1. DESIGN EXAMPLESVersion 14.0AMERICAN INSTITUTEOFSTEEL CONSTRUCTION

2. Design Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONiiCopyright 2011byAmerican Institute of Steel ConstructionAll rights reserved. This publication or any part thereofmust not be reproduced in any form without thewritten permission of the publisher.The AISC logo is a registered trademark of AISC.The information presented in this publication has been prepared in accordance with recognized engineering principlesand is for general information only. While it is believed to be accurate, this information should not be used or reliedupon for any specific application without competent professional examination and verification of its accuracy,suitability, and applicability by a licensed professional engineer, designer, or architect. The publication of the materialcontained herein is not intended as a representation or warranty on the part of the American Institute of SteelConstruction or of any other person named herein, that this information is suitable for any general or particular use orof freedom from infringement of any patent or patents. Anyone making use of this information assumes all liabilityarising from such use.Caution must be exercised when relying upon other specifications and codes developed by other bodies andincorporated by reference herein since such material may be modified or amended from time to time subsequent to theprinting of this edition. The Institute bears no responsibility for such material other than to refer to it and incorporate itby reference at the time of the initial publication of this edition.Printed in the United States of AmericaFirst Printing, October 2011 3. Design Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONiiiPREFACEThe primary objective of these design examples is to provide illustrations of the use of the 2010 AISC Specificationfor Structural Steel Buildings (ANSI/AISC 360-10) and the 14th Edition of the AISC Steel Construction Manual.The design examples provide coverage of all applicable limit states whether or not a particular limit state controls thedesign of the member or connection.In addition to the examples which demonstrate the use of the Manual tables, design examples are provided forconnection designs beyond the scope of the tables in the Manual. These design examples are intended to demonstratean approach to the design, and are not intended to suggest that the approach presented is the only approach. Thecommittee responsible for the development of these design examples recognizes that designers have alternateapproaches that work best for them and their projects. Design approaches that differ from those presented in theseexamples are considered viable as long as the Specification, sound engineering, and project specific requirements aresatisfied.Part I of these examples is organized to correspond with the organization of the Specification. The Chapter titlesmatch the corresponding chapters in the Specification.Part II is devoted primarily to connection examples that draw on the tables from the Manual, recommended designprocedures, and the breadth of the Specification. The chapters of Part II are labeled II-A, II-B, II-C, etc.Part III addresses aspects of design that are linked to the performance of a building as a whole. This includescoverage of lateral stability and second order analysis, illustrated through a four-story braced-frame and moment-framebuilding.The Design Examples are arranged with LRFD and ASD designs presented side by side, for consistency with theAISC Manual. Design with ASD and LRFD are based on the same nominal strength for each element so that theonly differences between the approaches are which set of load combinations from ASCE/SEI 7-10 are used fordesign and whether the resistance factor for LRFD or the safety factor for ASD is used.CONVENTIONSThe following conventions are used throughout these examples:1. The 2010 AISC Specification for Structural Steel Buildings is referred to as the AISC Specification and the14th Edition AISC Steel Construction Manual, is referred to as the AISC Manual.2. The source of equations or tabulated values taken from the AISC Specification or AISC Manual is notedalong the right-hand edge of the page.3. When the design process differs between LRFD and ASD, the designs equations are presented side-by-side.This rarely occurs, except when the resistance factor, , and the safety factor, , are applied.4. The results of design equations are presented to three significant figures throughout these calculations.ACKNOWLEDGMENTSThe AISC Committee on Manuals reviewed and approved V14.0 of the AISC Design Examples:William A. Thornton, ChairmanMark V. Holland, Vice ChairmanAbbas AminmansourCharles J. CarterHarry A. ColeDouglas B. DavisRobert O. DisqueBo DowswellEdward M. EganMarshall T. Ferrell 4. Design Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONivLanny J. FlynnPatrick J. FortneyLouis F. GeschwindnerW. Scott GoodrichChristopher M. HewittW. Steven HofmeisterBill R. Lindley, IIRonald L. MengLarry S. MuirThomas M. MurrayCharles R. PageDavis G. Parsons, IIRafael SabelliClifford W. SchwingerWilliam N. ScottWilliam T. SeguiVictor ShneurMarc L. SorensonGary C. VioletteMichael A. WestRonald G. YeagerCynthia J. Duncan, SecretaryThe AISC Committee on Manuals gratefully acknowledges the contributions of the following individuals whoassisted in the development of this document: Leigh Arber, Eric Bolin, Janet Cummins, Thomas Dehlin, WilliamJacobs, Richard C. Kaehler, Margaret Matthew, Heath Mitchell, Thomas J. Schlafly, and Sriramulu Vinnakota. 5. Design Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONvTABLE OF CONTENTSPART I. EXAMPLES BASED ON THE AISC SPECIFICATIONCHAPTER A GENERAL PROVISIONS.........................................................................................................A-1Chapter AReferences ......................................................................................................................................................A-2CHAPTER B DESIGN REQUIREMENTS .....................................................................................................B-1Chapter BReferences ...................................................................................................................................................... B-2CHAPTER C DESIGN FOR STABILITY.......................................................................................................C-1Example C.1A Design of a Moment Frame by the Direct Analysis Method ........................................................ C-2Example C.1B Design of a Moment Frame by the Effective Length Method ...................................................... C-6Example C.1C Design of a Moment Frame by the First-Order Method .............................................................C-11CHAPTER D DESIGN OF MEMBERS FOR TENSION...............................................................................D-1Example D.1 W-Shape Tension Member ..........................................................................................................D-2Example D.2 Single Angle Tension Member ....................................................................................................D-5Example D.3 WT-Shape Tension Member ........................................................................................................D-8Example D.4 Rectangular HSS Tension Member ............................................................................................D-11Example D.5 Round HSS Tension Member ....................................................................................................D-14Example D.6 Double Angle Tension Member .................................................................................................D-17Example D.7 Pin-Connected Tension Member ...............................................................................................D-20Example D.8 Eyebar Tension Member ............................................................................................................D-23Example D.9 Plate with Staggered Bolts .........................................................................................................D-25CHAPTER E DESIGN OF MEMBERS FOR COMPRESSION...................................................................E-1Example E.1A W-Shape Column Design with Pinned Ends ............................................................................... E-4Example E.1B W-Shape Column Design with Intermediate Bracing.................................................................. E-6Example E.1C W-Shape Available Strength Calculation .................................................................................... E-8Example E.1D W-Shape Available Strength Calculation .................................................................................... E-9Example E.2 Built-up Column with a Slender Web........................................................................................ E-11Example E.3 Built-up Column with Slender Flanges ...................................................................................... E-16Example E.4A W-Shape Compression Member (Moment Frame) .................................................................... E-21Example E.4B W-Shape Compression Member (Moment Frame) .................................................................... E-25Example E.5 Double Angle Compression Member without Slender Elements ............................................... E-26Example E.6 Double Angle Compression Member with Slender Elements .................................................... E-31Example E.7 WT Compression Member without Slender Elements ............................................................... E-37Example E.8 WT Compression Member with Slender Elements .................................................................... E-42Example E.9 Rectangular HSS Compression Member without Slender Elements ......................................... E-47Example E.10 Rectangular HSS Compression Member with Slender Elements ............................................... E-50Example E.11 Pipe Compression Member ........................................................................................................ E-54Example E.12 Built-up I-Shaped Member with Different Flange Sizes ............................................................ E-57Example E.13 Double WT Compression Member ............................................................................................. E-63CHAPTER F DESIGN OF MEMBERS FOR FLEXURE.............................................................................. F-1Example F.1-1A W-Shape Flexural Member Design in Strong-Axis Bending, Continuously Braced ................... F-6Example F.1-1B W-Shape Flexural Member Design in Strong-Axis Bending, Continuously Braced ................... F-8Example F.1-2A W-Shape Flexural Member Design in Strong-Axis Bending, Braced at Third Points ................. F-9Example F.1-2B W-Shape Flexural Member Design in Strong-Axis Bending, Braced at Third Points............... F-10 6. Design Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONviExample F.1-3A W-Shape Flexural Member Design in Strong-Axis Bending, Braced at Midspan..................... F-12Example F.1-3B W-Shape Flexural Member Design in Strong-Axis Bending, Braced at Midspan..................... F-14Example F.2-1A Compact Channel Flexural Member, Continuously Braced ...................................................... F-16Example F.2-1B Compact Channel Flexural Member, Continuously Braced ...................................................... F-18Example F.2-2A Compact Channel Flexural Member with Bracing at Ends and Fifth Points ............................. F-19Example F.2-2B Compact Channel Flexural Member with Bracing at End and Fifth Points ............................... F-20Example F.3A W-Shape Flexural Member with Noncompact Flanges in Strong-Axis Bending ...................... F-22Example F.3B W-Shape Flexural Member with Noncompact Flanges in Strong-Axis Bending ...................... F-24Example F.4 W-Shape Flexural Member, Selection by Moment of Inertia for Strong-Axis Bending ............ F-26Example F.5 I-Shaped Flexural Member in Minor-Axis Bending .................................................................. F-28Example F.6 HSS Flexural Member with Compact Flanges ........................................................................... F-30Example F.7A HSS Flexural Member with Noncompact Flanges ..................................................................... F-32Example F.7B HSS Flexural Member with Noncompact Flanges ..................................................................... F-34Example F.8A HSS Flexural Member with Slender Flanges ............................................................................. F-36Example F.8B HSS Flexural Member with Slender Flanges ............................................................................. F-38Example F.9A Pipe Flexural Member ................................................................................................................ F-41Example F.9B Pipe Flexural Member ................................................................................................................ F-42Example F.10 WT-Shape Flexural Member ..................................................................................................... F-44Example F.11A Single Angle Flexural Member .................................................................................................. F-47Example F.11B Single Angle Flexural Member .................................................................................................. F-50Example F.11C Single Angle Flexural Member .................................................................................................. F-53Example F.12 Rectangular Bar in Strong-Axis Bending .................................................................................. F-59Example F.13 Round Bar in Bending ............................................................................................................... F-61Example F.14 Point-Symmetrical Z-shape in Strong-Axis Bending................................................................. F-63Chapter FDesign ExampleReferences .................................................................................................................................................... F-69CHAPTER G DESIGN OF MEMBERS FOR SHEAR...................................................................................G-1Example G.1A W-Shape in Strong-Axis Shear ....................................................................................................G-3Example G.1B W-Shape in Strong-Axis Shear ....................................................................................................G-4Example G.2A C-Shape in Strong-Axis Shear .....................................................................................................G-5Example G.2B C-Shape in Strong-Axis Shear .....................................................................................................G-6Example G.3 Angle in Shear .............................................................................................................................G-7Example G.4 Rectangular HSS in Shear ............................................................................................................G-9Example G.5 Round HSS in Shear ..................................................................................................................G-11Example G.6 Doubly Symmetric Shape in Weak-Axis Shear .........................................................................G-13Example G.7 Singly Symmetric Shape in Weak-Axis Shear ...........................................................................G-15Example G.8A Built-up Girder with Transverse Stiffeners ................................................................................G-17Example G.8B Built-up Girder with Transverse Stiffeners ................................................................................G-21Chapter GDesign ExampleReferences ....................................................................................................................................................G-24CHAPTER H DESIGN OF MEMBERS FOR COMBINED FORCES AND TORSION ............................H-1Example H.1A W-shape Subject to Combined Compression and BendingAbout Both Axes (Braced Frame) ...............................................................................................H-2Example H.1B W-shape Subject to Combined Compression and Bending MomentAbout Both Axes (Braced Frame) ................................................................................................H-4Example H.2 W-Shape Subject to Combined Compression and Bending MomentAbout Both Axes (By AISC Specification Section H2) ..............................................................H-5Example H.3 W-Shape Subject to Combined Axial Tension and Flexure......................................................... H-8Example H.4 W-Shape Subject to Combined Axial Compression and Flexure ..............................................H-12Example H.5A Rectangular HSS Torsional Strength .........................................................................................H-16 7. Design Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONviiExample H.5B Round HSS Torsional Strength ..................................................................................................H-17Example H.5C HSS Combined Torsional and Flexural Strength .......................................................................H-19Example H.6 W-Shape Torsional Strength ......................................................................................................H-23Chapter HDesign ExampleReferences ....................................................................................................................................................H-30CHAPTER I DESIGN OF COMPOSITE MEMBERS................................................................................... I-1Example I.1 Composite Beam Design ............................................................................................................... I-8Example I.2 Composite Girder Design ........................................................................................................... I-18Example I.3 Concrete Filled Tube (CFT) Force Allocation and Load Transfer .............................................. I-35Example I.4 Concrete Filled Tube (CFT) in Axial Compression .................................................................... I-45Example I.5 Concrete Filled Tube (CFT) in Axial Tension ............................................................................ I-50Example I.6 Concrete Filled Tube (CFT) in Combined Axial Compression, Flexure and Shear ................... I-52Example I.7 Concrete Filled Box Column with Noncompact/Slender Elements ............................................ I-66Example I.8 Encased Composite Member Force Allocation and Load Transfer ............................................ I-80Example I.9 Encased Composite Member in Axial Compression ................................................................... I-93Example I.10 Encased Composite Member in Axial Tension ......................................................................... I-100Example I.11 Encased Composite Member in Combined Axial Compression, Flexure and Shear ................ I-103Example I.12 Steel Anchors in Composite Components ................................................................................. I-119Chapter IDesign ExampleReferences ................................................................................................................................................... I-123CHAPTER J DESIGN OF CONNECTIONS...................................................................................................J-1Example J.1 Fillet Weld in Longitudinal Shear ................................................................................................. J-2Example J.2 Fillet Weld Loaded at an Angle .................................................................................................... J-4Example J.3 Combined Tension and Shear in Bearing Type Connections........................................................ J-6Example J.4A Slip-Critical Connection with Short-Slotted Holes ....................................................................... J-8Example J.4B Slip-Critical Connection with Long-Slotted Holes ................................................. J-10Example J.5 Combined Tension and Shear in a Slip-Critical Connection ................................................. J-12Example J.6 Bearing Strength of a Pin in a Drilled Hole ................................................................................ J-15Example J.7 Base Plate Bearing on Concrete................................................................................................... J-16CHAPTER K DESIGN OF HSS AND BOX MEMBER CONNECTIONS...................................................K-1Example K.1 Welded/Bolted Wide Tee Connection to an HSS Column ...........................................................K-2Example K.2 Welded/Bolted Narrow Tee Connection to an HSS Column .....................................................K-10Example K.3 Double Angle Connection to an HSS Column ...........................................................................K-13Example K.4 Unstiffened Seated Connection to an HSS Column ...................................................................K-16Example K.5 Stiffened Seated Connection to an HSS Column .......................................................................K-19Example K.6 Single-Plate Connection to Rectangular HSS Column ..............................................................K-24Example K.7 Through-Plate Connection .........................................................................................................K-27Example K.8 Transverse Plate Loaded Perpendicular to the HSS Axis on a Rectangular HSS.......................K-31Example K.9 Longitudinal Plate Loaded Perpendicular to the HSS Axis on a Round HSS ............................K-34Example K.10 HSS Brace Connection to a W-Shape Column ...........................................................................K-36Example K.11 Rectangular HSS Column with a Cap Plate, Supporting a Continuous Beam ...........................K-39Example K.12 Rectangular HSS Column Base Plate ........................................................................................K-42Example K.13 Rectangular HSS Strut End Plate ...............................................................................................K-45Chapter KDesign ExampleReferences ....................................................................................................................................................K-49 8. Design Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONviiiPART II. EXAMPLES BASED ON THE AISC STEEL CONSTRUCTION MANUALCHAPTER IIA SIMPLE SHEAR CONNECTIONS.......................................................................................IIA-1Example II.A-1 All-Bolted Double-Angle Connection ...................................................................................... IIA-2Example II.A-2 Bolted/Welded Double-Angle Connection ............................................................................... IIA-4Example II.A-3 All-Welded Double-Angle Connection .................................................................................... IIA-6Example II.A-4 All-Bolted Double-Angle Connection in a Coped Beam .......................................................... IIA-9Example II.A-5 Welded/ Bolted Double-Angle Connection (Beam-to-Girder Web). ...................................... IIA-13Example II.A-6 Beam End Coped at the Top Flange Only .............................................................................. IIA-16Example II.A-7 Beam End Coped at the Top and Bottom Flanges.................................................................. IIA-23Example II.A-8 All-Bolted Double-Angle Connections (Beams-to-Girder Web) ........................................... IIA-26Example II.A-9 Offset All-Bolted Double-Angle Connections (Beams-to-Girder Web) ................................ IIA-34Example II.A-10 Skewed Double Bent-Plate Connection (Beam-to-Girder Web)............................................. IIA-37Example II.A-11 Shear End-Plate Connection (Beam to Girder Web). ............................................................. IIA-43Example II.A-12 All-Bolted Unstiffened Seated Connection (Beam-to-Column Web) ..................................... IIA-45Example II.A-13 Bolted/Welded Unstiffened Seated Connection (Beam-to-Column Flange) .......................... IIA-48Example II.A-14 Stiffened Seated Connection (Beam-to-Column Flange) ........................................................ IIA-51Example II.A-15 Stiffened Seated Connection (Beam-to-Column Web) ........................................................... IIA-54Example II.A-16 Offset Unstiffened Seated Connection (Beam-to-Column Flange)......................................... IIA-57Example II.A-17 Single-Plate Connection (Conventional Beam-to-Column Flange) ..................................... IIA-60Example II.A-18 Single-Plate Connection (Beam-to-Girder Web) .................................................................... IIA-62Example II.A-19 Extended Single-Plate Connection (Beam-to-Column Web) .................................................. IIA-64Example II.A-20 All-Bolted Single-Plate Shear Splice ...................................................................................... IIA-70Example II.A-21 Bolted/Welded Single-Plate Shear Splice ............................................................................... IIA-75Example II.A-22 Bolted Bracket Plate Design ................................................................................................... IIA-80Example II.A-23 Welded Bracket Plate Design. ................................................................................................ IIA-86Example II.A-24 Eccentrically Loaded Bolt Group (IC Method) ....................................................................... IIA-91Example II.A-25 Eccentrically Loaded Bolt Group (Elastic Method)................................................................ IIA-93Example II.A-26 Eccentrically Loaded Weld Group (IC Method)..................................................................... IIA-95Example II.A-27 Eccentrically Loaded Weld Group (Elastic Method) .............................................................. IIA-98Example II.A-28 All-Bolted Single-Angle Connection (Beam-to-Girder Web) .............................................. IIA-100Example II.A-29 Bolted/Welded Single-Angle Connection (Beam-to-Column Flange).................................. IIA-105Example II.A-30 All-Bolted Tee Connection (Beam-to-Column Flange) ........................................................ IIA-108Example II.A-31 Bolted/Welded Tee Connection (Beam-to-Column Flange) ................................................. IIA-115CHAPTER IIB FULLY RESTRAINED (FR) MOMENT CONNECTIONS............................................... IIB-1Example II.B-1 Bolted Flange-Plate FR Moment Connection (Beam-to-Column Flange) ............................... IIB-2Example II.B-2 Welded Flange-Plated FR Moment Connection (Beam-to-Column Flange) ......................... IIB-14Example II.B-3 Directly Welded Flange FR Moment Connection (Beam-to-Column Flange). ..................... IIB-20Example II.B-4 Four-Bolt Unstiffened Extended End-PlateFR Moment Connection (Beam-to-Column Flange) ............................................................. IIB-22Chapter IIBDesign ExampleReferences ................................................................................................................................................. IIB-33CHAPTER IIC BRACING AND TRUSS CONNECTIONS.......................................................................... IIC-1Example II.C-1 Truss Support Connection........................................................................................................ IIC-2Example II.C-2 Bracing Connection ............................................................................................................... IIC-13Example II.C-3 Bracing Connection ............................................................................................................... IIC-41Example II.C-4 Truss Support Connection...................................................................................................... IIC-50Example II.C-5 HSS Chevron Brace Connection ............................................................................................ IIC-57Example II.C-6 Heavy Wide Flange Compression Connection (Flanges on the Outside) .............................. IIC-69CHAPTER IID MISCELLANEOUS CONNECTIONS.................................................................................. ,ID-1 9. Design Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONixExample II.D-1 Prying Action in Tees and in Single Angles ............................................................................ IID-2Example II.D-2 Beam Bearing Plate ................................................................................................................. IID-9Example II.D-3 Slip-Critical Connection with Oversized Holes...................................................................... IID-15PART III. SYSTEM DESIGN EXAMPLES ................................................................. III-1Example III-1 Design of Selected Members and Lateral Analysis of a Four-Story Building .............................III-2Introduction..................................................................................................................................III-2Conventions .................................................................................................................................III-2Design Sequence..........................................................................................................................III-3General Description of the Building ............................................................................................III-4Roof Member Design and Selection ...........................................................................................III-5Floor Member Design and Selection ........................................................................................III-17Column Design and Selection for Gravity Loads ..................................................................... III-38Wind Load Determination ........................................................................................................III-46Seismic Load Determination .....................................................................................................III-49Moment Frame Model ...............................................................................................................III-61Calculation of Required StrengthThree Methods ..................................................................III-67Beam Analysis in the Moment Frame........................................................................................III-77Braced Frame Analysis ..............................................................................................................III-80Analysis of Drag Struts..............................................................................................................III-84Part III ExampleReferences ...................................................................................................................................................III-87PART IV. ADDITIONAL RESOURCES..................................................................... IV-1Table 4-1 Discussion................................................................................................................................... IV-2Table 4-1 W-Shapes in Axial Compressions, Fy = 65 ksi ........................................................................... IV-5Table 6-1 Discussion................................................................................................................................. IV-17Table 6-1 Combined Flexure and Axial Force, W-Shapes, Fy = 65 ksi .................................................... IV-19 10. A-1Chapter AGeneral ProvisionsA1. SCOPEThese design examples are intended to illustrate the application of the 2010 AISC Specification for StructuralSteel Buildings (ANSI/AISC 360-10) (AISC, 2010) and the AISC Steel Construction Manual, 14th Edition(AISC, 2011) in low-seismic applications. For information on design applications requiring seismic detailing, seethe AISC Seismic Design Manual.A2. REFERENCED SPECIFICATIONS, CODES AND STANDARDSSection A2 includes a detailed list of the specifications, codes and standards referenced throughout the AISCSpecification.A3. MATERIALSection A3 includes a list of the steel materials that are approved