4,96 in. SBOT OF STEEL = =,46 in..1 48.88 Composite Section; n=8 Component A d Ad Ad I o I Steel Section 76.81 70.00 61,95 Concrete Slab 9" x 100.5"/ 8 11.1 4.50 4,807 04,87 76. 05,050 189.9 4,77 67,00 4.94(4,77) = 118,141 4,77 I = 148,86 in. 4 NA dn = = 4.94 in. 189.9 dtop OF STEEL = 5.50 4.94 = 10.56 in. dbot OF STEEL = 5.50 + 4.94 = 60.44 in. 148,86 148,86 STOP OF STEEL = = 14,097 in. SBOT OF STEEL = =,46 in. 10.56 60.44 Calculate F nc using the smaller section at the transition: 0 rt = = 4.95 in. 1 8.9(0.565) 1 1 + 0(1) 1.0(4.95) 9,000 Lp = = 8.40 t 1 70 π(4.95) 9,000 Lr = = 1.5 t 1 49.0 Page 78 o 1 May, 004

LRFD rd Edition Determine R h : Determine R b : D n = 8.9 in. A n = 0(1) = 0.0 in. n = 70.0 ksi (8.9)(0.565) β = =.154 0.0 ρ = 50.0/70.0 = 0.714 1 +.154 (0.714) (0.714) Rh = = 0.971 1 + (.154) Dc (8.9) = = 16.1>λ rw = 116.0 t 0.565 w (8.9)(0.565) a wc = =.154 0.0.154 R b = 1 ( 16.1 116.0) = 0.977 100 + 00(.154) Since L p = 8.40 eet < L b = 0.0 eet < L r = 1.5 eet, 49.0 0.0 8.40 Fnc = 1.0 1 1 ( 0.977 )(0.971)(70) = 57.11 ksi < 0.977(0.971)(70) = 66.41 ksi 0.971(70.0) 1.5 8.40 \ F nc = 57.11 ksi Obviously, there is a signiicant discontinuity (reduction) in the predicted lateral torsional buckling resistance when a lange transition is moved beyond 0.L b rom the brace point with the smaller moment, and the preceding approximate procedure is applied to determine the LTB resistance o the stepped lange. A more rigorous approximate solution or determining the LTB resistance or this unbraced length is presented or consideration in Appendix C (to this design example). However, the results rom this procedure are not utilized in this example. Local Buckling Resistance (Article 6.10.8..) Calculate the local buckling resistance o the bottom lange at Section -. Determine the slenderness ratio o the lange: b c λ = Eq. (6.10.8..-) tc 0 λ = = 5.0 Determine the limiting slenderness ratio or a compact lange (alternatively, see Table C6.10.8..-1): ( ) E λ p = 0.8 Eq. (6.10.8..-4) F yc Since λ < λ p, 9,000 λ p = 0.8 = 7.7 70 F = R R F nc b h yc Eq. (6.10.8..-1) May, 004 Page 79 o 1

LRFD rd Edition F nc = (0.990)(0.984)(70.0) = 68.19 ksi Calculate the local buckling resistance o the bottom lange in the smaller section at the lange transition. Determine the slenderness ratio o the lange: 0 λ = = 10.0 1 Since λ > λ p, determine the limiting slenderness ratio or a noncompact lange as ollows: () E λ r = 0.56 Eq. (6.10.8..-5) F yr Fyr = 0.7Fyc Fyw Fyr = 0.7(70) = 49.0 ksi < 50.0 ksi F yr must also not be less than 0.5F yc = 0.5(70) = 5.0 ksi ok. Thereore: And: 9,000 λ r = 0.56 = 1.6 49.0 F F 1 λ λ 1 R R F ok yr p nc = b h yc RF h yc λr λp Eq. (6.10.8..-) 49.0 10.0 7.7 Fnc = 1 1 (0.977)(0.971)(70.0) = 59.6 ksi 0.971(70.0) 1.6 7.7 At Section - and at the lange transition, F nc is governed by the lateral torsional buckling resistance o 57.11ksi, which is less than the local buckling resistance o 68.19 ksi at Section - and 59.6 ksi at the lange transition. Thereore, F nc = 57.11 ksi at both locations. Stress Check As speciied in Article 6.10.1.6, or design checks where the lexural resistance is based on lateral torsional buckling, bu is to be determined as the largest value o the compressive stress throughout the unbraced length in the lange under consideration, calculated without consideration o lange lateral bending. For design checks where the lexural resistance is based on yielding, lange local buckling or web bend-buckling, bu may be determined as the stress at the section under consideration. Thereore, For STRENGTH I: Section - Top lange: Bot. lange: Flange transition (Span 1) bu = 54.57 ksi (computed earlier) = -55.64 ksi (computed earlier) Top lange: 1.5(,656) 1.5( 7) 1.5( 58) 1.75(,709) bu = 1.0 + + + 1 = 5.86 ksi 1,700 1,945 1,945, 448 Bot. lange: 1.5(,656) 1.5( 7) 1.5( 58) 1.75(,709) = 1.0 + + + 1 = 57.54 ksi 1,789 1,86 1,86 1,975 \ Bot. lange: bu = -57.54 ksi Page 80 o 1 May, 004

LRFD rd Edition For STRENGTH III: Section - 1.5( 4,840) 1.5( 690) 1.5( 664) Top lange: bu = 1.0 + + 1 = 1.66 ksi,94,194,194 1.5( 4,840) 1.5( 690) 1.5( 664) Bot. lange: = 1.0 + + 1 = 0.01 ksi,149,08,08 Flange transition (Span ) 1.5(,718) 1.5( 78) 1.5( 64) Top lange: bu = 1.0 + + 1 = 0.7 ksi 1,700 1,945 1,945 Bot. lange: 1.5(,718) 1.5( 78) 1.5( 64) = 1.0 + + 1 = 9.5 ksi 1, 789 1,86 1,86 \ Bot. lange: bu = -0.01 ksi For STRENGTH IV: Section - Top lange: Bot. lange: Flange transition (Span ) Top lange: Bot. lange: 1.5( 4,840) 1.5( 690) 1.5( 664) bu = 1.0 + + 1 = 7.4 ksi,94,194,194 1.5( 4,840) 1.5( 690) 1.5( 664) = 1.0 + + 1 = 5.6 ksi,149, 08, 08 1.5(,718) 1.5( 78) 1.5( 64) bu = 1.0 + + 1 = 5.65 ksi 1,700 1,945 1,945 1.5(,718) 1.5( 78) 1.5( 64) = 1.0 + + 1 = 4.5 ksi 1, 789 1,86 1,86 \ Bot. lange: bu = -5.6 ksi For STRENGTH V: Section - Top lange: Bot. lange: Flange transition (Span 1) Top lange: Bot. lange: 1.5( 4,840) 1.5( 690) 1.5( 664) 1.5( 4,040) bu = 1.0 + + + 1 = 49. ksi,94,194,194,70 1.5( 4,840) 1.5( 690) 1.5( 664) 1.5( 4,040) = 1.0 + + + 1 = 49.78 ksi,149,08,08,10 1.5(,656) 1.5( 7) 1.5( 58) 1.5(,709) bu = 1.0 + + + 1 = 47.55 ksi 1,700 1,945 1,945, 448 1.5(,656) 1.5( 7) 1.5( 58) 1.5(,709) = 1.0 + + + 1 = 50.96 ksi 1,789 1,86 1,86 1,975 \ Bot. lange: bu = -50.96 ksi May, 004 Page 81 o 1

LRFD rd Edition In this example, lateral bending in the bottom lange due to wind-load eects is considered at the strength limit state. For simplicity in this example, the largest value o within the unbraced length will conservatively be used in all design checks. is to be taken as positive in sign. Eqs. C4.6..7.1-1 and C4.6..7.1-, presented earlier, are again used to compute the actored wind orce per unit length, W, applied to the bottom lange, and the maximum lange lateral bending moment due to the actored wind load, M w, within the unbraced length, respectively. Again, the wind load acting on the live load (WL) is assumed transmitted directly to the deck and is thereore not considered in the STRENGTH V load combination in this example. The overturning eect o WL on the wheel loads is also not considered. According to Article 6.10.1.6, lateral bending stresses determined rom a irst-order analysis may be used in discretely braced compression langes or which: CR Eq. (6.10.1.6-) b b Lb 1.Lp bm F yc bm is the largest value o the compressive stress due to the actored loads throughout the unbraced length in the lange under consideration, calculated without consideration o lange lateral bending. In this case, use bm = bu = -50.96 ksi, as computed earlier or the STRENGTH V load combination (which is the controlling load case with wind included or this particular computation). Thereore: 1.0(0.977) 1.( 8.40) 11.68 t Lb 0.0 t 50.96 70 = < = Because the preceding equation is not satisied, Article 6.10.1.6 requires that second-order elastic compression-lange lateral bending stresses be determined. The second-order compression-lange lateral bending stresses may be determined by ampliying irst-order values (i.e. 1 ) as ollows (assuming an elastic eective length actor or lateral torsional buckling equal to 1.0, which should not be modiied since the lange is stepped within this unbraced length): 0.85 1 F cr = 1 1 bm or: = (AF) 1 1 Eq. (6.10.1.6-4) where AF is the ampliication actor and F cr is the elastic lateral torsional buckling stress or the lange under consideration speciied in Article 6.10.8.. determined as the smallest resistance within the unbraced length as: CRπ E = Lb r b b cr F 1.0(0.977) π (9,000) cr F = = 119.0 ksi 0.0(1) 4.95 t Eq. (6.10.8..-8) Note again that the calculated value o F cr or use in Eq. 6.10.1.6-4 is not limited to R b R h F yc as speciied in Article 6.10.8... The ampliication actor is then determined as ollows: Page 8 o 1 May, 004

