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2018 IBC®SEAOC
STRUCTURAL/SEISMIC
DESIGN MANUAL
VOLUME 2
EXAMPLES FOR LIGHT -FRAME,
TILT-UP, AND MASONRY BUILDINGS
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Copyright
Copyright 0 2020 Structural Engtneers A~iation ofCaltfornia AJI rights reserved. This publicalJOI\
or a.1y pan thereof must not bl!' reproduced t.n any fonn without the written pemllssion of the Struc:turnl
Engineers Assoctalion of California.
..The IntemalJOnal Buildmg Code'" and the ..JBC"' are registered trademarks oflhe InteJnalional Code
Council.
Publisher
Structural Engtneers Assocaalion ofCahfornaa (SEAOC)
92 1 lllb Street, Suue 1100
Sac.rameruo. Califomja 95814
Telepoone: (916) 447-1198; Fax: (916)444- 1501
E-mad: seaoc@seaoc.or~ Web address: \VWw.seaoc.org
The Structural Engineers ASSOCiatton of CalifOrnia (SEAOC) is a pmfesstonaJ assocaauon of four regtonal
member organizatttm.>(j (Southem CalifOrnia. Northem CahtOmia. San Diego. and Central CalitOmia).
SEAOC represents lhe structural e-ngu~erutg c:ommuntty ll\ C.ahfornta. nus document IS published in
keepang walh SEAOC's stated miSSion:
To advance lhe structural engtneering profession~ to provide the public with strucrures of
depe-ndable pe.rfom'taoce through the application of state-of-the-an structural engineering
pnnc.tples; to assist the public m obtaining professional strocturaJ engineering servtees; to
promote natural hazard mttigation; to provtde conunutng educauon and encourage research;
to provide structural engtneers wtth the most current tnformauon and tools 10 improve their
practice; and to mamtatn the honor and dtgntty of the pfOfess.on.
Editor
lnrernattonal Code Council
Disclaimer
\Vhtle the mfocmation prese1ned tn this document is believed to be.COO'ect, neither SEAOC nor its ll'k!'.mbe-r
org.anizauons, committees, writers, editor~ or indtviduals \'-ilo have contrabuted to this publication make
any warranty, expressed or tmplted, or assume any legal habtlny or responslbdity for the use, application
of, and/or retere-nce to optnaons. findtng~ ooncluston.'i., or recommendanons mcluded mthis publication.
The material prese-nted tn thts publication should not be used for any s~t fic application wuhout compete-nt
e.xamination and verification of its ac.cumcy, suitabthty, and applicabdity. Users ofinfocmation from this
publtcation assume all liability arising from such use.
F1rs1 Pnn11ng: July 2020
ISBN: 978-1-60983-997-0
T0251t2
ii
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Suggestions for Improvement
Comme-ms and suggesttons for improvements are welcome ru\d should be sent to lhe following:
Structural Engineers Association ofcaltfbmta (SEAOC)
Don Schtnske, ExecutiVe Dtree-tor
921 I lib Street, SUite I 100
Socromenro, Califbrma 95814
Telephone: (916) 447- 1 198; Fax: (916) 444- 1501
E-mad: d,.,hinske@seooc.org
Errata Notification
SEAOC has made a substantral etfon to ensure that the information tn tlus docwnent is accutate. In
the event that cocrecuons or elardicauons ilte needed, these will be posied on the SEAOC website at
~-w-w.staot.org and on the ICC websne at ~w.k:tsart.org.
SEAOC. at
it~ sole discreuon. may 1ssue wr itte1l
errata
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Table of Contents
Preface to the2018/BC SEAOC !>intcturol/Sei.smic Design Mtmual . .....• . ... . . .. . ... . ........
VII
Preface to Volwne 2 .
IX
AcknO\vledgmenL~
.. •...•. .
XI
References .. ...... ...... .
Xtli
t-10\V to Use Th1s Document. . . . . . . . . . . . . . • • • • • • • • • • • • • . . . . . . • • • • • • . . . . . . . . . . . . . • • • • . . . . x.xi
Design E.xsunplt I
Four-Story Wood Ltght-Frame Structure.... .
Design E.xamplt 2
Flexible Diaphragm Design. . . . . . . . . . . . . • . • • • . • • • . . • . . • . • . • . . . . . . . . . . . . . . . . . . . . . 133
Design E.xamplt J
Three-Story Light-Frame Mulufanuly Buildlng Design Ustng Cold-Formed-Steel
Wall Framing and Wood Floor and Roof Fram1ng . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . 155
Design E.xamplt 4
Masonry Shear Wall B01ld1ng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273
Design E.xamplt S
Tih·Up BU1Id1ng . . . . . . . . . . . . . . . . . • . . . • . • • • . • • . . • • • . • . • . • . . . . . . . . . . . . . . . . • . . . . 299
2018 IBC SEAOC SttuctutaVSeismic Design Manual. \obi. 2
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Preface to the 2018 IBC SEAOC Seismic/Structural Design
Manual
The IBC SEAOC Selsmic/Srructural Dtsign ManU(J/, throughouttl"i many edtuons. has served the purpose
of iHusuaung good seismtc destg.n aoo the correct appl icat1011 of buildtng-code provtsions. 1be l&ltmual has
bridged the gap betv.'een lhe discursive ueaunent of topics in the SEAOC Blut Book (Rtcommtntkd La/eru/
Force Requiremenu and Commentary) and reai-YI'Orld dectstons that desagners face t.n thetr pracuce.
The examples tllusttate code-complianr designs enguleered to achttve good pe,rformance under se\•ere
sei.snhc load mg. In some c.ases simply complytng walh butlding-code reqUirements does not ens·ure good
seismic response. This Manual takes the approach of e.xceedtng the mtntmum code require-mems in such
cases. wnh diSC-ussion of the reasons for doing so.
llus manuaJ comprises four volumes:
Volume 1: Code Application Exomples
Volume 2: Examples for ltght-Frame. Tllt-Up. and Ma.:;onry Budd1ngs
Volume 3: Examples for Concrete Buildings
Volume 4: Examples for Steel-Framed Buildings
In general. the provisions for developing the des1gn base shear. d1sttibut1ng lhe base-shear-forces \'ertically
and hori2onmlly. dled:ing for irregu1arit1es.. etc .• are illosuated m Volume I. 11te olher volumes contaJn
more extens1ve design examples that address tJ1e requarernents of the mater1al standards (for example. ACI
318 and AISC 341) that ate adopted by the lBC. Bulklingdes&gn examples do not illustrate 1nany of the
1tems addressed 1n Volume I in order to perm at the inclusaon of less--redundant contenL
Each volume has been produced by a small g(Oup of authors under the direction of a manager. lbe
managers have assembled te\'lewers to e.nsure coordmation wnh other SEAOC work and publications. most
notably me Blue Book. as well as numenca1 ac:curac:y.
ThiS rnanua1 can serve as \1aluable tool for eng.i~n seektng £O design buildings and build1ng components
fOf" good seismtc response.
Rafael Sabello and Katy Broggs
Project Managers
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Preface to Volume 2
Volume 2 oflhe 20 18/BC SEAOC SJructuro//Se,snuc De.slgn Mamw/ addresses the design of light-frames
ooncrere nit-up, and masonry shear wall buildmg systems for seismic loodmg. These ln('.fude the tllus.trarjon
oi the design requirements for the shear \\ails ru\d diaphragms, ao; were illustrated in previous editions, and
also tmponaru mtertaces with the rest of the structure.
The de:s1gn example.-; ln lhis volume represent a range of structural systems and seismtc syst·ems. The
design of each of lhese systems is go,•e.rned by standards developed by the American Concrete Jnsutute
(ACI) and the Amencan Wood Council (AWC). The methods tllustrated hereto represent approaches
conststent w1lh the ducuhty expec.tauons for each system and wnh the des&red se1smtc response. In nlost
cases there are se\'eraJ dem.Jls or mechanisms thal can be uti!ized to achieve the ducul ity and resistance
requtred, arld the author or each example has selected an apptopriate opdon. In many cases alternadves are
discussed. Th1s .Mauua/ tS oot imended to serve as a buildi11g code or to be an exhausuve catalogue of all
valtd approaches and de-rolls.
llus ,\fa.nuli/IS prese-nted as a set of examples tn whtch lhe engineer has considered the butldtng-code
requtremems in conjunction v.ith lhe opumal setsmic respon.o;e oflhe system. The examples follow the
guidelines oflhe SEAOC Blut Book and other SEAOC rec.ornmendatioos. The exampl e.~ are intended to atd
conscie-nlJOUS destgnetS 1n c.-raftmg destgns that are likely to achieve good sttSJtuc perfOrmance cons.stent
v.1th expectations inhere1U tn lhe requtrements for the syste:.ns.
Douglas Thompson
Volwne 2 Manager
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Acknowledgments
VoJume 2 oflhe 20 18/BC SEAOC ~wmdStrucwral De.slgn .M.tmubl was wnnen by a group of highly
qualified srructutal enganeers, chosen for lhear know1edge and experience with structural engu-.eertng
practice and seismic destgn. The authors are:
DouglasS. Thompson, S.E., S. E~C. B-Vo lumt M.anagtr and E~:amplt- I
Doug Thompson has over 40 years of expenence 1n designang of wood st.rucrures. He as the author of
severaJ pubhcauons an umber desagn including the WoodWotks publications: Four-story IV()()(/{ranw
StruCIUrt Ol·er Pcdlum Slab and Fn-e-story U'IHK/-frame StniCtuf? O'rtr Podmm Slab. Doug has instruc.ted
ltcense review·classes in umber destg.n for t11e PE and SE e.xams tor 20 years. He 1S a past prestdent of the
Structural F..llgtneets Association of Southern C.a1iforma and holds Jteenses tn SlX states. vN;W.stbse.com
John
la~'So n~
S.E..-Examplrs 2 11nd S
Professor John Lawson has provtded st.ruclutal engineenng consultlng se:rv1ces tOr 0\'et 30 years, mdudjng
overseeing more than 100 mill1on square feet ofiM\'-sloped roof and uh-upconcrete engt.neenng. He now
teaches m the Architectural El'lg1neenng depanment at Cal1fomia Polytechnic State Universuy 1n San ltJjs
Obispo. John is lhe recipient oflhe 2006 Tth-Up Conctete Associauon·s David L Kelly DisungUtshed
Engu~er Award. www.arce.calpoly.edu
Mithatl Cochran, S.E., S.E~C. B.- Examp lt 3
Michael Cochran is a Vice President wnh Thonuon Toma<;en.i.. lnc... 1n Los Angeles, Cahtbmia, with over
25 years of design experience. He has an extenswe background in the design ofmulustory hght-frame
commerctaJ and m.ultJfami1y res1denuat wood and cold-fOrmed steel-stud buddjngs. He lS a regtstered
SU\Icrural engmeer in Cahfomia. an active member of the AISC Connection Prequahfic.ation Review
Panel. a pa.<n. presldent of the Structutal Engineers AssociatiOn of Southern Cal1fomja (SEAOSC) and lhe
SU1Jctural Engineers Association ofcahfornla, ru'ld a SEAOC fellow.
Jtff EUis, S.E., S.E.C.B.- Eumpl< 3
Jeff Elhs., Director of Codes and Compliance for Simpson Strong-lie Company, Inc., ha'i more than 28
years of expe:nence as a professional engineer and manages the company code and compliance efforts.
Addtl!onally, he ts tn\•olved tn research and de-vclopm.ent and provides suppon for existing product hnes.
induding technical guidance fOr connectors. fastemng S)'S[em~ and lateral-force.reststing syste.t\S. He
was a pracucing dts1gn englOet-t for c.om.merctaJ, residential. and forenstc projects tbr more than nu~
years prior to joanmg Simpson at the end of2000. He currently serves on lhe Inte-rnational Code Council
E\•aluauon Service Boord and has served as president of the Struc.tural EngtneetS Assoctation of Southern
California (SEAOSq SEAOC secretuy, chainnan of !he AISI COFS Lateral Design Subcommittee, and
pres idem of the Cold-Formed Sleel Engineers Jnsutute (CFSEI).
2018/SC SEAOC Structural/Seismic Design Manual. Va'. 2
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Cbuk\\Urtli G. E.kwutmt, PhD.SE, LEED AP-E.umplt 4
Dr. Ekwueme is a Pnncipal with TilorntOil Tomasetu 1n Los Ange-les, california He ha.~ an extensive
background tn the destgn and anatys1s of a wade vattety of stn.ICtures, indudmg concre-te and masonry
eonsuucuon, steel and aluminum structures. and light-frame wood buildtngs. He is a reglstered
structutal enganeer m Caltfomja and Ne\•ada and is an aCU\1t member of the matn commiuee, the seismtc
subcommittee, and the axial fte.xutalloads and shear subcommittee of the Masonry Standatds Joint
Committee (MSJC).
AddittonaUy. a number ofSEAOC rnembef'S and other SltUCtutal enganeef'S helped chec.k the examples tn
th1s \'Olume. Ounng llS development, drafts of the examples \vere sent to these Individuals. Thelr help was
sought m review of code tnterpre.tations as well as detailed checking of the numencal computations. The
reviewers tnclude:
James La1, S.E.
Alan Robinson, S.E.
Tim Slafford, S.E.
Doug TOOmpson. S.E.
Tom VanDotpe, S.E.
Close collabormjon w1th the SEAOC Setsmology Comm.tttee ·was m.auuamed dunng the development
of this document "The Sels:mology Comminee has re\•iewed the document and provtded many helpful
comments and suggesrjons. Their assistance is gratefully acknowledged.
Productton and art was provided by the lnte.rnat.ional Code Council.
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References
Standards
ACI 318~ 2014. Bulldmg Code R~quirementsfor Stnu:turol Concnue, Amel'tcan Concrete lnsuune,
Farmington
~hils,
Mich1gan.
AISI S 100- 16, 2016. North American Specificalionfor the Destgn ofCold-Formed Steel Slructwu/
.f&lemMrs. American lton and Steel lnsutute, Washtngton, DC.
AISI 5240- 15, 2015. North American Sltindtlrdfor Cold-Formed ~ie~l StruCturtll Frammg.
Amer1can Iron and Steel Institute, Washington, DC 20036.
AlSJ 5400- 15,2015. North Amencan Sumdardfor &wnic De.tign ofCold-Formed Stul Stn1CturaJ
SJ·.vtems. Amencan Iron and Steellnsutute. Washington. DC 20036.
American Wood Counc1l, 2018, NattOtJ(J/ Design Specification for U't?od Con..tfruction lncludmg
Suppltm(mts. NDS-18. American Wood Coui'ICII, Waslungton, OC.
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Wood Council, Washington, OC.
ASCEISEI 7. 2016, Mimnwm I:kstgn Load! tmJ A.uoctated Cnt~riafor Buildings andOth~r
Structlll?s, Amencan Soctecy ofCtvd Engtneers, Structural Engineermg lnsutute, Reston,
V1rginia
lntemanonaJ Code Council, 201 8,/ntemtllitma/ Bmlding Ctxlt (IBC). lnternadonal Code Council,
Washtngton. DC.
TMS 402- 16, 2016. But/ding Code Requit?mentsfor A1asonry Structurtl, 1be Mason')• Society.
Boulder, Colorado.
TMS 601- 16,2016. Sp€cijicaumrfor MasomySiructUJ?s, The Masonry $OCtet)', Boulder, Colorado.
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ACI 551 .2R- 15, 2015. Guitk for the IJI!:ugn ofnlt-up Concrttt Panel.t. American Concrete
Institute, Fanninglon Htlls., Michtgan.
ACI 551.1 R- 14, 2014. Guidt 10 Till-up Concrett Construcuon. Amertcan C.oncre-te lnstttute,
Farmington ~hils, Mich1gan.
AISI 0100-08, A/Sf Mtuwal. Cold-Farmed Stu/ Design. American Iron and Steellnsutute,
Washt.ngton_. DC.
AJSI 0110-07, Cold-Fonned Sue/ Frammg Destgn GUide, Second Edition.Amertcan Iron and Steel
lnstnute, WashtnguMt, DC.
2018 fBC SEAOC StructuraJISeismic De.sigJ Manual. \tll. 2
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American FOfes[ and Paper Associauon, 1996, HOOd Con£/ruCIIOn Manual. Amencan Forest ruld
Paper A<SOCiauon, Washington, DC.
American P1)'Wood Assoctalion, 1997~ t:ksignl Om.uructfon Guide- Diaphragms and ShMr Walls.
From L350, Engu1eered Wood Associanon, Tacoma. Washtngton.
Amencan P1ywood Association, 2007, Diaphragms and Shear lf't11/s. Engmeered Wood A~tatton,
Tacoma. WashJngton.
A.tnertcan Plywood.A'>Sociation, 1993, revised, HOOd Struc.turol Pa.nel ShMr Wa/l.s. Repon 1S4,
Engineered Wood Assoclalion, Tacoma,. Wa'iliington.
Amencan P1ywood A'>SOCtation, 1994, Northrid~. Califorma Earthquake. Repon T-94-5.
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Amencan P1ywood Association, P~rformance SJandard For Heod-Based Strucwral Use Panels.
PS2-04. National 1nsutute of Standards and Technology, \Vashmgton, OC.
A.tnertcan Plywood Association, 1997, Ply..-ood Design Spectjications, From Y510. Engtnee:red
Wood Associauon. Tacoma, Wa.<mmgton.
Amencan P1ywood A'>SOCtation, 1988, PIJ,.-t)od Diaphrogmt, Researc.h Re-pon I38. Amencan
Pl)'\\-ood Associatton, Tacorna, Washtngtoo.
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American P1ywood Associario~ 2005, Using NariVIr Piues ofW()()(f Structural Panel Sheathing in
Wood Sh~ar Walls. APA T 1005-08~ The Enginee.red Wood Association, Taootna, Washington.
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APA, Tacol'na~ Washtngron.
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Applied Technology Councol, Redwood City, Cali1ornoa.
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Fall 2012, Forest ProdllctsSociecy, Madison, Wtsconsin..
Bend'iten, B.A. and W.L. Galltgan. 1979, Mean and Toltrtmu Limit Stresses and Stre.s.s Motklmg
for Comp~?SJion ~rJMnthcular to (iraln in Hardwood and Softwood Species. Rese,arch Paper
FPL 337. US Department of Agnculture, Forest Service, Forest Products Laboratory, Madison,
Wisconsm.
xiv
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Bendsten, B.A. and W.L Galligan, Vol. 29, No.2: Pg.. 42-48. 1979. Moddmg and Stre$$Comprt!$Jion R~latk>nslu'ps tn U~ in Compn!SSIOn Perpendiodar to Grow. U.S. Oepan.m.enl of
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Cal1forma
Cook. R.A., I999...Suength Oestgn ofAnchorage to Concre.te." Ponland Cement As:soctauon,
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Openmgs and Bast! Re.s1rai111 Co,(tgurotions. Timber Engtneering Repon No. TE-1997-00 I.
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Dolan. J.D. and Heine-, C.P.~
1991b.Seq~Nnlttll Phastd Displaument Cyclic TRstsofWood Frame
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Virgtnja.
Dolan. J.D., and Heine.• C.P.• 1997c. Sequential PIJtJStd Displacement Ttst ofJJOOd Fran~ Shear
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and State Un1vers•ty. Blac-ksburg. Virgin1a.
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Reconnai.uanct Report. Earthquake Spectro. Vol. II, Supplement C. Earthquake Engineermg
Research lnstuute, Oakland. CaJ1fom~-a.
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Stntctun magazine. National Council of Strucmtal Engineers Assoctauons (NCSEA).
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Rtcommended Pll'Jl•tstons for Mtsmlc Regultwonlfor New Bmldmgs and Other Structures and
Commentary. FederaJ Emergency Managemem Agency. Wa<ili1ng10n, DC.
F1ccadenu. S.K.• T.A. Caslle. D.A. Sandercock.,and R.K. Kazanjy. 1996, ..Laboratory Tesung to
lmrestigate Pneurnatically Driven Box Na1ls for the Edge Nall1ng of 3/8'. Plywood Shear Walls,.·•
Proceedmgs: AMual SEAOC COIWention. Sttuctural Eng.meers Association ofCal1fomu1,
Sacramento, Cal•fornia.
Fohe1Ue, Greg C., 1994. AnlllyJis. J:Msig" and Testing of Timber S tructurts under Seilnuc Loads.
University of California Forest ProducLo; Labor.uory, Richman~ California.
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Wisconsm.
Ghosh. A. S. Pryor, and R. Arevalo, June 2006...Mulli.StOI')' L1ght-frame Construction:
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(NCSEA).
Goers R. and Associates. 1976, A Methodologyfor Seismic Design and ConltnKtiOII ofSmgltFamll)' Dwellings. Applied Technology Counc-.11, Redwood City, California.
Gupta, R.. H. RedJer. and M. Clauson. 2007. "Cychc TestS ofErlguteered Shear Walls with Diffe-rent
Bouom-plaxe and Anchor-boll sizes (Phase 11):• Depan:ment of Wood Science and Engineering.
Oregon State University, Corvants.. Oregon.
Haygreen, J.G. and Bowyer, J.L. 1989. Forest Products (Jild Hhod Science- An IIIJrQduction.
Un1ve.rsiry of Iowa Press, Ames, Iowa.
XVi
20t8 IBC SEAOC Structurai/Seis;c Desig'J Manual,
tAll 2
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Hess, R, 2008, "For What Planet Is Thts Code Wnuen?,.. Str11cturt magazine, November. National
Councd of Structural Engtneers Associatiol'ls (NCSEA).
Hohbach, D., S. ShiotanJ, 2012. lmpro,·~d Seumu: Analysis ofW()(}(/ Liglu-Framed Muflt-Story
Residtnllaf Buildlngs, Wood Oestg.n Focus, Fall20 12, Foresl Produels Socrel)', Madtson.
Wisconstn.
Ju, S. and Lt.n., M., 1999, ..Companson of Building Analysts As.~umang Rigid or Flexible Floors.,·•
Journal ofStrucrutaJ Engmeenng. American SoctetyofCIVII Engineers. Washrngton, DC.
Knight, Bnan. June 2006. High-Ri£~ Wood Frome Construe/ion. S tntetmY magazme. NCSEA.
Kong, II, Ea1henon, M, aod Schafer, B, 2019. "Design of F1xed Base Hollow Stru<rural Sect1on
Columns Subjected to Large Setsmic Dnft," Proceed1ngs of the Annual Stabdtty Conference,
Apnl20 19, Strucrural Stabihry Research Council, SL louis. Mtssoun.
Lawson, J., 2019, "hnprovang the Accuracy of Wood Diaplmlgm Deflection Compul3uons and It<
lmpacr on ASCE 41 Pseudo-Lateral Force E.c:Hmates; ' SEAOC 2019 Convenuon Proceedings.
Strucrural Eng.tneers Assoctauon ofCahfom~a, Squaw Valley, C.alafornta.
Lawson, John, 2007~ "DtRection Limits for Tilt-up Wall Serviceabtltty;• Coucrtlt lnrertJIJJional,
Ameracan Concrete lnstuute. September.
Lawson. J, Kol1ou, M, Flllatrauh~A.aod Kelly. D. 2018 ...The Evalu:auon of Current Wall-to-Roof
Aochorage. For'Ce Provisions tOr Sangle-stocy Concrete and Masonry Build1ngs wirh lightweight
Flextble Diaphragms,.. SEAOC 2018 Con,'ention Proceedtngs~ SU1Jctural Engineers Association
of California, Palm Desen. California.
Mauera, Domintc, 2009, 5 Owr I
High-Ri£~ Podwm StruCfllrFS,
Wood Soluuons Faar Presentation.
Maneson, Thor, 2004, Wood-Frtuned Shear Wall ConstructtOtJ. lnternarjonal Code Council, Couru.t')'
Club Hills, lllincis.
Mayo. John l., 200 I, ..Metal Roof Construction on Large Warehouses or DIStrtbmion Centers;·
Steel lips. SLruCtural Education Councal, I4 I Greenbriar. Moraga. CA 94556, June.
Mendes, S., I987, ..Rtgid versus Flex1ble: lnappropn:ue Assumpuons C-3.11 Cause Shear Wall
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CaJit'omla, Sacramento, Caltfornia.
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Murphy. Michael, 20 12, Shrmkag~ Challenges wi1h Mid-Rt.tl! CoustnJCiion, Wood Destgn Focus.
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MulliS[Qty SLruCtures."' Struclurt Magazine~ March. NCSEA.
201 8 IBC SEAOC Structural/Seismic Design M anual. \lbl. 2
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xvii
Rose, J.D., 1998, Prelimintll)' TeSJmg ofWood Structural Panel Shear Walls und~r Cyclic (Ret·ersed)
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and Jf17ndon' [SJreathmgStandard. Sec.1.3.3j. Research Report 158. APA- The Engineered
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Bwldmgs C:ase Swdies Projtct, Setsm.tc Safety Cotntnts:ston, Smte ofcaJitbrnlil. Sacramento,
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In Light-Frame Co-muroctlo-n. SutJctuml Engu-.eers Assoctatton of California, Sacramento,
Caltfornaa.
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Sacramento, Calafornia.
SEAOC Blue Book. 1999, Rtcommendtd Lattral Fore-e Rtquirements lllld Commentary. Structural
Engineers Association of CaJtfornta, Seve.nth Edition. Sacramento. Caltfornia
SEAOC Seismology Commiuee,l009, "Anchor Bolts m Ltght-fi'ame Construcuon at Small Edge
Distances.,·• June~ MS. M6, M7 m the SEAOC Blut Book: Seismic Destgn RecommendaJions~
StruCtural Engineers Association ofCaltfbrnia, Sacramento, California. http:l/www.seaoc.org/
bluebook/index.hunl
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Blut Book: Sei$mic Deslgn Recommendations. Structural Engtneers Association of California,
Sacramento, Caltforn1a. www.seaoc.org/bluebookltnde.x.html
SEAOC Seismology Commiuee,l008. "Tilt-up Butldings,·· The SEAOC Blue Book: Seismic Design
Recommendations. Suuctural Engineers Association ofCaJifornia. Sacramento, Cahfonua ac
W\~tw.seaoc.orgfbluebookllndex.html
SEAOC Se1smology Commmee, 2007, "Wood-ti'amed Shear Walls \\11h Open.ng.<; May, tn the
S£.40C Bllte &>ok: Set.t nuc Dtsig11 Recomme,datiom, Structural EngtneersAssociauon of
Caltfbrnaa. Sacramento, Cal1fomia. \\'\\'W.seaoc.orglbluebook/tnde:x.htm1
SEAOSC, 1979. Rtoommtndtd ntt-up Uh/1 Design, Structural EngineetS Assoctatton of Southern
Caltfbrnaa., Los Angeles., Cahtbmta. June.
xviii
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20 18 IBC SEAOC Structural/Seismic Design Manual. \obi. 2
SEAOSC/COLA, 1994. 199./ Northridge EarthqutJkE (StroCfllrtJI Engineers Astociatitm ofSouth~m
Califomia!Ctt)' ofLos Angeles) Special Im·estigatlon Task Fore~. Tilt-up Subcommittee. Final
repon dated September 25, I994.
Shjotani. S.• D. Hohbach, J. Robens. 2011, L(JitrtJI S)'!temfor A·!Jtlti-Umt ConsirUCtton. Wood
Ptoduct.<i Couoctl Wockshops.
Shipp, John and D. Thompson, 200 1 Timber Design/. II. and Ill. JO/ume.t VIII. IX. and X.
Professtonal Engineertng Development Publtcauons, Inc .• lrvane, CaltfomJa.
Simpson, Wtlham T. 1998. Equilibrium MQislm"t Content ofJJ'Ood ln OUidoor LocaltOtLt in the
United States and Worldwide. Res. Note FPL-RN-0168. Forest Prodoc.rs Labomtory. Madtson,
Wisc.onstn.
Skaggs, T.D. and Z.A. Martin, 2004. "Estimating Wood StruCtural Panel Diaphragm and Shear Wall
Deflection... Practic-e Periodict~l an Structural Design and Constntclion. ASC~ May 2004.
Steinbrugge. J., 1994, "'Standard of Care m Structural Engtnee..-i11g Wood Frante Multiple Hou'iang.Proceedings: Annual SEAOC Convennon. StruCtural Enguteers Assoctauon of California,
Sacramenlo. Ca1ifornia.
Techntcal Coordtnaung Committee for Masoruy Research (TCCMAR), 1985. James NolandChauntan, U.S. -Japan Coordintued Pmgramf()r MQS(Jnry Buildlng &s~tJrch, U.S. Research
Plan.
Thompson, O.S.• 2009, Four-ltory Wood-frame Structure m·er Ptxllw, Slt1b. Woodworks, Tacoma,
Washington.
Thompson, D.S.• 2012, Fl\·~-stary W(J{)(/-framl! StruciUn ot·l!r Ptx/ium Slab, Woodworks, Tacoma,
Wao;hJ.rtgton.
Thompson, D.S.• 2012. 2009 JBC StniCturtJJ/Mt:tmlc ~sign Manual. 10/ume 2. ~sign E.tample.t I.
l & 3 Structural Eng.neers Assodauon of caJ 1fornta Sacra..nemo. Cahfornia.
VanDorpe, Tom and Andy Fenoell. 2010,2010 Building Code UptkJJe R<-roolmg Your Office for
Changes to the 2010 California Butldlng Regulattons that A/foct Light-Frame Slructure.'l. Orange,
C31tfotl\l3.
Washington Assoctauon of Butldmg Offictals and Structural Ettg.ineets Association ofWashtngron
(WABOISEAW) liatson Commtttee, 2013, Whtr< Paper 9-2013: nwaded Rod Holdown
Systems In Wood Frame Buildlngs. SeanJe, Washmgton. www.wabo.«g/waboseaw-white-pa.pe-rs
Weslem Wood Prodocts Association (WWPA). November 2002. Ttch Note! Refl(>rt No.
I 0-Shrmkb~ Calculations for Afttltt-St()ry H't'>od Frame Construc.llon, Porlland, Oregon.
WWPA, 1990, Dtmensional Stability of Western lumber, Portland. Oregon.
Yousefi, Ben, Son, James, and Sabel h. Rafael, 2005. Structural Engineering Re1·iew Manual (2005
Edition), BYA Publtcations, Santa Monica, Caltfomta.
2018 fBC SEAOC StructuraJISeismic De.sigJ Manual. \tll. 2
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2018/SC SEAOC Structura/ISf!;smic Design Manual. Vol. 2
How to Use This Document
Equation numbers in the nglu-han<l margtn refer 10 the one of !he slandards (e.g.,ACI J 18, ASCE 7, or
IBC). The default S[andard ts g.JVen in the headtng of each sec-tion of each e.xample~ equation numbers 10
that secuon refer to that slru\dard un.less another standard is explicitly cited.
Abbrevlanons used 1n the ..Code Refere-nce" column are
§ - Section
T - Table
F - Ftgure
Eq - Equation
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2018/BC SEAOC Slructurai/Seismic Design Manual, 1.01. 2
Design Example 1
Four-Story Wood Light-Frame Structure
OVERVIEW
ThiS design example &llusrrates lhe seismic design of selected e-lements for a four-story \vood-frame hotel
structure. The gravity-load framtng system conslSIS ofwood-fiame be~mng walls. The lateral-load-resisung
system consists of wood-frame beating shear walls (common box-type system). A typical building ele-vauon
and floor plan of the structure are shO\m tn F&
gures 1- l and 1 -2~ respecu,.-ely. A typtcal section showtng
the heights oflhe struc.ture is shown m Ftg_ure I-3. The wood roof IS framed with pre-manufacwred
wood ttusses. The floor tS framed ·with prefabricated wood 1-JOiSt(i. The floors have a 1Y!-inch lightweight
ooncre.te topplng. The roofing ts composiuoo shtngles.
When des1gn1ng thjs l)'pe of mtd-nse wood-frame srruct-ure. there are several un1que des1g.n elements to
cons1de.r. The foUoo•ing steps prov1de a detailed analysis of some oflhe tmponant setsmic requ1rement'i
of the shear "ails per lhe 20 18 JBC. Th1s des1gn e.xample represents a very s1mple wood-frame wood
struc.ture~ most wood-frame structures have severn! u1tique features requ1nng engmeer~ng design and
detailing not sltown m th1s des1g.n example.
ThiS design example is l'lOl a comple.te bwldtng destgn. Many aspects have not been tocluded, spectfically
the gravtcy-lood framing system, and onJy cenatn steps of the seismic design related to pontons of a
selected shear wall have been tllustrated. In addltton. the lateral requircmenLc; for Wind desagn related to
the se.lected shear wall have not been iUustrated (only se&srruc). The steps that have beentllusumed may
be more detailed than what ts neces:sar)' for an acrual buildtng destgn btn llte presented tn thls manner to
help lhe design engtneer undet"Stand the process. For a ntote detatled Jisung of the items not addressed, see
Section 10.
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Design Example 1 • Four-Story~ Light-Frame StTucture
OUTLINE
I. Building Geomeuy and load<
2. CalcuJarion of the Design Base Shear
3. Location of Shear Walls ruld Honzonml Distr1bution of Shear
4. Mec.ha.ntcs of Multistory Segm.emed Shear Walls and Load Comblnauons
5. Mec.hantcs ofMuJtistory Shear Walls with Force Transfer around Opemngs
6. Tile Enve.lope Process
7. Design and Demoling of Shear Wallmlone C
8. Diaphragm De.fle<-tions 10 Dete.rmine if the
Ou~phragm
Is Flexible
9. Disconunoous System Constderation."> and the 0\•e.rstrength Factor
10. Special Jnspecuon and Structural Obsetvation
II . hems No1 Addressed 1n ThJs Eumple
1. Building Geometry and Loads
ASCE7
1.1 GIVEN INFORMATION
roof is '1-1>-onch-thick DOC PS I· or DOC PS 2-rated wood structlll111panel (WSP) shemhu1g, with a
32/16 span raung and Exposure I adhestve or \\<lterproof adhestve.
TJ-.,
The ftoor is ll.l.c-inc:h-thic.k DOC PS 1- or DOC PS 2-nued Srurd-1-Fk>or 24 mches o.c. mnng. with
a 48n4 span raung(40/20 span mung with topping is also accepmble) and Exposure I adhesove or
waterproof adhesive.
DOC PS I and DOC PS 2 are the US Deparunent of Commerce (DOC) presc.roptive and performance-based
standards for plywood and onented strand board (OSB), respecuvely.
Wall fram tng. 1S a ..anodtfied balloon framing•· where the jotSL(i hang tfom the walls tn JOISt hangers.
(See Ftgure 1-7 detail of lhis and an explanation of Olher common framing conditions.)
Franung lumber f()( studs and posts
2
NDST4A
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Design Example 1 • Four-Story VKiod Ught Frame Structure
4
Dou.gllls Fir-Lartb-No. I G radt unadj us tt-d dtsigu \'tlut-s:
F• = I ,000 pso
Ft = 1,500 psi
Fa =625 psi
F,= 675 psi
E = 1,700,000 psi
Em,.= 620,000 psi
C"'= J.Odry ln-servtce condttions assumed
C, = 1.0 normal te:mpetature conditions assumed
Frammg lumber used for studs and posts is desogned per lhe National Design Specofication"' (NOS")
for Wood ConstrucLJon and NOS Supplement: Destg.n 'hllues for Wood Construc.uon.. Only l'-''<> end-use
adjustment factors are shown here. Others Will be de-fined and ~'1\ Later tn me design example.
Commott wire nails ate used for shear waJJ~ diaphragms. and straps. Whe.n specifymg nails on a proJect.
specttication of the penny wetgh~ type. diame-ter, and length (e.xample IOd cotntnon = 0.148 inc.h. x 3
inches) are recommended.
~'
t
~'
';..---...,;,
Figw·e I- I. Buildtng el~ l"lltilm
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Design Example 1 • Four-Story~ Light-Frame StTucture
~
NORTH
FigwY 1-1. 7jpicalformdarlonplan
4
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Design Example 1 • Four-Story VKiod Ught Frame Structure
4
4
23/32" APA R~TED
SH£A1WNO 48/24 SPAN
RATING 10d SCREW SHANK
HAllS 0 6" ojc EOct'S
PANEL EDGES UNBlOCKED
SHE"AlHING GI.UED TO
fRAU lNG
----10
HORIZONTAl. TIE
METAL STRAP AND
BLOCKING
I~
----10
·~
NORTH
Figunt 1-3. TJ.picaljloor frammg plan
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Design Example 1 • Four-Story~ Light-Frame StTucture
'
I
I'
I'
J - ------0
l
I
I
-----0
g
-----®
15/32. APA RATtO
••
SHEAll-!II+G J2/16 SP
RAnNG lOG 0 6• o}c
EOGES PANEt. EDGES
----®
UN9LOCKt0
ED
'
·-o
2'-0
~~
~
NORTH
----0
-- --0
~
ll
----®
-------®
FtgiJTl' /-4. T)p•ca/ roofframing plan
Notes for Ftgures 1-2 through 1-4:
I. Nonstructural ..pop.outs" on the exterior '"ails at hnes I. 4 need special detading showmg the
YI'OOd structura1 panel sheat.lung ruruung c:onunuous at hnes I. 4 and lhe pop-outs framed after
the shealhing is u\stalled.
2. All walls stack from the foundauon to lhe fourth floor.
3.
6
<) Desognates shealhed "all per shear-wall schedule (see Table 1-35).
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Design Example 1 • Four-Story VKiod Ught Frame Structure
4
PROIANUTAC'l'UREO
WOOD ROOF' lRUSS£S
0 24" 0/¢
TOP Pt.ATE.S + lfS,M'
4TI{ FlOOR +28.32'
~
FlOOR +18.86•
PREMANUFA.CT\JRED
"r"
.asTS 024"o/c
tsr n.OOR ..o.oo·
Figu~?
1-5. Typical bmldit,g secti()uS
Notes fOr Figure 1-5:
The cemer of mass of the roof lS higher than one-third of the ..tnal'l.gle., shape plus lhe blocking he.aght
over the top plates d:ue to the weight of the roofing re roof. and sheathing is heavie.r lhan the wetght of the
cetling. A eonsef\1ative height equal to the center of the roof diaphmgm (average hetg,ht of the sloped roof)
has been used tn this desagn example.
4
1.2 FACTORS THAT INFLUENCE DESIGN
Prior to smrung the se.ismtc design of a sLructure. the following must be considered:
Cood Sbtar-\Vall Consl:ruuion Dttailing GuidanceUse ortented strand board (OSB) sheathmg mther than plywood for 1mproved suffness.
Use dtst:rtbuted nallang to end studs and plates. Th1s reduces the need to sutch multi-ply end studs 3Jld
redoces potenual framing spin damage due to putung all nalls tn a single 2x end framtng mernber. ln a
typical Jugh-strenglh wall test. placing aJI the nalls m the ootermost stud tends to increase compression
perpendtcular to grain deformation tn the bottom plate that is flush with the e.nd stud
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Design Example 1 • Four-Story~ Light-Frame StTucture
Provide greater than minimum nail edge distaoc.e at osn panel edges. Putting nails at% inch from the
panel edge 1ncreases the potenual for nonconforming consuucuon. even m testing.. Because some nails will
be closer than% 1nch., tt places panel edge tasten.trtg in the outermost stud of a multi· ply stud pack. whtch
might lead to sphtungdamage even if the outermost rowofnatls is staggered (per above).and tn some
cases promotes earlier occ-urrence of panel edge tear out relatl\'e to dtstribure<l nailing.
Use concentric hold-downs or c.onllnuou.~ ue-d0\111\ systems rather than ec:centtic hofd.downs. Data from
bolh shear-wall testing with conunuous rods and from prior wood shear-wall tesung with conventional
hold-down dev1ces on both stdes ofa butlt-up 2 x 6 end stud pertbrmed better than tests \\ith a sangle
convenuonal hold-down on the tnboard stde of the wall.
Provide at least equivalent strength and stiflhe.ss (induchng the partictpauon of finishes) at first.fevel Lateral
system elements as compared with the second Jevel. Consjder nonstructural walls (especiaHy when such
walls do not stack down to the butldmg base) when evaluaung lateraJ system layout for weak and/or sofl
story con£huons.
Sptclu of lumbtr
The species of lumber used tn lhjs design example IS Douglas Fir-Larch (DF-L). which is common on the
west coast The autJ'IOf does not Intend to imply that this species can or should be used in all areas or for all
market~.;. Species that are both appropnate for this (ype of construction and locally available wry by region
and commonly 1nclude (among olhers) Soothern Pine (SP) and Spruce Pme Fir (SPF).
Gradt or Lumbtr
The lower l\vo stories of the wood-frame structure carry higher gravity loads than the upper two stones.
One approach is to use a higher grade oflumber for lhe lower two stories than the upper two stories. This
approach can produce designs that yield a consiStent wan construcuon over the hetght oflhe buildlng.
Anolher approach is to c.hoose one grade of lumber for all tOur wood-frame stones. Thts approach produces
the need to change the size and/or spacang oflhe studs based on the loadtng requirements. Sill-plate
crushing may control stud sizing a[ lower levels. For simplicny. thts design e.xample tllustrates the use of
one lumber grade for all floor levels.
k td)
m!S.
Figure I -6. Typical gradf! .s1amp
Notes for F1gure 1-6:
a. Certrfi.catJon Mark: Cert&ties assoctalion quality supetviSJon
b. M1ll ldentificauon: Firm name. brand. or assigned mall number
c. Grade Designat.ion: Grode name. number, or abbreviation
d. Species ldentificaljon: Jochcates species by individual species or species cotnbinatton
e. Cofldttjon ofSeasomng: Jndicau.~ conditton of seasomng at the nme of surfacmg
8
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Design Example 1 • Four-Story VKiod Ught Frame Structure
4
Moisturt Stn·it:t Condition of Lumbtr
l11ere rue three levels of wood seasoning (drying)~ which denote the moisture coment of the lumbe.r at the
ume of surfacing. The identification stamps are as fOllows:
S-GRN =over 190.4 moisture content (unseasoned)
S DRY, KD or KD-fff = 19"1ct max1mum moisrure content (seasoned)
MC 15 or KD 15 = 5% maximum molsrureOOJ\tent
4
ll\ese designru.ioi'IS may be fOund tn lhe grade stamp.
Unseasoned lumbe.r (S-GRN) IS manufactured overstzed so that when the lumber reaches 19 percent
moisture content (MC), it will be approximately the same size as the dry (seasoned) size.
Heat-tteated (l-IT) lumber lS lwnber that has been plac-ed u\ a closed chamber and heated unul tt anams a
m1nimum core te-mperature of 56<;oC for a minimum of 30 mtnutes.
The wntd ..DRY" tndic.ates that the lumber was e-Ither kiln or air dned to a ma.'<imum motsrure COJ\tent of
19 percenc
Kt!J1-dried (KD) lumber ts lumber that has been seasoned 1n a chamber to a predetermmed m01srure content
by havong heat applied.
Kdn-dried heat-treated (KD-Hn lumbe.r has bee-n placed in a c.losed chambe.r and heated wuil n ac.hieves
a minimum core temperature of56°C for a minimwn of JO minutes ru\d is dried to a ma\:tmum moisture
contem of 19 percen[ or less.
Moisture-content restrictions apply at tune of s.h1pment as well as tune of dressing tf dressed lumber is
Involved. and at time ofdelivery to the buyer wtless shipped exposed to lhe weather.
Platrorm Fnllmtd ~l. Modlfitd Balloon-fran\td floo rs
Eng~r.eered 1-JOlsts framed in the mochfied ba1loon-tfame fash1on were used for thJs design e.'<ample (see
Figure 1-7); ho\'lC"\'et, g.JVen lhe shon span on the floor jo1sts. sawn lwnber could have been used. Many
engmeers and developers use lhe sawn lumbet JOlsts in these cypes of structures because lhey are cheaper
than lhe engu-ured l-jo1sts. In tJus case~ [he joiSt shrinkage perpendicular to the gram would need to be
utduded in the overall shrinkage calculation. Also. sawn lumber joJstS can be supponed in joist hange.rs so
as to nol conlnbi.J[e to the overall bu1ld1ng shnnkage. The metl\Od of framing the floor joists and tl'le type of
floor joists used vanes. depending on engineers· and developer's preferences. For th1s design example. sawn
Jwnber IS used for the stud- fi'amed walls and premanufacturered roof truSses.
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Design Example 1 • Four-Story~ Light-Frame StTucture
VtrCtul Dis:plactmut (Shrinkage) ia .:\1ultiltvtl \Vood Framing
IBC §2303.7
Vernc.aJ diSplacement can be a stgruficam problem m rnulule-v-el wood frammg unless spec1al con.o;tderattons
are accowued for during destgn and con..">trucllort An estimate of shnnkage and consolidation per
floor is requtred tn order tOr the conttnuous ue-down rod system supplier to detemune the amount of
ttavel requtred for the shrtnkage-device components. Verttcal dtsplace-ment may be caused by one or a
combination of the following:
Mol~turt
Content and \Vood S hrinkage
From a se.rviceabllny and performance perspecuve-, lhe most stgntficant potenlial 1ssue related 10 mulustory
\\'nOd-frame con..wucuon ts wood shnnkage. whic-h tS tmpacted by the moiSture content and, more
spec:lfic:ally, whe-ther the. wood used as green or kdn drted. Wood shnnkage can be mallgated by proper
detail.ng.. For all wood-frame structure~ the areas requtring anentiotl are most co1nmonly ensunng that the
plumbtnglelectrical lines and the exterior finishes (sn.tec.o~ veneer) acoommodate the buddtng shrinkage/
se-ttlement Building shrtnkagelstnle.ment for the shear-wall hold-downs as accomod;ued by compensating
dev1ces that are dJscussed 1n Secuon 4 .4C.
A mote serious problem can be differential shnnkage bet\\'een full-he1ght steel frames and the wood-frame
system or fuJI-height concrete or masonry walls and the wood-frame system. Both of these condatiotts
teqUtrt special detaHing.
Tt.> availabilil)' of both types of lumber (grren and kiln-dried) is largely dependent on the region and
a10sociated market C01tditions. Typteally~ wood used ln construction ut the US southwest has both green
(S-GRN) and lc1l~tted (KD) \100<1. while other parts of !he country have more kil ~ried wood. K1IMitted
lumber is aYatlable~ and when the corutaclor need~ lO prov1de the product, thl!> may mean a longer lead ume
ln some areas lO obtain 11.
Wood Shrinkagr
Shr~nkage occurs when wood drtes~ chang1ng the structural dlmensJons of the lumber. llus 1s a faelOr in any
wood structure, and the compounding (adduive) effect of multiple stories can mc.rease the potenual lO have
a shrinkage pmblent. Shrinkage problems can be crac.ktng of finishes~ ptnchmg of door and ·wutdow frame~
or bucldtng of piping and conduits.
Both the me and NOS requare cons1derat:Jon be gaven to the etfeels of eross-gratn dunensJonal changes
(shnnkage) when lumber IS fabrica[ed in a green conduion. In addJtton, lBC Section 2304.3.3 requires that
bearing walls supponmg more than £WO floors and a roof be analyzed for shttnkage of the wood framutg
and that possible <llhrerse effects on the structure be satisfactonly remediated. This analysis and remediation
ts submmed to the butld1ng offictal.
The rotal shrinkage in wood-frame buildtngs can be calculated by summing the estimated shrittkage of the
hcrt=outallwnber me-mbers ln \\o<l.lls and ftoors (\\'all plates, stlls. and floor joists). Most of the shrtttkage
is cross grain. The amount of shnnkage parallel to grain (length of studs) IS approXJ1nately 1140 of the
shnnkage perpendicular to graln (cross gra1n) and can be neglected.
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Design Example 1 • Four-Story VKiod Ught Frame Structure
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ThiS design example illu.">ltates two melhods for determtntng lhe amount of wood shrinkage:
Comprtbtn.lih 't Shria.kagt Eslhnalioo
For a dime.nstonal change with the motsrure cornent IUnits of6 to 14 percent. the fom1uJa IS
where
S
= shnnkage (m 1nches)
D, = intlial dtmenston (tn 1nches)
C1 = d 1me nsion c.hange coe tfic ient. tange.ntial direction
c, = 0.00319 for Douglas Flf-l.arch
C1 =0.00323 for Hem-Fit
Cr = 0.00263 for Spruee-Pine-Fir
·".fF= final moi:S1ure content (o/o)
M1 = tnitial molsture content (%)
The formulas ace from lhe lf'c)oo' Handbook: Wood as an Engineering M(JJtnal. Dimens.tonal Stability ofJit..ttern
Lumlwr Pnxlucu. and AcctJmmodatiug Shnn~ In Multistory Wood-Frtune $Jru<:tut-es (W\1/W.WOod\\'Orks.org/
~tentluploodslwood_soluuon_paper-Accomodating-Simnkage.pdf).
11te final motsture content ( MF) for a buddi1tg ts referred to as the equdibrtum motsture content {EMC).
The final equdtbrium motsture coment can be h1gher 1n coastal areas aod lowe.r tn tnlan.d or desert areas.
These ranges are nonnally from 6to 15 perce.m (low to h1gh).
For this destgn example. a final motsture content AJ,.. ( Et-.<tC) of 12 percent is used.
Project specifications c~ll tor all top plates and s1ll (sole) plates to he Douglas F.r-L.arch kiln dried (KD) or
surfuced dried (S- Dry). Kiln-dried or surfuced-dned lumber has a ma.ximwn moisture conten[ of 19 percent
and an average of IS percent.
It mtght be more realiStiC to use a lower number lhan 19 percent ut the calculauoo so as to oot overest1ma[e
the shrinkage.
Typ1cal floor fram1ng has a 4 x4 top plate and two 2 x 4 sole plates(see Figure 1-7).
Ftnd the tndividual shnnkage of the two members:
Detenntne shnnkage of two 2 x 4 top plates and sill plates:
Smce the iniual MC (M1) is 19 percent and the final MC (Mp) is 12 percent., the equation is
S=
D,(MF - M, )
30(100) _ 30+M
S1
•
_ (2xl.5)x(l2 -1 9) _ _ . in
0 06
30(100) _ 30+19
7.ns
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Design Example 1 • Four-Story~ Light-Frame StTucture
The final size of the t\\'0 2 x 4s 1s
(2 X 1.5) - 0.06 = 2.94 In
Quitk Sbrinkllgt Estim.ation
A dose approxunauon that lS much more easily u.')ed to determine amount of shr1nkag.e is
where
S =shrinkage (1nches)
C =average shnnk.ag.e constant
C=0.002
M~= final
EnOlSture content (o/o)
M 1 = iniHaJ moisture cont.tm (%)
Determene shn.nkage of the 4 x4 top plate and two2 x 4 sell plates:
For !he two 2 x 4 soli plates, sonce the onotial MC (A!,) os 19 percent and the final MC (M,) is 12 percent, !he
equation ls
S= CD~M,- M1)=0.002 X (2 X 1.5)x ( 12- 19) = -o.4 1n
The final stze of the l\\'0 2 x4s is
(2 x 1.5) - 0.04 = 2.96 in
Thls queck esumatJon ts \\1th&n 0.5 perceru of the actuaJ calculated dlmenston of2.94 mches using the
eomprehens1ve tbrmuJas.
For the 4 x 4 top plate
S=
CD~MF - .If,)= 0.002 X
(3.5) X ( 12 - 19) = -o.05 on
The final size of ti-e 4 x 4 is
(3.5) - 0.05 = 3.45 1n
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Design Example 1 • Four-Story VKiod Ught Frame Structure
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AT UPPER LEVELS SHALL
STACK AND UNE UP DIRECTLY
ABOVE STUDS BELOW
DSL 2•4 SILL
p,,, ,.-~ -~
SCREWS f OR SHEAR TRANSFER
+x+
HANGER OVER SHEATHING.
HANGER SHALL BE RAlEO FOR
11-iiS TYPE or II>ISTALLATlON
CONT. TOP I(
STUDS PER PLAN/SCHED.
Figure I -7. Floor f1tllning til n·all
Nmes for Ftgure 1-7:
1. The double sdl plate as to pcovtde a nadiJtg surface tOr the fintshes. An altemauve detail could use
one sill plate and bloc-kmg between the studs for finish nathng.
2. Web stiffe1ters at joist hange-rs ma)• be required. depending on joast size and ananufacturer.
3. Hangers ror lhe floor jo1st are ms!alled over me shealhing (gypsum, plywood, or OSB) and must be
rated/approved for thts tnstallation (e.g .• Technical Bulletin frotn joist hange.r manufacturer listjng
reduc.ed aUoY..able hanger loads).
4. For conditton at shear waJI wath lhe tloor joistS parallel to the waJI. see Ftgure 1-24.
Total shrinkage per floor Je\•el wath the 4 x 4 top plate and two 2 x 4 sill plates:
S
=0.05 + 0.04 =0.09 1n
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Design Example 1 • Four-Story~ Light-Frame StTucture
Stftltmtni-Undt.r-Consrruttion
Gap~
(Consolidation)
An esUJnate of shrinkage and consol idation per floor lS requtred m order for the eonunuous u~own rod
system supplier to determine the amount of travel requited fbr the shnnk:age-device componerns. Thts may
be calculated on each project, or a typical value may be used fol' buildings havtng Slmtlar con.muction and
similar tnitial mo1·sture coote.nt in the lumber.
Small gaps can occur between plates and studs, caused by (among other lhmgs) m1s-curs (shoo Studs)
and the lack of square-cut ends. These gaps can account for up to !h indl per StOt')'. where ..pe,rt"tcc·
workmanshtp would be 0 1nches, and a mote ..sloppy" workmanshtp would beY, inch. Some senlement
can be attributed 10 compresston at jotst and berun-hanger top flanges. This case study factors 1n gaps and
comptess10n at hanger flanges of~ tnch.
OtJormatlon undtr S ustaintd Loading
Wood beams that suppon waJis can creep fi'om the su.'itaaned loading. The rate of cree-p c-an be higher
for beams that are loaded whde dry1ng under load because the modulus ofelasttctty IS 10\\•er for higher
mOlstute contents. Apperub.x F of the NOS provades commenmry !'elated to creep in wood and recommends
a (cree.p) de.flection runplicanon factor of betwee-n 1.5 and 2.0 for computing deflections under sustained
loads.
Table 1-1.
l~rtical di!placemtnt.S
Venic:al Displacement
Des.gn
Per Floor
Cwnulauve
Di.o;placement
(on)
(on)
(in)
41h Floor
0.22
0.65
3/4
3rd Aoor
0.22
0.43
L/1
2nd Aoor
0.22
022
1/4
Level
where
shnnkageof0.09 in+ senlement of0. 125 m =0.12 in
Mtthods to Rtduct Vutiut Displactmtnl
I. Usektln-droed pla1es(MC< 19percem)orevenMCI5(MC< 15percenl)lumberor
engmeered lwnber tbr plaxes.
2. Cons1der a stngle top plate instead of a double top plate.
3. Cons1der balloon tiamtng or a mochfied balloon framang..
4. Place floor JOISts 1n metal hangers beanng on beams or top plates tnstead ofbeanng on the top
plates.
5. lllesite storage oft11e maxenal stock can negate all destgn and p1anmng when the matenal is
not prope-r1y stored. lumber should be kept away from mo1sture sources and rain.
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Design Example 1 • Four-Story VKiod Ught Frame Structure
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Mtlbods 10 Atcounl for Vtrriul Oisplactmtn:l
I. Use c:ottunoous tie-dov.n systMJs with shrtnkage c:ompe1lsaung da•tces in shear walls.
2 . Arc:htteelural finish details near the floor Iines need to account for verttcal d1splaceme.ru..
3. Provtde a ~tnch gap between windO\\' and door tops to the Framtng lumber.
Ust of Prtmaaufac•urtd ~tttai-Piate-Connt"cttd \ \'ood Trusse:s
§2303.4
The pe.rmanent individuaJ truss me-mber restramtlbracing. of wood tru:Sses from individual truSS to mdividual
uusses lS the responstbtlny of the truss designer/truss tnanufacrurer. The responsibility of whtch party is
requtred 10 desig.nldemH lhe rrstraml of the lateral bracing. lS the problem. The truss submittal drawings
(ANSiffPI I St.ruldard Sec.uon 2.32.4) are required to show the approximate location oflhe reqmred lateml
btac1ng on the indlt·iduu/ truss members. ANSiffPJ I Standard Section 2.3.3.1.3 sHues that as an optton.
the building. destgner/engineer of record may design the nwraml of tJ'us lateral bractng. Otller options for
the design ofpermanent latemJ bracing are use of BCSI-B3. PermtJtWIII Restramt/Brac/ng ofChords tmd
Web Members or BCSI-B7. G11ide for Htmdll11g. luslillling. and Braci11g of3Xl a.nd 4x2 Paralltl Choni
Trusses. These doc.uments are available from the Truss Plate h'lsurute. JBC Section 2303.4.1 .2 references
Uldusuy stattdards for bractng without dtre<-lly referencing the nest docume-nts. These bracing forces
are accumulated along the length of the brace and need to be detailed as to how they are resisted by the
buildtng. The problem ts that when lhe build1ng design is completed, the locations of the truss bracing are
not known to lhe buddtng des1gner because the truSS desJgns done by the truss destg.nc-r.hruss manufacturer
have not been cOt:npiNed. The current praeuce wtth building destgne-rs •s tJuu the truss destgn and truss
brncing is by others.
It should be acknowledged lhat many prefabrtcated truss design subnuuals do oot seem to take
responstbthty fOr cenaan de.stgn items lhat are shown m them. such ao; sizmg trUss-to-truss cooneeuons,
setung cambers. or the bracmg and connec.tions of tntemal \\~bs that may othe:n.vtse be slender. Often,.
notes are placed tn tl'le submittaJ documents uxhcaung lhat another pany. such as the ••buJid1ng destgner~"'
should be the one to take responsibiht)' for these des1gns eve.n though the design originated OUlSide the
office of the "'butldittg designer."
See Section 4.1 tbr further discussion about shear tran.sfer (drag) trusses.
Efftds or Box Nails on \Vood Struttural Paotl Sbur Walls
ThlS design example uses common nails for t3stening wood structural panels. Based on cycliC testlng of
shearwaJisand performance m p351. eanhquak:es,. the use of common nalls is preferred. The 20181BC
no longer lists allov.'3ble shears fOf" wood suuc1ural pane-l s.hei.U" walls using nails; IBC Table 2306.3 lists
allov.--able she.ars for wood structural panel shear walls for staple.s. The lBC direc.tly references the Spee:ta.l
De.tfgn Pro-.·tsionsfor W'ind and St>simJc (SOPWS) by the Amencan Wood Councd (AWC) for wood
sU\Ictural panel shear capaciues wnh nails as fa~teners.
SOPWS Table 4.3A hsts nominal unH shear capacittes for wood structural panel shear walls fot common
or gal\•antZed box nails. Foomote 7 of Table 4.3A stues that lhe galvanized nails shall be ..ho[-dipped"' or
twnbled. but these nails will not work'" some power-drive-n fa~ner devtces. Most c:onuactors u.-.e po\vt:rdriven fasteners for diaphragm and shear-wall installations.
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Design Example 1 • Four-Story~ Light-Frame StTucture
Box natls have a s:ma11er diameter shank and a smaller head size than common nails. Using IOd box
nalls resuhs in a 19 percent reduction in allowable load for diaphragms and shear Y.'3lls compared to IOd
common rta.~ls. Ustng 8d box nails \llOuld result m a 22 percent reducuon in allowable load for dtaphragms
and shear walls compared to 8d common natls. Thls ts based on comparing allo\\'able shear values listed
tn Table 12N 111 tlte National Design Spectficuion {NOS) for Wood Construct1on for ¥.1-it'lCh. side member
thickness,tJ• and Douglas Ftr-Larch ti'amang. In addttion to the reduc.tion oflhe shear wall and diaphragm
c.apacuies, where box nails are u'ied,lhe walls will also dnft 11l0re lhrut where common nails are used.
Alternanvely,lhe englOee-r can con.o;ader usang nail and sheath•ng thtcknesses not listed tn the me or
SDPWS by ustng !he value< listed tn ICC-ES report ESR- 1539, available from lhe International Staple,
Nail, and Tool Association (I SANTA).
A contributor to lhe problem IS lhat when contractors buy large quanuues of nruls fOr nad gw\S, lhe words
"box" or "common"' do oot appear on the canon label. Nail length and dtameters are the most common
lisung on the labels. Thus. 11 is u crtmtly hnporta..nc to list the required nad lengths and diameters on lhe
structum.l drawtngs fbr all dlaphrogms rutd shear \\'3lls.
Some manufacturers of nads have developed a melhod of color coc:hng and stampt.ng of the nrul head that
allows the 1nspection of the nail stze and Lenglh in the mstalled condition as opposed to someone either
physically wit.ness1ng the mstaJiation of the nails or re.movmg the Installed natls for vertfication. These
color-c,oded and stamped nails add only a nominal cost to the. n.atls and are gainu1g tn popularity with
engmeers, special inspectors. and building officlals.
Another pt"oble.m is lhat contractors prefer smke.r and shon natls because the.ir use reduces splitting, eases
drtvtng. and costs less.
To tllusuate a potnt, tf an engmeer de~:tgns tOr dry lumber (as discussed earlter) and common nads, and
subsequently green lumber and box nads ate used in the construcuon, lhe result lS a compoundmg of the
reducLiOn.(i. f« example.• for IOd nails tnstalled tnto green lumber, the reduction would be 0.8 1 umes 0 .7 or
a 43 percent reduction tn capacity.
\Vo-od StudJC in Firt-Rtsistanct·Rattd \Valls
IBC T504.4 and T60!
The four-story buddmg ln th.is design example has rated \1/alls and ce.lhngs and is con(iidered a Type V
building. When \\'OOd-frame structures exceed lhe hmito; for Type V c,onstruct.ion (e.g .• five stonesofwoodfi"atne consm~c.tion),lhe code requires enher Type Ill or Type IV construction.
IBC Section 602.3 defines Type Ill construCtion as butldings walh exten()( \WIIs made from nottc:ombusttble
marenaJs. Therefore,. as an excepuon, use of fire-retardant-treated wood (FR1W) ts allo\1/ed for the extertor
load-bearing wall assemblies.
Fire-rated assemblies can be found 1n a number ofsources.lnclud•ng the IBC, the UndernTuers
Laboratories (UL) Fire-Re.tt.tUlnct· Rated Systenu und Pr(){hu:ts, the UL Flrt Resistance Dirtctorys the
Gypswn AssociatJon's Fu~ Re.tt.stanct Destgn Mtuwal, 3.00 AWC's Destgn forCodeAcuptanu 3 (DCA 3)
- Flrt· Re.rt.sumu-Rated Wt>Od-Framt Wall and Floor/Ceiling Assemblies.
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Design Example 1 • Four-Story VKiod Ught Frame Structure
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Table n 141(2) of the me ltsts fire raungs for var1ous wall oonst.ruC'tion types.. Many of the wall
consU\ICtion types using wood COJ'ISUUc.tion reference Footnote m. Foomote m of the table requlfes the
reductton ofF; to be 78 percent of the allowable whe.re the slenderness rotio 1/d > 33. For sruds w1th
a slenderness ratio lid< 33, the design stress shall be reduced to 78 peicent of the adjusted stress F;
calculated for studs hav1ng a slet~demess ra110 1/d of 33.
TheAWC has tesl"ed a nwnber of wood-frame fire.-mred waJI assemblies to 100 percent design load.
There is a d1sparity betwee.n the JBC and publ ic~ujons such as AWC"s DCA 3. Fire-R~slstanuRated UOoti-Frame. lf'(J/1 and Floor!Ctllmg A.u~mblie!. wh1ch does not requtre the reducuon tn
allovro.ble suess. l1le bUlldtng·s archuect and/or engtneer shoukf check with lhe local JU..flsdtchon to
determine the accepted appmach. The AWC procedure is drouled at hnps://w"'"v.awc.otg/faqslfirelfi.rel
what-IS-.00-backgrotmd-behind-footnote-m-<>f-ibc-20 12-table-721.1-{2).
11te cakufated suesses for F; used in th1s design example w1JI not use the reducuons for Foomote m.
1.3 WEIGHTS
Roor l\'tig.bts:
Floor wtights:
Roofing+ re-roof
5.0 psf
Flooring
She-athing
2.5
Lt. wt. concrete
Trusses
3.0
Sheathing
2.8
Insulation+ sprinklers
2.5
1-JOlst
4.0
2 layers gyp + m•sc
7.0
2 layers gyp+ m1sc
8.2
1.0 psf
14.0
Dead load
20.0 psf
30.0 psf
Live load
20.0 psf
40.0 psf
lntertor and ex1erior \\-all weigtm; have not been tnduded tn the above l oads~ they have bee.n tncluded tn the
diaphragm weights shown in Table 1.2. Typical interior and exter1or paruuon '"~-tg.hts (for determming the
building we.ights) can "'at)' between 15 psf to 20 psf (based on the horiZOntal plan dtmensJon). depe-nding
on room sizes, number of layers of gypsunt board on wans. S[agge.red studs, etc.
Weights of roof d1aphragrns are typically determined by taku1g one-half the he.ght of the 1\~lls from the
founh floor le\'eJ to the roof. Weights of floor dtaphragms are typtcally determu'led by taking one-half
the walls above and below for the.fOurth. th1rd. and second floor diaphragms. The weighlS of all walls,.
indudmg interior nonbearsng panjtions.. are included m the respectl\'e weights of tl1e vartous leve.ls. The:
weigh! of parapets (where they occ:ur) ha..~ been mcluded tn tl1e roof weight.
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Design Example 1 •
Four-Story~ Light-Frame StTucture
Table 1-1. Weigh1s ofrooftmdfloor diapl~ragms
Assembly
(psi)
Area
(It')
Weight
(ktps)
Roof
20
5288
105.7
Ex1 "all
15
1350
20.25
Jnt wall
15
5288
39.66
Floor
30
5288
158.6
Ext wall
15
3 100
46.50
Jnt wall
15
5288
79.32
Floor
30
5288
158.6
Ext wnll
15
3 100
46.50
Jnt wall
15
5288
79.32
Floor
30
5288
158.6
Ext wall
15
3100
46.50
lnl v.-all
15
5288
79. 32
Unit Wt
Level
Roof
4th Aoor
3rd Floor
2nd Aoor
Story\Vt
(kips)
165.7
284.5
284.5
284.5
W= 3(284.5 ktps) + 165.7 ktps= 1,019 ktps
2. Calculation of the Design Base Shear
ASCE7
2.1 CLASSIFY THE STRUCTURAL SYSTEM
Frocn ASCE 7 Table 12.2-1 tOr beanng-\'r'all systems ustng hght-frame wood walls sheathed Walh wood
strucnual pan~ls mted tor shear resistance:
I R=6.5
!l,=3.0 C,=4.0
2.2 DESIGN SPECTRAL ACCELERATIONS
The specual acc:elerauons lObe used 1n design are denved from sod profile and stte locauon:
Ss= 1.540g
s, =0.568g
2.3 RESPONSE SPECTRUM
Determine the app..-oxt.mate fundamental butldmg pertod using Sedion 12.8.2. I:
C,=0.02 andx=0.75
T., = C,h; = 0.02 x4J0·" =0.34sec (see the tOIIowtngdtscusston)
I T..=0.34 sec I
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Tl2.8-2
Eq 12.8-7
Design Example 1 • Four-Story VKiod Ught Frame Structure
4
r. =0.2~=0.1
s,
568
°·1.026
=O.II soc
s. =S""(0.4+0.6;.)= 0.4+5.6T forT< T,
Eq 11.4- 5
=~=~=0.55soc
§ 11.4.6
Sm ; 0.568
, T>T
.
- 10r
S ; 1
•
T
T
Eq 11.4-6
T
J
SA~
1.026
§ ,
<n'
e
g o.a
~
1
06
...{!
S.,=OA+5.6T
OA
ioJ
g
I
I
I
o L-----~-----------------------------J
0
05
1,5
Penod (Sec)
Flg~.~n
1-8. Design respon.'it spectntm for Jhe aample building
The long period equation for S., does not apply he.re because the Jong period tranSition occurs at 12 seconds
(from Figure22- 14).
As shown in F1gure 1-8, the des1gn spectral acceferanon 1S between T~and Ts. so the de~agn spectral
acce-lerationS., ts I.Og.. It IS not requ1red to const.nJCtlhe design response spectrum when using the
equl\•alentlate.mJ tbrce procedure, Since the response spectrum is Implicit tn the calculation of cl In Secuon
12.8.1.1.
11te response spec.trwn demoosuates the eft'ect oflhe assumpuons used m tJ1e c.aJc:ulaoon ofbu1lding
penod. Values ofC,; 0.02 and :c= 0.75 were selected as spectfied for shear wall bui1d1ngs. which result 10
an approximate period, 7;,;. 0.34 sec. As shown later tn this example, dnfts are dose to, if not at, the story
drift limns ofSect1on 12.12.
The penod oflhe structure can be established through struCtural analysis. Secuon 12.8.2, however,limlls
the pertod that C3J\ be used 10 calculate spectml accele:raLion 10 a value ofT_= C., x 7;.. Yl'here C, is a
factor found tn Table 12.8-1. In thtSC<>Se, T_ = 1.4 x 0.34 =0.48 soc.
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Design Example 1 • Four-Story~ Light-Frame StTucture
2.4 HORIZONTAL IRREGULARITIES
T12.3-1
By inspection, the buildmg does not quality for any of the honwntal structutal
Ia. to 5.
UTegulanues.
I NO HORJZONTAL STRUcnJRAL IRREGULARITIES
T12.3-2
2.5 VERTICAL IRREGULARITIES
I a. to 5b.
By tnspec.uon. lhe buiJdmg does not qualify for any of the vertical st.rucmral
irregularittes.
NO VERTICAL STRUCTURAL IRREGULARITIES
2.6 LATERAL FORCE PROCEDURE
T12.6-1
Design checklist:
1. Oetermtne occupancy category and 1mponance factor.
2. Dere-rmane s.. S1• and sod profile from sne location.
3. Test for SetSJntc Destgn Cotegory (SOC) E.
4. Oerermine SMand SDI.
5. Oete.mune T and test for short period e.xception on SOC. Dec ermine if equjvalent lateral force
analysis ts allo,-..-ed.
6. Determine SOC (tf not E).
7. Oere.rmine Rand verity heiglu..
8. Test for SJ < 1.5 Md calculate c. ba.~ shear.
9. Determine C",.
• Detemu.ne risk category and lmporta.tl(".e factor:
Rc<k Category II
T l.5·1
I~=
Tl.5·2
1.0
• De-termine SJ• 5 1• and soli profile:
Site Class D (based on g.eotedmtcal trrve.sugallon)
Per geotechnical mvest.iganon: Site Class 0
Locauon: I rv t~ CA
Longitude= - 117.8320, Latnu<le =33.6800
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Design Example 1 • Four-Story VKiod Ught Frame Structure
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Therefore.• from USGS \\'e-bsite applicauon:
S,= 1.540g»0.15
11\erefore. not SOC A
s, = 0.567g »
0.04
11\erefore. not SOC A
s, = 0.567g < 0.75
Therefore. not SOC E
s, = 0.567g < 0.6
Therefore, &]uatioo 12.8-6 applies.
Determme SOC:
Based 0 11 the geotechnical mvesugatton, SOC= D.
The above addmonal checks weJe added tbr reference when an mvestigauon has nm
detemuned the SOC.
Oe.termtne Rand verity heaght:
For ltgh!-fmme walls with wood structural panels that are both shear walls and beanng walls:
R=6.5
Tl2.2- l
Maxtmum he1ght pemmted tn SOC D lS 65 feet
Tl2.2- l
The budding structure is less than 65 feet; tl'lerefore. IllS permined.
In addition. Equauon 12.8-6 requ1res an addtuonal chedc for C,. the mintmwn for strucrures that
are located where S 1 is equal to or greater than 0.6g:
C~=
o.ss,
0.5(0.567)
oR)= 065 )
li
.
-0.04~therefore, Jtdoesnolconrrol
li.O
I C,=O. I58
I·=C,W:O.l58W
Design base shear 1n ASCE 7 is based on a strength design ba(jis.
Since the 20 12 IBC, all tables for "natled"• wood diaphragms and shear walls ha\'t been removed
from the IBC; the onJy tables that are left are fOr ••staph.>d .. \\"OOd diaphragms and shear \\o-alls
and have allowable she.ars. All tables 1n the SDPWS are ..dual format,.. and the nomu\31 values
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Design Example 1 •
Four-Story~ Light-Frame StTucture
must be adjusted for both sttength and alhm'ilble stress destgn (ASD) loads. Since ASD is still
predorrunantly pracuced. it ha<i been decided to have this design example tn ASO fonnat. In
adduton, aJI the manufacturers of meml hard\\o-are connectors currenlly pubhsh ASD values.
I. S1mpJified alternative structural design criterta-Ac:cotding to Section I2. 14. I.I.
this analysis procedure can be used for bearmg wall systems~ but not for bulldtngs
over 3 stories-NOT PERM!lTED.
2. Equwale-ntlateral force analysts--Accordmg to Table 12.6-1, since T < 3.5T_,
(0.34 sec< 1.93 sec), and the butldtng is regular and !Usk Category 11PERMITfED.
3. Modal response Sp«ll\lm analysts-PERMilTED.
4. Seismic respol'tse htstOI')'
procedure~PERMilTED.
USE EQUIVALENT LATERAL FORCE ANALYSIS
2.7 BASE SHEAR
IC,=O.I 58
r= C,W= 0. 158 X 1019 = 160.8 ktps
&! 12.8-1
1
v = 160.8kips
2.8 VERTICAL DISTRIBUTION OF SHEAR
Tab/~
w,
Level
1-3.
J.~rttcal dtStributtOn
h,
(tl)
o[.fhearfrom SecttOII 11.8.3
w:Jt.
(k-ft)
c.,
F,
(k)
F,
F~
"'•
(k)
Roof
165.67
43.07
7135
0.3 1
49.37
0.298
49.37
4th Floor
284.46
28.32
8056
0.35
55.14
0. 196
105.1
3rd Floor
284.46
18.88
537 1
0.23
37. 16
0. 131
142.2
2nd Floor
284.46
9.44
2685
0. 12
18.58
0.065
160.8
Total
1019.1
23,247
1.00
(k)
160.8
457.47
The te-rms used in Table 1-3 are defined tn Section 12.8.3. Since the period= 0.34 < 0.5 sec, the \1alue fork
lS 1.0.
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In thts design example.• k = 1.0. The distrlbUlion of SlOI)' shear as earned OUl usang
ll' ""
F,.=C""V.whereC.,. =~
&j 12.8- 11 and &j 12.8- 12
w,h:
0
i ...
2.9 REDUNDANCY FACTOR
According to Sectton 12.3.4. lhe redwtdancy fuctor should be calculated for each prtnc.apaJ a•us. The
redw'dancy factor as 1.3 unles.~ eather Section 12.3.4.2(a) or 12.3.4.2(b) as shown to be true, 1n which case
the redundancy factor can be taken as 1.0. Section 12.3.4.2(b) requares that there be t'Yt'O bays of shear walls
on each penmeter lane. Therefore, the redundancy factot is 1.0 for bmh the east-west dtrection and the
nonh-south dtrection.
p = 1.0 FOR EAST- WEST DIRECTION
p = 1.0 FOR NORTH-SOUll·l DIRECTION
3. Location of Shear Walls and Horizontal Distribution of Shear
ASCE7
3.1 LOCATION OF SHEAR WALLS
The lateral-f«ce-restSting system in Lhts design e.xample uses bolh intenor and exterior v.'3Jis for shear
walJs (see Figures 1-2 and 1-3). The seismic-force-restsung system for the uansve.rse dlrection (northsouth) utihzes the mteraor walls between the horel guest rooms. A seismtc design of a selected tnterior shear
wall an the transverse dJrection is Illustrated in thts design e.xampfe-. The seisJnic-fotee-resisting system for
the loogatuchnal direction (east-west) uuhzes lhe long mtenor corridor '"'3-lls located at the center of the
SU\Icture, walh shear v.'<llls on bmh sides of the corridor ul addttton to shear \valls on rhe extenor walls Md
shear waHs at the bathroom walls.
In a lateral-force-res:isttng system in the longuudtnal direction for structures similar to this design e.xample,
some structural enganeers wall utilize only the tnterior corridor '"ails and not place shear walls on tlte
e.~enor \\'31ls. In prevtOus editJons of the SDPWS,lhts type ofdestgn used a ngad diaphragm approach
for the dtstnbution oflateml forces to lhe shear walls. Also, previous editions of ASCE 7 and lhe SDPWS
were not clear or spectfic on what was allowed by the code. The 2015 SOPWS has added language and
requtre-me.nts fot a semlrigid a.'ld envelope diaphragm analysis for open-froatt structures. These added
requtremenLS indude le.nglh to width (UW) Jjmits wilh a tn<Damwn cantilever distanc.e of 35 feet. In
addtttoo, in Seasnuc Oestgn Categories B, c. D, E, and F the e-ng.uteer must con.~der torsaon., accidental
torsion, and StOt')' dnfts in conformance with ASCE 7. and dtaphragm shear a.'ld benchng deformation
contnbutions to story drift at the open front.
3.2 FLEXIBLE VS. RIGID DIAPHRAGM ANALYSIS
ASCE 7 conunues to allow the asswnption offtexable dtaphragms tbr most wood structures wttl'lOut a
rigorous analysis if the COt:ldittons tn Sectiott 12.3.1.1 are sansfied. The enganeer should use Judgment
in applying this code requirement in detenn1ntng shear dislfibmions to the shear walls since, in real tty,
all diaphragms behave as semangtd. Flexable dtaphragm assumpuons may be ju.<rufiable from a codecompliance perspective fbr this destgn example due to the O\'eralluruformity of shear wall lengths and
spactng for the building's tran.~\'erse direction (nonll-south).
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Design Example 1 • Four-Story~ Light-Frame StTucture
ASCE 7 and prior edttions of the SOPWS dtd nm have a definjtion of an emlelope: analysts approach for
late.ral analysts. The 2015 SOPWS states that the distnbution of shear to the ven~eaJ resisung element~ shall
be ba~ on an analysts where tJte daaphragm IS modeled a~ semmgtd, tdeallzed as fte.xable, or tdeahzed
as rigtd. As an alternauve to a se-mirigtd analysis, SDPWS also describes permissible use of an envelope
analysJs.
Recognizmg that honzontal wood diaphragms ate sem1rig1d and are netlher pt.trely ftexable nor purely
ngtd, an eng.meer ussng the envelope method analyzes the dastnbution of horizontal stoty forces tWICe,
wnh d~aphragrns separately idealized as flexable and rigid. Where.by~ the daaphmgm shear to eac-h ve.rucal
res:isung efe.ment IS tJ1e larger of the shear fOrces resultmg from the analyses.
Commutny on Diaphragm Fle-xibility for ~1uhistory Ug:ht- Fnmc- Structurts
In current practice~ usually to acrommodate Jarge.r opemngs and windows at the buildmg perimeters,
some destgn professionals have opted to omit or limulatecal-force-resisung elements from extenor wall
lines and rely on the interior lines oi lateml resistance~ suc.h as corridor walls. Some junsdicuons have not
allo\1/ed. lhe destg.ner to disregard lhe resuJts tfom a tradniona.l flextble d1aphragm approach. Tins prnc:uce
l~ not recommended unless the engineer explicnly constders bwldtng performance, incluchng the control
oflocaJIUd honzonta1 diaphragm de-flections that could lead to insrabtlity. Regardless of the approach~
the design shall comply With code hmtts for horrzontal diaphragm cantilevers (shown in code reference
SDPWS), defonnatjoo compatibtlit}·~ stabihty, ruW seismtc dnft requtrements. Ne\\1 research is undern,ay
that looks dtrectly imo lhe efficacy ofth1s approach., whtch should be avatlable to destgners of mid-rtse
wood fi'ame bwldtngs 1n the near future.
For this design example, an e.nvelope method of destgn is ut•ltzed.
Btha\'ior or Cantilt\'trt-d Diaphragms
Although allowed by Standards. use of cantilever diaphragms do not appear to have suffictent test data to
jusuiy Wide-spread use. Prevtous \\Tite-ups have urged cau1.1on on the use of C3Jltilever diaphragms without
additional study and val•dattngphystcal tesung. Reportedly somejuttsdiCuons, mcludtnglhe Cny of LA.
are not allowing or are restncting cantilever dtaphragm use.
3.3 USE OF CANTILEVER DIAPHRAGMS
The 2015 SDPWS had stgrufi.cant changes to the docUJnent related to the use ofcanulever djaphragms.
2015 SOPWS open-front requtrements have changed from lhe2008 eduion regatdingallowable cannlever
length. TI1e 2008 ednion l1m1ted the max1mum cantt.lever le.ngth of an open-front structure to 25 fee~
however, it had an excepuon that a.Jiowed an tncrease ln the cantilever length where calc.uJations shov.•lhm
d1aphragm de-flecuons can be tolerated. Consequently~ there did 110t extst a hard hm ll on the canulever
length ptovided that the aspect ratio and all other requtrements could be met. The 2015 edition limit~ lhe
cantilevered length~ L', to 35 feet wtth no exc.eption provided. Section 4.2.5.2 of the 2015 SDPWS also
has stgntfi.cantly changed from the requ1reme11ts of the p(t:\' IOUS 2008 edition, requmng the foiiO\v.ng for
c-antilever diaphragms with se-istntc loading:
Calculation of story drtft ax the edges of the structure.
Ver~llcauon that the budding is not torsionally
trregular.
Calculauon..:; to JUstify that the dtaphragm can be idealized as rigid (or modeled as sem1rigad).
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The 2015 SDPWS and buildlng codes prtor had provlstons for use of cantilever diaphragms; however,
deflect:Joo equations tOr the cantilever c-ase were never provtded. The proposed 2021 SDPWS equations
are to address t\\"0 types of loading conditions of cantilever diaphrag.ns (Case 2 and Case 3). The new
deflectioo calculations utilize the shear stiffness propeJ1ies, G... Loading C.ase 2 represe.nts a wufonnly
loaded canuJeve.r diaphragm as could be represented by dtstrtbuted loads uansferred from floorsfroofs and
walls. Loadtng case 3 represent" a oondttion where forces may be concentrated at the canulever end such
as from pertmeter \.111\lls at the oulSide edge of the diaphragm. CaJculattons of diaphragm deflecnon acoount
for bendtng and shear deflections, faste.ner deiOrmauon. chord sphce shp. and other contrtbuting source.s
of deflecuon. The d~aphragm deflecuon, s...,, shall be pe.nn.ned co be calculaced by use oftl1e follow1ng
equations !hacare proposed lor !he 202 1 SDPWS.
T hru-Ttrm Diaphragm DtOt"ction Equations
Eqwujon
Loachng Case
I. Mid-span deftecuon of a smgle-span Simply
S
supported d~aphragm \\i lh uniformly
distr1buted load.
""
2. End deftection of a cantde\•er diaphragm wnh
()
untformly disrr1buted load.
s,·L3
SEAlY
0.25FL
IOOOG.
I6.c:tt
lvL'l
O.StL'
!.6c:xl"
s,·L ' 1
\"L'
r.aa
E4W'
IOOOG"
JV'
= - - + - - - +- 2JY
. = - - + - - - +- EAW' IOOOG11
w·
dit
3. End deflection of a canttlever diaphragm wnh
concentrated lood at the end.
() . ; - -+ - - - +- '
""
Four-Tum Diaphrag-m Dd1t-ttioo Equatioa..s
Loochng case
Equation
I. Mtd-span de-flec.tJon of a single-span s1mply
supported diaphragm with uniformly
dtstributed Joad.
Svi!
8£4W
1·L
4 ,·tr
3vL'1
EAJfl'
rL'
2G\•J'\•
I6.al"
211'
S..., = - - +G+0.188Le +- -
2. End detlecuon of a canulever diaphragm with
untformly dismbuted load.
3. End deftectton of a cantilever diaphragm wtth
concentrated load at the end.
"
S = - - + - - +0.316L'e
...
II
!..6a
+-IV'-'
81·L'1 ¥L'
I:6cx
&""' = -- + -- +0.15L'~ +- - '
W'
EAW' Gl·h·
"
where
E =modulus of elasucity of diaphragm chords. psi
A =area of chord cross-section, tn2
G(f = appatMt diaphragm shear sttffl'tess from nail slip and pru\el shear deformatioo. kipshn
L = diaphragm length, ft
L' = canulever djaphragm lenglh. tl
v = 1nduced unit shear in diaphragm, lblft
Jfl = d 1aphragm \\i dlh, ft
W' =- canttla-e.r diaphragrn width, ft
-'r =-distance from chord spl1ce (0 neare.~t support, ft
M =-diaphragm chord sl1ce slap (in) at the mduced unn shear 1_
1\ draphragm
S.t.a =- maxtmum mtd-span dlaphragm deftecuon detecmtned by elastte at'lalysts. tn
t, =nail sl1p per SDPWS C4.2.2D for !he load per fastener ac,,
G,•fl· = panell'igtdity through the thickness
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Design Example 1 • Four-Story~ Light-Frame StTucture
\Vood\Vorks has rece-ntJy released a new publicat10J'I., &tsmic Design Example ofa Canti/n·~r Ubod
Diaphrugm. This publication has a comprehensive design example for calculating cantilever dtaphragm
deflecuon.s 1n 1ight-frame construcuon.
3.4 WEAK AND SOFT STORJES IN LIGHT·FRAME STRUCTURES
A recent Applied Technology Project, ATC- 11 6 Proje-ct: Solutions to the Issue of Shon-Period Building
Perfomlance has additJona1 gutdance. The projec.t·s purpose wa~ to:
I. ldenllfy key misstng elements of current modeling practice related 10 shon-penod buildings.
2. Develop a merhodology to inlprO\'e analyucaJ modehng of shon-penod bulldmgs.
3. Caltbrate lhe melhodology wnh obse,rved per1brmanc:e ofshon-penod butldings tn recent
eanhquakes.
4. S1mpltfy lhe methodology 1n1o pracucal solutions that can be tmplemented m codes and
slandards.
As pan oflhe ATC-116 shon-penod project, numertcal studtes identified that when finish matenals are
tncluded ln anaJysis models, the)• can dra.rnancally change bolh the stre.ngth and stifthess of light-frame
shear wall struCtures and appear to result tn the oonststent occurre.J'l(:e of a first-story collapse mechantsm.
Several factors are thought to be dnving. the first-story mechanism:
I. Whtle the c:apaCit)' of the destgned shear wall varies over the hetgh1 of the structure w1th
demand. the capacity added by the finish material remain... the same. whtch results tn lhe
overstreng_th (designed shear waJis plus finish matenals). However. these shear walls plus finish
materials. due to common architectural and occupancy issues, oont:rtblfl.e to pl'oviding the least
Sltength al lhe first story. In other words. shear walls at the lowe.stleveJ will have tl'le h1ghest
demand-lo-capacity ratio (OCR) and will therefore y1eld first.
2. Once inelastic behavior begins m tlte numerical srudies, addinonal 1nelastic behavtor tends to
fOCtts m that story and doe.~ not have lhe capac tty to push ytelding up tnto lhe uppe.r levels as
would be expected wnh a well-distributed and ductile lateral system.
ThiS potential tssue is thought to be amphfied in multiStory bUildings with stacking shearwaJis. where due
to the desire. for more open space at the base leve-l (e.g... lobby. e.xe.rcise room. meeting rooms. retail space).
there are often fewer nonstructural beartng walls, resulting m a somettmes significant reductiOn tn the
acrual st1f'rness and suength oompared 10 the upper levels.
The wood mdusuy is currently undertakmg tesung to expand the data available fOr nunlerical mode-hng~
th1s could result tn additional numerical modehng being available fOC" c.onside:ration. It ts theretbre
recommended that the designer evaluate the effects of the lower level layout of the struCtural and
nonstructural fin1shes as well as lhe structural and nonsuucnual \!r-alls and where possJble. ptovide extra
strength m lhe first-Slory shear walls.
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4. Mechanics of Multistory Segmented Shear Walls and Load
Combinations
4.1 OVERTURNING EFFECTS OF SHEAR·TRANSFER (DRAG) TRUSSES
The structurnl destgn an th.is design example uses premanufacrured wood rooftflJSses. Under seismic forces,
tl1ese mustlmJlsfer the lateral forces from the roof diaphragm 10 the tops of the tnlenor shear walls. To
accompltsh thLS.. specull oons!derattons must be made in the destg.n and detailed on the plans. In parttcular,
any rru:sses that are to be used as collectors or shear-uru'ISfer (drag) truSses should be clearly 1ndtc~ued oo
the srrucruraJ framing p1ao(s). The magnitude of the forces, the means by Whl('.h lhe fOrces are applied to
the trusses, and tl1e means by '"'htch the forces are tmnSferred from the trusses to the shear walls must be
shown on the plans. In addtlion, if the roof sheathing at the htpends breaks above the joiJll bttween the
end JOCk uusses and the supponmg gatder truss, the lateral forces to be restSted by the end jacks should
be specified so that an appropnare connection can be provided to resL~ these forces. The drawtngs tor the
buddtng also must specify the load comb1nalions and whetl1er or not a Sltess ~.nerease 1s pennmed. If r1dge
verus are being used, special detailing for shear transfers must be tncluded because normal diaphragm
continuity IS disrupted.
Mtthod of Load Tra.a..\Jtr
The Lrus.s~ that uansfer drag loads actually only t:ranster the loads from their top chord to thear bottom
chord4 llus IS a fine poutt~ but u is lhe engmeer ofrecord's (EOR) responstbilny to getlhe Joods from
the diaphragm to the top chord and then from the bottom chord to the top plate (or whew.·er 1t goes from
there). Also,lhe trusses do l'lOl. in general, ttanSfer loods to the shear walls~ they transfer lhe Joads to the
bonom chotd and from theu they may be transferred to the upper top pJare, another truss, a strap at the
corner. or many other types of mec.hanlsms. llle shear walls are usually at least one other link down the
load path, someumes many more than one.
ASD ,.s, Strtngth L.oads for Trusst-S
Truss designers are typtcally g01n.g to assume ASD loads. It is not recornmended that strength-level forces
are g1ven to the uuss des1gners.
Ridgt Vtats
With currem e.nergy standards, ndge \'etl.ts are bec.omang the new norm and are almost alv.-ays used.
At ridge vents, the forces on dtaphmgms that tranSfer loads to the eaves are usually small, but If there are
inteti()( shear waHs (hke 1n thiS design e.xample) witlt blocking pane.ls bet\\>een the trusses up to the roof,
these forces can become much higher.
The truss metal connector plates are not destgned for loads in this d.irecuon. OnJy Iimited testtng on this
c.ondiuon has been done, and the capacity of the plates tn this direction ls generally large enough to carry
tl1e forces that result in tl1e lowrangeexistingatlhe rooflevet If the peak is not near the loc.auon of zero
diaphragm shear, 11 is prudent to bwnp up the tntSS peak metal connector plate size.
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Design Example 1 • Four-Story~ Light-Frame StTucture
0\o'trturning or Drag Tru..~s
The roofuusses that uansfer lateral loads have overrurn1ng I'Orces at their ends. These ovenurrung forces:
are lhe only result:.lnt vertical fbn~es imposed on the struCture. The lateral forces are U'ansferred to the shear
waJI(s) by tension drag ties and/or framtng chps.. Where the shear wall is at the end of the uuss, the tension
and eompress:ton forces are addttive to the overturning forces in the shear \\'all. Where the shear wall is in
the interior part of the truss and does not e.xtend to the end of the truss, there are no venical effect'i on the
shear wall.
ThiS des1gn example is only the stmplest of cases. Most aetuaJ design conditions are more complex. When
the cond1t.ton of a bearmg wall and/or shear wall is below the truss. the truss design software used when
tdeahzed as an mfinJtely rigtd conunuous suppon can yteld bogus results. Qutte often these results: are
deeply htdde-n ln the cotnputer-genetated outpt.Jt and requare close scruuny. Many tames the computer modeJ
needs: to be mampulated to remove the erroneous uplift numbers the computer model yields because of the
tnfinitely rigtd co.uinuous s:uppon model.
w...
1:.&JJI1
J. • 4lUI n
115
T
\
Figure 1-9.
Total horizomal fOfe< to be
(esistcd at sh<:3r w-all(s)
c
o~·~rturniug at shear-tTtJnifer (drug)
truss
NOles for F1gure 1-9:
T =uplift in lb
Slope= 3, 4, 5, 6,etc., 1n mcrements of 12 (e.g., for a 6: 12 roof slope, slope =6)
Spactng of roof trusses IS 2.0 ft. on center
Ot."'ad load on roof truss (tncludes both top chord and bottom chord loading):
IV"' = 20.0 psf x 2.0 fl = 40.0 plf
Slope oflruss = 6
Span of truss= L=48.0 ft
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Design Example 1 • Four-Story VKiod Ught Frame Structure
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Otttrmint uplift fon:t Tat lint I dut to sd s mit ovtrturning fortts on roof sbtt r-tnn.sftr (drtg) truss
(stt Figurt 1-9).
l: M, =0
H Lsine
2oose
~~~ -
W"'(L)'
- TxL=O
2
H Lsine _ w.,(L)'
T=C= 2oose
2
L
For gable-type 1russes:
~
w..,•.,,.x_ L
T = H(slepe) _ _
24
2
C = H(>lope) _ .::
w.E.·
I>I.':x=
L
24
2
For single-slope-type trusses:
......----1
~
I
T = H(slepe) _ W01. x L
24
2
C = H(•lope) _ _
w_.·
"'.,_x_ L
24
2
From Table 1-1 3, the total hori2ontal (strength) shear force for the wall at hne C is:
H= 12,5481b
Otttrmint strtngrh lt\'tl of \'t rtiuJ fortt:s.
For tension force:
=
T = H(slepe) _ W0 , xL _ 12,548x6 _ 27.8x48.0 _
_
lb
3137 667 2470
24
2
24
2
where
(0.9 - 0.2Sm)D + pQ,
ASCE 7 §2.3.6, Eq 7
H= pQ£= 12,5481b
Eq 12.4-3
IV"'= (0.9 - 0.2S..)IYD<
Wm =0.9 - (0.2 X 1.026)1Vm = 0.691Voc
IVDL = 0.69 X 40.0 plf = 27.8 plf
L= 48.0 ft
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Design Example 1 •
Four-Story~ Light-Frame StTucture
For compression force:
C = H(slope) +IV"'- x L = 12,548x6 + 56.0 x48.0 _
+
=
lb
3 137 1344 4481
24
2
24
2
where
(I .2 + 0.2SIJS)D + pQ,
ASCE 7 §2.3.6, E<j 6
E<j 12.4-3
H = pQ,= 12,548 lb
IV.,_= ( 1.2 + 0.2S.os)IVDL
1Voc=(l.2+0.2x J.026)1V,._= 1.4Wm
IVIlL= 1.4 X 40.0 plf = 56.0 plf
L =48.0 ft
Ot:ttrmint ASD ltvtl or Vtrdr.a l fortes.
For lension force.:
T _ H(.tlope) _
24
w"' xL = 8784x6 _ 18.4x48.0 _ 2196 _ 442 = 1754 lb
2
24
2
where
(0.6 + 0. 14Sos)D + 0 .7pQ,
ASCE 7 §2.4.5, E<j 10
E<j 12.4-3
H = 0.7r>Qc=0.7 X 12,548 = 8784 lb
Wm =(0.6 + 0.14S.os)WOI.
IV.,_ =0.6- (0.14 X 1.026)1V,. =0.4611'0 ,
IV"' =0.46 x40.0 plf= 18.4 plf
L =48.0 ft
For compresston force:
T _ H(.tlop•) + W"' x L _ 8784 x6 + 44.0 x48.0 _
24
2
24
2
2196
+
1056
=
3252
lb
where
(I .0 + 0 . 14SI>S)D + 0 .7pQ,
ASCE 7 §2.4.5, E<j 8
E<j 12.4-3
H = 0.7r>Qc=0.7 X 12,548 = 8784 lb
IVDL = ( 1.0 + 0. 14S.os)WDI.
IVoc= 1.0+(0.14x 1.026)Wm= I.IIVDL
IV.,_= 1.1 X 40.0 plf = 44.0 plf
L =48.0 ft
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Design Example 1 • Four-Story VKiod Ught Frame Structure
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4.2 SHEAR WALL CUMULATIVE OVERTURNING FORCES AND WALL STABILITY
Shear \Vall Cumulatin 0\'trtuming fortt:s:
When des1gn1ng 0\1ertumang tbrces in muhJievel structures. shear and the respecllve O\'enurnang fOrces due
to se-ismic (or wand) must be carried down to the foundatio11(or base where a pcxhwn slab exists). While
shear is resisted by the sheathing. ovenurmng compress1on and tension fooces are typtcalJy designed to
be res1sted by the boundary studs and continuous tiMO\vn system, respective-ly. These forces are not only
cwnulattve ove.r the he1giU of the budding. but are also dependent on the hetght of the lood appi1C3t10il;
shear fOrces applied at the upper levels will ge.nerate much larger base ovenurmng momer'lt..(llhan if the
same sheat forces were applied at the lower stoty.
The overturntng tOr~es (M07~ for the. honzomalloads on lhe shear waJI (Figw-e 1-14) can be oblatned by
summmg forces abmn the base of the wall for the level betng des1gned.
The height m whtch the load is applied to the wall for detetmaning ovenummg forces depends on the type
of construction and the method used to apply the load to the shear wall.
For a sheathed wall that does not have a nbbon board at the floor line but has top plate(s) below the floor
sheathtng (modified platlOrm framed. see Fagures 1-10 and 1- 11 ). the height of the walliS the drstance
betw"'n the bonom of !he Sill plate and the top of !he top plate(s) (same as story height).
bl
h2
Ftgure I - I 0. o,·ertumingforus a1 floor lt,·el
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Point of application of
load to chord/collector
Point of application
of load to shear wall
Ft:- 1-11.
Fronung s«uon or.floor
Notesonftg.ures 1·11 and 1-12:
I. Ftg.ure 1-11 as a wall ele'\'~JltOn " tth a mochfied baUoon. f.ranung. S) stem.
2. FtgW'e 1~12 lS a SC<.:tton eut an Ftgure 1-11 taken atlhe shear wall sl\0\\mg the 4 x 4 top plate (see
also F•gure 1-7 j(>r ad<hi•OI131 n01es).
Where a shealhed wall has a nbbon boord • • !he 600< line (see F1gures 1-12 nnd 1-13) wnh the 10p plale(s)
below the nbbon board (plrufol'm framed). and where the sheathtng ts noc placed tu the ttbbon board.
the hetght of the walliS not thl! same as the story he1ght but rather •s the dt.stance bet\\~n the bottom
of!he sdl pial< and lop oflhe top pla1e(s) below !he ribbon board (wall hc•glll~ In 1h•s cas;>,!he pomt
ofapphcauon or !he hon>onml lood "' the shear wall JS ol\en considered U> be below the floor level nnd
thereby SOI"Mtames constdertd 10 reduce O\erturrung moment as opposed to constdtrang that the loads are
apphed at lhe floor Ime. AS dus percentage offoroe vanes from shear "~llO&hc31 wall \\llhtn the buddmg.
!he potenual reductiOn ofth< Cl'ertunung fcxces M3) be a small percei'Ul!•· Tlus melbod of oo.enummg
reducuon &S not recommended as tt can be tune-coo:swrung to account for all the O\Mt.rnutg forces a:od
1113) be IIIOC<Ur.ll<, depending on !he 3CtWJ bendmg S"ffness and ckWh'l! of !he nbiJon bootdllloo<
franung and !he kalJOn of !he '"'II (l!l!he end ofbwldlng. adptall U>a "'~or largeopenmg. and so
on). Pranclples or SWlCS also appear to be "' odds wllh reducong me "'en..-n•na h<•g)>• "'that be""- me
nbllon boord (wall h<tgllt) .. opposed to floor h<tght (st"') height).
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Design Example 1 • Four-Story VKiod Ught Frame Structure
4
I~
'
r
"
.'
',
v
!'- ·
W(l/1 tlb()\'(!
II
\
II
II
II
II
II
II
II
II
II
II
!
bl
h2
".~;_
·,
I I
I
I
11· ,I!..II Jr I
II
Ftgurt! 1-12.
