2012_IBC_SEAOC_SSDM_VOL4

2012

IBC

SEAOC STRUCTURAL/SEISMIC DESIGN MANUAL

Volume 4 EXAMPLES FOR STEEL STEEL-FRAMED -FRAMED BUILDINGS

®

Copyright Copyright © 2013 Structural Engineers Association Association of California. All rights reserved. This publication or any part thereof must not be reproduced in any form without the written permission of the Structural Engineers Association of California.

Publisher Structural Engineers Association of California (SEAOC) 1400 K Street, Ste. 212 Sacramento, California 95814 Telephone: (916) 447-1198; Fax: (916) 444-1501 E-mail: [email protected]; Web address: www.seaoc.org The Structural Engineers Association of California (SEAOC) is a professional association of four regional member organizations (Southern California, Northern California, San Diego, and Central California). SEAOC represents the structural engineering community in California. This document is published in keeping with SEAOC’s stated mission: To advance the structural engineering profession; to provide the public with structures of dependable performance through the application of state-of-the-art structural structural engineering principles; to assist the public in obtaining professional structural engineering services; to promote natural hazard mitigation; to provide continuing education and encourage research; to provide structural engineers with the most current information and tools to improve their practice; and to maintain the honor and dignity of the profession. SEAOC Board oversight of this publication was provided by 2012 SEAOC Board President James Amundson, S.E. and Immediate Past President Doug Hohbach, S.E.

Editor International Code Council

Disclaimer While the information presented in this document is believed to be correct, neither SEAOC nor its member organizations,, committees, writers, editors, or individuals who have contributed to this publication make organizations any warranty, warranty, expressed or implied, or assume any legal liability or responsibility for the use, application of, and/or reference to opinions, findings, conclusions, or recommendations included in this publication. The material presented in this publication should not be used for any specific application without competent examination and verification of its accuracy, suitability, and applicability. Users of information from this publication assume all liability arising from such use. First Printing: August 2013

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Vol. 4 2012 IBC SEAOC Structural/Seismic Design Manual, Vol.

Suggestions for Improvement Comments and suggestions for improvements are welcome and should be sent to the following: Structural Engineers Association of California (SEAOC) Don Schinske, Executiv Executivee Director 1400 K Street, Suite 212 Sacramento, California 95814 Telephone: Tel ephone: (916) 447-1198; Fax: (916) 444-1501 E-mail: [email protected] [email protected] g

Errata Notification SEAOC has made a substantial effort to ensure that the information in this document is accurate. In the event that corrections or clarifications are needed, these will be posted on the SEAOC web site at www.seaoc.org and on the ICC web site at www.iccsafe.org . SEAOC, at its sole discretion, may issue written errata.

2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

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Table of Contents Seismic Design Manual . . . . . . . . . . . . . . . . . . . . . . . . . . . . Preface to the 2012 IBC SEAOC Structural / / Seismic

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Preface to Volume 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Acknowledgementss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Acknowledgement

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References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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How to Use This Document. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Design Example 1 Special Moment Moment Frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Design Example 2 Special Concentrically Concentrically Braced Frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Design Example 3 Buckling-Restrained Braced Frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Design Example 4 Special Plate Shear Walls Walls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Design Example 5 Eccentrically Braced Braced Frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 Design Example 6 Multi-Panel OCBF. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 Design Example 7 Metal Deck Diaphragm Diaphragm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211 a. Bare Metal Deck (Flexible) (Flexible) Diaphragm Diaphragm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215 b. Concrete-Filled Deck (Rigid) Diaphragm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239 Design Example 8 Special Moment Frame Base Connection Connection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 Design Example 9 Braced-Frame Base Plate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281 Appendix 1: General Building Information. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

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Preface to the 2012 IBC SEAOC Structural / Seismic Design Manual The IBC SEAOC Structural / Seismic Design Manual, throughout its many editions, has served the purpose of illustrating good seismic design and the correct application of building-code provisions. The manual has bridged the gap between the discursive treatment of topics in the SEAOC Blue Book ( Recommended Lateral Force Requirements and Commentary) and real-world decisions that designers face in their practice. The examples illustrate code-compliant designs engineered to achieve good performance under severe seismic loading. In some cases simply complying with building-code requirements does not ensure good seismic response. This manual takes the approach of exceeding the minimum code requirements in such cases, with discussion of the reasons for doing so. Recent editions of the IBC SEAOC Structural / Seismic Design Manual have consisted of updates of previous editions, modified to address changes in the building code and referenced standards. Many of the adopted standards did not change between the 2006 edition of the International Building Code and the 2009 edition. The 2012 edition, which is the one used in this set of manuals, represents an extensive change of adopted standards, with many substantial changes in methodology. Additionally, this edition has been substantially revised. New examples have been included to address new code provisions and new systems, as well as to address areas in which the codes and standards provide insufficient guidance. Important examples such as the design of base-plate anchorages for steel systems and the design of diaphragms have been added. This expanded edition comprises five volumes: • • • • •

Volume 1: Code Application Examples Volume 2: Examples for Light-Frame, Tilt-Up, and Masonry Buildings Volume 3: Examples for Reinforced Concrete Buildings Volume 4: Examples for Steel-Framed Buildings Volume 5: Examples for Seismically Isolated Buildings and Buildings with Supplemental Damping

Previous editions have been three volumes. This expanded edition contains more types of systems for concrete buildings and steel buildings. These are no longer contained in the same volume. Volumes 3 and 4 of the 2012 edition replace Volume 3 of the 2009 edition. Additionally, we have fulfilled the long-standing goal of including examples addressing seismic isolation and supplemental damping. These examples are presented in the new Volume 5. In general, the provisions for developing the design base shear, distributing the base-shear-forces vertically and horizontally, checking for irregularities, etc., are illustrated in Volume 1. The other volumes contain more extensive design examples that address the requirements of the material standards (for example, ACI 318 and AISC 341) that are adopted by the IBC. Building design examples do not illustrate many of the items addressed in Volume 1 in order to permit the inclusion of less-redundant content. Each volume has been produced by a small group of authors under the direction of a manager. The managers have assembled reviewers to ensure coordination with other SEAOC work and publications, most notably the Blue Book, as well as numerical accuracy. This manual can serve as valuable tool for engineers seeking to design buildings for good seismic response. Rafael Sabelli Project Manager 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

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Preface to Volume 4 Volume 4 of the 2012 IBC SEAOC Structural / Seismic Design Manual addresses the design of steel building systems for seismic loading. Examples include the illustration of the design requirements for braced frames and moment frames, as were illustrated in previous editions, and also important interfaces with the rest of the structure. The design examples in this volume represent a range of steel structural systems. The Manual includes a set of examples that illustrate a more complete design: the design of diaphragms and collectors is illustrated, as are the design of base plates and anchorages for moment-frame and braced-frame columns. With the addition of these items this edition of the Manual offers more extensive guidance to engineers, addressing the design of these critical components of the seismic system. The design of each of these systems is governed by standards developed by the American Institute of Steel Construction (AISC). AISC produces its own Seismic Design Manual to illustrate the correct application of the AISC Seismic Provisions (AISC 341) and the AISC Prequalification Standard (AISC 358). The AISC Seismic Design Manual is a valuable resource for designers, and this volume is not intended to duplicate AISC’s efforts. This manual, for example, does not include the detailed range of options for gusset-plate design, as the AISC Seismic Design Manual addresses this design aspect thoroughly. Nevertheless, there is a fundamental difference in purpose and approach between this manual and the AISC Seismic Design Manual. The AISC Manual illustrates the code requirements, while the SEAOC Structural / Seismic Design Manual illustrates SEAOC’s recommended practices, which traditionally have gone beyond the code (or in advance of it). The design examples for base plates are important examples of design methodologies not explicitly defined by building codes. Building code provisions for these connections are difficult to apply and do not correspond well to the mechanisms of resistance. The examples herein provide a convenient and valuable alternative methodology, one that is not an illustration of explicit code requirements. The methods illustrated herein represent approaches consistent with the ductility expectations for each system and with the desired seismic response. In most cases there are several details or mechanisms that can be utilized to achieve the ductility and resistance required, and the author of each example has selected an appropriate option. In many cases alternatives are discussed. This Manual is not intended to serve as a building code or to be an exhaustive catalogue of all valid approaches and details. The Manual is presented as a set of examples in which the engineer has considered the building-code requirements in conjunction with the optimal seismic response of the system. The examples follow the recommendations of the SEAOC Blue Book and other SEAOC recommendations. The examples are intended to aid conscientious designers in crafting designs that are likely to achieve good seismic performance consistent with expectations inherent in the requirements for the systems. Rafael Sabelli Volume 4 Manager


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Acknowledgements Volume 4 of the 2012 IBC SEAOC Structural / Seismic Design Manual was written by a group of highly qualified structural engineers, chosen for their knowledge and experience with structural engineering practice and seismic design. Kevin S. Moore, S.E., SECB, Principal, Simpson Gumpertz & Heger—Examples 1 and 8 With multiple state licenses, Kevin has more than 18 years of experience in structural engineering design, analysis, and evaluation. He is the Chair of the SEAOC Structural Standards Committee, Past Chair of the SEAOC Seismology Committee, and Chair of the Seismic Subcommittee of the NCSEA Code Advisory Committee. He has written multiple papers and design examples associated with steel design, seismic forces, and structural systems. Kevin is also a member of the AISC Connection Prequalification Review Panel. www.sgh.com Rafael Sabelli, S.E., Principal, Director of Seismic Design, Walter P. Moore—Volume 4 Manager and Example 2 Rafael Sabelli is a member of the AISC Task Committee on the Seismic Provisions for Structural Steel Buildings, Chair of the AISC Seismic Design Manual committee, a member of the ASCE 7 Seismic subcommittee, and a member of the BSSC Provisions Update Committee and Code Resource Support Committee. He is the coauthor (with Michel Bruneau) of AISC Design Guide 20: Steel Plate Shear Walls as well as of numerous research papers on conventional and buckling-restrained braced frames. He has served as Chair of the Seismology Committee of the Structural Engineers Association of California and as President of the Structural Engineers Association of Northern California. Rafael was the co-recipient of the 2008 AISC T.R. Higgins Lectureship and was the 2000 NEHRP Professional Fellow in Earthquake Hazard Reduction. Anindya Dutta, S.E., Ph.D, Simpson Gumpertz & Heger—Example 3 Dr. Dutta has over 12 years of experience in structural and earthquake engineering. He has provided analysis and design of a variety structures in high seismic zones. Dr. Dutta’s experience also includes seismic evaluation and strengthening of low-rise to high-rise structures. He has taught graduate and undergraduate level courses on concrete design and structural analysis at the State University of New York at Buffalo and is a regular lecturer at the San Francisco State University’s graduate program and at the University of California at Berkeley’s extension program. He has authored a number of technical reports and journal papers as well as served as a member of the review board for ASCE’s Structural Engineering Journal. Kenneth Tam, Simpson Gumpertz & Heger—Example 3 Kenneth has more than 17 years of experience in the field of structural and earthquake engineering. His experience includes structural design and evaluation of variety of structures in high seismic zones. H e has co-authored various papers on design and analysis of buckling-restrained braced frames and has served on the ASCE41-13 Steel Subcommittee. Matthew R. Eatherton, Ph.D., S.E., Assistant Professor, Virginia Tech—Example 4 Matt has seven years of experience as a practicing structural engineer conducting high-seismic design in the San Francisco Bay Area. Now he serves on the faculty at Virginia Tech where he teaches classes on steel design, structural dynamics, and earthquake engineering. His research program includes both experimental and computational investigations of steel-plate shear walls, self-centering seismic systems, steel connections, and more. www.eatherton.cee.vt.edu


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Scott M. Adan, Ph.D., P.E., S.E., SECB, Principal, Adan Engineering—Example 5 With over 21 years of experience, Dr. Adan specializes in the investigation and design of buildings and structures. He is also actively involved in the research and development of steel moment-resisting connections. For the Structural Engineers Association of Northern California, he chairs the Steel Subcommittee. For the American Institute of Steel Construction, he is a member of the Seismic Design Manual Subcommittee, the Connection Prequalification Review Panel, and the Seismic Design Task Committee. www.adanengineering.com Anna Dix, S.E., Associate, Liftech Consultants Inc.—Example 6 Anna has eight years of practice in design and analysis of steel and concrete structures. Her focus is on special-use and marine structures including cranes, wharves, and heavy-lift and container-handling equipment. She has specialized experience with ductile tie-down systems for cranes, seismic design and analysis of steel structures, seismic crane-wharf interaction, designing ductile steel frames, and investigating fatigue cracking for various structures. In her spare time, Anna introduces engineering to inquisitive young minds. www.liftech.net Katy Briggs, S.E., Project Engineer, Thornton Tomasetti—Example 7 A licensed S.E. in the State of California, Katy Briggs has seven years of experience in structural analysis and design. She has worked on new buildings and seismic retrofits of existing buildings utilizing wood, steel, concrete, and masonry construction. These projects include education, healthcare, government, correctional, and commercial facilities. She has been involved with writing and editing design examples for steel diaphragms and special concentrically braced frames. Amit Kanvinde, Ph.D., Associate Professor of Civil and Environmental Engineering, University of California, Davis—Example 8 Amit’s research heavily focuses on the seismic response of steel structures and connections through experimentation and simulation. Pertinent to the design example, he has conducted 28 large-scale tests on column base connections and is the author of two major technical reports and several journal and conference papers on the topic of base plates. His other recent research has addressed the fracture of seismic column splices in moment frames and braces in SCBF systems. He is the recipient of the 2008 ASCE Norman Medal and the 2003 EERI Graduate Student Paper award addressing the collapse of structures. David A. Grilli, M.S., E.I.T., Graduate Student Researcher, University of California, Davis— Example 8 David is a doctoral student in the Department of Civil and Environmental Engineering at UC Davis. Through large-scale experimentation, his work addresses the seismic response of embedded and exposed column-base plates. Pertinent to this example, he is co-author of a journal article that characterizes the rotational flexibility of exposed column base connections. David was the recipient of the AISC Structural Steel Education Council scholarship in 2009, and the Farrer/Patten Award for outstanding student in Civil Engineering at UC Davis in 2012. Lindsey Maclise, Associate, Forell/Elsesser Engineers Inc.—Example 9 Lindsey is currently an Associate with Forell/Elsesser Engineers specializing in seismic design for both new construction and retrofit. She received her B.S. and M.S. from the University of California, Berkeley and is an active member of SEAONC, SEI, and EERI. She is currently serving as a Housner Fellow for her work in Sustainable Seismic Design. www.forell.com

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Additionally, a number of SEAOC members and other structural engineers helped check the examples in this volume. During its development, drafts of the examples were sent to these individuals. Their help was sought in review of code interpretations as well as detailed checking of the numerical computations. The reviewers include: Geoff Bomba Mike Cochran Andrew Cussen Tom Hale Walterio López Sara Jozefiak Ryan Kersting Benjamin Mohr Carrie Leung Thomas Nunziata Patxi Uriz Laura Whitehurst Close collaboration with the SEAOC Seismology Committee was maintained during the development of the document. The Seismology Committee has reviewed the document and provided many helpful comments and suggestions. Their assistance is gratefully acknowledged. Production and art was provided by the International Code Council.


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References Standards American Concrete Institute. ACI 318: Building Code Regulations for Reinforced Concrete, Farmington Hills, Michigan, 2011. American Institute of Steel Construction. AISC 341: Seismic Provisions for Structural Steel Buildings, Chicago, Illinois, 2010. American Institute of Steel Construction. AISC 358: Prequalified Connections for Special and Intermediate Steel Moment Frames for Seismic Applications, Chicago, Illinois, 2010. American Institute of Steel Construction. AISC 360: Specification for Structural Steel Buildings, Chicago, Illinois, 2010. American Society of C ivil Engineers. ASCE 7: Minimum Design Loads for Buildings and Other Structures. ASCE 2010. International Code Council. International Building Code (IBC). Falls Church, Virginia, 2012.

Other References American Institute of Steel Construction. Manual of Steel Construction, Chicago, Illinois, 2012. American Institute of Steel Construction. Seismic Design Manual, Chicago, Illinois, 2013. Anonymous, 1977. “Shear walls and slipforming speed Dallas’ Reunion project” Engineering News Record , 20–21, July 28. Anonymous, 1978a. “Patent problems, challenge spawn steel seismic walls” Engineering News Record , 17, January 26. Anonymous, 1978b. “Quake-proof hospital has battleship-like walls” Engineering News Record , 62–63, Sept. 21. Astaneh-Asl, A. 2005. “Design of Shear Tab Connections for Gravity and Seismic Loads,” Steel Technical Information and Product Report . Structural Steel Educational Council, CA. Basler, K. 1961. “Strength of Plate Girders in Shear” Journal of the Structural Division, ASCE, Vol. 87, No. ST7 October. Berman, J. W. and Bruneau, M. 2004. “Steel Plate Shear Walls are Not Plate Girders” AISC Engineering Journal, Third Quarter. Berman, J. W. and Bruneau, M. 2008. “Capacity Design of Vertical Boundary Elements in Steel Plate Shear Walls” AISC Engineering Journal, First Quarter.


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Bozorgnia, Y., Bertero, V., 2004. Earthquake Engineering: From Engineering Seismology to Performance-Based Engineering. CRC Press, LLC, Danvers, Massachusetts. Bruneau, M., Uang, C.M., and Sabelli, R. Ductile Design of Steel Structures. McGraw-Hill, 2011. CAN/CSA S16-09 2009. “Limit States Design of Steel Structures,” published by Canadian Standards Association. Cheng, J.J.R., and Kulak, G.L. 2000. Gusset plate connection to round HSS tension members. Engineering Journal, AISC, 4th Quarter, 133–139. Clifton, C., Bruneau, M., MacRae, G., Leon, R., Russell, A., 2011. “Steel Structures Damage from the Christchurch Earthquake of February 22, 2011,” NZST, Bulletin of the New Zealand Society for Earthquake Engineering, Vol. 44, No. 4. DeWolf, J. T., and Ricker, D. T. 1990. AISC Design Guide 1—Column Base Plates, Published by the American Institute of Steel Construction, AISC. Engelhardt, M., and Popov, E., 1989. “On Design of Eccentrically Braced Frames ,” Earthquake Spectra, EERI, Vol. 5, No. 3, 495–511. Englehardt, M. Personal correspondence and notes. 2012. Fisher, J.M. and Kloiber, L.A. 2006. “Base Plate and Anchor Rod Design,” 2nd Ed., Steel Design Guide Series No. 1, American Institute of Steel Construction, Inc., Chicago, IL. Gomez, I.R., Kanvinde A.M., and Deierlein G.G. 2010. “Exposed Column Base Connections Subjected to Axial Compression and Flexure,” Report Submitted to the American Institute of Steel Construction (AISC), Chicago, IL. Gomez, I.R., Kanvinde, A.M., and Deierlein, G.G. 2011. “Experimental investigation of shear transfer in exposed column base connections,” Engineering Journal, American Institute of Steel Construction, 4th Quarter, 246–264. ICC/SEAOC 2006. “Design Example 4—Steel Plate Shear Walls”, 2006 IBC Structural / Seismic Design Manual, Volume 3, Structural Engineers Association of California, Sacramento, California. Imanpour, A., Tremblay, R., and Davaran, A. “Seismic Evaluation of Multi-Panel Steel Concentrically Braced Frames,” 15th World Conference on Earthquake Engineering, 2012. Lehman, D., Roeder, C. 2, Johnston, S. 1, Herman D. 1, and Kotulka, B. 1 2008 “Improved Seismic Performance of Gusset Plate Connections”, ASCE Journal of Structural Engineering, Vol. 134, No. 6, 181–189. Luttrell, Larry D. 1967. “Strength and behavior of light-gage steel shear diaphragms”, Cornell Research Bulletin 67-1, sponsored by the American Iron and Steel Institute, Ithaca, NY. Moehle, Jack P., Hooper, John D., Kelly, Dominic J., and Meyer, Thomas. 2010. “Seismic design of cast-in-place concrete diaphragms, chords, and collectors: A guide for practicing engineers,” NEHRP Seismic Design Technical Brief Number 3, produced by the NEHRP Consultants Joint Venture, a partnership of the Applied Technology Council and the Consortium of Universities

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for Research in Earthquake Engineering, for the National Institute of Standards and Technology, Gaithersburg, MD, NIST GCR 10-917-4. Moore, Kevin S., Feng, Joyce Y., June 2007. “Design of RBS Connections for Special Moment Frames,” Steel Tips. Structural Steel Educational Council, Moraga, California. Myers, A.T., Kanvinde, A.M., Deierlein, G.G., and Fell B.V. 2009, “Effect of Weld Details on the Ductility of Steel Column Baseplate Connections,” Journal of Constructional Steel Research, Volume 65, Issue 6, June 2009, 1366–1373. Porter, D.M., Rockey, K.C. and Evans, H.R. 1975. “The collapse behavior of plate girders loaded in shear”, The Structural Engineer , London England, Vol. 53, No. 8., Aug. Prasad, Badri K., Thompson, Douglas S., and Sabelli, Rafael. 2009. Guide to the design of diaphragms, chords and collectors based on the 2006 IBC and ASCE/SEI 7-05, International Code Council Publications, Country Club Hills, IL. Purba, R. and Bruneau, M. 2009. “Finite-Element Investigation and Design Recommendations for Perforated Steel Plate Shear Walls” Journal of Structural Engineering, Vol. 135, No. 11, 1367–1376. Purba, R., and Bruneau, M. 2007. Design Recommendations for Perforated Steel Plate Shear Walls Technical Report MCEER-07-0011. Qu, B., and Bruneau, M. 2010. “Capacity Design of Intermediate Horizontal Boundary Elements of Steel Plate Shear Walls” Journal of Structural Engineering, Vol. 136, No. 6. Ricles, J., and Popov, E., 1989, “Composite Action in Eccentrically Braced Frames,” Journal of Structural Engineering, ASCE , Vol. 115, No. 8, 2046–2065. Roberts, T. M. and Sabouri-Ghomi, S. 1991. “Hysteretic Characteristics of Unstiffened Plate Shear Panels” Thin-Walled Structures, Elsevier Science Publishers, Great Britain, 1991. Rogers, C.A. and Tremblay, R. 2008. “Impact of Diaphragm Behavior on the Seismic Design of Low-Rise Steel Buildings”, AISC Engineering Journal, First Quarter. Sabelli, R. and Bruneau, M. 2006. AISC Design Guide 20—Steel Plate Shear Walls, Published by the American Institute of Steel Construction, AISC. Sabelli, Rafael, Sabol, Thomas A., and Easterling, Samuel W. 2011. “Seismic design of composite steel deck and concrete-filled diaphragms: A guide for practicing engineers,” NEHRP Seismic Design Technical Brief Number 5, produced by the NEHRP Consultants Joint Venture, a partnership of the Applied Technology Council and the Consortium of Universities for Research in Earthquake Engineering, for the National Institute of Standards and Technology, Gaithersburg, MD, NIST GCR 10-917-10. Schumacher, A., Grondin, G.Y. and Kulak, G.L. 1999. “Connection of Infill Panels in Steel Plate Shear Walls” Canadian Journal of Civil Engineering, Vol. 26. SDI 2004. Diaphragm design manual, Third Edition (SDI DDMO3), Steel Deck Institute, Fox Grove, IL.


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SEAOC Blue Book, Recommended Lateral Force Requirements and Commentary. Structural Engineers Association of California, Sacramento, California. SEAOC Seismology Committee 2007. “Development of System Factors,” May 2007, The SEAOC Blue Book: Seismic design Recommendations, Structural Engineers Association of California, Sacramento, CA. SEAOC Seismology Committee 2008. “Concentrically Braced Frames,” August, 2008, The SEAOC Blue Book: Seismic Design Recommendations, Structural Engineers Association of California, Sacramento, CA. Accessible via the world wide web at: http://www.seaoc.org/bluebook/index. html SEAOC Seismology Committee, FEMA 350 Task Group, 2002. “Commentary and Recommendations on FEMA 350—Appendix D,” Structural Engineers Association of California, Sacramento, CA. Stoakes, C.D., Fahnestock, L.A. “Influence of Weak-axis Flexural Yielding on Strong-axis Buckling Strength of Wide Flange Columns,” Proceedings of the Annual Stability Conference, Structural Stability Research Council, April 2012. Structural Engineers Association of California (SEAOC) Seismology Committee, 2008. SEAOC blue book: Seismic design recommendations, Structural Engineers Association of California, Sacramento, CA. Thornton, W.A., and Fortney, P. 2012, “Satisfying Inelastic Rotation Requirements for In-plane Critical Axis Brace Buckling for High Seismic Design.” Engineering Journal, AISC, Vol. 49, No. 3, 3rd Quarter. Tremblay, R. 2001, “Seismic Behavior and Design of Concentrically Braced Steel Frames,” Engineering Journal, AISC, Vol. 38, No. 3, Chicago, IL. Tremblay, R., Archambault, M.-H., Filiatrault, A. “Seismic Response of Concentrically Braced Steel Frames Made with Rectangular Hollow Bracing Members,” December, 2003, Article 2003. 129:1626–1636, Journal of Structural Engineering, American Society of Civil Engineers. Tremblay, R., et al. “Seismic Design of Steel Structures in Accordance with CSA-S16-09,” July 25– 29, 2010, Paper No. 1768, Proceedings of the 9th US National and 10th Canadian Conference on Earthquake Engineering, Toronto, Ontario, Canada. Vian, D., and Bruneau, M. 2005. “Steel Plate Shear Walls for Seismic Design and Retrofit of Building Structures” Technical Report MCEER 05-0010. Vian, D., Bruneau, M., Tsai, K.C., and Lin, Y.-C. 2009. “Special Perforated Steel Plate Shear Walls with Reduced Beam Section Anchor Beams 1: Experimental Investigation” Journal of Structural Engineering, Vol. 135, No. 3, 211–220. Wong, Alfred F. “Multi-tier Bracing Panels within a Storey,” Advantage Steel, Canadian Institute of Steel Construction, No. 43, Summer 2012. Zayas, V., Mahin, S., Popov, E. “Cyclic Inelastic Behavior of Steel Offshore Structures,” August 1980, Report No. UCB/EERC-80/27 to the American Petroleum Institute, Earthquake Engineering Research Center & College of Engineering at University of California, Berkeley.

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How to Use This Document Equation numbers in the right-hand margin refer to the one of the standards (e.g., AISC 341, AISC 358, AISC 360, ASCE 7). The default standard is given in the heading of each section of each example; equation numbers in that section refer to that standard unless another standard is explicitly cited. Abbreviations used in the “Code Reference” column are § – Section

T – Table

F – Figure

Eq – Equation


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Design Example 1 Special Moment Frame

OVERVIEW Structural steel special moment frames (SMF) are typically comprised of wide-flange beams, columns, and beam-column connections. Connections are proportioned and detailed to resist internal forces (flexural, axial, and shear) that result from imposed displacement as a result of wind or earthquake ground shaking. Inelasticity and energy dissipation are achieved through localized yielding of the beam element outside of the beam-column connection. Special proportioning and detailing of this connection is essential to achieving the desired inelastic behavior. The anticipated seismic behavior of the SMF system is long-period, high-displacement motion, with well distributed inelastic demand shared by all participating beam-column connections. System yielding mechanisms are generally limited to frame beams with the intent to invoke yielding at the base of frame columns. In many cases, engineers may model a SMF system with pin-based columns as significant stiffness is required to yield the base of large wide-flange members. If yielding at the base of the frame is desired to occur within the column section, the column might be extended below grade and tied into a basement wall or a ground-level beam, which is added to create a beam-column connection. Economies of construction usually limit the size of beam and column elements based on imposed displacement/drift limits. Design regulations for steel SMF are promulgated in a series of standards: ASCE/SEI 7, ANSI/AISC 341, ANSI/AISC 358, and ANSI/AISC 360. AISC 358 provides specific regulations related to prequalification of certain SMF connection types that obviate project-specific testing required by AISC 341. This design example follows the provisions of AISC 358 for the RBS connection type for the steel SMF seismic-forceresisting system. The six-story steel office structure depicted in the figure above has a lateral-force-resisting system comprising structural steel special moment frames. The typical floor framing plan is shown in Figure 1–1. A typical frame elevation is depicted in Figure 1–2. This design example utilizes simplifying assumptions 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

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Design Example 1



Special Moment Frame

for ease of calculation or computational efficiency. Because bay sizes vary, the example frames can be designed with different participating bays in each direction, which will result in different sizes of beams and columns for each frame depending on location. This example explores the design of a single frame and a single connection of that frame. Assumptions related to base-of-column rotational restraint (assumed fixed), applied forces (taken from the base example assumptions), and applied wind force (not considered) are all incorporated into the example in “silent” consideration. Beam and column element sizes were determined using a linear elastic computer model of the building. These element sizes were determined through iteration such that code-required drift limits, element characteristics, and strength requirements were met. While this example is accurate and appropriate for the design of steel SMF structures, different methodologies for analysis, connection design, and inelastic behavior can be utilized, including the use of proprietary SMF connection design. This example does not explore every possible option, nor is it intended to be integrated with other examples in this document (i.e. Base Plate Design, Passive Energy Dissipation).

OUTLINE 1. Building Geometry and Loads 2. Calculation of the Design Base Shear and Load Combinations 3. Vertical and Horizontal Distribution of Load 4. SMF Frame 5. Element and RBS Connection Design 6. Detailing of RBS Connection

1. Building Geometry and Loads 1.1 GIVEN INFORMATION • Per Appendix A 

Office occupancy on all floors



Located in San Francisco, CA, at the latitude and longitude given



Site Class D



120 feet × 150 feet in plan with typical floor framing shown in Figure 1–1



Frame beam and column sizes for lines 1 and 5 (Figure 1–2) 



Beam and column sizes will vary from those on lines A and F

Six-stories as shown in Figure 1–2

• Structural materials

2



Wide-flange shapes



Pates



Weld electrodes


ASTM A992 (F y = 50 ksi) ASTM A572, Grade 50 E70X-XX

Design Example 1

A

B

C

E

D




F

5 @ 30' – 0'' = 150' – 0'' 5

4

3

2

1

Figure 1–1. Typical floor framing plan

A

B

E

D

C

F

TOP OF PARAPET

ROOF

W21 X 150

W30 X 99

W30 X 99

W30 X 99

W21 X 150

6th FLR

W21 X 150

W30 X 116

W30 X 116

W30 X 116

W21 X 150

5th FLR

W21 X 150

W30 X 132

W30 X 132

W30 X 132

W21 X 150

4th FLR

W21 X 150

W30 X 148

W30 X 148

W30 X 148

W21 X 150

3rd FLR

W21 X 150

W30 X 173

W30 X 173

W30 X 173

W21 X 150

2nd FLR

W21 X 150

W30 X 191

W30 X 191

W30 X 191

W21 X 150

1st FLR Figure 1–2. Frame elevation – line 1 (line 2 in background) 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

3

Design Example 1




1.2 FLOOR WEIGHTS For development of seismic forces per Appendix A:

Table 1–1. Development of seismic forces per Appendix A

Unit W t (psf)

Area (ft2)

Weight (kips)

Floor

78

15,220

1187

Ext Wall

19

6990

133

Roof

36

15,220

548

Ext Wall/Parapet

19

5700

108

Level

Assembly

Typical floor

Floor W t (kips) 1315

Roof

656

W = 5(1320 kips) + 656 kips = 7256 kips

2. Calculation of the Design Base Shear and Load Combinations

ASCE 7

2.1 CLASSIFY THE STRUCTURAL SYSTEM AND DETERMINE SPECTRAL ACCELERATIONS Per ASCE 7 Table 12.2–1 for special steel moment frame: R = 8.0

Ωo = 3

C d = 5.5

2.2 DESIGN SPECTRAL ACCELERATIONS The spectral accelerations to be used in design are derived in Appendix A: S DS = 1.00g

S D1 = 0.60g

2.3 DESIGN RESPONSE SPECTRUM Determine the approximate fundamental building period, T a, using Section 12.8.2.1: C t = 0.028 and x = 0.8 x

T

t

= 0 028

T 12.8–2 08

= 0 86 sec

(see discussion below)

Eq 12.8–7

T a = 0.86 sec

=

4

S

02

06 1 00

sec


§11.4.5

Design Example 1

 S

+06 0 60 10

S =

=

06

T  T

0

=

+50



For T < T o


Eq 11.4–5

sec

§11.4.5

For T > T s.

Eq 11.4–6

The long-period equation for S a does not apply here because the long-period transition occurs at 12 sec (from ASCE 7 Figure 22–12).

1.2 TS = 0.60 sec

) g ( 1 a S , n o 0.8 i t a r e l e 0.6 c c A l a r 0.4 t c e p S n 0.2 g i s e D 0

SDS = 1.0g

SMF Building Period Ta= 0.86 sec, Sa= 0.70g

To= 0.12 sec

Tmax= 1.20 sec, Sa= 0.50g Sa= 0.4+5.0T

Sa= 0.60/T

0

0.5

1

1.5

2

Period (Sec) Figure 1–3. Design Response Spectrum for the example building

Figure 1–3 depicts the design spectral acceleration determined from T , which is greater than T S , so the design spectral acceleration S a is 0.70g. ASCE 7 Section 12.8.2 indicates that the fundamental period of the structure “can be established using the structural properties and deformational characteristics of the resisting elements in a properly substantiated analysis,” which might allow a linear elastic modal analysis to suffice. Section 12.8.2, however, limits the period that can be used to calculate spectral acceleration to a value of T max = C u × T a , where C u is a factor found in Table 12.8–1. In this case T max = 1.4 × 0.86 = 1.20 sec. For preliminary design, the approximate period, T a , will be used to design the SMF. As SMF designs are heavily dependent on meeting drift requirements, the initial value (usually found to be much lower than the period found through mathematical modeling) will suffice for the first design iteration. 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

5

Design Example 1




2.4 HORIZONTAL IRREGULARITIES 1a. and 1b.

T12.3–1

Torsional Irregularity—A torsional irregularity exists when the maximum story drift computed including accidental torsion is more than 1.2 times the average story drift. Extreme torsional irregularity exists when the maximum story drift computed including accidental torsion is more than 1.4 times the average story drift. A static linear elastic three-dimensional computer analysis is used to obtain the displacement at the corners of the building. This building has no torsional response, so the difference between the maximum drift and average drift is 1.0. Table 1–2 provides an example of how one would evaluate the presence of a torsional irregularity for the earthquake load case in the longitudinal direction with positive accidental eccentricity and differences between maximum and average drift. Table 1–2. Story displacements, line 1 and line 5, torsional irregularity check

Story

δ x at Line 5 (in)

δ x at Line 1 (in)

δavg (in)

δmax / δavg

Roof

15.8

11.00

13.4

1.18

5

14.3

10.5

12.4

1.16

4

12.0

10.7

1.12

9.44

3

8.86

7.22

8.04

1.10

2

5.26

4.51

4.89

1.08

1

1.85

1.80

1.83

1.01

NO TORSIONAL IRREGULARITY: δmax / δavg < 1.2 2.

Reentrant corner irregularity exists where both plan projections of the structure beyond a reentrant corner are greater than 15 percent of the plan dimension of the structure in the given direction. The plan projections in longitudinal and transverse directions are 30 feet. The plan dimensions are 150 feet and 120 feet in the longitudinal and transverse direction respectively: 30 feet/150 feet = 20 percent in the longitudinal direction; → REENTRANT CORNER IRREG 30 feet/120 feet = 25 percent in the transverse direction → REENTRANT CORNER IRREG

3. to 5.

By inspection, the building does not qualify for any of these horizontal structural irregularities.

REENTRANT CORNER IRREGULARITY EXISTS Per Section 12.3.3.4, forces for the connections of diaphragms to vertical elements and collectors and the design of collectors and their connections must be increased by 25 percent. If forces for the design of collectors and their connections are calculated using seismic load effects including the overstrength factor, the 25 percent increase is not required.

6


Design Example 1

2.5 VERTICAL IRREGULARITIES 1a. to 5b.




T12.3–2

Calculation for type 1a, 1b, 5a, and 5b may be required for less experienced engineers. These irregularities consider the stiffness and strength of one story relative to another. If the story under consideration is less than 70 percent (1a) or 60 percent (1b) of the story above or 80 percent (1a) or 70 percent (1b) of the average of the three stories above, a soft-story irregularity will exist. This irregularity does not exist based on the element sizes/ stiffness, identical sizes of frames in each principal direction, the height of the stories, and the similarity of bay sizes. If the story under consideration is less than 80 percent (5a) or 65 percent (5b) of the story above, a weak-story irregularity will exist. This irregularity does not exist based on the element sizes, identical sizes of frames in each principal direction, the relative height of each story, and the similarity of bay sizes. By inspection, the building does not qualify for any of the vertical structural irregularities, but the engineer is encouraged to calculate the conditions identified above and described in Table 12.3–2 for this problem. Other sections of the SEAOC Structural / Seismic Design Manual cover general analysis and irregularities in more detail.

NO VERTICAL STRUCTURAL IRREGULARITIES

2.6 LATERAL FORCE PROCEDURE

T12.6–1

1. Simplified Alternative Structural Design Criteria—According to Section 12.14.1.1 this analysis procedure cannot be used for buildings over 3 stories—NOT PERMITTED 2. Equivalent Lateral Force Analysis—According to Table 12.6–1, since the structure is less than 160 feet and has only Type 2 horizontal irregularity—PERMITTED 3. Modal Response Spectrum Analysis—PERMITTED 4. Seismic Response History Procedures—PERMITTED

USE EQUIVALENT LATERAL FORCE ANALYSIS

2.7 BASE SHEAR C

=

S

10

=

125

 R   8  I  1 0

Eq 12.8–2

but need to exceed C

= T

R  I



=

06 0 86



8



= 0 087 for T ≤ T L

Eq 12.8–3

10


7

Design Example 1




but shall not be less than 0.

e

=

.

> 0 01

Eq 12.8–5

and, for structures where S 1 is equal to or greater than 0.6 g, C s shall not be less than: C s

=

0 5 S 1

   I e 

=

(0 5)(0 6)

 8  1 0

= 0.038

Eq 12.8–6

C s = 0.087 V = C sW = (0.087)(7256) = 631 kips

Eq 12.8–1 V = 631 kips

2.8 REDUNDANCY FACTOR According to Section 12.3.4, the Redundancy Factor should be calculated for each principal axis. The Redundancy Factor is 1.3 unless either 12.3.4.2.a or 12.3.4.2.b is shown to be true, in which case the Redundancy Factor can be taken as 1.0. 12.3.4.2.a and Table 12.3–3 requires that for each story resisting more than 35 percent of the base shear, loss of moment resistance at the beam-to-column connections at both ends of a single beam would not result in more than a 33 percent reduction in story strength, nor does the resulting system have an extreme torsional irregularity. Section 12.3.4.2.a can be satisfied by showing that each story does not resist more than 35 percent of the base shear (a taller, well-distributed frame design). Indeed, this is the case for this structure as evinced later in the example; therefore, the Redundancy Factor can be taken as 1.0. Section 12.3.4.2.b can be considered as follows. There are total of six bays of moment frame in the longitudinal direction and four bays of moment frame in the transverse direction; thus, by inspection, removal of an individual frame beam and its rigid connections will not result in more than a 33 percent reduction in story strength (1/4 = 25 percent). The second condition needs to be confirmed by rigid diaphragm analyses by removing individual moment beams and checking whether an extreme torsional irregularity is produced. Frames can be designed with different elements in each direction, reducing the influence of this provision. From the three-dimensional linear elastic static computer analysis, single-beam elements can be changed to “pinned” to confirm that an extreme torsional irregularity does not exist. Since the example satisfies the requirement promulgated in Section 12.3.4.2.a, Table 1–3 is superfluous and not an actual representation of the building response, but it provides an example for how one might check for extreme torsion with the analysis results from a model with a single moment beam removed:

8


Design Example 1




Table 1–3. Story displacements, lines 1, 5, A, and F, extreme torsional irregularity check (illustrative)

Story

δ x at Line 1 (in)

δ x at Line 5 (in)

Story

δ y at Line A (in)

δ x at Line F (in)

(in)

δmax / δavg

(in)

δmax / δavg

STORY6

10.7

7.65

9.20

1.17

STORY6

15.8

11.00

13.40

1.18

10.5

12.4

1.16

10.7

1.12

δavg

δavg

STORY5

9.82

6.48

8.15

1.20

STORY5

14.3

STORY4

8.27

5.35

6.81

1.21

STORY4

12.0

STORY3

6.15

4.89

5.52

1.11

STORY3

8.86

7.22

8.04

1.10

STORY2

3.69

3.01

3.35

1.10

STORY2

5.26

4.51

4.89

1.08

STORY1

1.32

1.25

1.29

1.03

STORY1

1.85

1.80

1.83

1.01

9.44

ρ = 1.0 FOR EAST-WEST DIRECTION ρ = 1.0 FOR NORTH-SOUTH DIRECTION 2.9 LOAD COMBINATIONS See Appendix 1 for the derivation of combinations based on ρ = 1.0 and 0.2S DS = 0.2. Load combinations of consequence for the design of the SMF are 12 0

+05 =0

+

5

§12.4.2.3 Load Combo 5 (modified)

0Q

Load Combo 7 (modified)

3. Vertical and Horizontal Distribution of Load

ASCE 7

3.1 VERTICAL DISTRIBUTION OF SHEAR §12.8.3

Table 1–4. Vertical distribution of shear

Level

wi (kips)

hi (ft)

C vx

F x (kips)

Roof

656

72

101,356

0.18

114

6th

1315

60

163,890

0.29

185

5th

1315

48

125,991

0.22

142

4th

1315

36

89,762

0.16

101

3rd

1315

24

55,662

0.10

62

2nd

1315

12

24,592

0.05

27

Total

7231

i i

561,254

631


9

Design Example 1




The terms used in Table 1–4 are defined in Section 12. 8.3 and are presented in a rounded form taken from spreadsheet calculations (values will not be duplicated by hand calculation of Table 1–4). Since the period = 0.86 > 0.5 sec, the value for k is interpolated between a value of 1.0 for T = 0.5 sec and 2.0 for T = 2.5 sec. In this example, k = 1.18. The distribution of story shear is carried out using F x = C vxV , where, C

w

=

Eq 12.8–11 and Eq 12.8–12 w

=

i

3.2 HORIZONTAL DISTRIBUTION OF STORY SHEAR

A

B

C

E

D

F

5 @ 30' – 0'' = 150' – 0'' 5 ±6' – 0'' ACC. TORSION 4

C.M. C.R.

3

2 TORSION

±7' – 6'' ACC. TORSION

1

Figure 1–4. Rigid diaphragm analysis to distribute story shear to SMFs

As shown in Figure 1–4, the center of mass and center of rigidity coincide at the middle of the building. (While not precisely true, the example building is very well behaved and torsional response is non-existent so the coincident C.M and C.R. is not an unrealistic assumption.) To distribute the load to each SMF, the story shear, F i , is applied in the X and Y directions. Per Section 12.8.4.2, the point of application of the story shear is offset 5 percent to account for accidental eccentricity. This example designs all SMFs in the X direction the same as the SMFs in the Y direction. However, for this illustration, a generic stiffness K is used for all frames, and the assumption is made that frames have different stiffness in each direction as shown. Each frame bay is described in Table 1–5 by using the gridline location of the frame (A, F, 1, 5) and the gridline that intersects the frame line of interest (23, 34, BC, CD, DE).

10


Design Example 1




Table 1–5. Center-of-rigidity calculation

Frame

Dir

x

A23

Y

A34

y

R x

R y

xR y

−75

1K

−75 K

0K

Y

−75

1K

−75 K

0K

F23

Y

75

1K

75 K

0K

F34

Y

75

1K

75 K

0K

1BC

X

−60

0.8 K

0K

−48 K

1CD

X

−60

0.8 K

0K

−48 K

1DE

X

−60

0.8 K

0K

−48 K

5BC

X

60

0.8 K

0K

48 K

5CD

X

60

0.8 K

0K

48 K

5DE

X

60

0.8 K

0K

48 K

0K

0K

Sum

4.8 K

4K

yR x

Center of rigidity: X r = 0 K/4 K = 0 ft, Y r = 0 K/4.8 K = 0 ft (from center of mass)

Table 1–6. Rigid diaphragm analysis to distribute shear to SMFs (X-direction) X-Direction R x y

d = x − xr

Y

1K

−75.0

−75 K

A34

Y

1K

−75.0

F23

Y

1K

F34

Y

1K

1BC

X

0.8 K

1CD

X

1DE

V acc. tors

V total

5625 K

0.00F i

0.00F i

−0.01F i

−0.01F i

−75 K

5625 K

0.00F i

0.00F i

−0.01F i

−0.01F i

75.0

75 K

5625 K

0.00 F i

0.00F i

0.01F i

0.01F i

75.0

75 K

5625 K

0.00 F i

0.00F i

0.01F i

0.01F i

−60.0

−48 K

2880 K

0.17F i

0.00F i

−0.01F i

0.16F i

0.8 K

−60.0

−48 K

2880 K

0.17F i

0.00F i

−0.01F i

0.16F i

X

0.8 K

−60.0

−48 K

2880 K

0.17F i

0.00F i

−0.01F i

0.16F i

5BC

X

0.8 K

60.0

48 K

2880 K

0.17 F i

0.00F i

0.01F i

0.17F i

5CD

X

0.8 K

60.0

48 K

2880 K

0.17 F i

0.00F i

0.01F i

0.17F i

5DE

X

0.8 K

60.0

48 K

2880 K

0.17 F i

0.00F i

0.01F i

0.17F i

A23

Sum

4.8 K

4K

Rd

Rd 2

V torsion

Dir

R y x

d = y − yr

V direct

Frame

39,780 K


11

Design Example 1




Table 1–7. Rigid diaphragm analysis to distribute shear to SMFs (Y-direction) Y-Direction Rd 2

V direct

V torsion

V acc. tors

V total

−75 K

5625 K

0.25F i

0.00 F i

−0.01 F i

0.24F i

−75.0

−75 K

5625 K

0.25F i

0.00 F i

−0.01 F i

0.24F i

1K

75.0

75 K

5625 K

0.25 F i

0.00 F i

0.01 F i

0.26F i

1K

75.0

75 K

5625 K

0.25 F i

0.00 F i

0.01 F i

0.26F i

−60.0

−48 K

2880 K

0.00F i

0.00 F i

−0.01 F i

−0.01F i

0.8 K

−60.0

−48 K

2880 K

0.00F i

0.00 F i

−0.01 F i

−0.01F i

X

0.8 K

−60.00

−48 K

2880 K

0.00F i

0.00 F i

−0.01 F i

−0.01F i

5BC

X

0.8 K

60.00

48 K

2880 K

0.00 F i

0.00 F i

0.00 F i

0.01F i

5CD

X

0.8 K

60.00

48 K

2880 K

0.00 F i

0.00 F i

0.00 F i

0.00F i

5DE

X

0.8 K

60.00

48 K

2880 K

0.00 F i

0.00 F i

0.00 F i

0.00F i

R x y

d = x − xr

Y

1K

−75.0

A34

Y

1K

F23

Y

F34

Y

1BC

X

0.8 K

1CD

X

1DE

Frame

Dir

A23

R y x

Sum

4.8 K

d = y − yr

Rd

4K

39,780 K

The following calculations find the force in SMF along line F for shear in the X direction: R y

×

M acc. tors F cc. ors

0 25

=

.05

. to

=

=

R

M Ftota

R

0

x

to to

2

= +F rect

.5F i ;

7 5F i × 75 K 39, 75

39,

K 0 00

= 0 01F i i

; . tor

t s

= 0 26F i

Table1–7 shows that the maximum design force for any frame bay is 0.26 times the force at that level. The rest of this example focuses on the design of the SMF along line F. Table 1–8. Story forces applied to SMF along line F

12

Level/Story

F Story (kips)

F SMF-F (kips)

V SMF-F (kips)

Roof/6th

114

30

30

6th/5th

185

49

79

5th/4th

142

38

116

4th/3rd

101

27

143

3rd/2nd

62

16

160

2nd/1st

27

3

163


Design Example 1

4. SMF Frame




ASCE and AISC

4.1 SMF LAYOUT The layout of the lateral system should be well distributed to avoid torsional behavior as well as provide reasonable redundancy. The layout for SMFs is constrained by the limitation of column alignment and uniaxial connections (SMF beams aligned along a single column cross-sectional axis), which can sometimes require discreet bays and frame lines. This example lends itself to the layout shown previously, which tends to obviate any torsional response and related forces and/or penalties for unbalanced layouts. In some designs, it may be desirable to design SMF connections in a bi-axial arrangement; only specific connection types (prequalified or tested) may be utilized for bi-axial applications. Treatment of this complicated condition is beyond the scope of this example.

4.2 DEFLECTION LIMITS SMF design is highly influenced by drift limits. Some augmentative structural elements can be used to help keep drifts low (i.e. dampers) or SMF can be used in conjunction with stiffer systems, increasing redundancy and seismic resilience while meeting strict drift limits (i.e. dual systems). However, SMF can be designed to meet required drift limits without the assistance of other structural elements. The engineer should consider cost, constructability, anticipated performance, and client influence when weighing the use of SMF with or without supplemental systems. This example examines only the design of SMF. The elastic story drifts are the difference in the elastic deflections at the floor above and below. The maximum predicted inelastic story drift (based on elastic story drift extracted from the linear elastic analysis model described earlier without any modification of ground motion) is given by the following for the second story along line F: x

Eq 12.8–15 and F 12.8–2

−1

The limit on story drift is given in Table 12.12–1 based on building system (SMF), > four stories in height and the Risk Category (II):

∆ = 0.020 ∆

T 12.12–1

∆

.

29i

T 1–9 and T 12.12–1

∆ ∆ → DRIFT IS OK When the engineer is calculating lateral deflections for drift analysis, Section 12.8.6.1 provides an exception that removes the lower limit on C s , specified in Section 12.8.1. 1 in determining seismic design forces. This exception allows a recalculation of T without consideration of the upper limit on the period (C uT a) as required by Section 12.8.2 (per Section 12.8.6.2). Therefore, the period calculated in the linear elastic model described above can be used without modification ( T x and T y below). Without the lower limit on C s , the design base shear per Equation 12.8–3 is Longitudinal (X-direction): T x = 1.35 sec S D1 R 

06

 8 0  1 35  1 0 

0 06

V x = C sW = (0.06)(7256 kips) = 403 kips

Eq 12.8–3

Eq 12.8–1 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

13

Design Example 1




Transverse (Y-direction): T y = 1.64 sec C

= T

R  I



06

=

1 64

 8 0

=

05

Eq 12.8–3

10

V x = C sW = (0.05)(7256 kips) = 332 kips.

Eq 12.8–1

Using these modified design base shears, the force distribution to each level is adjusted and applied to the output from the computer model. The structure displacement, drift ratios, and stability coefficients are derived as shown in Table 1–9.

4.3 STABILITY COEFFICIENT The following calculation is used to develop the stability coefficient:

=

V

Eq 12.8–16 x

where P x = the total vertical design load at and above Level x

∆ = the design story drift as defined in Section 12.8.6 occurring simultaneously with V x V x = the seismic shear force acting between Levels x and x − 1 hsx = the story height below Level x C d = the deflection amplification factor in Table 12.2–1.

The stability coefficient, θ, shall not exceed θmax as determined from 0.5

βC

≤ 0 25

Eq 12.8–17

where β is the ratio of shear demand to shear capacity for the story between Levels x and x − 1, which can be taken conservatively as 1.0. When the stability coefficient θ exceeds 0.10 but is less than θmax, the incremental factor related to P–∆ effects on displacements and member forces shall be determined by rational analysis, or displacements and member forces can be multiplied by1.0/(1 − θ). If θ is greater than θmax, the structure is potentially unstable and shall be redesigned. A review of drift ratios tabulated in Table 1–9 shows that all interstory drift ratios are less than 0.020 using seismic forces corresponding to the actual period, T , in base shear Equation 12.8–1. Stability coefficients are all less than 0.10, so P– ∆ effects need not be considered. However, AISC 358 indicates that the global drift consideration of reduced beam section (RBS) SMF design must consider the reduced stiffness

14


Design Example 1




Table 1–9. Interstory displacements and drifts

Longitudinal (X-direction) Displacements and Drifts Height h (in)

δ x = δ xe

∆ Drift

Story

C d (in)

(in)

Drift Ratio (∆ / h)

Stability Coefficient θ

6th

144

9.52

0.82

0.006

0.01

5th

144

8.69

1.29

0.009

0.01

4th

144

7.41

1.67

0.012

0.02

3rd

144

5.74

2.35

0.016

0.03

2nd

144

3.39

1.61

0.011

0.02

1st

144

1.78

1.78

0.012

0.03

Transverse (Y-direction) Displacements and Drifts Height h (in)

δ x = δ xe

∆ Drift

Story

C d (in)

(in)

Drift Ratio (∆ / h)

Stability Coefficient θ

6th

144

14.66

2.48

0.017

0.04

5th

144

12.18

2.16

0.015

0.02

4th

144

10.02

2.42

0.017

0.03

3rd

144

7.60

2.55

0.018

0.03

2nd

144

5.05

2.55

0.018

0.03

1st

144

2.50

2.50

0.017

0.04

associated with the reduced beam section. The actual reduction in stiffness could be calculated and included in the mathematical model to determine “accurate” story drifts based on increased beam flexibility. However, a general value of system stiffness can be used to simplify the design and is allowed to be taken as 1.1 times the elastic story drift for a 50 percent flange reduction per AISC 358 Section 5.8, Step 1. Using this approach for the maximum drift ratio, the revised drift ratio is (∆ / h)(1.1) = (0.018)(1.1) = 0.0198 < 0.020 . . . ΟΚ (θ)(1.1) = (0.04)(1.1) = 0.044 << 0.25 . . . ΟΚ. Most published reference documents (Bozorgnia, 2004; Moore, 2007) recommend a mathematical model using centerline-to-centerline dimensions of framing members while also allowing for realistic assumptions for the stiffness of panel zones or modification of the effective span length for beams and columns to more accurately model frame stiffness. Many different analyses using different computer software, beam-column joint models, and other parametric considerations indicate that a centerline analysis (assuming no panel zone stiffness) represents a reasonable assumption for the majority of steel special moment frame designs. This assumption allows for balanced connection behavior. This simplified analysis assumption (center-tocenter modeling of frame elements) with a 10 percent increase on expected elastic story drift is generally acceptable for the design of steel SMF in the static linear elastic analysis/design regime and has been used to determine the displacements and forces extracted for this example problem. 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

15

Design Example 1




5. Element and RBS Connection Design

AISC

5.1 ELEMENT SIZING In this part of the design example, representative SMF beam and column members of Frame 1 are designed in accordance with ASCE/SEI 7-10. SMF structures assigned to Seismic Design Category D are to conform to the requirements of the 2010 AISC Seismic Provisions for Structural Steel Buildings (AISC 341). This example problem uses the RBS SMF connection type. Steel moment frame designs are typically drift controlled. Frame members are chosen to provide sufficient stiffness to meet drift limits imposed by the model building code as described above, in this case IBC 2012, which relies upon limits promulgated in ASCE/SEI 7-10. The selected members are then checked for SMF member design requirements per AISC 341. The selection of beam-column combinations must also conform to the connection limitations set forth in AISC 358 and AISC 360 as appropriate. After confirming that drift limitations were reasonably satisfied through selection of beam and column members to determine acceptable system performance, the W30 × 132 beam at the fifth floor, line 1, and the W33 × 221 column that comprises an interior connection at line C (see box on connection in Figure 1–2) were chosen to illustrate this design example.

5.2 MATERIAL SPECIFICATIONS AND STRENGTH PROPERTIES AISC 341 Chapter A indicates the material specification and strength properties for steel members used in seismic-force-resisting systems. Where the term “expected strength” is used, the value is to be taken as F ye = ( R y)(F y).

The “lower bound” strength is the minimum specified yield strength, F y. For this design example, material strengths are taken as W30 × 132 beam, A992 F y = 50 ksi F u = 65 ksi

W33 × 221 column, A992 F y = 50 ksi F u = 65 ksi.

Per AISC 341, Table A3.1, for ASTM A992 R y = 1.1.

5.3 DESIGN TYPICAL BEAM The fifth-floor beam in Frame 1 is selected to illustrate SMF beam design. This W30 × 132 beam is shown in Figure 1–5.

16


Design Example 1

C




C 30' – 0''

W30 x 132 VL

5TH FLOOR

VR MR

ML

Figure 1–5. Typical beam at fifth floor of Frame 1-CD

From a review of the computer output prepared in support of this design example, the moments and shears at the right end of the beam are greatest. The unfactored moments and shears at the face of the column at line D are M DL = 502 kip-in M LL = 288 kip-in M seis = + / −3208 kip-in V DL = 5.42 kips V LL = 3.23 kips V seis = + / −17.8 kips.

Seismic forces identified above include both vertical and horizontal components E v and E h. The vertical component, E v, is added to the dead load in Equation 12.4–1 and subtracted from the dead load in Equation 12.4–2: (1.2 + 0.2S DS ) D + ρQ E + 0.5 L + 0.2S

ASCE 7-10, §12.4.2.3

E = ρQ E = + / −0.2S DS ( D)

Eqs 12.4–1, 12.4–3, 12.4–4

E h = ρQ E = 1.0Q E E v = 0.2S DS D = 0.2(1.0g) D = 0.2 D.

Using the Basic Combinations of Section 2.3.2: 1.2 D + 1.6 L: M D+ L = 1.2(502) + 1.6(288) = 1063.2 kip-in V D+ L = 1.2(5.42) + 1.6(3.23) = 11.7 kips.

Using the Seismic Load Combinations, Basic Combinations for Strength Design of Section 12.4.2.3: (1.2 + 0.2S DS ) D + ρQ E + 0.5 L + 0.2S : M D+ L+ E = [1.2 + 0.2(1.0)](502) + 1.0(3208) + 0.5(288) + 0 = 4055 in-kip


17

Design Example 1




V D+ L+ E = [1.2 + 0.2(1.0)](5.42) + 1.0(17.8) + 0.5(3.23) + 0 = 27.0 kips

(0.9 − 0.2S DS ) D − ρQ E − 1.6 H : M D− E = [0.9 − 0.2(1.0)](502) − 1.0(3208) − 0 = −2857 in-kip V D− E = [0.9 − 0.2(1.0)](5.42) − 1.0(17.8) − 0 = −14.0 kips.

Prior to evaluating demand/capacity ratios for member strength, SMF beams must be checked for stability and proportions per AISC 358, 5.3, w hich references AISC 341 for limiting width-thickness ratios for elements subjected to compression forces. Check the flange width-thickness ratio:

=

≤0

2

29,000 k si

=

50 ksi

y

For W30 × 132:

b

2t

=

=7

.

AISC 341, T D1.1

7 22 . . . OK

However, if the section did not pass b / t criteria, then per AISC 358, Section 5.3.1 (6) for RBS connections, b f may be measured at the end of the center two-thirds of the reduced beam section, provided that gravity loads do not shift the plastic hinge location a significant distance from the center of the reduced section. Check the web width-thickness ratio, confirming that the beam is a highly ductile member: /F 1 0.

for

C a

=

P c

Py

=

66 0 9(50 ksi 38.8

≤

=

)

C )

AISC 341, T D1.1

=

.125, so o

a

f otnote

])

0

w

=

For W30 × 132:

0 . . . OK

w

Check beam depth, weight and span-to-depth ratio limits per AISC 358, Section 5.3: Maximum beam depth: W36 > W30 . . . OK Maximum beam weight: 300 lb/ft > 132 lb/ft . . . OK Maximum flange thickness: 1.75 in > 1.0 in . . . OK Minimum span-to-depth ratio: 7

(30)(12) 30

= 12

. . . OK.

Check beam lateral bracing requirements per AISC 358, Section 5.3.1 (7), which references AISC 341, Section D1.2b. Note that per the exception in AISC 358 Section 5.3.1 (7), beams supporting a concrete structural slab that is connected between the protected zones with welded shear connectors spaced at a maximum of 12 inches on center do not require supplemental top and bottom flange bracing at the RBS.

18


Design Example 1




As this is the case in most conditions (and assumed for this example), bracing will be considered under the auspices of AISC 341, Section D1.2b: Maximum brace spacing, Lb Lb = 0.086r y E / F y = (0.086)(2.25)(29,000/50) = 112 in or 9.35 ft.

Place minimum bracing at quarter points: L = 30/4 = 7.5 ft on center

= 0 02

P r



M r

AISC 360, Eq A-6–7

o

M r = R yZeF y, where R y = 1.1.

AISC 341 §D1.2a and T A3.1

Use Z x instead of Z e for a conservative estimate of bracing

= 1.1(437 in3)(50 ksi) = 24,035 kip-in = 2003 ft-k Prb = 0.02 M r C d / ho Prb = ( .02

 ( 

)( .0)  (

.3 1. ) in

= 16.4 k . 

The length of the brace is assumed to be measured from the centerline of the W30 × 132 to the centerline of the adjacent gravity beam. Assuming 10-foot beam spacing, the length of the brace is L =

(

)

 12 in   ft

=

=

.3 f

Based on the Manual of Structural Steel Construction, a L4 × 4 × 3 ⁄ 8 as an eccentrically loaded single angle will be examined:

φc Pn = 16.3 kips ~ 16.4 kips (using 11-foot length and Z x instead of Z e)

AISC Manual T 4–12

AISC 360 states that a minimum stiffness is also required to provide adequate lateral bracing (A-6–8). This type of brace is considered a nodal brace, providing a rigid brace support, so the required brace stiffness is

β r =

1 10 M C L

AISC 360, Eq A-6–8

o

where

φ = 0.75 M r = R y Z eF y (however, Z x will be used to estimate bracing)

= 1.1(437 in3)(50 ksi) = 24,035 kip-in = 2003 ft-k C d = 1.0 Lb = 10.3 ft = 124 in ho = d b − t f = 30.3 in − 1.0 in = 29.3 in

10(24 35 r

)(1 0

0 75(124 in)( 29.3 )

88 2 k/in


19

Design Example 1




The brace stiffness can be calculated as E

= θ= K =

cos2

 30 3 in  = 13.7  124 in 

an −



29,000 k si )  ( 124 in

( .86

.7

= 631 k /in

K > βbr . . . OK.

L4 × 4 × 3 ⁄ 8 kickers provided at 7.5 feet on center to brace the beam bottom flange to the top flange of an adjacent steel beam meet lateral bracing requirements. The W30 × 132 beam meets stability and proportion criteria; next, check the design flexural strength (LRFD) per AISC 360. From AISC 360 (Chapter F2) for W30 × 132: L p

F

=

.95 t

n

=

.

Design Flexural Strength (conservative assumption): φb M p = 19,680 kip-in.

AISC 360, Eq F2–5 AISC, T 3–6

Demand-capacity ratio: D / C = 4055 k-in 19,680 k -in

=

.0 . . . OK.