LRFD rd Edition For STRENGTH III: 0.85 AF = = 1.14 > 1.0 ok 0.01 1 119.0 For STRENGTH V: 0.85 AF = = 1.49 > 1.0 ok 50.96 1 119.0 Note that irst- or second-order lange lateral bending stresses, as applicable, are limited to a maximum value o 0.6F y according to Eq. 6.10.1.6-1. The largest section within the unbraced length will be conservatively used to compute W, and the smallest bottom lange will conservatively be used to compute. Thereore, For STRENGTH I: Wind loads: For STRENGTH III: Wind loads: M 9.04(1) ( ) Not considered 1.0(1.4)(0.05)(.0 + 69.0 +.0) /1 W = = 0.6 kips / t 0.6(0.0) Mw = = 9.04 kip t 10 w = = = 1.6 ksi * AF = 1.6(1.14) = 1.86 ksi < 0.6F y = 4.0 ksi S 1.0 0 6 For STRENGTH IV: Wind loads: For STRENGTH V: Wind loads: M.56(1) ( ) Not considered 1.0(0.4)(0.05)(.0 + 69.0 +.0) /1 W = = 0.064 kips / t 0.064(0.0) Mw = =.56 kip t 10 w = = = 0.46 ksi * AF = 0.46(1.49) = 0.69 ksi < 0.6F y = 4.0 ksi S 1.0 0 6 Now that all the required inormation has been assembled, check Eqs. 6.10.8.1.1-1 and 6.10.8.1.-1, as applicable: Bottom Flange bu 1 + φf nc Eq. (6.10.8.1.1-1) ok ok May, 004 Page 8 o 1

LRFD rd Edition For STRENGTH I: For STRENGTH III: For STRENGTH IV: For STRENGTH V: Top Flange For STRENGTH I: Section -: Flange transition: For STRENGTH III: Section -: Flange transition: 1 bu + = -57.54 ksi + 0 = 57.54 ksi φ F = 1.0(57.11) = 57.11 ksi nc 57.54 ksi > 57.11 ksi say ok (Ratio = 1.008) bu 1 1 bu + = 0.01 ksi + ( 1.86) = 0.6 ksi φ F = 1.0(57.11) = 57.11 ksi nc 0.6 ksi < 57.11 ksi ok (Ratio = 0.56) 1 bu + = 5.6 ksi + 0 = 5.6 ksi φ F = 1.0(57.11) = 57.11 ksi nc 5.6 ksi < 57.11 ksi ok (Ratio = 0.617) 1 1 bu + = 50.96 ksi + ( 0.69) = 51.19 ksi φ F = 1.0(57.11) = 57.11 ksi nc 51.19 ksi < 57.11 ksi ok (Ratio = 0.896) = 54.57 ksi h y φ R F Eq. (6.10.8.1.-1) bu h φ R F = 1.0(0.984)(70.00) = 68.88 ksi 54.57 ksi < 68.88 ksi ok (Ratio = 0.79) bu = 5.86 ksi φ R F = 1.0(0.971)(70.00) = 67.97 ksi h y 5.86 ksi < 67.97 ksi ok (Ratio = 0.778) bu = 1.66 ksi φ R F = 1.0(0.984)(70.00) = 68.88 ksi h y 1.66 ksi < 68.88 ksi ok (Ratio = 0.460) bu = 0.7 ksi φ R F = 1.0(0.971)(70.00) = 67.97 ksi h y 0.7 ksi < 67.97 ksi ok (Ratio = 0.445) y Page 84 o 1 May, 004

LRFD rd Edition For STRENGTH IV: Section -: Flange transition: For STRENGTH V: Section -: bu = 7.4 ksi φ R F = 1.0(0.984)(70.00) = 68.88 ksi h y 7.4 ksi < 68.88 ksi ok (Ratio = 0.541) bu = 5.65 ksi φ R F = 1.0(0.971)(70.00) = 67.97 ksi h y 5.65 ksi < 67.97 ksi ok (Ratio = 0.54) bu = 49. ksi φ R F = 1.0(0.984)(70.00) = 68.88 ksi h y 49. ksi < 68.88 ksi ok (Ratio = 0.716) Flange transition: bu = 47.55ksi φ R F = 1.0(0.971)(70.00) = 67.97 ksi h y 47.55 ksi < 67.97 ksi ok (Ratio = 0.700) Finally, it should be noted that or continuous span lexural members that satisy the requirements o Article B6. to ensure adequate robustness and ductility o the pier sections, a calculated percentage o the negative moment due to the actored loads at the pier section under consideration may be redistributed prior to making the preceding checks (Article 6.10.6..). The moments may be redistributed using the optional procedures o Appendix B (to LRFD Section 6 -- speciically, Articles B6.4 or B6.6). When the redistribution moments are calculated according to these procedures, the lexural resistances at the strength limit state within the unbraced lengths immediately adjacent to interior-pier sections satisying the requirements o Article B6. need not be checked. At all other locations, the provisions o Articles 6.10.7, 6.10.8.1 or A6.1, as applicable, must be satisied ater redistribution. 10..1.. Shear (6.10.6.) Article 6.10.6. reers to the provisions o Article 6.10.9 to determine the nominal lexural resistance at the strength limit state. Separate calculations similar to those shown previously or the interior panels in Field Section 1 are used to determine the spacing o the transverse stieners in the interior panels o Field Section, and will not be repeated here. The resulting stiener spacings are shown on the girder elevation in Figure. Note that although larger spacings could have been used in each panel in Field Section, the stieners in each panel were located midway between the cross-rame connection plates in each panel or practical reasons in order to help simpliy the detailing. 10... Service Limit State (Article 6.10.4) 10...1. Permanent Deormations (Article 6.10.4.) Article 6.10.4. contains criteria intended to control objectionable permanent deormations due to expected severe traic loadings that would impair rideability. As speciied in Article 6.10.4..1, these checks are to be made under the SERVICE II load combination speciied in Table.4.1-1. These criteria were discussed previously under the service limit state checks or Section 1-1. For members with shear connectors provided throughout their entire length that also satisy the provisions o Article 6.10.1.7, Article 6.10.4..1 permits the concrete deck to also be considered eective or negative lexure when computing lexural stresses acting on the composite section at the service limit state. Earlier calculations were made to ensure that the minimum longitudinal reinorcement satisied the May, 004 Page 85 o 1

LRFD rd Edition provisions o Article 6.10.1.7 or both the actored construction loads and the SERVICE II loads. Thereore, the SERVICE II lexural stresses will be computed assuming the concrete deck to be eective or loads applied to the composite section. Determine R h : Section -: D n = 54.97 in. (conservatively use the short-term composite section) A n = 0() = 40.0 in. n = 70.0 ksi (54.97)(0.565) β = = 1.546 40.0 ρ = 50.0/70.0 = 0.714 1 + 1.546 (0.714) (0.714) R h = = 0.977 1 + (1.546) Flange transition: D n = 59.44 in. (conservatively use the short-term composite section) A n = 0(1) = 0.0 in. n = 70.0 ksi (59.44)(0.565) β = =.44 0.0 ρ = 50.0/70.0 = 0.714 1 +.44 (0.714) (0.714) R h = = 0.960 1 + (.44) Check the lange stresses due to the SERVICE II loads at Section - and at the lange transition within the unbraced length in Span 1 adjacent to Section -. η is speciied to always equal 1.0 at the service limit state (Article 1..). For the example bridge, is taken equal to zero at the service limit state: For SERVICE II: Section - Top lange: Bot. lange: Flange transition Top lange: Bot. lange: Bottom Flange Section -: 1.0( 4,840) 1.0( 690 + 664) 1.( 4,040) = 1.0 + + 1= 6.96 ksi,94 6,148 1, 77 1.0( 4,840) 1.0( 690 + 664) 1.( 4,040) = 1.0 + + 1= 9. ksi,149,594,875 1.0(,656) 1.0( 7+ 58) 1.(,709) = 1.0 + + 1 =.51 ksi 1,700 4,96 14,097 1.0(,656) 1.0( 7+ 58) 1.(,709) = 1.0 + + 1 = 8.88 ksi 1,789,46,46 h y + 0.95R F h y 0.95R F = 0.95(0.977)(70) = 64.97 ksi 9. ksi + 0 < 64.97 ksi ok (Ratio = 0.604) Eq. (6.10.4..-) Page 86 o 1 May, 004

LRFD rd Edition Top Flange Flange Transition: Section -: Flange Transition: 0.95R F = 0.95(0.960)(70) = 6.84 ksi h y -8.88 ksi+0<6.84 ksi ok (Ratio = 0.609) h y 0.95R hfy 0.95R F = 0.95(0.977)(70) = 64.97 ksi 6.96 ksi < 64.97 ksi ok (Ratio = 0.415) 0.95R F = 0.95(0.960)(70) = 6.84 ksi h y.51 ksi < 6.84 ksi ok (Ratio = 0.68) crw Eq. (6.10.4..-1) Under the load combinations speciied in Table.4.1-1, Eqs. 6.10.4..-1 and 6.10.4..- do not control and need not be checked or composite sections in negative lexure or which the nominal lexural resistance at the strength limit state is determined according to the provisions o Article 6.10.8 (see Article C6.10.4..). Nevertheless, the checks are illustrated above or the sake o completeness. Web bend buckling must always be checked, however, at the service limit state under the SERVICE II load combination or composite sections in negative lexure according to Eq. 6.10.4..-4 as ollows: c F Eq. (6.10.4..-4) where c is the compression-lange stress at the section under consideration due to the SERVICE II loads calculated without consideration o lange lateral bending, and F crw is the nominal bend-buckling resistance or webs determined as speciied in Article 6.10.1.9. Determine the nominal web bend-buckling resistance at Section - and at the lange transition within the unbraced length in Span 1 adjacent to Section - according to the provisions o Article 6.10.1.9.1 as ollows: but not to exceed the smaller o R h F yc and F yw /0.7, where: F 0.9Ek = D t crw w ( D D) c Eq. (6.10.1.9.1-1) 9 k = Eq. (6.10.1.9.1-) According to Article D6..1 (Appendix D to LRFD Section 6), or composite sections in negative lexure at the service limit state where the concrete deck is considered eective in tension or computing lexural stresses on the composite section, the depth o the web in compression in the elastic range, D c, is to be computed rom Eq. D6..1-1 as ollows: = c 0 Eq. (D6..1-1) c Dc d t c + t where t is the sum o the various tension-lange stresses caused by the actored loads, calculated without considering lange lateral bending, and d is the depth o the steel section. Eq. D6..1-1 recognizes the beneicial eect o the dead-load stress on the location o the neutral axis o the composite section (including the concrete deck) in regions o negative lexure. Thereore, Section -: ( 9.) Dc = 7.0.0 = 41.7 in. > 0 ok 9. + 6.96 May, 004 Page 87 o 1