for use with the AISC Specification. Thecomplete ASTM standards for the most commonly used steel materials can be found in Selected ASTM Standardsfor Structural Steel Fabrication (ASTM, 2011).A4. STRUCTURAL DESIGN DRAWINGS AND SPECIFICATIONSSection A4 requires that structural design drawings and specifications meet the requirements in the AISC Code ofStandard Practice for Steel Buildings and Bridges (AISC, 2010b).Design Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONReturn to Table of Contents 11. A-2CHAPTER A REFERENCESAISC (2010a), Specification for Structural Steel Buildings, ANSI/AISC 360-10, American Institute for SteelDesign Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONConstruction, Chicago, IL.AISC (2010b), Code of Standard Practice for Steel Buildings and Bridges, American Institute for SteelConstruction, Chicago, IL.AISC (2011), Steel Construction Manual, 14th Ed., American Institute for Steel Construction, Chicago, IL.ASTM (2011), Selected ASTM Standards for Structural Steel Fabrication, ASTM International, WestConshohocken, PA.Return to Table of Contents 12. B-1Chapter BDesign RequirementsB1. GENERAL PROVISIONSB2. LOADS AND LOAD COMBINATIONSIn the absence of an applicable building code, the default load combinations to be used with this Specification arethose from Minimum Design Loads for Buildings and Other Structures (ASCE/SEI 7-10) (ASCE, 2010).B3. DESIGN BASISChapter B of the AISC Specification and Part 2 of the AISC Manual describe the basis of design, for both LRFDand ASD.This Section describes three basic types of connections: simple connections, fully restrained (FR) momentconnections, and partially restrained (PR) moment connections. Several examples of the design of each of thesetypes of connection are given in Part II of these design examples.Information on the application of serviceability and ponding provisions may be found in AISC SpecificationChapter L and AISC Specification Appendix 2, respectively, and their associated commentaries. Design examplesand other useful information on this topic are given in AISC Design Guide 3, Serviceability DesignConsiderations for Steel Buildings, Second Edition (West et al., 2003).Information on the application of fire design provisions may be found in AISC Specification Appendix 4 and itsassociated commentary. Design examples and other useful information on this topic are presented in AISC DesignGuide 19, Fire Resistance of Structural Steel Framing (Ruddy et al., 2003).Corrosion protection and fastener compatibility are discussed in Part 2 of the AISC Manual.B4. MEMBER PROPERTIESAISC Specification Tables B4.1a and B4.1b give the complete list of limiting width-to-thickness ratios for allcompression and flexural members defined by the AISC Specification.Except for one section, the W-shapes presented in the compression member selection tables as column sectionsmeet the criteria as nonslender element sections. The W-shapes presented in the flexural member selection tablesas beam sections meet the criteria for compact sections, except for 10 specific shapes. When noncompact orslender-element sections are tabulated in the design aids, local buckling criteria are accounted for in the tabulateddesign values.The shapes listing and other member design tables in the AISC Manual also include footnoting to highlightsections that exceed local buckling limits in their most commonly available material grades. These footnotesinclude the following notations:c Shape is slender for compression.f Shape exceeds compact limit for flexure.g The actual size, combination and orientation of fastener components should be compared with the geometry ofthe cross section to ensure compatibility.h Flange thickness greater than 2 in. Special requirements may apply per AISC Specification Section A3.1c.v Shape does not meet the h/tw limit for shear in AISC Specification Section G2.1a.Design Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONReturn to Table of Contents 13. B-2CHAPTER B REFERENCESASCE (2010), Minimum Design Loads for Buildings and Other Structures, ASCE/SEI 7-10, American Society ofDesign Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONCivil Engineers, Reston, VA.West, M., Fisher, J. and Griffis, L.G. (2003), Serviceability Design Considerations for Steel Buildings, DesignGuide 3, 2nd Ed., AISC, Chicago, IL.Ruddy, J.L., Marlo, J.P., Ioannides, S.A. and Alfawakhiri, F. (2003), Fire Resistance of Structural Steel Framing,Design Guide 19, AISC, Chicago, IL.Return to Table of Contents 14. C-1Chapter CDesign for StabilityC1. GENERAL STABILITY REQUIREMENTSThe AISC Specification requires that the designer account for both the stability of the structural system as awhole, and the stability of individual elements. Thus, the lateral analysis used to assess stability must includeconsideration of the combined effect of gravity and lateral loads, as well as member inelasticity, out-of-plumbness,out-of-straightness and the resulting second-order effects, P- and P-. The effects of leaningcolumns must also be considered, as illustrated in the examples in this chapter and in the four-story buildingdesign example in Part III of AISC Design Examples.P- and P- effects are illustrated in AISC Specification Commentary Figure C-C2.1. Methods for addressingstability, including P- and P- effects, are provided in AISC Specification Section C2 and Appendix 7.C2. CALCULATION OF REQUIRED STRENGTHSThe calculation of required strengths is illustrated in the examples in this chapter and in the four-story buildingdesign example in Part III of AISC Design Examples.C3. CALCULATION OF AVAILABLE STRENGTHSThe calculation of available strengths is illustrated in the four-story building design example in Part III of AISCDesign Examples.Design Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONReturn to Table of Contents 15. C-2EXAMPLE C.1A DESIGN OF A MOMENT FRAME BY THE DIRECT ANALYSIS METHODGiven:Determine the required strengths and effective length factors for the columns in the rigid frame shown below forthe maximum gravity load combination, using LRFD and ASD. Use the direct analysis method. All members areASTM A992 material.Columns are unbraced between the footings and roof in the x- and y-axes and are assumed to have pinned bases.Solution:From Manual Table 1-1, the W1265 has A = 19.1 in.2The beams from grid lines A to B, and C to E and the columns at A, D and E are pinned at both ends and do notcontribute to the lateral stability of the frame. There are no P- effects to consider in these members and they maybe designed using K=1.0.The moment frame between grid lines B and C is the source of lateral stability and therefore must be designedusing the provisions of Chapter C of the AISC Specification. Although the columns at grid lines A, D and E donot contribute to lateral stability, the forces required to stabilize them must be considered in the analysis. For theanalysis, the entire frame could be modeled or the model can be simplified as shown in the figure below, in whichthe stability loads from the three leaning columns are combined into a single column.From Chapter 2 of ASCE/SEI 7, the maximum gravity load combinations are:LRFD ASDwa = D + L= 0.400 kip/ft + 1.20 kip/ft= 1.60 kip/ftPer AISC Specification Section C2.1, for LRFD perform a second-order analysis and member strength checksusing the LRFD load combinations. For ASD, perform a second-order analysis using 1.6 times the ASD loadcombinations and divide the analysis results by 1.6 for the ASD member strength checks.Frame Analysis Gravity LoadsThe uniform gravity loads to be considered in a second-order analysis on the beam from B to C are:Design Examples V14.0wu = 1.2D + 1.6L= 1.2(0.400 kip/ft) + 1.6(1.20 kip/ft)= 2.40 kip/ftAMERICAN INSTITUTE OF STEEL CONSTRUCTIONReturn to Table of Contents 16. C-3LRFD ASDwu' = 2.40 kip/ft wa' = 1.6(1.60 kip/ft)= 2.56 kip/ftConcentrated gravity loads to be considered in a second-order analysis on the columns at B and C contributed byadjacent beams are:LRFD ASDDesign Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONPu' = (15.0 ft)(2.40 kip/ft)= 36.0 kipsPa' = 1.6(15.0 ft)(1.60 kip/ft)= 38.4 kipsConcentrated Gravity Loads on the Pseudo Leaning ColumnThe load in this column accounts for all gravity loading that is stabilized by the moment frame, but is not directlyapplied to it.LRFD ASDPuL' = (60.0 ft)(2.40 kip/ft)= 144 kipsPaL' = 1.6(60.0 ft)(1.60 kip/ft)= 154 kipsFrame Analysis Notional LoadsPer AISC Specification Section C2.2, frame out-of-plumbness must be accounted for either by explicit modelingof the assumed out-of-plumbness or by the application of notional loads. Use notional loads.From AISC Specification Equation C2-1, the notional loads are:LRFD ASD = 1.0Yi = (120 ft)(2.40 kip/ft)= 288 kipsNi = 0.002Yi (Spec. Eq. C2-1)= 0.002(1.0)(288 kips)= 0.576 kips = 1.6Yi = (120 ft)(1.60 kip/ft)= 192 kipsNi = 0.002Yi (Spec. Eq. C2-1)= 0.002(1.6)(192 kips)= 0.614 kipsSummary of Applied Frame LoadsLRFD ASDReturn to Table of Contents 17. C-4Per AISC Specification Section C2.3, conduct the analysis using 80% of the nominal stiffnesses to account for theeffects of inelasticity. Assume, subject to verification, that Pr /Py is no greater than 0.5; therefore, no additionalstiffness reduction is required.50% of the gravity load is carried by the columns of the moment resisting frame. Because the gravity loadsupported by the moment resisting frame columns exceeds one third of the total gravity load tributary to theframe, per AISC Specification Section C2.1, the effects of P- upon P- must be included in the frame analysis. Ifthe software used does not account for P- effects in the frame analysis, this may be accomplished by addingjoints to the columns between the footing and beam.Using analysis software that accounts for both P- and P- effects, the following results are obtained:First-order resultsLRFD ASD1st = 0.149 in.1st = 0.159 in. (prior to dividing by 1.6)Design Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONSecond-order resultsLRFD ASD2nd = 0.217 in.210.217 in.0.149 in.ndst== 1.462nd = 0.239 in. (prior to dividing by 1.6)210.239 in.0.159 in.ndst== 1.50Check the assumption that Pr Py 0.5 and therefore, b = 1.0:Return to Table of Contents 18. C-5PPDesign Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONPy = FyAg= 50 ksi(19.1 in.2)= 955 kipsLRFD ASD1.0(72.6 kips)955kipsPPry== 0.0760 0.5 o.k.1.6(48.4 kips)955kipsry== 0.0811 0.5 o.k.The stiffness assumption used in the analysis, b = 1.0, is verified.Although the second-order sway multiplier is approximately 1.5, the change in bending moment is small becausethe only sway moments are those produced by the small notional loads. For load combinations with significantgravity and lateral loadings, the increase in bending moments is larger.Verify the column strengths using the second-order forces shown above, using the following effective lengths(calculations not shown):Columns:Use KLx = 20.0 ftUse KLy = 20.0 ftReturn to Table of Contents 19. C-6EXAMPLE C.1B DESIGN OF A MOMENT FRAME BY THE EFFECTIVE LENGTH METHODRepeat Example C.1A using the effective length method.Given:Determine the required strengths and effective length factors for the columns in the rigid frame shown below forthe maximum gravity load combination, using LRFD and ASD. Use the effective length method.Columns are unbraced between the footings and roof in the x- and y-axes and are assumed to have pinned bases.Solution:From Manual Table 1-1, the W1265 has Ix = 533 in.4The beams from grid lines A to B, and C to E and the columns at A, D and E are pinned at both ends and do notcontribute to the lateral stability of the frame. There are no P- effects to consider in these members and they maybe designed using K=1.0.The moment frame between grid lines B and C is the source of lateral stability and therefore must be designedusing the provisions of Appendix 7 of the AISC Specification. Although the columns at grid lines A, D and E donot contribute to lateral stability, the forces required to stabilize them must be considered in the analysis. For theanalysis, the entire frame could be modeled or the model can be simplified as shown in the figure below, in whichthe stability loads from the three leaning columns are combined into a single column.Check the limitations for the use of the effective length method given in Appendix 7, Section 7.2.1:(1) The structure supports gravity loads through nominally vertical columns.