II
II
II
II
II II
II
II
II
II
II
Ortttumingforc~s tllj/()()r Jn·t/
Point of application of
load to cbord/collcclor
Framing
Ftgur~ 1-13. Frtmung s~ction tJJjloor
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Design Example 1 • Four-Story~ Light-Frame StTucture
Thos design example will apply the lateral loads ill the roof and ftoor levels (see Figure 1-14):
Cumulative overturntng tbrce for the roof le\•el:
Mar= F,(H,)
Cumulative 0\•erturntng tOrce for the fowlh-floor lt\•el:
Mar= F,(H, + H1) + F1(H1)
Cl.unulauve O\•enurning tbrce fot tl'le third-floor level:
Mor= F,(H, + H1 + li
2)
+ F1(H, + H2) + F2(H2 )
Cl.unulauve O\•enurning tbrce fot tl'le-second·flOQ( level:
Mur= F,(H, + H1 + li, + H 1)+ F,(H1 + H2 + H 1) + F2(H 2 + H 1) + F1(H1)
For ove:nurning forces for the ovenumtng effects of the shear-uansfer (drag) truSses. see Secuon 4.1.
In shear waJis with continuous tie-down systel'ns, the overturning resistattce in lhe shear waJi ts res:tsted by
the postS and/or end studs reststtng the compression forces and the reruaoo rods resiSttng the teno;ton fOrces.
In shear walls wtth c:onventtonal hold-down systems. tl'le overtummg resistance tn lhe shear wall is reststed
by the posts and/or end studs resisting the compression forces and the lension rods resisttng the tenston
forces.
F4
4-----ROOF)
...
:c
..,+
:c
+
.....
:c
+
:c
.F3 )
-'f<,...-----,4TH FLR
..,
:c
;!<
:c
+
:c
F2
-'l<---:;3;m'"° FLR)
...
STORY
HEIGHT H2
:c
+
::r::
STORY
HEIGHT HI
R~
34
STORY
ltEIC IIT lf3
1-14. Cumultww• ot·erturnmgforcts
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Design Example 1 • Four-Story VKiod Ught Frame Structure
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Commrntary on Tens ion a nd Compmsion Fortts in S btar \Valls
If a shear wall were an infinitely ng..d body. all of the Joads acllng on the ·wall (dead loads tncludtng selfweight and live loads. ba~d on load combinatio~ tn additi-On to lhe ovenurmng forces) would contrtbute
to the compression force at t11e shear wall boundary4 Recogn121ng that shear walls consuucred out of \\'OOd
sU\Icrural pane-ls are.not infinitely ngid bodies, as soon as defOrmations occur. the loads to the compresston
posts change to the lttbutary loads on the compression members. Certatnly a ''ery taU aspect rauo shear
waJI can act as a rigid body and can have all the loads acting on Ute \\-all contttbuung to the compression
force at the Yl'311 boundary. Ltkewtse. very long walls do tlot act as a ngad body and wlll only see tributary
loads to the v.'all boundary posts. h can be argued that if only mbutary loads are used on lhe compress ton
posts. then only tributary loads should be used at the opposite end for dete.rmirung the tensiott force 1n lhe
ue-dO\\n rods.
For de.tenntng me tenston force m the ue-down rod.. current t.ndusuy prnctice 1S to use the net ovenurnlng
moment dtvtded by the resisting arm. where the net 0\•enurnang force is gross ovenuming moment mtnus
the resisung nloment. The resisung moment ls the swn of any dead toad suppone<l by the wall. tncludang
self-weight tunes the distance from tl'le wall end to tl1e center ofgravtty of the dead loads. For a untformly
distrtbuted dead load, the lever arm is stmply one-halflhe shear wa11length
where Wut. or .HN have the apploptiate load combinations ofASCE 7. and b is the overall
shearwilll (seeFagwe 1- 14).
length oflhe
The uph fi force T lS de.temuned by
T = _A_.
far"-:- _,_
11.._,
d
where d ts the d1stance between the ue-down rod and the center of mass oflhe boundary metnbers(see
F1gures 1-1 5 and 1-1 6).
4.3 LOAD COM BINATION S
Load Combinations Using the 2018lBC
IBC Section 1605.3.2 has altemauve baste load combinations lO ASCE 7. For allowable stress design. the
earthquake lood combinatjons are
D+L+S+ -
E
IBC Eq 16-2 1
1.4
Since S l'i not present, the simphfied load combination is
E
D+L+ 1.4
where E =-the honzontal selsmic force (F)
0.90+.!..
1.4
IBC Eq 16-22
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Design Example 1 • Four-Story~ Light-Frame StTucture
Load Combinations Using ASCE 7
§2.4.3
Per Section 2.4.5, the follmving load oombtnauons shall be used for baste comb1nauons for allo'"<lble stress
design:
( l.O + 0 .14S..,)D + H + F + 0 .7pQ,
ASCE7 Eq8
(l.0+0.105S..,)D+H+F+0.525pQ,+0.75L+0.75(L, or S or R)
ASCE 7 Eq9
(0.6 -0.14Sos)D +0.7pQ,+ H
ASCE7 Eq 10
where the dead load D ts 1.nerea~ed (or decreased) tbr ven1cal accelerauons by the SM coeffic1enL
Since H, F. S,and Rare not present,lhestmplt1ied load combinations are
(l.O + 0 .14S..,)D+0.7pQ,
ASCE 7 Eq8
( 1.0 + O. IOSS,.)D + 0.525pQ, + 0.75L + 0 .75L,
ASCE 7 Eq9
ASCE 7 Eq 10
(0.6- 0 .14SI>S)D + 0 .7pQ,
where Q£= the honzornal seism1c fOrce F
§12.4 .2.1
O. I05Sos= 0.105(1.026) =0.11
0.14SIJS = 0. 14( 1.026) = 0.14
4.4 DETERMINE MECHANICS OF SEGMENTED SHEAR WALL C
To help in the illustrauon of a des1gn e.xample. the shear wall located at grid line C will be used. Reti!r to
other secuons in thts destgn example for the denvauon of the fOrces to th1s wall.
4.4A SHEAR WALL CHORD (BOUNDARY) MEMBERS
The vertic.a1 members at the end oflhe shear walls are the walls' c.hords (boundary members). As in a
dtaphragm. the chords res:1st flexure-. and the she.athtng (web) resistS the shear. The ovenumu1g moment
is resolved mto a T-C couple, creaung axtaltens1on and compression forces. When o11Jy the horizontal
component of the seismic fotces are constdered.lhe ten.~on and compressi011 forces are equal and opposite
for each load case. The ovenuming compressive force IS dete.nnined by dividing the O\'Murntng mo1nent
by the distance bo:fbetween lhe center of the te-nsion rod rutd the center work point of the compressiOn poStS
(Figure 1-15). However. tn most designs.• the sae and number of chords (boundary members) change from
story to stoty~ as shown 1n Figure 1- 15, whtch can necessitate iterations to derive lhe actua1 diStance b,.
Many engineers Will appropriately take a coJ'tservatiVe average distat\C.e b4 and use the same value tbr aJI
c-ases to mimtnize tteJ"ations. The dismnce h4 someumes c-an be justified as betng closer to lhe wall end
where stresses are IM\1 in the boundary members.
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Design Example 1 • Four-Story VKiod Ught Frame Structure
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Four4 x 4s
H
---.--.----.-----.,---.----
Centroid o f
Centroid
boundary
oflension
pOSIS
rod
Rgur~
I- I 5. Tension and comprtssionforc~s
r-
t-t-
II
....
d
!!!!• ,.----belt
d"
d
L
Rgure I-15A. ROiation at 'roll base
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Design Example 1 •
Four-Story~ Light-Frame StTucture
Figure 1-16 tllusttates muluple boundary me.mbers that are common to multtlevel wood-frame shear waHs.
d' =distance from edge ol wall to center of compression post
••
'TlE-00~
ROO
BOUNDARY POST
PER SC><EOUL£
.
I
I
I'
\I
j
!
I
1/2w
I
CENlER OF MASS OF BOUNDARY
l.lO.l~ERS FOR COMPRESSION
I
I
w/2- ! -1
x+J/2w
1~-----'-
w+x+ w/2
f'-,,----.;.._~
~+7/2w
w+x+w+w+w/2
__;__ _
_ _;.__II
K+ 5/2w
_
i'IAJIJNG TO COMPRESSION
PO~TS PER SCHEOUL£
w+x+w+w/2
3x+16/2w
(4) posts
(2) posts
(7) posts
(6) posts
d' ~A"d/A
d' ~A'd/A
d' ~A"d/A
d' ~A"d/A
~ A"(3x • 16/2W)/{4A)
~ A'(x • 212W)/(2A)
~ A'(Sx • 36/2W)/(6A)
~ A"(6x • 49/2W)/(7A)
~ 516x +3W
~3/4x•2W
~ 112x•
~617x• 712W
w
Figure 1-16. Example pltm section at boJmdary~ members
Notes forFtgure 1-16:
I.
d' = [(1t- I )'.t • n'•n/2 j/1t
n is the number of posts
x ts the gap dtstanc:e for lhe threaded rod (in)
w is the width of the post~ (tn)
d'""(= [{4 - I )' 6 + 4''2.512¥4 = 9.5 in
n = 4•.t
=- 6 lr\, w =- 2.5 in
tf11Mf 1 =9.5 i.n+5.5 in = 15tn= 1.25 tl
2. Edge natling to boundary member posts ts detenruned by dividu'lg. tlte d1ftt:renHalload per tloor (see
Table 1-6) by the story he1gllt dovtded by the numbe.r of boundary members. For the second-Hoor
level (doff<rential load):
VenicaJ shear force:
6373 Lb = 675 plf
9.44 ft
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Design Example 1 • Four-Story VKiod Ught Frame Structure
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The numher ofboondaty posts is four (see Table 1-5); !he verueal shear force per boundary poSl ts:
675 plf = 169 plf
4 postS
An edge nailiJtg of6 inches on center will be adequate for the four boundal)' posts. Havtng edge nailing
equal to the nailing ax the panel edges can be done. but the overnruhng of the wall at the jantbs can
produce undesirable failure modes due to the eooceruration of Sl!ength at the jamb. It is common to have
a boundary post naihng schedule that is placed on lhe drawmgs that spectfies this spectaJ nailing panem.
3. A gap of 6 mches is used at the ue-down rod. This d1mension should be at least as w1de as the
bearing plm~ For this de:stgn e.xample, the widest beanng plate IS 5.5 inches (see Table 1-8). Where
the gap is narro\\o'er than the beanng plate-width, u will force the ..dapptng... of the boundary posts
for the bearmg plates; in additton [0 the tncreased Labor, it also produces reduced bearing area'i for
the boundary members.
4. The distanced" rematns a contant dtstance from the v.'311 end. For the shear wall at hne C in this
destgn example, the d1stanc:e d" IS de.tennmed at the second-floor level where there are four 4 x 6
boundary posts(seeTable 1-5).
For this she-ar wall, the dlstance d":
. . =).)+2=
- - 6·0 8 .)11'1=
0 •71 ft
Otttrmint chord mtmbtrs ror tht sbtar wall at liot C.
The axtal loads lO the beating waJJ and boundary tnembets are determined from lhe tOIIowtng loods:
Dead loads:
"'""f=(20.0 psf)(2.0 ft) = 40.0 plf
••,,.,.= (30.0 psf)(2.0 ft) = 60.0 plf
For the roofle\'el:
••• ..,= IO.Opsf{8.21 ft)=82 .1 ft
For the 2nd- through 4th-floor levels:
" '•"' = 10.0 psf{9.44 ft)= 94.4 plf
ll\e '"all ·weig.ht of 10 psflsdtlferem than the 15 psfused tn delemttningthe building weighl based on plan
dimenstons (see Section 1.3).
Ltve loads:
"'""f= (20.0 psf)(2.0 ft) = 40.0 plf
••,,.,.= (40.0 psf)(2.0 ft) = 80.0 plf
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Design Example 1 • Four-Story~ Light-Frame StTucture
Dead + live loads:
"'""~'= (20.0
psf + 20.0 pst)(2.0 ft) = 80.0 plf
,..,...,. = (30.0 psf + 40.0 psf)(2.0 tl) = 140.0 fi
For lhe roofleve-1:
w,.,.= IO.Opsf(8.21 f\)=82. 1 plf
For the 2nd- through 4th-floor levels:
"'•w= 10.0 psf(9.44 ft) =94.4 plf
For ASD compresston on the chord members, lhe alternale baste load oomblnatlon wdl be used:
E
D+L+ 1.4
IBC Eq 16-2 1
For strength-level compression on the d\OC'd members, the ASCE 7 seJsmie load combtnauon will be used.
The strength-level compresston loads are used later m this example to determme the shear v.oall deflection
at strength loads (stll plate crushmg,). Per ASCE 7 Sections 12.8.6 and 12.12. I. strengt11-leveJ tbrces are
requlred for lhe detemHnalJOn of shear wall deft~uons:
( 1.2 + 0.2SIJS)D + pQ• + L + 0.2S
ASCE 7 §2.3.6
where
St.l'lce S 1s not present, the stmpl ified load combtnauon is
( 12 + 0.2S..,)D + L + E
ASCE 7 §2.3.6
where
( L2+0.2Sos)=(l.2+0.2x 1.016)= 1.4
Per ASCE 7 Secuon 2.3.6,lhe load factor on L is perrnttted to be 0.5 Slnce the live load is equaJ to or less
than 100 psf and oot of publtc assembly. llte 0.5 facmC' tS used in the followtng Iwe load detennmauons:
"'""~' =((L4x20.0 psf)+(0.5x20.0)](2.0 ft) = 76.0 plf
•• , . =[(1.4 X 30.0 psf)+(0.5x 40.0)] (2.0 ft) = 124.0 plf
For the roo fie-vel:
w,.,.= 1.4 X 10.0 psf(8.2 J tl) = 115 plf
For the 2nd- through 4th-floor levels:
"'•w= 1.4 x 10.0 psf(9.44 tl) = 132 plf
40
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Design Example 1 • Four-Story VKiod Ught Frame Structure
4
For shear wall chord me.mber force-:
E=Mor
d
Tabl~
1-1. Dtttrmmatio" ofshtar wall d'ord member forces at line C
Strength Demand
Compression
ASD Demand
Compres:swn
Morcz•
ASD
Pv+t
(l1p-ft) (kips)
Level
-
d,.,~.
'
-
(ft)
b
(6)
•§
M
1.4d
C= -f!!.+P
I>-L
.tlor
C : d + (1.2 +0.2.~llS)PO.L
(tl)
(kips)
(k1ps)
-
329
4.49
Drag Truss
-
Roof
51.5
0.365
1.25
19.54
5.58
7.60
4th Floor
ln.6
0.892
1.25
19.54
10.78
14.69
3rd Floor
348.3
1.630
1.42
19.38
17.92
24.81
2nd Floor
541.2
2.812
1.75
19.04
26.63
37.57
Notes tor Table 1-4:
I. Drag truss compres.o;ton force tS determtned tn Secuon 4.1 and tsappl1ed at lheoneend of the shear wall.
For Slmpl1c-ny 1n design and det31lin~ the fOrce will be apphed to both e.nds of the shear waJL
2. Mo1nent O\•enurning fOrces (Mm) are based on hortzontal fOrces from Table 1 13 tOr shear wall at hne
C. Since there are two shear \~ails at hne c. the horl20ntal fOrce from Table 1-13 tS divided by 2.
3. Compression forces for ASD deJ'n.and aJ\d strength dernand are detennined lfomload combinations hsted
4
plus the compress1on fOrce tram the drag truss.
4. The overall out-to-<>Ul wall length is 21.5 feet (see Table 1-13). The d' dJmension is U.: dJstanee trom the
approxlmate cenler of mass of the multiple boundary posiS to the conunuous tie-dO\'rTI rod (see Ftgute 1-12).
5. The dislaJ'IC.e df. shown tn Table 1-4, uses a (conservarjvely) larger distance of 5.5 tnches more than the
actual distance. Unless the engtneer rnanually computes lh1s dtstance or uses a oornputer program that
automaucally computes th1s dtstan<:e,lhe destgn process would tn\'olve rec-aJculaung lhe d1stance d' and
then redetermimng/ched.::ang the boundary post, bearang plates.. tenston rods, e-longattons, and c.rushing
effectS for m1nute adjustment-; 1n post sizes, which can be an nerative process ro say the Je.as:L Also, see
Footnote 4 for Table 1-36 about possible adchtion.al iterauon of shear wall loads due to the rigidltlexible
rarjo betng greater lhan I00 petcent. The initial esumate of d' usang an addiuonal 5.5 tnches is only used
for detemuntng lhe force 1n lhjs example. Once the posts have been chosen, the actual tf is used.
6. The distance b'.((diStance from the center of mass of the boundary posts 10 the tens ton rod. see Figure
1-15) is determmed by
b_,=b -d' -d·
where
b is the length of the shear wall
d' is the djstance from the wall e.nd to the center of compression poslS
d" is lhe distance from the wall end to the tens1on rod (see Figure 1-16)
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Design Example 1 • Four-Story~ Light-Frame StTucture
7. DetemuoeASD compression lood.
PO.L = w(d')2
For ASD demand:
PO.L ,..,.= (80 plf + 82. I plf)( I. I3 X 2) = 0 .365 ktps
Po.L ,,.,,-=(140 plf + 94.4 plf+ 80 + 82.1)(1.13 x2)=0.892 kips
Po.L ..,,,_ = ((( 140 plf + 94.4 plf)x 2) + (80 + 82.1)](1.29 x 2) = 1.630 ktps
Pt><L """"' = ( (( 140 plf + 94.4 plf) x 3) + (80 + 82. I)]< 1.63 x 2) = 2.8 I2 ktps
For strength demand:
( 1.2 + 0.02S..,)D + L = 1.40 + L
Pt><L...,.=(76 plf + 124 plf)( l. ll x2) =0.452 ktps
Po..,,.,,-=(124plf+ 132 plf+76+ 124)(1.13x2)= J.OJOkips
PI><L '"''""" = [ ((124 pU·+ 132 plf)X 2) +(76 + 124)](1.29 X 2)= 1.837 ktps
Pt><L """"' = ( ((124 plf + 132 pi I) x 3) + (76 + 124)
Table I -5.
j(J .63 x 2) = 3. I 55 ktps
~termuJfJJ/()n ofshear wall chord memberS tJJ lin~
Bearing
Chord
S tab~uy
c,
Capacity
(kips)
ASD
Demand
(ktps)
Capacity
(kips)
D/C
Ratio
1.1 5
0.258
21.88
5.58
24.93
0.26
8.84
1.1 5
0. 192
21.88
10.78
18.54
0.58
~9.0
8.84
1.1 5
0. 192
30.63
17.92
25.96
0.69
77.0
8.84
1.1
0.200
48.13
26.63
40.70
0.65
1,
POS(S
Total
Area
(ft)
CF
Roof
Four 3 x 4s
35.0
7.82
4 th Floor
Four 3 x 4s
35.0
3rd Flooc
Four4 x 4s
2nd Floor
Four 4 x. 6s
Level
C
Nates for Table I-5:
I . See lhe following example for detemunation of compression member capacity.
2 . The typiCal tnlerior stud \'t'3Ji ts framed with 4-inc:h nominal framing studs.
3. Interior beanng walls tbr thts design example are nonrated (fire) and, as such. wouJd not reqture the
reduction m allowable stresses required for FRnv.
4 . The shear wall ha~ sheathmg on one stde oflhe studs (nm both stdes)~ some engmeers constder the
ovenurntng compression tbrce wnJ1 sheathing on one stele as ecc:entncaHy loading the posts. Some
engineers will combine the compressive force with the 5 psf laterol: loading (IBC Section 1607. 15). but
lhis load combined v.'ith seismtc fOrces does not appear to be the tnlent of the rode. h ts recooc:}'Jzed that
combining tn-plane compresston loods with OUl-of-plane lateral loads or eccentric loads (NOS Secuons
3.9 .2 and 3.9 .3) woukl rtduce the ventcal c.apac:tty of the boundal)' posts. The possible eccenmc loadtng
from the sheathang on one stde or oul-of..plane loadtrtg on the compres.~ton posts has not been constdered
ln this desJgn example.
5. The beanng capacity tOr the 3 '< 4 postS at the roof leve-l controls over the stabthty capac tty.
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Design Example 1 • Four-Story VKiod Ught Frame Structure
4
E.umplt Compmsioa Mr.m btr Capat.il)' Dttrrmlnatioa
4
x
4 post - Dooglas Fir-Larch No. 1:
where
A= 12.25 tn'
C0 = 1.6
Em• = 620s()()() pst
d 1 =3.5 in
The tbllowang coefficients for Cuand C, are not referenced tn lhe NOS foonulas (fOI' sampltcity).
CM= 1.0
C,= 1.0
K11
= 1.0
ll\e members• span betwee-n the lop of the run 2 x 4 Sill plates and the underside of lhe 4 x 4 lop plate (see
Figure 1-7) is
(1.5x2)+(23/J2)+3.5 _ _ fi
0 60
12
I= 9.44- 0.60 = 8.84 fl
1,1 = 8.84 x 12 = 106 tn
Assuming WSP sheathing IS nailed to all postS to prevent lateral bucklangoflhe posls parallel to lhe shear
wall~ the post dimension perpendicular lO lhe-shear wall will be used for buckhng calculations (see NOS
A. I I.3). In all cases.. this ls the 3.5-lnch di.nenston.
Compression
C
"
par~dld
=I+(
t)
lc
to grain (only applicablt adju.ftmrnl rattOri shown.):
t J]' __;. =0.192
[I+(
2c
c
NOS E<j 3.7-1
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where
<=0.8
Fd:-
F«
,...
0,8'
..2£_
tH
~
0.8llx620,000
•
___
-)))psi
NDS§3.7.15
JO.J'
=~•0201
2760
•
•
F; =Fc{:,f:,C,•ISOOx 1.6x 1.15x0.192=530po•
For a 4 x 4 post:
P.,,.. =Ax F; •12.25 x530=64931b
Compress1on petpenc:hcular to gmm:
NOS §3.10.4
Fora 4 x 4 pos1.
P,._=Ax F;. •12.25 x625=76561b
4.48 DETERMINATION OF RESISTING MOMENTS AND UPLIFT FORCES
The: res•stmg momen1 \Ia IS dete:rmmed from the follootng de:3d loads;
w...,=20.0 psl\2.0 ft)=40.0 pU'
= 30.0 pSI\2.0 ft) = 60.0 plf
w.....,
For the roofle>el:
....... 10.0 P>q8.21 ft) = 82.1 plf
For the 2nd-flOO< through 41Mloor 1.-els:
..... = 10.0 psl\9.44 ft)= 94.4 plf
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Design Example 1 • Four-Story VKiod Ught Frame Structure
4
Table I -6. Dtltnnin~ £h~ar wa/1 uplift forces u:rmg ASCE 7 lood combinations atlm~ C
Strenglh
ASD Uplift
(M0 r X 0. 7)-(0.6- 0 .14S"')MR
Level
Drag
Truss
b_,
M•
(ft-lb)
(ft)
-
-
Mm
(fl-lb)
-
M,(0.6- 0.14S,.)' 11
(ft-lb)
-
Dttlerent•al
b<l
(lb)
1758
Load
per Floor
(lb)
-
Roof
31,063
19.54
5 1,511
14,176
2878
0
41h
Floor
66,749
19.54
177,607
30,462
6561
3684
102,435
19.38 348,283
46,747
11,928
5367
138,120
19.04 541,249
63,033
18,345
6416
3rd
Floor
2nd
Floor
Notes tOr Table 1-6:
I. Where (0.6 -0.14SI>S)= (0.6 -0.14 x 1.026)=0.46
2. Drag truss uplift (tension) tbrce Is determined m Section 4.1 and IS applied at the ooe end oflhe shear
wall. For simplicity 1n design and de.tathng. the force will be applied to bothe,n dsofthe shear wall.
3. Ovenurning forces (Af01) are from Table 1-4.
4.4C SHEAR WALL TIE-DOWN SYSTEM COMPONENTS
Tit-Down Rod
Tie-down rc.xls are usually made from A36/A307 steel. This IS called standard rod strength. Unless marked,
rods should be constdered standard rod strenglh. H tgb-strenglh rods are ASTM A449 or A 193- 87 and are
usually marked on the end w•th an embossed stamp, though some rod manufacturers stamp the rod grade
on the side. If the rod is stamped at lheend and 1s cut. it needs to be re-marked. There should be a spec.al
tnSpection ofhag.h-streng.th rods to coofirm the rod type since the ends of these rods may be embedded 1nto
a c.oupler where the marks cannot be ~,n after installation. High-strength rods are usually not weldable
·withoUl special welding procedures. Pro1>rietary systems have spec:Jal rod colors and marklng.~ oo the stdes~
some rods are propnet.ary, the manufacrured components are proprietary.
Rod Uongatlon
The net te.ns•le area. A~· is used tn the rod strength ru'ld elongation calculauons. However, to tacilitate rod
suength des1gn. AISC 360 Tab1e J3.2 tabulates the nomtnal tensale strenglh, F"" so the destgner may use
the nominal bolt area,A6. TheCommentruy states ..F..,= 0.75F,.- and "The factorof0.75 iocluded in thiS
equation accounts fonhe approximate ratio of the effecuve tension area of the threaded portion of the bolt
to the area of the shank oflhe bolt for common sizes.'· Somejunsdtctions have ltmttson the amount of rod
elongation that can occur betw~n restrauus and/or stone~ and some tequtre that the allowable sues:s area
(A~ vs. A ) be used an rod elongatton caJculruions. As suc-h, local butldmg depanment requirements should
1
always be checked. This design example uses A~ for rod elongation and .41 or A" fO! rod capacity.
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Design Example 1 •
Four-Story~ Light-Frame StTucture
Many conttnuous rod system manufactuters rry to de.termine lhe rnost C-Ost-effecuve solutton fOr a given
ttnston load considering rod strength and dtameter. The use of a htgher-strength rod wdl tncrease lhe drift
of the shear \valltf used in place of a standard-strength rod due to tncreased rod elongauon from the smaller
rod dtameters.. and the modulus of elast1city of lhe steel. whtch does not change, l~ the same forboth
standard- and high-strength rods.
Tie-down rod elongation 1s computed bel'ween bearing plates (restraints). This design example has bearang
plates located at each fl()()(. Table 1-7 computes the rod capac1ties and elongat10l'IS (per floor) bem~n t11e
bearing plates.
Table 1-7.
Derennin~ rod si:~s.
capaCilltS. and e/onga11on.t 01 lint C
ASDRod
Capacity
ASD
Rod
Elf.
Plate
Height
Tension
De-m.and
Dia.
Dia.
A,
(fi)
(kops)
d,
(in)
A,
Level
d
(on)
(in')
(in')
F,
(ksi)
F,
(kso)
2
(kops)
ASD
Rod
Elong.
(in)
Roof
8.1 1
2.88
0 .625
0.527
0.307
0.216
58
36
6.68
0.043
4th
Floor
9.44
6.56
0 .625
0.527
0.307
0.216
58
36
6.68
0.113
3rd
Floor
9.44
11.93
0 .875
0.755
o.60 1
0.462
58
36
13.07
0.10 1
2nd
Floor
9.44
18.34
1.00
0.865
0.785
0 .606
120
105
35.33
0.118
0.7sxr: xA~
Notes for Table I-7:
I. Tension demand (ASD upli11) values are computed tn Table I.<>.
1
3.14<1
2. Rod area: .4., :: - 3.
•
4
Net te.nsiJe area:
A, =0.7854 x(d• -0 ·9743
n )'
where n; lhe number of threads per 1nch per AISC Table 7-1 7
4. For ASD desig~~, the allowable rod capacny for the AISC 360-10 Secuon Jl.6
F.,
X
A•
•
and
A4' =0.75A•
Q = 2.0 for ASD
Therefore.
0.75 X F_ X .411
2
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IS
Design Example 1 • Four-Story VKiod Ught Frame Structure
4
54 Standard rod tsASTM A36 rod with mtntmwn F. = 58 ks1, Fy= 36 kst. High-strength rod isASTM
A 193-87 rod with minimum F. = 125 ksi; Fy = 105 ksi for rods up to 2!1.! inches tn dtamerer and A449
rod with minimwn F.,-= 120 ksi; Fy-= 92 ks1 for rods up to I tnch 111 diameter then drops to F. = 105 ksi,
Fy= 81 k.(jj (per ASTM A449) for rods from Ito
I~
tnches in diameter and drops to F.,= 90 ksa. F1 =
58 kst (pe-r ASTM A449) fOI' rods from 1¥.. 10 3 tnche:s 111 d~ame1er.
.
PL
6. Rod elongauon: A=--
A,E
where
A= lhe elong:nton of the rod tn inc.hes
P = lhe accumulated uphft tension force on the rod tn kaps (tenston demand)
L =the length of the rod 111 inches from beartng restratnt to bearing restraint, With lhe beanng
restr.unr betng where the load is transfened to the rod
E = 29,000 kSI
A~ = the effective area of the
rod tn square 1nchei. Where smooth rods are used, the area l'i
equal to the gross area (A8). Where threaded (all-thread) rods are used, the area 1s equal to
the tc-ns1on area (A,) of the threaded rod. Since many of the proprtetary sys1ems that have
smooth rods have long ponions threaded at the ends. tt is recommended that A~ be used
when cakulating rod e-longation.
7. Rod elongauon from allowable Stress destgn fOrces is based on ustng the etl'ecuve area (A,) and the
following lengths:
a. For the firsl level, the anchor bolt is projecting 4 incites above the foundadon (the height of
lhe coupler nut to the anchor bolt at the tiOOf slab~ The snlall runoum of elongation of the
rod anchored tn the concre.te foundation and projecting 4 mc.hes above IS negJecte.d in lhe
computauon.
b. For the framed floors, the rod from below ts proJectlng 6 tnches above lhe sole plate.
8. Rod elongations for determ1nauon of she<~r wall dtJfts will he fac.tored up by 1.4 (see Table 1-12).
Rod Couplers
Couplers are used to connect the rods. Couplers can e11her be Sltaight or reducing and can be supphed
tn different strengths or grades. Couplers for htgh-strength rods need (0 be ofhig,h-srre.ngth steel and are
marked walh notches or matks on the C-oupler. Also. high-strength nuts or couplers are not weldable. For a
rod to develop its full stteng.~ the rod must be e-mbedded a set amount into the coupler (usually lhe depth
of a standard nut). It 1s recom.mended that v.1tere couplers are used. they have ptlot or Wttness holes in the
side so lhe Inspector can witness the threads of the rods in the holes to ensure proper embedmenL
Reducing couplers are used where the rod s12e is changed. In reductng coupler~ the st.ze of the lhreadtng
cl'langes at the m1ddle of the coupler devtce. It IS tntended that the rods be embedded unttlt.hey bottom our
atlhe cente-r of the coupler. If the rod.~ are installed in this fashton.. witness holes will not be necessary~
hov..-e\•er. IllS recommended that couplers w1th Witness holes be used so that proper tnstallation can be
confirmed by M tnspec-tor. Some couplers have rod stops tn them to ensure the proper length of bolt is 111
the coupleJ'. Reducing couplers shouJd have the same notches and identtf)ting maries as Straight couplers
when used wath h1gh-strengt11 rods.
St-aring Platts
Bearulg plates transfer lhe dttferenual overturning forces into the conunuous ue-d<mon system through
bearing and compression in the floor framtng. through lhe stU plate(s) or the top plates and ribbon board
(when used) tnto the rod (see Ftgure 1-17). Premanufactured bear111g plates are usually tdenufied by patnt
color or by a number marked on the plate. However, patnt colors or unpauued plaxes Wlf)' among dtfferent
rod system manufacturers.
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Design Example 1 •
Four-Story~ Light-Frame StTucture
Tublt 1-8. De1erml11t bearing-piau Jl=ts and tXlpadllts tllline C
Bear1ng Plate
Beanng
Beanng
Factor
(tn')
c,
Load
(ktps)
AIIO\I;'able
Capacity
(ktpS)
0.8125
15.98
1.07
2.878
10.69
0.4
0.8125
9.98
1.11
3.684
6.92
5.5
0.6
1.0625
15.61
1.07
5.367
10.44
5.5
0.6
1.1875
15.39
1.07
6.416
10.29
Level
Width
(m)
Length
(m)
Th1ckness
(m)
Holed1a
(m)
Roof
3.0
5.5
0.6
4th
Floor
3.0
3.5
3rd
Floor
3.0
2nd
Floor
3.0
A ...
Notes for Table 1-8:
I. llearang pi are L< ba<ed on ASTM A36 steel witlt F1 = 36 ksi.
(/ +0.375)
2. Be.ati1tg area factor fOr /11 < 6 tncl\es C• = '"''--:----'1,
NOS §3.10.4
3. Beanngarea factor for 111 ~6 incites: C11 = 1.0
4. Be,aring plate thicknesses shall be c.hecked for bending us1ng lengths governed by the area sausfact.ion
check and the a.<SOCiated hole an the plate (see Figure 1- 17).
Eunaplt Btnding Cbtt.k of Btaring PlAit II Sttond Aoor
Beanng plate s1ze = 3.0 in x 5.5 in"' 0.6 1n thick
Beanng load =63731b (Table 14>)
Bear1ng area for wood: subuacung for ~~~o-1nch overs12ed hole ln \\"OOd plates
( 16.5 - 0.887) = 15.6 m'
r
- 6.416 - 411 .
pSI
15.6
Ia -
F:"=F"C,=625x i.07=669psi>4 11 pst .. . OK
Steel plate bendmg check:
•5.5)'
(411 X 3.0) X
l~
= 4662 tn-lb
_ lxl' _ (3.0 -1.0625)x0.6'
. ,
ZI"M-44
- 0. 174 tn
M
4662
- = - - = 2 6.8 ksa .. . OK
z 0. 174
5. Bearing load= differential load from Table 1-6.
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Design Example 1 • Four-Story VKiod Ught Frame Structure
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Bolte-d Tit-Down Dn'it t Eltmtnts
Anmher type of tie-d0\1111 devtce uulizes bolts tnstead of bearing plates to uansfer the overturning forces
to the contanuou:s rods. In thts system, posts need to transfer tension forces. Although this type of system
ts still avatlable, most fi'a.ning cont.raetOC'S prefer the beanng plate dev.ces due to qUJcker and ea~t et
installauon in the field.
4
Sbrink.agt Comptnsating orTakt-Up Dtvicts
Most conunoous rod S}'Ste.ns have methods of compensaung tor shnnkage Wtth propr1ewy e.xpanding or
contracung de\•tees..
The purpose of these devtces is to mintmize the clearance created between the hold-down, te.nsion ne
con..neclOr, or plate wa.'\her and the anchor bolt/nut due to butlding se-ttlement or wood shnnkage (resulung
from a reducuon tn moisture content). The nut tS rotated down (or compress1on spring IS used) on the rod so
the hold-down, tension tte, ot beartng plate remain.~ tight to the wood surface.
There are se\·eral ditferem types of devices 3\'atlable, as defi1'led in Acceptance Critena (AC) 3 16:
Comprusioa-Controlltd Shrinkage- Comptnsa ting De-viet (CCSCD): A device controlled by
compression loadtng Yl'here the rod passes uninterrupted through lhe device, and that is used to
compensate for localrzed wood shrinkage.
Ttnsion--Gontrolltd Sbriukag_t Comptnsaring Coupling Ot\'ict (TCSCCD) : A devtce controlled
by tens.ton loading that coonecrs rods or anchotS togethe,r. and that tS used to compens:a[e for localtzed
wood shrinking. ThL~ product t~ a conunuous1y varying uavel device.
Ttnsion-Coatrolltd Sbrinkag_t Compt.nsating On•irt (TCSCD): A device controlled by tens ton
loading where the rod is anaehed to or e1\gaged by the devtce-, and lhat 1s used to c:ompensrue for
loc:alt2ed wood shrinkage. The product may be an in.cre.mental or ratcheung travel devic-e or a
continuously varytng mwel device.
O!her definitions defined by AC3 16:
Ot' ice- An ragt Tnwt l and Stating lntrtmtnt (4 1t) tS the average of the moveme.nt cequued to cause
uxremen.tal motton from a seated posttion and the opposne movement requtred to reseat the device after
the ac.tuation (or ratcheung).
Allowablt Otfttttion Limit (A,J of Dt,•ice- Loadt-d at Allowablt Ot fttction Limit is the tnaxuuwn
allowable de-ftec.uon limit ofa shnnJ.."age dev1ce.
ICC-Evaluation Service (ICC-ES) has acceptance ernena (AC3 16) for slmnkage compen.<ating (take-up)
llte destgn enginet"r should check to see that the proprietary devic~ conform to these criteria.
device.~.
The use of take-up devices IS highly des1rable 111 mululevel \\'00d frame c.onstru("tlon. Stnce the total
shrinkage of the buildtrtg has to be accounted fOr m the tie-do\tm dtsplooement (d41), u is very difficult to
meet the code dnft requJtements fbr most shear \\'ails wtlhout take-up da•tces, espe:clally for shon-len.gth
4
shearwa.Jis.
Take-up devtces de-flect under load just hke the con,•entional hold-down. The reponed take-up device
(connector) deflectioo found in lhe C\'aluation repo11 is per AC l 55 the computed defl.ecuoo and fastener sltp
at the maximum rated Jood for the device. The deflec.t1on can be reduced based on actual da •tce loads that
are less than the 1nax.imwn tOr the device.
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Design Example 1 • Four-Story~ Light-Frame StTucture
The deformation or inttial slack of these devtces needs to be considered tn the overall ue-down
dtsplacement (d,).
Take-up devices have moving pans and may jam tf not properly tn.<a:alled. Jammtng typically occurs as a
resuh of e.xcesstve continuous LJe rod angle (ottt-of-plwnb). See the manufacturer's tnstn.K:tions for proper
tnstallation.
Wlule shrinkage compensattng or take-up devtces do not dtstinguash between the gaps created by tmttal
shnnkage or senJement or the gaps created at other tJmes or by other causes. ICC-ES AC316 for shrinkage
compensating dt\•tces states they are used to ..compensate for movement tn a connection due lO senlement
or wood shrinkage. stnce the codes do not specify procedures for qualtlytng and tnstall ing such produCL"i:'
Sundar statements are in shnnkage compensattng devace ICC-ES evafuauon repons. The destg.nc-r could
spec:tfy what seulement should be c:onstde-red on lhe plans. Se.nJement cannot be calculated., only estimmed.
TCSCO (e.g.• ratchetingdevices) move along a tension rod "s length with no maxtmum gap hmitauon.
1-JO\\'eVer. the devic.e average travel and seating increment (llJ (deflection) of these devices increases
signtficamly as the tension rod dtamete:r ux:reao;es, maktng u difficult lO comply With deflectton lunnations.
Whtle there is no current sysle.m lO evaluate the acceptable long-term perfonnance of shnnkage or take-up
device.s,ICC-ES AC316 Section 4.5 requires that ..LdJoc.umentation shall be submmed showang 110\'.• the
CCSCD~ TCSCCO or TCSCD will be protecled from ntOlsture and contaminatioo during cot\.~ruction, or
data shaH be submmed~ demonsuatlng lhm exposure to tnOISlUte at'l.d contaminauon during oof'\SU1JCtton
will have no long-term adverse effects on the devace." Mat\)' of the de\•tce manufucrure-rs perform corrosion
testing in otder to provade tnformation to the de.~igner regatdJttg the devtce ·s corrosjon-resistanc:e capability.
-
r/
/
'
B"""G .!AlE
/'5 OCCR!£ ·
/
I ,,
'
Ill Ill I
I
2X4 mt.IIIER Sflfl TO
UN£ t.ifl • / 4~6 8Cl0*'
-... ~ ~
"""'
4X4 COW
I'<JSTS
'
\
"'lssl&"l
...
..
'
19.9"
Figurr 1-17. Betmng =one lhrOughfrumlngfrom uplifting posts 10 bttmug de'\·ice
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Design Example 1 • Four-Story VKiod Ught Frame Structure
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4.40 BEARING ZONE THROUGH FRAMING
Compression loads to the boundary members (posts) are ac.hteved by naihng the shear wall sheathing to
each boundary me-mber. thus transfernng the overturning (compression) forces, ru\d are accumulative to the
stories bei O\V. As the shear wall ua.nsters the overrumutg (tensaon} forces to the boundary me.nbers.lhese
forces collect at each Je\'el between restmtnt devices and uansfet the ditferenttalloads (see Table 1-6) to
the beanng plates atlhe level above (see Ftgure I -1 7). The engmee:r should c.on.o;Kier how the differential
uphft forces are transferred from the bow\d.ary .nembe.rs to the beara.ng plate. As a general rule. \~,otlere the
dift'e-ren.ual uplift forces c.an be transferred within a bearmg area located withJn a 45-degree plane from the
beanng plate, oo further inve.~ig.atton ts necessary (see F1gure 1- 17). Where the transfer of fOrces reqUJres
an area larger than the 45-degree plane. some son of funhe,r uwesugauon tS necessary (e.g.• beochng and
shear checks of top plates).
Examplt During Chttk (Stt Figurt 1-17)
Difierentml Load at third-ftoor le,•el = 5367 lb (from Table 1-6)
llucknessoftfamingat floor= (2 x 1.5) + (23/32)+ 3.5 = 7.2 in
Bearing plate width= 5.5 an(from Table 1-8)
Beanrlg wtdth at bottom of4 x4 top plate=(5.5 + 7.2 + 7.2)= 19.9 1n
Negle<:n.ng the trimmer stud, there are dtree 4 x 4 c.o.npres.sjon postS withtn the be.aring area:
Bearang area= (3.5 x 3.5) x 3 = 36.7 an1
Bearing stress= 5367136.7 = 146 psi < 625 pso ... OK
S1nce the beartng area lS more than 3 inches from the end of the top plate(s). an mcrease m beanng stress
(NOS 3.10.4) c.an be considered. HO\\~ver. the length of the bearing area is greater than 6 inches. so the
be.anng factor lS C6 = 1.0.
4.4E Sill PLATE CRUSHING
F;..
Per NOS Section 4.2.6, where compression perpendicular to grainJ;l.ls less than 0.73
crushtng will
be .approxamately 0.02 mch. Where.{;..!.=
crushtng is approximately 0.04 tnch. The efti!ct of sill plate
crushing is the downward effect at lhe oppos1te end of the wall (resulting from the boundary chords) and
has the same rotational effect as the ue-down dts:placement (d,.). Shon \~t'alls that have no (net) uplift forces
will sttll have .a cruslung etfect at wall boundartes and contribute to rotatton of the waJJ.
F;.
The crushing effect on wood ts not ltnear~ a graph of load versus deformation is s:hown in Ftgure 1-1 8. llte
vaJues of0.02 1nch and 0.04 toc.h are based on a meillf pltll~ beanng on wood perpendicular to tJ'Ie gram.
These \'alues are llmit-st.ate values and are not adjustable for the duration of load (Co).
NOS Comme,ntary Sectton C4.2.6 states that when a joint made of two y..'QOd me-Jnbers are both 1ooded
perpendtcular to grain, the amount of deformation will be apptoxirnately two and one-hal fumes that ofa
metal plate bearingjmnt. Table l-91tStS the detbrmauon adjustment factors for d•tll!cent be.armg C.OJ'Iditions.
Exeepung post caps and ba.~ most connections in wood eonstrucuon do not have .neml plates for beanng_
In tJ1e case of the shear wallm thtS design ex.ample. the only metal plales an the wall construc.uon are the
bearing plates at tJte continuous tje-down rods. Accordtngly, t11e cruslung values of the boundary posts
should be inc:re.ased by lhe deforanauon adjustment factor shown in Table 1-9.
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Design Example 1 • Four-Story~ Light-Frame StTucture
Table 1-9. Defonnatton adjustn~ntfactor for bearing ctmdittou
Bearing Condmon
DefOrmation AdJustment Factor
I. Wood-to-wood (both P"rpend•cular to grain)
2.5
2. Wood-lo-wood (one parallel to gram and one perpendicular to
gram)
1.75
3. Meml-lO-\\.'OOd (wood loaded perpen<hcular to gram)
1.0
Fc.J.. Load Oeformatjon Curve
1.8
1.6
1.4
1.2
-
Fu. 1.0
0.8
0.6
'
o.•
0.2
-
0
'
---
'
0
0.02
0.04
0.06
0.08
0.1
0. 12
0 .14
0.16
Deformation, in
-- - Eq. 1.0 -
- Eq. 2.0 - - Eq. 3.0
Figure 1-18. Fti./()(ld deformation cun·~ (Eq. 3.0 dtril·edfrr>m B~udtun-Gallrgau. 1979)
For lhe lhtee ditlfte1\l region.~ oftlte load versus defonnauon curve sh0\\1\ ln Figure I- I8. equauons for
detemuning compression perpendkular to gratn def'ormatiorl (t1) may be calculated as follows:
where fu S FtJAIIl..
& = 0.02
X(___&_)
Eq 1.0
~J.OI.l2"
where Fclb.lrt' <f u < Fcll)J)J..
& = 0.04-0.02 X
52
(1-fa)
FcJJl,.,.
0.27
Eq2.0
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Design Example 1 • Four-Story VKiod Ught Frame Structure
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where
Eq3.0
where
a = deformatton. in
Ia =induced stre~. psi
F~·=Fe~= reference design ,,ruue at 0.04-tn<".h deformauon, psi (Fu)
FtJJ.IJil. =reference destgn \'alue at 0.02-inch deformat.~on. psi (0.73Fa)
In the case of !he shear wall m tlus design e=ple (Figure 1- 17), tl>e boundary posts bear on the top plate
(bearing conditaon 2). and the Sleel beartng plate reststs this upward movement and bears on the sill plate
and floor she.alhtng. whtch ul rurn bear dov.·n on the understde oflhe top plate (bearing rond.ttion 2). The
crushing. effect IS comang from two dtrecttons at the same ume, thus doubling the amount of deformation.
In addition, there is the crushtng effect of the floor s.ht.'>3thing. S1nce lhere are wood-t~wood bearing
conchtions,the-defom'lauon adjustment factor (Table 1-9) is I. 75. NOS does not have a crushtng wJue for
the \\'OOd structural panel floor sheathing. and it lS assumed to be higher than for sawn lumber. As a way
of account.ittg tbr thts crushing effect of the floor sheathmg. a deformatton adjl.ISU1"'em fbctor of2.5 wdl be
used in heu of the 1.75 factor,producinga C01npoundmgeffect of2 x2.5 = 5.0 tunes the \1alues computed
in Eq. 1.0, Eq. 2 .0, or Eq. 3.0.
C rushiog Efft(IS of Uplift Boundar')' Mtmbtr!l
Differentl3J sue.ngth uplift forces fOr the boundary chords transfer the slory upl lft forces to the metal
be.artng plate at the floor abo\1e {Figure 1-17); however, these dttTereniial uplift forces are less than lhe
cumulatwe strenglh compression downward forc-es. Since the crusJung effects have already been considered
for the tugher downward fOrces, there IS no need to consJder the lesser crushang effecL(i oflhe upltft forces.
Table 1- 10. De1ermint Sill piau C'rU.fhing tJIIine C
ASD
Stre.ngth
Total
Demand
(kaps)
Area
(in')
Fa
(ksi)
0.73F;.
Posts
Demand
(kips)
(k-<i)
Crush
(in)
Roof
Four3x4
5.58
7.00
35.0
0.2 17
0.456
0.048
4th Floor
Four3x4
10.78
14.69
35.0
0.420
0.456
0.092
Jrd Floor
Four4x4
17.92
24.8 1
49.0
0.506
0.456
0.130
2nd Floor
Four4x6
26.63
37.57
77.0
0.488
0.456
0. 11 9
Chord
Level
Notes tor Table 1-10:
I. ASD de.mand and strength (LRFO) demand ,,aJues are obtained fi'om Table I-4.
2. Allowable compresston perpendicular lO gratn \'alues FtJ. are-not allowed lObe ancreased for the dutauort
of the load~ therefore, F;. = FtJ..
For Douglas Ftr-larch: FtJ.. = 625 psi
For He.m-Fir: Fu = 405 psi (a 35% red~Xtion from Douglas Far).
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Design Example 1 • Four-Story~ Light-Frame StTucture
For both Douglas FIf and liem·FIf, the allov.'able compression perpendicular to the gram IS the
same for the various grades within the species.
For Southern Pine, the aJiowable compression perpendicular to lhe gratn var1es by grade Within
the species from 480 to 660 ps1.
For lam1nated veneer lumber (LVL). the values vary between 1nanuJ'acturers and ha\'e a value for
allowable oompressiOE\pe;rpe,ndEcular to lhe gram of approxunately 480 psi.
For lanunated strand lumber (LSL). the values vary between manutacrurers Md have a ''alue for
allowable oompress1on perpend1cular to lhe grain of approxtmately 435 ps1.
3. Crushing value ranges from 0.00 to 0.02 1nch when/d ranges from 0.0 ps1 to 0.73F; and ranges from
0.02 to 0.04 mch when);.J. ranges from 0.73F;. to F; .Values are Interpolated lO obmm lhe crushing
values IL<te<l (crush).
4. Crushing values have been mutuplied by 2 x 2.5 = 5.0.
Table 1-11.
Det~rmi11e
INuring plate ,·ruthmg 01/me C
Bearing
Plate
Load
(kipS)
Strength
Bearing
Load
(kipS)
!<J.
(ksi)
0 .73F;.
(ksi)
Crush
(in')
Roof
2.878
4.111
15.98
0.257
0.456
0.011
4lh Floor
3.684
5.262
9.98
0.527
0.456
0.028
3rd Floor
5.367
7.667
15.6 1
0.491
0.456
0.024
2nd Floor
6 .416
9.166
15.39
0.595
0.456
0.037
ASD
Bearing
level
A...
(1n)
Nmes tbr Table I· I I:
I. ASO beanng lood \1alues are obta1ned from the dd'teremjalloads ofTable 1-6.
2. Strength (LRFD) beanng loads are obtained by dividing ASD be<lfing loads by the conversion fae!Ol' of
0.7.
F;.
3. The allowable
may be exc.eded; however, th1s design example uses strength (LRFD) loads where
lhe bearing resistance is
4. ASO bearing-plate capacities and bearing are from Table 1-8.
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Design Example 1 • Four-Story VKiod Ught Frame Structure
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4.4F DETERMINE TIE-DOWN ASSEMBLY DISPLACEMENT
Table 1-11. Dettrmme tie-down a.ssemhl): displacements at /Jne C
Seattng
Increment
le\·el
Take-Up Device
lnttial A_, &
Strength
(Venical
Bearulg-
Destgn
Total
Rod
Displacement
t.,)
Chord
Plate
Displacement.
Crushing
(in)
(in)
Crushmg
(tn)
De-ftectJon.
t.,
Elong.
(tn)
(in)
d.
(tn)
Roof
0.061
0 .031
0.048
0.011
0 .030
0.181
4th Floor
0.159
0 .031
0.092
O.o28
0 .030
0.340
3rd Floor
0.141
0 .031
0. 130
0.024
0 .030
0.356
2nd Floor
0.166
0 .031
0. 119
0 .037
0 .030
0.382
N01es for Table 1- 12:
I. Dtsplaeement vatoes are taken out to three figures 10 avotd oowanted compoundmg of rounchng up for
each term.
2. Rod elongation values are obtalned from Table I-7 and factored up by 1.4 for suength-level (LRFO)
elongaoons.
3. Seating increment values (venical dJsplaceme.nt) ate obtauled from Table 1- 1, for most devices. this
1
1
\ alue IS zero a~ take-up de\ tces att- used aJ'ld mitial (seaung). All> take-up de\•ice deflection is take.n into
account as well as the device destgn deflecuon. ~_,.In lhe de\•tce ddltc:tton value. These \1alues should
be obtruned from the manufacturer's e\'aluation report evaluated to the ICC- ES AC316 cmeria F« this
desagn example, a vaJue of Yl! tnch l'i used, recogntzang that the devtce used wtll ha\•e to travel a dtstance
before the device get.~ to the next: groove in the dev1ce to readjusc
4. Chord c.rushing (crusb) values are obta1ned from Table 1- 10.
5. Be-aring-plate (c-rush) values ate obtatned from Table I- ll.
6. Without shnnkage co.npe.nsalOrs (Table 1-1). the ue-down assembly displacements are
accumulatl\<e from ft()()(-lo-tloor level.
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Design Example 1 • Four-Story~ Light-Frame StTucture
4 .5 SEGMENTED SHEAR WALL DEFLECTION
SDPWS
A considerable amount of monotomc and eyeIic teSting has been done (and conunues to be done) on
canulever wood structural pane.l shear '"ails 1n the Last ""'o decades. Tests results and tesrjng protocols used
for the testing can be found 1n the references l1sLed m Commentary Section C4.3 oflhe SOPWS. To date,
there has been ''el')' l1mited testing on mutustOI)' shear walls and therefore there is not a sigmficant amount
of intOrm.adon on the defleeuon chatacterisuc of multistory shear walls, specifically on the contribution
of the tloor ttanung on the wall deftecuon and whether to model the. wall caJculat.ons on a structure that
is assumed to have-signtficant rotation a1 each floor level (no1 perfectly ngid). have no (or insignificant)
rOlation at each level (has fixity), or IS somewhere tn between the two. Cenainly, the cype and de.tailing
of framing at the floor and the mag.mudes of spans of the floor clements would be expected to affect the
floor's conmbutton to lhe wall fixity.
Past phystcal tesung of sheat willis and current state of practice tend.~ 10 tgnore the fin.ish matenals and
nonstructural walls' contrtbutton to srjffiless~ however, Iimited whole building shake table testjng on wood
and cold-formed stei'llight-frame sttuc.tures confit•m.s that the finjsh materials and other nonslfUCtuml
elements can conmbute stgruficandy 10 the deftecuons and can have 3 mea~urable eft'ect (bmh p<)$1ll\'e and
negative) on lhe overall lateral perfOrmance of multtstory butldangs..
Curre.ntly the,re ate t\\'0 pnmary methods used to determine shear waJI deftecuons tn mtd·rtse v.'OOd·
ti'ame structures. One method is to asswne fixny only at the base of the building and consider the shear
wall to canule"er up from the base wtth no fixuy a1 e.ach floor. Essenually c.onstdenng the wall as one
tall canttlever element with applied lateJaJ loads al each ftoor. The second me.thod lS to assume that the
wall elements have fi.:<ity at each floor leYel and canulever up 10 the floor leve-l direc.tly above, and thereby
detemune the wall deftecuon on a stOI')'-by-story basis. This design example uses the second method~ which
ts to assume fixtty at the bottom of each floor level.
A well-known e.'(pression for determantng (segmented) shear wall deflecuon using four sources of
deflecliOt\ ts tn IBC Section 2305.3; however~ the 20 18 IBC now requites only the four-term equation for
5tllpl~d shear walls. For determinu1g the calculated deflection for a nailed shear wall. the SDPWS defines a
three-term equattOn for calculattng shear wall de-flecuon. as shown here.
8t·h;
1·h
EAb
JOOOG.
S= - - +- - - +d -
h
Eq 4 .3-1
• b<ff
This equation ts a sunpli:fied form oflhe tOur-term equatiOf'l~ whtch adds lhe e.ftet'lS of four difthent sources
c.ontnbuung to the deflection: the canulever (beam) ben£hng of the \'erttcal wall elernent, 1nclucltng the
contflbution of the boundary members-; the shear deformation of the wood s-uucrural panels; the bendtng
and slip of the fasreners; and the additional detlection due to the anchorage (tte-dO\m) deformation. The
ortgtnaJ fc)(lt-term shear \\o'all de-flec.tion formuJa ts sho,-.n here.
g,·h 3 ,.h
h
S= - - + - +0.7:>/Je +d EAb
Gt
"
EqC4.3.2-J
• b<l
The simpldioo expression using "'""' terms (Eq 4.3- I) combines the second and lh~rd terms of me foor.
tenn equation into one le-nn. Computed deflection-; by using either the tOur-term equatiOil or the three-tenn
equarjon produce nearly Identical resuJt~ at the crtrtc.al stte.ngth leveJ ( 1.4 tunes the allo\\'3ble shear values
for setsmtc). Thus, euher eqwujon may be used for c:ompuung the deflection of 3 she.ar wall.
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For determmauon of shear wall deflecttons with sheathmg on lhe enure wall. see Sect100S 5.4 and 5.5.
Although Equation 4.3 1 is easier to use. the deflec.uons computed w1JI be larger than the actual deflecuons
since the apparent sh<M wall suflness (G.,) listed in Tables 4.3A, 4.38, 4.3C, and 4.30 os based on !he sh<M
in the wall be1ng at 1ts capacny for lhe gtven na1Hng.
4
For the calculated shear-wall deflection to be more accurate. the computaiton.~ fot shear \\'all deflections m
lh1s des1gn e."'mple will use the four-tenn equauon (SDPWS Eq. C4.3.2- l ).
Htigbt of lbt Sbuthtd Wall (lr)
See Section 4.2 for s.hear-wa11cumulative ovenuming fotc.es for determination ofhe-tght.
For a shealhed wall lhm does not have a nbbon board at the floor hne but ha.o:o lop plates belov.• the floor
(or modtfied platfonn framed, see Figure 1-1 9), the height of the wall is the distance between the
Slll plate and !he top plate(s) below the nbbon board (story height).
sheath~ng
For a sheathed walllhat has a nbbon board at the floor hne watha top plate(s) below lhe rtbbon board
(plarform framed), and !he sheathing is not placed at 1h< nbbon board (see F1gure 1-20), the he1g)ll of the
wan ts oot lhe story height but the distanee between the sill plate and the top plate(s) be.low lhe rtbbc)Jl
board.
·rr~- !
A.
6 ~/
'b
/
•.
'
•
'!
(
1111
111111
11
11111111
,'
..".,
'
lr
•
''
~
'·,
II
;
- /i
~
I I II I II I I II I II
'
'
i
II
II
II
II
II
II
II
II
II
II
Ftgw-e 1-19. Shear wall height- modiflt!d ba/loo,froming
Note-:
For a sectio11 cut at the floor hne at Ftgure 1-1 9, see Ftgure 1· 11.
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Design Example 1 • Four-Story~ Light-Frame StTucture
~~.
.·,\.
'
h ·'
1).
''
'
'
II I II
.'
~
'
\,·
.....
'
\
I
'
'
II
II
II
II
II
II
II
II
II
II
II
II
II
II
II
h
II
I
F,
..L
II
'
/I
!
II
'
'
II
II
II
Figwi! 1-20. Shear wall height- plaifonnframlng
Nme:
For a section cut at the floor line in Figure 1-20. see Figure 1- 13.
fJa.sttntr Slip/Nail Dtformation Valuts (r..)
The two basic equations for fastener sl1p from SDPWS Table C4.2.2D for IOdc:ommon nails used tn lh1s
e.xample are shown here.
Whett nails are driven. into green lwnbe:r: e.= ( V,.f977) 1.J9A
When. nails are drtve.n tn.to dry lumbe.r: ~.. =(I ~1769)J.!?6
TC4.2.2D
TC4.2.2D
where
v.. = fastener load in pounds pe:r fa.(itener
e.,= t3stener sltp tn tnches
Values off'asterer sltp from the above formulas are based on the performance with StnJ(:t:urol-1sheathang
and must be increased by 20 percent for other sheathing types.
Many engineers have expressed concern that lf the corurac.tor 1nstaHs the nails at a dttferen.t spacang (too
many or too few). the ragidities wtll be sigmficantly dtfferent from those calculated.. However. based on. past
tesung. a n.omtnaJ changing of the nail spacing ln a given \\nil does not appear to significantly change the
suffness.
For decades.. the Uniform Buildmg Code, lnt~rnatJt>,Ptll Buildmg COlk, APA docurnen.L<~., text books. earlier
ed1UOJ\S of the SlnJCtlutJIISI!lslfuC Desig" Mtmual, and earlier editions ofSDPWS used a nooltlon. of d.. for
the Lie-down assembly displacement. In the 2008 SDPWS~ notation c.hanged to tJ.#. llus design e.xampfe
uses both terms as most eng1neers wtll take a whi1e lO change what ha~ become tngrained.
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Design Example 1 • Four-Story VKiod Ught Frame Structure
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Conlinuou.s Tit-Down A~stmbly Displattmtul
11te continuous tie-down assembly displacement (d4 ) tS a colle-cuve accumulauon of the deformation of
ue-down elements. Each of these elemems within the assembly deforms. elongates. and/o( compresses. See
Figure I-21A for the efifct of the ue~own assembly displacement on lhe shear waH deflection.
The 2018 JBC and the 2015 SDPWS defined, as follows:
d4 =Total ven1cal elongation of wall anchorage system (such as fastener slip. device elongation.
rod elongation) at the mduced unn shear in the wall (l·)
It lS uncertatn whether lhe d4 factor is mtended to tnclude y.'OO<J floor venical deformation and crush1n.g
due to shear wall rotation. as lhe rode JS not spec1fic. Ba.~d on test resultsofwood-fmme shear walls. the
crushing. effect of\\'OOd at the panel ends can be sign{/ictmt. ThiS design example Includes \'emcal wood
deflection and c.rush1ng in the d4 factor. The inclusion of ve:ntcal wood defonnatJon and crushing 10 the d4
factor can be signdi:cant especiaHy for higher aspect ratio shear walls, while less sjgn1ficant for lower aspect
ratio she.ar walls4
Some eng1neets Will argue that only the ma.xamum value of e-longauon at the tension side (ue~own
a.~mbly dis-placement) versus-the max1mum vaJue of the wood deflection/crushing on tJte compress1on
s1de a~eed be used. A~ the shear wall rotates from the lnduced shear, the tensaon s1de moves up and the
compression side mO\'es down, producang an addiu,•e effect on the shear wall rotation (d..).
11te net e.ffecl oflhe ue-down assembly dtsplacement plus the wood deflecuonfcrushmg IS an addJttonal
roomon of the shear wall, as a ngad body, with the addnionallaxeral diSplacement at the top of the wall (.6}
equal to the aspect ratio (hid) of the \1/aU multiplied by lhe calculated ue..down assembly dio;placement (d4 )
projec1ed 10 !he o<.nsideedge of me wall. See F1gures l -ISA and 1-21.
h
~ =da-­
bt/1
F
......
-
I
I
I
I
I
'
'
I
I
!
I
I
Tic-down
displacemenl. d,.
L
II ; '
I
i'
\
..
-
I
I"
'
'
'·
l ~
Figure 1-21. PrOJtCtd.
NmestOrf&gure 1-21:
h = he1g.ht of shear wall
b =the out-to-out dimension of the shear wall
b11 =the distance between centtoid of tension hold-down and cenuoad of boundary element
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Design Example 1 • Four-Story~ Light-Frame StTucture
M.uimum Compultd d. Ttrm
Over the last I0 years, several mumdpal JUrtsdicuons adopted pol1c1es luruHng the allO\\o<lble conmbutton
of the Lie-down displacement to lhe overall shear \vall lateral drift. Ahhough siJnilar 10 magnnude, 1n
many cases lhe hmHaUOI'lS adopted val'led. In order to provide for a more untform approach and to prov1de
consistency for both the engtneets des1gning the systems and the manufacturers,ICC-ES modified AC3 16
to tncludea ma.xtmum design hmit of0.2 i.nch for the calculated vert1cal deformation oflhe continuous
ue-down system, Including the shrmkage compensaung devices., when restsung Lateral overruming forces
(ICC-ES 2012c). This 0.2-1nch ltmittS defined at ASD forc<S. The ltnlll is for each stooy or between
resmunt devices, regardless of the wall aspect ratiO. Nw.•caJculations must be submuted that confirm that
the total t1e-down assembly (ventcal) displacement (rod elongation, connector dt\•ice elongation. wood
crush1ng. etc.) 1s not e.'Cceeded. It ts noted lhat the reqUJrement IS for ASD fOC'ces even tJ~ough the eng.u'leers•
deJieclJon calculatJOns need to be based on strength-level fOrces when check1ng building drtfts.
AC316 aliO\\'S the 0.2-inch limit to be exceeded pro,•lded lhat the total shear waJI story dr1ft is tn
compliance with the requirements of the buildtng code. Note that W1th regard to added lateroJ drift,lhe 0.2·
tnch Hm1t may be very conservauve for low as-pect ratio shear walls. but may be u.nc-anset\•ative tor narrow,
h1gh aspect ratio walls. For example, fot a shear waJI With a 2-t~l (he•ght-to-length) aspect rauo, 0.2 inch
of vertical displacement at ASO le,•el will result in an addiuonal 2.3 tnches of lateral drift when converted
to strength level and calculated with the C11 factor. For a 10-foot story hetght., this resuJts 1n 1.9 percent
adchtional drift (not acc.ounttng for n<lll slip, wa11bending. and shear defOnnation).. where the code level
dnft tOr 5-story wood construCtion ts hm1ted to 2 pe.rcem.
Tables 1-1 5, 1-20, 1-24, and 1-28 liSt the total ue-do\\-n assembly displacement~. (d.) using strenglh-lel•el
forces. The ASD-Ievel diSplacements are not hsted but can be apprmamated by multtply1ng the values by
0.7. Many of the values exceed the hmu of0.2 1nches; however. as allowed m the AC316 cmena, drift
checks of the shear walls at the roofle.•el are hsted in Table 1-1 9.
F
.....
•
••'
I
•I
1\.
I
I
I
I
I
Tic-down
displacement. tl.
L
l
'
!
•·,,
,.
E
"
•
• '
' !
\
b
•.
~
Figure 1-1/A. Effect ofd. on drift
Note:
h =height ofshear wall
b = the out-to-out dimension of the shear wall
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Design Example 1 • Four-Story VKiod Ught Frame Structure
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Stiffntss or Plywood
l 'S.
Orluttd Strand Board (OSB)
For shear walls with Y..'OOd structuml panels. there can be a significant d1tfe-rence in the shear wall
deflection. depending on which type of strucrural use paoels are used tOr tl'le construction of the shear wall.
The difference c.omes from t11e apparent shear wall suffness vaJue (GII'). This apparent shear wall stJft'hess lS
a combination of nall sl1p and panel shear deformation. Most e-ng&neers are not aware of the d1frerence and
can mislakenly allow or approve a change to lhe different wood structural pane-ls Without knowledge of the
true 1mpacL The mtio oflhe apparent shear wall stiflhess value vane~ depend1ng on the fastener spacing,
where the closer the fustener spac-i~ the greater the d1fference. ln addmon to shear walls, the d1tltrence
in G11 bemreen plywood and OSB can have a measurable effect on diaphragm deflection. llus could
significantly aftfct the expecred perfOrmance of canuleve,r or ngid diaphragm..-..
The followmg IS a comparison of apparent shear wall sufthess values tbr 15k -1nch Suucturall wood
sEnJctural pane-ls us1ng IOd common nails:
G.
Faste,oer Spacmg
PLY
OSB
RauoOSB/PLY
6"
16
22
1.4
4"
20
29
l.S
3"
22
36
1.6
2"
28
51
1.8
For other thicknesses of sheathing applied d!fectJy to stlJd framutg.. different fas£e-ne,r sizes. and so on, refer
10 SOPWS Table
E~ample
4.3A.
using three-tenn equation:
SOPWS Eq 4.3-1
where
r=750plf
"= 9.0
ft
£=1.7xl06
2
A-= 5.26 X 2-= 10.52 1n
b= 8.0 ft
d4 -=0.188m
Sheath1ng thickness-= •Yc 1n
Fastenets = IOd common nails at 2-tn spacmg
G.=28 PLY
G.=Sl OSB
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Design Example 1 • Four-Story~ Light-Frame StTucture
USing Strucmrall plywood she:nhing:
8=
1
8x750x(9.0)
+750x9.0+ 0.IS 8 9.0= 0 .48 "'
1.7x 10' x J0.52x8.0 1000x28
8.0
Using Structural I OSB sheathing:
8=
1
8x750x(9.0)
+ 750x9.0 + 0. 188 2£=0.JJin
1.7x l0 6 xl0.52x8.0 1000x5 1
8.0
Using JOd OOx nads would result In a 19 percent reducuon tn allo\vable-load fordlaphragms and shear
walls compared to IOd co1nmon nails. Using 8d box nails would resuh 1n a 23 percent re<l:uc:tton in
allov.-able load for dlaphragms and shear walls compared to 8d common nails. This is based on comJXlflng
aiiO\Wble shear values l1sted tn Table 12N m the NOS-1 5 for %-tnch side member thickness.t., and Ooug.la.~
F•r-Larch frammg. In addtuon to the reducuon of the shear \\'all and diaphragm capacities. where box nails
are used,lhe walls wtll also dritl more than where common nails are used. Alternatively. the engtneer can
consider ustng nail and shealh1ng lhjcknesses not listed m the JBC or SDPWS by ustng lhe va1ues liSted 111
ICC-ES repon ESR-1539, available from the International Staple, Nail, and Tool Assocoatton (ISANTA).
A conmbutor to lhe problem IS lhat whe.n c.ontractors buy large quan11ues of nru.ls (for nail guns). lhe words
box or common do not appear on the canon label. Nail length and diamete.rs are the most common lisung on
the labels. Thus. n is ~.-r.tremtly imporumt to liSt the requtred nail lengths and diameters (nQI J"Sf s.per•f)'ing
Sd c~mmon or IOd comn~on natls> on the structural drawmgs for all diaphragms and shear walls.
5. Mechanics of Multistory Shear Walls with Force Transfer around
Openings
5.1 DESIGN OF WALL FRAME WITH FORCE TRANSFER AROUND OPENINGS
SDPWS
Section 4.3.5 of the SOPWS liSL(i the three basics types of shear walls: segmented method, force-transfer,
and perlOrated. The followang method iUusuates force-uansfer shear walls, which is more commonly
referred to 1n the 1ndusuy as fOrce ll'3nsfe-r arouod opentng~ or FfAO. Sect.ioo 4.3.5.2 states that this type
ofdesign shaH be based on ..a rational analysJs," leaving the type of analysis to be chosen by the huildmg
engtneer/enginee-r of record. There are basically four types of analysis techmques that are g_e.nerally
accepted as rational analysts: the drag strut, the cantdeve.r beam, the ..Diekmann method.'" and the SEAOC/
Thompson melhod. The engineer as encouraged to refer to the SEAOC Btu~ Bcok antele ""Wood-framed
Shear Walls walh Opening_.~...
Unul rece11t1y, there has not been adequate testjng of these types of shear walls to JUStify the rational
analysis methods available, but APA and the Forest Products laboratcuy have now tested full-size walls
W1th open1ngs. The tesu11g focused on determmang lhe horizontal tensaon strap forces at the opening
corners. One of the conclusions m the testing repon ts that etther the Diekmann method or lhe SEAOC/
Tho1npson method most accurately estimates the actual tension suap force.~ ln the shear waJJ~ while
the other two me.thods can either produce overly nonconsen•ative or overly conservauve tensaon str~
force esumates. The-method dlustrated ln this design example. lhe SEAOCfThompson me-t~ uses
the Otekmann method with "'anauons. Many englneets wall arburarlly add lie-downs al the window
jamb members (Fagure 1-40). However. wnh lhjs type ofdesjgn. the ue-ckwms at these locations are not
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Design Example 1 • Four-Story VKiod Ught Frame Structure
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necessary and may result 1n higher shear stresses atxwe and below the wtndow. However. addtng tie-downs
at the window jambs Will ine-rease the wall frame perfOrmance and help prevem sill plate upltft at the
window jambs,. which occurs (to some degree) when they are nm prov1ded.
It lS poss1ble to get the m&staken tmpresston from SOPWS F1gure 4E that all a des.:tgner needs to do to
reduce the hlw ratio IS add some blocktng and straps. This design example has a structure with 9-foot plate
he1ghts, whtch makes ustng a wall frame feasib1e. HO\\'e\•er, when the plate height is 8 feet, there are chotd
development and panel nailing capacity problems. Most often. the wall shears above and below the opening
wdl be higher than m the wall pters. These destgn examples analyze the wall frame and neglect gravity
loads. Although correct from a techntcal sta.ndpotnt, some engtneers wdl argue that vet1tcalloads need
to be constdered when dete-rmJnutg wall shears. The standard practice of neglecung gravtty loads when
oonsidertng wall shears is deemed appropru.lte.
11tree-damension.al fintte element analystS compute-r programs are nov.• bemg used to model structural-use
panels on honzontal dulphragms and lateral-resisung elements. These programs model the strOCwral-use
pane-ls as eonunuous without panel spltces. Depending on the-si2e and location of the panel spltces, the
oomp..ner model may not accurately model the field condit1011S. Rece-nt fuJI-scale tests ha\•e sho\vrt that
even locating a paneJ splice at the window corners versus away from the Window comers (cut comers) can
prodoce dlll'erent test resultS.
The 2012 and earlier editions of SEAOC Srn~t:turuUSeismic Design Manual Volume 2. Design Example
I illustrated a me.thod where the Inflection pointo; 1n tl1e \\o'all piers as well as the wall segments above and
below the window openmg.~ were a~umed to be at the mtd·potrtt of the openmg. The methods illustrated
in this editton of the Design Afanual proportion the lnflteLJon potnt locauon based on walllenglhs of
the adjacent waJI segments. All of the full-scale testtng to date of shear walls with force transfer around
ope.nings has yet to deternune where the inflecuon potnts occur, thus. these methods of analysis are
cott..c;tde.red approximate and within lhe intent of providing a rational analysis. Proponioning the Inflection
po!rtt locauon.~ produces untforrn shears at the wa11 p1ers and Yo>all see__rnents abo\'e and below the opemngs.
To some degree, the 1nftectjon polrtts should be based on suft'h~ which would tend to attract more Jood
to the wall segment above the ope:rt1ng than the segment below the opening due to the header beam that is
above the opening.. Likt\\~se. the full-scale tesung to date h.as not determined whether the shear stresses in
wall pters is proponionaJ to wall le.ngths or based on the sufthesses of the wall pters.
Due to the indete-rmJrtate nature of calculaung inflection points and wall suffn~ the current pmcttce fot
force transfer around opemngs is to d1stnbute lhe forces to the wall piers based on \'t>aJI Ieng.th as opposed to
waJ I stiffness.
Fig.ure 1-22 illustrates a genertc wall frame with four wall pie.rs. Where there are more than three \\'311 piers,.
wall pier ..U" (the waJJ p1er between two adjacent openings) repeats as 1\eC'essary. WheJe there are two
wall piers, wan pier ..L2" does not exist. 1'he wall frame used 1n this design example has more wall area
Md less window are.a than tnany projects and has been chosen to illustrate force ttaJt.'\fet around openings
m large muJtistory walls. Budding struc.rures that have shd1ng glass doors to balconies ot \\'all· mounted a•rcondtuoner un.its below the wlrtdow~ such as those tn hotels. usually cannot uuhze fot"Ce transfer around
ope:rt~ngs bocau.~ie there IS not enough wall length above and beiO\\' the openings (dimen.<tion ..HI" and
..H2'"). In these cases. a segmented shear waJI with conunuous ue-downs may be necessary.
A more g.lobal approach for whether a wall face should be destgned wtth the segmented method or
the tOrce-uansfe-r method should be based on capac:nies. In other words. 1fthe \'-'311 shear forces are
significantly h1gher abo,•e and below tlte opemngs. then tl1e use of force tranSfer may not be the appropriate
cltot-ce.
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Design Example 1 • Four-Story~ Light-Frame StTucture
5.2 THOMPSON METHOD
A new method of force transfer IS tllustr.ned in thts destgn example and is dtfferent than the SEAOCI
Thompson method Illustrated tn pre\•tous edttions. This method first appeared in the 2012 SSDM editton
and has been Illustrated tn more detail tn the 20 15 and 20 18 edhionsand will be referred to as the
Thompson method.
Due to the relatively uniform. wall le-ngths on the extertor waH~ thJS destgn example uses a uniform
dtsutbution of wall shear fotces in the wall frame. This me.thod may not be appropnate on other buildtng
\vaJI fac~ and tts use will be up to the engineenngjudgment of the building designer/eng.tneer of recotd.
The lbompson method assumes the followiJ'I.g:
Tile untt shear at the sides of the opentng Lli untform.
ll1e locauon of lhe inflection points above and below the openings lS propontonal to the wall
pter lengths adjacet~tto that specific: opemng..
The locatton of the tnflection poinlS at the s1des of the window opentngs is proporuonal to the
\\'all lengths above and below lhat spectfic: openang..
l1le verttcal shear forces above and below t11e open1ng are determu-.ed by stattc: free body
equiJtbnwn of the free- body of that wall pier element.
Otttrmiat Shur Fortts around Optn.ings for lht Roor Lt"tl
Uslng stauc:~ detenmne the shears and forces m each free body panel.
Figure 1-22 IS genenc in nature Md deptets three windoo•openings. For this design example. tl'le shear
wall des1gn for the piers between grid lines B at'ld D Wtll be ilii.ISlrated where lhe wall frarne continues for
several opentngs. maktng 1he r1ght stde of the figure Look ltke the center of the figure.
Laleral (strength le\•el) fOrces to y,'alJ l1ne I (see Table 1-14):
= II ,223 lb
= 23,893 lb
Thtrd. ftoor le,.el: Fw = 32,340 lb
Rooflevel: Fw
Fourth-ftoor 10\•el: Fw
Secon<l. ftoor level: F~ = 36,563 lb
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Design Example 1 • Four-Story VKiod Ught Frame Structure
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I(
::t:
"'
t
La
Lb
Lc
Ld
20.69'
15.4'
13.5'
Hvsa
1l V4a
-
F~
a : b
- E--
VI
:...<'1 0
1i
V4::::.. ..=:"'
[X [X [X
V2
V3
-
«:
"!::t:
-<i
00
1ivv6a
e ; f
c :d
4444.4444
'
'
tv-~
1J~v~~
.
)I
1 l~V(
-
-;r.~
-.....=:
~
N
..,
::t:
b
'
Ll
8.0'
L5
9.5 '
(
L2
12.75 '
Figurt! 1-12.
L6
L3
4.0'
I 1.5'
lf'tll/forc~
L?
4.0'
L4
11.5'
determmation
Firsl: Otttrmint \\'all dilllt-osious ror the \\all framt.
Ll = 8.0
f~
L2 = 12.75 ft, L3 = 11.5 ft, lA = 11.5
f~
L5 = 9.5f~ L6 = 4.0 ft, L7 = 4.0 ft
Total waiJ length tncludingopemngs on grid hnes 1and 4 (see Fagure 1 1):
4
L=96.0 ft
Total wall length tOr waJI ptets on gnd lines 1 and 4:
! L=8.0+ 12.75 + 11.5+ 11.5 + 12.75+8.0:64.5 ft
Total length otwall piers shown an Figure 1-16:
! L = 8.0 + 12.75 + 11.5 + 11.5 =43.75 ft
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Total shear to \\o'all p1ers sho\\'11 1n F1gure 1-22 IS propontonal co the torol wall length tn lhe same \\':111 hne:
r.L
To<al IL
=(~)11,2l31b=76121b
64.5
H= 8.21 II, HI =2.54 II, f/2 =4.0 ll,l/3 = 1.67 ft
a=(- L_I _ )L4=(
S.O
)9.5=l.66ft
Ll + L2
8.0+12.75
b= U - a= 9.5 - 3.66= 5.84 ft
<=(_g_)L5=(
12 75
·
)4.0 =2.10 a
12.75+11.5
d=L5 - c=4.0 - 2.1
=1.90ft
r=(___!]_)L6=(
11 5
·
)4.0 =2.0 !
11.5+11.5
L2+L3
Ll+L4
!= L6 - • =4.o - z.o= z.o n
La=LI +o=8.0+3.66= 11.66ft
Lb= b + Ll +C= 5.84 + 12.75 + 2.1 • 20.69 ft
Lc=d+ L3 +t = 1.9+ 11.5 +2.0 = 15.4 ft
Ld= L4 +f= 11.5 +2.o = 13.5 n
Check !he sum of !he lengths:
11.66+20.69 + 15.4 + 13.5= 61.25 no:>cllec.ts
Oeterm1ne the inflection pomcs althe Sides of the opc mn~ :
3
Ha= ( H I: H 3
)f/ =(2_~:~.67 )8.21 = 3.26
ft
Hb=H - Ha=8.21 - 3.26= 4.95 ft
ha= Ha - 113 = 3.26 - 1.67= 1.5911
hb=fl2 - ha=4 - 1.59=2.41 ft
Label the upper and IOI>tr free-body poruons or the woll
66
r...,., ($« flgutt 1-23).
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Design Example 1 • Four-Story VKiod Ught Frame Structure
4
20.69'
15.4'
....-----
Flgure I-13.
J17JI/fran~ .showing UJ~r and IUII'er fru-body portiou.t
St(Ond: Fi.ad rortt-S attlng on wall pius and shur i.a-waU pitrs.
Shear forces lO the v.oall pl<!'rs are propomoned based on the length of the pter 1n relauon lO the lomllength
oftl1e piers (Figure 1-22).
11
=[~ )v =[~05 } 1,223= 1392 lb
1'1 1392
.
••l=-=-- =174pll
Ll
8.0
1'2
l
=[~ )v= ~ 7:)11,223=221S ib
1'2 2218
••2 = - = - - = I 74 plf
L2 12.75
1'3=[~}-=[~:~} 1,223=2001 lb
V3 2001
1'3=- = - - = 174 plf
LJ
I 1.5
1'4 =
•L4)
lu v = •11.5)
l64.5 11,223= 2001 Jb
··4= ~=~ =174pl(
L4
I 1.5
Check the sum of the shears 1n the piets:
11 + V2 + V3 +I 'I= 1392 +22 18 +2001 +2001 = 76121bs=>ch•ck.•
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Design Example 1 •
Four-Story~ Light-Frame StTucture
Third: Find tht \'trtital shtar fortt:s arting abovr and brlow tht optnings and tht unit shur fortt:S
(s.. Figu,.. 1-24).
11.66'
Vl = l392#
G) ~V4
VI::::..
-N
---------
~
'
00
V(
G) ~V4b
Vl = l392#
Figut? 1-24. J'l>rtica/ .sh~llr forct.s acting abm~ turd below opemng
1'4a+ V 4b= l'lxH _ 1392x8.21_ 9801b
La
I 1.66
1'4a=(V4a+V4b)(
v4a =
67
3
H
)=980x(
1.
)=3891b
H I+ H3
2.54 + 1.67
!:::!!!
= ~ = 233 plf
H3
1.67
V4b=(l'4a+JI4b)(
HI
HI+ H3
)=980x(
2 54
·
2.54 + 1.67
)=5921b
t•4b =.!::!!!. = ~ = 233 plf
HI
2.54
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Design Example 1 • Four-Story VKiod Ught Frame Structure
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Fourth: Otttnnlnt shtar tnd tit fortt:S (stt- Figure 1-lS).
Fr<t body Ill:
I
(
~9#
11.66 ft
853#
.....::
..., ,....-
~
)I
0
1392#
~389# 1.67ft
3.26 ft
,
~
0
-'--
..
,,
Tie Force
~
ft
1.5 9ft
3.66 ft
a
Flgmv 1-15. Frett-bodyd1agram U l
Determ1ne honzontal shear fOrce at the topoft11e \WI! above-"a'":
I'<>= V4ax
3 66
(.!!...)
= 389x( · )= 8531b
HJ
1.67
Oetemune horJzonml shear three at upper left oorner:
V: Vl -1&= 1392 - 853:5401b
v
540
·· =-=-=67plf
L i 8.0
Detemune hor1zonml ue force from header to wall p1er V1 :
Tie= V4ax.!!.... = 389x~= 8531b
HJ
1.67
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Design Example 1 •
Four-Story~ Light-Frame StTucture
Fru body LI :
38911
~
8.0 ft
3.66 ft
!}92#
a
-:r-
@
1 81ft
Tie Fo rce
4.95 ft
0 0
~
~592#
~
2.54ft
...
539#
85311
11.66 ft
ll
20211
F•gr•.-. 1-16. F.-.•-bodydwgram Ll
Determine honz.ontal shear force at lhe bonom of lhe wall at ..a":
J& =
3 66
F4bx(~)=
592 x( · ) = 853ib
Hi
2.54
Oetermme honzocual shear force at lhe lower left come.r:
I"= ~1 - il> = 1392 - 853= 539 Jb
v 539
·· = - = - =67pif
L i 8.0
Oetermme hortzontal ue three from wmdow Sill lOwall p1er V1 :
Tte= V4bx~ = 592x~=853 ib
Hi
2.54
Determ1ne venical tte fOrce mlo\\o't"r corner of wall pier J-1:
Tie T1 = V4b - 1'4a= 591- 389=202 Jb
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1392#
1"4o+l'4b•l'Sa+l'5b: nxH- 2218x8JI =880 lb
l.b
20.7
1'4a•(l'4a+ 1'4b{
Hl~lHl )=880x( 2.~·:~.67 )=3491b
J'Sa•(l'4a+l'4b)(
67
Hl )=88ox(
1.
)= 349Jb
HI+ H3
2.54 + 1.67
1' 4a
t•4o • vSt1 • -
H3
349
=-
1.67
-= 209 plf
20H
3A9•J
3.2Sft
4400
..::::: 558 If
...::::: 1220#
0 .. ,_ 8
' ,.
8
nEFORCE
\
..:::::
O...rnune horl.lOflbJ <l..aT force a1lbe top ofthe 1.:1llabo\.. .,~
II> • 1'4a)((.!....) = 349x(~) = 1220 lb
1.67
IC• J'Sa"(..!...)=349x(1.1..)=440
lb
If)
1.67
Detenmnc hon1'..ontal shear ~Orce at upper left 1'2. poc11on:
, •• l'b- · ~·1220 - 440=558 Jb
I'
SS8
.
··=- · --=44 pll
Lb
2 10ft
c
b
H)
T1EFORCE
221U
12.75 ft
584ft
-
4o --,
8
12.75
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221
l 3Au
l
1.8H
59ft
Design Example 1 •
Four-Story~ Light-Frame StTucture
Determine hortZOntal ue force from header to v.'<lll p1er J? (left stde):
Tte= 1' 4ax ~= 349x 5·84 = 1220 lb
H3
1.67
Determine honzomaJ ue fOrce from header to \\'1111 pter • 2 (right side):
Tte = V5a X..£_= 349 X 2J_ = 440 lb
H3
1.67
Fru body diagram ll:
V4b=r5b=(V4a+l"4b)(
2 54
·
HI )=880x(
)=5lllb
HI+ H3
2.54 + 1.67
V ' b 53 1
•·4b = vSb= ~= --= 209 plf
HI 2.54
20.7 fl
t. 2218.
8
8
8
..,
llE FORCE
3.26 ft
...
1-
~.. J
1.
'
'
1220~
1.8 1ft
,# 1 ...
~
ttE FORCE
lr
780#
12.75ft
-5,84 ft
8
-140#
2. 10 tl
c
b
Figun! 1-18. Fll'•-badydwgram LZ
Determine honzomaJ shear force at the bouom of the waH below ..b..,:
4
lb = V4b x(_!_)= 53 1x(; ) = 1220 lb
HI
... >4
!
Oetermtne honz.ontal shear force at ..c..:
I"C = I'Sb
72
X(~)=
531 X(2.2....)= 440 lb
HI
2.54
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.I
...
2.
2.218.
Design Example 1 • Four-Story VKiod Ught Frame Structure
4
Detennme horizontal shear fOrce at lower center I'2 portion:
V= Vb- Vc= 1220-440=7801b
,.
780
.
,, = - = - - = 61 pll
Lb 12.75
Detennane h«tzonml tie force from wu\dow sill to wall pier V2 (left side):
5 84
Tie=V4bx...£...=531x · =12201b
HI
2.54
Detenn1ne honzomal ue force fi'om Wlndow stllto wnU pier V2 (rtght side):
Toe= V5b X_:_= 531 X.!:!.._= 440 lb
HI
2.54
~ ~
G0
G) ~
e
e
G)
e
G) ~ ~ <2:) ~
Detennme strap forces and boundary forces around ope,nings.
Des1gn honzornalue sttaps above and below WLndows.
Detennane the LJe forte for the horizontal strap fbi' window header to wall pier (from Figures 1-25 through
1-28). lie force (strength level) os maxomum 31 tl1e header beam.
F,,= 1220 lb
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Design Example 1 • Four-Story~ Light-Frame StTucture
Determine the ue force fOC' the hor120ntal strap for wtndow sail to wall pie.r (from Figures 1-25 through
1-28). Tie force (suength level) ts max1mum at the wtndow sill.
F.,= 1220 1b
Coosult ICC EvaJuarjon Repons f« the a11owable load capacity of premanufuc.tured straps.
Check manimum penetratiOn for fuJI ruul capacny ut acc«dance wtlh NOS Table 12P Footnote 3.
For IOd nails through 16-gage strop and ~inch sheathing penetration; 3.0 - 0 .060 - 0.5 ;2.4 in.
Requ.red penetration for full value= IOD= 10 x 0.148= 1.5 <2.4 in . .. OK
AJIOYo•able load per IOd common nads wuh 16-gag.e memJ stele plate= 116 Jb
NDST 12P
Z'= 11 61blnali x C0 = 11 6x 1.6= 185 1blnail
For ASD des1gn:
Number of IOd nailS requ.red each end =(1220 x 0.7)1184 =4.6 nalis
Use a conunuou.o;: 16-gage by I Y..-anch sU'ap across me openang head and saJI co blocktng.
Alloo·abfe strap lood is me area umes the allowable suc.'SS:
(1.25)(0.06)(0.6
X
33,000) = 14851b > 1220
X
0.7 = 8541b ... OK
An allowable stress increase has not been used for the metal strap.
5.3 THE DIEKMANN METHOD
Thts method first appeared an the 2018 SEAOC Convention Pmceedtngs and was presemed at mat
convention.
The method assumes the tOIIowang:
The unh shear above and below lhe ope.ung L<;: umform.
The corne.r forces are bao;ed on the shear above and belcm• the openJn_gs and only the piers adjacent
to that specafic ope,ni1tg.
The trtbutary length of the opening ts the basis for c.alculaung the shear to each pier. This trtbutary
le.ngth IS the ratio of the le.ngth oflhe pier multtplied by the length of the openjng it is adjacent
10, divided by the sum oflhe leng<h of !he pier and !he lenglh of !he p ier on !he otller side of !he
opening..
For example, T1 = (Li • L.,)I(Li
+ L,)
The shear an each pier is the toulJ sheardtvKied by the L of the wall. muluplied by the surn of the
lenglh of 1he pier and ns rribmary length, divided by !he lenglh oflhe pier:
(V/L)(L i + Ti)/Li
The.unJI shear of the corner zones lS equal to subuactJng the comer forces from lhe panel
resistance, R. R js equaJ to the shear of the pier multiplied by the pter length:
lfi 1 =(,11L 1 - F 1)/L 1
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Design Example 1 • Four-Story VKiod Ught Frame Structure
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Lateral (sttength level) forces to wall !ones I 3lld 4 (see Table 1-14):
Rooflevel:
F~=
11,2231b
Fourth-floor level: F~ = 23,893 lb
llurd-floor level: F~ = 32,340 lb
Second-floor level: Fw= 36,5631b
Des1gn lhe waJI wnh the first four wall piers on hne I belween grid hnes B and E (see Figures I I and 1-2).
4
First: Determt.ne wall dimensions for lhe waH frame {Ftgure 1-30).
L 1 =8.0 ft, L, = 12.75 ft, 1.1 =I L5 ft, L, = 11.5 ft, 1.0 1 =9.5 ft,L.,=4.0 ft, L.,=4.0 ft
Total waiJ length for wall p1ers and openmgs on g.rtd lines I and 4 (see Figure 1- 1):
I.= 96.0 ft
Total wall length tOrwaJI ptets shown m Figure 1-30:
rL = 8.0 + 12.75 +
n.s + 11.5 = 43.75 ft
The uxal shear to wall ptets shown in F1gure 1-30 IS proponional to the toml \\'alllenglh m the same wall
hne:
r=
rL
To!al H
t
- ·
43 75
· )11,223 1b=76121b
64.5
v
;..,
N
oO
Figure 1-30. Wall e/C~·a.tlon
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Design Example 1 •
Four-Story~ Light-Frame StTucture
Figure 1-31 IS genenc in nature Md deptets three windoo•openings. For this design example. tl'le shear
wall des1gn tOr the piers between gnd hoes Band 0 will be tllustrated where the wall frame COiltinues fot
several open1ngs. maktng lhe nght stde of the figure Look like the center of the figure.
For lenglhs 1n F1gure 1-31:
L 1 =8.0 ft. L,= 12.75 ft, L,= 11.5 1\.L, = 11.5 ft
L,1 = 9.5 ft, L., = 4.0 fl, L., = 4.0 ft
h =8.21 fl,h,= 2.54 ft,h,= 4.0 1\, h,= 1.67 ft
Ftgure- 1-31. Tte--downforr:e.t
Second: Cafc.ulate tie-d0\\'1'1 forces.
H = I"JL :7612lbx 8.21 tl/61.25 ft = 1020 lb
Solve fOf shear above and below openi11gs.:
,., = v, = Hl(h, + h,) = 1020 lb/(1.67 ft + 2.54 ft) = 242 plf
F~rod
total boundal)' forces:
0 1 = ,., x(L.,) = 242 plfx 9.5 ft= 2299lb
0 2 = •·, X (L,2) = 242 plfx 4.0 ft = 968lb
o, = v, x(L,,)=242 plfx 4.0 ft= 9681b
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Thtrd: Oetermtne comer forces.
ll\e cnrner forces are based on the shear above and below lhe ope,n1ngs and only lhe piers adjacent to that
spec1fic opening.
1---::r
Figure / 31. Corner forces
4
Calculate come-r forces:
F 1 = 0 1(L 1)/(L 1 + L2) =22991bx 8.0/(8.0 + 12.75) = 886 1b
F2 =01(l-,)I(L 1 +L2)=2299 1bx 12.75/(8.0+ 12.75)= 1413 1b
F3 = OJL,)I(L, + L,) =968 lbx 12. 75/(12.75 + 11.5) = 509 1b
F, = 0:/.L,)I(L, + L3) = 968 lbx 11.5/(12.75 + 11.5) =~59 1b
F, = O;;{L3 )1(L3 + L,) = 968 lb x 11.5/( 11.5 + 11.5) = 484 lb
F, = O:!,L,)I(L3 + L,) = 968 lb x 11.5/( 11.5 + 11.5) = 484 lb
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Design Example 1 •
Four-Story~ Light-Frame StTucture
Fourth: Calculate trtbutary lengths.
The tnbut.ary length 1S the product of the pter length and the opentng length, dJV1ded by the swn of the pter
lengths on e.ach side of the: opentng..
Figwe 1-33. Tributary lengths
=L, x L.,t(L, + L,l =8.0 x 9.5/(s.o + 12.75) =3.66 n
T =L X L. 1(L + L,) =12.75 X 9.5/(8.0 + 12.75) =5.84 ft
r,
1
1
1
1
T3 = L 1 x L.,I(L1 + L3)= 12.75 x 4.0/(12.75 + 11.5) = 2.10 ft
=L X L.,I(L, + L =11 .5 X 4Jl/( 12.75 + 11.5) =1.90 ft
T, =L X L./(1., + L,) =11.5 X 4.0/( 11.5 + 11.5) =2 .0 fi
T, =L, x L.,I(L + L,) =11.5 x 4.0/( I 1.5 + 11.5) =2 .0 fi
T,
3)
1
3
3
Fifth: D.!lemune un11 shears at sades of or)(•nmgs.
The shear force an each pier is lhe total shear divtded by the length oflhe wall. multiplted by the sum of the
pter length and tts tributary length, divided by the length of !he pier.
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v
Figure 1-34. Umt sh~arsdtten11inauon
1·, =(VILXL, + T1)/L 1
v, =(7612161.25X8.0 +3.66)18.0= 181 plf
V, = ( VILXT, + L, + T1)1L,
v, =(7612161.25X5.84 + 12.75 + 2. 10)1 12.75 = 202 plf
v, =(VILXT, + L, + T,)IL,
v, = (7612161.25X 1.90 + 11.5 + 2.0)111.5 = 166 plf
V, = (VILXT, + L,)IL,
v, =(7612161.25X2.0 + 11.5)111.5 =
146 plf
0
.c:
.,
.c:
Ftgure 1-35. Sumforc-esmpitrs
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Design Example 1 • Four-Story~ Light-Frame StTucture
SL~th:
Check sum of shears in the piers.
V1 X L 1 = I8 I X 8.0 = 1448 lb
v,xL,=202x J2.75=2576Jb
v,xL,= J66x 11.5= 1909 Jb
r,xL,= J46x 11.5= 1679 Jb
Sum forces:
r, + 1; + v, + v, = 1448 + 2576 + 1909 + 1679 = 7612 lb=>checks
Design horizontal ue straps above and below wandows.
Oetermme the ue fOrce fOI' the hon:zontal strap for wtndow header to wall pter (Figure 1-32). Tie force
(strength level) is maxtmum at lhe he.ader beam.
F,..= 1413 Jb
Oetermtne the tte tbrce fOI' the honzornal strap for wmdow stll to wall pter (Ftgure 1-32). Tie force
(strength level) Js maxtmum at the window sill.
F,..= 1413 Jb
Coosult ICC Evaluation Repon.s for the allowable load capac1ty of premanufactured straps. Check
minimum penetration for full nail capacity tn accordance with NOS Table 12P footnOte 3.
For JOd naal through 16-gage strap and ~inch shealhingpeneltalton =3.0 - 0.060 - 0.5 =2.4 tn.
Requ.redpene.tmtion forli.li l value= JOel= 10 •0.148= 1.5 <2.4.n ... OK
AIIO\\•able load per IOd common nail with a 16-gage me-tal side plate= 116 Jb
Z' =I 16 Jblnail x C0 =I 16' 1.6 = 185 Jblna•l
For ASD design:
Number of JOd nruls requ.red each end =(1413 x 0.7)/185 = 5.3 na1ls
lJse a continuous 16-gage by I Y.--tnch suap across lhe opentng head and sill to blocktng.
Allov.·able strap load is lhe area umes the allowable suess:
( 125)(0.06)(0.6
X
33,000) = 1485 Jb > 1413
X
0.7 = 989 Jb . .. OK
An 3110\vable stress t.nc-rease has nm been used fot the metal strap.
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Design Example 1 • Four-Story VKiod Ught Frame Structure
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5.4 COMPARISON OF THOMPSON METHOOWITH DIEKMANN METHOD
ll'l<>mpson method shears ar sides ofope-nings and above and below opening.~:
G
0
0
0
G)
0
®
0
0
0
Figure 1-36. Sumforus in piers-Thompson me1hod
..
L_
F•852 1b
F = 852 1b
F•1220 1b
F = t~ lb
...
r
F•440 1b
F__;.,.440 lb
r
Flgure 1-37. 71eforces- Thomp.t()n me1hod
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Design Example 1 • Four-Story~ Light-Frame StTucture
Die-kmann Method shears at sides of ope1li1lgs and above and bel®• openings:
Figm-e 1-38. Shear fon.·es m s~ment~D1d.mann method
Flgu" 1-39. Shear forc.es in segmentJ-DitkmantJ method
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Design Example 1 • Four-Story VKiod Ught Frame Structure
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5.5 SHEAR WALL DEFLECTION USING WINDOW STRIPS (UNIT STRIP METHOD)
ll\e de-flecuon for a shear \\'all can be approximated by ustng an analysts sim1lar to computmg the stiffness
for a ooncrete wall walh 3.11 opentng 10 it The deflectjon tOt the solid v.'all is oomputed,lhe.n a deftecuon for
a honzomal Wtfldow SU!p IS subuacted.. and the deflectton for the wall pters added back in.
Eng&r-.eeringjudgmem may be used to stmpJif)• thts approxtmation. Htm•ever,tlte method shO\'rl\ in Figure
1-40 is one method to approximate the deflection.
v
L
'
r'l
'"'"!
N
~
c:
..,.
0
o\
1"'-
~
N
T~
-v
I> 3.0'
4.0'
~
..).
o·
'1 ...
c
10.0'
Fi'gu« I -.fO.
Et~·auon
ofwallframt wilh o~nmg
The method lltusunted here neglecL~ any romtion of the wall piers due to any displaceme-nts of the sail plate
below the window jambs. These amounts can be calculated but are extremely lime-consuming..
Ftrst, determtne deftectJon of the enure wall, Without an ope:mng_
&! 4.3- 1
Second, determine deflection of window strtp.
S1nce the boundary e-lements are connected to cominuous posts lhat extend above and below the openang,
the value of a., equals: the sheathtng-naiJ deformauon \•alue calcuJated above (boundary clement chord
e-longation as neglected).
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Design Example 1 • Four-Story~ Light-Frame StTucture
Thls de-ftectjoJt lS ne-gative because u JS subtracted from the sum of the de.tlectJOI'tS, as sl\0\\n later.
Th1rd, determme deftecuoo of wall plers.
Since the boundary elements are connected to continuous postS that extend above and below the opening.,
the value of~~~ equals the sheathang-nall defomwion value calculated for the wall pie.rs.
Last. determine the swn of the de-flec.uon..:;..