LRFD rd Edition 9 k= = 5. ( 41.7 69.0) 0.9(9,000)(5.) F = =4.71 ksi<min(r F, F 0.7)=R F =0.977(70)=68.9 ksi ok crw h yc yw h yc 69.0 0.565 9. < 4.71 ksi ok (Ratio = 0.898) Flange transition: ( 8.88) Dc = 71.0 1.0 = 4.5 in. > 0 ok 8.88 +.51 9 k = =.9 ( 4.5 69.0) 0.9(9,000)(.9) F = =9.7 ksi<min(r F, F 0.7)=R F =0.960(70)=67.0 ksi ok crw h yc yw h yc 69.0 0.565-8.88 <9.7 ksi ok (Ratio = 0.979) 10... Fatigue And Fracture Limit State (Article 6.10.5) 10...1. Load Induced Fatigue (Article 6.6.1.) Fatigue o the base metal at the connection-plate weld to the top (tension) lange at the intermediate crossrame in Span 1, located 0.0 eet to the let o Section -, will be checked or the FATIGUE load combination (Table.4.1-1). The stress range due to the atigue live load modiied by the corresponding dynamic load allowance o 15 percent will be used to make this check. The lateral distribution actors or the atigue limit state, computed earlier, are also used. From earlier computations, the (ADTT) SL was calculated to be 1,600 trucks/day. The provisions o Article 6.6.1. apply only to details subject to a net applied tensile stress, which by inspection is the case at this particular detail. As stated previously, the concrete deck is assumed eective in computing all stresses and stress ranges applied to the composite section in the atigue calculations. Thus, the stress range γ( ) at the connectionplate weld to the top lange due to the actored atigue load (actored by the speciied 0.75 load actor) at the cross-rame under consideration is computed using the properties o the short-term composite section as: 0.75(4)(9.56)(1) 0.75 86 (9.56)(1) γ ( ) = + = 0.675 ksi 148,86 148,86 Determine the atigue detail category rom Table 6.6.1..-1. Under the condition o illet-welded connections with welds normal to the direction o stress, the atigue detail category or base metal at the toe o transverse stiener-to-lange welds is Category C. According to Eq. 6.6.1..-1, γ( ) must not exceed the nominal atigue resistance ( F) n. Both the resistance actor φ and design actor η are speciied to be 1.0 at the atigue limit state (Article C6.6.1..). From Eq. 6.6.1..5-1, the nominal atigue resistance is determined as: Page 88 o 1 May, 004

LRFD rd Edition 1 ( F) = ( ) n A 1 N F Eq. (6.6.1..5-1) TH Table C6.6.1..5-1 shows the values o (ADTT) SL or each atigue detail category above which the atigue resistance is governed by one-hal the constant amplitude atigue threshold, ( F) TH. (such that the detail will theoretically provide ininite atigue lie). The values in the table assume a 75-year design lie and one stress cycle n per truck passage. For continuous spans with span lengths greater than 40.0 eet, the number o stress cycles per truck passage n is equal to 1.0 at sections away rom the pier (Table 6.6.1..5-). Sections away rom the pier are deined as sections greater than a distance o one-tenth the span on each side o the interior support. For other values o n, Table C6.6.1..5-1 should be modiied by dividing the values in the table by n. Thereore, rom Table C6.6.1..5-1, the 75-year (ADTT) SL equivalent to ininite atigue lie or a Category C detail is 745 trucks per day < 1,600 trucks per day. Thereore, For a Category C detail, ( F) TH 1 ( F) ( F) = n T H = 1.0 ksi (Table 6.6.1..5-). Thereore: 1 F = 1.0 = 6.00 ksi n ( ) ( ) ( ) F ( ) n γ 0.675 ksi < 6.00 ksi ok (Ratio = 0.11) 10... Special Fatigue Requirement or Webs (Article 6.10.5.) cr p Eq. (6.6.1..-1) As discussed previously, interior panels o stiened webs must satisy Eq. 6.10.5.-1 in order to control elastic lexing o the web so that the member is assumed able to sustain an ininite number o smaller loadings without atigue cracking due to this eect. where: Vu V Eq. (6.10.5.-1) V u = shear in the web at the section under consideration due to the unactored permanent loads plus the actored atigue load V cr = shear buckling resistance determined rom Eq. 6.10.9..-1 In this check, the actored atigue load is to be taken as twice that calculated using the FATIGUE load combination (Table.4.1-1), with the atigue live load taken as speciied in Article.6.1.4. Again, the atigue live load is modiied by the dynamic load allowance o 15 percent and the lateral distribution actors or the atigue limit state are used. In this example, the panel adjacent to Section - will be checked. The transverse stiener spacing in this panel is d o = 10.0 eet (Figure ). The shear at Section - to be used in this check is computed as ollows: Vu = 159 + + + (0.75)( 56) = 88 kips The shear buckling resistance o the 10-inch-long panel is determined as: V = V = CV n cr Eq. (6.10.9..-1) C is the ratio o the shear buckling resistance to the shear yield strength determined rom Eq. 6.10.9..-4, 6.10.9..-5 or 6.10.9..-6, as applicable. First, compute the shear buckling coeicient, k May, 004 Page 89 o 1

LRFD rd Edition Since, yw 5 k= 5+ do D 5 k= 5+ = 6.65 10.0 69.0 Ek 9,000(6.65) D 69.0 1.40 = 1.40 = 86.9 < = = 1.7 F 50 t 0.565 ( 1.7) V p is the plastic shear orce determined as ollows: 1.57 Ek C = D Fyw t w 1.57 9,000(6.65) C = = 0.40 50 V p yw w w Eq. (6.10.9..-7) Eq. (6.10.9..-6) = 0.58F Dt Eq. (6.10.9..-) Vp = 0.58(50)(69.0)(0.565) = 1,16 kips Thereore, Vcr = 0.40(1,16) = 45 kips > Vu = 88 kips ok (Ratio = 0.66) 10..4. Constructibility (Article 6.10.) 10..4.1. Flexure (Article 6.10..) In regions o negative lexure, Eqs. 6.10...1-1, 6.10...1- and 6.10...-1 speciied in Article 6.10.., to be checked or critical stages o construction, generally do not control because the sizes o the langes in these regions are normally governed by the sum o the actored dead and live load stresses at the strength limit state. Also, the maximum accumulated negative moments rom the deck-placement analysis in these regions, plus the negative moments due to the steel weight, typically do not dier signiicantly rom the calculated DC 1 negative moments. The deck-overhang loads do introduce lateral bending stresses into the langes in these regions, which can be calculated and used to check the above equations in a manner similar to that illustrated previously or Section 1-1. Wind load, when considered or the construction case, also introduces lateral bending into the langes. When applying Eqs. 6.10...1-1, 6.10...1- and 6.10...-1 in these regions, the bottom lange would be the discretely braced compression lange and the top lange would be the discretely braced tension lange or all constructibility checks to be made beore the concrete deck has hardened or is made composite. The nominal lexural resistance o the bottom lange, F nc, would be calculated in a manner similar to that demonstrated above or Section - at the strength limit state. However, or loads applied beore the deck has hardened or is made composite, F nc would be computed ignoring any contribution rom the longitudinal reinorcement. For the sake o brevity in this example, the application o Eqs. 6.10...1-1, 6.10...1- and 6.10...-1 to the construction case or the unbraced lengths adjacent to Section - will not be shown. Web Bend-Buckling For critical stages o construction, web bend-buckling should always be checked in regions o negative lexure according to Eq. 6.10...1- as ollows: Page 90 o 1 May, 004

LRFD rd Edition φ F rw Eq. (6.10...1-) bu c where bu is the compression-lange stress at the section under consideration due to the actored loads calculated without consideration o lange lateral bending, and F crw is the nominal bend-buckling resistance or webs determined as speciied in Article 6.10.1.9. In this example, check Eq. 6.10...1- or the noncomposite section at Section - and at the lange transition within the unbraced length in Span adjacent to Section -. By inspection, the STRENGTH IV load combination governs this check. The sum o the accumulated unactored negative moments during the deck casts plus the unactored moment due to the steel weight is 4,918 kip-eet (versus the unactored DC 1 moment o 4,840 kip-eet) at Section -, and,796 kip-eet (versus the unactored DC 1 moment o,718 kip-eet) at the lange transition (Table 4). For STRENGTH IV: Section - 1.5( 4,918) Bot. lange: bu = 1.0 1 = 8.11 ksi,149 Flange transition (Span ) Bot. lange: 1.5(,796) bu = 1.0 1 = 8.1 ksi 1, 789 Table 4: Moments rom Deck-Placement Analysis Unactored Dead-Load Moments (kip-t) Span -> 1 Pier Span -> Length (t) 10.0 15.0 140.0 0.0 15.0 0.0 4.0 47.0 74.0 87.5 Steel Weight -1-414 -777-777 -48-1 9 67 64 88 SIP Forms (SIP) -18-18 -9-9 -185-14 5 1 118 19 Cast 1 17-10 -90-90 -90-90 -90-90 -90-90 -1684-1754 -1965-1965 -108-79 69 91 1911 05-17 -7-917 -917-15 -48 79 79 79 79 Sum o Casts + SIP -1777-1 -4141-4141 -68-1850 -17 01 178 151 Max. +M 4 0 0 0 0 0 5 01 178 151 DC + DW -549-71 -154-154 -74-564 69 179 551 597 Deck, haunches + SIP -1709-4 -406-406 -90-177 64 8 1459 1594 Determine the nominal elastic web bend-buckling resistance according to the provisions o Article 6.10.1.9.1 as ollows: but not to exceed the smaller o R h F yc and F yw /0.7, where: F 0.9Ek = D t crw w ( D D) c Eq. (6.10.1.9.1-1) 9 k = Eq. (6.10.1.9.1-) At Section -, D c or the steel section is equal to.6 inches. At the lange transition, D c or the steel section is equal to.59 inches. From separate calculations, R h or the steel section is equal to 0.98 at Section - and 0.970 at the lange transition. Thereore, May, 004 Page 91 o 1