(2) The ratio of maximum second-order drift to the maximum first-order drift will be assumed to be nogreater than 1.5, subject to verification following.From Chapter 2 of ASCE/SEI 7, the maximum gravity load combinations are:LRFD ASDwa = D + L= 0.400 kip/ft + 1.20 kip/ft= 1.60 kip/ftPer AISC Specification Appendix 7, Section 7.2.1, the analysis must conform to the requirements of AISCSpecification Section C2.1, with the exception of the stiffness reduction required by the provisions of SectionC2.3.Design Examples V14.0wu = 1.2D + 1.6L= 1.2(0.400 kip/ft) + 1.6(1.20 kip/ft)= 2.40 kip/ftAMERICAN INSTITUTE OF STEEL CONSTRUCTIONReturn to Table of Contents 20. C-7Per AISC Specification Section C2.1, for LRFD perform a second-order analysis and member strength checksusing the LRFD load combinations. For ASD, perform a second-order analysis at 1.6 times the ASD loadcombinations and divide the analysis results by 1.6 for the ASD member strength checks.Frame Analysis Gravity LoadsThe uniform gravity loads to be considered in a second-order analysis on the beam from B to C are:LRFD ASDwu' = 2.40 kip/ft wa' = 1.6(1.60 kip/ft)= 2.56 kip/ftConcentrated gravity loads to be considered in a second-order analysis on the columns at B and C contributed byadjacent beams are:LRFD ASDDesign Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONPu' = (15.0 ft)(2.40 kip/ft)= 36.0 kipsPa' = 1.6(15.0 ft)(1.60 kip/ft)= 38.4 kipsConcentrated Gravity Loads on the Pseudo Leaning ColumnThe load in this column accounts for all gravity loading that is stabilized by the moment frame, but is not directlyapplied to it.LRFD ASDPuL' = (60.0 ft)(2.40 kip/ft)= 144 kipsPaL' = 1.6(60.0 ft)(1.60 kip/ft)= 154 kipsFrame Analysis Notional LoadsPer AISC Specification Appendix 7, Section 7.2.2, frame out-of-plumbness must be accounted for by theapplication of notional loads in accordance with AISC Specification Section C2.2b.From AISC Specification Equation C2-1, the notional loads are:LRFD ASD = 1.0Yi = (120 ft)(2.40 kip/ft)= 288 kipsNi = 0.002Yi (Spec. Eq. C2-1)= 0.002(1.0)(288 kips)= 0.576 kips = 1.6Yi = (120 ft)(1.60 kip/ft)= 192 kipsNi = 0.002Yi (Spec. Eq. C2-1)= 0.002(1.6)(192 kips)= 0.614 kipsReturn to Table of Contents 21. Return to Table of ContentsC-8Design Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONSummary of Applied Frame LoadsLRFD ASDPer AISC Specification Appendix 7, Section 7.2.2, conduct the analysis using the full nominal stiffnesses.50% of the gravity load is carried by the columns of the moment resisting frame. Because the gravity loadsupported by the moment resisting frame columns exceeds one third of the total gravity load tributary to theframe, per AISC Specification Section C2.1, the effects of P- must be included in the frame analysis. If thesoftware used does not account for P- effects in the frame analysis, this may be accomplished by adding joints tothe columns between the footing and beam.Using analysis software that accounts for both P- and P- effects, the following results are obtained:First-order resultsLRFD ASD1st = 0.119 in.1st = 0.127 in. (prior to dividing by 1.6) 22. C-9 = = + K K P EI EIr H HR P L HL L HLDesign Examples V14.0r r moment frameAMERICAN INSTITUTE OF STEEL CONSTRUCTIONSecond-order resultsLRFD ASD2nd = 0.159 in.210.159 in.0.119 in.ndst== 1.342nd = 0.174 in. (prior to dividing by 1.6)210.174 in.0.127 in.ndst== 1.37The assumption that the ratio of the maximum second-order drift to the maximum first-order drift is no greaterthan 1.5 is verified; therefore, the effective length method is permitted.Although the second-order sway multiplier is approximately 1.35, the change in bending moment is small becausethe only sway moments for this load combination are those produced by the small notional loads. For loadcombinations with significant gravity and lateral loadings, the increase in bending moments is larger.Calculate the in-plane effective length factor, Kx, using the story stiffness method and Equation C-A-7-5presented in Commentary Appendix 7, Section 7.2. Take Kx = K2( )2 22 0.85 0.15 2 2 1.7xL r(Spec. Eq. C-A-7-5)Calculate the total load in all columns, PrLRFD ASD2.40 kip/ft (120 ft)288 kipsPr ==1.60 kip/ft (120 ft)192 kipsPr ==Calculate the ratio of the leaning column loads to the total load, RLLRFD ASDr r moment frame( )288 kips 71.5 kips 72.5 kips288 kips0.500LrP PRP = +==( )192 kips 47.7 kips 48.3 kips192 kips0.500LrP PRP = +==Calculate the Euler buckling strength of an individual column.Return to Table of Contents 23. C-10Design Examples V14.0( )( )2 2 29,000 ksi 533 in.42 20.85 0.15 0.500 72.4 kips2,650 kips 0.000496 in./in.0.576 kips2,650 kips 0.000496 in./in.0.85 0.15 0.500 1.6 48.3 kips2,650 kips 0.000529 in./in.2,650 kips 0.000529 in./in.AMERICAN INSTITUTE OF STEEL CONSTRUCTION( 240 in.) 2,650 kipsEIL==Calculate the drift ratio using the first-order notional loading results.LRFD ASD0.119in.240in.0.000496in./in.HL==0.127in.240in.0.000529in./in.HL==For the column at line C:LRFD ASD288 kips( ) ( )( )( )1.7 7.35 kips3.13 0.324Kx + = = Use Kx = 3.13( )( ) ( )( )1.6 192 kips( )0.614 kips( )( )1.7 1.6 4.90 kips3.13 0.324Kx + = = Use Kx = 3.13Note that it is necessary to multiply the column loads by 1.6 for ASD in the expression above.Verify the column strengths using the second-order forces shown above, using the following effective lengths(calculations not shown):Columns:Use KxLx = 3.13(20.0ft) = 62.6ftUse KyLy = 20.0 ftReturn to Table of Contents 24. C-11EXAMPLE C.1C DESIGN OF A MOMENT FRAME BY THE FIRST-ORDER METHODRepeat Example C.1A using the first-order analysis method.Given:Determine the required strengths and effective length factors for the columns in the rigid frame shown below forthe maximum gravity load combination, using LRFD and ASD. Use the first-order analysis method.Columns are unbraced between the footings and roof in the x- and y-axes and are assumed to have pinned bases.Solution:From Manual Table 1-1, the W1265 has A = 19.1 in.2The beams from grid lines A to B, and C to E and the columns at A, D and E are pinned at both ends and do notcontribute to the lateral stability of the frame. There are no P- effects to consider in these members and they maybe designed using K=1.0.The moment frame between grid lines B and C is the source of lateral stability and therefore must be designedusing the provisions of Appendix 7 of the AISC Specification. Although the columns at grid lines A, D and E donot contribute to lateral stability, the forces required to stabilize them must be considered in the analysis. Thesemembers need not be included in the analysis model, except that the forces in the leaning columns must beincluded in the calculation of notional loads.Check the limitations for the use of the first-order analysis method given in Appendix 7, Section 7.3.1:(1) The structure supports gravity loads through nominally vertical columns.(2) The ratio of maximum second-order drift to the maximum first-order drift will be assumed to be nogreater than 1.5, subject to verification.(3) The required axial strength of the members in the moment frame will be assumed to be no more than50% of the axial yield strength, subject to verification.From Chapter 2 of ASCE/SEI 7, the maximum gravity load combinations are:LRFD ASDwa = D + L= 0.400 kip/ft + 1.20 kip/ft= 1.60 kip/ftDesign Examples V14.0wu = 1.2D + 1.6L= 1.2(0.400 kip/ft) + 1.6(1.20 kip/ft)= 2.40 kip/ftAMERICAN INSTITUTE OF STEEL CONSTRUCTIONReturn to Table of Contents 25. C-12Per AISC Specification Appendix 7, Section 7.3.2, the required strengths are determined from a first-orderanalysis using notional loads determined below along with a B1 multiplier as determined from Appendix 8.For ASD, do not multiply loads or divide results by 1.6.Frame Analysis Gravity LoadsThe uniform gravity loads to be considered in the first-order analysis on the beam from B to C are:LRFD ASDwu' = 2.40 kip/ft wa' = 1.60 kip/ftConcentrated gravity loads to be considered in a second-order analysis on the columns at B and C contributed byadjacent beams are:LRFD ASDDesign Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONPu' = (15.0 ft)(2.40 kip/ft)= 36.0 kipsPa' = (15.0 ft)(1.60 kip/ft)= 24.0 kipsFrame Analysis Notional LoadsPer AISC Specification Appendix 7, Section 7.3.2, frame out-of-plumbness must be accounted for by theapplication of notional loads.From AISC Specification Appendix Equation A-7-2, the required notional loads are:LRFD ASD = 1.0Yi = (120 ft)(2.40 kip/ft)= 288 kips = 0.0 in. (no drift in this load combination)L = 240 in.Ni = 2.1(/L)Yi 0.0042Yi= 2.1(1.0)(0.0 in./240 in.)(288 kips) 0.0042(288 kips)= 0.0 kips 1.21 kipsUse Ni = 1.21 kips = 1.6Yi = (120 ft)(1.60 kip/ft)= 192 kips = 0.0 in. (no drift in this load combination)L = 240 in.Ni = 2.1(/L)Yi 0.0042Yi= 2.1(1.6)(0.0 in./240 in.)