Check nominal shear capacity for W30 × 132:

AISC 341, T I–8–1

φV n = 0.6F y AwC v

AISC 360, Eq G2–1

C v = 1.0

AISC 360, Eq G2–3

43.

.24

F

=

.

=

0

AISC 360, Eq G2.1(a)

φvV n = 1.0(0.6)(50)(0.615)(30.3)(1.0) = 559 kips D C =

27 k 559 k

=

. 48 1 . . . OK.

USE W30 × 132 BEAM Note: The W30 × 132 beam is larger than required by strength considerations as calculated from prescriptive load combinations. However, given the constraints of meeting frame drift limits, it is a reasonable choice for this design. Limitations related to beam-column connection design and reduced beam section parameters guided the size of beam selected.

20


Design Example 1




5.4 DESIGN TYPICAL COLUMN The column to be designed is the second-lift column of Frame 1 (line C) as shown in Figure 1–2 and Figure 1–6. The maximum strong axis moments occur at the bottom of the column and are taken at the top flange of the fifth-floor beam. For brevity, the example will look at critical conditions affecting design, omitting the many different inconsequential demand values that are determined for the various analyses, including out of plane bending loads, external wind forces, etc.

4 5TH FLOOR

W33 X 221

4TH STORY

4TH FLOOR W33 X 318

Figure 1–6. Typical second-story column at Frame 1

For the fourth-story column at line 4, the maximum unfactored column forces generated by the frame computer analysis are M DL = 15.4 kip-in M LL = 13.2 kip-in M seis = + / −5559 kip-in V DL = 0.22 kips V LL = 0.22 kips V seis = + / −66 kips P DL = 167 kips P LL = 116 kips Pseis = + / −2.23 kips.


21

Design Example 1




Seismic forces identified above include both vertical and horizontal components E v and E h. The vertical component, E v, is added to the dead load in Equation 12.4–1 and subtracted from the dead load in Equation 12.4–2: (1.2 + 0.2S DS ) D + ρQ E + 0.5 L + 0.2S

ASCE 7-10, §12.4.2.3

E = ρQ E = 0.2S DS ( D)

Eqs 12.4–1, 12.4–3, 12.4–4

E h = ρQ E = 1.0Q E E v = 0.2S DS D = 0.2(1.0g) D = 0.2 D.

Using the Basic Combinations of Section 2.3.2: 1.2 D + 1.6 L: P D+ L = 1.2(167) + 1.6(116) = 386 kips M D+ L = 1.2(15.4) + 1.6(13.2) = 39.6 kip-in V D+ L = 1.2(0.22) + 1.6(0.22) = 0.6 kips.

Using the Seismic Load Combinations, Basic Combinations for Strength Design of Section 12.4.2.3: (1.2 + 0.2S DS ) D + ρQ E + 0.5 L + 0.2S : P D+ L+ E = [1.2 + 0.2(1.0)](167) + 1.0(2.23) + 0.5(116) + 0 = 294 kips M D+ L+ E = [1.2 + 0.2(1.0)](15.4) + 1.0(5559) + 0.5(13.2) + 0 = 5587 in-kip V D+ L+ E = [1.2 + 0.2(1.0)](0.22) + 1.0(66) + 0.5(0.22) + 0 = 66 kips

(0.9 − 0.2S DS ) D − ρQ E − 1.6 H : P D− E = [0.9 − 0.2(1.0)](167) − 1.0(2.23) − 0 = 115 kips M D− E = [0.9 − 0.2(1.0)](15.4) − 1.0(5559) − 0 = −5548 in-kip V D− E = [0.9 − 0.2(1.0)](0.22) − 1.0(66) − 0 = −66 kips.

Try W33 × 221, ASTM A992 column. Check the flange width-thickness ratio:

=

≤03

2

E

29,000 k si

=

50 ksi

y

For W33 × 221:

=

2

= 7 22

AISC 341, T D1.1

7 22 . . . OK.

Check the web width-thickness ratio: C

=

P c

=

294 09⋅

−

22

⋅ 65 3 in .

10 29 000 k si s

−

.


.

53.5

Design Example 1



for W33 × 211: h / t w = 38.5 < 53.5.


AISC 341, T D1.1

Check column depth, weight, and span-to-depth ratio limits per AISC 358 Section 5.3: Maximum column depth: W36 > W33 . . . OK Maximum column weight: unlimited . . . OK Maximum flange thickness: compact per AISC 341 . . . OK. Unbraced column height (taken from top of framing at bottom to mid-depth of beam at top): L p = 12.7 ft, Lr = 38.2 ft

AISC Manual, T 3–6

H = 12.0 − (2.5/2) = 10.8 ft Lb = 10.8 ft < L p.

As SMFs are usually governed by stiffness criteria, most of the frame stability checks are superfluous related to the large members required to provide adequate stiffness. Much effort can be expended on frame stability checks and column capacity checks, but in general these systems fall well within limits of stability and strength. Other documents (e.g. AISC Seismic Design Manual) provide exhaustive treatment of these issues, but generally checking the columns for combined flexure/compression is adequate. Therefore, the combined stresses for the critical load combination is 386 k 2939 k P

2P

+

M M

=

= 0 13

02

386 (2 0. (2939

+

5587 38,520

=

22 < 1 .

AISC 360, Eq H1–1b

Check column shear strength:

≤ w

Since

F y

=

.24

29,000 k si 50 ksi

= 54.0

= 38 5 V n = 0.6F y AwC v

AISC 360, Eq G2–1

C v = 1.0

AISC 360, G2–2

φvV n = 1.0(0.6)(50)(26.3)(1.0) = 788 kips V u = 66 kips < φvV n = 788 kips . . . OK.

Column flange bracing must also be considered and is addressed in Part 6 below. USE W33 × 221 COLUMN Note: The W33 × 221 is not necessarily an optimum size based on strength and local or global stability. The size was chosen to meet drift requirements with consideration toward connection design limitations and costs associated with the beam-column connection. A more refined optimization could result from a number of detailed iterations, but this size is adequate for illustration of the design example. 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

23

Design Example 1




5.5 RBS CONNECTION LIMITATIONS The column-beam relationship must satisfy limitations identified in AISC 358 Sections 5.4 through 5.7. Section 5.4 requires that panel zones conform to the requirements of AISC 341 and that the column-beam moment ratios are limited to conform to the requirements of AISC 341, with some prescriptive conditions associated with calculating the ratio of column-beam moments. Section 5.5 requires that SMF beam flanges are connected to column flanges using complete-joint-penetration (CJP) groove welds and that the welds and access hole geometry conform to the requirements of AISC 341 and AISC 360 respectively. Section 5.6 requires that the beam web provide the required shear strength according to Equation 5.8–9 and that the SMF beam web be connected to the column flange using a CJP groove weld extending between weld access holes. Section 5.7 defines some very specific conditions associated with the fabrication of flange cuts and repair of gouges and notches. These requirements should be clearly identified in contract documents to ensure that the quality of the connection is attained and maintained consistent with the AISC provisions.

5.6 RBS CONNECTION DESIGN The RBS connection is a pre-qualified connection type per AISC 358. This design example follows the procedure outlined in AISC 358 with reference to AISC 341 and AISC 360. The following calculations comprise a design methodology that is inherently iterative and requires some experience to gain proficiency. After considering code drift limits and evaluating several combinations for strong column-weak beam and panel zone strength criteria, the combination of a W30 × 132 beam and W33 × 221 column was selected for use in this design example. Note that the “allowance” for using deep columns is a relatively recent development based on various research (Ricles et al, 2004; Uang et al, 2001). The chosen frame members were shown to have adequate strength to resist the factored load combinations (Parts 5.3 and 5.4), and this combination of beam and column sizes in the computer analysis results in overall frame drifts within the code limits (Part 4). The W33 × 221 column was chosen to elucidate consideration of deep columns, while providing an efficient frame design (based on drift limits). However, an increase in column size might be warranted to obviate the need for doubler plates and possibly reduce the cost of the frame considering fabrication costs. The connection detailing might also give rise to consider different frame elements. When given the option, steel fabricators often request to use heavier columns in lieu of installing doubler plates for economy. But, in light of these requests, the designer must consider the actual realized savings when replacing fame elements. For example, if a W14 shape is selected for this frame (instead of the W33 proposed), the comparable column is W14 × 665, an element more than three times heavier than the selected column. In addition to connection detailing, necessary preheat for all welds, required inspection, and the fabrication and inspection associated with column-splice details would all drastically increase. Column splices in SMFs must comply with AISC 341 Section D2.5 (which is beyond the scope of this example). The detrimental effects of using very heavy columns might be most evident when executing a very complicated and expensive CJP weld at the column splice. These types of system considerations are important when determining the final sizes appropriate for a steel SMF. The focus for the following section will be on the design of the reduced beam section (RBS) connection and associated design details that are appropriate for the chosen sizes. Establish plastic hinge configuration and location

AISC 358, §5.8, Step 1

The fundamental design intent expressed in the AISC 358 design procedure for RBS is to move the plastic hinge away from the weld between the beam flange and column flange. This physical relocation of the plastic hinge region is accomplished by providing a defined reduced beam section located at a predetermined distance away from the column flange (Figure 1–7).

24


Design Example 1




AISC 358 Section 5.8 identifies specific limitations for the dimensions associated with the radius cut for the reduced beam section. 0.50bbf ≤ a ≤ 0.75bbf

AISC 358, Eq 5.8–1

0.65d ≤ b ≤ 0.85d

AISC 358, Eq 5.8–2

0.10bbf ≤ c ≤ 0.25bbf

AISC 358, Eq 5.8–3

Sh

COL

HINGE

bf FOR FLANGE COMPACTNESS, SEE PART 5a

center 2/3 a

b

Figure 1–7. RBS geometry

W30 × 132 0.5b f = 0.5(10.5) = 5.25 in 0.75b f = 0.75(10.5) = 7.88 in Use a = 7 in and 0.65d = 0.65(30.3) = 19.7 in 0.85d = 0.85(30.3) = 25.8 in Try b = 24.0 in The initial estimate for the depth of the cut, c, should be made such that 40 to 50 percent of the flange is removed. This should limit the projection of moments at the face of the column to within 90 to 100 percent of the plastic capacity of the full beam section.


25

Design Example 1




With a 45 percent reduction in the flange area, c = 0.45 b f /2 = 2.36 in

= 23 ⁄ 8-in cut, (2.38 in/10.5 in) = 0.23b f < 0.25b f . . . OK. Use c = 2.38 in R =

2

4 8

=

4⋅

2

8 2.38)

= 31.5 in (radius of the flange cut).

The plastic hinge may be assumed to occur at the center of the curved cut such that S h = a + b /2 = 7 + (24/2) = 19 in

and L = 30.0 ft = 360 in L′ = L − d c − 2(S h); (this assumes columns are the same size at each end of beam) L′ = (360 in) − 33.9 in − 2(19 in) = 288 in.

The length between the plastic hinges, L′ (see Figure 1–8), is used to determine forces at the critical sections for connection analysis.

L' PLASTIC HINGES

DRIFT ANGLE

L

Figure 1–8. Plastic hinge locations

Determine plastic section modulus at the reduced beam section

AISC 358, §5.8, Step 2

The plastic section modulus at the center of the reduced beam section is calculated as Z RBS = Z x − 2ct bf (d − t bf )

2c = 2(2.38) = 4.75 in Z RBS = 437 − [4.75(1.0)(30.3 − 1.0)] = 298 in3.

26


AISC 358, Eq 5.8–4

Design Example 1

Determine probable maximum moment at the reduced beam section




AISC 358, §5.8, Step 3

Next, the probable plastic moment at the reduced beam section, M pr , is calculated as M pr = C pr R y F y Z RBS

AISC 358, Eqs 2.4.3–1, 5.8–5

The factor C pr accounts for peak connection strength, including strain hardening, local restraint, additional reinforcement, and other connection conditions.

C

=

C

=

≤12

F

AISC 358, Eq 2.4.3–2

50 ksi + 65 ksi 2 ⋅ 50 ksi

= 1 15

M pr = (1.15)(1.1)(50)(298) = 18,849 kip-in.

The value for M pr must be such that the projected moment demand at the face of the column M f , is less than the expected strength of the full beam section; this condition is verified in Part 6e below. Compute the shear force at the center of each RBS

AISC 358, §5.8, Step 4

Determine the shear force at the center of the reduced beam sections at each end of the beam. AISC 358 Section 5.8 requires that the shear force at the center of the reduced beam section is determined by a freebody diagram of the portion of the beam between the centers of the reduced beam sections assuming the moment at the center of each RBS is M pr , and shall include gravity loads acting on the beam based on a specific load combination: 1.2 D + f 1 L + 0.2S (where f 1 = 0.5). (Figure 1–9 and Figure 1–10)

L'

Mpr

Mpr Vpr

Vpr

Figure 1–9. Beam equilibrium under the probable plastic moment M pr

( )( M )

(

)

L ′ V

+V

288 in V RBS =

V

131 ki

V p = V D+ L( L′ / L) V p = 11.7 kips (288 in/360 in) = 9.36 kips

′ =

V V

V

−

36 k ipss

= 122 k ips. 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

27

Design Example 1




Compute the probable maximum moment at the face of the column

AISC 358, §5.8, Step 5

M f = M pr + V RBS (S h)

AISC 358, Eq 5.8–6

M f = 18,849 kip-in + (140 kips)(19 in) = 21,509 kip-in (positive sense) M

=

−

=−

,531 kip-in (ne ativ sense)

RBS Vu

Vu

Mf

Mf

VRBSMpr

Sh = a + b 2

Figure 1–10. Free-body diagram between center of RBS and face of column

(AISC 358, Fig. 5.2)

In this example (as in many actual applications), the moment attributable to the gravity load applied between the plastic hinge and the face of the column flange is negligible ( < 0.5%): therefore, it is omitted and only briefly considered when comparing M pe to M f . Compute the expected plastic moment of the beam

AISC 358, §5.8, Step 6

To compute the plastic moment of the beam based on the expected yield stress of the beam material, compute M pe as M pe = Z b R yF y = (437)(1.1)(50) = 24,035 kip-in.

Check that M f does not exceed  d M pe

AISC 358, §5.8, Step 7

Check the value of M f against φd M pe as follows M f ≤ φd M pe = 21,509 < 24,035 . . . OK.

If M f exceeds φd M pe, the depth of cut at the reduced beam section ( c) should be increased, but not to exceed a 50 percent total reduction of the beam flange. The difference between M pe and M f is greater than 0.5 percent, so the omission of the gravity load between the plastic hinge and the face of the column flange is acceptable.

28


Design Example 1

Determine the required shear strength, V u




AISC 358, §5.8, Step 8

Determine the required shear strength of the beam and beam web-to-column connection from the following equation: ( )( M )

+

AISC 358, Eq 5.8–9

ty

where V u = required shear strength of beam and beam web-to-column connection L′ = distance between the centers of the reduced beam sections V gravity = beam shear force resulting from 1.2 D + f 1 L + 0.2S (where f 1 = 0.5) V u = V pr + (V D+ L) = 131 kips + (11.7 kips) = 143 kips.

The design shear strength of the beam is checked in accordance with AISC 360, Chapter G. This calculation was performed as part of the beam design (Part 5c). The shear strength of the W30 × 132 beam was determined to be 559 kips. Therefore, the beam is adequate to resist the shear demand at any location along the beam length, as the calculation considers the beam web only. Design the beam web-to-column connection according to AISC 358 Section 5.6 AISC 358, §5.8, Step 9 This check references AISC 358 Section 5.6. Section 5.6 indicates that the strength of the beam web-tocolumn connection strength must be determined in accordance with Equation 5.8–9. Furthermore, the following description (from AISC 358) identifies the only allowable detailing for the beam web-to-column connection: For SMF systems, the beam web shall be connected to the column flange using a CJP groove weld extending between weld access holes. The single plate shear connection shall be permitted to be used as backing for the CJP groove weld. The thickness of the plate shall be at least 3 ⁄ 8 in (10 mm). Weld tabs are not required at the ends of the CJP groove weld at the beam web. Bolt holes in the beam web for the purpose of erection are permitted. Because the beam web-to-column connection is made with a CJP groove weld, the shear capacity of the weld is greater than or equal to the shear capacity of the beam, so no further checks are required to verify the adequacy of this condition. Check continuity plate requirements according to AISC 358 Chapter 2

AISC 358, §5.8, Step 10

This check references AISC 358 Chapter 2, 2.4.4. Section 2.4.4 identifies the following equation (via AISC 341 Section E3.6f) for use in determining the need for continuity plates and the thickness of the plates if they are required by calculation. When the beam flange connects to the flange of a wide-flange or built-up I-shaped column having a thickness that satisfies Equations 2.4.4–1 and 2.4.4–2, continuity plates need not be provided. R F

t

=04 c

≥

6

=

10.5 6

1 8(10. 5 1. )

(1 1 1)( 50) (1 1)(50)

1 7 in

AISC 341, Eqs E3–8, E3–9

1 75 i 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

29

Design Example 1




where t cf = minimum required thickness (inches) of column flange when no continuity plates are provided. bbf = beam flange width, in t bf = beam flange thickness, in F yb = specified minimum yield stress of the beam flange, ksi F yc = specified minimum yield stress of the column flange, ksi R yb = ratio of the expected yield stress of the beam material to the specified minimum yield stress, per AISC 341 R yc = ratio of the expected yield stress of the column material to the specified minimum yield stress, per AISC 341 t cf = 1.28 in < 1.74 in and 1.75 in.

CONTINUITY PLATES ARE REQUIRED

Per AISC 358 Section 2.4 (AISC 341 Section E3.6f) the thickness of the plates is determined as follows: (a) For one-sided (exterior) connections, continuity plate thickness shall be at least one-half of the thickness of the beam flange. (b) For two-sided (interior) connections, the continuity plate thickness shall be at least equal to the thicker of the two beam flanges on either side of the column. Continuity plates shall also conform to the requirements of Section J10 of the AISC specification. The requirements of AISC 360 Section J10 pertain to detailing/sizing of the continuity plates. The thickness requirements listed above combined with the detailing requirements of AISC 360 Section J10 ensures that the continuity plate has adequate strength. The detailing for the continuity plates dictates some of the verifying calculations associated with the continuity plate, including the welded connections between the continuity plate and the column flanges and web. The following detailing provisions (AISC 358 Section 3.6) affect plate design: Along the web, the corner clip shall be detailed so that the clip extends a distance of at least 11 ⁄ 2 in (38 mm) beyond the published k det dimension for the rolled shape. Along the flange, the plate shall be clipped to avoid interference with the fillet radius of the rolled shape and shall be detailed so that the clip does not exceed a distance of 1 ⁄ 2 in (12 mm) beyond the published k 1 dimension. The clip shall be detailed to facilitate suitable weld terminations for both the flange weld and the web weld. When a curved corner clip is used, it shall have a minimum radius of 1 ⁄ 2 in (12 mm). Using these requirements, the projected contact area between the edge of the continuity plate and the column flange and column web are A pb = (W pb)(t cont−pl) W pb = bcont−pl − (“k 1col” + 0.25 in).

30


Design Example 1




The continuity plate width ( W pb−flange) can be determined considering the required strength at the projected bearing surface as identified in AISC 360 Section J7, using φ factors for nonductile limit states ( φ = 0.9) per AISC 358 Section 2.4.1:

φ Rct = (0.9)1.8F y A pb

AISC 360, Eq J7–1

≥

R t

p

≤

734 (0 9 (1 8)( F )

=

t

=

21509 , 30 3 1

734 (0 9 (1 8)(50)

= 734 kips

= 9 06 in 2 .

If the width available within the column section (between the column web and the edge of the column flange) limits the total available width for bearing, the following can be used to size the thickness of the continuity plate:

p m

=

t c

= 7 51in

2

W pb = W pb(max) − (“k 1col” + 0.25 in) = 7.51 − (1.1875 + 0.25) = 6.07 in p p

−

cont− pl

W p

=

9 06 6.0 7

= 1.4 9 in; say t cont−pl = 1.5 in.

Therefore, use two pairs of 1 1 ⁄ 2-inch × 71 ⁄ 2-inch continuity plates in the column aligned with the top and bottom beam flanges. In addition to the size of the continuity plate, the attachment/welding of the continuity plate shall meet the criteria established in AISC 341, E3.6f(3): Continuity shall be welded to column flanges using CJP groove welds. Continuity plates shall be welded to column webs using groove welds or fillet welds. The required strength of the sum of the welded joints of the continuity plates to the column web shall be the smallest of the following: (a) The sum of the design strengths in tension of the contact areas of the continuity plates to the column flanges that have attached beam flanges (b) The design strength in shear of the contact area of the plate with the column web (c) The design strength in shear of the column panel zone (d) The sum of the expected yield strengths of the beam flanges transmitting force to the continuity plates No design calculations are required for continuity plate to column flange portion of the connection. However, the connection between the continuity plate and column web should be calculated to determine an appropriate weld for this connection. The maximum contact area between the continuity plate and the column web is A pw = [d c − 2t cf − 2(k + 1.5 in)](1.5 in) = [33.9 − 2(1.28) − 2(2.06 + 1.5)](1.5 in) = 36.3 in2.


31

Design Example 1




The tension strength of the continuity plate is limited by the connection strength between the edge of the continuity plate and the face of the column flange. This condition is expressed in the limitations for welding required between the continuity plate and the column web. The following calculations identify the controlling case (minimum value of the four cases) for the required strength of the continuity plate-tocolumn web welds.

ΣφF y A pb = 2(0.9)(50)(9.06) = 815 kips

AISC 341, Eq 6f.(3)(a)

φvV nw = φ(0.6)F y A pw = (1.0)(0.6)(50)(36.3) = 1090 kips

AISC 341, Eq 6f.(3)(b)

φ Rv = 870 kips (see calculations in Step 11 below)

AISC 341, Eq 6f.(3)(c)

p

t

=2

(0 9 (24, 35)  30 3 − 1 0

= 1477 kip s 

AISC 341, Eq 6f.(3)(d)

The smallest value is 815 kips, which will be used to design the welds between the continuity plate and the column web. The minimum required double-sided fillet weld size to develop 815 kips follows. Dmin

=

=

(

.392

.5 in]

t

815 (2 1.392

33.9 − 2)(1 28

−

2)( 2 06

+ 1 5 in

= 10 19 (16ths fillet).

The designer could use double-sided 5 ⁄ 8-inch fillet welds (within 2 percent of required size) to connect the continuity plates to the column web. However, PJP preparation costs are not significant and often less expensive than filler metal placement, so it may be more economical to use an equivalently sized PJP weld or a CJP groove weld between the continuity plate and the column web. The most economical solution can be determined through a conversation with the project structural-steel fabricator. Check column panel zone according to AISC 358, Section 5.4

AISC 358, §5.8, Step 11

The panel zone strength is calculated below using the provisions of AISC 341 SectionE3.6e. The panel zone shear calculation is derived by projecting the expected moments at the beam plastic hinges to the face of the column, assuming points of inflection at the column mid-height between floors. (Figure 1–11)

Figure 1–11. Panel zone forces

32


Design Example 1



Special Moment Frame Frame

d b = 30.3 − (1.28) = 29.3 in M f = 21,509 kip-in

16,531 kip ip -i

+ M ′ = V c

=

=

,531 = 38,040 k ip ip--in

38, 40 k ip ip--in 144 in

t

= 264 kips

Figuree 1–12. Column shear Figur

R

=

M

(

−

=

38,04 040 0 k ipp-ii

−

(30.

.0 ) in

− 264 kip s

kips

The panel zone shear strength is determined from AISC 360, J10.6 for Pr less than 0.75 Pc

 

+

3

c

t 2

 AISC 360, Eq J10–11

c

where:

φ = 1.0 per AISC 341, Section 9.3a bcf = width of the column flange d b = depth of the beam d c = column depth t cf = thickness of the column flange t w = total thickness of the panel zone, including doubler plates.


33

Design Example 1




For the W33 × 221 column, the panel zone shear strength is

 +  = 870 kips < 1034 kips

R

3

.8)(1. .

)

3

 

870 ki

N . Doubler plates or thicker column web indicated.

The W33 × 221 column panel zone strength (without doubler plates) is not adequate when matched with the W30 × 132 beam. Howeve However, r, the panel zone is 84.1 percent of the required strength without doubler plates. This difference may be adequate considering the anticipated behavior of the connection (balanced yielding in beam and column panel zone), but it does not strictly comply with code regulations. If the panel zone must be strengthened, and if doubler plates are used in lieu of increasing the column size, then compliance with AISC 341 Section 6e(3) is required. The minimum panel zone thickness, t z, is also checked per AISC Section 341, E3.6e.(2). T

(

w )

AISC 341, Eq E3–7

90

where: d z = d − 2t f of the deeper beam at the connection w z = panel zone width between column flanges t = thickness of column web or doubler plate

=

≥

30.3 − 2(1 0) 90

=

.314 in in..

Check moment ratio according to AISC Section E3.4a

AISC 358, §5.8, Step 12

The moment ratio is checked in accordance with AISC Section E3.4a, with special attention paid to the definition of * and how the engineer derives the summation of nominal beam and column flexural * . The difference between M and strengths. M pc is projected to the centerline of the beam to compute M pc pc * is the column shear multipled by the distance from the flange of the beam to the centerline of the beam (Figure 1–17). The cruciform is defined by the assumed inflection point in the column(s) and the center of the beam(s) RBS (Figure 1–13).

35.95''

35.95'' RBS

RBS

Figuree 1–13. Beam-column dimensions Figur

34

2012 IBC SEAOC Structural/Seismic Design Manual, Vol. Vol. 4

Design Example 1




V'RBS = 122k Mpr = 18,849in-k

Mpr = 18,849in-k VRBS = 140k

35.95''

35.95'' RBS

RBS

Figuree 1–14. Demands from beam Figur

M'pb-r

M'pb-I M'RBS

MRBS

V'RBS

VRBS

35.95''

35.95'' RBS

RBS

Figure 1–15. Free-body diagram for M p*

M p* − M p* −r

p

+ (V

S

p

+ (V

S

 2

+

 2

=

+ (122

19 in +

=

+ (14 140 0

) 19 in +

 

33. 2 33 9 2

in



= 23,235 kip ip- -in

in  = 23,882 kip ip- -in

+ 23 88 882 = 47,117 kip ip-i -in n

* p

M pc−t −

−

2 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

35

Design Example 1

=

V




M pc −t

− M *

−

2

+ V ol

pc t

pc

*

t

−

 = 2 

+ M *pc

−t

M

   +  2 − 2   M   d  + pc −  2 − 2    t

d

+

*

*

 = 2 

  H

Mpc = 37,776 in-k M'pc M'pc Mpc = 37,776 in-k

Figure 1–16. Fr Free-bod ee-bodyy diagram for

386 kips 65 2 in2



M pc *

36

yc

=(

−

5 92 ksi P

 

=

− 5 92 

144 44 −



=

= 37 776 ki kipp-iin

,687 kip-in


*

pb

Design Example 1




Vcol-t

ht – db 2

db

Mpc – b

hb – db 2

Vcol-b

Figuree 1–17. Development of Figur

Is

* M pc *

*

1. ?

p

*

M p*

=

5,68 687 7 k -i 47,11 117 7 k -in

= 2.0 3

. . . OK.

Evaluate lateral bracing of columns

AISC 358, §5.3.2

Per AISC 358 Section 5.3.2, lateral bracing of columns shall conform to requirements of AISC 341. To preclude SMF column members from lateral torsional buckling AISC 341 Section E3.4c.(1)(1) specifies requirements for column flange bracing. The W33 × 221 column has a perpendicular beam framing into it at each level providing out-of-plane joint restraint. At the beam top flange, the concrete slab effectively effectively provides bracing for the column flange. The column flanges therefore need to be laterally braced at the beam bottom flange only if AISC 341 Equation 3–1 is not greater than 2.0. *

M p*

2.0

From Step 12 of this design example: * M pc * p

5,68 687 7 k -in 4 ,

k-i

2 03 . . . OK.


37

Design Example 1




The column flanges, therefore, do not need lateral bracing at the beam bottom flange. If bracing were required, it may be provided by perpendicular beams connected to full-depth stiffeners stiffeners and column continuity plates. When bracing is provided, it must be designed for a required strength that is equal to 2 percent of the available beam flange strength, as appropriate. Welding parameters Parameters for pre-qualified welded joints are presented in AISC 358 Chapter 3. The requirements identified in AISC 358 reference AISC 341. The requirements for welding are clearly outlined in AISC 341 Chapter I Section I2.3, which references AWS D1.8. Specific protocols associated with weld backing and weld tabs are described in AISC 358 Chapter 3 and AISC 341 Section I2.3. Quality control and quality assurance Quality control and quality assurance requirements are outlined in AISC 358 Section 3.7, which refers to AISC 341. AISC 341 Chapter J covers quality control and quality assurance related to SMF fabrication and erection. Specifications regarding the quality of RBS fabrication are provided in AISC Section 5.7. A “protected zone” is established for the connection (AISC 358 Section 5.3.1(8)). The zone is defined as the portion of the beam between the face of the column flange to the end of the RBS farthest from the column.

6. Detailing of RBS Connection The details shown in Figures 1–18 through 1–22 are representative SMF RBS details.

38


Design Example 1




Figure 1–18. RBS connection 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

39

Design Example 1




Figure 1–19. RBS connection


40


Design Example 1






41

Design Example 1




Figure 1–22. Continuity plate

42


Design Example 2 Special Concentrically Braced Frame

OVERVIEW This example shows procedures for the design of Special Concentrically Braced Frame (SCBF) buildings. It is intended to provide specific methods for the design of braced frames that comply with the International Building Code and the AISC Seismic Provisions for Structural Steel Buildings (AISC 341), guiding designers toward the careful consideration of the performance of concentrically braced frame structures under severe seismic loading. Certain recommendations provided are considered best practice; nevertheless, the calculation methods illustrated are applicable to a wide range of designs. The SCBF system has been developed over several cycles of building codes as a moderately ductile system that can withstand moderate inelastic drift while maintaining strength. In order to provide this performance, SCBF braces must accommodate significant compression buckling demands. In addition, the system must be able to realize the strength and stiffness of braces subject to tension as the strength and stiffness of buckled braces in compression diminishes. Thus SCBF are intended to have post-elastic behavior that differs significantly from the elastic distribution of forces. A simple linear analysis of force distributions is insufficient, and an amplification factor is, in many cases, insufficient as well. Design rules for SCBF have thus always contained some form of requirement for consideration of post-elastic conditions.


43

Design Example 2



Special Concentrically Braced Frame

The 2010 edition of AISC 341 has gone further than previous codes in making this latter requirement an explicit requirement. The provisions require that beams and columns have sufficient strength to withstand forces corresponding to two different conditions: the maximum forces the frame can resist (with frame forces corresponding to braces reaching their expected buckling strength and expected tension strength), and the post-buckled condition (with frame forces corresponding to braces reaching a low estimated post-buckling strength and expected tension strength). These are essentially plastic-mechanism analysis requirements and are illustrated in this example. As important as determining design forces for frame members is the detailing of connections to accommodate building drift and ductility demands. To accommodate building drift, the effect of gussets on the beam-to-column connection and the column base-plate connection must be considered. These gusseted connections should be considered fully rigid unless special detailing is used to allow for relative rotation. (The use of typical “simple” connections in combination with a gusset plate is insufficient to guarantee adequate rotation capacity.) Connections considered rigid will develop large moments at the design story drift, and AISC 341 requires that the connection have flexural strength corresponding to the strength of the beam or of the column. In this example a fully rigid beam-to-column connection is employed. This connection is a combination of the gusset plate with a Special Moment Frame WUF-W (Welded Unreinforced Flange-Welded Web) connection. The design of alternative connections (including the accommodation of rotation) is illustrated in the AISC Seismic Design Manual. Accommodating brace compression ductility demands entails detailing the gusset plate to allow for brace rotations or designing the connection as a fixed end for the brace. In this example the former approach is taken, with a hinge plate oriented to allow for in-plane rotation. Alternative designs are illustrated in the AISC Seismic Design Manual. This example does not include the design of a base-plate connection. However, Design Example 9 illustrates a base-plate design for a Buckling Restrained Braced Frame and can serve as a guide for SCBF base plates. For more information on the SCBF system see Blue Book article 08.03.050: “Concentric Braced Frames”; August 2008.

44

2012 IBC SEAOC S tructural/Seismic Design Manual, Vol. 4

Design Example 2




OUTLINE 1. Building Geometry and Loads 2. Design Base Shear and Load Combinations 3. Vertical and Horizontal Distribution of Load 4. Brace Sizing 5. Plastic Mechanism Analysis 6. Beam Seismic Forces 7. Column Seismic Forces 8. Detailing and Design of Connections 9. Additional Considerations 10. Items Not Addressed in This Example

1. Building Geometry and Loads 1.1 GIVEN INFORMATION The building is a six-story office building located in San Francisco. The Seismic Design Category is D. See Appendix 1 for the following information: • Building dimensions, • Floor and roof weights, • Latitude and longitude, • Soil type, • Spectral accelerations, and • Load combinations including the vertical seismic-load effect.

1.2 FRAME LAYOUT Early design decisions that should be made include • Location of frames, • Configuration of frames, and • Relationship of braces to the architecture.


45

Design Example 2




1.2.1 LOCATION OF FRAMES In this example the frames are located at the building perimeter, which is more efficient in controlling building torsion and ensuring redundancy. The plan layout of frames at floors 1–4 is shown in Figure 2–1.

Figure 2–1. Plan

46


Design Example 2




1.2.2 CONFIGURATION OF FRAMES Frames are configured in a two-story X configuration, which is advantageous in limiting beam flexural demands in the post-buckled condition. Additionally, the frames are offset at floors 5 and 6. This reduces the column overturning demands, which is especially beneficial for column size (and consequently columnsplice demands), base-plate demands, and foundation demands. This constitutes an irregularity, but this irregularity does not represent a dramatically different collector beam demand from those in similar regular configurations. A typical frame elevation is shown in Figure 2–2.

B

C

D

E

ROOF

6th FLR

5th FLR

4th FLR

3rd FLR

2nd FLR

1st FLR

Figure 2–2. Frame elevation


47

Design Example 2




1.2.3 RELATIONSHIP OF BRACES TO THE ARCHITECTURE Braces in this example are exposed. Thus, no special wall detailing is required to allow for brace buckling deformation. However, braces are located within 8 inches of the exterior glass, potentially leading to a falling hazard should braces buckle out of the plane of the frame, causing damage to the façade. Such a hazard could be mitigated in several ways, including increasing the physical separation or treating the glass to prevent falling debris. In this example the hazard is eliminated by configuring braces to buckle in the plane of the frame by using a round section in combination with end detailing that favors in-plane over out-of-plane rotation. The detail employed at the beam-to-column connection is shown schematically in Figure 2–3.

Figure 2–3. Typical gusset configuration

This detail has many advantages over more typical single-gusset-plate details. In addition to avoiding out-of-plane brace deformation, the detail avoids potential conflict between the composite deck and the hinge plate as the latter bends as a result of brace buckling. The detail is also highly adaptable to special conditions, including connections to sloped beams.

48


Design Example 2




2. Design Base Shear and Load Combinations

ASCE 7

2.1 CLASSIFY THE STRUCTURAL SYSTEM

T12.2–1

There are two options in Table 12.2–1 for using SCBF: a pure SCBF, and a SCBF/SMF dual system. For the purposes of this example, a pure SCBF without the dual system will be used: R = 6.0

Ωo = 2.0

C d = 5.0

2.2 DESIGN RESPONSE SPECTRUM The spectral accelerations to be used in design are derived in Appendix 1: S DS = 1.00g

S D1 = 0.60g

2.3 HORIZONTAL IRREGULARITIES

T12.3–1

The building has a type 2 horizontal irregularity. This requires an increase in diaphragm-to-collector transfer forces (as illustrated in Design Example 7), but does not affect the SCBF.

2.4 VERTICAL IRREGULARITIES

T12.3–2

The building has type 3 and 4 vertical irregularities. The type 4 irregularity requires that the elements supporting the braced frame be designed for the special seismic-load combinations (including the “amplified seismic load”). This is a requirement for SCBF columns regardless of irregularity. It also requires an increase in diaphragm-to-collector transfer forces, which is also required by the horizontal irregularity. The type 3 irregularity precludes the use of the equivalent lateral force (ELF) procedure per Table 12.6–1.


T12.6–1

Because of the type 3 irregularity, the modal response spectrum (MRS) procedure is employed. For preliminary design, the ELF procedure is employed, with the design to be confirmed using MRS. Using the MRS procedure, ASCE 7 allows for a reduction in design base shear, with a minimum of 85 percent of the ELF base shear. Thus it is advantageous to utilize MRS. For preliminary ELF design, the MRS base shear will be assumed to be scaled to 90 percent of the ELF base shear.

2.6 BASE SHEAR Determine the approximate fundamental building period using Section 12.8.2.1: C t = 0.02 and x = 0.75 T

x t

0 02

T 12.8–2

= 0 49se


Eq 12.8–7

T a = 0.49 sec


49

Design Example 2




The calculation of seismic response coefficient C S is performed using four equations. These equations, when presented together in a graph of the base shear coefficient versus building period, constitute the response spectrum for the site. (See Volume 1 of the 2012 IBC SEAOC Structural / Seismic Design Manual for further explanation.) The first equation defines the coefficient as: C s =

S DS R I

=

1 00 0

= .167 g .

Eq 12.8–2

The coefficient need not exceed: C =

1

( R I T

=

60 6 0 / 1 0) 49

= 0.204 .

Eq 12.8–3

The value of seismic response coefficient, C s, computed in Equation 12.8–1 must be at least equal to the following

= 0 044(1 0

C s

.

Eq 12.8–5

In addition, for buildings in Seismic Design Category E or F, and those buildings for which the 1.0-second spectral response, S 1, is equal to or greater than 0.6 g, the value of the seismic response coefficient, C s, must be at least equal to the following value: C =

05

=

0 5( 0 6 g) 6 0 / 1 0

=

05 .

Eq 12.8–6

Equation 12.8–2 controls. C s = 0.167

V = C sW = 0.167(7231 kips) = 1205 kips

Eq 12.8–1

V = 1205 kips

2.7 REDUNDANCY FACTOR The structure qualifies for a redundancy factor of 1.0 through calculation. See Volume 1 of the 2012 IBC SEAOC Structural / Seismic Design Manual for the calculation methodology.

2.8 LOAD COMBINATIONS Load combinations are addressed in Appendix 1 of this volume.

50


Design Example 2





ASCE 7

3.1 VERTICAL DISTRIBUTION OF SHEAR

§12.8.3


Level

wi (kips)

hi (ft)

i i

C vx

Roof

656

72

70,906

0.174

210

6th

1315

60

116,413

0.286

344

5th

1315

48

91,177

0.224

269

4th

1315

36

66,539

0.163

197

3rd

1315

24

42,683

0.105

126

2nd

1315

12

19,982

0.049

59

Total

7231

407,700

F x (kips)

1205

The terms used in this table are defined in Section 12.8.3. The period is assumed to be 1.4 times the approximate period: 1.4 * 0.49 sec = 0.69 sec; this should be confirmed or adjusted after member sizing. Since the period = 0.69 > 0.5 sec, the value for k is interpolated between a value of 1.0 for T = 0.5 sec and 2.0 for T = 2.5 sec. In this example, k = 1.095. The distribution of story shear is carried out using: F x = C vxV , where, C vx =

w

∑= w

Eq 12.8–11 and Eq 12.8–12 i

1

3.2 HORIZONTAL DISTRIBUTION OF STORY SHEAR The horizontal distribution of forces requires consideration of diaphragm torsion. This can be approximated at this stage of design (prior to selection of brace sizes) by making rough assumptions to be confirmed at a later stage. The frames on grids A and F are each assumed to resist 50 percent of the north-south seismic forces, plus 80 percent of the accidental eccentricity. This results in a share of the north-south forces equal to: R A = 0.5V + 0.8(V [0.05 L]/ L) = 0.54V .

The frames on Grids 1 and 5 are designed with a similar assumption, such that each frame is designed for 54 percent of the base shear. This overestimates the effect of accidental eccentricity slightly. This factor is applied to the force at each story. A more complex layout might require a more in-depth analysis—possibly modeling—to distribute the forces to each frame for design purposes.


51

Design Example 2




All four frames will be designed to resist the forces in Table 2–2:


Level

F x (kips)

Frame Force (kips)

Frame Shear (kips)

6th

210

113

113

5th

344

186

299

4th

269

146

444

3rd

197

106

551

2nd

126

68

619

1st

59

32

651

4. Brace Sizing

AISC 341

4.1 REQUIRED STRENGTH The required strength of braces is calculated from the seismic forces in Table 2–2 above. It is common for designers to neglect the gravity forces in braces, although that approach is not explicitly permitted for SCBF. Where gravity forces are significant they should be considered. In this case the gravity forces, shared by four braces and the beam, have been determined to be 4.5 kips for the typical floor condition for the seismic load combinations. The tension effect of this gravity force is conservatively neglected for design. Converting the frame shear into a brace shear considering the four braces and the brace angle of ATAN(12/15) = 38.7 deg, the brace forces are calculated and presented below. The required strength in compression is calculated adding the gravity force.

Table 2–3. Brace design forces for Frame A

Level

Brace Seismic Force (kips)

Brace Gravity Force (kips)

Pu (kips)

6th

36

0

36

5th

96

4.5

100

4th

142

0

142

3rd

176

4.5

181

2nd

198

0

198

1st

208

4.5

213

The same is done for Frame 1. At the top two stories there are only two braces, instead of the four in Frame A.

52


Design Example 2




Table 2–4. Brace design forces for Frame 1

Level

Brace Seismic Force (kips)

Brace Gravity Force (kips)

Pu (kips)

6th

72

0

72

5th

191

2

193

4th

142

0

142

3rd

176

2

178

2nd

198

0

198

1st

208

2

210

4.2 SECTION SELECTION The effective length of the braces is assumed to be the work-point length (19.2 feet), minus some distance at each end to account for connection size. For member selection purposes the effective length will be assumed to be 16 feet. To conform to the limit on brace slenderness, the minimum radius of gyration is: r ≥ 16 ft (12 in/ft)/200 = 0.960 in.

Round HSS will be used. See Blue B ook article 08.03.050 for a discussion of the behavior of brace section types. The availability of sections should be verified. Round HSS must meet the requirements of AISC 341 for highly ductile members: D / t ≤ 0.038 E / F y = 26.2.

The following brace sizes are to be used for Frame A, based on design strengths tabulated in the Manual (Table 4–5):

Table 2–5. Brace designs for Frame A

φPn (kips)

Level

Brace Size

r (in)

kL / r

D / t

6th

HSS5.563 × 0.258

1.88

102

23.2

80

5th

HSS6 × 0.312

2.26

85

20.6

113

4th

HSS7 × 0.375

2.35

82

20.1

183

3rd

HSS7 × 0.375

2.35

82

20.1

215

2nd

HSS7.625 × 0.375

2.58

74

21.8

215

1st

HSS7.625 × 0.375

2.58

74

21.8

215


53

Design Example 2




The following brace sizes are to be used for Frame 1:

Table 2–6. Brace designs for Frame 1

Brace Size

r (in)

kL / r

D / t

φPn (kips)

6th

HSS7 × 0.375

2.35

82

20.1

183

5th

HSS7.625 × 0.375

2.58

74

21.8

215

4th

HSS7 × 0.375

2.35

82

20.1

183

3rd

HSS7 × 0.375

2.35

82

20.1

183

2nd

HSS7.625 × 0.375

2.58

74

21.8

215

1st

HSS7.625 × 0.375

2.58

74

21.8

215

Level

The sixth-floor brace is substantially oversized to limit unbalanced forces on the intersected beam.

5. Plastic Mechanism Analysis

AISC 341

Once the braces are sized, the maximum demands on the rest of the system can be calculated. Two conditions are considered: the maximum forces that braces can deliver (in either tension or compression), and the condition after some braces have buckled (which can be critical for certain members). In order to perform these analyses, three values are required for each brace: the expected tension strength, the expected compression strength, and the approximate post-buckling strength. The expected tension strength is: R y F y Ag The expected compression strength is: 1.14 F cre Ag, where F cre is calculated using the expected yield strength, R y F y in the AISC 360 Chapter E equations in Section E.3. The post-buckling strength is taken as 0.3 times the expected compression strength (as required in Section F2.3): 0.3(1.14 F cre Ag) = 0.342F cre Ag. The values for these forces for the sections used are shown in Table 2–7.

Table 2–7. Brace capacity forces

Expected Tension Strength (kips)

Expected Compression Strength (kips)

Post-Buckling Strength (kips)

HSS5.563 × 0.258

236

110

29

HSS6 × 0.312

307

161

42

HSS7 × 0.375

429

275

72

HSS7.625 × 0.375

469

332

87

Brace Size

54


Design Example 2




5.1 CONDITION 1: MAXIMUM TENSION FORCE AND MAXIMUM COMPRESSION FORCE Figure 2–4 shows the maximum-force condition for Frame A, with the braces removed and the capacity forces substituted. Assuming a first-mode deformation, all braces in tension are assumed to reach their full expected tension strength ( R y F y Ag), and all braces in compression are assumed to reach their full expected compression strength (1.14 F cre Ag). Lateral forces indicated do not correspond to the calculated base shear.

B

C

D

Ry Fy Ag

1.14Fcre Ag

1.14 Fcre Ag

Ry Fy Ag 1.14Fcre Ag

Ry Fy Ag 1.14Fcre Ag

Ry Fy Ag

E

Ry Fy Ag

1.14 Fcre Ag

Ry Fy Ag

1.14Fcre Ag

Ry Fy Ag

1.14Fcre Ag

Ry Fy Ag

1.14Fcre Ag

1.14Fcre Ag

Ry Fy Ag

Ry Fy Ag

1.14 Fcre Ag

Figure 2–4. Maximum-force condition for Frame 1

5.2. CONDITION 2: MAXIMUM TENSION FORCE AND POST-BUCKLING COMPRESSION FORCE Figure 2–5 shows the post-buckled condition for Frame A, with the braces removed and the capacity forces substituted. Assuming a first-mode deformation, all braces in tension are assumed to reach their full expected tension strength ( R y F y Ag), and all braces in compression are assumed to have degraded to their nominal post-buckling strength (0.342 F cre Ag).


55

Design Example 2




B

C

D

Ry Fy Ag

0.342 Fcre Ag

Ry Fy Ag 0.342 Fcre Ag

0.342 Fcre Ag

Ry Fy Ag

Ry Fy Ag

0.342 Fcre Ag

0.342 Fcre Ag

Ry Fy Ag

0.342 Fcre Ag

Ry Fy Ag

E

Ry Fy Ag 0.342 Fcre Ag

0.342 Fcre Ag

Ry Fy Ag

Ry Fy Ag

0.342 Fcre Ag

Ry Fy Ag

0.342 Fcre Ag

Figure 2–5. Post-buckled condition for Frame 1

6. Beam Seismic Forces

AISC 341

6.1 INTERSECTED BEAM (SIXTH FLOOR, Frame 1) The seismic forces on this beam for Conditions 1 and 2 can be calculated considering the brace-capacity forces from the table above. Condition 1 will have larger axial force; Condition 2 will have a larger flexural force. Figure 2–6 shows a free-body diagram of the beam.

Condition 1: 1.14 F cre A g

R y F y A g

Condition 2: 0.342 F cre A g

R y F y A g

Condition 1: 1.14 F cre A g Condition 2: 0.342 F cre A g

Figure 2–6. Free-body diagram of the sixth-floor beam

The diagonal forces are converted to vertical and horizontal forces in order to obtain beam shear, flexure, and axial forces.

56


Design Example 2




6.1.1 CONDITION 1 The vertical force on the beam is QV 1 = [ R y F y Ag − 1.14F cre Ag]5 sin (θ5) − [ R y F y Ag − 1.14F cre Ag]6 sin (θ6)

= 12.5 kips. The horizontal force on the beam is Q H 1 = [ R y F y Ag + 1.14F cre Ag]5 cos (θ5) − [ R y F y Ag + 1.14F cre Ag]6 cos (θ6)

= 76 kips. Given the symmetry of the condition, the axial force in each segment of this beam can be taken as Pu = 1 ⁄ 2Q H 1 = 38 kips.

6.1.2 CONDITION 2 The vertical force on the beam is QV 2 = [ R y F y Ag − 0.342F cre Ag]5 sin (θ5) − [ R y F y Ag − 0.342F cre Ag]6 sin (θ6)

= 15 kips. The horizontal force on the beam is Q H 2 = [ R y F y Ag + 0.342F cre Ag]5 cos (θ5) − [ R y F y Ag + 0.342F cre Ag]6 cos (θ6)

= 42 kips. Given the symmetry of the condition, the axial force in each segment of this beam can be taken as 1 ⁄ 2Q H 1 = 21 kips. Typically Condition 2 governs. In this case it is unclear which condition governs due to the selection of similar brace sizes above and below the beam. U se of a significantly smaller brace size at the sixth floor would result in a much larger vertical force on the beam and may be uneconomical. The forces above combined with gravity forces are applied to the beam for design. The governing combination from Appendix 1: 12+

+ 0 5 L =

+

5L + Q

Load Combination 5 (modified; see Appendix 1)

The gravity moment is based on

= 1.4 D + 0.5 L = 1.4 (6 ft * 67.7 psf + 12 ft * 19 psf) + 0.5 (6 ft * 100 psf) = 1.19 klf M u = wg L2 /16

= 66.8 ft-kip. All braced-frame beams are W18 × 50 to provide for uniformity of connections. For brevity, the design of this beam will not be shown. 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

57

Design Example 2




6.2 REDISTRIBUTION BEAM (THIRD FLOOR, Frame 1) For this beam, Condition 2 governs by inspection. Figure 2–7 shows a free-body diagram of this beam. The story force is assumed to be delivered two-thirds via shear along the beam and one-third via a collector force on the right-hand side.

R y Fy A g

0.342 Fcre A g

0.342 Fcre A g

R y Fy A g

Figure 2–7. Free-body diagram of the third-floor beam

In order to calculate the axial force in the beam, the story force delivered to this portion of the frame corresponding to the Condition 2 forces must be calculated: F 3 = [ R y F y Ag + 0.342F cre Ag]2 cos (θ2) − [ R y F y Ag + 0.342F cre Ag]3 cos (θ3)

= 43 kips. At the right-hand connection, one-third of this force is delivered: F 3 R = 1 ⁄ 3F 3 = 14 kips.

Figure 2–8 shows a free-body diagram of the right-hand connection.

0.342

Fcre A g

F3R

Pu R y Fy A g

Figure 2–8. Free-body diagram of the right-hand connection of the third-floor beam

58


Design Example 2




The axial force in the beam is Pu = F 3 R + [0.342F cre Ag]3 cos (θ3) − [ R y F y Ag]2 cos (θ2)

= 296 kips. The moment is the same as for the beam in the previous section: M u = wg L2 /16

= 66.8 ft-kip. The compressive strength can be calculated considering the restraint provided at the top flange by the slab. As described in Example 7, constrained-axis flexural-torsional buckling is the limiting buckling mode for this condition. The AISC Seismic Design Manual gives this expression for use in calculating constrainedaxis flexural-torsional buckling for a beam restrained at the top flange (Equation 8–3):

 E Cw F

=

+ I y

( K )

2

 d   2

 1

+ GJ

2

x

  + 2

Using this equation in conjunction with Equation E3–1, the constrained-axis flexural-torsional buckling strength for the W18 × 50 with top flange continuously braced and the bottom flange braced at 15 ft is

φPn = 359 kips Pu / φPn = 296 kips/359 kips = 0.83.

From Table 3–10 of the Manual, the flexural strength is

φ M n = 258 ft-kips. For this moment diagram and bracing condition C b = 3.0. Thus the flexural strength is

φ M n = 3.0 * 258 ft-kips ≤ φ M p = 379 ft-kips M u / φ M n = 66.8 ft-kip/379 ft-kips = 0.18 Pu / φPn + 8/9 M u / φ M n = 0.99 . . . OK.

6.3 FIFTH-FLOOR BEAM (Frame 1) The two examples above (the sixth-floor and third-floor beams) are representative of all the beams in Frame A, with the even-numbered floors similar to the sixth floor and the odd-numbered floors (and roof) similar to the third floor. Frame 1 is similar, except for the in-plane offset at the fifth floor. Due to this, the three beams at the fifth floor in Frame 1 should be analyzed in a single free-body diagram, as shown in Figure 2–9.


59

Design Example 2




Condition 1: 1.14 Fcre A g

Ry Fy A g

0.342 Fcr e A g

Condition 1:

Condition 1: 1.14 Fcr e A g

Condition 2:

Ry Fy A g

1.14 Fcr e A g

Condition 2:

Condition 2:

0.342 Fcr e Ag

0.342 Fcr e A g

Ry Fy A g

Figure 2–9. Free-body diagram of fifth-floor beams (Frame 1)

The seismic forces on this beam for Conditions 1 and 2 can be calculated considering the brace-capacity forces from Table 2–7. Condition 1 will have larger axial force in the end beams; C ondition 2 will have a larger axial force in the middle beam. For brevity, the design of these beams will not be shown.

6.3.1 CONDITION 1 The force entering the frame at level 5 is Q H 1 = 2[ R y F y Ag + 1.14F cre Ag]4 cos (θ4) − [ R y F y Ag + 1.14F cre Ag]5 cos (θ5)

= 474 kips. This force is distributed axial force on each beam of 474 kips/3 = 158 kips. The right-hand beam has an axial force at the right end of Q H = [ R y F y Ag]4 cos (θ4) = 335 kips.

At the left end of the right-hand beam, the axial force is 335 kips − 158 kips = 177 kips. This axial force controls for the two outside beams (considering the load to be applicable in either direction).

6.3.2 CONDITION 2 The force entering the frame at level 5 is Q H 2 = 2[ R y F y Ag + 0.342F cre Ag]4 cos (θ4) − [ R y F y Ag + 0.342F cre Ag]5 cos (θ5)

= 348 kips. This force is distributed axial force on each beam of 348 kips/3 = 116 kips. Recognizing the equal forces in each bay, the center beam has an axial force at each end of 1 ⁄ 2(116 kips) = 58 kips.

60


Design Example 2




7. Column Seismic Forces

AISC 341

7.1 CONDITION 1 Column seismic forces for Condition 1 are shown in Table 2–8. Four columns are shown, with compression shown as positive forces and tension as negative. The forces are, of course, for one direction of loading.

Table 2–8. Column seismic forces for Condition 1

Level

Column B/1 Force (kips)

Column C/1 Force (kips)

Column D/1 Force (kips)

6th

−172

268

5th

−177

263

Column E/1 Force (kips)

4th

−172

−202

298

268

3rd

−172

−202

298

268

2nd

−647

263

−177

733

1st

−647

263

−177

733

Base

−941

470

−470

941

These column forces are obtained considering the brace forces from Figure 2–4. In addition, where a beam supports an unbalanced vertical seismic force, the reactions from this force can be included. (If they are not included, the summation of vertical seismic reactions may not equal zero.) Here the only beam with an unbalanced vertical force is the sixth-floor beam, which has a small unbalanced vertical seismic force. (Because of repetition of brace sizes and equal story heights, other beams have balanced seismic vertical forces.) The vertical unbalanced force on the sixth-floor beam was calculated in the previous section on beam forces. Q = [ R y F y Ag − 1.14F cre Ag]4 sin (θ4) − [ R y F y Ag − 1.14F cre Ag]5 sin (θ5)

= 10 kips. This results in a slight adjustment to the Column 2 and Column 3 forces below the sixth floor.

7.2 CONDITION 2 A similar process is used for Condition 2. (The vertical unbalanced force is shown in the section on beam forces.) The resulting seismic forces are presented in Table 2–9.


61

Design Example 2




Table 2–9. Column seismic forces for Condition 2

Level




6th

−45

268

5th

−37

276


4th

−45

−62

285

268

3rd

−45

−62

285

268

2nd

−368

276

−38

606

1st

−368

276

−38

606

Base

−661

330

−331

661

7.3 OVERSTRENGTH FACTOR Columns should also consider the overstrength factor. These seismic forces are presented in Table 2–10.

Table 2–10. Column seismic forces considering the overstrength factor

Level





6th

−91

91

5th

−91

91

0

4th

−178

−152

152

178

3rd

−178

−152

152

178

2nd

−646

316

−316

646

1st

−646

316

−316

646

Base

−906

576

−576

906

7.4 DESIGN SEISMIC FORCES Columns B/1 and E/1 may use the overstrength forces as a maximum (and thus neglect Conditions 1 and 2). However, in this case the mechanism forces are lower. Thus Columns B/1 and E/1 should be designed for the maximum of Condition 1 and 2, with the overstrength forces as a maximum. Columns C/1 and D /1 must be designed for the maximum of Condition 1, Condition 2, and the forces considering overstrength. The seismic forces considered for design are shown in Table 2–11.

62


Design Example 2




Table 2–11. Column seismic forces used for design

Level




6th

−172

268

5th

−177

272


4th

−172

−202

298

172

3rd

−172

−202

298

172

2nd

−646

316

−316

646

1st

−646

316

−316

646

Base

−906

576

−576

906

Column sizes will be governed by compression design. The governing combination from Appendix 1: 12+

+

=

+

+Q

Load Combination 5 (modified)

The design forces for interior columns and the selected sizes (using Table 4–1 of the AISC Manual) are shown in Table 2–12.

Table 2–12. Column design forces and sizes

Level

Exterior Column Force (kips)

6th

21

5th

69

4th

294

3rd

341

2nd

856

1st

903

Exterior Column Size

Interior Column Force (kips)

Interior Column Size

300 W12 × 40

385

W12 × 45

491 W12 × 45

570

W12 × 96

640 W12 × 96

8. Detailing and Design of Connections

719

W12 × 96

AISC 341

The connection of the HSS7 × 0.375 to the roof will be designed. Brace connections are designed to meet three conditions: the brace yielding in tension, the brace reaching its maximum compression strength, and the connection rotation associated with a brace buckling and undergoing large axial deformation. Additionally, the connection must be detailed to accommodate a frame drift of 2.5 percent, either by providing relative rotation capacity or moment capacity to allow beam or column yielding.


63

Design Example 2




Figures 2–3 and 2–10 show the details of the connection. The gusset plate is slotted to receive the knife plate. Also, the weld of the knife plate to the brace is wrapped around the end of the brace. This has been shown to be effective in forestalling net-section rupture (Cheng and Kulak, 2000).

Figure 2–10. Connection detail

8.1 CONDITION 1: BRACE YIELDING The required tension strength of the connection has been calculated previously: Ru1 = R y F y Ag = 429 kips.

64


Design Example 2




8.1.1 SELECT KNIFE PLATE Assume a 7 ⁄ 8-inch knife plate thickness ( t KP). The required width is: Ru1 /[(φF y t KP)] = 11 in.

A 7 ⁄ 8-inch × 12-inch knife plate will be used. The plate may be rectangular or shaped as shown in the figure.

8.1.2 BRACE TO KNIFE-PLATE WELD Assume a 5 ⁄ 16-inch weld size ( s). The length ( Lw) is Lw ≥ Ru1 /[4φ0.6F EXX (0.707 s)] = 15.4 in.

A 16-inch-long weld will be used for both welds. This is greater than 1.3 times the brace diameter (1.3 * 7 in = 9.1 in), which permits U = 1.0 (AISC 360).

8.1.3 KNIFE PLATE TO GUSSET WELD The weld of the knife plate to the gusset transfers the brace axial force into the gusset; it also provides continuity of the gusset across the slot. Thus in lieu of the smaller weld size selected for the knife plate to the brace, a size sufficient to provide gusset continuity will be used. Assume a 1 ⁄ 2-inch gusset ( t g). Based on recommendations by Lehman et al. (2008), the ratio of weld size to plate thickness is derived considering the expected plate tension strength and the strength of fillet welds loaded transverse to their axis: s ≥ ( R y F y t g)/(2 * 1.5φ0.6F EXX 0.707) s ≥ 0.82t g s = 7 ⁄ 16 in.

The length ( Lw) is Lw ≥ Ru1 /[4φ0.6F EXX (0.707 s)] = 11 in.

A 12-inch weld will be used, providing a moderate over-capacity.

8.1.4 BRACE CHECKS The brace wall may be checked for shear rupture. However, as the weld size is less than the brace thickness, by inspection this will not control. Because the knife plate is welded around the end and U = 1.0, the limit state of brace net section rupture does not apply.


65

Design Example 2




8.1.5 GUSSET THICKNESS, WIDTH, AND DIMENSIONS At the end of the weld, the required gusset width is calculated using the assumed thickness: Ru1 /[(φF y t g)] = 19.1 in.

This is less than the maximum Whitmore width: [2 * Lw tan (45°)] = 24 in. 45 degrees is used for the Whitmore width. To achieve this width, a rectangular gusset 21 inches horizontal by 13.5 inches vertical will be used. A 3-inch corner clip is made to provide a roughly orthogonal condition at the lap with the knife plate. This plate provides for the required length of knife-plate-to-gusset weld. If the gusset is adequate in tension at the end of the knife plate, checks of the vertical and horizontal sections through the gusset at the beam and column face will show adequate demand-to-capacity ratios.

8.1.6 GUSSET ANALYSIS The gusset is analyzed in order to obtain design forces for checking web yielding. (Welds are designed to match the gusset plate capacity.) The gusset may be analyzed using any convenient methodology. Here the Ricker method is used for simplicity; the resulting welds at the beam-to-column connection are small and can be neglected in the case of the WUF-W connection. Using the Ricker method, the following forces are obtained: V ub = 193 kips V uc = 75 kips H ub = 241 kips H uc = 94 kips.

8.1.7 GUSSET TO BEAM AND COLUMN WELDS These welds are designed to ensure that any ductility demands occur in the gusset plate (which has substantial ductility) rather than in the welds. Based on the proportioning rule discussed above for the knife-plate to gusset weld: s ≥ 0.82t g

Two 7 ⁄ 16-inch fillet welds will be used.

8.1.8 BEAM WEB YIELDING φ Rn = 1.0t wF y( N + 2.5k ) = 461 kips ≥ V ub = 193 kips . . . OK.

66


Design Example 2




8.1.9 COLUMN WEB YIELDING φ Rn = 1.0t wF y( N + 5 k ) = 317 kips ≥ H uc = 94 kips . . . OK.

8.2 CONDITION 2: BRACE BUCKLING (MAXIMUM COMPRESSION FORCE) The required compression strength of the connection has been calculated previously: Ru2 = 1.14F cre Ag = 275 kips.

This force is used to check web crippling in the beam and column. (By inspection, gusset buckling will not occur for the gusset plate stiffened by the knife plate.) The forces from the gusset analysis can be reduced by the factor: R y F yg = 275/429 = 0.641. 1.14F cre Ag /

Thus: V ′

′

=

= 124 kips = 60 kips.