LRFD rd Edition Section -: 9 k= = 8.7 (.6 69.0) 0.9(9,000)(8.7) F = =67.1 ksi<min(r F, F 0.7)=R F =0.98(70)=68.81 ksi ok crw h yc yw h yc 69.0 0.565 φ F = 1.0(67.1) = 67.1 ksi crw 8.11 < 67.1 ksi ok (Ratio = 0.419) Flange transition: 9 k= = 8.0 (.59 69.0) 0.9(9,000)(8.0) F = =65.91 ksi<min(r F, F 0.7)=R F =0.970(70)=67.90 ksi ok crw h yc yw h yc 69.0 0.565 φ F = 1.0(65.91) = 65.91 ksi crw 10..4.. Shear (Article 6.10..) 8.1 < 65.91 ksi ok (Ratio = 0.47) For critical stages o construction, Article 6.10.. requires that interior panels o stiened webs satisy the ollowing requirement: V φ V u v cr Eq. (6.10..-1) where: φ v = resistance actor or shear = 1.0 (Article 6.5.4.) V u = shear in the web at the section under consideration due to the actored permanent loads and actored construction loads applied to the noncomposite section V cr = shear buckling resistance determined rom Eq. 6.10.9..-1 In this example, the panel adjacent to Section - will be checked. The transverse stiener spacing in this panel is d o = 10.0 eet (Figure ). Since shear is rarely increased signiicantly due to deck staging, the actored DC 1 shear at Section - will be used in this check (the STRENGTH IV load combination governs by inspection): ( Vu ) = 1.0(1.5)( 159) = 9 kips DC 1 The shear buckling resistance o this 10-inch panel was previously determined to be V cr = 45 kips. Thereore, 10.4. Shear Connector Design (Article 6.10.10) φ V = 1.0(45) = 45 kips v cr 9 kips < 45 kips ok (Ratio = 0.58) Shear connectors are designed according to the provisions o Article 6.10.10. According to Article 6.10.10.1, continuous composite bridges should normally be provided with shear connectors throughout the entire length o the bridge. In regions o negative lexure, shear connectors must be provided where the longitudinal reinorcement is considered to be a part o the composite section. Both stud and channel shear connectors are permitted in Article 6.10.10.1.1. Stud shear connectors will be utilized in this example. Page 9 o 1 May, 004

LRFD rd Edition 10.4.1. Stud Proportions Terminating the studs at approximately the mid-thickness o the concrete deck will place them well within the limits or cover and penetration speciied in Article 6.10.10.1.4 and will also clear the reinorcing steel. Thereore, 9.0 (.5 0.875) 7.15 in. + = Use 7/8 x 7 studs. Check that the ratio o the height to the diameter is not less than 4.0, as required in Article 6.10.10.1.1. 10.4.. Pitch (Article 6.10.10.1.) h 7.0 = = 8.0 > 4.0 ok d 0.875 According to the provisions o Article 6.10.10.1., the pitch o the shear connectors along the longitudinal axis o the girder is to be initially determined to satisy the atigue limit state. The resulting number o shear connectors is then to be checked against the number required to satisy the strength limit state. 10.4.. Fatigue Limit State As speciied in Article 6.10.10.1., the pitch, p, o the shear connectors must satisy the ollowing: where: p nz V r Eq. (6.10.10.1.-1) sr n = number o shear connectors in a cross-section Z r = shear atigue resistance o an individual shear connector determined as speciied in Article 6.10.10. V sr = horizontal atigue shear range per unit length V sr is to be computed as ollows: where: V VQ I sr = Eq. (6.10.10.1.-) V = vertical shear orce range under the FATIGUE load combination speciied in Table.4.1-1 with the atigue live load taken as speciied in Article.6.1.4 Q = irst moment o the transormed short-term area o the concrete deck about the neutral axis o the short-term composite section I = moment o inertia o the short-term composite section The parameters I and Q should be determined using the deck within the eective lange width. Article C6.10.10.1. does permit I and Q in regions o negative lexure to be determined using the longitudinal reinorcement within the eective lange width, unless the concrete deck is considered to be eective in tension or negative lexure in calculating the range o longitudinal stress, as permitted in Article 6.6.1..1. Since the minimum required one-percent longitudinal reinorcement is provided in the deck according to the provisions o Article 6.10.1.7, the concrete deck is considered to be eective in tension or negative lexure when computing longitudinal stress ranges in this example. Thereore, I and Q must be determined using the short-term area o the concrete deck along the entire girder. The atigue resistance o an individual shear connector, Z r, is given in Article 6.10.10. as: Z 5.5d r = αd Eq. (6.10.10.-1) α = 4.5 4.8 Log N Eq. (6.10.10.-) May, 004 Page 9 o 1

LRFD rd Edition where: d = diameter o the stud N = number o cycles speciied in Article 6.6.1..5 determined as: N = (65)(75)n(ADTT) SL n = number o stress cycles per truck passage taken rom Table 6.6.1..5- Eq. (6.6.1..5-) From earlier computations, the (ADTT) SL was calculated to be 1,600 trucks/day. For continuous spans with span lengths greater than 40.0 eet, n is equal to 1.0 at sections away rom the pier and 1.5 at sections near the pier (Table 6.6.1..5-). Sections near the pier are deined as sections within a distance o onetenth the span on each side o the interior support. Thereore, N = (65)(75)1.0(1,600) = 4,800,000 cycles (away rom the pier) 5.5 α = 4.5 4.8 Log(4,800,000) = 1.79 < =.75 N = (65)(75)1.5(1, 600) = 65, 700, 000 cycles (near the pier) 5.5 α = 4.5 4.8 Log(65,700,000) = 1.04 < =.75 ( ) 5.5 0.875 \ us e Zr = =.11 kips (at all locations) The number o shear connectors in a cross-section, n, will be assumed equal to three (). Requirements or the transverse spacing o shear connectors across the top lange are given in Article 6.10.10.1.. The vertical shear orce range, V, is determined or the actored atigue load (actored by the 0.75 load actor speciied or the FATIGUE load combination) including the speciied dynamic load allowance o 15 percent. The lateral distribution actors or shear or the atigue limit state are used. A sample calculation o Q or the transormed short-term area o the concrete deck about the neutral axis o the short-term composite section at the interior-pier section (Section -) is given below: Q = (9.0 x 100.5/8.0)(.0) =,491 in. Based on Eqs. 6.10.10.1.-1 and 6.10.10.1.-, Table 5 summarizes the required stud pitch along the girder to satisy the atigue limit state. As speciied in Article 6.10.10.1., the pitch must not be less than six stud diameters = 6(0.875) = 5.5 inches or more than 4.0 inches. Table 5: Required Stud Shear Connector Pitch Fatigue Limit State Length (t) V (kips) Q (in ) I (in 4 ) V sr (kips/in.) p (in./row) 0.0 4.8 1,766 18,081 0.590 10.7 14.0 7.5 1,766 18,081 0.517 1. 8.0. 1,766 18,081 0.445 14. 4.0 1.5 1,766 18,081 0.44 14.6 56.0.,104 161,518 0.41 15.0 70.0.,104 161,518 0.41 15.0 84.0.8,104 161,518 0.440 14.4 98.0 6.0,104 161,518 0.469 1.5 11.0 7.5 1,985 148,86 0.500 1.7 16.0 9.8,491 0,760 0.449 14.1 140.0 45.0,491 0,760 0.508 1.5 157.5 4.0 1,985 148,86 0.560 11. 175.0 8. 1,985 148,86 0.511 1.4 19.5 5.,018 151,50 0.470 1.5 10.0 4.5,018 151,50 0.459 1.8 7.5 4.5,018 151,50 0.46 1.7 Page 94 o 1 May, 004

LRFD rd Edition 10.4.4. Strength Limit State (Article 6.10.10.4) The resulting number o shear connectors will now be checked against the number required to satisy the strength limit state. According to Article 6.10.10.4.1, the actored shear resistance o a single shear connector, Q r, at the strength limit state is to be taken as: Q = φ r sc Q n Eq. (6.10.10.4.1-1) where: φ v = resistance actor or shear connectors = 0.85 (Article 6.5.4.) Q n = nominal shear resistance o a single shear connector determined as speciied in Article 6.10.10.4. As speciied in Article 6.10.10.4., the nominal shear resistance o one stud shear connector embedded in a concrete deck is to be taken as: where: Q = 0.5A E A sc F u Eq. (6.10.10.4.-1) ' n sc c c A sc = cross-sectional area o a stud shear connector E c = modulus o elasticity o the deck concrete determined as speciied in Article 5.4..4 (=,644 ksi or this example as determined previously) F u = speciied minimum tensile strength o a stud shear connector as speciied in Article 6.4.4 = 60.0 ksi π ( ) Asc = 0.875 = 0.60 in. 4 A F = (0.60)(60.0) = 6.00 kips sc u Qn = 0.5(0.60) 4.0(,644) = 6. kips > 6.00 kips \ Q n = 6.00 kips Q r = 0.85(6.00) = 0.60 kips At the strength limit state, the minimum number o shear connectors, n, over the region under consideration is to be taken as: P n = Eq. (6.10.10.4.1-) Q where P is the total nominal shear orce determined as speciied in Article 6.10.10.4.. According to Article 6.10.10.4., or continuous spans that are composite or negative lexure in the inal condition, the total nominal shear orce, P, between the point o maximum positive design live load plus impact moment and an adjacent end o the member is to be determined as: P = P p Eq. (6.10.10.4.-1) where P p is the total longitudinal shear orce in the concrete deck at the point o maximum positive live load plus impact moment taken as the lesser o: ' P 1p =0.85cbsts where b s and t s are the eective width and thickness o the concrete deck, respectively. r Eq. (6.10.10.4.-) or: P p =Fyw Dt w +Fyt btt t +Fyc bct c Eq. (6.10.10.4.-) The point o maximum positive live load plus impact moment in Span 1 is located 60. eet rom the abutment. May, 004 Page 95 o 1