The deflection is the sum oflhe solid wall (first step) mmus the deflection oflhe window strip (second step)
plus lhe deflecuon of the wall piers (third step).
Using lhe wtndow slfip method for determming shear wall deftecuon for the shear wall tllusrrated in Ftgure
l-20,lhe y,aJJ has 92 perce-nt ofsuffness of the same y,aJJ wnhout the opening..
5.6 SHEAR WALL DEFLECTION USING PERFORATED SHEAR WALL METHOD
SDPWS
The deftecuon tbr a shear wall can be approximated by usmg the method of perforated shear \valls m
SOPWS Section 4.3.2. 1. where'' tn Equation 4.3-1 is equal to '~--obtained 1n Equation 4.3-9 and b IS
taken as r.L1•
F1rst, de-termine ,.1114\"
&l4.3-9
where
C0 =shear capacny adjusunem facror fro1n Table 4.3.3.5 or can be calculated \Vlth Equauons 4.3-5 and
4.3-6:
r
)L~
&l4.3-5
Co = ( 3 -2r ll
,
__
1
I
..:..,_
&l4.3-6
I+~
h!.L,
where
= sheatJung area rauo
L. = tota1 length of a pe.rforated shear wall tncluding the lengths of pe.rforoted shear ·wall segments
and the lengths of segments containing openings
A0 =total area of openmgs in the perforoted shear wall uxltVJdual opening areas calculated as the
opening wadth times the clear opening hetght
h = height of the perforated shear wa11
IL, = sum of perforated shear wall segment lengths~ ft
r
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Design Example 1 • Four-Story VKiod Ught·Frame Structure
Ustng the pe-rforoted '"all method tbr determanmg shear waJI deftecuon for the shear wall illustrated m
Figure 1-20, the waJI has 59 percent of sttffness of the same waJI wnhoul the ope,mng_
The window stnp method for detemuntng shear wall deflecuon appears to overesumate the stiffness of lhe
walJ and IS more tune-consumtng than the perforated wall method.
For detemumng shear \\'all deflections and sufrnesses., tJtis desjgn example wtll use the pe.rforared waJI
me-dtod. Testtng has shown the shear walls with opemngs have good stiffness dt..~radatton and performance
compared to shear walls wtthout ope-mngs.
&. The Envelope Process
ASCE7
6.1 ASSUMPTION OF FLEXIBLE DIAPHRAGMS
§12.3.1 .1
ASCE 7 Secuon 12.3.1.1 allows wood structural panel dtaphragms to be ideahzed a" flexible tf any oflhe
following oondhions exisr.:
I. In strucnues where the 're:rucal clements are steeJ.braced frames~ steel and concrete compositebraced frames; or concrete. masonry~ steel, or steel and concrere composite shear walls.
2. In one- and two-tamil'y dwellutgs.
3. In struclures oflight·frame construction where aU of the following oonditjons are met.:
a Topping of concrete or slmtlar materials ts not placed over wood structurnl panel
dtaphmgms except tbr nonsuucturaJ topplng not greater lhan 10.: inches th1ck.
b. Each hne ofventcaJ elemenlS of the setsmic-forte-resistingsystem complie.~ wilh the
allowable stcxy dr~t\ of Table 12.12·1.
In thts destgn example. oondtt•on 3 is nh~t s11tce the strucrure does nm exceed the
concrete.
I~
inches of l&gtuwetght
Section 6.4 of this destgn example fOr dr1lt check of typical shear wall complies with the allowable story
drift.
6.2 LATERAL FORCES ON SHEAR WALLS AND SHEAR WALL NAILING IDEALIZED AS
FLEXIBLE DIAPHRAGMS
In thts ste-p. fOrces on shear walls due to sets:mic forces wdl be determined. As was customary 1n the
ponion oflhe example assumes fle.~1ble dtaphragms. ASCE 7 Section 12.8.4. I does not require
torstonal eOects to be considered Cor flex&ble diaphragms. 1be effects of torsion and wall rig.idioes will be
considered 1n Pan 6.5 of this design example.
pas~ th1s
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85
Design Example 1 • Four-Story~ Light-Frame StTucture
Under diaphragms tdealized as flexible. loads 10 shear walls are determtned based on ttlbutary areas w1th
simple spans between support.-;. Another method of detenntntng k>ads to shear walls assumes a connnuous
beam. This destgn example uses the total buddtng wetght W applted to eac-h respe-cuve dtrectton. The
results shO\\n wdl be slightly conser"\•ative su\Ce the buddtng_ wetght IV tncludes the wall \'/eights tbr lhe
dtrecuon of load, whtch can be subtracted out. This example conven.s the story forces mto seismtc forces
per square foot of floor or roof area. This may result ut the loss of a cettatn amount of precision. but it also
result.<; ln much simpler calculations. This approach is genemlly conside;red acceptable unless there appears
to be a concenttatton of dead load tn a parucular area (e.g.> a mechantcal penthouse).
A detailed analysis wtll tnc:lude the derivation of these Lributar)' weights,. whtch tnclude the tributary
exterior and tnterior wall \\'e-tghts.
Using fOrces from Table 1-3 and the area oflhe floor plan= 5288 square feet. calculate trtbutary \\~ttghts.
For roof diaphragm:
Roof area= 5288 ft'
f.
, • ..,.
- 49.37 X 1000- 9 .34 f
5288
ps
For founh-ftoor diaphragm:
Fl{)()( are.a = 5288 ftJ
f.
,~.
_ 55.74 x 1000 -IO.S4 sf
5288
p
For lhird-floor dtaphragm:
Fl{)()( are.a = 5288 ttl
f.
plr-d
= 37.16 x 1000 _ 1.03 sf
5288
p
For secood-ftoor dial'hragm:
Floor area= 5288 n-
f.'""-
86
18.58xl000
_ f
5288
- 3.> 1ps
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Design Example 1 • Four-Story VKiod Ught Frame Structure
4
Table I-11. Forees to walls tmd reqwred pamd nailmgfor eaJt-wf!Jt dinCttOJ1 1 '~ 3
Wall
Tnb.
Area
IF....,,
IF,
F~
(I\')
(I b)
(I b)
( Ib)
If''
(ft)
F
,. ;.....!!!..
b
F
(1.4)b
(plf)'"
~· = --!!!!.-
Sheathed
I or2
Sides
Allowable
Edge
Nail
Sberu'"
Spacing
(pi I)
(in)
Sht-ar \Vall~ a1 Roof Lt~tro•
A
170
0
1587
1587
12.5
127
91
I
340
6
B
746
0
6965
6965
22.0
317
226
I
340
6
c
1344
0
12,548
12,548
43.0
292
208
I
340
6
E
1344
0
12,548
12,548
43.0
292
208
I
340
6
F
960
0
8963
8963
43.0
208
149
I
340
6
G
554
0
Sin
5172
22.0
235
168
I
340
6
12.5
127
91
I
340
6
H
170
0
1587
1587
1:
5288
0
49,372
49,372
Shur \Valls al Fourlh- Ftoor Ltvtl
A
170
1587
1792
3379
12.5
270
193
I
655
3
B
746
6965
7864
14,829
22.0
674
481
510
4
c
1344
12,548
14, 167
26,7 15
43.0
62 1
444
I
I
510
4
E
1344
12,548
14,167
26,7 15
43.0
62 1
444
I
510
4
F
960
8963
10,119
19,082
43.0
444
317
I
655
3
G
554
5172
5840
11,012
22.0
501
358
I
510
4
H
170
1587
1792
3379
12.5
270
193
I
655
3
1:
5288
49,372
55,741
105, 112
870
2
665
3
Sbur \Valls al Thlrd- VIbor Ltvtl
A
170
3379
1195
4574
12.5
366
261
B
746
14,829
5242
20,071
22.0
912
652
I
I
c
1344
26,715
9445
36,160
43.0
841
601
I
665
3
E
1344
26,715
9445
36, 160
43.0
841
601
I
665
3
F
960
19,082
6746
25,829
43.0
601
429
I
655
3
G
554
11,012
3893
14,905
22.0
678
484
I
655
3
H
170
3379
1195
4574
12.5
366
261
I
870
2
1:
5288
105,112
37,161
142,273
A
170
4,574
597
5171
12.5
414
295
I
870
2
B
746
20,071
2621
22,692
22.0
1031
737
I
870
2
c
1344
36,160
4722
40,882
43.0
951
679
I
870
2
E
1344
36,160
4722
40,882
43.0
951
679
I
870
2
Sbur \Valls at S«ond- Floor Lt\•el
F
960
25,829
3373
29.202
43.0
679
485
I
870
2
G
554
14,905
1947
16,852
22.0
766
547
I
870
2
H
1:
170
4,574
5171
12.5
414
295
I
870
2
5288
142,273
597
18,580
160,853
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Design Example 1 • Four-Story~ Light-Frame StTucture
Nores tbr Table 1-13:
I. In SOC D, E, or F, SDPWS See1ion 4.3.7.1 requires Jx nominal lhickness Slud framing a1 abumng
panels or two 2x members \'ltlere lhe required nomtnal shear exceeds 700 plf. or the nail spactng is 2
tnches on ce-nter or less at adjoining panel edges. or IOct common na1ls havtng penettatjon into framing
members and blocking of more than I ~ inches are 3 inches on cenrer or less.
2. Refer to Section 7.3 m this design example for sill-plate anchorage.
me Secuoo 1705. 12 requites spec1al inspection where the nad spactng ts4 tnches onc-emer or closer
3.
wilh SOC C and higher.
4. The shear wall length used for wall shears JS lhe ..out-to-our• wall length.
5. Fotces are suength leve-l and lhe shear lR wall is divided by 1.4 lOcooven to allowable streSs destgn.
6. APA « TECO performaoce-ra!ed SlfUCtllral-1-ra!ed wood ''"""ural panels may be enher plywood
or OSB. The allowable shear values are fro1n SDPWS Table 4.3A using IOd common nails Wllh a
minjmum I ~inch penetratiot\ and 1-MHnch panel thickness and divided by the ASO reduction factor of
2 .0 .
7. Shear \\ails al hnes c. E. and F extend totlte bonom of the prefabncared wood trusses at the rooflevel.
Shear t.tanstfr ts oblalOed by framing chps from the bottom chord oftl1e trusses to the top plates of the
shear walls. Projec1 1>lans call tOr trusses at these hnes to be destgned fOr lhese horizontal forces (see
also comments in Secuon 1.2). Roof shear forces are also uansttrred to lanes A. n. G. and H.
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Design Example 1 • Four-Story VKiod Ught Frame Structure
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Table 1-14. Foru.t 10 walli and rtqutred panti iJIJIIIngfor north-south du?Ciion1•2•3
F
Tnb.
If''
F
~· = --!!!!.-
(l.4)b
(plf)'"
Edge
Sheathed
I or2
Allowable
Sberu'"
Spacing
Sides
(pll)
(in)
6
Area
IF....,,
IF,
F~
Wall
(I\')
(I b)
(lb)
( Ib)
I
1202
0
11 ,223
11,223
64.5
174
124
I
340
13,463
(ft)
,. ; .....!!!..
b
Nail
Shea r \Valls a1 RoofLn't:l
0
60.0
224
160
I
340
6
3
1442
1442
13,463
0
13,463
13.463
60.0
224
160
I
340
6
4
1202
0
11 ,223
11 ,223
64.5
174
124
I
340
6
49,372
Shu r \Valls a l Fourlh- floor Ltvtl
I
510
4
2
1:
5288
0
49,372
I
1202
11,223
12,670
23,893
64.5
370
265
2
1442
13,463
15,200
28,663
60.0
478
34 1
I
340
6
3
1442
13,463
15,200
28,663
60.0
478
341
I
340
6
4
1202
11 ,223
12,670
23,893
64.5
370
265
I
510
4
1:
5288
49,372
55,74 1
105,112
I
1202
23,893
844 7
32,340
64.5
50 1
358
I
510
4
2
1442
28,663
10,133
38,797
60.0
647
462
I
510
4
3
1442
28,663
10,133
38,797
60.0
647
462
I
510
4
64.5
50 1
358
I
510
4
I
655
3
Shur \Valls at Third- Floor Ltvtl
~
1202
23,893
8447
32.340
1:
5288
105, 112
37,16 1
142,273
I
1202
32,340
4223
2
1442
38,797
5067
43,864
60.0
73 1
512
I
665
3
3
1442
38,797
5067
43,864
60.0
73 1
522
I
665
3
64.5
567
405
I
655
3
Shtar \ Valls at Sttond-Fioor Lt-nl
36,563 64.5
567
405
4
1202
31,340
4223
36,563
1:
5288
142,273
18,580
160,853
Notes rorTable 1- 14:
I. In SOC D. E. 01' F, SDPWS Section 4.3.7. 1 requires lx nonunal thickness stud franungat abuntng panels
or two 2x members where the requ1red nom mal shear exceeds 700 pit: or lhe nail spacmg ts 2 ioc,h eson
center or less at adjommg panel edges. or JOd common nails havtng penetration mto framing members and
blocking of more than I ~ inches are 3 indtes 01'1 center or Jess.
2. Refer to Sectton 7.3 tn this Design Example tor sill-plate anchorage.
3. IBC Sec.tton 1705.12 requires special inspeeuon when the nail spacing is 4 tnd'!eS on center ot closer with
SOC C and h1gber.
4. 1l1e shear wall length used for wall she-ars lS the ..out-Lo-ouf' wall length4
5. Forces are strength level and shear in y,aJl ts divided by 1.4 to convert to allowable suess desagn.
6. APA or TECO performance-rated StructuraJ-1-rated wood struc.tural panels may be etthc-r plywood or OSB.
The allowable shear values are from SDPWS Table 4.3A using IOd common nails wtlh a mtnimum I'15:-mch
penetratiOn and 1Y.c-mch panel thidmess and dt\•tded by the ASD reductJOO factor of2.0.
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Design Example 1 • Four-Story~ Light-Frame StTucture
6.3 CALCULATION OF SHEAR WALL RIGIDITIES
In thts example-. shear wall ng_~dtues are calculated ustng the three- or fOur-term code deftecuon equauon.
These calculation.. are facilitated by the use of a spreadsheet progra.m., which eltminates posstble amhmelic
errors from the many repeuuve compulanons that must be tnade.
The first step lS to c-31culate lhe displacement (i.e.• ven1cal elongation and deflecuon) of the ue-down
assembly elements and the crus.ltutg effect of the boun.dary elemem. ThHl is lhe term dr The force
conside-red w act on lhe ue-down assembly is the net uphft force determined from the fte.xtble diaphragm
analyses. These forces are summarized '"Tables 1-1 5, 1-20, 1-24, and 1-28 fonhe roof, the fourth floor,
the third floor. and the second tloor. respectively.
After the tJe-down assembly dtsplacements are determtnOO.. the-four-term defleeuon equation ts u.<1ied to
deternune lhe de-fleeuon..D.. of each shear wall. These are summartzed ut Tables 1-1 6 and 1-1 7 for the roof
level, on Tables 1-21 and 1-22 for the fourth-floor level, tn Tables 1-25 and 1-26 fonhe lhird-floor level, and
tn Tables 1-29 ru1d 1-30 fOr the second-floor level.
Finally. the rigidities of the shear walls are swnmartzed in Tables I- IS, 1-23. 1-27, and 1-3 Lfor lhe roof,
fourth floor. thlrd ftoor. and second floor. respectively. For both strength and allowable stress design.
ASCE 7 now requtres buildtngdnfts to be determined by eanhquake forces without muluplying by 0.7.
Using strength-level forces for wood destgn utihZutg ASCE 7 and the IDC meaJ'ls lhat lhe engtneer wtll use
fOrces and allowable stress fOrces. This can create some confusion because the code
requires drifl checks to be strength-level forces. HO\\ever. most engu-.eers destgn \vood using aJiowable
stress design. Drtft and shear Y.>all ngidmes should be cakulated from the strength-level fofces. Remember
that the structuml system faClO( R tS based on ustng strength-leYel fOrces.
both strength-level
Rigiditlts of Sbur \\'ails
Determination of wood shear wall rtgidities is om a sinlple mslc In practtce-. appro.xtmate methods are ofte11
used. The method tHustrated ln lh1s example is by flu lhe most ngorous melhod used. There are other. more
simphfied methods, and their use is often appropnate.
It must be emphastzed that. at the present ume. ~Yf!IJ' method ts approxamate. parucularly for mutustory
struc.tures suc.h as those lR thts exrunple as there has been only very limned testing on petformance of
multistory wood-sheathed shear walls. Untd more defin1te general procedures are establtshed through
funher testing and research. the designer must exefcise judgment tn selecting dle appropriate method to be
used for a given structure.
When in doubt, consult With the local buddutg official regardmg methods accepmble to the JUfiSchction. At
Lhe time oflhis public.auon. the type of seism1c design requtred fOr a projec.t of this type vanes greatly from
Ot'le jurisdictiOn to anothe-r.
Wall ngidities (stiffness) are appro.xtmate. lbe inaual ng.idny oflhe structure can be significantly higher
because of stucco. dryv.>all. bnck. and stone venee,rs. stifferung effects of walls not COI'IStdered. and areas
over doors and wtndows. During an eanhquake, some loo•-stressed walls: may malntajn their sttft"ness
while others degrade in suflhess. Some walls and thetr collectors may anract signaficandy mote lateral Jood
than anticipated tn semt-fte.x:ible diaphragm analysis. The method of analyzing a struc(ure us:tng inflexible
90
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Design Example 1 • Four-Story VKiod Ught Frame Structure
4
dtaphragms takes signdicruuly more eng~neering eftOn. However, use of the ngid-djaphragm method
uldicates that some lateral-resisung elements can aruact sagntficantly htgher se1smic de-mands than tttbutary
are-a (a.e., flexible diaphragm) analysis methods. Yet with the ltmited testjng data on multistory wall
performance and the ofteat approxtmate methods available for determuung wood-she.athed wall and flOOI'
dtaphragm ddJecnons, engtneers are as.ked to design more complex lateral systems that include mulustory
narrO\\' shear Y..alls, cantlle\•ered diaphragms, and/or rigid diaphragms to accommodate open-front desigruL
p
- +I-
I
'•
I
I
I
I
l
i
•'1
I
I
I
Tie-down
dis.,laeement d.,
L
I
h
r'.
II
I
''
'
'
•
•••
l ~
Figure /_./.
In llus example, shear wall ng.tdities, ~are computed ustng the baste st.iffness equatiOn
Of
F
k =;;
Estimation of Roor-Lt\'tl Rigiditits
To estimate roof-level wall ngtdities, roof-le\•el dtsplacemems must first be detemllned. The followtng are
a series of calculauons m table form to esumate the roof:tevel drifts, 6, 1n eac.h shear \\o'all. Ftrst, the shear
wall tie-down assembly displacements are determined (Table 1- 15). lbese and the parameters gwen m
Table 1- 15 are used to arnve at the drifts.~ for each s:hear \1/all at the root level (Tables 1- 16 and 1- 17).
Rigtdities are estirnated 10 Table 1-18 tOr waJis in both dtrections. Once the drifts are known, a drift check
ts performed. Thts ts summanzed 1n Table 1-19.
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Design Example 1 •
Four-Story~ Light-Frame StTucture
Table I -I 5. !Htermine tit-down anembly displaumtnts at the IT)()f/en!/1
S[(et~gth
ASD
Design
T1e-Down Assernbly D1splaceme-m
Dovm
Devtce(6•
Uphftm
(lb)
Uphft"'
(I b)
Elo1\gation
(1n)
(1n)
(1n)
(in)
Bear.ng
Plille
Crush
(1n)
0
0
0.000
0.030
0.03 1
0.003
0.000
0.064
1922
2691
0.040
0.030
0.03 1
0.009
0.008
0.118
H
Rod
Rod
Rod
Rod
Rod
Rod
Rod
Rod
Rod
Rod
Rod
Rod
I
2a
Takeup
Tie-Down<31
TieWall
A
81
82
Device
Chord
Eloogation Shnnt<"• Crush($)
6m
'
(In)
1922
2691
0.040
0.030
0.031
0.009
0.008
0.118
2878
4029
0.061
0.030
0.031
0.048
0.01 1
0.340
2878
4029
0.061
0.030
0.031
0.048
0.01 1
0.340
2878
4029
0.061
0.030
0.031
0.048
0.01 1
0.340
2878
4029
0.061
0.030
0.031
0.048
0.01 1
0.340
1848
2588
0.039
0.030
0.031
0.005
0.007
0. 113
1848
2588
0.039
0.030
0.031
0.005
0.007
0. 113
1922
2691
0.040
0.030
0.031
0.009
0.008
0. 118
1922
2691
0.040
0.030
0.031
0.009
0.008
0.11 8
0
0
0.000
0.030
0.031
0.009
0.000
0.064
None
0
0
0.000
0
0.000
0
0
0
None
0
0
0.000
O.oJO
0.031
0.003
0.000
0.065
2b
None
0
0
0.000
0.030
0.031
0.004
0.000
0.065
2c
None
0
0
0.000
O.oJO
0.031
0.013
0.000
0.065
3a
None
0
0
0.000
0.030
0.031
0.003
0.000
0.065
Jb
None
0
0
0.000
0.030
0.031
0.004
0.000
0.065
3c
None
0
0
0.000
0.030
0.031
0.013
0.000
0.065
4
None
0
0
0
0.000
0
0
0
Cl
C2
El
E2
Fl
F2
Gl
G2
92
@Seismicisolation
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2018/SC SEAOC Structura/ISf!;smic Design Manual. Vol. 2
Design Example 1 • Four-Story VKiod Ught Frame Structure
4
Notes for Table 1-15:
I. Tte-down assembly displacements for the rooflewl are calculated for the tie-downs ailhe founh-floor
level.
2. Uphft fb(ce is deteanined by using the net overturning momem (Mot-MR) dtvided by the dJstance
between the c-enlrvids oflhe boundary ele.tneots with 4x members at the ends of the shear wall where
MR uses load combmatiOI'IS oudtned tn Pan: 4.2 ofthjs design example. Usittg allowable stress des1gn,
ue-<I0\111\ devices need ooly be-sjzed by ustng the ASD uphft fo(ce. 1he strength design uplift force Is
used to detemune ue-down assembly displacement and then to detemune suength-level displacements.
3. The conunuous tte-down (rod) system selected fo( this structure wlll have a shnnkage compen.wtng
system. Most of these systems have shrinkage compen.o;atton by e.ither p(e-tenstonmg ofcables ot a
self-rarcheting hardware ronnectot and are propnetary. The device selected in tJus design e.xample has
adjusting grooves at Yl(l-tnch incremenu:, meantng the most the system w111not have compensated for in
shnnkage and crushlng will be Yt. inch. If the selected devic.e does not have a shnnkage compensaung
de'\' lee, then shnnkage of floor ftamtng, sill plates, compress ton bridges, crushing of bridge suppon
studs, and collector studs will need to be considered.
4. Woodshnn.kage IS based on a change ln matsrure content (MC) fi'otn 19 percem to 12 percent, wnh 19
percent MC being assumed for S-Ory lwnber per J)(OJect spedfications. The MC of 12 percent iS the
assumed finaJ MC at equahbrium with ambient hwnidity for the project location. The final equlllbrtum
value can be higher in coastal areas and low-er tn tnland or desen areas. Thjs equates to (0.002)(d)( 19
- 12), where dIS the dianenston of the lwnber (see Ftgure 1-7). Pressure.rreated lumber has a moisture
content of less than 16 percent at ueaunem completion. Shrinkage of the 4x top plate+ 2x DBL sJII
plate= (0.002)(3.5 + 2 x 1.5)(19 - 12)= 0.078mch. Stnceshrinkage compen."'ting devices are us~ a
\'3.lue ofYl! lOCh is used. recognjzing that most devtces have to travel a distance bef«e they get to the
ne.xt groove tn the device to readjust.
5. Per NOS Secuon 4.2.6, when compression perpendicular to grainfu js less than 0.73Fw crushmg wtll
be 3pP(OXtmately 0.02 inch. When.fc.L = 0.13Fcl• crushing ts esttmated at 0.04 inch. The efttct of sill
plate crushing is the downY1ard efti!-cl at the opposite end of the wall with uplift force and has the same
rotauonal etfec.t as me tte-dmvn dJspfacement. Shon walls thai have no uphft fo(tes Will sull have a
crushing effect and oonrnbute to rotation of the wall.
6. If bo1ted-type ue-down devtces are used, the displace.anent for the bolt slip can be determiJ'Ied by NOS
Sectioo 11.3.61oa<l/sltp modulus y = (270,000)(D"), plus an addttoonal \-1, onch for the overst:>.e<l hole
for bolts. For nruls, values for~,. can be used.
7. !111 Is the total ue-down a~mbly displacement. Th1s also could lnclude misculS (short swds) and Jack of
square-cut ends.
2018/BC SEAOC StroctUtai/Sei.smic Design Manual.
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2
93
Design Example 1 •
Four-Story~ Light-Frame StTucture
Table 1- 16. lkj/tctionJ of!hear wallt attht roofln·tl in the nut-west direction
Strength
v
Wall
(pit)
127
A
Nail
E
b
( ft)
h
( II)
A
(in')
8.2 1
17.5
I, 700,000 12.5
(pSi)
Bl
317
8.2 1
17.5
I, 700,000 11.0
B2
317
8.2 1
17.5
I, 700,000 11.0
B
22.0
Cl
C2
b-s
( I\)
G,t,
S)Xlcjng
r·
(in)
(lb)
'·
(in)
c...
t.
(ps~
(tn)
(tn)
11.6
45,500
6
63
0.0003
0.064
o.on
10.1
45,500
6
158
0.0056
0.118
0.19
10.1
45,500
6
158
0.0056
0.118
0.19
•
-
292
8.2 1 35.0 1,700,000 21.5
20.3
45,500
6
146
0.0043
0.180
0.15
292
8.2 1 35.0 1,700,000 21.5
20.3
45,500
6
146
0.0043
0.180
0.15
c
43.0
-
El
292
8.2 1 35.0 1,700,000 21.5
20.3
45,500
6
146
0.0043
0.180
0.15
E2
292
8.2 1 35.0 1,700,000 21.5
20.3
45,500
6
146
0.0043
0.180
0.15
Fl
208
8.2 1 35.0 1,700,000 21.5
20.3
45,500
6
104
0.0014
0. 11 3
O.Q9
F2
208
8.2 1 35.0 1,700,000 21.5
20.3
45,500
6
104
0.0014
0. 113
0.09
F
43.0
Gl
235
8.2 1
17.5 1,700,000 11.0
10.1
45,500
6
118
0.002 1 0.118
0.16
G2
235
8.2 1
17.5 1,700,000 11.0
10.1
45,500
6
118
0.002 1 0. 118
0.16
45,500
6
63
E
43.0
G
22.0
H
127
8.2 1
17.5 1.700,000 12.5
11.6
0.064
0.0003
o.on
Table 1- 17. DejltcttOu!l t>Jshtar wai/J· at the roof/n·el m the nm·th-s.outh dt~YCtlon
"
E
b
(II)
Wall (pit) (ft)
A
(in2 )
1,4
11.55 1,700,000 64.5
I'
(psi)
b-s
(II)
-
G~t.,
(poi)
Nail
Space v.
(in) {Ib)
••
(tn)
d.
(in)
Ao
r
c.
·-
t.
(tn)
45500
6
a,3a 224 8.21
38.50 1,700,000 18.0 16.6 45500
6
112 0.0018 0.07
0.08
b,3b 224 8.21
38.50 1,700,000 24.0 22.6 45500
6
112 0.0018 0,07
0.08
c~3c
38.50 1,700,000 18.0 16.6 45500
6
112 0.0018 0.07
0.08
174 8.21
I, 4
1, 3
94
87 0.0008 0.00 11 2 0.83 0.9 1 191 0.04
IL1 =64.5
224 8.21
60.0
@Seismicisolation
@Seismicisolation
2018/SC SEAOC Structura/ISf!;smic Design Manual. Vol. 2
Design Example 1 • Four-Story VKiod Ught Frame Structure
4
Table 1-18. Shear wall ng1dl1its 011h~ ll)()f/~nd1
F
Jlf!)
F
(I b)
k=•
0
-
(kltn)
K
(klin)
1587
22.1
22.1
0.19
3483
18.2
0.19
3483
18.2
Wall
(Ill)
A
0.072
Bl
B2
6965
36.2
Cl
B
0.15
6274
40.9
C2
0.15
6274
40.9
12,548
82.0
c
El
0.15
6274
40.9
E2
0.15
6274
40.9
12,548
82.0
4482
48.3
E
Fl
0.09
F2
0.09
F
4482
48.3
8963
97.0
Gl
0.15
2586
16.7
G2
0.15
2586
16.7
36.2
81.7
81.7
97.0
33.4
5172
33.4
H
O.Q7
1587
22.1
22.1
I, 4
0.04
11,223
281.0
281.0
2a, 3a
0.08
4039
47.5
2b,3b
0.08
5385
70.8
2c, 3e
0.08
4039
47.5
13,463
166.0
G
2, 3
166.0
Notes for Table 1-18:
I. Deflections and fOrces are based on sue.ngth-force Je\'els.
2. <1s are me deSign-level displacements from Tables 1- 16 aod 1- 17.
2018/BC SEAOC StroctUtai/Sei.smic Design Manual. \.t)J. 2
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95
Design Example 1 •
Four-Story~ Light-Frame StTucture
Es1in111tion of fourth-Floor ltvt-1 RigidWts
Shear wall ngtdittes at the fourth floor are esumated tn the same manner as lhose at the roof. The calculanons
are summanzed tn Tables 1- 19, 1-20, I-2 J.and 1-22. A drift chec.k ts not shown.
Table I -I 9. ne-down assembly displactments atth~ fourth-floor ln·el
ASD
Strength D<sign
Tie-Down Asse-mbly Otsplacetnent
Beartng
Takeup
Upltft"1
(I b)
Uplift<''
(I b)
Tie- Om.\nm
Elongation
(in)
8
II
0.000
0.030
0.031
0.009
0.002
0.072
7 11 7
9964
0. 172
0.030
0.031
0 .04 7
0.057
0.337
7 11 7
9964
0.1 72
0.030
0.031
0.047
0.057
0.337
6:%1
9 186
0. 159
0.030
0.031
0 .092
0.028
0.340
6:%1
9 186
0.1 59
0.030
0.031
0 .092
0.028
0.340
6:% 1
9 186
0. 159
0.030
0.031
0 .092
0.028
0.340
6:% 1
9 186
0.1 59
0.030
0.031
0 .092
0.028
0.340
424 1
5938
0. 103
0.030
0.031
0.012
0.015
0. 191
424 1
5938
0.1 03
0.030
0.031
0 .012
0.015
0.1 91
7 11 7
9964
0. 172
0.030
0.031
0 .047
0.057
0.337
7 11 7
9964
0. 172
0.030
0.031
0 .047
0.057
0.337
Tie-
Down
Device
Elongation
(1n)
Chord
Plme
Shnnk<",
Crush(')
6(11
(1n)
(in)
Crush
(m)
'
(1n)
Wall
Oevtce
A
H
Rod
Rod
Rod
Rod
Rod
Rod
Rod
Rod
Rod
Rod
Rod
Rod
8
II
0
0.030
0.031
0 .009
0.002
0.072
I
None
0
0
0
0
0
0
0
0
2a
None
0
0
0
0.030
0.031
0 .009
0 .000
0.074
2b
None
0
0
0
0.030
0.031
0 .012
0 .000
0.074
2c
None
0
0
0
0.030
0.031
0 .013
0 .000
0.074
Bl
B2
Cl
C2
El
E2
Fl
F2
Gl
G2
3a
None
0
0
0
O.o30
0.03 1
0 .009
0 .000
0.074
3b
None
0
0
0
0.030
0.03 1
0 .012
0 .000
0.074
3c
None
0
0
0
O.o30
0.03 1
0 .013
0 .000
0.074
4
None
0
0
0
0
0
0
0
0
Notes for Table 1-19:
I. Tie-down assembly displacements for the lht.rd-floor leve1 are calculated for the tie-downs at the second-
floor level.
2. Footrores 2 througll7, see Table 1-1 5.
96
@Seismicisolation
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Design Example 1 • Four-Story VKiod Ught Frame Structure
4
Table 1-10. DeflectionS ofshear wallt aJ the fourth-floor /n·e/ m tht I!(I$1-wtJt dlrrction
Srrenglh
,.
h
A
E
b
b.,
v,
(pit)
(ft)
( in')
(pSi)
(ft)
(ft)
G,l.,
(pSi)
Spa«
Wall
(in)
(lb)
(tn)
••
6.
(tn)
6
(in)
A
270
9.44
17.5
1,700,000
12.5
11.6
45,500
3
68
0.0003
0.072
0. 12
Bl
674
9.44
17.5
1,700,000
11.0
10. 1
45,500
4
225
0.0 178
0.332
0.59
B2
674
9.44
17.5
1,700,000
11 .0
10. 1
45,500
4
225
0.0178
0.332
0.59
Cl
621
9.44
35.0
1,700,000 21.5
20.3
45,500
4
207
0.0136
0.338
0.39
C2
621
9 .44
35.0
1,700,000 21.5
20.3
45,500
4
207
0.0 136
0.338
0.39
B
22.0
c
43.0
El
621
9 .44
35.0
1,700,000 21.5
20.3
45,500
4
207
0.0 136
0.338
0.39
E2
621
9.44
35.0
1,700,000 21.5
20.3
45,500
4
207
0.0 136
0.338
0.39
E
43.0
Fl
444
9.44
35.0
1,700,000 21.5
20.3
45,500
3
Ill
0.00 18
0. 182
0.20
F2
444
9 .44
35.0
1,700,000 21.5
20.3
45,500
3
Ill
0.00 18
0. 182
0 .20
F
43.0
Gl
SOl
9 .44
17.5
1,700,000
11.0
10.1
45,500
4
167
0.0067
0.332
0 .48
G2
SOl
9.44
17.5
1,700,000
11 .0
10.1
45,500
4
167
0.0067
0.332
0 .48
270
9.44
17.5
1,700,000
11 .6
45,500
3
68
0.0003
0.072
0 .12
G
22.0
H
12.5
Table 1-21. Deflections ofshear wallt at the ftmrth-jloor le1·e/ in the north-s.ouJh dt~YCtion
,,
A
E
Wall (pll) (ft)
(in')
(psi)
1,4
11.55 1,700,000
"
b
(ft)
-
b,,
G;,
(fi)
(psi)
Space v
•
(tn) (I b)
••
(tn)
d,
( in)
A,
r
c, ,.
~
6
(tn)
45,500
4
16.6 45,500
6
239 0.02 17 O.Q7
- -
-
-
0.30
2b,3b 4 78 9.44 38.50 1,700,000 24.0 22.6 4 5,500
6
239 0.0217 O.Q7
-
-
0.29
2c,3c 4 78 9.44 38.50 1,700,000 18.0 16.6 45,500
6
239 0.02 17 O.Q7
- - -
1,4
370 9.44
0
112 0 .84 0.96 386 0. 10
I L1 =64.5
2a,3a 4 78 9.44 38.50 1,700,000 18.0
2,3
123 0.0025
-
-
0.30
60.0
2018/BC SEAOC StroctUtai/Sei.smic Design Manual. \.t)J. 2
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97
Design Example 1 •
Four-Story~ Light-Frame StTucture
Table 1-12. Shear wall rigid/lies a11h~ fourth-jl()Or ltl-t/1
!J.fl•
Wall
(in)
F
(lb)
F
k, = ;(k/in)
A
0.12
3379
27.6
Bl
0.59
7414
12.5
B2
0.59
7414
12.5
14.829
24.9
B
Cl
0.39
13,358
34.5
C2
0.39
13,358
34.5
26,715
69.0
13,358
34.5
c
El
0.39
E2
0.39
E
13,358
34.5
26,715
69.0
Fl
0.22
9541
48.7
F2
0.22
954 1
48.7
19,082
97.4
F
Gl
0.48
5506
11.5
G2
0.48
5506
11.5
11,012
23.1
G
....
K
(k/in)
27.6
24.9
69.0
69.0
97.4
23.1
li
0.13
3379
27.6
27.6
240.0
1,4
0.10
23,893
240.0
2a,3a
0.30
8599
28.9
2b,3b
0.29
11,465
40. 1
2c~3c
0.30
8599
28.9
28,663
97.9
2 ~3
97.9
Notes tbrTable 1-22:
I . Deflections and tbtces are based on strength levels.
2. Mare the design-level displacements from Tables 1-20 and 1-2 1.
98
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Design Example 1 • Four-Story VKiod Ught Frame Structure
4
Es tirnatioa ofTbird- floor Lt"vtl Rig:iditits
Shear wall ngidities at the secood-ftoor level are esumated in the same manner as those for lhe roof and
third ftoor. The calculations are summarized in Tables 1-2.3. 1-24, 1-25, and 1-26. A drift check is not
shown4
Table 1-23. Tte-down tLUemblydJJplacemenu a.r th~ thu-d-floor lev~/1
ASD
Strength Des•gn
Tie.-dO\m Ass...~bly Displac.eme.nt
Tie-
wau
down
Device
Uplifta1
(lb)
Upllfi"'
(Ib)
576
806
Bearmg
Takeup
Device
Tte-downm
EJongatioo
( in)
Elongation
0.007
Chord
Plate
Shrmk1">
Crush'$)
Crush
(in)
(10)
(10)
(in)
t.•"'
0.030
0.031
0 .010
0 .002
0.080
(m)
H
Rod
Rod
Rod
Rod
Rod
Rod
Rod
Rod
Rod
Rod
Rod
Rod
I
None
0
0
0
2a
0
0
0 .019
0 .030
0.031
0 .010
0 .006
0.1 0 1
0
0
0 .033
0.030
0 .031
O.QI5
0 .006
0. 11 5
0
0
0 .019
0 .030
0.031
O.QI5
0 .006
0.1 0 1
0
0
0 .019
0.030
0.031
0 .010
0 .006
0. 10 1
.lc
Rod
Rod
Rod
Rod
Rod
Rod
4
None
A
Bl
B1
Cl
C2
El
E2
Fl
F2
Gl
G2
1b
2c
3a
3b
11,85 1
17,991
0.152
0.030
0 .031
O.QI8
0 .029
0.260
12,85 1
17,991
0.152
0.030
0 .031
O.QI8
0 .029
0.260
11 ,928
16,700
0.141
0 .030
0.031
0 .130
0.024
0.356
11 ,928
16,700
0.141
0.030
0 .031
0 .130
0.024
0.356
11 ,928
16,700
0.141
0 .030
0.031
0 .130
0.024
0.356
11 ,928
16,700
0.141
0.030
0 .031
0 .130
0.024
0.356
783 1
10,963
0.093
0 .030
0 .031
O.QI5
O.QI5
0.1 84
783 1
10,963
0.093
0.030
0 .031
O.QI5
0 .015
0. 184
12,85 1
17,991
0.152
0.030
0 .031
O.QI8
0 .029
0.260
12,85 1
17,991
0.152
0.030
0 .031
O.QI8
0 .029
0.260
576
864
0.007
0 .030
0 .031
0 .010
0 .002
0.080
0
0
0
0
0
0
0
0 .033
0.030
0 .031
O.QI5
0 .006
0.11 5
0
0
0 .019
0.030
0.031
O.QI5
0 .006
0. 10 1
0
0
0
0
0
0
0
0
N01es for Table 1-23:
I . Tte-dov•.-n assembly dtsplacements for the second-floor level are calculated tbr lhe tte-downs at the first-floor
level.
2. See Table 1-1 5 for Fo01notes 2 tlltougl\7.
2018/BC SEAOC StroctUtai/Sei.smic Design Manual. \.t)J. 2
@Seismicisolation
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99
Design Example 1 •
Four-Story~ Light-Frame StTucture
Table 1-2-1. Dejlutlons ofthear n·al/s attht tlurd-j/oor In·~/ in tht east-1~e.st dirrct/(}IJ
Strenglh
,.
b
(ft)
b_,
G,t~
Space
v.
(poi)
(ft)
(p<i)
(in)
(lb)
(in)
'·
(tn)
"·
(in)
17.5
1.700,000
12.5
11.6
45,500
2
69
0.0002
0.080
0.15
49.0
1,700,000
11.0
9.5
45.500
3
228
0.0 187
0.257
0.59
49.0
1.700,000
11.0
9.5
45,500
3
218
0.0187
0.257
0.59
\\\ill
(plf)
h
(ft)
A
(in')
A
366
9.44
Bl
912
9.44
82
9 12
9.44
E
B
Cl
C2
22.0
84 1
9.44
84 1
9.44
49.0
1.700,000 21.5
20.0
45,500
3
210
0.0 143
0.351
0.45
49.0
1,700,000 21.5
20.0
45.500
3
210
0.0 143
0.351
0 .45
c
43.0
El
84 1
9.44
49.0
1,700,000 21.5
20.0
45.500
3
210
0.0 143
0.351
0 .45
E2
84 1
9.44
49.0
1.700,000 21.5
20.0
45,500
3
210
0.0 143
0.351
0.45
Fl
60 1
9.44
49.0
1.700,000 21.5
20.0
45,500
2
100
0.0013
0.176
0.22
F2
60 1
9.44
49.0
1,700,000 21.5
20.0
45.500
2
100
0.0013
0.1 76
0 .22
Gl
678
9.44
49.0
1,700,000
11.0
9.5
45.500
3
169
0.0070
0.257
0.45
G2
678
9.44
49.0
1.700,000
11.0
9.5
45,500
3
169
0.0070
0.257
0.45
11.6
45,500
2
61
0.0002
0.080
0.15
E
43.0
43.0
F
G
...
22.0
366
9.44
17.5
1.700,000
12.5
Table 1-25. Dtjkcuons ofsh~ar walls at the third-floor ln·el In t/t(! north-south diret:110 n
l'
h
\\lll (plf) (ft)
.,4
A
( in')
E
(psi)
501 9.44 11.55 1.700,000
.,4
b
(ft)
-
b,Jf
(ft)
••
Glr
(p<i)
Spaee
v.
(in)
( Ib)
45.500
4
167 0.0067 0.00 112 0.84 0.96 523 0.1 5
(in)
d.
(tn)
U 1 =64.5
16.0 45.500
4
2 16 0.0 155 0.10
).Jb 647 9.44 77.00 1,700,000 24.0 22.0 45.500
4
2 16 0.0 155 0.12
:.•3c 647 9.44 77.00 1.700,000 18.0
4
2 16 0.0 155 0.10
>,3a 647 9.44 77.00 1.700,000 18.0
~.
"
3
100
16.0 45.500
60.0
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2018/BC SEAOC Slructurai/Seismic Design Manual, 1.01. 2
Ao
r
c.
~·-x
- - - - - - - - - -
"
(in)
0.31
029
0.31
Design Example 1 • Four-Story VKiod Ught Frame Structure
4
Table I -26. 1*~11 rigidttie.s tJttlurd1foor /t,·ei
F
l!;,(l)
Wall
(in)
F
(I b)
A
0.15
Bl
B2
k=(k/tn)
K
(kiln)
4574
30.6
30.6
0.59
10,036
17.1
0.59
10,036
17.1
20,071
34.2
B
'
Cl
0.45
18,080
40.5
C2
0.45
18,080
40.5
c
36,160
80.9
El
0.45
18,080
40.5
E2
0.45
18,080
40.5
E
36,160
80.9
Fl
0.25
12,914
58.1
F2
0.25
12,914
58.1
25,829
116.1
7453
16.4
F
Gl
0.53
G2
0.53
G
"""
0
7453
16.4
14,905
32.9
34.2
80.9
80.9
116.1
32.9
H
0.15
4574
30.6
30.6
1,4
0.16
32,340
204.0
204.0
2a,3a
0.31
11,639
38.1
2b,3b
0.29
15,519
52.7
2c,3c
0.31
11,639
38.1
38,797
129.0
2,3
129.0
Notes for Table 1-26:
I. Deflections and forces are based on strength-force levels.
2. L\s are the design-level displacements from Tables 1-24 and 1-25.
Estimation or St-tond-floor
Uv~l
Rigiditit:s
Shear wall rigidities at the second-floor level are esumated m the same manner as lhose tOr lhe roof. tburth
ftoor. and third floor. The calculauons are summartZed in Tables 1-27. 1 28. 1-29, and 1-30. A dnft c.heck JS
4
n01 sho\~Tl.
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101
Design Example 1 •
Four-Story~ Light-Frame StTucture
Table 1-27. Tttt-down (I$$tmblydisplattmenu at the second-floor ln·e/1
ASD
Strength Destgn
Tie-Down Assembly Displacement
Wall
TieDown
JRvice
Tie-Down11)
Elongation
(in)
Tilkeup
Devtce
Elongauon
Shnn.k'',
(in)
(•n)
Chord
Crush(!)
(in)
Bearing
Plate
a"'
•
Uplift"'
(lb)
Uplift"'
(I b)
1320
1848
0.016
0.030
0.031
0.015
0.003
0.095
20,188
28,264
0.182
0.030
0.031
0.040
0.047
0.331
20,188
28,264
0.182
0.030
0.031
0.040
0.047
0.331
18,345
25,683
0.166
0.030
0.031
0.119
0.037
0.382
18,345
25,683
0.166
0.030
0.031
0.119
0.037
0.382
18,345
25,683
0.166
0.030
0.031
0.119
0.037
0.382
18,345
25,683
0.166
0.030
0.031
0.119
0.037
0.382
14,184
19,858
0.128
0.030
0.031
0.017
0.036
0.242
Crush
(in)
(•n)
14,184
19,858
0.128
0.030
0.031
0.017
0.036
0242
20,188
28,264
0.182
0.030
0.031
0.040
0.047
0.331
20,188
28,264
0.182
0.030
0.031
0.040
0.047
0.331
1320
1980
0.016
0.030
0.031
O.QIS
0.047
0.095
0
0
0
0
0
0
0
0
0
0
0.096
0.030
0.031
0.015
0.013
0.189
0
0
0.109
0.030
0.031
0.024
0.013
0207
0
0
0.096
0.030
0.031
0.019
0.013
0.189
0
0
0.096
0.030
0.031
0.015
0.013
0.189
0
0
0.109
0.030
0.031
0.024
0.013
0.207
3c
Rod
Rod
Rod
Rod
Rod
Rod
Rod
Rod
Rod
Rod
Rod
Rod
None
Rod
Rod
Rod
Rod
Rod
Rod
0
0
0.096
0.030
0.031
0.019
0.013
0.189
4
None
0
0
0
0
0
0
0
0
A
Bl
82
Cl
C2
El
E2
Fl
F2
Gl
G2
H
I
2a
2b
2c
3a
3b
Notes for Table 1-27:
I. Tie-dov.·n assembly displacement.<~ for tJ1e second-floor level are calcuJated ror the ue-downs at the fitst-Hoot
lt\•el.
2. S<e Table 1- 15 for Footn<lles 2 lhrough 7.
8. Rotations of the foundilltOn are assumed to be 0 tnches. The engineer should consider possible rouuionaJ
effect~ on shear waJis due 10 lhe grade beam rotations. especially ln mller Slltlctures.
102
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Design Example 1 • Four-Story VKiod Ught Frame Structure
4
Tab/~
1-28. Dejlectiont ofshear wall..~ tllthe second-floor ln·e/ m 1h~ ttJ.SI-t~tsl dt~ction
Strength
,.
"
E
b
b'l
(psi)
(fi)
( ft)
G,
(ft)
(psi)
Space
(in)
(lb)
"·
(in)
••
tJ..
(m)
( on)
11.6
12.5
45.500
2
69
0.0004
0.095
0 .17
1,700,000 11.0
9.5
11.0
45,500
2
0.0074
0.327
0 .60
1,700,000 11.0
9.5
11.0
45.500
2
m
m
0.0074
0.327
0 .60
21.5
45.500
2
158
0.0057
0.377
0 .42
21.5
45,500
2
158
0.0057
0.317
0 .42
Wall
(pi f)
( ft)
A
( in2)
A
414
9.44
17.5 1,700,000 12.5
Bl
1031
9.44
49.0
B2
1031
9.44
49.0
Cl
951
9.44
17.0 1,700,000 21.5
19.5
C2
951
9.44
17.0
1.700.000 21.5
19.5
B
22.0
c
Gl..
IJ.
22.0
43.0
43.0
El
951
9.44
17.0
1.700.000 21.5
19.5
21.5
45,500
2
158
0.0057
0.317
0 .42
E2
951
9.44
17.0 1,700,000 21.5
19.5
21.5
45.500
2
158
0.0057
0.377
0 .42
E
43.0
43.0
Fl
679
9.44
17.0 1,700,000 21.5
19.5
21.5
45.500
2
11 3
0.00 19
0.24 1
0 .27
F2
679
9.44
17.0
1.700,000 21.5
19.5
21.5
45,500
2
11 3
0.0019
024 1
0 .27
43.0
F
43.0
Gl
766
9.44
49.0
1.700,000
11 .0
9.5
11.0
45,500
2
128
0.0028
0.327
0.51
G2
766
9 .44
49.0
1,700,000 11.0
9.5
11.0
45.500
2
128 0.0028
0.327
0.51
414
9 .44
17.5 1,700,000 12.5
45.500
2
0.095
0.17
22.0
G
11
22.0
11.6
12.5
69
0.0004
Table I -29. INjlt!Ctions ofshear walls Ulthe stcond-jloor ltnd in the- north-.Wulh d 1t?ction
,.
"
A
(in2)
E
b
b,l
G;,
\IIlii (pi f) ( fi)
(psi)
(A)
(A)
(psi)
567 9.44
11.5
1.7E6
-
1,4
Spaoe v
•
( on) ( Ib)
'·
(m)
45,500
3
142 0.0039
d.
(in)
0
r
c.
-
¥
IJ.
(m)
11 2 0.84 0.96 591 0. 15
r.L,= 64.5
1,4
2.a,3a 731 9.44
135
1.7E6
18.0
15.3 45,500
3
183 0.0090 0. 19
2b,3b 731 9.44
115
1.7E6
24.0 21.3 45,500
3
183 0.0090 0.2 1
2c,3c 731 9.44
135
1.7E6
18.0
3
183 0.0090 0. 19
2, 3
Ao
15.3 45,500
-
-
- -
0.33
-
0.31
-
- -
0.33
60.0
2018 IBC SEAOC Structural/Seismic Design Manual. \lbl. 2
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103
Design Example 1 •
Four-Story~ Light-Frame StTucture
Tab/~ 1-30. Wall riguliues attht second-floor /~rt/1
Ji(l)
F
k, -= ;(klin)
(k/in)
29.8
Wall
(in)
F
(I b)
A
0.17
5171
29.8
Bl
0.60
11,346
18.8
82
0.60
11,346
18.8
B
22,692
37.6
Cl
0.42
20,441
48.1
C2
0.42
20,441
48.1
40,882
96.3
0.42
20,441
48.1
0.42
20,441
48.1
40,882
96.3
14,601
53.5
c
El
E2
E
Fl
0.27
F2
0.27
F
14,601
53.5
29,202
106.9
Gl
0.51
8426
16.4
G2
0.51
8426
16.4
16,852
32.8
G
K.-
37.6
% .3
% .3
106.9
32.8
H
0 .17
5171
29.8
29.8
1,4
0.15
36,563
238.0
238.0
2a,3a
0.33
13,159
39.4
2b,3b
0.31
17,545
57.4
2c,J.c
0.33
13,159
39.4
43,864
136.0
2,3
136.0
NOtes for Table 1-30:
I . Oeflecuons and forces are based on strength-force levels.
2. lis are tl\e desogn-level displaeemeniS from Tables 1-28 and 1-29.
104
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Design Example 1 • Four-Story VKiod Ught Frame Structure
4
6.4 DISTRIBUTION OF LATERAL FORCES TO THE SHEAR WALLS USING RIGID
DIAPHRAGMS
ASCE 7 §12.8.4
The base shear was distributed ro the fOur levels ln Part 2.8. In this step. lhe SlOf)' forces are d1stributed to
the shear \valls suppomng each level ustng the ngtd dlaphragm assumptton. See Pan 8 fOr a coofirmatton of
tlus assumption.
For many years It has been common e.ngtneertng practice to assume flexible diaphragm.~ and to distnbure
loads to shear walls based on tflbutaty areas. This has become a well-establtshed conventional destgn
assumptton. In thts destgn e.xample-. the ng~d dtaphrag.m a-;sumpt.ton wtll be used. Thts IS not Intended to
unply that se1smic design of wood light-frame consuuction in the past shouJd have been perl'onned ln lhts
manner. Howe,·er, recent earthquakes and testing of \\'OOd panel shear walls have mdicared thar dnfls can be
COt'L~iderably higher than what \\"aS known or assumed in the pasL This knowledge of the increased drifts of
narrow wood-pane.! shear '"-ails and the factlhat me dtaphragms tend to be much more ngid man the shear
walls has tncreased the nee.d for the engu-.eer to consider the relattve rigtdities of shear walls.
ASCE 7 requtres that the story force at the ce-nter of mass be dtsplac.ed from the cakulated cettter of mass
(CM) a distance of five
percent of the building diJ'ne.nston at that level perpendtcular to the direction of
force. Thts IS to account for acctdentaltorston. ASCE 7 requlres the nlost severe load combinatton to be
conside-red and also pemuts the negative wtsionaJ shear to be subtracted ffom the direct load shear. The net
effect of this is to add five percent acctdental eccentrteity to the cafc.ulate<l eccentrtcity.
However, lateral forces must be constdered to act m each dtrecuon of the 1\\'0 pnnc1pal axes. Thts destgn
example does not OOttStder eccentriCities betw·een the centers of mass between leve-ls. In thlS destgn
example-. these eccentriCities are smaJI and are therefore deemed insignificanL The engineer must t~\:e:rctSe
good judgment in determining when those effects need 10 be consu:lered.
Section 12.8.4.3 exempts structures that do not have Type Ia or Type I b trregulariues from amplifymg the
acctdentaJ torsional mome-nt; the torstonal amplification fac.tor A~ has not been calculated in th1s design
example.
The dtrect shear fOr~e F.,1s determined from
R
•R
F=F-
•
and
the torsional shear force F, ts determined from
F =TRxd
'
J
where
J=
r.Rd: + r.R<J,'
R =shear wall rigidny
d =distance from the lateral reststmg element (e.g.~ shear wall) to the center of ng_idity (CR)
f;Fxe
4
F = 49,370 lb (for roof diaphragm)
e = eccentttdty
2018 IBC SEAOC Structural/Seismic Design Manual. \lbl. 2
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105
Design Example 1 • Four-Story~ Light-Frame StTucture
DttttmiM Ctntr.r of R.igidiry, Ctnltr of ~lass., and E«tnlricitirs for Roof Diaphragm
Forces m the easl-west (.\1 d•recuon:
Usang lherigadtty valuesk from Table 1-18 and the distance.v from line 1-1 lO the shear wall:
)',(22.1 + 36.3+ 82+82 +97 + 33.5+ 22.1) = 22.1(116)+ 36.3(106) + 82(84.2) + 82(5o.o) + 97(26.0)
+ 33.5( 10.0) + 22.1 (0)
Distance to calculated CR
;, =
2
~~:
3
= 53.6 fi
The bu1lding ts symmetrtcal about the X axas (Ftgure 1-22). and the c:enler of mass (CM) ts dete-nntned as
:v. = 11:.0 = 58.0 t\
The mtmmum five percent accidental eccentricity for east-west forces. ~P is computed from the length of
the structure perpendicular lO the applied story fOrce:
•,=(0.05x 116 t\):±5.8 n
The new J.to the displaced CM = 58.0 ft ± 5.8 t\ = 63.8 It or 52.2 fl
The toull eccenUtcity l~ the d1slance between the displaced center of mass and the cenler of rig1dny:
y,= 53.6 t\
:.•,=63.8 -53.6= 10.2!\ or 52.2-53.6=-1.4ft
Otsplacing the c:emer of mass five percent can resuJt lR lhe C:M beutg on eilher side of the CR and c:an
produce added lOrstonal shears to all walls.
The fi,•e percent may not be conservmive. The content~ to-structure weight rauo can be higher tn '"'OOd
framang than tn heavier types of c:onstruction. Also,lhe locauon of the cafruJated center ofrigadtty is Jess
reliable than in olher structural systems.
Use eng.ineenngjudgment when selecting lhe eccent:ricny e.
Amplificauon of accidental tors1o.nal momenl requjred tn ASCE 7 Section 12.8.4.3 is exempted for
struc.tures that do not have Type Ia or Type Ib 1treguJanues.
106
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Design Example 1 • Four-Story VKiod Ught Frame Structure
4
Fo rte~
in tht North-South ( Y) Dirntion
ll\e building JS S)'mmeltlcal aboul the y-.axis (F1gure l-42). TherefOle. lhe dismnce to the CM and CR JS
·---2-- .
-
- 48·0 - 240ft
t;= (0.05X 48 ft) = ±2.4 ft
Because the CM and CR locations coinc1de,
•
t.\ =e~
4·. ,., = 2.4 ft or
- 2.4 ft
Detennme total sheatS on ·walls at roof level.
The total shears on the v,-alls 31 the roof le,-el are the direc1 shears. F,. and the shears due to totsJon
(combined actual torsion and accJdental torsion). Fr Torsioo on the roof diaphragm is computed as
T, = Fe1 = 49,370 lb(I0.2 ft)= 503,722 ft-lb for walls A, B, and C
or T, = 49,370 lb(l.4 ft)= 68,989 ft-lb tbr walls E, F, G, and H
r, =Fe,= 49,370 lb(2.4 ft) = 118,492 ft-lb
Because the build1ng is sym.metricaJ for forces in the nonh-south direction, tl\e torsional forces can be
subuacted for those walls located on the opposite s1de from the dtsplaced CM. The criucaJ force will then
be used for the des1gn of these \valls. Table 1-32 summanzes t11e spreadsheet fot determintng c.ombtned
forces on the roof 1evel walls..
4
Determme the center of ng1dity, center of ma-cos, and eccenttictttes for the fOurth-, lh1rd and second-Boor
4
d~aphmgms.
Since the \\'3lls stae:k with umform nailing. It can be assumed that the cenre.r of rtgtdity fot the lhird-floor
and the second-floor diaphragms Will ooinC-ide with the center of ngid1ty ofthe roofdiaphragm.
Tors1on on the fow~ftoor diaphragm:
F= (49,370+ 55,740)= 105, 11 0 lb
T, =Fe,.= 105,110 lb(I0.6 fl)= 1, 11 8,643 ft-lb for walls A, B, and C
or 105, 11 0 lb( I.O ft) = 100,660 ft-lb for walls E, F, G, and J-1
r, =Fe,= 105,110 lb(2.4 ft) = 252,270 ft- lb
201 8 IBC SEAOC Structural/Seismic Design M anual. \lbl. 2
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107
Design Example 1 • Four-Story~ Light-Frame StTucture
Level
=?cof
4th
5rd
2nd
.
,.
e, .
10.2'
10.6'
10.2'
9.6·
1.1,•
1.0'
1.4'
••
2.0'
~
NORTH
Figure 1-11. C~nrer ofrigtdtry and !octJ.rion ofduplaced centers ofmass for
Jecond-. rhJrd-. andfout1h-jloor /~1·e/1 and roofdiaphragm
108
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Design Example 1 • Four-Story VKiod Ught Frame Structure
4
Results for the founh-ftoor le.•el are summanzed in Table 1-32.
Torsio11on the third-floor dlaphtagm:
F= (49,370 + 55,740+ 37,160)= 142,270 Jb
r. =Fe,= 142,270 Jb(J0.2 ft)= 1,446, 183 ft-Jb tor walls A, B,and c
or 142,270 Jb(J.4 ft) =204,182 ft -Jb for walls E, F, G, and H
r,. =Fe,= 142,270 Jb(2.4 ft) = 341,455 fi- Jb
Results for the third-floor level are summanzed in Table 1-33.
Torsi011on the second-floor diaphragm:
F= (49,370+ 55,740+ 37,160 + 18,580) = 160,850 Jb
T, = Fe1 = 160,850 Jb(9.6 ft)= 1,551 ,003 ft-Jb for walls A, B,and C
or 160,850 Jb(2.0 ft) = 314,894 ft -Jb for walls E, F, G, and H
r,. =Fe,= 160,850 lb(2.4 ft) = 386,047 fi- lb
Results for the second-floor level are sum.man.zed 1n Table 1-34.
Table 1-31. Distribu110n offorc.t.f to .'lh~ar walls /Nimv the T(J()f/e,·~l
~
-ij
"'
3
0
'1
-€0
z
d,
Ref
Total
Force
F,
Force
F,+F,
R,
A
22.1
62.4
1379
86,054
2919
893
38 12
B
36.2
52.4
1896
99,353
4779
1227
6007
c
81.7
28.4
2322
65~945
10,798
1503
12,301
d,
Rd
Tors1ooal
Wall
E
81.7
3.6
294
1058
10,798
190
10,989
F
96.5
27.6
2664
73,524
12,752
1725
14,477
G
33.4
43.6
1454
63,395
4406
941
5347
H
22.1
53.6
1184
767
3686
1:
.<:
R,
Otrecl
Force
F,
373.8
63,482
2919
452,811
49,372
I
280.6
24.0
6735
161,633
15,516
1026
16,542
2
165.8
2.5
415
1036
9170
63
9233
3
165.8
-2.5
-415
1036
9170
-63
9106
4
280.6
-24.0
161,633
15,516
-1026
14,491
1:
892.9
325,338
49,372
1:
-6735
778,149
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Design Example 1 •
Four-Story~ Light-Frame StTucture
Table 1-31. Di.sJrihutlon offorces to s~ar walls below the fourth-j/()()r llt1·el
ii
:!:
,;,
t!l
R,
d,
Rtf
Dlrec.t
Force
F,
Torsional
Force
F,
Foo:e
Fr+F,
8828
2763
11 ,59 1
Total
Wall
R,
A
27.6
62.4
1n4
107,605
B
24.9
52.4
1306
68,437
7962
2093
10,055
c
69.0
28.4
1960
55,668
22,046
3141
25, 186
E
69.0
3.6
248
893
22,046
398
22,443
F
88.6
27.6
2445
67,466
28,300
3918
32,2 18
G
23. 1
43.6
1006
43,867
7373
1612
8986
H
26.8
53.6
1436
76,959
8559
2301
10,860
1:
329.0
420,895
105,112
5748
137,956
37,303
2077
39,38 1
dr
24.0
Rd
I
239.5
-"
2
97.9
2.5
245
6 12
15,253
88
15,341
$).
3
97.9
-2.5
-245
6 12
15,253
- 88
15,164
"2
4
239.5
-24.0
-5748
137,956
37,303
-2077
35,226
l:
674.9
277, 137
105,112
3
z0
1:
698,032
Table 1-33. DlJtnbutiou offoreesto shear wa/IJ below the thwdfh>orln·el
-"'
~
.!.
~
R,
Rtf
D1rect
Force
F,
Torsional
Force
F,
Total
F«ce
F,+F,
Wall
R,
A
30.6
62.4
1907
119,028
11,185
3760
14,946
B
34.2
52.4
1794
94,00 1
12,526
3537
16,063
c
80.9
28.4
2298
65,275
29,609
4531
34, 140
d,
d,
Rd
E
80.9
3.6
291
1047
29,609
574
30, 183
F
104.5
27.6
2885
79,6 12
38,25 1
5687
43,939
G
28.0
43.6
1222
53,285
10,259
2410
12,668
H
29.6
53.6
1587
85,04 1
10,833
3128
13,961
497,289
142,273
I
203.7
24.0
4888
117,322
43,574
2275
45,850
-5
2
128.8
2.5
322
805
27,562
150
27,7 12
0
3
128.8
- 2.5
- 322
805
27,562
- 150
27,4 12
4
203.7
- 24.0
117,322
43,574
-2275
41,299
1:
665.0
236,255
142,273
l:
:z"
1!."
388.8
-4888
l:
110
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Design Example 1 • Four-Story VKiod Ught Frame Structure
4
Table 1-3-1. Distribution offorc~.t 10 J.'htar walls btlow second1foor /e,·e/
Otrecl
~
"'
oll
~
5
~
0
z
Force
F
Torsional
Force
F,
Total
Force
Wall
R,
d,
Rd
RJ
A
29.8
62.4
1860
11 6,081
11 , 160
3587
1~ .748
B
37.6
52.4
1973
103,384
14,095
3805
17,900
c
96.3
28.4
2735
77,683
36,052
5275
41,326
R,
d,
•
F.,+ F,
E
96.3
3.6
346
1246
36,052
668
36,720
F
106.9
27.6
2952
8 1,454
40.~
5692
45.732
G
32.8
43.6
1432
62,421
12,295
2761
15,056
li
29.8
53.6
1598
85,633
11 , 160
308 1
14,241
!
429.7
527,902
160,853
I
238.4
24.0
5122
137,317
5 1, 189
2746
53,935
2
136.2
2.5
340
851
29,238
163
29,401
3
136.2
- 2.5
- 340
851
29,238
- 163
29,075
4
238.4
- 24.0
- 5122
137,317
5 1, 189
- 2746
48,442
!
749.1
276,336
160,853
!
804,239
6.5 COMPARISON OF LOADS ON SHEAR WALLS USING FLEXIBLE DIAPHRAGM
ASSUMPTIONSVS. RIGID DIAPHRAGM ASSUMPTIONS
Table 1-3 1 summarizes wall l'btces determined unde.r lhe-separate flexible and rigid djaphragm analyses.
Stnce nailtng requtreme-nls were establtshed tn lhe flexible diaphragm analysis of Part 6.2. they must be
checked for result< of1he r~gtd diaphragm analysis and adjusled if neces.<ary (also given in Table 1-35).
201 8 IBC SEAOC Structural/Seismic Design Manual. \lbl. 2
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Design Example 1 •
Four-Story~ Light-Frame StTucture
Table 1-35. Compan.wn ofloads 011 shear walls usingjlexlblt n-. rigid diaphrtJgm analysts
and ~chtck ofnailmg in walls
F.
Rigid/
Wall
FftmbJ<
(lb)
F,,l)4
Fle.xible
(lb)
(%)
b
(ft)
ll=~
1.4bm
(pll)
Plywood
I or2
Sides
Allowable
Shear
Edge Nail
Spacmg
(plf}'""
(in)
I
340
6
6
Roof level
A
1587
3812
140
12.5
2 18
B
6965
6007
-14
22.0
226
I
340
c
12,548
12,301
-2
43.0
208
I
340
6
E
12,548
10,989
-12
43.0
208
I
340
6
F
8963
14,477
62
43.0
240
I
340
6
6
G
5172
5347
3
22.0
174
I
340
tl
1587
3686
132
12.5
2 11
I
340
6
I
11,223
16,542
47
64.5
183
I
340
6
2
13,463
9233
-31
60.0
160
I
340
6
60.0
160
I
340
6
64.5
183
I
340
6
3
13,463
9233
-31
4
11 ,223
16,542
47
Fourth-Floor Level
A
3379
11,594
243
12.5
663
I
665
3(11)
B
14,829
10,054
-32
22.0
48 1
I
510
4
c
26,715
25,159
-6
43.0
444
I
510
4
E
26,715
22,391
-16
43.0
444
I
510
4
F
19,082
32,182
69
43.0
535
I
665
3
G
11,012
8982
-18
22.0
358
I
510
4
tl
3379
11,201
231
12.5
640
I
665
3(11)
I
23,893
39,374
65
64.5
436
I
510
4(8)
2
28,663
15,341
-46
60.0
341
I
340
6
6
4(11)
3
28,663
15,341
-46
60.0
341
I
340
4
23,893
39,374
65
64.5
436
I
510
(cominued)
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Design Example 1 • Four-Story VKiod Ught Frame Structure
4
Table 1-15. Comparison of/()(J(/s ou shear wallt usingjlexibl~ ,..t rigid diaphragm antJI)'SI.'f
and ~VC~ck ofnailing in walls-continued
Wall
Fftmlll~
(lb)
Frigid
Rigid
Flexible
( lb)
(%)
F
,.=~
b
(II)
1.4b<'l
(plf)
Plywood
I or2
stdes
Allowable
Shear
(pi f)"~'
EdgeNa1l
Spac1ng
(1n)
Third-Floor Level
A
4574
14,879
225
12.5
850
I
870
2"'
B
20,071
15,915
-21
22.0
652
I
665
3
c
36,160
33,370
-8
43.0
601
I
665
3
E
36,160
28,978
-20
43.0
601
I
665
3
F
25,829
47,691
85
43.0
792
I
870
2"'
G
14,905
14,643
-2
22.0
484
I
665
3
1-1
4574
14.291
212
12.5
817
I
870
2"'
I
32,340
45,787
42
64.5
507
I
510
4
2
38,797
27,708
-29
60.0
462
I
510
4
3
38,797
27,708
-29
60.0
462
I
510
4
4
32,340
45,787
42
64.5
507
I
510
4
Second-Floor Level
A
5171
14,748
185
12.5
843
I
870
2"'
B
22,692
17,900
-21
22.0
737
I
870
2
c
40,882
4 1,326
I
43.0
686
I
870
2
E
40,882
36,720
-10
43.0
679
I
870
2
F
29,202
45,732
57
43.0
760
I
870
2
G
16,852
15,056
-II
22.0
547
I
870
2
li
5171
14.241
175
12.5
814
I
870
2'~
I
36,563
53,935
48
64.5
597
I
665
3'~
2
43,864
29,401
-33
60.0
522
I
665
3
3
43,864
29,401
-33
60.0
522
I
665
3
4
36,563
53,935
48
64.5
597
I
665
3"'
Notes for Table 1-35:
I. In SOC D. E, or F. SDP\VS Section 4.3.7.1 requires 3x nomtnaJ thickness stud frammg at abutting
panels or two 2x members where the reqUJ.red nom mal shear exceeds 700 plf, ot lhe n.ad spactng ls
2 anches on center or less at adjoantng panel edges. or JOel common nails ha\'ing penetration into f:ram1ng
members and blocktng of more than 10.: tnches are 3 mches on center or less.
2. Refer to Pan 7.3 m this design example for sill-plrue anchorage.
3. IBC Section 1705.11 requires special inspection where lhe nail spacing lS 4 ioc,hes on center or closer
\Vlth SOC C and higher.
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Design Example 1 • Four-Story~ Light-Frame StTucture
4. The shear wall length used for \vall shears is the ..out-to-out" wall le.ng.th.
5. Forces are suength leve-l and shear in wall tS d1vided by 1.4 to conve.n to allowable stress destgn.
6. APA « TECO perfonnance-rated Strucruml-1-rated wood structural panels may be either plywood
or OSB. The allowable sht."ar values are from SDPWS Table 4.3A usmg IOd common nails w1th a
mmtmum I~toc.h penetration and ·~Hnch panel thickness and d1vided by the ASD reduction tactor of
2.0.
7. Where the fOrce used was the highe-r tOrce tOr the same wall at the oppos1te side of the st:ru('ture, the
h1gher force was used.
8. The shear forces due to the rig1d dtaphtagm analysis exceeded me force us.ng flexible diaphragm
assumpuons requlnng the naihng tn the wall to be inc.reased to mee1 the dernand. Whe-re fOrces ti'om
rigid diaphragm analySts are htgher than those from the flexible diaphragm analysis, wall stabtl ity and
anchorage must be reevaluated. However. eng,neer~ngJudgmem may be u~ to determine 1f a complete
r~gid diaphragm analysis should be repeated due to changes in wall r~gidity. If ng1d d1aphmgm load~
are used, me dtaphragm shears should be rechecked tOr [Oial she.ar load for the shear \valls at that hoe
dl'\'lded by diaphragm length along the u~hvidual y,'3JII anes..
6.6 DETERMINATION OF SEISMIC DRIFTS USING DIAPHRAGMS
IDEALIZED AS RIGID
ASCE 7 §12.8.6
The shear \Yall deflections used to determine the shear wall rig1d1ttes tn Tables 1-16 through 1-30 are-based
on suength-leve-1 seismic fOrces us1ng diaphragms idealiZed as fte.x1ble. Since lhe shear waJI designs are 1n
the ela~ic and lmear range. the seismiC dnfls for diaphragms tdeaJtZed as r1g1d can be propon10ned lO lhe
se.•stmc dnfl..; obu11ned from d1aphrngms 1dealized as flexible.
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Design Example 1 • Four-Story VKiod Ught Frame Structure
4
Table 1-16.
Det~rmmation ofsttvnlc drifls
using diaphragms ldealt:ed as rlgid
g
!J.rig1d
ll
Frigid
~
c
Wall
Frigid
Fftex
F Hexible
"-flexible
( F ngid
Ftlex
)t.
flex
Roof level
~
~
~
0
"'
€
~
A
3870
1587
2.438
0 .07
0.17
B
ti083
6965
0.873
0 .18
0.16
c
12,244
12,548
0.976
0 .15
0.15
E
10,757
12,548
0.857
0 .15
0.13
F
12,906
8963
1.440
0 .09
0.13
G
4633
5172
0.896
0 .15
0.13
H
3061
1587
1.928
0 .07
0.13
I
16,363
11,223
1.458
0 .04
0.06
2
9,385
13,463
0.697
0 .08
0.06
3
9,385
13,463
0.697
0.08
0.06
4
16,363
11,223
1.458
0 .04
0.06
FotJnh-Fioor Level
~
~
Ul
~0 ~0
Z"'
A
10,409
3379
3.080
0. 1~
0.43
B
10,792
14,829
0.728
0.56
0.41
c
26,0ti0
26,715
0.975
0 .38
0.37
E
22,495
26,715
0.842
0 .38
0.32
F
29,3n
19,082
1.539
0 .21
0.32
G
8064
11,012
0.732
0 .45
0.33
H
7998
3379
2.367
0 .14
0.33
I
34,146
23,893
1.429
0 .15
0.21
2
20,143
28,663
0.703
0 .29
0.21
3
20,143
28,663
0.703
0 .29
0.21
4
34,146
23,893
1.429
0 .15
0.21
{continued)
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Design Example 1 •
Four-Story~ Light-Frame StTucture
Table 1-36. DeJerminaJu:m of.teismlc drifl.s u.tmg diaphragms ideali:ed a.r ngt'd-a:Jn.Jmmtd
.2
1l
.~
c
6ngjd
Fngid
Wall
Fngjd
Fftex
F fle.XJble
6 ftexible
(Frigid )6 flex
F flex
llurd-Foor Level
~
~
-5
~
0
~
~
A
11 ,538
4574
2.523
0.21
0.52
B
18,326
20,071
0.913
0.55
0.50
c
37,548
36,160
1.038
0.43
0.45
E
32.552
36, 160
0.900
0.43
0.39
F
34,927
25,829
1.352
0.29
0.40
G
12, 109
14,905
0.812
0.49
0.40
H
8928
4574
1.951
0.21
0.40
I
45,696
32,340
1.413
0. 16
0.22
2
28,062
38,797
0.723
0.30
0.22
3
28,062
38,797
0.723
0.30
0.21
4
45,696
32,340
1.413
0.16
0.22
Second-Floor Level
~
~
"'
.,!. ~
c~
0
A
10,7 14
517 1
2.012
0.26
0.54
B
21 , 134
22,692
0.931
0.55
0.52
c
46,262
40,882
1.132
0.40
0.46
E
39,457
40,882
0.965
0.40
0.39
F
39,005
29,202
1.336
0.30
0.40
G
13,105
16,852
0.778
0.52
0.40
H
8074
5 171
1.56 1
0.26
0.41
I
48,065
36,563
1.315
0.20
0.26
2
35,087
43,894
0.800
0.32
0.25
3
35,087
43,864
0.800
0.32
0.25
4
48,065
36,563
1.315
0.20
0.26
0
Zv>
Nmes tbr Tab!e 1-36:
I. Values for F ngid and FfJex are the fOrces ro the respecuve walls from Table 1-35.
2. The d flexible values are from Table 1-18 for the roof level, Table 1-22 for the fourth-floor level, Table
1-26 for the !lurd-ftoor level, an<! Table 1-30 for <he secon<l-ftoor level.
3. Values ford ngtd are obtained by mulupJytng the rario ofF ng.id/F flex limes d tlexable.
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Design Example 1 • Four-Story VKiod Ught Frame Structure
4
SDPWS §4.2.5.1
6.7 DETERMINATION IF A TORSIONAL IRREGULARITY EXISTS
Per ASCE 7. a Type Ia torsional irreguJarity exast(i whe1llhe ma>amwn story drift exceeds 1.2 umes lhe
average story dnll A Type I b torstonaJ trregularny ex1sts when the maxtmum stoty drift e.xeeeds 1.4 umes
the average swry dnfi (see Figure 1-43).
The ratios of the maxtmwn 10 the average story dtdls (see Table 1-37)are close to exceedtng lhe hmtt
of 1.2 lin1.es lhe 3\'erage story dnft; when they do, a Type Ia torstOnaJ trregularny e.'<ISts. It IS notlhat
uncommon 10 exceed the hmit of 1.4. and when they do. a Type 1b torsional trregulamy ex1sts.
Avo rago Story Drift
Flgu" 1-13.
Tab/~
le\td
fourth
Floor
llurd
Floor
Typ< Ia
Typ< lb
Extsts?
ll..-
~
IJ.,...Ill...,
(> 1.2)
£.'(1Sts?
I> 1.4)
0.1 5
0.17
1.13
North-South
0.06
0.06
1.01
£a'it Wesa
0.37
0.43
1.15
Nonh-Sou!h
0.21
0.21
1.01
E.ast Wes1
0.46
0.52
1.15
North Soutb
0.22
0.22
1.02
East Wesa
0.47
0.54
1.16
Nonh-Soulh
0.26
0.26
1.01
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
Dtrccuon.o;
l!ast Wes1
4
4
4
Sc:rond
floor
oftn·erog~ SIOI'Y drift
1-37. Dett,.mmation ofa1·ero~ and ma:C1m11m s1ory drifts
4
Roor
Det~rminatitm
4
Notes rorTable 1-37:
I. Values for ~·cn~ are the average story drifts obtained from Table 1-36.
P<r ASCE 7 S«<ioo 12.3.3. 1:
Fora Type latorsiol'lal trreg.uJartty:
• Does not e.x1st with flexible diaphragms
o Appl tes only 10 r1gad or sem1ngtd diaphragms
For a Type Ib extreme torsional irregulamy:
Does not e.'HSt with fte."<&ble diaphragms
Appltes only to ngtd or semirigtd diaphragms
Shall not be permitted ut Setsmtc Design Category E or F
o
o
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Design Example 1 • Four-Story~ Light-Frame StTucture
The tests for diaphtagm irregulamy in ASCE 7 pe:ttatn to diaphmgms classified as semirtgid or idealized as
rtgid. For diaphragms tdeahzed as flexible. lhese requtrements do not apply stnce dtstributton is by lttbutary
area, and the diaphragms are not able to distribute torstonal forces. Refer to Secuon 3 ln dus destgn
example for further dlscusston on the use of diaphragms idealized as ftextble.
ASCE 7 requires structures with Type Ia or Type Ib hortzontal itregulariues to be amplified per the
equouion be Iow:
•l s_ )'
A • - l.2Sms
S...n =maximum displacement using Ax= 1.0
8...,.!. =average displac-eme.nt at the extreme points ustng Ax= I.0
where the amphficatton fuctor A.. is multiplied by the accide,ntaJ torsion length. e (see Section 6.4).
Tors1onalt.r1"egulanty can potenttally be a cause for failure tn an eanhquake. In suuctures classtfied as
tors tonally irregular. SDPWS now has added specUlJ requtremems that must be met m order tbr the
struc.ture to have a beuer pertbrmance. These 1ndu.de Iim1ts on dtaphragm matenals. dtaphragm aspect
ratios. and bullding drtft.
Ptr SDPWS Strtioo 4.2.5.1
Where a torsional irregulanty e.\:lsts in suuctures asstgned tn SetSmtc Destgn categories B, C. D. E, and F,
dtaphragms shall meet the tbllow1ng requirements:
I. The d~aphragrn conforn1s to Sections 4.2.7. 1. 4.2.7.2. Md 4.2.7.3.
2. The UW mtio is not greater than 1.5:1 when sheathed i.n conformance to Secuon 4.2.7.1 or
not grearer than 1: I when sheathed tnconfonnance to Sectl0114.2.7.2 or 4.2.7.3.
3. l11e maxintum story drtft at each edge of the structure shall nOt e.xeeed the ASCE 7 allawable
story dnft when subject to seismic destgn forces. mdud1ng torsion and accidental tOI'Sion.
Checks:
I. l11e dlaphragm assemblies used m thts design e.xa.mple are constructed wtth \\"OOd structural
panels~ therefOre. requirement I is met.
2. The maximum L/W ratio tn this design example (see Figure 1-2):
Therefore. requtrement 2 ts met.
3. Building dnfts Will be checked tn Secuon 6.8.
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4
6.8 DETERMINATION OF BUILDING DRIFTS
ASCE7
Drifl Cbt<ks
To establish dnfl. the story drift, 8...-. must be detemuned a(i follows:
§12.8.6
where
C4 =4.0
I= 1.0
5X
=4 "I06 =4.0d
The ca1c:ulated story drtft ustng 8x shalJ not exceed the maxamum d.v whtch IS 0.025 umes the story he1ght
The draft check is swnmanzed 1n Table 1-38.
Ptr ASCE 7 S<ttioa 12.8.6
Stoty dnfts shall be computed as the difference of the defteetions at the centers of mass (or a ve:nacal
ptojecuon where the centers of mass do not align).
In Setsmic De:stgn Caregones C through F havtng Type Ia or lb honzontaltrregulanty. the design stoty
drift shall be lhe largest difference oflhe deftections of vertically aligned points along any of the edges
of the strucu.ar~
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Design Example 1 •
Four-Story~ Light-Frame StTucture
Tab/~
Wall
1-38. Dnfi check at ttJCh /e,·~/
tJ.
He1glu
8,
(1n)
(It)
(1n)
Max. A"
(In)
SllUUS
Roofle\•el
A
~
~
"'
~
0
~
0.17
8.21
0.67
2.46
2.46
B
0.1 6
8.2 1
0.64
c
0.1 5
8.21
0.58
2.46
E
0.13
8.2 1
0.51
2.46
F
0.13
8.21
0.52
2.46
G
0.13
8.2 1
0.52
2.46
H
0.13
8.21
0.53
2.46
1,4
0.06
8.2 1
0.23
2.46
2a, .la
0.06
8.21
0.23
2.46
2b,3b
0.06
8.2 1
0.23
2.46
0.06
8.2 1
0.23
2.46
z" 2c.Jc
ok
ok
ok
ok
ok
ok
ok
ok
ok
ok
ok
Founh-Fioor Level
~
,;.
<ll
~
0
~
A
0.43
9.44
1.72
2.83
B
0.4 1
9.44
1.64
2.83
c
0.37
9.44
1.47
2.83
E
0.32
9.44
1.27
2.83
F
0.32
9.44
1.29
2.83
G
0.33
9.44
1.31
2.83
H
0.33
9.44
1.32
2.83
1,4
0.21
9.44
0.85
2.83
2a, .la
0.2 1
9.44
0.85
2.83
2b,3b
0.2 1
9.44
0.83
2.83
0.2 1
9.44
0.83
2.83
z" 2c.Jc
(conti11ued)
120
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ok
ok
ok
ok
ok
ok
ok
ok
ok
ok
Design Example 1 •
Four-Story~ Light-Frame StTucture
Tab/~
Wall
1-38. Dnfi check at ttJCh /e,·~/
tJ.
He1glu
8,
(1n)
(It)
(1n)
Max. A"
(In)
SllUUS
Roofle\•el
A
~
~
"'
~
0
~
0.17
8.21
0.67
2.46
2.46
B
0.1 6
8.2 1
0.64
c
0.1 5
8.21
0.58
2.46
E
0.13
8.2 1
0.51
2.46
F
0.13
8.21
0.52
2.46
G
0.13
8.2 1
0.52
2.46
H
0.13
8.21
0.53
2.46
1,4
0.06
8.2 1
0.23
2.46
2a, .la
0.06
8.21
0.23
2.46
2b,3b
0.06
8.2 1
0.23
2.46
0.06
8.2 1
0.23
2.46
z" 2c.Jc
ok
ok
ok
ok
ok
ok
ok
ok
ok
ok
ok
Founh-Fioor Level
~
,;.
<ll
~
0
~
A
0.43
9.44
1.72
2.83
B
0.4 1
9.44
1.64
2.83
c
0.37
9.44
1.47
2.83
E
0.32
9.44
1.27
2.83
F
0.32
9.44
1.29
2.83
G
0.33
9.44
1.31
2.83
H
0.33
9.44
1.32
2.83
1,4
0.21
9.44
0.85
2.83
2a, .la
0.2 1
9.44
0.85
2.83
2b,3b
0.2 1
9.44
0.83
2.83
0.2 1
9.44
0.83
2.83
z" 2c.Jc
(conti11ued)
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Design Example 1 • Four-Story VKiod Ught Frame Structure
4
Table 1-18. Dnft chedi tJI ~ad' /n-~1-t•ontmued
t.
Height
s,
M~a..
Wall
(tn)
(ft)
(to)
(tO)
Status
A
0.52
9.44
2.09
2.83
13
0.50
9.44
2.00
2.83
c
0.45
9.44
1.80
2.83
E
0.39
9.44
1.56
2.83
F
0.40
9.44
1.59
2.83
G
0.40
9.44
1.61
2.83
H
0.40
9.44
1.62
2.83
ok
ok
ok
ok
ok
ok
ok
ok
ok
11tird-Aoor Level
~
-§
~
Ul
,s
~
0
~
0
z
1,4
0.22
9.44
0.90
2.83
2a, 3a
0.22
9.44
0.90
2.83
2b, 3b
0.22
9.44
0.86
2.83
2c, 3c
0.22
9.44
0.86
2.83
ok
ok
Second-Floor Level
~
.
~
ell
,s
A
0.54
9.44
2.16
2.83
13
0.52
9.44
2.06
2.83
c
0.46
9.44
1.83
2.83
E
0.39
9.44
1.56
2.83
F
0.40
9.44
1.59
2.83
G
0.40
9.44
1.62
2.83
H
0.41
9.44
1.63
2.83
9.44
1.04
2.83
1,4
0.26
2a, Ja
0.26
9.44
1.04
2.83
2b, Jb
0.25
9.44
1.01
2.83
2c, 3c
0.25
9.44
1.01
2.83
~
0
~
0
z
ok
ok
ok
ok
ok
ok
ok
ok
ok
ok
ok
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Design Example 1 • Four-Story~ Light-Frame StTucture
7. Design and Detailing of Shear Wall at Line C
7.1 DETAIL OF SHEAR TRANSFER AT ROOF
Edge nailing from the roof sheathtng to tJ1e collector trus.<; may need to be closer than lhe roof sheatJ\Ing
edge nailing or "field natling., because ofshe3fS being collected from each s1de oflhe truss. It is also
c.ofnmon to use a double colle-ctor uuss at these locauons. 11\e 2 x 4 btaces at the top of lhe shear waJI need
to be designed for compression or to provide tenston bracing on each side of the waJI (Figure 1-44).
ROOF EDGE NAILING
ROOF SHEA THING
COLLECTOR TRUSS
MFRD ROOF
TRUSSES
FRAMING CUP
II
8'-i/c
2x4 BRACE AT
EDGE NAILING
PER SCHEDULE
2x CONT. w/
16d SILL NAILING
PER SCHEO.
2
X
4
STLIDWALL
Figure I -·1-1. Dt1all ofshear lftmsftr tl/100/
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Design Example 1 • Four-Story VKiod Ught Frame Structure
4
7.2 DETAIL OF SHEAR TRANSFER AT SECOND FLOOR
ll\is detail uses the double top plates at the underside of the floor sheathing (Figure 1-45). This IS
adnJUageous for shear transfer. Another often-used detail is to bear the floor joists d1rectly on lhe top
plates. However, when the floor JOists are on top of the top plates, shear t.ra.nSfer IS requJted lhrough the glue
jomt in the Ylebs and heavy nailJng from the joJ:St chord to the top plate.
The r.culers for tlte drywall ceding need to be installed after the \\'all sheath1ng and gypsum board have been
Lnstalled.
1ST SPACING
DOUBL E 2 x 4
SILL V !TH
PROGRESSIVE
NAILING
4
X
4
TOP
PLATE
Figure I -45. !Niatl of.'lhtar lftllufir OJ stCOtfd-jloor Jerel
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Design Example 1 • Four-Story~ Light-Frame StTucture
7.3 DESIGN SHEAR TRANSFER AT FOUNDATION SILl PLATE
SiD-Platt Anchoragt for Sbur \Valls
\Vas btr Platts
~ tnch oflhe edge oflhe
bottom plate on the stde(s) wnh sheathang wheu lhe nomlna1 shear capacity is greater than 400 plf for both
WU'Id and seismic forces. The SDPWS has an exce-ptton to the 3-inch x 3-tnc:h x 0.229-inch plate \'l<lShers.
Section 4.3.6.4.3 pe:rmtts standard-cut washers tn tndividual fuJI-height shear wall segments (Section
4.3.5.1) when three condttions are met:
The SDPWS requtres lhe 3-inch x 3-inch x 0.229-inch plate washers to within
I. 1lle shear wall anchorage at the \vall ends u.o;ed to resist ovenuming is designed neglecung all
dead loads for ovetturning resistance.
2. ll1e shear wall aspect rauo is a maximum of2: I.
3. 1lle ASD setsmic demand is a maxtmum of 490 plf.
The most stgmficant tssue wnh sill-plate anchorage is the requtrement ofha\•&ng the washer extend to
wtthin ~ inch of the bottom plate edge on the stde of the sheathing.
I. For a 4-lnc:h-nominal-wtde sill plate and ustng 3-tnc:h x 3-i.nch x 0.229-inch washer plates. this
presents no problem.
2. For 6-mc.h-nomtnal and larger stll plate~ lh1s will reqmre lhe anchor bolts to be otrset from
the center of the sail plate when 3-inch x 3-t.nch x 0.229-mc.h washer plates are used. h is also
possible to use a larger rectangular plate.
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Design Example 1 • Four-Story VKiod Ught Frame Structure
4
~
~SQUARE PLATE WASHERS
v /'""~
Wj
3X4 PT SILL PLA'JE ~
W/ M.B. PER SCHED.
0
2 ROWS E.N. \
~~
••
PL
s
DRI~ PIN
•
I
/
•
Flgrn I -16. ShMr transfer at the foundatlo"
Anchor Bolli
IBC Section 1905.1 .8 exemptsACJ318 Chapre:r 17 requt.rementson anchor bolts in wood stU plates. The
lateral design strengtlt is determined using the values spec1ti.ed in NOS Table 12E, provJded that the anchot
bolts comply w1lh all of the following condttions:
I. AIIO\'<able in-plane shear suength IS determined in accordance with NOS Table 12E.
2. The anchor bolts must be a ma:<Jmum of~-tnc.h diame.ter.
3. The aochor bolts are embedded a mtnunwn of7 mches imoooncrere.
4. The anchor bolts rue located a mmimum of I~ inches from the concrete edge that1s paralJel to
the sill plate.
5. The anchor bolt IS Ioc.ated a mmimum of I 5 anchor diameters from a concrete edge that is
perpendicular to the stll plate.
6. The sill plate is of2 inch or l-inch nomtna1thtckness
4
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Design Example 1 • Four-Story~ Light-Frame StTucture
It IS the opinion of the SEAOC Seismology Committee that %-inch-diameter bolts can be tncluded tn those
condtlions, and tlte Committee is worktng on code change proposals for thts. Outside of Calitbrnia. the
destgner sull has tO meetlhe reqwremenl'l ofACI 318 Chapter 17. In Sectton 17.2.3.5, there IS a prO\•tsJOn
for srud beartng "ails. his the opinton of the SEAOC Setsmology Commmee that additional code changes
need 10 be implemenred tnto ACI allowing the exemptton ofACt 3 18 Chapter 17 for sill plate botts.
Dtsig:n S ill
Pltlt An thor Bolls
See the discussion about fasteners for preservative-treated wood.
From Table 1-36:
1"=806plf
For a side me.mber, thickness= 2.5 utches tn Hem-Fir wood (nme that designing for Hem-Fir vt11l require a
Ughter nallru\d bolt spacing) and using a ~-inch bolt:
Z,= 1070 lb I boll
6
Requtredspacin•=
Z,Co
- 255in
~
,. = III70xl.
8 0 6212fi
-..
Use 5~tneh-dtame-ter anchor bollS at24-tnch spacing.
Corrosion or Fasttntrs in Trtaltd Sill Pltlts
With the EPA removing CCA as a wood preserv~uive, the chemtcals that are 1'10\.\' used for wood
presel"\•atives oontru.n htgheJ levels of copper. nus h1gher level of copper ln the presef\•auve-t!eated lumber
has a dJSSimtlar metals reacrion wtlh carbon steel. IBC Secuon 2304.10.5. I requires faste-ners in contact
with preser\'aUve-ueated wood to be hot-dipped 2inc:-cooted galvaruzed steel. stail\less steel. sthcon bronze.
or copper. This requirement applies 10 not onJy the anchor bolts but to the washers and nails. E\'en the toe
t~a~ls from lhe stud to the sill plate would require lhe prOtection. 1be na1hng of lhe shear-v.oall sheathing to
the Sill plate would need 10 meet thts requirement Most nail guns used by carpenters cannot use gal\•ant.Zed
nails. mea.ung thm the carpe-nter need.~ 10 enher hand nail the nails from the sheathing ro the sill plate or
change nrul guns to make these fastenings. Elecltogalvanaed natls (silve-r s.huty natls) do not meet the
above requ1remt11ts. The IBC does give 3.1\ e.xception to the.above requirementS when two conditaons are
meL:
I. The presen'ative-tte-aunent process uses SBXIDOT or tlnc borate.
2. Fasteners are used in an tnterior, dry environment.
Many of the pc-eservative-treatmetU plants do nm use the above ptocesses, so dependtng on lhe locat10n of
the prOJect, this type of treated wood may not be available. There can be no e.xcuse that the f'hlmer/carpeatter
dtd not know \\'hich type of prese.f\'31t\'e treatment was used for the lumber purchased; these are stated
on a label thattSStapled omo theendsofeach ptece of lumber. Many of these labels will have statementS
··corrosive to Fasteners.. « sam alar wordmg.
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Design Example 1 • Four-Story VKiod Ught Frame Structure
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Sptcits or Pru e.rvativt-Trrattd \Vood
An addiuona1 caut.ion for sill plates lS the type of wood used4 The JllO:St oommoo species used on the west
coast tor pressure treatment is Hem-Far. whtch has loo·er tasuner \1alues (spectfic gravity G = 0.43) for
natls and boltS than Oougla(i Ftr-Larch. A tighter nad spacutg to the stU plate may be necessary. depending
on the loads.
Gap at Bouom or S bratblng
lnvesugauons into wood-frame construCtion have found that plywood or OSB sheathtng that bears on
concrete 3[ perimeteJ ex1erior edges can wtck up moisture from the concrete and cause corroston of the
fasteners and rott.ing. in the sheathmg. To help prevent th1s problem. the sheathtng can be placed with a
gap above the conctete surface. A Y
...-lnch gap is recorrunended for a Jx sill plate? and a ~tnch gap is
recomme.nded for a 2x siJI plate (F1gure 1-47).
STANDARD CUT
00 1-tEAVV PLATE WASiiERS
z
~
'
HOT·OIPPEO ZINCCOATED NAILS AT E N.
TO SILL PLATE
Flgr.tn I-./7. Sill p/(lJt at theJaundation ~dg~
11te SDPWS only requlres a mintmum edge diStance of Ys mch fOI' natls tn sheax.hiJtg. Tests have shov.n that
sheathtng wtth greater edge dtstances has performed bener. In additJon_.lhe ~- utch gap shown at the bonocn
of the sheathing should be a tUtntmwn. Rece-tU cyclic lesung has shown that when this gap is Y! toc,h? the
shear wall has performed beuer.
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Design Example 1 • Four-Story~ Light-Frame StTucture
8. Diaphragm Deflections to Detennlne If the Diaphragm Is Flexible
Thls step is shown only as a reference tbr hoo• to calculate horizontal diaphragm deflections. Su-.ce
the shear \\'all forces were detemuned using both fte.xlble and r1g1d diaphragm idealization. there JS no
requ1remem to venfy the diaphragm ts actually rigtd or tlextble.
The roof diaphragm has been selected to 1llusuare the methodology. The design sets:mec force in the roof
dtaphragm usingASCE 7 Equatton 12.10-1 mus1 first be dete.rmined. The design selsmic focce is then
d1vtded by the diaphragm area to determu'le-lhe horizontal loading in pounds per square foot. These values
are used for dete-nmning diaphragm shears (and also collector forces). The destgn se1smac force shall be not
less than 0.2SJJ$/w14 nor greater than 0.4Smlw,..
8.1 ROOF DIAPHRAGM CHECK
The roof diaphragm Will be checked in two steps. Firs~ the she,ar in the diaphragm Will be detem1jned
and compared to alloooable. Next, the d1aphmgm deftecuon will be calculated. In Pan 7b. the diaphragm
deJte(:t:ion is used to detennine whether the diaphragm lS flextble or rig1d.
Cbttk Diaphragm Shur
The roof diaphragm consiStS of 1¥.c-tnch-thtck sheathtng nailed wnh IOd common na1ls at 6 inches on
cente-r, and pane-l edges ate unblocked. Lood1ng oo the segment be.tv."een lines C and E. where
v- (8.31)x48.0ftx(32.0ft)x0.7 _
If
93
p
48.0 fix2
Dtapbragm span= 32.0 f\
Diaphragm dep1h = 48.0 ft
Diaphragm shears are convened to aiiO\\'Ilble stress des1gn by muluplying by 0.7.
From SDPWS Table 4.2C, the allo\\~ble shear of380/2 = 190 plf ts based on % -.nch APA orTECO
performance-rated wood struetural pan~ls wnh unblocked edges and JOd common nails spaced at 6 utches
on center at boundaries and supponed panel edges. APA or TECO pe-rformance-rated wood struetural
panels may be either plywood or OSB. However, !his design example will use ply\\ood \\; lh five plys.
Cb~tk
Diapb.-.g·m Dr:Ortlioa
The code spec1fies that the de.tlectjon ts calculated on a unit load basts. In other words. the diaphragm
deflection should be based on the same lood as the load used for the lateral-resjsting elemen[S, not F,. total
force at the level cons:1dered. S1nce the code requ1res buildmg dnfts to be de.temuned by the Strength-level
forces spec: died in Secuon 12.8. strength loads on the buildmg diaphragm must be determined.
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Design Example 1 • Four-Story VKiod Ught Frame Structure
4
The basic equation to de.tennine seismic forces on a diaphragm tS
•
0
"'
F
.
F,
,.
=~w
0
w
ASCE 7 Eq 12.10-1
'
_ (43.92xl65.67)
,k.
- 439
. • tps
165.67
~fy~
For s unpllc~ty. the wan wetght..(l parallel to the d1recuon ofload1ng are 1nduded '" ...,.....
For the uppermost level. the abtwe calculation WJU always pnxJuee the same fOrce as compmed m ASCE 7
Equation 12.8- 11. Then divide by the area of the d1aphragm to find the equivalent uniform force.
f
= 43.92x l000 _ 8_3 1 sf
5288
p
-
In this e.x.ample. the roof and floor diaphragms spanntng bet\\•een ltnes C and E Will be used 10 Illustrate the
is
mt-d lod. The basic SOPWS equation to de.temune the deHecuon of a dtaphragm
ft.= 5t·L' + 0.25•L + !(xtt., )
S£AW IOOOG.
211'
SDPWS Eq 4.2-1
For the purpose of thts destgn e.xampJe. the djaphragm IS as."'umed to be a simple span supported at l1nes C
and E (refer to Figure 2-4). In real ity~ \v;th conum.uty. the actual deflection wdl be less.
With n.ads at 6 lnches on center. the stre-ngth load per n.ad ts 93 x 1.4(61 12) = 651b/nad = V..-Other terms 1n
the deflection equauon are
r:93plf
I.= 32.0 ft
JV:48.0 ft
SDPWST4.2C
G.= 10 x 1.2 = 12.0 kops/on
£ = I,700,000 psi
A2xactr.or.:h -=5.25 in2 x2 = 10.50 1nl
Assume a chord spltce at the m1d-span oflhe dtaphragm that will be nailed. The allowable loads tbr
fasleners are based on limn state destg_n4 In other words, the defOrmation is set at a limn rather lhan the
Sltength of the fastener. The defonnation hmn l'i 0.05 diameters of the fastener. Fot a 16d common nail, a
conservauve shppage of0.0 1 tnch wtll be used.