(192 kips) 0.0042(192 kips)= 0.0 kips 0.806 kipsUse Ni = 0.806 kipsReturn to Table of Contents 26. C-13Design Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONSummary of Applied Frame LoadsLRFD ASDPer AISC Specification Appendix 7, Section 7.2.2, conduct the analysis using the full nominal stiffnesses.Using analysis software, the following first-order results are obtained:LRFD ASD1st = 0.250 in.1st = 0.167 in.Check the assumption that the ratio of the second-order drift to the first-order drift does not exceed 1.5. B2 can beused to check this limit. Calculate B2 per the provisions of Section 8.2.2 of Appendix 8 using the results of thefirst-order analysis.LRFD ASDPmf = 71.2 kips + 72.8 kips= 144 kipsPstory = 144 kips + 4(36 kips)= 288 kipsRM = 1 0.15(Pmf Pstory ) (Spec. Eq. A-8-8)1 0.15(144kips 288kips)0.925= =H = 0.250 in.H = 1.21 kipsL = 240 in.Pmf = 47.5 kips + 48.5 kips= 96 kipsPstory = 96 kips + 4(24 kips)= 192 kipsRM = 1 0.15(Pmf Pstory ) (Spec. Eq. A-8-8)1 0.15(96kips 192kips)0.925= =H = 0.167 in.H = 0.806 kipsL = 240 in.Return to Table of Contents 27. C-14LRFD ASDP = R HL= = Design Examples V14.01 1AMERICAN INSTITUTE OF STEEL CONSTRUCTIONP = R HLestory MH(Spec. Eq. A-8-7)1.21 kips (240 in.)0.9250.250 in.1,070 kips== = 1.0021 1= 1 storye storyBPP(Spec. Eq. A-8-6)1 1= ( )1.00 288 kips11,070 kips1.37=estory MH(Spec. Eq. A-8-7)0.806 kips (240 in.)0.9250.167 in.1,070 kips== = 1.6021 11 storye storyBPP(Spec. Eq. A-8-6)( )1.60 192 kips11,070 kips1.40=The assumption that the ratio of the maximum second-order drift to the maximum first-order drift is no greaterthan 1.5 is correct; therefore, the first-order analysis method is permitted.Check the assumption that Pr 0.5Py and therefore, the first-order analysis method is permitted.0.5Py = 0.5FyAg= 0.5(50 ksi)(19.1 in.2)= 478 kipsLRFD ASDPr = 1.0(72.8 kips)= 72.8 kips < 478 kips o.k.Pr = 1.6(48.5 kips)= 77.6 kips < 478 kips o.k.The assumption that the first-order analysis method can be used is verified.Although the second-order sway multiplier is approximately 1.4, the change in bending moment is small becausethe only sway moments are those produced by the small notional loads. For load combinations with significantgravity and lateral loadings, the increase in bending moments is larger.Verify the column strengths using the second-order forces, using the following effective lengths (calculations notshown):Columns:Use KLx = 20.0 ftUse KLy = 20.0 ftReturn to Table of Contents 28. D-1Chapter DDesign of Members for TensionD1. SLENDERNESS LIMITATIONSSection D1 does not establish a slenderness limit for tension members, but recommends limiting L/r to amaximum of 300. This is not an absolute requirement. Rods and hangers are specifically excluded from thisrecommendation.D2. TENSILE STRENGTHBoth tensile yielding strength and tensile rupture strength must be considered for the design of tension members.It is not unusual for tensile rupture strength to govern the design of a tension member, particularly for smallmembers with holes or heavier sections with multiple rows of holes.For preliminary design, tables are provided in Part 5 of the AISC Manual for W-shapes, L-shapes, WT-shapes,rectangular HSS, square HSS, round HSS, Pipe and 2L-shapes. The calculations in these tables for availabletensile rupture strength assume an effective area, Ae, of 0.75Ag. If the actual effective area is greater than 0.75Ag,the tabulated values will be conservative and calculations can be performed to obtain higher available strengths. Ifthe actual effective area is less than 0.75Ag, the tabulated values will be unconservative and calculations arenecessary to determine the available strength.D3. EFFECTIVE NET AREAThe gross area, Ag, is the total cross-sectional area of the member.In computing net area, An, AISC Specification Section B4.3 requires that an extra z in. be added to the bolt holediameter.A computation of the effective area for a chain of holes is presented in Example D.9.Unless all elements of the cross section are connected, Ae = AnU, where U is a reduction factor to account forshear lag. The appropriate values of U can be obtained from Table D3.1 of the AISC Specification.D4. BUILT-UP MEMBERSThe limitations for connections of built-up members are discussed in Section D4 of the AISC Specification.D5. PIN-CONNECTED MEMBERSAn example of a pin-connected member is given in Example D.7.D6. EYEBARSAn example of an eyebar is given in Example D.8. The strength of an eyebar meeting the dimensionalrequirements of AISC Specification Section D6 is governed by tensile yielding of the body.Design Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONReturn to Table of Contents 29. Return to Table of ContentsD-2EXAMPLE D.1 W-SHAPE TENSION MEMBERGiven:Select an 8-in. W-shape, ASTM A992, to carry a dead load of 30 kips and a live load of 90 kips in tension. Themember is 25 ft long. Verify the member strength by both LRFD and ASD with the bolted end connection shown.Verify that the member satisfies the recommended slenderness limit. Assume that connection limit states do notgovern.Solution:From Chapter 2 of ASCE/SEI 7, the required tensile strength is:LRFD ASDDesign Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONPu = 1.2(30 kips) + 1.6(90 kips)= 180 kipsPa = 30 kips + 90 kips= 120 kipsFrom AISC Manual Table 5-1, try a W821.From AISC Manual Table 2-4, the material properties are as follows:W821ASTM A992Fy = 50 ksiFu = 65 ksiFrom AISC Manual Tables 1-1 and 1-8, the geometric properties are as follows:W821Ag = 6.16 in.2bf = 5.27 in.tf = 0.400 in.d = 8.28 in.ry = 1.26 in.WT410.5y = 0.831 in.Tensile YieldingFrom AISC Manual Table 5-1, the tensile yielding strength is:LRFD ASD277 kips > 180 kips o.k. 184 kips > 120 kips o.k. 30. D-3Tensile RuptureVerify the table assumption that Ae/Ag 0.75 for this connection.Calculate the shear lag factor, U, as the larger of the values from AISC Specification Section D3, Table D3.1 case2 and case 7.From AISC Specification Section D3, for open cross sections, U need not be less than the ratio of the gross area ofthe connected element(s) to the member gross area.Design Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONb tA= 2U =2 f fg2(5.27 in.)(0.400 in.)6.16 in.= 0.684Case 2: Check as two WT-shapes per AISC Specification Commentary Figure C-D3.1, with x = y = 0.831 in.U =1 xl=1 0.831 in.= 0.9089.00 in.Case 7:bf = 5.27 in.d = 8.28 in.bf < qdU = 0.85Use U = 0.908.Calculate An using AISC Specification Section B4.3.An = Ag 4(dh + z in.)tf= 6.16 in.2 4(m in. + z in.)(0.400 in.)= 4.76 in.2Calculate Ae using AISC Specification Section D3.Ae = AnU (Spec. Eq. D3-1)= 4.76 in.2(0.908)= 4.32 in.2224.32 in.6.16 in.AAeg== 0.701 < 0.75; therefore, table values for rupture are not valid.The available tensile rupture strength is,Return to Table of Contents 31. Return to Table of ContentsD-4Pn = FuAe (Spec. Eq. D2-2)P =Design Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTION= 65 ksi(4.32 in.2)= 281 kipsFrom AISC Specification Section D2, the available tensile rupture strength is:LRFD ASDt = 0.75tPn = 0.75(281 kips)= 211 kips211 kips > 180 kips o.k.t = 2.00281 kips2.00nt= 141 kips141 kips > 120 kips o.k.Check Recommended Slenderness Limit25.0 ft 12.0 in.1.26 in. ftLr= = 238 < 300 from AISC Specification Section D1 o.k.The W821 available tensile strength is governed by the tensile rupture limit state at the end connection.See Chapter J for illustrations of connection limit state checks. 32. D-5EXAMPLE D.2 SINGLE ANGLE TENSION MEMBERGiven:Verify, by both ASD and LRFD, the tensile strength of an L442, ASTM A36, with one line of (4) w-in.-diameter bolts in standard holes. The member carries a dead load of 20 kips and a live load of 60 kips in tension.Calculate at what length this tension member would cease to satisfy the recommended slenderness limit. Assumethat connection limit states do not govern.Solution:From AISC Manual Table 2-4, the material properties are as follows:L442ASTM A36Fy = 36 ksiFu = 58 ksiFrom AISC Manual Table 1-7, the geometric properties are as follows:L442Ag = 3.75 in.2rz = 0.776 in.y = 1.18 in. = xFrom Chapter 2 of ASCE/SEI 7, the required tensile strength is:LRFD ASDP =Design Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONPu = 1.2(20 kips) + 1.6(60 kips)= 120 kipsPa = 20 kips + 60 kips= 80.0 kipsTensile YieldingPn = FyAg (Spec. Eq. D2-1)= 36 ksi(3.75 in.2)= 135 kipsFrom AISC Specification Section D2, the available tensile yielding strength is:LRFD ASDt = 0.90tPn = 0.90(135 kips)= 122 kipst = 1.67135 kips1.67nt= 80.8 kipsReturn to Table of Contents 33. D-6Tensile RuptureCalculate U as the larger of the values from AISC Specification Section D3, Table D3.1 Case 2 and Case 8.From AISC Specification Section D3, for open cross sections, U need not be less than the ratio of the gross area ofthe connected element(s) to the member gross area, therefore,U = 0.500Case 2:U =1 xP =P =80.8 kips > 80.0 kips o.k.Design Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONl=1 1.18in.= 0.8699.00 in.Case 8, with 4 or more fasteners per line in the direction of loading:U = 0.80Use U = 0.869.Calculate An using AISC Specification Section B4.3.An = Ag (dh + z)t= 3.75 in.2 (m in. + z in.)(2 in.)= 3.31 in.2Calculate Ae using AISC Specification Section D3.Ae = AnU (Spec. Eq. D3-1)= 3.31 in.2(0.869)= 2.88 in.2Pn = FuAe (Spec. Eq. D2-2)= 58 ksi(2.88 in.2)= 167 kipsFrom AISC Specification Section D2, the available tensile rupture strength is:LRFD ASDt = 0.75tPn = 0.75(167 kips)= 125 kipst = 2.00167 kips2.00nt= 83.5 kipsThe L442 available tensile strength is governed by the tensile yielding limit state.LRFD ASDtPn = 122 kips122 kips > 120 kips o.k.n 80.8 kipstReturn to Table of Contents 34. D-7Design Examples V14.0Recommended LmaxUsing AISC Specification Section D1:Lmax = 300rzAMERICAN INSTITUTE OF STEEL CONSTRUCTION= (300)(0.776in.) ft 12.0 in.= 19.4 ftNote: The L/r limit is a recommendation, not a requirement.See Chapter J for illustrations of connection limit state checks.Return to Table of Contents 35. Return to Table of ContentsD-8EXAMPLE D.3 WT-SHAPE TENSION MEMBERGiven:A WT620, ASTM A992 member has a length of 30 ft and carries a dead load of 40 kips and a live load of 120kips in tension. The end connection is fillet welded on each side for 16 in. Verify the member tensile strength byboth LRFD and ASD. Assume that the gusset plate and the weld are satisfactory.Solution:From AISC Manual Table 2-4, the material properties are as follows:WT620ASTM A992Fy = 50 ksiFu = 65 ksiFrom AISC Manual Table 1-8, the geometric properties are as follows:WT620Ag = 5.84 in.2bf = 8.01 in.tf = 0.515 in.rx = 1.57 in.y = 1.09 in. = x (in equation for U)From Chapter 2 of ASCE/SEI 7, the required tensile strength is:LRFD ASDDesign Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONPu = 1.2(40 kips) + 1.6(120 kips)= 240 kipsPa = 40 kips + 120 kips= 160 kipsTensile YieldingCheck tensile yielding limit state using AISC Manual Table 5-3.LRFD ASDtPn = 263 kips > 240 kips o.k. ntP= 175 kips > 160 kips o.k. 36. D-9Tensile RuptureCheck tensile rupture limit state using AISC Manual Table 5-3.LRFD ASDtPn = 214 kips < 240 kips n.g. ntP= 142 kips < 160 kips n.g.The tabulated available rupture strengths may be conservative for this case; therefore, calculate the exact solution.Calculate U as the larger of the values from AISC Specification Section D3 and Table D3.1 case 2.From AISC Specification Section D3, for open cross-sections, U need not be less than the ratio of the gross areaof the connected element(s) to the member gross area.Design Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONb tA= 2U = f fg8.01 in.(0.515 in.)5.84 in.= 0.706Case 2:U =1 xl=1 1.09in.