. .

8.2.1 BEAM WEB CRIPPLING The beam web crippling strength is:



 +3

N 





  y



= 124 kips

V ′

R

2

. . . OK.

8.2.2 COLUMN WEB CRIPPLING The column web crippling strength is:



 +3

N 

 R

H ′



2

  y



= 60 kips

. . . OK.

8.3 CONDITION 3: BRACE BUCKLING (ROTATIONAL DEMANDS) This condition is satisfied by the detailing of a hinge zone in the knife plate equal to three times the gusset thickness. This approach has been demonstrated in recent tests (Thornton and Fortney, 2012).

8.4 ACCOMODATION OF FRAME DRIFT Additionally, there are rotation-capacity requirements for gusseted beam-to-column connections. F2.6b requires that gusseted beam-to-column connections be detailed to accommodate large drifts. The option 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

67

Design Example 2




taken in this example is to use a fully welded moment connection (option “b” in the provision). The beam-to-column portion of the moment connection follows the prescriptive requirements for the Welded Unreinforced Flange-Welded Web connection of AISC 358. The gusset, welded to both the beam and the column, provides additional resistance but this is not explicitly considered. Figures 2–11 and 2–12 show the completed connection detail.

Figure 2–11. Completed connection detail

68


Design Example 2




Figure 2–12. Connection detail

9. Additional Considerations 9.1. BRACE TRANSVERSE DISPLACEMENT Braces can be expected to undergo significant transverse displacement as a result of buckling and axial deformation (Tremblay, 2001). In this design, the brace displacement is in the plane of the frame. Thus, brace buckling will not affect the cladding system. The braces are exposed, and thus no special detailing of partition walls enclosing the braces is required.

9.2 PROTECTED ZONES To minimize the chances of premature fracture, braces and gussets should be designated as protected zones, with a restriction on fasteners and low-toughness welds. (There are regions of the brace that need not be so designated; see AISC 341 for specific requirements.) The effect of fasteners in the protected zone is currently being investigated.


69

Design Example 2




9.3 DEMAND-CRITICAL WELDS Certain welds must be designated as demand-critical, with the corresponding toughness requirements per AISC 341. These welds include: • Beam-flange to column-flange welds, • Column splice welds, and • Column-to-base-plate welds.

9.4 QUALITY ASSURANCE A quality assurance plan is required for this structure. For the SCBF system, the plan should include verification of the hinge-plate offset dimension complying with the detailing requirements and tolerance.

10. Items Not Addressed in This Example The following items are not addressed in this example but are nevertheless necessary for a complete design of the seismic load resisting system: • Comparison of wind and seismic forces, • ASCE 7 stability check (ASCE 7 Eq 12.8–16), • Design of the column base plate connection, • Design of the column splice, • Design of lateral bracing of the bracing connection at beam midspan, • Design of the foundations, and • Design of the diaphragm system.

70


Design Example 3 Buckling-Restrained Braced Frame

OVERVIEW Buckling-restrained braced frame (BRBF) is a relatively new system. It was developed in the late 1990s and has since gained acceptance in the United States as engineers look for a system with high ductility and energy dissipation. Unlike conventional steel braces, buckling-restrained braces consist of a steel core that is restrained from buckling by outer casing of concrete, steel, or both. As a result, the steel core will yield in tension as well as compression. The purpose of the outer casing is to prevent global buckling of the core only, and it must be kept free of axial forces. The methods to prevent transfer of axial forces to the casing as well as to provide confinement to the core are fairly specialized. As a result, BRBF in the United States consists of proprietary braces provided by a steel fabricator or a specialty manufacturer. The 2005 edition of AISC 341 and ASCE 7 contained the first U.S. code provisions for the design as w ell as testing requirements for BRBF. Subsequent editions of ASCE 7 (2010 edition) and AISC 341 (2010 edition) introduced additional updates. For example, the 2010 edition of ASCE 7 added provisions to estimate period (Method A) for BRBFs. In addition, the differentiation between BRBFs where beam-column joints are fixed or pinned has been eliminated and replaced by a single value of R = 8. AISC 341 included updates of beam-column joint requirements where the connection must either be able to allow rotation (0.025 radians) or to be designed to resist expected flexural strength of the beams or sum of the columns.


71

Design Example 3



Buckling-Restrained Braced Frame

OUTLINE 1. Building Geometry and Loads 2. Calculation of the Design Base Shear and Load Combinations 3. Vertical and Horizontal Distribution of Load 4. Preliminary Sizing 5. Preliminary Analysis and Brace Deformations 6. Iteration and Final Design 7. Detailing and Design of Connections 8. Items Not Addressed in This Example

1. Building Geometry and Loads 1.1 GIVEN INFORMATION • Per Appendix A:

72



Office occupancy on all floors



Located in San Francisco, CA, at the latitude and longitude given



Site Class D



120 feet × 150 feet in plan with typical floor and roof framing shown in Figure 3–1



Six stories as shown in Figure 3–2.


Design Example 3




Figure 3–1. Typical floor and roof framing plan

Figure 3–2. Braced-frame elevations


73

Design Example 3




1.2 FLOOR WEIGHTS Per Appendix A:

Table 3–1. Floor weights per Appendix A

Level

Unit Wt (psf)

Area (ft2)

Weight (kips)

Floor

77.7

15,220

1183

Ext Wall

19

6990

133

Roof

36

15,220

548

Ext Wall/Parapet

19

5700

108

Assembly

Typical floor

Floor Wt (kips) 1315

Roof

656

W = 5(1315 kips) + 656 kips = 7231 kips.


ASCE 7

2.1 CLASSIFY THE STRUCTURAL SYSTEM AND DETERMINE SPECTRAL ACCELERATIONS Per ASCE 7 Table 12.2–1 for steel buckling-restrained braced frame: R = 8.0

Ωo = 2.5

C d = 5.0

2.2 DESIGN SPECTRAL ACCELERATIONS The spectral accelerations to be used in design are derived in Appendix A: S DS = 1.00g

S D1 = 0.60g

2.3 DESIGN RESPONSE SPECTRUM Determine the approximate fundamental building period using Section 12.8.2.1: C t = 0.03 and x = 0.75 t

T 12.8–2

=

4

(see discussion below) T a = 0.74 sec

74


Eq 12.8–7

Design Example 3

S D1

T

02

S



S S D1

0 60

06

T  T o

0


1 00

6

10 S D1

06



=

§11.4.5

0T for T < T o

Eq 11.4–5

§11.4.5

for T > T s.

Eq 11.4–6

The long-period equation for S a does not apply here because the long-period transition occurs at 12 sec (from Figure 22–12).

1.2 ) g (

a S

TS=0.60 sec SDS=1.0g

1

, n o i t a 0.8 r e l e c c A l a 0.6 r t c e p S n 0.4 g i s e D

Approximate BRBF Building Period, Ta=0.74 sec, Sa=0.81g

To=0.12 sec

Calc'ed BRBF Period in y-direction, Ty=0.87 sec, S a=0.69g

Sa=0.4+5.0T Calc'ed BRBF Period in x-direction, Tx=0.97 sec, Sa=0.62g

0.2

Sa=0.60/T 0 0

0.5

1

1.5

2

Period (Sec)

Figure 3–3. Design response spectrum for the example building

As shown in Figure 3–3, the design spectral acceleration is greater than T S , so the design spectral acceleration S a is 0.81g. It is not required for the engineer to construct the design response spectrum when using the equivalent lateral force procedure, since the response spectrum is implicit in the calculation of C s in Section 12.8.1.1. The response spectrum demonstrates the effect of the assumptions used in the calculation of the building period. Values of C t = 0.03 and x = 0.75 were selected as specified for steel buckling-restrained braced frames, which result in an approximate period T a = 0.74 sec. As shown later in this example, drifts are close to, if not at, the story drift limits of Section 12.12.


75

Design Example 3




The period of the structure can be established through structural analysis of the BRBF. Section 12.8.2, however, limits the period that can be used to calculate spectral acceleration to a value of T max = C u × T a, where C u is a factor found in Table 12.8–1. In this case, T max = 1.4 × 0.74 = 1.04 sec. For preliminary design, the approximate period, T a, will be used to design the BRB F, and the period of the building will be verified later in the design process. The first mode period of the final design shown in Figure 4–14 was calculated to be 0.97 sec and 0.87 sec in the longitudinal and transverse direction respectively; therefore, the initial assumption of T = 0.74 sec is shown to not only be valid, but still quite conservative.

2.4 HORIZONTAL IRREGULARITIES 1a. and 1b.

T12.3–1

Torsional Irregularity—A torsional irregularity exists when the maximum story drift computed including accidental torsion is more than 1.2 times the average story drift. Extreme torsional irregularity exists when the maximum story drift computed including accidental torsion is more than 1.4 times the average story drift. From the 3-D computer analysis later in this example we can obtain the displacement at the corners of the building. The following shows the calculation for earthquake load case in X-direction with positive accidental eccentricity:

Table 3–2. Story displacements, line 1 and line 5, torsional irregularity check

Story

δ x @ Line 5

δ x @ Line 1

δavg

δmax / δavg

Roof

1.35

1.10

1.22

1.10

6th

1.19

0.97

1.08

1.10

5th

0.94

0.76

0.85

1.10

4th

0.68

0.55

0.62

1.11

3rd

0.44

0.36

0.40

1.11

2nd

0.22

0.18

0.20

1.11

δmax / δavg < 1.2 → NO TORSIONAL IRREGULARITY 2.

Reentrant corner irregularity exists where both plan projections of the structure beyond a reentrant corner are greater than 15 percent of the plan dimension of the structure in the given direction. The plan projections in longitudinal and transverse directions are 30 feet. The plan dimensions are 150 feet and 120 feet in the longitudinal and transverse direction respectively: 30 ft/150 ft = 20% in the longitudinal direction; 30 ft/120 ft = 25% in the transverse direction → REENTRANT CORNER IRREG. EXISTS

76


Design Example 3

3. to 5.




By inspection, the building does not qualify for any of these horizontal structural irregularities.

REENTRANT CORNER IRREGULARITY EXISTS Per Section 12.3.3.4, forces for the connections of diaphragms to vertical elements and collectors and the design of collectors and their connections must be increased by 25 percent. If forces for the design of collectors and their connections are calculated using seismic-load effects including the overstrength factor, the 25 percent increase is not required.

2.5 VERTICAL IRREGULARITIES 1a. to 5b.

T12.3–2

By inspection, the building does not qualify for any of the vertical structural irregularities.



T12.6–1

1. Simplified Alternative Structural Design Criteria—According to Section 12.14.1.1 this analysis procedure can be used for BRBF, but not for buildings over three stories—NOT PERMITTED 2. Equivalent Lateral Force Analysis—According to Table 12.6–1, since the structure is less than 160 ft and has only Type 2 horizontal irregularity—PERMITTED 3. Modal Response Spectrum Analysis—PERMITTED 4. Seismic Response History Procedures—PERMITTED


2.7 BASE SHEAR 10

C

0 125

 R   8 0     1 0 

Eq 12.8–2

but need to exceed C

=

S T

R  I



06

= 07

 8 0 

= 0.101 for

Eq 12.8–3

10


77

Design Example 3




but shall not be less than

=

0.

. 44 ≥

01

Eq 12.8–5

and, for structures where S 1 is equal to or greater than 0.6 g, C s shall not be less than 05

1

   I

=

5 06

 8 0   1 0 

= 0.038

Eq 12.8–6

C s = 0.101

V

=

.101 ×

= 730 k ips

Eq 12.8–1 V = 730 kips

2.8 REDUNDANCY FACTOR According to Section 12.3.4, the Redundancy Factor should be calculated for each principal axis. The Redundancy Factor is 1.3 unless either 12.3.4.2(a) or 12.3.4.2(b) is shown to be true, in which case the Redundancy Factor can be taken as 1.0. 12.3.4.2(a) and Table 12.3–3 require that for each story resisting more than 35 percent of the base shear, removal of an individual brace or connection thereto would not result in more than a 33 percent reduction in story strength, nor does the resulting system have an extreme torsional irregularity: There are total of six BRBs and eight BRBs at each level in the longitudinal and transverse direction respectively; thus, by inspection, removal of an individual BRB would not result in more than a 33 percent reduction in story strength (1/6 = 17%). The second condition needs to be confirmed by rigid diaphragm analyses by removing of individual braces and checking whether extreme torsional irregularity exists. By inspection, since brace sizes will be designed to be identical for frames in each direction, the following two conditions will result in highest torsion: 1. Removal of brace along line 4, or 2. Removal of brace along either line A or line F. From 3-D computer analysis later in the example, the braces per above were removed and it was confirmed that extreme torsional irregularity does not exist in either of the above cases. The following shows check of extreme torsion:

78


Design Example 3




Table 3–3. Story displacements, lines 1, 5, A, and F, extreme torsional irregularity check

Story

δ x at Line 1 (in)

δ x at Line 5 (in)

δavg (in)

δmax / δavg

Roof

1.40

1.11

1.25

1.11

6th

1.25

0.98

1.12

1.12

5th

1.01

0.79

0.90

1.12

4th

0.77

0.59

0.68

1.13

3rd

0.54

0.41

0.48

1.14

2nd

0.32

0.23

0.28

1.16

Story

δ y at Line A (in)

δ y at Line F (in)

δavg (in)

δmax / δavg

Roof

0.89

1.07

0.98

1.09

6th

0.80

0.96

0.88

1.09

5th

0.64

0.77

0.70

1.09

4th

0.47

0.56

0.52

1.09

3rd

0.31

0.37

0.34

1.09

2nd

0.16

0.19

0.17

1.09

ρ = 1.0 FOR EAST-WEST DIRECTION ρ = 1.0 FOR NORTH-SOUTH DIRECTION Alternatively, the designer can elect to use ρ = 1.3 for simplicity.

2.9 LOAD COMBINATIONS See Appendix A for the derivation of combinations based on ρ = 1.0 and 0.2S DS = 0.2. Load combinations of consequence for the design of the BRBF are

0 9 D

+05 = 0 7D

+ 1 0Q .

5

§12.4.2.3 Load Combo 5 (modified) Load Combo 7 (modified)

Load combinations with overstrength are not used for the BRBFs (although they apply for collectors and at other conditions outside the BRBFs).


79

Design Example 3





ASCE 7

3.1 VERTICAL DISTRIBUTION OF SHEAR


Level

wi (kips)

hi (ft)

i i

C vx

F x (kips)

Roof

656

72

78,907

0.18

128

6th

1315

60

128,960

0.29

210

5th

1315

48

100,442

0.22

164

4th

1315

36

72,775

0.16

118

3rd

1315

24

46,213

0.10

75

2nd

1315

12

21,262

0.05

35

Total

7231

448,560

730

The terms used in Table 3–4 are defined in Section 12.8. 3. Since the period = 0.74 > 0.5 sec, the value for k is interpolated between a value of 1.0 for T = 0.5 sec and 2.0 for T = 2.5 sec. In this example, k = 1.12. The distribution of story shear is carried out using F x = C vx V , where,

h

.

n

∑=

i

1

80


Eq 12.8–11 and Eq 12.8–12

Design Example 3




3.2 HORIZONTAL DISTRIBUTION OF STORY SHEAR

Figure 3–4. Rigid diaphragm analysis to distribute story shear to BRBFs

As shown in Figure 3–4, the center of mass and center of rigidity coincide at the middle of the building in the X direction, but the center of rigidity is 10 feet below the center of mass in the Y direction. This is due to the layout of the braced frames. To distribute the load to each BRBF, the story shear, F i, is applied in the X and Y directions. Per Section 12.8.4.2, the point of application of the story shear is offset 5 percent to account for accidental eccentricity. For simplicity, all the BRBs in the X direction are the same, and all the BRBs in the Y direction are the same. Since they are the same, a generic stiffness, K , is used for all braced frames.


81

Design Example 3




Table 3–5. Center of rigidity calculation

BF No.

Dir

x

A1

Y

A2

y

R x

R y

xR y

−75

1K

−75 K

0K

Y

−75

1K

−75 K

0K

F1

Y

75

1K

75 K

0K

F2

Y

75

1K

75 K

0K

1

X

−60

1K

0K

−60 K

3

X

0

1K

0K

0K

4

X

30

1K

0K

30 K

0K

−30 K

Sum

3K

4K

yR x

Center of rigidity: Xr = 0 ft, Yr = −30 K/3 K = −10 ft. Table 3–6. Rigid diaphragm analysis to distribute shear to BRBFs (X-direction) X-Direction R y

d = x − xr

Y

1K

−75.00

−75 K

A2

Y

1K

−75.00

F1

Y

1K

F2

Y

1K

1

X

1K

3

X

4

X

BF No.

Dir

A1

Sum

R x

d = y − yr

Rd 2

F direct

F torsion

F acc. tors

F total

5625 K

0.000 F i

−0.028F i

−0.017F i

−0.045F i

−75 K

5625 K

0.000 F i

−0.028F i

−0.017F i

−0.045F i

75.00

75 K

5625 K

0.000 F i

0.028F i

0.017F i

0.045F i

75.00

75 K

5625 K

0.000F i

0.028F i

0.017F i

0.045F i

−50.00

−50 K

2500 K

0.333 F i

−0.019F i

−0.011F i

0.303F i

1K

10.00

10 K

100 K

0.333 F i

0.004F i

0.002F i

0.339F i

1K

40.00

40 K

1600 K

0.333 F i

0.015F i

0.009F i

0.357F i

3K

Rd

4K

26,700 K

Table 3–7. Rigid diaphragm analysis to distribute shear to BRBFs (Y-direction) Y-Direction R y

d = x − xr

Y

1K

−75.00

−75 K

A2

Y

1K

−75.00

F1

Y

1K

F2

Y

1K

1

X

1K

3

X

4

X

BF No.

Dir

A1

Sum

82

R x

d = y − yr

2

F direct

F torsion

F acc. tors

5625 K

0.250F i

0.000F i

−0.021F i

0.229F i

−75 K

5625 K

0.250F i

0.000F i

−0.021F i

0.229F i

75.00

75 K

5625 K

0.250 F i

0.000F i

0.021F i

0.271F i

75.00

75 K

5625 K

0.250 F i

0.000F i

0.021F i

0.271F i

−50.00

−50 K

2500 K

0.000F i

0.000F i

−0.014F i

−0.014F i

1K

10.00

10 K

100 K

0.000F i

0.000F i

0.003F i

0.003F i

1K

40.00

40 K

1600 K

0.000 F i

0.000F i

0.011F i

0.011F i

3K

Rd

4K


Rd

26,700 K

F total

Design Example 3




The following calculations find the force in BRB along line 4 for shear in the X-Direction:

F r ct

M acc. tors

i

R

= 0 33F i =

i

=

M Ftota

rect

; F acc. tors

M tor

; F o

+ F cc. tor

i

R x 2

Rd

tors n

. to

= =

=

R i

× 40 K

26,700 K

=

6 F i × 40 K 26,700 K

.009 F

. 15F

= 0 357 F .

Table 3–6 shows that the maximum design shear for any frame line is 0.357 times the story shear. The rest of this example focuses on the design of BRBF along line 4. Table 3–8. Story forces applied to BRBF along line 4

Level

F Story (kips)

F BF4 (kips)

V BF4 (kips)

Roof

128

46

46

6th

210

75

121

5th

164

58

179

4th

118

42

222

3rd

75

27

249

2nd

35

12

261

4. Preliminary Sizing 4.1 BRB LAYOUT The layout of the lateral system should be well distributed to avoid torsional behavior as well as have reasonable redundancy. However, many times the layout is constrained by architectural requirements. For this example, the layout is to illustrate such constraint. Also, the layout of the BRBFs is used for the diaphragm design example (Example 7). To illustrate the diaphragm design where the lateral system is not symmetrical, the BRB frames in the X direction are located asymmetrically. Typical BRB capacity ranges from a minimum of around 50 kips to a maximum capacity upwards of 1400 kips. While using larger BRB can potentially reduce the number of braced bays, larger BRB sizes result in large overturning forces, which drive up the sizes of the columns, beams, and foundations. Thus, aside from redundancy considerations, the layout of the BRB frames is often driven by these considerations. In addition to well-distributed braced bays, AISC 341 commentaries also note that “engineers should consider effects of configuration and proportioning of braces on the potential formation of building yield mechanism.” Unlike conventional steel braces, the axial yield strength of BRBs can be set more precisely by specifying core area as well as limiting the range of material yield strength, which are established by coupon test. Thus engineers have more control in their design to distribution of yielding over the building height. 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

83

Design Example 3




4.2 BRB CONFIGURATION Similar to the configuration of concentrically braced frames, configuration of the BRB frames can be V-type, inverted V-type, or single diagonal configuration. Neither X-bracing nor K-bracing is an option for BRBF. For V-type and inverted-V-type braced frames, AISC 341 Section F4.4 lists special requirements. This will be discussed further below. For certain situations, e.g. tall story heights and w ide bays, brace configuration may depend on maximum lengths that the BRB manufacturers have tested. AISC 341 requires that the design of the BRB be based on results from qualifying cyclic tests. Existing test data by manufacturers are often used for design, but interpolation and extrapolation of existing test results needs to be justified. Where the length of the BRBs exceeds the range of typical tested specimens, project specific tests may need to be performed.

4.3 REQUIRED BRACE AREA

AISC 341

Required steel core area of a BRB is given by P ysc = F ysc Asc with φ = 0.9

Eq F4–1

where F ysc = specified minimum yield stress of the steel core, or actual yield stress of the steel core as determined from a coupon test. Asc = net area of steel core.

For this example, an average F ysc = 42 ksi was used with a tolerance of + / −4 ksi. Thus for the required steel core area for strength consideration, the lower bound F ysc = 38 ksi was used. Preliminary core area can be designed. Pu = V u /2 braces/cos (θ).

For first-floor brace, Pu = 261 kips/2/ cos (38.7) = 167 kips. Required core area Asc req’d = Pu / φF ysc = 167 kips/(0.9 × 38 ksi) = 4.9 in2. Use Asc = 5 in2. Table 3–9 summarizes the preliminary core area for BRB frame along line 4:

Table 3–9. BRB core area along line 4

84

Level

V u

Pu

Asc Req’d

Asc Used

Roof

46

29

0.9

2

6th

121

77

2.3

2.5

5th

179

115

3.4

3.5

4th

222

142

4.2

4.5

3rd

249

159

4.7

5

2nd

261

167

4.9

5


Design Example 3




4.4 A NOTE ABOUT MATERIAL PROPERTIES The engineer should consult with the BRB manufacturer regarding typical range of yield stress for the steel core. If the BRB manufacturer is pre-selected during the design phase, it is sometimes possible to procure the material ahead of time and thus have less variability of the yield stress. However, more commonly, the project is to be competitively bid by various BRB manufacturers; thus, a reasonable range of yield stress should be used to permit procurement process for brace manufacturers. The range of 38 ksi to 46 ksi is fairly standard for the industry. As discussed later in Section 6.2, this material variability needs to be accounted for when the engineer specifies requirements for the BRBs.

4.5 COMPUTATION OF ADJUSTED BRACE STRENGTH The adjusted brace strength in tension and compression is ω R y P ysc and βω R y Psyc respectively, where ω and β, (the strain hardening adjustment factor and the compression strength adjustment factor) are determined based on measurements from BRB qualification tests corresponding to two times the design story drift or 2 percent of story height, whichever is larger. Brace deformations at the design story drift can be obtained from computer analysis as discussed in the section below. Prior to running a computer analysis, the engineer can estimate initial values for ω and β assuming braces are sized such that φP ysc = Pu, (assuming redundancy factor ρ = 1.0). The deformation of brace at design level forces would then be ∆bx = φ(∆by), where ∆by is the deformation of the brace at first significant yield; 2∆bm = 2C d ∆bx = 2C d (φ)(∆by) = (2)(5.0)(0.9)(∆by) = 9∆by. The corresponding core strain is approximately 9(ε y) = 9(F y / E ) = 0.013. This is an approximate value only since the core strain is based on multiple factors such as the length of core, the effective stiffness of braces, etc. Prior qualification tests by BRB manufacturers show that β = 1.04 and ω = 1.54 for a core strain of 1.3 percent. For preliminary beam and column design, these values will be used. After the computer analysis has been performed, the core strain and whether a minimum 2 percent inter-story drift requirement would control will be confirmed.

4.6 BEAM DESIGN The beam in a BRB frame acts both as an axial load carrying member and as a flexural member. The axial load arises because the beam acts as a collector transferring the diaphragm forces to the braces. The flexure arises because the beam carries gravity loads from the floor and in special geometrical configurations (chevron) carries the unbalanced load from the compression and tension braces. The beam at the second floor in the braced-frame bay will be designed in this illustrative example.

4.7 CHECK WIDTH-THICKNESS RATIO The beam in the BRB frames needs to be seismically compact. The beam size at the second floor will be assumed to be W18 × 65 and its adequacy will be checked.

= 7 22 for W18 × 65

5 06 . . . OK.

T D1.1

Since the axial load in the collector is not known yet, it is conservative to assume that C a is greater than 0.125 and hence using the minimum limit on the h / tw ratio Per Table I-8–1 ps

y

=

88 for W18 × 65

ps

=

w

= 35 7

. . . OK.

T D1.1

BEAM WIDTH-THICKNESS RATIOS ARE OK 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

85

Design Example 3




4.8 FIND MOMENT IN BEAM The flexure in the beam arises because it carries the tributary gravity loading. In addition, for a chevrontype configuration, Section 16.4(1) requires that the beam be designed for the net vertical component due to the BRBs in tension and in compression. For the current case, the tributary gravity load is w D = 5 ft × 77.7 psf = 388.5 plf. Add 15 psf of loading from the elevator door as well as the self-weight of the beam. Thus the total dead load on the beam is 388.5 + 15 psf × 12 ft + 65 plf = 633.5 plf. Similarly, the live load is estimated to be 40 psf × 5 ft = 200 plf. The ultimate uniformly distributed load under Load Case 5 on the beam is calculated as

= (1.

S ) D

×1 ×

+ 0.

= 1 k ip-ft .

If the brace is considered as a location of support for the beam, the ultimate beam moment at mid-span using the beam formula for a two-span continuous beam is M

w (

=

/

=

=

.

.5 ki -in = 28.13 k-ft.

The unbalanced load due to the unequal compression and tension capacity of the BRBs are calculated as Pun a an e

R P

sin ) .

Since the maximum yield is limited to 42 ksi + 4 ksi = 46 ksi and will be confirmed from a coupon test, R y P ysc = 46 ksi × Asc. Also, assuming β = 1.04 and ω = 1.54 as presented earlier with Asc = 5 in2 above and below as shown in Table 3–9: Punbalan ed

× 1.

−

× 46 × 5

sin

.66))

.85 kips.

This produces a moment of 8.85 kips × 30 ft/4 = 66.38 kip-ft. The unbalanced load will act in opposite to the gravity loading. Since the brace was considered as a location of support, the negative support moment due to gravity will be additive to this moment. Thus the net design moment is M

=

.13

.

=

.1 kip-in.

4.9 CHECK BEAM FOR COMBINED AXIAL AND FLEXURE DUE TO COLLECTOR ACTION A capacity based approach to calculate the axial force in the beam in the BRB bay will be followed. This is done by assuming that the brace capacity in tension and compression are mobilized in the lowermost story producing a net horizontal shear capacity of R

os ) =

+ 354 2) cos(38.66) =

+ 276 4 = 564 2 kips

Part of the axial load in the brace at the lowermost story is due to the seismic inertia of the second floor mass, while the large majority is due to the inertia of the upper floors, which is carried down through the braces. If the same capacity-based approach is adopted for the upper floors, the shear immediately above the second floor is also 564.2 kips since the same braces are used between the second and third level. This would signify that no inertial load is generated at the second floor, which is incorrect.

86


Design Example 3




While there could be various combinations as to how much of the total shear of 564.2 kips is due to the inertia of the second floor, it will be seen that for the beam within the braced frame bay only, the maximum axial load in the beam is unaffected since the axial load based on the brace capacity will finally need to be delivered by the beam. For beams outside the braced frame bay, this ratio is important and needs to be substantiated adequately so that the collectors are not under-designed. For this illustration we will assume that the proportion of shear arising out of the inertia of the second floor is in the same ratio as the distribution of the equivalent static lateral load presented earlier. Other distributions such as a uniform distribution or that based on response spectrum analysis should be considered to bound the axial force outside the braced fame bays. Hence, the total capacity of 564.2 kips can be split into two parts. The first part has a magnitude of (249/261) × (276.4 + 287.5) = 263.7 + 274.3 = 538 kips. The second part has a magnitude of 26.2 kips, where 249 kips is the total shear above the second floor and 261 kips is the total shear below the second floor obtained from Table 3–8. The 249 kips shear stays within the braced frame bay at the second floor while the 26.2 kips is collected over the entire diaphragm width at the second floor. The resultant collector force diagram is shown in Figure 3–5.

600

Force input from upper braces 400

538.0 kips

276.8 k = 263.7 k + 0.17 k/ft x 75 ft

200

Collector Force Dgm. 0

Shear dragged from 2nd floor @ 26.2k/150 ft = 0.17 k/ft

-200

287.4 kips Force coming out from lower braces

-400

564.2 kips -600 0

30

60

90

120

150

Distance along Collector (ft) Figure 3–5. Collector force diagram along line 4 at second floor

From Figure 3–5 it can be seen that the maximum collector force is 287.4 kips. The beam for this axial load and a net moment of 1134.1 kip-in presented earlier will be checked.


87

Design Example 3




Determine beam flexural buckling stress (assume that the beam will be braced transversely at the location of the chevron brace): 49 = 48 06

K x x / rx K

/

/1 69

106 51 . . . governs.

From AISC Table 6–1 for W18 × 65 with an unbraced length about the Y axis equal to 15 ft:

= 374 5 k ip

p

.

The torsional buckling length equals the minor-axis buckling length. Under these circumstances, torsional buckling need not be checked for an I-shaped member as minor axis buckling has a lower value for design strength. Also since the flange and web are seismically compact, Qs = Qa = 1.0 and thus will not affect the nominal compression strength. AISC design methodology utilizing constrained axis flexural-torsional buckling is illustrated in Examples 2 and 7. The reader is referred to those examples for alternative design approaches. Calculate the flexural strength of the beam: Check beam yielding:

=

9

50

5985 k ip-in .

From Table 6–1 of the AISC:

=

. 4

−

.

Strength for members under combined loading is calculated in accordance with Chapter H, “Design of Members for Combined Forces and Torsion” of AISC. /374 5 =

0 2.

Therefore, using H1-1a Pu

P

=p

+

x

u

= (287. ×

.1 12

.

09

1.0 . . . OK.

4.10 SUMMARY OF BEAM DESIGN A W18 × 65 section can adequately resist the combination of axial and flexural loads. The member is assumed to have adequate shear strength.

4.11 COLUMN DESIGN Columns strength is based on following load combinations as listed in Appendix A: 12 0

+05 =0

+

5

0 E .

In determining amplified seismic load, E mh is taken as the forces in all braces corresponding to their adjusted braced strength. For this example, the column axial forces are based on the BRB reaching adjusted brace strength at all levels.

88


Design Example 3




Adjusted brace strength in compression is Pbr = βω R y P ysc. Since the maximum yield is limited to 42 ksi + 4 ksi = 46 ksi and will be confirmed from a coupon test, R y P ysc = 46 ksi × Asc. The vertical component of this adjusted brace strength is Pbr sin (θ). For the chevron configuration, the unbalance load at the mid-span of the beam is βω R y P yscsin(θ) − ω R y P ysc sin (θ) = (β − 1)ω R y P ysc sin (θ) acting upwards. Half of this unbalanced load is resisted at each end of the beam and reduces the axial force in the column in compression (and increase the tension load in the column in tension). Thus the column loads can be reduced by ( β − 1)/2ω R y P ysc sin (θ). Table 3–10 summarizes the design axial forces in the columns.

Table 3–10. Summary of brace core strain calculation

Level

Asc in2

Roof

2

P ysc (From Brace Above) (kips)

P Emh = Pbr sin θ (kips)

ωβ R y P ysc (kips)

Reaction From Unbal. Load at Beam (kips)

Trib. Area (sf)

DL (psf)

0

0

0

−1.8

900

31

L (psf)

Cum. ΣPu (kips)

Lr (psf)

P(1.4 D+0.5 L) (kips)

P(1.4 D+0.5 L+ Emh) (kips)

20.0

39.1

37.3

37.3

6th

2.5

92

147

92

−2.2

900

67.7

65.0

114.6

204.5

241.8

5th

3.5

115

184

115

−3.1

900

67.7

65.0

114.6

226.6

468.4

4th

4.5

161

258

161

−4.0

900

67.7

65.0

114.6

271.8

740.2

3rd

5

207

332

207

−4.4

900

67.7

65.0

114.6

317.4

1057.6

2nd

5

230

368

230

−4.4

900

67.7

65.0

114.6

340.4

1398.0

4.12 CHECK WIDTH-THICKNESS RATIO For a column at the first floor, select a W14 × 211 trial section. In accordance with AISC 341 Section F4.5a, BRB columns shall comply with the width-to-thickness requirements of Section D1.1 for highly ductile members. Flanges of rolled I-shaped sections:

≤

= 7 22

t = 5 06 . . . OK.

T D1.1

Webs of rolled I-shaped sections: C

=

P c

≤

P

=

1398 2790

= 05 −

T D1.1 45

t

6 . . . OK.

T D1.1

COLUMN WIDTH-THICKNESS RATIOS ARE OK


89

Design Example 3




4.13 FIND MOMENT IN COLUMN As discussed in later sections, a brace connection consists of gussets that are welded to beams and columns. Additionally, beams flanges are fully welded to columns to resist high collector forces. For this example, the beam-column joints are modeled to be momently connected. Per AISC, it is permitted to neglect flexural forces resulting from seismic drift. Thus the moment in the columns will be from gravity and vertical seismic ( E v) only. From previous sections under beam calculations, the uniform load from gravity plus E v at the second-floor beam: w

= (1 2 +

S )

×1 ×

+05

= 1 k ip-ft .

Calculate negative moment at beam ends (fixed-fixed) condition:

=

M

= 75 k-ft =

00 kip-in

For the “inverted-V” configuration, the unbalanced load from the BRBs counteracts the gravity loads. For the column moments, this effect will be ignored. For a column at the first floor, the beam end moment will be resisted by columns above and below. The column base is pinned, and assuming zero moment at the mid height of second-floor column, the column shear and moment can be calculated as follows: 75 k-ft

=

V M u

u

12 6 1

= ( 17

= 4 17 kips

12

50 k-ft .

4.14 CHECK COLUMN FOR COMBINED AXIAL AND MOMENT Check axial strength: Per AISC Steel Manual, Column Table 4–1, for W14 × 211, kL = 12 ft: P P

= 2550 kips P

/

)=

.0 . . . OK.

Check flexural strength: Per AISC Steel Manual, for W14 × 211 with unbraced length = 12 ft: M n = 1463 k-ft

/

)=

<1 0

. . . OK.

Using H1-1a for combined axial and moment: P

P

=

.0 . . . OK.

4.15 SUMMARY OF COLUMN DESIGN Column design is summarized in Table 3–11.

90


Design Example 3




Table 3–11. Summary of column check

Size

φPn (kips)

Pu / φPn

M u

φ M n

M u / φ M n

Pu / φPn + 8/9[ M u / φ M n]

37

W14 × 68

700

0.06

37.8

431

0.09

0.13

6th

242

W14 × 68

700

0.35

37.5

431

0.09

0.42

5th

468

W14 × 132

1568

0.30

37.5

878

0.04

0.34

4th

740

W14 × 132

1568

0.48

37.5

878

0.04

0.51

3rd

1058

W14 × 211

2550

0.42

37.5

1463

0.03

0.44

2nd

1398

W14 × 211

2550

0.56

50.0

1463

0.03

0.60

Level

Pu (kips)

Roof

5. Preliminary Analysis and Brace Deformations 5.1 BRACE EFFECTIVE STIFFNESS

AISC 341

The steel core of a BRB typically consists of a yielding segment, transition zones, and connection zones where the core exits the casing. The effective stiffness of the BRB is based on these sections in series. The yielding length, transition zones, and connection length vary among different manufacturers. It is possible for the designer to control the stiffness by varying the proportions of yield length of the core to the overall length of the brace. However, a shorter yield length, while increasing the stiffness of the brace, also forces inelastic strain to occur at a shorter segment thus increasing the ductility demand and reducing the cumulative energy dissipation capacity. Given the variation discussed above, the structural engineer should consult with the manufacturer to obtain the effective stiffness of the braces. For this example, the effective Lwp , where K eff is the stiffness modification factor obtained from BRB brace axial stiffness is K eff EAsc / manufacturer, and Lwp is the distance between work points.

5.2 ANALYSIS

ASCE 7

A three-dimensional computer model is created to analyze the building and to obtain story deflections. Beam ends are modeled as fixed (see the discussion in a later section). Column bases are modeled as pinned. The BRBs can be modeled as truss elements with area equals to the BRB core area ( Asc) with a stiffness modification factor ( K eff ) = 1.43 to capture the effective axial stiffness described above.

5.3 CHECK DEFLECTION The resulting elastic deflections are given in Table 3–12. The elastic story drifts are the difference in the elastic deflections at the floor above and below. The predicted inelastic story drift is given by the following for the fourth floor: C

x

I

5

28 10

14i .

Eq 12.8–15

The limit on story drift is given in Table 12.12–1 based on building system and the occupancy category:

∆ =

=

020

= 2 88 in.

T 12.12–1

δ x < ∆a . . . DRIFT IS OK 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

91

Design Example 3





Story

Elastic Deflection (in)

Elastic Story Drift, δ xe (in)

Inelastic Drift, δ x (in)

Story Drift Limitation (in)

Roof

1.48

0.17

0.9

2.88

6th

1.31

0.27

1.4

2.88

5th

1.03

0.28

1.4

2.88

4th

0.75

0.26

1.3

2.88

3rd

0.49

0.25

1.3

2.88

2nd

0.23

0.23

1.2

2.88

5.4 ITERATE TO SATISFY DEFLECTION The drifts were found to satisfy the story drift limits. However, if the inelastic story drift had exceeded the story drift limit, it would have been necessary to either increase the size of cores to increase the stiffness of BRBs or discuss with the manufacturer regarding shortening the yield length.

5.5 BRACE DEFORMATIONS AND CONFIRMATION OF EXPECTED BRACE STRENGTH Depending on the analysis software, brace deformations ∆b can be obtained directly from computer analysis for seismic-load cases to compute drift. The seismic load for drift may be different than for strength since the redundancy factor ρ = 1.0 for drift and there is no upper limit for period ( C uT a) used to calculate drift. Since the analysis performed is linear elastic, ∆bm = C d ∆b where C d = 5.0 for BRBFs. Corresponding strain in the core at 2 ∆bm can be calculated based on ε = 2∆bm / L ysc = 2C d ∆b / L ysc or 10∆b / L ysc. If braced deformations are not part of the computer analysis output, BRB deformation can be easily EAsc, where Lwp is the distance between work points and K eff calculated from brace forces ∆b = Pbrace Lwp / Keff is the stiffness modification factor (discussed in the previous section). Pbrace are brace forces under load combinations for drift. BRB yield length, L ysc, and effective stiffness vary among BRB manufacturers, so the structural engineer should consult with the BRB manufacturer prior to calculating BRB strains. For this example, the yield length L ysc is based on 0.7 times length between work points, and the stiffness modifier is K eff = 1.43. For = L ysc, thus, ε @ 2∆m = 10[Pbrace Lwp / Keff EAsc]/ L ysc = 10Pbrace / EAsc. our example, Lwp / Keff AISC also requires that story drift be based on a minimum of 2 percent of story height. This condition needs to be checked with the results from computer analysis. Where two times the design story drift is less than 2 percent of story height, one method to estimate ε at 0.02 times story height is to scale the brace strain accordingly, i.e. scaled ε = 0.02/(2 × design story drift) × ε at design story drift.

92


Design Example 3





Asc (in2)

ε at 2× Design Story Drift

Story Drift (δ xe) from ETABS (in)

2C d × δ xe (in)

29.9

2.0

0.5%

0.17

5th

72.0

2.5

1.0%

4th

108.6

3.5

3rd

135.0

2nd 1st

Story

Pbrace from ETABS (kips)

6th

Drift Ratio

Is Drift Ratio at Least 2%

Scale Factor

Scaled

1.73

1.2%

No

1.67

0.9%

0.27

2.73

1.9%

No

1.05

1.0%

1.1%

0.28

2.84

2.0%

OK

1.00

1.1%

4.5

1.0%

0.26

2.63

1.8%

No

1.09

1.1%

151.0

5.0

1.0%

0.25

2.51

1.7%

No

1.15

1.2%

159.2

5.0

1.1%

0.23

2.35

1.6%

No

1.23

1.3%

ε

β = 1.04 and ω = 1.54 based on core strain of 1.3 percent is conservative for all floors. 5.6 BRACE CONFORMANCE DEMONSTRATION Brace testing requirements are stated in Section K3 of AISC-341. There are two applicable tests: the first is a subassemblage test that incorporates end rotations of the brace; the second test may be a uniaxial test. Displacements are based on brace yield displacement ∆by. The actual yield displacement of the brace should be determined from the test, although an estimate of ∆by can be calculated from properties of the brace (yield strength, yield length, transition zone length, etc.). For this example, ∆by is calculated to be 0.2 inches. ∆bm is brace deformation taken from analysis and scaled to two times the design story drift or 2 percent, whichever is larger. See Table 3–14.


∆b

2∆bm (in)

2C d × δ xe (in)

Drift Ratio


2.0

0.083

0.83

1.73

1.2%

No

1.67

1.39

72.0

2.5

0.160

1.60

2.73

1.9%

No

1.05

1.69

4th

108.6

3.5

0.172

1.72

2.84

2.0%

OK

1.01

1.75

3rd

135.0

4.5

0.167

1.67

2.63

1.8%

No

1.09

1.83

2nd

151.0

5.0

0.168

1.68

2.51

1.7%

No

1.15

1.93

1st

159.2

5.0

0.177

1.77

2.35

1.6%

No

1.23

2.17

Story

Pbr (kips)

Asc (in2)

6th

29.9

5th

Scale Factor

Scaled 2∆bm


93

Design Example 3




Table 3–15 summarizes the test protocol for the uniaxial test for brace at first floor.


Cycle

Deformation as Ratio of ∆by

Deformation (in)

2 @ ∆by

0.2

1.0∆by

2 @ 0.5∆bm

0.54

2.7∆by

2 @ ∆bm

1.08

5.4∆by

2 @ 1.5∆bm

1.63

8.2∆by

2 @ 2∆bm

2.17

10.9∆by

Aside from brace axial deformations, the building drifts impose rotations to the ends of the BRBs. The amount of rotational demand is required for sub-assemblage tests per AISC 341. Brace-end rotations may be obtained from the computer model. Maximum end rotations of the lower and upper ends of the brace are summarized in Table 3–16.

Table 3–16. Summary of brace end rotations

Story

Rotation @ δ xe (radians)

Rotation @ 2× Design Story Drift (radians)

Story Drift (δ xe) from ETABS (in)

Drift Ratio


Scale Factor

Rotation at 2× Design Drift or 2% radians

2C d × δ xe (in)

6th

0.0006

0.006

0.17

1.73

1.2%

No

1.67

0.010

5th

0.009

0.009

0.27

2.73

1.9%

No

1.05

0.009

4th

0.0012

0.012

0.28

2.84

2.0%

OK

1.00

0.012

3rd

0.0011

0.011

0.26

2.63

1.8%

No

1.09

0.012

2nd

0.0011

0.011

0.25

2.51

1.7%

No

1.15

0.013

1st

0.0011

0.011

0.23

2.35

1.6%

No

1.23

0.014

6. Iteration and Final Design The period of the building can be obtained from the computer model and compared to the period based on method A. Based on the computer analysis, the first-mode periods in longitudinal and transverse directions are 0.87s and 0.97s respectively. As shown, the period of the building is higher than that computed by method A. Since the building is at the descending branch of the spectrum, the resulting base shear will be slightly less. The BRBs in the transverse direction can also be reduced in size based on the results from the computer analysis. Iterations can be performed to reduce brace sizes and to optimize the design.

94


Design Example 3




6.1 MEMBER SIZES After optimization, the final brace design is shown in the figure below. (The numerical values adjacent to braces represent the steel core area, Asc, i.e. BRB2 has a core area of 2 square inches):

Figure 3–6. Final braced-frame elevations

6.2 BRACE SPECIFICATION Since BRBFs are proprietary items specified by the engineer and manufactured by the supplier, the engineer must clearly specify the design requirements and assumptions to the supplier. Construction documents need to clearly specify the following: BRB required strength. There are several options to specify BRB. Option one involves specifying a steel core area ( Asc) and a range of acceptable F ysc. Option two involves specifying brace strength ( P ysc) , thus allowing the brace manufacturer to adjust Asc based on F ysc. While option two results in lower overstrength in the brace, it results in a variation in brace stiffness from the one the engineer has in the computer model. This may result in different load distribution than in the analysis. For this reason, in the opinion of the author, it is preferable to use option one. In addition to brace strength, construction documents should also show i) the permissible range of core yield strength; ii) the permissible variability of brace stiffness (since brace stiffness is related to core yielding length and can be manufacturer specific); and iii) the maximum strength adjustment factors ( β, ω ). 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

95

Design Example 3




7. Detailing and Design of Connections Connections are to be designed such that yielding occurs at the BRB cores. Thus, per AISC 341 Section F4.6c, the diagonal brace connections are designed for 1.1 times the adjusted braced strength (Pu = 1.1 R yβω P ysc).

7.1 BRB-TO-GUSSET CONNECTION Since BRB-to-gusset connections are typically manufacturer-specific and can vary from welded, bolted, or pinned-type connections, BRB manufacturers will typically design and detail the brace-to-gusset connections. However, the gusset plate dimensions and subsequently the connections to beams and/or columns will depend on this detail; thus, some amount of coordination will be required between the design professional in responsible charge and the BRB manufacturer. At a minimum, the design professional in responsible charge needs to convey information such as work-point locations and the beam/column connection configuration in the design documents.

7.2 GUSSET PLATE REQUIREMENTS The design of gusset plate detail needs to consider local and overall buckling. In addition, lateral bracing of the gusset plate needs to be consistent with that used in the specimen used in the assemblage test.

7.3 GUSSET CONNECTION TO BEAM AND COLUMN

Figure 3–7. Brace connection details

96


AISC 360

Design Example 3



Buckling-Restrained Buckling-Restr ained Braced Frame Frame

The design of the gusset plate and the connections to column/beams are sometimes delegated to the brace manufacturer.. This is because different manufacturers manufacturer manufacturers typically have different BRB end-connection details that affect the geometry of the gusset. The design of the connection between the beam and the column is usually the responsibility of the structural engineer of record. Since the design forces on the beam/ column joint typically depend on the assumptions in gusset plate design, the engineer of record needs to coordinate with the brace manufacturer as well as review calculations by the manufacturer to make sure the assumptions are consistent. The gusset connection to beam and columns can be designed using various methods. The Uniform Force Method is commonly used to design gusset connection and is not part of this example. The reader can refer to the AISC Steel Manual Part 13 for a discussion of Uniform Force Method. Per AISC 341 Section F4.6b, where a brace or gusset plate connects to both members at a beam-to-column connection, the connection shall either be a) a simple connection that allows unrestrained relative relative rotation of 0.025 radians between the framing elements being connected, or b) a fixed connection designed to resist a moment equal to the lesser of 1.1 (LRFD) times expected beam flexural strength ( R y M p) or 1.1 (LRFD) times the sum of expected column flexural strengths ( Σ R y F y Z ). ). Unless collector loads are relatively small, it is often difficult to design the beam-to-column connection connection without restraining rotation. In addition, typical welded gusset connection would not allow the relative rotation described in AISC; thus, the connection is considered fixed. The connection for this example between the beam and columns consisted of complete joint penetration welds at both beam flanges. Additionally,, the stiffener at the top of the gusset plate increased flexural capacity of the connection. Additionally

7.4 COLUMN SPLICES Per AISC 341 Section D2.5b, the required strength of column splices shall be greater of a) the required strength of the columns, including that determined from Chapters E, F, G, and H and Section D1.4a or b) the required strength determined using the load combination with an amplified seismic load. The required strength need not exceed the maximum load that can be transferred to the splice by the system. In addition, per AISC 341 Section F4.6d, column splices shall be designed to develop at least 50 percent of the lesser available availa ble flexural strength of the connected member. Column splices occur above the third floor where columns change sizes from W14 × 211 to W14 × 132. The required column strength between the third and fourth floors is 740.2 kips. For W14 × 132: B f = 14.7 and t f = 1.03. φ A f F y = 681 kips per flange. The splice consists of a complete penetration weld of the columns flanges. By inspection, the column’s splice is capable of resisting the required axial load as well as 50 percent of the flexural strength of W14 × 132. H c Per AISC 341 Equation F4–2a, the required shear strength of the splice is determined by V u = Σ M pc / using LRFD. M pc for W14 × 211 = 50 ksi × 390 in3 /12 = 1625 k-ft M pc for W14 × 132 = 50 ksi × 234 in3 /12 = 975 k-ft H c = Story height − beam depth − slab thickness = 12 ft − (18 in + 4.5 in)/12 = 10.13 ft V u = (1625 k-ft + 975 k-ft)/10.13 ft = 257 kips.

Check shear strength of web of W14 × 132:

φvV n = (φv)(0.6)(F y)(t w)(d ) = (1.0)(0.6)(50 ksi)(0.645 in)(14.7 in) = 284 kips > V u . . . OK. 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

97

Design Example 3



Buckling-Restrained Buckling-Restra ined Braced Frame Frame

7.5 ST STABILITY ABILITY BRACING

AISC 360

For V-type or inverted-V-type brace configurations, as a minimum, one set of lateral braces is required at the point of intersection of the braces, unless the beam has sufficient out-of-plane strength and stiffness to ensure stability between adjacent brace points. The design of stability bracing is beyond the scope of this example.

8. Items Not Addressed in This Example The following items are not addressed in this example but are nevertheless necessary for a complete design of the seismic load resisting system: • Comparison of wind and seismic forces, • ASCE 7 stability check (ASCE (ASCE 7 Eq 12.8–16), 12.8–16), • Design of the column column base plate plate connection, • Design of the foundations, and • Design of the diaphragm system.

98


Design Example 4 Special Plate Shear Walls

OVERVIEW Special Plate Shear Walls Walls (SPSW) consist of a plate bounded at the sides by columns, also referred to as vertical boundary elements (VBE), and at the floor levels by beams, also referred to as horizontal boundary elements (HBE). The alternate terms for beams (HBE) and columns (VBE) emphasize the boundary element’ss role of resisting the tension field developed in the plate. element’ Steel plate shear walls have been used as early as the 1970’s in the United States (Anon 1977, Anon 1978a, Anon 1978b, Troy 1979). The early implementations of these walls in the U.S. were designed so that the web plates did not buckle under shear loading. To To prevent buckling, the web plates had to be thick and stiffened at intermediate intermediate points across a shear panel. This type of system is commonly referred to as a stiffened steel plate shear wall and is still the preferred behavior in Japan where buckling is not permitted in the lateral-load-resisting system. Another manifestation manifestation of the stiffened steel plate shear wall is the Composite Steel Plate Shear Wall Wall (C-SPW), which uses concrete to brace the plate. On the other hand, the fact that web plates have post-buckling capacity has been known and applied for decades when it comes to plate girders (Basler 1961, Porter 1975). Upon buckling, a diagonal tension field develops in the plate, which, along with the boundary members, creates a truss-like system that can resist significant shear. shear. In some ways, SPSW are similar to a plate girder turned vertical. The major difference between plate girders and SPSW is the greater strength and stiffness of the SPSW column compared to the plate girder flange (Berman and Bruneau 2004). Unstiffened SPSW design is focused around keeping the boundary elements elastic while the web plate yields along an inclined tension field.


99

Design Example 4



Special Plate Shear Walls

Since the 1980’s, tension field action has been utilized in steel plate shear walls and is the basis of the AISC 341 provisions for SPSW. Since these systems allow thinner plates and do not require stiffeners, they are commonly cheaper, and thus have become more popular than their stiffened counterparts in North America. Steel plate shear walls that utilize the post-buckling capacity of the shear plate are commonly referred to as thin or unstiffened. U.S. code provisions apply to unstiffened steel plate shear walls that utilize the post-buckling capacity of the web plate. Engineers can choose whether to design the SPSW as a dual system with special moment frames (SMF) capable of resisting 25 percent of the lateral forces ( R = 8.0) or as a SPSW with ordinary moment connections ( R = 7.0) (ASCE 7). AISC 341 requires moment connections between the HBE and VBE in both cases, because they fill out the hysteresis loops (dissipate additional seismic energy), provide lateral lateral resistance during load reversal (while the tension field in the web plate reverses direction), direction), and add lateral stiffness. One of the defining characteristics of SPSW is the significant amount of strength that can be developed in a short length of wall. SPSW can be especially useful in building configurations with open floor plans and little room to hide braced frames. Of course, with more slender lateral-resisting elements comes correspondingly higher overturning overturning forces and a greater likelihood that deflection will control the design. This example is tailored to the use of unstiffened SPSW in high-seismic regions with all the ductile detailing requirements implied therein. Stiffened SPSW, SPSW, composite steel plate shear walls, and low seismic applications for SPSW are not addressed. See the AISC Design Guide 20 (Sabelli and Bruneau 2006) for more information on these topics. Other references of note are the MCEER report on design of perforated steel plate shear walls (Purba and Bruneau 2007), the MCEER report by Vian and Bruneau (2007), which gives test results for panels with perforations and low-yield-strength low-yield-strength steel, the paper by Berman and Bruneau (2008) which discusses capacity design of VBE’s, and the paper by Qu and Bruneau (2010) which discusses HBE capacity design.

OUTLINE 1. Building and Special Plate Shear Shear Wall Wall Layout 2. Calculation of the Design Base Base Shear and Load Combinations 3. Vertical and Horizontal Distribution Distribution of Load 4. Preliminary Sizing of the Web Web Plate and Boundary Members 5. Analysis and Design Design (Iterative) (Iterative) 6. Detailing and Design of Connections Connections 7. Items Not Addressed in This Example Example

100


Design Example 4

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1. Building and Special Plate Shear Wall Layout Layout 1.1 GIVEN INFORMATION The building is a six-story office building located in San Francisco. The Seismic Design Category is D. See Appendix A for the following information: • Building dimensions • Floor and roof weights • Latitude and longitude • Soil type • Spectral accelerations • Load combinations combinations including including the vertical seismic-load effect

1.2 LA LAYOUT YOUT OF SHEAR WALLS Figure 4–1 shows the plan location of the shear walls. There are four shear walls oriented in each direction, making eight walls in total. Figure 4–2 shows a typical wall elevation. A

B

C

E

D

F

5 @ 30'-0" = 150'-0" 150'-0"

5C

5D

5

4

A4

3

F4

" 0 ' 0 2 1 = " 0 ' 0 3 @ 4

Deck Span

A2

F2

2

1

1C

1D

Figuree 4–1. Plan layout of shear walls Figur 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

101

Design Example 4




A

B

C

D

1 5 '-0 "

E

F

1 5 '-0 "

" 0 ' 4

PARAPET ROOF

6th 6th FL R

5th 5th FL R

" 0 ' 6 7 = " 8 ' 2 1 @ 6

4th 4th FL R

3rd FLR

2nd FLR

1st FL R

Figuree 4–2. Ty Figur Typical pical wall elevation

The floor and roof weights are given in Appendix A. The seismic weight is tabulated in Table Table 4–1.

Table 4–1. Building seismic weight calculation

Level

Assembly Floor

Unit Wt (psf)

Area (ft2)

Weight (kips)

75

14,400

1080

Second to Sixth

1218 Ext Wall

20

6885

138

Roof

30

14,400

432

Ext Wall/Parapet

20

5603

112

Roof

544 Total Building Weight, W = 5 × 1218 + 544 = 6634 kips

W = 6634 kips

102

Floor Wt (kips)


Design Example 4

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ASCE 7

2.1 CLASSIFY THE STRUCTURAL SYSTEM AND DETERMINE SPECTRAL ACCELERATIONS There are two options in ASCE 7 Table 12.2–1 for using special plate shear walls: a SPSW, and a SPSW/ SMF dual system. The two options are shown in Table 4–2 along with performance factors and height limitations.

Table 4–2. System factors for SPSW from ASCE 7 Table 12.2–1

Seismic-ForceResisting System

Response Mod. Coef., R

OverStrength Factor, Ωo

Defl. Amp., C d

Height Limits by Seismic Design Category B

C

D

E

F

B26

SPSW2

7.0

2.0

6.0

NL

NL

160

160

100

D13

SPSW—SMF Dual System1, 2

8.0

2.5

6.5

NL

NL

NL

NL

NL

1. Dual system with Special Moment Frames capable of resisting at least 25 percent of prescribed seismic forces. 2. IBC Section 2205.2 references AISC 341 for detailing requirements (note that IBC 2012 does not adopt ASCE 7 Chapter 14 as stated in IBC Section 1613.1). In Seismic Design Categories A, B, and C, it is not required to follow the provisions of AISC 341 if using R = 3, Ωo = 3, and C d = 3 from part H of Table 12.2–1. For the purposes of this example, a special plate shear wall without the dual system will be used. The height of the structure is 76 feet, 0 inches, which is less than the limit for Seismic Design Category D. The system factors are: R = 7.0

Ωo = 2.0

C d = 6.0

The spectral accelerations to be used in design are calculated in Appendix A based on S S = 1.50g and S 1 = 0.60g to be S DS = 1.00g

S D1 = 0.60g

2.2 RESPONSE SPECTRUM Determine the approximate fundamental building period using Section 12.8.2.1: C t = 0.02 and x = 0.75 T a

t n

= 0

T 12.8–2

=

.51 sec (see discussion below)

Eq 12.8–7

T a = 0.51 sec 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

103

Design Example 4




Develop the design response spectrum using Section 11.4.5: T =

S



S

T =

S

S

= 1

T

06 1 00

6

DS

S

= 02

=

0 60 10 06 T

 T

=0

=

s c

=

§11.4.5

0T for T < T o

Eq 11.4–5

s c

§11.4.5

for T s < T < T L

Eq 11.4–6

The long period equation for S a does not apply here because the long period transition occurs at T L = 12 sec (from Figure 22–12).

1.2 ) g (

a S

Approximate SPSW Building Period, Ta=0.51 sec, Sa=1.0g 1

SDS=1.0g

, n o i t a 0.8 r e l e c c A l 0.6 a r t c e p S n 0.4 g i s e D

TS=0.60 sec To=0.12 sec

Tmax = 0.71 sec

Sa=0.4+5.0T Calc’ed SPSW Period, T = 0.92 sec

0.2

Sa=0.60/T 0 0

0.5

1

1.5

2

Period (Sec)

Figure 4–3. Design response spectrum for the example building

As shown in Figure 4–3, the design spectral acceleration is between T o and T S , so the design spectral acceleration S a is 1.0g. It is not required to construct the design response spectrum when using the equivalent lateral force procedure, since the response spectrum is implicit in the calculation of C s in Section 12.8.1.1. The response spectrum demonstrates the effect of the assumptions used in the calculation of building period. The approximate fundamental period was computed above to be T a = 0.51 sec. The period of the structure can also be established through structural analysis of the SPSW. Section 12.8.2, however, limits the period that can be used to calculate spectral acceleration to a value of T max = C u × T a, where C u is a

104


Design Example 4

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factor found in Table 12.8–1. In this case T max = 1.4 × 0.51 = 0.71 sec. As shown later in this example, drifts are close to, if not at, the story drift limits of Section 12.12. Therefore, the maximum period, T max, will be used to design the SPSW and the actual period of the building be verified later in the design process. The first mode period of the final design shown in Figure 4–15 was calculated to be 0.92 sec; therefore, the initial assumption of T = 0.71 sec is shown to not only be valid, but still conservative.


T12.3–1

Determine if the structure has horizontal irregularities. 1a. and 1b.

Torsional Irregularity—A torsional irregularity exists when the maximum story drift computed including accidental torsion is more than 1.2 times the average story drift. An extreme torsional irregularity exists when the maximum story drift is more than 1.4 times the average story drift. From the rigid diaphragm analysis presented later in this example:

= 0.26

F K

25

ma vg

=

26 0 25

i

K F

= 1.

.2 → NO TORSIONAL IRREGULARITY

where: F i = story shear K = stiffness of one SPSW (assuming all SPSWs are identical).

2.

Reentrant Corner Irregularity—This plan irregularity exists when a reentrant corner has plan dimensions in both directions that are greater than 15 percent of the overall plan dimension. The reentrant corner in this structure is 30 feet/150 feet = 20% of the plan width and 30 feet/120 feet = 25 percent of the plan depth.

→ REENTRANT CORNER IRREGULARITY EXISTS.

3.–5.

By inspection, the building does not qualify for any of these Horizontal Structural Irregularities.

HORIZONTAL IRREGULARLY TYPE 2

According to Section 12.3.3.4, because the structure has a horizontal irregularity of Type 2, the forces in connections of diaphragms to vertical elements, connections of the diaphragm to collectors, connections of collectors to the seismic-force-resisting system, and the collectors themselves shall be increased by 25 percent unless designed for seismic-load effects including overstrength factor.


105

Design Example 4





T12.3–2

By inspection, the building does not qualify for any of the Vertical Structural Irregularities.



T12.6–1

Determine which lateral force procedure to use: 1. Simplified Alternative Structural Design Criteria—According to Section 12.14.1.1, this analysis procedure can be used for SPSW, but not for buildings over three stories—NOT PERMITTED 2. Equivalent Lateral Force Analysis—According to Table 12.6–1, since the building is less than 160 feet tall and only has plan irregularity 2—PERMITTED 3. Modal Response Spectrum Analysis—PERMITTED 4. Seismic Response History Procedures—PERMITTED


2.6 BASE SHEAR Compute the seismic base shear using Section 12.8. C

=

C

≤

S

=

1 00

 R   7 0  I  1 0 1

  T I 

=

06

=

 7 0 71  1 0

14

Eq 12.8–2

= 0 12.

Eq 12.8–3

Also, C s

e

C s ≥ C

≥

= 0 044(1 0 )(1 0) = 0 044

01 5S 1



=

0 5 × 0 60

 7 0  1 0

= 0.044

Eq 12.8–6

C s = 0.12

106

Eq 12.8–5


Design Example 4

V

=

12 ×




= 796 k ips

Eq 12.8–1 V = 796 kips

2.7 REDUNDANCY FACTOR According to Section 12.3.4, the Redundancy Factor should be calculated for each principal axis. The Redundancy Factor is 1.3 unless either Section 12.3.4.2(a) or Section 12.3.4.2(b) is shown to be true, in which case the Redundancy Factor can be taken as 1.0. Since Section 12.3.4.2(a) is satisfied, the Redundancy Factor is 1.0 for both directions.