Ustng suength-level dtaphragnl shear:
!(<leX)= (0.0 1) 16.0 ft(2) = 0.32 in-ft
6
=
5(93x 1.4)32.0 1 + 0.25x(93x 1.4)x(32.0) + 0.32 =0.0 in
9
8(1.7£6)10.50(48.0)
IOOOx 12.0
2(48.0)
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Design Example 1 • Four-Story~ Light-Frame StTucture
From Table 1-21, the shear wall deftecuonal hoe C and hne E is 0.33 inch. For flexible diaphragm
as.sumpuons (one of tluee me-lhods), the SOPWS and ASCE 7 require the diaphragm to de.flect t\vice as
much as the laleml-res1sung elements, or in other \llOrds, [0 be cons1dered flexible the diaphragm would
need to de-flect 0.33 x 2 = 0.66 lnch. By the definition tn the rode, the dtaphragm is considere.d to be rigid.
9. Discontinuous System Considerations and the Overatrength (fl0 ) Factor
9.1 ANCHOR FORCES TO PODIUM SLAB
For over 20 years, the butld1ng codes ha\'e had requtrements to use amphfied setSintc forces tn the
design of e-lementS supporung diSCOntinuous systems. Earlier e.dllions of the codes used the term 3Rw/8,
while current codes use the tenn 0.0· ASCE 7 Section 12.3.3.3 reqUires ampJificauon of seiSmic loads
tn the design of structural eleme.ms supponing disconunuous walls. Previous editions of the IBC and
the 1997 Uniform Budding Code e:<empted concrete slabs supporting hght-frame construction from
these require-ments. However, ASCE 7 does not ha\'e th1s exce.puon, thus adding slabs to tl1e Iist of
elemenlS needing the design strength to resist the maxlmum a>< tal force lhat can be de-levered per the load
combination.~ with the overstrength (actor (Clo) tn ASCE 7 Secuoo 12.4.3.2.
Thls means that the shear y,oaJI boundary overturning forces (axial uplift and axial compresston) need to
have the Q factor of 3.0 apphed to the support1ng slab destg.n. Footnote g ofASCE 7 Table 12.2-1 states
that for struCtures wnh Hexible diaphrag.nt~ llus \•alue may be 2.5. ll\e overstrength factor does not need
to be applied 10 tl'!e shear wall's connec.tions. ASCE 7 Sec.tion 12.3.3.3 stues that the cottnecuons of the
dtscotttinoous wall to the supporting element need only be adequate to resist the forces for which the
dtscontinuou.~ wall \llas destgned. 1be expanded commentary ( 3rd prtntmg ofASCE 7-10) ofSecuon
12.3.3.3 provides further explan<llion:
..For wood Jight-frame shear wall construction, the final sentence of §12.3.3.3 results 11'1 the shear atld
overtum1ng connections at the base of a dtscont.inued shear wall (i .e .~ shear fasteners and tie-downs) betng
destgned us1ng the load combinauons of§2.3 or 2.4 rather than the load combu'lat.ions Y.~th overstreng.th
faeoor of§ 12.4.3.IIO\Iltver, O'!.apler 17 of ACI 318. Buildmg Code Requirements for Structural Concrete, does apply a fac.tor
similar to an oversuength factor to brittle conc.rete breakout failure modes tflhey govern the anchorage
destgn. It tScommon to have anchorage to the podtum slab not fall Wlthtn the scope of Chapter 17 because
ofedge distances or available embedmen[ lengths. Othe-r means of bolt anchorage co1nmonly used mclude
through bolting or sleeves tbr post-tnstalled through-bolts, embedtng plates with we.lded studs, beartng
pla[e washers at tl1e bolt nut, or spectaf Sl:eel re.inforc:ing bars used ln conjunction wtth the anchor bolrs/
bearing plates. Recent tesung ha~ sho\\Ttlhat adding retnfOrctng cages allhe ue"ov.·n rods mcrease lhe
strength sigmfican~y.
As d1sc:ussed 10 ASCE 7 Secuon 12.4.3.1, one poss:tble route to reduce the calculated overstreng.th load
oocurs when it can be shown that yieldlng ofother eleme1US(tie-down rod, shear \liall. diaphmgm. collec.tor,
etc. ) wtll oc.cur below the overstrength-level forces. When tlus 1S the case, the setsJntc load effect~ includtng
ovetstre.ngth c-an be reduced w a lower value. ASCE 7's commentary on Section 12.4.3 pro,rides furt.her
e.xplanat'ion:
'"The S[andard penntt~ the seasmtc lood effeclS, tncluding overstrength fuctor, lObe taken a~ less than the
amount computed by applytng n., to the destg.n seismic fOrces where It can be de-termined that yieldtng of
othe-r elements in the struc(ure hmits the amount of load that can be de-hvered to the element and, therefore~
the amowu of force that can develop an the ele-me.nt...
130
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Design Example 1 • Four-Story VKiod Ught Frame Structure
4
10. Special Inspection and Structural Observation
Sptdallnt p«tion.s
IBC Secuon 1705.12.2 requlres special lnspection. when the nad spacing is 4 tnches on cen.ter or closer ·w1th
SOC C an<l higher.
Cu.rrently, me SectJon. 1705.12.2 is nm clear on how far up the load path to go with the specUll u1Spectiorts.
The IBC States: ..pe,riodic spectal tns:peclion. shaJI be requued for nathn~ bolung. anchonng and olher
fa(lten.ing of elements of the seismic force-resisung system. including wood shear \'-'ails. \\'OOd dmphmgms,
drag struts, broces,. shear pane-ls and hold-do\"ns.r The code then. exempts specialtnspecoon. requtreme.nt<t
for the above ire.tns when the fasteoer spacittg to the sheathmg is more Jhan 4 lnches on cettter.
The code also e:.:empts specialtnspect:ions when eu.her lhe SJJS does oot exceed 0 .5 and the budding he-tght
of the strucwre does not exceed 35 fee.t. In the case ofthis design e.xample.lhe value of SAS'e.xceeds 0.5,
and the height exceeds 35 ite.t.
An itemtzed list of spec tal inspection tte.ms placed on the structural draY.tngs can help t11e process of whic.h
ttems should be inspected. but from experience. the uems that the enganeer of record and the buHdmg
official agree to being placed on t11e permn set of plans often ge.ts either agnored or amended by the city
tnspeclO(
at the St£1!.
Struttural Observations
l11e code ha(l been. ame1lded to tndicate a structural observation as follows due to t11e building otlicial at the
e.nd of the work, not prior to c.aJiing for each tnspection:
1704.6.1 Structural obsrrvatioas for struttuns. Structural observations shall be provtded for those
suttctures whert one or more of the followang oottdttions e.xist:
I. The structure 1s classified as Risk Category IV.
2. The-structure is a high-rise buildmg.
3. Such observation is required by the registered design profe:sstona1 responstble fbr the structural
desJgn.
4. Suc-h obsel'\'ation IS specifically required by the buildtng official.
1704.6.2 St:rut-tural obsrn·ations for stbmit ruistantt:. Structural obsetvattons shall be pr0\1tded
fo( those structures assigned to Setsmtc Design C-ategory D. E or F where 0 1te or m.o(e of the following
condations ex1st
I. The structure is classified as Risk Category III odV.
2. The struc.tu(e ts ass1gned to Setsmac DesJgn Category E. is classified ao; Rtsk Category I or u.
and tS greater lhan two stories above the grade plane.
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131
Design Example 1 • Four-Story~ Light-Frame StTucture
11. Items Not Addressed In This Example
The fbllow1ng item.c; are not addressed 1n this example but are nevenhe-less necessary for a complete design
oftlte seismtc-load-resisling system:
Compatt<;OO of wind and seismic tOrce.s.
Anchorage to concrete for tie-dO\m bolts.
Budd1ng separauons.
Design of foundauons.
Rotational effects of foundarions (if any) on shear ·wall defleclions.
Des1gn of the venteal load-carrytng sysle..n, u\cludtng beanng sruds.
Des1gn of load-bearing \\ails 10 mee-l !-hour tire-ranng requlremems.
Shear transfer at roof ltusses to lnterior cotrtdor walls.
Determinruton of redundancy factor p.
VertficattM of shear wall drifts at all levels.
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Design Example 2
Flexible Diaphragm Design
OVERVIEW
ThL'i example illusuates lhe design or a large fle.xable diaphragm in a big-box retatl store subJected to lateral
seisrniC load~ng. ll\e (()Of structure c.on.otists of a panelized hybrid roof system. which as very common
m large-diaphragm roo& in the seJsmicaJiy actave western United States. Thts roof syste.m compnses
SU\JCtural wood-panel sheathmg wnh hght dtme.nslonallumber suppons, resttng on open-w-eb stee1 JOISts
and joist girders. While this example dlust.rates the design of a "'OOd diaphragm. a samilar methodology is
applicable to untopped steel deck diaphragms.
OUTLINE
I. Buildtng Geotneuy and Loads
2. Roof Diaphragm laterol Loading
3. Shear Nathng of the RoofDtaphragra (North-South)
4. Constderottons for Plan lrreg,u1ant•es
5. Dtaphragra Chords (NOrth- South)
6 . Dtaphragra Collectors
7. Diaphragm Oeflec.tion
Figure 2-1. 1jplca/ bwldmg withjle.dble dtaphragm
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133
Design Example 2 • F/Wble ()japhragm Dt!sign
9
<f <f
r
80'. 0"
I
'
'
I
I
I
{
cp cp
0
I
I
0
320' . 0"
I
I
I
-'l-Ir--,--
A
b
6
•
®-
b
b
•b
(~}§
-
I
0
I
I
cp
}
I
I
b
b
•
®
b
®-
b
•
STEEL BEAM
COLLECTOR
OPE.N WEB STEEl.
JOISTS@ S' O.C.
@
7 1/4" PRECAST
CONCRETE PANELS TYP
ROOFPlAN
Figure 1-2. ExampleS T(J()fplan
Figure 2-3. E:wmple :t building sec•tion
134
STE:El JOIST GIRDER
FRAMJNGTYP
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Manual,
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@Seismicisolation
Design Example 2 • Aexible Diaphragm Design
ASCE7
1. Building Geometry and Loads
1.1 GIVEN INFORMATION
Seismic·force-Resistlng System
Bearing-\~,o-all system consistJng of mtermedtate preca(it concrete shear walls suppotttng a tlexlble
dtaphmgm of wood sEnJctural panel.
Seismic and Site Data
Mapped spectral aecelerauons for tl>e siteS,= 1.5 (shon period) s, = 0.6 ( l· secood P'"'od)
Risk Category II (Occupancy)
Site Class 0
Seismic respon.<e coefficoent R = 4 (T 12.2· 1)
Overstrength fac.torl>., = 2.5 (T 11.2·1)
Cs=0.25 (ASCE 7 Section 12.8.8.1)
r=0.25Jff(ASCE 7 Secuon 12.8.1)
Setsmje Destgn Categocy D
Sos= 1.0
Wind
Assumed not to govern
Roof
Dead load = 14 psf
Live load (roof)= 20 psf(reductble)(IBC Table 1607.1 )
Walls
Thtckness = 7.25 tnches of concre-te
Height= 23 teet
Nonnal wetglu concrete= 150 pcf
Roof Structure
Structural-) s:heathtng (oriented strand board wood SlfUCturaJ panel)
Pre-engmeeredlp«"manufactured open-web steel JOtsts and JOISt g~rders wath fuJI-\Vtdth \\"OOd natlets
All '"'od is Douglas for. Larch
2. Roof Diaphragm Lateral Loading
ASCE7
2.1 ROOF DIAPHRAGM SHEAR COEFFICIENT
11te roof dtaphragm mu.st be destgned to resist seismic forces tn each direcuon. The foJiowtrtg formula IS
used to determme lhe total seismtc force F'* on lhe djaphragm at a g~venlevel of a butkling..
.. .... ' ,.,
o • F
F =~w
0
w
,..
Eq 12.10· 1
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135
Design Example 2 • F/Wble ()japhragm Dt!sign
Asgtven, the base shear for this butldtng ts 1'=0.2SW. Because ll is a one-story butldtng. Equation 12.10-1
simply becomes the fOIIowang:
F,. shall be not less than
=
=
Eq 12.10-2
0.2S"'l,w,. 0.2( LOX I.O)w,. 0.2w,_
but need not exceed
Eq 12.10-3
0.4S,.J,w,. = 0.4( 1.0)( I.O)w,. = 0.4w,.
Based on the cmeriagtven in Secuon 12.10.1.1, FJW=0.2s,..p·
Alternauvely, JnS[ead of des:igmng a diaphragm (includmg chords a~td collectors) to Secuons 12.10.1 and
12. 10.2. ASCE 7- 16 tnuoduced Sec.uon 12.10. 3 a.~ an aJternauve methodology. whtch eonstders unique
dtaphragm design force reducuon factors, R~ and a dtaphmgm design acceleratton coefficient, c,.. The
setSmtc force on the diaphragm iU any given Je\<el of a budding is:
F
P
c
=::...J!!..w
R~
Eq 12.10-4
,_
For a one-s tOI)' butlding(N = 1), Figure 12.10-2 ut<hc:ues c,.= c,.., where
c'* = J<r .,n6C1 f' +(r..,2C11 )
=
2
Eq 12.10-7
=
For a one-story building(r,..1 1.0, r _, 0, CSl =0). C,.. bec.ot'ttes
c,.. =0;{.:, = 2.5(0.25) = 0.625
(Note: Foomo<e b to Table 12.2-1 shall not apply.)
However, C,_ shall be not less than Cli' whete c, is the greater of
c,. = O.SC,. = 0.8(0.4S,..,) = 0.8(0.4)( 1.0)( 1.0) = 0.32 < 0.625
... OK
and
c,. =0.9r.,n,c,= o.9( 1.0)(2.5)(0.25) = o.5625 <0.625 . .. OK
In practice, c,..wdl never cot\trol the \1a1ue of C,_ for ooe-stocy butldin~ thus makmg this check trrelevant.
The-refore, ci"X = 0.625.
The diaphragm desogn forre factor, RP ts 3.0 per Table 12.10- 1. Therefore
c,.. 1vpr = -0.625 wJ~J =0.208wP
.
FP. = -R
30
.
Eq 12.10-4
But F,...sball be not less than
F,_ = 0.2S,.J,w,. = 0.2( 1.0)( I.O}w,. = 0.2w,..
136
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@Seismicisolation
Eq 12.10-5
Design Example 2 • Aexible Diaphragm Design
Thus, when using Section 12.10.3
Ffit; 0.208up,.;. In COJtjunc.tJon With using 0o; 2.5
Alternativd)', when using Sections 12.10.1 and 12.10.2
Fp=0.15w,.,tn OOt\junction with using ilo = 2.5-0.5 ;2.0 (FoolllOte b to Table 12.2- 1)
nus example wlil use ASCE 7 Section 12.10.3.
TherefOre, for diaphragm destgn use F,; 0.208,,,.
2.2 ROOF DIAPHRAGM SHEARS
The v.'OOd structural panel roof system lS permnted lObe tdealtzed as a flexible diaphragm per ASCE 7
Section 12.3. 1. 1 or SDPWS Section 4.2.5. Seismtc forces for the roof are computed from lhe trabutary
weigh! of the roof and lhe walls oriented pe-rpendu~ular to lhe dlrecuon of the seismic forces. \valls parallel
to the direction of seismic fOrces do not load the fte.xlble diaphragm. The distrtbuted lateral loading to the
d1aphragm wdl be computed tn each onhogonal d1rection.
Easl-\\'tst dirtttion
Because the panehzed wood roofdtaphragm. m th1s buildmg ts 1dealrzed a.~ fleXIble, lines A. B, and E are
cons1dered lines of resistance tOr the ea.~-\\""tSl selSJntC forces. A coJiector ts needed along I ine B to drag
the trtbtttary east-west dtaphragm force.~ tnlO the-shear wall on lute B. lOe load1ng and shear dtagrams are
shown in Fag.ure 2-4.
@rll
I~
I
b
1:::
1:::
~
E
w,•,F41:A!
8.C 4 k
W.i: !401r,j
~Uft
"""
"""'""
Figur.l-4.
Eas1-w~st diaphrogm
/oadmg
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137
Design Example 2 • F/Wble ()japhragm Dt!sign
The uniform loads H,.1 and W:! 1n the east-west direction are computed using the diaphragm lengths and ·wall
hetghts.
Roof dead load= 14 psf
Wall dead load •
7 25
· 150 pcf = 90.6 psf
12
Roofhetght = 21 feet average
Parapet height= 2 feet average
w; = 0.208( 14 psl)(240 ft) + [ 0.208(90.6 psf)(2J>( 2; ) ;1]2 = 1174 plr
w, = 0.208(14 psf)(320 ft)+ [ 0.208(90.6 psf)(23>(~) ;I}= 1407 plf
In lhts example-. the effect of any wall openings reduc-tng the wall weight has been neglected. Thts is
considered 3.1' acceptable stmplificatton because the opemngs usually occur tn lhe bottom half of the wall.
In addmort, sagnificant changes in parapet hetght should also be cons1dered tf lhey occur due to stgntfic:ant
roof slope.
Diaphragm shear at lane A and on tl1e north stde of lane B lS
23,500 lb _ . If
97 9
240ft
p
Diaphragm she.ar at the south side of line Band at ltne E is
S4,400 lb - 264 If
320ft
p
Nortb-soulb dlrtction
Diaphragm forces for lhe nonh-soulh dtreclion are computed usmg the same procedure and assumptions as
the ea<n-wesr dtrec6on and are shown m Ftgure 2-5.
w, = 0.208(14 psf)(l20 ft)+[0.208(90.6 psf)(23>( ~3 ) ;I}
W,= 824 plf
w, =0.208(14 psf)(l60 ft)+[0.208(90.6 psf)(2J)(~) ;I}
W,=941 plf
Diaphragm unit shearal lu\e I and lhe west side ofhne 3 is
33,000 lb ,r r
120ft - - >pf
138
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Design Example 2 • Aexible Diaphragm Design
I~-~ I1.--_-. ".-_- ,I
1----'
t•t• t tit t t t t t titt
~=a24,.tf
W,r• ~11)(1
"''"""
, ,,. k
..
, ,
Figure 2-5. North-JOJith dlaphrogm loading
Diaphragm untt shear at lhe east s1de of line 3 and at line 9 is
113,000 lb _
If
706
160ft
p
3. Shear Nailing of the Roof Diaphragm (North-South)
SDPWS
The du:tph.ragm loaded 1n the north-south direction has been selected to Illustrate the design of a wood
struc.tural panel roof dtaphragm. A similar design ts requtred in the other onhogorlal dtrec.tton, ea(it-west..,
but IS not iUusuated here~ Allowable sues:s destgn (ASO) wall be used. The baste loadmg eombtnauons are
gwen in IBC Secuon 1605.3.1, and those tn\'olvtng eanhquake loading have been simpltDed 1n ASCE 7
Section 2.4.:5.
The govenung seismic load combi.natJoo ror allowable stress design is (8)
i.OD+ 0.7£, +0.7£,
§2.4.5
where£,= pQ,
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139
Design Example 2 • F/Wble ()japhragm Dt!sign
When destgntng lhe st.ru&::tural diaphragm. the engineer need nm coosider the venicalloading in conjunction
with the lattral diaphragm shear stresses. Therefor~ the dead load D = 0 and the vemcal earthquake effec-t
E, = 0 tn this load rombmation.
For diaphragms of buildings tn Setsmtc Destgn Category D~ E, or F~ the redundancy factor p is requtred
withtn lhe seismtc-foad combination, but is typical f)• set top= I .0 for diaphragm loads per Section 12. 10.3
Item 3. In untque mulustocy sttu.auons where the diaphragm tS acung 10 transfer forces hort20f'ltally due
to ventcal syste.m offsets or due 10 changes in venteal element stiffnesses, the redundancy factor p Will
conform to Sections 12.3.4 and 12.1 0.3.3. In thtsexample, p = 1.0 for the diaphragm destgn. Thus, the
applicable basic lood oomblnatton reduces to stmply 0.1Qt·
IS to be constructed wnh ·~!-inch $t.fU('tural-l OSB sheathing (wood struc-tural
panels) with all edges supported (blocked). Referto the 2015AWC SDPWS Table 4.2A for nailing
requirements. The sheathing arrangement (sh0\\1\ tn Figure 2-2) for north-south seismic forces is Ca'ie 3
Assume the dtaphragm
with the long pane.l direction parallel to the supports.
Because open-web stte-1joist purl ins tn a hybnd roof structure have full-wtdth wood nail er~ the conunuous
sheath1ng edges loaded tn this nonh-south direction are supponed by framing_ greater tJ"IaJt a 3-tneh nominal
width, allowing nail S1Jac1ngs of2% inches or less at adjouung panel edges (AWC SDPWS Sectjon
4.2.7.1.1 ). HO\vever, in the east-west direcliOJ\ the sheathtng edge.~ are supported by only 2x subpurlin
franung. and the strength IS there tOre Iunited by the nad spac tngs as.:soc rated with 2-tnch oom uta! franung
width. Although not applicable 1n lhis example, 3x framing ts required \\'here adjoin1ng panel edges are
fastened wnh IOd common nails at 3-tnch spacang tfthe nail peneuauon is greater than I% tnches. In these
large panelized roof systems. the nailtng contrac-tor lS often instructed to order custom 2-tnch length n.atls to
obtam the I %-tnch penetrauon. allowtng 2x franung m ce-rta.tn loc:auons.
Various nail spactngs al sheathtng panel edges and their respective seismtc shear capac:iues for Case 3
(nonh-soulh seismic loadtn_g) are given in Table 2- I. Mtn1mum tntetmedtate (field) nailing is 10d common
rwls at 12-inch spacing. and IOd common nails require 1Y:-tnch member penetration. A stmtlar calcularjon
(nOl shown) mUSt be done for east-west setsnuc fOrces.
Tab/~
1- 1. Allolt'a.ble diaphragm !ltea.r c.a.pacillts
Edge Nailingl'· 11
Eas!-WeSl Edge
Nrultng''- 31
Nominal Unu Shear
Zone
Capacity (pi f)
ASD Allowable
Shear {plf)
Boundaty and North-South
A
10d @2Y.ttno.c.
10d @ 4 '"o.c.
1280
640
B
IOd @ 4 1n o.c.
IOd @ 61n o.c.
850
425
c
IOd @ 6
10d @ 6 '"o.c.
640
320
10 o.c.
Notes for Table 2- I:
I. The nonh-south runntng sheel edges are lhe ..conttnuous paneJ edges paralle1 to load" men[toned 1n
SDPWS Table 4.2A
2. The east-weSt sheet edges are the "other panel edges" tn SOP\VS Table 4.2A. The naJitng for east-west
runntng dtaphragm boundaftes 1S per the ughter boundary spacing.
3. Natls are common smooth-shank n.atls ( IOd = 0.148-tnch dtam.e-ter). Screw-shank nails (deformed
shank) are ofte-n used tn some regton."i of the counuy where spec tal concerns e:<.tst of natls backing out tn
withdra\\'3.1 due to wind upltft loads or framing member drytn_g shrmkag_e.
The diaphragm boundartes i.U lanes 3 and 9 have a shear de.mand of1·= 706 plf(see Pan 2.2). Converung
to all<m'ilble streSS destgn, 1"A.So = 0.7(706) == 494 plf. which is less than natling zone A·s allowable su-ess of
640 plf.
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Design Example 2 • Aexible Diaphragm Design
At a pruticular location across lhe dtaphmgm. natltng zone B (425 plf) will become acceprable as lhe
diaphragm shears reduce farther from the diaphragm boundary. llte demarcatton between nruling zones A
and B may be.located as follov.•s using allowable sLress design:
Shear demand (ASD) s Shear capacny (ASD)
0.711 13,000 Jb- (941 plf)xj S: 425 plf( 160 fi)
where
.1:
=the demarcauoo distanc:e from the dtaphragm boundary
Solvtng for x obtams
x=l6.9ft
Because a paneHzed wood roof system typically ronststs of8-fom-wtde panel modul~ the demarcauon is
increased to the next 8-foot increment or to .1: = 24 feeL
A stmdar process is unde.rtaken to de.termine lhe de.tnarcation betwee.n zones B and C. In this suuauon.
x =- 42.4 teet, and the dem.arcatioo is tnc.reased to 48 feet from the diaphragm boundary. The resulting
dtaphragm shears at these demarcatiOn boundartes are shown 1n Table 2-2.
Table 1-1. Emluation ofnailing ::on~ dillt111Ce~·
Allowable Shear
Natling
Zone
DislaOCe from
llonndary
Maxtmum Shear
A
(pi f)
ASDSheat
(pit)
Capacity
(pi f)
Ofeet
v-=106
,.ASo= 494
640
B
24 feet
v-=565
'·.uo= 396
425
c
48 feet
''-=424
l'ASu =297
320
The resulting nailing zones for the nonh-south loading are shown tn Figure 2-6.
11tese demarc;~ujon calculauons asswne the full depth of the d1aphtag.m. IS <Wallable for shear capac tty.
However. ryptcal warehouse c:onstruclJon often cootatns sl-ylighL(i and smoke vents that can substanually
perforate the st.ruCturaJ diaphragm. In these situauons, the designer must account for these diaphragm
intem.tptjons. resulttng tn la.rger shear stresses.
Commentary
Plywood and other wood SliUCti.Jfal panels are common diaphragm tnatena1s in the west and parts of
the south. Other pans of the Unned States commonly use steel deck for d~aphragms tn conjUJ'ICuon wnJt
steel roof framjng. Stei'-1-deck diaphragms are approached in the same manner wtth a simtlar dtaphragm
table assigntng varJous deck gauges and anachmetlts to specttic diaphragm zones. dependmg on tl'le shear
de.mands. In this example-. the wood structural panel edge wiJI coincide with the collector l1ne. For a metaldeck dtaphragm. th1s m1ght not be the case. Therefore, speclal detads wtll be requtred to define adequate
shear transfer between the steel deck and the collector hnes.
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Design Example 2 • F/Wble ()japhragm Dt!sign
1
3
9
240' - 0('
80' -0"
24' - 0 11
24' - 0"-
'I
c
A
144' . 0"
24' - 0"
i-:-'24' - 0''
'I ' I
'I
c
B
B
A
Flgun 2-6. 11/ustrtuion ofnailmg : ont loctJIIOn.t
Thts wood diaphragm res1sting seism1c fOrces must have 1Lo:; aspec.t rarjo checked agauwlhe Iimitauons in
SDPWS Table 4.2.4. For blocked diaphragms usmg \I/OOd structural panels, the max1mum aspect mtio is
LIW=4:1.
For<his example, LIW = 240/160 = 1.5 < 4 . . . OK
Commentary
Aspect rarto hmitations for metal-deck d&aphtagms ate fOund under the s:pectfi.c deck manufacturer's
ICC-ES Evaluation Repon. Wnlun these reports. a table utled .. Diaphragm Flexibiluy llmitation"' provides
gwdance on hmiung diaphragm ftexlbtllty in coojW'lCUon \'11th diaphragm aspect ratios.
4. Considerations for Plan Irregularities
ASCE7
Because tl'lfre IS a ree11trant corner at the tntersection ofl ines 13 and 3, a chec.k for Type 2 honzontal
structural irregularuy must be made. Requtrements for horizontal strucwral l!regularities are g1ven tn ASC.E
7 Table 12.3-1. 1be buildtng projecttons beyond the reentrant comer are checked dtme.nsionally against a
IS percent plan dtmensJon cmeria.
Ea~t-west dtrection
check:
0.15x320ft=48ft
80 ft projec<ion >48ft
North-south direction check:
0.15x 160 ft=24 ft
40 ft projec<ion > 24 ft
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Design Example 2 • Aexible Diaphragm Design
Because both proJecuons are greater than I5 percem of the plan dtmension 10 the dlrection consJdered
and the structure is SOC 0 Of higher. a Type 2 horiZontal structural ttregularity exist-.. ASCE 7 Secuon
12.10.3.3 direct-. the destgn engineer to the require.me-nts of Section 11.3.3.4. resulting tn a 25 percent
ioc.rease in se,ismtc forces for connections ofdiaphragms to the venkal elements. connections of
dtaphragms 10 collectors. and the collectors themselves.
Accord tog to Awe. the 25 percent force tr.crease is not tntended to require tncreased diaphragm sheathing
nailing. Instead~ Section 12.3.3.4 mcreases the forces tOr lhe shear transfer from ledgers at1d woOO nailers
10 shear walls and collecto..-s. The design of this shear tranSfer tS not a pan of thlS example. Th1s horizontal
plan trregularlly also affects the collector destgn, as w~U be shown 10 Pan 6. The 15 percent force tncrease
IS not applied 10 out-of-plane wall anchoro.g.e fOrces connected lO the dtaphragms.
5. Diaphragm Chords (North-South)
ASCE7
Chords are required 10 carry the reruaon forces developed by the tn-plane bending moments tn the
diaphragm. In th1s buildittg. the chords are conunuous retnforcement loca£ed tn the wall panels at or near
the roof level as shown 1n Figure 2-7. In this example,lhe chord reinforcemem IS below the roof ledger to
rac.d itate the c.hot'd splice connection at the wall panel joints.
Note: Wall anchorage is a very important aspect or destgn when concrete or masonry walls mterface with
wood d1aphragms. The necessar)' members and connec6oi'IS for anchorage are tlot shown in Ftg.ure 2-7 for
stmplicity, and the des1gn of wall anchomge is tllu:strated in Oestgn Example 5 ofthts publication.
Figure 2-7. lntetfau ofdtaphrugm at wall
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Design Example 2 • F/Wble ()japhragm Dt!sign
The oonh-south dtaphragrn spans between lines I and 3 and hnes 3 and 9. llte diaphragm 1s ideahzed a-.
flexible. and the moments in segme.nts 1-3 and 3-9 can be computed independently. assuming a simple span
for each segment. In thls example-. the chord remforcement between lanes 3 and 9 will be detenntned for an
applied dtstrtbuted Jood w omo the diaphragm. This reinfotttment is em.bedded in the \Yall panels on lines
A and E.
"'= 941 plf from Part 2 .2
0.94 1 klf(240)' _ TIS kip-ft
6
8
The chord forces are computed from
p = 6775 k-ft - 42.3 kl
•
160ft
ps
The chord will be de.-.igned ustng strength design wtth ASTM A706 Grade 60 reinforctng steel A706
reinforcing ts used here m amjc:tpatton that the chord steel will be welded at the splice belwee.n wall panels
(see ACI Secuon 26.6 .4. 1). The load fuctor is 1.0 for seismic forces (ASCE 7 §2.3.6).
A;~=
'
•
~
42.3ktps -0.7S3inl
0.9(60 kso)
The-refore. use a mtnimum of two #6 reinforctng bars. A, =0.88 in2 > 0.783 tn2
•••
OK
The: chord destg.ned above consists of two #6 bars. These must be sphced at the jotnt between adJacent wall
panels, typically usmg details that are htg,hly depende.nt on the ac:c:uracy an placang the bars and the quality
of the field weldtng.. The we-lded reinforcing splice connection mu!il develop at least 125 percent!, per ACI
318 Sec.tion 25.5.7.1.
Chords can also be combll'led with the ledger \\'he-re steel channels O( angles are used. Whe-re steel shapes
are used, YlOod nailers can be bolted to the ledgers to provide a \\'OOd substrate for the diaphragm nail utg.
Alternatt\•ely,lhe structural wood panel may be direct1y fa(itened to the steel ledger/chord with propriemry
pneuanattcally dnven stee1 pm fasteners (see mal\ufacrurer's ICC-ES evaluatton reports for requtrements
and desogn values).
In some cases, concrete \Vall pane-l shrankage and any thermal e.xpansionlconuacuon should be cons1dered
tn the chon:l detaihng, but is not a pan of th1s design example.
6. Diaphragm Collectors
ASCE7
Collectors (or suttts) are provided tn d1aphragms 10 uansmat the dJaphmg.m reaction-. to venacallateralresisung elements suc.h as shear walls or frames. Where the enme d~aphtagm's len_glh parallel to the lateral
load is uniformly attached to a shear \vaJI, the diaphragm shears stmply transfer unttbrmly into the she-ar
wall. However. where the supponang shear walls are shorter in le-ngth than the diaphtagm, or whe.re a frame
ts placed tOr late.ml diaphtagm support, the shears must be collected aJld the.n transmitted to the shear ,~,-all
or frame. The collector ts a criucal me.mber 1n the k>ad path and IS thus subjected to an adchuonal de~ilgn
considerauon tn Se-Ismic Design Categones c. D. E. and F (ASCE 7 Section 12. 10.3.4).
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Design Example 2 • Aexible Diaphragm Design
6.1 DESIGN THE COLLECTOR ALONG LINE 3 BETWEEN UNES BAND C
ll\e collector a.'ld shear wall ledger along line J cany one-half of the north-south roof d1aphragm setsm1c
force. The fOrce in the collector is ..collected- from the tributary shear length between lines B and E and
uans:mined to the shear waJI on l11'!e 3.
6.2 DETERMINE THE COLLECTOR FORCE IN THE STEEL BEAM COLLECTOR
From the dtaphragm shear dtagram for noM-south setsm1c fOrces: (Figure 2-5). the accumulattng unn shear
e.merUlg the collector on line J JS: calculated:
On a hybnd roof system, lightly loaded collectors can uultze the steel jolSts and jo1st gtrders if properly
destgned by the JO ISt manufacturer. For heavter-loaded collectors. rolled steel beams are more common. In
tlus e.xample-, a \VIS x 50 wtth \1/000 natle:r is a~umed to be the collector and assumed adequate to suppon
dead and live loads. ASTM A992,Fy; 50ksJ will be used tn design. Not knowing whether the mode of
fa1lure would be yielding of the cross-section ot buckl1ng at mtd-span.. the engineer will e"Yalume-the destgn
collector force at the highest magrutude along tts length.
P = ql = 0 .98 1 kit{ 120 ft) = 11 8 ktps ten.<ton or compression m beam
6.3 DETERMINE THE WOOD NAILER ATTACHMENT ON THE STEEL BEAM COLLECTOR
In a hybrid roof system. the wood structural panel diaphragm tS fa(itened to the steel roof frnrrung with the
use offtat wood nailers attached to the top chords oftl1e steel joists and the top flanges of the steel beam
collectors. ll\e an.achment can be 10 the form of steel bolung or threaded we-ld studs.
To prevent a confhct with lhe wood struCtural panels' placement and lhe weld studs' nuumg. the wood
nailers are countersunk so the nuts recess below the wood structtl(aJ panel. In this example, ~ · lOCh·
diameter threaded we-ld studs wdl be asswned to be welded to the oollectot's top flange a.'ld bolted to a 4x
wood nailer with 1-inch·deepcowuersunk recesses.
The connection shear capacity of the wood nailer faste.ned wtth threaded we-ld studs is based on the 2018
NOS. The threaded weld studs have a fixed end support on tlte steel flange. and the reference design
shear value could be calculated by applying the y teld limit equations of NOS Table 12.J. IA. A stmpler
a1'ld slightly cooservattve method is for the designer to use the vaJues associated with bolts that are fixed
in concrete fOund ln NOS Table 12E. It ts 1mponant to note that these threaded fasteners have a reduced
diameter D, compared wnh standard wood bol~ and thts reduc-uoo w1ll affect lhetr desig,n shear value
(NOS Section 12.3. 7~ Using NOS Appendix l, a fully threaded lr.·utch shank wtll eftectively have aD,=
0.627 tnc.h, which ts approxtmately% tnch.
For a -%-tnch-dtamete.r fully threaded dowe-l shank ( _$,-i-lnch daamele.r umh.readed) in 2YJ inches of rtel wood
tluckJ'Iess
Zj = 1180 lb (ASO)
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Design Example 2 • F/Wble ()japhragm Dt!sign
Applying 1he adjusunenl faclors of NOS Table I 13.1:
Zi= I ISO lbx CoX CJJx C,xc,x CAx c..,xcdtx c,.
Cu= 1.6
C"'= 1.0
NOS §11 .3.2
NOS §11 .3.3
C,= 1.0
NOS §11 .3.4
c,= 1.0
NOS §11 .3.6
c.= 1.0
NOS §12.5. 1
c,. = 1.0
NOS §12.5.2
c.= 1.0
c.= 1.0
Zj = 1180 lb X 1.6 = 1888 lblsrud (ASO)
NOS §12.5.4
NOS §12.5.3
The shear demand under allowable sl!ess des1gn isq,= 0.7(0.98 1 x 1.25) = 0.858 kif, where 1he 1.25 factor
IS per Sectioo 12.3.3.4 as directed by Secuon 12.10.3.3. Tile requtred weld stud spacing IS
s = 1.88/0.858 = 2.19 fllstud
Thus, the de.~igner should provide -%-I.OCh-diameter x 3Yi-mch-long fully threaded weld studs at 24-tnch
spacing with nuts c.ounlersunk I tnch.
6.4 CHECK STEEL BEAM COLLECTOR AS REQUIRED BY SECTION 12.10.3.4
ASCE7
The govermng setsmtc load combinauon for LRFD under Secuon 2.3.6 JS
(6) 1.2D+E,+E,+L+0.2S
For this example, L = o. S = 0, £~ = 0.25~. £ 11 = pQ£> and Sa;= 1.0. Because collectors are cons1dered a
pan of the diaphragm system, the redundancy factor p = 1.0, as wa~ d1scussed pt""eviously 1n Part 3. Thus,.
the apphcable baste load eomblnation for LRFO redoc.es to the followi1lg:
(6) 1.4D+Q,
The unfactored dtstnbuted gravity lood and bendtng moment are as fOliO\\~:
"'•= 8 ft(l4 psf) +50 plf= 162 plf
M0
=
162 pJQ40 ft)'
-32,400 11>-ft or 32.4 k1p-ft
8
As shown 1rt Part 4, thi~ building contams a Type 2 horizontal sm~c.tutal irregularity, and the reqUirements
ofSecuoo 12.3.3.4 apply. Th1s result..~ 1n a 25 percent tncrea~e Ln selsnuc fotces fot coUectors and the1r
oon.nections. The collectOr's aJoal seismic force becomes Qs:= 1.25 x 11 8 kaps= 148 kips. Additionally,
collectors m SOC C through F requlre a 1.5 muluplier per Section 12. 10.3.4 because ofthe1r criucal role.
ThUS, Q,= 1.5 X 148 kips :2221ups.
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Design Example 2 • Aexible Diaphragm Design
AISC 360 Sect100 HI contains the equauons tbr combined a.xiaJ compression and bending. Because the
bc!.ndlng ts not biaxial. it is ad\'antageoos tbr lhe e:ngtneer to use Sectton H1.3 by checking failure about
each axis independently. In this example. the collector's top flange JS continuously supponed with closely
spaced dlaphtagm nailing. thus pt'e,·enllng lateral-torsional buckhng. The colleclOr·s bouotn flange will be
larerally braced at the member's equal third pcHnL~ Y.'lth use of an angle brace (destgn not sho,vn). resuJtmg
in an unbraced lenglh of L., = 40/3 = I 3.33 1\. The strong axis unbraced length is Simply lhe span L, = 40 fl
As a condition of AJSC 360 Section H 1.3. the efttcuve lateral-totSJonal buckhng length Le: must be less
than the effecuve out-of-plane weak-axis bockhng length LC? and thts 1s confirmed.
Le S: L9'
o n,; 13.33 n
Frulure will be checked sepatately about each axis per Secuon H1.3.
X-a:ds limit statt ( in-plant instability)
AISC 360 Secuon H J.J(a) provides lhe approach to check tn-plane stability oflhe loaded Wl8 x 50
collector. F1rst. the designer must compute the available tn-plane stre-ngths Pex and M0 for use in Equarjon
H 1-1 . PaIS the available axial strength and is a funcuon of the collector's strong-axis w1braced length.
La = KL, = 1.0(40.0)12 _ .
65 0
r:r
r.7.38
§E3
Eq E3-4
r,
Bocau~
FJF, s 2.25. Equatton EJ-2 is applicable.
Fa=
(o.658~ ~,. = (o.6586~1 ~o = 36.7 hi
P~= F,A,= 36.7(14.7) =539 kips
Eq EJ-2
Eq E3- l
P0 = ~' Pm = 0.90(539) = 485 kips
§EI
Mn is lhe avaJlable fte.xural strength for tn-plane bendtrtg.
Ma = 0,M. = 0,F,Z,= 0.90(50 kstX IOI) =4545 ki(>-tn
Ma = 379 kl(>-ft
Second, the destgner mus[ de[ermme the reqUJred axiaJ and flexural strengths P, and M, usang tlte baste
LRFD load c<>mbtnaiiOn (6) 1.4D + Q,:
P, = Q£= 222 k1ps (mcludes the 1.25 tncre-ase for plan irregularity ru'ld the 1.5 increase tbr oollecror)
M, = 1.4M0 = 1.4(32.4) = 45.4 lip-ft
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Design Example 2 • F/Wble ()japhragm Dt!sign
Per Section Ji 1.3(a). the in-plane stability check uses Equation H1-1. Pa tS the appropriate m-plane
buc.ldtng strength.
~
P0
=
222
= 0.46• 0.20
485
The.refore, E<juation H 1-la
~
s•M,)
1S applicable
for checking combined forces.
s•45.4)
•
Po +9l Mo =0.46+9t 379 =0.>7!> 1.0 ... OK
\'-axis linlil st.att (out-of-plant butkll.ng a.11:d lattral-torsional budding)
AlSC 360 Section lll .3(b) provides the approach to check out-of-plane buckhng and late.ral-torsiOilal
buc.ldang oflhe loaded collector. First. the destg.nc-r must oompute the available out-of-plane strength P9
and Lateral-torsional budding strength Ma For use 10 Equation H 1-3. Pq tS theavadablea-xtal strength and
is a funcuon of the collector's weak-a.·os unbmc.ed length.
§E3
Eq E3-4
Because F/F,,; 2.25,AISC E<juanon E3-2 is applicable.
F~
=[0.658*1F, =10.658.:.]50 = 25.2 Jcsi
P.,= F~A1 =25.2(14.7) = 370 kips
P'>'= $,P., =0.90(370) = 333 k ips
Eq E3-2
Eq E3-l
§EI
With the top flange ful ly supported laterally:
Ma = 0_.11. = 0,F,Z,=0.90(50 Jcsi)(JOI) =4545 kip-In
Ma = 379 kip-ft
Per AISC Section li J.J(b), the out-of-plaoe buckling check uses E<juanon H 1-J.
P •
P )
P:l i.5 - 0.5P:
• M
+lc.A~o
)'
222 •
222) • 45 4 )'
= m li.5 - 0.533J +t(I.0)~79 =0.79
0.79 s 1.0 ... OK
Thus. theW 18 x 50 collector ts adequate for combined a.x1al compress ton and bending.
Evaluattng the WI 8 x 50 collector for combined axial tension and beOOing per Section H 1.2 is not
necessary because axlal suength P~ is less For compress1on and thus more cnucal than for tension.
Evaluallng the WL8 x 50 collector under a load combinauon walh overstrength (Secuon 12. 10.2.1) as not
necessary because the alternattve desagn provtstons ofSecuon 12.10.3 were followed instead.
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Design Example 2 • Aexible Diaphragm Design
6.5 COLLECTOR CONNECTION TO SHEAR WALL
ll\e design of the connection of the steel be.am to the shear \\'all ooltne 3 is nm given. but lhe unfac:tored
force is calculated as FP = 0.981 klfx 120 tt = 118 kips. This is an imponam connection because it is the
sole avenue to t.ronSfer the large "collected" sets:mec force imo the shear \~;-all. The oonnecuon must be
designed to carry the yav&ty and seismic forces from lhe collector beam, induchng the 1.5 muJuplier on
axull loods per ASCE 7 Secuon 12.10.3.4 ln Sets:mic Oes:tgn Categortes C, D. E, and F. Ali. shown in Pan 4,
tJus buildtng has a Type 2 horizontal struc.tural irregulanty. and lhus the coUec.tor connection fOrces inc.reao;e
by 25 percent In addttion to the 1.5 multtpller.
Because there is also a collector aloog line B, there is stmtlarly an Important coonection of the gtrder
betwee.n ltnesl and 4to the shear wall on ltne B. Havtng locarry two large tension and/orcompressiott
forces tJuough the intersection of hoes B and 3 (but not simultaneously) requires care.ful design
constderauon.
7. Diaphragm Deflection
ASCE7
Diaphragm deftecttons are estimated to dete:nnu-.e tl'lf' displacement~ 1m posed 01'1 attached structural and
nonstructural eleme.nts, at'ld to evaluate the significa.nee oflhe P-delta effects.. Under ASCE 7 Section
12.12.2, dtaphragm detlecuons are linuted to the a.nount that wtll pennu theanached elemetus to maintain
SU\JCtural integrity and to eonunue suppott.iJtg lhelt prescnbed loads. For suuctural elemems. the tntent here
is to ensure structural stability by avoidtng fonn:u:ion of oollapse mechanisms in the venieaJ suppon system
and avoidmg excessive P-delta loading etfects. For non$trucwral elements, the tnt"ent of this section is to
prevent fatlute of connections or self-integrtty that could result 1n a localized falling hazard.
Trad.Juonally~
in-plaoe diaphragm deflecuons are ()'pically e\•aluated sunply by compatlng the span-to-
Wtdth ratio Ub to the acceptable code hmits. For this example. SDPWS Table 4.2.4 hmn.s the spruHowJdth ratio for blocked wood structural panel dtaphragms to a maximum of four, wbich is greater than
t11e dtaphtagtn.(i contained tn this bulldt.ng.. Nevenheles..-c;., there are tmportani reasons to estimate the actual
diaphmgm de.flecuons. such as to detemune whether a dtaphragm may be tdealt2ed as fte.xtble or not and
when determtnt.ng buildtng setbacks from property ltnes or other ad;acent bulldt.ngs. Another unponant
reason tha[ Will be Illustrated nex1lS to evaluate exces.<>t\'e P-delta loading effects.
For the purpose of e\'aluatmg honzonml diaphragm deftecuons. the dtaphragrn loading u<;ed is that
associated with the setSIUIC stocy force F11 ofSecuon 12.8.3 tnstead of the dlaphragrn destgn force FJW
from Sec.tJOf'l 12. 10.1 or 12.10.3. It ts imponantthat both the venicaJ seJSmic-force-resiSltn:g system's:
dtsplacement and horiumta1 diaphragm displacements share the same basis of F1 , a.Jiowtng both
displac.eme.nts to be dire<:-.d y added toget11er for eompuung the total structure."s maximwn dJsplactJnel\t.
Addmonally~ when evaJuat.i.ng whethe.r a diaphragm ts flexible by calculation (Section 12.2.1 .3). it is
approprtate that the dtsplacements ofbolh the vert1cal and honz.onta1 systems also share the same basts F11•
For a one-story butkhng. F1 = C.~IJl, and thus in tltis e:umple btulding the diaphragm de-flection is evaluated
under loodtng F11 = 0.25 W.
7.1 DEFLECTION OF NORTH..SOUTH DIAPHRAGM
SDPWS
An accepmble me.thod ofde.temttning the honzonml deflection of a blocked wood structural panel
diaphmgm under lau~ral forces: is given in SDPWS Sectjon 4.2.2.
<
0
...
5•·L'
0.25rL l:(x6,)
=--+---+--8EAW IOOOG.
2W
&] 4.2- 1
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Design Example 2 • F/Wble ()japhragm Dt!sign
Thls e.xample wl11 compute tl'!e deflection of the OSB diaph.mgm spanning belwee.n hoes 3 and 9. Values fOr
each of the parameters in Equation 4.1-1are given here:
,,= 706 plf for suenglh design (see Pan 2.2)
,. =706 x 0.25/0.208 =849 plf for defleeuon evaluauon
L=240fl
£=29x IO'psi
A= 2 116 bars= 2 X 0.44 = 0.88 m'
IV= 160ft
G41 = 20.0 klin Zone A (see Part 3 for n.amng zones)
T4.2A
15.0 klin ZoneD
24.0 klin Zone C
6-.:o == 0 (Assume no slip ln stee.l chord connectaons)
- L'
The fte:<ural deformation ponton of the equation~ assumes a unjformly Joaded diaphragm ru'ld tS
computed as follows:
SEAW
s•.......,.""""' =
1
SEA•·L' • =
5(849 ptf)(240 fl)
8 11 8(29x 10' ps~)0.88(160 ft)
The shear deformation portion of the equatton
0.2Sv~
1000(,.,
_
I. SO in
is dertved from a uruformly loaded dlaphragm with
uruform shear stiftite:s:s. Because Uus example has various nailing zones. and the apparent shear suttness G
varies b)' nalltng zone. this ponion of the equauon must be modt:fi:ed. Using \•tnuaJ work methods, the shear
defonnauon of a unaformly loaded d1aphragrn wtth vartous shear stt.tfness zones JS
Q
where
•·;.~ =the average dlaphragm shear wuhln each shear stUfness zooe ba(led on F_.. oot F,.
L, =the length of each stiffness zone measured perpendicular to the loading direcuon
G111 =the apparent shear suffness of each shear stiffness zone betng considered
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Design Example 2 • Aexible Diaphragm Design
Worktng across the diaphragm from grtd 3 to 9, Table 2-3 l~ helpful using infonnation from Pan 3.
Table 1 -3. IJOrkshtet computing Jhtar deformatldn
O.Sl·,.VLI
Zone
1.1ttl
,.,.
1",_.
L,
c.
IOOOG"
A
849
679
764
24ft
20
0.46 in
B
679
509
594
24ft
15
0.48 tn
0.38 in
c
c
509
0
255
72ft
24
0
509
255
72ft
24
0.38 tn
B
509
679
594
24ft
15
0.48 in
A
679
849
764
24ft
20
0.46 tn
l:=2.64tn
Because the chord re.lnfbrcu'lg bars are dlfectly welded together at their splice, no chord slip tS as....;wned to
occur:
- l:(.H~,) - 0 00 .
2W - . m
•
odlclfd..,-
84• = s~~ = sd.w.:.r + sdlool.p = 1.so + 2.64 + o.oo = 4.44 1n
To compute the ma.xtmum expected diaphragm deflec.tioo Sv ASCE 7 Equatton 12.8-15 is used;
5 = c,5w
'
I
•
8u = 4.44 in (UStng an elastiC anili)'SIS under Strength forces, 04..._)
C4 =4
4( 4.44 ) 17.8 in
5
~
1.0
=
Eq 12.8- 15
T 12.2- 1
=
Note: The defJecuon amplificauon factor C11 from ASCE 7-16 Table 12.2- 1 is primanly associated with
reversing the efTeels of appl1ed response mod,ficatton coefficient R used lR detenninmg the base shear J•=
0.251V.
Instead of using SOPWS Equatton 4.2-1 to compute the elastic deflection. the designer could u:se Equation
C4.2.2-l in the SDPWS Cammenuuy . Although th1s method IS a hn1e more complex, it will be more
accur.ue if property applied. Additional mfonnatton lS a\1adable in Lawson. 2019. and Skaggs. 2004.
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Design Example 2 • F/Wble ()japhragm Dt!sign
7.2 LIMITS ON DIAPHRAGM DEFLECTION
ASCE7
lim1ts are placed on d1aphmgm deftecuon pnmanly fOr rwo reasons. The first 1s to separate the btuldtng
from adjacem Structures and propeny l1nes in accordance With Sectton 12.12.3. In thtssituation, 0-' l~
computed fOf the diaphragm and then added to lhe d1splacement contrtbution oflhe vemcal lateralforce-resisting system (shear ·walls~ frames, etc.) to oOOun the total maximum bu1ld1ng djsplaceme-nt
S.v- Buildi1tgs at propeny li1'!eS are usually set back a distance of8JJ in ac:corda.lce with Sectjon 12.12.3.
AdJaCent structures shall be separated from each othe.r by a dlstanee O,vr 10 accordance wnh Equat1on
12.12-2 whe,re 0,.,. and O,w.are the maximwn structutaJ displacement<; for structure I and struCture 2 at their
adjacent edges.
&..,. =
Jcs,,>' +C&"' >'
The second reason for linuung dtaphragm deflection IS to matntatn structutaJ integrity under design load
condttions. Otaphragrn deflectjons are limhed by Sectton 12.12.2. and f()( wood dlaphmgms. SDPWS
Section 4.2. I:
Penmss1ble deftecuon shall be that deflection that wtll permit the diaphragm and any attached
eleme.tts to maintain their SLruCtural mtegmy and conunue to suppon tJ'!e1r prescnbed loads as
de-termined by the applicable butld1ng code or standard.
The language ofth1s secuon 1s 1menuooaJiy not well defined~ wnh the approach left much to lhe engtneer's
own mtional JUdgment
The diaphragm •s ddlecuon resull~ tn lhe col umn.~ and perpendicular watts romting about their bases
because of the dtaphragm's uanslauon at the top (see Ftgure 2-8). Assuming the oolumns and v.oalls were
modeled with pmned bases dunng their tndiv1dual des1gn. tJus ba10e rotatton IS pennmed to occur even 1f
some umntent10nal fixity e.xiSL(i.
Unintenuonal fixity may be the result of standard column base pla[e anchorage or waJI.to-slab anc.h()(3ge.
The assumption of plastte hmges tbrmmg at the base may be acceptable. provtded that these hmges do not
result in Joss of supporL Research conducted b)' Kong et aJ. (20 19) provides seve-ral destg_n approaches for
HSS column.o; subject to large dnft.
A possible source of instabil it)' IS the P-de1ta effect resutnng Jn added diaphragm loading due to a
hor120ntal thrust component from the leanang a-xaally loaded gravity columns and \'o'aliS.
~
~
<f
®
I
0
I
H:f±jj;t
Ftg11nt 1-8. B11ilding section n•tlh diaphrogm deformation
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Design Example 2 • Aexible Diaphragm Design
Although 1t was not orig1nally mtended to be use.d to evaluaxe diaphragm deformations. Section 12.8. 7 can
be used a.~ a gUJde lO investigate stab1lny of the roofsystem under diaphragm P-deha effects. The stability
coetlkient 9 is defined as
&] 12.8- 16
P. is the venical load acung on the translating system and has lwo component~ m this example. P!ilfWI/lS
t.he translating roof load. and because load oombinanon 6 ofSectjoJt 2.3.6 is applicable. no roof live load
is conside-red. P}l •"'' lS the lranslating concrete waH dead load and comprises lhe upper half of lhe waJI plus
parapeL Lood factors need not e.xceed 1.0.
P,_,-= 14 ps(\240 ft)( 160ft)= 538 kips
P, ... =
7
-~~i"(l5o pcf)C~ ft +2 a )240ft(2sides)=544 kips
P~- = Puo.~+ P::a_,= 538 + 544 = 1082
k1ps
6. = the average hortZOntal story drift
Because th1s IS a flex1ble dtaphtag.m W1th an approximately parabolic deflected shape, the average
translation is
2
2
3 s. = 3(17.8) = 11.9 in
V::a =the seismtc shear fOfCe act.ng on the transJaung systtm ut1der considerat.ion
RecaJJ that dlaphrag.m deflections are computed With F}l = 0.25W, not F14 = 0.208W. Thus.
V,=(941 plfx0.25/0.208)(240 ft)=271 kipS
hJ;T=21 fix 12=252.n
c,=4
T 12.2- 1
TherefOre:
9 = I082( I I. 9)1.0 _ O.O-I? < O.IO
271(252)4
Thus. P-de-lta effects on story shears, momenL'i. and story dnfts are not reqwred to be oons1de.red.
Nole: The stOI')' drift l1mnations ofSecuon 12.12.1 are not Intended to apply to fle.xible diaphragm
de-flectiorts, buttnstead are intended to apply to the acting lateral-resisting Y.>all or frame systems. The.o;e
ltmltiltiOI'lS on budd1ng dnft were pnmarlly deYeloped for the class1c flexible frame sysr:e..n w1th ng1d
d1aphragm. Story dnft limits are designed to ensure that the frames and walls do not excess1vtly dJston U\
plane. Stmilarly. the P-delta hmitations ofS~uon 12.8.7 are also intended to restrict in-plane movements
of the ventcal seJsmie-rtslstmg system. especta1Jy tn fte.x1ble frames reslsttng venieaJ and lateral forces
together while subjected to potenua11y Large secondary mo1nerus. sua· concrete and masoory shear \\all
butldtngs generally are not unpacted by secondary moments from tn-plane P-delta effects.
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Design Example 2 • F/Wble ()japhragm Dt!sign
7.3 DEFORMATION COMPATIBILITY ISSUES
Thls example building provtdes a very stmple tllu.strauon to desJ_gn a flexible diaphragm. While ftexlble
dtaphragm analysis tS \'1!-f')' suatghtforward on stmple butldt.ngs, the 3nal)'Sis can become quite dtfficuJt to
accurately model more comple.'< budding shapes. Often small changes in the outltne of a buildtng can create
sjgni:ficant deformation comparibility lSSues. Small offsets or J<>gs along wall lanes Introduce suffe-lements
that are not long enough to act a.;; shear V.'aJIS, yet \Viii inad"\•enemly resist d~aphtagrn movemem due to their
std'fness.
These elements are not a pan of the lateral-force-resisting systems. but dependtng on where ll~ey are located
within a buildtng. they may be requtred to translate 31 the dtaphragm level or else the diaphragm may tear
3\\'i.l)' from the element. llus concept of defonnauon oompaubd ity must be considered by the eng.u~r
when designing budd1ng.~ with flexible diaphragms.
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Design Example 3
Three-Story Light-Frame Multifamily Building Design Using
Cold-Formed-Steel Wall Framing and Wood Floor and Roof
Framing
OVERVIEW
ThL'i example will be revi~Ntng the design of a three-story aprarunent budding sitting on a raised structural
retnfbreed concrete podju.m. deck, wnh ooe le\•el of subterranean parking.
I. The destgn ofth1s example wtU be hmited to the lateral destgn of one of the lhree,.story
apanment buildtngs and does nOl mvesugare the destgn of the raiSed first-floor structural
concrete podium deck that the three-story apartment buildings sit on.
2. The adequacy of the building's gravtty system also ts not evaluated e.xc:ept where requtred as
pan of the buddtng·s lau:ml-force-restsrjng system.
OUTUNE
I. Building Geometry and Seismic Crtteria
2. Roof and Floor Gravny Loads
3. Laternl Loading: Setsmic
4. D1ophtagm Flexobolity
5. Flextble Diaphragm Condttton
6. Buddmg Classificauon: Regular or Irregular
7. Redundancy Factor
8. Redundancy Check tor Build1ng B
9. Sefecled Anai)1Jcal Procedure
10. Distnbulion ofSe1smic Forces to Shear Walls
II. She.athed CFS-Stud Shear Walls: Franung Marenals
12. Shear Wall Design Example: Building ll
13. She~r Wall Deflecuon
14. Disc:ossion: Framtng with Coki-Formed Steel
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15. Dt~USSIOO: Se1.srnte Joants
17. lt<mS No< Addressed
111 ThiS
E>ampl<
18. Refenmc<s
DESIGN EXAMPLE BUILDING
Ftgures J.J th(()tlgllJ-S tllustmle lhe apMment bu~dtng comple.x·s front elevallon. Hrsl Ooor, second
floor. tJ11rd floor, and roof' plans used '" lh•s destgn example. F1gure 3-6 shows an elevation cut through
the butldoo>g of smeked cold·fom>ed steel (CFS) framed shear 11~lls. The buoldmg complex os cut tnto
a sJopmg grade, wht(:h tS vr:ry cotnlnolt for tJtese types ofbutldtngs. The fir.n Ooor, podtwule\el. is an
elevated retnforced concrete Strt.J(,1ural slab oo;er the pamal subtetro.nean pad.tns, lt\'CI. There are two ltgh1frame lhree-story bUIIdangs thai s1t on thas podtum deck: Bu1ldmgs A and B. Uutldtng A has a sutgle-story
communtl)' room w•ng,. \\hteh has a green roof. ThtS roof. as \\ell as the dtrect•on ofche decoratrve wood
slat sadmgon the ~tenor '''ails oftht b.uldtng. can be seen 1n the fron1 ele-.11t1oet off1gure J..l.
II•
I
,._.
·-~Ji
,...igtJrt J-1. Apartmtnl complex jm111 (#/notion
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'
I
Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
The two three-story BuddUlg5 A and Bare Interconnected by a pedestrian brtdg_e lhm ha') a seism1c joint
(see Figures 3-3 and 3-4). Access rome residentjaJ units is by means of exte-nor stairs and cantilever
walkways. The walls of the reside-ntial unns at each floor level staCk verttcally. so there are no honzonta1
ofrsets of the load-bearing. \\ails and shear waJL~ be-tween ftoor levels. Engineered wood 1-Joists are used
fo( the fl()()( system, and pre.fabncated metal-plate-connected wood trusses are used fo( the roof memhets.
Rated 'vood strucnnl panel (WSP) sheath1ng is used for the floor and roof diaphragms.
Figu" 3-2. Ftrtr-jloor pltm--Bmldings A and B smmg on a common podmm deck
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
Flgrmt 3-J. Second-floor plan
Figure J-4. Tlurd-jloor plan
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
?
ii ! i
I
0-----+
I
~----h'.~--+---+--,J,
j
<p
i
i
e-=---'--=--~==<i==="11=='-i
o em
on
- - - --C-1-
- - - -"""-t::)
--- -0
------0
--- -€>
----0
Figure 3-5. Roofplan
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
"f'CDQH B=f!
SEMJQH Q=G
Figw¥! 3-6. Shear wall de-t·aJion u.ting mixedfroming materla/~old-formed Steel (C FS)
ltglu{romt sMar wall and wood-frt:tme jloor and rOO/
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
The Wl.tque a~c.t of this butlding is that CFS v.'all framing is used fOr the budding's wall gravtty and
lateral fi'amtng (wood structural panel-sheathed shear walls). whale the roof and floor systenl is framed wnh
wood membe.rs. The use of conttnuous wood ram joists along the top ofCFS stud beanng walls an y.'QOd
floor and roof framing systems aJiows for flexibility 1n lhe floor- and roof-jolst layout, regardless of where
the CFS studs ahgn. The wood floor and roof ttaming ts supported by hangers from the wood rim JOist,
which acts as the venrcaJ load distribuuon member to lheCFS studs.IJlhe floor or roof framing is to bear
darectly on the wall, tlten lhe CFS swd wall top track or wood top plates must be destgned for flexure and
shear to support the wood 1-jotst reacuons and transfer them to the CFS studs.
Where CFS floor JOistS and metal-plate-connected wood roof trusses are tJSed wtlh CFS-frruned beattng
walls. the alignment of the CFS studs wttlti.n the wall wnh the CFS floor JOists and wood roofU\Jsses is
much more criucal. Regardless oflhe floor system selected. CFS floor framtng or wood floor fi'atning, the
design oflheCFS stud waUs'top track or wood double plate, or combtnauon of both. must be ck.~igned to
uansfe.r the floor grawry loads to the Yl'3ll studs.
MIXED-USE, MULTIFAMILY, MULTISTORY PROJECTS
The apartment complex is no longer just a relatively ssmple structure buah fi'om wood or CFS studs sining
on a concrete slab-oo-gmde wnh ronttnuous and tso1ated retnforced concrete pad fooungs.
The apartment buddtng is now typically just one component of a Larger mixed-use. anulufanuly, multistory
complex that tncludes retail spaces, office spaces,loadtng doc.ks, and above- and be-low-grade parkmg tn
addirjoo to the apartment un1tS. The apatunent/rental units typically ocwpy the space above the uppem1ost
podium deck, wh1le the concrete podium structure conta1ns the parking. reuul, dehvery. and commercial
funcuons. 'The concrete pcxilum deck can QC(:Uf at grade level over multiple levels of subterranean parktng
or be several stories taU above grade. These mc<ed-u.<ie proJects commonly oontam any\\<here from Just five
or six mdjvidual rental units to severaJ hundred apamnenlfrentaJ untts.,. and the supponing concrete podium
footpnnt can extend fOr hundreds of feet m each dtrecuon.
Apanment bt.llldjngs consuuc.ted during the 60s, 70s, and 80s typically were smaller, and the desagner
onJ)' womed about the Slfeet-fi'ont appeamnce since the sides were obscured by adjacent buildings, and
not much emphasis ·was placed on the alley elevations. These: Larger mixed-use projects often take up lhe
maJOnty of the block. stretchmg from st~l to adjacent street, to taktng up the entlre block. requtring very
e-laborate street-front appearances on all exterior elevations of the bualding complex as opposed to jLL">l the
s1ngle street-front elevation oftlte apartment buildtng of past years.
FIRE-RESISTIVE CONSTRUCTION
The demand for residenual housing (apartment rentalslcondomintums) st.nce lhe 1980s has led to an
increase in the typtcal number of residential floors being provided tn lhese residential projects from three
stones ru'ld less. to typtcally four ruld th1e stones, and tn some Jurisdictions to SIX stories using fightframe consliUCUon. Along wtlh inc.reastng lhe number of floor levels betng uttlt.Zed in reside-ntial houstng
have come additional building code reqUirements regardtng use of ooncombusttble COt'l..">truc.tJon framtng
members (CFS studs. fire-treated lumber, etc.). Ltght-frame butldutgs lhat are three stones or less typically
fall into the category of Type V coosttuction. but oa'lCe lhe d~igner goes above three stori~ the building
may now faJI tnto the c.ategory ofType Ill construction, whtch has more stnngent fire-ratJng requtrements.
In Type IU construction. the fire mung of the extertOC' '"''all assembly has now become more unportant. often
dictating lhe d•rectaon of the. floor fi-aming to mtntmlU extertor wall fire-raung requtrements, depending on
whethe.r the extertor Y.allts a bearing wall Of a nonbearing wall.
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
For light-frame structures five stones and taller. it ts recommended mm the destgne.r review the requtrements
of Type: lilA orB con.'itrucuon provided tn the l11ttrnauonal Buildmg Code (IBC) Chaprer 6~ since they
sigrufi.cantly impact framtttg-destgn requtremems. This would mclude the engtneer being familiar wtm
fire-separation distances. since l.hey lmpact the required fire rating of the extenor walls. St.nce me buildmg
e.xampfe beang presented here is just three stories in height. the requirements of Type IUA orB construction
will not be dJscussed.
TERMINOLOGY
The light-frame-mulufamdy buildmg sttucture abcwe the structural retnforced c,oncrete podtum deck
at me first-floor level will typically be referred to as the ..superstrocture" for dtscusston purposes. The
de.finjtion is to divtde the budding \renically t.nto two zones: the hghr- ttame smtCture above the podtum deck
(superstruCture zone) and the tirst-ftoor podtum deck down to the lowest parking le\•el (subsl.n.lcture 20t1e).
DESIGN EXAMPLE OUTLINE-CODES
ThiS destgn example follo\\'S the provisions of the 2018 IBC and the appltcable standards referenced tn IBC
Chapter 35. These 111CiudeACI318- 14,AWC NOS 2018,AIVC SDPWS-2015,AISC 360- 16,AISI 5201 - 17,
A lSI S240-15,AISI 5400-15/SI-16, ASCE 7- 16,and applicable ASTM SI<Uldards. Some groupsolfer thetr
electronic standards for free~ such as AISI through CFSEI. AISI 5240 Commentary Sectjon A2 re-ferences
AISI 520 I s~uon A 1. whic.h defines cold-formed stee-l hght-frame construction as that whe.re the CFS
memberth1cl-ness is between 0.0179 inch (25 gauge)and 0.118 inch (10 gauge).IBC Section2211 requires
the design ofCFS light-fi'ame const:ruction to be ln accordance with AISI 5240 and that seistruc-tOrceresisl.lng syslems shaJJ be 1n accordance with AJSI S4-00.
While Chapter 22 of the 20181BC references SI00- 16 for CFS design, for CFS hght-lrame const:rueuoo,
5240-15 and S4-00- 15/S 1-16. whtch i.n turn reference the prevtous CFS specificauon edtltoo.
S 100- 12. The authors recommend using S 100- 16 along with tlS supplementS 1- 18 as n represents the lalest
CFS research and understandtng. Note that the secuons m 5100- 16 and suppleme.nt Sl-18 are reorganized
and some c:onlent was added/dele.ted. S 100-16 contains a c.hapter/sectiorl cross reference back to the S 10012 edttJon to assist tn understanding me changes.
tt references
TheAJSI stru\dards consist of the followtng:
AISI SI00-16: Specification
AISI 5201-17: Product Standard
AISI S202-15: Code of Standard Pmcuce
AISI 5220-15: Nonstruc!Urnl Members
AISI 5230-15: Prescnptive Method for One- and Two-Family Dwellings
AISI 5240-15: Structural Framing
AISI $310-16: Profiled St"'l Diaphragm Panels
AISI 5400-15/SI-16: Smrruc Des1gn of Cold-Fanned Steel Struc!Ural Systems
The tblhm•mg has been updated SU'Ice the adopt ton by the 2018 IBC and the authots recommend lhal 11 be
considered as n contains update<Vdeleled information:
AISI SI00-16/SI-18: Supplement
Where standards are referenced tn the folhm•mg des1gn exarnple, such a.~ AJSI S400,lhe author"s expecraoon
l~ the reader wiU be usmgthe adopted or latest supplement lO that standard (example: AISI 400-15/SJ-16)
where tt occurs.
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
Only Butldmg B lS betng evalurued in this design example stnce the design approach tbr Budding A would
besimtlar.
1. Building Geometry and Seismic Criteria
1.1 GIVEN INFORMATION
I. The btuld1ng fi'aming c:onstSt(i oflig.ht-frame c:.onstroclion utthZtng CfS-fi'amed beanng wall~
wood structural panel (WSP) shemhing, engtneered wood 1-joists and Slructural composite
lumber (SCL) beams fOr the floor, and engineered metal-plate-connected wood trusses for the
root: The superstrucrure·s type of c:onstructjon IS constdered 10 be Type V-A. R-2 (apanment
budding).
2. The floor height< are (finish floor to finiSh floor)
a first to second floor:
10 tl (actual= 10ft II in)
b. second 10 thtrd floor:
10 tl (actual= 10tH\ 1n)
c. third floor to roof.
II tl (actual= 10 tlll 1n)
3. Butldtng Setsmtc Destgn category (SOC): D
4. Buildtng Risk Category: II
5. Buildtng superstructure·s lateroJ.resisung syStem: Beanng lYaii System (ASCE 7 Table 12.2· 1):
a Shear walls: light-frame (cold-fonned steel) walls sheathed ·wtlh ste.!l sheet(i or wood SliUCtuml
panels ra1ed fot shear testsmnc:e
b. R=6.5,0.,=3.0,C1 =4.0
L
If diaphragm is con.<tdered to be flexible, t.hen 0.= 2.5 (Footno<e b,ASCE 7 Table 12.2· 1)
c. Maximwn allowed buildtng height SOC 0::65 ft.
1. A.re-hiteclutal and fire requirements may reduce allowable building height Building
occupancy is R-2.
ii. Ftre spnnklers are f'equJred by the bt.nlding code, type of fire spnnlde-rs 1mpacts height lunns
• IBC Table 504.3: Type Y·A construction: Standard sprinkler (S) = 70 ft (S 13R sprinklers=
60ft)
• IBC Table 504.4: l)'pe Y·A construcuon: Standard sprinkler (S) or (S 13R sprulklers) = 4
s1ory levels ma:(tmum above podiwn
ui. 65 ft>60ft>butldtngheight=( 10ft+ 10ft+ II ft)=31 ft
6. Sefecled analytical procedure: To be delenntned
7. Floor and roofdtaphtagms are conslJ'UCted ofWSP sheathing and are blocked
8. Solis report: Soil Sue Class D
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
2. Roof and Floor Gravity Loads
The folloo•ing is a description of the buildlng supemructure·s dead Joods and the supenmposed hve loads
for the roof. thtrd-. and secOild-floor levels of the buddmg. The applic.ab1e live loads to use ror lhis destgn
e.xample are defined in the bu.ildtng code-. depending on the func:tioo of the space (roof~ I iving unit. balcony,
COITtdor, parktng. publtc access. etc.).
The gravity loads are d1vided between horizontal fromlng elemenLo~t (floors) and venicaJ fmm tng elements
(walls). l11e gravtty loads are also used ln calculaung lhe design seismtc forces for lhe butldtng strucrure.
2.1 ROOF LOADING
love load: 20 psf Dead load: 16.0 psf
Table 3-1. Roofdead-lood nraJent1/S
Matena1s
Weigllt (pst)
Roofing
2.0
~-inch ot %-tnch plywood/OSB span-rated sheath1ng
2.0
Wood
roofttus.~sal
24 inches on center
2.0
1\vo layers of ~-tnch gypswn boartt•• cethng
5.0
Spnnklers
1.0
lnsulatton
1.0
Miscellaneous (ducts, coodwts, ptpes)
3.0
16.0
Total:
Note tor Table J- 1:
I. Gypswn board, Gyprock, Sheetrock, plasterboard, wallboard ruKI drywall
are all W'Ords used to descnbe the same product.
Additaonal roof dead loads:
I. Some designers include the we-tght for a future re-roof where new roofing is placed over the e.xisung
roofing m:uenaJ. Thls can be a good design practice to account tOr a future re-roof. but tncludtng the
re-roorwetght also lmpacts the seismic destgn of the budding. It JS ac;sumed for this example lhat
when new roofing is required, the exisung roofing will be rem<:wed down to the existlng wood roof
sheathing, and thu.o; the weight of a re.roof is not tncluded.
2. Roof sheathi1tg thtclcness vanes, depending oo prOJe<-t·des&gn reqUJreme-nls. The thicker the roof
sheathing, the fe-y.•er issues w1th the de-bond&ng of the roofing mateJial adhes1ves from the'"~
sheathing as a result of matntenance people or others walkmg on the roof over the years. A two-ply
roofing membrane system lS assumed for this project
3. Rooftop air-oonditJoning/heaung unns are not being used on this project, and a cust:om atrcondttioning unit ha.o~t been installed in tndlviduaJ uniK
4. Solar panels: The owner would hke the roof to aecommodare future solar panels m predesignated
areas. An add1tional 8.0 psf of dead load will be oonsuJered tn those designated areas for the weight
of the solar panels and franung. The furure solar panel design wtll have to won: withu1 these deadload hmtts.
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
5. No root areas for the lhree-story bulldtngs are destgnated to be green roof areas reqmring additional
dead load tbr rooftop landscape. Only the stngle-srOI')' w1ng has a green roof. butn is not part of the
three-story build1ng chose-n for the design example. and thus the additional weiglu lS not accounted
tor in the roofdead-load mater-ials rable.
2.2 THIRO·FLOOR AND SECOND-FLOOR DEAD LOAD AND SUPERIMPOSED LIVE LOADS
The live loads at the second and lh1rd ftoors are hmited to use withm the unn and walk1ng along corridors
or stattS to get to indtvidu:.al units. The liVe lood used Wtthin the umts acoounts for a normal exp«:ted
fumJture and cabtneuy Hoor Layout. The destgne,r suit must revtew whether spectal ftOOI' framu1g wdl be
required under any special permanent fh.1ures to be Joc:ated withtn the unit An example might be a heavy
spa tub.
Tab/~
3-2. Third- tmd St!COud-/foor lh·e /()(ld!
Ltve-Load Requirement~ by locatioo
Design Load (pst)
Wtthin unns
40.0
01
BaJoontes
60.0
Corridors
100.0
Exterior stairs
100.0
Note for Table 3-2:
I. Balcony ltve load (per ASCE 7 Table 4.3·1) is a minimum of 1.5 umes tl1e
una served live load, but IS not teqUJred to exceed 100 psf.
Tab/~
3-3. Third- and second-j/()()r d~ad /()(ldt
2nd- and 3rd-Fioor
Wetght (psf)
Floor Dead-Load Matenal
Ftntsh floortng (carpet/ltnoleum)
1st-Floor
(pochum lev<!)
Wetght (psf)
1.0
H\-tnch hgh1weightconcre1e ( 120 pet)
15.0
1.0
-
%-inch or ~-mch plywood/OSB span-rmed floor sheathmg
2.5
-
Wood ftQ()( JOISIS at 16 tnches on center
2.5
-
Spnnklers
1.0
1.0
MJscellan.eous (ducts. condutts, papes)
3.0
3.0
2 Layers of ~ t nch gypsum board cethng
12-anch-tbick retnforced concrete SltU(:tUtal slab
5.0
-
150.0
6-tnch exre-nor concrete topptng slab (where occurs)
-
o.on5.o
Subtotal
30.0
1551230.0
lntenor paruuons (see part1tton ealculauon)
14.0/15.0
14.0
To!al
44.01~5 .0
169.01244.0
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
Addiuonal floor dead loads.:
1. The type ot'fin1sh floonng be1ng used should be venfied w1th the archnect/o\mer to see 1f 11 t<;:
going to be stone/ma-c;onry tile, wood, linoleum ttle, or carpe-t. For this example, the finish floor
is asswned to be carpet throughoUl the budd1ng and hnoleum tn the kuchen and bathrooms wath
a 1\~iglu of 1.0 psf.
2. l1le thickness of the OOJ'tcrete topptng o\'et the floor JOistS. whe.re provaded, ranges from Y., 1,
or I~ inches w1th1n the unn. and can be in the range of 3 inches in corr1dor trnnsiuon zones
from intenor to e.xterior locations along the corridor. which has a sagnaficant tmpac.t on the
floor weight. The conc.rete topping \\-eaght, dependtng on thickness and denstl)', typtcally vanes
be.t\veen 8.0 psfto 15.0 psfwnhin lhe mdivtdual units. For thiS bulldmgexample,the wetg.hl iS
assumedtobe IS.Opsf.
3. lne reqmred thtckness of the \VSP floor sheathing can be de.pendent on the requared sound
rransnuss.on control (STC) rati11g. llus design example assumes that ~-inc.h WSP sheathing
wdl be used along w1th two layers of ~tnch gypsunt board on the bottom of the jol$1S.
4. The ftOOf wood 1-jOl'it weight wtll \'31)', depending on actual 1-jotst depth, span. spac1ng. and
the size otl-jo1st flanges and webs. The weight of com.merc•ally available ftoor wood 1-joists
9~ tnches to 14 inches deep typically ranges between 2.3 to 4.2 pounds per linear foot of
JOist. The average mulufumtly, multistory project Will ()'pic.ally use Olle wood 1-JOISts "depth"
throughout the project. but will hkely include three or tOur dd'fe,rent types of wood l·joists
for the selected joist depth to address different JOISt-span condtuons. S1nce different types of
wood 1-jotsts and wood 1-joist spacings often occur throughom a floor level of the building. an
..average" floor JOISt we1ghtts used. The average ftoor JOtst wetght for th1s des•gn example IS
asswned as 2.5 psfwhen spaced at 16mches on center.
5. lne first-floor (podiwn level) weights are included for comple.teness of the building weight.
There are often addittona1 concrete topping slabs on the podium deck tOr waterproofing..
Adchuonalloads for planters, cantilever masonry securny waHs., and other pennanent
archuectural fe.atures (shade structures., trelhs. colonnades, etc.) must be included tn the
coocrele podJwn de:s1gn.
2.3 INTERIOR PARTITION WALL LOADS (GRAVITY DESIGN)
The 1ntertor part1tion walls 1nclude both bearing walls and nonbearing walls, some ofwhic.h are shear walls.
IBC Section 1607.5 reqmres floors on which partitions can be rearranged or moved to use a mtnimwn
partition live lood of 15 pst: Oft.en people h~we u...-;ed 20 psffor the pamtion gravity load an oommerctaJ
buddtngs that are constdered to have moveable walls., and a lesser amount tn restdenual buddtngs. Since
the reside.ntiaJ.untt walls are.considered fixed. the actual a\re.rage partttion wall w-eight wtll be used. The
tndt\•tduaJ unh demising waJis often have WSP sheathing or sreel sheets, but this is not typical for the unns,
Interior walls. The uppe.r floor shear v.alls wlll typically have ooly WSP sheathtng. or stee.l slle.!ts on one
side. while the lower floors are likely to have \VSP she.athtng or steel sheel~ on both sides.
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
Table 3--1. Partltit>" dead-load mtJierial
Non-shear Wall
Weigh! (psi)
Matenal
1ft....
ioc.h \VSP sheathing (oonservatwely asswne both fac.es)
Shear Wall
Weight {psf)
0.0
3.0
5.0
5.0
4-ineh or6-mch CFS studsal 16 inches on cenrer
1.5
1.5
Moscellaneous (pipes, conduits)
1.5
1.5
Total
8.0
11.0
~inc.h gypsum
board each face
The average parutton ftoor load was detemuned by summing up tl1e the walJ weighl of aJI the walls within
an tnd!VIdual unit (length of walls x height of walLs x wall weiglu pst) and then dtvtdJng the toml weaghl by
the square fOotage of the tndivtdual unu. This \\'aS done tor several w1its, resulting in average pattuion wall
weoghL< of 12+ and 13+ psf of floor area.
l11e ftoor-lo-floor heighl as typ1cally 10 feet 0.: tnch. Wall clear height is aboul 9 feet 0 inches at the first and
second floor, 3.Rd n is aboul 9 f-eet 6 tnc.hes from tl1e third floor to the unders1de of the roof tfaming.
For this destgn example, a paruuon gravity load of 14 psf (consen•atively used stnce tt excee.ds 13+ psf
found for some units) shaJI be used utufOrm.ly across the floor at the second floor~ while a 15 psf parution
gravity k>ad is used at the thtrd floor due to the taller wall hetght to the root:
2.4 EXTERIOR WALL WEIGHTS
The exteraor waJI \!fetght needs to be accouttted for wtten determtmng the buildmg's gravtty Joods imposed
on the first floor slfliClural concrete podtum. The waJI weight is tmportant as well for calculation oflhe
buddtng's setsmte base.shear. The extenor wall weight calculation should also lrtclude tnterior eorrtdor wall
wetglu..o; tfthey we·re not 1ncluded in the pat'ttuon loadlng. The wall hetghts ryptcally used are the f1oor-t~
floor height., since this will account for the \1/e'.ight ofrtm joist and blocktng tn lhe wall that is not included
itt the normal floor dead loads. Some destgners choose to define the wall height as the. clear height betwee11
tl1e top of the floor sheathing and the top of the platform waJI cor1strucuon (underside of floor or roof
JOtst). lrt whtch case the weighl of the floor Ime nm j<>lst and blocktng are tncluded as pan of the floor JOlst
framtng Y.'ttght.
Table 3-5 summanzes the e.xterior wall m.ateriaJ and its representative wetght based on the wall verucal
surface area. Wall weightS are typtcally determined based on vertical surfuce area
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
Tablr 3-5. Exterior wall dtad-load mattrial
Wall Surface Area
We1ght (psf)
Matenal
Wall S1d1ng or Ys-inch-thiC-k StUCCO
10.0
\'\- inch plywood (assume pl)~l·ood both laces)
3.0
l'\- 1n<h gypsum board (mtenor face)
2.5
6-1nch CFS studs at 16 1nches on center
1.5
Jnsulauon
0.5
Miscellaneous (p1 pe.~. eonduu)
1.5
Total
19.0
2.5 EXTERIOR WALL AND INTERIOR WALL PARTITION GRAVITY AND LATERAL LOADS
The extenor wall and imenor wall pamuon we1ght is accounted for ditl'erently, depe.ndtng on the purpose
of the \liall weight calculation:
J. Budd1ng stlsmic lateral calculations.
2. First-ftoor OOJ'lcrete podmm tmposed design loads.
Thus~
two diff'erem waH weighl calrulations are typ1cally done.
E.xttrior \\'alh: Buildiag Stismit lattral Calculations
The butld1ng•s exterior wall weight tOr this de.~ign example is summed based on a floor-~floor analysts.
Extenor wall we1ght = (tnd1vidual waJI Ieng.th) x (wall he1ght) x (wa11 we1ght in pst)
All extenor wall weighlS m the buildmg are then summed together for each floor la•el, and this toW. v.-eight
is used m calculating the bu1ld1ng·s seismic base shear. For stmplicity of desi_gA door and window ope:ni.ng.s
are often omitted. and lhe walls are consuJered solid. Some designers may account for openings by acrually
calculaung the area of lhe opemng.-; or subtracung some pon1on of the e.xteraor wall weight to account for
the opemngs.
Si.nce lhe butld1ng \\ails span between floor levels. half oftl'le wall weighl JS distributed 10 the floor level
abcwe and half is distributed 10 the floor lt!\•el below. llus division is used to determine the tributary mass
f« each buddjng floor level. Atlhe upper mOSl floor level of the budding. the upper half oftl'lat floor level 's
wall ,~,--eight is associated wuh the tnbutaty roof mass. If the floor-t~floor heights vary. then lhe tributary
waJI weight a~iated With each floor level wall vary over lhe bujldtng height.
This total weight fOr each floor level may then be divided by the total floor area for that particular level of
the bu1lding. Th.is can be helpful1n accounung for the average wetght Of mass assoc.Lated W1th each floor
level of the buildtng foorpnnt when calrulaung the seismic des1:g,n force distributJon over the hetght of the
bwldlllg.
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
E.tttrior \Valls: Firs-t-Floor Contrtlt Podium lmpostd Dtsign Loads
ll\e design of the concrete podiwn ts oot pan ofthtsdesjgn example.• bot the supersuucrure design loads
are shown for complete.ness to demonstrate how lhey would be calculated and unplemenred lJ'l the conc.rete
podium desogn.
The desogn loods 1m posed on !he podium ore typically divided ornong:
I. Umfonn surfuee load.' (example: live loads).
2. Line loads (example: "all loads).
3. Point load(i (example: column reactions).
Current destgn practiCe ullhzes finjte element analysis sotm'ate to evaluate lhe elevated structural concrete
podium and deternune the resulu.ng design forces based on the 1m posed design loads. The design enginee.r
determjnes how they \W.Otto apply the design loads. Some design practiuoners just sum up all of the
verucal loads and apply thts as a umform surfuce load over the slab. while others use a more refined
approach to consJder all the untform, I i.ne. and pOtntloods separately. The appfoprtateness of dttferent
design approaches used for applying loads (surface, line, point) for podium desogn IS left to the design
engtneer.
Pod1um-unpos:ed wall desJgn loads are generally accounted tOr uxhvtdually. The ' 'en1cal Y~'3llloads are
usually c:aJculated on a per-tbot-lengtJ\.of-waJI basJs to make them easter to enre.r into the pod tum analys1s
software.
Ex1erior wall \VCighl (per foot)= [(total wall he1ght) x ("all weight in psf)J +[individual ftoor tributary
loads to extenor wall (per ioot of\•alllength)J
The same appt"oach would be used fOf calculaung t.ntenor beartng wall loads to be imposed on the podiwn
de<-.k as we.ll.
3. Lateral Loading: Seismic
ASCE7
3.1 LATITUDE AND LONGITUDE
The .setsmtc spectral response acceleraLionsSJ and S1 fOr use specrficaJiy with the IBC and ASCE 7 are
oblail'led ffom the Unned States Geolog•cal Survey (USGS) based on tJ\e latttude and longatude coordtnates
of the build1ng sue (hnps.://earthquake.usgs.go\'/hazards/designrn.apslusdesign.php) using third-party web
GUisfrom:
I. ASCE: hnps:/lasce7ha:zardtool.onhne/
2. SEAOC/OSHPD: hllps:l/se1Sm1cmaps.org/
3. ATC: hnps:llh32ards.atcouncil.org/
USGS pre,•toUS1)1 provided online tools to detem1.ine s. and s,. but it ha<i now dlscontinued this service.
USGS sull provtdes the set.srruc data that lhJtd-pany software deve.lopers retraeve ln deterrninmg the s. and
S1values for a given locaLion.
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
The,re are programs avatlable on the internet that allow for the conversion of the butldtng"s suee.t address
to latuude and longnude ooordmates. A good program will check severa1 dtffe.rent uuernet sources to
detenn10e the laurude and longuude coordtnates for a g.l\'en address stnce each tntemet source calculates
Lhe latitude and longttude coordanates ddfereml)1• For example, a good program would check severnl
sean:h engine< hke Google, Tamu, Geocooer, Yahoo, and Vinual Earth. Generally, !he re.pcned lautude and
longuude results fot the bu1lding site from eac.h source are very close to each other. The three tlurd-party
web GUis refereoc.ed earlier gh•e the user the option of using either latitude and lo11.gitude or the stt~t
address ofthe building s!le.
lftl1e project site is large. or tfthe e.nganeer wants a ge.neral idea of the seismicny around the project sue.•
they may \\'3nt to check several ditferet\t latitudes at\d longarudes across the project sne or surrounding area
by va.cying the program Jautude and long11ude coordmates in IncrementS of 1/ IOOth degree plus or mmus.
Latitude: 1/IOOth degree= 0.6 minutes= 36 see= 36 ( - 100 fttsee) = 3600 ft
Longitude (at laurude= 30°): 1/ IOOth degree= 0.6 minutes= 36 see= 36(-88 fttsec)= 3168 ft
Thls gtves the e.ngtneer a se,nse of the variation m ~~ aoo S1 across the area surrounding the butlding site.
3.2 BUILDING SITE SEISMIC DATA
The apartment buildlng compte:~ is located in los Angeles. wh1ch is a high se.asmlc region. The following
destgn tnfbnnauon wa~ determmed based on the lamude and longnude enre.red mro the SEAOC/OSHPD
seesmu:· destgn maps website program tbr Lhe buildmg sate. The geotechnical re'JXm for the buddang site
should include some of this information as well to confirm the engtneer"s innial design asswnptions
reg.ardtng site. sod pmpenies used to determme soil site cla'iS, S,. and S1•
§II A. I
Ntar-Fauh Silts
Site is not a Jlear-fault site (per soils repon)
1. Site l'i located ihnher than 9.5 mile.-; from a surface proJectioo of a known acti\·e ralllt capable
of producing M.7 or larger e\'ents.
2. Site is located farther than 6.5 mdes from the surfac-e projecuon of a known active fault capable
of producing M_.6 or larger eve-11ts.
Table 3-6. Spectral accel~ratJOn! Umm SEAOCIOSHPD program)
Penod (seeoods)
s.(sJ
0.2
1.675 (S,)
1.0
0.600(S1)
Sptctnl RtsponSt Act:tltratioas Ba.Std on Sitt-SpttUit Soil Class.ificatlons
Stte sot! properues: Site Class D--sutr sot! (confirmed tn the projoct's sods repon)
Long_-penod transition period TL = 8.0 sec
Site coefficient: F, = 1.0 ( 1.0 per soils repon; otherwise, F, = 1.2. by default per §11.4.4)
S11e coefficient: F, = I. 7 (also see § 11 .4.8 that follows)
170
F 22-14
T ll.4- l
T 11.4-2
s..=213S,,.=213F,'>,= 1.117g
§ 11.4-1 and §11.4-3
s.. =213 s"' = 211 F,:>, = o.680g
§ 11.4-2 and § 11.4-4
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
Sitt-Sptcitic Ground Motion Prottdurt:s
§11.4.8
Per Item 3 ofthis sect.iOI\ a ground motion hazard analysis shall be performed m accordanc.c with Secuon
21.2 for structures located on etlher Sotl Site Class 0 orE where S1 as greater than or equal to 0 .2. Tile
ground motton hazard analysls l~ not required for struCtures 1ocated on Soli Site Class 0 when the
followang are satisfied:
I. C,(per equation I2.8-2): Use 1.0 x C, (when T,; 1.5T,) where T, = S0 ,1S.o<= 0.68011.1 I 7 = 0.609
2. C,(per equation I2.8· 3): Use 1.5 x C, (when T,;, T> l.5T,)
3.
(.~(per equation
12.8.4): Use 1.5 x C, (when T> T,)
A response spectrum analysis ts not required tOr this butldtng structure.
Tile eqUJvalent lateral force procedure shall be used fOr thiS analysts.
T 12.6· 1
3.3 BUILDINGS A AND B-SEISMIC DESIGN REQUIREMENTS
Structural system: l1ght- ftame CFS walls sheathed Wtth wood structuml panels
Number offtoors: 3
Buildtng hetgllt (above tl\e podium level)= 11, = 31 ft
Risk Category: II
IBC T 1604.5 aod T 1.5· I
Sod Site Class: 0-sutT sotl
Seismac Design Category (SOC): 0
lmponance. factor: 1, = 1.0
Bu1ldU''I;g response modification coefficient: R = 6.5
From Solis Repon
T I 1.6- I and T I 1.6· 2
Tl 1.5· 2
T 12.2· 1
Building period detemunat.ion:
Approximate period parametet, C~ = 0 .02
Approxunate period parameter. x = 0.75
Approxamate fundamental period~ =C,H'.= 0.263 sec
Coefficient C.= 1.4
Butldtng period (1) used= CJT.,) = 1.4 x 0.263 =0.37 see< 1.5T, = 1.5(0.609)
T
T
T
T
12.8· 2
12.8·2
12.8· 7
12.8· 1
ASCE 7 §I I .4.8 (Exeepuon 2)
Nme-: Seismic Design category E is not re-quare<l since S1 is less than 0.75.
Stl~mic
Latt.ralloads
V=C..W
Eq 12.8· 1
Base shear ( Jl) equation checks (ustng No Ground Mouon Hillard study) T & T, per ASCE 7 §I I .4.8:
V(when T < 1.5TJ= (S,J/R)W= [(1.1 I 7XJ.0)/(6.5)JIV=O.I 12W
Eq 12.8· 1 ond Eq 12.8· 2
I'(when 1.5T, < T < T,) = (Sb 11/R1)W = [(0.68X I.OJYI(6.5)(0.37)]( 1.5)11' = 0.4241V
I"(when T> T1 ) = (S01 I,T.fRT')W = [(0.68)( 1.0)(8.0)]1{(6.5)(0.37}']( 1.5)11' = 9. I 7W
Eq 12.8· 3
v•• = 0.044Snsf,W= [(0.044)(1.1 17)(l.O)JIV= 0.0491V~ 0.01 IV
v•• = (0.5511/R)W = [(0.5)(0.60)( I.OJY(6.5)W = 0.04611'
Eq 12.8· 5
Eq 12.8-4
Eq 12.8· 6
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
The.refore, use V = 0. I 7211:
Vernc.al diStribution coeffictent, k = 1.0
§ 12.8.3
Author's Distussion: Equanons I2.8-3 and 12.&-4 .are shown fbr completeness. as destgners often u.c;e a
genencspreadshee.t showing alllhree base shear Equations 12.8-2, 12.8-3, and 12.8-4 so they can use the
spreadsheet fOr every project. Since the calculated butlding pe-riod ( is less lhan 1.5 T,.. EquatiOns 12.8-3
and 12.8-4 do no1 need to actually be chec.ked tor thts destgn example. If the butldtng pertOd (1) was longer
than l.5T.., then Equa11ons 12.8-3 and 12.8-4 would need to be evaluated. as ap)X"opriate.
n
I fa soils re.port hao; not been provided, then by default I'"= 0. 172 x 1.2W = 0.2061V a" Ftt = 1.1 tnstead of
I .0. For longer penod (7) buildings, a ground bazard analysis woold likely be perfonned by a ge01ecluucal
eng1neer as the 1.5 multtplter for C, IS a Large penalry for the butldang base shear ('1· Su'tCe a soils report
was performed, tltis would include the Site-S~1 fic Ground Motion Procedures per Seetions I I.4.8 and
21.2; thus, the multiplie.r for C, = 1.0 for Equations 12.8-3 and 12.8-4 instead of 1.5 whe.n no stte-specific
ground motion procedures are checked.
M.a.d mum S 0 s Va.lut in Dttrrmi.aation of C, and£.
ASCE 7 §12.8.1.3
c, to be calculated ustng a value of s&..equal to 1.0 . but oo1 less than 70 percent
1nASCE 7 Section I 1.4.5. whe.n the fbtlO\ving occurs:
The budding code allows
ofs~~ao;defined
1. The buildmgdoes not have trregulanties, as defined m SecLion 12.3.2.
2. ll'!e buildmg IS thre stones. or le~ above the lower of the base or grade plane, as defined tn Section
11.2. Where present, e..ach me:zzarune level shall be considered a story for lhe purpose oflhas 1tm11.
3. The buildmg period (7) is 0.5 second or less, as determmed using Section 12.8.2.
4. ll1e strucrure meelS lhe requtrement.'i necessary tbr the redundancy factor~ p, to be permitted to
betaken as 1.0, tn acoordance wnh Secuon 12.3.4.2.
5. The s ite sod propentes are not dassstied as Site Class E or F,asde,fined tn Secuon 11.4.3.
6. ll1e strucrure is C'lassi:fied a" Risk category I or 11. as defined m Secuon I.5.1.
As Will be de.mon.'ilrated late.r in lhis design example, Budding B due to tts "l-shaped,.. plan cottfiguration is
elass1fied a'i ..Irregular'', and thetefore lS not permitted to use the Sa; hmn. SDS used in this design example
shall rematnSbS= 1.117.
Author's Distussion: The number of stones is measured from lhe grade plMe, so for purposes of
detemuning SD$> thlS bu.ildtng wooJd be cons1deJed a(i being four stories (one level of podium and three
levels ofhght-frame construction), \\-hich is Jess than the five-story limit.
3.4 BUILDING MASS
Ctntrit- Building Footprinlfl.ayour or Building§
The schematic OUI.lines oflhe tw"O three-story buildings on the first floor podium are shown tn Ftgure 3-7.
As noted prevtously. only the seasnuc fOrces tOr Budding B are betng calculated.
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
,.----------- ...
.-------1 Zooe 82
Zone 81
: Bldg. B
"'-----------..1
Figure 3-7. Layout of Building$ A and 8 on larger podf11m deck
Building 8
The ftoor area tOr each floor level will be determined utiltZ.ing a sumJnation. of rectangular ateas that are
slightly la~ger than the actual ftoot footptin.t. Floor open1ngs,. such a~ stairs or duct/pipe c.hao;es, wlll not
be subuaeted out and will be considered as being solid. Recesses and notches along the penme-ter of the
butldtng wiU be considered as nonexlstent
8UJid1ng B is d1v1ded uuo two rectangles for calculating the building weaght. Zones B I and B2. This t.s to
account fat' add1Uonal point loads/masses atlhe floor and roof levels and the fact tha[ Roof Zones B 1and
82 are almost essentially detached from each other. as can be seen in. Figure 3-5...Roof plan.""
Though there ts some conservatisrn 1n this approach. using a more accurate floor area cah:::ulauon does oot
usually lead to a Significant decrea~ in the seismic mass at each ftoor level of the buildtng. If the opemngs
are signtficantJy large in the ftoor or roof. then the designer may warn to account for these openings in
reducang the mass assoc1ated wath that ftoor level The other reason to use a slightly target footprint IS to
account fof pmenual adjustme-nts 1n the build1ng footprmt by the architect as the construeuon documents
are developed. It is always benef for the designer to be conservative on the \'-'eight of the building than to
discover larer that the bujlding weight/mass has been Wlderestimared.
Two SurfaceArtu (81. 82) and Unreal Fttt or Exttrior \Vall
As noted above. the actual bu1kling dimensions have been tncreased shghtJy for calculating the 8001' and
roof areas.
Seool'ld- and thtrd-ftoor are.as:
Bl =45 X 135 =6075 fl'
B2 =45 x96 =4320 t\'
Roofatea:
Bl =40 X 135 = 5400 fl'
B2 =40x 96 = 3840 t\'
The calculated budding aiea is larger at the flootS than the-roof to account tor the e:aerior canulever
walkways.
Perome1er \\rul lengills (second and llurd):
B I = 40 + 135 + 40 + 135 = 350 lineal fe<1
B2 = 40 + 96 + -10 + 96 = 2721meal feel
Perinteler wall lengills (root):
Bl = 40 + 135 + 40 + 135 = 350 lineal feet
B2 = 40 + 96 +40 + 96 =2721ine<ll fee1
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
sn
390
:; I==~
II
0
¢::~
zoneBl
Z01~0
Bl
11
g~ L----------~'L-'-----J
~
1340
(13511 USED)
390
(40 R USED)
Figure 3-8. Bm'ldmg B joo1prm1
3.5 SEISMIC WEIGHT
The weig.htsln~ass assoctated with each ftoor le,•e.l of the buddtng are determu'led and are used to calcu1ate
the budding seismic base shear.
The paniuon wall weights fOr seism1c destg,n ate dtstnbuted hal r to the floor le,·el above and half to the floor
level below. The coJUtlbutton of half oflhe pattttion wnll weight to the roof mass needs to be constdered.
Tablt 1-7. Building B ,.-et'ghlsfor building base shear (V) calculatlon.t
(psi)
Total Floor
We1ght
(lb)
Wall
Length
(ft)
Tnbutary
Wall
Heigllt
(ft)
Floor
Area
(II')
Weight
Wall
(psi)
Tolal Wall
Weigilt
(lb)
7.0
19.0
46,550
7.0
19.0
36,176
Weight
Level
Zone
Roof
Bl
5400
(16 + 1.5)
126,900
350
B2
3840
(16+7.5)
90,240
272
2 17, 140
622
350
5.5+5.0
19.0
5.5 + 5.0
19.0
9240
Total
3rd
Bl
B2
Total
(30 + 14.5)
270,338
4320
(30 + 14.5)
192,240
272
462,578
67?