= 0.93216.0 in.Use U = 0.932.Calculate An using AISC Specification Section B4.3.An = Ag (because there are no reductions due to holes or notches)= 5.84 in.2Calculate Ae using AISC Specification Section D3.Ae = AnU (Spec. Eq. D3-1)= 5.84 in.2(0.932)= 5.44 in.2Calculate Pn.Pn = FuAe (Spec. Eq. D2-2)= 65 ksi(5.44 in.2)= 354 kipsReturn to Table of Contents 37. D-10From AISC Specification Section D2, the available tensile rupture strength is:LRFD ASDP =PPDesign Examples V14.0 5.44 i n. 0.75 5.84 in.AMERICAN INSTITUTE OF STEEL CONSTRUCTIONt = 0.75tPn = 0.75(354 kips)= 266 kips266 kips > 240 kips o.k.t = 2.00354 kips2.00nt= 177 kips177 kips > 160 kips o.k.Alternately, the available tensile rupture strengths can be determined by modifying the tabulated values. Theavailable tensile rupture strengths published in the tension member selection tables are based on the assumptionthat Ae = 0.75Ag. The actual available strengths can be determined by adjusting the values from AISC ManualTable 5-3 as follows:LRFD ASDtPn = 214 kips A e 0.75Ag 5.44 i n.2 0.75 5.84 in.2= 214 kips ( )= 266 kipsnt= 142 kips A e 0.75Ag22= 142 kips ( )= 176 kipsThe WT620 available tensile strength is governed by the tensile yielding limit state.LRFD ASDtPn = 263 kips263 kips > 240 kips o.k.nt= 175 kips175 kips > 160 kips o.k.Recommended Slenderness Limit30.0 f t 12.0 in.1.57 in. ftLr= = 229 < 300 from AISC Specification Section D1 o.k.See Chapter J for illustrations of connection limit state checks.Return to Table of Contents 38. D-11EXAMPLE D.4 RECTANGULAR HSS TENSION MEMBERGiven:Verify the tensile strength of an HSS64a, ASTM A500 Grade B, with a length of 30 ft. The member iscarrying a dead load of 35 kips and a live load of 105 kips in tension. The end connection is a fillet welded 2-in.-thick single concentric gusset plate with a weld length of 16 in. Assume that the gusset plate and weld aresatisfactory.Solution:From AISC Manual Table 2-4, the material properties are as follows:ASTM A500 Grade BFy = 46 ksiFu = 58 ksiFrom AISC Manual Table 1-11, the geometric properties are as follows:HSS64aAg = 6.18 in.2ry = 1.55 in.t = 0.349 in.From Chapter 2 of ASCE/SEI 7, the required tensile strength is:LRFD ASDDesign Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONPu = 1.2(35 kips) + 1.6(105 kips)= 210 kipsPa = 35 kips + 105 kips= 140 kipsTensile YieldingCheck tensile yielding limit state using AISC Manual Table 5-4.LRFD ASDtPn = 256 kips > 210 kips o.k. ntP= 170 kips > 140 kips o.k.Tensile RuptureCheck tensile rupture limit state using AISC Manual Table 5-4.Return to Table of Contents 39. D-12LRFD ASDtPn = 201 kips < 210 kips n.g. ntP= 134 kips < 140 kips n.g.The tabulated available rupture strengths may be conservative in this case; therefore, calculate the exact solution.Calculate U from AISC Specification Table D3.1 case 6.P =Design Examples V14.04.00 in. 2 2 4.00 in. 6.00 in.AMERICAN INSTITUTE OF STEEL CONSTRUCTIONx =2 24( )B +BHB +H=( ) +( )( )( )4 4.00 in. +6.00 in.= 1.60 in.U =1 xl=1 1.60in.= 0.90016.0 in.Allowing for a z-in. gap in fit-up between the HSS and the gusset plate:An = Ag 2(tp + z in.)t= 6.18 in.2 2(2 in. + z in.)(0.349 in.)= 5.79 in.2Calculate Ae using AISC Specification Section D3.Ae = AnU (Spec. Eq. D3-1)= 5.79 in.2(0.900)= 5.21 in.2Calculate Pn.Pn = FuAe (Spec. Eq. D2-2)= 58 ksi(5.21 in2)= 302 kipsFrom AISC Specification Section D2, the available tensile rupture strength is:LRFD ASDt = 0.75tPn = 0.75(302 kips)= 227 kips227 kips > 210 kips o.k.t = 2.00302 kips2.00nt= 151 kips151 kips > 140 kips o.k.The HSS available tensile strength is governed by the tensile rupture limit state.Return to Table of Contents 40. D-13Design Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONRecommended Slenderness Limit30.0 ft 12.0 in.1.55 in. ftLr= = 232 < 300 from AISC Specification Section D1 o.k.See Chapter J for illustrations of connection limit state checks.Return to Table of Contents 41. D-14EXAMPLE D.5 ROUND HSS TENSION MEMBERGiven:Verify the tensile strength of an HSS60.500, ASTM A500 Grade B, with a length of 30 ft. The member carries adead load of 40 kips and a live load of 120 kips in tension. Assume the end connection is a fillet welded 2-in.-thick single concentric gusset plate with a weld length of 16 in. Assume that the gusset plate and weld aresatisfactory.Solution:From AISC Manual Table 2-4, the material properties are as follows:ASTM A500 Grade BFy = 42 ksiFu = 58 ksiFrom AISC Manual Table 1-13, the geometric properties are as follows:HSS60.500Ag = 8.09 in.2r = 1.96 in.t = 0.465 in.From Chapter 2 of ASCE/SEI 7, the required tensile strength is:LRFD ASDDesign Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONPu = 1.2(40 kips) + 1.6(120 kips)= 240 kipsPa = 40 kips + 120 kips= 160 kipsTensile YieldingCheck tensile yielding limit state using AISC Manual Table 5-6.LRFD ASDtPn = 306 kips > 240 kips o.k. ntP= 203 kips > 160 kips o.k.Tensile RuptureCheck tensile rupture limit state using AISC Manual Table 5-6.Return to Table of Contents 42. D-15LRFD ASDtPn = 264 kips > 240 kips o.k. ntP= 176 kips > 160 kips o.k.Check that Ae/Ag 0.75 as assumed in table.Determine U from AISC Specification Table D3.1 Case 5.L = 16.0 in.D = 6.00 in.P =Design Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTION16.0 in.6.00 in.LD== 2.67 > 1.3, therefore U = 1.0Allowing for a z-in. gap in fit-up between the HSS and the gusset plate,An = Ag 2(tp + z in.)t= 8.09 in.2 2(0.500 in. + z in.)(0.465 in.)= 7.57 in.2Calculate Ae using AISC Specification Section D3.Ae = AnU (Spec. Eq. D3-1)= 7.57 in.2 (1.0)= 7.57 in.2227.57 in.8.09 in.AAeg== 0.936 > 0.75 o.k., but conservativeCalculate Pn.Pn = FuAe (Spec. Eq. D2-2)= (58 ksi)(7.57 in.2)= 439 kipsFrom AISC Specification Section D2, the available tensile rupture strength is:LRFD ASDt = 0.75tPn = 0.75(439 kips)= 329 kips329 kips > 240 kips o.k.t = 2.00439 kips2.00nt= 220 kips220 kips > 160 kips o.k.Recommended Slenderness Limit30.0 ft 12.0 in.1.96 in. ftLr= = 184 < 300 from AISC Specification Section D1 o.k.Return to Table of Contents 43. D-16See Chapter J for illustrations of connection limit state checks.Design Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONReturn to Table of Contents 44. D-17EXAMPLE D.6 DOUBLE ANGLE TENSION MEMBERGiven:A 2L442 (a-in. separation), ASTM A36, has one line of (8) -in.-diameter bolts in standard holes and is 25 ftin length. The double angle is carrying a dead load of 40 kips and a live load of 120 kips in tension. Verify themember tensile strength. Assume that the gusset plate and bolts are satisfactory.Solution:From AISC Manual Table 2-4, the material properties are as follows:ASTM A36Fy = 36 ksiFu = 58 ksiFrom AISC Manual Tables 1-7 and 1-15, the geometric properties are as follows:L442Ag = 3.75 in.2x = 1.18 in.2L442 (s = a in.)ry = 1.83 in.rx = 1.21 in.From Chapter 2 of ASCE/SEI 7, the required tensile strength is:LRFD ASDDesign Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONPn = 1.2(40 kips) + 1.6(120 kips)= 240 kipsPn = 40 kips + 120 kips= 160 kipsTensile YieldingPn = FyAg (Spec. Eq. D2-1)= 36 ksi(2)(3.75 in.2)= 270 kipsReturn to Table of Contents 45. D-18From AISC Specification Section D2, the available tensile yielding strength is:LRFD ASDP =Design Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONt = 0.90tPn = 0.90(270 kips)= 243 kipst = 1.67270 kips1.67nt= 162 kipsTensile RuptureCalculate U as the larger of the values from AISC Specification Section D3, Table D3.1 case 2 and case 8.From AISC Specification Section D3, for open cross-sections, U need not be less than the ratio of the gross areaof the connected element(s) to the member gross area.U = 0.500Case 2:U = 1 xl= 1 1.18in.21.0 in.= 0.944Case 8, with 4 or more fasteners per line in the direction of loading:U = 0.80Use U = 0.944.Calculate An using AISC Specification Section B4.3.An = Ag 2(dh + z in.)t= 2(3.75 in.2) 2(m in. + z in.)(2 in.)= 6.63 in.2Calculate Ae using AISC Specification Section D3.Ae = AnU (Spec. Eq. D3-1)= 6.63 in.2(0.944)= 6.26 in.2Calculate Pn.Pn = FuAe (Spec. Eq. D2-2)= 58 ksi(6.26 in.2)= 363 kipsFrom AISC Specification Section D2, the available tensile rupture strength is:Return to Table of Contents 46. Return to Table of ContentsD-19LRFD ASDP == = 248 < 300 from AISC Specification Section D1 o.k.Design Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTIONt = 0.75tPn = 0.75(363 kips)= 272 kipst = 2.00363 kips2.00nt= 182 kipsThe double angle available tensile strength is governed by the tensile yielding limit state.LRFD ASD243 kips > 240 kips o.k.162 kips > 160 kips o.k.Recommended Slenderness Limit25.0 ft 12.0 in.Lrx 1.21 in. ftNote: From AISC Specification Section D4, the longitudinal spacing of connectors between components of built-upmembers should preferably limit the slenderness ratio in any component between the connectors to a maximumof 300.See Chapter J for illustrations of connection limit state checks. 47. D-20EXAMPLE D.7 PIN-CONNECTED TENSION MEMBERGiven:An ASTM A36 pin-connected tension member with the dimensions shown as follows carries a dead load of 4 kipsand a live load of 12 kips in tension. The diameter of the pin is 1 inch, in a Q-in. oversized hole. Assume that thepin itself is adequate. Verify the member tensile strength.Solution:From AISC Manual Table 2-5, the material properties are as follows:PlateASTM A36Fy = 36 ksiFu = 58 ksiThe geometric properties are as follows:w = 4.25 in.t = 0.500 in.d = 1.00 in.a = 2.25 in.c = 2.50 in.dh = 1.03 in.Check dimensional requirements using AISC Specification Section D5.2.1. be = 2t + 0.63 in.Design Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTION= 2(0.500 in.) + 0.63 in.= 1.63 in. 1.61 in.be = 1.61 in. controls2. a > 1.33be2.25 in. > 1.33(1.61 in.)= 2.14 in. o.k.Return to Table of Contents 48. D-214. c > a2.50 in. > 2.25 in. o.k.From Chapter 2 of ASCE/SEI 7, the required tensile strength is:LRFD ASDP =P =Design Examples V14.03. w > 2be + d4.25 in. > 2(1.61 in.) + 1.00 in.= 4.22 in. o.k.AMERICAN INSTITUTE OF STEEL CONSTRUCTIONPu = 1.2(4 kips) + 1.6(12 kips)= 24.0 kipsPa = 4 kips + 12 kips= 16.0 kipsTensile RuptureCalculate the available tensile rupture strength on the effective net area.Pn = Fu(2tbe) (Spec. Eq. D5-1)= 58 ksi (2)(0.500 in.)(1.61 in.)= 93.4 kipsFrom AISC Specification Section D5.1, the available tensile rupture strength is:LRFD ASDt = 0.75tPn = 0.75(93.4 kips)= 70.1 kipst = 2.0093.4 kips2.00nt= 46.7 kipsShear RuptureAsf = 2t(a + d/2)= 2(0.500 in.)[2.25 in. + (1.00 in./2)]= 2.75 in.2Pn = 0.6FuAsf (Spec. Eq. D5-2)= 0.6(58 ksi)(2.75 in.2)= 95.7 kipsFrom AISC Specification Section D5.1, the available shear rupture strength is:LRFD ASDsf = 0.75sfPn = 0.75(95.7 kips)= 71.8 kipssf = 2.0095.7 kips2.00nsf= 47.9 kipsBearingApb = 0.500 in.(1.00 in.)= 0.500 in.2Return to Table of Contents 49. D-22Rn = 1.8FyApb (Spec. Eq. J7-1)Pn == 16.2 kipsP =Design Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTION= 1.8(36 ksi)(0.500 in.2)= 32.4 kipsFrom AISC Specification Section J7, the available bearing strength is:LRFD ASD = 0.75Pn = 0.75(32.4 kips)= 24.3 kips = 2.0032.4 kips2.00Tensile YieldingAg = wt= 4.25 in. (0.500 in.)= 2.13 in.2Pn = FyAg (Spec. Eq. D2-1)= 36 ksi (2.13 in.2)= 76.7 kipsFrom AISC Specification Section D2, the available tensile yielding strength is:LRFD ASDt = 0.90tPn = 0.90(76.7 kips)= 69.0 kipst = 1.6776.7 kips1.67nt= 45.9 kipsThe available tensile strength is governed by the bearing strength limit state.LRFD ASDPn = 24.3 kips24.3 kips > 24.0 kips o.k.Pn= 16.2 kips16.2 kips > 16.0 kips o.k.See Example J.6 for an illustration of the limit state calculations for a pin in a drilled hole.Return to Table of Contents 50. D-23EXAMPLE D.8 EYEBAR TENSION MEMBERGiven:A s-in.