ρ = 1.0 FOR EAST-WEST DIRECTION ρ = 1.0 FOR NORTH-SOUTH DIRECTION

2.8 LOAD COMBINATIONS See Appendix A for the derivation of load combinations based on ρ = 1.0 and 0.2S ds = 0.2. Load combinations of consequence for the design of the SPSW are 12 0

+05 =0

+

§12.4.2.3 Load Combo 5 (modified)

5

0Q E .


Load combinations with overstrength are not used for the SPSW frame (although they apply for collectors and at other conditions outside the SPSW frame).


ASCE 7

3.1 VERTICAL DISTRIBUTION OF SHEAR The equation for distributing the shear along the height of the building is shown below. Since the period = 0.71 > 0.5 sec, the value for k is interpolated between a value of 1.0 for T = 0.5 sec and 2.0 for T = 2.5 sec. In this example, k = 1.105. The other terms are defined in Section 12.8.3. F x = C vxV , where, C

=

w

Eq 12.8–11 and Eq 12.8–12 w

=

i


107

Design Example 4





Level

wi (kips)

hi (ft)

i i

C vx

F x (kips)

Roof

544

76.0

65,148

0.16

127

6

1218

63.3

119,248

0.29

232

5

1218

50.7

93,189

0.23

181

4

1218

38.0

67,812

0.17

132

3

1218

25.3

43,324

0.11

84

2

1218

12.7

20,141

0.05

39

Total

6634

408,862

1.00

796

3.2 HORIZONTAL DISTRIBUTION OF STORY SHEAR As shown in Figure 4–4, the center of mass and center of rigidity coincide at the middle of the building. To distribute the load to each SPSW, the story shear, F i, is applied in the X and Y directions. Per Section 12.8.4.2, the point of application of the story shear is offset 5 percent to account for accidental eccentricity. The 5 percent offset is 6 feet for the N-S plan dimension of 120 feet and 7.5 feet for the E-W plan dimension of 150 feet.

5C

A4

5D

F4

" 0 '

STORY SHEAR, Fi

6

ACC. T OR S IO N 7'-6"

CENTER OF MASS AND CENTER OF RIGIDITY

A2

F2 STIFFNESS, K, TYP

Y X

1C

1D

Figure 4–4. Rigid diaphragm analysis to distribute story shear to SPSW

108


Design Example 4

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Since all of the SPSWs are the same, a generic stiffness, K , is assumed for all walls. Example calculations of direct shear, V direct, the shear in each SPSW due to torsion, V tors, and the total shear, V total, are given below SPSW 5D for shear in the X-Direction. The values associated with horizontal distribution of shear for other walls and for forces in the Y-Direction are tabulated in Table 4–4. The distance from the center of mass in the N-S direction, R y, and E-W direct, R x are used in these calculations. R

V rect

V tors

R y

=

i

.0 5

M

R x

Vtota

= 0 25F

R

=

=6 i 6 F i × 60 K 36900 K

= 0 01 F i

+ Vtors

rect

F

i

Table 4–4. Rigid diaphragm analysis to distribute shear to SPSW X-Direction Rd 2

V direct

V tors

−75 K

5625 K

0.25 F i

−0.012F i

−75

−75 K

5625 K

0.25 F i

F2

75

75 K

5625 K

F4

75

75 K

Wall

x

A2

−75

A4

y

R y x

R x y

Y-Direction V total

V direct

V tors

V total

0.24F i

0

−0.015F i

−0.015F i

−0.012F i

0.24F i

0

−0.015F i

−0.015F i

0.25 F i

0.012F i

0.26F i

0

0.015F i

0.015F i

5625 K

0.25 F i

0.012F i

0.26F i

0

0.015F i

0.015F i

1C

−60

−60 K

3600 K

0

−0.01 F i

−0.01F i

0.25F i

−0.012F i

0.24F i

1D

−60

−60 K

3600 K

0

−0.01 F i

−0.01F i

0.25F i

−0.012F i

0.24F i

5C

60

60 K

3600 K

0

0.01 F i

0.01F i

0.25F i

0.012F i

0.26F i

5D

60

60 K

3600 K

0

0.01 F i

0.01F i

0.25F i

0.012F i

0.26F i

Σ Rd 2 = 36,900 K

Table 4–4 shows that the maximum design shear for any wall is 0.26 times the story shear. The rest of this example focuses on the design of SPSW 5D. The story shear and cumulative shears for SPSW 5D are given in Table 4–5.

Table 4–5. Story forces applied to SPSW 5D

Level

Incremental Shear (kips)

Cumulative Shear (kips)

Roof

33

33

6th

60

93

5th

47

140

4th

34

175

3rd

22

197

2nd

10

207


109

Design Example 4

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4. Preliminary Sizing of the Web Plate and Boundary Members 4.1 PRELIMINARY WEB-PLATE DESIGN AND CONSIDERATIONS

AISC 341

The web plate of a SPSW can develop especially large shear capacity even with thin web plates. However, to promote uniform yielding in web plates along the height of the structure, it is preferred to proportion web plates to have shear capacities that match the shear demands at each floor. This can result in very thin web plates. For example, the thickness of the sixth-floor web plate was computed based on a solid web plate, the computed shear demand, and the equation for shear capacity given in AISC 341-10: y w

c

=

w

with φ = 0.9

Eq F5–1

where, F y is the web-plate yield strength (assumed to be 36 ksi) t w is the thickness of the plate to be computed Lcf = clear distance between VBE = 15 ft × 12 in/ft − 17 in = 163 in (assuming d c = 17 in)

α = angle of inclination of tension field relative to vertical (assumed to be 40°) (

) 0.015 in required thickness for a solid web plate at the sixth floor.

It is shown above that a web plate as thin as 0.015 inches thick would be sufficient to resist the design shear forces at the sixth floor. However, there are constructability issues associated with using plate material that thin including weldability, handling, and obtaining suitable steel material. More information about using thin web-plate material is included in Sabelli and Bruneau (2006) and ICC/SEAOC (2012). AISC 341 Chapter F5.7 presents two alternatives to the solid plate SPSW including a perforated SPSW and a corner cut-out SPSW. These alternatives, especially the perforated SPSW, represent a method for reducing the strength of the plate to allow the use of thicker plate that can be made with more commonly available material. Furthermore, tests have shown larger buckling loads and improved energy dissipation in perforated SPSWs as compared to similar strength solid-web SPSWs (Vian and Bruneau 2005). Perforated SPSWs are selected for this design example. Design of a perforated SPSW begins with selecting a hole diameter, D, the diagonal hole spacing, S diag, and the angle of hole lines, α. The only restriction given is that the diagonal spacing, S diag, shall be at least 1.67 D [Section F5.7a(2)]. However, there are some important considerations when selecting these variables. The ratio of S diag / D is an important parameter as it defines the amount of strength reduction compared to the D result in closer-spaced holes and thus more strength reduction. For this solid plate. Smaller values of S diag / D = 1.67. example, use the minimum ratio to maximize the plate thickness, S diag / D = 1.67 USE S diag /

(OR SLIGHTLY LARGER AS NECESSARY)

110


Design Example 4

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Three criteria are considered here for selecting the hole diameter: the hole diameter used in experiments, the cost of cutting the holes, and how well the holes can be configured in the wall geometry. The hole diameter used in the one large-scale test (approximately half scale) of a perforated SPSW was approximately 8 inches (200 mm) (Vian et al. 2009). Finite element models have been made of a perforated SPSW with perforation diameters between 2 inches and 12 inches (Purba and Bruneau 2009). Tests on smaller plates (12 inches × 12 inches or 18 inches × 12 inches) with single perforations were conducted with hole diameters between 2.4 inches and 6 inches (Roberts and Sabouri-Ghomi 1992). The cost of cutting the perforations is directly related to the length of cut. Figure 4–5 shows the variation in the length of cut normalized to area of the web plate for varying hole diameter and spacing ratio. It is clear that larger hole diameters result in shorter length of cut and thus more economical fabrication. However, the layout of larger holes ( D ≥ 20 inches) in the typical 15-foot-wide × 12-feet-8-inches-tall bay used in this example can result in configurations that do not develop the tension strip concept illustrated in Figure C-F5.7 in the commentary. Figure 4–6 shows the perforation layout for four different hole diameters in the D = 1.67, and satisfy required typical bay using a hole orientation angle of 45 degrees, spacing ratio of S diag / distance [Section F5.7a(2)] between edge holes and web-plate connection to boundary elements of between D and D + 0.7S diag. With the goal of maximizing the hole diameter to improve economy, while selecting a perforation layout capable of developing well-defined tension strips, a hole diameter of 16 inches is selected for this example. The 16-inch diameter is also considered reasonable relative to large-scale testing performed at half-scale on panels with 8-inch-diameter holes.

USE D = 16 in

Figure 4–5. Effect of hole diameter and S/D ratio on length of cut


111

Design Example 4




(a) D = 24"

(b) D = 20"

(c) D = 16"

(d) D = 12"

Figure 4–6. Four possible hole arrangements for the perforated SPSW with α = 45°

Finally, the angle of the hole lines, α, needs to be selected. Although the specification does not explicitly restrict the possible hole line angles, there are some good reasons to set the angle equal to 45 degrees. First, the tension strip will be most effective when the angle is near the angle of inclination of a solid plate as given by Eq F5–2. The web plate will naturally form a tension field at that orientation, which is usually between 35 degrees and 45 degrees.

112


Design Example 4




Second, the effective tension strip width, S eff , will reduce as the angle varies away from 45 degrees. This is demonstrated in Figure 4–7 for angles of 30 degrees and 60 degrees, both of which have an effective tension strip width, S eff , equal to 87 percent of the diagonal hole spacing, S diag. The effective tension strip width is used in the literature (Purba and Bruneau 2009, Vian et al. 2009) to develop the strength and stiffness of the panel, so if the angle is not 45 degrees, the effective strip width, S eff , should be used in place of S diag to reduce the strength, stiffness, and expected tension stress in the specification. This is supported in the literature (Purba and Bruneau 2009, Vian et al. 2009, Purba and Bruneau 2007) in which S diag is used to represent the width of the effective tension strip in the finite element analyses. If the angle is 45 degrees, the tension strip width is equal to S diag. The equation relating the effective tension strip width, S eff , to the D and angles in Figure 4–8. diagonal hole spacing, S diag, is given below and plotted for a range of S diag /

 D





sin [ 2

−

)]

To maximize the efficiency of the perforated plate, use α = 45° for this example.

USE α = 45°

Sdiag

Seff Seff

Sdiag

60°

α

= 60°

30°

α

= 30°

Figure 4–7. Possible variations in the angle of holes


113

Design Example 4




Figure 4–8. Effect of varying the angle of holes on effective strip width

The layout of the perforations is now defined and is shown in Figure 4–6(c). Compute the required thickness of the plates using the following equation. The web plate is designed to resist the full story shear without consideration of the moment connections, and the equation below implicitly assumes an angle of the hole lines of, α = 45°. V n

 

−

7

 S diag

with φ = 0.9

Eq F5–3

where, F y and t w are the yield strength and thickness of the plate respectively Lcf = clear distance between VBE = 15 ft × 12 in/ft − 17 in = 163 in (assuming d c = 17 in) D and S diag are defined above as the hole diameter and the spacing of the holes along the diagonal

Plate and steel sheet materials that are allowed in AISC 341-10 Section A3.1 include A36 and A1011 Gr. 55. However, use of A1011 SS Gr. 55 may be undesirable because of the larger yield stress. A1011 SS Gr. 30 and A1011 SS Gr. 33 provide a good alternative even though they are not explicitly allowed in the specification. A1011 SS Gr. 30 and A1011 SS Gr. 33 have been used in construction of SPSWs (Eatherton 2004) and are discussed further in Sabelli and Bruneau (2006) and SEAOC/ICC (2006). For this design example, it is assumed that A36 steel sheet is available in 1 ⁄ 16 inch (0.0625 inch), 14 gage (0.0747 inch), 12 gage (0.1046 inch), 1 ⁄ 8 inch (0.125 inch), 10 gage (0.1345 inch), and 3 ⁄ 16 inch (0.1875 inch) as suggested in Sabelli and Bruneau (2006).

114


Design Example 4




Size the web plate at the first floor. Try 3 ⁄ 16-inch-thick A36 plate with hole diameter, D = 16 inches, S diag = 28 inches, and angle α = 45°. V u = 207 kips

=

0 42

3 16

− 0.7

16 28

From T 4–5

= 250 k ips → OK

φV n = 250 kips The rest of the web-plate designs are tabulated in Table 4–6.

Table 4–6. Preliminary web-plate design at each level

Level

Plate Mat’l

Plate Thick.

Design Thick., t w (in)

Hole Diameter, D (in)

Hole Spacing, S diag (in)

Shear, V u (kips)

Shear Capacity, φV n (kips)

DemandtoCapacity Ratio

6th

A36

1/16

0.0625

16

28

33

83

0.40

5th

A36

14 ga

0.0747

16

28

93

99

0.94

4th

A36

12 ga

0.1046

16

34

140

156

0.90

3rd

A36

10 ga

0.1345

16

28

175

179

0.98

2nd

A36

10 ga

0.1345

16

34

197

200

0.98

1st

A36

3/16

0.1875

16

28

207

250

0.83

4.3 GENERAL DESCRIPTION OF BOUNDARY ELEMENT DESIGN For boundary element design, AISC 341 Section F5.3 states that boundary members are to be designed for the forces corresponding to the expected yield strength, in tension, of the web calculated at the angle α. The definition of the expected yield strength of the web plate is F y R y and the expected flexural strength of the HBE is 1.1 R y M p (for LRFD). The values of R y for different materials are given in Table A3.1 and is R y = 1.3 for A36 plate and sheet. Figure 4–9 shows the forces in the HBE and VBE due to the application of the expected yield strength of the plate at angle α and the expected moment strength of the HBE. F

F R t → Expected web plate yield strength at angle α

→ Expected moment strength of the HBE


115

Design Example 4




Figure 4–9. SPSW divided into parts to show forces

116


Design Example 4




∆ /sinα

Figure 4–10. Components of expected web strength applied to HBE and VBE (based on Sabelli and Bruneau 2006)

For the design of the HBE and VBE, it is necessary to separate the force, F , into components. Figure 4–10 shows a representative strip of web plate acting at an angle α. Sabelli and Bruneau (2006) present a derivation of the resulting horizontal and vertical forces acting on the HBE and VBE: R y F y

Force Length R F t

∆ cos α ∆ os α

∆ sin α = ∆

2

t

F t

Ft

R Ft

1

sin (

)

sin ( 2

)

2



os F1

∆ ∆

=

os

=R

1



sin F 2 =

R F

∆ sin

= R y F y w sin α .

sin α


117

Design Example 4




For this design example, the perforations have the effect of reducing the web-plate strength and fixing the angle of inclination, α. Since the angle of hole lines is 45 degrees, the trigonometric terms in the above equations simplify to: cos

1

=sn

2

2

.

The effect of the perforations on web-plate strength is to reduce the expected yield strength as follows:



F t → perforated plate,

Solid plate,



−

7 D  . a



So the components of the yielding web-plate tension simplify to

=F 12

F

= 2

1 2

R y Fy

w



7





S iag



1 −

.

For the sixth floor, the forces F and F ij are calculated below. All tension field components are given in Table 4–7.

 1

0.7





− 0. 7

−

1.3 36 0. 625 1 0.

a

=

1

16  2



 

(1 3)(36 0. 625) 1 0 7

−

1 76 ki -in 16 28

=

88 kip-in

Table 4–7. Components of the expected strength of the tension field

Level

F (k/in)

F 11, F 12, F 22, F 21 (k/in)

6th

1.76

0.88

5th

2.10

1.05

4th

3.28

1.64

3rd

3.78

1.89

2nd

4.22

2.11

1st

5.27

2.63

4.4 OVERVIEW OF ROOF AND SIXTH-FLOOR HBE DESIGN The HBEs are designed for the moments due to vertical force associated with the web-plate yielding, F ij, and vertical forces associated with the gravity loads. The HBE resists axial compression due to the inward pull applied to the columns by the yielding web plate in addition to the axial force due to collector forces. The HBE resists shear associated with the web-plate yielding, expected plastic moment capacity at both

118


Design Example 4




ends, and gravity loads. Figure 4–11 shows the forces acting on the roof HBE and the sixth-floor HBE. Note the following: • The roof HBE resists the force of the tension field on one side only. In some cases this will result in a heavier top HBE than in floors below. • At intermediate floors, if the plate is the same thickness above and below, the HBE does not resist vertical forces other than gravity loads. When the plates are of different thickness above and below, the HBE is designed for the forces associated with the difference. • The HBE is not required to be designed for end moments unless using a dual system where the moment frame is to be designed to resist 25 percent of the lateral forces.

Figure 4–11. HBE design forces

For this example, the calculations for the roof HBE and sixth-floor HBE will be presented, then design information will be presented for all the HBEs. The design process will consist of computing the required flexural strength, M u, calculating the required axial strength, Pu, calculating the design flexural strength, φ M n, calculating the design axial strength, φPn, checking the interaction equation of AISC 360-10 Section H1.1, then checking that the design shear strength, φV n, exceeds the required shear strength, V u.

4.5 FIND THE HBE MOMENT Assume that the HBE will be laterally braced at mid-span with a beam as shown in Figure 4–1. Find the moment on the roof HBE for SPSW 5D:

 

=

1

w



 × 

25 ps 20 ps L

 = 

w F

0 13 kip-ft 10 kip-ft L

Uniform load due to roof gravity loads D Uniform load due to exterior wall

Q . 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

119

Design Example 4




Using load combination number 5, the required design strength for the roof HBE is

w

u

L2c . − )] 4

+w


The equation M u = wL2 /4 to calculate the required moment due to a distributed load is based on work by Bruneau et al. (2011) that shows that HBEs subjected to end moments and distributed load can develop in-span plastic hinging unless the beam is designed to resist a maximum moment of wL2 /4. The required design strength for the roof HBE is therefore M

=

+ 0.

)2

(

)+

= 511 kip-f f .

M u = 511 KIP-FT ROOF HBE

The required design strength for the sixth-floor HBE is similar. The differences are the inclusion of floor live load per load combination 5 (no floor live load for the roof HBE) and the vertical force associated with web-plate yielding. The vertical force associated with web-plate yielding is the difference in vertical force applied by the sixth-floor web-plate yielding and the fifth-floor web-plate yielding. The required design strength for the sixth-floor HBE is therefore

w

 

1

w L



+

+ 65 ps D

=

65 ps L

  

Lc

4

0 33 k kip-ft D 0 33 kip-ft L

w M

in/f

D

=

+ 0.

)+

+ 1.

2.

)]

)2

( 4

.

M u = 139 KIP-FT 6TH FLOOR HBE

4.6 FIND THE HBE COMPRESSION FORCE

ASCE 7

Next, find the compression force in the roof HBE for SPSW 5D. The roof HBE resists two components of compression: 1. Compression from the story shear collected to the wall. This can be estimated as the larger of the diaphragm collector force, Pcollector, or the maximum force that the web plate can develop, P12. 2. Horizontal reaction from the VBE that resists the inward pull of the tension field, P22. The compressive load from collecting load to the SPSW is limited by the shear that can be developed at the top of the panel: c

120

88

= 143 k ip

.


Design Example 4

Compare this load to the collector force:




§12.10.1.1

n i x n

oo

∑

127 ki

Eq 12.10–1

i

x

F −

e

F − r of 4 frames

=

ect

544

109 ki

127 ips

F px −roof P

=02

w

=

2 0 × 127

63 kips .

4

The design collector load, Pcollector, is less than the force required to fully yield the web plate (neglecting the moment connections), P12. The intent of the AISC 341 provisions is clear, in that the boundary elements are to be able to resist the effects of the expected yield strength of the web plate. The HBE should, therefore, be designed for the larger of Pcollector or P12 plus the horizontal reaction from the VBE calculated below. The horizontal reaction from the VBE is the load required to resist the component F 22 for the height of the column tributary to the top beam. This force accounts for the tendency of the columns to bow inward from tension field action. The HBEs brace the column against this inward pull. F −6

= 0 88 k ip-in from Table 4–8

P

F 2

=

2

×

12 67 ×

− 18

2

=

+ 59

= 59 kips (assuming a W18 beam) 202 kips

Pu = 202 KIPS ROOF HBE

The process for finding the axial compression force on the sixth-floor HBE is similar. The beam axial force due to web-plate shear acting along the axis of the beam, P12, is found using the difference between the web-plate forces above and below. Similarly, the beam axial forces due to the inward pull of the columns, P22, has a portion due to the web plate above and the web plate below the sixth-floor level. −

−

=

−

=

0.88)

= 28 kips

= 248 kips

i x

Eq 12.10–1

x

F − F px −6

e

w

= 0 2×

× 1218 = 244 kips

= 248 kips

Pcollector

=

−

4 frames

=

20

248 4

−

P

×

2 P2

=

=

= 124 kips 1.05)

12.67 × 2

− 18

129 ki (assume W18 Beam)

+ 129 = 253 kips 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

121

Design Example 4




Pu = 253 KIPS 6TH FLOOR HBE

4.7 SELECT PRELIMINARY SECTION FOR THE ROOF HBE

AISC 341

W18 sections are used for the preliminary HBE sections in this example. Choose a W18 × 86 for the roof HBE. Since the calculations are identical for the remaining HBEs, they are not repeated for the sixth floor here. For this example it is desired to only have one point of lateral bracing between the columns. Check the maximum spacing for lateral bracing for this section. Section F5.4c references Section D1.2a for the lateral bracing requirements associated with moderately ductile members. L

y

=

F

§D1.2a

1 67

29,000

21

50

7i

The HBE is laterally braced at 7 feet, 6 inches . . . OK.

ROOF HBE LATERAL BRACING IS OK

Since it is necessary for the HBEs to develop plastic hinges for the full mechanism to occur, Section F5.5a requires that the HBEs meet the section compactness requirements associated with highly ductile members. The following calculations verify that the roof HBE satisfies those compactness criteria.

 F y



202

C

w

=

1265

≤

22



= 7 2

. . . OK

0 16

T D1.1

− C ) ≥ 1 4 

≤ 51 4

T D1.1

=

.9] = 51.

 . . . OK.

T D1.1

ROOF HBE WIDTH-THICKNESS RATIOS ARE OK

Another consideration for the HBE suggested in the AISC Design Guide for Steel Plate Shear Walls (Sabelli and Bruneau 2006) is to provide an HBE with web that is stronger than the SPSW web plate. The suggested equation is

w

≥

t w Ry F y F

.

For A36 web plate and A992 roof HBE, this reduces to w

122

for the W18 × 86: 0.48 > [0.94 (0.0625) = 0.06].


Design Example 4




ROOF HBE WEB THICKNESS IS OK

Verify that the stiffness of the HBE is sufficient to develop uniform yielding in the web plate per Section F5.4a: 0.0031

I

t L



1530

4

0.0625(15 * 12 (12.67 * 12)

= 1338



.

ROOF HBE STIFFNESS IS OK

4.8 CHECK COMBINED FLEXURE AND AXIAL FOR THE ROOF HBE

AISC 360

For a W18 × 86 roof HBE: KL

10×

× 12

r y

12

r

)

7 77

=

2

2

=

 

29000

34.22

Limit on inelastic buckling, 4 71

Because

= 34 for weak-axis buckling

2 63 10

F

)

K

= 244 ksi

F

 < 4 71

n

F

=

Eq E3–4

= 113 y

 = 5 9 ksi

, F

r

= 20 for strong-axis buckling

45

= 1045 ki

Eq E3–2 Eq E3–1

φPn = 1045 KIPS

L p = 9.3 ft Lr = 28.6 ft

Eq F2–5 and Eq F2–6 (can also be found in AISC 13th T 3–2)

Lb = 7 ft-6 in, so Lb is less than L p; therefore, M n

x

=

y

= 775

Eq F2–2

× 775 = 698 k ips. φ M n = 698 KIP-FT


123

Design Example 4




Since Pr / Pc > 0.2: P

8 M

P

M

P P

+

8

+



M r

10

M c

Eq H1-1a

 202 8 511 = + Mn  1045 9 698

M

84 . . . OK.

COMBINED AXIAL AND FLEXURE = 0.84 The calculations for combined flexure and axial force for the sixth-floor HBE is similar to the roof HBE.

4.10 CHECK SHEAR FOR THE ROOF HBE AND SIXTH-FLOOR HBE

AISC 360

The shear at the face of the column is the sum of the shears associated with plastic hinging in the beam, gravity loads, and the effect of the expected yield strength of the tension field. The shear associated with plastic hinging in the beams is taken from Equation E2.1 in AISC 341. The following calculations show that the factored beam shear in the roof beam is less than the factored nominal shear capacity.

=

V u

2 11

o

c

+

−

2

c

2 1.1 1.1)( 775 × 12) 163

=

V u

+ 72 + = 213

163

+

2

+

4

2

(0 13 + 0 21 163 12

2

ips .

For the W18 × 86:

=

.24

w

F y

= 53.

so:

=1 0

w

§G2.1 50

= 265 kip

0 48

. . . OK.

Eq G2–1

V u− R = 211 kips < φV n = 265 kips

ROOF HBE SHEAR IS OK The calculation for the sixth-floor HBE shear load and capacity is similar and is given by the following: V−

2 11

p − o

c

F − ) 2

c

2 1.1 1.1)( 554 12) −

V −

163

+

− 0.88)

+ [1 4(w 163 2

+

.4 0.33 +

+ 14 + = 119 kips.

Use a W18 × 65 for the sixth-floor HBE: V u−6 = 119 kips < φV n = 248 kips

6th-FLOOR HBE SHEAR IS OK

124


w )]

c

2

0.5 0.33)

163/ 12 2

Design Example 4




4.11 SIZE THE REMAINING HBEs Table 4–8 gives the preliminary sizes for the HBEs with key design information. The same calculations as presented above were carried out for each HBE. Section compactness requirements, lateral bracing, stiffness, web thickness, and shear were checked, but are not given in Table 4–8.

Table 4–8. Summary of preliminary HBE designs

Level

VBE Reaction (kips)

Max of Pcollector or P12 (kips)

Pu (kips)

M u (k-ft)

(k-ft)

Combined Axial and Flexure1

Shear Demand to Capacity

Roof

59

143

202

1045

698

0.84

0.80

6th

129

124

W18 × 65

699

476

0.62

0.48

5th

180

373

W18 × 86

1045

698

0.76

0.73

4th

358

182

W18 × 71

766

524

0.77

0.49

122

389

168

W18 × 71

766

524

0.79

0.48

122

439

334

W18 × 86

1045

698

0.85

0.71

φPn

φ M n

Size

(kips)

511

W18 × 86

253

140

122

302

236

122

3rd

267

2nd

317

1. Using Eq H1–1a of AISC 360-10.

4.12 VBE DESIGN Similar to HBEs, the VBEs should be designed to resist the forces associated with the expected yield strength of the plate ( F y R yt w) acting at an angle α (equal to 45 degrees for this example). A key difference is that the VBE is subjected to six floors of web plates at their expected yield strength. This is not dissimilar to the eccentrically braced frame or buckling-restrained braced frames that require the columns to be designed for the loads associated with each floor’s energy-dissipating element at its capacity. Unlike those systems, however, columns in SPSW are required to resist considerable moment due to the web-plate tension field pulling inward, in addition to the overturning forces. In this sense, the columns are doing more work in SPSW than in other systems, and larger structural sections are often required. For VBE design, the moment and compression forces are separated and examined individually, and then interaction is considered in accordance with Chapter H of AISC 360. The design checks for the VBE include the following: 1. Combined axial and flexure considering (Section F5.3) • Moment due to expected yield strength of the web plate acting at the angle α in addition to plastic hinging of HBEs • Compression/tension due to overturning loads 2. Compactness criteria (Section F5.5a) 3. Minimum moment of inertia requirement (Section F5.4a) 4. Strong-column/weak beam check (Section F5.4b, which references Section E3.4a).


125

Design Example 4




The VBE is subjected to axial force, moment, and shear. The commentary for AISC 341 suggests three methods for determining the forces acting on vertical boundary elements: 1. Nonlinear push-over analysis—Develop a nonlinear 2D model and conduct push-over analysis to find maximum axial forces, moments, and shear forces. 2. Indirect capacity design approach. 3. Combined Plastic and Linear Analysis—A capacity-design approach in which the maximum forces that can be delivered by the HBE and web plates are applied to a model of the column. The Combined Plastic and Linear Analysis method will be used in this example with some simplifications. Forces and moments are calculated based on a complete mechanism with all web plates yielding and plastic hinges at each beam-column connection. The column compression calculated by the capacity-design approach consists of: downward components of the web-plate tension field, F 21; beam shear due to web-plate tension field, V 11; and beam shear due to plastic hinging, V p, (shown in Figure 4–12). The forces applied at the roof and sixth floor are given by F −

1 6

= F 2−6 =

F −

Lc

88 k ip-in

0 88 163

72 ki

−

M pe− V p

r

y

=

2

y

=

Lc

x

=

1 1(1 1)(50)186 12 in/f

2

12 in/f 163

=

38 kip-f t

= 138 kips.

The beam shear due to the web-plate tension field at the sixth floor is due to the difference in web-plate thickness above and below the HBE and is given by

V

=

× Lc

( F − 2

=

× 163

(1 05 2

= 14 kipss .

The rest of the applied loads and moments used for VBE design are given in Table 4–9. Table 4–9. Loads on VBE due to web plate and HBE

126

Level

Tension Field, F 21 and F 22 (k-in)

HBE Plastic Hinge, M pe (k-ft)

Beam Shear Due to Hinge, V p (kips)

Beam Shear Due to Plate, V 11 (kips)

Roof

—

938

138

72

6th

0.88

671

99

14

5th

1.05

938

138

48

4th

1.64

736

108

20

3rd

1.89

736

108

18

2nd

2.11

938

138

43

1st

2.63

—

—

—


Design Example 4

F

V 11-r

21-6

V 11-r V p-r

Vp-r

F

21-6

21-5

V p-6

21-4

V p-6

21-5

V11-4

F

21-3

V p-4 F

V

V p-3 F

21-1

V

21-4

V p-5 F V p-4

21-3

F 22-3 M pe-3

V 11-3 F V p-3

21-2

F22-2 M pe-2

F

11-2

(a) Axial Loads on C olumn E xperiencing T ens ion

F 22-4 M pe-4

V 11-4

V 11-2

V p-2

22-5

F

11-3

21-2

F M pe-5

V 11-5 V p-5

22-6

F V 11-5

F

F M pe-6

11-6

F


M pe-r

V 11-6

V



V p-2

21-1

(b) Axial Loads on C olumn E xperiencing Compression

F 22-1

(c) Loads C ontributing to S hear and Moment

Figure 4–12. VBE free-body diagrams for the capacity-design approach


127

Design Example 4




4.13 DETERMINE VBE AXIAL FORCES, SHEARS, AND MOMENTS A 2D model of a single VBE as a continuous beam over interior supports at each floor was used to compute shears and moments in the VBE. Figure 4–13 shows the resulting shears and moments associated with the given loading on the 2D model, and Table 4–10 gives the values for moments at various heights along the column.

M pe-r

163 F

938

22-6

M pe-6 129 F

390

29

-281

22-5

-266

M pe-5

186

F 22-4

-31

710 -164

M pe-4

200 -64

F 22-3 M pe-3

676 -206

221 -87

F22-2

699 -268

M pe-2

268 -99

864

F 22-1

-275 -132

2D MODE L

COLUMN

COLUMN

S HE AR (K IP S )

MOME NT (K-FT)

Figure 4–13. VBE shear and moment due to capacity-design forces

128


Design Example 4




Table 4–10. Maximum moments at three locations for each floor

Level

Moment at Top of Column, M uT (k-ft)

Max Moment In-Span, M uM (k-ft)

Location of Max In-Span Moment (fraction of L)

Moment at Bottom of Column, M uB (k-ft)

6th

938

−281

0.00

−281

5th

390

−266

0.20

−228

4th

710

−164

0.25

−60

3rd

676

−206

0.31

−37

2nd

699

−268

0.31

−73

1st

864

−275

0.33

0

The axial compression forces can be computed as the sum of the applied axial forces shown in Figure 4–12b. Table 4–12 tabulates the compression forces at the top of the column, the bottom of the column, and at the location of maximum moment identified in Table 4–11.

Table 4–11. Axial forces applied to the VBE

Level

Column Compression from Plate, P21 (kips)

Compression at Top of Column (kips)

Compression at Location of Max Moment (kips)

Beam Shear, V p + V 11 (kips)

Incremental Column Compression (kips)

Compression at Bottom of Column (kips)

6th

117

210

327

210

326

327

5th

140

113

253

440

552

580

4th

219

186

406

766

930

985

3rd

252

129

381

1114

1288

1366

2nd

282

126

408

1492

1687

1774

1st

352

181

532

1955

2190

2306

4.14 SELECT VBE SECTIONS

AISC 341

With the moments given in Table 4–10 and the compression forces given in Table 4–11, wide-flange sections are chosen that satisfy the compactness criteria, that satisfy the minimum moment of inertia, and that can resist the combined forces. The column is to be spliced above the fourth floor. Choose a W14 × 283 for the first three stories and a W14 × 159 for the upper three stories. The compactness requirements for the


129

Design Example 4




W14 × 283 are verified below to satisfy the criteria for highly ductile members as specified in Section F5.5a (the procedure for the W14 × 159 is identical):



C

E

=

= 7 22

2306

=



= 3 89

. . . OK

T D1.1

= 0 62

3749

T D1.1

− C ) = 46 1

= 8 84

. . . OK.

T D1.1

VBE WIDTH-THICKNESS RATIOS ARE OK Section F5.4a places a minimum on the moment of inertia of the VBE. This limit is intended to require a stiff enough VBE such that the tension field can develop uniformly across the panel.

= min

0 031t

I m n

4

0.

= c

(3 16 12. (

=

)

)

n

. . . OK.

§5.4a

VBE MOMENT OF INERTIA IS OK

4.16 CHECK COMBINED FLEXURE AND AXIAL FORCES ON VBE

AISC 360

The column must be able to elastically resist the combined moments and axial forces computed above. The design axial strength, φPn, and design moment strength, φ M n, are computed, and then the interaction equation is checked. This process is conducted for the top of the first-floor column (W14 × 283), and then a summary of all the interaction equation values is presented in Table 4–13. K

=

r y

1 0×

× 12 17 29 00

 KL

2

36 5 2

Limit on inelastic buckling, 4 71

 Because

r

)

< 4 71

F y

= 36.5 for minor axis

215 ksi

E

= 113 F y



, F cr

=

Eq E3–4

y

45 4

=

.4 ksi

= 3404 kips φPn = 3404 kips

130


Eq E3–2 Eq E3–1

Design Example 4

L

F

=




7 ft

Eq F2–5 (can also be found in AISC 13th T 3–2)

Lb = 12 ft-8 in, so Lb is less than L p; therefore,

=

M

× 542

09

12 in/f

= 2033 kip-ft

Eq F2–1

φ M n = 2033 kip-ft The worst case combined axial and flexure might occur at one of three locations: the base of the column, the maximum moment in the in-span, or at top of the column. All locations were checked and are tabulated in Table 4–12. The worst case was found to be the top of the first-floor column: Since Pr / Pc > 0.2: P

8 M

Pc P P

M cx

+

8

+

M u M

M r



M y

10

Eq H1–1a

 1955 8 864  = 3404 + 9 2033 = 0 95

. . . OK.

COMBINED AXIAL AND FLEXURE = 0.95

Table 4–12. Interaction equation values

Level

Design Compression Capacity φPn (kips)

Design Moment Strength φ M n (k-ft)

Interaction at Top of Column (kips)

Interaction at Location of Max Moment (kips)

Interaction at Bottom of Column (kips)

6th

1891

1076

0.89

0.40

0.40

5th

1891

1076

0.55

0.51

0.49

4th

1891

1076

0.99

0.63

0.57

3rd

3404

2033

0.62

0.47

0.42

2nd

3404

2033

0.74

0.61

0.55

1st

3404

2033

0.95

0.76

0.68


131

Design Example 4




4.17 STRONG COLUMN—WEAK-BEAM CHECK Section F5.4b states that the strong-column/weak-beam provisions in Section E3.4a associated with special moment frames shall be met neglecting the web plate. The capacity-design approach for the VBE will develop more moment capacity in the VBE than required to satisfy the strong-column/weak-beam provisions. However, a typical check for the connection of the first-floor HBE to the VBE is provided below for completeness. The strong-column/weak-beam check can be more critical when column forces are determined using methods other than the capacity-design approach. The sum of the projections of column flexural strength to the beam centerline is given by

 pc

c

yc

−



P

=



−

1955   83.3 

28,760 k-in.

The shear on the first-floor HBE is given in Table 4–12 as 181 kips. The eccentricity between the shear force and the column centerline is half the depth of the column because the moment connection selected is the WUF-W connection, which uses a plastic hinge at the face of the column.

p

* M pc p

y

28,760 12 764

y

+ M uv )

.

.

+

 16 7  181 2  

12,764 k -in

2 3 1 0.

The column and beam satisfy the strong-column/weak-beam moment ratio.

5. Analysis and Design (Iterative) 5.1 CHECK DEFLECTION

ASCE 7

Calculating deflection requires the use of a 2D computer model. There are multiple ways to model the web plate, of which two of the most useful are shown in Figure 4–14. The Canadian Steel Code (CAN/CSA S16-01) recommends the tension strip model. In this model, the web plate is divided into at least ten tension strips per panel, oriented at the angle α. The tension strip section is merely the tributary width of plate times the effective thickness of the plate and is pinned at each end. CAN/CSA S16-01 uses this model to determine the design capacity of the wall as the shear that causes a strip to reach its factored tension capacity. The elastic 2D model is not appropriate, however, for boundary element design. In order to use the 2D model to design the boundary elements, it would be necessary to account for nonlinearities such as the plate yielding and plastic hinging of the beams. Another modeling option is the orthotropic membrane model. The web plate is meshed into a number of membrane elements. The local axes of these elements are rotated to the angle of inclination, α. The modulus of elasticity along local axis 1 and the shear modulus are set to zero (or as near zero as the modeling software allows). The modulus of elasticity along local axis 2 is set to 29,000 ksi for steel. The membrane thickness is then set to the effective web-plate thickness. Advantages to this type of modeling include: 1. The angle of inclination, α, can be adjusted without reconfiguring the model geometry. 2. The membrane can be meshed automatically whereas the tension strips have to be added individually.

132


Design Example 4




ME S H O F MEMBRANE ELEMENTS

ROTATE LOCAL AXES T E NS IO N ONLY STRIP (CAN/CSA RECOMMENDS 10 MIN)

2 1

SET TO ZERO STIFFNESS

ORTHOTROPIC MEMBRANE MODEL

T E NS IO N STRIP MODEL

Figure 4–14. Modeling options to check deflection

For this example, the tension strip model will be used. Because the angle of inclination of the hole lines is fixed at 45 degrees, it is possible to develop a model that does not need to be adjusted for changes in the angle of inclination. Also, the 2D tension strip model facilitates nonlinear pushover analysis if desired. Each panel was split into ten tension strips with the following properties: Width = spacing between strips = 20.75 inches Angle = 45°


133

Design Example 4




The effective plate thickness is given by



1− 1−

4 S

 D  4  S di

D



1−



a

N D sin α 

.

Eq F5–4



H

For the first floor, the value of the effective plate thickness can be found as 1−

16  4 28

 16   5 16 1− 1− 4  28   134

45

1875

 

0 139 in

The effective plate thickness was calculated for the first through sixth floors as 0.139 inch, 0.112 inch, 0.100 inch, 0.087 inch, 0.056 inch, and 0.046 inch. The resulting elastic deflections are given in Table 4–13. The elastic story drifts are the difference in the elastic deflections at the floor above and below. The predicted inelastic story drift is given by the following for the fifth floor:

x

=

C I

x

=

60

43 10

= 2 58 in

ASCE 7 Eq 12.8–15

The limit on story drift is given in Table 12.12–1 based on building system and the occupancy category:

∆ =

.

=

. 20

=30

in

ASCE 7 T 12.12–1

δ x > ∆a → DRIFT IS OK

Table 4–13. Calculated deflection for preliminary design

Level

Elastic Deflection (in)

Elastic Story Drift, δ xe (in)

Predicted Inelastic Story Drift, δ x (in)

Story Drift Limitation, ∆a (in)

Roof

2.26

0.34

2.04

3.04

6th

1.92

0.43

2.58

3.04

5th

1.49

0.43

2.58

3.04

4th

1.06

0.42

2.52

3.04

3rd

0.64

0.36

2.16

3.04

2nd

0.28

0.28

1.68

3.04

134


Design Example 4




5.2 ITERATE TO SATISFY DEFLECTION The drifts were found to satisfy the story drift limits. However, if the inelastic story drift had exceeded the story-drift limit it would have been necessary to either increase the size of HBEs and VBEs to increase the stiffness of the moment frame, or increase the web-plate thickness. If the capacity-design approach is used, the column moments are the only parameter not easily calculated by closed-form equations (although they can be estimated). It is possible, therefore, to create a spreadsheet or Mathcad sheet to carry out most, if not all, of the required calculations. Using one of these computer tools can significantly simplify the iteration process.

5.3 FINAL PLATE THICKNESSES AND BOUNDARY MEMBERS Since all design checks were satisfied, the final design for the web plate, HBEs, and VBE’s are shown in Figure 4–15.

W18x86 1/16" A36 Plate and Sdiag=28" W18x65 9 5 1 x 4 1 W

9 5 1 x 4 1 W

14 ga A36 Plate and Sdiag=28"

W18x86 12 ga A36 Plate and Sdiag=34"

Sdiag Typical

" 0 VBE Splice ' 6 7 = " 8 ' 45° Angle 2 1 Typical, Center @ 6 Pattern on Wall

W18x71 10 ga A36 Plate and Sdiag=28" W18x71

3 8 2 x 4 1 W

3 8 2 x 4 1 W

10 ga A36 Plate and Sdiag=34"

W18x86 3/16" A36 Plate and Sdiag=28"

16" Diameter Holes Typical

15'-0" Figure 4–15. Final plate and boundary-member design 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

135

Design Example 4




6. Detailing and Design of Connections 6.1 PANEL ZONE

AISC 360

AISC 341 Section F5.6b(2) references Section E3.6e for a check of the panel-zone shear for the panel zones next to the top and base HBE of the SPSW. The roof HBE and an HBE at the foundation are expected to experience larger shear forces because there is web plate on only one side of the HBE. For this design example, check the column panel zone at the roof HBE (W18 × 86) connection to the VBE (W14 × 159) including a 1 ⁄ 2-inch-thick doubler plate in the column panel zone.

− V c

Vu

= 580 −

+ 72 + = 529 kips

where, M

F

=

=

=

R Z F

163

Lc

Since

+

w

Pu

210

P y

2335

18.

−

= 580 kips

.77

= 126 kips

=0

due to web plate

= 3 k ips

)

due to flange forces

due to plastic hinges

ki

2 V

1 1(186 50

2 1 ,230

c

F

=

due to gravity loads.

.75, then

+

t

3

LRFD §J10.6

c

t c

LRFD Eq J10–11

c w

6×

R



× 15.0 ×

5 1+

3 15 6 (1 19)2 18.4 15 0)( .745 + 0 5)

= 601 kips

and the minimum panel zone thickness is determined from Section E3.6e as

≥ min

(

w )

0

=

(

.4 −

.

+

.

−2

1.

)

0

0 33 in

Eq 3–7

The doubler plate and the column web are both thicker than the minimum thickness.

V u = 529 kips

φ Rn = 601 kips

t min = 0.33 in

t wc = 0.745 in

Use 1 ⁄ 2-inch doubler plates following the requirements of Section E3.6e(3) at the panel zone of the VBE at the connection of the roof HBE.

136


Design Example 4




6.2 HBE-TO-VBE CONNECTION Section F5.6b requires that the HBE-to-VBE connections satisfy the requirements of Section E1.6b, beam-to-column connections for ordinary moment frames. Welds shall be in accordance with AISC 358-10 including criteria for removal of backing bars, backgouging, and reinforcement with fillet welds. Weld access holes shall be in accordance with AWS D1.8 Section 6.10.1.2. Additional requirements for the connection are given in Section E1.6b. For this example, the HBE web and flanges will get complete joint-penetration welds to the VBE flange. Since the HBE web will get a complete joint-penetration weld to the VBE flange, a check of the HBE shear will suffice. The shear in the HBE is checked in Section 4.10 of this design example.

6.3 PLATE-TO-BOUNDARY ELEMENTS

AISC 360

In general, tolerances make it impractical to weld the plate directly to HBEs and VBEs. Either a plate or angle is required to provide a flat surface against which the plate can be placed and welded. For this example a plate will be used. The connecting plate is sometimes referred to as a “fish plate.” Section F5.6c specifies that the connection of the plate to the HBE and VBE shall resist the forces associated with the expected yield strength of the plate in tension, oriented at the angle α. In this case it is most convenient to express this force in terms of the effect of the tension field F 1 and F 2 acting on the connection at the angle α. Figure 4–16 and the following equations demonstrate the decomposition of the tension field force into these components.

Figure 4–16. Web-plate-to-boundary-element connection


137

Design Example 4






1−

F 1

=

fo length



=

7 D  S dia

∆ 

∆

= R F t

1−



7

cos

S i

os



−

0 7 D 

force

∆

a

t

∆

lengt



1



0 7 D  sin



si Because the angle is 45 degrees, the two components of force, F 1 and F 2, are the same. For the first-floor plate, this becomes 13

1875

−

0 7(16

s(45

28

3. 2 ki -in.

Try a 3 ⁄ 16-inch continuous weld on one side and a 3 ⁄ 16-inch weld intermittent at 2 inches every 6 inches on the other side. The equation for weld capacity comes from Section J2 of the LRFD specification and includes a factor for the angle of loading relative to the longitudinal axis of the weld, θ, which effectively increases the weld capacity. ) R

.6 ×

×

Eq J2–5

e

 3 6 3 2  × + × (1 + 0.5 2  16 6 16 6 

1

V u = 3.72 kips-in

45 5

7.22 kip-in → OK

φ Rn = 7.22 kips-in

At the corners of the panels, it must be decided how to treat the intersection of the two fish plates. Schumacher, et al. (1999), studied the effect of different ways of joining the fish plates. Figure 4–17 shows the four types of corner connections tested. All four corner details exhibited similar response and were found to be acceptable for use in the SPSW system.

138


Design Example 4




Figure 4–17. Corner details tested by Schumacher et al. (1999)—All were found to be acceptable

6.4 COLUMN SPLICE AISC 341 Section F5.6d references Section D2.5 for splices in the VBE. As stated previously, columns in the SPSW system do more work than other lateral resisting systems. As shown in Figure 4–13 there is considerable moment that is developed in the VBE. Figure 4–13 also shows that the maximum moment occurs at the floor levels, and although it is far from zero at a typical splice height, it should not be subjected to inelastic demands. For this example the column splice is accomplished with complete jointpenetration welds at the flanges and web, so the connection will have at least the strength of the smaller section.

6.5 BASE OF WALL TO FOUNDATION

AISC 360

There are different approaches that can be used to anchor the first-floor web plate to the foundation. It is sometimes preferable to use a wide-flange beam. See the AISC Design Guide 20 for an example (Sabelli and Bruneau 2006). For this example, a WT is utilized to transfer the significant uplift and shear to the foundation. The anchor bolts will be designed to resist the combined shear and tension. The AISC Design Guide 1 (DeWolf and Ricker 1990) suggests that if anchor bolts are used to resist shear, that they either be designed for friction or include a weld washer. The weld washer allows for the use of oversized holes in the WT and tolerances in anchor bolt placement, while still encouraging uniform bearing on each bolt.


139

Design Example 4




For an anchor bolt spacing of 12 inches determine the loads to each bolt: 1

=

F

2

.1875 1 −

1

0 7(16) 

= 2 63 k ip s-in  F × 12 in spacing

28

Bolt tension without prying action, Rut = Bolt shear, V u =

12 in spacing

F 12

2 bolts

2 bolts

= 15.8 kips

= 15.8 kips

Prying action will amplify the bolt tension. Chapter 9 of the AISC Manual 13th Edition gives the method to determine the added bolt tension due to prying action. A WT section with a 10-inch-wide flange is assumed (such as the WT sections between WT7 × 30.5 and WT7 × 41). Bolt gage, g = 6 in WT flange width, b f = 10.0 in, so a = 2.0 in, and b = 2.8 in, F u = 65 ksi

AISC Manual Part 9

Using 11 ⁄ 8-inch bolts, a′ = 2.6 in and b′ = 2.2 in, p = 5.6 in. The minimum WT flange thickness to neglect prying action is

min

4T

=

=

4×

.

9

.2 65

=

65 in.

Choose a WT7 × 34 that has a flange thickness of 0.72 inch. Prying action can therefore be neglected. Try two 11 ⁄ 8-inch F1554 Gr. 55 bolts at 12-inch spacing. Ab = 0.99 in2, F u = 75 ksi. The nominal tensile and shear stress are obtained from AISC 360 Table J3.2 for threaded parts with threads not excluded from the shear planes. Tn

=

F u

× 0 45 ×

n



.75 ×

2



 15.8  1 8

× 0 = 8 kips × 0 9 = 25 1 kips

T J3.2

2

+ 2

+

Eq C-J3–5a

10 15.8

2

 25 1 

=

54

COMBINED TENSION AND SHEAR = 0.54 Figure 4–18 shows the detail at the base of the wall.

140

T J3.2


Design Example 4

W E B PL 3/16




3/16 WT7X34

2-6

1/8" P3"X3" L

T W O 1-1/8" F 15 54 G R 55@ 12"

1/4

FOUNDATION

Figure 4–18. Web-plate connection to foundation

Additional design checks for the WT that should be examined include ensuring the stem is stronger than the SPSW web plate, bolt bearing, bending of the WT between bolts, and local bending of the flange.

6.6 COLUMN CONTINUITY PLATES Section E3.6f gives requirements for continuity plates. Check whether continuity plates are required at the connection of the first floor HBE: t

R F

= 1 6 and

c

6

= 1 85 in

Since the flange thickness of the first floor VBE is, t cf = 2.1 inches, no continuity plates are required at the first floor. If continuity plates are required, they should conform to the requirements of Section E3.6f .

6.7 COLUMN BASE CONNECTION

AISC 360

The VBE, in addition to resisting considerable compression, will experience large uplift forces associated with seismic forces. Uplift forces can be computed by the capacity-design approach similar to that employed to determine compression and as shown in Figure 4–12a. It is shown in the AISC Design Guide 20 that tension forces are best calculated using this type of capacity-design approach because it can yield less uplift than other methods (Sabelli and Bruneau 2006). Vertical forces on the tension column include the beam shear from plastic hinging, V p; beam shear from web plate yielding, V 11; and the vertical component of the expected web-plate strength acting on the column, V 21. The capacity design approach produces the uplifts given in Table 4–14. 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

141

Design Example 4




Table 4–14. Column uplift computed by the capacity design approach

Beam Shear due to Plate, V 11 (kips)

Beam Shear due to Hinge, V p (kips)

0

72

−138

−66

−66

6th

−117

14

−99

−202

−268

5th

−140

48

−138

−230

−498

4th

−219

20

−108

−307

−805

3rd

−252

18

−108

−342

−1147

2nd

−282

43

−138

−377

−1524

1st

−352

0

0

−352

−1876

Level

Plate Uplift on Column, V 21 (kips)

Roof

Incremental Uplift (kips)

Cumulative Uplift, T (kips)

The uplift at the base of the column is found to be 1876 kips. A similar method can be used to estimate the maximum shear at the base of the column as: ( −

) 2

+

M

= 2 63

(152

− 18.4 in 2

+

38 k-ft 12 67 ft

= 250 kips .

Try twelve 13 ⁄ 4-inch-diameter F1554 Gr. 105 bolts. The nominal tensile and shear stress are obtained from AISC 360 Table J3.2 for threaded parts with threads not excluded from the shear planes. Tn

.75 ×

F u N 2

× 0 45 ×

× 2 41 × × 2 41 ×

= 2033 k ips = 1220 kips

T J3.2

T J3.2

2

  + 10   2 2  1876  250 + 0 89 .  2033   1220 

Eq C-J3–5a

COMBINED TENSION AND SHEAR = 0.89 The column base plate is designed not to resist moment as reflected by the zero moment at the base of the column in Figure 4–13. Design checks for the base plate are required, but they are not shown here.

142


Design Example 4




7. Items Not Addressed in This Example The following items are not addressed in this example but are nevertheless necessary for a complete design of the seismic-load-resisting system: • Comparison of wind and seismic forces, • ASCE 7 stability check (ASCE 7 Eq. 12.8–16), • Design of the column base plate connection, • Design of the foundations, and • Design of the diaphragm system.


143

Design Example 5 Eccentrically Braced Frame

OVERVIEW This example presents design procedures for a six-story Eccentrically Braced Frame (EBF) office building. Given its substantial capacity for inelastic energy dissipation, an EBF seismic-force-resisting system is often used in buildings designed to resist severe earthquakes. This system can be considered a hybrid, combining the stiffness of a concentrically braced frame with the ductility and energy dissipation capacity of a moment-resisting frame. The distinguishing characteristic of an EBF is that at least one end of every brace is connected so as to isolate a segment of a beam called a link. The link is designed and detailed to sustain large inelastic deformations without loss of strength. In a well-designed EBF, under severe seismic shaking, inelastic activity is restricted primarily to the link. It acts as both a ductile fuse and an energy dissipator, limiting the forces transmitted to the braces and other frame members and permitting development of stable and predictable hysteretic behavior. Design requirements for EBF systems are contained in a series of standards. ASCE/SEI 7, Minimum Design Loads for Buildings and Other Structures, sets the basic loading criteria together with associated lateral drift limits. ANSI/AISC 341, Seismic Provisions for Structural Steel Buildings, provides detailed design requirements relating to materials, framing members, connections, and construction quality assurance and control. AISC 341 is applied in conjunction with ANSI/AISC 360, Specification for Structural Steel Buildings, and AISC 303, Code of Standard Practice for Steel Buildings and Bridges. AISC 360 is the main specification that provides the design and detailing requirements for all steel buildings. In addition to these standards, American Welding Society (AWS) standards D1.1, Structural Welding Code and D1.8, Structural Welding Code Seismic present requirements for welding and fabrication that pertain to EBF systems. Another useful document is the AISC Seismic Design Manual, which presents EBF design aids and examples. The International Building Code or IBC refers to ASCE 7 by reference for the determination of seismic loading and building drift determination. 2012 IBC SEAOC St ructural/Seismic Design Manual, Vol. 4

145

Design Example 5



Eccentrically Braced Frame

Single bay EBF configurations can position the link at the ends of the beam as shown in Figure 5–1(a) and (b), or at the center, as shown in Figure 5–1(c) and (d). For links positioned at the ends of the beam, AISC 341 Section F3.2 requires the link-to-column connection be designed in accordance with Section E3.6e. This example positions the links in the center using an inverted-V configuration (Figure 5–1(c)) in one of the primary orthogonal building directions and for comparison, a two-story X (Figure 5–1(d)) configuration in the other. The title figure on the previous page shows an isometric elevation of the selected EBF configurations.

(a)

(b)

(c)

(d)

Figure 5–1. EBF configurations: (a) end-link V; (b) end-link two-story X; (c) center-link inverted-V; (d) center-link two-story X

146


Design Example 5




To accommodate the expected inelastic drift, beam-to-column attachments are fully restrained moment connections. At locations with an attaching diagonal brace, the connection also includes welded gusset plates. Although not explicitly illustrated in this example, a similar beam-to-column connection configuration for a Special Concentric Braced Frame (SCBF) is included in Design Example 2. In order to reduce the moment in the beam outside of the link, the diagonal braces are connected to the link with a fully restrained Bolted Flange Plate (BFP) moment connection. The flange plates facilitate installation of the braces in the field. At the opposite end, away from the link, the brace attachment is idealized as a pin with bolted connecting elements attaching braces to gussets. To economize their size and facilitate erection, the columns are spliced at the third and fifth levels. At the base, the column attachment to the foundation is also idealized as a pin. The column connects to the foundation with a bolted base plate. Although not illustrated in this example, a similar base-plate connection configuration for a buckling restrained brace frame is included in Design Example 9. The beams, braces, and columns used in this example are all ASTM A992 rolled wide-flange sections. Plates and built-up I-shaped sections are all ASTM A572 Grade 50 material. The beam segments outside of the links are composite with the concrete slab. Within the link, the slab is not composite.

OUTLINE 1. Building Geometry and Loads 2. Calculation of the Design Base Shear and Load Combinations 3. Vertical and Horizontal Distribution of Load 4. Preliminary Member Sizing and Analysis 5. Analysis Verification and Member Final Design 6. Design and Detailing of Connections 7. Items Not Addressed in This Example

1. Building Geometry and Loads 1.1 GIVEN INFORMATION The subject building is a six-story office located in San Francisco, California. See Appendix 1 for the following information: • General information including location latitude and longitude, site class, and risk category; • Building geometry including plan dimensions, elevation, and axonometric view; • Assembly weights; • Floor and roof weights;

2012 IBC SEAOC St ructural/Seismic Design Manual, Vol. 4

147

Design Example 5




• Design spectral response acceleration parameters; • Seismic Design Category (SDC); and • Load combinations including the vertical seismic load effect.

1.2 PLAN OF FRAME LOCATIONS AND TYPICAL ELEVATIONS The figure on the first page of this example and Figure 5–2 show the isometric and plan locations of the frames, respectively. As shown, there are two frames oriented in each direction, for a total of four frames. Figure 5–3 shows the typical frame elevations.

Figure 5–2. Typical floor and framing plan

148


Design Example 5

3

2 Roof




C

D 30’-0”

30’-0” BM-5

12’-0”

BM-4

BM-5

6th

12’-0”

BM-3

5th

12’-0”

4th 12’-0”

BM-6

3rd

12’-0”

BM-1

2nd

BM-2

J-1

12’-0”

BR-1 C-1 -Indicates Hinge

Base

-Indicates Splice

(a)

(b)

Figure 5–3. EBF elevations: (a) inverted-V along Grids A and F; (b) two-story X along Grids 1 and 5


ASCE 7

2.1 CLASSIFY THE STRUCTURAL SYSTEM AND DETERMINE DESIGN COEFFICIENTS AND FACTORS As indicated in Table 12.2–1, “Design Coefficients and Factors for Seismic Force-Resisting Systems,” there are two options for using steel eccentrically braced frames. The first is listed under line B1, “Building Frame Systems,” and the other is under line D1, “Dual Systems.” For this example, an EBF without a dual system is used. The following are the listed EBF Response Modification Coefficient, R, Overstrength Factor, Ωo, and Deflection Amplification Factor, C d : R = 8.0

Ωo = 2.0 C d = 4.0 2012 IBC SEAOC St ructural/Seismic Design Manual, Vol. 4

149

Design Example 5




2.2 DESIGN SPECTRAL RESPONSE ACCELERATION PARAMETERS As derived in Appendix 1, the following are the example design spectral response acceleration parameters for short periods, S DS and 1-sec period, S D1: S DS = 1.00g

S D1 = 0.60g

2.3 RESPONSE SPECTRUM The approximate fundamental building period, T a, is determined in accordance with Section 12.8.2.1 as follows: C t = 0.03 and x = 0.75 (steel eccentrically braced frames) T a

= 0

t

=

.74 se (See discussion below).

T 12.8–2 Eq 12.8–7

T a = 0.74 sec

For assistance in illustrating the design response spectrum curve shown, the following parameters are provided in accordance with Section 11.4.5: S D1

=



S

T =

S

02

1

S DS S

= =

06

6 06 1 00 0 6

=

 T

=

§11.4.5

0T for T < T o

Eq 11.4–5

sec

§11.4.5

for T > T s.

Eq 11.4–6

As indicated in ASCE 7 Figure 22–12, the long period transition period, T L, occurs at 12.0 sec. Therefore, the long period equation for S a(T > T s) does not apply. As indicated in Section 12.8.2, the fundamental period of the structure, T , can be established using the structural properties and deformational characteristics of the resisting elements in a properly substantiated frame analysis, or it is also permitted to use the approximate building period, T a. Furthermore, the maximum fundamental period, T max, shall not exceed the following: C u = 1.4(S D1 ≥ 0.4)

T 12.8–1

T max = C uT a = 1.4(0.74 sec) = 1.04 sec.

§12.8.2 T max = 1.04 sec

As shown in Figure 5–4, the approximate fundamental building period, T a, is greater than T s, but less than T max. At T max, the corresponding design spectral acceleration, S a, is 0.58g. In accordance with Section 12.8.1.1, when using the equivalent lateral force (ELF) procedure, it is not required to construct the design response spectrum curve because the accelerations are implicit in the calculation of the seismic response coefficient, C s.

150


Design Example 5




1.2 ) g ( a

TS = 0.60 sec SDS = 1.0g

1

S

, n o i t 0.8 a r e l e c 0.6 c A l a r t 0.4 c e p S n 0.2 g i s e D

To= 0.12 sec

Approximate EBF Building Period, Ta= 0.74 sec, Sa= 0.81g Tmax= 1.04 sec, Sa= 0.58g

Sa= 0.4+5.0T

Sa= 0.60/T

0

0

0.5

1

1.5

2

Period (Sec) Figure 5–4. Design response spectrum curve for the EBF building

For this example, the fundamental period, T , as computed in the elastic model analysis (Section 4.8), exceeds the maximum fundamental period, T max. Therefore, T max is used in the determination of the seismic response coefficient, C s.

2.4 HORIZONTAL IRREGULARITIES In accordance with ASCE 7 Section 12.3.2.1 and in conjunction with Table 12.3–1, structures having one or more of the following shall be designated as having a horizontal structural irregularity: Type 1a or 1b (Torsional Irregularity): The irregularity is defined to exist if the maximum story drift, δmax, including accidental torsion, is greater than 1.2 times the average story drift, δavg. For this example, the maximum story drift is assumed to conform with the requirement and then subsequently confirmed in the rigid diaphragm analysis of Section 3.2. Type 2 (Reentrant Corner Irregularity): The irregularity is defined to exist if the plan projection, pi is greater than 15 percent of the plan dimension, Li, in the given direction as indicated: p x max < 0.15 L x = 0.15(150 ft) = 22.5 ft

T 12.3–1.2

p y max < 0.15 L y = 0.15(120 ft) = 18.0 ft.