LRFD rd Edition P1p = 0.85(4.0)(100.0)(9.0) =,060 kips For the steel section yielding the smallest orce in this region: P p = (50.0)(69.0)(0.5) + (50.0)(18.0)(0.875) + (50.0)(16.0)(1.0) =,1 kips \ P = P p = P 1p =,060 kips P,060 n= = = 100 studs Q 0.60 r Compute the required pitch, p, in this region at the strength limit state with studs per row: No. o rows = 100 4 rows = 60.(1) p = = 1.9 in. ( 4 1) The total nominal shear orce, P, between the point o maximum positive design live load plus impact moment and the centerline o an adjacent interior support is to be determined as: P= P + P Eq. (6.10.10.4.-4) p where P n is the total longitudinal shear orce in the concrete deck over an interior support taken as the lesser o: P1n = Fyw Dtw + Fyt bttt + Fyc bctc n Eq. (6.10.10.4.-5) ' or: P = 0.45 b t Eq. (6.10.10.4.-6) n c s s Eq. 6.10.10.4.-6 is a conservative approximation o the tension orce in the concrete deck to account or the combined contribution o both the longitudinal reinorcement and also the concrete that remains eective in tension based on its modulus o rupture. A more precise value may be substituted, i desired. The distance between the point o maximum positive live load plus impact moment in Span 1 and the adjacent interior support is (140.0-60.) = 79.8 eet. For the steel section and eective concrete deck yielding the smallest orces in this region: P 1n = (50.0)(69.0)(0.5) + (50.0)(18.0)(1.75) + (50.0)(16.0)(1.0) =,76 kips Pn = 0.45(4.0)(100.0)(9.0) = 1,60 kips \ P n = 1,60 kips P = P p + P n =,060 + 1,60 = 4,680 kips P 4,680 n = = = 15 studs Q 0.60 r Compute the required pitch, p, in this region at the strength limit state with studs per row: No. o rows = 15 51 rows = 79.8(1) p = = 19. in. ( 51 1) The distance between the point o maximum positive live load plus impact moment in Span and each o the adjacent interior supports is 87.5 eet. Using calculations similar to the above: Page 96 o 1 May, 004

LRFD rd Edition P p =,060 kips P n = 1,60 kips P = 4,680 kips n = 15 studs No. o rows = 51 rows p = 1.0 in. The inal recommended pitches are governed by the atigue limit state and are shown in Figure 17. The eective width o the concrete deck is larger or the interior girders, which in conjunction with dierent atigue shear ranges, may result in slightly dierent recommended pitches. However, or practical purposes, unless the dierences are deemed signiicant, it is recommended that the same pitches be used on all the girders. 10.5. Exterior Girder: Field Section 1 Figure 16: Shear Connector Pitch 10.5.1. Transverse Intermediate Stiener Design (Article 6.10.11.1) Intermediate transverse stieners are designed according to the provisions o Article 6.10.11.1. In this example, each intermediate transverse stiener consists o a plate welded to one side o the web. The distance between the end o the web-to-stiener weld and the near edge o an adjacent web-to-lange or longitudinal stiener-to-web weld must not be less than 4t w or more than 6t w. Stieners not used as connection plates must be tight it at the compression lange (and are generally illet welded to the lange), but need not be in bearing with the tension lange. However, it should be noted that the aorementioned Guidelines recommend calling or a tight it o these stieners to the tension lange. Stieners used as connecting plates or cross-rames or diaphragms must be connected by welding or bolting to both langes. Welded connections are generally cheaper. Also, as noted earlier, a Category C' detail still exists at the termination o the connection-plate weld to the web just above (or below) the tension lange even when the stieners are bolted to that lange. The design o the intermediate transverse stieners or Field Section 1 (not serving as cross-rame connection plates) will be illustrated in this example. The same size stieners will be used within the ield section or practical purposes. Grade 50W steel will be used or the stieners (i.e. F ys = 50.0 ksi). 10.5.1.1. Projecting Width (Article 6.10.11.1.) Size the stiener width, b t, to be greater than or equal to b /4 as required in Eq. 6.10.11.1.-. b is to be taken as the ull width o the widest compression lange within the ield section under consideration in order to ensure a minimum stiener width that will help restrain local buckling o the widest compression lange. 16.0 bt = 4.0 in. Eq. (6.10.11.1.-) 4 Stieners are commonly made up o less expensive lat bar stock. Flat bars are generally produced in whole-inch width increments and 1/8-in. thickness increments in Customary U.S. units. May, 004 Page 97 o 1

LRFD rd Edition Check that: Use b t = 5.0 in. > 4.0 in. ok d bt.0+ Eq. (6.10.11.1.-1) 0 (1.75 + 69.0 + 1.0).0 + = 4.4 in. < 5.0 in. ok 0 Try a stiener thickness, t p, o 0.5 inches. The Guidelines recommend a minimum thickness o 7/16 or stieners and connection plates, with a minimum thickness o 1/ preerred. Check that: 16tp bt Eq. (6.10.11.1.-) 16(0.5) = 8.0 in. > 5.0 in. Finally, it should be noted that according to Article 6.10.11.1.1, or webs which satisy: D.5 t w E F yw J ok Eq. (6.10.11.1.1-1) stieners used as connection plates need only satisy the projecting width requirements o Article 6.10.11.1.. 10.5.1.. Moment o Inertia (Article 6.10.11.1.) The moment o inertia requirement ensures that the transverse stiener has suicient rigidity to maintain a vertical line o near zero lateral delection along the line o the stiener in order or interior web panels to adequately develop the shear buckling resistance. In this particular example, separate calculations show that the 07-inch panels govern the required moment o inertia or the intermediate transverse stieners in Field Section 1. The moment o inertia o the stiener, I t, must satisy: where: I d t t o w Eq. (6.10.11.1.-1) d o = transverse stiener spacing I t = moment o inertia o the transverse stiener taken about the axis in contact with the web or single stieners and about the mid-thickness o the web or stiener pairs J = required ratio o the rigidity o one transverse stiener to that o the web plate 1 10.5.1.. Area (Article 6.10.11.1.4) o D J=.5.0 0.5 d 69.0 J =.5.0 = 1.7 < 0.5 07.0 \ J = 0.5 d t J = 07.0(0.5) (0.5) = 1.94 in. o w 4 4 I t = (0.5)(5.0) = 0.8 in. > 1.94 in. 4 ok Eq. (6.10.11.1.-) For web panels required to develop a tension ield, the transverse stieners need suicient area to resist the vertical component o the tension ield and must also not be subject to local buckling. This includes the irst stiener in an end panel i it adjacent to a tension-ield shear panel. The local buckling resistance o a transverse stiener is combined with the area requirement o the stiener as ollows: Page 98 o 1 May, 004

LRFD rd Edition D V F As 0.15B ( 1 C) 18 t t u yw w w vv φ n Fc rs Eq. (6.10.11.1.4-1) where: φ v = resistance actor or shear = 1.0 (Article 6.5.4.) B = 1.0 or stiener pairs; 1.8 or single angle stieners;.4 or single plate stieners C = ratio o the shear buckling resistance to the shear yield strength determined rom Eq. 6.10.9..-4, 6.10.9..-5 or 6.10.9..-6, as applicable V n = nominal shear resistance determined as speciied in Articles 6.10.9. and 6.10.9. or unstiened and stiened webs, respectively V u = shear due to the actored loads at the strength limit state F crs = elastic local buckling stress or the stiener F crs 0.1E = F ys Eq. (6.10.11.1.4-) b t t p 0.1(9,000) Fcrs = = 89.9 ksi> F ys = 50.0 ksi 5.0 0.5 \ F crs = 50.0 ksi In this particular example, separate calculations show that the 01-inch panel adjacent to the end panel governs the required area or the intermediate transverse stieners in Field Section 1. From earlier calculations or this panel: Thereore, C = 0.67 V u = 45 kips φ v V n = 475 kips 69.0 45 50.0 As 0.15(.4) (1 0.67) 18 0.5 =.11 in. 0.5 475 50.0 A = 5.0(0.5) =.50 in. >.11 in. s ( ) Note that in some cases, Eq. 6.10.11.1.4-1 can give a negative result, which indicates that the web alone is suicient to resist the tension-ield load. In these cases, the required area is zero and the stieners need only be proportioned to satisy the moment o inertia and projecting width requirements. The area requirement need not be investigated or web panels not required to develop a tension ield. 10.6. Exterior Girder: Abutment 1 10.6.1. Bearing Stiener Design (Article 6.10.11.) Bearing stieners are designed as columns to resist the reactions at bearing locations. According to Article 6.10.11..1, bearing stieners must be placed on the webs o built-up sections at all bearing locations. At bearing locations on rolled shapes and at other locations on built-up sections or rolled shapes subjected to concentrated loads, where the loads are not transmitted through a deck or deck system, either bearing stieners must be provided or else the web must be investigated or the limit states o web crippling or web local yielding according to the provisions o Article D6.5 (Appendix D to LRFD Section 6). It should be noted that the provisions o Article D6.5 should be checked whenever girders are incrementally launched over supports. ok May, 004 Page 99 o 1

LRFD rd Edition Bearing stieners must extend the ull depth o the web and as closely as practical to the outer edges o the langes. Each stiener is to either be inished-to-bear (allowing the option o milling or grinding) against the lange through which it receives its load and attached with illet welds, or else attached to that lange by a ull penetration groove weld. The Guidelines recommend using inish-to-bear plus illet welds to connect the bearing stieners to the appropriate lange, regardless o whether or not a crossrame or diaphragm is connected to the stieners. Full penetration groove welds are costly and oten result in welding deormation o the lange. The design o the bearing stieners or Abutment 1 will be illustrated in this example. Grade 50W steel will be used or the stieners (i.e. F ys = 50.0 ksi). Assemble the bearing reactions due to the actored loads at Abutment 1. The STRENGTH I load combination controls. r [ ] R = 1.0 1.5(87 + 1) + 1.5(1) + 1.75(19) = 88 kips 10.6.1.1. Projecting Width (Article 6.10.11..) The width, b t, o each projecting stiener element must satisy: E bt 0.48tp Eq. (6.10.11..-1) F Try two 7.0-inch-wide bars welded to each side o the web. Rearranging Eq. 6.10.11..-1 gives: ( ) t ( t p) = min. p min. 7.0 b ys E 0.48 F t = = 0.61 in. 9,000 0.48 50.0 F ys \ Try t p = 5/8 10.6.1.. Bearing Resistance (Article 6.10.11..) According to Article 6.10.11.., the actored resistance or the itted ends o bearing stieners is to be taken as: ( R ) = φ ( R ) sb r b sb where: φ b = resistance actor or bearing = 1.0 (Article 6.5.4.) (R sb ) n = nominal bearing resistance or the itted end o bearing stieners ( ) R = 1.4A sb n pn ys n Eq. (6.10.11..-1) Eq. (6.10.11..-) A pn = area o the projecting elements o the stiener outside o the web-to-lange illet welds but not beyond the edge o the lange Assume or this example that the clip provided at the base o the stieners to clear the web-to-lange illet welds is 1.5 inches in length. Thereore, ( ) A = (7.0 1.5)(0.65) = 6.88 in. pn (R ) = 1.4(6.88)(50.0) = 48 kips sb n R = (1.0)(48) = 48 kips > R = 88 kips sb r r ok Page 100 o 1 May, 004