-thick, ASTM A36 eyebar member as shown, carries a dead load of 25 kips and a live load of 15 kips intension. The pin diameter, d, is 3 in. Verify the member tensile strength.Solution:From AISC Manual Table 2-5, the material properties are as follows:PlateASTM A36Fy = 36 ksiFu = 58 ksiThe geometric properties are as follows:w = 3.00 in.b = 2.23 in.t = s in.dhead = 7.50 in.d = 3.00 in.dh = 3.03 in.R = 8.00 in.Check dimensional requirements using AISC Specification Section D6.1 and D6.2.1. t > 2 in.Design Examples V14.0s in. > 2 in. o.k.AMERICAN INSTITUTE OF STEEL CONSTRUCTION2. w < 8t3.00 in. < 8(s in.)= 5.00 in. o.k.Return to Table of Contents 51. Return to Table of ContentsD-24P =Design Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTION3. d > d w3.00 in. > d(3.00 in.)= 2.63 in. o.k.4. dh < d + Q in.3.03 in. < 3.00 in. + (Q in.)= 3.03 in. o.k.5. R > dhead8.00 in. > 7.50 in. o.k.6. q w < b < w wq(3.00 in.) < 2.23 in. < w(3.00 in.)2.00 in. < 2.23 in. < 2.25 in. o.k.From Chapter 2 of ASCE/SEI 7, the required tensile strength is:LRFD ASDPu = 1.2(25.0 kips) + 1.6(15.0 kips)= 54.0 kipsPa = 25.0 kips + 15.0 kips= 40.0 kipsTensile YieldingCalculate the available tensile yielding strength at the eyebar body (at w).Ag = wt= 3.00 in.(s in.)= 1.88 in.2Pn = FyAg (Spec. Eq. D2-1)= 36 ksi(1.88 in.2)= 67.7 kipsFrom AISC Specification Section D2, the available tensile yielding strength is:LRFD ASDt = 0.90tPn = 0.90(67.7 kips)= 60.9 kips60.9 kips > 54.0 kips o.k.t = 1.6767.7 kips1.67nt= 40.5 kips40.5 kips > 40.0 kips o.k.The eyebar tension member available strength is governed by the tensile yielding limit state.Note: The eyebar detailing limitations ensure that the tensile yielding limit state at the eyebar body will controlthe strength of the eyebar itself. The pin should also be checked for shear yielding, and, if the material strength isless than that of the eyebar, bearing.See Example J.6 for an illustration of the limit state calculations for a pin in a drilled hole. 52. D-25EXAMPLE D.9 PLATE WITH STAGGERED BOLTSGiven:Compute An and Ae for a 14-in.-wide and 2-in.-thick plate subject to tensile loading with staggered holes asshown.Solution:Calculate net hole diameter using AISC Specification Section B4.3.dnet = dh + z in.= + from AISC Specification Section B4.3.Line A-B-E-F: w = 14.0 in. 2(0.875 in.)2 2 2.50in. 2.50in.w= + += 11.5 in. controls2 2.50in.2 2 2.50in. 2.50in.w= + += 12.1 in.Design Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTION= m in. + z in.= 0.875 in.Compute the net width for all possible paths across the plate. Because of symmetry, many of the net widths areidentical and need not be calculated.2w d s14.04 netg= 12.3 in.Line A-B-C-D-E-F: ( ) ( )( )( )( )14.0in. 4 0.875in.4 3.00in. 4 3.00in.Line A-B-C-D-G: ( ) ( )( )14.0in. 3 0.875in.4 3.00in.w= += 11.9 in.Line A-B-D-E-F: ( ) ( )( )( )( )14.0in. 3 0.875in.4 7.00in. 4 3.00in.Therefore, An = 11.5 in.(0.500 in.)Return to Table of Contents 53. D-26Design Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTION= 5.75 in.2Calculate U.From AISC Specification Table D3.1 case 1, because tension load is transmitted to all elements by the fasteners,U = 1.0Ae = AnU (Spec. Eq. D3-1)= 5.75 in.2(1.0)= 5.75 in.2Return to Table of Contents 54. Return to Table of ContentsE-DesignExamples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTION1Chapter EDesign of Members for CompressionThis chapter covers the design of compression members, the most common of which are columns. The AISCManual includes design tables for the following compression member types in their most commonly availablegrades. wide-flange column shapes HSS double angles single anglesLRFD and ASD information is presented side-by-side for quick selection, design or verification. All of the tablesaccount for the reduced strength of sections with slender elements.The design and selection method for both LRFD and ASD designs is similar to that of previous AISCSpecifications, and will provide similar designs. In this AISC Specification, ASD and LRFD will provide identicaldesigns when the live load is approximately three times the dead load.The design of built-up shapes with slender elements can be tedious and time consuming, and it is recommendedthat standard rolled shapes be used, when possible.E1. GENERAL PROVISIONSThe design compressive strength, cPn, and the allowable compressive strength, Pn/c, are determined as follows:Pn = nominal compressive strength based on the controlling buckling modec = 0.90 (LRFD) c = 1.67 (ASD)Because Fcr is used extensively in calculations for compression members, it has been tabulated in AISC ManualTable 4-22 for all of the common steel yield strengths.E2. EFFECTIVE LENGTHIn the AISC Specification, there is no limit on slenderness, KL/r. Per the AISC Specification Commentary, it isrecommended that KL/r not exceed 200, as a practical limit based on professional judgment and constructioneconomics.Although there is no restriction on the unbraced length of columns, the tables of the AISC Manual are stopped atcommon or practical lengths for ordinary usage. For example, a double L33, with a a-in. separation has an ryof 1.38 in. At a KL/r of 200, this strut would be 23-0 long. This is thought to be a reasonable limit based onfabrication and handling requirements.Throughout the AISC Manual, shapes that contain slender elements when supplied in their most common materialgrade are footnoted with the letter c. For example, see a W1422c.E3. FLEXURAL BUCKLING OF MEMBERS WITHOUT SLENDER ELEMENTS 55. Return to Table of ContentsE-DesignE3-2InelasticbucklingTransition betweenequations (locationvaries by Fy)KL/rFig. E-1. Standard column curve.Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTION2Nonslender sections, including nonslender built-up I-shaped columns and nonslender HSS columns, are governedby these provisions. The general design curve for critical stress versus KL/r is shown in Figure E-1.The term L is used throughout this chapter to describe the length between points that are braced against lateraland/or rotational displacement.E3-3ElasticbucklingTRANSITION POINT LIMITING VALUES OF KL/rCriticalstress,ksiFy , ksi Limiting KL/r 0.44Fy , ksi36 134 15.850 113 22.060 104 26.470 96 30.8E4. TORSIONAL AND FLEXURAL-TORSIONAL BUCKLING OF MEMBERS WITHOUT SLENDERELEMENTSThis section is most commonly applicable to double angles and WT sections, which are singly-symmetric shapessubject to torsional and flexural-torsional buckling. The available strengths in axial compression of these shapesare tabulated in Part 4 of the AISC Manual and examples on the use of these tables have been included in thischapter for the shapes with KLz = KLy.E5. SINGLE ANGLE COMPRESSION MEMBERSThe available strength of single angle compression members is tabulated in Part 4 of the AISC Manual.E6. BUILT-UP MEMBERSThere are no tables for built-up shapes in the AISC Manual, due to the number of possible geometries. Thissection suggests the selection of built-up members without slender elements, thereby making the analysisrelatively straightforward. 56. Return to Table of ContentsE-DesignExamples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTION3E7. MEMBERS WITH SLENDER ELEMENTSThe design of these members is similar to members without slender elements except that the formulas aremodified by a reduction factor for slender elements, Q. Note the similarity of Equation E7-2 with Equation E3-2,and the similarity of Equation E7-3 with Equation E3-3.The tables of Part 4 of the AISC Manual incorporate the appropriate reductions in available strength to accountfor slender elements.Design examples have been included in this Chapter for built-up I-shaped members with slender webs and slenderflanges. Examples have also been included for a double angle, WT and an HSS shape with slender elements. 57. Return to Table of ContentsE-DesignExamples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTION4EXAMPLE E.1A W-SHAPE COLUMN DESIGN WITH PINNED ENDSGiven:Select an ASTM A992 (Fy = 50 ksi) W-shape column to carry an axial dead load of140 kips and live load of 420 kips. The column is 30 ft long and is pinned top andbottom in both axes. Limit the column size to a nominal 14-in. shape.Solution:From Chapter 2 of ASCE/SEI 7, the required compressive strength is:LRFD ASDPu = 1.2(140 kips) + 1.6(420 kips)= 840 kipsPa = 140 kips + 420 kips= 560 kipsColumn SelectionFrom AISC Specification Commentary Table C-A-7.1, for a pinned-pinned condition, K = 1.0.Because the unbraced length is the same in both the x-x and y-y directions and rx exceeds ry for all W-shapes, y-yaxis bucking will govern.Enter the table with an effective length, KLy, of 30 ft, and proceed across the table until reaching the least weightshape with an available strength that equals or exceeds the required strength. Select a W14132. 58. Return to Table of ContentsE-DesignExamples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTION5From AISC Manual Table 4-1, the available strength for a y-y axis effective length of 30 ft is:LRFD ASDcPn = 893 kips > 840 kips o.k. ncP= 594 kips > 560 kips o.k. 59. E-DesignExamples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTION6EXAMPLE E.1B W-SHAPE COLUMN DESIGN WITH INTERMEDIATE BRACINGGiven:Redesign the column from Example E.1A assuming the column islaterally braced about the y-y axis and torsionally braced at the midpoint.Solution:From Chapter 2 of ASCE/SEI 7, the required compressive strength is:LRFD ASDPu = 1.2(140 kips) + 1.6(420 kips)= 840 kipsPa = 140 kips + 420 kips= 560 kipsColumn SelectionFrom AISC Specification Commentary Table C-A-7.1, for a pinned-pinned condition, K = 1.0.Because the unbraced lengths differ in the two axes, select the member using the y-y axis then verify the strengthin the x-x axis.Enter AISC Manual Table 4-1 with a y-y axis effective length, KLy, of 15 ft and proceed across the table untilreaching a shape with an available strength that equals or exceeds the required strength. Try a W1490. A 15 ftlong W1490 provides an available strength in the y-y direction of:LRFD ASDcPn = 1,000 kips ncP= 667 kipsThe rx /ry ratio for this column, shown at the bottom of AISC Manual Table 4-1, is 1.66. The equivalent y-y axiseffective length for strong axis buckling is computed as:30.0 ft1.66KL == 18.1 ftReturn to Table of Contents 60. Return to Table of ContentsE-DesignExamples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTION7From AISC Manual Table 4-1, the available strength of a W1490 with an effective length of 18 ft is:LRFD ASDcPn = 929 kips > 840 kips o.k. ncP= 618 kips > 560 kips o.k.The available compressive strength is governed by the x-x axis flexural buckling limit state.The available strengths of the columns described in Examples E.1A and E.1B are easily selected directly from theAISC Manual Tables. The available strengths can also be verified by hand calculations, as shown in the followingExamples E.1C and E.1D. 61. E-DesignFP =Examples V14.0AMERICAN INSTITUTE OF STEEL CONSTRUCTION8EXAMPLE E.1C W-SHAPE AVAILABLE STRENGTH CALCULATIONGiven:Calculate the available strength of a W14132 column with unbraced lengths of 30 ft in both axes. The materialproperties and loads are as given in Example E.1A.Solution:From AISC Manual Table 2-4, the material properties are as follows:ASTM A992Fy = 50 ksiFu = 65 ks