T 12.3–1.2

or,


151

Design Example 5




As indicated in Figure A1–1, the building has a plan projection of 30 feet in both orthogonal directions. Therefore, the building is irregular and the provisions of Section 12.3.3.4 and Table 12.6–1 apply. For this type of irregularity, Section 12.3.3.4 requires that diaphragm design forces be increased 25 percent for diaphragm and collector connections. Also, Table 12.6–1 restricts the permitted structural analysis analytical procedures. Type 3 (Diaphragm Discontinuity Irregularity): The irregularity is defined to exist if an open area is greater than 50 percent of the gross enclosed diaphragm area or if a change in effective diaphragm stiffness varies by more than 50 percent form one story to the next. As indicated in Appendix A, the gross enclosed diaphragm area is 15,220 square feet. Per Figure A1–1, the open area at the building’s core is approximately 600 square feet, much less than half the gross area. As indicated in Appendix A, the floor is comprised of 3 1 ⁄ 4-inch-thick light-weight concrete fill on a 2-inch metal deck (rigid), while the roof is simply metal deck (flexible). By inspection, the stiffness of the flexible roof deck is less than 50 percent that of the rigid floors. Therefore, the building is irregular at the roof, and as with the Type 2 irregularity, the provisions of Section 12.3.3.4 and Table 12.6–1 apply. Type 4 (Out-of-Plane Offset Irregularity): The irregularity is defined to exist if there is a discontinuity in the lateral-force-resistance path. As indicated in Figure 5–2 and 5–3, the EBF configuration provides vertical continuity with no out-of-plane offsets. Therefore, the building does not have an out-of-plane offset irregularity. Type 5 (Nonparallel System Irregularity): The irregularity is defined to exist where vertical lateral-resisting elements are not parallel to the to the system’s major orthogonal axes. As indicated in Figure 5–2, the lateral-resisting frames are positioned parallel to the major orthogonal axes. Therefore, the building does not have a nonparallel system irregularity. TYPE 2 AND 3 HORIZONTAL STRUCTURAL IRREGULARITIES

2.5 VERTICAL IRREGULARITIES In accordance with Section 12.3.2.2 and in conjunction with Table 12.3–2, structures having one or more of the following shall be designated as having a vertical structural irregularity: Type 1a or 1b (Stiffness-Soft Story Irregularity): The irregularity is defined to exist if the stiffness of any story is less than 70 percent of the story above or less than 80 percent of the average stiffness of the three stories above. For this example, the vertical stiffness of each story is assumed to conform to the requirement and then subsequently confirmed in the story drift analysis of Section 4.8. Type 2 (Weight Irregularity): The irregularity is defined to exist where the effective mass (weight) of any story is more than 150 percent of any adjacent story. A roof that is lighter than the floor below need not be considered. As indicated in Appendix 1, the weight is the same on each story except the roof. Therefore, the building does not have a weight irregularity. Type 3 (Vertical Geometric Irregularity): The irregularity is defined to exist where the horizontal dimension of the lateral-force-resisting system at any story is more than 130 percent of that for an adjacent story. As indicated in Figure 5–3, the horizontal dimension (30-feet) of the frame is the same at each level. Therefore, the building does not have a geometric irregularity.

152


Design Example 5




Type 4 (In-Plane Discontinuity in Vertical Lateral-Force-Resisting Element Irregularity): The irregularity is defined to exist where an in-plane offset of the vertical lateral-force-resisting element results in overturning demands on supporting elements. As indicated in Figure 5–3, the elevation of the frame is not discontinuous or offset in any manner. Therefore, the building does not have an in-plane discontinuity irregularity. Type 5 (Discontinuity in Lateral-Strength-Weak-Story Irregularity): The irregularity is defined to exist if the lateral strength of any story is less than 80 percent of the strength of the story above. For this example, the lateral story strengths are assumed to conform with the requirement and then subsequently confirmed in the link strength comparisons of Section 4.4. NO VERTICAL STRUCTURAL IRREGULARITIES

2.6 REDUNDANCY FACTOR In accordance with Section 12.3.4, the Redundancy Factor, ρ, is calculated for each principal axis and is 1.3 unless either Section 12.3.4.2(a) or 12.3.4.2(b) is shown to be true, in which case it can be taken as 1.0. For braced frames, Section 12.3.4.2(a) in conjunction with Table 12.3–3 requires that the removal of an individual brace would not result in more than a 33 percent reduction in story strength, nor does the resulting system have an extreme torsional irregularity (horizontal structural irregularity Type 1b). Alternatively, Section 12.3.4.2(b) requires at least two bays of framing on each side of the structure in each orthogonal direction at each story resisting more than 35 percent of the base shear. As indicated in Figures 5–2 and 5–3, one bay of framing is provided on each side of the structure. In each bay, two braces resist seismic forces. For this example, the removal of one out of the four braces in either of the two orthogonal directions will only result in a story strength reduction of approximately 25 percent. Furthermore, the resulting system does not have an extreme torsional irregularity as determined by rigid diaphragm analysis (Section 3.2). Therefore, in accordance with Section 12.3.4.2(a), the redundancy factor is 1.0 for each orthogonal direction.

ρ = 1.0 2.7 ANALYSIS PROCEDURE SELECTION In accordance with Section 12.6 and based on the Seismic Design Category (SDC), structural system, dynamic properties, and building regularity, the structural analysis shall consist of one of the following types as permitted in Table 12.6–1: 1. Equivalent Lateral Force (ELF) Analysis—Structural Characteristics: SDC-D, no applicable structural irregularities, and structural height not exceeding 160 feet. 2. Spectrum Analysis 3. Seismic Response History Procedures. USE ELF ANALYSIS


153

Design Example 5




2.8 SEISMIC RESPONSE COEFFICIENT Using the importance factor, I e, determined in Appendix 1, the seismic response coefficient, C s, shall be determined in accordance with Section 12.8.1.1, as follows: C

=

S

=

10

 R   8 0   I   1 0 

= 0 125.

Eq 12.8–2

The coefficient need not exceed the following: S D1

06

R 

072.

 8 0 1 04  1 0

Eq 12.8–3

Also, the coefficient shall not be less than the following: C s = 0.044S DS I e = 0.044(1.0)(1.0) = 0.044 ≥ 0.01.

Eq 12.8–5

And when S 1 ≥ 0.6g, the coefficient shall not be less than the following: C s = 0.5S 1 /( R / I e) = 0.5(0.6)/(8.0/1.0) = 0.038.

Eq 12.8–6

C s = 0.072

2.9 SEISMIC BASE SHEAR Using the effective seismic weight, W , determined in Appendix 1, the seismic base shear, V , shall be determined in accordance with Section 12.8.1 as follows: V = C sW = 0.072(7231 kips) = 521 kips.

Eq 12.8–1 V = 521 kips

2.10 SEISMIC LOAD COMBINATIONS As derived in Appendix 1 and in consideration of ρ equal to 1.0, the basic seismic load combinations in accordance with Section 12.4.2.3 are simplified as follows:

0

+ ρQ =

+ 0 5L E

10

.

Load Combo 5 (modified) Load Combo 7 (modified)

Also, as derived in Appendix 1 and in consideration of Ωo, the basic seismic load combinations with overstrength factor in accordance with Section 12.4.3.2 simply as follows: 1 4 D 0

154

+

+ 0 5L =

7 D

2 0Q

2 0Q .



Design Example 5





ASCE 7

3.1 VERTICAL DISTRIBUTION OF LATERAL SEISMIC FORCE In accordance with Section 12.8.3, the lateral seismic force, F x, and the vertical distribution factor, C vx, induced at any level shall be determined from the following equations: F x

vx

vx

=

w h

Eqs 12.8–11 and 12.8–12

∑= w

i i

i 1

where the vertical weights, w, and heights, h, are summarized in Table 5–1 and the distribution exponent, k , is interpolated in accordance with Section 12.8.3 as 1.27 for T = 1.04 sec. The corresponding vertical distribution factors and lateral seismic forces are also summarized in Table 5–1 for each level.

Table 5–1. Vertical distribution of lateral seismic force

Level

wi (kips)

hi (ft)

Roof

656

72

149,870

0.188

98

6th

1315

60

238,330

0.299

156

5th

1315

48

179,516

0.225

117

4th

1315

36

124,575

0.156

81

3rd

1315

24

74,438

0.093

49

2nd

1315

12

30,866

0.039

20

Total

7231

797,595

1.000

521

i

C vx

F x (kips)

3.2 HORIZONTAL DISTRIBUTION OF FORCES In accordance with Section 12.8.4, the seismic design story shear in any story, V x shall be determined from the following equation: Eq 12.8–13 i x

where the portion of seismic base shear, F i, is distributed to the frames based on the relative lateral stiffness of the frames and the diaphragm. With respect to the diaphragms, Section 12.3.1.2 permits a concrete-filled metal deck to be idealized as rigid when there are no horizontal irregularities. As determined previously in Section 2.4, the building does have a Type 2 reentrant corner horizontal irregularity. However, given the symmetrical configuration and equivalent frame stiffness, a rigid diaphragm idealization is considered acceptable. In Figure 5–2 the center of mass and center of rigidity, by inspection, coincide at the middle of the building. Therefore, in consideration of Section 12.8.4.1, there is essentially no eccentricity and no corresponding inherent torsional moment, M t . However, per Section 12.8.4.2, to account for accidental torsional moments, 2012 IBC SEAOC St ructural/Seismic Design Manual, Vol. 4

155

Design Example 5




M ta, the point of application of the story shear shall be offset a distance equal to 5.0 percent of the dimension of the structure perpendicular to the direction of applied forces, Li, from the center of mass, as follows: M tai = 0.05 L y(F i) = 0.05(120 ft)F i = 6.0F i

§12.8.4.2

M tai = 0.05 L x(F i) = 0.05(150 ft)F i = 7.5F i.

§12.8.4.2

or,

To distribute the moment loads to each frame, the seismic base shear, F xi, is applied in the building’s X and Y directions. For any given story, each bay of framing has equivalent stiffness, and a generic stiffness, R, is used to represent the rigidity. A fixed distance, d i, is then measured from the respective centroidal axis to each frame. The seismic design story shear, V xi, can be determined by summing the direct story shear force, V i, inherent torsional force, V ti, and accidental torsional force, V tai, as follows: V xi = V i + V ti + V tai

Eq 12.8–13 (modified)

where, i

, V ti = 0, and

ai ai

i

R

i i 2

.

The corresponding story shear rigid diaphragm distributions are shown in Table 5–2. Table 5–3 shows the corresponding design shear forces for each typical frame. Table 5–2. Story shear rigid diaphragm distribution

Grid

Direction

d i

Rd i

V i

V tai

V xi

A

X

−75

−75 R

5625 R

0.50F i

−0.024F i

0.48F i

F

X

75

75 R

5625 R

0.50F i

0.024F i

0.52F i

1

Y

−60

−60 R

3600 R

0.50F i

−0.024F i

0.48F i

5

Y

60

60 R

3600 R

0.50F i

0.024F i

0.52F i

R

Σ Rd 2 = 18,450 R Table 5–3. Typical frame seismic design story shears

156

Level

Design Story Shear, V xi, (kips)

Roof

51

51

6th

81

132

5th

61

193

4th

42

235

3rd

26

261

2nd

10

271

1st

0

271


Cumulative Shear, (kips)

Design Example 5




With respect to torsional irregularity and using the values from Table 5–2, the maximum and average story drifts can be determined as follows: (

)

.52F

R a

=

[(

R

+

xi

]

xi

2

=

( .52

0.

)

i

2

=

05

i

Therefore, the maximum to average story drift ratio can be determined as follows: ma

=

avg

52 05

= 1.

.2.

T 12.3–1 Type 1a

NO HORIZONTAL TYPE 1a TORSIONAL IRREGULARITY In accordance with Section 12.3.4.2(a), with the removal of an individual brace, in this case along Grid A, the resulting frame stiffness, R, is reduced 50 percent. With that reduction, the corresponding modified story shear rigid diaphragm distributions are shown in Table 5–4. Given the symmetrical nature of the frame, it can likewise be shown that similar distributions occur with the removal of an individual brace along Grids F, 1, or 5. Table 5–4. Modified story shear rigid diaphragm distribution

Grid

Direction

d i

Rd i

V i

V tai

V xi

A

X

−75

−38 R

2813 R

0.50F i

−0.014F i

0.49F i

F

X

75

75 R

5625 R

0.50F i

0.029F i

0.53F i

1

Y

−60

−60 R

3600 R

0.50F i

−0.029F i

0.47F i

5

Y

60

60 R

3600 R

0.50F i

0.029F i

0.53F i

R

Σ Rd 2 = 15,638 R With respect to torsional irregularity, using the values from Table 5–4, the maximum and average story drifts can be determined as follows: (

= =

)max R

= +

[(

.53F i R xi

]

2

=

( .53 + 0 2

)

i

=

05

i

Therefore, the maximum to average story drift ratio can be determined as follows:

= a

.5 3 05

= 1 06

1 .

T 12.3–1 Type 1b

NO HORIZONTAL TYPE 1b TORSIONAL IRREGULARITY


157

Design Example 5




4. Preliminary Member Sizing and Analysis 4.1 MATERIAL SPECIFICATIONS

AISC 341/360

In accordance with AISC 341 Section A3.1, the specified minimum yield stress of steel to be used for members in which inelastic behavior is expected shall not exceed 50 ksi. For this design example, steel material strengths are taken from the AISC Manual Table 2–4 as follows: Rolled wide-flange sections:

ASTM A992, F y = 50; F u = 65 ksi

Built-up I-shaped sections and plates:

ASTM A572 Grade 50, F y = 50; F u = 65 ksi.

F y = 50 ksi; F u = 65 ksi

The material ratio of expected yield stress to the specified minimum stress, R y, is taken from AISC 341 Table A3.1 as follows: ASTM A992 or A572 Grade 50, R y = 1.1. R y = 1.1

Bolt material strengths are taken from the AISC 360 Table J3.2 as follows: Shear tab bolts

ASTM A325, F nv = 68 ksi (threads excluded from shear planes)

Flange plate bolts

ASTM A490, F nv = 84 ksi (threads excluded from shear planes). GROUP A: F nv = 68 ksi; GROUP B: F nv = 84 ksi

4.2 PRELIMINARY LINK SIZE AND LENGTH

AISC 341

In an EBF system the primary design focus is the link. Optimizing the link can be a challenge due to member local buckling requirements, geometric constraints, and considerations associated with the strength of the beam outside of the link. Of primary importance is the link length, e, and its relationship to the inelastic yielding behavior of the frame. The design provisions set forth in AISC 341 Section F3 are intended to provide reliable and ductile link behavior when the link is subjected to seismic loading. In the provisions, the nominal shear strength of the link, V n, is determined as the lesser of the shear yielding strength, V p, or the shear associated with the flexural yielding strength, 2 M p / e. In general, when e = 2 M p / V p , the yield condition is balanced between shear and flexure. For values less than 1.6 M p / V p , the link behavior is generally controlled by shear, whereas for values greater than 2.6 M p / V p it is controlled by flexure. For link lengths between 1.6 M p / V p and 2.6 M p / V p , a combination of shear and flexural yielding occurs. Because shear yielding is much more reliable than flexural yielding, it is generally considered advantageous to keep link lengths short enough to be controlled by shear. With this in mind, a target value of 1.6 M p / V p is used for the link length. To achieve this, a lower design value of 1.3 M p / V p is recommended by Engelhardt

158


Design Example 5




and Popov (1989). This lower value allows some flexibility in changing the link beam size and frame V p target. geometry while still maintaining a final link length consistent with the 1.6 M p / V To derive the preliminary link beam shear force, V r , refer to the invert inverted-V ed-V frame beam, BM-1, shown in Figure 5–3(a). As shown in the frame free-body diagram of Figure 5–5 and ignoring the nominal effects of gravity loads, the preliminary shear force, V r , is estimated in consideration of the cumulativ cumulativee design seismic shear force, V i, story height, hst , and frame width, L, as follows: V r = V 2hst / L = (271 kips)(12.0 ft)/30.0 ft = 108 kips.

For the two-story X configuration and BM-2 shown in Figure 5–3(b), the same preliminary force includes the cumulative shear is estimated from the two associated stories as follows: V r = (V 2 + V 3)hst / L = (271 kips + 261 kips)(12.0 ft)/30.0 ft = 213 kips.

L/2 e/2 Vi /2

Vr

hst Vi /2 Vr Figuree 5–5. Inverted-V free-body Figur free-body diagram (cut through through the link centerline)

The preceding methodology is repeated for all the corresponding links with the resulting design shear force summarized in Table 5–5. Table 5–5. Preliminary link design shear force

Shear, V r , (kips)

Level Inverted-V

Two-Story X

Roof

20.4

No Link

6th

52.8

73.2

5th

77.2

No Link

4th

94.0

3rd

104

2nd

108

171 No Link

213


159

Design Example 5




For the beam BM-1, assuming a short link governed by shear yielding, the required link area, Atw, is estimated in consideration of the link design shear strength, φvV n, as follows: V r

108 kips

w

9( 6)50 ksi

4 0 in2 .

Eq F3–2 (Modified)

Excess capacity in the link segment is an important economic consideration, as the other elements in the frame are designed to develop the full inelastic link capacity. The beam-to-brace geometric constraint is also a consideration in that the beam flange must be approximately equal to or wider than the brace flange. In this example, the braces will be W10 rolled sections. Therefore, in consideration of the required link area, flange width, and buckling requirements summarized in AISC Seismic Design Manual Table 1–3, evaluate a W10 × 68 ASTM A992 wide-flange for the beam BM-1 as follows: Alw = (d − 2t f )t w = [10.4 in − 2(0.77 in)](0.47 in) = 4.16 > 4.00 in2.

Eq F3–4

Typically, the most efficient links are deeper sections. However, given the geometric constraints associated with the fully restrained beam-to-brace moment connection in this example, only shallow W10 and W18 rolled sections have an adequate flange width for the brace attachment. Therefore, in consideration of W10 rolled sections for braces, the minimum required flange width in the second through fifth floors is b f > 10.0 inches. Above Above the fifth floor, where lighter braces are adequate, the minimum required flange width is b f > 8.0 inches. Per AISC 341 Section F3.5b(2), the nominal shear, V p, and plastic flexural strengths, M p, are determined assuming Pr / Pc < 0.15, as follows: V p = 0.6F y Alw = 0.6(50 ksi)(4.16 in 2) = 125 kips

Eq F3–2

M p = F y Z = 50 ksi(85.3 in 3) = 4265 kip-in.

Eq F3–8

The preliminary link length, e, which will optimize the W10 × 68, is determined as follows: e = 1.3 M p / V V p = 1.3(4265 kip-in)/125 kips = 44 in. . . . say 48 in.

The preceding methodology is repeated for each link for both frame configurations with the selected link sizes, nominal shear strength, V p, and calculated link lengths, e, summarized in Table 5–6.

Table 5–6. Preliminary link sizes, shear strengths, strengths, and link lengths

Inverted-V

Two-Story X

Level

160

Link

V p (kips)

e ( (iin)

V p (kips)

e (in)

Roof

W10 × 6 68 8

125

48

No Link

—

—

6th

W10 × 6 68 8

125

48

W10 × 6 68 8

125

48

5th

W10 × 6 68 8

125

48

No Link

—

—

4th

W10 × 6 68 8

125

48

W18 × 8 86 6

243

50

3rd

W10 × 6 68 8

125

48

No Link

—

—

2nd

W10 × 6 68 8

125

48

W18 × 8 86 6

243

50


Link

Design Example 5




As indicated in Table 5–6, for the inverted-V, a preliminary link size of W10 × 68 is selected for all of the floors. With a corresponding increase in floor height, somewhat lighter link sections might be expected. However, Howeve r, for this example, in consideration of buckling width-to-thickness ratios, a lighter W10 × 60, W10 × 54, or W10 × 49 rolled section cannot satisfy the highly ductile requirement of Section F3.5b(1).

4.3 PRELIMINARY ADJUSTED LINK SHEAR STRENGTH

AISC 341

As noted in AISC 341 Section F3. 5(a), the diagonal brace and beam segment outside of the link are intended to remain essentially elastic. Therefore, in accordance with Section F3.3, the link shear strength is “adjusted” to include the amplification effect of both material overstrength and strain hardening. For the shear-governed shear-gov erned link of beam BM-1, the adjusted shear strength, V mh, is determined as follows: V mh = 1.25 R yV p = 1.25(1.1)(125 kips) = 172 kips.

§F3.3

The preceding methodology is repeated for each story and each frame configuration with the adjusted link shear strengths summarized in Table 5–7. As a measure of the link shear strength capacity, Table 5–7 Vr . In order to also shows the ratio of the adjusted link shear strength to the design shear force, V mh / V economize the link member selection, the ratio should be as close to 1.25 R y as possible, which in this case is approximately 1.4.

Table 5–7. Preliminary adjusted link shear strength strength and shear ratio

Inverted-V

Two-Story X

Level V mh (kips)

V mh / V Vr Ratio

V mh (kips)

V mh / V Vr Ratio

Roof

172

8.4

No Link

—

6th

172

3.3

172

2.4

5th

172

2.2

No Link

—

4th

172

1.8

334

2.0

3rd

172

1.7

No Link

—

2nd

172

1.6

334

1.6

As indicated in Table 5–7, several of the links significantly exceed the optimum ratio. This could unnecessarily increase the size of the associated braces and columns. G iven the limited selection of highly ductile rolled wide-flange sections with a corresponding compatible flange width, this example considers the use of built-up I-shaped sections as an effecti effective ve alternative for these locations. However, However, the use of built-up link sections is not without both code and economic considerations. For example, AISC 341 Section F3.5b.(1) requires Complete Joint Penetration (CJP) groove welds to connect the link web to the flanges. In addition, Section F3.6a(5) classifies these CJP groove welds as demand critical. The AISC 341 demand-critical designation designation requires specific filler metals (Section A3.4b) and among other requirements, a weld procedure specification in Section I2.3. In terms of economy, in consideration of both the plate material and fabrication, the cost of the built-up shapes can be three to four times that of a rolled wide-flange section (B. Manning, personal communication). With With these considerations in mind, built-up I-shaped link beam sections are deemed necessary only where the indicated capacity ratios in Table 5–7 exceed 2.0. 2012 IBC SEAOC St ructural/Seismic Design Manual, Vol. 4

161

Design Example 5




4.4 PRELIMINARY BUIL BUILT T-UP LINK BEAM SIZE

AISC 341

As with rolled sections, the built-up I-shapes must satisfy both local buckling and geometric constraints. For beam BM-3 shown in Figure 5-3(a), the required flange width for geometric compatibility with the brace is approximately 10.0 inches. In accordance with the exception in AISC 341 Section F3.5b.(1), the required flange thickness, t f , in consideration of a moderately ductile 10.0-inch flange width, b f , is determined as follows:

=

2

= m

10 in

=

.2 9. 15)

55 in . . . say 5 ⁄ 8 in.

T D1.1 (modified)

For the same beam, assuming a short link governed by shear yielding, the required link area, Atw, is estimated in consideration of the link design shear strength, φvV n, as follows:

lw

=

6

=

77.2 kips 0 9 0.6 50

)

= 2 86 in2


Using a web thickness, t w, of 1 ⁄ 4-inch, the required depth, d , can be determined as follows:

= ( A d = Alw / t tw ) + 2t f = (2.86 in2 /0.25 in) + 2(0.625 in) = 12.69 in . . . say 13.0 in.


The preceding methodology is repeated for each targeted link in both frame configurations. The preliminary built-up link beams designations are shown in Figure 5–3, and their associated parameters are summarized in Table 5–8:

Table 5–8. Preliminary built-up link beam beam parameters

Designation

Beam Size

Depth, d (in)

Web, t w (in)

Flange b f (in)

Flange t f (in)

Area, A (in2)

Weight (plf)

Z (in3)

BM-3

BU13 × 53

13.0

0.25

10.0

0.625

15.4

53

86.0

BM-4

BU12 × 3 37 7

12.0

0.25

8.0

0.50

10.8

37

53.6

BM-5

BU9 × 34

9.0

0.25

8.0

0.50

10.0

34

38.0

Additional built-up link beam design parameters including the link shear strength, V p, moment strength, M p, and preliminary link length, e, are summarized in Table 5–9:

Table 5–9. Preliminary built-up link beam parameters and length

162

Beam Size

Shear, V p ( (k kips)

Flexure, M p ( (k kip-in)

V p (in) Length, e 1.3 M p / V

BU13 × 5 53 3

88

4299

64

BU12 × 3 37 7

83

2678

42

BU9 × 3 34 4

60

1900

42


Design Example 5




Table 5–10 shows the location and designation of the built-up link beams. In addition, the table indicates the adjusted link shear strength, V mh, and associated shear ratios. Although the inverted-V roof link beam shear ratio exceeds 2.0, given the nominal plate sizes, a further adjustment is deemed unnecessary. unnecessary. Table 5–10. Preliminary adjusted link shear strength and shear shear ratio

Inverted-V Level Beam Size

V mh (kips)

Two-Story X V mh / V Vr Ratio

Beam Size

V mh (kips)

V mh / V Vr Ratio

Roof

BU9 × 3 34 4

82.5

4.0

No Link

—

—

6th

BU9 × 3 34 4

82.5

1.6

BU12 × 3 37 7

114

1.5

5th

BU13 × 5 53 3

1.6

No Link

—

—

121

With respect to the Type 5 vertical irregularity provision outlined in the Section 2.5 base shear calculations, Tables 5–7 and 5–10 show the adjusted link shear strengths either remain the same or increase when compared with that of each story above. Therefore, for this example, there is no vertical discontinuity or weakness in the frame’s lateral strength (Type 5 vertical irregularity).

4.5 PRELIMINARY NON-LINK BEAM SIZE (2-STOR (2-STORY Y X)

AISC 341/360

As shown in the two-story X configuration of Figure 5–3(b), the beams at the third, fifth, and roof levels have no associated link. Therefore, they are classified as a beam outside of the link. In accordance with AISC 341 Section F3.5a, beams outside of the link need only satisfy the width-to-thickness limitations in Section D1.1 for moderately ductile members. For beam BM-6, shown in Figure 5–3(b), evalua evaluate te a W16 × 26 ASTM A992 wide-flange in consideration of deflection and serviceability requirements. Determine the minimum beam deep, d min, as follows: d min = L /24 = 30.0 ft(12.0 ft/in)/24 = 15 in < 15.7 in.

The beam is designed as composite (except at the roof) in accordance with AISC 360 Section I5 and is braced at the third points in accordance with AISC 341 Section D1.2a(a). USE A COMPOSITE W16 × 26 FOR NON-LINK BEAMS IN THE TWO-STOR TWO-STORY YX

4.6 PRELIMINARY BRACE SIZE

AISC 341

To derive the preliminary brace axial force, Pr , refer to the inver inverted-V ted-V brace, BR-1, shown in Figure 5–3(a). Using the beam shear and moment diagrams of Figure 5–6 and ignoring the nominal effects of gravity loads, the downward vertical brace reaction, Rmh, and the moment at the end of the link, M mh, are determined as follows:

m

M m

V m

 

L

2

  30 ft  = 172 kips = 198 kips e   30 ft − 4 ft  4 ft  = 172 kips = 344 kip-ft kip-ft.. 2  2012 IBC SEAOC St ructural/Seismic Design Manual, Vol. 4

163

Design Example 5




(L-e)/2

Vmh

M1

Mr

R1

Rmh

R1

e/2

(L-e)/2

e/2

Vmh

Rmh

V2

V2

V1

V1 Shear

Shear M1 Mmh

Moment

Moment

Mr

(a)

Mmh

(b)

Figure 5–6. Beam shear and moment diagrams (cut through Figure through the link centerline) for (a) pinconnected beam and brace connections (b) fully restrained beam and brace connections

Using the geometry shown in Figure 5–5 and the compatible beam and column depths, d b and d c, respectively respectiv ely,, the diagonal brace unbraced length, Lb, and angle, β, are calculated by trigonometry as follows:

−

L

)

 L + 

2

 = 

e−

2

.

− 1.

)

+

30

4

0 83 ft

2

= 16.7 ft . . . say 17.0 ft

β = tan−



 L − e

2



=

 an



12 ft 30 ft 2



 ft 

42 7 d g .



The required preliminary diagonal brace axial force, Pr , for the brace is calculated using trigonometry as follows: P

=

=

m

φ

β

198 kips .90 sin 42. )

= 324

ips.. ips

In this example, the brace and brace end connections are fully restrained (fixed). Therefore, the moment at the end of the link will be distribute distributed d in proportion to the relative stiffness stiffness of the diagonal brace and the beam outside of the link as shown in Figure 5–6(b). For a preliminary estimation, this example uses approximately 50 percent of the adjusted end moment for brace resistance. The exact brace end moment is subsequently confirmed using an elastic computational analysis. Therefore, the preliminary required brace flexural strength, M r , in the brace BR-1 is determined as follows: 50

m

50(344 kip-ft 09

164

191 ki -f -ftt .


Design Example 5




Using the AISC Manual Table 6–1 for combined loading with unbraced length, Lb = 17 ft, and AISC Seismic Design Manual Table 1–3 for local buckling requirements, a W10 × 77 ASTM A992 wide-flange section is deemed adequate for the preliminary diagonal brace size. In the selection process, the flange width of the brace, t bf , is confirmed geometrically compatible for attachment to the beam. The preceding methodology is repeated for each story and each frame configuration with the selected brace sizes summarized in Table 5–11.

Table 5–11. Preliminary brace sizes

Brace Size Level Inverted-V

Two-Story X

6th

W10 × 45

W10 × 45

5th

W10 × 45

W10 × 45

4th

W10 × 54

W10 × 54

3rd

W10 × 77

W10 × 68

2nd

W10 × 77

W10 × 68

1st

W10 × 77

W10 × 68

As indicated in Table 5–11, the corresponding first-story brace for the two-story X is a size smaller than that of the inverted-V frame. At this location, the correspondingly larger second-level link beam (W18 × 86) reduces the moment distribution to the brace, allowing for the reduction in size.

4.7 PRELIMINARY COLUMN SIZE

AISC 341

In accordance with AISC 341 Section F3.3(1)(b), when a column receives force from at least three links, the seismic force is permitted to be multiplied by 0.88. The reduction represents the diminished likelihood of simultaneous strain hardening. Given the configuration shown in Figure 5–3(a) for the inverted-V column C-1, the strain hardening reduction factor is applicable. As with the downward vertical reaction of the first-floor brace, the first-floor upward force, Rmh, from the second-floor link is determined as follows: Rm

V m



 L

e

= 172 kips





4 ft 30 ft − 4 ft

= 26.5 k kips (upward).

As shown in the mechanism model of Figure 5–7, the simultaneously downward column force resulting from the overturning of the levels above column C-1 is determined from the adjusted link forces at those levels. Therefore, the total first-floor required column axial load, P Emh, in consideration of the strain hardening reduction factor, is determined as follows: Pm

oo m

−

m

=

+ 82 5 +

+ 172 +

− 26.5) kips = 531 kips.


165

Design Example 5




Roof

Vmh

6th

Vmh

5th

Vmh

4th

Vmh

3rd

Vmh

2nd

Vmh Rmh

roof 3

Vmh

Figure 5–7. Column mechanism model (cut through the link centerline)

The seismic forces are combined with the appropriate proportion of gravity forces to determine the required column axial load, Pr , as follows: Pr = 1.4 D + 0.5 L + 1.0Q E = 1.4(286 kips) + 0.5(98.0 kips) + 1.0(531 kips) = 980 kips.

As mentioned previously, the applied free-body-diagram methodology is reasonable when pin-ended members are used. In this example, the beam-to-column connection is detailed as fixed, so for preliminary estimation purposes, assume approximately 15 percent of the link end moment will be distributed to the column. The exact column moment is subsequently confirmed with the elastic computational analysis. Therefore, the required column C-1 flexural strength, M Emh, in consideration of the 0.88 reduction factor, is determined as follows: 0 15 0. 88) m

166

m

0 15 0.88)344 kip-ft 09

50 4 kip-ft .


Design Example 5




Using the AISC Manual Table 6–1 for combined loading with unbraced length, Lb = 11.0 ft, and AISC Seismic Design Manual Table 1–3 for local buckling requirements, a W12 × 96 ASTM A992 wide-flange section for the preliminary column size is selected. In the selection process, the flange width of the column, t cf , is confirmed to be geometrically compatible for the beam attachment. The preceding methodology is repeated at each splice location with column sizes summarized in Table 5–12 for both frame configurations. Table 5–12. Preliminary column sizes

Level

Column Size

5th to roof

W12 × 50

3rd to 5th

W12 × 96

1st to 2nd

W12 × 96

The preliminary frame configurations with the selected sizes are shown in Figure 5–8.

3

2

C

30’-0” Roof

30’-0” W16x26

BU9x34 W12x50 W10x45

6th

5th

D

TYP.

BU9x34

W10x45

12’-0”

BU12x37

W10x45

W10x45

BU13x53

W16x26

12’-0”

W12x96 W10x54 4th

TYP.

W10x68

2nd

W10x68

W10x68

W10x77

12’-0”

W16x26

W10x68

W10x77

12’-0”

W18x86

W10x77 3rd

W10x54

W12x96 TYP.

W10x68

12’-0”

W18x86

W10x68

12’-0”

Base

(a)

(b)

Figure 5–8. Preliminary EBF member sizes: (a) inverted-V along Grid A; (b) two-Story X along Grid 1 2012 IBC SEAOC St ructural/Seismic Design Manual, Vol. 4

167

Design Example 5




4.8 ELASTIC MODELING AND STORY DRIFT DETERMINATION

ASCE 7

Using the preliminary beam, brace, and columns sizes, a two-dimensional computer model is generated for the entire frame. An elastic computational analysis of the model is used to confirm the fundamental structural period, and the link end moment distribution assumptions, and to determine the elastic deflections of the frame. From the analysis of the model, the fundamental period, T , is computed as 1.16 and 1.25 seconds for the inverted-V and the two-story X, respectively. Both computed periods exceed the maximum fundamental period, T max, confirming that it be used in the determination of the seismic response coefficient, C s. In accordance with AISC 341 Section B1, the design story drift and the allowable story drift limit are those required by the applicable building code. From the analysis of the model, the elastic deflection, δ xe, between the first and second level is computed as 0.348-inch. The allowable story drift limit, ∆a, as indicated in ASCE 7 Table 12.12–1 is calculated in consideration of the story height, hst , as follows:

∆a = 0.020hst = 0.020(12.0 ft)(12 in/ft) = 2.88 in.

T 12.12–1

In accordance with ASCE 7 Section 12.8.6, the story deflection, δ x, is calculated as follows:

x

=

−

= 1

C I

x

=

4 0(0.348 10

)

− = 1.

88 in n.

Eq 12.8–15

δ x < ∆a → DRIFT LIMIT IS SATISFIED The preceding methodology is repeated for each level and each frame configuration with the story drifts summarized in Table 5–13. Also included in Table 5–13 is the soft story vertical irregularity check.

Table 5–13. EBF Story drift and soft story parameters

Inverted-V Level

Two-Story X

Story Drift, δ x (in)

δ x /0.70 (in)

Story Drift, δ x (in)

δ x /0.70 (in)

Roof

1.04

n.a.

1.57

n.a.

6th

1.66

2.37

1.85

2.64

5th

1.88

2.69

1.66

2.37

4th

1.66

2.37

1.44

2.06

3rd

1.60

2.29

1.61

2.30

2nd

1.39

n.a.

1.09

n.a.

As indicated in Table 5–13, the story-drift analysis confirms that there is no soft-story irregularity (Type 1a) as set forth previously in Section 2.5 (i.e. δ x < δ x+1 /0.7). In accordance with ASCE 7 Section 12.3.2.2 exception 1, the top two stories need not be evaluated.

168


Design Example 5




If the story drift were to exceed allowable limits, ASCE 7 Section 12.8.6.1 provides an exception removing lower limits on the seismic response coefficient, C s. The exception as outlined in Section 12.8.6.2 allows the use of seismic design forces based on the computed fundamental period, without the upper limit specified in Section 12.8.2.

4.9 P-DELTA EFFECTS

ASCE 7/IBC

In accordance with ASCE 7 Section 12.8.7, P-delta effects shall be considered on story shears, moments, and drifts when the ratio of secondary moments to primary moments, defined as the stability coefficient, θ, is greater than 0.10. The stability coefficient is determined as follows:

=

P I V h x

≤01

.

Eq 12.8–16

Where, in accordance with Table 5–1, IBC Table 1607.1, and live load reductions specified in IBC Section 1607.10 and 1607.12, the total second floor vertical design loads are determined as follows: P2 = Σ D + Σ L = 7231 kips + [5(0.035 ksf)(15,220 ft 2) + 0.012 ksf(15,220 ft2)] = 10,077 kips.

Therefore:

=

10,077 k ips 2 88

1. 0 )

730 kips(144 in)(4 0

= 0.07 ≤ 0 1

.

Eq 12.8–16

θ ≤ 0.10 → STABILITY COEFFICIENT IS SATISFIED The preceding methodology is repeated for each level with the total unfactored design loads and the corresponding stability coefficient, θ, summarized in Table 5–14.

Table 5–14. Total unfactored design loads and stability coefficient

Level

Design Loads, P x (kips)

Stability Coefficient, θ

Roof

7413

0.05

6th

7947

0.05

5th

8480

0.06

4th

9013

0.06

3rd

9363

0.06

2nd

10,077

0.07

4.10 SECONDARY EFFECTS

ASCE 7

As the stability coefficient is less than 0.10 on all floors, in accordance with Section 12.8.7, P-delta considerations (secondary effects) are not required.


169

Design Example 5




5. Analysis Verification Member Final Design Using the preliminary selected member sizes and output from the elastic model, specific design provisions of the EBF are investigated.

5.1 LINK BEAM REQUIRED STRENGTHS

AISC 341

For this example, refer to beam BM-1 shown in Figure 5–3. As determined in the preliminary analysis, the link segment is a W10 × 68 wide-flange section. From the computational analysis, the applied loads for the beam are shown in Table 5–15.

Table 5–15. Link segment loading for BM-1

Dead, D

Live, L

Seismic, Q E

Axial, P

—

—

—

Shear, V

1.20 kips

1.00 kips

102 kips

12.4 kip-ft

7.6 kip-ft

216 kip-ft

Load

Moment, M

From the applied loads, the required axial, Pr , shear, V r , and moment design strengths, M u, can be determined using the applicable load combinations as follows: Pr = Pu = 1.4 D + 0.5 L + 1.0Q E = 0.0 kips V r = V u = 1.4 D + 0.5 L + 1.0Q E = 1.4(1.2 kips) + 0.5(1.0 kips) + 1.0(102.0 kips) = 104 kips M u = 1.4 D + 0.5 L + 1.0Q E = 1.4(12.4 kip-ft) + 0.5(7.6 kip-ft) + 1.0(216 kip-ft) = 237 kip-ft.

5.2 LINK BEAM WIDTH-TO-THICKNESS RATIOS

AISC 341

In accordance with AISC 341 Section F3.5b(1), the link shall comply with the width-to-thickness requirements of Section D1.1 for highly ductile members. However, for the flanges of short, shear dominated links there is an exception for link lengths less than 1.6 M p / V p . In this case, the flanges need only satisfy the width-to-thickness requirements for moderately ductile members. The width-to-thickness ratio for the flanges can be determined as follows: m

=

=

.38 29 00 k si/ 50 ksi =

15.

T D1.1

Because b f /2t f = 6.58 < λ md , the flanges meet the local buckling requirements. The width-to-thickness ratio for the web can be determined as follows: C a = 0.0 < 0.125 m

170

=

T D1.1 1−

.4

2


. .

T D1.1

Design Example 5




t w = 16.7 < λ hd , the web satisfies the local buckling requirements. Because ht / t

LINK BEAM WIDTH-TO-THICKNESS RATIOS ARE SATISFIED SATISFIED Alternatively and in subsequent calculations, the width-to-thickness ratios are inve Alternatively investigated stigated using AISC Seismic Design Manual Table 1–3.

5.3 LINK BEAM STRENGTH DESIGN

AISC 341

Per AISC 341 Section F3.5b(2), the nominal shear, V p, and plastic flexural strength, M p, are determined in consideration of the required axial strength as follows: Pc = P y = F y Ag = 50 ksi(19.9 in 2) = 995 kips

Eq F3–6

Pr / Pc = 0.0 < 0.15.

Therefore, use V p and M p as determined in the preliminary calculations. calculations. In accordance with Section F3.5b(2), the link design shear strength, φvV n, shall be the lower of the value obtained in accordance with the limit states of shear yielding in the web and flexural yielding in the gross section. Because for this example V p < M p, shear yielding of the link controls and the design shear strength is determined as follows:

φvV n = φvV p = 0.9(125 kips) = 113 > 104 kips.

Eq F3–1

φvV n > V r , SHEAR YIELDING CONTROLS 5.4 LINK ROT ROTA ATION ANGLE LIMITA LIMITATION TION

AISC 341

The link rotation angle is the primary variable used to describe link inelastic deformation. AISC 341 Figure C-F3.4 defines the link rotation angle as the inelastic angle between the link and the beam outside of the link under design story drift conditions. In accordance with Section F3.4a, the link rotation angle shall not V p or less and 0.02 radians for link lengths of 2.6 M p / V V p or exceed 0.08 radians for link lengths of 1.6 M p / V greater.. Linear interpolation is required for link lengths between the two limits. As previously determined, greater the second-floor link length for this example is approximately 1.3 M p / V V p . Therefore, the link rotation angle, γ p, in consideration of the plastic story drift angle, θ p, is determined as follows:

γ

L

where: θ p = ∆ p / hst .

F C-F3.4

In accordance with Section F3.3, the inelastic link rotation angle is determined from the inelastic portion of the design story drift as follows:

∆ p = δ x − δ xe = 1.39 in − 0.348 in = 1.04 in. Therefore,

θ p = 1.04 in/144 in = .007 rad γ p = (30 ft/4 ft)(0.007 rad) = 0.05 < 0.08 rad. γ p < 0.08 rad → LINK ROTATION IS SATISFIED 2012 IBC SEAOC St ructural/Seismic Design Manual, Vol. 4

171

Design Example 5




5.5 BEAM OUTSIDE OF THE LINK REQUIRED STRENGTHS

AISC 341/360

According to AISC 341 Section F3.3, the required strength of beam outside of the link must be “adjusted” for both material overstrength and strain hardening. Howeve However, r, with respect to this segment, the adjusted shear strength is allowed to be taken as 0.88 times the force to account for the increased strength provided by a composite slab and recognizing the fact that limited yielding is unlikely to be detrimental to frame behavior.. If there is not a concrete composite slab, a strain-hardening factor of 1.25 should be used as behavior recommended by Section C-F3.3. As discussed in the overview, overview, for this example, the beam outside of the link is composite with the concrete slab. Additional lateral bracing along the length of the beam, if required, is designed per AISC Specification Appendix 6. If the beam outside of the link is a different section than the link, then it must also satisfy the width-to-thicknesss requirements. width-to-thicknes For this example, refer to beam BM-1 shown in Figure 5–3. As determined in the preliminary analysis, the beam is a W10 × 68 wide-flange section. From the computational analysis, the applied loads for the beam are shown in Table 5–16.

Table 5–16. Beam outside of the link loading for BM-1

Load

Dead, D

Live, L

Seismic, Q E

Axial, P

—

—

138 kips

Shear, V

7.80 kips

6.70 kips

11.6 kips

11.6 kip-ft

7.60 kip-ft

115 kip-ft

Moment, M

In accordance with Section F3.3, the “adjusted” required shear strength, V mh, is calculated as follows: V mh = 0.88(1.25) R R yV p = 0.88(1.25)(1.1)(125 kips) = 151 kips.

As outlined in the overview, overview, the brace-to-beam connection is detailed as a fully restrained connection. Using the methodology described in AISC Seismic Design Manual Example 5.4.3, the adjusted link end moment will be distributed into the beam outside of the link and the diagonal brace based on amplification factors. The amplification factor, α, is determined by dividing the adjusted shear strength, V mh, by the required link shear force, V QE , reported in Ta Table ble 5–15 as follows:

α = V mh / V VQE = 151 kips/102 kips = 1.48. The seismic axial, shear, and moment forces in the beam outside of the link are multiplied by the amplification factor as follows: P Emh = αPQE = 1.48(138 kips) = 204 kips V Emh = αV QE = 1.48(11.6 kips) = 17.2 kips M Emh = α M QE = 1.48(115 kip-ft) = 170 kip-ft.

172


Design Example 5




From the applied loads, the required axial, Pr , shear, V r , and moment design strengths, M u, can be determined using the applicable load combinations as follows: Pr = Pu = 1.4 D + 0.5 L + 1.0Q E = 1.4(0.0 kips) + 0.5(0.0 kips) + 1.0(204 kips) = 204 kips V r = V u = 1.4 D + 0.5 L + 1.0Q E = 1.4(7.80 kips) + 0.5(6.70 kips) + 1.0(17.2 kips) = 31.5 kips M r = M u = 1.4 D + 0.5 L + 1.0Q E = 1.4(11.6 kip-ft) + 0.5(7.60 kip-ft) + 1.0(170 kip-ft) = 190 kip-ft.

5.6 BEAM OUTSIDE OF THE LINK WIDTH-TO-THICKNESS RA RATIOS TIOS

AISC 341

Because the beam outside of the link is the same section as the link, no additional local buckling checks are required. BEAM OUTSIDE OF THE LINK WIDTH-TO-THICKNESS RATIOS RA TIOS ARE SA SATISFIED TISFIED

5.7 BEAM OUTSIDE OF THE LINK STRENGTH DESIGN

AISC 341/360

Due to the large axial force and bending moment, the beam outside the link is designed as a beam-column in accordance with the user note of AISC 341 Section F3.5a. The beam outside of the link will be braced at the column and at the link. Therefore, the unbraced length, Lb, in consideration of the column depth, d c, is determined as follows: Lb = ( L L − e − d c)/2 = [30 ft(12 in/ft) − 48 in − 12.7 in]/2 = 150 in.

For the W10 × 68 beam, in accordance with AISC 360 Section F2.2, the limiting yielding unbraced length, L p, and the limiting lateral-torsional buckling unbraced length, Lr , in consideration that the coefficient, c = 1, are determined as follows: L p

y

r t

F

= . 76 76(2 59 in)

E

Jc

0 7F

S h

+



29 00 ksi 50 ksi

Jc



S h

= 110 in

Eq F2–5

2

+ 6 76

 0 7 F y  

E

Eq F2–6



where r ts =

I yC w

=

134 in4

x

75

n ) in

2 92 in .

3

Eq F2–7

Therefore: 1 95 2.92

)

29,000 ksi 7(50 ksi

3 56 75 7

1. ) (9 63

)

+

3 56 75.7

2



1.0 ) 9.63

)

+ 6 76

7(50 ksi

2

Eq F2–6

29 000 si

Lr = 487 in

Alternatively and in any subsequent calculations, L p and Lr , are determined from the AISC Manual Alternatively Table 3–2, as follows: L p = 9.15 ft(12 in/ft) = 110 in

Manual T 3–2

Lr = 40.6 ft(12 in/ft) = 487 in

Manual T 3–2 2012 IBC SEAOC St ructural/Seismic Design Manual, Vol. 4

173

Design Example 5




In accordance with Section C3, the effecti effective ve length factor, K , shall be taken as unity. The member slenderness is determined in accordance with AISC 360 Section E2 as follows: KL / r r y = 1.0(150 in)/2.59 in = 57.9 < 200.

§E2

In accordance with Section E3, the elastic buckling stress, F e, is determined as follows: /(KL / r F e = π2 E r) 2 = π2(29,000 ksi)/(57.9) 2 = 85.4 ksi.

Eq E3–4

In accordance with Section E3, if the yield to elastic buckling stress ratio is less than or equal to 2.25, the column buckling behavior is considered elastic. The stress ratio is determined as follows: F y / F Fe = 50 ksi/85.4 ksi = 0.58 < 2.25.

The critical stress, F cr , is determined as follows: Fe ) F cr = [0.658(F y / F ]F y = (0.6580.58)50 ksi = 39.2 ksi.

Eq E3–2

Alternatively and in any subsequent calculations, F cr , is determined from the AISC Manual Table 4–22, as Alternatively follows: F cr = 35.3 ksi/ φc = 35.3 ksi/0.9 = 39.2 ksi.

Manual T 4–22

The nominal compressive strength, Pn, is determined as follows: Pn = F cr Ag = 39.2 ksi(19.9 in 2) = 780 kips.

Eq E3–1

In accordance with AISC 341 Section A3.2, the strength of the beam outside of the link can be increased by the expected yield stress ratio, R y, given the link and the beam are the same member. In accordance with AISC 360 Section H1.1, the design axial strength of the beam, Pc, is determined as follows: Pc = R yφcPn = 1.1(0.9)(780 kips) = 772 kips

§H1.1

Pr / Pc = 204 kips/772 kips = 0.26.

§H1.1

In accordance with AISC 360 Section F2.2, when L p < Lb < Lr , then the nominal flexural strength, M n, in consideration that the lateral-torsion lateral-torsional al buckling modification factor factor,, C b = 1.0, is determined as follows:



−

.7

)

 L

p

   ≤

Eq F2–2

where M p = F y Z x = 50 ksi(85.3 in 3) = 4265 kip-in.

Eq F2–1

Therefore:

=

− [4265 kip-in − 0 7(50 ksi

75 7 in 3 ]

M n = 4094 kip-in.

174


150 in − 110 in 487 in 110 in



Eq F2–2

Design Example 5




In accordance with Section H1.1, the design flexural strength, M c, is determined as follows: M c = R yφb M n = 1.1(0.9)(4094 kip-in)(1.0 ft/12 in) = 338 kip-ft.

§H1.1

Because Pr / Pc > 0.2, the combined beam strength is limited as follows: Pr / Pc + 8/9( M M r / M c) = 0.26 + 8/9(190 kip-ft/338 kip-ft) = 0.76 < 1.0.

Eq H1–1a

BEAM OUTSIDE OF THE LINK C OMBINED FLEXURE AND COMPRESSION STRENGTH STREN GTH RA R ATIOS ARE SA SATISFIED TISFIED Alternatively and in subsequent calculations the combined strength equations are determined using the Alternatively AISC Manual Table 6–1, with Lb = 12 feet and R y = 1.1, as follows: p = 1.40 × 10−3 (kips)−1 /1.1 = 1.27 × 10−3 kips−1

Manual T 6–1

b x = 2.88 × 10−3 (kip-ft)−1 /1.1 = 2.62 × 10−3 (kip-ft) −1

Manual T 6–1

Pr / Pc = pPr = 1.27 × 10−3 kips−1(204 kips) = 0.26.

Because pPr > 0.2, the combined strength is limited by AISC Manual Equation 6–1 as follows: pPr + b x M rx = 0.26 + 2.62 × 10−3(190 kip-ft) = 0.76 < 1.0.

5.8 DIAGONAL BRACE REQUIRED STRENGTHS

Manual Eq 6–1

AISC 314

For this example, refer to beam BR-1 shown in Figure 5–3. As determined in the preliminary analysis, the brace is a W10 × 77 wide-flange section. From the computational analysis, the applied loads for the brace are shown in Table 5–17.

Table 5–17. Diagonal brace loading for BR-1 BR-1

Load

Dead, D

Live, L

Seismic, Q E

Axial, P

10.8 kips

8.30 kips

182 kips

Shear, V

1.00 kips

0.50 kips

5.00 kips

4.10 kip-ft

3.20 kip-ft

101 kip-ft

Moment, M

As with the beam outside of the link, in accordance with AISC 341 Section F3.3, the required strength of brace is “adjusted” for both material overstrength overstrength and strain hardening. The adjusted required shear strength, V mh, is calculated as follows: V mh = 1.25 R yV p = 1.25(1.1)(125 kips) = 172 kips.

Using the amplification method described previously previously,, the factor α is determined by dividing the adjusted shear strength, V mh, by the required link shear force, V QE , reported in Ta Table ble 5–15 as follows:

α = V mh / V VQE = 172 kips/102 kips = 1.69. 2012 IBC SEAOC St ructural/Seismic Design Manual, Vol. 4

175

Design Example 5




The seismic axial, shear, and moment forces in the brace are multiplied by the amplification factor as follows: P Emh = αPQE = 1.69(182 kips) = 308 kips V Emh = αV QE = 1.69(5.00 kips) = 8.45 kips M Emh = α M QE = 1.69(101 kip-ft) = 171 kip-ft.

From the applied loads, the required axial, Pr , shear, V r , and moment design strengths, M u, can be determined using the applicable load combinations as follows: Pr = Pu = 1.4 D + 0.5 L + 1.0Q E = 1.4(10.8 kips) + 0.5(8.3 kips) + 1.0(308 kips) = 327 kips V r = V u = 1.4 D + 0.5 L + 1.0Q E = 1.4(1.0 kips) + 0.5(1.0 kips) + 1.0(8.45 kips) = 10.4 kips M u = 1.4 D + 0.5 L + 1.0Q E = 1.4(4.10 kip-ft) + 0.5(3.20 kip-ft) + 1.0(171 kip-ft) = 178 kip-ft.

5.9 DIAGONAL BRACE WIDTH-TO-THICKNESS RA RATIOS TIOS

AISC 341

In accordance with AISC 341 Section F3.5a, EBF braces shall comply with the width-to-thickness requirements of Section D1.1 for moderately ductile members. As indicated in AISC Seismic Design Manual Table 1–3, the W10 × 77 satisfies brace local buckling requirements. BRACE WIDTH-TO-THICKNESS RATIOS ARE SATISFIED

5.10 DIAGONAL BRACE STRENGTH DESIGN

AISC 341/360

As with the beam outside the link, the diagonal brace is designed as a beam-column in accordance with the user note of AISC 341 Section F3.5a. As mentioned previously, previously, the brace is hinged at the column and fixed at the link. The combined strength equations are determined using the AISC Manual Table 6–1, with Lb = 17 feet (as determined in the preliminary design), as follows: p = 1.54 × 10−3 kips−1

Manual T 6–1

b x = 2.65 × 10−3 (kip-ft)−1

Manual T 6–1

Pr / Pc = pPr = 1.54 × 10−3 kips−1 (327 kips) = 0.50.

Because pPr > 0.2, the combined strength is limited by AISC Manual Equation 6–1 as follows: pPr + b x M rx = 0.50 + 2.65 × 10−3 (kip-ft)−1 (178 kip-ft) = 0.97 < 1.0.

Manual Eq 6–1

BRACE COMBINED FLEXURE AND COMPRESSION STRENGTH RATIOS ARE SATISFIED In accordance with AISC 360 Section G2.1, the brace shear strength is determined in consideration of the web width-to-thickness ratio. The width-to-thickness ratio is determined as follows: h / t tw ≤ 2.24( E E F y )0.5 = 2.24(29,000 ksi/50 ksi)0.5 = 53.9 > 14.8. / F

176


§G2.1

Design Example 5




Therefore, with the web shear coefficient, C v = 1.0, the design shear strength, φvV n, is determined as follows:

φvV n = φv0.6F y AwC v = 1.0(0.6)(50 ksi)(10.6 in)(0.53 in)(1.0) = 169 kips > 10.4 kips.

Eq G2–1

φvV n > Pr , SHEAR STRENGTH IS SATISFIED Alternatively, the design shear strength, φvV n, is determined using the AISC Manual Table 3–6 as follows:

φvV n = 169 kips > 10.4 kips.

Manual T 3–6

5.11 COLUMN REQUIRED STRENGTHS

AISC 360

For this example, refer to column C-1 shown in Figure 5–3(a). As determined in the preliminary analysis, the column is a W12 × 96 wide-flange section. From the computational analysis, the applied loads for the column are shown in Table 5–18.

Table 5–18. Column loading for C-1

Load

Dead, D

Live, L

Seismic, Q E

Axial, P

286 kips

98.0 kips

339 kips

Shear, V

1.00 kips

0.120 kips

3.0 kips

4.20 kip-ft

2.20 kip-ft

38.0 kip-ft

Moment, M

As with the beam and brace, in accordance with AISC 341 Section F3.3, the column strength must resist the forces generated by the sum of the adjusted link shear strengths. In Section 4.7 of the preliminary design, the adjusted column required axial load, P Emh, was determined to be 531 kips. The computational analysis confirmed that 15 percent of the link end moment, M Emh, or 50.4 kip-ft is distributed to the column. From the applied loads in axial compression, the required axial, Pr , shear, V r , and moment design strengths, M u, can be determined using the applicable load combinations as follows: Pr = 1.4 D + 0.5 L + 1.0Q E = 1.4(286 kips) + 0.5(98.0 kips) + 1.0(531 kips) = 980 kips V r = V u = 1.4 D + 0.5 L + 1.0Q E = 1.4(1.0 kips) + 0.5(0.12 kips) + 1.0(3.0 kips) = 4.50 kips M u = 1.4 D + 0.5 L + 1.0Q E = 1.4(4.2 kip-ft) + 0.5(2.2 kip-ft) + 1.0(50.4 kip-ft) = 57.4 kip-ft.

From applied loads in axial tension, the required axial, Pr , design strength can be determined using the applicable load combination as follows: Pr = 0.7 D + 1.0Q E = 0.7(286 kips) + 1.0(−531 kips) = −331 kips.

By inspection, the governing load combination is that for axial compression. However, foundation and anchorage design must also consider the applied loads in axial tension.


177

Design Example 5




5.12 COLUMN WIDTH-TO-THICKNESS RATIOS

AISC 341

In accordance with AISC 341 Section F3.5a, EBF columns shall comply with the width-to-thickness requirements of Section D1.1 for highly ductile members. As indicated in AISC Seismic Design Manual Table 1–3, the W12 × 96 satisfies column local buckling requirements. COLUMN WIDTH-TO-THICKNESS RATIOS ARE SATISFIED

5.13 COLUMN STRENGTH DESIGN

AISC 360

As mentioned previously, the column is idealized as a pin base. The combined strength equations for the W12 × 96 column are determined using the AISC Manual Table 6–1 ( Lb = 11.0 ft), as follows: p = 0.901 × 10−3 kips−1

Manual T 6–1

b x = 1.61 × 10−3 (kip-ft)−1

Manual T 6–1

Pr / Pc = pPr = 0.901 × 10−3 kips−1 (980 kips) = 0.88.

Because pPr > 0.2, the combined strength is limited by AISC Manual Equation 6–1 as follows: pPr + b x M rx = 0.88 + 1.61 × 10−3 (kip-ft)−1 (57.4 kip-ft) = 0.97 < 1.0.

Manual Eq 6–1

COLUMN COMBINED FLEXURE AND COMPRESSION STRENGTH RATIOS ARE SATISFIED

6. Design and Detailing of Connections 6.1 LINK END STIFFENER REQUIREMENTS

AISC 341

In accordance with AISC 341 Section F3.5b(4), full-depth, double-sided web stiffeners are required at the diagonal brace ends of each link. The minimum required width, wmin, of each end stiffener is determined as follows: wmin = (b f − 2t w)/2 = [10.1 in − 2(0.47 in)]/2 = 4.58 in.

§F3.5b(4)

The minimum required thickness, t min, of each end stiffener is determined as follows: t min = 0.75t w ≥ 3 ⁄ 8 in = 0.75(0.47 in) = 0.35 in < 3 ⁄ 8 in.

PROVIDE FULL-DEPTH, 43 ⁄ 4-INCH-WIDE, DOUBLE-SIDED END STIFFENERS In order to simplify detailing and fabrication, use the same stiffener thickness as required for the intermediate stiffeners.

178


§F3.5b(4)

Design Example 5




6.2 LINK INTERMEDIATE STIFFENER REQUIREMENTS

AISC 341

In accordance with AISC 341 Section F3.5b(4), where e > 5 M p / V p , full-depth intermediate (within the link) web stiffeners are not required. For all other conditions, intermediate stiffeners are required, spaced at intervals in accordance with Section F3.5b(4)(a through c). For this example, with e < 1.6 M p / V p , the required intermediate stiffener spacing, s, for a link rotation angle, γ p, of 0.08 rad, is determined as follows: s = (30t w − d /5) = 30(0.47 in) − (10.4 in/5) = 12.0 in.

§F3.5b(4)(a)

For a link rotation angle of 0.02 rad or less, the required spacing is determined as follows: s = (52t w − d /5) = 52(0.47 in) – (10.4 in/5) = 22.4 in.

§F3.5b(4)(a)

Interpolating between these limits using the calculated link rotation angle of 0.05 rad, the maximum spacing between stiffeners is 17.2 inches. For links where d ≤ 25.0 inches, the stiffeners are required only on one side of the link web. Where d > 25.0 inches, the stiffeners are required on both sides of the web. The minimum required width, wmin, of each intermediate stiffener is determined as follows: wmin = b f /2 − t w = (10.1 in/2) − 0.47 in = 4.58 in.

§F3.5b(4)

The minimum required thickness, t min, of each intermediate stiffener is determined as follows: t min = t w ≥ 3 ⁄ 8 in = 0.47 in.

§F3.5b(4)

For simplification purposes, both the end and intermediate stiffeners will use ½-inch-thick material. PROVIDE FULL-DEPTH, 43 ⁄ 4-INCH-WIDE × 1 ⁄ 2-INCHTHICK, SINGLE-SIDED STIFFENERS SPACED AT 17 INCHES ON CENTER

6.3 STIFFENER WELD REQUIREMENTS

AISC 341/AISC 360/AWS D1.8

With respect to the fillet welds connecting link stiffeners, AISC 341 Section F3.5b(4) requires the strength in proportion to the horizontal cross-sectional area of the link stiffener, Ast . For welds connecting the stiffener to the link web, the required strength is F y Ast . For welds connecting the stiffener to the link flange, the required strength is F y Ast /4. In accordance with AISC 360 Section J2.4, the connecting fillet weld strength, φ Rn, is determined, in consideration of the weld size (in sixteenths of an inch), D, and weld length, l, as follows:

φ Rn = φ0.60F EXX (0.707) Dl /16 = 0.75(0.60)(70 ksi)(0.707) Dl /16 = 1.39 Dl kip-in. Eq J2–4 (Modified) Alternatively, and in subsequent calculations, the fillet weld strength is determined using the AISC Manual Part 8 as follows:

φ Rn = 1.39 Dl kip-in.

Manual Eq 8–2a


179

Design Example 5




In accordance with AISC 341 Section C-F3.5b(4), welds in the k-area of the beam should be avoided. Appropriately detailed stiffener corner clips avoid the k-area. In accordance with AWS D1.8 Clause 4.1.1, the required minimum corner clip length along the link w eb, lcw, is determined as follows: lcw min = 1.5 in + k des − t f = 1.5 in + 1.27 in − 0.77 in = 2.0 in.

AWS D1.8 Clause 4.1.1

In accordance with AWS D1.8 Clause 4.1.2, the required maximum corner clip length along the link flange, lcf , is determined as follows: lcf max = k 1 − 0.50t w + 1 ⁄ 2 in = 0.88 in − 0.50(0.47 in) + 0.50 in = 1.14 in.