LRFD rd Edition 10.6.1.. Axial Resistance (Article 6.10.11..4) For computing the axial resistance o bearing stieners that are welded to the web, Article 6.10.11..4b states that a portion o the web is to be included as part o the eective column section. For stieners consisting o two plates welded to the web, the eective column section is to consist o the two stiener elements, plus a centrally located strip o web extending not more than 9t w on each side o the stieners, as shown in Figure 18. As speciied in Article 6.10.11..4a, the radius o gyration o the eective column section is to be computed about the mid-thickness o the web and the eective length is to be taken as 0.75D, where D is the web depth. The area o the eective column section is computed as: s [ ] A = (7.0)(0.65) + 9(0.5)(0.5) = 1.5 in. The moment o inertia o the eective column section is computed as: Figure 16: Eective Column Section or Bearing Stiener Design 0.65(7.0+0.5+7.0) 4 I= s =158.8in. 1 The radius o gyration o the eective column section is thereore computed as: I 158.8 s r= s = =.46 in. As 1.5 The eective length o the eective column section is computed as: Kl=0.75D=0.75(69.0)=51.75 in. Check the limiting slenderness ratio o 10 speciied or main members in compression in Article 6.9.: K 51.75 = = 15.0 < 10 r.46 s As speciied in Article 6.10.11..4a, calculate the actored axial resistance, P r, o the eective column section according to the provisions o Article 6.9..1 using the speciied minimum yield strength o the stieners, F ys. P = φ P r c n ok Eq. (6.9..1-1) where: φ c = resistance actor or axial compression, steel only = 0.9 (Article 6.5.4.) P n = nominal compressive resistance determined (or this case) as speciied in Article 6.9.4.1 May, 004 Page 101 o 1

LRFD rd Edition Calculate λ. K λ = rπ s F ys E Eq. (6.9.4.1-) 51.75 50 λ = = 0.09.46π 9,000 Since λ <.5, then: P 0.66 λ F A Eq. (6.9.4.1-1) 10.6.1.4. Bearing Stiener-to-Web Welds r n = ys s 0.09 Pn = 0.66 (50.0)(1.5) = 65 kips P = 0.9(65) = 587 kips > R = 88 kips As speciied in Article 6.1...4b, the resistance o illet welds in shear which are made with matched or undermatched weld metal is to be taken as the product o the eective area o the weld and the actored resistance o the weld metal. For a illet weld, the eective area is deined in Article 6.1.. as the eective weld length multiplied by the eective throat. The eective throat is the shortest distance rom the root o the joint to the ace o the illet weld (equal to 0.707 times the weld leg size or welds with equal leg sizes). As speciied in Article 6.1..5, the eective length o a illet weld is to be at least our times its nominal size, or 1½ inches, whichever is greater. As speciied in Article 6.1..1, matching weld metal (i.e. with the same or slightly higher minimum speciied minimum yield and tensile strength compared to the minimum speciied properties o the base metal) is generally to be used or illet welds. Undermatched weld metal may be speciied by the Engineer or illet welds when the welding procedure and weld metal are selected to ensure sound welds, and is encouraged or illet welds connecting steels with speciied minimum yield strengths greater than 50.0 ksi. For ASTM A 709 Grade 50W steel, the speciied minimum tensile strength is 70.0 ksi (Table 6.4.1-1). Thus, assume the classiication strength o the weld metal is also 70.0 ksi. The classiication strength o the weld metal is expressed as EXX, where the letters XX stand or the minimum strength level o the electrode in ksi. According to Table 6.1..4-1, the minimum size illet weld is ¼ inch when the base metal thickness (T) o the thicker part joined is less than ¾ inches. The actored shear resistance o the weld metal is taken as: R = 0.6φ F r e r exx ok Eq. (6.1...4b-1) where: φ e = resistance actor or shear in the throat o the weld metal = 0.8 (Article 6.5.4.) F exx = classiication strength o the weld metal = 70.0 ksi in this case R r = 0.6(0.80)(70.0) =.6 ksi The resistance o a ¼ inch illet weld in shear in kips/inch is then computed as: v =.6(0.707)(0.5) = 5.94 kips /in. The total length o weld, allowing.5 inches or the clips at the top and bottom o the stiener, is: L = 69.0 (.5) = 64.0 in. The total actored resistance o the our ¼-inch illet welds connecting the stieners to the web is thereore: 4(64.0)(5.94) = 1,51 kips > 88 kips ok Page 10 o 1 May, 004

LRFD rd Edition 10.7. Exterior Girder: Design Example Summary The results or this design example at each limit state are summarized below. The results or each limit state are expressed in terms o a perormance ratio, deined as the ratio o a calculated value to the corresponding resistance. 10.7.1. Positive-Moment Region, Span 1 (Section 1-1) 10.7.1.1. Constructibility (Slender-web section) Flexure (STRENGTH I) Eq. (6.10...1-1) Top lange 0.844 Eq. (6.10...1-) Top lange 0.85 Eq. (6.10...1-) Web bend buckling 0.697 Eq. (6.10...-1) Bottom lange 0.55 Flexure (STRENGTH III Wind load on noncomposite structure) Eq. (6.10...1-1) Top lange 0.61 Eq. (6.10...1-) Top lange 0.170 Eq. (6.10...1-) Web bend buckling 0.085 Eq. (6.10...-1) Bottom lange 0.7 Flexure (STRENGTH IV) Eq. (6.10...1-1) Top lange 0.955 Eq. (6.10...1-) Top lange 0.977 Eq. (6.10...1-) Web bend buckling 0.86 Eq. (6.10...-1) Bottom lange 0.60 Shear (96-0 rom the abutment) (STRENGTH IV) 0.447 10.7.1.. Service Limit State Live-load delection 0.4 Permanent deormations (SERVICE II) Eq. (6.10.4..-1) Top lange 0.469 Eq. (6.10.4..-) Bottom lange 0.775 10.7.1.. Fatigue and Fracture Limit State Base metal at connection plate weld to bottom lange 0.99 (7-0 rom the abutment) Stud shear connector weld to top lange (100-0 rom the abutment) 0.08 Special atigue requirement or webs (shear - 7 - rom the abutment) 0.618 10.7.1.4. Strength Limit State (Compact Section) Ductility requirement 0.4 Flexure Eq. (6.10.7.1.1-1) (STRENGTH I) 0.741 Flexure Eq. (6.10.7.1.1-1) (STRENGTH III) 0.87 Flexure Eq. (6.10.7.1.1-1) (STRENGTH IV) 0.4 Flexure Eq. (6.10.7.1.1-1) (STRENGTH V) 0.68 Shear (End panel) (STRENGTH I) 0.995 May, 004 Page 10 o 1

LRFD rd Edition 10.7.. Interior-Pier Section (Section -) 10.7..1. Strength Limit State (Slender-web section) Flexure (STRENGTH I) Eq. (6.10.8.1.1-1) Bottom lange 1.008 Eq. (6.10.8.1.-1) Top lange @ Section - 0.79 Eq. (6.10.8.1.-1) Top lange @ Flange transition 0.778 Flexure (STRENGTH III) Eq. (6.10.8.1.1-1) Bottom lange 0.56 Eq. (6.10.8.1.-1) Top lange @ Section - 0.460 Eq. (6.10.8.1.-1) Top lange @ Flange transition 0.445 Flexure (STRENGTH IV) Eq. (6.10.8.1.1-1) Bottom lange 0.617 Eq. (6.10.8.1.-1) Top lange @ Section - 0.541 Eq. (6.10.8.1.-1) Top lange @ Flange transition 0.54 Flexure (STRENGTH V) Eq. (6.10.8.1.1-1) Bottom lange 0.896 Eq. (6.10.8.1.-1) Top lange @ Section - 0.716 Eq. (6.10.8.1.-1) Top lange @ Flange transition 0.700 10.7... Service Limit State Permanent deormations (SERVICE II) Eq. (6.10.4..-1) Top lange @ Section - 0.415 Eq. (6.10.4..-1) Top lange @ Flange transition 0.68 Eq. (6.10.4..-) Bottom lange @ Section - 0.604 Eq. (6.10.4..-) Bottom lange @ Flange transition 0.609 Eq. (6.10.4..-4) Web bend buckling @ Section - 0.898 Eq. (6.10.4..-4) Web bend buckling @ Flange transition 0.979 10.7... Fatigue and Fracture Limit State Base metal at connection plate weld to top lange 0.11 (0-0 to the let o the interior pier) Special atigue requirement or webs (shear at interior pier) 0.66 10.7..4. Constructibility (Slender-web section) Flexure (STRENGTH IV) Eq. (6.10...1-) Web bend buckling @ Section - 0.419 Eq. (6.10...1-) Web bend buckling @ Flange transition 0.47 Shear (at interior pier) (STRENGH IV) 0.58 Page 104 o 1 May, 004