10,395
Toml
2nd
6015
82,726
69,825
54,264
124,089
Bl
6015
(30 + 14.0)
267,300
350
5.0+5.0
19.0
66,500
B2
4320
(30 + 14.0)
190,080
272
5.0+5.0
19.0
51,680
457,380
622
10,395
118,180
1. Half oflhe thtrd-floor pot~Ju on load ( 15n = 7.5 psi) and half oftl~e third-floor e.xtenor wall we.ght
( 19.0 psi) is dtstributed to the roofle>·el.
2. The hetght of the thtrd floor is lO teet I I tnches. and half the wall hetght 1S 5 feet 5~ 1nche:s.. wh1ch 1s
rounded to 5.5 ttet. ll\e parape.t height tS I foot 6 1nches, so the lOlaJ \'-'all he1ght tnbutary to the roof IS
s.5 + 1.5= 1.0 teet.
3. The trtbutary panuion wall load at the thjrd floor is ( I5.0 + 14.0)12 = 14.5 psfto account for the different
ftoor he.ights. The ttiblltary partiuon wall load at the second floor is 14.0 psf.
4. The wall he1ght for eac-h of second and third floors IS 10 feet~ lnch. Half ofthjs wall he1ghl is 5 teet
~ tnc.h.,. which IS rounded to 5.0 feeL
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
AdditJonal \Vtigbts/MIIISSt:~ to Considrr
Often there are additiOnal individual masses lhat should be <K.--cotJnted for ln the boddtng design that do not
fit into a floor surface area or are limned tocenain areas of the butldtng. These loads are ollen added tn as
addtltonal floor loads at lhe floor le\-el where they occur.
I. Solar ponels.
A poruon ofBulldtng B's roof will support solar panels tn the future. The weighl of the lOdJvtduaJ
solar panels(often rangmg between 2.0 to 3.5 pst) and the suppon framing above lhe roof is
assumed 10 equal 8.0 psf.
The future roof area for solar panels ts 14 fee-l x 100 feet= 1400 square feet.
The assume<! weigh! tor !he solar ponels lS 1400 x 8.0 = LL,200 powl<Ls.
2. Rooftop equipmen1.
Individual rooftop utUl~ typically rat\ge tn weight between 100 to 500 pounds. The rooftop
eqUJpmem wetg.lu should also tnclude we1ght for the curb or house-keepu"tg pad the eqUJpment sits
on. There are often other hg_hter \\'t-ight fans and mher mtscellaneous equipment on the roof as well.
Some designers will add an addittonal 1.0 psfto the roof-wetg.ht load to account for mechanical
unit~; others will asswne a number of rooftop units of a gtve.n weight range and add tt to the roof
weighL
There are no rooftop unns on this butldU\g. and the mt:SCellaoeous roof dead-load weight of 3.0 psf
can be considered to cover the wetght of the few pieces of mechanical equipment that are installed
on the roof.
3. Exterior stairsfwa.U,"\\>ays.
The majority of exter1or st:atrs are mtegral With buildtng.c;. unless there lS a seismte jotnt, and
therefore lhe1r mass as added. lialfofthe slatr weaght between floors wtH be added to the floor
above. and half w1ll be added to lhe floor below. The stair we-1glu needs lO include the guard-raJ!
we-Ight. which can often be \•ery heavy. The dead load of a sta,jr considering a combinauon of
concrete infill treads. stnngers. conc.rete-filled pans.,. guard ralls and stanc.htons, and infill sr.eel
picke·ts can be extremely heavy. When lhe totaJ stair dead lood is divtded by the footprtnt of a flight
of staJts belween floor le\-el~ the wetghl can often ronge bel\veen 50 to 100 psf.
The added staJr Wtl gl'll at the second floor IS 9 fee-t x 12 feet x 75 psfx2 stalrs= 16,200 pounds.
The adde<L weagh1 a! !he !hard ftoor is half oftl1e second ftoor= 16,200/2 = 8 LOO pounds.
4. Ele--'ated walk\vay/bndge between buildings.
Stnce there is a seismic joint, the mass of the bridge needs to be Included with the mass of the
aJlC-hor buiJd1ng from which it ts cantilevering. When the bridge moves ln a direction parallel to the
bndge span. lttnduces a load on the anchor buildtng by uytng to pull a\\-ay or push into the anchor
budding. When the bridge mm·es perpendicular to the bndge span. an Independent lateral-resisting
system probably should be provided (such as moment frames at the end where tlte seismic jouu
occurs) or some mher system created that ties the bridge back 1nto the anchor buildmg.
The added mass for Buildtng 8 at the third floor and second floor to account for the bndge IS 13 feet
x 8 fee1 x 50 psf(assumed brtdge dead Load)= 5200 pounds.
5. Planters.
Wuh green build1ng codes and stormwater retention systems be-ing required in these butlding
project~ there are often planters on upper floor levels to provide green space as well as remporanly
store rainwater (0 be released at a later date into the public stonn drain systems.
There are no planters on Butldtng B in thts design e.xample. lnere 1S a planter/stonnwater retention
system that is pan: or Buildtng A that would have to be addressed in the design of Budd1ng A.
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
Summary of Building 8 \Vti.g:bts:
= 184.7 kops
= 353.5 kops
RoofareaBI = 126,900+46,550+ 11,200=
184,650 pounds
353,463 pounds
Th.rd-ftoor area Bl = 270,338 +69,825 + 8100 + 5200=
Secon<l-floor area 81 = 267, 300+ 66,500 + 16,200 + 5200 = 355,200 pounds
Tmal =
893,3 I3 pounds
Roof area B2 = 90,240 + 36,176 =
Th.rd-ftoor area B2 = 192,240 + 54,264 + 8100 =
Secon<l-floor area 82 = 190,080+ 51,680 + 16,200 =
Tmal =
=355.2 kips
= 893.4 kops
126,416 pounds
254,604 poun<ls
257,960 pounds
638,040 pounds
= 126.5 kops
=254.6 kops
=258.0 kops
=639. 1 kops
3.6 BUILDING BASE SHEAR (11}-STRENGlll LEVEL
Building 81 = 0 . I72W = 0. 172(893.4) = 153.7 kips.
Buildong 82 = 0 . I72W = 0. I 72(639. I)= 109.9 kips
Use 160.0 kips.
Use I 16.0 kips.
To help stmplif)' the math, the numbers are rounded up to an even number and tncreased slightly to Jl(OYJde
a srnall buffer for future smaJi nems that might be added during the design stages of the building. Thus. the
engtneer will nm have to go back and check the lateral analysis. ln this case the author has added a cushion
of about 5 percent to the calculated building base shear. The building code does not requ1re this round-off.
and the destg.ner can use the calcuJated base.shear 1f they wtsh.
Building Bl : I·= I53.7 x 1.05 = 161.4 k ops, whoch is approxomately 160 kops.
8uildJng 82: V= 109.9 x 1.05= 115.4 kips, whoch is approxi ma~e ly 116 kops.
3.7 VERTICAL DISTRIBUTION OF SEISMIC DESIGN FORCES, STORY SHEARS, AND
DIAPHRAGM FORCES
Tables 3-8 and 3-9 show the distrtbutlon of design fOrces. story shears. and djaphragm design forces for the
two zones of Building D.
Table 3-8. Buildlng :.one 81
Stooy
Diaphragm
(kops)
Stoty
I'
(kops)
0.350
56.0
7070.0
0.433
3552.0
16,347.7
IV,
H,
...evel
(kops)
(fi)
IV,H,'
Roof
184.7
31
3rd
353.5
2nd
355.2
check
893.4
(kops)
1:11;
(kops)
~
m;
56.0
184.7
184.7
1.0
69.3
125.3
353.5
538.2
0.217
34.7
160.0
355.2
893.4
1.000
160.0
%
I'
5725.7
20
10
F,
w,.
nmat'ion check (V)= 160, same as 160 hand calcuJation. tlterefore okay.
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F,.
IF,
F,.
(kips)
(kops)
(mon)
(kops)
56.0
56.0•
41.3
0 .657
125.3
82.4'
79.0
0 .398
160.0
63.6
79.4'
Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
Table 3- 9. Bs1ildmg =tme 82
S10ry
Dtaphragm
Story
v
w,.
r.wl
I'
F,
(kips)
( kips)
(kips)
(kips)
rw,
(ktps)
F"'
(kips)
F,.
(min)
(kips)
0 .337
39.1
39.1
125.5
125.5
1.0
39.1
39.1'
28.1
20
5092.0 0.440
51.1
90.2
254.6
380.1
0.670
90.2
60.4•
56.9
10
2580.0 0.233
27.0
117.2
258.0
638.1
0.405
117.2
47.5
57.7•
IV,
H,
Level
(kips)
(fl)
W}"f/
Roof
125.5
31
3890.5
3rd
254.6
2nd
258.0
l: ch«k
638.1
%
11 ,562.5
1.010
IV
__!!!_
r.F,
117.2
Swnmauon check (J?= 117.2 > 116.0 hand calcularion, therefore okay.
Nores tor Tables 3-8 and 3-9:
I. % V = C,. = W,H,'I!i.JV,H,") and k = 1.0 (ASCE 7 Secuon 12.8.3).
2. Fl"=l:F,(W,.fl:ll',~
3. Diaphragm ltmits (ASCE 7 Socriolt 12. 10.1.1 ):
a. F,. need nat e.xceed 0.4S,.t,WI"= 0.4{ 1.11 7)(J.O)W,_ = 0.447W,.
b. F,. shall not be less rhan 0.2S.o.<J,JV,... = 0.2(1.117)( I.O)JV,... = 0.223W,.
Eq 12. 10-3
Eq 12.10-2
4. f« samplicity, the weaght mbutary to the diaphragm at each level, JV,..... and the wetght trtbutcuy to
the level, W,. are considered equal m the two onhogonal dtreclions of the bui1dlng. The weight of
shear \WIIs 1n the dtrection under consideration could be subtracted from the djaphragm weag.ht., but
this refineme-nt is cypically not warranted in hght-frame con.~truction. However, 1n some cases, such
as a loog narrow buildang walh a Jarge number of shear Y~<llls in the building's transverse direction,
lhe des1gner may want to oonsJder the refinement to reduce daaphtagm design fOrces.
As nOted by !he symbol(' ) tn Tables 3-8and 3-9 for Buildongs Bland B2, the dtapbragm design force at
the second floor is governed by the buddmg code mtntmum des1gn requtrements.
4. Diaphragm Flexibility
ASCE 7 Secuon 12.3. I requtreS t.hal the.struCtural analys1s fOr all buildtngs tnclude the relam-e suft'ness
of the floor as'ld roof diaphragms aod the vemcal elements of the seasmic-fbrce-reststtng syste~ whkh ts
ltght-frame shear \valls tn thjs design example. If the d~aphragm cannot be tdeali.Ud as betng e-1ther flexible
or ngld. the sltuctural analysiS is to e.xplicitly ioc-lude cooside:rarion of the stttiness of the diaphragm as
betng semingid.
The commentary for ASCE 7-05 Section C 12.8.4. I ..Inherent Torsion·· stated the followtng:
Most diaphragms of light-frame ronsrruclion llte somewhere-between ng.id and flextble tbr
analysis purposes, that tS. seming_id. Such diaphragm behav1or is difficult to analyu when
considenng torsion of the sttucture. As a resul~ tt is belteved that constderouon of ampltficaljon
oflhe torsional momMt is a refinement lhat is not \llatranted for light-frame construction.
TI>e comme.mruy tbr ASCE 7-16 Section Cl2.3.1 "Dtaphragtn Flexobiluy" srares:
The diaphragm tn moSl buildangs braced by wood light-frame shear \WI Is are semlrtg.td. Because
semmgtd diaphragm modeling ts beyond the capabilities of available software for wood lightframe buildings. u is anticipared that thiS requ1rement wdl be met by evaluating force distl'lbution
us1ng both ngid and fteXJbJe d1aphragm models and talang the worse case oflhe two.
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
Section C 12.3. I. I ..f le:oble Diaphragm COtldttton" nores that compltance with story dr1ft llmits along each
I me of shear walls is intended as an tndtcator that the shear \\o'3.1ls are substantial enough 10 share load on a
tnbutary area basts and nm requJte torstonal force redtSlftbutJon.
ASCE 7 Section C 12.8.4.1 .. Inherent Torsion"' states that for structures with fteXtble diaphragt~ ventcal
eleme-nts of the seismic-force-resisung system are assumed to resist the aneruaJ forces from the mass that IS
tribt.naty to the elements with no explicitly computed torston.
Therefore, the v.'OOd diaphragm tn hght-t!ame c:onstruction ls COJ'IStdertd a sem1rtgid diaphragm. even tfthe
common practice has been uaditJonally to consider it as ftexable. The semirigid diaphragm classaticauon l'i
a result of a large number of mtenor nonseismac bearing and nonbeartng V.'3.1l parutaons. as \Veil as interior
and exten or pertmeter setsmtc shear walls c:onrnbutiJtg to the st.iftiless of the diaphragm. These part1t10ns
and walls form a honeycat:nb gnd of cross walls that tnterloc.k the adjacent flO()( dtaphragms lOgether
over nluluple le\'els of the buildmg. so Lhe building structure behaves more as a rigtd box struCture than a
c.ollectjoJt of stjffvenical Lateral-resisting eleme-nts with a flextble diaphragm spanning bet\\"eelllhem.
Thls semmg.1d d1aphragm be-havtor has led some hght frame destgners to use an envelope method for the
design of the diaplltagms, which is discussed m ASCE 7-16 Seeuon Cl2.3.1 as ooted previously.
I. Ftrs~ consider the dtaphrngms as ftextble and de-term1ne the distttbuuon of lateral fOrces to the
shear walls.
2. Second. conside-r the daaphragms as rigtd and determine the dastnbution of lateral forces to the
shear walls.
3.
Thtr~
the actual tndtvtduaJ shear wall destgn wiU the-n be based on the diaphragm analysts that
produces the larger design forces.
The en\•elope method results 1n some of the andividuaJ shear wa11designs using the forces from the tlextble
diaphragm analysi~ whjle lhe othe.r shear walls are designed using the forces from the rigid diaphragm
analysis. Checkmg both e.'1remes ofideal12ed diaphragm behavior. ngid and flextble. provides lxlunds
that will encompass the ma:<tmwn shear wall destgn fOrces, as tf a semmg.id diaphragm anal)•sts approach
was utthze<l. The envelope methtxl has 00\\' been included as an altemanve method to semirigid horrzonml
analysis for horizontal distnbut10n of shear to shear walls an Section 4.2.5, AWC SOPWS 2015. (Note
thal the authors recommend refernng to lhe 2018 SOPWS as u is lhe most recen1 published standard even
though 11 will not be adopted unul the 2021 IBC.)
Due to the gene.rally g_ood performru'lCe of one- to tOur-story light-frame buildings sheathed with WSP.
ASCE 7 ha.<i allowed for a tew presc.nplive design requiremenl.(i which, if satisfi~ allow the diaphragm to
be tdeaJtzed as tlex.tble tbr des1gn purposes, whtle the actuaJ be-h.avaor lS more likely between se:ming.id and
rtgad m behav1or.IJthe prescriptive section requlrementscannot be met., then lhe buildtngc:ode aJiows tOr
a more rigoroos approach 10 determtne if the diaphragm behavior is fte.xtble. Build1ng designers typically
prefer the "flextble dtaphragm" approach su-,ce Jess destgn tiJne is requ1red for the budding design than
for a ngid diaphragm approach. There are some design offices though that have auto1nated electronic
spreadsheets for do1ng a ngtd diaphragm analysiS with tors1on and use that tOr the shear wall designs.. and
do not c:ons1der the flextble diaphragm approach. So IllS left up to t.he designer as to the design approach
they \l;'aJ\t to take and to determine the ng.idny of the diaphragm.
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1.01. 2
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
5. Flexible Diaphragm Condition
Building B IS an L-shaped bualdinS> and the roof d1aphmgm IS not continuous around the tntersection of the
two roofs. A flexible dtaphragm will be a~umed for disttibuuon of setsrnic forces to the vatiouo; shear \Valls
at each floor level ofButlding B. This is permissible since per ASCE 7 Section 12.3.1.1(c):
I . F« ltght-frame construction where the venaca.l seis.ntc-force-resisting syste.Jn elements c.ocnply
With the allowable srory dflft tn Table 12.12-1 and the non.mucrural concrete. or stmiJar
material. ftoor topptng thickness over the wood structural panel diaphragm does not exceed 1.5
tnches.
a. A possible exception might incJOOe that topptng slab th1ck:ness can 1ncrease to 2.5 tnches along.
edges where topptng slopes down to 1.5 inches or less at floor drainage systems at breezeways
and bo.lcooJes sance they represent a relauvely mu\Or area of the btukling's toW. square footage.
Since the ftoor and roof dtaphragm.s are considered to be flexible by defimtion of Section 12.3.1.1(c).
the addittonal ..ca1culated.. ftex1ble dtaphragm oondU10n (ASCE 7 Sectjon 12.3.1.3) is not reqUtred to be
checked. This sect1on requtres the tn-plane defl~lioos of the.horizontal diaphragm between shear \valls be
compared to the top of shear wall de.flections. If lhe average horr.zontal diaphragm deflecuon exceeds two
times the top of shear wall deftecuon. the.n the honzontal diaphragm ts constdered to be flexible.
6. Building Classification: Regular or Irregular
Bujldlng B needs to be classified as being either a regular or Irregular bualdtng based on whether thete are
horiZOI\tal andfor vertical structural irregularities present wathin the budding.
I. Regular buddtng: No honzontal or \le.rtical i.rregulariues.
2. Irregular bttildtng: One or more horizontal or venical irregularities.
1-Jorizontallrrtgularil)' Cbtck
ASCE 7 § 12.3.2.1
Honzontal sttuCturaltrreguJanues are defined 10 ASCE 7 Table 12.3-1. Table 3- 10 indiC:3tes hO\Y these
hoti2011tal irregulantles apply to Butfdjng B.
Tublt 3- 10. AppltcabJ!lty ofhori=Oni(J/ structural irregularities 10 Building B
ASCE 7 Table 12.3-1-Horizontal Structural lrregulanues
Descraption
Type
Design Check for Buildmg B
Ia
Torsionallrregulanty
N.A.- Fiexible floor and roof diaphragms
lb
Extreme Torsaonal lrregulanty
N.A.- Fiexible floor and roof diaphmgm.s
2
Reentrant Comer Irregularity
Oetatled check required-See below
3
Diaphragm Discontinuity JrreguJaray
OK- No large diaphragm openings
4
Om-of-Plane Offset Irregularity
OK- Shear walls stack
5
Nonparallel System Irregularity
OK- Shear '"ails oriented onhogonal to each
other and parallel to buddmg axis
I. N.A.: Not Applicable-No check requtred (forreason stated).
2. OK: BuiJdtng does not meet condittOI'IS of noted trregularaues requir1ng a detaded evaluauon (for reason
stated).
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The only honzontallrregulanl)' that requtrts a deta.ded ched for Ouddang B IS "Typt 2~ntra.ru corner
arregutarny.- sance Bualdln,g 0 tS L·.sh:'lped and an SOC D. The reentrant comer as consader~ lObe a
po<cnual honzonlal trrq;ulo.rll) ,.,.,.. lh< two legs oflh< bwld111g come togecher. The prOJ<Cuon of lh<
reerurant..,._ "'lh< dtr<<l""' betog considered cannot be more than IS pcttaU oftht M<fllll hutldmg
ltnglh llllhill dtr..,_ IO a>Ot<Ja horuontal ~1)-
Roofs:
81:5 ft/(134 ft + S ft)= 3.6% < IS%
81: 6IVJ9 ft = 15.Wo-IS%(soy OK)
82: (95 ft - J9 ft - S ft)I9S ft =53. 7"'. > IS%
81: 1341V(I34 ft + S ft + J9 ft) = 7S.J% > IS%
82: 5 IV(J9 ft + 5 ft)= 11.4% < IS%
B2: 6 IV9S ft = 6.3% < IS%
Floors: liave Type 2 trregulanty (reentrant corner)
Roof: Cons•derc..-<1 regulor smce the roofs are essentially separote.
sn
39ft
Zone B2
loneB2
.,
c
.,
"'
"'
"'
Zone 81
13<4 II
...
511
39ft
=
c
...,
=
"'"'
c
39ft
Zone 81
"'"'
I I
I I
;-,
u•n
39ft
Schemadc Floor Plan
Schomarlc Roof Plan
Flgurt J.9, Bulldmg B .fChtmtJJtC plans/Or rtt#lt/J'tJtll comer de101l ch«k
When a Table 12.3.·1 T}pe 2 hott2Mtaltnegulamy occurs. lh<n lh< dest.go forcn for lh< franung elemems
oflh< floor dtaphrosm (example: collecl<lfS and collector CO<IIIecttons must be •ncreased 25 percetll per
ASCE 7 S«uon 12.J.H~ In lhiS.,...,, Bwldmg B's second- and lhtrd-lloor d~ framtngelement
destgn forces "wid ha\e 10 be tncuased 25 perceu. hut lh< dtaplr.t@m framtng <Iemen• destgn fon:es
,.wJd 110( ha\0 to be,,.....,... 01 lh< roof. ThiS destgn (.,..,., tneteaSe IS Ill O<idtt""' 10 011) redund:lnc)
destgn requu...,.,..._
ASCE 7 § 12J.2.2
Verucal suucrumltrrq;ulanues are defined tn ASCE 7 Table 12.3-2. Table l-11 tndiCiltcS how these \ertJ<:al
lrregulanues apply to Butldu-.g B.
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
Table 3-11. Applicabillly of•·~rttcal structural irngulantie.v to Buildmg 8
ASCE 7 Table 12.J-2- VerticaJ Struc-tural lrreguJaritaes
Type
Descr1ption
Ia
Stlflhess--Soft-story trreguJamy
lb
Stifthess-E.xtreme soft-story 1rreguJamy
2
We1ght (mass) 1treguJamy
3
Ventcal geomeuic 1rregutariry
4
ln-pla.ne dtsconLinuity in venu::a1 Late:ral-foreeresist1ng element irregularity
5a
DLSCOntinuhy in lateral suength-Weak-story
uregu1arity
Desogn Check for Budding B
OK. Shear walls stack; lateral sljfliless of shear
walls at each floor level1s more than 70% of
shear \\o'alls at floor level alxwe.
OK. Shear \valls stac.k; lateral stjffness of shear
walls at each floor level 1s more than 6()0/e of
shear \\o'alls at floor level alxwe.
OK. Weight of any floor level1s not more than
lsoe'lo of floor level above or below.
OK. Shear \\'alls stac.k; Individual shear wall
lengths at a given ftoor level are not more than
1JO-Io oflhat in the adjacent story.
OK. Shear \\'alls stac.k; there are no in-plane
offsets between shear walls ofadjacent floor
levels in the same hne that are more than the
length of the shear wan above.
OK. Shear walls stac.k; stor)' lateral strength J:S
more than 8<>-.4 of floor Jevel above.
OtsconLinuity in lateral suenglh-Exueme
OK. Sheill' wall strength e.xcee<ls 65% of the
weak-slory 1rreg.uJaruy
lateral strength :u the adjacent floor above.
I. OK: BuJidtng does not mee-t condJttons of noted trregulanues requinng a detailed el•aluauon (f« reason
5b
staled).
2. Mass Irregularity exce.ption: When a roof is lighter than the floor below n, n need not be considered for
mass ttregularity.
When a vertica1 ttregularny occurs per Table I2.3-2. then the I:JujJdlng des1gn mu'» satisfy lhe requtremMts
of the applicable referenced sections in Table 12.3-2. The destgn example for Buddtng B Illustrates
three different shear 'Wall configuratiOl'L"" that could occur along a given waU ltne. In the case of shear
wan c.onfigutauon opuons 1 and 3. the-re 1s no ventcal 1rregulartty. Shear waH configumuon opuon 2 is
cons1de.red to have a l)'pe 4 \'ertical trreg.ulamy since the shear wall tS shone-rat lhe first floor than at the
floor level above and would have to be funhe.r evaluated per the requtrements of Table 12.3-2.
7. Redundancy Factor
The concept of redundancy in buiJdjng selsmic-focce-resJSung systeJns is to provide enough ele.anents
res1stmg Lateral forces so cata'itrophic failure of the build1ng does not occur lf one eleanent is overloaded
or fails. The redundancy factor, P~ tmposes a penalty on the design of ce.nain components of the lateralforce-resisting system by requir1ng design f« higher forces when adequate redundancy is not prl)Vided.
RedW\dancy IS typtcally prov1ded in buildlngs by the use of mulliple lateral-force-resisting elements
ASCE 7 Secuoo 12.3.4 reqwres that a redundancy factor be ass1gne<l to the buddmg·s selsnuc-tbrceresistmg system in each of the two onhog.onal direcuons relauve to the budding's footprint. A value of p
equal to 1.3 is used unless the condiuons of either Section 12.3.4.1 or 12.3.4.2 are satisfied. tn which case
tlle value of p = 1.0 may be used.
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
Conditions \Vhtrt Valut ofp l.s 1.0
ASCE 7 §12.3.4.1
ASCE 7 SecLJon 12.3.4.1 tS a ge.neral catchall seaton thm recognizes that it 1s not apptopnate for the
engu-.ee.r 10 automatically apply a redundancy factor greater than L.0 to all seismic regions (se.is.mic desig,n
categories) or aspects of the seismic lateral-force-resisting system evaluation and c.omponent design. A
1
\ alue of p equal to 1.0 may be used for the stx destgn condtttons related to lighl-frame c.on.wuction listed
below. The six tndiv1dual destg.n condtuons have been sorted according to apphcauon lnstead ofllsung
Lhem indi\•tdu:.ally. as oc:curs tn Section 12.3.4.1. There are three olher condit1ons not mcluded as they are
n01 genernlly applicable oo light-frame building construcl'ion.
I. Setsmtc Des1gn Category
a. Building structures asstgned 10 Seismic Design Categories (SOC) B. C.
1.
Structures assigned to SOC A do not require-consideration of seismic fOrces. so the
redundancy factor, p, does not apply.
2. Buddtng Dtsplacements
a. Odft c.a1culat10ns and P-delta etlOCts.
3. Buddlng Component Des1gn
a. Design of nonstructural componenL">.
b. Design of collec.tor elements. spltces, and thetr conneclions when the overslfength fac tor.~ is
tncluded in the seismic lood combtnation being used for de.~ign oft.he element.
c. Design of sttsmtc-forc:e -resist~ng systern members or connecuons where the se1smte load
combinations requtre the 0\'e:rstrength factor. D.o. be tncluded.
d. Diaphragm inertta11oads derived from using Equatton 12.10-1 tor distribution of dtaphragrn
setsmtc forces to the vartous floor/rooflevels of the butldtng_
[Note-: p:: 1.3 is requtred for fOrces ttanSferred through a diaphragm due to an otlSet of the
lateral-force-resisung system belween floor levels (e.'<ample: Oi.ll-of-plane otiset of shear walls
between ftoor levels)].
Rtdundanty Factbr, p, rbr Sti:sRlit DHign Cattgorit.s D through F
ASCE 7 §12.3.4.2
ASCE 7 Section 12.3.4.2 SJk""Cifically addresses Se1smic Design Categortes (SOC) o. E and F and requtres
p equal to 1.3 for the desagn of the setsmtc- tbrce~reststtng S)'Ste-m unless the conditions tn e1ther Section
12.3.4.2.a or 12.3.4.2b are sausfied, •n which case a valueofp equal to 1.0 may be used.
ASCE 7 Section 12.3.4.2a
For each story level resislU"'g more than 35 perce~\l oflhe buddmg base shear m the direetton under
consideration. the seJsmtc-force-resisting system complies wnh ASCE 7 Table 12.3-3 ..Requtrements fOr
Each Story Resisung More Than 35% oflhe Base Shear....
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
Table 12.3-3 pro\'Kies requtremems for dttferent types oflaternl fOrce-resisting elemenlS. The
category of interest for most light-frame buildings is ..shear walls or wall pters with a heig,hl-tolength rario of greate-r than 1.0." The requtremellts to be evaluated are
I. Removal of either a shear wall or wall pie-r wnh a height-to-length ratio greater than 1.0
w:ithirl any story level()( collector connection.-; there.to would not result in more than a
33 percent reduction tn story sue.ngth.
2. Removal of a shear wall or pter does not result in an extreme totstonal trregulanty (Type
lb horizontal irregularity).
ASCE 7 Strtioa 12.3.4.2b
Section 12.3.4.2b also applies to each story level reslstmg more than 35 percent of the building base shear,
and IS summanzed as follows:
I . Structure tS ..regular" tn plan at all floor levels.
2. Seismic.-forc.e-resisting system requtremellt<;:
a. At least two bays, atlhe pertme.ter. on each side of the building.
b. Seismic-fOrce-resisting systems are onhogonal to each other.
A s&ngle bay IS commonly denoted as the dtsta.nee between two adjacent colwnns of a moment frame Of
btaced frame. Since shear wall behavtor is dtfferent from a momem frame-or btaced frame. an equtvalent
definioon for shear wall "bay" has been develope<l. Tite number of shear wall bays JS defined as folio"~ for
tl1e vartous wall construction matertals:
Concrete, ma..~nry, or steel shear \vall bay quanmy: \vaJI Iengthlstory he1ght
Light-frame shear wall bay quanuty: (2) x (\\~lllengthlstory height~
Therefore, a single long shear wall could be equivale.nt to one or more shear wall bays, whereas a stngle
shon she-ar wall could be equivalent to less lhan one bay. The sum of all shear wall bays along the
buildtng's perinleter,on each side of the bullding. must e~\:Ceed two in order to use p = 1.0.
Autbor•s Dlstussio~Equinlrnl Sh<'tr \Vall Bays
Ltght-fra.me shear \\'3-lls typically have openings such as doors. windO\\'S, and dueL"'. The decisio1' of which
shear \vall bays IS left to the engtneer,
\\'all lengths should be used m calculating lhe number of equrvaleru
bul two options are dtscussed here for conside.ration_
Opuon I: Shear wall length defined from vemcal edge of operung to venkal edge of ope-ning.
Conser,•atively, lhe e1lg.inee-r mtght use only the shear ·wall lenglhs Ot\ each side of the \~tall
openings. Th1s wouJd be appropnate where lhe ope tung ts a door that interrupts lhe pe-rimeter
boundary edge of the shear '"-aiL
Option 2: Shear wall element" designed to trarlsfer design fOrces around lhe opentng in the shear wall.
In the author's opinion, a wall opentng length (wiOOow,duct, etc.) can be ignored and the wall
consideted as betng one single longer wall that tncludes the length of the opening and the shear
wall elements on each s1de of the wall openi1tg. 1fthe shear ·wall has been destgned to uansfer
the wall design forces around the open1ng.
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
Some e-ngmeertngjudgmem should be apphed when ustng this option. and the fbllow1ng
crtteria are suggested.
I. The length oflhe wall opening shall not exceed 30 percent of the sum of the individuaJ
shear \Vall lengths on each side of the opening; otherwise, thts condition should be
lteated as separate Individual shear walls on each side of the ope1ling.
2. The widtl1 of a wall pier tn a shear waJI on e-nher stde of an open1ng shall be not less
!han 24 inches (AJSI 5240-1 5 Section 65.2.2. 1 "Type I Shear Walls"'), half !he height
of lhe opening.. or half the lenglh of lhe open1ng. \\tlichever is more; otherwise, this
condauon should be ueated as separate tndividual shear \\ails on each side of the
openang_
8. Redundancy Check for Building B
Components of the seismic-force-resisting-system m this des1gn example w11l be evah.tared using a
redundancy factor, p, equal to 1.0 based on !he enndnions noted tn ASCE 7 Section 12.3.4.2 and diaphragm
Section 12.1 0.1. 1
ASCE 7 ~ttion 12..3.4.1 {applirablt tonditions whtrt p
= 1.0 in Building 8)
I. Ortft calculations: lne shear wall drtfts and dtaphragm drifts ·would be evaluated using p = 1.0
since p = 1.3 ts not applicable 10 drift calcuJations and P-delt3 eftk1s.
2. Diaphragms: The destgn of !he floor and roofdtaphragms shall usep = 1.0 as tl1e dtaphragm
fO!ces are dtstrtbuted to the various floots and roof per Equmion 12.10-1 .
a. In ac:cordaJlce with Section 12.10.1.1, diaphragm design, p applies to diaphragm desagn in SOC
o. E. and F. a.nd stnce the 1ne-nial forces we-re calculated 1n accordance with Equatton 12.10-1,
the value of p = 1.0 is penmssible fOr diaphragm design.
b. In accordance with Secuon 12.3.3.4, tfthere is a honzontal or \'e.ntcal otfstt trregulatity, then the
destg.n f«ces shall be tncreased 25 percent tOr the follov.•mg setsmtc-fbfce-resisting eJeme-ms:
1.
Diaphtagm conoection ro collector (1 .25 x diaph.rn_gm. destgn force at collector)
u. Collector membe-.rs
111.
Collector member c:onnecuons
3. Collectors (collector elements, spltces, and collector sphces)
a. ASCE 7 Section 12.10.2.1 has an exemption for strucwres braced enurely by hght-frame shear
walls. The collectot elements and collector conn~tions for buildings in SOC C, 0 , E, and F
need 10 be designed only to resist fOrces for Section 12.4.2.3 seismic lood combinattons whe1l
dtaphragm destgn forces are calc-ulated per Secuon 12.10.1.1. Stnce Section 12.4.3.21oad
combinations with 0\'erstrenglh factor flo are not requited to be tncluded, p used fot the structure
ts tequlred tbr collector design (equal to 1.0 tn Lhts destgn e.xample).
184
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
b. ASCE 7 Section 12.10. 1. 1 requires the value of' p for elements tran.-.fe:rnng forces between
Ytnical elements above and below the diaphragm (example: collectors at an tn-plane offsetType 4 \'ertical iiTeguJarity) be the same as for the O\'erall buildtng system. If the requirement~
of enher Section I2.3.4.2a or 12.3.4.2b cannot be met, the.n p =- 1.3 \\'Ould be required for the
structure and therefOre aJso for the colleclOr destgn.
c. The value of p tor coUeclOr design for this example is 1.0 m accordance with Secuoo 12.3.4.2,
and thts detenntnatton IS shO\\n below.
ASCE 7 Stttioa 12.3.4.2 (Building 8 rtdundanty fat tor)
Smce the budding tn thts design example ts located in SOC D. the requiremenl of Section 12.3.4.2 has to be
evaluated. Enher AISC 7 Section 12.3.4.2a or 12.3.4.2b must be sausfied to allow for p= 1.0 for the design
of the selsmic-force-reststmg system compone-nts. For completeness of the des1gn problem. both secuons
shall be checked.
ASCE 7 S.ctioa 12.3.4.2•
Determtnauon of where story leve-l shear exceeds 35 perce.m of the base she.ar IS based on calculations fOr
the complete bttildtng footpnnt
Table 3-12. Bwldmg B SIOIJ' shear swtmJatiOII chtd
Floor Le'•el
ZooeBI
Story Shear
(kips)
Zone B2
Story Shear
(ktps)
BuildJng B
Story Sbeats
(ktps)
Perc-ent Stot')'
Shear per Floor
Levei(J!)
Summation
Percem Story
Shear(l1
3rd
56.0
39.1
95.1
0 .345
0.345 < 0.35
2nd
69.3
51.1
120.4
0.436
0.78 1 >0.35
1st
34.7
27.0
61.7
0.223
1.004 > 0.35
l:
160.0
117.2
277.2
1.004
At the first and second stories where the shear exceeds 35 percent of the building base shear, there are
suffic:tent numbers ofshear walls 1n each onhogonaJ direction, so that:
I. Remova1 of a shear wa11or wall pier with a height-to-length rauo greate-r than I .0 does not
result tn a 33 percent reductton in story strength iU each floor level. In the eve.nt all of the shear
walls have a helghHo-length ratiOless than I .0. then the LMeral-fotce-resiSttng elemerli noted
as "Other'' tn ASCE 7 Table 12.3-3 v.'Ould be used. In thts case. there are no requirement~. so
the buildtng would be deemed to sausty this section without funhe:r investigatton.
2. Removal of a shear wall does not result in extreme tors1onal Type lb horizontal trregularity,
whtch does not apply where the ftOOI' and roof dJaphragm are tdeahzed as bemg tle.xable. a~ IS
the case tbr Butldtng B.
TherefOre, p = 1.0 is permissible for the l1ght-fmme shear wall setsmtc-force-resisting system tn thts design
example bec-ause the condtlion 11'1 Section 12.3.4.2a IS satisfied.
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
ASCE 7 Stttion 12J.4.2b
Thls secuoo would not normally be checked if the build1ng successfUlly complied wnh Secuon 12.3.42a
for allowtng p = 1.0. However, tOr comple-ten~ the requjred checks are dtscussed. Also, in some cases
th1s may be a qutcker check than can be performed viSl.lalty, so Secuon 12.3.4.2b may be checked befOre
Section 11.3.4.2a.
The shear exceeds 35 percent of the bwlding base shear at the first and second stories. The redundanc)'
faclOf p = 1.0 1s alloo•ed based on the following requtrements:
1. Have the equtva1ent of at least t\\'0 bays ofltght-fmme shear walls at all penmeter sides of the
bUilding at tath Door level of the bUild mg.
2. Shear wall setstutc-force-resisttng systems are onhogona1 to each other.
Hov.-ever, the re<IW'ldattcy factor, p. would have to be inere.ased to 1.3 Stnce the budding ts not regular in
plan as it has a l)•pe 2 honzontal trregulamy and therefOre does not satJsty the reqUJrements of Secuoo
12.3.4.2.b.
Thts sectjon cannot be used to permtt p = 1.0 s1nce the bmldmg is not regular 10 plan. As nmed previously,
the requtrem.e1n ofSecuon 12.3.4.2 ts to satisfY either Section 12.3.4.2a or 12.3.4.2b, and they are
c.onsidered to be mdependent checks. Sance Build1ng B passed the requlre.mentS ofSecuon 12.3.4.2a. p =
1.0 is permlssjble tOr the design of Building B.
Author's DiKu.ssion
A redundancy (actor of' p = 1.0 is commonly used for many multistory, light-frame. shear Y<'all struCtures
per Seetion 12.3.4.2.3 due to the large number of individual shear \valls that occur tn these buHdmg.o;.
However, m the event a structure does not satisfy the requuemem.s ofSecuon 12.3.4.2a. 11 seems hkely a
redundancy fuctor of p = 1.3 ·would be requited for many multistory. Iight-ftame~ she,ar wall buJiding.o; per
Section 12.3.4.2b as these bulld1ngs often are not ••re-gular" in plan and have irreguJarttJes noted m ASCE 7
Tables 12.3-1 and 12.3-2.
One weakness of Section 12.3.4.2a IS that it does nm take into account the layout of the shear walls m
plan when the engineer encounters a fle.xtble diaphragm. since the engineer does not have to constder
buddlng torston. Section 12.3.4.2b, on the Olher hand, specifically addresses the requuement of shear walls
on the perimerer of the butldtng. While some engineers have desig,ned. light-frame nlultistOry structures
using just the mtertor comdor walls for the shear resistance tn one direc.uon of the building walhout any
parallel buildtng pertmeter she.ar walls, the SEAOC Se1smology Light-Frame Commmee has stated "'this
prac11ce IS not reromme;nded wtthout expl1cttl)' COilSldering bulldiJtg performance, mclud1ng the control
of localaed honzontal dtaphragm de-flections lhru. could lead to instabality," and ..lim nations on cantilever
spans, diaphragm deflections at butldtng comers and the corresponchng methods of structutal analysis are
unde-r tnvesugarjon by tl'le Structural Engtnee.rs Assocuujon of Cal ifom1a (SEAOC) and the Arnerican Wood
Coundl (AWC).- The desired seismic performance of the building is expected to be achieved where there
are per1meter e.xtert()( shear walls. If perunete.r laght-frame shear walls are not included tn the dlrectJ:On
unde.r consideration m the buildmg destgn. then cons1derauon should be given to using p = 1.3.
186
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
E\<t-O if the engtneer saus:fies Sectjon 12.3.4.2a and uses p = I.0. n tS common m hght-fmme mulus.tOI)'
buildtngs to have vertical irregularittes suc.h a~ out-of-plane and in-plane oftSets of shear walls from story
to story. It is left to the designer to determane the nwnber of acceptable out-of-plane or in-plane offsets after
which the redundancy factor ofp = 1.3 mtght be used for the laght-fratne s.hear wall system. The u\-plrule
offSets and out-ot:plane offsets of ltght-ti'ame shear walls do resolt tn destgn penalues tbr connecting and
supportmg frammg members, but the butkhng's overall stiffness is sull less than when the hghr-frame
shear waJis stack from floor to floor. The tnctease of p = 1.3 becomes a seismte brute srrength approach to
address that the buildmg has structural trregularities and is not COI'!Stdered to be regular tn plan.
By the same token, just because an t.rregulatil)' does occur once tn a light-frame multtstoty butlditlg. the
enure buddtng should not neces.-.artly be penahzed by a strict unerpretatton of the building code. It would
be. tnappropriately penahztng a large nru.llistory light-frame bulldmg that has many shear v.'<llls to requare
p = 1.3 just because only one or two shear \\>ails do not stack or are offset at one floor level.
The suggested nwnber of m-plane and out-of-plane shear \\oall offsets between floor levels should OOl
exceed tl'!e foiiO\\olng when the engtneer ts using a flexible diaphragm and p = 1.0. after whjch consideration
should be g_jven to using a value of p = 1.3:
I. 15 percent of the total number of shear walls 1n that directaon for that floor level
2. 20 petcMt of the total number of shear walls tn that directaon for all floor levcls of the building.
3. 20 percent oflhe tota..l number of shear walls for both onhogonal directJons tbr aU floor Jeve.ts
ofd>e b!Jildtng.
It tsalso recomme.nded mat p = 1.3 If a mjnimum of two light-frame shear waHsare not being provtded
along or near each perimete-r race of the buiJdtng.
9. Selected Analytical Procedure
ASCE 7 Table 12.6- 1 hsts the permmed analytteal procedures that can be used to e,·aluate the bu1lding..
The tquwalent latera.J force procedure per A.SCE 7 Sect10n 12. 8 is permmed to be u.-.;ed since the suuctural
characteristk of the btuldmg superstructure is ltght-fiam.e construction.
10. Distribution of Seismic Forces to Shear Walls
Seismic des1gn forces are dasltlbuted to lhe individual shear walls ba~ on their tributary area s1nce lhe
floor and roofdtaphragms are constdered to be flexible. ln th•s desagn example. stnce p = 1.0. the calculated
seismac design forces do not have to be adjusted re1ated to redundancy issues..
I. Redundancy factor, p = 1.0.
2. Inherent tors1on consKJerattons are omitted per flextble dulphragm assumptaons.. so the designer
does not have to co11sider building lOtsion.
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
Table 3 -13. Buildmg B ~·~rtlcal !tiJnuc-foru-t?sislmg sys1em !eismic !IOry/<m:f!£ (powrd! per square foOl)
Zone
Level
Roof
3rd
81
2nd
I
Area
(fi')
Seismic
Design force
(F,)(kips)
(F/Area)
(psi)
5400
56.0
10.4
6075
69.3
11.4
6075
34.7
5.7
Area
(ft')
Zone
82
Selsmic
Design Force
(F,) (kips)
(FjArea)
(psi)
3840
39.1
10.2
4320
5 1.1
11.8
4320
27.0
6.3
160.0
117.2
Olitu.tSion
The distribuuon of the seasmtc forces between t11e two zones of Buildtng B are Slmtlar, and to further
sintplify tl'lf' lateral force design, u IS suggested to use the larget design forces ofdle two buildang w1ngs for
the entire building.
If the setsmic de:Sig.n forces were Slgtuficanrly d1fferent between the two bulldh'lg wings (zones), then
the designer may want to use lhese destgn forces for lhe tv.'O dtfrerent butlding zone desagns. Stnce lhe
bwldtng is l -shaped. n L<; suggested the shear wall design forces for those \\'3lls located at tl'le \'erte:< ZOI'Ie
(antetsection) oflhe two budding w1ngs should be the larger of the two zone design forces.
So tbr lh1s design example, the vertex zone setsmic desagn forces would be:
Zone B2
-Zooe Bl
I
Vertex
Zone
I
FiguJ? 3-10. But/ding 8 =ones (plain wew)
Tublt 3-14. l'erlex =one sei!mtc deJignf{)f'Ce.S
Conttollmg
Desagn Force
Level
Zone
(psi)
Roof
B2•
10.2
Jrd
82
11 .8
2nd
82
6 .3
• Zone B2 go\rerns at the roof su-.ce the roof diaphragms are separate~ tfthey ·were COJmec:ted., then Zone BI
would have governed the vene.x zone seismic design forces.
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
11. Sheathed CFS-Stud Shear Walls: Framing Materials
The tbllowang lS a brief review of the fi'aaning materials utilized for CFS-framed sheathed shear waJ I~ as
well as mtnt.mwn buJJdu"tg code desagn requireme,ms. The gravity load of the buildlng 1S supported by CFS
bearing stud~ whic.h are shealhed With a strucrural material to pro\•tde lateral stabiJuy fbr lhe budd mg. The
pr1mruy elements of the shear wall construCtion tnclud.e shealhi.ng screws, CFS St\Jds, CFS tra.cks, CFS
block.inglbridgtng. shear transfer fasteners at the top and bottom of walls. an overturning resttatnl (holddown) system., and structural sheathing.
The American Iron and Steel lnsutute (AJSJ) publishes the North American Spec!ficalion for llre Design of
Cold-Formtd Steel StniCtural .tfembers (A lSI SIOO), a senes of standards for cold-formed steel framing
(AISI 520 1, S240, S400),as \\~II as !he Cold-Formed Steel Cksign Mmmai(DIOO). These pubilcauons
cover the destgn requirements for CFS member, c.onnecuon., aod system desagn.
11 .1 SCREWDESIGN
JUSIS100,S240,S400
The des1gn require:menlS for screws used tn CFS materials are covered in
I. AlSI SIOO North Amencan Specification for the I:Hsig" ofCold-Fol"mtd Sttel Stntcturtll
.f&femMrs. Se-cuon E4 and AppendiX A.
2. AISJ S240 North American Standard for Co/d-Rmntd Stu/ Strucwral Frammg. Sections
B 1.5.1 , 85.2.2.3.2, and 85.2.2.3.3.
3. AISJ S400 North.4menctm Standardfor Seismic Dengn ofCo/d- Form~d S tu/ Stnlt"tw·al
~):uems (with Supple.ment I).
ScrtwSius
Some.of the [)'pical screw sizes used fot attaching CFS studs to CFS studs or structural steel are shown in
Table J-15 (AISI 5240 Tables C-8 1.5.1.1 -1 and C-8 1.5.1.1-2).
Table 3-15. Sc"w si:esandprope,.tles
Sc.rewStze
Screw Diameter (in)(l)
AvaJLable Tip Pouu
Styles•2)
#6
~
# 10
#12
~ln
0.138
0. 164
0.190
0.216
0.250
2
2,3
2, J
2, 3. 4. 5
1,4, 5
I. Screw shank dulJneter ts measured out-to-our (threads are ettt tnto the screw shank).
2. The higher the potnt-slyle number, the th•cker lhe
Table C-8 1.5.1.1-2).
st~l
materaal the screw can peoeuone (AISJ S240
Successful tnsmllauon of a screw also depend-; on the screw manufacturer and maximum allowed drill
motor speeds. Some screws may fracru.re dunng 1nstallatton tf insmlled at lOO high of a drill speed.
Installers cypically want to use h1gher speed drtlls sance they can tnstall more screws 1n a ga"\·en ume pe,riod.
Generally th1s is not a problem when sc:rew1ng thin CFS studs together. but 1fthe 1n'itallers are screwtng to
struc.tural steel members (wide flange beams. HSS sect.iotts. etc.). they can encounter a potenttal problem.
CFS framers may ha\'e to try several different screws from d11Terem nlanufacturers before finding
t.he requ1red. sctew that works we.IJ wuh t.hear haJld dnHs. Screw values can vary significantly. son
is recOtnmealded to use a screw t.hat is ltsted tn an evaJuation repon by M ANSI-accredtted product
cemfication company.
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
5(.ttw- to-S(rt~1
Spacing
AISI SIOO §J4.1 and J4.2
Centec.to-center spacing (m1mmum):
AISI 5200 §B 1.5.1.3
a. 3 screw diameters: 100% screw shear \1alue
b. 2 screw dtameters: 80% screw shear value
Edge and End Distance (minimum) Wllh lhe destgn force perpe-ndicular to edge:
a. 3 screw d1amerers: 100% screw shear \f3Jue
b. 1.5 screw diameters: Mirumum edge distance
AISIS IOO §1U
c. Shear rup~ure
AISIS IOO §16.1
f',.
= 0.6F.,A..,..
A lSI SIOO Eq 16.1- 1
For a connection where the screw pulls through the steeJ toward the limning edge:
A..,.=bue_
AISI SIOO Eq 16.1-2
n = Number of screws along cntical cross-secuon
1 = Base saee.l thtc.kness
e.,= Clear distrutce from end of member and edgeofsc.rev.• hole
Three screw dtameters is [)'pically recommended for use m design so tlte tear-out dtstance to the
perpendtcular free edge lS not less than the tear-out d.islance betwee.n the center-tt>-eente.r spacmg of screws.
Edge Distance (m1mmum) with lhe design fotce parallel to edge:
a. 1.5 SCte\\' diameters: I00% screw shear \'alue
AI 51 5240 §B 1.5.1.3
b. 1.5 screw diameters: Mirumum edge distance
AISI 5 100 §1U
St.rtw Otsigu Valuu
A lSI SIOO §J4
Screwdestgn values tor steel to steel are to be ca.Jculated per AISI S 100 Secuon J4, and requuemenL(j for
c.onnecuons to other matenals are found ln Secuon 16. When screws are tnstaJied through material other
lhlln steel. such as steel to wood or steel to d.rywaJI. the screw destgn \•alues Will have to be evaluated based
on smndards for that material or published in an ANSI-at..--credited p!oduct-certtficarjon company evaluation
report.
The Steel Stud Manufacturers Association (SSMA) ..Product Technic.a11nfonnauon" and the Steel Framing
Industry Associauon (SFIA) "'TechnicaJ Guide'" pro\'tde a conveniet\t screw design table that ioc.ludes
shear and pullout destgn \'aJues f« var1ous screw s.zes. CFS stud thtcknesse~ and steel strength values.
The serew design \'alues tn this mble are for Allowable Stress De:stgn (ASD). lfth«!' designer Intends to use
st.re.ngth dessgn. lhe screw ..nom mal shear strength·· \'aiUt..~ can be found tn Pan IV of the A/Sf AftmwlCold-Formed Steel EMsign (D IOO).llowever. the tabuJated values may have slightly dift'e-rent thicknesses
for framing me..nbers than typically used. The notnanaJ shear strength values have to be adjusted by either
the restsmnce factor, q,, or J3ctor of safety. 0, to detemlute the screw des1gn values.
The CFS-framed shear wall shear strength does not have to be calculated usmg screw con.t'l«tion ''alues
aod scaled to those tabulated values •n AISI S240 and S400, tn accordance with AISI S240 Secuon D1.2.6,
when the enginee,r chooses an a'iSembly wnh tabulated nominal values 1n AISI S240 Tables 85.2.2.3-I and
85.2.2.3-2 and S400 Tables E6.3- l, El.l- 1, and E2.3- l for wtnd, seismic., and othefln-plane loads.
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
Scnws to AU.ath Sbur \Vall WSP Sbtathing to CFS Framing
AIS I S240
ll\e design requirements fOr CFS-fi'amed shear walls are covered in AISI S240. AJSJ S240 Sect1on
85.2.2.3.3 lirnns the shear waJI WSP sheathjng anaehme.ru to lhe CFS framing 10 be either #8 or# 10
screws. For WSP shea!hill£. AISI 5400- 15/S 1-16 Table E 1.3-1 (shear wength for seismic) specrfies !he
screw size to be used wnh a cenam fram1ng thickness 10 oy 10 preclude early shear fa1lure of the screws.
When using sreel sheet sheathed shear v.'311~ a mlntmum of a #8 screw is requtred for the stee-l sheet
sheathtng attachment to !he CFS franung members, AISI 5140 Table E2.3-l.
The screw sJzes typtcally used for the attachment ofWSP or Sleel sheet sheath1ng to the CFS framing are
#8 scre\\o'S:
# I0 scre\li'S:
#8 screws:
For attachment ofWSP sheathtng 10 CFS studs Wllh a desigilaUon lhick:ness
ofS4 mils (16 gauge) and thmner.
For attachment of WSP sheathing to CFS studs with a designation thid11ess
..
of68 mils (14 gauge).
Mtnimum screw sae for the auachment of steel sheet sheathtng 10 CFS fmming.
ForWSPsheathing. the #8 screws and # IOscrews used to lnstall the sheathing are requlfed to have a
m1nimum head diameter of 0.285 inch and 0333 tnch, respecuvely. CFS-framed shear walls have nominal
shear stte.n.gths ba<iied on sheath1ng l)'pe. fram1ng thu::kness. and scte\V spaetng 1n the tOIIowtng seyen
dttferent destgn mbles:
AISI 5240 Table 05.2.2.3-1
Untt Nominal Shear Strength [Resistance] (V.) for Shear Walls'""' Steel
Sheel Sheath1ng on One Stde of Wall
AISI S240Table 05.2.2.3-2
Umt Nominal Shear Strength [Resismocej (V.) for Shear Walls'""' Wood
StruCtural Panel Shea!hing on One Side of Wall
AISI S240 Table 05.2.2.3-3
Untt Nom1nal Shear Strength [Resistance] ( V.) lor Shear Walls'""' Gypswn
Boord Panel Sheathing on One Side of Wall
AISI 5240 Table 85.2.2.3-4
Untt Nominal Shear Strength [Resistance] ( V.) for Shear Walls \\1!h
Fiberboard Panel Sheathing on Ot\e S.ide of \Vall
AISI 5400 Table E 1.3-1
Unu NominaJ Strength [Resistru\Cej (VIt) per UtlU length tOr Setsmic and
Other In-Plane Loads tor Shear Walls Shea!he<l wi!h Wood Structural Panels
on One Side of Wall
AISI 5400 Table E2.3-l
Un1t Nomtnal Strength [Reststancej (V,.) per Untt length tbr Se.is:mic and
Other In-Plane Loads for Shear WaJls wnh Steel Sh~t Sheathing 0 11 Orle
Side of Wall
AISI 5400Table E6.3- l
Nomtnal Shear Strength (I',.) per Unttl.ength for Se1smic Loads for Shear
Walls She.alhed with Gypsum Board Panels: or Fiberboard Panels on One
Side of Wall
AISI 5240 Tables 05.2.2.3-1 !hrough 85.2.2.3-4 may only be used lor seismic design \\!leo !he sersmic
respon<e modification factor, R, 1s 3 in SOC 8 and C ct for SOC A (AISI 5400 Secuon A 1.1.3), so n does
not apply to this design example. It is t.tnponan.t to nme tha[ AJSJ now requtres one to design seismtc fotceresisung systems in SOC B through F tn accordance with S400, no matter the R-factor, unless an R of 3 is
used 1n SOC 0 and C.
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191
Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
Author 's Distu.ssion
The msmllauon of sc.rews for eonnecung CFS studs logether or sheatJung to CFS studs generally takes
longer than naals inro v.'OOd framing. Nails typ!C.ally art shot-installed into v.ood fra.mtng members using a
nrul gun. whereas a screw requires a drill and the ume fOf the installer 10 manua11y push the screws through
the sheatlung matenal and engage the CFS stud ftange. Add•tionally,AISI S240 Commentruy Section
85.2.3 states that overdriven sheath1ng screws will result in IOY.'ef strength, stitli\es:s, and duc.ulity in
sheathed CFS-fcruned assemblies., and therefore the sctt\\o"'S should be dnven Hush wnh lhe sheathing.
AISIS100, S24~S400
11.2 CFS STUD DESIGN
The des1gn requirements tOr CFS 1n sruds are conred by the followtng documents, including supplements:
I. AISI S 100 North Amencan Specification for 1he Design ofCold-Formed S11~e/ StruCfl»·a/
Membert. Secuon C and AppendiX A.
2. AlSI S240 North Amencan Standardfor Cold-Formtd S1ul Structw·al Frammg. Secuons A. B.
and C.
3. A lSI S400 North Amencan Standardfor Seismic Design ofCold-Formed Steel Strm•turol
Systems. Section E.
CFS studs are 3\'3ilab1e 1n various dephs. Eac-h stud depth has se\'eraJ different flange widths a(i well as
mruenal thicknesses (gauge, m•ls) available. Structural CFS studs used 10 bwlding design are typieally
manufactured us1ng steel material With y1eld strengths of either 33 ksi or 50 ks1, with 54 mil and th1cker
franung t)'JllcaUy u:s1ng a yteld st.re.ngth of 50 ks1.
The material thickness used tbr CFS framing was pre.vtous.ly gt\•en a~ a gauge llttCkness (example: 16
gauge). The CFS 1.ndusuy has moved 10 measunng the material thickness now as ..mils*" where I mil =
1/1000 of an lnch. Table J-1 6 shO\\o'S the conversioo for structucal studs. Nonstruc.twal studs ate available in
lhJnner matenal.
Table 3- 16. CFS thickne.ss properties
ThickJ>e<s'' 1(in)
Des•gn
Thickness ( in)
Design lns1de Bend
ThickneSS (m!IS)
Designation
M1rumum
Radius''' (in)
Reference Only
(gauge no.)
33
0.0329
0.0346
0.0764
20 - struCtural
43
0.0428
0.045 1
0.0712
18
54
0.0538
0.0566
0.0849
16
68
0.0677
0.0713
0.1069
14
97
0.0966
0. 10 17
0 .1525
12
118
0.11 80
0.1242
0 .1863
10
Notes for Table 3- 16:
I. M1n1mwn lhickness represents 95 percent of des.gn thick.nes:s and the m~nimaJ acce-ptable thickness
for deh,rery to the job Stte.
2. Des1gn 1nsK1e bend rnd1us 1n acc«dance wnh AI SI S20 I Product Dam standard Table C3-l . 11us •s
used by SSMA and SFIA to calc.ulate the CFS stud propen1es shoy,.n in lhetr pubhcauons.
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
Author 's Distu.ssion
The msmllauon of sc.rews for eonnecung CFS studs logether or sheatJung to CFS studs generally takes
longer than naals inro v.'OOd framing. Nails typ!C.ally art shot-installed into v.ood fra.mtng members using a
nrul gun. whereas a screw requires a drill and the ume fOf the installer 10 manua11y push the screws through
the sheatlung matenal and engage the CFS stud ftange. Add•tionally,AISI S240 Commentruy Section
85.2.3 states that overdriven sheath1ng screws will result in IOY.'ef strength, stitli\es:s, and duc.ulity in
sheathed CFS-fcruned assemblies., and therefore the sctt\\o"'S should be dnven Hush wnh lhe sheathing.
AISIS100, S24~S400
11.2 CFS STUD DESIGN
The des1gn requirements tOr CFS 1n sruds are conred by the followtng documents, including supplements:
I. AISI S 100 North Amencan Specification for 1he Design ofCold-Formed S11~e/ StruCfl»·a/
Membert. Secuon C and AppendiX A.
2. AlSI S240 North Amencan Standardfor Cold-Formtd S1ul Structw·al Frammg. Secuons A. B.
and C.
3. A lSI S400 North Amencan Standardfor Seismic Design ofCold-Formed Steel Strm•turol
Systems. Section E.
CFS studs are 3\'3ilab1e 1n various dephs. Eac-h stud depth has se\'eraJ different flange widths a(i well as
mruenal thicknesses (gauge, m•ls) available. Structural CFS studs used 10 bwlding design are typieally
manufactured us1ng steel material With y1eld strengths of either 33 ksi or 50 ks1, with 54 mil and th1cker
franung t)'JllcaUy u:s1ng a yteld st.re.ngth of 50 ks1.
The material thickness used tbr CFS framing was pre.vtous.ly gt\•en a~ a gauge llttCkness (example: 16
gauge). The CFS 1.ndusuy has moved 10 measunng the material thickness now as ..mils*" where I mil =
1/1000 of an lnch. Table J-1 6 shO\\o'S the conversioo for structucal studs. Nonstruc.twal studs ate available in
lhJnner matenal.
Table 3- 16. CFS thickne.ss properties
ThickJ>e<s'' 1(in)
Des•gn
Thickness ( in)
Design lns1de Bend
ThickneSS (m!IS)
Designation
M1rumum
Radius''' (in)
Reference Only
(gauge no.)
33
0.0329
0.0346
0.0764
20 - struCtural
43
0.0428
0.045 1
0.0712
18
54
0.0538
0.0566
0.0849
16
68
0.0677
0.0713
0.1069
14
97
0.0966
0. 10 17
0 .1525
12
118
0.11 80
0.1242
0 .1863
10
Notes for Table 3- 16:
I. M1n1mwn lhickness represents 95 percent of des.gn thick.nes:s and the m~nimaJ acce-ptable thickness
for deh,rery to the job Stte.
2. Des1gn 1nsK1e bend rnd1us 1n acc«dance wnh AI SI S20 I Product Dam standard Table C3-l . 11us •s
used by SSMA and SFIA to calc.ulate the CFS stud propen1es shoy,.n in lhetr pubhcauons.
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
The CFS stud tndustry has also adopted a dttferent namtng nomenclature standard to specifically tdenufy
the individual CFS studs. An example of the naming nomenclature 1S as follows.:
Tab/~
3-17. CFS pn>duct tkslgnatkms
Style ofS«:non< 1•
(S, T, U, F)
Flange Width
(x/100) (in.)
Thickness (m1ls)
Material Yield
Strength (k.<i)
600
s
162
54
50
600
T
150
54
33
Stud Name
Deplh
(x/I00)(1n)
6005162-54 (50)
600TI50-54
Material
Note larTable 3- 17:
I. S =-Stud, T =Track, U =Channel, F =Furring Channel.
The CFS swds are aulOmatically assumed lO ha\'e mm1mwn yteld suength of 33 ksatfthe matenal
strength is not identified_ It tS recommended that the engineer reviey,•AISI 5201 and lhe SSMA and
SFIA publications lO funhe-r Utlderstand the nanung nomenclmure u..-;ed to tdenuty CFS fmm1ng.
AJSI 5400 Stttion El.4.t.t- Limitations: forTtbuhutd Sys:ttms (Sbur\\'aiJ Studs)
The tbllowang are the minimum CFS srud dtmensional requirementc; for shear \~t'alls ulihzutg CFS studs.:
Table 3-18. ShttJthtd shear wall mmlnuun CFS stud dimensional Nqm~nu!niS
Style ofSecuon
Mtnimum
Thickness
M injmum Flange
Widd1 (x/IOO)(tn)
MtnimumWeb
Depdl
Minimum Flange Edge
Silffener Width (tn)
C-shape
33 mil s
162
350
14
T-shape
33 mil s
125
350
N.A.
The CFS stud tnaxunwn-aJiowed spacing ts limited to 24 inches on c.ente.r tor shear waHs.
AJS I S400 Tobles EI.J- 1 and El.J-1
Table El.3-l requtres the following additional material requirements for shear wall CFS studs and ttacks~
Tablt 3-19. Shtatlwd shear wall nummum CFS ltud ma.ttnal rtquirtmenl.t
Struc.tural Grade
SleeJf2'
Stu<l Thickness'" (mtls)
ASTM
33,43
Al003
33
Ty pe I-I
54 and lh1cker
Al003
50
l)'pel-1
Notes for Table 3-19:
I. FOC' \\rsp sheathed shear \\'all S400 table values, the thtd:es[ CFS stud thatts petmined ts 68 mils.
2. Type H stee.l is a htgh~uchlny steel.
AISI 5240, 5400
11.3 SHEAR WALL SHEATHING
The des1gn requtrements for shear \valls utillzittg CFS Studs are OO\'e-red by AlSI S240 North American
Sttmdardfor Cold-Formed Steel Structural Framing, Seetion Band Commenmry Section B.
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
The shear \vall nonunal shear suength de-pends on sheathing type.• screw size and spactng. and lhe thickness
oflhe CFS framang. 1l1e follow&ng are sheathed shear \1/all m1nimum requtrements per AISI S400 Sect.ion
E l.4.1.1'"limlrottOns for Tabulated Systems•·:
1. Fastener location from edge of sheathing (mlrumum edge disronce):
Minch
2. Minimwn width dimension of sheathing in any direction:
12 tnches
a. Panels less than 12 inches wide shall not be used for shear walls.
3. All sheathing edges shall be anached to framang or blocking.
a. Suap blocking shall be a muumum of I~ Lnches wade and nOf less than 33 mtls 1n lhtcL::ness.
b. Slrap block:mg. can be mstaJied e,ither under or ove-r lhe sheathing.
c. For sheathiJ\g Other lhan steel sheet sheathang.lhe screw shaH be tnslalled through the shemhmg
to the blocktng.
Stttion A lSI S400Stt:tion E1.4.1.1- \\'ood Struttural Pantl Sbtathing_
Nominal shear suength shaJI be as given 1n AJSI 5400 Table E 1.3- 1. Jncreao;es of nomu'lal strengths shown
tn thts table. as allowed by olhe.r stru'lda!ds, shall not be permmed.
Addittonal requirements for wood structural panel sheathtng are smted m AISI notes. Those reqmred for
desJgn are noted here:
seiSmiC
(n) Wood structutal panels shall be manufactured uStng e."tterior glue and shall comply wnh DOC
PS I or PS2.
(o) Wood struclural panels are pe-rmaned lO be applied e1lhe.r parallel lO or pe.rpend1cular to
franung.
(p) Wood structural panels shall be anached with m1mmum No.8 countersunk tapptng sc.rews that
have a minimum head dtruneter of0.'285 lnch or No. 10 oountefSUJ'tk tapping screw'S that have a
mtnimum head d1amele.r of0.333 tnch tn accordance with AISI S400 Table E 1.3-1.
(q) Screws used toanach \\"OOd sLructutal pane.ts shall be tn ac:cotdanc.e WllhASTM C l513.
As noted prevtously, there are seven shear wall shear strength tables provtded tn AISI (AISI 5240 Tables
85.2.2.3-1 through 85.2.2.3-4 and AlSI 8400 Tables E 1.3- 1, E2.3- l, and E6.3- l) tabulating the nominal
shear strength for various sheathed CFS-framed shear wall assemblies. AlSI S400 Tables E 1.3- 1 and E2.3- l
reproduced here a'i Table 3-20. is fbr seismic desjgn_ A review oflhe table tndicates:
1. l1le maxhnwn shear \...-all aspect rauo (h/w) is typically hmiled to 2: I.
a. There IS an exception for Identified shear wall assemblies wnh an aspect ratto beru·ee-n 2: I and
4: I tflheir shear suength is reduced by 2wlh (2 x shear wall length/shear wall height).
2. The shear wall sheathtng can be % -tnch WSP Strueruml I sheathing (4-ply), %.-tnch OS8, or
sleet sheet sheathing.
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
The nominal shear strength values form-plane lateral loads other than setsmic (example: wind) for shear
walls are pro\rided tn AJSI 5140 Sectaon 85.1.3. The nomtnal shear strength values tOt setsmic load.~ tOr
shearwallsare pmvaded tn AISI 5400 Tables El3-l and E2.3- l. The nominal s.heat stteng_th values shown
in Tables 3-20A and 3-208 are reduced by enher tbe resistance factor (t = 0.60) for LRFD or !be factor
of safety (!1= 2.5) for A~D per AISI 5400 SectiOns El .3.2 and E2.3.2. Per AISI 5240 Secuon 85.2.3 and
AISI S400 Sections E1.3.1.1.2 and F2.3.1.1.3, when the same sheathing and fastenet stze and spactng are
used on both sides of the \\'all, then the shear Willi design values can be doubled.
Tab/~ 3-2QA. AJS/ S40Q Tab/~ £1. 3-1:
mut nommal Sll?ngth [~Ysisumctj (l',) per umtl~ngthfor str.smic
and othtr in-plant foods for shear u·atl.t shtatlwd with w()()(/ stnu:·rurat panels on one std~ of wall
Fastener Spac.an~ at
Descnption
Max.
Aspeet
Rarjo
(h/w)
*%!"'
2: ll))
Assembly
6
4
780
990
3
2
-
-
Struc.tural I
sheathing
Designatlorl
Thtck.ness•S."J
Panel Edges'
(tn)
2:1
890
1330
1ns
2: 1(3)
700
9 15
-
1:1(3)
825
1235
1545
2:1
940
14 10
2:1
1230
1850
2 190
ReqUired
Sbe:lllung
Screw S12e
33 or 43
8
~3
8
or 54
68
10
33
8
2060
43 or 54
8
1760
2350
54
8
2310
3080
68
10
(4-ply)
%~·osa
ofSrud,
Track and
Blocktog
(mils)
-
Tobit 3-208. A IS/ SIOO Tablt £2.3-1: unll nommal Slrtugth [resisumcej (V,) per writ ltngthfor
s~t.smlc and other m-plane /()(J(/sj()r .t~ar walls with sJeel shut sheathing on on~ sldt ofwat/9•
Max.
Assembly
Description
0.0 18" steol
sheet
o.o2r steel
sbeet
Fastener Spacing at
Designation
Panel Edges'"
(in)
llucknes:s'"-"1
6
4
3
2
ofSwd,
Track and
Blockmg
(mils)
2: 1
390
-
-
-
33 (mon.)
8
2: ICJJ
-
1000
1085
1170
43 (mon.)
8
2: 1Cl)
647
7 10
n8
845
33 (min.)
8
Aspeet
Rau o
(/r/w)
ReqUJred
Sbealhong
Screw Si:ze
Notes for Tables 3-20A and3-208:
I. Nominal strengzh shall be muJuplied by the rtsistance foctor (4) to detenn1ne design strtngth or
divided by the safery filctot (0) to detennine allon·able strt,gzh. as set fbl'th in AISI S400 Sect:tons
E 1.3.2 and E2.3.2.
2. Sctews in the field of the panel shall be tnstalled 12 1nches (305 mm)o.c. unless otherwise shown.
3. Shear wall he-aght to \\1dth aspect rouo (hhv) greatet than 2: I, but nOt e.xceedJng 4: I, shall be
permmed pmvided that the nominal strength values ate muJuplte.d by 2wlh. See AJSI S400 Sections
El.l.l.l and E2.3.1.1.
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
4 . SeeAISI S400 Section.< EI.3.LL2, El.l. I. 1.3, E2.3. I. L2,and E2.3. 1.1.3 for r<qu~rement< forshea~hing
applied to bolh stdes of wall at1d more than a single sheaLhing marerial or fastener configuration aJong
the same wall line, respectively.
5.
Unless noted a~ (mln.), subslhulion of a stud ot uaek ofa diff'erem dtsignaJton thickne.ss is not
pe-l'mined.
6 . Wall swdsan<l track shall be ofASTM A l003 Strocrural Grade 33 (Grade 230) Type H steel for
members with adRsignaJton thtclmes.sof33 and 43 mils. and A 1003 Structural Grade 50 (Grade-340)
Type H steel \Vith a designatton thtd.ness equal to or g.re.ater than 54 mils.
7. Fot v.'OCXI sU\Ictural panel sheathed shear walls. tabulated R,. values are applicable for shan-term load
duration (seismtc loads). Fot other in-plane lateral loads of normal or permanent load duration as defined
by the AWC NOS, me values'" Tables 3-20A and 3-20B for wood strucrural panel shea!hed shear '"'lis
shall be muluplted by 0.63 (normal) or 0.56 (pennanem).
8. For Sl: I"= 25.4 mm, I
fo01 = 0.305 m, I lb = 4 .45 N.
9. AddJ!tonal steel sheet micknesses can be found '" A lSI S400 Table E2.3. 1.
The original table was updated and split tnto two separate tables(EI .3-I and E2.3-l) inAISI S400 based on
sheathing type.
11.4 SPECIAL SEISMIC REQUIREMENTS
AISI S240, S400
The design requirements tbr shear \\"ails ulili2.1ng CFS st-uds are covered by AISI S400 Secuons E1.4 and
E2.4: Syste.m Requiremems.
These AISI seismiC destgn standard sections apply to shear ·walls sheathed with WSP or steel shee.t
sheathang when the buildtng S)'Stem response mochfic~uion ooeffictent R is other than 3.0 . WSP and steelshee.t-sheathed CFS-framed shear walls are typacally designed usmg an R greater than 3.0. However. 11 JS
lmponant to note that the setsmic destgn standard does apply to any lateral system.reslsting. seismic forces.
e\<e.n w1th an R lower than 3.
I. AISI S400 Sections E 1.4. 1.2 and E2.4. 1.2 "Chord Studs, Anchorages, Collec.tors (a•KI
COntlt'C[IOJ'IS)"
a. El .4.1 . 1 and E2.4 . 1.1: Chord studs or olherventcal boundary members at the endsof\vall
segme.nts that resist seismic loads, braced with sheathing. shall be attc.hored such that the botto1n
rmck is not requiJ'ed to resJst uplift by bendtng of the track web.
b. El .4 . 1.2 and E2.4. 1.2: The available strength (destgn strength or factored resistance) of
connections and me.mbers nm pan of the destgnated e1'1ergy-disstpa6ng mechanisrn (DEDM)
shall exceed the expected strength of the shear \vall, but need not exceed the load de-termtned
from the setsmtc load cornbtnattOns with O\'erstrength. Thts tncludes collectors, collector
connections, chord studs. hold-downs.. hold-down anchorage.• and bottom track shear anchorage.
2. A lSI S400 Sections E 1.4. L I and E2.4. 1.1 "Ltmit:ruons for Tabulated Systems"
a. 33 mtl an<l43 mll wall sruds and uack shall be of ASTM A 1003 Structillal Grade 33,
Type li.
b. 54 mtl or lh1cker wall studs and track shall he ASTM A 1003 Structural Grade 50, Type H.
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
AI51 S240 &ctlon 85.2: Sh"• Wall Dcsign
There are two types of shear walls: Type I, whtch is coveted tn AISJ S240 Sec.tion 85.2.1 .1 and AJSI
5400 Secuon 1.4.1. and Type n. "hich is covered in AISI 5240 Section 85.2. 1.2 a•KI AISI S400
Section I .4.2. The l)tpe: I shear wall IS the convenuon.al she.athed shear wall wnh a hold-down at each
end~ it is typtcally referred to as a segmented shear wall tn wood construcuon. l1le Type U shear wall
i.~ typically referred to as a perfOrated shear \\'311 t.n wood consuuction. Type II shear \\ails must meet
the requiremeniS oFType I shear walls except where noted •n AISI S240 Section B5.2.1.2 and AISI
5400 Secuon E 1.4.2. This desogn example is ba.<ed on us•ng l)'pe I shear walls. The engineer shoold
revtew AISJ S240 Section 85.2. I.2 and AISI S400 Secuon El .4.2 for the design requiretnems fOr l)tpe
II shear wall<.
12. ShearWBII Design Example: Building B
One uuenor lhree-story. st.acktng shear wall tn Butlding B~ Zone B I ha~ been selected for this design
example. The CFS-framed shear wall des•gn wtll follow the requtremems of AISI 5240 and 5400.
Shear Yo>aJI patametets:
I. Classificauon: Type I.
2. Wall shealhong; Wood structurnl panels (DOC PS-I and PS-2): ply"ood or OSB.
3. Wall studs: CFS studs(54 nul [16 gauge]): 6005162-54 (F,= 50 ksi)and (68 nul[l4 gauge]):
6005200-68 (F1 =50 kst).
4. Continuous rod tie-doo1\ (hoh~doo1n) system: Resists tension (upltft) from O\renurnmg and
comprises conunuous rods, coupler nuts, nuts. bearing plates, take-up devtces, cnpple studs,
and bridge blocks.
5. Chord Studs: RestSt compression from gravaty and overtumiJlg; CFS studs.
6. Wall base connection: Elevated floots: screws.
Podiwn level: anchor boiL<; in a 12-U'lCh-thick structural concrete slab.
Conlinuou.s Rod Tit-Dowa (llold-Down) System
ll\e shear wall design w111 uuhze a conunttous rod tie-down system that is cenu~red. between tl'!e chord studs
at each end of the shear wall and extends from the podium deck to appro.~imately halfway bet\lteen lhe th1rd
floor and the roof. where Jttemunates at a hodge block. Each lod.JVtduaJ story's shear ·wall uphft forces are
resisted by a bearing plate on top of the floor framtng above lhe shear wall or by a bearing plate on top of a
brtdge block. This story shear \Vall upltft fOrce is someumes referred to as tncremental upltfi. The beartng
p1ares transtCr the O\'e:rturning force for each story of the buddtng to the conttnuous rod. The conunuous rOO
restslo; lhe cumulauve overturning tel\Sion (uphft) fOrce from each stocy·s shear wall and uansfers n to t.he
concrete pod tum level of the butlding.. The shear wall CF$ chord studs rest.St the upl1ft tbrce at eac.h story tn
compression and be.ar agamst the \\'OOd floor or roof system or bridge block, but are typically governed by
the ovenurning and gravity oompresston load.
In mulustory budding.-;~ take-up (shnnkage compensatmg) devtces are often used lO ensure the ovetturntng
rest.ramt system is ught even after wood sh.nnkage or settlement occurs. Use of engu~ered wood framing
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
heJps reduce the wood shttnkage effect. AISI S240 Sectjon C3.5.3 requu-es that structural Yl'<lll studs are
seated ught against the tracks and states that tight means a maximum gap tolemnce of~ 1nch is acceptable
between the.stud and the [op and bonom uack Thts could pos.s1bly lead to up to ~ -tnch ve111cal senJement
per story. Since the steel rods 3l'e a fixed length 10 a continuous rod ue-dov..n system, shrinkage and
se.ttle.ment accumulates over the he1ght of the building, possibly rtwJting 1n ¥...-tnch venicaJ settlement at
the top of a three-stOI)' srructure.
AISI S100-16 Nonh Am~nctm Specrficallonfor lht IJI!:ugn ofCold Fonn~d Sled SfniCtural Mtm~rs
AISI 520 1-1 7 Norlh American Sumdardfor Cold-Formed St~~l Fronung- Produt.1 Data
AISI 5240- 15 Nonh American Sumdardfor Cold-Formtd Stu/ Stmcturol Frtmung
AISI S400- 15/S 1-16 Nor1h Amen can Standardfor Seismic Design of Cold-Formed Stu/ StniCturol S>·sttms
LRFD andASD
The AISI spectficauons and smndards aiJowthe use ofetther LRFD or ASD tor the design of the hgJuframe-CFS struc.tures. Both LRFD and ASD loads are shown in the beg.innang sections oflhis design
example as an aid in transtuorung for those more familiar with ASD than LRFD. When deflections are to
be calculated for eitl1er v.'OOd or CFS light-frame WSP-sheathed shear walls,lhey are to be based on usang
strength-level load combinatioos (LRFD). AI SI 5240 and S400 require !he CFS light-frame shear wall
uphft anchorage and venteaJ boundary members (c.I'IO(d sruds) have the nom1nal strength to resist whatlhe
system c-an deHve:r, but need not exceed the amphfied seismic load. The amphfied loads are based on using
the LRFD seismic load coanbination.o; with the oversttength fac tor. Since ASD amphfied setsmic fO!ces are
nOl addressed b)• AJSI.lhe remainder of the design example is done ustng LRFD.
12.1 SHEAR WALL SHEAR AND OVERTURNING REQUIRED STRENGTH
(BUILDING B, ZONE B1)
The lfibutary area and destgn shear force to the shear wall at each floor le\•el. ustng the Zone B I desJgn
force~ tS as follows. The overall shear wall length IS de[enntned from the waH out-t~out dm\ension of
39 feet and subt.rneting the two exterior 6-tnch CFS stud c:ross-v.'alls' thickness. The cross-v.all overall
thickness equaJs 8 tnches (6-tneh CFS Cstud + 1-Jnch-Uuck stucco finish+ y;.,nch plywood one face+
~inch gypsum wall board). The shear wall length is approximately 39 feet minus two 8-tnch c.-ross waH~
which equals 371eet 8 inches. To stmplit)' the math, the shear wall destgn length is shortened to 37 feet
0 mc.hes. The ltlbutary area to the shear waJI though IS sull bao;ed on U1e 40-foot length used in detemun1ng
the seismic mass of Building B. The shear \\'1111 se.ismic ttlblltary area width IS 23 feet8 1nches. ll\e story
wall heights are pr0\•1ded tn Table 3-21.
Table 3-21.
Sh~ar
wall de.stgn mformallo-n (l.RFD)
Story Level
Tributary Area to Shear
Wall Ltne (II')
Level
Design FOC'ce
"(psf)
Shear Wall Story
Forces(lb)
3rd Floor
(23 118 tn) x 40 ft = 947
10.4
947 X 10.4 = 9849
2nd Floor
(23
n 8 •n) x 45 n =
1065
11 .4
(23 118 tn) x 45 ft = 1065
5.7
1st Floor
l: Shear
ASD
Wall Forc.e
Forces
( Ib)
(0.7F)
9849
6895
1065 X 11.4 = 12,141
21,990
15,393
1065 x5.7 =6071
28,06 1
19,643
The locatjon of the e.xample shear ,~,-all can be seen on lhe tollowing framtng plans as \veil ao; the other
shear walls for Bu1lding B.
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
Figu" 3-//_ Buildmg ~ThmJ-ji()(Jr !hear n·al/s
Ftgw-e 3-11. Buildmg B-Second-jloor shear wall!
OENOT£5
SHEAR WALl
l OCATION IN
8Uil.01Nn6
FlguTP 3-13. Buildbrg ~Flrs1-j1oor shear n·all.s
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
11
10
10
-----
:v-c
I
I
I
I
ontlnuous-rod
IJ
I
I
I
l
r
37'
Figure 1-14. OptiOn 1: Fuii-/Rngth, .Slacked shear wall
11
10'
10
---
-
I
I
I
I
I
I
v-- s•eel gravity beam
I
I
I
L
r--- ..- ~
I
'4 -'- - : r .
I
,...-:..
1
25'
12'
'
vone-
story steel columns
I
T
Rgun! 3-I 5. Option 2: Short finii-j/oor shear wall
-
COLLECTOR
(DRAG)
11'
-
10'
10'
OPENING IN WALL
25'
Flgw~
200
3-16. Option 3: Short. Stllcked shear u·a/1
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
12.2 SHEAR WALL CONFIGURATION AND TYPE
AISI S400 SECTIONS E1 AND E2
Often 1n mulufamily. mufhstory res1denLtal butldtng.s there are many archhectural restrictions as to where
shear waJLs can be located, how much oflhe wall length can be utilized fOr shear waJis., and ho\"' s.hear Y..alls
align from floor to floor. Three opoons are shown for the shear wall configurauon at this localJon.
Optton I: Full-length, stacked shear walls
a. Opwnwn condiuon~ smaller wall element desJg.n fOrces
Opuon 2: Full-length upper floor waJis, shan wall length 31 first floor
a. 0Jsconttnoous she-ar wall sys.tem at second floor
b. SeOOfld-floor beam and first-floor column design to c.heck
c. High shear design forces on first-floor shear wall
Optton 3: Shon-length shear wall
a. Wall elements ha"e htgher design forces tJ'IaO Option I
b. Htg.h overturning forces on wall
c.. Do not have to design for discontutuny
Other multistoty shear wall eontigwauons are also poss1ble, and the selected shear \\'311 configuration Will
uJtimately depend on the layout of the restdenual units at each floor level. OesJgn fOrces tor the three shear
wall options are provided for compart'iOn.
Shear walls are classified as one of two types:
I. Type I (segmented)
a. Hold-downs required at each end ofeach Type I shear wall
b. Design for force uansfe.r around openlngs
2. Type II (perforated)
a. Hold-downs required at eac.h end ofeach Type II shear wall
b. Opentngs permitted betw·een the ends of a Type II shear waJI wtthout designing for force transfer
around openings
c. Design for Ultifbrm uplift as well as increased unit shear between wall ends
A Type I shear wall ,.;u he used lbr this design example.
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
A IS I S400 Sttlioo EI.J.I.I- Typt l ShtarWaUs
The shear walls m th1s design example are reststjJ'Ig setsmtc fOrces and the building 1s located tn SOC D;
therefore. the destgn has 10 be tn accordance Wllh AISI 5400.
The requu-emems of a Type I shear wall are as fOllows:
I. Fully sheathed and 1\avtnghold-downs at each et\d of the wall segmeru.
2. When sheathed wnh etther wood or steel sheet, then an openutg is permitted m the \vall
between the hold-downs as long as details are provided for lareml-tbrce transfer around the
opemngs.
a. Heig.ht-£0-Width aspect mtio (lrhf') of shear wall pie-r on each side or a window shall be linuted 10
2:1~ where Jr =-Window height and'"= v.'idth of pier adjacem to the wu\dow.
b. Minimum pte.r Width=- 24 1nches.
3. Shear wall heiglu·tO.IVldth aspect ratio (h/w) is limited to AISI S400 Tables EI.J·I, E2.3·1, and
E6.3-l , with a tnrunmum a.~1 ratio of 4: I and walls not less than 24 tnches wide. Mtn1mum
screw s tze reslfictjons for selsmtc destgn are nmed in the shear wall stten.glh tables tn AJSI S400.
4. AISI S240 can be used for seismic destgn only when the building is located m SOC B or C and the
seismic response modtficat:ion coefficie.tt R is equaJ to 3. AISI S400 does not apply to SOC A~ so
AISI 5240 requtrements would apply.
12.3 SHEAR WALL SHEATHING AND SCREW SELECTION
AISI 5400 SECTIONS E1 AND E2
The design requirements tbr shear \\'311 sheathtng and screws are CO\'ered by AJSI S400:
I. Sectaon E 1.3.2 "Available Strength (Factored Reststance)....
2. Sectton El.3. 1.1 ..Nomanal Strength (Resistance)-- tor v.'OOd structural panels.
The \VSP-s:heatlled shear wall nomu1al she.ar strength values are taken from AISI S400 Table El .3- l . The
shear wall can be e''alurued using e-tlhe-r S((ength design (lRFO) or a11owable stress design (ASD). The
avatlable strength is determined tn accordance with AISJ S400 Sectton E1.3.2 where the seistnic resistarl<:e
fuctor, 9. equals 0.6 for LRFD and the factor of safety, !l~ equals 2.5 for ASD (see AISI 5240 Section
85.2.3 for nonsetsmic reststanee faclOrs and factor of safety ''alues). Table 3-22 shows the aYallable
sheathmg strength destgn values for one sheathmg and stud combtnation.
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
Tab/~
3-11. Shtar wall strengths
Fastener Spac:tng at Panel Edges
(inches)
Wall Sheathang
6
I
4
I
3
I
2
SttJ<I
lluclcness
Sheathing
(mols)
SctewSJze
Table E 1.3- 1: Nominal Shear Strength (pounds per 1001)-Sheathong One Sode Only
¥» Structural I (4 ply)
0
I
890
I mo I ms I
2190
54/68
I
8/10
Table EI.J- 1: Available Strength Design Values (pounds) q, = 0 .6
05
1» Structural I (4 ply)
I
534
I
798
I
1065
I
1314
54/68
I
8/10
I
110
I
876
54/68
I
8/10
Table EU- 1: AvailableASD values(pounds) ll = 2.5
% StruCtural I (4 ply)
I
356
I
532
Sbnr \Valll~ngtb
Earlier in lhedestgn example: the width ofBuildtng Bl \lias shown as 39 feet, but a width of 40 feet ·was
used for detennining the seismic mass of the buildmg.. The 39 feetts lheout-to-out dtmensaon of the
buildtng. and lhe shear wall sheathing is not considered to be extended to the outs1de face of the exterior
waJJs; thus. the shear wall length will be shoner than 39 feet. The extenor ,~,·aUs are 6-tnch CFS studs, so
after accounrjng for 1-inch-thtck stucco exterior finish. I.A.:- tnch plywood, and ~tnch gypsum board intenor
siding. the length of the shear Y.'all is reduced as fOllows:
Shear wall full length= 39ft- (2)(6 in+ I in+\<! in+ \<\on) = 39 fi- 16 on= 37 ft 8 on
To keep the design calculauons easy. the full-length shear wall length is reduced lO 37 teet 0 1nches for
destgn purposes. Acrual shear wall coosuucuon length wtJI be 37 ftet8 mches. so there wdl be a hnle more
shear waJI capacity than is bemg accounted For tn the destgn.
Sbur \Vall Shealbing Design
The WSP sheatho.ng selecuon process n>ay uuhze lRfD and ASD, and both are shoiVn on Tables 3-23, 3-24,
and 3-25 for the thtee shear wall options p~e\'Jously lisled. Shear wall des1gn is based on using 1 ~-utc.h
Structural I (4 ply) WSP sheathong.
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
Tablt 3-13. Opli<>ll I: Shear wa/1/engtht: Full-length !hear n·tdl.s each floor ltl·el (L =- 37feet)
LRFD
r
Le\•e-1
S~ear
Wall
Force<••
( F) ( lb)
s~ear
(FIL)
(pll)
ASD
#8or#IO
O..ign
Sc-rew
Capacotym Spacong
(in)
(pit)
EShearWall
FotcesASD
= (0.7F )(I b)
Oestgn
#8 or # 10
Screw
Shear
(FIL)
(pit)
Capacuy'"
(pit)
Spacing
(on)
3rd Floor
9849
267
534
6
6895
187
356
6
2nd Floor
2 1,990
595
798
4
15,393
416
532
4
lsl Floor
28,061
759
798
4
19,643
531
532
4
I. Shear wall fon:es are from Table 3-21 .
2. Oe:stgn capaciLies are from Table 3-22.
Table 1-14. Option Z: Shear,..a/1/mgt/u (L =thudfloor. s.co..djloor= 17 fw: L =firstft>or= Z5fut)
LRFD
Level
3rd Floor
2nd Floor
1st Floor
r Shear
Wall
Force<•,
(F)( Ib)
9849
21,990
28,061
Shear
(FIL)
(pit)
267
595
1123
ASD
#8or#IO
O..ign
Screw
Capaeuy"> Spactng
(pit)
( on)
534
6
798
1314
#8 or# IO
r Shear Wall
Forc-esASD
= (0.7F)(Ib)
Shear
(FIL)
(pit)
Destgn
Capacity'''
(pit)
Screw
Spacing
187
356
6
4
6895
15,393
2
19,643
416
786
532
876
4
2
(on)
I. Shear wall forces are from Table 3-21 .
2. Design capac.utes are from Table 3-22.
Table 3-25. Op1ion 3: Shon-lenglh Jhear walls each floor leu/ (L =- 25fut)
LRFD
Le\rtl
ESheM
Wall
Forcet•)
(F)(Ib)
3rd Floor
2nd Floor
9849
2 1,990
1st Floor
28,061
ASD
#Sor#IO
Screw
Spacong
( on)
rshearWall
ForcesASD
=(0.7F)(Ib)
Shear
{FIL)
(pit)
Design
Capacoty'21
(pit)
1065
6
3
6895
15,393
276
6 16
356
710
6
3
1314
2
19.643
786
876
2
Shear
{F/L)
(pit)
394
O..ign
Capacotytll
(pit)
534
880
1123
#8or# IO
Screw
Spactng
(on)
I. Shear wall forces are from Table 3-21 .
2.
~ •gn capac.tlies are
from Table 3-22.
The shear \Yall sheathing is requtred only on ooe stdeof the shear wall and is assumed to be InStalled
vertically since lhe floor-tc)..ftoor heights are eithe,r 10 or II feet, and the clear CFS srud height between
floor fram1ngs is e.nher 9 or 10 fee~ respectively. As an altemauve design for an Optioo 2 and Option 3
shear waH, the first-floor sheatlung could be installed on both sides of the wa11 us1ng a sheathing edge-screw
spaclng of4 u'tChes on center. \\'hen shealhlng JS used on both mdes of the srud wa11, tt is reromme-nded that
Lhe sheath1ng panel jotnts should be offset to fall on d.-fti!rent framing membe.rs, the same as tbr light-frame
wood- frame shear walls.
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
S1nce the wall sheathmg is being Installed venically. no horizontal CFS flat-strap blockang as requtred
because all sheathing edge screwtng will be to the CFS studs (vertical panel edges). bonom track. and
top track (honzonta1 pru'lel edges). Ifthe wall sheath1ng is to be tnstalled horizol'llall)'. then a mlntmwn
I ~-inch-wide CFS flat-strap blocking. the same lh1d:ness as the CFS wall fram1ng. w-ould be required at
the sheathJngs• horizontal pa.'lel edges. VerucaJ panel edges are asstJmed 10 a11gn with lhe wall studs.
Whtle honzontal CFS Rat-strap bracutg as nOl required 111. this case for blocking when the "all sheathsng
IS installed \'emcally. t11e honzontal CFS flat-strap bracing is sull required W\der the wall sheathing for
de\•elopang the axtal capac tty of the CFS beanng stud$. as will be discussed Later.
Full-scale shear wal.l testu'lg c.ompanng Yl'inged-ttp to non-winged-up screws anaching WSP sheathutg to
CFS fmming has shown the peak loads for the shear wall assembly within 5 to 10 perce,m. but also a 20
percent design strength reducuon for the wtnged-up screw~ probably due to the wings creating a larger
hole m the WSP sheathing_ Also, testmg has shown stgnificantJy reduced strength, suffnes:s, and duc.tility
when self-tappmg screws were overdrtven 1n the WSP sheathing of a CFS-framed shear wall assembly, as
noted an AISI S240 Co•nme1ttary Section 85.2.2.
12.4 01/ERTURNING RESTRAINT (TIE-DOWN) SYSTEM REQUIRED STRENGTH
AJSI S400 requires a hold-dO\m (aftemauve terminologies used: hold-down ancho! or tte-down or uplift
anchorage) at each end of a shear \\all to prevent the shear y,'alJ from ovenurmng as a result of applied
honzonmJ loads dirutb..ned ven1cally over the buildtng structure's height The hold-dO\\nS are considered
uphft anc.horage tor the shear waJJ chord studs (ven.cal boundary members) and must comply wuh AJSI
S400 Seeuons E 1.4.1.2, E2.4.1 .2, and E6.4.1.2. AISI S240 Seeuon F I reqUJres tests for hold-downs 10
be in accordance w1thAJSI $100 Section Fl (K J fOr SI00-16) where requjred to delermane their strength
and stiftne<s and pemtits the use of test standard A lSI 5913 to be uS<!<! for hold-downs. Figure 3- 18
shows a three-story staeked CFS-fmmed shear wall, including the ovenurnmg restraint syste-m, which IS
representauve of the Opuon I and Opuon 3 shear walls. as the Opuon 2 shear wall ts dtscon.ttnuous.
AISI 5400 Stttion EU.I.l " Rtquir<d Strength [Efft<l of Faclort<l Loads[ for Chord Studs,
.~ ncbl)ngt:, and Colltt'tors"'
Whtle collectors tn butldmgs braced enttrely by wood l1glu-frame shear walls do no1 have 10 be destgned
for the seiSJ11.ic lood combinations 1nclud1ng the overstrength factor l!t, per ASCE 7 Seetion 12.10.2.1,lhe
collectors and collector connecUOJ\S for bulldulgS us1ng CFS frame shear \vaJis are requ1red to have the
available (design) strength 10 resist tlte Jesser of the expected strength of the shear wall and the amplified
seismic load per A lSI S400 Sections El .4. 1.2 and E2.4. 1.2.
AJS I S400 Sections El.4.1.2 and E2.4.1.2 "Chord Stud.~"' (nrtital boundar)' mtmbus)
S400 Sections E 1.4.12 and E2.4.1.2 require tha!tbe shean,all CFS chord studs (verucal boondary
members) rutd the uphft anchorage have the a...allable strength to res1st the amplified se1sm1c load. For ease
of reference tn this design example. the term amplified :seismic food will mean the Jesser of
I. The load detenn&ned from the ASCE 7 seismte load combinations wtth lhe overstrength factor Oo2. The e.xpected Sl!englh ofthe shear wall (1'.
walls per AJSI 5400 Section El .3.3).
X
n,), where n,= 1.8 (for WSP-shealht<l shear
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
Ftgun 3-17. Ovtrlummg nsJramJ in stacked CFS-fromtd shear wall
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
S1nce the floors and roof diaphragms are considered to be flextbfe 1n lhis example-, the overstrength faeroe,
IJo, is equal 102.5 perASCE7Table 12.2 -1 F001n01e b.
The available/reqUired stre1lgth of the collectors. chord studs, other venical boundary eleme.us., holddowns and anchorage connected to the shear v,-an. and all other compooenLo:; and oon.nectJons of the shear
wall shall be greater than or equaJ to the amplified seismtc load.
Typ1cally, designing for the expected strength of the shear waJI will not govern over the seism1c load
combtnauons., utdudltlg !lo- llus will be shown later ln the des•gn example.
Sbnr \Vall On.rturnlng f ortt-ll
Tables 3-26 and 3-27 show the amplified se1smtc ovenurmng forces for the three shear \\'all opt1on.~. The
overstrength facror, flo. lS 2.5 su'ICe Bmldtng B I has ftex1ble floor diaphragms and roof diaphragm. The
overstreng.t.h factor,~ lS used to determine the requtred suength that the CFS chord studs and continuous
rod tie-down system ·s nominal strength must equal or exceed.
Tablt 3-16. Shtar wall Ol'trturning moment (OTli/) Oplitm I and OpliOII 3
LRFD
Level
l:Shear
Wall
Wall
Heigh! (h) Force (F)
(11)
(I b)
ASD
S1ory-
l..e\•el
0,
OTM
OTM
(k- ft)
(k-fi)
Wall l: Shear Wall
Hetglu
Force
(h)
ASD=(0.7F)
(I b)
(ft)
II
9849
108,339
2n<l Floor
10
21,990
328,239
820,598
10
1st Floot
10
28,061
608,849
1,522,123
10
Jrd Floor
270,848
II
6895
S!OryLevel
0.
arM
OTM
(k-ft)
(k-fi)
75,837
189,593
15,393
229,767
574,418
19,643
426,197
1065,493
L OTM th~d ftoor =It xF= II x 9849 = 108,339 ft-lb = 108,339 ktp-fl
arM seron<l ftoor=CJTM third ftoor + (ltx F)= 108,339 +( 10 x21,990) = 328,239 ft- lb
= 328,239 ktp-ft
arM fits! ftoor= OTM secon<l ftoot + (ltx F)= 328,239 +(10 x28,061) = 608,849 ft-lb
= 608,849 ktp-ft
2. The story LRFD seismic Ioree is 9849 pounds, 12,141 poun<ls, ruld 6071 poun<lsa1 the 3rd-, 2nd-, and
lst-ftoor levels, respecuvely, per Table 3-21.
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
Tab/eJ-17. S/war wal/OTMOption 1:
(JI~l/length
L: 3rd. 2ndfloor= 37 ftet. 1stfloor= 15fttt)
LRFD
Level
l: Sheat
Wall
Wall
Height(h) Force (f)
(ft)
( Ib)
ASD
Wall l: Shear Wall
lleogho
Force
(Jo)
ASD=(0.7F)
(ft)
(I b)
Story-
Level
0.
OTM
(k-ft)
arM
(k-ft)
StoryLevel
OTM
(k-ft)
0.
OTM
(k-ft)
3rd Floor
II
9849
108,339
270,848
II
6895
75,837
189,593
2nd Floor
10
21,990
328,239
820,598
10
15,393
229,767
574,418
L-R:
608,849
L- R:
1,522,123
L- R:
426,197
L-R:
1.065.493
R-L:
280,610
R-L:
701,525
R-L:
1%,643
R- L:
491,608
1st Floor
10
28,061
10
19,643
A:(laJ
AXJal
Level
1st Floot
Col. Ht.
(ft)
l: Shear
Force
(P)
(I b)
10
0
8872
Force
( Ib)
Col.
Ht.
(ft)
(P)
( Ib)
IV
1: Shear
22,180
10
0
6011
15,028
IV
I. L-R = setsnuc lateral forc.e applied left to rtghl
2. R-l = seismtc lateral fOrce applied right to left
3. OTM third floor= loX F= II X 9849= 108,339 fi-lb
OTM second floor =OTM thord floor+(h xf)= 108,339+ (10 x 21,990)
= 328,239 tl-lb
OTM first floor= L- R: OTM second floor +(h x f)= 328,239 +(10 x 28,061)
= 608,849 ft-lb
OTM first floor= R-L: (h x f) =(10 x 28,061)= 280,610 ft-lb
First-floor column axoal = (OTM second floor/L) = (328,239/37) = 8872 lb
-
,.