AWS D1.8 Clause 4.1.2

Based on the AWS recommendations, 2.0-inch and 1.0-inch-wide clips are provided along the w eb and flange, respectively. The link stiffener horizontal cross-sectional area, Ast , in consideration of the flange clip is determined as follows: Ast = (w − 1.0 in)t = (4.75 in − 1.0 in)(0.50 in) = 1.88 in2.

§F3.5b(4)

The minimum thickness of double-sided fillet welds connecting the stiffener to the link web and flange are determined as follows:

Web: Dmin Flange:

=

F

2 .39 ki -in )

=

50 ksi .88 in ) 2 .39 ki -in [10. in

F mn

( .

in

4.

50 ksi .88 in )

4 )( .39 ki -in

4 )(1 3

4.75 in − 1 0

)

n

= 6 96 sixt enths

2 25 sixteenths .

In accordance with AISC 360 Table J2.4, the minimum fillet weld size for a 1 ⁄ 2 inch thick plate is 3 ⁄ 16 inch. PROVIDE DOUBLE-SIDED, 1 ⁄ 2-INCH and 3 ⁄ 4-INCH FILLET WELDS ALONG THE LINK WEB AND FLANGE, RESPECTIVELY

6.4 BRACE-TO-LINK CONNECTION

AISC 360

For this example, refer to joint J-1 shown in Figure 5–4. As determined previously, the applied loads for the connection are shown in Table 5–19.

Table 5–19. Diagonal brace connection loading for Joint J-1

Load

180

Required Strength

Axial, Pu

327 kips

Shear, V u

10.4 kips

Moment, M u

178 kip-ft


Design Example 5




As discussed in the overview, the diagonal braces are connected to the link with a fully restrained bolted flange plate (BFP) moment connection. The BFP moment connection is prequalified per AISC 358 Chapter 7. In accordance with AISC 360 Section F13.1, the flanges of the brace shall have adequate strength to avoid the limit state of tensile rupture. The brace tension flange gross area, A fg, is determined as follows: A fg = b f (t f ) = 10.2 in(0.87 in) = 8.87 in2

§B4.3a

Since F y / Fu = 50/65 = 0.77 < 0.8, then Y t = 1.0. Using two rows of bolts, the maximum hole size, d max, is determined as follows:



 − g

y

g

  

1.0 (

8 87 in

F

8.87 in 65 ksi

2

2 .87 in)

 i .

§F13.1(a)

Therefore, a bolt nominal diameter, d b, of 1.0-inch with a standard hole is used. The maximum hole size, in consideration of the hole diameter, d h, is determined as follows: d max = d h + 1 ⁄ 16 in = 11 ⁄ 16 in + 1 ⁄ 16 in = 1.13 in < 1.17 in.

Table J3.3

In accordance with the Manual Table 7–1, a nominal bolt area, Ab, of 0.785 square inches is used. In accordance with the preferred spacing requirements of AISC 360 Section J3.3, a longitudinal center-tocenter bolt spacing (pitch), s, of 3d b is used. Two rows of bolts are used with a 4.0-inch transverse center-tocenter spacing (gage), g, between rows. In accordance with Section J3.10, a longitudinal bolt edge distance, Lc, of 2d b is used. Finally, in order to provide installation clearance, from the face of the link beam flange to the first row of bolts, a 7.0-inch clear distance, S 1, is used. In consideration of brace and beam flange compatibility, a flange plate width, b fp, of 10.0-inch is used. The flange plate is full penetration welded to the beam and must be less than or equal to the flange width. In consideration of the brace flange thickness, a flange plate thickness, t p, of 1.0-inch is used. In this case, it is preferable that the plate thickness is at least greater than or equal to the thickness of the brace flange. PROVIDE 10.0-INCH-WIDE, 1.0-INCH-THICK BOLTED FLANGE PLATES The flange plates resist both axial and moment loading. The required flange plate strength due to the axial load, P fa, and moment, P ff , are determined as follows: Axial: P fa = Pu /2 = 327 kips/2 = 164 kips Moment:

M t

178 kip-ft(12 in/ t 10.

.0 in

1 4 ki .

The corresponding total required flange plate strength, P f , is determined as follows: P f = P fa + P ff = 164 kips + 184 kips = 348 kips.


181

Design Example 5




In accordance with Section J3.6 and J3.10, the controlling shear strength, φ Rn, per bolt considering bolt shear and bearing is determined as the minimum of the following:

φ Rn = φ1.0F nv Ab = 0.75(1.0)(84 ksi)(0.785 in 2) = 49.5 kips

Eq J3–1

φ Rn = φ2.4F ud bt f = 0.75(2.4)(65 ksi)(1.0 in)(0.87 in) = 102 kips

Eq J3–6a

φ Rn = φ2.4F ud bt p = 0.75(2.4)(65 ksi)(1.0 in)(1.0 in) = 117 kips.

Eq J3–6a

Therefore, the number of bolts can be determined as follows: n=

P Rn

=

348 kips 49.5 ips

= 7 03 . PROVIDE EIGHT 1-INCH-DIAMETER A490X BOLTS WITH STANDARD HOLES

In accordance with Section D3, the net effective area of the flange plate, Ae, in consideration of a shear lag factor, U = 1.0, is determined as follows: Ae = AnU = t p[b fp − 2(d h + 1 ⁄ 16 in)]U = 1.0 in[10.0 in − 2(11 ⁄ 16 in + 1 ⁄ 16 in)](1.0) = 7.75 in2.

Eq D3–1

In accordance with Section J4.1, the trial flange plate thickness for tensile yielding and tensile rupture is confirmed as follows:

Tensile Yielding:

Tensile Rupture:

= min

P F P

mn

=

348 kips 0. (50 ksi 10 i ) 348 kips( .0

p

A

= 0 77 in.

0.

65

7 75 in 2

0 92 i .

Eq J4–1

Eq J4–2

With respect to block shear in the W10 × 77 brace flange, in accordance with Section J4.3, the gross and net areas subject to shear, Agv, Anv, respectively, and the net area subject to tension, Ant , are determined as follows: Agv = 2[S 1 + (0.5n − 1)s]t f = 2{7.0 in + [0.5(8) − 1]3 in}(0.87 in) = 27.8 in2

§J4.3

Anv = 2[S 1 + (0.5n − 1)s − (0.5n − 0.5)(d h + 1 ⁄ 16)]t f

= 2{7.0 in + [0.5(8) − 1]3 in − [0.5(8) − 0.5](1.13 in)}(0.87 in) = 21.0 in2 Ant = [g − (d h + 1 ⁄ 16)]t f = [4 in − (1.13 in)](0.87 in) = 2.50 in2.

§J4.3 §J4.3

For the limit state of brace flange block shear rupture, the available strength, φ Rn, for the limit state of brace flange block shear rupture in consideration of a uniform tension stress, U bs = 1, is determined as follows:

φ Rn = φ(0.60F u Anv + U bsF u Ant ) ≤ φ(0.60F y Agv + U bsF ut Ant ) = 0.75[0.60(65 ksi)(21.0 in2) + 1.0(65 ksi)(2.50 in2)] = 736 kips ≤ 0.75[0.60(50 ksi)(27.8 in2) + 1.0(65 ksi)(2.50 in2)] = 747 kips φ Rn = 736 kips > 348 kips.

182


Eq J4–5

Design Example 5




Although not shown, some the preceding methodology is repeated to determine the block shear rupture strength in the 1.0-inch-thick flange plates. For this example, the flange plate block shear rupture strength is assumed to be more than adequate. BLOCK SHEAR RUPTURE STRENGTHS ARE SATISFIED In consideration of the flange plate slenderness, the radius of gyration, r , is determined as follows: 1 0 in 12

0 2 in.

12

In accordance with Section E2, the member slenderness, KL / r , using an effective length factor, K = 0.65, and a lateral unbraced length, L = S 1, is determined as follows: KL / r = 0.65(7.0 in)/0.29 in = 15.7 < 25.

§E2

In accordance with Section J4.4, the available strength, φPn, for the limit state of compression yielding is determined as follows:

φPn = φF y b fpt p = 0.90(50 ksi)(10.0 in)(1.0 in) = 450 kips > 348 kips.

Eq J4–6

FLANGE PLATE COMPRESSION STRENGTH IS SATISFIED The vertical component of the flange force, V f , is determined as follows: V f = P f (sin β) = 348 kips(sin 42.7) = 236 kips.

In accordance with Section J10.2, the web local yielding and crippling strengths of the beam at the flange plate connection are determined as follows: Yielding: φ Rn = φF y t w(5k + lb) = 1.0(50 ksi)(0.47 in)[5(1.27 in) + 1.27 in]

Eq J10–2

= 179 kips < 236 kips

 +3

Crippling:

 =

15   y w       1 27 in  +3  10.4 in  

Eq J10–4 47 in 77 in

15

 

29 000 ksi(50 ksi)( 0 77 in) 0 47 in

= 240 kips > 236 kips. As indicated, the web local yielding strength is insufficient to satisfy the required flange-.plate strength. Therefore, web stiffeners are required adjacent to the flange plates. The component portion of the flangeplate force directed through the stiffener is determined as follows: Ps = 0.50(V f − φ Rn) = 0.50(236 kips − 179 kips) = 28.5 kips.


183

Design Example 5




As determined previously, the link end stiffener width is 4.75-inches. In consideration of 1.0-inch-wide stiffener corner clips, the minimum thickness of the stiffener to satisfy the required local yielding strength is determined as follows:

min

P

=

F

=

28 5 ki

− 1 0 in)

0. (50 si 4 75

0.17 in

0 5 in .

The minimum thickness of double-sided fillet welds connecting the stiffener to the link web and flange are determined as follows: 28.5 kips

Web:

mn

2 .3 ki -in

Flange: Dmin

=

2 .3 ki -in 1 4

Ps

=

2 .39 kip-in

4. in

28 5 kips 2 .39 kip-in )(3.75 in)

=

=

8 00 sixt

h

.00 sixteenths .

LINK END STIFFENERS AND ASSOCIATED WELDS SATISFY REQUIRED STRENGTHS Although not shown, some the preceding methodology is repeated to determine the bolted brace shear tab attachment. For this example, two A325X bolts are determined adequate for the shear tab-to-brace web connection. In addition, a double-sided, 1 ⁄ 4-inch fillet weld is determined adequate for the shear tab-to-beam flange connection. The final brace-to-beam connection geometry is shown in Figure 5–9.

Wide flange beam

PL ”x4 ” full depth stiffeners each side of web

Link, e

PL ”x4 ” full depth intermediate stiffeners at 17” on center max., near side side of web

Link, eeff

Beam centerline typ.

3/16 3/16

1/2 1/2

typ.

PL ”x4”x0’-6”with (2)-1” diameter A325X bolts (bolts not shown)

typ. 1/4

Shims, if required

Brace centerline

1/4

PL 1”x10”x1’-6” with (8)-1” diameter A490X bolts, both sides, (bolts not shown) Wide flange brace Figure 5–9. Brace-to-beam connection, J-1, at the link

184


Link centerline

Design Example 5




6.5 FINAL LINK DESIGN CHECK

AISC 341

With the design of the brace-to-beam connection complete, the effective link length, eeff , can be distinguished from that used in the preliminary design. As shown in Figure 5–9, the preliminary link length, e, is based on the frame centerline offset and represents the distance between the intersection of the brace and beam centerlines. The effective link length represents the distance between the brace connections measured from the edge of the end stiffener. For beam BM-1, shown in Figure 5–3a, the effective link length, eeff , is determined as follows:

e

+

(

)

ta

−

t p

si

=

+

10.4 in tan 4 .

10.

(1. sin 4 .

)

40

in

Say 41 in.

The preceding methodology is repeated for each link for both frame configurations with the preliminary link sizes, e, effective link sizes, eeff , and calculated link capacity ratios summarized in Table 5–20.

Table 5–20. Preliminary link lengths, effective link lengths, and link capacity ratios

Inverted-V

Two-Story X

Level e (in)

eeff (in)

Roof

42

35

6th

42

5th

e

e (in)

eeff (in)

1.1

—

—

No Link

35

1.1

42

38

1.1

64

61

1.2

—

—

No Link

4th

48

41

1.2

50

52

1.3

3rd

48

41

1.2

—

—

No Link

2nd

48

41

1.2

50

52

1.3

As indicated in Table 5–20, the link capacity ratios are all less than 1.6. In accordance with AISC 341 Section F3.3, a ratio less than 1.6 indicates shear yielding will dominate the inelastic response. If the links were subject to flexural yielding, the required flexural strength would be calculated at the link end rather than at the centerline intersection. As all the links are shear-governed, no further flexural redesign considerations are required. Commentary Section C-F3.5b further discusses considerations associated with the centerline model analysis.

6.6 ALTERNATIVE BRACE-TO-LINK CONNECTIONS

AISC 341

Although this example uses a BFP, there are other options to provide a fully restrained brace-to-link connection. Alternatives for other connections include a Welded Flange Plate (WFP) or a Welded Unreinforced Flange (WUF). Unlike the BFP, a WFP connection is not prequalified per AISC 358. In addition, the SEAOC Seismology Committee, FEMA 350 Task Group has documented a number of potential concerns with the connection’s inelastic performance (SEAOC, 2002). 2012 IBC SEAOC St ructural/Seismic Design Manual, Vol. 4

185

Design Example 5




With respect to a WUF connection, geometric compatibility for the CJP groove welds requires that the linkbeam flange width be greater than or equal to that of the brace. In many cases, including this example, to provide adequate flange width, the link beams would either require oversized rolled wide-flanges or custom built-up I-shaped sections. Both possibilities would have significant economic implications. The CJP groove welding associated with a WUF connection also presents an overhead field-weld challenge that most fabricators would preferably avoid (B. Manning, personal communication). In some cases, to avoid this condition, a short segment of the brace can be shop welded to the link and subsequently field spliced (Bruneau, Uang, and Sabelli, 2011). With respect to WUF connection inelastic performance, several notable link connection fractures occurred in the 2010 to 2011 Christchurch series of earthquakes. At the time of this publication, the exact cause of the fractures was still being investigated (Clifton et al., 2011).

6.7 BEAM-TO-COLUMN CONNECTION

AISC 341

As indicated in Design Example 1, beam-to-column connections in a Special Moment Frame (SMF) must satisfy beam plastic hinge strength requirements, because the hinge serves as the frame’s structural fuse. In accordance with the user note in AISC 341 Section F3.5a, these stringent requirements are not needed in an EBF because the link, not the beam hinge, is the structural fuse. However, beam-to-column connections in an EBF are required to accommodate significant inelastic drift in accordance with AISC 341 Section F3.6b. Therefore, the connections must satisfy ordinary moment frame beam-to-column connections requirements as outlined in Section E1.6. For this example, as discussed in the overview, the moment connection is fully restrained, conforming to option (b) of Section F3.6b. When in combination with the brace gusset, the connection configuration is similar to the SCBF beam-to-column connection illustrated in Design Example 2.

6.8 DIAGONAL BRACE-TO-GUSSET CONNECTION

AISC 341

As indicated in Design Example 2, the gusset plates in a SCBF must accommodate the inelastic rotation due to brace buckling. However, as with the beam-to-column connections, the user note in AISC 341 Section F3.5a indicates this stringent requirement is not needed in an EBF because the link, not the diagonal brace, is the structural fuse. Therefore, in accordance with Section F3.6c, it is considered unnecessarily conservative to design the connection for the brace buckling strength. The connection is simply required to resist the adjusted brace strength corresponding to link yielding and strain hardening in accordance with ordinary concentric braced-frame requirements as outlined in Section F1.6. For this example, as discussed in the overview, the brace attaches to the gusset with bolted connecting elements. Although not illustrated in this example, several bolted brace-to-gusset connection design examples are included in the AISC Seismic Design Manual.

6.9 LINK STABILITY BRACING

AISC 341/360

In order to provide stable inelastic behavior, lateral restraint against out-of-plane displacement and twist is required at each end of the link in accordance with AISC 341 Section F3.4b. The lateral restraint stabilizes the diagonal bracing and the beam outside of the link. In accordance with Section F3.5c, the link is considered a protected zone and as such, headed stud anchors cannot be used between the ends. As determined by Ricles and Popov (1989), the concrete slab alone cannot be relied upon to provide adequate lateral bracing at the ends of the link. Therefore, the provisions ignore any slab composite action effects and require supplemental bracing.

186


Design Example 5




For I-shaped sections, Section F3.4b requires bracing for both the top and bottom link flanges. In accordance with Section D1.2c, the bracing required strength is the same as that required for SMF expected plastic hinge locations. In addition to requirements for strength, Section D1.2c(a)(3) also refers to AISC 360 Appendix 6 stiffness requirements. For this example, a schematic representation of the bracing is shown in plan on Figure 5–2. Although the link bracing design requirements are not explicitly illustrated in this example, a similar lateral restraint example for a SMF beam is included in Design Example 1.

6.10 INELASTIC STRAIN AND QUALITY CONSIDERATIONS

AISC 341/360

As an expected area of inelastic strain, the link is considered a protected zone (AISC 341 Section F3.5c). In a protected zone, the members are subject to fabrication and attachment limitations as defined in Section I2.1. These limitations include among other items the attachment of headed stud anchors and decking that penetrate the flanges of the link. Inelastic strain is also likely in the column base plate, column splice, and beam-to-column moment connection welds. Therefore, CJP welds in these regions are to be treated as demand critical in accordance with AISC 341 Section F3.6a. As outlined previously, CJP groove welds associated with built-up link beam fabrication are also treated as demand critical. Although not specifically mentioned in Section F3.6a, for this example, the CJP groove welds connecting the bolted flange plates to the link are also treated as demand critical. To provide a reliable ductile seismic response, the EBF is required to meet the quality requirements outlined in both AISC 341 Chapter J and AISC 360 Chapter N. The building code requires a specific Quality Assurance Plan (QAP) that is typically prepared by the engineer of record and is part of the contract documents. Chapter J provides the minimum acceptable requirements for the plan. As the requirements may incorporate economic and scheduling considerations, the QAP should be provided to the fabricator and erector as part of the bid documents.

7. Items Not Addressed in This Example The following items are not addressed in this example but are nevertheless necessary for a complete design of the seismic load resisting system: • Comparison of wind and seismic forces, • Design of the collector elements, • Design of the column splice, • Design of the column base plate connection, • Design of the foundations, and • Design of the diaphragm system.


187

Design Example 6 Multi-Panel OCBF

OVERVIEW A multi-panel braced frame is a braced frame that has multiple tiers of bracing in one story. It is best suited for buildings with tall stories that have restrictions on the width of the frame, where steep brace angles or very long braces lead to a less efficient design. The behavior of a single-story multi-panel braced frame is different from a multi-story braced frame, since it does not have out-of-plane supports at each beam-column-brace intersection. The designer must take special considerations for design beyond those covered by AISC 360, AISC 341, and ASCE 7. This example demonstrates one method for designing the main components in an ordinary, multi-panel, concentrically brace frame, but it could be adapted for other multi-panel frames. The method consists of first sizing the members using AISC and ASCE design requirements for a steel OCBF, then iterating the design considering the inelastic behavior of the frame. The frame is designed to control which braced panel will experience concentrated yielding. Another aspect of the design will consider out-of plane frame buckling.


189

Design Example 6  Multi-Panel OCBF

OUTLINE 1. Building Geometry and Loads 2. Calculation of the Design Base Shear and Load Combinations 3. Vertical and Horizontal Distribution of Load 4. Preliminary Sizing 5. Analysis 6. Final Member Sizes 7. Discussion Topics 8. Items Not Addressed in This Example

1. Building Geometry and Loads 1.1 GIVEN INFORMATION The building is a one-story warehouse that will be used for wine storage, public tours, and events. It is located in San Francisco, CA. The Seismic Design Category is D. The following information is provided: • Building geometry: 200 feet × 300 feet open floor plan with flat roof. • Multi-panel braced frames consist of three panels (tiers), 20 feet × 20 feet, with X-brace configurations, stacked vertically. Building height is 59 feet at the perimeter wall from finish floor. • Braced frame layout is doubly symmetric. • The roof diaphragm qualifies as flexible. • Storage racks are braced independently from the lateral-force-resisting system. • There are no mezzanines attached to the exterior walls • There will be fewer than 300 occupants in the building at a time (Risk Category II) See Appendix A for the following information: • Latitude and longitude, • Soil type, • Spectral accelerations, and • Load combinations including the vertical seismic load effect.

190



1.2 LAYOUT OF BRACED FRAMES Figure 6–1 shows the plan location of the braced frames. There are four braced frames oriented in the north-south direction and six braced frames oriented in the east-west direction, making 10 braced frames in total. Figure 6–2 shows a typical braced frame elevation.

Figure 6–1. Plan layout of braced frames


191


Figure 6–2. Typical braced frame elevation (frame height is measured from the top of the footing.)

WEIGHTS Table 6–1. Assembly weights

Roof Gravity Load

Effective 2 Seismic Weight

Metal roof deck

4 psf

4 psf

Insulation

2 psf

2 psf

8 psf

8 psf

6 psf

6 psf

20 psf

20 psf

Dead Loads

Steel framing Ballasted photovoltaic solar system Total Dead Load

1

Roof live loads3 Ordinary Pitched Roof

192


Live Load 20 psf


Exterior Wall Gravity Load


Cladding

7 psf

7 psf

Metal studs

2 psf

2 psf

Insulation

2 psf

2 psf

5

⁄ 8 in gypsum board

3 psf

3 psf

Miscellaneous

5 psf

5 psf

19 psf

19 psf

Dead Loads

Total Dead Load

1. From Borrego Solar, News & Press article “Roof Load s and Solar Energy,” James E. Trant, June 24, 2010. 2. ASCE 7 Section 12.7.2 describes the loads that are included in the Effective Seismic Weight. 3. From ASCE 7 Table 4–1. Table 6–2. Roof weights

Assembly

Unit Wt (psf)

Area (ft 2)

Ext wall

19

30,000

570

Roof

20

60,000

1200

Weight (kips)

W =

1770


ASCE 7

2.1 CLASSIFY THE STRUCTURAL SYSTEM AND DETERMINE SPECTRAL ACCELERATION PARAMETERS Multi-panel braced frame systems are not specifically addressed in ASCE 7 Table 12.2–1. For this example, the systems fit into the concentrically braced frame category. Under footnote j, the building qualifies as an OCBF, since it is one story with a 59-foot roof height (60 feet from top of the foundation), and the roof dead load is no more than 20 psf: R = 3.25

Ωo = 2.0

C d = 3.25

If the footnote j requirements are not met, the frame is not permitted to be designed as an OCBF. The frame can then be designed to meet the special concentrically braced frame requirements.

2.2 DETERMINE SEISMIC RESPONSE COEFFICIENT The spectral accelerations to be used in design are derived in Appendix A: S DS = 1.00g

S D1 = 0.60g


193


2.3 DESIGN RESPONSE SPECTRUM Determine the approximate fundamental building period using Section 12.8.2.1: C t = 0.02 and x = 0.75

t

T 12.8–2

= 0 43

0 02


Eq 12.8–7

T a = 0.43 sec S

T =

S DS



S

= S

1

D

S D1 T

=

=

2

06 1 00

+06 0 60 10 06 T

T  T

=0

=

=

s c

+ 5 0T for T < T o

§11.4.5 Eq 11.4–5

s c

§11.4.5

for T > T s.

Eq 11.4–6

The long period equation for S a does not apply here because the long period transition occurs at 12 sec (from ASCE 7 Figure 22–12).

Figure 6–3. Response spectrum

As shown in Figure 6–3, the design building period T a = 0.43 seconds is between T o and T S , so the design spectral acceleration S a is 1.0g. It is not required to construct the design response spectrum when using the equivalent lateral force procedure, since the response spectrum is implicit in the calculation of C s in Section 12.8.1.1. The response spectrum demonstrates the effect of the assumptions used in the calculation of building period. Values of C t = 0.02 and x = 0.75 were selected as specified for other structural systems, which result in an approximate design building period, T a = 0.43 sec.

194



The period of the structure can be established through structural analysis of the OCBF. Section 12.8.2, however, limits the period that can be used to calculate spectral acceleration to a value of T max = C u × T a, where C u is a factor found in Table 12.8–1. In this case T max = 1.4 × 0.43 = 0.60 sec. It is recommended that this period, T max, be used to design the OCBF and that the period of the building be verified later in the design process, because often the period will be governed by T max. The designer should be aware that for a short-period building, such as this one, the ASCE ductility factor, R, may not be a good representation of the ductility of the system. Some research has shown that shortperiod structures tend to exhibit less ductile behavior than predicted by R, although further research is needed to determine whether or not the overall code design procedure accounts for this by requiring higher design forces in other ways. Refer to The SEAOC Blue Book article “Development of System Factors.”


T12.3–1

By inspection, the building does not qualify for any of these Horizontal Structural Irregularities. NO HORIZONTAL STRUCTURAL IRREGULARITIES


T12.3–2

By inspection, the building does not qualify for any of the Vertical Structural Irregularities. NO VERTICAL STRUCTURAL IRREGULARITIES


T12.6–1

Since the warehouse is Risk Category II and is only one story above the base, all three analysis options in Table 12.6–1 are permitted: Equivalent Lateral Force Analysis, Modal Response Spectrum Analysis, and Seismic Response History Procedure. In addition, the building meets all applicable requirements for the Simplified Alternative Structural Design Criteria in Section 12.14.1.1. All options are available to the designer. The Simplified Lateral Force Analysis Procedure per Section 12.14.8 will be used in this example to determine the seismic base shear of the building. USE SIMPLIFIED LATERAL FORCE ANALYSIS

2.7 SEISMIC BASE SHEAR V =

FS DS R

§12.14.8.1

W

Eq 12.14–11

F = 1.0 for buildings one story above grade plane.

1

0 3 25

1770

545 ki

V = 545 kips


195


2.8 SEISMIC LOAD EFFECT AND LOAD COMBINATIONS The equations for the Seismic Load Effect for the Simplified Alternative method are the same as for those of the Equivalent Lateral Force method, except that the Redundancy Factor, ρ, is not included in the equations (equivalent to ρ = 1.0). Since the derivation of the load combinations in Appendix A of this text are based on ρ = 1.0 and 0.2S DS = 0.2, they are applicable for this example. Load combinations of consequence for the design of the OCBF are 12

= 09

=

+ 0 5 L

= (1 2 +

+Q

0 5L

§12.14.3.1.3 Load Combo 5 (modified)

5 L for this example (0 9 −

+


− Q E for this example.

Load combinations with overstrength are not used for the OCBF main frame elements, although they apply for collectors and at other conditions outside the OCBF frame. Since the braces in one panel of the three vertically stacked panels will tend to yield before the others, it could be argued that after braces in this panel yield and buckle, the other members in the braced frame become collectors. Commentary Section C12.4.3 discusses that the overstrength factor is intended to be used for these elements, unless yielding of other elements relieve its load. When the braces in one panel yield and buckle, the shear transmitted to the remaining elements in the frame is then governed by the moment capacity of the columns at the yielded panel and the remaining capacity in the yielded braces, if any. Therefore, the overstrength factor could apply to the columns. Consideration will be given later to the behavior of the vertically stacked panels, including determining the governing panel and the moment demand on the column. Load combinations with overstrength factor for column design are 1 2 D 0

+ 0 5L = 0 7 D

+

o

0 5L

.


§12.14.3.2.2 Load Combo 5 (modified) Load Combo 7 (modified)

ASCE 7

3.1 VERTICAL DISTRIBUTION OF SHEAR F r = V r = V = 545 kips

3.2 HORIZONTAL DISTRIBUTION OF STORY SHEAR For flexible diaphragms, the shear can be distributed to the braced frames based on their tributary area (Section 12.14.8.3.1). Since the braced frames all have the same tributary area, the horizontal distribution is as follows: F E −W = F r /6 = 90.8 kips (frames in east-west direction) F N −S = F r /4 = 136.2 kips (frames in north-south direction)

The rest of this example focuses on the design of braced fame BF-1 at A-8.

196



4. Preliminary Sizing Preliminary sizing of the braces, columns, and beams w ill be based on AISC 341 and AISC 360. The members will be resized based on a non-linear approach in the “Analysis” section below where a computer program will be utilized to determine the distribution of seismic and gravity forces in a 3-D framing model of the building.

4.1 DIAGONAL BRACE DESIGN

AISC 341 AND 360

AISC 341 Section F1, which refers to Section D1.1, requires OCBF braces to meet the limiting width-tothickness ratios for moderately ductile members. Preliminary sizing will use a seismically compact, square HSS; however, as discussed in Section 7, research shows improved performance of HSS sections with b / t ratios lower than code minimum requirements, as the code minimum is not necessarily conservative. The force in each brace for the seismic load is shown in Figure 6–4. Brace force =

= 68 1 2 = 6 3 kips, tension or compression.

Figure 6–4. Frame BF-1 brace design force


197


The diagonal brace end connections will be pinned in-plane, and fixed out-of-plane. Refer to AISC 341 Chapter F for connection design requirements. The center connections where the diagonal braces cross will be fixed in-plane and out-of-plane. The length of the compression brace is L = 14.1 feet, since the tension brace tends to provide lateral resistance for the compression brace at the center connection. K = 1.0 from Table 3 in the SEAOC Blue Book article “Concentrically Braced Frames.” Preliminary brace size is selected as HSS 4 1 ⁄ 2 × 41 ⁄ 2 × 3 ⁄ 8 based on the design aid Table 4–4 in the AISC Steel Construction Manual. Fulfillment of the local buckling requirement is confirmed in the design aid Table 1–5b of the AISC Seismic Design Manual. The brace properties are Ag = 5.48 in2 I x = I y = 15.3 in4 r x = r y = 1.67 in KL / r = 101 < 200 (recommended criterion in E2, User Note) F y = 46 ksi, F u = 58 ksi (A500, Gr. B).

The compression strength of the brace is calculated as F

=

2

= 28.1 k si

 KL

=

< 4 71



y

AISC 360 Eq E3–4

= 118

F

 23 2 ksi

r

=0

× 5. =

AISC 360 Eq E3–2 .4 kips.

AISC 360 Eq E3–1

The HSS 41 ⁄ 2 × 41 ⁄ 2 × 3 ⁄ 8 is adequate.

4.2 COLUMN DESIGN

AISC 360

The preliminary column design considers axial load only. Column in-plane moments due to brace yielding in tension and brace inelastic buckling in compression will be addressed in the iteration and final design. Frame BF-1 has a significant overturning moment, due to the frame’s aspect ratio (3:1) and the relatively small amount of dead load. Neglecting the dead load of the frame for now, the maximum design compression force is based on modified Load Combination 5 above with overstrength, and the maximum design tension force is based on modified Load Combination 7 above with overstrength. The basic design forces are shown in Figure 6–5, where the cladding loads are all applied conservatively as point loads at the top of the column.

+ 22.8 = 26.8 k ips

D

= P P

198

= 4 k ip +

4

+

.

856

ki

, maximum compression

= 798.4 kips, maximum tension


Load Case 5 Load Case 7


Figure 6–5. Frame BF-1 column design forces

The column unbraced length in the out-of-plane direction is 60 feet, and the compression load is very high. The unbraced length in the in-plane direction, however, is only 20 feet. The columns could buckle out-ofplane before the braces reach their intended loads. This idea will be discussed in more detail later in this example. In order to achieve desirable behavior from the frame, the designer must carefully consider a few options. Two possible solutions are shown below: 1. Add out-of-plane bracing to the column: This could be tricky, since a large, open floor plan is required to maximize storage space. One option is to use W12 or W14 frame columns and build up a relatively shallow vertical truss perpendicular to the frame for the height of the columns to stiffen the columns against out-of-plane buckling. A complication with this method is that once the diagonal braces yield and buckle, the orthogonal stiffener truss will need to deform with the columns without becoming ineffective.


199


2. Column Orientation: Select a wide flange column and orient it such that the in-plane KL / r is larger than the out-of-plane KL / r so buckling will occur in-plane of the frame. This option may require a relatively deep wide-flange section for out-of-plane buckling resistance due to the height of the column. This option also complicates the connections between the columns and the beams, and the columns and the braces, since the connections would frame into the column webs. Even with this option, it is desirable to add out-of-plane stability bracing at the panel intersections. For this example, a relatively deep wide-flange section will be used for the preliminary design, oriented with the column strong-axis for out-of-plane bending moments. Bracing options are discussed in Section 5. Design for wind is not addressed in this example, but may govern the column out-of-plane requirements. In a real-life building, the designer should consider wind in the preliminary column sizing, or in the column out-of-plane bracing. Preliminary column size is selected assuming that the 20-foot unbraced length governs, since support will be provided between the panels. The preliminary size is selected as a W21 × 111 based on the equations of Chapter E. The strong axis of the columns is oriented in the out-of-plane direction for the preliminary design. Columns of OCBFs are not required to be seismically compact, but the W21 × 111 qualifies. The column properties are Ag = 32.7 in2 I x = 2670 in4 I y = 274 in4 r x = 9.05 in r y = 2.90 in KL / r y = 82.8 F y = 50 ksi, F u = 65 ksi (A992) E

KL r

41. ksi

2

 KL  r

E

<

F y



AISC 360 Eq E3–4

= 113

F y

F r

y

t

y

=

= 09× = g

.3 ksi

× 32 7 891 k ips .7 = 1472 kips

AISC 360 Eq E3–2 AISC 360 Eq E3–1 AISC 360 Eq D2–1

The ratio r x / r y is 3.1 > 60 feet/20 feet = 3.0, which confirms that the column weak axis direction (20-foot unbraced length) governs the KL / r. Additional out-of-plane column stability requirements will be discussed later.

200



4.3 BEAM DESIGN The beams in this structure below the roof level are redundant and do not carry more than self-weight, provided that the compression braces do not buckle. There are no provisions in the AISC 341 to address multi-panel OCBFs, but the Canadian Standards Association (CSA) S16 standard has some requirements that can be considered. CSA requires the redundant beams for multi-panel X-braced frames to handle the unbalanced brace forces resulting from the compression brace buckling. For the preliminary design, the beams will be sized to transfer the entire lateral load from one tension brace to the next, as if the compression brace does not exist. The cladding is not supported by the beams. The force in each beam for the seismic load is shown in Figure 6–6.

Figure 6–6. Frame BF-1 beam design force

136.2 k ip , compression M u

=

4

=14

20 f 2

0.025 8

. 25 klf × 20 f 2

= 1 75 kip-ft

= 0 35

ips


201


A preliminary size of HSS 6 × 6 × 3 ⁄ 8 is selected from AISC Steel Construction Manual Table 4–4, using an unbraced length of 20 feet. The beam properties are Ag = 7.58 in2 I x = I y = 39.5 in4 r x = r y = 2.28 in KL / r = 105 F y = 46 ksi, F u = 58 ksi (A500, Gr. B) Pn M

= 149 kips = 54.6 kip-ft

T 4–4 in the Steel Construction Manual T 3–13 in the Steel Construction Manual

Checking the requirements of AISC 360 H1.1: 136.2 149 P P

+

8 M M

0 91

=

91 +

0 2, 8 1 75

× 54 6

=

.

AISC 360 Eq H1–1a

Shear strength exceeds the required strength by inspection. The HSS 6 × 6 × 3 ⁄ 8 is adequate.

5. Analysis After preliminary sizes are selected in accordance with AISC 341 and AISC 360 minimum requirements, the designer can evaluate the performance of the preliminary structure with a 3-D computer program to model the building and resize the members as required. A non-linear analysis of the building frame is recommended to capture the performance of the braced frame members.

5.1 DETERMINE CRITICAL PANEL The geometry of the vertically stacked multi-panel brace is conducive to brace tension yielding and inelastic buckling being concentrated in one panel, which is called the critical panel. That is because when the braces yield and buckle in the critical panel, the tier level shear resistance in that panel is limited, and the braced frame can no longer develop the shear necessary to yield and buckle the other panel braces. When inelastic deformations occur in the critical panel under high enough lateral load demands, the column begins to develop a moment at the interface between the critical panel and the adjacent unyielded panel. The other panels in the system may not yield. The critical panel is the panel w ith the lowest shear resistance. Since all three stacked panels in this example are identical in geometry and member size, it is not obvious which one will yield and buckle first. One way to control the location of the critical panel is to arbitrarily boost the brace sizes in the other two panels, to force the inelastic deformations to occur in the preferred panel. In this example, the third (top) panel will be designed as the critical panel because the columns in this panel carry the least gravity and overturning compression forces. During an earthquake the braces in the critical panel will yield, and the columns will begin to develop moment, as discussed above. Therefore,

202



less compression in the columns is more desirable to allow for more flexural capacity in the columns at this panel level. Typically the top panel of vertically stacked multi-panel braced frames will be designed as the critical panel. In order to make the top panel yield first, the lower two panel braces w ill be increased to HSS 41 ⁄ 2 × 41 ⁄ 2 × 1 ⁄ 2, and the top panel braces will be kept as HSS 41 ⁄ 2 × 41 ⁄ 2 × 3 ⁄ 8. See Figure 6–7 for the preliminary member sizes.

Figure 6–7. Frame BF-1 preliminary member sizes (critical panel at top of frame)

5.2 COLUMN MOMENTS AND ANALYSIS Since the in-plane moment demands on the column due to inelastic brace deformations cannot be determined from an elastic analysis, an alternative must be considered. CSA has some guidelines that the columns should be designed to resist the axial load plus 20 percent of the column plastic moment capacity. This approach is prescriptive and is not based on the brace deformations for individual systems, nor does it consider the consequences of the columns yielding after the braces become ineffective.


203


It is important to understand the failure mechanisms of this type of frame in the final design iteration. When the braces in the critical panel buckle and yield in an earthquake, their strength may degrade rapidly after a few cycles. If the braces become ineffective, the columns in that panel must alone carry the shear demand of that panel. In this example, the axial and moment demands in the columns at the bottom of the critical panel are very high. If the columns yield in the in-plane direction, an undesirable collapse mechanism can develop; the columns can become unstable in the out-of-plane direction. (See Figure 6–8 for a graphical representation of this concept.) A multi-panel braced frame is particularly vulnerable to this mechanism, since it does not have floors framing in at each level to stabilize the frame at the yield locations. Some research has been performed on offshore platforms, where the columns are very large, but many full-scale tests have not yet been performed on multi-panel braced frames in which the columns are allowed to yield.

Figure 6–8. Column yielding and out-of-plane instability

204



There are a few ways to address this undesirable failure mode. In this example, the columns are designed to remain elastic after the braces yield and buckle. This results in very large columns, such as for offshore platforms. A more efficient design may be obtained by allowing the columns to yield, but considering the strength degradation of the HSS braces, the residual in-plane post-buckling brace strength, and the P-delta behavior of the columns for out-of-plane stability. The designer must also decide how to effectively brace the columns after they yield, which is a challenge. This design approach is beyond the scope of this example. In order to determine the axial and moment demands in the columns, a structural analysis program is used. The diagonal braces in the critical (top) panel are assumed to be ineffective; thus, the columns in the top panel are acting as cantilever elements above the middle panel, as shown in Figure 6–9. The foundation is assumed rigid, as permitted by ASCE 7 Section 12.7.

Figure 6–9. Cantilever behavior of column


205


The forces in this analysis model include the frame self-weight, and loads in Combination 5 with overstrength. The resulting column moment for the preliminary design between the critical (level 3) and the level 2 panels is 3130 k-ft, with a corresponding axial load of 46 k. The column section W21 × 111 is checked for beam-column action as follows: c cr

g

= 891 kips

Previously Calculated

09

68 2

255 8 k-ft.

AISC 360 Eq F2–1

The required moment exceeds the weak axis bending strength significantly, which is a consequence for orienting the column in this direction. The design must be iterated. Another analysis is run using a W30 × 211 column oriented in strong axis bending in the plane of the frame, which results in a revised column moment of 2750 k-ft and a revised axial load of 46 k occurring at the bottom of the top panel. The column axial load is higher in the lower portion of the frame due to overturning forces, but the moment is small, only 200 k-ft. The column strength is rechecked for the W30 × 211 for the maximum moment location at the bottom of the top panel: Ag = 62.2 in2 Z x = 751 in3 r x = 12.9 in r y = 3.49 in KL / r x = 37.2, using an unbraced length of 40 feet in the column strong axis direction, based on the deflected shape in Figure 6–9, when the top panel diagonal braces are considered ineffective. The column will be braced out of plane in the weak axis direction so strong axis column bucking governs. F y = 50 ksi, F u = 65 ksi (A992, Gr. B) F

=

π2 E  

= 207 ksi

AISC 360 Eq E3–4

r

KL

E



y

= 113

 45 2 ksi

r

= 0

AISC 360 Eq E3–2

× 62 2 = 2530 k ips 09

751

AISC 360 Eq E3–1 2820 k-ft

2 50 k-ft

AISC 360 Eq F2–1

Checking the requirements of AISC 360 H1.1:

=

46 2530

P

M

2 P

M

= 0 02 < 0 2, =0

2750 2820

=

99

1.0

AISC 360 Eq H1.1b

The column section is adequate for strength. H owever, the columns must now be braced in the weak axis direction for out-of-plane buckling.

206



5.3 OUT-OF-PLANE BEHAVIOR In the previous section, the out-of plane frame behavior was briefly addressed. Ongoing research in this area indicates that critical panel yielding and buckling does not produce significant out-of-plane moments; however, the multi-panel braced frame is vulnerable to twisting at the panel intersections even if the columns remain elastic. The frame is especially vulnerable to this situation now that the column w eak axis direction is oriented in the out-of-plane direction of the frame. For this example, out-of-plane twisting is addressed by providing “relative” bracing or stiffening along the columns. This can be in the form of a shallow truss or a stiffener intermittently welded into the column web. The bracing element should comply with AISC 360 Appendix 6 as a minimum. As mentioned previously, the designer of a real building should consider that the out-of-plane column bending or bracing requirements may be governed by w ind.

Figure 6–10. Out-of-plane column stiffening (not to scale)

5.4 DRIFT CALCULATION

ASCE 7

According to footnote c in Table 12.12–1, there is no story drift limit for single-story structures that have non-structural components designed to accommodate the drift; however, structural separation requirements must still be met. As an experiment, “story” drifts are calculated for each panel as if they were in a threestory building, and the results are compared to the Table 12.12–1 limit of 0.025 hsx. The drift calculation is based on the procedure in Section 12.8.6. δ xe is calculated by elastic analysis in a structural analysis program. The results are presented in Table 6–3 for information.

=

C I

x

= 3 25

Eq 12.8–15


207


Table 6–3. Calculated and maximum story drifts

δ xe (in)

Level

δ x (in)

Level 1

0.7

2.2

6

Level 2

1.5

4.9

12

Roof

2.58

8.4

18

6. Final Member Sizes The final member sizes are shown in Figure 6–11.

Figure 6–11. Frame BF-1 final member sizes

208

Limit (in)



7. Discussion Topics The following topics are discussed, but are not supported by calculations in this example. Brace Shape and End Conditions: The hysteretic behavior of the braces is closely related to the shape and the end conditions. In this example, square braces with pinned ends w ere selected. However, cyclic loading tests of braced frames demonstrate that braces degrade rapidly after they buckle. Round braces tend to perform better than rectangular or square braces, but buckling leads to member tearing after a few cycles. More ductile hysteretic behavior can be achieved with lower brace KL / r values. Some research suggests that KL / r values of close to 25 provide good ductility. The designer should consider designing beyond the code minimum with more stout braces and fixed connections to inhibit buckling. In a frame with long brace spans, this leads to heavy braces. Wind vs. Seismic Loads: Buildings in some parts of the country may need to be designed for high winds and high seismic demands. The building in this example has a particularly large projected wind area, and the forces could be significant, possibly governing the column design. Consideration must be given to the frame behavior for both loading conditions. The critical panel framing members shall not yield for w ind, as is intended for the seismic response. Foundations: Multi-panel braced frames tend to have high aspect ratios. They also can have very little dead load, since they are only supporting loads essentially at the top of the frame. The combination of those two characteristics results in large uplift forces on the foundation. The designer should consider this, if possible, when selecting a foundation type. Foundation types such as piles or combined footings may be more suitable than the isolated pad footings shown in the design example. The frame footings could also be tied to the adjacent column footings to help resist uplift. Two-panel or Four-panel Frames: This example uses a three-panel frame with equal panel heights. The same procedure could apply to multi-panel braced frames with a different numbers of panels or unequal panel heights. However, the designer should consider some additional ideas when selecting a load-resisting system and laying out a multi-panel frame. First, the higher the frame aspect ratio, the more vulnerable the frame is to twisting, and the bracing requirements may be higher. The uplift demands on the foundation may also be very high. A somewhat free-standing, multi-panel system may not be appropriate where the frame is required to have a very high aspect ratio. A more economical lateral-force-resisting system could be selected based on the building height and geometry, the frame layout, the architectural restrictions, and the soil conditions. Second, different panel heights can complicate the frame in-plane and out-of-plane behavior. Some additional iteration may be required to determine the critical panel and the appropriate column moment demand. One-story or Two-story X-Bracing: In the case where the foundation type and uplift capacity is limited, but a multi-panel frame is still justified, a wider frame may be selected. The designer may decide to use a double-wide or two-story, X-brace configuration with three columns instead of the one-story configuration used in this example. See Figure 6–12. The two-story configuration may not be economical for a situation where the layout is not governed by the capacity of the foundation, due to the increased slenderness of the braces.


209


Figure 6–12. Two-story X-brace configuration (left), and (right) one-story X-brace configuration

Mezzanine: Since the multi-panel braced frame is best suited for a very tall story, there is a chance that a mezzanine level will be incorporated into the design. The designer needs to decide if the mezzanine level will be tied in with the multi-panel frame or if it will be independently braced. If it is tied in with the frame, the designer needs to then consider the mezzanine level in the layout of the panels. This may govern the number of panels, the panel height, and the frame base. The designer should also consider if the mezzanine will provide any out-of-plane stability to the frame.

8. Items Not Addressed in This Example The following items are not addressed in detail in this example but are nevertheless necessary for a complete design of the seismic load resisting system: • Comparison of wind and seismic forces, • ASCE 7 stability check (ASCE 7 Eq 12.8–16), • Design of the column base plate connection, • Design of the foundations, • Design of the roof members and connections, • Design of torsional bracing for the multi-panel frames, and • Design of brace and beam to column connections.

210


Design Example 7 Metal Deck Diaphragm OVERVIEW Diaphragms are the horizontal components of a building—typically the roof and floor elements—that transfer lateral loads to the vertical elements of the Seismic Force Resisting System (SFRS), which are Buckling Restrained Braced Frames (BRBF) in these design examples. Diaphragms also • Provide gravity support for the structure • Transfer dead and live loads to beams and/or bearing walls • Provide lateral support to walls subject to out-of-plane loading • Provide lateral stability to vertical elements, decreasing the unbraced height of these members and consequently increasing their capacity to support axial loads • Allow horizontal framing members supporting gravity loads such as beams and joists to carry larger loads by providing lateral stability to the compression flanges of these members. Diaphragms comprise three distinct components: the metal deck or concrete fill and metal deck (in these examples); the chord elements, which are perpendicular to the applied lateral loading and carry axial loads caused by the moment in the diaphragm; and the collector elements (also known as drag struts or diaphragm ties), which are parallel to the applied lateral loading and transfer diaphragm forces along a lateral-load-resisting line to the vertical elements of the SFRS. These three components are identified for loading in the east-west direction at the roof in Figure 7–1 and for loading in the north-south direction at the roof in Figure 7–2. All bays are 30 feet, 0 inches × 30 feet, 0 inches.

COLLECTOR BEAM, TYP ALONG GRIDLINES 1, 3, AND 4

METAL DECK DIAPHRAGM

CHORD BEAM, TYP ALONG GRIDLINES A, B, E, AND F

Figure 7–1. Diaphragm components for east-west loading 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

211

Design Example 7



Metal Deck Diaphragm

CHORD BEAM, TYP ALONG GRIDLINES 1, 2, 4, AND 5

COLLECTOR BEAM, TYP ALONG GRIDLINES A AND F

METAL DECK DIAPHRAGM

Figure 7–2. Diaphragm components for north-south loading

The stiffness of the diaphragm, like the stiffness of any structural element, determines how much it will deflect. Stiffness is affected by many things, including the size, spacing, and connections of the framing members, but the factor that contributes the most is the stiffness of the metal deck or concrete fill and metal deck. A bare metal deck diaphragm is not as stiff as a concrete-filled metal deck diaphragm, and thus a bare metal deck diaphragm will have higher deflections than the same one filled with concrete. The diaphragm stiffness also determines how lateral loads are distributed to the vertical elements of the SFRS. ASCE 7 has three categories for diaphragms depending on their stiffness: flexible, rigid, and semirigid. The more flexible a diaphragm is, the less ability it has to transfer torsional forces. When a diaphragm can be classified as flexible due to the construction type per ASCE 7 Section 12.3.1.1 or by calculation in accordance with ASCE 7 Section 12.3.1.3, it will transfer lateral load to the vertical elements of the SFRS based on tributary area. Rigid diaphragms are stiff enough to transfer torsional forces to the vertical elements of the SFRS. A diaphragm can be classified as rigid only if it is one of the construction types identified in ASCE 7 Section 12.3.1.2. A rigid diaphragm distributes lateral loads to vertical elements of the SFRS based on its relative rigidities. When a diaphragm cannot be classified as flexible or rigid, it is said to be semi-rigid. A building with semi-rigid diaphragms must be modeled with the stiffness of each diaphragm taken into account. Per ASCE 7 Section 12.3.1.2, semi-rigid modeling will most likely be required when any horizontal irregularity (per Table 12–3.1 of ASCE 7) is present or the span-to-depth ratio of the diaphragm exceeds three and the diaphragm is constructed of a type that otherwise qualifies as being rigid. The presence of horizontal irregularities and a span-to-depth diaphragm ratio greater than three impact the stiffness of the diaphragm in such a way that the rigid idealization is no longer accurate.

212


Design Example 7



Metal Deck Diaphragm

The following examples will illustrate how to design diaphragms in accordance with the 2012 IBC to adequately distribute applied lateral loads to the vertical elements of the SFRS. The first example, 7A, goes through the design of the roof diaphragm, which is constructed of bare metal deck and can be idealized as a flexible diaphragm. The second example, 7B, goes through the design of the second-floor diaphragm, which is constructed of 3 1 ⁄ 4 inches of concrete fill over a 2-inch metal deck and can be idealized as a rigid diaphragm. Discussions for semi-rigid diaphragm design will be included in Example 7B.


213

Design Example 7A Bare Metal Deck (Flexible) Diaphragm OUTLINE 1. Given Information 2. Determination of Diaphragm Forces and Code Discussion 3. Diaphragm Analysis without Openings 4. Diaphragm Analysis at Opening and at Re-entrant Corner 5. Diaphragm Design 6. Chord and Collector Design 7. Collector Connection Design 8. Items Not Addressed in This Example

1. Given Information 1.1 INFORMATION PROVIDED IN APPENDIX 1 AND EXAMPLE 1 • S 1 = 1.0g • S DS = 1.0g • S D1 = 0.6g • I e = 1.0 • R = 8.0 • ρ = 1.0 • Ωo = 2.5 • Six stories with heights as shown in Figure 7A–1 • W roof = 656 k • W floor = 1315 k for all floor levels • SFRS = Buckling Restrained Braced Frames (BRBF)


215

Design Example 7A



Bare Metal Deck (Flexible) Diaphragm

Figure 7A–1. Building elevation with story heights

1.2 METAL DECK TYPES AND DIAPHRAGM CAPACITIES Per XYZ-ES Report #1000 for ACME Steel Deck Company, Roof Deck Diaphragm Capacities: • vall = 800 plf for 16-gage deck with 1 1 ⁄ 2-inch rib height, four-weld pattern per sheet to supports with 11 ⁄ 2-inch top seam welds at side laps at 24 inches on center for beam spans up to 10 feet. • vall = 1500 plf for 16-gage deck with 1 1 ⁄ 2-inch rib height, seven-weld pattern per sheet to supports with 1 1 ⁄ 2-inch top seam welds at side laps at 12 inches on center for beam spans up to 10 feet.

2. Determination of Diaphragm Forces and Code Discussion

ASCE 7

2.1 CALCULATION OF DIAPHRAGM FORCES In order to calculate the diaphragm forces in accordance with Section 12.10, the story shears must be calculated in accordance with Section 12.8. These are based on the seismic response coefficient, which is also calculated in accordance with Section 12.8. This is the procedure for calculating diaphragm forces regardless of the type of analysis used for the vertical elements of the SFRS (Equivalent Lateral Force Procedure or Modal Response Spectrum Analysis). Determine the approximate fundamental building period in accordance with Section 12.8.2.1: C t = 0.03 and x = 0.75 a

216

t

= 0 03

T 12.8–2

= 0 74 se


Eq 12.8–7

Design Example 7A




The approximate building period per Equation 12.8–7 will be used in these examples to determine the seismic response coefficient and thus the story shears. These numbers will differ from the ones used in Example 3, which utilizes the calculated building period to determine these values. Determine the seismic response coefficient in accordance with Section 12.8.1.1: C

=

S DS

  I

=

1 00

=

≤



1

T

10

( )



I

60

= 0.7 4

=

 .1 0 .

1.0 

Eq 12.8–2 and Eq 12.8–3



Also, C ≥

05

01, and C

 R   I

6

=

=

80

375.

Eq 12.8–5 and Eq 12.8–6

10

Therefore, Equation 12.8–3 controls, and V

S

= 0.100 ×

= 723 k .

Eq 12.8–1

Determine the vertical distribution of base shear in accordance with Section 12.8.3:

Table 7A–1. Vertical distribution of shear

Level

wi (kips)

hi (ft)

i i

C vx

F x (kips)

Roof

656

72

78,907

0.18

130

6th

1315

60

128,960

0.29

210

5th

1315

48

100,442

0.22

159

4th

1315

36

72,775

0.16

116

3rd

1315

24

46,213

0.10

72

2nd

1315

12

21,262

0.05

36

Total

7231

448,559

723

The terms used in Table 7A–1 are defined in Section 12.8.3. Since the period = 0.74 > 0.5 sec, the value for k is interpolated between a value of 1.0 for T = 0.5 sec and 2.0 for T = 2.5 sec. For this building, k = 1.12. The distribution of story shear is carried out using F x = C vxV , where, C vx =

w x

∑= w

x

.

Eq 12.8–11 and Eq 12.8–12

i

1


217

Design Example 7A




Determine diaphragm design forces in accordance with Section 12.10.1.1:

Table 7A–2. Diaphragm design forces

Level

w px (kips)

F x (kips)

F px (kips)

F px,min (kips)

F px,max (kips)

F px,des (kips)

Roof

656

130

130

131

262

131

6th

1315

210

227

263

526

263

5th

1315

159

200

263

526

263

4th

1315

116

176

263

526

263

3rd

1315

72

153

263

526

263

2nd

1315

36

132

263

526

263

F px,des is the force the will be used for the diaphragm design. All other terms used in Table 7A–2 are defined in Section 12.10.1.1. n

∑

i x

∑

Eq 12.10–1 i

i x

F px shall not be less than: F F px need not exceed: F px ,max

,m

S

I w

S DS Ie w px .

Eq 12.10–2 Eq 12.10–3

2.2 DISCUSSION OF DIAPHRAGM FORCES VS. STORY FORCES As illustrated in Figure 7A–2, the diaphragm forces calculated in accordance with ASCE 7 are distributed differently than the story forces calculated in accordance with the Equivalent Lateral Force Procedure of ASCE 7. The vertical elements of the SFRS are designed for the story forces, which are typically less than the diaphragm forces (the roof is the only location where they are not). D iaphragm forces are larger because they are calculated to account for the peak modal response of each story. Story forces do not account for this because it would be overly conservative to design the vertical elements of the SFRS for the peak response at each floor, as each floor will experience its peak response at a different time.

218


Design Example 7A




                     

   

  













 

Figure 7A–2. Comparison of story and diaphragm forces

The minimum diaphragm design force was developed to account for responses due to higher mode effects. The Response Modification Factor, R, has the most effect on the lower modes of the building, so buildings with higher R values would unconservatively be designed for low diaphragm forces that would not capture all of the demands if this minimum were not provided. In this BRBF building, the minimum diaphragm forces govern at every level because the R value for this type of building is so high. The maximum diaphragm design force was developed to prevent overly conservative demands on diaphragms with low R values.

2.3 DISCUSSION OF THE USE OF THE REDUNDANCY FACTOR,  Per Sections 12.3.4.1 and 12.10.1.1, ρ = 1.0 for inertial forces when diaphragm forces are calculated in accordance with Equation 12.10–1. Engineers should exercise caution when following this code prescription. In some situations, typically at the roof and occasionally at higher floors, F px / F x < 1.3. If ρ = 1.3 for the design of all other elements of the SFRS, the diaphragm would actually be designed for a lower force. Although it is not a code requirement, it is recommended that F px be factored up to ρ*F x. For this building ρ = 1.0, so the calculated values of F px will be used without any alteration.

2.4 INCREASE OF DIAPHRAGM AND COLLECTOR FORCES PER SECTION 12.3.3.4 Per Section 12.3.3.4, if Horizontal Irregularity Types 1a, 1b, 2, 3, or 4 per Table 12.3–1 or Vertical Irregularity Type 4 per Table 12.3–2 are present in Seismic Design Categories D or higher, diaphragm and collector forces must be increased by 25 percent. The only exception is if the overstrength factor, Ωo, is used and these forces exceed the 25 percent increase. In the situation where Ωo > 1.25, the overstrength level forces govern. Per Section 12.3.3.4, the connections of diaphragms to vertical elements and collectors in buildings where these irregularities are present will always be designed for the 1.25 percent increase. In this building, Horizontal Irregularity Types 1a and 2 are present (per Example 3), so this increase will be applied. Collectors in buildings having any of these irregularities with only light-framed shear walls acting as the vertical elements of the SFRS meet Exception 2 of Section 12.10.2.1. Therefore they will be designed for the 1.25 percent increase, as the overstrength factor does not apply. 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

219

Design Example 7A




2.5 DISCUSSION OF DIAPHRAGM FLEXIBILITY Per Section 12.3.1.1, untopped steel deck diaphragms in buildings with steel braced frames acting as the vertical elements of the SFRS can be idealized as flexible. The roof is bare metal deck, and the SFRS utilizes only BRBF, so this diaphragm will be analyzed using flexible diaphragm assumptions. As briefly discussed in the Overview, it is assumed that flexible diaphragms cannot transfer torsional forces. Lateral forces are distributed to the vertical elements of the SFRS based on the tributary area of the diaphragm to each vertical element. Flexible diaphragm analysis can easily be used to determine the forces in the vertical elements of the SFRS, as the distribution can be done by hand (this will be illustrated in Section 3.1 of this design example). Rigid diaphragm analysis, as discussed in the Overview, accounts for torsional forces imposed on the diaphragm and the vertical elements of the SFRS. The analysis requires a more in-depth calculation for the distribution of lateral forces. Typically when rigid diaphragms are present, an analytical model of the building is used. The lateral forces in the vertical elements of the SFRS can be extracted from this model and used to calculate the diaphragm forces. This procedure will be illustrated in Design Example 7B.

2.6 OTHER APPROACHES TO DIAPHRAGM DESIGN There are other approaches to designing diaphragms, such as non-linear response history analysis and capacity-based design. The National Building Code of Canada, in fact, explicitly requires that diaphragms be designed to not yield, which means that diaphragms must be designed for the capacity of the vertical elements of the SFRS.

3. Diaphragm Analysis without Openings A flexible diaphragm can be idealized as a beam with the vertical elements of the SFRS acting as reaction points (the BRBFs in this example). The seismic-force tributary to that level is treated as the distributed load along the beam, calculated by dividing the seismic force by the area of that level. Figure 7A–3 below illustrates this idealization for the roof diaphragm in the north-south direction, and Figure 7A–4 below illustrates this idealization for the roof diaphragm in the east-west direction. Diaphragm force at the roof = F px = 131 k per Table 7A–2. Roof Area = A = 15,220 ft2 per Appendix A

=

F

=

131 k × 1000 lbs/f 15,200 f

8 62 ps

1034.4 plf 517.2 plf

517.2 plf

RF

R A

Figure 7A–3. Diaphragm loading in north-south direction

w w

220

= = 2

60 f

pf .4 plf


Design Example 7A




1293 plf

775.8 plf R1

775.8 plf

R4

R3

Figure 7A–4. Diaphragm loading in east-west direction

w w

x

1

= =

. plf 150 f

plf

where d = depth of the diaphragm. The loading along the beam takes into account the distribution of the seismic mass based on the configuration of the building.

3.1 DIAPHRAGM SHEAR The shear diagrams for the “beam” in each direction are shown in Figures 7A–5 and 7A–6.

65.5 k RF

R A

65.5 k 65

Figure 7A–5. Diaphragm shear in north-south direction

23.3 k

18.1 k

22 k R1

R3 40.1 k 40.

R4

20.7 . k

Figure 7A–6. Diaphragm shear in east-west direction

The maximum shears occur at the reaction points, which are the locations of the BRBF. The diaphragm is designed for the maximum unit shear, which is calculated as the shear along a line divided by the length of the diaphragm at that line. These diaphragm shear demands are calculated in Table 7A–3.


221

Design Example 7A




Table 7A–3. Diaphragm shear demands

Gridline

V (kips)

V corrected (kips)

L (ft)

v (plf)

A

65.5

65.5

1.1

72.1

60

1201

F

65.5

65.5

1.1

72.1

60

1201

1

22.0

23.2

1.12

26

90

289

3

58.2

61.4

1.07

65.7

150

438

4

44.0

46.4

1.13

52.4

150

350

e

V des (kips)

V = Calculated diaphragm shear V corrected = Diaphragm shear corrected so the summation of the reactions in each direction = F px = 131 k e = Eccentricity V des = V *e L = Diaphragm length v = V des / L

The summation of the reactions, V , in the east-west direction does not equal F px = 131 k. This is due to rounding of the distributed loading. The variable V corrected has been introduced to account for this discrepancy. The variable e is included to account for accidental torsion. Though ASCE 7 Section 12.8.4.2 does not require that this be accounted for when designing flexible diaphragms, it is recommended that accidental torsion be considered for the same reason that it is required to be considered in the design of non-flexible diaphragms, which is to provide capacity in the diaphragm to resist redistribution of loading due to nonuniform inelastic behavior of the elements of the SFRS. For the simply supported beam condition in the north-south direction, shifting the mass 5 percent of the building width to account for accidental torsion creates a 10 percent increase in the reactions. For the more complex condition in the east-west direction, shifting the center of mass 5 percent of the building length to account for accidental torsion creates a 12 percent in the reaction at gridline 1, a 7 percent increase in the reaction at gridline 3, and a 13 percent increase in the reaction at gridline 4.