LRFD rd Edition Appendix A: Elastic Eective Length Factor or Lateral Torsional Buckling By Proessor Donald W. White, Georgia Institute o Technology Michael A. Grubb, P.E., BSDI, Ltd. The equations or determining the nominal lateral torsional buckling (LTB) resistance o the compression lange in Articles 6.10.8.. and A6.. (Appendix A to LRFD Section 6) assume an elastic eective length actor o K = 1.0 or the critical unbraced length. When adjacent unbraced lengths are less critically loaded, substantial restraint can exist at the ends o a critical unbraced length such that K may be less than 1.0 or the critical length. Should the unbraced length under consideration end up being the critical unbraced length or which K is less than 1.0, the lower value o K can then subsequently be used to appropriately increase the elastic LTB resistance o the compression lange, F cr. A lower value o F cr will in turn result in a lower value o the ampliication actor (speciied in Article 6.10.1.6) that may be applied to calculated irst-order compression-lange lateral bending stresses within the unbraced length, should they exist. The unbraced length, L b, also can be modiied by the eective length actor K < 1 to determine a larger nominal LTB resistance or the compression lange within the critical unbraced length. Article C6.10.8.. reers to Galambos (1988) and Nethercot and Trahair (1976) or a practical design procedure or determining elastic eective length actors associated with LTB, applicable or the case where a member is continuous with adjacent unbraced lengths. The procedure is based on the analogy between the buckling o a continuous beam and the buckling o an end-restrained column. As such, the alignment chart or nonsway columns given in the AISC LRFD Speciications (1999) can be used to determine the eective length actor or the critical unbraced length. The procedure is conservative because the moment-envelope values in adjacent unbraced lengths are assumed to be the concurrent loadings associated with LTB o the critical unbraced length. The application o this procedure is demonstrated or the unbraced length in Span 1 o the example bridge containing Section 1-1. This unbraced length is in a region o positive lexure and spans between the cross-rames located 48.0 eet and 7.0 eet rom the abutment. Thereore, L b is equal to 4.0 eet. The LTB resistance o the top (compression) lange o the noncomposite section is computed or this unbraced length in the example in order to check the top lange or the construction condition. This unbraced length is identiied herein as Segment M. The equal 4-oot-long unbraced lengths immediately to the let and to the right o Segment M (Figure ) are identiied as Segments L and R, respectively. STEP 1: Determine the moment gradient modiier, C b, or each segment. Segment L: Segment L contains a bottom-lange transition 4.0 eet rom the abutment (Figure ). Since the transition is located at a distance greater than 0 percent o the unbraced length rom the brace point with the smaller moment, C b is taken equal to 1.0 (as recommended in Article C6.10.8..) 1. Segment M: Since mid / > 1 within this segment, C b must be taken equal to 1.0 according to the provisions o Article 6.10.8... Segment R: Since the member is prismatic within Segment R and since mid / is less than 1.0 and is not equal to zero, calculate the moment gradient modiier, C b, according to Eq. 6.10.8..- 7 as ollows: 1 The procedure outlined in Appendix C (to this design example) may be used to obtain a more precise estimate o the LTB resistance o unbraced lengths with stepped langes. May, 004 A1

LRFD rd Edition = + 1 1 Cb 1.75 1.05 0.. o Eq. (6.10.8..-7) is generally taken as the largest compressive stress without consideration o lateral bending due to the actored loads at either end o the unbraced length o the lange under consideration, calculated rom the critical moment envelope value. is always taken as positive or zero. I the stress is zero or tensile in the lange under consideration at both ends o the unbraced length, is taken equal to zero (in this case, C b = 1 and Eq. 6.10.8..-7 does not apply). The value o 1 is given by Eq. 6.10.8..-10 as: = 1 mid Eq. (6.10.8..-10) where mid is the stress without consideration o lateral bending due to the actored loads at the middle o the unbraced length. o is the stress without consideration o lateral bending due to the actored loads at the brace point opposite to the one corresponding to. Both mid and o are to be calculated rom the moment envelope value that produces the largest compression at the respective points, or the smallest tension i that point is never in compression, and both are to be taken as positive in compression and negative in tension. In this particular example, the STRENGTH IV load combination governs the constructibility check. The stresses below are computed rom the results o the deck-placement analysis (Table 1): For STRENGTH IV: Top lange: STEP : Identiy the critical segment. 1.0(1.5)(,706)(1) = = 0.81 ksi 1,581 1.0(1.5)(, 7)(1) mid = = 5.88 ksi 1,581 1.0(1.5)(1,585)(1) o = = 18.05 ksi 1,581 = (5.88) (0.81) = 0.95 ksi > = 18.05 ksi = 0.95 ksi 1 o 0.95 0.95 Cb = 1.75 1.05 + 0. = 1.17. 0.81 0.81 < ok The critical segment is deined as the segment that buckles elastically at the smallest multiple o the design loadings based on the largest moment envelope value within each segment, and with F cr calculated using the actual unbraced lengths L b as the eective lengths. The multiple o the design loadings associated with the buckling o the critical segment is denoted as γ m, and the multiples o the design loadings associated with the buckling o the adjacent segments (should they exist) are denoted as γ rl and γ rr, respectively. For all o these segments, the ollowing equation applies: F cr γ = (A1) where bu is the largest value o the compressive stress throughout the unbraced length in the lange under consideration and F cr is the elastic LTB stress or the lange speciied in Article 6.10.8.. determined as: bu CRπ E = Lb r b b cr F t 1 Eq. (6.10.8..-8) A May, 004

LRFD rd Edition For checking constructibility, the web load-shedding actor, R b, is to be taken equal to 1.0 (Article 6.10.1.10.) since web bend buckling is prevented during construction by a separate limit state check. The eective radius o gyration or LTB, r t, is taken as the value within the unbraced length that produces the smallest buckling resistance. Thereore, Segment L: Segment M: Segment R: Separate calculations show that bu is controlled by the section at the lange transition and F cr is controlled by the larger section within the segment (r t is smaller). Thereore, bu 1.0(1.5)(,677)(1) = =.45 ksi 1,485 1.0(1.0) π (9,000) cr F bu = = 5.49 ksi 4.0(1).90 5.49 γ = γ rl = = 1.6.45 1.0(1.5)(,889)(1) = =.89 ksi 1,581 Fcr = 5.49 ksi 5.49 γ = γ m = = 1.60.89 bu (governs) 1.0(1.5)(,706)(1) = = 0.81 ksi 1,581 1.17(1.0) π (9,000) cr F = = 61.41 ksi 4.0(1).90 61.41 γ = γ rr = = 1.99 0.81 STEP : Calculate a stiness ratio, α, or each o the segments. The stiness ratio, α m, or the critical segment is determined as: and or each adjacent segment is determined as: 1 bt + Dt r 6 α m = L c c c w t bcr 1 nbt + Dt r α r = L c c c w t 6 γ m b 1- γ where n = i the ar end o the adjacent segment is continuous, n = i the ar end o the adjacent segment is pinned, and n = 4 i the ar end o the adjacent segment is ixed. These equations are a generalization o the procedures outlined by Nethercot and Trahair (1976) and Galambos (1998) to allow or consideration o the more general case o singly-symmetric I-sections, which are the most common type o section used in steel-bridge construction. I one end o the critical segment is a simply supported r (A) (A) May, 004 A

LRFD rd Edition end, α r = at that end. In this case, the ar ends o the adjacent segments are both continuous; thereore, n = or both segments. Also, or cases involving singly-symmetric I-sections and reverse curvature bending in any one o the above segments, the area (b c t c + D c t w /6) and r t terms are the corresponding values that produce the smallest buckling resistance. Segment L: Segment M: Segment R: 1 16(1) + (8.6)(0.5) (.90) 6 1.60 α r = α rl = 1 = 0.05 4.0(1) 1.6 1 16(1) + (8.6)(0.5) (.90) 6 α m = =.0 4.0(1) 1 16(1) + (8.6)(0.5) (.90) 6 1.60 α r = α rr = 1 = 0.98 4.0(1) 1.99 STEP 4: Determine the stiness ratios, G = α m /α r, or each end o the critical segment. Let end: Right end: α.0 = = = > α 0.05 m G 81. 50.0 rl α.0 = = = α 0.98 m G 5 rr.10 use 50.0 STEP 5: Obtain the eective length actor, K, rom the nonsway restrained column nomograph. From Figure C-C. o the AISC LRFD Speciications (1999), or the sidesway inhibited case: K = 0.96 Thereore, or the critical unbraced length, the elastic lateral torsional buckling resistance may be computed as: * 1 1 F cr =F cr =5.49 =56.96 ksi K ( 0.96 ) an 8.5% increase which will result in a slightly smaller ampliication o the irst-order lateral lange bending stresses in the compression lange within this unbraced length according to Eq. 6.10.1.6-4. O course, the beneit is relatively small in this particular example, but it may be a signiicant beneit in some cases. A slightly smaller unbraced length o KL b can also be used in this case, i desired, to determine the nominal LTB resistance o the compression lange within the critical unbraced length, F nc. Once the eective length actor or the critical segment has been determined, the eective length actor or the adjacent segments should be computed as: K γ = (A4) γ r r * m The sidesway inhibited nomograph shown in AISC (1999) does not label values or G larger than 50. The top o the nomograph actually corresponds to G = ; however, eectively the same results are obtained using G = or G = 50. A4 May, 004

LRFD rd Edition * where γ m is the multiple o the design loadings associated with the buckling o the critical segment based on the reduced K value. For this case, * 56.96 γ m = = 1.7.89 1.6 KrL = = 0.97 1.7 1.99 KrR = = 1.07 1.7 Note that the eective length actor or the adjacent segments may actually exceed 1.0, but these segments are always less critical segments. For all remaining unbraced lengths not adjacent to the critical segments, K should be taken equal to 1.0 or the condition under investigation. The procedure is ocused on a local subassembly composed o the most critical segment and the unbraced lengths adjacent to this segment. The Engineer may assume that more remote unbraced lengths are not aected signiicantly by buckling interaction with the critical segment. Note that the same procedure may also be applied when the optional provisions o Appendix A (to LRFD Section 6 -- Article A6..) are used to compute the nominal LTB resistance, and when Eq. 6.10.1.6-5 is used to compute the ampliication actor or lange lateral bending. Reerences AISC. 1999. Load and Resistance Factor Design Speciication or Structural Steel Buildings and Commentary. American Institute o Steel Construction, Chicago, IL, Third Edition, December 7, 1999. Galambos, T. V., ed. 1998. Guide to Stability Design Criteria or Metal Structures. 5 th ed. Structural Stability Research Council. John Wiley and Sons, Inc., New York, NY. Nethercot, D. A. and N. S. Trahair. 1976. Lateral Buckling Approximations or Elastic Beams. The Structural Engineer, Vol. 54, No. 6, pp. 197-04. May, 004 A5