-
10'
~Column
10'
T
2s·
1 1
12'
Figr.ut! 3-18. Optitm 2 sh(!ar n·a/1 R-L OTM and reactions
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(lb)
Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
Sbur \Vall Onrturnlng Rtsisting Fortt
ll\e building. dead load counters a porttOn or some-limes all of the she-ar wall uplift. lbe shear waH resisung
force will use lhe floor dead lood if the framing is perpendicular to the ·wall~ and n wdl also use the ·wall
self-we1ght. The metaJ-pJate-eonnected wood roof trusses are framed parallel to thi: shear wall, so the roof
dead Joad does nm cootribute significandy to reducmg the shear wall upl ift.
ASCE 7 Stdion 12.4.3 Sti..limic Load Efrttts, intluding Ovtrstrtngtll; Stttions 2.3.6 and 2.4.5
Basit Combinations with Stismit load Eff«ts
Only a portion of the dead load c-an be used for resist1ng shear wall ovenurn1ng forces. and for lightframe wood-frame construction, the o•.rerturning. resisung forc-es would be based on Secuon 2.3.6
equauons.
ASCE 7 S•<tions 2.3.6 and 2.4.5
LFRD:0.90 - £,+£.,.
[(0.9- 0.2S.,.)D+ pQ,+ 1.6HJ -> (0.9 - 0.2S.,.)O= [0.9- 0.2{1.117)]0 =0.6760
ASD: 0.60-0.7£, + 0.7£"'
[(0.6- 0.14S,..)D+0.7pQ,+ H)-> (0.6 - 0.14Sos)D = [0.6- 0.14(1.117)]0 =0.4440
Per AJSJ S400 Sec:uon EJ .4. 1.2. the chord studs and uphft anchorage must have the a\1allable
strength to resist the lesser of( I) the e.xpeete<l strength of the sheM \\all or (2) the load determined
us1ng the ASCE 7 se1smic load c-ombu-uuions. including lhe overstten.gth fac-tor~ Uo.
ASCE 7 S•<tion 12.4.3
LRFD: [(0.9 -0.2Slfi)O+ Q,Q,+ 1.6HJ -+ (0.9- 0.2SbS)D = [0.9- 0.2(1. 117)]0 =0.6760
ASD: [(0.6 -0.14Sns)D+0.7(!loQ,) +H)-+ (0.6 - 0.14S"')O= [0.6- 0.14( 1.117)]0 =0.4440
In e1ther case. the dead load used to restst the overturning ts the same, but the uplift loads w1U be
more mgni6cam when the ove.rstrength factor. t1o. 1S lOc:luded Of uplift fOrces are derrved u.,c;ang the
expected shear s1rength of the shear \\'ails.
Aulhor's Distussion--load.li tht Systtm Can Oelivtr
In ptevious ediuons,. ..loads lhe System Can Deh,•e-r" v.a~ conside.red as one load rombinauon. That
procure has been disconunued. The shear wan. ilS compot~ents~ and its collectors shall be able to resist the
lesse-r of the shear v.<l.ll expected stte.ngth or loads detemuned USlng the seism1c load oombitlaUons \\1th
the overstre.nglh factor. It is no longer allowed to be designed to a lesser cap-acny~ as ut the case. where the
butldtng system fram1ng e-lement (say dtaphragm) cannot de-liver lhat large of a design force to the shear
wall~ 1ts component.;;, or its collectors.
Aulhor's Distussion--Otad Loads
There are three methods destgners u.-.e to detemune the amount of dead load along the shear wall that can
be used to C:OU11ter the shear \\<l.ll uplift fOrce.
I. Neglect the dead load for conditions when 11 is mtn1maJ (example: ftoor fmming parallel to
shear ·wall), wh1ch 1S lhe most conservative approach for the uplift stde, but not the c:ompresston
s1de oflhe shear wall.
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
2. Cons1der the shear wall as a ngid body where the dead load is used to reslstlhe uplift and is
used as part of tlte gta\'ltY load demand m addttion to the ovenumjng compression load. Some
JUStifY thts usmg a deep-beam analogy.
3. Constder the rtm and top track as a beam on an eJa.o;tic foundat1on. Such an analysis shows that
typically only the dead load W1tl11n a few srud bays can be mobtlrzed to resist uplift. Further
discusston c.an be found in FEMA 451 NEHRP Recommtnded Pro1·islon.t: /Hslgn Examples
Chapter 10 as weU as m the book Structural De.ttgn t(Lon--Rt..te Buildmgs. Section 6.7.
ThiS destgn example uses the rtgid body method although good argument.c; could be made in suppon of
the beam on elastiC fOundation approach. Therefor~ the dead load of and ttibutary to the wall is used to
resist the uplift. ru'ld tt ts also used as pan of the gravny-load demand to the cl\ord-stud a~mbly oo the
c.ompressK>n stde of the she.ar wall.
Table 3-28 summanzes the wall and floor weights to be used for calculating the shear wall resisung
moments.
Table 3-28. Shttu 1.-all OTMdestgn mformatJOn
Resisung Dead loods
Trib.
Wtdth to
Dead
level
Wall
Height (/z)
(f\)
Shear Wall
(ft)
load
(psi)
3rd Floor
II
2.0
16
Self-wetght
Total We1ght
:!: Tolal
Wetght
(pi I)
(IV_,=hxwt)
(pi t)
(IV.,..,)
(pi I)
(Wro)
(pi t)
32
llxll=l21
153
617
1081
Dead load
to Wall
(W,._)
Shear Wall
2nd Floor
10
11.8
30
354
lOx II= 110
464
1st Floor
10
11.8
30
354
lOx II= 110
464
153
I. Shear Ylall wetg.ln = II psf of wall surt'llce area.
2. Me.tal-plare-cont'lected wood rooflniS:SeS at24 lnches on center.
Tables 3-29, 3-30. and 3-31 show the res1sting dead-load mome-ms (Rl\•t) for the Lhtee shear wall options.
Table 3-29. Optto11 I shear wall (wall length= 37 fl}
Reststing Dead Load Mome.ms (Rl\1)
:!: Shear Wall
Level
3rd Floor
Wall Height
(h)
(ft)
II
l:DeadLoad
(fV,;)
(pll)
Restsung Moment
(RM)
(ft- lb)
LRFD
0.676RM
(ll- lb)
104,728
285,499
187,517
500,202
328,535
2nd Floor
10
617
1st Floor
10
1081
739,944
I. ResiSting momem = (W,)(L x L)fl = ( 153)(37)(37)(1 = 104,728 lb.
210
~6,499
153
422,336
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70,796
ASD
0.444RM
(f\-lb)
Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
Table 3-30. Op110n 2 .~h~ar wall (wall l~ngth = 37fi and 15/t)
Res1sHng Dead Load Mo1nents
:!: Shear Wall
Wall Height
Level
(h)
(fi)
Resisting Moment
:!: Dead Load
(11',.)
LRFD
0.676RM
(ft- lb)
ASD
0.444RM
(ft-lb)
104,n8
70,796
46,499
(RM)
(ft- lb)
(plf)
3rd Floor
II
153
2nd Floor
10
617
422,336
285,499
187,517
1081
L-R: 337,812
R-L : 567,336
228,360
383,519
149,988
251,897
1st Floor
10
Table 3-31. Op110n 2 .~h~ar wall (wall l~ngth = 37fi and 15/t)
Res1sttng Dead Load Column Axmllood
Colwnn Axlal
le\•el
1st Floor
Col. HI.
(ft)
10
I. ResiSting mome.nt
(lb)
0.676
Dead Load
( Ib)
0.444
Dead Load
(lb)
3702
2502
1644
:!: Dead Load
Force
(plf)
617
= (W,.)(Lx L)/2 = (153)(37)(37)12 = 104,n81b
= (617)(37)(37)12 = 422,3361b
2. Resostlng moment (L - R) = (W,.)(L)(L/2)
= [(1081)(25)(2512)= 337,8121b
3. ResiSting moment(R - L) = [(W10)(L X L)/2] + [(( 1212) + 25)(617)( 12)]
= [(1081 )(25)(25)12)] + (229,524) = 567,3361b
4. Column dead load
= (W,.)( I2)12 = (617)(12)12 = 3702 lb
-.-------.
11 '
-1-------t
10'
-
1----.----t
10'
1
Agu~
25'
l
12·
T
3-19. Opt1on 1 shear wall
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21 1
Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
Table 3-32. Op1ion 3 shear wall (wa/1/englh =25ft)
Reststing Dead Lood Moments (RM)
l: Shear Wall
Level
Walllletght
(h)
(ft)
l: Dead Load
(W,;)
(pit}
Res:tSung Mome-nt
LRFD
ASD
(RM)
(ft- lb)
0 .676RM
(ll- lb)
0.444RM
(tl-lb)
3rd Floor
II
153
47,8 13
32,321
21,229
2nd Floor
10
6 17
192,8 13
130,341
85,609
1st Floor
10
1081
337,8 13
228,361
149,989
I. ResiSting moment= (Wro)(L x L)fl = ( 153)(25)(25)(1 = 47,813 lb.
Sbur \Vall Ovtrturning fort-t~Using Expt('ltd Sbur \Vall Capadty Desjgn
The ovenurning forces destgn c.heck is hmited to lRFD only as lhe expected capacny of shear wall
sheathing considers factored loods. To dernonsuate that lood combtnations:-~ incluchng ilo, g_enemte design
forces less then lhe sheathing capacity~ tncludi.ng nt~ see Table 3-32A. The provtded wall sheathing shear
c.apacuy selected by lhe designer wtll always equal or exceed the seismic applied forces fOr ASD or lRFD
to sausfy bu.tld1ng code reqmremenL'>. Basing the collector and boundary element design forces on the shear
waH shealhjng shear capacity, expected suength will always be greater (more consen•ative) than the design
setSmtc forces resulung from load combinattons. ineludu'lg OoTablt J-JZA. OpliOn I -<hear wall-comparison ofd-slgnforees (LRFD}
\Vllh Overstrength
Wllh Expected Strength
v. x flc = Shear force
Shear(F/L)
(pit)
Nail Specmg
(in)
V. x !l, = Shear fcrre
level
(pi f)
(pit)
3rd Floor
267
6
267 x2.5 =667.5
890 X 1.8 = 1602
2n<l Floor
595
4
595 x2.5 = 1487.5
1330x L8=2394
1st Floor
759
3
759 X 2 .5 = 1897.5
1775x 1.8=3195
0, = overstrength factor= 2.5
.QJ:' = expecte-d shear wall factor = 1.8
Thus, the arnphfied de.•ign loads, using the lesser of the two loading types, w~l be controlled by the load
combtnations Wllh Oct. a.~ thts wtll always be the smaller of the two. h v.'OUld be conservative to design the
ampli"fied design loads us.ing the expected strength of the shear waJI sheathing.
Sbu.r \\'all Condnuous Rod Tit-00\\U (Hold-Down) Fort:ts
AlSI S240 and S400 requtre shear waJis to have hold-downs (also referred to as ue-downs or uphft
anchorages) at the ends. Hold-dcmn.~ prevent the shear wall from uphtUng when the ovenurnang forces
e.xeeed the eftectave reststtng forces. The summation of lhe shear waJI 's effecuve ovettumtng resasung tbrce
and hold-down uphft capacny has ro exceed the ovenurnmg fo1'Ces acttng on the shear wall.
212
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
The amplified sets:mic ovenuming moments shown m the following tables are provided in terms of strength
Je\•el design or LRFD frocn thL~ point fOrward, as thL~ example de-fines the amphfied selsmic forces as the
LRFD setsmic load combinations Wtth the overstreng.th factor OoThe upltft fOrces are calculated using the dtstance between the conunuous rods (hold-downs. uplift
anchorages) that are centered 1n chord studs (venicaJ boundat)• members) at each end of the shear \\o'all.
The measured center of the compresston CFS chotd-stud pack from the end of the shear wa11will vary
de.pending on the number of chord studs, which typically lS governed by the magnitude of the ovenuming
compresstve axtaJ load. The chord Slud.S are spin symmemcally on etther side of the conunuous rod tiedown, typtcally wtth the gap width staning at 6 1nches so the components of the ue-down system may
be installed, to foi'M a concentric O\•enurning resuainL system for the shear Yo>aJI. The CFS chord studs
are placed first near and on each side of the conttnoous rod and then,. as the overtllt'ning compressioo
force 1ncreases in each lower story. CFS chord studs are added symmetrically to the out(iide of the CFS
chord studs on euher side oflhe conunuous rod. The center of the conunuous rod IS also the center of the
compression chord-stud assembly.
ll\e width oflhe c.hord-stud assembly at the lowest level of the stacked shear waH assembly will [)'pically
govern the locauon of the continuous rod and compression chord studs at the upper-floor le\•els. Howe,·er.
the upper-Hoot bndge-block demd may govem the gap wtdth by influencing the chord-stud gap width used
between the chord studs in IO\\'eJ stories tfthe enganeer lSaligning CFS chord studs from srory-to-slOry.
W11en the force that tl\e cont'muous rod tie-<10\vn resists tS relatively small, 11 may be reasonable to asswne
the center oflhe oonunuous rod and CFS chord-stud pack is located 6 mches tn from the end of the wall.
When the force that the conllnuous rod ue-down resJsts ts Large. asswn1ng the ceme.r of the CJ-""S chord sruds
1s 12 inches m from lhe end of the \\o-'<111 is probably more reasonable. TI\e width of the CFS chord-srud pack
varies., depe,ndtng on lhe location and tnelhod of uphft load transfer from the continuous rod to the CFS
chord studs and the required strength c.onsidenng O\rerturning and gravny loads. The uplift trano;fer me.thod
utali2es ea~r a steel beanng plate nested 1n the stud-wall bottom track. or a Slee.l beartng plate on a bndgeblock element placed bew.--een the CFS studs several feet above the stud-wall CFS bonom rrack. The CFS
chord-stud assembly w1dth often ranges between 12~ to 30 tnche.o; from outside edge 10 outside edge. A
wide-r cho!'d-stud-pack Width typically occurs where a bndge block is used fbr the uphft load transfer.
Some engtneers may want to tnstallthe steel bearing plate below the CFS bouom uac.k so as nm to have to
cut the CFS chord studs at a ddferent hetght than. other CFS wall SlUds. but then the floot sheathtng would
need ro have adequate depth for the stee-l beattng plale, and shear transfer from the diaphragm to the shear
waJI below around the stee-l beanng plate would need to be detailed.
Tables 3-33. J -34, 3-35A, and 3-358 show the amplified seismic uphfi forces for the three shear waJI
opttons using different end distances 10 the center of the CFS chord studs. The non-amplified seismic uplift
forces are included fOr the Option 3 shear wall since they are required for calculating the CFS-framed shear
wall deftecljons, which are based on oon-runpli'fied LRFD loads.
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
Tablt 3-33. Option I shear wall
Amplified Uplift Forces (LRFD)
level
Wall Length
(L)
(II)
Wall Length
(L,)
(ft)
l: Shear Wall Oo
OTM
(ft-lb)
0.676RM
(ft-lb)
Story-Level Uphft
(I b)
3rd Floor
37
36
270,848
70,796
5557
2nd Floor
37
36
820,598
285,499
14,864
1st Floor
37
36
1,522, 123
500,201
28,387
I. L£= L Etfecuve = Dtstance be1weeneenterhne ofhold-dO\\'nS =(37ft - 6 m- 6 m =36ft)
2. Uphft = (OTM -0.676RM)I(L,) =(270,848 - 70,796)136 = 5557 1b
3. Uphft = (OTM - 0.676RM)I(L,) = (820,598 - 285,499)136 = 14,8641b
4. Uphft =(OTM- 0.676RM)I(L,) =( 1,522, 123 - 500,202)136 = 28,387 1b
Table 3-34. Option 2 .Wear wall
Amplified IJploft Forces (LRFD)
Wall Length
Level
Wall Length
(L)
(ft)
(ft)
l: Shear Wall 0o
OTM
(ft-lb)
3rd Aoor
37
35.5
270,848
(L,)
0.676RM
(fl-lb)
Story-Level Uplift
( lb)
70,796
5635
2nd Floor
37
35.5
820,598
285,499
15,073
1st Floor
25
25
23
23
L-R: 1,522,123
R-L: 70 1,515
228,360
383,519
56,25 1
13,827
I. L-R =Seismic force
1n the
le-ft to rtg.ht directron. R-L =Seismic force tn the rtght to left direction
2. L, = L Effective= 37ft - 1.0 fl- 0.5 ft = 35.5 flat third and second floors; L, = 25 ft - I f\ - I ft = 23 ft
:u first floor
Table 3-35A. Optiou 3 shear wall
Amplified IJploft Forces (LRFD)
Wall Length
Level
Wall Length
(L)
(ft)
3rd Aoor
25
23
2nd Floor
25
1st Floor
25
l: Shear Wall 0o
OTM
(ft-lb)
0.676RM
(fl-lb)
Story-Level Uplift
( lb)
270,848
32,321
10,371
23
820,598
130,341
30,011
23
1,522,123
228,361
56,251
(L,)
(fl)
I. Uphf\ = (OTM - 0.676RM)/(L,) = (270,848 - 32,321)123 = 10,371
2. L, = L Eftective = 25 fl - I tl- I ft = 23 fl
214
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
Table 3-3j8. Option 3 shear wall
Non- Amplified Uplift Forces (LRFD) lor Shear Wall Deflection
Level
Wall Length
(L)
(ft)
Wall Length
(L,)
(ft)
l: Shear Wall n,
OThl
(ft- lb)
3rd Floor
25
23
108,339
32,32 1
3305
2nd Floor
25
23
328,239
130,341
8604
1st Floor
25
23
608,849
228,36 1
16,542
0.676RM
(ft-lb)
Story-Level Uplift
(I b)
I. Uplift= (arM - 0.676RM)I(L.,) = ( 108,339 - 32,32 1)123 = 3305
2. L,=LEtlective=25 fi -1 ft-1 fi=23 fi
Author's Distussion
The uplafi forces are rather large due to the use of lhe load combinations wnh the system oversuenglh
factor. 0,. The upltft forc~es on tJle bmlding's mteriot transverse walls are generally smaller than forces on
the perimeter shear waHs due to the Interior transverse walls typically being longer than the penmeter walls.
The bullding•s peruneter shear walls are typicaHy shorter due to the presenc.e of doors and windows. The
aspect rauo (hetg.htlloogth) of the penmeter shear walls ryptcally ranges be1ween 4: I and I:2.
Where the shear wall aspect ratio is approximately I: I or greater (example: 2:1. 4: I). the upl1ft fOrces at
the \\o'all ends can be sigmficant m a multistory light-frame sttuctute. If the Option 3 stacked shear walls
had been only 12 feet long (a<;pect ratiO = 10 feet: 12 feet. sheathing screw spacmg now at 2 mches on
cenrer, bolh Stdes a t first floor), the uplift design force at the podium le•'tl would ha•'t been [(I ,552,123 (77,832 x 0.676)]/10 = 149,950 pounds. This wotild likely be a difficult anchorage force to design and
develop tnto the structural oooc.rete poc:hwn. thereby most likely precluding the Option 3 shear \\'all bemg
just 12 fe<t long.
It lS very common to have shear wall upIift forces m the rnnge of 30,000 to 50,000 pounds for ASO in
mulustory light-frame wall systems. especially where the bwld1ng height mcreac;es from three stories to
five or SlX stories. The uplift loads are even greater where the setsmic-load combinauoos wtth the system
overstrength facror. flo. are used. Where the shear wall upl1ft loads become Large, the use of conunuous rods
\\1JIIIkely be reqUJred. as opposed to ustng cold-formed stee.l hold-downs screwed or bolted to lhe shear
wall CFS cbord studs.
12.5 OVERTURNING RESTRAINT (liE-DOWN) SYSTEM AVAILABLE STRENGTH AND
DISPLACEMENT
The upltR anchorage (also referred to a"i ovenuming restratnt, hold-do\\--n,. ot ue-down) system at each end
of the shear wall in tlus des1gn e.umple ts a conunuous rod tie-down system that e.xtends from the podiurn
le\•el to the top ftoor level. The conunuous rod ue-down system consists of a continuous rod and multiple
CFS chord studs placed on e1ther Side oflhe continuous rod at each floor leve-l. The conunuolJS rod ac.ts
as the tension component of the system and holds the uphn-tng shear wall down at each floor leo.•el and
holds n to the podJwn-level base. The mutuple CFS chord studs resi'il lhe compressi,•e axial uphft load
and typically bear against the wall top track and ftoor framing on the uphft end of the shear wall unless a
bt1dge block at ne.ar mid-heaght of the shear walliS used. In that cao;e, the CFS chord studs are fa~tened to
CFS ct~pple sruds lhat be-ar agatnst lhe br1dge block. as sho\.\1\ 1n F1gure 3-20. The chord studs also res 1st
the combined gravny and seismic ovenurntng compressive axial down load 011the compression side of the
shear wall.
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215
Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
:SHEAI' llr~~ ~O.floWIG
t:OOC SUICWS
Sl!n HU1
Slt(L PV.l"f 'I'IA.Sti(R
rAKt.-t.ll'" D£VIet
'>f(fl. g£AA;~:. F\lot£
...n..r;'lo<-f:J~!}~ --rm1+-ilttlt
\toiT-' IW}tJ"\.IIfftNJ
~
Flgunt 1-10. Deuu/ ofthe Jlurd-jloor bndge-hlock tenmnatl on oflh~ oontim.J()u.t rod ti~-down system
216
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
J" WN
CONTINUOUS
STEEL ROD
CFS CHORD STUDS
ATTACHED FACE TO F"ACE
-~--TIIn-,..,IF"'rnrlil''"~-c!FS CRIP<'L£ STUDS
ATTACHED FACE TO f ACE
STEEL COUPLER NUT
WITH WITNESS HOLES
WOOO <LOOR- - -,
SHEATHING
SlEEL S£ARING PLATE IN
CFS BOTTOM TRACK
lRO FLR
2ND FLR. ~-};~;;;;;til~
SCL RIM ---+-~
SCflEW CFS TOP TRACK
THROUGH 2X WOOD PLATE
TO SCL CONllNUOUS Rlld
8
CF"S CHORO
Sl UOS
9'" GAP Fort CONTINUOUS STEEL ROO
Cf S CHORD STUDS
SfCliOH B - A
Figure 3 -2/. 1)pica/ chord-sltld a.u tmbly oflht j/()()r lint. Con1muou.s rod Jut-down (hold-down) S)'Stem
chmt/ s1uds at 1Jw !tcond and third floor!.
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217
Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
The conunuous ue-down system design JS based on the Opdon 3 shon-smcked shear wall forces.
since it has the largest uplift fOrces. TI'Ie conunuous rod extends abo\'e the lturd floor~ btU terminates
at a bridge block several feet below lhe lh1rd-ftoor wall top track.
In a contu'lllous rod tie-d<mn system~ upl1ft caused whe.n a lateral load acts on a shear waJI is
typically resisted by a steel beanng plate attached to the conunuous rod at the floor above the
shear waJI or a bear1ng plate on top ofa bridge block wnhin the height of the shear \\"all~ as shown
tn Figure 3-21. l1\e be-aring plate or load transfer point m each fl()()( level resists the overrumJng
(uphft) for~e of the shear \WII below (someumes referred 10 as lncrenumJa/ bearing). wh1le the
continuous rod resistS the cumulative ovenuming force from the top of the continuous rod tie-d<m'n
system run down 10 the floor level under conside:muon. The lowest le...el m the des1gn example is
the concrete podJUm deck 10 whtch the conunuous rod 1s anchored and that resists the cwnulauve
ovenurn1ng force.
AlSI S400 requ1res in SOC B through F that component,. of the seismic-force-restsung systems (SFRS)
00{
pan of the designated e.nergy-dJSSip-ating system (e.g... chord stud~ tie-downs. shear a1tchorage.
and collectors). regardless of R-f3ctor used unless exempted otherw1se, have the available strength to
resist the lesser of( I) the expected strength of !he si"e:lr "all (O,V.) « (2) the load determined ustng
the seismic load combinauon.o; with the overstrength factor n.,. llus JS different from light-frame wood
structures where lhe shear wall chord (end) sruds, hold-down or 11e-down system, shear anchorage. and
eollectOfS are designed OJ'IJ)' tOr the ASD or LRFO seismic load. However. buildings are exempt from
havlng to cOt:npJy Wllh 5400 \\'here located 1.n SOC A. or located in SOC B or C and ustng an SFRS
w1lh an R equal to 3, or an SFRS \vith a higher R-fact()( but assigned an Rof 3 1n SOC B or C. But these
e.><empte<l boddtngsmust sttll comply wtth SIOO orS240.
The a railablt strength tS defined by the AlSI standard as the matenal nommal strength ( V.J mod1fied by a
resistance factor(~) when ustng LRFD or dtvtdtng by a fac tor ofsafety (!l) when ustng ASD. The values
for~ and (!l) are defined tn the AISI spectfications.
Available strength (LRFD) = V,.(9)
Avadabl< strength (ASD) = V)(!l)
Table 3-36 shows the continuous rod area required and the provtded strength and rod s.12es. The
continuous rod ts requlred to have the available strength to reslst the amplified seismic fOrce per AISI
5400 Secuons E1.4. 1.2 and E2.4.1.2. D1splacemeru of the overrurn1ng resuaint system (incluchng tension
rod e-longatJon., take-up device deflectiOn or the gap distance from senlement and/or shrinkage 1flake-up
devices are not used. wood crush.i11g. and bearing plale bertdmg) also needs robe determined lR order tOr
the engutee.r to determine 1flhe horizontal deflection at the top of the shear walliS lO accordance with the
ASCE 7 drift hm11 as well as the ICC-ES and/or local jurisdichon story "e:rucal displact"meJU hntit
218
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
Table 3-36. Continuou:r rod tie-down sy.tt~m n'!qwntd. pronded .\'tre11gth. and rod St.:.es
(Op11on 3: Shear wa/1)-LRFD
Level
S10ry-Level
Rod Net Tensale
31
FloorAmplified
Rod Gross Area S10ry-Level Atea Requtred' /
Se.ismtc Area Pro\•tded (A,.)
to-Floor
Seismtc
Reqwred'''IArea
Height Uphft ForceHI Prov1ded (A,)
Uplifi
for Venacal Story
Foreef.!' (P)
(L)
for Strength
Dls:placement(6'
(P..,.)
(In')
( Ib)
(in')
(It)
( Ib)
Rod Diameter
Provided ( 1n)
(rod tensile
strength l ksi J)
3rd Floor
II
10,371
0.31&10.442
3305
0.063A:U34
~(F.= 58
2nd Floor
10
30,011
0.92010.994
8604
0. 15010.763
ll-I(F,= 58 ks1)
8604
0.15010.334
Y. (F, = 125 ksi)
16,542
0.28810.763
1% (F.= 125 ks1)
2nd Floor
10
30,011
0.42710.442
1st Floor
10
56,251
0.80010.994
ksi)
I. StOI')'-Ievel amplified seastnic upltft force ts from Table 3-35A.
2. StOf')'-lt\'el .seismtc uphft force is from Table 3-358.
3. Rod matenal:
ASTM A36
ASTMAI93B7
F, = 36 ks1
F, =58 ksi, E = 29,000,000
F,= 105ksl
F,= 125kso,£=29,000,000
4. Rod 3\'llllable strength (LRFD) = 0. 75(0.15F,)A., so A.,.= P...,/(0.56F,) from AISC 360 Equation JJ-1 ,
Table J.3.2, and corresponding commentary.
5. System ven1cal story displacemem lamtt = 0.170 = 0.20 tnch- 0.030 ioc.h take-up devace 1nitiaJ seatmg.
t.., and design deftection, t..., per manufacturers evaluation report (ASD) x 1.4 = 0 . 170 x 1.4 = 0.238
lOCh (LRFO). System vertical story dtsplaceanent hmit 8 0.238 inch.
==
f>=PUA,_.E
A,.= PL/0.238(£)
6. Rod ne-t let'!Sile area tabulated tn the 15th Editton ofAISC St~l Con.tfTUCiiOII Mam1al Table 7-17 or per
the followtngequation: .4,. = 0.7854 x [d-(0.9743/,)f \vbe.r en =threads per tnch and d= rod d1ame-ter.
Author's Discussion
Conttnuous rod Lie-down systems are not only designed to have the stte.n.gth to re.s1st lhe setsantc
ovetuJmtng force but must also be destgned to ensure compliance. wath the code se.is:mte-drtft Iimtt and the
local budding jurisdiction ·s story-elongation ltmu. The allowable story drift hmit is shown ln ASCE 7 Table
12.12- I and is taken as 0.02Shu for lhts design example. Devices that e.xpand to take up the gap between
tl1e bearing plate and the continuous rod nut are caUed wkr-up (shrinkage compe.nsating) device.t~ and lhese
are often used in mult.is[Qry buddtngs to ensure the O\'erturmng resuauu system JS tig!U.
The JCC-Evaluauon Servace·s publtcaJiy developed acceptance criterta tbr shnnkage compensaung devices,
AC316, has a 0.20-mch total vertical d&splacementlinltl, including the steel rod elongalion a.t'ld the take-up
device deflection, us:ang the more restricti,•e rod length at each stoty or between restraims at ASD. The
0.20-tnch hmn needs to be confirmed with lhe local junS(hcuon where the buddtng project occurs as tt may
have a more stnngent hm11 such as 0.125 tnch. Conunuous rods should be cons1dered standard strenglh
unless clearly ident:ifi:ed with markmgs as htgh-suength, and specialtnspecnon should be required to
ensure high-sue.ngth rod IS: used where specified. Also, high-suength rods and coupler nuts are usually not
weldable, but may be tf cenatrt supplementary requlreme;nts are followed (e.g., ASTM F I554 Grade 55
wnJ1 Soppletnentary Requ1remem S I).
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219
Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
The use ofsmallerdiameter high-strength rods. SU('.h asASTM Al93 87 with F. = 125 ksi. will elongate
more than larger-diameter standard-strength rods, such as ASTM A36 with F. = 58 ksi,. when sized for the
same load and so Wlllmcrease the top of sltear wall dnft wath the1r use.
Coudnuou.s Rod Tir.Ooy; a Slttl Buri.og Plait Dtsi~rn
As the shear wall attempts to uplift, the uplift force is transferred through compression of the CFS studs
located under the wood framu"tg (floor blockulg_ floor ram jot~ wall bndge block) and steel beanng plate.
A nut Of lake-~a•tce nut is used to tmnsfer the uphft fofce from the bearing plate to the eonltnuouo;: rod.
The steel beanng plate is design«! for bolh bendtng and shear.
The "-ood fra.antng often becomes the weakest elemem of the conunuous rod tie-down system si.n<:e atlS
typtcally governed by compression perpendtcular to gratn. The wood run JO ISt and blocking wall fi·aaning
is used as a load transfer eJeme:ntto help spread the co1npression loads under the steel beanng plate to the
ends of the CFS chofd-stud pack under tl'!e floot/roofwood franung.. The steel beattng plate width and
Length ate des1goed and fabt'icated to tit within the CFS wall uack. accowutng tOr the CFS bottom-uadc
md1us and some consttucuon erectton tolerance. When a bndge block liS used wtth.tn the shear wall, the
steel plate is simdarl)' designed and fabncated for some consU'llction erection tolerance and to not extend
o,-er the edges of the bridge block.
The steel bearing plate tS destgned to have the a\•ailable strength to resist the amplified overturning uplift
load (differe1lttal or story-uplift load) in the sltear wall below a1 each floor le,-el or load uansfer point So
for this design e.xample-~ the followtng occurs:
I. ll'!e steel beanng plate at the th1rd-floor wood brtdge block resists the ovenumtng uphft from
the th1rd-floor shear wall.
2. ll1e steel beanng plate tn the third-ftoof CFS-fi"amed.wall botto1n track resists the ovenurmng
uplift fbfce from the second-floor shear wall.
3. The steel beartng plate m the second-8001' CFS-framed-\\'311 bottom track reststs the
ovenumu1g. uplift force ffom the first-floor sltear \vall.
Table 3-3 7. Oplion 3 Wt~pltjied design forces (OJ and tie-rod /Maring piau St:mg- LRFD
Shear
Wall
Level
Floor-to-Floor
Hetgllt (L)
(ft)
Story-Level
Uplift (J'!4)
( Ib)
Uplift Loads
( lb)
Difi"brentlal
Beartng Plate
Level<'•
Rod Diameter
Pmvtded ( ut) (nominal
d1ameter. rod area)
Jrd Floor
II
10,371
10,371
3rd Hr. bridge blk.
Y. (0.442) (F. = 58 ksi)
2nd Floor
10
30,011
19,640
3rd flr. b. uack
HI (0.994)(F. =58 k.<i)
2nd Floor
10
30,011
19,640
3rd ftr. b track
y. (0.442) (F. = 125 k.<i)
1st Floor
10
56,251
26,240
2nd fir. b track
I V.(0.994)(F.= 125 ksi)
I. b lt3Ck = CFS stud-wall bottom track
220
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
The steel bearing plate at the second-floor CFS bonom track is shown below. The des~gn is simiLar for lhe
other steel beartng plates at the lhird-floor CFS bonom trac.k and third-floor bridge block.
I. Bearang plate wod:: space:
a. CFS stud stze and spacing.: f:H.nc.h CFS studs at 16 inches ott ce11ter
b. Clear ruea between CFS chord studs ond traek ftanges = ( 16 1n- 1.625 in)(6 m- 0.25 in)=
82.66 in2
c. Wood block or nm joist: Douglas Fir-Larch, Fu == 625 pst
d. Conunuous rod st:ze = 1.125-tnch diameter
2. Beartng plate rorees--second-floor Jeve.l:
As wood bearing perpendiCular to gnun and the steel bear1ngplate are part of the CFS-
framed shear wall overtum1ng-restr.unt (hold-do"n) system, AISJ S400 Secuons El .4. I .2 and
E2.4.L2 require that they must have lhe available strength to resist the minunum of( I) lhe
expected Sltength of the shear wall and (2) the load determined us1ng the ASCE 7 seism1c load
combtnaltons wnh the 0\'e:rstrength factor. AWC NOS Section 4.2.6 nmes lhat wood reference
compressiott design values perpendicular to gram (Fa) are based on an average compression
resistance at a deformatton of0.04 anch. NOS Equation 4 .2 -1 is used 10 calcuJare the
compresston des1g.n \'alue perpendicular 10 gr.:un for a reduced defonnauon level of0.02 tnch:
FclA&l =0.73Fu
The ASD value IS then tnuluplted by Kp equal to 1.67 and 9 equal1o 0.9 to obtajn the desagn value
tor LRFO, as can be seen 1n AWC NOS Table 4.3. I and sbo\\n here:
DouglaJ Flr-Uuch P(*rpendicular-lo-Grtun ~formatJo" and Fcl.. Dt.ttgn r&Jues
Approxunately 0.02 mch of crusl'\fdetOrmation
Approxamately 0.04 tnch of crushfdefonnataon
ASO:
625 pso
625 X 1.67: 1044 ps1
456 psi
Nominal: 456x 1.67=762pso
LRFO:
456x 1.67x0.9:685ps1
625 pSI X 1.67 X0.9:939 psi
The quest10n arises regarding tl'le app(Oprtate appl ic:ation of the AISI S400 required strength
pr0\1isions to the wood bearmg design values based ott a serviceability limit stue.
While AISI requires that the shear wall chord studs and uphft anchorage have the available strength
to ~ist the amplified setsmie loads (primarily to prevent CFS open-shape. sangly symmetric studs
from buckltng under compress1on loadtrtg. but also helpful1o preclude premarure failure oftlte
tens1on anchorage). th.is is not a requtrement for Y..'OOd design. If the ftoor jotst framing. or \vall
beanng bndge had been a CFS member 1nstead of wood framing or Slt\Jctural stee.l, these me.mbers
would ha\'e to develop the amplified seistnic loads.
Using e.1ther the ASO or LRFD seismte loads. the bearing plate in this desagn example shouJd be
st2ed such that the resuJung beanng pressure is equal to or less than the selected wood-compressiooperpendicular-t<>-gr.un value (F" )that has a stanmgASO base value of625 psi. Usmg LRFO, tl1e
wood ASO Fa. value tS multiplied by 1.67 and 0.9 per AWC NOS Table 2.3.5 and Table 2.36 (or
Table 4.3. 1). Per AWC NOS Appendix Table N3, time-effect !actor.; (A)equal 1.0 when used in load
combtnaltons Including eanhquake, so there is no additionaltncrease allow·ed for wood beartng
capacity when desig.n forces ine.lude earthquake.
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
Neithel' load-duratton (Cv) inc:l'tases fol' ASD nor 11me-etTect fac:tots (h) for LRFD ate permiued
when evaluattng wood framing members for pcrpendicular-to-gratn loading per NOS Table 4.3.1.
However~ a be.artng area factor~
is appltc.able tor be.anng lengths up to 4 tnches measured parallel
c..,
to gratn per NOS Section 3.10.4.
E.xamplt Forte:s
I. LRFD differential non-amplified "'"smic uphft force= 16,542-8604 = 7938 lb
a. Non-amplified seismic uphft faeces are from Table 3-358 ofthtS d~ign e.x.ample
2. ASD difiercnual non-atnplified seism1c uphft force= 7938(0.7) = 5557 lb
a. 11te use of 0.7 ao; a fac:tor to oonven from LRFD to ASD is not exact, but it is considered close
enough for design purposes because the diffe1'e1'1Ce tn differenual ampli'fied seismic loading
between ASD and LRFD ts expected to be Jess lhan three percent.
3. Beanng-plate hole size tOr continuous rcxl: ntf-14
=n(I'A + Yl•)z/4 =1.11 tn
2
a. A standard hole is used in the bear1ng plate whose diameter is de-fined as the conunuous rod
diameter plus y., tnc.h (0.0625 tnch)-see NOS Section 12.1.3.2.
b. llte use of EquaiJon NOS 3.10-2 is permisstble, but is tgnored tn this calcuJaLton as plates often
ha\'e dunension.o; equaJ to or larger than 6 1nehe~ e\1ett though this e.xample shows a plate size of
J m x 5 1n. The 1ncrease ln CilJJacny per Equation 3.10-2 is not that muc.h..
S.oriog-Piatt Si2lng for Wood (ASD)
I. Plate area (m1nimwn): 5557/625 = 8.89 in2 < trJal plate size= 3.0 in x 5.0 tn = 15.0 in2
2. Wood bearing suess: PIA~,= 55511(15.0 - 1.11) =400 pSI< 625 ps1
3. Plate shear. V = P(A-)12 = 400 X ( 15.0- 1.11 Y2 = 2718 lb
4. Plate bending: M=400 psi x (3.0 inXS.Ofl) x(5.0/4) = 3750 in-lb
Bt~riog-Piatt Sizing
for Wood (LRFD)
K,= 1.67,,=0.9, 1.= 1.0
I. Plate area(m1n1mutn): PJ(F,_K,~) =7938/(625 x 1.67 x 0.9) = 8.45 in'
Trial plate st.ze = 3.0 in x 4.0 in= 12.0 inz > 8.45 in2
2. Wood bearing stress: PJA- =7938/(12.0 - 1.1 I)= 729 psi< 625 x 1.67 x0.9 = 939 psi
3. Platesbear: V.,=P.(A-Y2=729x( I2.0 -1.11 )12=3969Jb
4. Plate bending: M.= 729 psi x 3.0 1n x(4.012) x(4.0/4)=4374 1n-lb
For the ASD and lRFO plate bending-moment calculaiJons. the area of t11e hole IS conservatjve.ly not
subuacted out to simply the caJculauon.
222
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
Stttl Btaring-Platt O.sign (AJSC 360)
ll\e steel-plate yaeld suength lS F1 = 36 ksi The plate design tbr be-nchng will C-onsider the oonttnuous rod
ho1e in plate. A standard hole is assumed, where the hole"s diameter is equal to lhe rod diameter plu.~ Y,.
inch (0.0625 .nch) per AISC 360 Table JJ.3.
I . Plate lhicknes:s-Fi e.~re
AISC 360 §FI I "Rectangular Bars and Roonds"
AISC360EqFII- I
ASD (0 = 1.67)
LRFD (~ = 0.9 AISC)
M,.IO>M-
qM.>M**-
Z reqwred = MQIF1
Z requ1red = (Miq)IF1
= 3750(1.67)136,000
= (437410.9)136,000
=0.135 inl
=0.174 in 1
Z= !xi'/4 b = 3.0- (1.125 + ~>~•)
Z= lxi'/4 b= 3.0- (1.125 + Y,,)
b= 1.813 in
b= 1.813 on
r=J4Zib
r=J4Zib
= J(4)(0.174)/(1.8 13)
= JI4XO. JJS)/(1.813)
= 0.620 tn
=0.546 In
Use 0.625-mch-t.htclc plate minamum.
Use 0.625-mch-tJuck plate mmlmum.
1
1.6M1 = 1.6Fy5,
1
S, = blr 16 = 3(0.625) /6 = 0.1953 on'
= 1.6(36)(0.1953)
= 11.25 1n-kop
M, =F,Z= (36)(3)(0.625)'14 = 10.55 in-ktp < 11.25 in-k1p ... OK
2. Plate lhickne~She<>r
AISC 360 §G I and G2
Shear is typically not a problem for the beatmg plate, butu is checked for completeness of the
bearang-plate desagn.
ASD (H = 1.67)
LRFD (~ = 0.9)
v, = 0.6F_,A..C.
AISC 360 Eq G2- l
Cr= 1.0 for flat plates
1
A,. = 3.0 in x 0.625 in=- I .88 in
v. = 0.6(36)( 1.5)(1.0) = 32.4 kips
~~~=
0.9(32.4)= 29.2 kipS> 3969~. 7 = 5674Jb
V,/0=29.2/1.67= 17.5klps>39691b
It is acceptable to use.: ASD: a 3.0- incll x 5.0-inch x 0.625-inch-thick bearing plate
LRFD: a 3.0-tnch x 4.0-tnch x 0.625-inch-lhick bearing plate
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223
Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
A smaller 3-inc.h x 3-tnch plate could be used if the c. increase
Jn allo\~o'able beanng \WS considered. The
piare would have been the same thickness and the deformation of the wood would be a linle more than 0.<»
tnch due to the smaller beanng area.
While LRFD allows for smaller beanng plates, the resulting sill plate crushing defonnauon is greater than
the 0.02 and 0.04 crush hmits established for ASD service values. LRFO uses KF to adjust the £-va1ue for
stre.ngth destgn, resulting in larger perpendteular-t~gr.un bearing "alues. The increase in crush assoc.lated
Wtth 939 psi and 1044 pst can bedeternuned tfom the load-dtsplacernent cut\'e shown in Figure 1- 18 and
associated equalioos of Design Example I ofth1s manual. The authors rec.ornmend that beating plates be
sized to hmn the compression perpendtcular to gram to 625 ps1 (crush= 0.04 t.nch) or less when us1ng
LRFD to reduce deftecuons oftlte tie-down assembly.
Podium Coone-erion (antboragt to concrttt)
The shear \vall continuous rod ue-dO\m tenston (uphft) base connection occurs al the first-floor podtwn
level. The amplified seismic uplift force is 56,2 15 pow'lds (Table J.-35A) a~td the LRFO seismtc uplift fbrce
ts 16,542 pounds (Table 3-358). The ampltfied compress ton force (OTM oompresston plus gravuy) tor rlus
design e.'ample ts 91,318 pounds(Table 3 -39~
The elevated podtum structural slab is curre.ntly cons1dered to be a disconttnuny for the shear '"all system.
I. Anchorage.: Similar to requtrements for light-frame wood structures. ASCE 7 Section 12.3.3.3
requires tha[ the connections of the discononoous ele-me.•u to the supporting structure be
desig.tled fot the forces used to des1gn the discontinuous elemenr. AISI S400 Sections El.4.1.2
and E2.4.1.2 requtre the uphft anchorage. which includt.."*S the anchor rod, ha\'e tl1e available
suet~gth to restst the amplified se1smic force. UstngAISC 360.lhe steel anchor rod matenal and
size are selected to comply ""'th AJSI S400. Then one designs the anchorage. tn accordance w1th
ACI 3 18 Chapter 17 usingrhe rod mateml and size selecred rosarist)• AlSI 5240,5400, and
the LRFD seismic-load combtnations, except where ACI318 Chapter 17 requires the use of the
seismic-load combuw.ions with the oversuength factor. This ensures the conc.rete ·will not be the
governing limn f« the rod matenal and s:•ze selected unJes:s 11 is des1gned ustng the seismic-load
combinations ·w1th the overstrength f3ctor.
2. Podtum slab design: l11e concre-te podtum deck, SU'lCe it represents the element the dlsoonunuous
shear wall ts s itting on. is to be designed to restst the load determined us1ng the ASCE 7 seismkload c.ombinauons wnh lhe overstrength factor flo. This appl1es to both wood and CFS lightframe strucn.tres.
The lateraJ-restsung system tbr the concrete podium sttucture ts hkely a load-bear1ng wall system w1th
spect.aJ retnfbrced concrete shear walls. spec&al re1nforced masonry beartng she.ar walls, or a combmauon of
the tw~ where R IS equal to 5.0. Pe.r ASCE 7 Secuon 12.3.3.3, when the lower podium base structure is ten
umes stjffer than the hght-ti'ame constructiott upper portion, the upper and IO\ver portions of the st.ructure
are to be analyzed as two different structutes, wh1ch JS Iikely the case for this butlding.. The design force V
(or the IO\\I'tt podium structure lateral-resisting elements ''"'ill be amplified by the ratio of R-''alues of the
ligtu-ftame buildtng above, dtvtded by rhe R-value ofrhe podium smtclure below (6.5/5.0 = 1.30).
The design of the concrete pod1um deck will be controlled by tl'le light-frame construction fOrces and liS
overstre.ngth factor, llo (the podtum sfab IS COnsidered 10 be a dtSC,c>nUnUOUS element), Stnee ~: 3.0 is
gre.uer than evaluating the podium deck uslng the 1.3 force ampJdication associated v.IJ.th the ratlO of the
uppe.r and lower late-ral-restsung system R-mlues.
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
The des1gn ofthe oommuous rod -t~podium eonnecnon is not pan ofthtS design example. and n is assumed
the engineer tS famd1arwnh ACI318 Chapter 17. The desagn of the pod1um slab is also not panoflhis
design e=ple.
Autbor•s Discussion
MultlstOI')' residential buildings cypicaUy have many shear walls. and typically a symbol mark IS shown
on the framing plans representing the locations of the connnoous rod tie-downs {hold-downs) for each
of the shear \'l'alls. These marks then refer back to a conunuous rod (hold-down) schedule that 1denufies
all oflhe c.omponents of the commuous rod ue-down sySiem. usuaJiy wnh a gene-r1c detatl show1.ng how
the conunuous rod ue-down system is assembled. Projects 1nay use a combanation of differe-nt hold-down
types such as CFS strap hold--downs, enlbedded CFS strap hold-d0\1/1\S, conve-ntional CFS hold-downs.
and eontinuouo; rod tie-down systems. As the budd1ngs become taller. conunuous rod tie-down systems are
used. as they typteaHy have more sueng.th than com'enuonal hold-downs.
When destgning multistory bulldmgs.lhe framing contmctor should be requjred 10 pr0\11de the engtneer
of record 1nstallauon drawmgs showing all the dillerent types ofoonunuous rod ue-down runs. along wnh
framtng plans that shO\v wheJ'e these conunuous rod tie-downs octur in plan. Some bu1lding departments,
or the junsdicuons havtng responsabll ny. Will reqUJte these conttnuou.s rod ue-down installatton drawtngs
be submiued for pemumng ao; a deterred submmal.
There are conunuous rod tie-dov•.-n component manufactuters that will produce these shear wall continuous
rod ue-down installauon drawtngs as part of their setvice ro the fram1ng contractor who purchases lhe
oomponems t:i'om them. The conu.nuous ue-down component manufacturer may also c.ons1der and su_ggeSI
ahernative materials (steel-rod material. CFS stud material, beartng-plate sizes. etc.) for substitution based
on what JS cUJTe!ltly avaiJable 1n the construction marketplace. Ar'r]r substitutioos by the conttnuous rod tjedown component manufacturer are to be submitted to the engineer of record as well as to the jurisdiCLion
for rev1ew and approval if the subsutuuon IS proposed durtng or after plan check.
12.6 SHEAR WALL CHORD STUDS
Chotd studs (verucal boWldary members, eJtd studs) are at each end ofa shear wall and prunarily res1st the
shear wall o-venumtng and gravny-oompressJon loads. The chord studs are placed symmetrtcaHy on either
side of the shear waH continuous rod tte-down use ro resist the uplift forces. In thls destgn example. there
are two oven:uming-restrrunt (tie-down) system configurations used.
I. The thtrd-ftoor shear wall conunuous rod rermmates at the bf1dge block several teet below the
lOp of the waJI. As the \\'3.11 hfts up from ove.nurntng. the faslener connection of the CFS chord
studs to CFS cripple stud~ located under the v.'Ood bridge block, uansfers the upld't force to the
CFS cripple sruds, which then uansfe.-s the force to the unders1de of the bridge block through
beanng. lne.n the sreel bearing plate on top of the \\'Ood bndge block resist.(i the uphft fotce
through beattng and JS supported by the co.umuous rod.
2. The second- and firsl-ftoor shear wall conunuous rods e.~tend through the floor above. and as
the wall lifts up from ovenumjng. tJ'le CFS chord studs resist the uplift force through bearmg
in the CFS top track. which in tum bears on the w'ldersade of the Vt'OOd floor rim jotst. The
conunuous rod steel bearing plate tS p1aced tn the CFS bonom uack of the wall at the floor level
above. and it holds dO\m the v.'Ood rim floor joiSt.
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225
Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
The chord studs are checked fOr ruu different compressive axial load1ng cooditjoJ'ts:
1. Chord studs used for uansferr1ng the overturnmg uplift load are located on each side of the
conHnuousrod and under the steel bearing plate m the bottom r.rac.k of the wall above and any
wood filler block or nm joist under the steel beanng plate.
2. Chord studs resistutg downward ovenumjng and gnwny compression forces are located on
each stde oflhe ~I beanng plate 1n t11e wall bouom track or posstbly bearoo top of the
beanng plate.
When R IS other than 3.0. the CFS chord studs are requtred to have the available strength to restst the
compresstve ax.LaJ force equal to lhe lesser of ( I) the setStniC load combtnauons wuh lhe overstrength
fac10r ot (2) !he expe<!ed strength of !he sheM wall. The expected sueng1h oflhe shear walliS !he notnu>al
shear strength. • ·,.. muJuplied by the expected su-e:ngth filctOt'", n£. The expected suength factor ts I .8 for
shear walls with wood structural panels pe.r AJSI S400 Secnon E 1.3.3. Thjs l)'pically governs the destg.n of
the CFS chord studs rather than lhe dtfferential load. as the overturnmg and gravity loads are cumulati\·e
downward.
Only the ovenurrung uplift load m each floor le"Vel (dift'C-rent:ial load) should be considered tOr the
compression destgn of the uplift. studs at lhat floor le"Vel. The difrerenual upltft loo.ds were derived earlier
using the amplified loads subt.ractutg out the resisung dead loads.
In tltis desig,n e.~ple~ the CFS chord Sluds resisung uplift bear on tl'le top Lmck., which bears on a wood
top plate. \\ohich bears on a wood run JOiSt, whtch is held down by the stee.l beart.ng plate tn the wall above.
AJSI 0 110..1 6 Co/d-FOI"med Sttel Fl"tJmmg Design Guidt AppendiX F has a destgn example for desagn1ng
an axtally load steel stud-to track-to concrete bearmg coodtuon, whtch may provide gutdance for a steel
stud-to steel track-to wood beanng condition. In additton, lhere will be crushing or bearang deformatiOn
at lhe steel chord stud to wood top plate, lhe wood top plate to the wood nm joist, and the wood nm
jo1st to the steel beanng plate. AWC NOS Comme.ntary Section C4.2.6 states the amount of deformatiOn
IS approximately two and one-ha1ftimes a metal plate to VI'OOd bearing joint when a JOint contatns two
wood members loaded perpendicular to yam. nus beartng detbrtnation should be added to the vertJcal
ovenurnang-restra.int-system diSplacement ponion of the shear wall deflectioo equation.
AJSI 0 II().. I 6 Co/d-FOI"med Sttel Fl"tJmmg Design Guidt AppendiX F provides an example of one
methodology for calcuJating the approximate bearmg area when destgning CFS compt""esston studs in a
bottom track bearutg on concrete.
Chord-Stud Assrmbly Compre:ss ion
Load~
The des1gn forces shown are for 1lle Optton 3 shear wall (LRFD).
Roof-Ltvtl Otsig.o Lo1ds (Third-Floor Sbur W•lls)
Wall ya\•tty-cmnpressJon Joads:
Dead load= 153 plf
Ltve load =0 plf(roof live load not used in load combinations)
Load onmbtnation = ( 1.2 + 0_2(S"'))D + 0.5L
= [ 1.2 + 0_2( 1.117)]( 153) + 0.5(0) = 2 18 plf
Chord srud boundary gravtl)' load= (25-ft-long wai112X2 18 pll) = 27251b
Assumed chord stud boundary \\1dlh = 2 ft
226
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2018/BC SEAOC Slructurai/Seismic Design Manual, 1.01. 2
ASCE 7 Eq 6 §2.3.6
Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
Wall seismic forces (amplified seismic loads - compression loads):
OTM (Including the seisrntc overstrength factor 0.0) = 270.848 ft-lb
Lever arm= centelitne ofhold-dO\\n to centerline of hold-down =25 ft- 2(1 ft) =- 23 ft
C= T=270,848123 =I 1,776 Jb
Expected shear waJI strenglh - compression loads:
Wall length= 25ft, #8 screws a! 61n o.c., V......,= v. = 890 plf,fl,= 1.8
OTM = (25 ft)(890)( I .8)( I I ftYIOOO = 440.55 kop-ft
Lever arm= cemeritne ofhold-do""ll to centerline of hold-down=- 25- 2( I ft) = 23 ft
C=440,550/23 = 19, 154 Jb >I 1,776 Jb; therefore,use I 1,776 Jb
Conunuous rod te.ftsile capacity (for up1ift comparison}:
C.onlinuous rod= .Yt-tnch diameter (0.442 in2)
R, = 1.2 (ASTM A36 matenal)
AISC 34 I TA3. I
Continuous rod nom anal tension strength= (0.442)(0.75 x 58 ksi)( 1.2) = 23.07 ktps
AISC 360 §J3.6
Summary of chord-stud assembly axlaJ loads:
23.07 kips (rod uplift capaci!y)
> 19.154(expected-compression)
> I 1.776 kops(amplified - compression)
> 10.371 kops(rod amplified dift"eren!ial uplift)
Third-floor O..igo Loads (Setood-Fioor Sh.. r Wills)
Wall gmvity-compc-ession loads:
Dead load= 153 + ~64 = 617 plf
Live load =(I 1.8 ft x 40 pst) =472 plf
ASCE 7 &j 6 §2.3.6
Load comb1na1ion = ( 1.2 + 02(SJJS)D + O.SL
= [1.2 +0.2(1.1 17)](617)+0.5(472) =I 115 plf
Chord stod boundary gravo1y load= (25-ft-long wall/2)( I I 15 pi f)= I3,938 Jb
Assumed chord-stud assembly houndruy \\idth = 2 ft
W3JI seismic forces (amphfied se1smie loads - comp(eSSIOO loads):
(Jf)V( = 820,598 fi-Jb
Leve.rarm = 23 ft
T=C=35,6791b
Expected shear waJI strenglh - c.ompression loads:
Wall length = 25 ft, #8 screws at 3 1n o.c., V__, = J~ = I775 plf; o, = 1.8
OTM = (25 ft)( 1775X I .8X 10 ft)IJOOO = 799 kop-l\
OTM 1otal (roof+ !hord floor)= 440.55 + 799 = 1239.55. Use 1240 k1p-ft
Leve-r arm= centerhne ofhold-dov..n to centerline-of hold-down= 25- 2( I fi) = 23 ft
C= 1,240,000123 = 53.913 lb > 35,679 lb; therefore, use 35,679 lb
Conttnuous rod renstle capacity (for uplitl c.omparison):
Continuous rod= ~-inch diameter (area= 0.44 tn2)(F.. =- 125 k.s1)
R, = I. I (assumed for ASTM A I93 !!7 matenal)
Conlinuous rod nom1nal lension strength
=(0.44)(0.75 x 125 ksiX I .I)
AISC 360 §J3.6
= 45.375 kips
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
Summary of chord-st\KI assembly axial loads:
45.375 k tps (rod uplift capacuy)
< 53.913 ktps (expected compresston)
> 35.679 kips (amplified compression)
> 19.640 ktps (rod amplified-differential uplift)
S«ond-floor Drsign Loads (First-floor Shur \Valls):
Wall yav&ty-compression 1oads;
Dead load= 153 +464 +464 = 1081 plf
L""' load= ( 11.8 fl x 40 psf) + 472 plf = 944 ptf
Load combination
= ( 1.2 + 0.2(S..,))D + 0.5L
ASCE 7 Eq 6 §2.3.6
= [1.2 +0.2(1.117)](1081) + 0.5(944) = 1539 plf +472 plf=2011 plf
Chord st\KI boundary gravity load= (25-ft-long wall/2)(2011 pit)= 25,138lb
Assumed chord-stud assembly boundary width= 2ft
Wall selsmic forces (amplified seism&c loads - compression loads):
OTM = 1522.123kip-fl
Lever arm = 23 ft
T=C=66,180 lb
Expected shear wall suength - oompress1on loads:
wau lenglh = 25 ft. #8 screws at 2 in o.c.• f-:._...., = v. = 2190 plf. fit= 1.8
OTM = (25 11)(2190)( 1.8)( 10)/1000 = 986 k1p-ft
OTM total (roof+ third floor+ second floor)= 44 1 + 799 + 986 = 2216 ktp-ft
Lever arm= cemerl tne of hold-down to centerline of hold-down= 25 - 2( I ft) = 23 ft
C=2226/23 = 96,7831b > 66,180 lb; d~erefore, use 66,180 lb
Cormnuous rod tensile capacity (for uplift c.ompanson):
Conunuous rod= I Ys-inc.h diameter (area= 0.994 in2 ) (F, = 125 ksi)
R, = 1.1 (asstuned for ASTM Al93 87 material)
Conunuous rod norntnal capacuy
= (0.994)(0.75 x 125 kst)( 1.1 )
= 102.51 ktps
AISC 360 §J3.6
Summary of chord-stud assembly axial loads:
102.51 kips (rod uplift C<lJ>l<lly)
> 96.783 kips (expected compression)
> 66.18 ktps (amphfied compression)
> 26.240 kips (rod amplified dtfterenLial uplift)
GM1trning loads ror Compression CFS Chord Sluds
Table 3-38 shows the governing CFS chord stud destg.n loads for both the story-level diffe(enttal uplift and
accumulative downward compression loads due to the shear wall ove·rturnmg for each story level of the
buddlng. ln a typiCal conunuous rod ue-dov.n system for CFS c.onsuuction,lhe chord studs are requtrtd to
I. Have the available strength to resist the comprcsstve axial load due to the cumulative
overturning do,~onward compresston design load.' 1.ncludtng the design gravtty lood~i.
2. Have the avrulable strength to res 1st the compress1ve a..;tal load when beanng agautst the ftoor
fromtng or wall above on the te.nston (uplift) side of the shear \WI I.
228
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2018/BC SEAOC Slructurai/Seismic Design Manual, 1.01. 2
Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
The chord stud des1gn on the compression stde of the shear wall governs the chord stud des1gn. as the
compress1ve axlal load 1S cumuJauve wnh each successi,•e SlOI)' and includes the gra\'aty Joods.
Table 3 -38. Opuon 3 shear wall: summtlr)t q[CFS dmrd s1ud differenlial uplift
and accrumt/(JJh·e dowm•·anl compre.uion dtsign loads
Level
arM DitrerenuaJ
Uplift Loads (k1ps)
3rd Floor
10.371
2od Floor
19.640
1st Floor
26.240
Load Type'''
OTM Compression
Downward Loads (k1ps)
Lood
Gravuy
Type•lJ
Loads (kips)
11.n6
amplified
2.725
35.679
amplified
13.938
66.180
amplified
25.138
amplified
differential
amplified
differential
amplified
differenual
I. The uplift loads are lhe amplified ddTerenual SUll')' loads from Table 3-37.
2. The downward comp!ession design loads ate-the smaller of the calculated oversuenglh and expected
loads.
Chord Stud Resisting Compression Out to 0\•trturning and Gnwity loads
The CFS-chord studs are designed for the governing compression forces uulizmg commercially available
CFS design software.
The selected CFS StlJd stzes for lhe three floor levels lhat have an avaalable compressive a-< tal stte11gth
10 res1s1 1he lesser of ( I) !he expee1ed streng1h of <he shear wall or (2) lhe load from lhe se1smic load
comb1nations w1th the overstrenglh factor (amplified se-ISrntc load).
The eng1neer may want to t.ncrease the CFS stud thickness over that used tn the story above lOkeep the CFS
chord studs aligned from lop lO bottom. However, the tncreased stud thickness; and corresponding requ1red
screw size are required to be tn accordance wnh AISI S400 Tables E I .3-1 and E2.3- l.
Tablt 3-39. Op1itm 3 shear u·al/: CFS chord s1ud n.:.e. mu11ber. tmd s1reng1h
Avatlable
Chord
Strength
(kips)
Amphfied
LRFD"'
Axial
Des1gn
Load (k•ps)
A\•adable Nwnber Assembly
level
(ft)
Stud S1ze
(kipS)
of
Chord
Sltlds
3rd
Floor
10
600SI62-54(50)
4.48
4
17.92
14.489
2nd
Floor
9
6005200-68(50)
9.34
6
56.0
49.617
1st
Floor
9
6005200-68(50)
10
110.8
91.3 18
Stud
Stud
Height11 ) lnd1v1dual Chord Strenglht'"•
11.1
AXIal
Chord
S~t~d
Assembly Width('1
(8 x 1.625 in)+ 3 in
= 16ln
(6x2.0 in)+9in
=211n
( 10x2.0 in)+9tn
=29tn
I. Stud hetghl uses floor he1glu manus 12 tnches for run joL-.t hetg.lu on SlUd Yo>all.
2. Amphfied LRFD a.xial design load= OTM oompressive load+ gravity= (e.g., fii'Sl floor= 66.18 +
25.138 = 91 .318 kips); <hese loads can be fowKI in Table 3-38.
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
3. In accordance wuh IBC Section 1607. 14. a horizontal live load of5 psfis applied wtiformly across lhe
ven1cal race of the mterior shear wan and is assumed to be acung concurrently w1th the seism1c: arM
oompres..'>IOn ru1d gravtty axtal load'> on the shear wall chord studs.
4 . This destgn example uses a 3-tnch space between the CFS cnpple studs bei<>Yo1 the bridge block on either
side of the continuous rod at lhe Llurd ft{)()( and a 9-lnch width betwee.n the CFS chord stud packs at
lhe second and lh1rd fi()()(S. The gap needs to be adequate to install the uphft resmunt systenl. which
mcludes the cootinoou.~ rod to the coupler nut., the take-up device. etc.
5. The shear wall CFS studs are mechamcally braced by steel sttaps on each side and periodlC tntermitte1n
blockmg at midpo.nt at the lhud and second floors and at thtrd po1ms at the first floor.
The compressive axial strength of the mdividual CFS studs
JS dependent on lhe presence ofbrac1ng. lf
mechan1ca1 steel btactng is used. the axial capacity also depends on lhe spacmg of the bfacmg. As is
dtscussed later 1.n Secuon 14. 1 on waJI-stud braci11g. the bracing must be adequately secured to prevent
Late-ral twisung and buc.klmg of lhe CFS studs and must have anchorage to resist the cumulath•e bractng
force fo< a"ally loa<led stu<ls.
Wh1le the lhird-ftoor shear wall CFS studs are brac-ed by steel straps on each stde and peJ'todlc intermment
blocktng a1 the midpotnt, the engineer mtght use wood struc.nwl panel she.attung on one s1de-and gypsum
board sheathing on the other Side. But the bracing strength of the gypsum board sheathing IS weaker and
therefore governs 1n a~ordance with A lSI 5240 Section B 1.2.2. 1(b}. If wall sheathing is used as stud
bracing. Section B 1.2.2.2 also requtres lhat the eng.ineer1ng construction docume.nts 1denuf)• the wall
sheathing as a structural elemettton the plans. AlSI 5240 Table 83.2- 1 hmus the nomjnal ax:ialload tor a
CFS stud to 6 .7 ktps when braced by ~-1nch--th1ck gypswn board s.he.rufung us1ng #8 screws and so would
ha\'e worked only tbr lhe thtrd floor.
When lhe engtneel' selects lhe CFS suap-brac.t.ng hetghts on the wall. tile)• should consider the location of
electncal boxes. Iig.ht swucltes, recessed wa11pockel'> for fire e>.'ttngutshers, and othe.r waJI opentngs. The
CFS continuous stmp bracing should not be used as badang for objects to be supported off the tace of the
shear wall, so the engineer should cons1der the location and elevatton of wall fixtures. wall-hung cab1nets,
and the tops offtoor-supported cabt.nets.
Stud Slrap Bradng for CFS Stud Axial Capacity (Bradng Forte- and Bntting Syfttm)
IBC Section 22 11 states tl1<1t structural \\1-alls shall be designed 1n accordance with AISI S240, whic.h
reference'> S 100. and lhat seisnuc-tbrce-res1sting systems shall be designed using.AISI S400. Th1s design
example Wtll confonn to AISJ 5240, ut wh1ch Section 83.4 requites that each stud brace be destgned tbr
2 percem of the des1gn compress&on tbrce 1rt the me.mber. The.retbre, lhe CFS stud bracing anchorage ts
designed 10 have lhe available strength (LRFO) to resiSt lhe maximum of:
J. ll'!e cumulative brac1ng force fbr the CFS chord studs on the compression side of the shear
wall determined ustng the rompress1on force of the lesser of ( I) load from lbe se1smic load
combut.lhOI'IS ·w1th the overst.rength f3ctor, {lo. or (2) the expeeted strength of the shear \Yall
v.'ith just the gravity load from that same load combination on lhe mtermediate studs for half
the length of the shear v.-all (c.onservauvely. as the ngid-body assumptiOn would have only a:<tal
compression on the chord studs).
2. ll1e cumulative bmclng force on all the CFS chord and
wall ba'ied on the LRFD gravity-load oombinauons.
230
intermediate studs along the en.ure shear
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2018/BC SEAOC Slructurai/Seismic Design Manual, 1.01. 2
Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
The Optton 3 shear \\o'allm the desagn example as 25 fee1 long with 15 tntermediate CFS stud~ at 16 inches
on c-enter. TI\e first-floor chord-stud assembly has a pack of five chord studs on eac-h side of the conunuou.~
rod ue-down (ren chord studs total) wath an overall wtdth of29tnches on eac.h end of the shear waH. The
first-floor shear wall CFS studs are mechanically braced at the third points using continuous horizontal
CFS suaps on each taceofthe wnll. The two ca~es noted above are checked tbr lhe honzonta1 CFS stud
strap brocing destgn uulaztng a venical bracing system ofCFS strongbad:: studs turned flat on each face
of the wall (see Figure 3-28) spaced uniformly along the length of the shear \\oallto transfer the \\'311-smd
honzontal bracing forces to the floor le\•el above and below the wall. Flat-strap bractng requtres access
to bothstdesofthe \\aU and may cause "bump outS'' an the wall sheathing. so bndgang dvough the srud
punchouL(O may be a more desamble stud brocang method.
Ct:st I: Otttrmiaing CFS Stud Compnssivt Axial Load tJsiag Stism.k Lotld Combinations with 0.
Fcc Case l,lhe amplified LRFD axial load for the ten 6005200-68(50) by 9- tllot-tall CFS chord
stud pack as 91.3 kips per Table 3-JS~and the amermedtate CFS '"all studs ha\'ean lRFD axial load
of2.68 ktps [201 t plfx 16-tnch spaciny( l2tnches/foot)J.
a.
Stra~to-stud
connection design force
P._~-·= 0.02P =
0.02(91.3 ktps/IOchords studs)
= 0.183 ktps/2 straps(one on each Stde of stud)= 0.092 ktps
P._,_ = 0.02P = 0.02(2.68 kips) =0.054 ktps/2 sttaps (one on each side of stud)= 0.027 kips
Attach the CFS strap to the CFS stud flange with a #8 self-tapptng screw connection havtng
an LRFD shear strength of 114 pounds at !he u><IJVtdual chord SllJds and 27 pounds at !he
interme'thate studs. Slrap thtck:ness does not have to be the same thickness as the waJI smd.
b. Slfoogback bmcang system destgn force
P_ .. = (0.092 kips x 10 chord studs)+ (0.027 kips x 8 Intermediate studs tn half of shear \IIlii)
= 1.136 ktps
Try tour palfs of800S 162-54(50) by 9 -foot-tall CFS strongback studs starttng with !he first pair
near the chord stud pac.ks oo each end oflhe wall and then equ:ally spaced aloog the length of the
wall between the chord Sluds and on each stde oflhe sud waJJ reststtng the load from the strnp
bmctng. The strongback studs are designed unbraced for lhetr full height and ate attached wnh
CFS cltp angles at tl'le top and bouom tracks. Because strap bracing only work.~ an tension, the
enganeer must take special care in deterrntntng at'lChorage JocatiOJ'IS to ensure there as an adequate
load path for bracsng force.s acting an either direction along the le.nglh of lhe wall.
P_.,.= 1.136 ktps/4 suongbackstud locanonsalong wall length
= 0.284 ktps at a 3-foot and 6- foot hetght along each 9- foot -tall CFS suongback stud
Auach the CFS strap to the CFS sttongback web witlt a #8 self-tapping sere\\' connection havmg
an LRFD shear strength of284 pounds. Use a mutimum of two screws for the suap connection
to the strongback web. Use CFS chp angles wllh an LRFD shear strength of284 pounds to artach
the str011gback to the top and bonom ttac.k..~oflhe shear wall. Use a mtntmum oftwosccewsor
other appropnate fastenet an each leg of the CFS clip angle (example: expansion anchors tn a
ooncrele podiwn).
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231
Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
The honzon<al CFS straps shall be designed for a m1nimwn LRFD load of284 pounds. The CFS
strap width and thickness ts sized to resist the required tenston load, but does nm have to be the
same thtckness as the shear waJI CFS tiam~ng members as long as it IS not less than lhe mlnlmum
structural thickness of 33 mils. Typtcally, the strap wtdths used by destgners ate I~ to 2 mc.hes
wide to accommodate the mtrumum two-screw attachment to lhe ventcal bractng system rmnung
member. For !his des1gn example, a I ~-10ch-w1de by 33-mil (20 ga) strap {F, = 33 ks1) could be
used wtth two #S self-tapping SCte'\\'S for the attachment to the CFS strongbac::k.
The four prurs of SOOS 162-54(50) CFS s1rongbaek st\Kis, wtlh po1nt loads at third po&n!S along the&r
hetght from lhe strap btae1ng. starung wnh the first pair near lhe chord-stud packs and then spaced
equally along the shear Y.>all length, are detennined to be adequate 1n ftexure using commercially
avaJiable CFS destgn software. llus example a~mes thatlhe suongback studs are not axially
loaded. Ifa d1agonal flat-strap a.J'lC.horage method lS used tnste.ad of st:rongbacks.. such as suap ..x··
bracing (Figure 3--29). then lhe waH stud that the ftat diagol'lal strap attaches to must also be checked
ror the addttional ax tal load from the dJagonal flar strap.
Cast 2: Ottnmi.ning CFS Stud Comprt.ss-ivt- As:ial Load Using Gravity l oad Combinations
For Case 2, the LRFD a"al load for each of the CFS chord st\Kis IS0.486 k1ps [20 11 plfx
(29 >nches/(12 1ncbes/foot))/IO chord st\Kis] and 2.68 1 kips for each of !he &ntennediate CFS
wall studs [(2011 plfx 16-u><:I1SI>acing)/( 12>nches/foot)J. The st\Kisare braced with steel-strap
bc"actng on e.ach stde of the studs at thjrd pomts (two srrap braces at 3 fee t on center) W1th CFS stud
sttong.bac.ks. as shown u"t Figure 3-28.
a. Strap-to-stud design force
= 0.02P = 0.02{0.486 k1ps) = 0.010 k1psltwo suaps (one on each stde of stud)
=0.005 kips
P.,~..,...~·
= 0.02P = 0.02(2.68 I k1ps) = 0.054 k1psltwo srraps (one on each sl<le of stud)
=0.027 kips
Attach tl'!e CFS strap to tl'!e CFS stud flang.e with a #8 setf-tapptng screw connection having an
LRFD shear strength of 5 pounds at the tndtvtdual chord studs and 27 pounds at the tntermedulte
studs.
b. Sttongback bracing system design force
P_. = (0.005 kips x 20 chord st\Kis) + (0.027 k1ps x 15 intermediate sruds) = 0.505 kips
Try two pair.s ofSOOS 162-54{50) by 9-foot-tall CFS strongback studs starting with the first pa11
n.....r the chord srud packs on each end of the wall and then "'luaU)' spaced along the length of the
wall and on each side of the stud y,:all resistJng the load from the strap brac.tng. The strongback
studs are des1gned unbraced for lhe1r full he.1ghi and are attached wtth CFS dip angles at the top
and bouom tracks. Because strap btac1ng only works'" tenston. the engineer must take special
care in detemumng anchorage k>cations to ensure there ts an adequate load path for bracing
forces acttng tn eitJ1er direction along the length of the wall.
PflldJJ.> = 0.505 ktps/2 suongback stud locations along wall length
= 0.253 k1ps at a 3-foot and 6-foot height along eac.h 9-foot-tall CFS sttongbaek st\1<1
232
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
Attach the CFS strap to the CFS suongback web w1lh a #8 self-tappang screw connecuon hav1ng
a LRFD shear strength of253 pounds. Use a nt1mmwn of two scre\\'S fOf the srtap connection lO
lhe S[(ongback web. Use CFS clip Mgles with an LRFD shear sue.ngth of253 pounds to attach
lhe strongback to lhe top and bonom t:mcksoflhe s.hear wall. Use a mintmum of two screws or
olher appropnate fa~te.ter (example: expanston anchors cmo a concrete podium) tn eac-h leg of
lhe CFS chp angle. The CFS straps shall be destgned tbr a mt.ntmwn LRFD tension load of253
pounds.
The two pairs of 800S 162-54(50) CFS s!fongback studs, wllh poin! loads at th1rd poin!S along melt
hetght from the suap brnc:ing. starung with the-first pau near the chord srud pac-ks and then spaced
equally along the shear wall length, are detenntned (0 be adequate 1n fte.xure using oommerctally
available CFS design soft\\'3te..
Stnce Case 2 requtres fewer sttongbac:ks than Case 1. the Case I b!actng destgn ustng the seismic
load oombtnauons governs, and !bur pairs of SOOS 162-54{50) CFS strongbaek studs are required.
The spactng of these strongbaclcs typically ranges between 6 to 16 feet on center along the length
of the stud wall and IS highly depende.nt on the floor system gravtty loads as well as the resultjng
ovenurnjng axialloods imposed on the chord-stud a~e-.mbl ies at lhe e.nd of the shear walls. ln
muJtistory buildings. the strong.back spactng WlllfYptcally get closer together frotn the upper-nlost
floor level 10 the lowest floor le\'el. Whe.re suongback spacing exceeds 8 feel on center. full-depth
stud bridg.ng block."' should be provtded at not more than 8 feet on center spacing tor the contlnuous
CFS strap-brac.&ng attachments. Where CFS horizontal bractng sttaps are 10 be spliced, it is
prefemble to splice the straps on fuJI-deplh stud bndgmg be.twee.n the y,'3JJ studs than on the venJcaJ
brac1ng system strongbad:.s.
Other anchorage solutions tbr the stud bractng mtg.ht work as wen. such as anchoring into a cross
waULwhere the wall studs. tn-wa.JI pos~ columns. or other \\all tnatenaJ (example: concre.te or
CMU wall) are destgned for lhe strap-bractng force] or ustng diagonal srrap bractng. taktng the
braclng force down to the bonom track from the lowest bractng hne and up to the lop uack from the
highest bractng line. This ts dtscussed li.trther m Secuotl 14.1.
Where taU s.healhed shear waJis are used (example: over 10 feet taJI).the engineer should consider
lhe ventcal s:pactng of conttnuous strap brncang wtlh the layout of the sheatht.ng so tt WtJI nm be
used. a.'i edge blocktng for the wall sheathing. This includes considering any honzontal stagger of
sheathing panel edges. Separate CFS fla1-stmp blockmg should he used for slleathing edge screwing
and not combtned with the continuous hori20fltal CFS straps used 10 brace CFS stud Ranges. The
autJ'lOr recommends a note be placed on lhe Str\JC.tural drawsngs that states sotnethmg like the
tblloo1ng:
The layout of the shear wall sheathing sball he such thai any CFS fla1-sttap blockong tbr enher
the sheatbittg edge screwing or for any items attached to the wall surface shall be placed separate
from the CFS stud waH conttnuous suap bracing.. Where a confttcl bet"~" blocking and CFS
stud wall continuous strap bracu1g placement OC(:UJ'S, contact the structural engineer tOr dtrection.
The assumption for CFS stud walls less than 10 feet !all is that tlle sheathing shall he placed
vertic.aHy so no horizontal CFS ftm-strap blocking tS required su'ICe sheathmg edges wall be anached
directly to the CFS studs (verucal panel edges), houom !rack, and top track (horizontal panel edges).
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233
Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
lne designer is a1so remtnded that a SI!UC[Utal note needs to be placed Oltthe engineering
construction drawangs staung somelhmg ltke the follow1ng:
All cold- formed-s!eel (CFS) srud wall suap brocmg and bridging shall be 111-<lalled and
adequately anchored per details shown on plans prior to applyu1g any axtal loads to the CFS
studs resulung from the placement of the floor/roof fram~ng level im.mediate.Jy above the CFS
studs. Shear \Vall sheathtng shall be u1stalled prtor to plactng of any tloor topptngs over the floor
sheathtng at the floor leveltmmedjareJy supported by the CFS studs and also before placemen.t of
the next tmmedJate floodroof level framing above betng supponed on CFS stud walls, unless the
bujldtng strutrure ts othe-rwise adequately laterally braced by the corn:racror. until the installation
of all sheathed shear '"-ails and any other lmeral-res1sung systems are completed.
Chord Stud AlignnJtn r bttY<t-tn Floors
In this design example. the third-floorchO!d-srud asse-mbly wtdth tncludes the CFS flange width ( 1%
tnches) of the crtpple studs under the bridge block. as the crtpple studs are on either side oftlte 3-tnch gap
width under the bridge block. The it!Stde fac-e of the chord-stud assembly usjng a 9-inch gap ru the first and
seoond Hoors below alagns almost directly wnh the utsJde face of the chord studs at lhe thltd floor where the
gap at the third floor. exduchng the crtpple studs. ts 9.5 inches; see F1gure 3-22. S1nce the ftoors are wood
framed. the depth of the wood nm joist or blocking sining on the ·wall can be used to horizontally transiuM
axiaJ loads resuhtng from any m1nor alignment ofrset<; between c.hord studs above the floor lane to the chord
studs below the floor ltne-.lfCFS Hoor ti'amtng was used.u would be more unponam 10 have the chord
studs ahgrung above and beiOY.' the floor hne. and this might change the nwnber of required chord Studs
for alignmettt between floors.. Also a larger beartng plate e:aen.ding across all the CFS c-hord studs v.'Ould
probably be requu'ed.
Flgmv 3-11. Chord .tllld ahgnmenl
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2018/BC SEAOC Slructurai/Seismic Design Manual, 1.01. 2
Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
loutJon Chetk of Conlinuou.s Rod Tit-Down from End of Shtar \Vall
As a last c.heck,the enganee.f reviews the origu~l asswnption of the continuous rod ue-d<m'n location from
the end of the shear waJI. The assumption used in lheOptton 3 shear wall was 12 inches frotn the e.nd of the
walJ. Table 3-40 shows the oompanson.
Table 3-40. Opllon 3
.th~bJ"
wall: continuous 11~-down rod IO<.YJ.tiOII chult jf"()m entl of wall
Chord-Stlld Assembly
Width, W
(in)
SheM Wall End
Distance, W/2
(in)
Corrected End
3rd Floor
16
2nd Fl001
1st Floor
Level
Distattce
Ong.inal Assumption
(tn)
(tn)
8.0
14.5
12
21
10.5
14.5
12
29
14.5
14.5
12
I. Su'l<:e the conunuous rod ue.-down IS to be 1nstaJied vettkaJiy \Vlth no offsets. the first-floor IOC31ton of
the continuous rod lie-down from the end of the shear wall diCtates the locauon at the h1gher floor levels.
but the bndge bloc.k may dtetate the saze oflhe chord stud pock gap width.
2. II'= individual chord stlld flange width+ gap(9 inches per Ftgure 3-22)
In thiS destg.n exatnple, at the thtrd- and second-floor levels. the locauon ofthe continuous rod ue-down
from the end of the waJI is less than the orig1naJ assumption. At the first-floor leve.l,lhe assumption was
slightly off ( 14.5 tn > 12 tn).
I.
When the actual cont1nuous rod ue-down Jocouion tS less than the as..:owned distance from
lhe end of the she.ar wall, thts IS oonsef\•auve, Since the uphft des1gn forces are less than the
calculated design force.
2. If the end dtstance tsgreaterthan the ongtnal a.~umpt.io1t., then the designer must detenntne if
the uphft forces need to be recalculated. In thts deStgn example, the change in dista.nee between
lhe cemerhneofthe tie rods is mintmal:
25 ft - 2( 14.5 tn/12) = 22.58 ft
23 ft/22.58 n = 1.019 The change is less than 2 percent, so I[ is deemed atteptable and
does not have to be recaJculated.
Su'tCe the rod lS centered 1n the Largest chord pacl; and the rod stacks from floor to floor,lhe.
co«ected end d1stance IS now 14~ inches.
Aulhor's Distus~ion
Jfthe end dJstance to the commuous rod tie-down 1S 6 inches or gre.ater from the e.nd of the shear '"-all
tl-.an the original assum.pt1on, and the shear wall aspeet ratio 1S 1: I or greater~ then perhaps the uphft force
should be recalculated. ll\is depends on the O\'erall le1\gth of the shear wall and wl'ltthe.r the shear \\'311
design forces are calculated ustng JUst a utbutary-area dtaphragm analys1s, a rigid-dt.aphragm analysis.
or an envelope procedure (c.omparison between tributary area and ng..td diaphragm). A ng1d-djaphragm
analysts depends on the shear wall st&
ffness and is dtrectly tmpacted by the distance between conunuous
rod tie-down/chord-stud a~semblies at the ends of the shear wall. Ustng a shear \\'all length longer or
shorter tl'Um the design ac.;sumpt.ions wtlllmpac:tthe accuracy of the rigid-diaphragm analysts. resulung 1n a
redistnbution of seismic destgn force.s be.tween shear walls.
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235
Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
If the ac.tual c.onunuous rod ue-down location lS less than 6 tnches beyond the origtr~l assumption from
the end of the shear wall to tlte conunuous rod tie-<10\\n, and the shear wall aspect rauo IS less than I: 1.
th1s probably will not result tn a large mc.reao;e in uplift and compresston des1g.n fotces.. Most hkely there is
enough resef\'e c.apacuy in the chord-stud assembly and c.otUmuous rod ue~0\\1'1 to be able to support the
shghtly increased destg.n forces resulung froJ'n a s.h1ft of6 inches or less from the ortgu'lal assumed location.
Similarly. if the conunuous rod tie-down location is less than tl'!e ong1naJ assumption from the end of the
shear waJI by 6 1nches or less. and the shear waJI aspect ratio IS le.'iS lhan I: I~ the uplift. a.'ld compression
des1gn fof'Ces should be less and oot lmpact the des1gn ofthe chord-stud assembly and conllnuous rod
ue-down.
12.7 BRIDGE BLOCK AND CHORD.STUD ASSEMBLIES
Bridgt Block
At tl1e third-floor wood bndge block~ shown tn Figure 3-23. t\\"0 CFS cr1pple studs under the bridge block
and on each side oftl1e continuous rod (tOur total) are anac.hed to lhe CFS chord studs and transfer the
lhlfd-ftoor shear waJI O\'enurnmg uplift force from the chord studs to the bndge block. The ru'O full-he1ght
CFS chord .studs on each side oflhe bridge block (four tmal) are also used to resist the th1rd-ftoor shear \vall
0\'enumlng compress1ve axial load. In lieu of a wood bndge block~ a steel HSS section or a steel channel
(turned ftat) could be deSigned to be used as the bridge block.
WOOD 9RIDCE BLOCK
Ftgw·' 3-23. Chord-stud assembly at wood bndge blodc
The bridge block needs to be the same \\1dlh as the CFS cripple studs under the br1dge block f« upliftforce uanster. If 111s not, the.n a tiller plate the same de-plh as the CFS c.r1pples needs to be provided
between tl1e bndge block and the CFS uack mto which the CFS cnpple studs frame. The bridge block also
needs to be sufr enough to uniformly transfer lhe uphft loads to the steel bear1ng plate on the top of the
bndge block.
In the case of a wood bndge block (spanning between CFS ctipple studs), the depth of the block needs to
be checked for shear just as would be done for a wood beam. As a guidehn~ the depth of the wood bndge
block should be not less than hal flhe gap distance or the chord-stud ao;sembly (clear distance between the
1nside tace of the chord studs). llus helps to assure a I: I slope for uplift-force rransfer from the underside
of the \\'Ood bridge block to tl1e stee.l bearing plate on the bndge block.. If a sreel bt1dge block IS used, then
deflectJon of the steel br1dge block end rela1we to the continuous. rod ue-dO\\n should be evaluated. ThJs
generally ls not a problem unless a ''et)' thin steel plate is used for the bridge block.
236
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2018/BC SEAOC Slructurai/Seismic Design Manual, 1.01. 2
Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
Sbuthing Artacbmtnl 111 Bridgt Blotk, C ripplt Studs, and C hord Sluds
ll\ere ate muluple v.ays to accomplish the uplift-force transfer from the wall sheathing to the bndge block
through the cripple studs aJ'ld adjac-e-nt chord stud.~. The one detatl common to all oflhe uplilt-tbrce transfer
methods IS to have lhe CFS chord stud directJy adJacent to each end of the bndge block anached to the first
CFS crappie stud under the bndge block. Thts spectfic oonnecuon detail be.tween the cl\ord Stud and crtpple
stud provides the load path for transfemng wall uphft fOrces to the crtpple studs. The required number
ofsheath1ng edge fasteners are divtded equally into the chord studs, not exceed tog 12 tnches on center.
c.on.necung the wall sheathtng to the chord stud~ to restst the required de:stg.n uphft force and downward
compress ton force. The desagner may use field screwspacut_gof 12 inches on center tOr the cnpple studs.
There are \'atiOUS sheathing-ta.~tener placement configuration opttons that designers can use-. but all still
requtre investigation to detennine the appcopnate force transfer design between the chord studs and crtpple
studs adJacent to the end of the brtdge block. Typically, lhe design-force transfer between the c.hord sruds
and ertpple s1uds IS accomplished ustng screws or \\'elds. The wall sheathtng should not be anached to
bndge blocks because wood bl ock.~ could shnnk and steel blocks restrain ventcal movement.
C hord Stud and Cripplt Stud Oritutatiou
One CFS chord stud-to-cripple srud connecLion option would be to have the CFS c.hord stud directJy
adjacent to the end of the bttdge block orte.nted so tlS web could be sc.re\\-ed 10 lhe web of the CFS cnpple
stud under the wood bridge block. If this option was used, the CFS chord studs and the CFS crtpple-sn.td
flanges would be rutached with tntennutent welds to tbrm a box shape, but lhe screw connection between
tJ1e webs of the CFS chords and the CFS crtpple studs would need to be completed prior to welding 10 make
the bo.x shapes. The e.ngtneer mtght dec1de to use an all-welded connectton approach wtth justtntemt~ttem
welds on either s1de of the chord studs 10 the cripple studs rnther than screw them together. Figure 3-24
shows an aU-we-lded chord and c.npple stud arrangement with individual chords and c.npple studs welded
100-t~toe and the back of the chord stud box welded to the back ofthecnpple stud box.
Figtmt 3-l.f. Chord bnd crtpple studs oriented tot-IO-toe (plan 1•inv)
Anod'ler CFS chord s1ud-u~cnpple. srud oon.necuon optton would be 10 have lhe CFS chord studs oriemed
bad:-to-back ustng screv.-'S or v.'tlds 10 attach them together and the CFS cnpple studs oriented back-t~back
ustng sc:re\li'S or \vtlds to attach them together. Then the CFS ch«d studs would be attached 10 tlte CFS
cnpple studs with mtermittent welds to the flanges between the CFS chord studs and the cr1pple studs.
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237
Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
Stttl Staring Platt at Floor Lint
The steel beanng plale thal ts ut the CFS bottom track of the lhtrd-floor shear \vall, whtch is uo;ed [0 resast
the overturning upllf\ fOrce from the second-floor shear wall below, is also designed to resast the overturning
compresssve fOI'Ce from the third-floor shear walL The third-floor CFS chord studs' nomtnal compresswe
strength and lhe beanng strength of the bonom track a.l'ld wood floor system are to be equal or greater than
the overtllrning ampltfied seismic compresstve load. The CFS-framed shear wall CFS chord studs align
from floor lO floor, centered on the conunuous rod Lie-dO\\'n. The CFS chord sruds may uansfe.r some of the
ovenurn1ng compressl\'e fo1'Ce lO the c:npp1e studs below the bridge block through the screw connection or
weJded connection, and this is taken through the steel bearang plate 1n the third-floor CFS bottom tmek and
the wood floor syste.m. so CFS chord studs directly be-low the thtrd-ftoor CFS cnpple studs do not need to
be added to the shear wall system below. The lh1rd-Hoor CFS cr1pple sruds migllt also be cut short so they
only transfer the O\'enuming uplaft force from lhe CFS chord studs lO the. wood bridge block..
Woll Bridgt Blotk
The \li-ood br1dge block for uplift tS used m the wall at the th1rd floor since a roof-drag, metal-plaleconnected wood truss will be connected to the shear wall double top plate/top track to transfer roof lateral
forces to lhe shear walJ. If the upl1ft conu.nuous rod Lie-dO\\'n system beanng plal'e was placed on the top of
the sl1e.ar wall double plate, it would likely interfere wuh the roof-drag metal-plate-connecled wood-truss
connecuon. The bridge block IS generaHy located betwee-n one-haJf to about two-th1rds of the VI'all hetght
above the floor leve-l.
Author 's Distu.ssion
The ventcaJ boundary element (chord-stud a.'\Sembly) at each end of the shear wall oons1sts of seveml
CFS studs. The CFS Studs sltould be placed symmetrically on each side of the connnuous rod. Each of
the tndiVid:uaJ chotd studs shall be screwed to the shear wall sheathing. Tile screw1ng of the shear wall
sheathtng to the chord studs does nOl have to be the same as f()( the olher CFS ttruning me.mbe-rs of the
shear waJI at the wood structural panel edges.
E.xarnple: The second floor has six CFS chord sruds. The second-floor shear waJI "-ood structural
panel sheathing has SCte\\'S at 3 tnches on center. Because there tS more than one member tn the
chord-slud asse-mbly, the edge screws may be divided equally bem-een them~ however. the edge
spacing should not exceed 12 tnc.hes on cenrer on any of the todivtdua1 chord or intennechate studs.
3-inches-on-center edge-screw spacing x 6 chord stud.~= 18-inches-on-center edge-screw spactng at
each chord stud. t-lowe\-e.r. the destgner shouJd oot e.xceed 12-inches-on-cente-r spacing.
Thls ts a savtngs tn ttme and matenaJ stnce the waJI framer does not have to tnstaJI screws at 3 U'l(:hes on
cente-r to each tnch\'tdual chord stW of the chord-stud assembly (boundary member).
Whe-re a bridge block ts used, each of the indh•idual chotd studs shall be screwed to the shear wall
sheathing wuh edge screws staggered equally between them. TyptcaHy. the bndge block CFS cripple studs
are not screwed lO the shear wall sheathittg with edge screws, but W1lh field screws at 12 lnches on center.
In this case. the total uplift Coree lS transferred from the welded or scre'lled anachment of the CFS chord
studs to tlte CFS cripple studs below the bndge block, which then bear on the uodeo;ide of the bridge block.
See Figure 3-20 for CFS cripple and chord-stud anachmenL">.
2 38
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2018/BC SEAOC Slructurai/Seismic Design Manual, 1.01. 2
Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
How-e\'er. afthe CFS chord Studs-to-crappie studs connecttons are not designed to transfer lhe e-ntire uphft
load from the c-hord studs as discussed previously. then the c-npple studs need to have shear waH s.Jtealbjng
edge SCJ'ews adequate to transfer the force not being transferred by the chOt'd studs. This might occur if the
enganeer considers that only the tndividuaJ chord stud immediate-ly adjacent to each e.nd of the bridge block
wdl transfer only a ponton of the ovetturni.ng upltft force to the adjacent cnpple stud.
StACCER SHEAR WAll
SHCATHt/'iC £DGE SCREWS
EOUAI_lV BElWEEN CHORD
STUD& {12~ M!.X S-DACtNC)
• '11000
SRtOCE:
stOCK
Jrr?i!P~'!fr--:- St!E,o.\fHING FA$.j£h[~$
CfS CRtPPLl SI\JI>Si----cm
AI tCNtf> rAt( TO
AI lj( O.C. Al C~ll"PL(
S1'UDS U.N 0.
(NO AnACHMtr...--:s
BtTWEEN
srvos)
/__J
i1"''---el'< CHVRO SiUOS
AUCNE.D rACt ·o rAC£
(NO AtTACitMOtr$
BfT\\'(fN S1UCS)
3 " 1/!N
Figure 3- 25. Wood bndge block tJIIhird-/foor chord-swd assembly
Cbord-StudAi.~tmhly
ar Bridgt Block; Shtatbing-Scrtw Uplifl Fortt Tranfftr Exampks
E.xamplt 1: The lhtrd-floor uplift bndge b1ock uses the chord-stud sheathing edge-screw des1gn but without
edge screws to the cripple studs. instead using field scte\VS to the crappie smds under the bridg.e block.
Uplift= 10,371 lb (LRFD amplified force)
Stud heigJu to wall nm joist= II ft- I ft = 10ft
Shear per foot of stud height= 10,371/10 = 10371blft
Shear per foot ofhe~ght per stud= 1037/4 chord studs= 260 lb
#8 screws 316 inches on c-enter tn plywood sheathing (nominal strength)= 890 lb
890 lb > 260 lb
However. shear wall edge fasteners are divided equally between chotd sruds 10 shear walls using c.ontinuous
rod tte-down systems (but they should not exceed 12 tnches on center). The third-floor shear wall uses #8
screws at 6 mches on .::enter. and there are four chord studs.
6-tnches-on-cemer edge-screw spac1ng x 4 chord studs= 24-tnches-o~nter edge-screw spacll'lg
at each chord srud. However. the designer nwsr not ex:ceed spactng of 12 tnches on center.
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
Use #8 SCI\."\VS at 6 inc.hes on c-e-nter for general wall-sheathing edge screv.'ing. except use 12 tnches
on center to each oflhe chord compression studs on each side of the bridge block. Stagger tlte
screws between adjacent chord studs. Attach sheath1ng to cnpple studs ustng field spactng of 12
inches on center.
Load Tran.sftr bttwttn lbt C hord Stud and tht Cripple Stud
Use #8 se.rew anachmenL~ bemreen the 54-mtl CJ-""S chotd srud web and the-54-mil CFS cnpple stud web for
uplift force transfer to the bndge block:
Tronsfertorce = 10,37Jn chord stud packs= 5186 lblchord pack
#8 screw capacity (•Pul = 0.5 x I278 = 639 Jb
(from Table 3-40)
(•Pu) <(•P~> fcrbeanng and ttlung
See Section 12.8 fOr screw des1gn infonnatton.
Qua:nuty of#S screYt'S = 5186/639::: 8.1 SCWr\'S; Lheref«e.• use etght #8 scre\li'S mimmum.
#8 screw spacong = (3.67 ft x 12)/((8n rows)- I)= 14.68 tn > 12tn
Since the required screw spacing IS greater than 12 tnches. it is recommended to nm exceed I2-inch
spacing.
Use two tO\\o'S of#8 screy..s at 12 i.oches on center.
Load Ttans ftr bttwttn lndh'idual Chord Stud.~ and Jndivid:ual Cripplt Stud.~
ll'!echord studs must be designed to tranSfer 10,371/4 =2593 pounds between lhem ussng weJds,
and the pairs of cripple studs at each end of the bndge bloc-k also need to be destgned to transfer
518612 crtpples = 2593 pounds between lhem ustng welds. Instead of screwmg the one chord stud
10 the one ct1pple stud. it may be easier for the fabric:arot to shop we-ld the chord stud 10 the crtpple
stud and bring_ it out as one assembly. since the welding between indtvtdual c.h«d studs (face to
face) and cripple studs (face to face) wtll likely be done tn the shop.
Ex.amplt 2: At the third-floor uphft bndg_e bloc I; the chord-stud sheathing edge-screw des1gn tncludes lhe
edge screws lOthe shoner cnpple studs under the bridge block. Otord and crtpple studs are n01 auached
togelher except tbr a sang_le chord stud we.b screwed to a single cripple stud web at each end of the bridge
block(see Ftgure 3-25).
Uplift= 10,371 Jb (LRFD ampltfied force). The bndge block is assumed to be 10 tnches deep.
The stud height to the underSide of the bndge block=4.5 ft - 10 tn=3
n 8 tn=3.67 ft
The uplift above the hotton1 of the bndge block= 10,37 1 x [(10ft- 3.67 ft)/10 1\j =6565 Jb
The shear per foot of stud height ahove the hottoro of the bridge bloc.k =65651(10 n- 3.67 fi)
= 1037 Jblt\
The shear per foot of chord stud ahove the bottom of the bridge block= 10371(2 chord stud<)
=519Jbll\
The upltft below tlle bottom of the bndge block= 10,371-6565 = 3806 Jb
The shear per foot of Stud hetght below the bndge bloc.k = 3806/3.67 = 1037 Jblt\
The shear pe.r foot of chord studlcnpple he1ght below the hottom of the bndge block:
Shear= 10371(4 cripple studs+ 2 chord studs)= 173 lb
#8 screv.-s at 6 tnches on center tn plywood sheathtng (nominal strength)::: 890 lb
#8 screws at 4 mches on center 1n plywood sheathing (nominal strength)= 1330 lb
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
Summary with screws at 6 1ncheson. center: 890 lb > 5191b> 173 lb
withscrewsat4 inches on. center: 1330lb> 1037lb
4-inch-on.-c-erue-r edge-S('.rew spac1ng (4 cnpple studs)= l()..inch-on.-cemer edge spacing
16 in.cltes 011ceme.r > 12-mch-on-center spac:•ng~ therefore. use 12 utches on cente.r.
Use #8 screws at6 1nches on center 10 all chOI'd studs s1.n.ce i.nd1v1dual chotds studs are not
mterconnec.ted by screws or \\'tlds.and use#8 screws at 12 1nches on center to the uxhvtdual
cripple studs below the bridge block.
load Transftr btt~tt-n tbt Cbord Stud and tbt Cripplt Stud
#8 screw anachment(i between the 54-mil chord stud we-b and the 54-mll cripple stud web for upl1ft
fbcce transfe-r:
Tran<fer force= (519 x 6.33 ft) + (173 x 3.67 ft)= 39211b
#8 Screwcapacil)' (qPu) = 0.5 x 1278 =6391b
(from Table 3-40)
(q>Pul < (qP.l tbr beanngand ulting
See Section 12.8 for screw design mfOl'matton.
Quant1ty of #8 screws= 39211639 = 6.1 screws~ therefofe-, use SIX #8 screvis minlmum
#8 screw spacmg = (3.67 II x 12)1(6 - I)= 8.8 in
Use #8 scre·ws at 8 tnches on center. staggered.