3.2 CHORD FORCES As briefly mentioned in the Overview, chord forces develop in boundary members perpendicular to the applied load as a result of the moment created in the diaphragm. The moment diagrams for the “beam” in each direction are provided in Figures 7A–7 and 7A–8. These diagrams have not been corrected to account for rounding as the forces in Table 7A–3 have been. The moment resolves itself as a tension-compression couple with an arm equal to the depth of the diaphragm at that point. The boundary members can be thought of as the flanges of the beam with the diaphragm acting as the web to provide shear resistance. Maximum chord forces will occur at locations of maximum moment and/or at locations where the diaphragm is fairly shallow.

222


Design Example 7A




R A

RF M = 2676.5 k-ft

Figure 7A–7. Diaphragm moment in north-south direction

349.1 k-ft

310.3 k-ft

R1

R3

R4

311.4 k-ft

Figure 7A–8. Diaphragm moment in east-west direction

Based on the moment diagrams for loading in each direction provided in Figures 7A–7 and 7A–8, the chord forces within the boundary members are: 2676.5 -ft 12 ft

22 3 k

311 4 k-ft 349 1 k-ft  0 t P

=

310 3 k-ft 150 t

0 t



x

5k

k

= 2 1 k

As shown in Figure 7–2, there are also chord members along gridlines 2 and 4, which develop because of the re-entrant corners. The forces in these members are calculated in Section 4.2 of this example. Distributed lines of lateral load resistance create shorter diaphragm spans, which lead to less moment developed within the diaphragm. Consequently, chord forces are lower w hen the lateral-load-resisting lines are located throughout the building than when they are located at the extremes of the building only. This concept is illustrated here by the difference in magnitude between the chord forces in the north-south direction and the chord forces in the east-west direction. There are no provisions for increasing the forces in chord members in ASCE 7 (i.e. designing for amplified Ωo level forces or increasing by 25 percent when certain irregularities are present). In Seismic Design Categories C or higher, collector elements are required to be designed for forces amplified by the overstrength factor Ωo per ASCE 7 Section 12. 10.2.1. Consequently, when members in a line of lateral resistance act as both chord and collector elements, collector forces almost always govern.

3.3 COLLECTOR FORCES As briefly mentioned in the O verview, collector members are parallel to the applied diaphragm loading. Their function is to drag load along the diaphragm into the vertical members of the SFRS. Per ASCE 7 Section 12.10.2.1, collector members and their connections must be designed for the maximum of Ωo*F x, Ωo*F px, and ρ*F px,min, but the force does not need to exceed ρ*F px,max. There is an exception for buildings that have their SFRS comprised entirely of light-framed shearwalls. Collectors and their connections in these buildings do not have to be designed for forces amplified by the overstrength factor, Ωo. 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

223

Design Example 7A




Table 7A–4 calculates the collector force for each story based on Section 12.10.2.1: Table 7A–4. Collector forces

Level

Ωo*F x (k)

Ωo*F px (k)

ρ*F px,min (k)

ρ*F px,max (k)

F collector (k)

Roof

325

325

131

262

262

6th

525

568

263

526

526

5th

398

500

263

526

500

4th

290

440

263

526

440

3rd

180

383

263

526

383

2nd

90

330

263

526

330

F collector is the force that will be used for the design of the collector beams and their connections. The force along each line of lateral resistance from Table 7A–3 will be scaled up to the collector force of 262 kips at the roof from the diaphragm force of 131 kips calculated in Table 7A–2.

Collector diagrams for each lateral load resisting line in the east-west direction are provided in Figures 7A–9, 7A–10, and 7A–11 for gridlines 1, 3, and 4, respectively. A collector diagram for gridlines A and F is provided in Figure 7A–12.

v = 262 vk =/ 131 k *k 26 ft =plf 578 plf 2.5*26 / 90kft/ =90 722

V = 2.5*26 k = 65 k

V = 262 k / 131*26 k = 52 k

) s p  i k (  e c r  o F  r o  t c e l l o C

26

17.3 21.7



32.5



17.3

26



21.7

32.5



(a)



Length Along Collector Line (ft) (b)

Figure 7A–9. Collector diagram along gridline 1



v1 v1 = 262 k / 131 kk */ 150 20.4ftk=/ 150 = 272 plf = 2.5*20.4 340 ft plf

v2 = 2.5*45.3 k / 150 ft = 755 plf v2 = 262 k / 131 k * 45.3 k / 150 ft = 604 plf V = 2.5*65.7 k = 164.3 k V = 262 k / 131 k * 65.7 k = 131.4 k

) s  p i  k (  e c  r o  F r   o t  c e  l l o  C 

65.882.1

52.6 65.7 26.323.9 





65.8 82.1



52.665.7

Length Along gC Collector Line ine (ft) (b)

Figure 7A–10. Collector diagram along gridline 3 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4



26.3

(a)

224



23.9



Design Example 7A






v1 =v1 262 k / 131 k k* /27.7 = 616 plf = 2.5*27.7 90 ftk =/ 90 769ft plf

v2 = 2.5*24.7 k / 150 ft = 412 plf v2 = 262 k / 131 k * 24.7 k / 150 ft = 329 plf V = 2.5*52.4 k = 131 k V = 262 k / 131 k * 52.4 k = 104.8 k

) s  p i k (  e  c r o F  r  o t c e l l o C

65.5 52.4

38.6 48.3

11.1 13.9 













11.113.9 52.4 65.5



38.6 48.3

Length th Along lon Collector lector L Line (ft) (b)

(a)

Figure 7A–11. Collector diagram along gridline 4

f l p 3 0 4 2 = V = 262 k / 131 k t f * 72.1 k = 144.2 k 0 6 / k 1 . 2 7 * k 1 31 k *k 26 90722 ft =plf 578 plf 3 .5*26 / 90kft/ = 1 V = 262 k / 131 k / k * 72.1 k = 144.2 k 2 6 2 V = = 2.5*26 k = 65 k 2 v k / 131*26 k = 52 k





e c r  o F  r o  t c e l l o C 

(a)

36.05 17.3 21.7

) s p  i k (

26



36.05

32.5



26

32.5

36.05



17.3 36.05



21.7


Figure 7A–12. Collector diagram along gridlines A and F

The (a) diagrams show how diaphragm forces are being dragged into the lateral resisting line. When lateral resisting elements are located at the interior of a building, load is dragged in from both sides of the line, but at lateral resisting lines located along the exterior of a building, the load along that line is dragged in from the interior side only. Figures 7A–10(a) and 7A–11(a) occur at the interior of the building, and the loads being dragged into these lines from either side of the diaphragm are taken from Figure 7A–6. The shear values used above have been corrected for use with V des from Table 7A–3. The (b) diagrams show the actual distribution of forces along the collector line, which takes into account the unloading of the diaphragm as the load goes into the vertical element of the SFRS. The jump in the collector force diagram occurs at the midspan of the BRBF at all of these collector diagrams because that is where the braces frame into the diaphragm. When braces frame into the diaphragm at columns (such as in Special Concentrically Braced Frames with X configurations) or in moment-frame buildings where the load gets transferred directly into the columns, there will be two jumps in the collector diagram—one at each brace/column within the frame. At shearwall buildings, the lateral load is considered to be uniformly transferred into the lateral resisting element, so collector diagrams for these buildings will have different shapes that the ones shown here.


225

Design Example 7A




The value of Ωo = 2.5 has been used in determining these collector forces. Per footnote g of ASCE 7 Table 12.2–1, when Ωo ≥ 2.5, it can be reduced by 0.5 for structures with flexible diaphragms. In this building, the roof diaphragm is flexible, but the floor diaphragms are not. Consequently, the reduction will not be used in this design example. When all diaphragms in a building are flexible, the 0. 5 reduction should be utilized when possible in order to avoid an overly-conservative design of the collector members and their connections. These collector diagrams are for design in accordance with ASCE 7 only. See Example 3 for collector diagrams within the BRBFs, where the collector forces shown here must be combined with additional axial and shear loads caused by brace yielding.

4. Diaphragm Analysis at Opening and at Re-entrant Corner 4.1 ANALYSIS AT OPENING Openings in diaphragms create a concentration of forces at portions of the diaphragm adjacent to them. Increased shear demand at members along the length of the opening and local chord forces along the top and bottom of the opening will be developed. Figure 7A–13 shows how forces in the north-south direction must be transferred around the opening between gridlines C and D just north of gridline 3, creating additional forces within in the diaphragm that are not accounted for when it is analyzed without the opening.

EXAGGERATED DEFLECTED SHAPE OF THE DIAPHRAGM AT THE OPENING

T

Vopn'g

Vopn'g

g ' n p o

d

C 3 wopn'g

lopn'g C

Figure 7A–13. Forces and deflected shape at diaphragm opening

226


D

Design Example 7A




The distributed load, wopn’g, is equal to the distributed diaphragm loading multiplied by the depth of the diaphragm at this location, which is 60 feet at this opening. The shear forces that must be transferred around the opening, V opn’g are calculated below based on the wopn’g and the opening length, lopn’g as shown below. Additionally, the local chord forces can be calculated from the moment created by w opn’g and the depth of the opening, d opn’g as shown below.

=

w

n

=

f

o n

=

2 w

=

plf

517.2 lf

f

2

= 7 76 k

.

1 1

= 5 82 k

The beams along gridlines C and D near the opening and their connections must be designed to drag this shear force around the opening. The beam along gridline 3.2 between gridlines C and D needs to be designed for this axial chord force in addition to gravity loads. The beam along gridline 3 between gridlines C and D is part of the B RBF, and this additional axial load from the local forces must be added to the axial forces from the collector analysis and from the yielding of the braces. The chord forces are conservatively based on a simple span moment, but the end connections of the beams along gridlines 3 and 3.2 should be designed for fixed-end moments created by the distributed loading along the opening. The shear demands along gridlines C and D will increase to account for the fact that the two openings located between gridlines 3 and 4 reduce the diaphragm length by 20 feet total. The new shear demands along these lines are calculated below:

= V , es

V

× 30 t =

w

V

25 6 k

− 20 ft

120 ft − 20 t

.

−

lf × 30 f

=

6 k

256 plf

Because these openings are not located near lateral-load-resisting lines in the north-south direction, they do not have much impact on the shear design of the diaphragm, and the shear demands along the lateral-loadresisting lines at gridlines A and F still govern. In the east-west direction, these openings occur along lateral-load-resisting lines. The method for determining the local chord forces will be the same as for loading in the north-south direction. The new shear demands along gridlines 3 and 4 are calculated below:

=

V ,de

150 ft − ,de

150 ft

=

65.7 150 ft − 30 ft 52.4 k 150 ft − 30 ft

= 548 plf 437 plf

These diaphragm shear demands that account for the effects of the openings are 25 percent larger than shear demands calculated in Table 7A–3, which did not account for the openings.


227

Design Example 7A




The collector force diagrams along gridlines 3 and 4 must also be modified to account for the transfer of the diaphragm force around the opening. Figure 7A–14 shows the actual collector force diagram along gridline 4, including the opening between gridlines C and D. The modified collector diagram along gridline 3 will be very similar.

v1 = 262 / 131 k * k27.7 ft = 923 v1 =k2.5*27.7 / 60kft/ =601154 plf plf

v2 = 2.5*24.7 k / 150 ft = 412 plf v2 = 262 k / 131 k * 24.7 k / 150 ft = 329 plf V = 2.5*52.4 k = 131 k V = 262 k / 131 k * 52.4 k = 104.8 k

)  s p  i k ( 

e c r  o F  r  o t c e l l o C 

(a)

46.9 58.6

52.4 65.5

11.1 13.9 





52.4 65.5







11.1 13.9



46.9 58.6


Figure 7A–14. Collector forces along gridline 4 with opening taken into account

Figure 7A–14a differs from Figure 7A–11a in that the distributed shear being dragged along the line in Figure 7A–14a is larger around the opening and zero at the opening on the side of the line where the opening occurs. The distributed shear along the other side of the line is unchanged from what was shown in Figure 7A–11a. The maximum collector forces in Figure 7A–14b are larger at gridlines C and D than those calculated in Figure 7A–11b. As illustrated above, openings in diaphragms can have a significant impact on the members surrounding them as well as on the diaphragm shear demands. While diaphragm analysis done without considering these openings is easier to calculate, it is important for the engineer to account for them in order to capture all of the demands on the elements that are part of the diaphragm.

4.2 ANALYSIS AT RE-ENTRANT CORNER Similar to openings in diaphragms, re-entrant corners create a concentration of forces at the portions of the diaphragm near them. Increased shear demand is created within the area of the diaphragm that has the narrower dimension. Chord forces are also developed at the re-entrant corner that must be dragged back into the main portion of the building. Figure 7A–15 illustrates how forces in the north-south direction are transferred at the re-entrant corner along gridline B, creating additional forces within the diaphragm in this area.

228


Design Example 7A




Figure 7A–15. Forces at re-entrant corner

From Table 7A–3 we know that R A = 72.1 k and from previous calculations we know that w1 = 517.2 plf. Utilizing these values and the dimensions of the re-entrant corner, l = 30 feet and d = 60 feet, the chord forces created are

− M r

c

r

= 72 1 k × = ×3 c M d

30 ft

1698 -ft 60 ft

56 6 k

1698 k-ft 28 3 k

These axial chord forces must be accounted for when designing beams along gridlines 2 and 4, and consideration must be taken to ensure that these axial forces are adequately dragged into the main portion of the diaphragm beyond the re-entrant corner.


229

Design Example 7A




In order for the BRBFs along gridline A to resist loading from their tributary portion of the diaphragm in the north-south direction, the diaphragm shear to the east of the re-entrant corner must be transferred to the west of the re-entrant corner. This creates a much higher shear demand along gridline B than was previously accounted for in the diaphragm analysis done without considering the re-entrant corner. The diaphragm shear demand along gridline B can be calculated as 56.6 k v =

=

56.6 k 60 ft

= 943 plf

Without accounting for the re-entrant corner, the distributed diaphragm shear along gridline B would be much lower, as it would be determined using the length of the entire diaphragm, which is 120 feet, instead of the length of the diaphragm at the re-entrant corner.

5. Diaphragm Design 5.1 DESIGN FOR SHEAR DEMANDS Based on the above calculations done for diaphragm analysis without consideration for openings or reentrant corners, with consideration for the openings, and with consideration for the re-entrant corners, the maximum diaphragm shear demands occur along gridlines A and F. vu,max = 1201 plf.

Diaphragm capacities must be obtained from a properly certified evaluation report produced by an accredited service. From the given information of this design example, per XYZ-ES Report #1000 for ACME Steel Roof Decks, vall = 1500 plf for 16-gage deck with 1 1 ⁄ 2-inch rib height, 7-weld pattern per sheet to supports with 11 ⁄ 2-inch top seam welds at side laps at 12 inches on center for beam spans up to 10 feet.

The diaphragm shear demand must be converted to an allowable value: vmax = 1201 plf*0.7 = 840.7 plf.

Per Section 2.4 of this design example, the connections of the diaphragm to vertical elements and collector members must be designed for a 25 percent increase to the calculated diaphragm force. Therefore the actual diaphragm shear demand is vmax = 840.7 plf*1.25 = 1050.9 plf < 1500 plf.

The capacity of the chosen deck, based on the attachments to the supporting members, can adequately transfer diaphragm forces to the collector members and vertical elements of the SFRS.

5.2 BARE METAL DECK DIAPHRAGM CAPACITY Bare metal deck diaphragm capacities are largely controlled by the type and spacing of fasteners used to attach the deck to the supporting structure. Just calculating the capacity of these fasteners, however, will not accurately provide the capacity of a bare-metal deck diaphragm. It is also influenced by the type

230


Design Example 7A




and spacing of fasteners used at side seams, the spacing of the supporting members, and the deck profile (rib height and width, rib spacing, and thickness of steel). When the diaphragm spans long distances between supports, buckling of the metal deck becomes an issue and may be the governing factor of the deck capacity. Deck capacities provided in approved testing agency reports are based on a combination of calculations used within the industry and testing data in order to determine capacities for a variety of configurations.

5.3 DISCUSSION OF DECK FASTENER DUCTILITY Bare metal decks can be fastened to the structural steel members supporting it by many different means. In this design example puddle welds are utilized, but self-tapping screws and powder-actuated fasteners can also be used as connectors. It is important to note that while welds do provide a stronger connection from the deck to the supporting members, they are not the most ductile form of attachment. Self-tapping screws and powder-actuated fasteners have been found to demonstrate more ductility than welds, and thus may be a more favorable method of attachment.

6. Chord and Collector Design 6.1 DISCUSSION OF O AND COMBINING FORCES FROM PERPENDICULAR LOADING As previously discussed in Section 3.2 of this example, chord members are not required by code to be designed for amplified, Ωo-level forces. This discrepancy between forces levels for chord and collector members is unclear, as the overstrength factor is provided to prevent non-ductile failure modes. ASCE 7 does not explicitly require that chord and collector forces be considered to act simultaneously, but there are cases, such as at cantilevered portions of diaphragms or at discontinuities in the diaphragm, where it is highly recommended to consider both of these forces in the design of these members.

6.2 COLLECTOR COMPRESSIVE STRENGTH The collector beam along gridline 4 between gridlines B and C will be designed, and for illustrative purposes its compressive strength will be determined for both the deck parallel condition and the actual deck perpendicular condition. Only the capacity based on the deck perpendicular will be used for the final design, as this matches the actual configuration of the building. When the deck is parallel to the beam, the beam can be considered to be completely unbraced for major axis buckling and braced only at points where perpendicular beams frame into it for minor axis and torsional buckling. The initial size chosen is a W18 × 35, and its compressive strength will be calculated in accordance with AISC 360 Chapter E, “Design of Members for Compression.” Per AISC 360 Table B4–1, check beam flanges for slenderness:

=

.38

=

.56

F

F

9 15

AISC 360 T B4.1b

13.49 .

AISC 360 T B4.1a


231

Design Example 7A




Therefore, the flanges of a W18 × 35 are compact for flexure and non-slender for compression. Next, check the beam web for slenderness:

h

=

.76

=

≥ 1 49

F

= 90.55

AISC 360 T B4.1b

33 88.

AISC 360 T B4.1a

E

Therefore, the web of a W18 × 35 is compact for flexure but slender for compression. The compressive strength of this member shall be calculated per AISC 360 Section E7, “Members with Slender Elements.” But first, F cr , the flexural buckling stress, must be calculated in accordance with Section E3 of AISC 360. K

= =

1 0 × 360 in 7 04 in 1 0 × 360 in 1 22 in

= 51 14 =

> 200 . This ratio should be less than 200 per AISC 360 E2 User Note.

Therefore, provide a nominal perpendicular beam with a full-depth shear tab connection to the W18 × 35 beam at its mid-span to cut the un-braced length in half: K L

=

r y F

=

 

1 0 × 180 in 1 22 in

=

2

=

.71

29,000 k si

= 13.1 ksi

(147.5

Therefore,

F y

= 113.4

11 5 ksi.

AISC 360 Eq E3–3

The W18 × 35 beam is composed of compact unstiffened elements and slender stiffened elements, so the compressive strength is calculated per AISC Section E7:

Qs = 1.0 and Q =

Because Therefore,

=

≤ 0 56

where Aeff is based on be E

= 0

=

29,000 k si 11 5 ksi

× 10

= 28 1 , Qs = 1.0. .6 k .

AISC 360 Eq E3–1

Because the torsional and lateral restraints are provided at the same locations, minor axis flexural buckling will govern the compressive strength of the collector. Torsional buckling need not be checked. As previously mentioned, the capacity of the W18 × 35 beam will be checked for the actual deck perpendicular condition. The deck running perpendicular to the beam can be assumed to act as continuous bracing for minor axis buckling, provided Constrained Axis Flexural-Torsional Buckling is checked.

232


Design Example 7A




AISC Equation E4–4 for elastic torsional buckling stress is modified as shown below to determine the compressive strength of a member accounting for Constrained Axis Flexural-Torsional buckling:

F

+

=

(K L

a ) 2

 + GJ

1

+a

where a = distance from the member centroid to the location of lateral restraint on the minor axis ( d /2 for wide flange beams). Constrained Axis Flexural-Torsional buckling governs over major axis flexural buckling, so

F

n

=

+

×

(1 0 × 180 in) 1

510 in

4

8.85in)2

19 76 ksi

and

8.85 in )

Therefore,

cr

g

2

)+

× 0.506 in

= 19 76 ksi

17 33 ksi .

=

s



AISC 360 Eq E3–3

× 10.

=

.7 k .

AISC 360 Eq E3–1

As calculated above, Constrained Axis Flexural-Torsional buckling allows for a higher compressive strength than minor axis flexural buckling (50 percent more capacity in this case), so it should be used to calculate the compressive strength of a collector beam when the deck runs perpendicular in order to allow for a more efficiently sized member.

6.3 COLLECTOR FLEXURAL STRENGTH The flexural strength of the W18 × 35 collector beam will be calculated in accordance with AISC 360 Chapter F, “Design of Members for Flexure.” Because the W18 × 35 beam has compact flanges and a compact web for flexure, local buckling need not be considered. Check beam yielding:

= 0

.5 in

= 2992.5 k -in.

AISC 360 Eq F2–1

Check beam lateral torsional buckling per Section F2–2 of AISC 360. For W18 × 35: L p = 4.31 ft

AISC Steel Manual T 3–2

Lr = 12.3 ft

Therefore Lb > Lr , so M

F S

M

AISC 360 Eq F2–3


233

Design Example 7A




Where, per AISC 360 Equation F2–4: C

F r =

E

   s 

=

1 0×

1 0.078

S

× 29,000 k si  180 in   1 51 in  2

 r ts 

L

c

1 + 0 78

 180 in

.506 × 1 0

3 in 1 52 in

57.

= 25 2 ksi

C b = 1 has conservatively been used in this calculation.

= 0 9

Therefore,

× 57 6 in

= 108 8 k-f f .

6.4 COLLECTOR DESIGN FOR COMBINED LOADING The capacities for compression and flexure have both been calculated. In order to determine the adequacy of the W18 × 35, these capacities must be checked in combination against the compressive and flexural demands. The axial load on the beam will be taken as a combination the full collector load, which is amplified based on Section 12.10.2.1 of ASCE 7, calculated considering the opening in Section 4.1 and 30 percent of the chord force calculated at the re-entrant corner in Section 4.2.

=

+ 0 3×

= 55 4 k .

The flexural demand on the beam is due to dead, live, and vertical seismic loads at the roof, which have been provided in Appendix A. This W18 × 35 beam has 10 feet of tributary roof. M

=

(( .2 +

. 0) ×

+

.5 200 p f ) × (

)

8

=6

1 k -ft

The maximum axial load and the maximum flexural load calculated above do not occur at the same location along the beam. The maximum axial load occurs at the left end, and the maximum flexural load occurs at midspan. These maximum axial and flexural demands will be checked in accordance with AISC 360 Chapter H. Pr

P

55.4 k

P

P

160.7 k

Therefore,

234

P P

8

M M



=

02.

8

60 1 k-ft

9

108 8 k-ft

+ ×

=

.84

1.0 .


AISC 360 Eq H1–1a

Design Example 7A




7. Collector Connection Design

AISC 360

This design example will check a simple shear tab connection for combined axial and shear loads. When horizontal and/or vertical loads are larger than the ones present at the end of this collector beam, a stronger connection will be required. Some commonly used connections for collectors with large forces are shear tabs with double rows of bolts or welded flanges. The reactions at this connection are T u = Pu = 55.4 k and V u =

( .4 ×

+ 0. × 2

× 30 ft

8.0 k .

Figure 7A–16 provides a detail of this connection.

1/2" SHEAR TAB WITH 5-7/8" DIAMETER A325N BOLTS

W18X35COLLECTOR COLLECTOR W18X35 BEAM

WF COL PER PLAN

5/16

Figure 7A–16. Collector connection detail


235

Design Example 7A




Check Shear Plate Yielding The 1 ⁄ 2-inch shear tab will be checked for combined loading in accordance with Section J4. The axial force causes a moment in the weak axis of the shear tab due to the eccentricity between the centerline of the beam web and the centerline of the shear tab.

=

M

1

1

5

2

2

16

=

.5 k -in .

The plate will be acceptable for the combined loading if



V V

  +

P

+

P

2

8

M uy



9

M



10.

Where the shear strength of the plate is

=

t

×06×

× 0 5 in ×

= 225 k ,

Eq J4–3

the compressive strength of the plate is F r KL

=

1 0 × 4 in

=

5 in

E

≤ 4 71

= 113 4

12 F

=

E

 

=

P

× 29,000 k si (27.7

= 373

si

Eqs E3–1, E3–2 and E3–4

r

 ∴

2

y



=



50 ksi s

=

=

 50 si

× 15 in

.3

47 3 k si

319 k si

and the flexural strength of the plate is n

t 2

 8 k  ∴ +  225 k 

× 15 in

= 4

k si

2

55. k

.5 -in 

+

k-in 

=

≤1

.

The 1 ⁄ 2-inch × 15-inch shear tab will not fail in yielding under the combined loading. Check Plate Rupture Plate rupture is checked based on the net area of the plate for combined axial and shear loading in accordance with Section J4.

 Vu  +  V = n t 236

T u T n 

10

n × 15 in −

n

n

/ 8 in)

= 5 in 2 .


Eq F2–1

Design Example 7A




Using Anet, the capacities of the plate in shear and tension are

× 0.

e

=

2 65 ksi = 243 8 k = 0 75  8 k  55 .0 ∴ + =  146.3 k   243.8 k 

.3 k

Eqs J4–2 and J4–4

The 1 ⁄ 2-inch × 15-inch shear tab will not fail in rupture under the combined loading. Check Bolt Shear Bolt shear for combined vertical and horizontal loading is checked in accordance with Section J3.6. The plate has been sized to be weaker than the bolts in order to provide a more ductile failure mode. R

u

+

=

+

56 k

and 5 DCR =

48 ksi

= 121 5 k

Eq J3–1

.0 .

The 57 ⁄ 8-inch-diameter A325N bolts will not fail in shear, and the plate will yield before the bolts fail. Check Bolt Bearing Bearing strength at bolt holes is checked in accordance with Section J3.10. Bolt bearing w ill be checked for loading in the horizontal direction only, as the vertical load is small enough to be neglected. The beam web, with a thickness of 5 ⁄ 16 inch will control over bearing at the 1 ⁄ 2-inch-thick shear tab. The holes are standard size, and deformation at the bolt hole under service loads is a design consideration, so t

≤ ×

4 t

Eq J3–6a

where Lc = 2 in − 1/2(7/8 in + 1/8 in) = 1.5 in t = 5/16 in F u = 65 ksi d = 7/8 in

∴φ Rn = 0.75 × 1.2 × 1.5 in × 5/16 in × 65 ksi ≤ 0.75 × 2.4 × 7.8 in × 5/16 in × 65 ksi

≤ 32 k

R

=

27 42 k

137 1 k

DCR = 0 41


237

Design Example 7A




Check Weld Strength The strength of the two 5 ⁄ 16-inch fillet welds will be checked in accordance with Section J2.4, and for simplicity, the vertical loading will be ignored. F y t

w

.6 ×

R

×

2

×2

2

= 209 k

Eq J2–4 and J2–5

02 Check Block Shear Block shear will be checked in accordance with Section J4.3. As in previous checks, the vertical load can be neglected, and the connection will be checked for block shear due to the horizontal load only. The beam web, being thinner than the shear plate, will govern.

≤

U F

F y

gv

+ U s F

nt

)

Eq J4–5

where

= × 5 /16 in × − 0 5 7 / 8 + = × 15 − × 1 5 in − nt = × 5 /16 in = 1 25 in2 n

U s

10

∴

= 0 75

0 9375 in

0 75 0.

≤ 195 R

.25 in

+

=0

375 in 2

+ 1 8 in ) = 3.

n

.

)

.

n )

k

195.7 k 0 28 .

Therefore, the simple shear tab connection detailed in Figure 7A–16 is adequate to resist the combined loading. The calculations were simplified by the fact that the vertical load due to gravity was small enough to be neglected. When the vertical reaction is significant, Astaneh’s “Design of Shear Tab Connections for Gravity and Seismic Loads” is a good resource.

8. Items Not Addressed in This Design Example The following items are not addressed in this example but are nevertheless necessary for a complete design of the bare metal deck roof diaphragm: • Design for gravity loads • Support for out-of-plane wall loading.

238


Design Example 7B Concrete-Filled Deck (Rigid) Diaphragm OUTLINE 1. Determination of Diaphragm Forces and Code Discussion 2. Diaphragm Analysis 3. Diaphragm Design 4. Chord and Collector Design 5. Discussion of Semi-Rigid Diaphragms 6. Items Not Addressed in This Example

1. Determination of Diaphragm Forces and Code Discussion

ASCE 7

1.1 DISCUSSION OF TORSION As mentioned in the Overview, concrete-filled metal decks are allowed to be classified as rigid per Section 12.3.1.2. This building has Horizontal Irregularity types 1a and 2, so per Section 12.3.1.2, it is required to model this building to account for the actual stiffness of the diaphragms (semi-rigid modeling). In order to provide a simplified calculation that can be presented in this design example, however, the semirigid modeling requirement will be ignored, and the concrete-filled metal deck diaphragms in this building will be assumed to be rigid. A discussion of semi-rigid diaphragm modeling will be presented at the end of this example. Rigid diaphragms must be designed to account for torsion, as they are assumed to have adequate stiffness to distribute torsional loads. Torsion is created when the center of mass (COM) and the center of rigidity (COR) do not coincide. A moment is induced in the diaphragm equal to the force applied at that level multiplied by the distance between the COM and the COR, e x. Per Section 12.8.4.2, all non-flexible diaphragms must include a 5 percent offset of the COM in each direction perpendicular to the applied load to account for any accidental torsion. This is not a minimum offset. For diaphragms where the COM and the COR are at the same location, such as in the north-south direction of this building, the value of the eccentricity due to accidental torsion, ea, which is equal to 5 percent of the building width, will be the total offset between COM and COR. Where the COM and the COR are not at the same locations, such as for loading in the east-west direction of this building, the offset from accidental torsion is added to the inherent torsion, so that the total offset between COM and COR becomes e x ± ea. Figures 7B–1 and 7B–2 illustrate the torsional effects for loading in each direction. The accidental torsion is included to account for any non-uniform inelastic behavior of the elements of the SFRS as well as any uneven distribution of loads. The components of the SFRS must be designed for the worst-case load effects due to accidental torsion applied in either direction of the actual COM.


239

Design Example 7B



Concrete-Filled Deck (Rigid) Diaphragm

V2nd

V2nd

COM COR

COR

-ea

+ea

V2nd

Figure 7B–1. Accidental torsion for north-south loading

COR V2nd ex +ea -ea

V2nd

COM

Figure 7B–2. Accidental torsion for east-west loading

240


Design Example 7B




Section 12.8.4.3 requires that accidental torsion be amplified when the story drifts at the extremes of the diaphragm exceed a certain ratio of the average story drift. This torsional amplification factor, A x, has been developed for buildings with vertical elements of the SFRS laid out in such a way that large amounts of torsion will occur. The factor A x is applied to the 5 percent required accidental torsion, increasing it in order to account for the fact that buildings with irregular layouts will have more non-uniform inelastic behavior in the elements of the SFRS.

1.2 FORCE DISTRIBUTION IN RIGID DIAPHRAGMS WITHOUT THE USE OF AN ANALYTICAL MODEL One way to design the second floor diaphragm, its chords, and its collectors is to perform a rigid diaphragm analysis by hand (or with the help of a spreadsheet). A rigid diaphragm is assumed to distribute lateral load to the vertical members of the SFRS based on the vertical members’ relative rigidities. By calculating the rigidity of each BRBF and their relative rigidities, the COR can be determined. The COM must also be calculated, as well the distance between the COM and COR including the effects of accidental torsion. This type of analysis would be done using the calculated diaphragm forces provided in Table 7A–2 of Example 7A, and no analytical model of the building would be required. The results of a hand distribution of forces in a rigid diaphragm for loading in the north-south direction would be exactly the same as the flexible diaphragm analysis done in Example 7A that included the 5 percent accidental eccentricity, as the BRBFs along gridlines A and F have the same stiffness. The only difference is that the moment created by the offset between the COM and the COR must resolved by the BRBFs in the east-west direction. For the flexible diaphragm analysis, resolving this moment was not required because including accidental eccentricity was not required, but because the accidental eccentricity must be included for a rigid diaphragm analysis, all forces must be resolved. As shown in Figure 7B–3 below, shear force is developed within gridlines 1, 3, and 4 when performing a rigid diaphragm analysis as a result of the moment created by the accidental torsion applied in the positive direction.

VT,4

COM COR

V2nd/2-VT,A

V2nd/2+VT,F

VT,3 +ea V2nd

VT,1

Figure 7B–3. BRBF forces due to positive accidental torsion in north-south direction 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

241

Design Example 7B




1.3 FORCE DISTRIBUTION IN RIGID DIAPHRAGMS WITH THE USE OF AN ANALYTICAL MODEL It is more efficient and usually more accurate (provided the modeling assumptions made are correct) to create an analytical model of a building with rigid diaphragms. When this approach is taken, the designer can use the forces in the vertical elements of the SFRS determined by the analysis program to calculate the diaphragm forces. These forces must be scaled up to the level of the diaphragm force calculated in accordance with Section 12.10.1.1.

2. Diaphragm Analysis 2.1 CALCULATION OF DIAPHRAGM SHEARS This example will use the results from the analytical model created for Example 3 to calculate the shears in the second-floor diaphragm. Table 7B–1 provides the section cuts from the ETABS model along each line of lateral load resistance in the north-south direction for the stories above and below the second floor with accidental eccentricity applied in each direction. The diaphragm shear along each line is then scaled up to the diaphragm force calculated in accordance with ASCE 7 Section 12.10.1.1 from Table 7A–2 of Example 7A. Table 7B–2 provides the same information as Table 7B–1 but for loading in the east-west direction. Table 7B–1. Diaphragm design forces for loading in north-south direction V 2 (kips)

V 1 − V 2 = F px (kips)

Line

+ / − e x

V 1 (kips)

F px,des (kips)

A

+

297.39

284.6

12.79

127.1

F

+

350.8

337.13

13.67

135.9

Sum

+

648.19

621.73

26.46

A

−

350.8

337.13

13.67

135.9

F

−

297.39

284.6

12.79

127.1

Sum

−

648.19

621.73

26.46

263

263

V px,des (kips)

263

263

Table 7B–2. Diaphragm design forces for loading in east-west direction

242

Line

+ / − e x

V 1 (kips)

V 2 (kips)

V 1 − V 2 = F px (kips)

1

+

174.62

169.7

4.92

43.9

3

+

200.42

188.95

11.47

102.3

4

+

211.38

198.27

13.11

116.9

Sum

+

586.42

556.92

29.5

1

−

190.71

183.19

7.52

67.0

3

−

197.08

186.11

10.97

97.8

4

−

198.47

187.46

11.01

98.2

Sum

−

586.26

556.76

29.5


F px,des (kips)

263

263

V px,des (kips)

263

263

Design Example 7B




The story force is much lower than the prescribed diaphragm force at the second floor. The difference between story force and diaphragm force becomes more pronounced lower in the building for reasons discussed in Section 2.2 of Example 7A. The fact that the minimum diaphragm force governs per Equation 12.10–2 of ASCE 7 further highlights this difference, as the high R value for this BRBF building allows for relatively low design forces in the vertical elements of the SFRS. Also, the story shears calculated from the analytical model are lower than the second-floor force calculated in Table 7A–1 of Example 7A. This difference is because the ETABS results are based on a R esponse Spectrum Analysis, which produce a different distribution of forces than the Equivalent Lateral Force Procedure distribution calculated in accordance with ASCE 7 Equations 12.8–11 and 12.8–12. The Response Spectrum Analysis is also why the story shears differ in the north-south and east-west directions. The building has different modal responses in each direction, which creates a different force distribution in each direction. Regardless of the analysis type used, however, the diaphragms must be designed for forces calculated in accordance with ASCE 7 Section 12.10.1.1. The internal diaphragm shears can be calculated from the scaled diaphragm shear reactions at each BRBF line in each direction for positive and negative accidental torsion. An efficient way to do this is to create a spreadsheet that calculates distributed loads based on the geometry of the diaphragm and the locations of the braced frames. Using the reactions at the BRBF locations and the distributed loads, the internal shear diagram can easily be calculated. Figure 7B–4 shows the input that would be used for loading in the north-south direction with positive eccentricity, and Figure 7B–5 shows the required input for loading in the east-west direction with positive eccentricity. For both loading directions, input for negative eccentricity is identical to the input for positive eccentricity except that the frame reactions are different.

  

     

     

 

    

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     

   

      

      

         

Figure 7B–4. Spreadsheet input to calculate diaphragm shears in north-south direction


243

Design Example 7B




  

     

       

 

            

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   

   

      



         

  

Figure 7B–5. Spreadsheet input to calculate diaphragm shears in east-west direction

The resultant internal shear diagrams for loading in the north-south direction for positive and negative eccentricity are shown in Figure 7B–6, and the resultant internal shear diagrams for loading in the east-west direction for positive and negative eccentricity are shown in Figure 7B–7.

      

      

127.1k

 

         

135.9k

 



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

 

      





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

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



135.9k

    



127.1k

    

Figure 7B–6. Diaphragm shears for loading in north-south direction

      

       

49.3k

43.9k

           















 

    

    







87.6k

67.6k

  

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  

     

64.5k      

Figure 7B–7. Diaphragm shears for loading in east-west direction

244

49.3k

33.3k



 

67k



14.6k


48.9k



Design Example 7B




As previously mentioned, in a rigid diaphragm analysis, forces are developed in the vertical elements of the SFRS perpendicular to the applied loading to resolve the torsional moment. These forces are presented in Tables 7B–3 and 7B–4 for the second floor based on the ETABS results from Example 3. Table 7B–3 provides the forces in gridlines A and F caused by loading in the east-west direction scaled up to the diaphragm-force level using the story shear from Table 7B–1, and Table 7B–4 provides the forces in gridlines 1, 3 and 4 caused by loading in the north-south direction scaled up to the diaphragm-force level using the story shear from Table 7B–2.

Table 7B–3. Forces created in north-south direction by loading in the east-west direction V 2 (kips)

V 1 − V 2 = V px (kips)

Line

+ / − e x

V 1 (kips)

A

+

51.82

51.4

.42

9.94

4.2

F

+

51.82

51.4

.42

9.94

4.2

A

−

13.27

13.38

−.11

9.94

−1.1

F

−

13.27

13.38

−.11

9.94

−1.1

Scale Factor

V px,des (kips)

Table 7B–4. Forces created in east-west direction by loading in the north-south direction V 2 (kips)

V 1 − V 2 = V px (kips)

Scale Factor

V px,des (kips)

Line

+ / − e x

V 1 (kips)

1

+

11.15

9.32

1.83

8.9

16.3

3

+

2.32

1.96

0.36

8.9

3.2

4

+

8.95

7.47

1.48

8.9

13.2

1

−

11.15

9.32

1.83

8.9

16.3

3

−

2.32

1.96

0.36

8.9

3.2

4

−

8.95

7.47

1.48

8.9

13.2

The forces calculated in Tables 7B–3 and 7B–4 will not be used to calculate internal diaphragm shears in each direction, but they will be included in the design of the diaphragm and its components when accounting for combined orthogonal effects in accordance with ASCE 7 Section 12.5.4.

2.2 CALCULATION OF CHORD FORCES The internal diaphragm moments can also be calculated from the scaled diaphragm shear reactions at each BRBF line in each direction for positive and negative accidental torsion. An efficient way to do this is to create a spreadsheet that calculates distributed loads based on the geometry of the diaphragm and the locations of the braced frames. Internal moments are calculated from one end of the building based on the BRBF reactions and the distributed diaphragm loading. Because of the torsional moments, the moment diagram created this way will not equal to zero kip-feet at the other end of the building. This non-zero moment at the end becomes the unbalanced moment shown in Figures 7B–8 and 7B–9, which is distributed throughout the diaphragm as the correcting moments based on the tributary diaphragm area along its length. 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

245

Design Example 7B




The internal moments are then summed with the corrected moments to create the final moment diagram for the diaphragm. Figure 7B–8 shows what input would be used for loading in the north-south direction with positive eccentricity and Figure 7B–9 shows the required input for loading in the east-west direction with positive eccentricity. For both loading directions, input for negative eccentricity is identical to the input for positive eccentricity except that the frame reactions are different.

  

     

     

 

  

    

           

     

   

      

  

   

               

Figure 7B–8. Spreadsheet input to calculate diaphragm moments in north-south direction

  

     

           

     

 

  

             

   

      

 

   



               

Figure 7B–9. Spreadsheet input to calculate diaphragm shears in east-west direction

The resultant moment diagrams for loading in the north-south direction for positive and negative eccentricity are shown in Figure 7B–10, and the resultant internal moment diagrams for loading in the eastwest direction for positive and negative eccentricity are shown in Figure 7B–11.

246


Design Example 7B




           

      

5671.2 k-ft





             

5671.2 k-ft







             

















 



















    

















    

Figure 7B–10. Diaphragm moments for loading in north-south direction

               

1076.6 k-ft

 

  

734.4 k-ft

 

   

    

             

     







  











             

381.4 k-ft



   











 







53.3 k-ft



903.4 k-ft



31.1 k-ft



          

Figure 7B–11. Diaphragm moments for loading in east-west direction

Based on the moment diagrams for loading in each direction provided in Figures 7B–10 and 7B–11, the chord forces within the boundary members are P

=

5671 2 -ft 120 ft

P

= max

P

= max

= 47.3 k

 903 4 k-ft 0 ft

 381.4 k-ft 150 ft

,

1076 6 k-ft 90 ft 31.1 k-ft  150 ft

max(

= max(

.5 k

12 k 2 k

= 12 k .5 k .

See Section 3.2 of Example 7A for a discussion of chord forces.

2.3 CALCULATION OF COLLECTOR FORCES Collector forces are calculated for rigid diaphragms in almost the same manner that they are calculated for flexible diaphragms. The only difference is that the shears caused by loading in the perpendicular direction due to the torsional moment must be included in the collector diagrams for 100 percent + 30 percent orthogonal load combinations. The collector diagrams along gridlines 1, 3, and 4 are shown in Figures 7B–12, 7B–13, and 7B–14 respectively. The collector diagram along gridlines A and F is shown in Figure 7B–15. These diagrams have been created using the maximum forces caused by positive and negative accidental eccentricity. They have also been created taking the openings between gridlines C and D into account, which is further discussed in Section 4.1 of Example 7A. At the second floor, the collector forces are scaled up to 330 kips per Table 7A–4 from the diaphragm force of 263 kips per Table 7A–2. 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

247

Design Example 7B






) 

v = 330 vk = /263 k * (67+k0.3*16.3k) + 0.3*16.3/ 90ft k) / 90 ft = 1002 2.5*(67k = 1997 plf plf

2.5*(67k+0.3*16.3k) = 179.7k V = 330Vk=/263 k * (67 k + 0.3*16.3 k) = 90.2 k (a)

k (  e c r o F  r o  t c  e l l  o C 

) k ( e c r o F r o  t c e l l o C

45.1 89.9k

59.9k 30.1







59.9k

89.9k 45.1







30.1


Figure 7B–12. Collector forces along gridline 1



v1 = 330k * (14.6k 0.3*0.5*3.2k) / 120 =ft 314 = 158 v1/ =263k 2.5*(14.6k ++ 0.3*0.5*3.2k) / 120ft plf plf

v2 = 2.5*(87.6k + 0.3*0.5*3.2k) / 150ft = 1468 plf v2 = 330k / 263k * (87.6k + 0.3*0.5*3.2k) / 150 ft = 737 plf V = 2.5*(102.2k+0.3*3.2k) 2 5* 102 2k+0 3*3 2k = = 258k 258k V = 330k / 263k * (102.2k + 0.3*3.2k) = 129k

) ) k k ( ( e e c  r c o r o F F  r r o o  t c t  e c l e l l l o  o C C

64.7 129k

53.7 106.9k 53.5k 26.8 















26.8 53.5k 129k64.7



53.7 106.9k

Length Along Collector Line (ft)

(a)

(b)


v1 = 330k / 263k * (49.3k + 0.3*0.5*13.2k) / 60 ft = 1072 plf v1 = 2.5*(49.3k + 0.3*0.5*13.2k) / 60ft = 2137 plf

v2 = 2.5*(67.6k + 0.3*0.5*13.2k) / 150ft = 1160 plf v2 = 330k / 263k * (67.6k + 0.3*0.5*13.2k) / 150 ft = 582 plf V = 2.5*(116.9k+0.3*13.2k) * * = 302.2k V = 330k / 263k * (116.9k + 0.3*13.2k) = 151.7k

133.7k 67.1

)  k ( e  c r  o F  r o  t  c e l l  o  C

75.9151.1k

34.8k 17.5 







75.9 151.1k



67.1



17.5

133.7k

Length Along Collector Line (ft)

(a)

(b)


t f l f 0 p 6 3 / 0 ) 4 k 2 2 . 330k 263k = VV==262 k //131 k* 4 t * f 0.3*4.2k) * 0.5 3 * (135.9k 72.1 k =+144.2 k . 0 0 6 = 85.5k + / k k 9 . 1 . 5 2 3 7 1 ( * * k V/ =90 330k /578 263kplf * k 1 / 131 ft =plf 3 3 k *k 26 = 2.5*26 / 90k(135.9k ft = 722 6 + 0.3*4.2k) * 0.5 1 2 / V = 262 k / 131 k / f = 85.5k l k * 72.1 k = 144.2 k k p 2 0 0 6 3 5 2 3 8 = 2 V = = 2.5*26 k = 65 k v 262 k / 131*26 k = 52 k v =



) s p  i k ( 

e c r  o F  r o  t c e l l o C

42.8 36.05 17.3 21.7

26



42.8 36.05



(a)

32.5



26

32.5

42.8 36.05



17.342.821.7 36.05


Figure 7B–15. Collector forces along gridlines A and F

248








34.8k



Design Example 7B




See Section 3.3 of Example 7A for an explanation of the collector diagrams and the use of Ωo for amplified force levels. Ωo has been applied to the collector forces caused by loading in each direction. Also, see Example 3 for the design of the beams within the BRBF. These beams must be designed to resist the collector loading provided in the diagrams above as well as the potential yielding of the braces within the BRBF.

3. Diaphragm Design 3.1 METAL DECK SELECTION For concrete-filled metal deck diaphragms, welded shear connectors are typically designed as the mechanism for shear transfer. The deck and its attachment to the structure, then, do not need to be chosen for shear strength as they would for a bare-metal deck diaphragm. When concrete-filled metal decks utilize welded shear connectors, the deck can be sized for gravity loads only. Typically construction loading will govern, as it is desirable to choose a deck that will not require shoring. Per Section I3.1b of AISC 360, the deck must adequately support all applied loading prior to the concrete reaching 75 percent of its specified compressive strength, ′ .

3.2 DISCUSSION OF CONCRETE PROPERTIES The welded shear connectors are designed to be the shear transfer mechanism, and so the concrete fill can be assumed to have adequate in-plane shear strength. The combined deck and concrete fill do need to be sized to support all applied gravity loads. Also, certain thicknesses of concrete fill help achieve one-hour and two-hour fire ratings. Reinforcing should be provided in the concrete fill to prevent cracking and also because manufacturer’s testing is often performed with a minimal amount of steel reinforcing in the concrete. In order to use the shear capacity provided by a manufacturer’s testing data, the designer must match, at a minimum, the concrete properties and the reinforcing steel used in the test specimens. Reinforcing in the concrete fill can also help increase the flexural capacity of the composite section when negative moments are applied to the beam. See Sections I3.2b and I3.2d of AISC 360 for discussions on the use of concrete fill reinforcing to support negative moments.

3.3 SHEAR STUD DEMAND AND CAPACITY The maximum shear demands along each line of lateral resistance can be calculated from Tables 7B–1 thru 7B–4: v

v1

F

= = =

=

+

135

× 4 2 k

60 t

67 k +

16 3 k

= 799 plf

2 k

= 686 plf

90 ft 102 k + 150 ft 116.

13 2 k 150 t

2286 plf

806 plf


249

Design Example 7B




The maximum diaphragm shear force occurs along gridlines A and F and is 2286 plf. This force must be increased by 25 percent per Section 12.3.3.4 of ASCE 7. See Section 2.3 of Example 7A for further discussion. The maximum diaphragm design shear becomes 2857.5 plf. Shear stud capacity can be calculated in accordance with AISC 360 Chapter I. Alternatively, the required shear stud spacing can be determined from testing provided by deck manufacturers. For composite beam design, shear studs are also the transfer mechanism between the steel beam and the concrete fill for gravity loading. When beams are simply supported, the shear force imposed on the shear connectors is applied in opposite directions from the center of the beam, so for half of the beam span, the shear demands on the welded studs are additive and for the other half, they act in the opposite direction. Figure 7B–16 illustrates the shear forces on shear studs supporting gravity loads and diaphragm loads.

CENTERLINE OF BEAM SPAN SHEAR FLOW FROM DIAPHRAGM FORCES

SHEAR FLOW FROM GRAVITY FORCES

DIAPHRAGM SHEAR IS DECREASED DIAPHRAGM SHEAR IS DECREASED BY GRAVITY SHEAR BY GRAVITY SHEAR

GRAVITY SHEAR AND DIAPHRAGM GRAVITY SHEAR AND DIAPHRAGM SHEAR AREARE ADDITIVE SHEAR ADDITIVE

Figure 7B–16. Shear loading on welded studs at composite beams

Because the maximum shear force for this example occurs along beams that run parallel to the deck, the gravity loads will be assumed to be fairly minimal. The required shear stud spacing will be determined by the diaphragm demands only. The metal deck and fill combination for the second floor has been determined to be 16-gage deck with 2-inch high ribs with 3 1 ⁄ 4-inch lightweight concrete fill having a compressive strength of 3 ksi. The deck can span 10 feet, 0 inches during construction without requiring shoring, and the 2-inch deck plus 3 1 ⁄ 4-inch concrete fill has a fire rating of two hours. The concrete fill is reinforced with #3 bars at 18 inches on center each way located at the mid-height of the concrete thickness. The shear studs will be 3 ⁄ 4 inch diameter and 4 1 ⁄ 2 inches long, which meets the requirements of AISC Section I3.2c. The load transfer between the steel beam and the concrete shall be taken as the minimum of concrete crushing, tensile yielding of the steel section, and the shear strength of welded studs. Concrete crushing and the yield strength of the beam are assumed not to govern, so the spacing of the shear studs will be designed for the maximum shear demand. E

250

≤

R R

F


AISC 360 Eq I8–1

Design Example 7B




where for deck with hr / wr ≥ 1.5, sc

= 0 44 in w1 5

= 2136 k si

15

Rg = 1 0

75 F Q

∴

65 ksi

= n

× 0 5×

×

= 13 2

3 si

×1

×

44

16 1 k

= 13.2 k

Provide shear studs at 24 inches on center for φvn = 13.2 k/2 ft = 6600 plf > vu,max = 2857.5 plf.

4. Chord and Collector Design 4.1 FLEXURAL AND AXIAL DEMANDS As in Example 7A, the collector beam along gridline 4 between gridlines B and C will be designed at the second floor. As determined in Figure 7B–14, the maximum axial collector load within this beam is 67.1 k. Additional axial force from loading in the north-south direction is created from the diaphragm reentrant corner. This additional load is 51.45 k, which was calculated using the methodology presented in Example 7A and the diaphragm shear loading from Figure 7B–4. Consequently, the total axial load imposed on this collector beam is Pu = 67.1 + 0.3*51.45 k = 82.5 k. The maximum moment is due to dead and live loads provided in Appendix 1 acting over a tributary width of 10 feet, 0 inches. M

=

( .4

+

.

( 8

)2

= 143.2 k-ft

A W18 × 46 beam with bracing at midspan will be used as the second-floor collector at this location. The axial and flexural demands are higher than in Example 7A, but the composite action from the slab provides larger flexural capacity and compressive strength of the beam is determined using Constrained Axis Flexural Torsion, so the beam does not need to be much larger than the one used at the roof.

4.2 COLLECTOR COMPRESSIVE STRENGTH As stated in the Commentary of AISC 360, there are no guidelines for computing the compressive strength of a composite beam, which is due to the lack of research and the fact that this would be a complex, rigorous calculation. In practice, the collector compressive strength is typically determined assuming noncomposite properties because of the reasons stated in the AISC 360 Commentary and because it would be potentially un-conservative to rely on the concrete fill once the diaphragm has been loaded and the fill has cracked, losing some of its compressive strength.


251

Design Example 7B




The deck braces the top flange of the beam, so the beam will be checked for Constrained Axis Flexural Torsion. A discussion of Constrained Axis Flexural Torsion is provided in Section 3.3 of Example 7A. When determining the compressive strength of a member using Constrained Axis Flexural-Torsional buckling, modify AISC 360 Equation E4–4 for elastic torsional buckling stress:

F

=

+ (K L



a )

1

GJ

2

+a

where a = distance from the member centroid to the location of lateral restraint on the minor axis ( d /2 for wide flange beams). Constrained Axis Flexural-Torsional buckling is assumed to govern over major axis flexural buckling, so

F

n

=

+

×

×

(1 0 × 180 in 1

+

9 in)

24 6

= 21 57 ksi.

712 in and

9 in)

= 0

Therefore,

.

n

 . .22 in

= 24.6 ksi AISC 360 Eq E3–3

× 13.5 in

262 k .

AISC 360 Eq E3–1

4.3 COLLECTOR FLEXURAL STRENGTH The flexural strength of the collector will be determined assuming composite action of the beam and the concrete slab with 3 ⁄ 4-inch-diameter × 41 ⁄ 2-inch-long shear studs spaced at 24 inches on center. Determine the horizontal shear force at the interface of the steel beam and the concrete fill over metal deck in accordance with Section I3.2d of AISC 360 based on the lower of the limit states of concrete crushing, steel yielding, and the capacity of the shear studs: Effective width of the concrete on each side of the beam is the minimum of: • one-eighth of the beam span = 30 feet/8 = 45 inches • one-half the distance to the center line of the adjacent beam = 10 feet/2 = 60 inches • distance to the edge of slab (will not govern for this beam). Therefore, the effective concrete width = 90 inches, as governed by one-eighth of the beam span. V

c

=

×3

×

× 3.

=

. k

AISC 360 Eq I3–1a

Only the thickness above the metal deck is considered, as concrete contained within the flutes is minimal and cumbersome to calculate.

= 50 ksi V

252

=

=

30 ft 2 ft

= 675 k ×

13 2 0 75

= 264 k .


AISC 360 Eq I3–1b AISC 360 Eq I3–1c

Design Example 7B




Therefore, the strength of the shear studs governs. Table 3–19 of the AISC Steel Manual will be utilized to determine the flexural strength of this composite W18 × 46.

−

/ 2

where

=

Y

=

n + 3 25 in = 5 25 in

=

0 85

∴ = 5

264 k 0 85

−

0i

1 15 in

= 4.675 in .

.

Knowing that the concrete has much less strength that the steel, and based on Qn vs. AgF y, assume that the PNA occurs in the web of the W18 × 46. 264 k

s

×6

n + 50 ksi

− 50 ksi

n

n

in

s

× 0 36 in ×

= 1 3 in. The PNA occurs 1.94 inches from the top of the steel beam, so Y 1 = 1.94. Using Table 3–19 and interpolating between Y 2 values of 4.5 and 5 and interpolating between Y 1 values of 0.57 and 2.08, the flexural compressive strength of this W18 × 46 is φ M n = 549.6 k-ft.

4.4 COLLECTOR DESIGN FOR COMBINED LOADING The maximum axial load and the maximum flexural load calculated above do not occur at the same location along the beam. The maximum axial load occurs at the left end, and the maximum flexural load occurs at midspan. These maximum axial and flexural demands will be checked in accordance with AISC 360 Chapter H. P

Pu

Pc

Pn

Therefore,

+

=

82 5 k 262 k

8

=



=

.2

+

8

141.6 k-ft 549 6 k-ft

=

54

1 0.

AISC 360 Eq H1–1a

Example 7A provides a design for the collector beam connection to the column. The connection at the end of this W18 × 46 beam will be similarly designed.

5. Discussion of Semi-Rigid Diaphragms Section 12.3.1 of ASCE 7 explicitly requires semi-rigid modeling of the diaphragm to be considered in the building analysis when the diaphragm cannot be classified as rigid or flexible. Semi-rigid diaphragm modeling is also required when diaphragms are of a type that be classified as rigid but horizontal irregularities exist such that per Section 12.3. 1.2 of ASCE 7 they cannot be considered rigid. The reason for semi-rigid diaphragm modeling is to accurately capture the behavior of the building, including the effects of the diaphragm. It is especially important when irregularities are present, as they alter the stiffness of the diaphragm.


253

Design Example 7B




Computer analysis programs will allow the designer to choose a semi-rigid diaphragm option. When this type of diaphragm is chosen, the program will require that the diaphragm in-plane stiffness be provided. This stiffness is typically based on the thickness of the concrete fill over the metal deck, neglecting the deck and the portion of the concrete within the deck. Cracked properties should be used for the concrete where diaphragm shears are large, and industry practice is typically to choose between 0.15 and 0.5 for the cracked section modification factor. Because this is a wide range of values and because the stiffness of the diaphragm can greatly impact that load distribution to other elements of the SFRS, it is advised either to envelope results based on reasonable minimum and maximum cracked section modifier values or to perform sensitivity studies to determine how much the diaphragm stiffness affects the transfer of forces to the vertical elements of the SFRS. As mentioned in the Overview of this example, the stiffness of the diaphragm is largely determined by the stiffness of the metal deck or concrete fill and metal deck, but it can be impacted by the stiffness of the other diaphragm components. When the designer is creating a semi-rigid diaphragm model, it may be important to include the actual axial stiffness of the chord and collector elements. Consequently, the designer may want to run a sensitivity check to determine whether or not this will impact the behavior of the building.

6. Items Not Addressed in This Example The following items are not addressed in this example but are nevertheless necessary for a complete design of this concrete-filled metal deck roof diaphragm: • Design for gravity loads, • Support for out-of-plane wall loading, and • Semi-rigid diaphragm analysis.

254


Design Example 8 Special Moment Frame Base Connection OVERVIEW This design example is a procedure for the design of a base connection for a Special Moment Frame (SMF). This example demonstrates a design using a possible anchorage configuration for columns with moderate expected anchorage forces. In practice, these connections generally consist of a plate welded to a first-story column at its base, supported by a grout pad, anchored to the foundation using anchoring bolts. When the base plate and anchor heads are positioned above the surface of the foundation, the connection is referred to as an Exposed Base Connection. In this case, as in this example, transferred loads are expected to be resisted entirely by base-plate bearing, plate bending, anchor rods in tension, and grout/concrete compressive strength. When large overturning moments or tensile forces are expected at the base of the structure (usually in high-rise buildings), columns can be embedded in the foundation to provide additional strength. This type of base connection is referred to as an Embedded Base Connection and is not considered in this design example. Laboratory testing (e.g. DeWolf and Sarisley 1980; Thambiratnam and Paramasivam 1986; Astaneh et al., 1992; Fahmy et al., 1999; Burda and Itani 1999; Myers et al., 2009; Gomez et al., 2010), has confirmed that exposed base connections are susceptible to various failure modes. As a result, efforts have been made to characterize the strength of Exposed Base Connections under various loading conditions. In this example, the results from these laboratory tests and pertinent design guidelines are utilized to identify limit states and determine the strength of each component of the base connection. This example assumes a typical configuration of an exposed base connection for seismic regions: anchor rods are placed along lines parallel to and outside the connecting column flanges. The following base connection design is based primarily on two sources (1) AISC Design Guide 1—Base Plate and Anchor Rod Design (Fisher and Kloiber, 2010) recommendations, along with ACI 318 Appendix A for anchorage design, and (2) a compilation of test observations from various studies.

OUTLINE 1. Building Geometry and Loads 2. Determine Required Base Connection Strength 3. Design Base Plate and Anchor Rods 4. Base Connection Shear Resistance 5. Design Column-to-Base Plate Weld 6. Shear Resistance Discussion 7. Anchorage


255

Design Example 8



Special Moment Frame Base Connection

1. Building Geometry and Loads

ASCE 7

1.1 GIVEN INFORMATION The building is a six-story office building located in San Francisco, CA, and the Seismic Design Category is D. See Appendix 1 for the following information: • Building dimensions, • Latitude and longitude, • Soil type, • Spectral accelerations, and • Load combinations including the vertical seismic load effect. The base connection to be designed connects a W14 × 211 ASTM A992 column section (F y = 50 ksi, F u = 65 ksi) to a reinforced concrete grade beam. The grade beam is 4 feet × 3 feet in cross section and extends longitudinally in both directions along the axis parallel to the column web (see Figure 8–1). The grade beam frames into a 4-foot × 4-foot concrete pedestal, level with the grade beam and concentric with the footprint of the base plate. The ultimate compressive strength of the concrete, ′ , is 4 ksi. The first-story height is 12 feet, and the column frames into a W24 × 76 beam above. Relevant section properties for the column and beam are listed in Table 8–1.

W14 x 211

Grout

Anchor Head Grade Beam Base Plate

4 ft

General Reinforcement Layers

Figure 8–1. General base connection profile

256


Design Example 8




Table 8–1. Section properties for the column and connecting beam above

Column

Beam

b f (in)

d (in)

t f (in)

Z x (in)

d b (in)

15.8

15.7

1.56

390

23.9

1.2 SERVICE LOADS AT COLUMN BASE Load Type

Axial Force (kips)

Shear Force (kips)

Moment X-X (k-ft)

Dead

185

0

−3

Live

59

0

−2

4

39

−348

Earthquake

Applied moments about the column weak axis are assumed negligible.

1.3 LOAD CASES See Appendix 1 for a derivation of load combinations based on S DS = 1.0. The overstrength factor, Ω0, is determined from Table 12.2–1:

Ω0 = 3.0 The load combinations of consequence for the design of the column base connection are as follows:

0

m

+0 5

+ 5 L

=

E

.

2. Determine Required Base Connection Strength

AISC 341

2.1 DETERMINE REQUIRED AXIAL STRENGTH Section D2.6a requires that the axial strength be the load calculated using the load combinations of the applicable building code, including the amplified seismic load as follows: Pu

=

+ 3 0(4 ips) = 142 kips.

2.2 DETERMINE REQUIRED SHEAR STRENGTH Section E3.6d requires that the shear strength of the base connection be based on the load combinations of the applicable building code, including the amplified seismic load. The factored earthquake load, E mh, is determined by approximating the expected shear corresponding to the plastic moment capacity of the column. Here, R y is the ratio of the expected yields stress to the specified minimum yield stress F y . M p is the


257

Design Example 8




nominal plastic flexural strength of the column, and L H is the distance from the column base to the bottom flange of the beam above: 211

m

where, taking H as the story height,

−

=

2

−

23.9 in 2

= 132 5 in.