LRFD rd Edition Unbraced cantilevers and members where mid Otherwise: ( ) ( ) 1 1 Examples: C b=1.75-1.05 +0.. = - 1 mid 0 Appendix B: Moment Gradient Modiier, C b > 1 or = 0: C b = 1 mid / = 0.875 1 / = 0.75 C b = 1.1 1 / = 0.75 C b = 1.40 mid > C b = 1 = 0 C b = 1 mid / = 0.75 1 / = 0.5 C b = 1. mid / = 0.65 1 / = 0.5 C b = 1.51 1 / = -0.75 C b =.19 Note: The above examples assume that the member is prismatic within the unbraced length, or the transition to a smaller section is within 0.L b rom the braced point with the lower moment. Otherwise, use C b = 1. B1 May, 004

LRFD rd Edition Appendix C: Lateral Torsional Buckling Resistance o Stepped Flanges By Proessor Donald W. White, Georgia Institute o Technology Michael A. Grubb, P.E. BSDI, Ltd. As speciied in Article 6.10.8.., or unbraced lengths containing a transition to a smaller section at a distance less than or equal to 0 percent o the unbraced length rom the brace point with the smaller moment, the lateral torsional buckling (LTB) resistance may be determined assuming the transition to the smaller section does not exist. For a case with more than one lange transition, any transition located within 0 percent o the unbraced length rom the brace point with the smaller moment may be ignored and the LTB resistance o the remaining nonprismatic unbraced length may then be computed as the smallest resistance based on the remaining sections. When all lange transitions are located at a distance greater than 0 percent o the unbraced length rom the brace point with the smaller moment, the LTB resistance is to be taken as the smallest resistance within the unbraced length under consideration. This resistance is to be compared to the largest value o the compressive stress due to the actored loads, bu, throughout the unbraced length calculated using the actual properties at each section. Note also that the moment gradient modiier, C b, should be taken equal to 1.0, and L b should not be modiied by an elastic eective length actor when this approximate procedure is used. As illustrated in the design example (i.e. in the design checks or Section -), this approximate procedure typically results in a signiicant discontinuity (reduction) in the predicted LTB resistance when a lange transition is moved beyond 0.L b rom the brace point with the smaller moment. In this particular example, an increase in the unbraced length, L b, adjacent to the interior pier rom 17.0 eet to 0.0 eet, with a single bottom-lange transition located 15.0 eet rom the pier, resulted in a drop in the predicted lateral torsional buckling resistance rom 68.19 ksi to 57.11 ksi (a 16 percent reduction). To help determine i the predicted drop in the nominal lexural resistance is reasonable, a more rigorous approximate procedure is presented herein or predicting the LTB resistance o the compression lange within an unbraced length containing a single lange transition. The procedure is based on work by Carskaddan and Schilling (1974), which attempted to address the general case o lateral torsional buckling o singly-symmetric noncomposite or composite girders in negative lexure subjected to a moment gradient within any given unbraced length. The calculations in this report are based on the ollowing ratio: χ = π P cr EI Lb where P cr is the elastic critical buckling load or a stepped column subjected to uniorm axial compression, and π EI L b is the corresponding elastic critical buckling load or a prismatic column having the larger o the two moments o inertia, I. This ratio is given by Figure C1, which is Figure rom Carskaddan and Schilling (1974). This igure is urther adapted rom Figure o Dalal (1969), but with changes in notation, where: L (C1) α = (C) Lb I and β = (C) I For an I-section in lexure, the above column analogy corresponds to lateral buckling o the compression lange, and thereore, 1 May, 004 C1

LRFD rd Edition t b β = c c tc1b (C4) c1 Based on Eq. (C1), the compression-lange stress at the maximum moment location at elastic lateral torsional buckling o the stepped unbraced length, normalized with respect to the yield strength o the compression lange at the maximum moment location, may be expressed as: F F F L r cr CbRbπ E =χ yc yc ( b t ) where the moment gradient modiier, C b, is calculated according to the provisions o Article 6.10.8.. (or Article A6.. as applicable) assuming the unbraced length is prismatic and based on the larger section within the unbraced length. χ is determined rom Figure C1 or the analogous equivalent stepped column. I available, other more rigorous estimations o F cr may be substituted or the value given by Eq. (C5). Carskaddan and Schilling (1974) show that or α = 0.5, Eq. (C5) is conservative relative to other more rigorous calculations o F cr. It is logical that χ would always be smaller or the case o uniorm axial compression within an actual or equivalent column versus the case o the same column subjected to an axial compression that increases toward the end with the larger lexural rigidity. Thus, it is conservative to apply the χ value rom Figure C1 as a actor that accounts or the reduction in the elastic critical stress level due to a single step in the geometry o a general member subject to moment gradient conditions. The elastic critical stress, F crs, at the smaller section within the unbraced length when the elastic critical stress, F cr, is reached at the maximum moment point can be computed as ollows (again normalized with respect to the yield strength o the compression lange at the smaller section): Fcrs F F cr yc bs Rb1 F F cr yc Ms Sxc Rb1 F F cr yc Sxc R b1 M 1 = = = 1+α -1 Fyc1 Fyc Fyc1 b Rb Fyc Fyc1 M Sxcs Rb Fyc Fyc1 Sxcs R b M where M 1 is a moment at the brace point with the lower moment, determined in general in the same manner that 1 is calculated when determining C b according to the speciication provisions. The expression within the square brackets in the inal right-hand side orm o Eq. (C6) is based on the replacement o the moment envelope associated with the unbraced length under consideration with an equivalent linear variation between M and M 1. The expression within the square brackets, multiplied by S xc /S xcs, is bs / b based on this equivalent linear variation o the moment along the unbraced length. Once the ratios o the elastic critical buckling stresses to the corresponding yield strengths are determined at Location and within the smaller section at the lange transition (denoted here as Location s), the corresponding F n /F yc values at each o the above locations may be calculated as ollows: I I I Fcr F π Fnc = RbRhFyc (C7) F F yc yr cr yc F F F F F yc < π cr yr < Fnc Fcr yc Fyc F F F 1 F 1 π 1 R R F yr cr yc nc = b h yc RF h yc π Fyr Fyc 1 (C8) = (C9) Eqs. (C7) through (C9) are obtained by writing the LTB resistance expressions given by Eqs. 6.10.8..-1 through 6.10.8..- in terms o the ratio o Fcr F yc (computed assuming R b is equal to 1.0), rather than in terms o the unbraced length L b. Eqs. (C7) through (C9) give exactly the same result as Eqs. 6.10.8..-1 through 6.10.8..- or the case o a prismatic member subject to uniorm bending moment. These (C5) (C6) C May, 004

LRFD rd Edition equations give a conservative representation o the inelastic LTB resistance o unbraced lengths with a single step in the cross-section. The equations, conigured in this manner, are based undamentally on a uniorm F cr /F yc within the compression lange o prismatic members. The compression lange in a stepped unbraced length is not stressed uniormly along its length, and thus the mapping rom F cr /F yc to F nc is conservative since the inelastic reduction in stiness is less or this case than i the compression lange were stressed uniormly. Since or the stepped member, the smaller cross-section may experience signiicant yielding within the middle regions o the unbraced length, Eqs. (C7) through (C9) are employed both at Location and at Location s to ensure that the result is still conservative or stepped members that experience signiicant yielding prior to reaching their maximum LTB resistance. The application o this suggested procedure is illustrated to determine the LTB resistance o the stepped bottom (compression) lange within the 0-oot-long unbraced length adjacent to the pier section (Section -) in the design example at the strength limit state (see Figure ). For this unbraced length, L 15.0 α = = = Lb 0.0 t b (0) 0.75 c c β = = = tc1bc1 1(0).0 χ = 0.9 rom Figure C1 Calculate the ratio o F cr /F yc at Location rom Equation (C5) and the ratio o F crs /F yc1 at the section transition (Location s) rom Equation (C6). The necessary data or these locations are obtained rom the design example calculations or this particular unbraced length: In both cases, Fcr 1.5(0.990)π (9,000) Fyc 70.0 0.0(1) 5. =0.9 =.50 ( ) F crs 70.0,10 0.977 8, 46 =.50 1 0.75 1 + =.79 F yc1 70.0 1,975 0.990 14,979 Fyr 49.0 Fcr = = 0.7 < <π = 9.87 F 70.0 F Thereore, inelastic LTB governs at both locations and: yc 49.0 π.50 1 Fnc = 1 1 (0.990)(0.984)(70.0) = 61.14 ksi (0.984)(70.0) π 0.7 1 49.0 π.79 1 Fncs = 1 1 (0.977)(0.971)(70.0) = 60.48 ksi (0.971)(70.0) π 0.7 1 The lange stress b at Location due to the actored loads is compared to F nc and the lange stress at Location s, bs, due to the actored loads is compared to F ncs, to determine that the unbraced length has adequate LTB resistance. Note, however, that the lange local buckling resistance o 59.6 ksi at Location s (as computed in the design example) would actually control in this case and would be taken as the nominal lexural resistance o the compression lange at Location s. The local buckling resistance o 68.19 ksi at Location would not control. In this particular case, the LTB resistance rom the more rigorous approach at Location s is only 5.9 percent greater than the single value o 57.11 ksi predicted or this unbraced length using the less rigorous approximate approach given in the speciications. The value o 57.11 ksi is calculated assuming the unbraced length is prismatic based on the section at Location s, with C b taken equal to 1.0. The increase yc May, 004 C

LRFD rd Edition in the LTB resistance may be more signiicant in other situations. The suggested method herein provides one possible approach or evaluating the calculated LTB resistance o a stepped lange (with a single step) in greater detail and or determining a larger resistance in situations where it may be desirable or necessary to do so. Note that similar logic can be applied to develop a set o equations to be used in lieu o the LTB equations given in Article A6.. (Appendix A to LRFD Section 6) or sections with compact or noncompact webs. The LTB equations given in Article A6.. include the eect o the St. Venant torsional rigidity, GJ. However, the more basic equations provided herein, ignoring the inluence o the torsional rigidity, may be conservatively used or these sections, i desired. Reerences Carskaddan, P. S. and C. G. Schilling. 1974. Lateral Buckling o Highway Bridge Girders. Research Laboratory Report -G-001 (109-), United States Steel Corporation, Monroeville, PA. Dalal, S. T. 1969. Some Non-Conventional Cases o Column Design. AISC Engineering Journal, American Institute o Steel Construction, Chicago, IL, January 1969. C4 May, 004

LRFD rd Edition Figure C1: c Ratio Chart May, 004 C5