\Vood Bridgt Bloc:k Horizootal Sbur and Btnding Cbtc:k
The \\'OOd bridge b1ock must also be designed. The br1dge b1ock was prev1ously checked for bearing when
the engineer was designing the steel beating plare. The bndge block also needs to be checked for h<>rrzonral
shear ru\d be:nchng. ASD is used for e\1aluating the bridge b1oc-.k. Cu is conserv.ujvely taken as 1.0.
Bndge length= (4 x 1.625) + 3-inch gap around c:onunuoos rod:: 9.5 in
Bndge reacuon (somple beam): R=(I0,371 x 0.7)12 = 3630 lb
Bridge ftexure (simple beam): M= PU4 = ( 10,371 x 0.7)(9.5)14 = 17,242 in-lb
2
Wood area reqUired= 1.5(3630)1170= 32.03 on < 41.25 in'= 6 x8
WoodS,= 17,24211350= 12.77 on'< 51.56 in'= 6 x8
Usea6 x 8 bndge block min1mum (Douglas Fir-Larch #I, F, = 1350 psi, F,= 170 pso).
12.8 SHEARTRANSFER
11te shear u-anster between floor levels uul1zes screws from the CFS wall uac:ks lO the wood fmnung
members (floor sheathtng and floor wood fram1ng members), as shown 1n F1gure 3-26.
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
y~~,.."""~---Cf'l.
~.­
......._ ... "'W"T
'V\-f\1..
....
'l.o(o\11 ..... ""-'-"'
~1!..("1100'. . ~·'Ill " ...._
r~
~1"1"-~fW..
....,
IJOCol>.'-'1
'
.,...,._
l•¥1'11
...""'~~If
...........
~·~ '""
>•1
~·
Ct
~~
1.011"11
cr;
Ql
......... ,,,tot
'1"1.
, •.,. ,,N.)p.
:
...-,:'lVI
·'
FJgur~
A
I
II
3-26. SMar Jrtmsftr through n·()()(/j/oorfrummg
The shear ltal'lSfer at the-podium uses anchor bolts embedded mto cone fete v.'1th standard cut washers over
the CFS t.me.k web. While standard cut washers may be. used for the shear anchor bolts, some designers use
plate washers in case the shear anc.hor bolt holes are oversized or if they want ro weld the plate washer to
the CFS t:raek w-eb. The de~ag.n f«ces for the Opuon 3 shear \\all are used for thlS destgn example.
Just as OCCtlts in lumber. the destgn values for screws and bolts are limited by stee.l sheet sheath1ng edge.
encL. and s:pacmg dismtt('.es. 1be location of the screws and bolts wtll be such that they are not controlled by
these minunum diStance.s .
Strtw St.ltctioo: A\VC !'iDS Stttlon 12.3 Rtftrtnct Lilt nil Ots:ign Val uts
(Tablts 12.J. I .A, 12J.I.B, 12M)
NOS Section 12. I.5.3 requlres lead holes for wood screws loaded laterally: .. For G S 0.6 (see Table
12.3.3A), the part of the lead hole receivtng the shank shall be about~ the dJamete'r oflhe shank and
that receiving the threaded portion shall be about %the diameter of the screw at the root of the thread."
Theufore. the designer should use a fag screw or wood screw installed through predlilled holes 1n the
CFS track-t().Wood cormecuon if a self-drilling tapping screw has a drill tip that drills a Larger lhan %-root
dlameter hole ullhe wood. If the lead hole is too large, the Jag screw or wood scre\v dveads do not have a~
muc-h ..bne," so lhe lateral values would be some reduced value, and the \\ithdrawal could be signdlcanll)'
less than the lag screw or wood screw des1gn value.
To 1nake the shear connection more econom1cal between floor levels and reduce the number of requtred
scre\''S. larger #12 screws should be used lnstead of#S screws. Typtcally, the screws are tnstalled ul two
parallel rows in the bottom uack of the CFS stud shear waJ11nto the wood floor sheathing and framing.
AWC NOS Table 12M does nm provide# 12 screw des1gn ''a lues (Z) for 0.048· inch or 0.060. mch-lluck
mill<flal, btn does have# 12 screw design values (Z) for 0.075· 1nCh ( 14 gauge) malerial of 147 pounds for
Douglas Flr-Larch.
The# 12 screw shear design values wlll therefore be calculated for use wnh CFS 54--md ti'amJng thtd:ness
marenal ( 16 gauge, 0.0566 lOC-h thick) attached to wood framing..
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
I. CFS Coontttions (A lSI SIOO Set lion J4 Srnw Coontttioos)
Screw notauons:
d = nomtnal screw diameter
d• :; screw head diameter
T1 = thtckness of matenal under sc.rew head or washer
7 2 = thtckness of matertal not ln contaCt with screw head or washer
F. 1 =tensile strength of member ln comact with screw head or washer
F. 2 = tensile. suength of member oot 1n contact wtth screw head or washer
Screw spac1ng::
Muumum screw spaciJtg. =3d
Mtnimum screw edge and end dtstance =3d
(force perpendtcuJar to edge)
Mtnimum screw edge and e.nd dtstance = I.Sd
(tbrc.e pamllel to edge)
AISI SIOO §14.1
A lSI 5240 §B 1.5.1.3
AISI 5100 §14.2, 5240 §BI.5.1.3
Sbnr (A lS I SIOO S.<tioa .14.3)
A. Conne-t:tion Sbtar-Um.ittd by Tilting and Bearing (P.-)
A ISI S IOO §J4.J.I
The designer must determane which of two sets of ltmll state equattons. based on the thtckness
oflhe two materials(T1• TJ beangJOUled by the screw. ts to be used to dere-tmine tlle screw nom anal
shear capacity. Shear U'al'ISfer connection design marenals are as follows:
T1 = trac.k matenal thickness: 54 mils=O.OS-66 ln (F1 = 50 kst. F.. :;65 ks1)
Tracks are to match shea! \Vall stud materiallhickness
and p!opentes.
T2 = plywood sheathing: Y. tn
f:/ 12 screwdtameter: shank= 0.216 an. heOO.d1amerer = 0.340 ul
Screw 1tm11 state equauons.:
T,IT0
=0.7510.0566 =I 3.3 > 2.5
T/T1 < 1.0 (P. = smaJiest oflhe three equauons)
I. Tilung:
2. Beanng.:
3. Beanng:
.=
P 4.2(t23d) 112F .a
PIU=2.1t,dF,.,
Pld=2.1t1dF.-J
AISI S IOO §14.3.1- 1
AISI 5100 §14.3.1-2
AISI 5100 §14.3. 1-3
TiT1 > 2.5 (P*=smallest oflhe two equations) ... governs
I. Tilung:
2. Bemng:
3. Beanng:
Not applicable
pM =2.7t,tJF., = 2.7(0.0566)(0.216)(65,000)
=2 146 1b
P... =2.1t2dFiil = NOl applicable for wood
AISI SIOO §14.3.1-4
AISI 5100 §14.3. 1-5
LRFD = ~pM = (0.5)(2146) = 1073 lb
ASD = P./0 =(2145/3.0)= 7151b
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
AI SJ SIOO §J4.3.2
l11e average screw shear \'alues (Pss> are taken fi'om CFSEI Techn1cal Note F70 1- 12. wh1ch prov•des an
average screw shear vaJue based on rev.ew of sever31 manufacturers' test reportS. The nominal scre\\o'S
shear values are s.hovm m Table 3-41. H1gher screw shear values are poss1ble based on the designer
specify1ng sc.rews that are evaluated 1n an evalwujon repon ln accordance with pubhcally developed and
publica11y available acceptance or evaluation criteria.
Tab/~ 3-./ I. Nommalscmv shetJr mlrNs (P,) from CFSEI Technical Note Fl0/-11
Sc.rewSize
Nomtn.al
LRFD (~ = 0.5)
ASD(0=3.0)
PM (Ib)
~PM(Ib)
P,/O(Ib)
~
1278
639
426
#10
1644
822
548
#12
2330
1165
777
y.•
3048
1514
1016
AISI S IOO §J 6.1
C. Sbtar R upturt (V..) if Scr ew Pulls tow11rd Limiting Edgt
AISJ SIOO Eq 16.1- 1
For a connection where lhe screw pulls through the steel toward lhe limiung edge:
AISI SIOO Eq 16.1-1
A_.=2nte'""
"=number of Scte\\'S along critical cross.-secuon (assume I screw)
1 = ba~ steel thtckness
e...,= c.lear distance !rom e·nd of member a~td edge ofscrew hole (assume 3 u1)
A.= 1(1)(0.0566)((3X02 16)J =0.07335 1n2
~~ =
o.6(65,oooxo.o73JS) = 2861 lb
LRFD=<pV. =(0.5)(2861 ) = 14311b
ASD = I )0 = (2861/3.0) = 954 lb
Cont:roUi.ag Stnw Valut (in CFS) for This Design Exam pit
ASD: #12 screw shear \•alue In 54-mtl (16ga) lll1Ck=7151b (PM)<
m
lb= p~
LRFD: g12 screw sheM value 1n 54-mil (16 ga) track= 1073 lb(P*) < 11651b = P"
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
2. Wood (AWC NOS) and Wood )ltmbtr "itb Stt<l Sidt )ftmbtr (ASD and LRFD)
ll\e wood to sreel stngle shear lateral destgn values are derennined from the AWC National Design
Specif~eatlon for JV()()(/ Constn1CIIO, (NOS) using a steel side member on ooe stde.
Screw size: #12 (AWC NDS Section 12.3.7.1 )
Nom1naJ dtameter D = 0.216tn
RootdiamererD,.=O.I71tn
Required screw length: Full destgn shear (Z) = IOd (only wf'len using tabulated values)
Minimum design vruue (Z) = 6d = 0.6V(NDS Secuon 12.1.5.6)
Length requ1red
= uack thJc.kness + sheathmg thu::kness + recet\1tng member penetration
=0.0566 + 0.75 +(10)(0.17 1) =2.517 in
Note: Use IOd fOt' penetration as a staning desjgn esttmate. Stnce
caJculating the capacity, one could use a shallower penettatioo, say 2.5 or
2 .0 inches. Use a #12 x 3.0-m-long screw mlnlmum > 2.517 indles.
Side member thtd::ness:
CFS track (Grade. 50): 54 mtl (0.0566 tn)
Wood member (Douglas F1r-Latch): thickness nm less than
screw length
AISI 5201 Table 1!2-1:
CFS track design thickness= 0.0566 in
(Mintmum base steel thickness= 0.0538 in)
AWCNDSTablei2M:
Stnce the shear wall design teqUlres the use of 16 gaugeASTM A653, (F.,.=50 kst) material, and
AWC NDS Table 12M deSign values (Z)are ba<ed on the steel Stde member material beingASTM
A653~ (Fy= 33 ks1), whtch is d1tl'erent from me CFS track belng used tn lh1s destgn example, lateral
destgn values wsll have to be calculated ustng NOS yteld equations.
ThedO\vel beanng strength is derived from AWC Technical Repon 12, Table AI and ts equal to
2.2F/ 1.6.
Dowel beanng strength (F,):
Assoc&ated with ASTM A653, Fy= 33 ksi, F.,= 45 kst, is established as betng 61,850 psi
As:soctated with ASTM A653, Fy =50 k_(ii, F.,= 65 k.o;a, is estabhs.hed as being 89,375 psi
The screw bending yield strength, F1•• ts dependem on the diameter of the screw. As the sctew
diameter becomes smallet, the bending yield strength typtcally increases. NOS Table 12M fOotnmes
provtde m.tnlmum screw bending yield strengths ranging from F,. = 70,000 pst tor sc-re\vs around
IA-ioc,h dtameter up to 100,000 pst for scre\VS Lhat are around %-inch diameter. Hagher screw
bendtng yield strengths are posstble by us1ng proprietary screws lhat have product evaluauon repons
(example: ICC, IAPMO) by an ANSI-accredited product certification oompany that could allow for
htgherscrew capacittes. See NOS Appcndoc E and AWC Technical Repon 12 for mote mfOrtnatton
on dowel beartng strength and bending yteld strength.
# 12 Screw Z values
tn
54-mtls CFS ( 16-gauge or 0.0566-mch destgn thickness)
# 12 serew tn 0.0566-mils ( 16-gauge) CFS (F, =50 ksi)
#12 scte\V length= 3.0-u\Ch-long screw
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
A\VC NOS \Vood Strtw Laltni.l Otsign V~tluts
D, =screw root diameter= 0. I71 tn
TI2.3.1A, §12.3.7.1
11 = bearang length: side member= 0.0566 1n
TI2.3. 1A
1.= bearing length =3.0 - 0.75 -0.0566= 2.19
F_ =- mam member 00\.\•el bearmg. sueng.th = 4650 ps1(Douglas Fir-Latt.h: G =- 0.50)
TI2.3.1A
Tl2.3.3
Fu =Side member dowel beanng strength= 89,375 psi (ASThl A653, Grade 50)
F,.= dowel bending yield s1rengdl (0.177 < D S0.236) = 80,000 psi (used)
TI2M FootnOie 2
=dowel bending yield strenglh (0.142 < D $0.177) = 90,000 psi
R,= reducuon term= K0 = 100+0.5 = 10(0.2 16) +0.5 = 2.66
TI2.3.1B
R, =F,./Fu =4650/89,375 =0.0520
R, =I./I,= 2. I9/0.0566 = 38.69
TI2.3.1A
TI2.3.1A
A\VC NOS Yitld Limit Equation Conslanu (CaJrulaltd from Tablt Jl.3.1.A Notes)
k, = 0.821, k,= 0.4769, k, = 14.977
Table 3-42. NOS Table 11.3./A ..Yield Lmut Et[uauou.t-
Yield Mode
Z lateral Value
(ASO)(Ib)
(Single shear)
NOS
Equation
Number
I.
654
12.3- 1
I,
325
12.3-2
II
267
12.3-3
m.
282
12.3-4
Ill,
123 (governs)
12.3-5
IV
168
12.3-6
Per NOS Table I 1.3. I, the calculmed lateral design value, Z, is to be adJUSted by applicable factors,
including !he load duration !actor for ASO, Ct> Therefoce, the adjusted lateral design value, Z' = I23( 1.6) =
197 lb for ASO. For LRFO, Z' = 123(3.32X0.65XJ.0)=265 1b.
Note: Per the AWC NOS Comme1ttaty Section Cl2.3.the design values for screws are uule.xed to average
shon-term propomonal li.mu test values div1ded by I.33. The NOS Table 12.3. 1A equations represent the
poss1ble screw fiulure modes, but do not account for the load duratioos ()'plcally assoc1ated with wood
des1gn. The I.33 factor IS based on the ongtnal reduclJon factor of I.6 tncreased 20 percent for normal
loading and expe.nence ( 111.6) x 1.2 =- 0.75. The 20 percen[ inc-rease wa~ tntroduced as pan of the Wotld
War II emerge.ncy 1ncreac;e 1n y..'QOd design values. After the war, the 20 percent1ncrease was codtfied as 10
percent f« change from pennanent to normal loadmg and I0 perce1tt fOr experience.
Instead ofcaJculanng the screw shear design values tn wood when corutecung CFS framing members. as
shown above. the des1gner can refer to the the Cold-Fol'med Steel Engineers Institute (CFSEI) Technical
Note F 101-12 (September 2012). which provides screw design values for both withdray..-aJ and shear.
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
Author's Distussion: A\VC, NOS, a nd A IS I Sttt.l Tbh:kntss and Connutioas
ll\e various st~l thtcknesses(inches) listed
tnAWC NOS Table 12M "Wood Sc.rews" are dtfferent from the
typical steel mtnimwn and design thicknesses (milsfinches) u.:oed by CFS stud tnanufacturers. AWC NOS
references steel up to 3 gauge (0.239 inc.h} in terms of gauge (inches), whereasAJSI and the CFS industry
have moved away from ustng the term gauge and refer to mili (Inches) for thtd::ness.
W11en using multiple matenal~ designers need 10 evaluate vanous matenal standards to detenn1ne which
material might control the destg.n for each element tn the load path. Both ASD and LRFD melhodologtes
are available for cornmon construc:uon matenaJs. C-are should be taken to ensure appropnate adjustment
factor's are used that maght be matenaJ spec.1fic. For example, NOS Section 11 .2.3 states the follow1ng:
11.2.3 Dtsi.gn of Mttal Parts
Metal plate~ hangers., fasteners,. and other metal pans shall be designed in accordance wtlh applicable
me.tal design procedure.~ to restst f3ilure in tension~ shear~ beanng (metal on metal), bendmg. and
buckI ing. When the capacity o( a connecuon IS con[foiJed by 1netal strength mther than wood strength.
me.tal stte11.glh shall not be multiplied by the adjusunent factors in thts Spec:tfication. In additiOJ\ metal
stl'ength shall not be 1ncre.ased by wind and eatthquake factors tf des&gn load..(O have already bee.n reduced
by load combination factors.
the wood members wiJI control screw design values as compared 10 the CFS jj·aming elemenl~ or
tndtvtdu.al screw shear capacny. When connecung CFS maxenals together. the screw sht."ar capactty can be
the governtng factor, depending oo the CFS matenal suength and thtdmess used. A con1panson example
ts shown below using 33 ks1 and 50 ksi CFS n1aterial and comparing 1110 the CFS tndlV1duaJ screw shear
TyptcaHy~
capacity per Technical Note F701-12.
fiS Screw: shank diameter = 0. 164 tn
54--mJI CFS fram1ng nuuertals: Fl'= 33 ksi. F. = 45 k.(OJ
Fy=50 kst~ F. =65 kst
Beanng: P* =2.7t1dF., = 2.7(0.0566)(0.164)(45,000)= 1128lb
pM = 2.1t,dF., = 2.7(0.0566)(0.164)(65,000) = 16291b
LRm = $P~ =(0.5)(1128) = 564 lb < 639 < qP-=(0.5)( 1629) = 8151b
ASD = P,/Q =(1128/3.0) = 376 1b < 426 < P,/Q =(1629/3.0)= 5431b
As shown alxwe, the #8 screw is v:eaker than the 54 mil matenal with a y1eld strength of F1 =50 kst~ but n
is stronger than Fy = 36 k.(ji.
Autbor•s Dlstussion: Scrtws in CF'S and \Vood
As l'l()(ed in the prevtous discuss•on. a companson of the SC-rew nonunal shear design vaJuei in thin CFS
members (P~land wood members (Z) shows that the shear design values'"" typically be governed by the
lower wood vaJues. A# 12 screw was selected for tts larger wood shear transfer design values as opposed to
usmg #S screws~ whtch was the fastener stze used for the CFS stud shear v.aU sheath1ng attachment. Table
3-43 shows a compan:son of ASD and LRFD shear design values tbr the #12 screw with AlSI-calculated
design values. Even when Jrtclud111g the sets:mic shon-tenn load duration (C0 ) InCrease of 1.6 for wood, the
#12 screw shear value is snll less than the shear strength 1n the CFS ftaming marenals.
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
Tab/~
1-·13. 5Cnrw shtar design ,·a/ut comparison behJ·ten wood and CFS
Screv.• Size
Type of Framing MatenaJ
ASD Destgn Values
LRFD Desigtt V..lues
# 12
54-mtl CFS (F.= 65 ksi)
P.J0=7 151b
1072
# 12
Wood (Douglas Fir-Larch)
Z' =( 1.6X 123)=1961b
Z'=2561b
Shear transfer belween fmmed floors is based on ASD loads sance lateral fotces are uansferrang anto a wood
member for !Ius example. LRFD loads could also be used d·1he destgtter prefers 00110 use AS D. AJSI
S400 requtres the bottom t:raek shear anchorage to have the available strength to reststlhe lesser of (I) the
e.xpected strength of the: shear wall or (2) the load de.termjJ\e'd using the seismic load comb1nalions with
ove(strength. Whe-n the latter governs. which is typical, ASCE 7 Sectjoo 2.4.5 permits the ASD capacity to
be multiplied by 1.2 when using the ASD seisantc load combinations with overstrength. So the #12 screw in
wood strength of I96 pounds then becomes 235 pounds.
Table 1-14. Shear \FOil sheathing shear design
o. Load
Shear
(0.7FIL)
(pU)
Oesagn
#8 or # 10
Capacil)'
Screw Spacing
Combinauons
Level
l: Shear Wall Forces
ASD=(0.7f)
(lb)
(pit)
(in)
(pit)
3rd Floor
6395
276
356
6
690
2nd Floor
15,393
6 16
7 10
3
1540
lSI Floor
19,643
786
876
2
1965
Shear from
Floor shear transfer:
Jrd Floor: #12 screws reqwred:
V= 1540 plf/(2 rowsx 235) =3.28 screws/ft and 12/3.28 = 3.66 in o.c.
Use two rows of# 12 scre\VS at 3 inc.hes on center, staggered.
2nd Floor: # 12 scre\\o'S required:
V= 1965 plf/(2 rowsx 235) =4.18 scre"<ift and 12/4. 18=2.87 tn o.c.
Use two rowsof#l2 screws at 2 inches on center. staggered.
The second-floor screw spacing should probably be three rows spaced at .t. 1nches on ce-tuer staggered. The
wood member would need to be v.l)de e.nough to accommodate three.row'S of fa<»e1lers.
3. Wood (AWC NOS) ond Wood Mtmbt.r w;lb Wood s;dt Mtmbt.r (ASD)
Dependtng on how the floor/wall framing assembly is consln.lcted, there may be several filler wood
plares-t()-bonom of wood nm joist connec:Lions usmg lags, screws. or \1•'3.11sheathmg natlmg to complere
the required shear transfer. The load path through the s.hear wall that is selected for the lateral design
detemhnes the coonect.ions.
Assumtng the: wall nm jo1st sits on a wood plate to distribute gta\'lty loads to the top of the CFS stud wall.
then a wood-t()-Wood s:hear ua.nsfer may be requ1red. ThlS would be a more trn<llti01lal \\-all sheathmg
anachmem us1ng na1lulg of the '"-all she:athmg to the wood plate wtth lags/scre\\'S for the shear U11J1Sfer
from the wood plate to the underside of the rim JOtst. The \ttall sheathtng should have edge nad mg to both
the wood plate and CFS top ttae.k. The shear ttai'ISfer to Lhe nm joist v.'Ould be through screws/lag.~ insmlled
through both the CFS top ttaek and wood filler plate to the understde of the run joist.
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
Scnw IH.sigu Sbur Valut (\Vood-to-\Vood Mtmbtr)
AWC NOS Tablel2l:
1Y.!-in side membe.r to main member
#12 screw=0.216-in diame-ter
Z= 1471b
The \\'OOd stde member-to-wood matn member shear value. Z. e.xceeds the calculated CFS 54-mll CFS s1de
member-to-wood main member connection where Z equals 123 pounds. llterefore, the SIJacing of the # 12
screws will be governed by Zequals 123 pounds.
If the wood rim JOist is lhe same width as lhe stud \WII.lhen the CFS top track would be screwed dtrectly to
the understde of the rtm joist.
Ancbor-Bolr Otsigu-Sbtar Tra.nsftr Sbur \Vall Bonom Track to Coutrttt BaSt
L Conc.rttt
ACI 318 CbJipttr 17 aad
me §1905.1.8
The anchor bolt tn concrete shear capacity io; determined ustng ACI 3 L8 Chapter 17. IBC Secuon I905.1.8
amends ACI 318 Sect1on 17.2.3.5.2 by a(khng exempuon.o; for the calculauoo of the Ln-plane shear strength
of anchor bolts used to attach wood sill plates or CFS tracks to coocrete. ll'll.s exc.epl:ion permns the
in- plane sbear suength to be determined
tn
aecordanee withAWC NOS Table 12E and AISI 510013.3.1,
respectively. rather than comply wnh ACI 318 Section 17.5.2, ..Concre.te breakout strength of anchor
m shear;• Secuon 17.5.3, ..Concre,te pryout strength of anc.hor tn shea(' Md Secuon 17.2.3.5.3. The
applicab1hty of e.ac.h of these revtsed proviSIOns to hght-frame CFS structures 1S dtscussed below.
2018 LBC Section 1905.1.8 Exceptions I, 2, and 3 d1scuss thedesignofshearanchorsanchonng lightframe wood mudsills or c:old-fbrmed S[eeJ ttaek.s to a concrete element near the edge oflhe concrete.
Excepuons 2 and 3 for CFS tracks fot both bearmg and nonbearlng walls may be used when the. following
condiuons are sausfied:
2. I. The maxtmum rutchor oomina1 diameter is% tnc.h.
2.2. Ail<'-hoi"S are embedded into c.<mctele a minimum of 11nches.
2.3. Anchors are located a minimUJ'I'I of I~ tnches from the edge of the concrete paralle1 to
tl'le length of the track.
2.4. Anchors are located a mmimUJ'I'I of 15 anchor diameters from the edge of the oone.rete
perpe.n(bcular to the length of the track.
2.5. The stll track is 33 to 68 nul designation thickness.
The last pan of the exception iMHcates that concrete anchors less than or equaJ to I tnch m diameter
that attach a y.'OQd sdl plate or CFS bottom traek to a foundauon need .~ot sausty ACI 3 IS- 14 Secuons
17.2.3.5.3(a) throogb (c) 1ftbe des1gn streng.h is determined using Section 17.5.2.1(c), nom1nal concrete
breakout sttenglh 1n shear parallel to an edge.
ACI 318 Stttion 17.5.2.1
"'Coarrt-t~
Brtakoul Sl:rtn.g th of Anchor in Sbtar'"
Stnce the Interior shear wall meets the requ1rements of JBC Seetion 1905. I.8, replacemMt Secdon
17.2.3.5.2. Excepuoo2, thts calculauon is not required but is 1ncJuded for completeness of the design
example. As will be seel\ the shear wall bonom track shear capacity will hmn the shear t:ransfe.r ao;
compared to the anchor strength tn concrete.
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
ACJ Section 17.5.2.1(c) states the method for determtrtlng the a.tl('.hor-boh shear strengdl when the shear
force is parallel to the slab edge. Since this is an mterior shear wall, the anchor-bolt slab-edge distance
ts fiOl a tactor, and anchor bolts wtUbe spaced far enough apan such that spacmg will not be a hmnauon
factor either.
Podaum slab:
Thickness: 12 ln
Concrete: F
; = 4000 psi
He.x-he.aded anchor bolt= d# = %-1rt diameter= 0.625 m (heaght of hex head= ~ tn)
End dtstance = 15d= ( 15)(0.625)= 9.375-in mintmum =minimum dismnce
from perpendicular edge of coocre.te slab to cente.rhne of anchor bolt
Bolt load beartng length ln concrete= 1, = 7 tn- ~an= 6.5 in
A= 1.0 (normal wetght concrete)
A
I'd,= -!t.'f',.,.vV"" 11'*'•• vr•
ACI318 Eq 17.5.2.1a
A_
1(1.,= 1.0
ACI318 §17.5.2.6
'-''I'l' = I.0 (conservative)
•'l'h = ~(1.5<',1)/!J,
ACI 318 § 17.5.2.7
ACI318 §17.5.2.8
= Jl4.06/12.0
= 1.082
I~= (7(/)d.)0 '
.Ji:p..fjjcc.,)..,
ACI318 §17.5.2.2
= (7(6.5/0.625)., Jo.625 )( I.O)J4000 (9.375)"
= 16,048
c.,,= 15d=9.375ln
1.5t'.,1 = 14.06 > 12-tn thickness of podium slab~ therefore., h.,= 12.0 ln
A~= 2(1.5c, 1)h, = 2( 14.06)( 12) = 337.44 m'
A-=4{<, 1)'= 4(14.06)' = 790.7 m'
,.,. = (337.4n90.7)( 1.0)(1.0)( 1.082)( 16,048) = 74091b
t= 0.70
,,,,=0.70(7409) = 51861b
ACI318 §17.3.3(c) Con<htion B
2. CFS Conne-ttioos (A lS I S l ~tt f ion J J '"Bolttd Conntttions"') Sbtar Tra.n.sftr Shtar Wall
Bottom Tratk to Coocrtlt BL~~
Bolt notations:
d =nominal boh diameter
1 =-uncoated
dlick:nes:s of thinnest c:onnected material
e =distance measured tn line of force frotn center ofSlandard hole to nearest edge ofadJacent
hole or to end of connected pan
F.= tensile strength of connected CFS member as defined trt AISI S JOO Secttons A3.1 and A3.2
C = be<lring factor (AISI 5100 Table 13.3.1-1)
"'I= modd\cauon factor for l)'pe ofbearmg connecuon (A lSI S 100 Table J.J.J.I -2)
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
Boll spa.cmg and bolt-hole size:
Mtnimum bolt spacing (ce-me.r hne to center hne) =3d
Mtnimum bolt edge and end dtstatlce from centerline of bolt = 1.5d
Bolt-hole SI.U: (bohs larger than !h-m d1ameter)
Using Slandard bolt hole= d + 'A. 1n
AISI S IOO §13. 1
AISI SIOO §13.2
AISI S 100 Appen<hx A TEla
Bolt beanng (AISI S 100 Section 13.3):
Track mater1al t111ckness: 1 = 68 rotls = 0.0713 in
Track nlate:nal F,-:: 50 kst, F.= 65 ksi
Anchor boh dianleter (D)=% in-= 0.625 m
A• =gross cross-sectional area of bolt= 0.3 1 tn2
Bolt type: A307 bolt. Grade A
Bearulg mength (res.smnce) WithOUt consideration of bolt hole defonnaLion (AISI S I00, Section 13.3 . 1):
AISI 5 100 §13.3.1- 1
P,.=-CmrJtF.
Q
= 2.50 (ASD)
~
= 0 .60 (LRFD)
D/r=(0.625/0.0713)= S.n < 10; lherefore, C = 3.0
AISI 5 100 §13.3. 1- 1
"'1-= 1.0 (w1th washer under anchor bolt nut)
AISI 5 100 §13.3.1-2
P. = (3.0)( 1.0)(0.625)(0.071 3)(65,000) = 8690 lb
ASD:
P,/11 = 869012.50 = 3476 lb
LRFD:
~P. =
0.6(8690) = 5214 lb
Author's Distussion; Since this 1s a she.ar wall.lhe desJgner should also consider the AJSI bo1t shear
equation~ wh1ch cons1ders bolt hole deformauon 10 the sheet meml (Al Sl S 100 Secuon J3.3.2). Shear
walls resisong seismic forces are deformation ltml[ed by the ASCE 7 story dnft limitation. Due to
poss1ble combu\at10ns of bolt dia.me[er and sheet metal th1ckness.,. AISl S 100 Section J3.3.2 requtres that
the calculated beartng strength ...with consideration of bolt hole deformation•• not exceed the calculated
beanng strength ••wnhout cons1derauon of bolt hole detbrmauon... The bolt hole cons1denng deformauon
calculation ts shown below.
P.=(4.64at+ 1.53)dtF,
Q = 2.22 (ASD)
~
AISI 5 100 §JJ.J. 1-2
= 0.6 (LRFD)
a= 1.0
P. = [4 .64(I.OX0.0713) + 1.53)(0.625)(0.0713)(65,000) = 5390 Ib
ASD:
P,111 = 5390/2.12 = 2428 lb < 3476 lb
LRFD:
~P.=
0.65(5390) = 3504 Ib <5214 Ib
AISI S IOO §13.3. 1
AISI S IOO §13.3.1
As can be seen~ the d1tference 1n shear capactty 1s S1grU6cant when the designer c.onsiders bolt hole
defortnation tn thts design example ( 1857 pounds vs. 2000 pounds). The value for n is larger and the value
fOt' q. is smaller to address the bolt hole e-longauon deformation limn of0.25 inch. which occurs before
reaching the hmued bearing strength of the sheet metal l11erefore, sheet metal beartng suength wtth
c-0ns1derauon ofboh hole deformat100 governs 1n thts destgn.
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
Controlling Sht11r Transrtr Duign (Con(rftt Embt-dmtnt and Cold-Formtd Stttl Tta(ks)
The bottom-track anchor-bolt attachment to the concre-te pod1um deck IS restriCted to strength destgn su1ee
2018 JBC Secuon 1901.3 requtres the anchor to ooncre.te to be macoordanc:e wuh ACI318. lneanchorboh govemtng strength design shear value IS controlled by the boltoJn lnlC-k's 68-mll thtck:ness and is
limited to 3504 pounds (LRFD).
Sbtar Transftr Summtl)'
Table 3-45. Opti011 3 shear n·al/: shear walljasf~11~r spacmg a1 floor /In~
Level
O,(F)
LoodCombo
(lb)
Anchor
Type
Connectot
Design
Number of
Capacny (Z') Fastenets
Max. Spacing
Allowed
(in)
Spacing U<ed
ASD-Superstrueture-abcwe Pod1um Deck
3rd floor
17,238
#11 screw
2351b
73
8.5
8
2nd floor
38,483
#12 screw
2351b
164
5.6
5 tn on center, 3 to\lt'S
1n on center, 2 ro\~t'S
35041b
20
15.8
LRFD-Coocrete Podium Deck
1st floor
70,153
* -Jn
Diameter
I5 in on centeJ
Anchor Bolt
I . Wall length= 25 ft, O. = 2 .5, shear foroes (F) taken from Table 3-25
2 . Max spacong = (25 ft x 12V(13n - I)= 8.5 in
=(25 fix 12V( 164/3 - 1)=5.6in
3. Max spacing (songle rowofanchor bolts)= (25ft x 12)1(20 - I)= 15.8 on
Author's DiSt":u.ssion: Spactng ofanchor bolts is calcuJated ba~d on applytng omega forces. Practtcal
spaclng of the rutcl\or bolts lS 16 Ul.Ches em center, whtch IS a module oflhe wall stud spactng. \Vlule the
suggested ma.x1mum spacing is I5 mches on center, 11 may be more practical to space the bolt-li at I2 inches
on center.
The shear wall coMect.ion to lhe concrete slab w1U typ1c.ally be controlled by lhe thickness of the coldfonned steel track If the dtameter of the anchor boli lS reduced, and the thickness of the cold-formed steel
track is increased. Jtl'l poss1ble that the anchor-bolt stre.ngth would c.onrrolthe OOJtneetion destgn to the
concrete.
In thts des.:tgn e.xample, the cootrolli.ng shear design value is per AJSI SIOO Secuon J3.3. 1 or the beartng of
the steel bonom track agaon.<t the anchor bolt (3504 pollll<is < 5186 pounds). A lower h<>ttom track bearing
capac1ty again.')~ the anchor bolt could be obtained by using a track with an F1 = 33 ksi (F.= 45 ksi). but
then 11 would not match tlte \\-all stud material stre1lgth used (F1 = 50 ksi. F. = 65 ks1).
12.9 DISCONTINUOUS SHEAR WALL
The Option 2 she.ar \\'all
is a diSCOntinuous shear wall lhat s.:it~ on a steel beam at the seOOJ'Id floor, which m
tum lS supponed by a first-floor steel column.
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
A disconhnuous she.ar \~t'all situation occurs when lhe shear waH directly below is eliminated and the
remaining shear walls a[ the Hoot below are offset either horizontally to a parallel grtd lt1'1f' (Type 4
honzontaJ 1rregulariry) or honzonmlly along the same grid line (Type 4 vemcal irregularity). In thts
design exam.ple.lhe Option 2 ~'Wall at lhe first floor is shorter than the shear v.-allammedt.ately above
at the second- and thard-tloor levels,. so thts would be cons1dered a vemealerregularity. The desagn of the
supportmg members of the d•sconHnuous shear ·wall has to c:omp1y with ASCE 7 Sect1on 12.3.3 ..Ele-ments
Supponing Disconunuous Walls or Frames."
11te steel beam and column supporting lhe dJSContinuou.~ shear wall are requlred to be destgned useng
the seas:mic load combi.nauons wnh the oversttength factOI'. The followmg elements supponing lhe
discotUmuous shear wall ate requt.red to be desig,ned to reslstlhjs amplafied seismic load:
I . The secottd-ftoor steel beam.
2. The first-floor steel columns.
3. The beam-to-oolumn connections.
4. The c:onnecuon of the column to the podium level.
The amphfied se1s:nuc load ts not used for the hold-down ot chord studs or their anaehmem to the
supportmg member for light-frame wood stud shear walls. AlSI S400 requires that the CFS-framed shear
wall hold-downs ( uphft andtorage) and CFS dtord studs (verucal boundary compresston elements) have
the available sttertgth to restSt the amplified se1smtc force.
13. Shear wan Deflection
AJSI 5400 §E1.4.1.4 "Design Deflection"
The de:stgner of the rigidity of a CFS-framed shear wall must constder seve.rat factors: bending detbnnatioo.
shear deformauon. fa.(itener slip, rutd hold-down movement. When there are multiple shear v.>alls along the
wall lme. the fOrces are to be distnbuted based on the rigtdtty oflhe shear walls aJong that waJIIine.
The rigidity of the indwidual shear wall ts de.termined from lhe deflection of the shealhed CFS stud shear
wall. AISI 5400 provides design equat.ioos to esumate the de-flection of either sheet steel or blocked wood
struc.tural panels (WSPs) on CFS stud shear walLs. The destgn equatJons can be found tn AlSI Section
El.4.1.4 "Destg~t Deflecuon."
The dtstribunon ofhOfi2ontaJ shear forces between shear walls tn a l1ne ts not covered 1n lhis design
example.
AJSI 5400
Deflection is an tmpottant cons1deratioo in design~ e.xcesstve deftection of lateral-force-res:isung syste.nts
may lead to uttdestrable build1ng performance or even collapse. AISI S400Sections E l.4.1.4 and E2.4. 1.4
state lhe detlecuon of a blocked wood strucTural panel or steel-sheet-slteathed CFS-srud-framed shear wall
may be determined by us1ng the four pan equat1on shown here.
+q "..,
I
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
The four pans of the equation take into account hnear elastjc: c.anulever bendmg. linear elastjc: sheathing
shear, overall nonltnear effects. and the top-of-\vall horizontal deftecuon contribution fl-om lhe ve.nical
dlsplacemenr or the holcklown or ue-down syste.m (uphft anchorage).
21·h1
Ltnear elastic c:anulever bending: S1 = - - 3E1A('b
.
~
L1near elasuc sheathtng shear. o1 =
0
(1)1 1
G
1-h
p ,........,
Overoll nonlinear effeCIS: S, = ro,'"• ; , • ,
(~)'
Lateral cocuribuuon from anchotagelhold-down defonnarjon: S4 = ~o..
ASCE 7 Table 12.12-1 hm1ts the story drtft for Rtsk. Category lt lhree-story bulldtngs to 0.025h.o~ where
hD. is lhe story he1ght. Th1s three-stocy hght-fi"ame buildtng 1s considered to comply with the Table 12.12-1
requ1rement lhat tnterior \valls. parution~ cei l 1ng~ and exterior walls have been des1gned to accommodate
the story dnfts. I r n did not. or the number of stones v.ti fi. ve stones or mote. the stol)' d.rd't would be
limhed to 0.020h.u-
The shear wall defteetion (S) os calculated per ASCE 7 Equauon 12.8-15:
Shear wall deftecuon cakulation:
2
Forst story: chord studsAc....,Ac=(IO) - 6005200-68 =lOx 0.7643 on =7.643 on'
AISI 0100 TI-2
Serond B story: chord studsAc.w,Ac= (6)- 6005200-68 = 6 x0.7643 in' =4.586 on'
A lSI 0100 TI-2
Second A story: chord studsAa.w,Ac= (6) - 6005200-68 = 6 x 0.7643 in'= 4.586 in'
A lSI 0100 TI-2
Third story: chord studs A"""-" A<= (4) - 6005162-54 = 4 x 0.5560 on'= 2.224 on'
AISI OIOOTI-2
Structural I (4 ply, 32/16) plywood sheathong shear modulu.s per AISI S400Commentary Section El .4.1 .4
and AWC SOPWS Commentary Table C4.2.2A:
G = G,J/t= (45,500 lb/on)/0.469 on =97,0 15 pso
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
Table 3-16.
Wall
lleogh~
Floor
Level
h
(in)
Op110n 3 shear ,.·a/1: deflection "ariables
LRFD
S!O<)'
LRFD
Level
Wall Requtred Uphft
Width, Strength. Force.
p
b
(on)
(lblon)
(Ib)
,,
Rod Net
TensjJe Area
Sheatlung
Shear
ThJckness. Modulus.
Are~ At"
I
G
(in')
(pso)
(in)
1
(A,)(on Y
LRFDRod
Take-Up
Elongauon(~H
Deftecuont•J
End
Studs
Gross
Sheathing
Device
d,,
d,.
(on)
(in)
0.030
132
(54f"
300
32.8
3305
2.224
0.469
97,015
0.334~).018
2nd A
120
300
73.3
8604
4.586
0.469
97,015
0.763/().046
0.030
2nd B
120
300
73.3
8604
4.586
0.469
97,015
0.33410.107
0.030
lSI
120
300
93.5
16,542
7.643
0.469
97,015
0.763/0.088
0.030
3rd
See 001es for Table 3-47.
Table 3-17.
Op11on 3 shear ,.·a/1: deflection "ariables
CFS
Frammg
Frammg
Modulus of Faste-ner Destg.nauon
Fastener Oestgnation
Elastieny, Spacing, Thickness Shealhong Spaeong
Thickness
Aspe<t Ratto Sheathing
Floor
£,
s
Factor.
Factor.
Factor.
Factor.
Material
(i)l ; i/6 ""= 0.033/t_, ""= ((hlbY2)"-' Factor. oo"
(in)
Level
(psi)
(on)
p
,...,
3rd
29.5E6
6
0.054
1.85
1.00
o.611
0 .469
I
2nd A
29.5E6
3
0.068
1.85
0.50
0 .485
0 .447
I
2nd B
29.5E6
3
0.068
1.85
0.50
0 .485
0.447
I
1st
29.5E6
2
0.068
1.85
0.33
0 .485
0 .447
I
Notes for Table 3-4 7:
I . Rod Jength at third-floor bridge block=- 4 ft 6 Ln =54 in.
2. Vertical rod elongatoon =a,.= PUA,£.
3. Modulus of elasucny (£)for tie rod =29,000,000 pso.
4. Reference.manufacturer"s evaluation repon e\'aluated to JCC-ES AC316 for take-up device tnittaJ
(seaung Lncremem) deftec.uon. !!.k> and allowable load deflectton. !!.,.,. to determine the total device
deflection,~,., whoch forlhtsdesign example osa,,. ~.=a,=~.+ ~A (PJP,). Shrinkage and
eslitnated build1ng settlement are to be taken as pan of the deflection computation afa take-up
device lS not used.
5. Total ven:tcal deformauon of uphft. anchorage system= 8, =-rod elongauon (0.,) +toW. take-up
device detlecuon (O,.J +stud beartng crushing at wood floor+ bearing plate crushing at wood tloor.
The wood crus.hmg terms are not shown tn this design example. but should be added.
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
For Jrd-floor shear wall:
2l·lr1
S, = - - = 0.0026 on
3£1.4~b
Sz = ro1 ~ 1 pGt~- 0.03 14 in
S,=ro,g'• f
,•
•(iJ =0.06non,~=67.5forpl)"'oodshealhing
AISI S400 §EI.4.1.4
h
132
S, =bS' =
(0.018+0.030)=0.0211 on
300
There are ackhtaonal contrtbutors to the deflection of multistory shear Y..alls that need to be taken into
accoun~ tnc:ludtng rotation at the top of the shear wall below; a-x tal shonentng and lengtl\ening of the
c.ompressK>n and tension chords, respecuvely. oflhe shear wall below~ and hold-down deflec.tion of the
shear waJis below.
2nd-floor shear \Vall addjtional top-of-wall deflection contttbutors:
1. Rotauon at top of 1st-floor shear wall due to canulever bending caused by seasmtc ovenuming
Forces. Angle of rotation for a fixed-base, cantde\•e-r beam wnh a concentrated load attl~ end:
a= PL11(2£1)
I for !he shear "~II =Ajl/2
a= PL1/(E.4,1>1)
a,.,~= 28,061 lb x(l20 in)' /(29,500,000 pso x 7.643 on' x(300 in)1 ) = 1.99E-5 radoans
Top of2nd-tloor shear waH defleeuon due to rotauon a1 top of 1st-floor shear v.-allas then
8•..._:: = h2 X a 1 = 120 tn X 1.99E-5 radjans = 0.0024 tn
2. Adduional rotation 31 top of 1st-floor shear wall caused by ten~ion and compressaon from 2nd-
floor shear wall chords:
T1 =86041b C2 =(328,1391b+ 130,341lb)/"_.3 fl= 19,9381b
Table 3-358
S...,..,=PU.4£
sdd:r;.IT=
8604 lb X 120 in/(7.643 ln2 x29,500,000 psi)= 0.0046 tn forT lst-ftoorshear wall chord
S,.,.,c = l9,9381b x 120 uv(7.643 on' x 29,500,000 psi)=O.O I06 on forC 1st-floor shear wall chord
Add.U10nal deflecuon at top of 2nd-floor shear wall due to lst-ftoor shear \WII chord a.xu1l
defom\ations:
S,....,.., =(S...,.,+ s..,.,c)(h,lb,) = (0.0046 in +0.0 106 in)(l20 on/300 on) =0.006 1 in
3. Addtt.ional2nd-ftoot shear wall deftecuon due lO Ist-floot tte-down system defOrmation. Note
that this accounts tOr only the rod elongation and take-up device deflection. It should also
1ndude stud beartng crushing and be.aring plate cru.~ 1.ng at wood floor when they oc-cur.
Deflection per Table 3-46 = S, = 0.088 in+ 0.030 on= 0.118 on
Top of2nd-ftoor deflection due to tie-down system deformauon is then
S,_1,._1 =S.x (lr,lb,)= 0.118 in x(l20 in/300 in)=0.0472 on(0.0304 in for 2nd-Floor Wall Optinn I)
256
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
For Jrd-floor shear wall:
2l·lr1
S, = - - = 0.0026 on
3£1.4~b
Sz = ro1 ~ 1 pGt~- 0.03 14 in
S,=ro,g'• f
,•
•(iJ =0.06non,~=67.5forpl)"'oodshealhing
AISI S400 §EI.4.1.4
h
132
S, =bS' =
(0.018+0.030)=0.0211 on
300
There are ackhtaonal contrtbutors to the deflection of multistory shear Y..alls that need to be taken into
accoun~ tnc:ludtng rotation at the top of the shear wall below; a-x tal shonentng and lengtl\ening of the
c.ompressK>n and tension chords, respecuvely. oflhe shear wall below~ and hold-down deflec.tion of the
shear waJis below.
2nd-floor shear \Vall addjtional top-of-wall deflection contttbutors:
1. Rotauon at top of 1st-floor shear wall due to canulever bending caused by seasmtc ovenuming
Forces. Angle of rotation for a fixed-base, cantde\•e-r beam wnh a concentrated load attl~ end:
a= PL11(2£1)
I for !he shear "~II =Ajl/2
a= PL1/(E.4,1>1)
a,.,~= 28,061 lb x(l20 in)' /(29,500,000 pso x 7.643 on' x(300 in)1 ) = 1.99E-5 radoans
Top of2nd-tloor shear waH defleeuon due to rotauon a1 top of 1st-floor shear v.-allas then
8•..._:: = h2 X a 1 = 120 tn X 1.99E-5 radjans = 0.0024 tn
2. Adduional rotation 31 top of 1st-floor shear wall caused by ten~ion and compressaon from 2nd-
floor shear wall chords:
T1 =86041b C2 =(328,1391b+ 130,341lb)/"_.3 fl= 19,9381b
Table 3-358
S...,..,=PU.4£
sdd:r;.IT=
8604 lb X 120 in/(7.643 ln2 x29,500,000 psi)= 0.0046 tn forT lst-ftoorshear wall chord
S,.,.,c = l9,9381b x 120 uv(7.643 on' x 29,500,000 psi)=O.O I06 on forC 1st-floor shear wall chord
Add.U10nal deflecuon at top of 2nd-floor shear wall due to lst-ftoor shear \WII chord a.xu1l
defom\ations:
S,....,.., =(S...,.,+ s..,.,c)(h,lb,) = (0.0046 in +0.0 106 in)(l20 on/300 on) =0.006 1 in
3. Addtt.ional2nd-ftoot shear wall deftecuon due lO Ist-floot tte-down system defOrmation. Note
that this accounts tOr only the rod elongation and take-up device deflection. It should also
1ndude stud beartng crushing and be.aring plate cru.~ 1.ng at wood floor when they oc-cur.
Deflection per Table 3-46 = S, = 0.088 in+ 0.030 on= 0.118 on
Top of2nd-ftoor deflection due to tie-down system deformauon is then
S,_1,._1 =S.x (lr,lb,)= 0.118 in x(l20 in/300 in)=0.0472 on(0.0304 in for 2nd-Floor Wall Optinn I)
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
3rd-floor shear wall additional to~f-waJI deflecuon contnbutors:
I. Rmation at top of 2nd-floor shear \~tall due to c:anulever bending caused by se-ismtc overturning
fOrces. Angle or rotation for a fixed-base. cantilever beam w1lh a c:ooc-entrated load at its e.nd:
1
a= PL 1(2£1)
I tor !be shear wall = 11,1>112
a= PL11(£4,b')
a,.'""= 21 ,990 lb x (120 tn)'/(29,500,000 psi x 4.586 m' x (300 on)')= 2.60E-5 radians
Top or3rd-ftoor shear wall deftectiott due to rotatton at top of 2nd-floor shear wnll is then
()lOW= h 2 X a 1 + ~~ = I20 lOX 2.60E-5 tadlat'IS + 0.0024 in= 0 .0055 U'l
2. Additjooal rotauon at top of 2nd-floor shear wall caused by te-nsion and compresston from Jrdtloor shear waJI chords:
r, = 3305 1b c,= (IOS,339 1b + 32,321 lb)/23 t\ =61 161b
Table3-35B
()<llds-t = PUAE
5,......,1. : 3305 lbx 120 tn/(4.586 in' X 29,500,000 psi) :0.0029 on for T2nd-ftoor shear wall chord
S,..._1c= 61 161b X 120 tn1(4.586tn1 x29,500,000 psi)= 0.0054 tn for C2nd-ftoor shear wall chord
Addttional deflecuon at top of3rd-floor shear wall due to 2nd-floor shear wall chOfd axial
defoonauons:
s.....,., = (S-.,., + S""'~"')(h,Jb2) + S,_, = (0.0029 tn + 0.0054 tn)( 120 tn/300 tn) + 0.0061 in
= 0.0094 in
3. Addll!onal Jrd-ftoor shear Vi-all deflection due to 2nd-floor ue-down system deformation. Note
lhatlhis accounts for ooly the rod elongation and take-up dev1ce deflecuon. It should also
1nclude stud beartng crushing and beanng plate c.rushing at \vood floor when they occur.
Deftec1ion per Table 3-46 = 1i, = 0.107 tn + 0.030 tn = 0.137 on
Top of Jrd-ftoor deflection due to tie-doY.n system deformation is then
S~tllO-> =
1i, x (h,lb2 ) + 11,_1 ,._1 = 0.137 tn x ( 120 in/300 on)+ 0.0472 tn = 0.102 in
Amphfy story dnft us:utg ASCE 7 Equauon 12.8-1 5 to ensure n lS Less than the aUO\'I'able story drtft, !!(If.
11. = 0.025/ou = 0.025( 120 in)= 3.0 on (firs! and second floor<)
= 0.025( 132 m) = 3.3 tn (third floor)
s•.= c;a.,;1,
ASCE 7 §12.12- 1
ASCE 7 Eq 12.8- 15
I,= 1.0
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Design Example 3 • Thre&-Siory Ught-Frame Mulifamity Buildng Design Using CokJ.Forrned.Ste/Wall Framing
and IMlod Floor and RoofFraming
Table 3-#J. Slunmaryoftop-tf-wa/1 dejlecuons
Floor
Level
s,
s,
s,
s,
(in)
(1n)
(in)
(1n)
s~
(in)
s.....
SnJs
s..,
8,
(tn)
( in)
(in)
(1n)
II,
(m)
SrOilll<Sl'
3rd I
0 .0026 0.0314 0.0677 0.02 11
0.0055 0 .0094 0 .1020 0.2397 0.9587
3.3
Yes, OK
2nd A
0 .0021 0.0254 0.1076 0.0304 0.0024 0 .0061 0.0304 0.2043 0.8172
3.0
2ndB
0 .0021 0.0254 0.1076 0.0548 0.0024 0 .0061 0 .0472 0.2455 0.9820
3.0
Yes, OK
Yes, OK
1st
0 .0016 0.0213 0.1041
-
3.0
Yes, OK
0.0472
-
-
0.1741 0.6970
I. 3rd-ftoor shear wall top-of-\WJI deftecuon calculauon ba<;ed on 2nd B rod size.
Each hne of vertical e-lements of the se1smic-force-resisung system complies wnh allowable story drift.(j per
ASCE ?Table 12.12-1.
Th.s des1gn example uses Type I shear walls, and the AISI 5400 deflecuon equauon .s for thou type of
wood structuraJ panel or steel-sheet-sheathed CFS-frarned shear Y<'all. However. while lhere are currently
no provisions g1ven tn AISJ S400 to calculate the deflectiOn of a Type II shear \vall. tl1e de:s1gner might
consider using a similar methodology as shown inAWC SDPWS Secuon 4.3.2.1 where v and b tnAISI
5400 Equat1on El.4.1.4-l are equal to,. tn AI 51 S400 Equation El.4.2 .2-l and the sum ofw1dths ofType II
shear wall se:g_~nems. IL,. respecuve-ly.
St:lsrn.ic St-paraHon
The story dr1fts are also used to detenmne the s1ze of any requJCed suuctural st!Sn'uc sepamuon between
adjacent budding strucTures as well as se1smic joints be1ween mdiv1dual build1ngs. ASCE 7 Secuon 12.12.3
requ1res lhe m1ntmUnl suucturaJ separation between buildings ofS,. =CdSm.Jl,. for each story level of
the bUilding where &n." mdudes bolh the uanslauonaJ and amp1ified torsional disp1acements of each floor
level and roof of the build mg. S,. is lhe minimum setback requtrement tl'om the property hne~ unless other
budd1ng code requu-emems stipulate a larger setback.
The requu'ed separauon distance (S.,) between adjacenl edges of structures on tl'le same property is
calculated using the equatton:
A5CE 7 Eq 12.12-2
8,.1 = Ma'<tmum displacement of adjacent structure I, as determined by an elastic analysis
8.w2 = Maxtmum dtsplacement of adjacent structure 2. as determu'led by an ela.'(i{!C analys:1s
The swnmatton of the BwJdmg B Opuon 3 shear wall ela.o;tic stoty drrfts equals 0 .240 + 0.246 + 0.174
= 0.660 1nches.. Assumlng this shear wall deflection is represental:ive of all the shear walls in ButldJng B and
account(i for amphfied totSion of the floor/roof diaphragl'ns, the roof~ level deflection-; are as fOllows:
I. Roof-level maximwn elasuc horizontal displacement (deflection)= 0.660 inches.
2. Roof-level mmumum inela.<tlc response horozontal displacement= (4.0)(0.660YI.O = 2.64
ifK'hes.
The calculated Buildtng B rooftop dnft of2.64 tnches IS Significantly less than the code default value hmit
of0.025H, or whm would be ( 10.0 + 10.0 + 11.0)( 12)(.025) = 9.3 1nches.
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Design Example 3 • Thre&-Story Ught-Frame Mulifamily Bu;lding Design Using Cold-Formed-Steel 'WaN Framing
andv.bod Floor and Roof Framing
The width of the se1smic joiJU at the br1dge between Buildings A and B wdl have to be determtned from the
calculated sei:sm1c hori?.on.tal de-flections of both Building A and BUIIdtng B 1n acco«lance with the budding
code requ1reme.nts for bulldtng separanon. Dm as can be seen for the Dudd1ng B shear wall calculation.
sizmg the seismic joirll width based on. the actual shear waJI deflecuons of both Building A and Buildtng B
\\1111 ikely be Significantly less than swng the setsmic joint Width ba'lied on using the maximum alJoo•able
shear wall deftecuons permuted by the budding code.
14. Discussion: Framing with Cold-Formed Steel
14.1 WALL STUD BRACING
Bec-ause CFS fi'amtng members are [)'ptcally made up of stn.gly symmetrtc (C-studs, uacks) or potmS)'mme.tric shapes (Z-purltn~ angles) with tlun elements. they often require suppleme.ual: btaclng. In the
case of ex1etior wall studs 111. beanng-wall construction, intermiuertt bracing is requtred 10 resist v.'eak-a.xis
buckJtng due to axialloods and torsion resulung from lateral loods (wu'ld or seismtc) nor applied through
the shear cel'lteJ oflhe member 11\ either the strong or weak ax1s. The foliOVtingdiscuss1on provides an
overvtew of stud-walJ bracmg des1gn.. For funher discusston, 111-ustrauons. and destg.n examples, the reader
ts re-ferred 10 Cold-Forn~ed Steel Frammg Destgn GUide (AISI 0 110). A more in-depth discussion of
bracing can also be found tn Bracing Cold-Form~d Stttl Strucltu-es-A Design Gmde~ by Thomas Sputa
and Jennifer L. Turner.
1\\'0 basic
melhods are recogntzed for btactng CFS studs: sheathing braced destgn and all-steel destgn (also
referred 10 as mechanical btactng). Sheathing braced design relies on the diaphragm action of the sheathing
anached (0 the studs and top at'ld bonom tracks. Mechanical broctng consist'li of t11e addttton of dlscrtle
bracing element~~ typic.aJiy comprised of a ootnbanatton of bridging channe-l, fta( stmpping,. connectors, and
blocking, that are de~ugned to prov1de lhe reqUlred restStance.
Sheathing Bractd Des-ign of CfS \Vall Studs
Sheathtng braced des1gn ts presc.npuve tn nature and 1S addressed m the Norlh American Slandurdfor
Co/d-Formtd Steel Structurol Frami,g (S240) and North American Suuulanifor Seismic DeSign ofColdFonntd Stu/ Strueturol Systems (S400). Since sheathing is typically reqUifed fOr architectural rea"'iins, use
of a sheathtng brac-ed design ts an.ractive for COSl reasons. However, studtes show that gypsum board that IS
commonly used ma>• not provide adequate bracing for members with a design thtckness greater than 0.0346
U'l(:h (20 gauge or 33 mtls).ln addiuon, many fee.t lhe bractng perfonnance of gypScwn board detertora[es
quickly under cyc.lic loadang and 1f exposed (0 motsture. Even when sheathing bt'aced design is used, IllS
standard prac'[ice for lhe engaoeer to provide some amount of mechantcaJ bractng to align members and
provide stabII iry durtng construction.
AISI 5240 Section 83.2.1.1 addressesax.alload design of the CFS S<u<l. The connecuon of !he s<ud ends 10
the bonom track and top track prevents twisl of the stud atlhe ends. The type of sheathing and fa.'litener stze
may limit the maxtmum axtaf capacity the stud can suppon unless supplemental steel bractng ts provided
be-tween the CFS studs. The ma.x1mum spactng of the sheathing fastener is 12 mc.hes on center. as typically
occurs in the field area oflhe sheath.tng attachment to the stud. Closer spactng of the sheathtng fasteners is
n01 c-redtted to tncre.ase the srud a.~ tal capactty.
In shear walls, it is oommon to have wood shealhing on just ooe side. while gypswn wall board JS prm•tded
on the opposite stde as the waJI finish ma(etial, wh1ch may be used to brace the CFS srud flanges. The
wood sheathmg fastener attachment, gypsum board fastener attachmem, and CFS Stud need to be checked
to detemune the l1miung a.~ial capacny of the CFS stud. The lesser \1alue of the three checks shaJI be used
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259
Design Example 3 • Jhree-Stoty Light-Frame Mu/6/amily Building Design Using Colt:J.Formed-Steelwall Framing
andYKiod Floor and Roof Framing
ll\e width of the seismic joint a1 the brtdge betweto-n Buddin~ A and B will have to be deternuned from the
calculmed sersmtc horizontal defteclions of both Butlding A and Buildmg B 1n 3ccordance v.ith the building
c-Ode requ1remems fOr buddmg sepa.ratton. But as can be see-n for the Buddtng B shear wall calculation.
siztng the seismic JOint width based on the actual shear wall deflecttons ofbolh BuildtngA ilnd Buildlng B
will likely be stgnificrunly less than stzing the seismic jouu 'A'idth based on using the maxtmum aJJoo·abfe
shear wall deflections permitted by the building code.
14. Discussion: Framing with Cold-Formed Steel
14.1 WALL STUD BRACING
Because CFS tronung members are typically made up of Stngly symmetroc (C-studs, tr:lcks) or pointsymmetric shapes (Z-purhns>angles) wnh thi1l elements. they often requjre suppleme.ntal btactng. In the
case of exterior wa11 studs in bearing-wall construc:uon. il'tterminent bracing is required to resist weak-axis
buckJtng due to ax tal loads at'l.d torsion resultmg from lateralloods (wutd or setsmtc) not appl ted through
the shear cemer of the member tn enher the strong or weak axis. The followmg discussion provides an
overview of stud. wall bracing destgn. For funher discusston> illustrations. and design eX3tnples> the reader
ts re..ferred to Cold-Fonned Stttl Froming Destgn Guide (AISI D 110). A more tn-depth dJsc.ussion of
brncmg can also be found tn Bracing CoJd. FormM Stttl Structures- A Design Gmde. by Thomas Sputo
and Jennifer l. Turner.
Two basic methods are recognized for bfactng CFS studs: sheathing braced design and aiJ.steel design (31so
referred 10 as mech3nical bracing). Sheathtng braced design relies on the dtaphragm action of the sheathing
anached to the studs and top and bonom track.-;. Mechruucal bractng oonststs of the additton of dtSCrete
btactng elemenL~ typically compnsed of a combinauon ofbrtdging channel>flat strapptng.. connectots, and
blodang> that are designed to provide lhe requ1red resistance.
Shu thing Bractd Dtsig:n of CFS \ \'811 Studs
Sheathing braced design l'i pre-sc-rtplive in nature and is addressed tn the North Amencan Standardfor
Cold-Fomrtd Sttel ~inteltual Fmming(S240) and Nw·th Amtrict.m Sumdardfor Selsmlc Desigtt ofCold·
Fontted Steel Structurul Sy$lems (S4-00). Since she3lhing is typically required tOr architectural reasons, use
of a sheathtng braced destgn lS an.mclive for c-OSt reasons. However, studjes show that gypsum board thmts
commonly used may no. provtde adequaxe brocutg for members wtth a design tJuckness gre.ater than 0.0346
tnch (20 gauge or 33 mtls).ln addinon> many t'i!el the bracing perfoonance ofgypswn board de-tenorates
quickly under cyc.hc loading and If exposed to nlotsture. Even when sheathing bmced design IS used. n is
sta1'ldard practjce for the engineer to provide some amount of mechantcaJ bracing to align members and
prov1de stab!I tty dunng construcuon.
AISI S240 Section 83.2.1.1 addrcssesaxial lo3d design of the CFS stud. The connecuon of the stud ends 10
the bottom track and top track prevents tWist of the stud at the ends. The type of sheathing and faste-ner stze
may lim1t the tnaximwn axial capacity the srud cat1 suppon unless supplemental steel braclng IS provided
between the CFS Studs. The maxtmum sp3Ctng of the she.ath1ng fastener is 12 snches on center> as typically
occurs in the field area of the sheathing attachment to the stud. Closer spacing of the sheathtng fastene.rs IS
not credited to mcrease the stud axl31 cap3City.
In she:lt Y.'illls. tt is oommoo to have wood sht.--athmg on just one stde, while gypswn wall bo3td IS provtded
on the opposite s1de as the wall firush material, whtch may be used to brace the CFS stud flanges. 1he
wood sheathtng fas1ener attaehme:nt. gypsunl board fa~ener illt3chmen[, and CFS stud need to be cheeked
tode.tenntne lhe limning axtal capacity oftheCFS stud. The lesser \'3.lue of the three checksshaJI be used
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259
Design E.xample 3 • Tluee-Sio<y Light-Name MultifamHy Buildng Design Using Cold-FOnn<d-Steelw..JI Framir"Q
and lo\bod Floor and Roof Framing
to determine the stud a:oaJ capacity. l1le des1gner may ooly use twice the bracing strength of the we,aker
auaehed shealhlng whe.n detennm.ing the CFS stld axaaJ capacny~ assuming sheathing occurs on both faces
of !he shear wal I.
Su'!Ce the desJg.n equauons for determining axu:tl capactty are based on a stngle fastener anached to one s1de
of the stud, when that sheatJung is attached to both Hanges of the wan stud, the nomtnal a.xial capacny of
the wall srud is twice the catcuJate<l capac.ny based on lhe single fastener attachment to JUSt one flange of
the CFS stud. The AlSI 5240 Commentary discusses how to detemune the sheatht.ng bracing streng~h of
the stud. Secuon B 1.2.2 of the new 5240 CFS hght-frame stru1dard tlOW permits load beanng and tlon- load
bearing walls to have sheathang braci1tg on ooe wall stud flange and d1screte bracing on the other flange
rather than requinng enher an aU-steel broced destgn ofstud flanges or an all-she.atlung braced design of
stud flanges for wall stud brocll1:g. Ho\\-eve.r, the authors recommend usmg enher an all-sheathtng braced
design or an all-steel braced design oflhe stud bracing for (I) load bearing and (2) n<>o>-load beanng walls
destgned for Lateral forces 1n excess of 6 psf unless research is 1denufied substanuaung waH srud broce load
shanng bet\•.'een the sheathmg on one face of the wall and dtscrete bracutg on the other face of the wall.
Thls CFS-framed shear wall design example ut.ihzes honzonta1 steel suap bractng to brace the CFS beartng
stud waJI flanges Y.'ith a full-depth block placed every 8 teet. with a minimum of two blocks per shear "all,
along the length of the. wall. The vemcal spac1ng of the c.onnnuous honzontal suaps at each floor level was
selected based on the engineer a.~urtng the chord-srud assembly would have sufficient capacity lD support
the tmposed compres:siye axu:d and ftexuta1 destgn loads.
As dtsc:ussed earher, sheathtng braced destgn could be used a t the third-floor shear wall level of the butldu'lg
su'ICe the axla1 1oads are s•nall. It 1s preferable to stay with a steel-braced des1g.n for two reasons:
I. For shear walls., u 1s prefemble to have the sheathing u..c;ed to resist JUst the seism1c latera1 forces
a.;; opposed to incluchng lhe sheathing to also stabilize lhe steel studs for axtal loods.
2. It the wall sheathing has to be removed for any reason (example: access a leakmg p1pe or the
gypsum \\1111 boatd beromes wet),lhe CFS studs are still braced by the continuous suappsng.
Stul (M « h aniui)-Brarrd D<slgn o f CFS WaU Studs
Rtqutrements for the design of aU steel bracing. or mecharucal bractng, are included i.n AISJ S 100 Secnon
C2. The destgn approach is somewhat ditlfrent for the bracing of axially and latera11y loaded studs. For
axial loading. bmc1ng is destgned for a percentage of the axial load 1n order to prevent buckltng of the stud.
sinular to AJSC 360 tbr compresston member nodal bractng. AISI S 100 provides des1gn reqmrements for
bolh the suength and suffness oflhe bracing. For bc'acmg of laterally loaded studs, bracing tS designed to
resist the torsion created due to lhe faa that the lateral load is not applied through the shear CMter. In thiS
case, AISI S 100 provides design requtrements for the stre1lgth of the steel bracing. but does not currently
tnclude requirement'i tor st.iftbess.
Design ofbrac.ng can prove challengmg for several reasons. Pr1or to lhe 2007 edttion, the AISI
Spectfication requ1red that members be braced~ but no design equations were provtded. As a result.
many engmeers designed bractng based on the ..2 percent rule," or other rules of lhumb, or prescriptive
guideIines. With the addiuon of explicn strength and stiffness requirements, destgners must now be more
ngorous wnh their bractng destgn. AISI OliO comains des1g,n examples lhm •nclude bractng des1g.n. Recent
testing of typical brocing detatls has been performed, and AlSI as worktng on improved bracing provisions
based on this research. AJso,lhe calculation of bracing system sufthess can prove to be quite complicated
and Inaccurate. Severnl manufa.c.turers h~we recently staned to pub!ish destgn values for propnetar)•
bndgtng connectors in accordance with AJSI 5915, T~st Sumdardsfar Through-11~-Wtb PundJOtd ColdF<>rn~d Steel n~l/ Stud Brulgmg
260
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Design Example 3 • Jhree-Stoty Light-Frame Multifamily Building Design Using Colt:J.Formed-Steelwall Framing
andYKiod Floor and Roof Framing
ll\e two most oommoo steel braclng systems consist of:
I.
l~mch U-channel bfidgtng lnstaJied through the web punch-out of the studs and attached to
the stud web wnh a connector. The connector generally extends across the stud web out to both
stud ftange.<.
2.
Flat-strap bmctng installed on each face of the stud in combmauon with intennmeot blocki ng
bemreen the CFS Studs.
Both of these bracing methods ha\<e their ad\13-tttag.es and disadvantage.~ and the declston of which to u.'ie l'i
often based on designer and/or installer preference.
U-Channt-1 Bridging
ll\e use of U-channel bridging IS probably more common. but requ1res c.oordinauon with other butld1ng
elemMts tn the srud bays. tncludmg plumbing pipes, electrical conduit-. and conrjnuous rods used for
ovenummg restraint.
I. For the design ofa..;:ialty loaded Sl~ tl1e bracing force accumulates 0\'er a number ofstuds,
requinng that the b!'aciJ1g be anchored to the structure pericxhcally. One bridgtng anchorage
option is to use suongbacks (CFS Stllds turned 90 degrees wnh1n the stud wall fol' strong axis
bending to res1st stra~braced loads) pe.nodically along the wall, and this ts shown in Figure
3-27. This aspect ofbmctng destgn is sorneumes ove-l'looked, bllt is critic-a1 to the perfoonance
of the btacang system.
2. For the destgn of late-rally loade-d sruds using U-channel bridging, tors1on in the stud 1s res1sted
by bending oflhe U-channel, so no addttional anchomge is required because rhe system ts in
equilebrtum wnhout rhe anchorage. The only horizontal shear l'eaction that needs to be- resolved
is at the ends of the U-channel run. or how the U-channe1 connects to a cross \\'311 Ol the last
stud in lhe waJI.
3. While not a code limit., tlus U-channe-1 bndgtng system IS typicall)•ltmued 10 CFS stud depths
of 8 inches or less.
r.
"'\
\
••
I
1...
\
a:s U·Channel
Bridging
\
~·
l ...
Strongbad< CFS
stud
I
~·
.
!It :
\
Bottom track
below
'enp
Figure 3-27. CFS s1ud-wall U-chtJJmtl bndgmg-s1rongback anc/l()r(Jge (plan
-.·}~f'}
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Design E.xample 3 • Tluee-Sio<y Light-Name MultifamHy Buildng Design Using Cold-FOnn<d-Steelw..JI Framir"Q
and lo\bod Floor and Roof Framing
Flat-Stra p Bradng
Flat-strap bracing on e.ach side of the CFS stud wall \\1th pencxhc stud blockmg ts gene-rally sttffe.r lhan
U-channe-1bridging and eliminates cortfl icts within lhe stud bays~ but insmllatton requires acc.ess to both
srdcs of the wa11.
I. Periodrc anchorage to the structure is required when tl\e engu\eer lS bracing a.xtaUy loaded CFS
studs. srmilar to U-channel bridgtng. Anchorage of bractng can occur several ways:
a. Srrongbac.ks (CFS studs rurned 90 degrees wnlun the stud \WII for strong axis bend1ng to resrst
suap-braced loads) spaced penodically along !he wall. l111s is shown in Figure 3-28.
b. Periodrc Oat-strap diagonal bracing attached to the flat-strap stOO bracing exte.rKhng and
aoc-hormg to the top and bouom tracks of the stud wall. as shown m F1gure 3-29.
c. Stud brnetng might also be ac.hteved usjng sheathing diaphragm ac.tJOn (0 del aver loads lO the top
and bonom uack.o; of the stud \\'all.
d. Stud btactng can also be achu~ved by anachtng bracit~g to some other structural member (Sleet
column/post, shear wall, etc.) that
is substantjally anchored itself between floor levels.
2. But unhke U-channel bridgang. Hat-strap bracmg only acts in renston. so thts system also
teqUJres tntermutent blocklng (bndgtng) between the CFS studs when bmctng laterally Joaded
studS. whtch may have an tmpact on tnsaalla[ton costs.
CFS blocking
at 8'-0" o.c.
Flat-strap bracing
each side
Strongback CFS
studs
along wall
Figure 3-18. CFS Slud· wallj/at-!lltlp bracing and blockmb),_Slrongbaclc anchorage (plan -.·ieu~
3. While flat-suap and blockmg btactng were used for this destgn example. some may prefer to
use a btidgtng and clip bracing sol utiot~ as discussed prev1ously. to avoid the wall sheathing
be.U1:g ••pushed ouf" by the tJuckness of the CFS strap and screw anachjng It (0 the CFS stud or
CFS trnck framong.
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Design Example 3 • Jhree-Stoty Light-Frame Mu/6/amily Building Design Using Colt:J.Formed-Steelwall Framing
andYKiod Floor and Roof Framing
.,
I
.,.,,..
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/
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~ IOl1'le IQUYJai.IOn
Figure 1-29. CFS :rtud-walljltJt·SilTJ.p brucmHiagona/ strap bracing tmd10r0ge (dt!ration new)
Detenmnauon of CFS stud capacities (axial and flexure) based on steel-brace spacing is ac:comphshed
most eastly ustng computer programs. Most commercially a\•a1lable computer programs for CFS srud
design allow the destgner to evaluare c:ombmattons of axlal loads and lateral loads based on various ventea1
spactngs of the CFS stud Hange braces to determine the nomu1al capacities oflhe selected CFS stud.
14.2 WALL STUD HEIGHT: BEARING AND NDNBEARING STUDS
Out-of-plane wind loads and axial gravuy loads are typically going to go\'em the des1gn of the-CFS waJI
studs. other than poss1bly the des1gn of the boundary studs at lhe end of shear walls for m-plane seism1c
loads. Design aids, such as the Steel Stud Manufacturers As:socun:ion (SSMA) and Steel Frammg Industry
Assoc1auon (SFIA) publication ..Product Techn1cal lnfonnauon." prov1de CFS srud-des1gn mformauon
fo( use 1n imenor nonbearang stud des1g.n., exterior nonbearang cunall'l·\\-"all des1g.n. and bear1ng stud waU
design.
The sae ofbeanng stud walls will depe-nd on the a.x1al load on the stud as well a.'i the out-of-plane loading..
SSMA and SFlA's "Product Technlcallnfonnauon•· pubhcat1on cypically prov1des desag.n values ror
bearing stud walls up to 16 feet tn height. This is probably a good cut-offheight. lftaller stud walls are
requ1red due to floor-to-ft()Q( heights, then an altemative gravny fram1ng system should be con.o;idered.
such as beams and C-olumns. to support the floor/roof gravity loads. The CFS studs would then be strictly
cot'l..~idered as nonbearing srud ""ails and only need to be designed for out-of-plane loads.
Taller beating studs may be possible-. but the stud depth will likely ha\'e 10 be deepe-r. or lhe thickness
will need to be tncteased to restst out-of-plane forces. As a rem1nder, CFS studs used 1n sheathed shear
walls have hmits on the CFS stud thickness. so at taller he1ghrs a deeper stud \Vlll be required to keep
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Design E.xample 3 • Tluee-Sio<y Light-Name MultifamHy Buildng Design Using Cold-FOnn<d-Steelw..JI Framir"Q
and lo\bod Floor and Roof Framing
the lhick:ness of srud walhtn thickness ltmttation.o;. Tile alternative solution as to make the sheill' \vall CFS
studs nonbeanng tn order to ac-hieve taller sn.Kl heights. The shear wall CFS studs would be infill studs
anstalled below a floor/roof beam supponed by columns or posts to transfe-r gravity loads to the foundatton
or podium level in th1s design example. Where irtfill studs ate used below a beam,lhey should be installed
only aft<r !he ftooc/roofbeam above, and after all (flbmary framing dead loads to the beam have been
placed.. to mmamize gravuy rudal1oad transfer to the studs. Bolh non-shear and shear ·wall mfill studs may
sttl11\eed 10 be c-hecked for tmposed lwe loods from the ftoor/rooffranung.
14.3 FLOOR SYSTEMS
ThiS design example intMttonaHy used enganeered wood 1-Joist.. rather lhan CFS joiStS for the floor
frammg syste.m to de.monstr.ue that wood joist-; c-an also be considered for use-. Generally the asswnption
when usang CFS beartng stud walls ts that the floor joist will also be ofCFS. which ts not ahvaysthe case.
Ftgures 3-30 and 3-31 shO\.\' platform and ledger CFS ligtu-fmme consU\.ICuon methods, respecu,,.ely, of
whic.h the ledger method currently seems popular for three-story and taller structures. CFS bearing studs
have also been ulied wllh other floor systems such as concrete topptng O\'er steel deck as well as precast
plank floors. Each of these floor systems has special coosiderations for the e1lg.i.Me.r to keep 1n m1nd when
designing the CFS bearing studs to support the floor framing.
I. CFS Floor J oist
A lS I Sl40 Stt'tion 81.2.3 " In-Lint fnuu.ing"
Su'ICe the CFS top track does not have much flexural capactty about its weak axts to suppon the reacuon of
horiZOntal fram1ng members. the centerline (m1d-W1dth) of the. CFS floor jo1st, CFS rafter. CFS ttuss, and
CFS structural wall stud (above or beneath) must be aligned vertically with1n the tolerances of AISI S240
Figure B 1.2.3- I.
As c.an be see.n m Figure 3-32. a fair amow1t of eoord~nauon IS required in placing the frammg members to
lunn lhe amoum of offse-t between them. Even w1th a %-inch offSet, there can be a fatr amount of reduction
1n the axial capacity of the srud.
Figure 3-30. FlOOr-JOist plaifomr-framtd CFS light-frame constnlCtion
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r"'"
~
~
~
pt:
. [..---
-
--=
--""
,_ ......
:.a::..
.,.....,ntu
II(JUOIIfA.
.............
-
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Figun J-J!. .~/SISNO Figurr BI.J.J./: " l,._u,. FromJng·· (fig"" 001111..,•ojAISl)
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Design E.xample 3 • Tluee-Sio<y Light-Name MultifamHy Buildng Design Using Cold-FOnn<d-Steelw..JI Framir"Q
and lo\bod Floor and Roof Framing
2. \Vood Floor Jois l
Si.nce the wall top lrnCk does not ha\•e much fte.xural capacity about tts weak a~lS., typically a wood single
plate or double plate is added to the top oflhe stud wall. On looge-r-span floor joists., where heavy floor Jive
loads occur or the destgn mcludes panttiooloods. the stngle or double wood rop plate may JlOl be adequatetn flexure about the weak axis either. In such cases. studs and jotSLs mUSt be altgned similar to m-hne
fram1ngdescribed pre\•iously. or a full-depth wood nm joist can be framed on top of the wood top plate
with the joists hung from the nm jotst using hangers. The shear strength. fte..xural Slfength. and stiffness
of the ""OOd run Joist allow the floor JOtst to occur anywhere along the length of the rim joist wtthout the
destgner needing to worry about al1gnme-nt w11h lhe waHstuds. This tS conststent wnh AISI S240 Ftgure
B1.2.3- 1 "In-line Fmmutg." whe.re the franung member ahgrunent toleranoe is not required when a
structural load distribt.Juon member IS specified tn accordance wnh an approved design or approved design
standard.
Since the oomtnal w1dths of available wood framing members used as nm JOists are typically narrower
than the depth of the CFS studs (4x rtm on 4-tnch-deep CFS suds, 4x or 6x rtm on 6-tnch-det:p CFS studs_
etc.), the wood stngle OC' double top plrue needs to be cut to be the same \\~dth as the CFS top track This
provides a load path from the natrO\vet rim joist to the fuJ I width of the CFS stud. Where beams frame into
the \\"all run JOist. t11en multtple sruds should occur under the rim jo1st where the beam connectton occurs,
a-; requtred to suppon the beam reactJon fO!ce.
3. Stttl Dt<'k with Conc.rttt- Topping and Prtt.as t Plank FloorS, " llh or widaout Contrttt Topping
CFS stud waHs are used 10 both res1denual and commerctal strucrures to suppon floor systerns constructed
from steel deck and precast planks that may or may not have concrete topping. Fagure 3-33 shows concrete
over a steel-deck floor system frarrung ove-r a CFS-framed wall. The use of these types of floor system..q IS
beyond the soope oflhis destg,n example.
Fi'gt.ll? 3-33. Concrete o,·er !ttel-dtckjloor
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Design Example 3 • Jhree-Stoty Light-Frame Mu/6/amily Building Design Using Colt:J.Formed-Steelwall Framing
andYKiod Floor and Roof Framing
Aulbor's Distuss-ion
Bare preca.~t planks and both steel-deck and precast planks with concre-te topping are floor systerns that ate
sigruficandy heavier than the floor jo1st systems used 1n ttadnionalltght-frame eonstrucLion. light-frame
construction is defined 1n Chapter 2 of the IBC as "'A type of construct1on whose ventcal and hotl2ootal
Slftlc.tural e-leme1\tS al'e primarily formed by a system of l'epelitive wood or cold-formed steel framing
members.... The use of bare pl'eca~t concrete planks and both steel-deck and precast planks wuh a concrete
topptng as a hortzontal fi"amul;g system dearly does not fit withm thiS de.finhton of hght-frame con..">tructJon.
Where CFS swd \\ails. or \\'00<.1 stud \\ails for that maner, are to be used wath tl'le:Se heavier floor/roof
framing syste.m.s, these bu1ldang structures should not be eonstdered ao; light-frame construction and should
not be des1gned to use a response modification coefficient R associated with hght-frame construction.
ASCE 7 Table 12.2.1 provides response mod1ficat"ion coefficients R based on the setsmie-force-restsung
system that the building uses, but does not differentiate a" to the mass oflhe ftoor system when utlliztng
light-frame constructlOf\. As a oomparison,lhe respon.o:;e modification coefficient R for the different Steel
moment-frame ()'pes used 1n high se1sm1c: zones vanes. and theJe are hetght and weight hmitattons. The use:
or hght-frame wood stud walls or CFS stud waJis sheathed wnh shear panels or CFS dtag.onaJ straps may
not be appropriate as the lateral-resisting system for this type of bu1ld1ng system 1n high-se1smic regions.
Late.ral-resisung systems us1ng suuctural steel frames (brace frames, moment frames. and stee.l-plate she.ar
'"''ails). conc.rere. «masonry shear walls are more appropmue to reslstthe se1sm1c lateral hor1zontal forces
induced ti'om the huger mass of these budd1ng structures.
In SOC D. both ordtnary and mtermed1ate steel moment frames v."Ould n01 be permiued tn these structUJ"CS
usmg precast plank or conCI'ete topplng over a steel dec.k, s1nce the ftoot mass hkely exceeds the 35 psf
we1glu hmuation and 1m plied expected ductihty performance of these steel frames associated wnh thetr
respoose modification coefficienlS R. Stmllatly. the implied ducultty performance of sheathed wood stud
shear walls., CFS stud shear walls. or CFS diagonal strap-braced \\'ails. based on the.ir hght-frame respon.o:;e
modttication ooetlkiem R-value.s tOr construction,. may be reduced related to the gre.atel' associated building
mas:s in buildtngs that have. flool'/roofsystems consuucted of precast planks Ol' metal decktng aod preca.~t
planks ·w1th a st.ruct:urol concrete topping slab.
AJSI S400 Section B 1.5 ..Sclsmtc Load Effec.ts from Other Concrete Ol' Masolll)' Compone1us" does pennn
CFS syste.ms to be destgned to l'estst setsmtc forces from other concre-te and masonry c.ompone:nts (other
than walls), 1ncludtng masonry veneers as v.'tll as ftool' systems uula1ng concrete toppmg ovel' a steel deck
or structurol l'etnfon::ed concrete ftoors. The design then has to also be in accordance wuh AISI S 100 and
required deflection ltmtl(i as specified 1n the oonCI'e.te aJld masonry standards or mode-l bujJdmg codes.
AISI S400 Section B 1.4 ...Setsrruc Load Effecu Contnbuted by Masonry and Concrete Walls" is similar to
wha11s permuted by AWC fol' ltght-frnme wood construction (s1ngle-story f1mn, two-story exceptions.,. e.tc.),
which are reasonable ltmats since a combmauon oflateraJ-res1sttng systems are being uul1zed (concrete or
masonry shear '"''ails with hght-frame con<ruucuon wood or CFS-framed shear walls). The aulhor's concern
IS for tal lei' buildtngs that have the heavtel' wetght floor systerns pl'eVJouotly dtscussed that de-pend on oul)'
ustng liglu-frame sheathed CFS or \\ood-frame sl>!ar walls or diagonally strapped CFS studs wuh the
associated response modification coe.fficie.nt<t R tOr the latc,ral-l'esisting system (no COU'Ibuwion of laterolresistmg systems). and achievmg the same expected ducttl ity perfOrmance of the l1ght-frame she-athed shear
,.,'311 system.
ll\e designer needs to use caution v.'he.n ustng a combination oflateral-resistmg systems (examples:
cooc.rete shear walls, steel-braced frames) with the light-frame sheathed she.ar walls or CFS diagonal strapbraced walls, pamcularly wtth these heavy flool' systems tn any se1snuc reg1on. The lateral loads go where
the stitl'ness occ.urs, and the light-frame sheathed or diagonally strap-braced walls al'e not going to have
near the same stttfness and suength as properly detruled masonry shear wall~ conc:rete shear walls. or steelbraced frames.
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Design E.xample 3 • Tluee-Sio<y Light-Name MultifamHy Buildng Design Using Cold-FOnn<d-Steelw..JI Framir"Q
and lo\bod Floor and Roof Framing
15. Discussion: Seismic Joints
As occurs with many large multistory restde.ntial light-frame buildtng project~ there can be sa'eral
budd1ngs on a st.ngle podtum slab. Where the butldtngs ate dtsunctJy separate., they must sat.isf)' the
mtnimwn separatiOn dtstMce between buildJng.c; a.o; reqUJted tn the buddtng code. In many cases, these
light-frame bluldmgs are typtcalty conunuous butldtngs arranged m a doughnut shape or \'tty long
buildings \v;th dtstinct areas where the floor plan ts greatly reduced tn width, often reducing down to 10 or
IS feet or poss~bly narrower. In some cases the buildings are tnterconne-ctOO W1th pedestrian bndges.
Historically most mulustory light-frame buddtngs have ignored the use of seismic joants m those areas
where the footpdm of the bulldmg significantly naJTows re-Lative to the building footprtnt on each side of
this narrowed area Depending on the actual wtdth and length ofthis nrurow buildtng strtp, the buildtngs
on each side oftlle narrow strip area may be behavang more a.;; separate iJldividu.al butldtngs as opposed
to a stngle bui1dlng. possibly movtng tn opposite dtrections during an eanhquake. The opposite-dtrecuon
movement of the butIdings could tear apan the natTow strap area diaphragm or cause 11 to pull away from
one of the butldtngs. This could be particularly true \\f'te.re pedesuian brtdges are used to connect the
''arious floor levels of individual bud dings.
Ifan engineer was destgnang a mullistocy aU-steel-frame build1ng wath concrete topping O\'er a steel deck,
or all-concrete cast-In-place or precast bmldings for a giVen footprtnt, and the engineer detemuned that a
setSmic jou\1 was required, then a multtstory light-frame blllldtng wnh the same footprtnt should probably
also have a seismtc jomt in the same location.
15.1 BUILDING SEISMIC JOINTS
Designe-rs need 10 give senous oon.><;;tde.ration to providing se.ismic jotnts at those locattons where the
ltght-frame butldtng foo1prtnt stgntficantly narrows tn wtdth. When lhe narrowed poruon of the butlding
l~OOI)• 10 to 15 feet wide or less and the rematning tbotpnnt on each side ofthtsnarrowed region is40 10
60 feet wide, the diaphragm cannot be realtSticaUy expected 10 transfer the design loads across this narrow
building wtdth. In such locations a seism1c jotnt should be introduced to allow the bualdtng to move as two
tndtvtduaJ struc.tures. Often the.se narrow are.as are paths of egress. so it may be prudent for the en.gu1eer to
destgn the narrow are.as so that they rematn funcuonal atler an earthquake. Consideratton \\111 also have to
be gtven 10 utjhties passjng across the seismic joint that have to remaln in service after the seismic t'\'ent.
15.2 PEDESTRIAN BRIDGES
In this design example there were t\\'0 separate buildings, A and B, and a pedesttian bridge was pro\•tded
between tl'lfm at the second· and third. floor levels. A se1smic joant was pro\•tded. between the pedestrian
bndge and D'uddtng A at bolh lhe second and third floors. A two-story steel J1l()ment frame was provided
ar lhe e.nd of the bridge at Building A to provide lateroJ stability, orthogonal to the direcuon of the bridge
span. Tile brtd.ge longttudlnal framing me.mbers were checked as drag members to t.ransfe-r the bridge
setSmu:· destgn forces bac.k tnro both the second- and thtrd-floor frami ng le,•els of Bwldang B when lhe
bndge seismic movement as in the direcuon of tJ'Ie brtdge span. As a path of egress, it was detennined that
the two-story bndge should have a seism1c jotnt 10 increao;;e the likelihood Lhe pedestnan bridge wtll re.m.am
fw'!Cttonal after an eatt.hquake.
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Design Example 3 • Jhree-Stoty Light-Frame Mu/6/amily Building Design Using Colt:J.Formed-Steelwall Framing
andYKiod Floor and Roof Framing
16. Discussion: Elevators
16.1 SEISMIC AND FRAMING CONSIDERATIONS
Mulustory liglu-ftaJne consuucuon Introduces issues with elevators and elevator guide-ratl supports.
particularly in high se,ismtc regions. In tigtu-frame c.on.-ruuction,lhe floor construetton io; typically nommal
wood-frame construc.uon. engineered 1-jotsts. or CFS jotsas. When the fl()()( constn.Jctlon IS conventional
dimension-lumber framing. there are issues with vertical shrinkage oflhe floor jotSL and to a muc.h
Jesser e.xrem with enganeered wood-framing products. CFS joists do not have the shnnkage issues bw do
experience senlement as studs settle tnto the tracks if they are not built completely seated. Since lumber
shnnk.age occurs over Hme as the wood franung dnes oUl.lhe elevator cab stops at each floor level will
hkety have to be reset a couple ofumes to account tOr the ventcaJ shrinkage of the wood ftoor fran11ng.
so thai when the ela•ator cab doors open. the ela•ator cab floor is flush with the finish floor 1a'el at each
floor Jeve1 of the building.. The amount of ven.caJ movement Vt'lll be greatest at the upper ftoor lf!\lefs of the
butJding as the amount of venical shnnkag.e is c.wuulatl\•e over muh:iple ftoor levels.
Elevator manutacturets also have concems regardtng attaching the elevator cab guKie-rad suppons tn the
elevaLOr shaft \\'ails to the ftoor framtng, typically not wanung lhe cab gu.tde-rad suppons to be anached
to wood framtng due to concerns wtth wood shnnkage and bec.ause the anac.hments of the wood framing
can become k>ose over time. The elevator cab guide. rail is attached by honzonml bracke.L~ ptovided by lhe
e.levator manufacrurer, to the elevator cab gutde-ratl suppon placed withtn the elevator shaft wall. TypteaUy,
a custom-fabricated steel face plate braclce.t lS destgned by the engineer of record (EOR) and mounted to the
elevatOt' cab gwde-rail support. An openjng ts cut1nto the shaft wall g.ypswn \vall board to fit flush around
the custom fabricated steel face bracke.t to whtch the elevator guide-rail horizontal bracket lS attached. The
custom-fabncated steel face plate allow.; for some flexthility tn locaung the gutde-ratl support in the shaft
wall.
ll\e elevator cab guide tatls themselves are otTse.t tnv.ar<L. away from the elevator shaft \~ails by adjustable
hottZOt\lal gutde-rail brackets, whtch are spac.ed vertically over the height of the ele,•ator cab guide rail. The
gUJde-rad offset from the tbce of the elevator shaft wall typically ranges between 4 and 18 t.nches.
One style of elevator that has become popular is cractton e.levators. whtch elt.mtnates the need fot an
elevalor machtne room. as tl\e hotst machlnery is now installed tn the ele\1ator shaft. SeiSO'Iic horizontal
forces for the attachment oflhe elevator cab gujde raJ! can be in the range of4000 to 6000 pounds tOr
passenger elevators (greater fOI'" fretght elevators) and need to be de\-eloped lnlO the floor-franung system.
The gutde-mll suppon deftectJon erttecfta are stringent. typically linuung the allowable deftection 10 Y. inch
onde.r normal elevator cab runnjog loads. and not .nore than V. tnch due to se.jsmtc forces.
11te applicauon of the ele--1ator c.ab gutde-rad honzontal destgn force~ being offset fi'otn lhe elevator shaft
wall. results tn a twist, or torque. on the gutde-ratl suppon tn lhe shaft wall tn addition to the d1rect forces.
Thts torque force is often overlooked by the destgn engineers. Typacauy. the be.o;t type offramtng members
10 support these honzon1al design fOrces is en.her a square or rectangular hollow struCtural secuon (HSS)
since each has a closed cross section. whach is good for resiSting torsaon forces.
Currently 10 hg.lu-frame eonSt.rUCuon it is common for the destgner to use wood rim beams on the four sides
of the elevator floor opening and then JUSt span the HSS gmde-rad supports \'ertically between ftoor 1e--'els.
Thts may not be acceptable to the elevator manufacturer since the connectiort of the gutde-rail suppon ts to
theft()()( wood nm beam. The load path for transfemng the gutde-rad design forces needs 10 be a •aluated.,
and the comers of the wood nm beam nng typtcaJiy need 10 he mechantcally connected 10 transfer the
design forces be.tween rtng nte:mbets. The ftoor sheathing nailed to the top of lhe wood beams should not be
considered as being suffictent to trano;fer loads acung perpe;ndjcutar to the wood beam as this puts lhe top
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Design E.xample 3 • Tluee-Sio<y Light-Name MultifamHy Buildng Design Using Cold-FOnn<d-Steelw..JI Framir"Q
and lo\bod Floor and Roof Framing
of the \\"OOd beam tn cross-grrun bending. Also. an e-levator cab guide.raiJ suppon attachment point often
occurs at the floor I ine, requirmg a bracket bolted to the Ylood rim beam, wh1c-h may require an addttional
strut framing member behind
n to extend back and away from the elevator shaft to transfer the design
loads tnto the surrounding ftoor.framtng system. A d1scus:sioo with lhe elevator manufacturer is needed to
detemune acceptable floor connecuons.
lfCFS JOists are ulied iltOund the elevator shaft lloor opening. then SU\Its extendtng back into the floorframing system av.>ay from lhe elt\•atot shaft will be reqUJred to transfer the elevator gujde-rail suppon
reacuons mto the surrouodtng floor frrunmg.
h t~ probably best for the e.nglneer to frame the fOur comers of the ele\'ator shaft W1th hollow strue:tural
section (HSS) columns and wnh a nng ofHSS beams around the e-levator shaft floor opening at each floor
level. tncluding lhe elevator overrun above the roof. The elevator guide-rail supports ln the shaft walls, set
to ahgn with the elevatot cab guide ralls, \\'Ould then be e-rected to span venically be.tw·een HSS Ooor and
roof ring beams. The steel rtngs prov1de a load path around lhe perlmeter of lhe elevmor shaft floor operung
to uansfer the gUide.rail des1gn forces mto the floor fram.mg. The steel beams are generally smaJier than
the wood member rings and can mote easily resist lhe design forces causing bending tn the ftoor beam,
posstbly leadtng to fewer strut members for transfernng forces into the surroundjng floor diaphragm.
A steel beam nng ts more expensiVe lhan an au. wood nm beam nng at the floor levels. An all.wood nm
beam nng, when acceptable to the elevator manufacturer. can lead to more complicated bolted conne(:t:ions
to wood framtng members. as well as additional strut nlembers Into the surrounchng lloor framing_ If
conventtonal2x lumbe.r is used for lhe floor~ this can be an tssue stnce the floor joJsts will shrtnk vertically
and the steel framirtg of the ele\'ator shaft wiU not shonen, which can lead to surrounding floors sloping
upv.'3td to lhe elevator shaft. the worst betng at the upper tloor le\·els since the jotst shnnkage tS cumulrujve
over the tloor le\•els of the butldtng_
Whether tlte elevator ts framed wid1 a steel rtng or wood framing aroood tlte elevator shafi floor opening.
the destgner should provtde separate structural drawtngs showtng full·hetg.ht buddtng franung elevations of
the elevator shaft venical suppon framing and gUide-mil suppon points. These drawiJtg.s should tnclude all
requlred frammg c.onnecttons to transfer the gutde· rail reaction fOrces into each floor level of the butkltng
for all elevators in lhe buddtng. Elevators in multistory liglu. fmme struc.tutes are complex, and additional
detailing is required.
The design oflhe elevator is typtcaJiy a deteCTed subminal to lhe buiJdtng depamnem, so something should
be shown oo the structuraJ drawing regardmg the elevaror fram~ng tor the permn set (ele\13tor gutde-ratl
suppons. roof hoist beam. floor.f.ramtng ring. comer steel columns l when used), etc.). This SlfUCtutal
drawing can be revised/updated once the elevator martufacturer and type of elevator is selected, but in the
meanwhtlt. the struCtural dmwtngs Jl(OYJde retttences for inrtial pnctng purposes by the buddtng general
contractor.
16.2 ELEVATOR SHAFT WALLS
Typtcally. CFS C studs are used tbr most stud wall framtng. Shaft walls tbr mechantcal docts or elevators
can be an exception where CFS J studs are uuli7.ed ro allow for dryv..nll placement from the opposite side of
the stud \~,onll. so no scaffolding has to be set up in the shaft to install the drywa11s.h.!athing. The fire raung
of the shaft may tntroduce furring c.hannels on the CFS swds as well. The designer needs to determine if the
shaft wall of the elevator or mechanical shaft should be tncluded as pan of the shear \"-all system as lhe-re is
no testing of a sheathed shear \~;nU ustng J studs or C studs w1th funing channels between the sheathing and
the
c stud<.
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Design Example 3 • Jhree-Stoty Light-Frame Mu/6/amily Building Design Using Colt:J.Formed-Steelwall Framing
andYKiod Floor and Roof Framing
17. Items Not Addressed In This Example
The tbllowang uems are not addressed m thts example but are nevertheless nettssa.t)' for a complete destgn
of the building's laterol-resisung system:
I. Wind desJgn to compare to setsmiC to determine which governs.
2. Design of the structural podium deek for shear wall O\•enurntng and upltft forces.
3. Design of shear wall connnoous rod anc.horages to a concrete pod tum deck.
4 . Speciallnspecr.ions and SutJctural Obse.l'\'ation ofbutlding construction
5. Dtstnbmion oflateral fOrces beru•een ltght-fra.me sheathed shear walls along the same wall line.
6. Destgn of setsmtc JOtnlS.
7. MIXed systems (when used wnh light. frame sheathed shear walls):
a. Rigtdny of coocrete/CMU walls Wtth light-frame walls at the same floor level.
1.
Otten there can be a couple of concrete or CMU shear walls that occur above the budding's
matn podium level for supporting a structutnJ slab at the ftoor level above. This can occur
over a drive\vay leading to a subterranean parking level below the main podium located
at grade le\•el or a high bay for a first-floor retad space where the seoond floor 1s lhe matn
pcxhum deck level fot the restdential frammg above. The SlnJCtural slab and concrete/CMU
walls are used to provide required occupancy and fire separruton bern•ee.n the residentl31 le\•el
above and the parkmg leveVretad space beiO\\'.
11.
The relatJVe ng1dJty of the concrete and CMU walls relarave to lhe light-frame sheathed shear
walls needs to be addressed concemmg how design forces are distributed between v.'aliS.
b. Rigtdny of sttuctural steel moment ttames\btaced frames with ltghl-frame walls at the same floor
leve.l.
L
Where strucrural steel moment frames or SlfUCruraJ steel-b!aced frames occur at the same
floor l!!vel as the liglu-frame. shear walls. the relarave ngidity betwee.n moment frames. btat..--ed
frames. and Iight-fi'ame sheathed shear walls needs to be addressed COilC.emang how design
forces are dtstributed berw~n frames and walls.
18. References
See the reterence Hstang at the from of this design volume.