Therefore, V u

m

358 ki

=

+ 2 1 1 1 1 50

)(390

) /132. 5 in

.

2.3 DETERMINE REQUIRED FLEXURAL STRENGTH Section D2.6c requires that the flexural strength of the base connection be at least equal to the lesser of 1.1 R y F y Z x or the moment calculated using the load combinations of the applicable building code, including the amplified seismic load. Assuming the column is adequately designed to resist all load combinations, the required flexural strength is calculated as follows:

−3

M

+

−384 k-ft = − 1155 k-ft.

3. Design Base Plate and Anchor Rods 3.1 DETERMINE INITIAL BASE-PLATE DIMENSIONS The base-plate dimensions are selected to be large enough for the installation of at least four anchor rods, as required by Occupational Safety and Health Administration (OSHA, 2001). An edge distance of 4 inches is selected for all sides. An additional 4 inches is added on all sides to provide adequate room between the column and the anchor washers for anchor installation. Based on these practical constraints, the required length in the direction of bending long the column web, N , and the required width, B, of the base plate are estimated as follows: N

+ 2 4. 0

+

= 15 7

B

+ 2 4.

+

= 15.

+

+8 + 8.

31 7 in

=

8 in .

Based on these practical considerations, the initial base-plate dimensions are selected to be N = 32 inches. and B = 32 inches, with an edge distance of 4 inches.

3.2 DETERMINE BASE-PLATE THICKNESS AND SELECT ANCHOR RODS The design bearing strength of concrete is first determined using ACI 318 provisions. Concrete bearing capacity can be increased using the factor , which accounts for the beneficial effects of concrete confinement. The area A1 is the area of the base plate, and the area, A2 is the largest area that is geometrically similar to the base plate that can be inscribed on the surface of the concrete footing without going beyond the edges of the concrete. Since the base plate is connected to a 4-foot × 4-foot concrete

258


Design Example 8




pedestal, the dimensions of A2 are taken to be 4 feet × 4 feet. The design bearing strength is determined as follows: 2

)

=

65

ACI 318.9.3.2.4

× B =

× (32

=

2

4 2304 in2

)

=

.

1024 in 2

) 12

)

2304 in

=15

1024 in

1 p

ACI 318.10.14.1

.

.

.

k si .

Determine the bearing intensity acting on the base plate as follows: q

)

( .315 si

1155 k-ft

N

(32 in)

−

k-i 7 6 in

142 k t

in

−

(142 k ) .

k-in

15 3 i .

AISC Design Guide 1 (Fisher and Kloiber, 2010) provides a “large moment” design procedure for the case when e > ecrit. The assumed force distribution is shown in Figure 8–2.

ƒ

e

x

Mu

Pu

Pu

T

q qy y/2

y/2

N

Figure 8–2. Force distribution at base plate


259

Design Example 8




Determine the distance from the centerline to the anchor rod line, f , as follows: N

2

− edge dista

32 in 

c

2

− 

12 in.

Determine if the base-plate dimensions are adequate to resist the applied loads as follows:



2

+

  784

N 

≥

2

2(

) 2

in +

32 in  2

Design Guide 1, Eq 3.4.4

2 13,860 k-in



in

142 k ips 12 in)]

106. 8 k -in n . BASE CONNECTION DIMENSIONS ADEQUATE

Determine the bearing length, Y , and tension in anchor rods due to base plate uplift, T u : Y

+

2

N

+

2

Y =

N 

2  .

2(

− n

− 293.

)


in

5 85 in T

P

.


.

−(

)

478 62 ki

.

AISC Design Guide 1 considers yield lines parallel to the edge of the column flange for both compression and tension side flexure. An examination of damage patterns during experiments suggest that yield lines may form at an angle to the column flange when anchor rods are placed near the corners of the base plate only (Gomez et al., 2010). The onset of inclined yield lines may act to reduce the capacity of the plate. To avoid the onset of inclined yield lines, and to reduce anchor rod and plate hole sizes, four anchor rods are used to resist tension ( n = 4) for a total of eight anchor rods. Figure 8–3 illustrates the differences between parallel and inclined yield lines.

260


Design Example 8




Anchor Rods

0.95d

Mu

Mu

d

Inclined Yield Lines

Figure 8–3. Comparison between parallel and inclined yield lines

The tension in each rod, T i , is given by 478.62 k ips 4

120 kips.

Strengths for commonly used anchor rod materials and sizes are tabulated in AISC Design Guide 1 Table 3.1. Anchor rods are selected based on the expected tensile force in each rod due to uplift: EIGHT 2-INCH ASTM F1554 GRADE 55 ANCHOR RODS (φ Rn = 133 kips) Recommended sizes for washers and base plate holes are listed in Table 2.3. Heavy hex nuts and standard washers are to be used and fastened to the plate washers. Holes are designed to accommodate setting tolerances, and washers are sized to cover the entire hole when the anchor rod is located at the edge of the hole. Sizes are selected based on an anchor rod diameter of 2 inches: EIGHT 3 ⁄ 4-INCH × 5-INCH ASTM A572 GRADE 50 PLATE WASHERS EIGHT 31 ⁄ 4-INCH HOLES The distance from the center of the holes to the edge of the base plate should be enough such that the entire washer will be on the base plate if the anchor is located at the edge of the hole. In addition, enough room between the column and the holes should be provided such that the washer will not come in contact with the column flange if the anchor is located at the edge of the hole. Considering the column section properties and the initial base plate size, the assumed edge distance is adequate: 4-INCH EDGE DISTANCE ON ALL SIDES


261

Design Example 8




In AISC Design Guide 1, the preferred material specification for base plates is ASTM A36. However, based on relatively large demands, the base plate material is selected to be ASTM A572 Grade 50. AISC Design Guide 1 recommends the designer take the lower of the capacity of the plate in bending on the tension and on the compression side when determining the strength of the base plate in bending. However, tests conducted on six large-scale base connection specimens (Gomez, 2011) demonstrated that characterizing base plate bending strength using the compressive side strength is highly conservative. More accurate strength predictions were obtained by considering yielding due to bending on the tension side only. The following strength calculation considers compressive side yielding for completeness, but only the tension side yielding is considered for design purposes.

Bearing interface: m= n=

32 in − 0 95 15 7 in

N

2

2

B

=

2

32 in −

8(15.8

)

2

=85

in

Design Guide 1, 3.1.2

= 9 68 in.


Since Y < max(m, n)



)−

Y max( )

2 11

Y 

2

( .315 ksi

Design Guide 1, Eq 3.3.15a–2

i

= 2.1 1 )

−

( .85 in 2

 

50 ksi

= 3 42 in .

Tension interface:

− +

x

2

=

−

p (

)

p (

)

15.

= 2 .11

= .


2

+

2

56 in 2

=

93 in

Tu x

Design Guide 1, Eq 3.4.7a

BF y

(

.62 k (

)(

.

) )

= 2.5 6 in .

Assuming thickness requirements are based on tension side yielding, the base plate design is as follows: 23 ⁄ 4-INCH × 32-INCH × 32-INCH ASTM A572 GRADE 50 BASE PLATE

262


Design Example 8




3.3 CHECK COMPRESSIVE STRENGTH OF THE BASE PLATE The following procedure is used to check that the base-plate thickness is adequate to resist factored compressive loads. It is based on recommendations from Design Guide 1 Section 3.1. In Appendix A, the governing load combination considering compression only is as follows: 12

.

Therefore,

=

P

+ 1 6(59

) = 317 kips.

Check minimum base plate area to resist compressive load:

(

)

=

1



P

2

0.

1

=

B =

 = 71.72 in si) 

317 k (0.

.85)(

=

(32

4

.

2

. . . OK

Check plate thickness required to resist base plate yielding, taking max ( m, n, λ n′) as the cantilever length:

=

=

=  (

317 k (32 in)(32 in)

b

 

)

= 0 3538 ksi

= 0 85

 4 15 7 in)(15. (

in) 

.7 in 15. in)

(

317 kips .65 ( .85 4 ksi 1024 in (1 5)

1

X = 0 093

=

2 X 1+

=

− 4

2 0.093 1+

− 0 093

= (0.269)

)

=

n

(15.7

313 15 8

)

4 2P

,

p

=

( .

in)

= 1 23 in 2(

)

( .9)(50 ksi 32 in 32 in)

2 75 in . BASE CONNECTION ADEQUATE TO RESIST FACTORED COMPRESSIVE LOADS

3.4 GROUT DESIGN Based on recommendations in AISC Design Guide 1 Section 2.10, the design thickness of the grout space between the base plate and the surface of the grade beam should be 1 1 ⁄ 2 inches to 2 inches. It is also recommended that one or two grout holes, typically 2 to 3 inches in diameter, be provided when


263

Design Example 8




plates are larger than 24 inches so that the grout can be placed properly beneath the plate. Based on these recommendations, the grout design is as follows: 2-INCH NON-SHRINK GROUT BENEATH PLATE TWO 2-INCH GROUT HOLES

4. Base Connection Shear Resistance This section will demonstrate the procedure for designing for shear loading based on recommendations in AISC Design Guide 1. See Section 6 for further discussion on shear resistance at column bases in Special Moment Frames.

4.1 SHEAR RESISTANCE THROUGH FRICTION As recommended in AISC Design Guide 1, the contribution of the shear should be based on the most unfavorable arrangement of factored compressive loads that is consistent with the lateral force being evaluated. Based on this, and referring to Appendix A, the following load case governs: 0

m

=

.

E

Therefore,

+

u

=

+

4 k

4 ki

.

Shear resistance through friction, V n , can then be calculated as follows: Design Guide 1, 3.5.1

′

ACI 349 11.7.5

c

= 0 75

ACI 349 11.7.5

= (

c

=

)

.2 in

The friction coefficient, µ, is recommended to be 0.4 in Design Guide 1. However, an experimental study (Gomez et al., 2010) has shown that µ should be taken as 0.45 for steel on grout. Therefore,

=

45

and,

=(

.75 0.

)(

)

. s → V

48 kips

.

48 kips .

112 ki

= 48 kips .

The revised shear load is then as follows: V

ota

−

V

=

−(

) = 310 kips

where V u,total is the factored shear load calculated in Section 2.2.

264


Design Example 8




4.2 PLATE WASHER BEARING STRENGTH The deformation at each washer hole at service load is not a design consideration. Thus, the bearing strength of each washer is as follows: 0 tF

AISC 341 Eq J3–6b

= 0 75

AISC 341 J3.10

where t is the thickness of the washer, F u is the specified tensile strength of ASTM A572 Grade 50 material, d is the nominal bolt diameter, and lc is the clear distance. The calculation of the clear distance, lc , is calculated using the width of the washer in the direction of loading, W washer , and the diameter of the anchor rod, d , as follows:

=

+

W was er

1 16

in

 

2

5 in

=

−

+

1 16

2

in

 

= 1.469 in.

Therefore,

≤

R

.75 3 0)(2 0 in (0 75

= 80 56 ki

65

)

.

The shear load at each washer is the revised shear load, V u , divided by the total number of anchors, n. Thus,

≤

R

310 kips 8

≤ 80 56 kips

38.

.56 kips. PLATE WASHER BEARING STRENGTH ADEQUATE TO RESIST SHEAR LOAD

4.3 SHEAR STRENGTH OF ANCHOR RODS The design shear strength of each anchor rod is as follows: F v

AISC 341 Eq J3–1

= 0 75 where F nv is the nominal shear stress when, threads are not excluded from the shear plane, obtained from Table J3.2, and Ab is the nominal unthreaded body area of the anchor rod. Thus, n

(0.450)(75

33.75 ksi


265

Design Example 8




where F u is defined as the lower limit of tensile strength specified for ASTM F1554 Grade 55 material. Therefore, R V u n

.

≤

.

=

.

.52 kips

Rn

38 75 kips ≤ 79 48 kips . SHEAR STRENGTH OF ANCHOR RODS ADEQUATE TO RESIST SHEAR LOAD

4.4 TENSILE STRENGTH OF ANCHOR RODS CONSIDERING COMBINED TENSION AND SHEAR Since anchor rods will be subjected to tension due to uplift as well as shear, the combined effect of tension and shear should be considered according to AISC 341 Section J3.7. The design tensile strength of a bolt subjected to combined tension and shear is as follows: AISC 341 Eq J3–2

t

= 0 75 where nt is the nominal tensile stress of each anchor rod modified to include the effects of shear stress, calculated as follows: F t

−

nt

nt

F

≤ F t .

AISC 341 Eq J3–3a

Here, F nt is the nominal tensile stress obtained from Table J3.2 of each anchor rod when threads are not excluded from the shear plane, and f rv is the required shear stress. Both values are calculated as follows:

= 0.75)(75

F nt

(V

= 56.25 ksi

) 38 75 kips

1

3 1415 in2

ksi

where nb is the total number of bolts. Therefore, F ′t

.

266

(

.

.

.25 ksi)

( .75 3 3 .

≤

nt

R

−

.

.

)

(

.34 ksi

→ nt′ = 45 76 ksi = .83 kips.

.25 ksi


Design Example 8




4.5 TENSILE STRESS DUE TO TENSION AND BENDING The stress induced in the anchor rods due to tension and bending is addressed in AISC Design Guide 1. The following calculations will be based primarily Design Guide 1 guidelines, with some deviations based on experimental testing (Gomez et al., 2010).

+

.

t

The anchor rod tensile stress is calculated as follows:

=

t

=

n

478 71 ki (4 3.1416 in )

= 38.09 ksi .

The anchor rod bending stress is calculated as follows: M t Z

.

The anchor rod bending moment lever arm, lb, is calculated using recommendations from Gomez et al., 2010 as follows: p

+

out

=

+

in)

5 in.

The anchor rod bending moment is calculated as follows: M t =

V u

=

(

)( .5)(5 in 8

= 6 56 k-in.

The anchor rod plastic section modulus, Z b, is calculated as follows: 3

=

6

=

(2 0 in 6

1 333 in .

Therefore, M t

=

Z

+

=

1.333 in

=

t t

96 56 k-in

.

= 72.42 ksi

+ .

) = 110.51 ksi

72.

= 4 2 1

k si

F t

111 ksi

42 ksi. ANCHOR RODS INADEQUATE TO RESIST STRESS DUE TO TENSION AND BENDING


267

Design Example 8




Bending stresses are introduced to the anchor rods due to sliding of the base plate on grout and deformation of the grout. As discussed in Section 6, this is not expected in connections for SMFs. However, a more conservative approach is to install a shear key as recommended by Design Guide 1. This approach would eliminate the need for anchor rods alone to resist shear, and prevent the sliding action that will introduce bending stresses in the rods.

4.6 SHEAR LUG DESIGN The lateral resistance can be expressed as:

′

− T

V u


= 0 60. Using this equation, the minimum bearing area of the lug, Al , can be determined as follows: V

F y

,m n

− T )

) − ( .2)[(4 3.1416

(

′

1.3

(0 6

)(55

478.72 k ips)]

1.3)( ksi)

in2 .

Assume the width of the lug, bl , to be 14 inches and the embedment depth past the grout bed, d l , to be 2.5 inches. Thus,

=(

)( .5 in = 35 in .

Since the shear lug is not bearing in the direction of a free edge of concrete, the shear strength of concrete in front of the lug is not considered. The bending moment acting on the lug, M ul , using a cantilever model is calculated as follows:

+

u

 = 2 

+

2 5 in 2

= 1161 k -in


where G is the thickness of the grout bed. The section modulus, Z l , and moment capacity, φ M nl , for the shear key is calculated as follows: Z =

4

= M u

n

F

=

4

(0 9)(50 ksi 14 in) 4

= 135t 2

→ = 2 72 in .

Based on Design Guide 1 Section 3.5.2, it is recommended that the base plate be of equal or greater thickness than the shear lug. The base plate thickness is 3 inches; thus, a shear lug thickness of 2 3 ⁄ 4 inches. is acceptable. Since using heavy fillet welds is the preferred method of attaching the lug to the base plate, fillet welds will be used along the length of the shear lug at the interface between the lug and the base plate. As in Example 4.10 in AISC Design Guide 1, the vertical forces are equal and are assumed to act on the welds act at the weld centers. The shear force per unit length acting on each weld, f v , can be calculated as follows:

=

268

2

=

(358

)

(2 14

)

= 12 79 k-in.


Design Example 8




Vertical forces are assumed to act at the weld centers due to the bending moment introduced at the connection between the lug and the base plate. If we assume a weld size, t w , of 11 ⁄ 4 inches, these forces per unit length, f c, may be calculated considering the distance between the forces, s, as follows:

+(

s

=

M u s

)

 1

( 3 

=

.25

=

( (

+( )

)( .58 in

)

 1  3

( .2 5

= 33 58 in

= 23.16 k-in .

The resultant of the forces per length acting on each weld, f r , is calculated as follows: 2

2

=

2

.

26 46 k-i .

The design strength of each 1 1 ⁄ 4-inch fillet weld, assuming the use of an E70 electrode, is calculated in accordance with AISC 341 Section J2.2 as follows:

=

.75)( .

)( .6 )(

.25

.84 k -in .

Therefore,

and the shear lug is designed as follows: 23 ⁄ 4-INCH × 41 ⁄ 2-INCH × 14-INCH-LONG SHEAR LUG 11 ⁄ 4-INCH FILLET WELDS, BOTH SIDES, ALONG LENGTH It is recommended in Design Guide 1 that the shear strength of concrete in front of the lug be considered when determining the lug embedment depth. The shear strength of concrete is evaluated as a uniform tensile stress acting on an effective stress area defined by projecting an 45-degree plane from the bearing edge of the shear lug to a free surface of concrete. In this example, the lug bears in the direction of the longitudinal axis of the grade beam in which there is no free surface to be examined. Thus, this strength criterion is neglected analysis. However, in cases where zero net axial load or net tension in the column is expected at service loads, a potential failure surface that extends at a 45-degree angle from the edge of the shear lug to the top surface of concrete must be considered. Since net compression is expected at the base in this example for all load combinations, concrete bearing strength was considered a sufficient design criterion for lug embedment.

5. Design Column-to-Base Plate Weld Experimental tests (Myers et al., 2009) conducted using base connection configurations similar to the connection designed in this example have shown that Partial Joint Penetration (PJP) welds in conjunction with fillet welds connecting the column to the base plate can sustain deformations that are sufficient for seismic design. Figure 8–4 illustrates a detail involving a combination of PJP and fillet w elds. The fillet weld and the PJP weld are sized such that the sum of the width of the PJP weld and the throat thickness of the fillet weld is 25 percent greater than the flange thickness; for an overmatched weld filler material, the flange may be fully developed in tension. The PJP weld is designed to be installed from the outside flange face through 80 percent of the flange thickness using E70XX electrodes, beveled at 45 degrees. The weld


269

Design Example 8




at the web is a fillet weld of the same size as the fillet weld on the inside of the flanges, determined below. These specifications are given to remain consistent with previously mentioned experimental setups and standard practices for these types of welds. The fillet weld size, wfillet, is determined as follows: 0 80 w

1 et

=

(0.707)

1 ( .707

(0 2

1.56 in) =

44 in.

The size of the PJP weld, wpjp , is determined as follows: wp p

=

0.80 )(1 56 in

= 1.248 .

Based on these calculations, the column-to-base plate weld design is shown in Figure 8–4.

1/2-in. Fillet Both Sides

1-1/4-in. PJP 1/2-in. Fillet

Base Plate

Figure 8–4. Column-to-base plate weld detail

Alternatively, a Complete Joint Penetration (CJP) weld detail may be designed to attach the flange to the base plate. However, experimental research (Myers et al., 2009) indicates that connections with CJP weld details do no exhibit superior performance when compared to PJP details, primarily due to stress concentrations introduced at the weld access holes that are required for CJP welds. In either case, the welds must be treated as demand critical, such that applicable toughness criteria outlined in the Seismic Provisions (AISC, 2010) are met.

270


Design Example 8




6. Shear Resistance Discussion Figure 8–5 illustrates the force distribution at the column base due to the applied loading presented in this example, including the frictional force. Since exposed based connections are generally used in low- to mid-rise buildings, net axial tension on these connections is highly unlikely. As a limiting case, an applied loading scenario is examined where zero axial load is present. It can be determined that, in general, shear resistance through friction provides adequate shear strength for base connections in Special Moment Frames (SMFs). Results are shown below.

Figure 8–5. Force distribution with zero net axial load

We can assume r = N , the length of the base plate in the direction of loading, to obtain an upper bound for our analysis. From Figure 8–5, α H is a fraction of the story height, H , which corresponds to the location of zero internal moment in the column. We can write the required shear strength to resist the applied forces, V u , as follows: V

=

T N

α H

.


271

Design Example 8




The shear resistance through friction, F r , can be computed by multiplying the resultant bearing force, qmaxY , which is equal to the total tension due to uplift, T u , by the coefficient of friction between the base plate and grout, µ. If we impose that the frictional force be adequate to resist the shear load, we can determine a constraint on α as follows: F

α

T u

H

≥ V u = =

(

T N H

)

( .45 (144 in

= 0.494 .

The result indicates that if the point of inflection or location of zero internal moment in the column is located at or above 0.494 times the story height for this limiting case of zero axial load, friction alone will adequately resist any applied shear load. As history has shown, the seismic response of a SMF at design loads generally results in first-story columns developing points of inflection at or above one-half the story height. The implication here is that, for the general case considering applied axial loads, moments, and typical base plate geometries at SMF base connections, friction between the base plate and grout alone will resist most, and likely all, of the applied shear load.

7. Anchorage The design for anchorage to concrete is generally dictated by provisions in ACI 318 Appendix D. This section will demonstrate an alternative methodology to design anchorages based on proven strut-and-tie concepts that have been used for similar structural configurations. This methodology may prove simpler in many cases, allows for simultaneous grade beam design, and may also provide more options for the designer. ACI 318 Appendix A provides a method of analysis using strut-and-tie concepts for deep beams. To perform the strut-and-tie analysis, certain geometric constraints exist. Anchors must be embedded to a greater depth than the centerline of the lower layer of reinforcement to engage the reinforcement in resisting anchorage forces. This requires that the lower reinforcement be elevated enough such that adequate cover is provided beyond the bottom of the anchors. The geometric constraints affect the design process as follows: the reinforcement locations, anchorage depth, and anchor-plate sizes must be chosen prior to analysis; then, ties can be sized, and the strength of the grade beam can be checked against the applied anchorage forces. If the strength of the grade beam is adequate, the design is successful. Appendix A defines a “D-region” as a region where the usual plane sections assumption is not applicable, and strut-and-tie design provisions may be employed. Based on the anchorage geometry and Section A.1 guidelines, D-regions exist through the entirety of the grade beam which resists applied anchorage forces. Figure 8–6 shows the assumed bottle shaped struts that are formed to resist the forces generated in the grade beam. An idealized, determinate truss can be constructed by considering struts as uniaxial compression members, and the longitudinal reinforcement between the struts as uniaxial tension members. Angles formed by struts are assumed symmetric at each node. The constructed truss is shown in Figure 8–7. The truss can be made determinate by considering the far ends as a roller and a pin support. Figure 8–8 shows the resulting force distribution and geometry.

272


Design Example 8




Figure 8–6. Struts formed in grade beam due to anchorage forces

Figure 8–7. Idealized truss formed in grade beam resisting anchorage forces 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

273

Design Example 8




Figure 8–8. Idealized truss layout and force distribution

7.1 DETERMINE DEMANDS Force demands are determined using standard truss analysis. To perform the analysis, as previously mentioned, reinforcement locations must be selected. One layer is chosen for both top and bottom grade beam reinforcement. The upper reinforcement is chosen to be 4 inches from the top surface based on cover requirements, and since the lower reinforcement must be elevated to provide enough cover for the extending anchor below, the lower reinforcement is chosen to be 9 inches from the bottom of the grade beam. Values h, a, and θ are calculated as follows: a

+

y

2

− =



2

12 in

− θ=

an −

a

+

– 4 in

=

an−1

32 in 2

5 85 in 

−

2

= 25.08 in

35 in

35 0 25.08

=

.

=

.4 .

Section A.2.5 requires that the angle between the axes of any strut and any tie entering a single node shall not be taken as less than 25 degrees. Since here θ = 54.4 degrees, the truss geometry is permissible. Reactions at the assumed supports can be determined through statics as follows: R

R

274

3 1 3

− P) =

3

+ 2P) =

1 3

− 142 kips) = 112.24 kips

.

.

− 2(142

)] = 254.24 kipss .


Design Example 8




Forces in each member can be determined through standard truss analysis, and are calculated as follows: 1

−

3 sin( )

=

)

3

− 3 35

)=

3 1

c

sin

0.949))

)

=−

3 sin( ) a

3

(25.07 in)

1

=

1

+

1 3

)

−138 ki

kips

+ 2(142

0. 49 )

)) = −313 ki

+ 2(142 kips ) = 182 kips

)

=−

− 142 ki

− 142 kips

.

(25. 7 in 3 35

(478.72 i

1 . 4

(2(478 72 i

s

kip

where compression is indicated by a negative value.

7.2 DETERMINE CAPACITY OF STRUTS, TIES, AND NODAL ZONES ACI 318 provisions require that the struts, ties, and nodal zones (truss joints) meet strength requirements given in Appendix A. This section will proceed by first determining reinforcement sizes required to meet tie demands based on the assumed geometry, then strut and nodal zone capacities will be calculated. In practice, each strut and nodal zone should be analyzed to ensure the strength of the entire truss system meets the required demands. Here, it is apparent that the most critical strut end and nodal zone exists at (c) in Figure 8–8. This location will be analyzed in detail, but it should be noted that the remaining nodal zones may be analyzed in an analogous manner. Figure 8–9 shows the geometry and applied forces at nodal zone (c).

Figure 8–9. Geometry and applied forces at nodal zone (c)


275

Design Example 8




Figure 8–9 shows that the nodal zone is split into two zones, each resisting a strut that acts on a face of the nodal zone. The node is split based on the factor α, which is based on the magnitude of the strut forces acting at the zone, and is calculated as follows:

=

= c

(

.17 kips

.17 kips) + (444 62 kips)

(

= 0 234 .

The height of the nodal zone, hs, is taken as twice the distance from the plate face to the centerline of the bottom reinforcement. The plate face is chosen to be embedded to a depth of 42 inches, resulting in hs = 6 inches. The length of the anchor plate, ls , is taken as 8.5 inches based on an initial anchor plate footprint sizing of 81 ⁄ 2 inches × 7 inches. Section 9.2.3.6 states that strut-and-tie models and struts, ties, nodal zones, and bearing areas in such models are to use a strength reduction factor of

= 0 75.

ACI 318 9.2.3.6

Strength of Ties According to Section A.4.1, the strength of a nonprestressed tie, F nt , shall be taken as ts

and t

where Ats is the area of the tie and f y is the yield strength of the tie. Thus, the area of each tie, assuming ASTM A615 grade 60 material, can be computed as follows:

ts ,ac

≥

F

F ts ,c

= =

(

)

( .75 6 ksi) (

)

( .75 6 0 ksi)

= 3.0 7 in 4 04 in2 .

Ties should be selected based on the required areas and so that the placement of the ties are compatible with the location of the anchors. Based on the anchorage configuration, one layer of five reinforcing bars spaced at 8 inches is selected. The resulting bar sizes are selected considering the practical advantage of having identical longitudinal reinforcement on top and bottom of the cross section. The area of each bar is calculated as follows:

n

=

ts ,c

5

=

4 04 in2 5

= 0.809 in .

Thus, USE #9 BARS (AREA = 1.00 IN2) AT 8 INCHES ON CENTER TOP AND BOTTOM LONGITUDINAL REINFORCEMENT FOR GRADE BEAM

276


Design Example 8




Strength of Struts The compressive strength of a strut without longitudinal reinforcement, F ns , shall be taken as the smaller value of the following equation evaluated at the two ends of the strut: F

e

cs

.

ACI 318 A–2

Here, Acs is the cross-sectional area at one end of the strut, and f ce is the smaller value of the effective compressive strength of the concrete in strut, given by

′

ACI 318 A–3

and the effective compressive strength of concrete in the nodal zone is given by

′.

ACI 318 A–8

According to Section A.3.2.2, for bottle-shaped struts, assuming shear reinforcement will be added to the grade beam, β is taken as 0.75. According to Section A.5.2.2, for nodal zones anchoring one tie, β ′ is taken as 0.80. The area of the strut is calculated considering the minimum width of the strut in the plane of loading, ws , shown in Figure 8–9, and the effective width in the transverse direction, bs. Conservatively, bs is taken as the sum of the widths of each of the anchoring plates. Anchor-plate footprints are initially sized at 81 ⁄ 2 inches × 7 inches. Thus, the compressive strength of the larger strut is calculated as follows: cos(

+ −

4 × 7 in

sin

=

+ 1−

.

.5 in [sin

}

= 246. 3 in

=

0.75)( 85 0.75)(4 k ksi

in

kips .

It follows that, c

n

. . . OK.

Therefore, the strength of the larger strut is adequate to resist the applied loads. The compressive strength of the smaller strut is analyzed as follows: cos

+

× 7 in)

=

+

0.

)( .5 in)[sin(54.4°)]}(

)

= 113.51 in

= Fc

0.75)( 85 0.75)(4 ksi

in

kips

F ns . . . OK.


277

Design Example 8




Strength of Nodal Zone The strengths of nodal zones at faces connecting to struts are equivalent to the strengths of the struts at those locations (excluding the β factor, which acts to increase the capacity at the faces). Therefore, it is apparent that the strengths at the faces of the nodal zone connecting to struts are adequate to resist the applied loads. The strength at the face connecting to a tie is analyzed as follows: F

ACI 318 A–7

n

=(

n

× ×

) = 168 in

= F

0.75)( 85 0.80 )(4 ksi 168 in

342 72 kips

F nn . . . OK.

The strength at the bearing face is analyzed as follows: ( .5 in)

n

7 in)

=

238 in 2 0.75)( 85 0.80)(4 4 ksi 238 in

485 52 kips

. . . OK. STRENGTHS OF TIES, STRUTS, AND NODAL ZO NE AT (C) ARE ADEQUATE TO RESIST APPLIED LOADS

7.3 DESIGN ANCHOR PLATES The anchor plates are expected to bend due to bearing against the concrete when anchors are in tension. Planar cantilever bending is assumed, and the strength of a plate is checked against the bending moment induced in the plate in a similar fashion as is demonstrated in AISC Design Guide 1 Section 3.3.2 for baseplate bending. Bending is assumed symmetric, and the cantilever length, x, is the distance that the plate extends beyond the anchor nut. A standard hex nut for a 2-inch anchor rod has a width, whex, of 3 inches. The following analysis, assuming A572 Grade 50 anchor plate material, is used to size the thickness of the anchor plate, t ap: 1

x

−

2

)

1

−3

2

2.75 in.

The moment per unit length to be resisted by the plate can be calculated considering the bearing pressure, f p , exerted on the bearing area of the plate, Abrg, as follows:

−

r

M p

x

=

2

=

= T rg

× (7.  

in) −

.

=

.61 in

 2

.

The nominal bending strength per unit length is as follows:

R =

278

F t

4

.


Design Example 8




Therefore, with φ = 0.9, M p

Rn

4 M F

2T x

=

F y

g

2(120

)(2 75 in

(0.9 50

60 61

)

USE 81 ⁄ 2-INCH × 7-INCH × 1-INCH ASTM A572 GRADE 50 ANCHOR PLATES

7.4 DESIGN SHEAR REINFORCEMENT Grade beam shear reinforcement is designed for the region in the beam where anchorage forces are resisted. ACI Appendix A requires that certain requirements are met that are consistent with the strut-and-tie analysis performed. An initial design will be based on minimum requirements for areas of shear reinforcement in deep beams specified in Section 11.7.4 as follows: v

s

g

ACI 318 11.7.4.1 ACI 318 11.7.4.2

g

,

where Av is the vertical reinforcement, Avh is the horizontal shear reinforcement, s is the longitudinal spacing for vertical reinforcement, and s2 is the vertical spacing for longitudinal shear reinforcement. The maximum spacing is as follows: g

= min 2,max

smax

5

,12 in



=

.

= 9 6 in.

,

By setting the spacing s and s2 at 9 inches, it follows that, v

v ,min

= 0.

(36 in )(9 in

.81 in

Appendix A dictates that when using a strut-and-tie model, the following equation for shear reinforcement must be satisfied:

g

s

sin( g

≥ 0 003

sn

s

ACI 318 A–4

where α1 is the angle between the strut axis and the vertical reinforcement, and α2 is the angle between the strut axis and the horizontal reinforcement. By setting the vertical reinforcement at a minimum, the horizontal shear reinforcement can be determined as follows:

 v

=

.

− −

s

in(

)



81 in2 (36

)(9

)

g

sin(

s2

)

sin(35.6°)

 (36

)(

sin(54.4°)

)

= 0.6 2 in

.


279

Design Example 8




It follows that minimum shear reinforcement is adequate in the vertical and longitudinal directions. Based on the reinforcement area and spacing requirements, it is determined that #3 stirrups (Area = 0.11 in2) can be used in the vertical direction spaced at 9 inches, and #3 strands can be used on either side in the horizontal direction in the configuration shown in Figure 8–10, the final base connection design.

Figure 8–10. Base connection design

280


Design Example 9 Braced-Frame Base Plate OVERVIEW This example shows procedures for the design of Buckling-Restrained Braced Frame (BRB) base plates. This example includes the design of the connection of the Buckling-Restrained Braced Frame gusset to the base plate and column, the column connection to the base plate, sizing of the base plate, and the connection of the base plate to the foundation. BRB connections are designed for the ultimate loads that can be delivered to the connections from the brace based on the brace expected yield strength, strain hardening, and, when applicable, compression overstrength. This is done such that yielding of the system will occur in the brace and not in the base plate. That is, a capacity design is performed.

OUTLINE 1. Design Parameters 2. Gusset Plate Design 3. Base Plate Design 4. Foundation Anchorage Design 5. Items Not Addressed in This Example

1. Design Parameters 1.1 GIVEN INFORMATION This design example is based on the information calculated in Design Example 3 (Buckling-Restrained Braced Frame). Design Parameters: Maximum brace size: As = 5.0 in2 Typical bay spacing = 30 ft Base story height = 12 ft F y (BRB) = 42 ksi + / − 4 ksi

Maximum brace yield strength = 46 ksi

β = 1.04 ω = 1.54 R y = 1.0

Column size = W14 × 211 S ds = 1.0 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

281

Design Example 9



Braced-Frame Base Plate

Material Properties: Concrete:

= 4000 psi

W-Shapes: ASTM A992; F y = 60 ksi Rectangular HSS: ASTM A 500, Grade B; F y = 46 ksi Plate: ASTM A 572, Grade 50; F y = 50 ksi Anchor Rods: ASTM F 1554 Grade 55; F y = 55 ksi Reinforcement: ASTM A615 Grade 60; F y = 60 ksi

1.2 PIN VS. FIXED BASE DESIGN This section will provide a discussion on different modeling and design options. Moving forward, the design of the plate will be done based on a “pin” assumption. Figure 9–1 shows the conceptual force diagram for the base connection. The vertical component of the brace force will be taken by the anchor bolts, while the horizontal shear component of the brace force will be resisted by a HSS drag strut.

Pu Pv

Pu

Figure 9–1. BRB connection force diagram

282


Design Example 9

2. Gusset Plate Design




ASCE 7

2.1 BRB CONNECTION TO GUSSET PLATE The connection of the BRB to the gusset plate varies among brace manufacturers. BRB connections can be bolted, welded, or pinned. In designing the BRB, all connection types should be considered during the design phase. The final design of the connection of the BRB to the gusset plate should be provided by the manufacturer. Typically the design of the gusset plate and its connection to the base plate is provided by the manufacturer as well; however, it is recommended that the engineer provide minimum weld lengths as part of the contract documents.

2.2 DETERMINE CONNECTION DESIGN FORCE The connection of the brace is designed for the maximum force that can be delivered by the brace. During a seismic event, strain hardening will occur in the brace, which increases the capacity of the brace. Therefore the capacity of the brace should be increased by the strain-hardening adjustment factor, ω . Additionally, the brace capacity should be increased by the compression-strength adjustment factor, β, which accounts for increased capacity of the brace when subject to compression loading. Both β and ω are determined based on testing of the brace and should be provided by the brace manufacturer. Finally, the brace capacity should be multiplied by R y which is the ratio of the expected yield stress to the specified minimum yield stress, F y of the brace material. For different member types, values of R y are given in Table A3.1 of the AISC 341-10 Seismic Provisions for Structural Steel Buildings. In this design example a range of material is considered from 38 ksi to 46 ksi. The upper value of 46 ksi will be used; therefore, R y is set equal to 1.0. The connection should be designed for the maximum tension and compression loads that can be delivered by the brace. The maximum connection design force under compression is Puc = βω R y F y As Puc = (1.04)(1.54)(1.0)(46 ksi)(5.0 in 2) Puc = 368 k.

The maximum connection design force under tension is Put = ω R y F y As Put = (1.54)(1.0)(46 ksi)(5.0 in 2) Put = 354 k.

2.3 SIZE CONNECTION WELDS Size connection of weld to the base plate: Brace Angle:

θ = tan−1(12/15) = 38.6°. Horizontal component of brace force: Ph = cos(38.6°)*368 k = 288 k.


283

Design Example 9




The work point for the brace is located at the center-line of the column at the top of the base plate; therefore, there is no eccentricity between the horizontal component of the force and design weld. Half of the horizontal force will be taken directly through the gusset plate to the HSS strut, while the other half w ill be transferred through the base plate to the HSS; see Figure 9–5 for the HSS connection to the base plate and gusset. Assuming a two-sided, 1 ⁄ 4-inch fillet weld, size weld length: (

(

)

× 2 sides 11 1  = (288

k/2)/11.1 = 12.9 in → 13-in minimum weld length required.

The vertical component of brace force transferred to the column is Pv = sin(38.6°)*368 k = 230 k.

Because the work point for the brace is located at the center-line of the column, and the resisting weld occurs at the face of the column, an eccentricity occurs between the force and the weld that is to be added to the design force of the weld. The eccentricity of the weld is equal to half of the column depth: e = 15.7 in/2 = 7.9 in.

Using Table 8–11 in AISC 360-05 for eccentrically loaded weld groups, where the force angle is equal to zero, determine the size and length of fillet weld required to resist the vertical force component. Since there is no return weld, the value of k is set equal to zero. Assuming a two-sided, 3 ⁄ 8-inch fillet weld, try a weld length of 20 inches. e = a. a = e /  = 7.9 in/20 in = 0.395 ∼ 0.4.

Therefore,

Using Table 8–11 for a = 0.4, then C = 1.33. Check the minimum weld length required for the determined C value to confirm the original length assumption. For LRFD: min

= Pu / φCC 1 D

where: φ = 0.75 C = 1.33

min

C 1 = 1.0

(per Table 8–3 for E70XX electrodes)

D = 12

( D is equal to the number of sixteenths-of-an-inch in the fillet-weld size; for a two-sided 3 ⁄ 8-inch fillet this is equal to 12)

= 230 k/(0.75 × 1.33 × 1.0 × 12) = 19.2 in < 20 in.

Therefore, the original assumption is acceptable. Use a two-sided, 20-inch-long, 3 ⁄ 8-inch fillet weld.

284


Design Example 9




Figure 9–2. Connection weld requirements

3. Base Plate Design 3.1 COLUMN CONNECTION TO BASE PLATE The connection of the column to the base plate should be designed for the maximum tension values that can be delivered to the system by the braced-frame system. Table 9–1 shows the forces to the column as determined in Design Example 3.


285

Design Example 9




Table 9–1. Column design forces P ysc (kips)

wβ R y P ysc (kips)

LLr (psf)

P(1.4 D + 0.5 L + E ) (kips)

20

48

48

20

20

65

207

255

−49

−30

67.7

65

230

484

−72

−102

900

67.7

65

276

760

−118

−221

207

900

67.7

65

322

1081

−164

−385

368

230

900

67.7

65

345

1426

−187

−572

368

230

230

1656

−230

−802

Level

Asc abv.

Roof

0

0

0

6th

2

92

5th

2.5

4th

P E = Pbr sinθ* (kips)

TA

DL (psf)

0

900

31

147

92

900

67.7

115

184

115

900

3.5

161

258

161

3rd

4.5

207

332

2nd

5

230

1st

5

230

LL (psf)

ΣPuc (kips)

P(0.7 D − E ) (kips)

ΣPut (kips)

*Pbr is equal to the brace force with overstrength For the connection of the column to the base plate, provide a CJP weld of the column flanges and a fillet weld for the column web. Per AISC 360 Table J2.5 for tension welds normal to the weld axis, the strength of the joint is controlled by the base metal. Therefore, the strength of the flange CJP welds will be equal to the strength of the column flanges:

φ Rn = φF BM A BM for W14 × 211: b f = 15.8 in and t f = 1.56 in = 0.9(50 ksi)(2 × 15.8 in × 1.56 in) = 2218 k. Therefore, the CJP weld of the flanges is adequate for the transfer of the column tension forces. Provide a minimum required fillet weld for the column web. For a W14 × 211, the thickness of the web is 1 inch. Per AISC 360 Table J2.4, the minimum required fillet weld for 1-inch material is 5 ⁄ 16 inch. Therefore, for the column connection to the base plate, provide a CJP weld for both column flanges and a 5 ⁄ 16-inch fillet weld for the column web. Alternatively, a PJP weld can be considered. If a PJP weld is used, it is recommended that it be toughness rated. The PJP weld should be sized for the tension demands of the column. From Table 1, the tension for the column is equal to 802 kips.

φ Rn = φF w Aw, From Table J2.5, for a groove weld in tension φ = 0.8, and F w = 0.6F EXX . Therefore the required area of the weld is Aw = 802 k/0.8/(0.6 × 70 ksi)

= 23.8 in2. The effective area of the weld is equal to the length of the weld times the effective throat thickness. Therefore, the required throat thickness is equal to t min = 23.8 in2 /2 flanges/15.8 in

= 0.754 in → 7 ⁄ 8-inch effective throat required.

286


Design Example 9




3.2 PLATE MATERIAL SELECTION For this design example, ASTM A572 Grade 50 steel will be used, which has a yield value of 50 ksi and an ultimate strength of 65 ksi. In areas of lower seismicity, ASTM A36 Grade 36 steel may be used.

3.3 DETERMINE BASE-PLATE THICKNESS The base-plate thickness is determined by the maximum tension and compression values that can be delivered to the plate. For sizing of the base-plate thickness, the values of B and N will be based determined assuming 11 ⁄ 2-inch-diameter anchor rods, which require 3 1 ⁄ 2-inch square washers minimum per AISC 360 Table 14–2. For this example, the value of N will be based on the area of plate directly below the column; see Figure 9–3.

Figure 9–3. Base-plate dimensions

The critical base-plate cantilever dimension is based on the maximum values of m, n, and λ n′. For the purpose of this design example, λ will conservatively be taken as 1.0. For W14 × 211, b f = 15.8 inches and d = 15.7 inches. m = ( N − 0.95d )/2 = (28 in − 0.95 × 15.7 in)/2 = 6.54 in n = ( B − 0.8b f )/2 = (22 in − 0.8 × 15.8 in)/2 = 4.68 in

=  = max(m,

t m n

×

= 3 93 in

n, λ n′) = max(6.54 in, 4.68 in, 3.93 in) = 6.54 in u

= 6 54 in 2 ×

. × 50 ksi × 22

2.26 in

Use a Grade 50 2 1 ⁄ 2-inch base plate. For the tension check of the plate, see Section 4.3.


287

Design Example 9




4. Foundation Anchorage Design 4.1 SIZE FOUNDATION THICKNESS Footing sizes will be sized based on the maximum force that can be delivered to the system. From ASCE 7-05 the maximum tension and compression cases for the foundation design will be Puc = (1.2 + 0.2S ds) D + 0.5 L + E , where S ds = 1.0 Put = (0.9 D − 0.2S ds) D − E .

Soil-bearing values to be used in the sizing of footings should be provided by the geotechnical engineer on the project for the specific site. Alternatively, default lower bound values can be used as provided in Table 1806.2 of the 2010 California Building Code. For the purpose of this design example we will assume a soil bearing value of 5000 psf under Dead and Live loads, with an allowable one-third increase under seismic loading. For determination of footing sizes and uplift forces to the foundation, ultimate brace forces are not required to be used. Instead, the brace demand forces are used to calculate the maximum compression and tension forces that are delivered to the foundation. Table 9–2 shows the forces as determined in Design Example 3.

Table 9–2. Foundation design forces

Level

Asc abv.

Pbr (kips)

Roof

0

0

6th

2

5th

P E = Pbr sinθ (kips)

LL (psf)

LLr (psf)

P(1.4 D + 0.5 L + E ) At Floor Level (kips)

20

48

48

20

20

ΣPuc

P(0.7 D − E ) At Floor Level (kips)

ΣPut (kips)

TA

DL (psf)

0

900

31

29.9

23

900

67.7

65

138

186

19

39

2.5

72

56

900

67.7

65

171

357

−14

25

4th

3.5

108.6

85

900

67.7

65

200

557

−42

−18

3rd

4.5

135

106

900

67.7

65

220

777

−63

−81

2nd

5

151

118

900

67.7

65

233

1010

−76

−156

1st

5

159.2

125

125

1135

−125

−281

(kips)

Size the footing based on the maximum compression force based on ASD loads. The controlling ASD load combination for this example will be 1.105 D + 0.75 L + 0.525 E , which is equal to 804 kips: 1

- -

f

tin .

Determine the foundation thickness based on the punching shear of the footing based on the maximum compression force per Table 9–2. The punching shear strength of the footing is determined by ACI Section 11.11.2.1 as ACI Eq 11–33 The critical perimeter bo is one-half the effective depth from a plane halfway between the face of the column and the edge of the base plate. Assume a footing depth of 3 feet with rebar clear cover of 3 inches and #8 reinforcing.

288


Design Example 9




Therefore, d = 36 in − 3 in − 1 ⁄ 2 in = 321 ⁄ 2 in

and bo = 2 x(0.95d col + m + d + 0.8b f + n + d ) = 207.5 in V =

× 0.

× 207.

=

.

steel column

m

c

m

2 m

2 base plate

critical section

d

2

c + m + d

critical perimeter

Figure 9–4. Punching shear

Alternatively, a grade-beam system can be used to facilitate transfer of loads in the foundation system and reduce the size of the square footings. It is recommended that grade beams be used to connect the foundations in a frame system and facilitate the transfer of the frame shear forces as detailed in Section 4. 2. 2012 IBC SEAOC Structural/Seismic Design Manual, Vol. 4

289

Design Example 9




4.2 TRANSFER OF SHEAR TO THE FOUNDATION SYSTEM The shear component of the brace force will be resolved in the foundation system through a strap. For this example a square HSS will be used to provide the strap element. The HSS section shall be sized based on the maximum horizontal brace force component Ph. Since the HSS will be slotted and welded to the gusset plate, the reduced section shall be checked assuming a gusset plate thickness of 1 1 ⁄ 2 inches. Determine area of steel required for the HSS section: As = Ph / φF y = 288 k/(0.9 × 46 ksi) = 6.95 in2 → HSS 6 × 6 × 3 ⁄ 8.

For HSS 6 × 6 × 3 ⁄ 8: As = 7.58 in2. Check HSS reduced section for fracture: Ae = As − 2 x((t gusset + 1 ⁄ 8 in) × t HSS)

= 7.58 in2 − 2 × (15 ⁄ 8 in × 0.349 in) = 6.44 in2 φPu = φ Ae × F u = 1.0 × 6.44 in2 × 58 ksi = 373 k > Ph. Size the weld of HSS to the gusset. Half of the horizontal load is being transferred directly from the gusset, while half is transferred through the base plate. Based on the gusset geometry, a starting weld length of 17 inches was selected. P

=

.6 F

=

D

× 70

2 / 2

)] where  = 17 in 16 17 × 4 sides

= 94.6 D D = 288 k/94.6 = 3.04 → 1 ⁄ 4-in fillet weld

Figure 9–5. HSS strap connection

290


Design Example 9




To transfer the shear loads to the concrete, welded headed studs will be used. Assuming 3 ⁄ 4-inch diameter WHSs on each side, determine the number of studs required based on the nominal horizontal shear capacity of the stud as given in Table 3–21 of AISC 360. Per Table 3–21 for normal-weight concrete with compressive strength of 4 ksi: Qn = 21.5 ksi.

Therefore, the number of studs required to transfer the shear to the concrete is Ph / Qn = 288 k/21.5 ksi = 13.4 studs → place studs at 12 in o.c.

The HSS strut beam can be ended after the forces have been transferred to the slab-on-grade diaphragm. Alternatively, the HSS strut beam can be extended to the adjacent column. In the case that the adjacent column is not a seismic column, it is recommended that a horizontal slotted connection be provided to the adjacent column to eliminate any horizontal force transfer to the column. The slab diaphragm connection should be checked to confirm transfer to the grade beam foundation through slab dowels; see Figure 9–7 for slab dowels. The number of dowels required to transfer the shear force through shear friction of the dowel bars, assuming #5 bars is

φV n = φ Avf × f y × µ where, φ = 0.75 Avf = 0.31 in2 f y = 60 ksi

µ = 1.0 (provide concrete shear key between grade beam and slab) φV n = 0.75(0.31 in2)(60 ksi)(1.0) = 13.9 k. Since Ph = 288 k, the number of dowel bars required is 288 k/(13.9 k/bar) = 21 dowel bars → provide two rows of #5 dowel bars at 12 in o.c.

4.3 TRANSFER OF TENSION TO THE FOUNDATION SYSTEM The anchor bolts shall be sized based on the maximum uplift of the system. For this case the maximum tension force that can be delivered to the connection is −802 kips as was determined in Table 1.


291

Design Example 9




Pu = 802k

10''

10''

Figure 9–6. Anchor-force distribution

Treating the base plate as a rigid body, the forces in the anchor rods will be Pu = 6P1.

Therefore, P = 802 k/6 = 133.6 k per anchor rod. Using Gr 105 rods then, As_rod = 133.6 k/(0.9 × 105 ksi)

= 1.41 in2 _

=

= .34 in → use 1½-in-diameter anchor rods.

Check base plate in bending to confirm rigid body assumption for distribution of tension force: M u = 2P1 × e e = 10 in − d /2 − tf /2 = 10 in − 15.7 in/2 + 1.56 in/2 = 2.93 in M u = 2(133.6 k) × 2.93 in

= 782 k-in. For a 21 ⁄ 2-inch base plate with anchor spacing of 20 inches: Z = (20 in)(2.5 in) 2 /4 = 31.25 in3

σ = M u / Z = 782 k-in/34.38 in3 = 25 ksi < 50 ksi.

292


Design Example 9




To transfer the anchor-rod loads to the foundation, an embedded plate will be used. The plate is extended below the bottom layer of reinforcing bars to engage the reinforcing. The reinforcing must extend beyond the embedded plate long enough to develop the full capacity of the reinforcing bar, or hooks shall be provided. The reinforcing bars must be adequate to resist the column uplift forces. Therefore the number of bottom bars required to resist uplift is 802 k / 2 planes / 0.75 / 60 ksi = 8.9 in2 → nine # 8 bars. Since the reinforcing bars are fully engaged, the Appendix D method is no longer applicable.

Figure 9–7. Foundation anchorage

The anchorage plate should be sized for the bending forces in the plate. The two rows of bolts are located 8 inches from the center line of the plate. The uplift will be resisted through the plate bearing on the concrete; therefore, the bearing of the plate is equal to the uplift force over the bearing w idth: wu = 802 k/16 in = 50.1 k-in M u = wu2 /8 = (50.1 k-in)(16)2 /8 = 1604 k-in.


293

Design Example 9




The width of the plate will be taken based on the spacing of the anchors plus 3 inches from the centerline of the bolt. Therefore, the width of the plate is equal to 2 feet, 2 inches. Using a 50 ksi plate, the required modulus for the plate is Z x = M u /0.9/ F y = (1604 k-in)/0.9/50 ksi

= 35.6 in3 Z x = bh2 /4 where b = 26 in.

Therefore, the required plate thickness is

=

=

.

34 in → use 21 ⁄ 2 × 22 × 26-in plate.

Checking the concrete for breakout in tension, with the plate width taken as the distance between the anchors: bo = 2 × (20 in + 32.5 in) + 2 × (16 in + 32.5 in) = 202 in V c =

× 0.

× 202 × 4000 psi

= 1245 k > 802 k.

5. Items Not Addressed in This Example The following items are not addressed in this example but are nevertheless necessary for a complete design of the seismic load resisting system: • Design of the foundations.

294


Appendix 1 General Building Information The following information is used for most of the design examples in this volume and is presented here to reduce repetition. Design Example 6, w hich is a one-story industrial building, uses only the location, site class, and spectral accelerations; it uses a different geometry and roof assembly weight. Each design example cites this appendix at the points in the design process at which the information is applied. This appendix contains the following information: 1. General Information 2. Building Geometry 3. Assembly Weights 4. Floor and Roof Weights 5. Design Spectral Accelerations 6. Seismic Design Category 7. Load Combinations

GENERAL INFORMATION • Office occupancy on all floors • Located in San Francisco, CA (site location: Latitude 37.783°, Longitude −122.392°). • Site Class D • Risk Category II

BUILDING GEOMETRY • 120-foot × 150-foot centerline dimension in plan with typical floor and roof framing shown in Figure A1–1. The edge of the deck dimension is approximately 1 foot, 6 inches from the gridline. • Six-stories as shown in Figures A1–2 and A1–3.


295

Appendix 1



General Building Information

Figure A1–1. Typical floor and roof framing plan

Figure A1–2. Building elevation 296


Appendix 1




Figure A1–3. Building axonometric view

ASSEMBLY WEIGHTS Floor Effective 3 Seismic Weight

Dead Loads

Gravity Load

Floor finish

5.7 psf

5.7 psf

2-in, 18-ga. deck

2.7 psf

2.7 psf

31 ⁄ 4-in light-weight concrete fill

39.0 psf

39.0 psf

Steel framing

10.7 psf

10.7 psf

Mechanical / Plumbing / Electrical

4.7 psf

4.7 psf

Ceiling

4.7 psf

4.7 psf

Under Live Load

10.7 psf 4

3.7 psf

3.7 psf

67.7 psf

77.7 psf

Partitions Miscellaneous Total Dead Load Office Building Live Loads1

Live Load

Lobbies and stairs = 100 psf Offices = 50 psf + 15 psf partition Corridors above 1st floor = 80 psf

100 psf 2

65 psf 80 psf


297

Appendix 1




Roof Dead Loads

Gravity Load


Built-up roof

6 psf

6 psf

Insulation

2 psf

2 psf

Metal roof deck

4 psf

4 psf

Steel framing

8 psf

8 psf

Mechanical / Plumbing / Electrical

4 psf

4 psf

Ceiling

4 psf

4 psf

4

Partitions

0 psf

5 psf

Miscellaneous

3 psf

3 psf

31 psf

36 psf

Total Dead Load Roof Live Loads1

Live Load

Ordinary Flat Roof

20 psf

Exterior Wall Gravity Load


Cladding

7 psf

7 psf

Metal studs

2 psf

2 psf

Insulation

2 psf

2 psf

5

⁄ 8-in gypsum board

3 psf

3 psf

Miscellaneous

5 psf

5 psf

19 psf

19 psf

Dead Loads

Total Dead Load

1. From ASCE 7 Table 4–1. 2. ASCE 7 Section 4.3.2 specifies a 15 psf live load where partitions will be erected or rearranged. 3. ASCE 7 Section 12.7.2 describes the loads that are included in the Effective Seismic Weight. 4. Per Section 12.7.2, 10 psf is included for partitions where partition load is required per ASCE 7 Section 4.3.2.

FLOOR AND ROOF WEIGHTS

Level

Unit Wt (psf)

Area (ft2)

Weight (kips)

Floor

77.7

15,220

1183

Ext wall

19

6990

133

Roof

36

15,220

548

Ext wall/Parapet

19

5700

108

Assembly

Typical floor

1315

Roof

656

W = 5(1315 kips) + 656 kips = 7231 kips.

298

Floor Wt (kips)


Appendix 1

DESIGN SPECTRAL ACCELERATIONS




ASCE 7

USGS web tools can be used to determine the maximum considered earthquake (MCE) spectral accelerations for 0.2 sec, S S , and 1.0 sec, S 1 based on the longitude and latitude of the site. The longitude and latitude can be entered into the web application found on the USGS website at https://geohazards.usgs .gov/secure/designmaps/us/application.php to get the S S and S 1 values interpolated for this site. The output values are S S = 1.50g

S 1 = 0.60g

The Site Class is D, so the factors to modify the MCE spectral accelerations are F a = 1.0

T 11.4–1

F v = 1.5

T 11.4–2

These values are used to modify the spectral accelerations: S MS = F a S S = 1.0 × 1.50 = 1.50g

Eq 11.4–1

S M 1 = F v S 1 = 1.5 × 0.60 = 0.90g

Eq 11.4–2

The spectral accelerations to be used in design are S S

2 3 2

MS

= =

2 3 2 3

×

=

Eq 11.4–3

.00 g

Eq 11.4–4

0 60

×

S DS = 1.00g

S D1 = 0.60g

SEISMIC DESIGN CATEGORY Risk Category

=

ASCE 7

II

Importance Factor

T 1.5–1/IBC T 1604.5 =

1.0

T 11.5–1

Short Period Seismic Design Category = D (for S DS = 1.0g)

T 11.6–1

1-sec Seismic Design Category = D (for S D1 = 0.60g)

T 11.6–2

According to Section 11.6, the Seismic Design Category is the more severe of the two results of Table 11.6–1 and 11.6–2. Both resulted in Seismic Design Category D, so

Seismic Design Category = D


§11.6

299

Appendix 1




LOAD COMBINATIONS E E E

Eq 12.4–1 and Eq 12.4–2

v

Eq 12.4–3

E

2D ;

=

E E

ASCE 7

therefore,

+02 Eq 12.4–5 and Eq 12.4–6

±

Eq 12.4–7

E E

± 0 2D.

Load combinations of consequence are 1 2 D

+ 0 5L

0

= 0 7 D

§12.4.2.3

5L

+


Q


Q

§12.4.3.2

1 09

300

Eq 12.4–4

m

+0 = 0 7 D

+

5 L

QE

Q .



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100+ WORKED-OUT EXAMPLES! B: SEISMIC AND WIND FORCES: STRUCTURAL DESIGN EXAMPLES, 4 TH EDITION This new edition by Alan Williams, Ph.D., S.E., F.I.C.E., C. ENG., has been updated to the 2012 IBC, ASCE/SEI 7-10, ACI 318-11, NDS-2012, AISC 341-10, AISC 358-10, AISC 360-10, and the 2011 MSJC Code. In each chapter, sections of the code are presented and explained in a logical and simple manner and followed by illustrative examples. More than 100 completely worked-out design examples clearly illustrate proper application of the code requirements. Problems are solved in a straight forward step-by-step fashion with extensive use of illustrations and load diagrams. (580 pages) Soft Cover #9185S4 | PDF Download #8804P FROM SFPE AND ICC! C: ENGINEERING GUIDE: FIRE SAFETY FOR VERY TALL BUILDINGS A new information tool from ICC and SFPE for designers, builders and all engineering professionals to better                              

covered include: risks, unique challenges, egress, smoke       

design, site infrastructure, commissioning, inspection, maintenance, and change of occupancy/use. (155 pages) Soft cover #9627S | PDF Download #8950P300

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NEW FROM NCSEA AND ICC! D: GUIDE TO THE DESIGN OF BUILDING SYSTEMS FOR SERVICEABILITY: IN ACCORDANCE WITH THE 2012 IBC AND ASCE/SEI 7-10 Provides practical information and design examples for design professionals to evaluate the serviceability performance of buildings in accordance with the requirements of the 2012 IBC and referenced standards. Author Kurt D. Swensson, Ph.D., P.E., LEED AP , provides detailed explanations and examples of proper application of code provisions and standards for a broad scope of materials, building systems, and building components, covering the vast majority of limit states encountered in the structural design of buildings. Each chapter contains basic information necessary for consideration of serviceability performance and several example problems associated with an example building type. (220 pages) Soft Cover #7071S12 | PDF Download #8950P293

E: STRUCTURAL DESIGN OF LOW-RISE BUILDINGS IN COLD-FORMED STEEL, REINFORCED MASONRY, AND STRUCTURAL TIMBER Author J.R. Ubejd Mujagic provides a concise, practical guide that authoritatively covers the primary aspects of structural design of low-rise buildings in the most common materials: cold-formed steel, reinforced masonry, and structural timber. (448 pages) Hard Cover #9252S F: REINFORCED CONCRETE STRUCTURES: ANALYSIS AND DESIGN Author David A. Fanella, Ph.D., S.E., P.E., covers the analysis, design, and detailing requirements in ACI 318-08         

you through code requirements, and worked-out examples demonstrate the proper application of the design provisions. (652 pages) Hard cover #9359S

ORDER YOURS TODAY! 1-800-786-4452 www.iccsafe.org/books

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G: DESIGN OF LOW-RISE REINFORCED CONCRETE BUILDINGS: BASED ON THE 2009 IBC ® , ASCE/SEI 7-05, ACI 318-08 This publication by David A. Fanella, Ph.D., S.E., P.E., F.ASCE, will help engineers analyze, design and detail low-rise cast-in-place conventionally reinforced concrete buildings in accordance with the 2009 IBC®. Section numbers and equation numbers from the 2009 IBC, ACI         

requirements are provided throughout the text. The guide’s straightforward approach makes it an ideal reference for practicing engineers, engineers studying for licensing exams, structural plan check engineers and civil engineering students. (384 pages) Soft Cover #7034S09 | PDF Download #8950P075

H: STEEL STRUCTURES DESIGN: ASD/LRFD Author Alan Williams, Ph.D., S.E., C. Eng. , introduces the theoretical and fundamental basics of steel design and covers the detailed design of members and their connections. The book provides clear interpretations of the 2010 AISC Steel Design Manual; ASCE 7-05; and 2009 IBC. Contains 250+ design examples with step-by-step solutions. (624 pages) Hard Cover #9363S FROM ICC AND NCSEA! I: INSPECTION, TESTING, AND MONITORING OF BUILDINGS AND BRIDGES Addressing both applications and emerging technologies related to the inspection, testing, and monitoring of civil engineering structures and structural systems, this book covers a variety of topics by recognized experts in    Written by the National Council of Structural Engineers Associations (NCSEA), the book considers visual, destructive and nondestructive evaluation. The organization of the book proceeds from primarily visual inspection and evaluation of structures to more in-depth analysis procedures. Many materials are addressed, including reinforced concrete, structural steel,       

is intended as a reference for engineers routinely faced with complex evaluation requirements. (180 pages) Soft Cover #7845S | PDF Download #8950P193 13-08300

2012_IBC_SEAOC_SSDM_VOL4

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