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Design Example 4
Masonry ShearWall Building
OVERVIEW
Whtfe masonry can be used to construct buildings of all types and s12e~ u is most ollen used m low-rise
construcuon. pa.rucularly tbr restdemial~ re-tail. ltg.ht oommerctal. and tnstllUuonaJ bu•ldtngs. Thts type of
c.onsu\tction has generally had a good earthquake pe:rfo.rmance record. The ware-house bwlding considered
ut this example is a l)'pical ooe-story masonry "box butldtng~· wtth a wood-ttame roof.
11te destgn of masonry tS based on the 20 I6 edttton ofTMS 402. But/ding Cod~ Requirem~11ts for M(ISt'Jnry
Structures. and TMS 602. Sptc!firolion for Ma.tonry' Strucllut!.t. v.tlich are de\•eloped by The Masonry
Socie[)' (Th1S). The code contatn.'> destgn provtsions tOr structural destgn using both strength-design and
aiiO\\o'3ble stress design procedures. Thts example wtll focus only on strength-design procedures.
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Design Example 4 • Masomy ~arWaN Building
OUTLINE
I.
BuHdmg Geometry and loads
2. Calculation oflhe Design Base Shear and Load Combinations
3.
Des1gn of Walls to Resist In-Plane Seismic Loads
4.
Des1gn of Walls 10 Resist Out-of-Plane Seismic Loads
5. Out-of-Plane Wall Anchorage
1. Building Geometry and Loads
1.1 GIVEN INFORMATION
60 tee-t x 90 fee-t in plan wnh typical floor and roof fi"aming shown in Figures 4- I and 4-2. An
elevauon oflhe wall on line A 1S shown in Figure 4-3.
Walls coosist ofS-inch-lhick (nominal), sohd-grOlned. medium-weight concrete masonry units
(CMU) laid 10 running bond with TypeS monar. Masonry has a specified compressive sue.ngth.
r_, of2000 ps1 With unns li:ud tn runntng bond. Steel remforcemem JS Grade 60 ({y = 60 ks1).
Figure 4-4 shows a secuon through the '"ails on lines I and 3.
The self-weight of the roof and roofframmg is 17 psf~ and the roof live load is 20 psf.
l1le buildtng 1s located 1n Oakland. California, on a s1te that JS classified as Sne Class D.
1.2 BUILDINGWEIGHTS
Su'!Ce the budd1ng has tnasonry shear walls and a roof With wood SU'UCtural pane-l (WSP) sheathtng, the
roof diaphragm may be 1dealozed as flex1ble as permmed by Section 12.3. 1.1 of ASCE 7. M1n1mwn Des1gn
Loads and Associated Criteria for BuHdings and Other Structures. This means that lhe dtaphragm spans
bet\l,"eell the load-resisling walls. which resist lateral load from the diaphragm in propon1on to tributary
width. Inertial forces generated by walls perpendicular to the dareclion otloachng are transferred to lhe
foundation.~ and to the diaphragm, which spans be-Eween wans parallello the apphed loads. In addttton to
the seismic load from the diaphragm and out-of-plane walls. walls parallel to the d1rection of loadmg also
res:ist the ine.nial loads generated by lhelr self-we1ghL
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Design example 4 • Masavy Shear
c
B
A
war Building
D
30'·1Y
r----------
~=======9========9========+-~ 1
------------~
~
I
t
I
HSS 6x6x1/4 COLI
rHSS 6>MI. Cr:A.
I
I
1
1
I
+-'-------~t----
·--~
~s·
_l!,; L;,;-±; ;-; ;-,;~ ; ,;~; ;-u; -'; ,;:Al; L;L~ ;,;-; ;-; -;,-; ; ;,;-; ; ;-; -; -;,;1; -; ; -; ;-; ;-; ; ; ; ; -; ;-; ;-dl. :_.:~~
SLAS.ON·GRADE
3
Ftgure 4-I. F/{)()r pla11
c
B
A
D
31Y-()'
30'·0'
30'-()'
1
I~ "'
I:
-~
0
1-;;;,.
S'l.oo24Gl8
I
~®
;•/.,.24 G.B
N
5'/,1<24 GlB
-<V
0!
·"' >-!'9":
e>'
"'';Or
'
"
~
N@
/
15132'PtVWOOO
ROOF SHEATI-ING
3
Ftgure 4-1. Roofframmg plan
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Design Example 4 • Masomy ~arWaN Building
cp
cp
cp
l
l
I
--------------------
l
30'-(}~
30'-a'
~
1
!
l
6'-0" .,
..
10' x 20' ODOR
IO' x 20' DOOR
l
1
..
.~ .
8'-0'
L
1
20'-!l'
L l
• 6'.().. .,
Ftgure 4-3. Elel-tltion on flu~ A
2Wx15"GI.B
JOISTS@24'
2!4'x15'GLB's
LEDGER
·8" CMJWALl - - - - + - f \
Figure .J-4. Sect10" through CMU wall a/o,g /i,rs I and 3
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Design example 4 • Masavy Shear
war Building
ll\e seismic weight of the roof is equa1to
Jv104)(
_ 17psf(60 ftx90ft)
-
1000
-
k
92 •ps
The masonry walliS constructed wnh 8-u'ICh-lluck, solid-grouted, me<hum-weight CMU, which has a
self-weig,Ju of 78 psf. TI'II!refore. for earthquake loads ln the nonh~south direc.non. the seismtc weiglu of
perpendicular walls ts equal to the we1ght of the east-west \Wlls and parapets that are tributary to lhe roof
diaphmgm. Conservaovely ignortng any openmgs tn lhe walls on lines I and 3:
w..,.,..,..,., =78 psfl"19')•
T l 16I)90 ftx2 =158 ktps
Taking mto account the tact that 35 percent of the walls 1n lhe nonh-south direction cuns•st(i of openings.
whtch can be asswned to have a we1ght of 10 psf:
w.,.," _u» =[0.35(10 psi)+ 0.65(78 pst)J(60 ftx 19 ft)2 =124 kips
Therefore, the toml se'lStnic weight 10 the north- south direction IS equal to:
2. Calculation of the Design Base Shear and Load Combinations
ASCE 7
2.1 DESIGN SPECTRAL ACCELERATIONS
The spectral accelerations to be used in de~agn are equaJ to
Sos = I.OOg
s.,=o.60g
2.2 CLASSIFY THE STRUCTURAL SYSTEM AND DETERMINE SEISMIC DESIGN
PARAMETERS
From ASCE 7 Tables 11.6-1 and 11.6-2. the buildtng IS assigned to Setsmic Design Category (SOC) D.
ASCE 7 Table 12.2- 1 utdJcates that only specta1 re1nforced masonry walls are permitted in SOC 0, E, and
F. The design coefficients tbr special reinforced shear walls tn beatutg wall systems are as foll<m!S;
R = 5.0
0,.= 2.5 CJ= 3.5
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Design Example 4 • Masomy ~arWaN Building
2.3 RESPONSE SPECTRUM
Determtne the approxtmate fundamental budding penod usulg Sect1on 12.8.2. 1:
C, = 0.02 and x = 0.75
T 12.8-2
J;,=C/1.: =0.02x 16a."=O.l6sec
r, =0.2 -SDI
SDS
Eq 12.8- 7
0.60
=0.2 - - =0.12sec
1.00
§11 .4.6
s. = sDS(0.4+ 0.6~)=
(o.4+0.6 ....!:....)=
0.4 + s.or for r < r.
T,
0.12
Eq 11.4-5
r, = s"' = 0.60 = 0.60 sec
s(Jf;
s
41
§11 .4.6
r.oo
= Sno = 0.60 lor T> T
T
T
s
Eq 11 .4-6
The long-period equauon for Sa does not apply he-re because the long-per.od tranSUJon occurs at I 2 seconds
(fr01n Figure 22-16).
2.4 HORIZONTAL IRREGULARITIES
Ia.
T 12.3-1
Torsionaltrregulartty--does not apply to diaplvag;m.~ that can be ideal ized as
ftexible.
lb. to5.
By tnspe<:t.ioo, the bwlding does not ha-..·e any of these- honzomal struCtural
trregulariues.
NO HORIZONTAL STRUCTURAL IRREGULARITIES
2.5 VERTICAL IRREGULARITIES
I a. to 5b.
T 12.3·2
By tnspe-ction, the one-story building does not quaJtJ}• for any of the vernc.al
suocruraJ trregulattties.
NO VERTICAL STRUCTURAL IRREGULARITIES
2.6 LATERAL FORCE PROCEDURE
T 12.6-1
I. Simphfied alte-rnati"\•e- srructuraJ design cmena-The- budding sausfies the reqwre-..nents of
Seclion 12.14.1.1- PERMJTfED
2. Equivalent latetal force analysts-Accordutg to Table 12.6-l.sinc.e T < 3. 5~ (0. 16 sec < 2.10
sec) an<l lbe building is regular and IS Occupancy C.legory II-PERMITTED
3. Modal response specuum analysis -PERM!TfED
4. Setsmtc response htstOI)' procedures-PERMilTED
USE EQUIVALENT LATERAL FORCE ANALYSIS
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Design example 4 • Mosavy Shear
2.7 BASE SHEAR
c
=
5
"" = I.OO = 0.20,
· l~)
l~:~)
r~
lrln
_
0 60
·
O.l 6 l~:~)
_
o. 75]
war Building
&] 12.8· 2 and Eq 12.8· 3
Also.
0.551 0.5 X 0.6
C,~O.OI,an<IC, ;, - ) =
• ) -0.06
<>
R
&] 12.8· 5 and&] 12.8· 6
~~>:>.0
l/, tJ.O
C,=0.20
&] 12.8· 1
1·= C,W=0.20 X 374 = 75 k1ps
V= 75 k1ps
2.8 REDUNDANCY FACTOR
According to Section 12.3.4, the redundancy f3ctor should be calc.ulated for each princ1pal axts. Tile
redundancy f3ctor is 1.3 unlessenher Section 12.3.4.2(a) or 12.3.4.2(b) lSsausfied, in which case the
redw'ldancy factor can be taken as 1.0. Tile waH segme.nts in the nonh-south bt.uldmg have a height-tolength rauo of greater than 1.0. By IO.'i:pection,. the removaJ of any wall segment does not result tn more
than a 33 percent reduction in story strength. Removal of any wall segment also does not result 1.n an
eXIJeme torsionaluregularity since the butldtng ha~ a flexible dtaphragm. Therefore. from Table 12.3-3, tlte
redundancy factor ts 1.0 m the north-south dU"eCuon. In the east-west d.irectton., the length of shear wall on
each stde of the structure consists or more than l\\'0 bays. where lhe number of bays 1s equal lOme length of
shear wall dJV1ded by !he SIO<)' he1ghl. Therefore, from Section 12.3.4.2(b),lhe re<lundancy factor is I.0 1n
the easr-west direc.tJon.
p = 1.0 FOR EAST· WEST DIRECTION
p = I .0 FOR NORTH~~OUTH DIRECTION
2.9 LOAD COMBINATIONS
For the one-slory budding. tlte applicable lood combinations for eanhquake design are as tOIIo·ws (Secuon
2.3.6):
(1.2 + 0.2Sus)D + PQc= ( 1.2 + 0.2 X I.O)D + I.OQc= 1.40+ Q,
Load Comb. 6 (modified)
(0.9 -0.2So.JD + PQr.= (0.9- 0.2 X J.O)D + I.OQ,= 0.70+ Q,
Load Comb. 7 (modified)
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Design Example 4 • Masomy ~arWaN Building
3. Design of Walls to Resist In-Plane Seismic Loads
TMS402
The procedures ror designing to resist ul-plane loads will be illustrated using the 8-foot-long \\all segrnent
on hne A. Su)Ce the roof dtaphragm can be tdeahzed as f1exible. the lateral load in the nonh-south dlrecuon
l~ resisted equally b)' the walls on lines A and D. Figures4-5 and 4-6 show the results of a computer
analysJs of the wall on hne A.
~- ---- - - --- -- -- -- --- ------- --- --
1r
~:3 5
13
lJ
5,f-
lJ
0
0
Figure 1-5.
~ad loads on
.
wall along lin~ A
f,;c::-------s:~-----------:~&1
1i.i
18.6ki:I\S
sn~.,
-..__,~ 120.6 ,,.,.,
~-
ld
~62~''""'
1UFtgu~
280
~
4-6. Earlhqullki! lotJds on wall along lm~ A
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Design example 4 • Masavy Shear
war Building
3.1 PRELIMINARY REINFORCEMENT LAYOUT
Ftgure 4-7 shO\\'Sa prel i.minary layout of retnforcement for the 8-foot-long wall segment.
64f.4VERT \
I ""
14
r
#4@ 24' o.c. HORIZ
Figure .f-7.
La)V'JUI
8~·nch (Nom'nal)
I
CMU
r. =2000 psi
-I
24"
16"
16"
r
16"
16"
4'1
ofn!mfercen~ntfor 8-foot-/ong wall stgmtntlmt A
In accordance with TMS 402 Section 7.3.2.6, tl:.e tnaxJ1num spac1ng (~) of venical remforcement 1n
specull retnlbrced masonry shear walls shall be the smallest of one-thtrd the length (L,.) of the shear wall,
one-thtrd lhe hetght (H) ofthe shear wall, and 48 inches (24 tnches for masonry oot latd in running bond).
s-.x=24in<4Stn . . . OK
H _ 10ft(12) _
OK
<3 3
- 40 U1. ..
<!:.= 8 ft(l 2 ) =32 in .. . OK
3
3
Horizontal reinforcement (A.) shall be at least 0.2 tn2 and have maxtmwn spactng of the smallest of
one-thtrd lhe length of the shear wall, one-third the height of the shear wan, and 48 inches (24 tnches fOr
masonry not laid in runntng botld).
A .. = 0.20 tn2 = 0.20 1n2 .. . OK
s-.u=24in<48in . . . OK
< H = 10 ft(12) _ 40 <n ... OK
3
3
<!:.= 8 ft(l2) =3211'1 ... OK
3
3
The m1n.1.mum cross-sectional area of reinforceme-nt in each dtteetion shaH be not less than 0.0007
muluphed by the gross cross-sec1tonal area of lhe wall:
6(0.lO tn') -0.0016 >0.0007 ... OK
l'!A·)=
tL,
7.63 in(96 <n)
0 20102
[·~~·)._
=
'
-0.0011>0.0007 ... OK
, _
24 on(7.63 1n)
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281
Design Example 4 • Masomy ~arWaN Building
where 1 is the wall thickness. The swn of the venteal and horaontal reinforcements shall exceed 0.002
umes the gross cross-secttona.J area of the wall:
•ll:.4,)+
IL,.
·l ~)=O.
OOI6+0.0011=0.0027>0.002
Sl
... OK
The venjcaJ retntOrcemem must also be at least one-third the shear reinforcement:
3.2 IN-PLANE AXIAL AND FLEXURAL LOADS
The most effective \\'<l.Y to evaluate the ability ofa masonry wall 10 resist a.xiaJ and tlexuraJ k>ads ts to
develop an .ntemcuon diagratn. An tnteractioo diagram ts a plot of the change in tle.xur31 sttengili of a
member W'lth axial load that is developed using the basic assumptions for strength destgn. Combinations
of factored axl3llood and bending moment that ran within the tmemct.ion diagram are acceptable. and
the me.mber can adequately res1st suc,h loads. If any eombtnation of axial load and bendtng moment faJis
olJISjde the intecactioo diagram. the cross seclton ts tnadequate. and the member must be redesigned.
To obtain an accurate interactiott diagram. the designe,r deteJ'mines lhe moment capacny for several values
of applted a-xtaJ load to obtatn the smooth curve. Sudl an analysts IS best dere.nnined with a spectaltzed
compute-r p!Ogrant. Howe\•er. an approximate tntemcrion diagram cun•e can be developed using selected
poi11ts on the tnteracuon diagram:
a. The axial k>ad capacity \\'hen there is no bendiJtg moment on the member.
b. The ftexuraJ strength when there l'i no axtal load.
c. ll1e balallCed strain oondttiott when lhe stratn in the e.xtreme compresston tiber is equal to the
maxtmum usable strain. and the strain 1n the extreme te-nsion steelts equal to the steel yield
strain.
The strength-reduc.Uort fac.tor. ' · for all combtnatiOI\."i of axial and flexuraJ load is equaJ to 0.9. From
Ftgures 4-5 and 4-6. the loads for the three applicable load combmmjons are as follows:
I. 1.20 + 1.6L,
P. = 49.1 ktps; M. = 0 ktp-ft; V, = 0 kaps (at the 10p of !he wall segment)
P. = 56.6 ktps; M. = 0 kip-ft; r: = 0 ktps (at the bouom of !he wall segment)
2. 1.20+£= 1.40+£.
P,= 40.5 ktps~ M,= 59.5 ktp-ft;
I~=
17.3 kips (at the top of the wall see__rnent)
P.= 49.3 ktps; M.= 120.6 kip-ft; V,= 18.7 kips (at tlle bottom of the w.>ll segment)
3. 0.90+£=0.70+£•
P, = 20.2 ktps; M, = 59.5 ktp-ft; V, = 17.3 kips (at the top of the wall segment)
P.= 24.6 ktps; .If.= 120.6 ktp-ft; 1',= 18.7 ktps(at the bouom of !he wall segment}
Check the capac.ny wtlh mdalloads alone rutd no ftexuralloads..
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Design example 4 • Mosavy Shear
war Building
ll\e effecuve hetght for compression loads. h. ts equal to the full story hetght, H. slnce Lateral suppon 1n
the out-of':pla.Jle dtrection occuts only at the ground and roof levels. As specified tn TMS 402 Section 9.3.2,
retnfOrctng steel IS not used to restst compresston because tt ts not supported by lateral lies.
7.63 tn
Jii = -:ti2
= 2.20 m
!! =16 11(12) =87.3 < 99
r
=
r
I
2.20
11te nom mal a.x1al strength. P,.. tS therefore equal to:
P,
=08[08J;(A, - A.)+
f,A.][1-c:~r
n
=0.8[0.8(2 ks.)(7.63 '" x96 m - (0)+60
TMS 402 Eq 9- 15
ksi(OJ)[I-(~:: )']= 573 kips
where A, ts the net area of the cross sectton and AM tS the area of latemUy ued long_~tudmal reantbrcement
(which <S equal
10 0
for !he 1Volll).
In accordance with TMS 402 Section 9.1.4, the strength reduc.non factor• .;.. for combu1auons of axial and
flexural load in reinfOrc.ed ma..(jonry is equal to 0.9. lberefore,
,P,= 0.9(573)= 5 16 klpS> P, . . . OK
For a gaven runal load (P..). the correspondtng moment (M,) on the mteractton diagram IS dete.nmned by
senmg the m.a.:;onry suatrlln the extreme oompresston tiber at r.... (wtuch IS equal to 0.0025 for cooc.rete
masonty; see TMS 402 S~lion of9.3.2) and selecting a neuual a-.:is depth (c). The eompresston force in
the masonry tS g1"\·e-n by:
c. = 0.64cbf:
where b ts wKflh of the compression block, whtch is equal to the wall thk:kness, 1 in this case.
Using Similar tnangl ~ the stram tn each remfbrc1ng steel bar (t..J IS given by:
d
-c)
£ • =£.... ( - ·t:- -
where d, IS the distance from the extreme oompression fiber to the reinforcing bat. The force tn each
remforc1ng bar ( TJJ) IS then g.l\'en by:
where£, is the steel modulus of elasucity, A,.. is the area of each reinforcing bar. and/y 1s the steel yteld
str~s. If the selected ne.utra1 axis locauon does not result in eqmlabrium of fOrces on the cross se<:rion. tlte
Jocauon of the neutral axis is modified unul equ11ibrium e.xlsts witJun acceprable limits:
'E.T11 +P,=C,.
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Design Example 4 • Masomy ~arWaN Building
Table 4-1 shows the forces on the cross sectaon after nerations have detennined tht depth of the neuual a'<is
to be 6. 14 inches when there ts no axtal load.
~M.0
. ft
=0.9 ( 3206.6)
- - - =240 kop12
Tahlt 4-1. Equilibrium ca/culatio,sfor flexural stnmgth with notuia/f(){JJ/ (c = 6.14 In)
A
(on1)
Masonry
••
d,
(on)
(oniln)
J.
(kso)
u..
p ..
(kops)
(kop-on)
60.0
2732.6
2.457
Steel
Bar I
0.20
4.00
0.0009
-
-
-
Bar2
0.20
20.00
...0.0056
-60.0
-12.0
-336.0
Bar 3
0.20
36.00
...0.0121
-60.0
-12.0
-144.0
Bar4
0.20
60.00
...0.0219
-60.0
-12.0
144.0
336.0
Bar 5
0.20
76.00
...0.0284
-60.0
-12.0
Bar6
0.20
92.00
...0.0349
-60.0
-12.0
521!.0
Total
1.2
-o.o
3260.6
A baJanced smun eonditton occurs when the strain tn the extreme compression fiber IS equal to the
maximum usable strain~£_. (whtch is equal to 0.0025 for concrete masonry~ see Th1S 402 Secuon 9.3.2).
and the suain tn the extre-me tension steel is equal to the steel yteld srtaln. The depth of the neutml axis at
the balanced stratn condition. c,. can the.refore be dettrmined from:
c, = (
d
f,
--+1
£ e
·-
) _ (
d.
) - 0.541d =0.547(92 on)= 50.Jon
60 kso
I
.
+
29,000 kso x0.0025
Table 4-2 shows lhe forces on the cross secuon at the balanced condition:
q,P,=0.9(469 kops)=422 kops
~M, = 0.9( 14•46: 2kip-tn) = 1085 kip-ft
Ftg.ure 4-8 shows the uueractJon dtag.ram and applted loads~ indtcattng the v.'aJI has suffic-tent flexural
strength.
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@Seismicisolation
Design example 4 • Mosavy Shear
war Building
Table .f-1. EquilibrilUn ca/cularionsfo r flexurol.wvngth at balanud conditlon (c = 50.3 m)
AI
(in·)
Mosonry
cl,
(1n)
••
P,
J.
(iniln)
(ksi)
20.13
(kips)
M,
(kip-in)
491.5
13,697.5
Steel
-
-
-
Bar I
0.20
4.00
0.0023
Bar 2
0.20
20.00
0.0015
Bar 3
0.20
36.00
0.0007
Bar4
0.20
60.00
...0.0005
- 13.9
- 2.8
-7.4
207.1
-12.0
528.0
469.3
14,466.5
-
-
-
-
Bat 5
0.20
76.00
...0.0013
- 37.0
Bar 6
0.20
92.00
...0.0021
-60.0
Total
1.2
33.5
460
---------------------------,
400
1
360
300
I
260
I...
s
!
~
,,
160
100
+
50
0
,'
,'
,'
,,
,,
,,
,,
,'
'
,'
- -
· VI~ Diagram
-o-AflptCllima Solulion
-60
100
0
100
500
Moment {kJp-ft)
Figure 4-8.
l.a)'()llf afreinforcement/or 8-f()()I-Ioug
n·a/1 legmen/
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2a5
Design Example 4 • Masomy ~arWaN Building
3.3 MAXIMUM REINFORCEMENT
TMS 402 §9.3.3.2
The ducttlit)' of a shear \ttall or ito;; ability to deform in the nonlinear range JS highly dependent on the
amount of reinforcement in lhe wall. Thus. TMS 402 hmits waJI ftexural reinforceme1n to levels that are
c.onsistem walh lheduetd ny 111herem in the desagn parameters (R and C.J spedfied u-.ASCE 7.
The max1mum fle.xural remforcemem is Iunited by ensunng that the strau'l m the ex[(e,me tensile
reinforcement exceeds steel yield strain by a specified amour'lt., depend1ng on lhe requtred duculity capacny.
Table 4-3 shows lhe m1nimum requtred smun at the uiUJnate state~ as a function of me.steel y1eld strrun.
where d .. is lhe total depth of the cross secuoo 1n the darectioo of shear betng constdered.
The verification oflhe mouomum reinforcement is performed using the load combanation D + 0.15L +
0.525Q~ where Q£ is the effect of the horizoma1 component of earthquake load. Reintblcement in
compressiOn may be used m evaJuaung equilibnum of the cross secuon when calculatang the max1mwn
reanfbrcemem even if II is not laterally supported by ties.
The prov1saons for maxamum reinforcement may be superseded by using TMS 402 Sectaon 9.3.6.6. which
provides an alternative approach to ensure that walls have sufficient ducul ity. Walls of short or moderate
he1ght with relatively low a.~ta11oads and shear suesses are assumed to possess suffic1en1 ducuhty.
Olhe.rwise~ the use ofconfined boundary eleme-nts is requtred to increase the masonry comp(eSSIOn sttain
capacny. Where special boundary elements are rtqU1red. their effectiveness must be confirmed by testtng to
verifY the strain capacity of the confined masonry.
Table 4-3. Minimw11 strom n!quiiY!dfor satufylng maxmwm remforeenumt m1io
Basic Se~,c- Force-Resisung
System
Special Reinforced Masoory
Shear Walls
ln[e.rmediate Remforced
Masonry Shear walls
Walls Loaded Out-of-Plane
All Others
.H.
R
5
3.5
-
v.d.,
>1.0
a
~
.•'
: ....!.
4.0
< 1.0
1.5
>1.0
3.0
< 1.0
1.5
-
1.5
All
>1.0
1.5
> 1.5
< 1.0
1.5
<1.5
< 1.0
None
At the base of the wan. the axial load tor checking the maxtmum re1nforcement ts equal to:
P_
= P0 +0. 75P, +0.525PQ, = 35.2 +0.75(0)+ 0.525(0) = 35.2 k1ps
Table 4-4 shows the equll1bflum calculations for deternuning the neuuaJ a-xis and steel stratn for the above
ax1aJioad. From lhe table, the strain in the extreme tension fibe( IS equal to:
<.....,=0.0223= 10.8<,.>4<,. .. . OK
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Design example 4 • Mosavy Shear
war Building
Tablt 4-1. Equilibrmm ctJit:ulationsfor maxmmm relnforumtnl (c = 8.93 in)
A,
••
d,
(tn)
(in')
Masonry
(in/tn)
f.,
P_
(ksi)
(ktps)
,\1_,.
(ktp-tn)
87.2
3873.7
3.57
S1eel
Bar I
0.20
4.00
0.0014
40.0
8.0
352.1
Bar2
0.20
20.00
-o.003 1
-60.0
- 12.0
- 336.0
Bar 3
0.20
36.00
-o.0076
-60.0
- 12.0
- 144.0
Bar4
0.20
60.00
-o.O I43
-60.0
- 12.0
144.0
336.0
BarS
0.20
76.00
-o.OI88
-60.0
- 12.0
Bar6
0.20
92.00
-o.0233
-60.0
- 12.0
528.0
35.19
4753.8
1.2
3.4 IN-PLANE SHEAR STRENGTH
TMS402§9.3.4.1.2
In order to reduce tl1e possibility of a non-ducule shear fa1lute~ TMS 402 Section 7.3.2.6.1 .1 stipulates that
shear strength. QV,.. of spec1al reinfoteed masonry shear walls must exceed the shear conesponding to the
development of 1.25 times the.nominal moment strength, M,.. Hcm·ever, the nom anal shear suenglh, 1:, need
n01 exceed 2.s1:. For shear, q. =0.8 perTMS Secuon 9.1 .-1..5.
From the tntemction diagram in Figure 4-8, the monlent strength u'lCreases \\ith axJal load t.n the rnnge of
a-xtal load con.~tdered The nom1nal momem suength at the maximum ax tal load combinatton (P. = 49.3
k1ps for 1.20 + £) is equal to -4 1I ktp-ft. Therefore, 1.25 times the nominal moment strength is g1ven by:
Then the shear de.mand correspoodtng to 1.25 umes the wall flexural .strength is equal to:
0
•
11 _,.,_ =
Howe\'er~
since
v.l1.2SM.) = 18.7
~
• 571 ktp-fi )
•
. 1ap-ft = 88.> ktps
k tpsl
120 6
+I: need not exceed 412.sJ.;..:
$2.5V,= 0.8(2.5)(18.7) = 37.4 kips
-+Governs.
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Design Example 4 • Masomy ~arWaN Building
The shear strength of ma.o;onry me-mbers ts a combination of the shear resastance provided by the masonry,
V__.. and the shear resistance provided by sheat reinforcement. V.u· The 001ninal sheat strength is therefore
equal to:
TMS 402 Eq 9-17
where y6 •s a grouted shear '"-all factor that tS equal to 1.0 for fully grouted \\"ails (Y, IS equal to 0. 75 for
partially grouted wal ls~ Shear strength IS htghly dependent on the shear span-to-deplh ratio of the member,
whic.h is defined by the term MJI ",//.,. 1n \\<Juch M.. is the moment de.mand cotrespondmg to the factoted
shear demand. V.,.and d.,. is the actua1 depth of the maSOI'U')' cross secuon tn the direction the s:hear is being
considered. The vaJue of MJI:P... must be pos1uve and need nm exceed 1.0. From Figures 4-5 and 4-6,lhe
shear span rauo of the wall segment is equal to:
M _ 120.6 ktp-H _ O.SI < I.O
18.7 kips(8 ft)
Vd,
Conservatively using the load combinatton wath the smallest axtal load at the base of the wall, the shear
strength provided by the masoory IS given by:
1;_ = [ 4 - 1.75( ;~;, )]A.!i. +0.25P,
=
TMS 402 Eq 9-20
[ 4 -1.75(0.81)](7.63 inX% in)J200o ps1
_
.
.
IOOO
+ 0.25(24.6 kipS)= 90.7 kips
The shear sttength c.ontnbuted by tl'le. steel 1s equal [O:
•o'"')
•A)
1:_ = O.SJ\ -'F,d, = 0.5l~
s
24 1n-
TMS 402 Eq 9-2 1
(60 ks1X96 tn) = 24ktps
Therefore,
When MJV.,d.,< 0.25, the nomtnal shear strength ts I1m ned by the folloY..'lng equation:
TMS 402 Eq 9-18
and when MjV,d, > 1.0,
J:s4(A.[r;)Y,
TMS 402 Eq 9-19
For values of MJV,fl. berwee1l 0.25 and 1.0, the nla.'Ctmum value of V., IS detennined by Interpolation
between Equations 9· 18 and 9-19. The max1mum shea! Slfength oflhe wall is lhus ga"\·e-n by:
,. s(
4 + (I- 0.81X6 - 4))A
lf./llil!(
(1-0.25)
rr; = 4.5IA rr: = 4.51(7.63 inx% in)J2ooo ps1
""'·
= 148 kips> 115 ktps _ .. OK
· "'·
1000
In accordance with TMS 402 Secuon 9.1.4.5, the strength reducuon factor for masotuy subJec-ted to shear lS
equal to 0.80. Tile des•gn shear suength oflhe wall segment ts the-refore equal to:
W.=0.8( 11 5)=92 kips>qo2.5l:=37.4 kips ... OK
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Design example 4 • Mosavy Shear
war Building
ll\e '"all segmem ha~ sufficient shear stre-ngth to resist the shear demands at the base of the wall. By
t.nspection. the wall segment ts also capable of resisting the shear at the top of the segment
3.5 SHEAR FRICTION STRENGTH
Masonry waJis with low walloods and low shear span ratK>S are susceptible to shear sliding when
subjected to u\-plane lateral load~. TMS 402 Sectton 9.3.6.5 provtdes procedures for the design of walls to
resist shear slidtng. The oomutal shear fricnon strength, V.,. lS obtained as tOIIows:
For .lfj( ~J:I,),; 0.5:
V'!f = )l(A..f,. + ~) ~ 0
TMS 402 Eq 9-33
where 1J is the eoefficienl of fnction. which ls equal to 1.0 fO! masonry on concrete wnh an unfinished
surfuce or a finJshed surface that has bee.n tntentionaJiy roughe.ned. and is equa1to 0. 70 for all other
condmons. A• is lhe area of steel (only web reinforce-ment in flanged walls) that is adequa£ely anchored
above and beJow the hortzontal shear plane oo develop the y teld S((ength of the reinforcement. P11 IS the
a.xull load on the wall and l~ jX>Sitive for compresston.
For M)(V,d,) ~ 1.0:
ThiS 402 Eq 9-34
where Aw.: is the net area of the ,.,'all cross secrion ( tndudtng applicable pontOJ\S of effective flange wtdth.o;)
be-tween the neuuaJ a·<ts and the fiber of maximum compressive stratn_. calculated at the nominal moment
strength of the cross secuon.
For values of MJ( V,d,) between 0.5 and 1.0, V.;-shall be determtned by linear tnterpolauon between the
values given by Equations 9-33 and 9-34.
11te load c.otnbanaoon0.9D + E(0.7D + £•) governs lOr the 8-foot wall because it provtdes the mtnimum
a.xull load to resisl sliding:
P, =24.6kips; M,= 120.6 ktp-t\; 1~=1 8. 7ktps
M, = 120.6 =O.SI
V,d, 18.7(8)
From Equatton 9-33:
1;, =)l(A.,J1 + P,) = 1.0[(0.2x6)60+ 24.6] =96.6 kips
And from Eqttatton 9-34, c:on:sef\'atively u~tng the neutml axas depth wath no axial load tn Table 4-1
(c=6.1 4tnches):
J;, = 0.42f:AN = 0.42J:ct = 0.42(2.0X6.14X7.63) = 39.4 ktps
By lt1terpolaung:
'-;,. =96.6+ (39.4 - 96:6) (0.81 - 0.5)=61.1 kips
(1.0 - 0.))
~F-(=0.8(61.1) =48.9
kips< V.= 18.7 ktps . .. OK
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Design Example 4 • Masomy ~arWaN Building
4. Design of Walls to Resist Out-of-Plane Seismic Loads
TMS402
A major constder.uton in lhe strength destgn of masonry walls to resist out-ofplane loads is the fact that
the lateral d1splacement lS otlen comparable to the walllhtclcness. Thts me-ans that secondary effects (also
known as P-de-lta etTects), whtch exist as a result of the application of loads on lhe wall·s deformed shape,
can be a stgnificant part of the v,.1l.Jl demand and must be considered tn lhe analysis.
TMS 402 rtqUJrements for out-{)f:plane destgn provide procedures for walls that sausfy one of the
followtng reqturernents:
I. P,
~0.05J.:A
8
1 ; 7$30
2. 0.05J.A1 < ~ S 0.20f~A
Eq 9·22
where Ptl is the factored axial load atlhe locatton of maximum moment, A1 is the gross cross-sectional area,
and f~ ts the masonry compressive stre-ngth. h is the effective hetght and 1 is the wall thickness. The-re are
no design requtreme.nts tOt- walls with a-<iaJ loads greMer than 0.2f;A, since there is no experimental data
on t11e om-of..plane response of nla..~nry walls wtth such large axial loads.
The out-of-plane demand may be determined either by a secol\d.order anaJysjs or by an alternative firstorder analysis that utdtzeS a moment magnifier to ancorpol'<lte P~e-lla effects. The alternative procedure is
applicable to waHs wtth different boundary c.ondtuons and loadmg.
4.1 DESIGN OF WALL ON LINE 1 (NO OPENINGS)
TMS 402 §9.3.5.4.2
The wall on hne I, whiCh has no openings, will be designed as a sunple span between the floor and roof
le-v-els. Thts IS conststent wnh computer analyses of out-of-plane wall respon..:;e, whtch 1ndtcate that dunng
dynamtc loodmg. the presence of a relati\1ely shon parapet does not reduce the momenttn the v.'all span
as much as durtng stauc loadmg. The procedure for a second-otder analysts provided tn TMS 402 IS
applicable to ptnned-pinned boundary condiuons and uniformly distrtbuted lateral loads and will be used to
evaluate the wall.
From Figure 4-2, the roofdead and live loads on the wall on Ime I are equal tO:
Pp= 17(15)=255 1blfi
P,._, =20(15)=3001blfi
At the \\all mtd-hetg.ht., whtch 1S the cnucal secuon. the self-weag.ht of the wall 1s equal to:
Ftgure 4-1 shows the con.necuon of tl1e roof to lhe waJI. lllC eccentrteity or the roof reactton, which occurs
at the face of the glulam ledger. JS gaven by:
t; I
- 7.63
- 6 .3 tn
+ 2..;,:;-+ 2.):;
2
2
Ftom ASCE 7 Section 12.11.1, the out-{)f:plane eanhquake load on the structural \\'all ts equal to:
F,= 0.4S,l,W1 =0.4{1.0)(1.0)(78) = 31.2 psf
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Design example 4 • Mosavy Shear
war Building
For the load combination 0.9D + £ =0.1D+ £11:
/'.;= 0.7(255) = I79 lblft
i'w =0.7(858)=60 1 lblft
P. = /'o~+ /'w= I 79 + 60 1 = 780 lblfi
!f
=
g
~~~3) = 8.5 ps• < o.os.r.; = 100 ps• ... OK
Try tl4 bats spaced at 24 lOC-hes on center (A.:: 0.10 in:!lft. d= 3.82 in) and detemune lhe wall properties.
The gross moment of tnenta is equal to:
1
= b(t)' = 12(7.63)
12
12
1
'
_
444
,
10
fi
1
From TMS 402 Table 9. 1 .9.~ the modulus of rupture,[~ for fully grouted ma.<Onry wnh TypeS monar with
flexural stresses nonnaJ to bed Joints ts 163 pst. The cracktng mome-nt IS thu.~ gaven by:
M~ = (.1;+
;:)s.
= (163 +8.5)
12 7 63
< ~ )' - 19,968 lb-inlfi = 1664 lb-ftlfi
The deplh oflhe neutral axis, c. is gtven by:
r = A.f. + 1', _ 0.10 tn2 /fi(60,000 ps~)+ 780 lblft _ O.<l4 10
0.64/;b
0.64(2000 pst)l2 in
TMS 402 Eq 9-3 1
From TMS 402 Table 4.2.2.1he elastiC modul t of steel and concretemasont)' are gtven by:
£$ = 29,000 ksi
E = 900f.• = 900(2000 psi) - I800 kSI
•
..
1000
11te modular ratio ts therefOre equal to:
,__ E,
E.
kst
-_ 29,000
- 16. I
1800 lest
and the cracked mome.m of tnerua ts g1ven by:
I
a
1' 1)
be'
=n A+~ (d - c)'+'
fy2d
3
(
= 16. l(o. 10 10'/ft +
TMS 402 Eq 9-30
780 lblft x 7.63 10 )(3.8 I- 0.44 tn)' + 12(0.44 tn)' - 2 1.0 m'
60.000pstx2x3.81 in
3
The wall deflection can be calculated directly (asswn1ng lhe Willi is uncracked) with the follwnng equauon:
wlr'
P_,e
(3 1.2psf(16ftx 12)' + 179 1blft(6.3tn))
8( 12 10)
2
.
,
-0.06tn
I' +I' ) 48(1,800,000 psiX444 in Itt) _
lblft
780
lw
.;
5(16xl2)'
- 8- + ?
3. =
£
48 1 _
____!!_!_
Sir'
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Design Example 4 • Masomy ~arWaN Building
Therefore, the moment demand.. including P-delta effects, is equal to:
w h1
M
II
P,r
= - "8 -+-+(PJ
+P
2
.,.-
)ll
TII1S 402 Eq 9-23 and Eq 9-24
II
_ 31.2 psf( l6 fi)' + 179 lblfi (6.3 in)+ 780 lb/11(0.06 in)= 1049 lb-lllfi
8
2
12
12
M. is less than Mu, so the assumption of an uncracked v.all 1s correct. The flexural strength of the \Yall is
gtven by (see oommeruary for TMS 402 Se<:uon 9.3.5.2):
M. =(A,[y+; )(\" )+A,f,(d-';)
where:
A.f.
a-
2
780 lb/11
O. IOtn lft(60,000pst) + - -
+PI~
11
r
-
0.8(2000pst)l2 in
0 .8J;,b
0 ·9
-0.36tn
Comm §9.3.5.2
Therefore,
( 0 10 111 , 1ft x60,000 psi+
78~-~b/ft )C-63 m ; 0.36tn)
9M =0.9
[ +0. 10in2 1ftx60,000ps{3.81in - 7.6;in)
•
12
= 1870 lb-fVfl> M.= 1049 lb-ftlft .. . OK
The above procedllre can be repeated to show the '"'all re1nfbrceme-m IS satJsfactory tor other appl1cable
load combinauons. The wall reinforcement needs to be checked against the maximum limits tn TMS 402
Section 9.3.3.2. From Table 4-3. tlte tensile stram in the extreme steel fiber must be at least 1.5 Limes lhe
y1eld stratn (a= 1.5) for the followmg load comb1nat1on:
P_ = P0 +0.75PL +0.525PQ, = [858+ 255]+0.75(0) +0.525(0)= I I 13 lb
From the commentary ofTMS 402 Section 9.3.3.2, tl1e maximum re-inforcement ratio for a fully grouted
member with only conce-ntrated te.nsion reinforcement is:
-0.64J;,(,_•:a.J£i
f.
64(
P-=
y
0_
2000
i)(
ps
113
0.0025
)- I
lb/11
0.0025+1.5x0.002 1 12 m x 3.8 1tn -0.009
60,000 psl
The wall re1ntbrcement ratio ts:
A
p = -'- =
bd
292
0 .10 in'
- 0.0022 < 0.009 ... OK
12 m(3.8 1m)
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Design example 4 • Mosavy Shear
war Building
ll\e \~tall horizontal deflectton at mtd-hetght under allowable stress design load combu..auoos~ Sp must
satisfy the fOllowing equation:
8Jo s: 0.007/J
TMS 402 Eq 9-32
For the allowable sttess load combu1atioo 0.6D + 0.7£:
"'• = 0.7(31.2) = 21.8 psf
P.r= 0.6(255) = I 53 lbltl
P, = 0.6(255 + 601)= 514 lblfl
And St.llce the dessgner can assume the \'r'allts not cracked because tt was uncrac.ked under suength-leveJ
loads. the deflectton at mtd-height l~ equal to:
1
wh
Pe
-1 - +...L
& 8
2
s
48£,.1,
5h2
(
21.8 psf(l6 fix 12)' + 153 lbltl(6.3 1n))
8(12)
2
48(1,800,000 psi)(444 in') - 514 lblfl
5(16 fiX 12) 2
- (P,)
= 0.04 '"< 0.007/r = 1.34 1n .. . OK
As wttll the determi.nauon oflhe strength oflhe wall. the wall can be checked tbr lhe other allowable sttess
design 1oad combinmion to verify that the deftecuon is acceptable.
4.2 DESIGN OF WALL SEGMENT ON LINE A
TMS 402 §9.3.5.4 .3
The 8-foot \WI! segment on line A tS adJacent to openings. TherefOre.• as shown sn Ftgure 4-9, the segment
suppons out-of-plane loods over a trtbutary width larger than its width. A conservative approach ts to ignore
the openings and destgn lhe 8-foot segment to resist a w'liformly distributed load mat corresponds to a 28foot tributary widlh over lUi entire height. Wnh this approach, the equations for pinned-pinned boundary
condmons and un1fonnJy dtstrtbuted loads. whteh were descnbed in the prevaou:s secuon. can be used.
Tributary Arta ror
·\r
Out-of Plaoe Load
<p
2~'·0'
cp
r
cp
----------- ------
1-----I
I
I
I
I
I
i
I
FiJ:tll? 4-9. Tnbuttuy lt~tdlh for 0111-bfpltmt loads for an 8-fooJ-/ong wall ltgmtnl
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Design Example 4 • Masomy ~arWaN Building
An altemau\•e, less consef\'ative approach is to use the moment magnification p(()cedurt\ whkh tS tncluded
tn TMS 402. The moment mag,nificatiott procedure can be used for ·walls wnh all boundary condJLtons
and loading and ts not subject to the limitations of provistons used 1n lhe prevtous secuon. To account tbr
P-delta eftUL<;, the f3cmred moment fi'om a first order analysts. .H.o. is modified by lhe followlng equatton:
TMS 402 Eq 9-27
where
I
'1'=-1- ~
TMS 402 Eq 9-28
~
and
TMS 402 Eq 9-29
When M., ts less than M",.1.,.1 shall be taken as 0.75 ttmes the gross momem oftnerua. Otherwtse, me wall
l~oons1dered cracked, and /#ff1s equal to /cr. The value of P,/P, must not exceed 1.0.
Figure4-IO shows the load and maximum moment on lhe waH segme.n t Ftgure 4-5 sh<m•sthat the axial
load at the top of the waJI segtnent is equal to 28.9 kips. For the load combmation witJt the 10\\-est axial load
(whtch results tn lhe smallest moment strength):
P.=0.7(28.9)=20.2 ktps
Af : 14.5 kip-It
Ftgurr .f-1 0. Out-ofpltme loads for an 8-foot-/ong wall segment
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Design example 4 • Mosavy Shear
war Building
ll\e cracktng momeruas equal to:
=
M
t:r
(1. + AP, )s• = (163 pso +
'
1000
1
20.2 kips
7.631t1X96tn
)96 on(7.63 on)' _ 178 ko on
p-
6
From Ftgure 4-7.A,:: 1.2 tn2 and b= 96 tn. Assurmng lhe wall iscrad::ed, the moment oflnenta is
calculared as follows:
c- A,[y+P, _1.2on'(60kso)+20.2 -0.75on
I = n(A + I'.t
t:r
f,2d
•
TMS402Eq9-31
0.64{2.0 ksi)96 in
0.64J..b
)(d-
c)'+ f><'
TMS 402 Eq 9-30
2
= 16.1(.1.2 on'+ 20.2 kipsx7.63 in)(3.81 in-0.75 in)'+ 12(0.75 m)' -233.4 on'
60 k-<ox2x3.81 in
3
The second order moment is equal to:
M,~
M, ='I'M..,o=
1
_
,
l',h-
rE.Ioff
14.5 kop-ft(12)
_ 20.2kops(16ftx12) 2
1
r(ISOO psip33.4 on')
= 1.22(14.5lap-ftx 12)=212kop-in
M. > Mtp so the assumption of a cracked wall is correc.L
A.f, +
a=
·
t
O.SJ;b
1.2 in 2 (60 ksi)+ lO.llaps
0 ·9
-0.61 on
0.8(2kso)96 on
= 0.9[( 1.2 on' x60 ksi
+20~
:ips)( 7.63on;0.61on
)+
1.2x60 k-si(O)]
= 298 kip-on> M, = 217kip-m . . . OK
Conservatively checking the wall de-flection at stre-ngth-level loads (anstead of servtce-level loads):
S,='I'S.,= 1.22(0.147)=0.18on<0.007h= 1.34on . .. OK
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Design Example 4 • Masomy ~arWaN Building
5. Out-of-Plane Wall Anchorage
TMS402
The om-of-plane anchorage of waJis IS a critacal aspect in the design of many ma..(ionty buildings.. Ourtng
past e.an.hquakes. the separauon of waHs from dtaphragms has resulted tn severe damage aod the partlal or
complete collapse of several buildings. ASCE 7 Section 12.11.2 provtdes p~oviSions for lhe anchorage of
\vaJis and ttansfer of the anchorage forces 1.nto dtaphragms. ln addh.ion to resistmg the prescribed forces.
continuous ties or struts must be prov1ded between diaphragm chords to dlstr1bute the anchorage forces
tnto the dtaphragms m SOC C, D, E, and F. Added chords may be used to create subd1aphragms w1th a
maximum length- to-Width rat.io of2.5 to Ito ttansrrut the anchorage forces to the matn continuous crossues. \Valls must be desagned to span bet\veen anchors 1f lhe spacing between anchors exceeds 4 feeL
5.1 CAlCULATION OF ANCHORAGE FORCES
ASCE7
The anchorage force is gi\·en by:
ASCE 7 Eq 12. 11-1
where k4 ts the anlplificauon f3ctor for d1aph.ragm flextbihty, whtch 1S gwen by:
L
k,
=1.0+~
ASCE 7 Eq 12. 11-2
L1 IS tl'le spaal between vertical elements that prov1de latera] suppon to the d~aphragm in lhe direcuon
c.OJ'ISidered. f« the wnll on line I:
90
k, = 1.0+ 100 = 1.9
The \velght of VI<ill tributary to the diaphtagm is equaJto:
w, = 78( ·; + 3)=858lblft
Theufore:
F, = 0.4Sn<kJ,W, = 0.4( 1.0)(1.9)( 1.0)(858) = 652 !bit\
The anchorage fbrce must e.xceed the following:
F,<: 0.2kJ,W,=0.2( 1.9)( 1.0)(858) = 3261blfi ... OK
5.2 DESIGN OF ANCHORAGE CONNECTION
TMS402
Figure 4-11 shows the out-ot-pJane anc.horoge connecuon for the wall on line I. For headed at\C-hor bolts,.
tensile strength IS de.tennined by e1tber masonry breakout or yteld and fracture of the bolt steel. For
nta'iOnt)' breakout,lhe anchor bolt strength IS given by:
TMS 402 Eq 9-1
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Design example 4 • Masavy Shear
war Building
Premanufaclured
Hold-Oown Each
Side w/518 Dia Bolls (I, = 36 ksi)
--Joist Hanger
SECT!ON A·A
Ledger
Figure .f- 1 I. Out..Qfpllme anchorage- COtrnRt-tion
11te strength reduction factor for masonry breakout JS equal to 0. 5~ and A,. 1s me proJected tenston area on
the masonry surface, which is equal to:
where 111 tS the efttcllve e.mhedment le-ngth. The projected tension area mUSt be reduced to account for the
overlap of the area'i or adJacent boltS so that no ponio11of rnasonry L~ included mol'e than once.
From Ftgure 4-12. the central angle of tlte overlapptng segment ts gsven by:
9 = 2cos·• (.!....)= 2oos· •
2/•
and the modified projected area
52
( 2X.)
~)= 2.036 rndians
ts equal to:
A~= nf,- ~f,ce-s.n 9) = 78.5 -~(5)'(2.036-SJn2.036) =64.2m2
The masonry tensale breakoU[ su-ength of each bolt lS therefore equal to:
For steel tensile )'ield. the strength of each bolt ts given by:
$B_ = ~.!,= 0.9(0.31 in2)36,000 psi= 10,().14 lb
TMS 402 Eq 9-2
The strength of the bolts lS detecmmed by masonry breakouL I f the two bolts are spaced at S feel on center:
F, = 8 ft(6521blft) _ _ < I.O
0 45
.. . OK
2(57421b)
$8_.
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Design Example 4 • Masomy ~arWaN Building
Venfy that lhe waJI can span 8 teet horaontally hem'ten the ane,hors. It the engtneer assumes a beam width
of6 feet (3-foot high parapet p1us an addJttona13 feet of waH below the root) with tl4 barsal24 mches on
cemer:
w.J'
M ~
= (652 lblftXS ft)' _ 52 16 lb-ft
•
8
8
'M
.,. •
- u
-YJy
A(d - 1.6/;b
/,A, )
f
=0.9(60,000X0.2 on' x3)(3.8 1tn -
102
60 000
psixO.l
xJ)..!_=98651b-fi .. . OK
•
1.6x2000psixnin 12
Radius=
Modified Projected
Tensile Area, Apt
Figw-e ./-12. 0 1·erlap ofprojuttd 1eu.ttl~ areas
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Design Example 5
Tilt-Up Building
OVERVIEW
ll\is e.xample prese11ts the seismic design of major components of a tih-up budding_ Many uh-up butld1ngs
have suffered severe structural damage 1n past eanhquakes. panicularly during the I 97 I San fe.rnando
and 1994 Nonhrtdge evenr..c:;. The most common problem has been wall-roof separauon~ w1th subsequent
partial collapse-of the roof. Since those events, the building codes have SlgnJficantly improved. yet a major
earthquake has yet to test the current uh-up code provlstons.
The example building is a ware.house, whtch has tih-up coocrete walls and a panehzed hybrid roof
systern. The hybrid roof, common in Ca1tfbrma. Nevada, Anzooa. and Oregon, COOSJSts ofa paneltzed
wood structural panel (pi)1Wood or onented st.rand board) system supponed on open-web steel jo1sts. The
butldtng's roofftanung.plan IS shown 1n Ftgure 5- 1, and a typtcal section through the budding tS given in
Figure 5-2. The emphasis tn lhis design example is on lhe seismic design oflhe slender \\o'all pane-l loaded
out-of-plane and the wall-lo-roof anchorage, wnh a dtscusston on shear wall destgn and computing base
shear in tiJt-up bt.nldings. A more complete e.xample 11lu.~ttaung general coocrele shear waH des1gn may be
found tn the 1018/BC Slrocturai/Selsmlc Destgn Manual Volume 3.
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Design Example 5 • Tlr-Up Building
OUTLINE
I. Building Geomeuy and Loads
2. Overv1ew of ACI Slender Wall Des1gn
3. Out-of-Plane Late,ral Design Wall Forces
4. Prtmary Moment fi'om lhe Out-of-Plane
Forces
5. Primary Moment from lhe Venical load Eccenwcity
6. Total Fac.tored Momen1 Including P-Del!a Effec!S
7. Nomtnal Mome-n t Strength fM.
8. Service-Load DefJecuon Considerations
9. Wall Anc.hornge a1 RoofPurlins (Nonh-Soulh Load1ng)
10. Wall Anchorage a1 Subpurilns (Ea.<t-WeSI Loading)
I 1. Subdiaphtagm Design(East-West Loochng_)
12. Conununy Ties across !he Mam Dmphragm (EaSI-WeSI Loading)
13. Shear Wall Des1gn Loads
14. References
1. Building Geometry and Loads
ASCE7
1.1 GIVEN INFORMATION
Setsmtc-tbrce-reststtng system
A bearmg-wall syslem conslsttng of intermediate ptecast concrete shear walls supponmg a fle.\;tble
diaphragm of wood SII\ICturaJ panels (WSP).
Seismtc and site dam
Mapped spectral accelerauons for lhe site:
s, = 1.5 (short period)
S1 =0.6 (1-second period)
Risk C.1egory II
SneCiass D
s..= 1.0
Wind
A'iSumed not to govern
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O.sigl Example 5 • Til-Up Butilg
Roof loading onto walls
Dead load = 12 psf
IBC T 1607.1
Live load (root)= 20 psf(reducoble)
Nosnowlood
Umfonn loa<hng from roof assumed
Roof load eccenmctty = 5 inches from tnstde face of panel
Walls
Thtckness = 9.25 tnches wtth penodJc ~-mch narrow honzont.al reveals
Heig)l! = 32 fee• (lop of wall)
!-leigh• = 28 fee• (roof hoe)
Normal weight concrete= 150 pc.f
=3000 pso
A615. Grade 60 stet-1 remforcang (F1 = 60 kst)
r;
Roof structure
Structural-! shealhing [orien1ed suand board (OSB) WSPJ
Pre-engineeredlpre.manufoctured open-v.-eb steel jOl'its and joist girde:rs wah fuJI-width wood ruulets
All'"'"" is Douglas For-Lareh (Dfl)
cp cp cp ~ <?
i i 24((:
~: ...;'...
+
9
cp
CD 0
'!::/111-rT
tI
·0'
I
"
I,--.-I I
I I
.C:O'· O''
I
+
'f;;:v.rv.\1'
ISI3:2"WOOO
STRUCWRAI.
PANS.
STEEl J.."'llSTG!ROER
@
AAAMII..IG TVP
tl tj4"1?RfCP..sT
CCNC~ PAII:fl.S T'IP
ROQFA:Aif
Ftgurt 5-/. Rbofframmg plan
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Design Example 5 • Tlr-Up Building
l.rJ. 0"
40 -<r
4.0'-tr
40'-a·
~
b
'
"'"'
'
I'
Ftgure 5 -1. Building section
2. Overvl- of ACI Slender wall Design
Concrete walls unde-r combined axial and out-of-plane lateral tOrces may be destgned under ACI 318
Chapter 6 by accounung for slendemesselfects usmg Sec'lion 6.6.4. 6.7. or 6.8. An altemauve wall destgn
procedure is provtded tn ACI 318 Section 1 I.8. wtuch allows \'tl)' slender concrete v.'3lls with certain
restncuons. This sectton made us debut i1l ACIJIS-99, and it JS generally based on lhe 1997 Uniform
Building C()(/t. wtuch tncorporated the equation~ conceptS. and full-scale tesung developed by the
Structural Engtneers Assoctation of Southern California and published 1n the Report oflhe Task Commiuu
"" S/t1tdtr ll'a/lttn I982 (SCCACIISEAOSC, 1982).
Walls designed un<ler the altemauve slender wall method ofACI 318- 14 Seeuon I 1.8 are typ•eally tilt-up
concrete panels that are sate-cast.. cured, and ulted into place. They are designed to withstand out-of-plane
forces and C-31T)' vettical loads at the same tune. These slende.r walls dlf'fe-r from concrete walls designed
under the stmphfied design method (ACI 318 Section 11 .5.3) and walls designed as compression members
(ACI 318 Section I I.5.2) ln lhat slender walls have greater re.strictions on axial loads and must be a
ten..:;ton-conltolled destg.n w~n subjected to combtned bendmg and axmllood. ln addtuon. secondary
effects of eccemricuies and P-delta moments play an important role in anaJysL" and design of these slender
tilt-up panels.
In this example-. the out-of-piMe latera1 destgn force..:; for a one-story lilt-up concrete slender \\all paoel ate
detemuned, and the adequacy of a proposed reinfOrced concrete secuon IS c.hec.ked. The example wall panel
has two door openings. a.:; shown in Figure 5-3. Tile p1er between the two opemngs ts analyzed ustng the
slender wall design method ofACI 3 I8 Section I I .8, as adopted by referenee through IBC Seetion I90 I .2.
AnaJysts oflhe wall panel for ltf'Ung stresses or other erection loads .s not a pan of this example and ts
usually completed by the subcontractor's specta1ty engineer.
3. Out-of-Plane Lateral Design Wall Forces
The concrete wall panels may be loaded out-of-plane by wand or seism1c forces. For eao;e of ana1ysis. the
waJI panel is subdivided into a design Slttp. TypicaJiy. a soltd pane-l is subdivtded tnto 1-foot-wide design
strtps for out-of-plane desagn. llowe\•er, for simplicity. wttere '"all openings are tnvol\'ed, tlte entire pter
wtdth between openmgs ts generally used as lhe design strip. The dtstr~buted loading accounts for the
strtp's self. weight as well as the mbutat)' loadmg from abo"e each opening.. Ftgure 5-4 Illustrates the
h«tzonml Ioad dism bunon on the wall"s profile.
302
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... .,
- I-
1'/ lf 1t' Ol)tlhi'IO
3"lt 7' cbow
[]_
t -cr
l" -.(
'1·0·
•··o·
..f
~
.r
l
-o·
-l'
Flguw j.J. £/e,·atlo, 11'ew ofwall ponal
_ _ _ .L__
I
I
......
12' x 14'
F/g11n S-4. Wa/1-/oadmg d10gram
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303
Design Example 5 • Tlr-Up Building
3.1 SEISMIC COEFFICIENT OF WALL ELEMENT
The wall panel is considered a beartng wall and a shear wall; lhus,ASCE 7 Sectton 12.11 .1 appltes tn
detennlntng lhe Lateral seismtc force.
but oot less than O.J ,..,.,
1,.= 1.0
s,.= 1.og
F=0.4( 1.0)(1 .0~''• =0.4-..
3.2 LOAD COMBINATIONS FOR STRENGTH DESIGN
ASCE7
For lhis example, the useof lBC Load combinauon Equatton 16-5 ofSect1on 1605..2 JS appltcable and
governs tor concrete strength design under seisn-uc loadmg. Because the axial Joad will create significant
secondary P-delta moments later tn lhe analysis, Equation 16-7 wdl nm govern O\'tt Equation 16-5. Wtth F
; 0 and H = 0, the load combinarjon becomes
1.20 + 1.0£ + f,L +J,S
IBC &] 16-5
where
D = self-wetght of wall and dead load of roof and equipment
L = 0 (floor live load )
S = 0 (snow load)
E = £, + £.= pQ,+ 0.2S0
p
&] 12.4- 1, &] 12.4-2 and&] 12.4-3
where
p = 1.0 for wall elements
§12.3.4. 1
IBC load combt.n.ation (Equation 16-5) reduces to
(12+0.2S..,)D+ I.OQ,or( l.2+0.2)0+ I.OQ,
orsimply 1.40 + I.OQ,
3.3 LATERAL OUT-OF-PLANE WALL FORCES
The lateral wall tbn~es Qe are determaned by multipl)'tng the walrs trtbutary weight by the lateral force
coeffic1em. Three different umform1y distrtbuted loads are determtned because of the presence of two door
open1ngs of dlfl'ering heights. See F tgure 5-4.
wan weoght = ••. = 9;~5 (150 lblft3 ) = 11 6 1blft'
F.,.,,= 0.4( 11 61blft') = 461b/ft
1
w,
304
W,
= 461blll'x 4 fl= 184 plf
: 46 lb/fl' X 3/2 ft : 69 pi f
w,
= 461bltl'x 12/2 ft=276plf
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O.sigl Example 5 • Til-Up Butilg
4. Primary Moment from the Out-of-Plane Forces
The objective is to check QM.. '2: M. where ·"111 =- .Mid+ P,.A.(ACI318 Equation 11.8.3.la). Mw.lS the
max1mum moment (usually ;.u mtd-he1ght) due to applied factored loads and constst~ of two compcments:
an out-of-plane loading mome.n.t (M.,«tt) and a venical eccenlricity loading moment (JH....«-)· PAts a
secondary moment created by P-delta effec-t~ and is investigated in Part 6.
To determtne M., ~the loading dtagram in Figure 5-5 should be u..-;ed.
b
Wj~ t841Q
;::: ~
1- ,_
.........
Rgure 5-5. Loodmg dtagrom
ACI 318 Secuon 11.8.2.1 s<ates, "The wall shall be analyzed as a Simply supported, axially loaded member
subject to an out-of-plane uniformly dtstrtbuted lateral load, with tnaxunum mome-nts ru'ld de.ftectjons
occurnng at mid-height.~ As evtdent from Figure 5-5, a pier between openings has netther a ul'tiform latera]
lood nor a max.1.mum mome.m occurrmg at mid-hetghL In thtS sattwujon, ttts acceplable to compute an
equl\1alent distrtbute.d load and the more accurate ma.xtmum mome-nt M., ... located shghlly away from mtdhetghL This ts then coJnbtned wnh M.#Ct and P,.A., as computed at mid·he&ghL
llte destgner then locates the point of zero shear for maxt.mum mome-nt AI.,~ They 1gnore the parapet's
negauve mome-nt benefits in reductng the pos11ive moment for s&mpltcJty ofanatysJs. If the des•gner dec1des
10 use the pampe.t's negauve moment 10 reduce the positive moment. spec tal care should be taken to use
lhe shortest occurnng parapet he.ighL FOC' this approach. the seis.nac coeffictent for the parapet shall be
the same a~ that for the wall below using forces based on ASCE 7 Secuon 12.11.1. The parapet should be
checked separately under Section 13.3.1 , but ts not a pan of th1s example.
llus example consef\•auvely assumes the max1mum moment occurs at a cn6cal secuon wtdth of 4 fee•.
In cases where the max.tmum mome-nt occurs well abo,•e the door~ a more comprehens1ve analysis could
corL.c;ider several criucal destgn sectjons, whtch v.ould account for a wider destgn section at the locauon of
maximum moment and for a narrower des1gn secuon wath reduced mome.nts near the top of the doors. ACI
318 Section II.S.I.Ia states thatlhe ••cross section is constant over the height of the wall." Therefbre,lhe
use of a comprehensive analysjs to account for a nonuntfonn design section wldlh ts beyond the scope of
Section 11.8.
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305
Design Example 5 • Tlr-Up Building
4 .1 DETERMINE THE SHEAR REACTIONS AT TOP AND BOTTOM WALL SUPPORTS
R,wt = shear reaction at grade la•el of design strap
ar r:~;:o)n auo:;:~)'<l ofd:s;::)su]·~
R....r= she
[
Rpok= 184 l
2
+69l
2
+276l
2
28 =40851b
R-1 =(184(28)+69(21)+ 276(14))-4085 = 6380 lb
Oetermtne the diStance of the ma-<unwn moment M11 1lo¥ as measured from the roof elevauon downward
(Figure 5-5):
X=
6380
+ +
- 12.1 ft!opotn!ofuroshear(maxtmummorrenl)
184 69 276
4.2 DETERMINE Muou•OI'oi'UIN•IOOPJ
ThiS is the pnmary momeru due lO factored OUl-of..plane forces. which e.xdudes P-deha effects and vertical
load ecc:entrictty effects. Using Figure 5-5. the following tS obtained:
M • .,., = 6380(1 2. I) - (I 84 +69 + 276) (l
2~ I)' = 38,473 lb-JI
u •.,., = 38.5 ktp.ft
5. Primary Moment from the Vertical Load Eccentricity
Any verucalloads that act at an eccenmc dtstance from the v.'alrs c:emer also apply a nlotnent to the design
wall section. In thts example only the roofloads are applied to the waJJ with an eccentricity.
P11»/= gravity loads from the roof acting on the destgn strip
P....r= (roof dead load) x (lrtbullll)' \\idlh of pier) x (1ribullll)' lenglh of roo I)
NOte: When concentrate.d g.ravtty loads, such as from a girder, are applied to slender \valls, the loads are
aiOSUmed to be distributed over an tncreasing wall width at a slope of2 units ''e.tticalto I untt horizontal
d0\\11 10 !he flexural design sec~ion hetghl (ACI 318 Section 11.82.2).
The applicable load comb1nauon detemuned in Pan 3 as 1.4D + I.OQ£ for se1smic consadecauons. Roof
live load Lo; not combined with seismic loads in the JBC strength design-load ooanbmanons. However. '"tten
tnvesugating loadoombinatiOt\S mcludtng wind design. a portion oflhetoofl1ve load is 1ncluded.
P• .w~= 1.4(2760) = 3864 lb
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O.sigl Example 5 • Til-Up Butilg
ll\e eccenuic load places an applted mome,nt at the roof level. With the base of the wall c.onstdered pinned,
the resulung moment at mid-hetght is approximau;·ly half of the applied moment Asswnang a 5-inch ledger
and Y.-anch reveals, the tOifowtng as obuuned:
M,~
•
= P.. n»f 2
where
-
t==>m+
9.25-0.75 925
-. m
2
0
M.~
9.25)
=3864lT
=17,871 11>-m.
-~~.~'¢
= 1.5 ktp-ft
6. Total Factored Moment Including P-Delta Effecta
11te total factored moment M,. lS the applied momem M., With an tncrea~ for P~elta efiOCL.;;. From Parts
4.2 and 5:
MIIIQ=M,.IXIfi+M"~
=38.5 + 1.5 =40.0 ktp-tl
M,.. is magnified usingACI 318 Equataon 11.8.3. ld
M_
M. = - - " '5"'Pc,
l', - 1_
Mt
(0. 75)48£,1~
The ca1c:ulation tbr .H, using ACJ Equation 11 .8.3.1 d provtdes a direct solution for second-order effectS,
indudmgP-delta moments, anstead of the iterative process of ACI Section 11.8.3.l(a). Various software
programs on lhe market today suit use an uerarjve second-otder approach or, m some cases. have
no second-order analysts. Software program resull~ can have significa1Uerrors when improper anput
assumpLIOJ'IS ate made. The desag.ner IS cauuoned lO e-1\.~re a pmper second-order analysts is utilized wnJ1
proper waJJ suffness assumptions.
To use ACI Equatton 11 .8.3.1d, tlte wall's venic:.al lood1ng and sec.u.on properties must be calculated.
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Design Example 5 • Tlr-Up Building
6.1 DETERMINE THE TOTAl VERTICAL LOAD
P~
=PI'IKI/+P_,..,
P_,. = 2760 lb (lrom Pan 5)
PwiiMIJI =the portion of the v.'alrs self we-Ight above the flexural design sectioo. It 1$ acc:eptable to
asswne the desig,n k"'C:tion is located mtdway beru-een the floor and rooflevels.
P....., =(116
psf>(4+~+ I~)(~ +4 )= 24,012 1b
P_,
=P...,.+P... ..,=2760+24,012=26,mlb
P.
= 1.4{26,7n) = 37,481 lb
:37.5 kipS
6.2 DETERMINE NECESSARY SECTION PROPERTIES
ACI318
For this pie;r, a uial double-cunatn reanrorcing arrangement of SL'< #5 bars Cor each face will be assumed,
as shown in Figure 5-6. The location of the re1nforc1ng around the pier is controlled by the necessary
conc-rete cover dimensions. For uh-up conc-rete, the reinforemg depth d can be based on ACI Secuon
20.6.1.3.3 tOr precast C0\1er dtmensions, provided that the construction is sunilar to that nonnaHy expected
under plant controlled coodtliOJ\S. Wnh the waH panels normally cast on the bulldutg•s conc:re-le floor
sJab. retnforcement placement on chairs. and shon-edge forms. nghter construCtion tolerances can be met
compared with tradiuonal monolithically poured concrete walls. For wall panels wtth #II bars and smaller,
the mutimum cover dimenston is% inch,
d = thickness - reveal - cover - tie diameter - ~ bar diameter
d= 9~ - Y. - Y. - %- (!h)(%)= 7.06 in
314" tevaal
Rgru~
5-6. Crou .ttclion
The cracked moment of tnenta /00 is necessary to dete-nnine the ?-delta etltc~ and ACI Equatton 11.8.3.1c
provides
P -h ) (d-e)2 +-I.e'
= -E, ( A+_..
I
N
308
£
'
'f.2d
y
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O.sigl Example 5 • Til-Up Butilg
where
E, : 29,000 ksi
E, :51.{j:: 3122 k<i
§19.2.2.1
a= P, + AJ, _ 37,500+6(0.31X60,000) - 1.22 on
o.85f; b
o.85(3000X48)
and thus per ACI318 Secuon 22.2.2.4,
c=~=
IIJ I
2
1.2 = 1.44 in
0.85
Eq 22.2.2.4. 1
and the efltctive steel area per ACJ 3 18 Sect.ion Rll .8.3. I considertngthe compression load 1s
k,W
I
~
:(A1', ..!!_)=
+
A
'
2d
; ;,
6(0.3 1)+(E222) ..2.2L =2.27 m'
60,000 2(7.06)
= 29•000 x2.27(7.06 - 1.44)' + 48(1.44 )' - 7 14 in'
3 122
3
The hl2d te.rm for A~,... moddi.es the tmpacl of cat:npression tbrce P, for conduJons wl\ere the relnforcing
steel 1s not a single cunatn at the center oflhe watrs thac.Jmess.
6.3 DETERMINE THE TOTAl FACTORED MOMENT MAGNIFIED FOR P·l!.. EFFECTS
Usong ACI 3 I 8 Equatton I 1.8.3. ld:
M_
M. = ---:5P~t';-l-
II
t
(0.75}'18£,/N
_
1
40 0
'
- 54.3 k ft
tp5(37.5X28x 12)'
(0.75)48(3 1l2X7 14l
7. Nominal Moment Strength ci>M.
The nominal moment strenglh 4'Af,. is given by the follov.·mg equation:
M.: A.J,(d -~
)=
2.27(60,000>( 7.06 -
1 2
~ ) = 878 ktp-in
M. = 73.2 kip-ft
~:
0.90 per ACI 3 I8 Table 2 I .2.2
: 0.90(73.2) = 65.9 kip-ft
M. : 54.3 kip-fi < 65.9 kip-ft
~M.
M 11 <<;1M, .. . OK
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Design Example 5 • Tlr-Up Building
7.1 CHECK FLEXURAL CRACKING MOMENT
~~> .Ha to determine the acce-ptability of the sle1lder wall design method (AC1318 Section
11.8.1.1c~ M~ lS defined in ACI 318 Secuon 24.2.3.5 \\i th !his use of/, defined tn Section 19.2.3.1.
Venfy lhati)M.
=
48(9.25)3
I
12
M_ =f.~=
7.5~3000-,.....-281,18711>-in = 23.4 kip-ft
.. 'y,
9 .25
2
Eq 24.2.3.53
Usang lhe ahernauve slender wall method.. tlte retnforctng IS sufficie.nt to provide adequate strength.
Nate: For the purposes ofAC1318 Secuon 11.8.J.Ic.• I,andy, areconsenoattvely based on lhe gross
thtdcness \vithout considemtion for archnectura1 reveal depth. This approach creates a \~t"Orst-ca~
compartson of Ma to q.Mtr ln addtuon. the exclusiOn of the revea1 depth tn the ,H" calculation hkely
produces a more accurate deflection \•alue when the re\'eals are narrow. sha11ow. ru\d few in number.
7.2 CHECK SECTION FOR TENSION-CONTROLLED RESTRICTION
ACI318 Secuon II.S. I.Ib requtre.s Y..alls lhat are designed usingthealtematjve design ofslerlder walls
provisions to be tenston eonltOlled. ACI 318 SectiOn R21 .2.2 defines tension-coJUtolled sections as those
whose net tensile stratn £, > 0.005 when the concrete in compression reaches ns assumed stram hmtt of
0.003. The net te-nsile strain hmits can a1so be stated in terms of the rat'io ctd,.. where c ts the depth of the
neutra1 axts at nominal strength and d, ts the d1stance from the extreme compression tiber to lhe extreme
tenston steel. A net ttt'Lo;ile strajn hmit of£,> 0.005 is equtvalent todd,< 0.375 for Grade 60 reinfon:ement
(ACI318 Ftgure R21.2.2a).
cld,
=1.4417.06 =0.204 <0.375 .. . OK
Therefote. the slender wall melhod
JS
acceptable.
7.3 CHECK THE MAXIMUM VERTICAL STRESS AT MIDHEIGHT
Check lhe factored venteal sues:s at the nud.height secuon to determute whether lhe alternauve slender
wall destgn melhod is acceptllble (ACI 318 Section 11.8.1.ld). With only dead load D and roofl.,·e lood L,
contrtbutjng toP.,. the IBC load combinations of Sec.tion 1605.2 with ASCE 7 Section 12.4.2 reduce to the
foUmvi11g:
IBC Equation 16- 1:
1.4D
IBC Equation 16-2 :
1.2D +0.5L,.
IBC Equation 16-3:
1.2D + 1.6L,+0.5W
IBC Equation 16-4:
1.2D + I.OW + 0.54
IBC Equation 16-5:
( 1.2+0.2S...)D+ I.OQ,= 1.4D+ I.OQ,
IBC Equation 16-6:
0.9D+ I.OW
IBC Equation 16- 7:
(0.9 -0.2S...)D+ I.OQ,=0.7D+ I.OQ.e-
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O.sigl Example 5 • Til-Up Butilg
From inspec.uon of the Jood combinations pre\•iously listed. only combination-. 16-1~ 16-3, and 16-5 can
govern the vertical load. As determaned trt Pan 6.1 .the total ven:ical dead load D is26~7n pounds. The
reduced roofltve load L, ts determtned from ASCE 7 Sectton 4.8 as follows:
40ft(4+l+
3 12)
=230ft'
2
A,= - 2
L,= L.,R 1R2
where
L. = 20 psf. R, = I.0 (fiat roof} and
R1 = 1.2 - 0.001.4,=0.97
L,= 19.4 psf
P..,
=19.4 psf x 230 n' =4462 lb
Load combinatiOilS
1~1 .
16-3, and 16- 5 result tn lhe fol lowt.ngP, verttcal loads:
IBC Equation I6- I
1.40 = 1.4(26,772) = 37,481 lb
IBC Equation 16-3
1.20 + 1.6L, + J.OIV = 1.2(26,772)+ 1.6(4462) = 39,266 lb (go\'erns}
IBC Equauon 16-5
1.40 + !:!.= 1.4(26, 772) = 37,48 I lb
Venical load P, = 39,266 <0.06(3000X48 x(9.25 - 0.75)) = 73,440 lb .. . OK
The comptesston load P,ts low e1lough to U.')e the alteJ'native sknde.r waJI
method~
otherwise a diftetet\t
method. such as the Slmphfieddesign method (ACI318 Secuon I 1.5.3)or the compresston me.mber
method (ACJ 3 I 8 Section 11.5.2). \'IOuld be required along wnh their restric.ttons on wall slenderness.
8. Service-Load Deflection Considerations
ACI318
Out-of.plane deflecuons are limned w:lder servic.e.Jeve.lloading to prevent pe.rmane11t wall defonnations
under frequent wtod and earthquake evenls. Based on the fuJI- scale slerw:ler wall tests conducted by
SEAOSC in the I980s (SCCACIISEAOSC, I982), u wa< obseJVed that pennanent set could occur pnor to
theoreucal yaeld of the reinforcing. and thus a servtce-level deflecuon lunn of //ISO \1/a:S adopted by lhe
Uniform Bw'/dlng Code and later the IBC. In addjtion, the fuJI-scale tesung revealed that cracking Initiated
at t; = 5{;;> .n.<tead of at t; = 7.5{;;> as gl\'en tn ACI Equation 19.2.3. I, and thus ACI Table I 1.8.4. I
ha-c; included a two-lhJtds adjustment factor on flw and .\fe r ThiS confhct be.tween observed behaVIor and
the ACt equation for modulus of rupture[, ts associated wnh re-mfOrcing restratnt to concrete shnnkage
(Lawson, 2007}.
Anolher depanure between the observed behavior a1\d tradittonaJ ACI equauons ts the load-deftection
graph. The full-scale tesr.mg program re,•e.aled a lood-deflecuon relattonslup tJuu was dearly btl i.near
li\Ste.ad offoiiO\\'lng Bran.'iOn's etfecttve moment of menia I, (ACI Equauon 24.2.3.Sa). lne reason behind
th1s deviarjon m sJender wall prutels lS associated wnh the unique flexural suffuess of thin bending members
(Lawson, 2007). TI1erefore, the follow.ng bihnear approach is p<O\'ided by ACI Table I 1.8.4. I.
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Design Example 5 • Tlr-Up Building
If the moment M. (tncludmg 1'-<lelro eftects) does oot exeeed (2/J)M,.:
where
Eq 11.8.4.la
Eq II.S.4.3b
Because M 0 must include P-delta effect'l based on ll... and~ depends upon .!&f.,. an nemt.ive process wdl be
necessary. To detennine M.,. service-level loads are lllil ized using the appropriate load combinations ln ACI
Section R 11.8.4.1.
The servtce-level setsmic load combination equals D + O.SL + 0.7£ where E = PQc+ 0.2SrJ] per ASCE 7
Sectioo 12.4.2. Thus, D + O.SL + 0.7(pQ,+ 0.2S,..) or ( I + 0.14S,)D+ O.SL + 0.1pQ, With L = 0, p =
1.0, ru'ld SJZ;= 1.0. the applicable load combinatjon for servtce-level seismtc loads reduces to the followtng.:
1.14D+0.1Q,
8.1 DETERMINE THE APPLIED SERVICE-LEVEL MOMENT
lntttally, M(J
tS
lhe applied servtee-Je\'el moment and comp(lses Af41 ~(out-of-plane) and M(J.-
(eccenlt1C·tty).
Because M41 oop 1S solely caused by the se1smtc loads Q£ 111: this example,
M• ..,=0.7M• .,=0.7(38.S) =27.0 ktp-ft
AdditaonaJJy, in an approach sun alar to Pan 5,
M.-= P..,.(et2) = 1.14(2760)9.2512 = 14,55211>-tn = 1.2 ktp-ft
M. = 27.0 + 1.2 =28.2 ktp-ft ("1lhout!'-6 eftects)
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O.sigl Example 5 • Til-Up Butilg
B-2 COMPUTE THE INITIAL SERVICE-LOAD DEFLECTION
11te 1ntttal M11 tS used to c.ompute the firsl Jterauon of P-delta e.tl'eru.
lnillal M, = 28.2 kip-ft (wuhout P-1!. efteciS)
Ma = 23.4 ktp-ft (from Part 7. 1)
(2/3)Ma = 15.6 ktp-ft
M11 > (213)!\1.,., so the cross sec:Lton is tnttially assumed to be a crncked sectton
UsingACI Table 11.8.4. 1 ForO,:
L\ = (2/3)<\
tr
.t
+
(M - (2/J)M )
'
a (!!.
(AI,. - (2/3).\fa)
•
- (2/3)1!. )
a
where:
!!.
a
=
5M.J! _ 5(23.4 x 12)(28x 12)1
48£,1,
11
!!. = )- .M.,
•
48£,Ja
48(3122t8(~:5J' )
_
0.
3310
5(73.2 X 12)(28 X 12)-'
4 63 10
48(3 122X714)
- .
(2/3)Ma=(213)23.4 = 15.6 kip-ft
28 2
!!., =(213)0.33+ (
· - 1 ~·6l(H3 - (2/3)0.33)
(73.2 - b .6)
b..,
= 1.1 8 tn (wnhout P-~ effects)
8.3 DETERMINE THE SERVICE-LOAD MOMENT M.,INCLUDING P-DELTA EFFECTS
For sel"\•tce loads including the effect~ of the vertical load eccentrtcity and P-deha etfeclS:
Use Ftgures 5-7 and 5-8 to detemune M41 ~1-Yand M.,p,:. fora de.flected shape.
Detenntne the force component H from stattcs summtng moment~ about the base of the wall. From Ftgure
5-7, and aoeosuming.a parabolic deftected shape:
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Design Example 5 • Tlr-Up Building
_;
I
I
!! '· ·
!
'j..-- -'Y'
2&·
1/ 3
Figurft 5-7. Verticalloadlng d1tJgrt11n
Because tl'le panel's ope~ung.~ are not posittoned symmetrically wtth the panel's mid-height, P..,..Nw- wtll
be less than P,.,llll"'ff" For ease ofcalcularion, COI'Iservatively assume PWJ~1 ,__ = P,...,,.. which ts sim1lat to
panels wtlhout openjngs. Sum moments about lhe base:
Using. F1gure 5-8, and substhuting for H. compute the moment component M from staucs to acoount for
eccentrtcity and P-delta eft-bets:
Therefore,
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O.sigl Example 5 • Til-Up Butilg
,..
__
Flgnre 5-8. Ftu-body diagram
Firsl iluation
1.1 8'"
M. =21.0+1.2+ 1.14(24.012+2.76)-12
M. = 31.2 kip-ft
Plug utto Equation 11.8.4.1b for a new t!J:
3 2 15 6
" =(213)0.33+ ( 1. - · >(4.63 - (2/3)0.33)
•
(73.2 - 15.6)
tt., = 1. 4 1 tn (first 1ttration wath P-tJ. e-ftkts)
Sttond ittration
M. = 27.0+ 1.2 + 1.14(24.012 +2.76)
1.~~ in
M0 = 31.8 kop-ft
Plug 1.nto Equation 11.8.4.1b for a new t!J:
3 8 1 6
IJ., = (2/3)0.33 + ( 1. - ~· ) (H3 - (2/3)0.33)
(73.2 - 1>.6)
6,
= I .46 in (second iteration with P-!l effect~)
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Design Example 5 • Tlr-Up Building
Third iltrt.lion
1.46'"
M. = 27.0+ 1.2+ 1.14(24.012+2.76)12
M.
= 31.9 ktp-ft
Plug into Equanon 11.8.4.1b for a new~
6
'
!:t.s
= 213)0.33 + ( 3 1.9 - 15•6) (4.63- 2/3)0.33)
(
(73.2 -15.6)
(
= 1.47 in
Fourth ittrat:ion
1.47 In
M.
= 27.0 + 1.2 + 1.14(24.012 + 2.76)1 2
M.
= 31.9 kip-ft (converged wttlt P-41 effects)
I
t:t., = 1.47 ins: ; = 2.24 tn ... OK
10
The:refote, the proposed slendeJ \Vall section is acceptable using the aJtemati\'e slender wall method.
Commentary
Instead of the lengthy tterauve approach. the designer could stmply have assumed a trial !>.1 = /jl50 = 2.24
tnches as a worsl case to compute."',. then use ACJ Equation 11.8.4.1 b for a new 6_,. If the new !:t., sausfies
the ljl50 Iimit, the slender wall sec.tioo is acc-eptable.
An example oflhe retnforcu1glay0ln fOr thee.ntire wall panellsshown in Ftg.ure 5-9.
~hO~I
remfQfting#S@
1frM
ft'!tltofeiOO
~n;Mincl Ql)tflng~
2·.S
Figure j -9. 1Jplca/pan~/ reinftJrcing
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O.sigl Example 5 • Til-Up Butilg
9. Wall Anchorage at Roof PurlIns (North-South Loading)
In single-story buildtngs., ult-up wall panels typically are supponed out-of-plane at the roofstructure and
floor slab. While Cailures of the conc:rett wall section ttself out of plane Me extremely rare, v.'alJ anchorage
failures at the rooftine have frequently occurred dunng earthquakes. In response to these failures. the
current anch«age des1gn forees and detailing requirements are significru1t1y more suinge.m than they were
under older codes (Lawson et a!., 20 18).
From a histoncal perspecuve. the most critical element i.n ult-up engtneered buJidtngs is the wall anchorage.
In the 1960s and up unttl the 1971 San Fernando, Caltfom~a, earthquake, engineers tn the west typic:.aJiy
prov1ded no posttive direct tie anchonng the pe.nmeter concre.te wall panels to the suppottiJtg \\'nOd roof
SU\Icrure. Jnsre.ad,lhe roof plywood sheathing was Slmply naaled to a Y..'OOd ledger that was bolted to lhe
tnSide face of the wall panels . The roars glued-laminated umber beams (glulams) were supported oo top of
concrete p1fa~ers and had ue connections wuh mintmal capactty. Thts tndirectLie arrang.eme,nt relied on tl'IC
wood ledger in cross-grajn be.ndtng. a very weak material propeny of wood.
In the 1971 San Fernando earthquake, many wood ledgers splittn half due to cross-grain ben<ltng loads,
and pi)~Vood edge natling pulled throogh plywood panel edges as the result of tension loads. Pattial roof
collapses and wall collapses were common in the areas of strong ground motton.
Beg.iMlng with lhe 1973 UBC. cross-grrun bend.ng ln wood \'o'aS e.xpressly pmhibtted. and spectfic wall
anchorage require-ment(i wece established. Over the years since then. tl'lf' waJI-anchomge design forces have
tncreased m response to conttnUJng poor perfocmru\Ce of wall anc.horage during earthquakes and additional
tnfonnation learned from tnstrumented ult-up buildtngs.
The current \\'llll-anchorage code requirements ate a result of the 1994 Nonhndge eanhquake. "The
unexpected wall-anchocage damage to newer buddtngs was prtmarlly anrlbuted to inadequate conned ion
oversttength for the roof accelerations. Research has sh<mn that roof-top acceletaitons may be three to
four tunes lhe ground acceleration. IBC Section 1604.8.2 specifies the need tor wall anchorage in the.se
structures. ASCE 7 Secuon I2.11 .2.1 de-termutes the anchorage destgn tbrces tbc lhe structural walls of
ult-up buildtngs. tncludtng the reqUJre.ments of Section 12.11.2.2. to ttanster waJI-anchornge forces mto
the diaphragm ln seismtcally active areas (Seismtc Design Categortes C through F). The wall-anchorage
forces in flexible diaphragms are amplified by the k# factor. whic.h \1attes from 1.0 10 2.0. depending on me
diaphragm•s span between Lru.eml-resJsting waJisand frames. Flexib1edlaphragms Wllh spans tn excess of
100 feet ha\'e wall-rutchorage loads that are double the normal wall design tbcce and lhree to tbur 11mes the
typicaJ ult-up buildtng base shear to account f()( the e.-<pected roof-top amphfic.anon associated wnh flexible
diaphmgms.
11te reqUirements ofASCE 7 Section I3.4 assoctated wtth anchoroge of nonsUttctural concrete componentS
do not apply here because all be.aring walls and shear walls are classified as structural walls under Sect1on
11.2. In additiOn, all oonstrueturalwalls Sllpported by ftextble diaphragms are also anchored per Section
12.11.2 per Table 13.5- 1 Footnote c. The de:sjgn forc-es associated wnh the concrete and ma'iOJ\ty \\'llll
attchorage at struc-tural walls ha\·e already been factored up to maximum expected le\•els ln comparison
wnJ1 matenaJ overstrengtllS.
9.1 FORCES ON WAll-ANCHORAGE TIES
ASCE7
In tlus example-. the sttucruraJ concrete v.'all-anchorage forces to the floable dtaphragm are governed by
ASCE 7 Equanon 12.11- 1:
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Design Example 5 • Tlr-Up Building
where FP shall not be taken less lhan 0.2kJ..W, and flexible diaphragm amphficauon factOI' k# IS
L
k, = 1.0+
I~
where kill need not be grecue.r than 2.0. L1 rs the fte.xtble dtaphragm ·s span m feet between the walls or frames
Lhat p(QVtde lateral suppon 10 the dtrecuon beang considered.
In lhls example btulding,.lhe reentrant corner provtdes diaphragm support reactions w1th the attached
collectotS, and thus the followmg d.aph.ragm spans are recogntzed:
North-south seismic. forces:
Gnd lutes I to 3
Grid lines 3 to 9
fr=80 tl
fr= 240 tl
k#= 1.8
k#=2.0
East-west setsmic forces:
Grid lines A to B
Grid hoes B toE
k,= 1.4
*.=2.0
In both the north-south and east-west orthogonal dlrections., the majority of the \\'all anchorages requtre
2 .0. To stmphfy lhe repetiuve \vall anchorage. ka = 2 .0 tor all designs 1n tJus example.
*• =
With SI>S = I .0, k, = 2.0, and 1, = I.0,
F, = 0.4SIJSkJ,W, = 0.80W,
Eq 12.11-1
For this budding's 9Y...-uK-h-thic.k concrete \\'3.11 panels ( 150 lblft 1), the urut wall load IS
F,= w,=0.8(1 16 psi)= 92.8 psf
Chec-k thjs umt load wnh the mtnimum "all-roof anchomge force per ASCE 7 Secuon 1.4.4 and Secuon
12.1 1.2.1.
5 psf <92.8 psf . . . OK
0.2( I I6 psf) = 23.2 psf < 92.8 psf .. . O K
0.2kJ,W,= 0.2(2.0)1.0( I 16 psf)= 46.4 psf < 92.8 psf .. . OK
Use F, = 92.8 psf
\Vlule thts bwlding has a c.ombtnauon of solid wall panels and panels wnh peneuarjons, the solid panel
oorxhtion wdl produce the worst-case wall-anchorog.e forces and will be used for design here-. lJsmg statics
to sum moments about the waWs base., the foiiO\Ytng calculation for the wall anchorage forc-e per foot of
roofhne iocludes the parapet·s cantilever effects (see Figure 5- 10).
F, = 92.8 ps~32 ft)l•J?ft)
T 28I ft = 1697 plfalong lherooftine
The anchorage force pe.r jolst is
F, =(8 fiX1697 lblfi) = 13,576 lb
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O.sigl Example 5 • Til-Up Butilg
b
N
"'
b
o.sw.
"'
N
Figure j-/0. Wall secuon wtth IObdmg
Commentary
When ue spactng exceeds 4 teet, Section 12.11.2.1 requires that suuctural \\'ails be des1gned to restst
bending between anchors. Instead ofus1ng ASCE 7 Equation 12.11-1. the des1gner may elect to use lhe outof-plane force for me wall tt•elf; per Section 12.11.1, F, = 0.4S,..f,W. The reasoning behind thts approach
IS that the wall's out-of-plane be.ndtng lS expected to have a mo..-e ductile behavior than lhe connectjoi\S and
thus are designed for lo,,rer forces. This design force also ha,. a Jo,ver bound of0.20WP per Secuon 1.4.4.
9.2 CHECK CONCRETE ANCHORAGE OF TYPICAl WAll-ROOF TIE
ACI318
Concrete anchorage design is'" accordance with Chapter 17 ofACI3 18 as modified by IBC Secuon
1905.1.8.
ACt's Table 17.3. 1.1 lists the various failure modes lhat must be considered tndetenniningthe required
anchor strength.
ll'!e first step IS to determme the resulung des1g.n loads acting on lhe \vaJI-roofanchorag.e sySlem. The wallroof anchoroge aJong the nonh and south walls consiSLli of a Steel JOtSl seat welded to an embedded plate
·with headed-weld studs (see Figure 5-11). Because the embed resists both the waH tie force and the \•erucal
gravny reaction of the steel joist. several loads must be comblned.
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Design Example 5 • Tlr-Up Building
&w>JrAng"b to
EmbedPiitto
J~ltoShof
)"
1' · 2"
1'-0"
, .. :
0
.
:· . .-·_.
;
:::.... '·
(.t} 112"0ia 11ead&d
t'~
}
\WidS~{S"I~}
....
~
;w'' fmhM I"JiH
.•··
Figurt 5-11. Steeljoist to wall-tit detail
The vertic.aJ gravity end reaction from the stee1 joast c.reates a prying force on the embedded plate's anchors.
h \\lll be assumed a force couple at the he.aded-wcld studs wdl rtS ISlihe eccentric gravny load.
Calculate the joist end reacuon It
R =(12 psf +20 psl)(8
ft>( ~ ft) = 1920 lb (dead)+ 3200 lb (In~)
Cooservatavely assummg the vemcaJ joist reaction is acung at lhe edge of the shelf angle, lhe reacuon
eccentnctty is 5 1nches, whtch matches the panel design assump£ion tn Part 5. With the 6-tnch \'etttcal
spacing between the two pairs of headed-weld studs, the fOllowing stud fOrces are determlned ustng the
load combinations ofiBC Sec1ion 1605.2 and ASCE 7 Sec1ion 12.4.2.
Loa d Combination IBC Equatioa 16--J
1.2(0 +f)+ 1.6(L, or S orR)+ 1.6H +({,Lor O.SJV)
GJVen S = 0, R = 0, L = 0, F = 0, H = 0, and wutd ts not being considered, thlS load comblnauon reduces to
1.20+ 1.6L,
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O.sigl Example 5 • Til-Up Butilg
.._::.A
• • !" ..
$ 187 1b
.,
tenston
-
.:-..
}
=1.2(1920) • 16(3200)
= 7424 1b
~!,;
~.·
·:··
;;· i l~
74241b Shear
v
Ftgure j - / 2. Lbad on embed
l oad Combination IBC E-quation 16-S ~'itb ASCE 7 St d ion 12.4.2
( I .2D + 0.2S"')D + P<Jt + L + 0.2S
GivenS()$= 1.0. L =O~S=O. and p = 1.0. this load oombtnatton reduces to 1.4D + Qe-
5"
.. ..
2688 lb
$tleilf
Ftgurt! 5- I 3. Lbad 011 embed
l oad Combination lBC EqWition 16-7 witb ASCE 7 S«tion 12.4.2
GivenSos= 1.0. H =O.and p = 1.0. th1s load comb&nation reduces lO 0.7D+ Qe-
5"
Ftgure 5-I 4. Load 011 embed
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Design Example 5 • Tlr-Up Building
The load combanauon from lBC Equatton 16-3 results tn only two weld studs loaded tn tension, while the
othe-r two load co•nbtoalions result tn all four weld studs tension loaded. The load c:ombinatton from IBC
Equarjon 16-3 is cOt\Sidered firsL
Load Cotubiaation Equation 16-J Analysi.'
ACI §17A.I
The nomtnal steel strength for a ~tnch-d1amerer steel headed stud anchor IS computed ustng ACt Equation
17.4.1.2:
Nsa=AMJ~f-
0.196 m1 (~tn-diameter srnooth shaft)
f-= 65,000 pSI (AWS 01.1, Type B)
A.,.JI=
Thus, N~ = 12.74 kips.
Steel headed studanchorscontbnnang toAWS 01.1 Type B quality as a ducule steel element, and thus the
strength reduction factor oj> tS 0.75 pe< Section 17.3.3:
oj>N~
=0.75(12.74 ktps) = 9.56 kips
The mOS( htghly stressed anchors 1n thts group are the top two headed Stud anchors. 11\e design tensJon
force for an mdtvidual anchor tn this group is Nw;= 6187/1 = 3094 1b.
'1>-iV~
= 9.56 kips> 3.094 ktps
... OK
Conrntt brtakout stnnglb in ttnsion Nt*r
ACI §17.4.2
The 'h-inch-diam.eh!'t studs have an after-weld length of 5 inches.. and considering their ~~tnch-thic:L: heads.
they have an etfec.uve embedment of htf= 4.688 tnc.hes. Tile plate's thickness may be added 10 h11 because
the fadure surface extends out past the outhne of the embed plate. Thts results in hrf = 4.688 + 0.375 = 5.06
tnches, rounded 10 hq = 5 1ncl\es.. For the dtft'erent ~sible failure modes beang evaluated, ACt Sec.tion
17.2.1.1 is consulted to de1emune whetJ1er group ac:uoal IS occurrtng. For concrete breakouttn tensjon_. the
cntica.J spactng IS 3hrf = 15 tnches, whteh is greater than the actual spacing and thete.fore group action is
()()C:urnng..
' ' _,
anchOr rod
/'
r
- -
,,
7.5"
1'. ()'
,.:
"•
/
~
"'
~
''
/
-:;,
'
7!S'
~
~----'
,,
Figurt 5-15. Projtctedfailure area
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~
O.sigl Example 5 • Til-Up Butilg
ACI E<juation 17.4.2.1 b:
Ndf -- AA,\'£ (•
"'
~,_vi\
wi,.V
)(• e-.\' "'
I\
q~.N
)N•
N«t
A"' = 2(7.5 tn)[2(7.5 m) + 12 mJ = 405 m'
A,v((lo = 9/J'!t/:::: 9(5)2 =225m2
F 5- 15
Eq 17.42.1 c
Per ACI Secuon 17.4.2.1 .Ax.- shall nm exceed nAft.'to·
The modificauon for any possible loadul£, ecceiUJJCJty is per Equauon I7.4.2.4:
where e~ JS the eccentrtclly of the resultant tens1le force from lhe cemro1d of the bolt group acung '"
tension. Because there is only one row of anchors acting 1n ten..:;ion tn lhis lood combination. the anchor
group's resultant tension force aligns wnh the anchor row and thus
t~ =0
In
I
'I' K.v =(I + 2(0))- 1.0 ma•
3(5)
"'"""= 1.0 (no adjacen1 edgeetfeciS)
§17.4.2.5
'Vt.N = 1.25 (uncrac.ked secuon due to shon parapet)
'1'.,.... = 1.0 (cas<- in- place anchor)
§17.4.2.6
§17.4.2.7
N, = k,>...[i'fr~' = 24(l.O)J3000(5)1·' = 14.7 k1ps
405
N,., = ,,__, (I.O)(I.OX I.25)( 1.0)14.7 = 33.1 ktps
Eq 17.42.2a
ll\e strength reduction factor 9 is 0.70 per Secdon 17.3.3 because oo supple.memary reanforcing is
provided. a~td thus
~N,..=0. 70(33 . 1Jcips)=23 .2 k ips > 6. 19kips
... OK
Pullout Slrtngth in ltnsion
AC I §17.4.3
N,. = v,.J'I,
'Vt.J>= 1.4 (assume uncmcked sectton due lO short pampet height)
Eq 17.4.3. 1
N, = 8.4.. f; (' mere t.!aded sruds or bollS are used)
A.. = (head area) - (shank area)=0.785 - 0. 196 =0.589 m'
E<J 17.4.3.4
§17.4.3.6
N,.= 1.4[8(0.589)(3000 psi))= 19.8 kips
nN,_ = 2( 19.8) =39.6 ktps
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Design Example 5 • Tlr-Up Building
The strenglh reduction factor I) is 0.70 per Se<:t:ion 17.3.3 because no supplementary reinforc1ng tS
provided-. and thus
<I>N,..=0.70(39.6kips)=27.7kips>6. 19kips ... OK
Cont rttt sidt-ract blowout strt.ngtb in ttnsion
ACI §1 7.4.4
In lh1s example~ 11 wtll be assumed that the coocrere anchorage is nO£ located near an edge~ thus~ N_.. Will
I'Kll control the des1gn. However, tJ'Ie purl in layout tS often not well ooordinated With the concrete panel jo1nt
Layout. and theufore conflicts between the tw'O can occur. Where purl an anchorage IS located near panel
JOints, Nlob mu.o;t be evaluated. Th1s ts true also for roof anchornge to wall panels with no parapet.
Govt.rning ttnsilt st.rtngtb
For this gravity-only load combination., the steel tensile strength at 19.0 kips for the top pair of anchors
is less tJ'tan the other \13fious concrete tensile failure modes checked, and thiS strength is greater than
the design te.ns1le load of6. 19 k1ps. HO\\•ever, acceptability IS pe1lding the combined tensile and shear
tnreract.ion check pre.sented below.
Sttt-1 slrtngth in shtar V8
ACI §1 7.5.1
The nommal stee-l strength fOr a ~tnch-dtamete.r steel headed stud anchor IS computed ustng ACI Equation
17.5. 1.2a. Bolls af presen1use ACI Equauon 17.5.1.2b.
~~=A~,yf-
Eq 17.5. 1.2a
A..-.1·=- 0. 196 Ln 2 (~-in djameter smomh shaft)
f-=65,000 pst (AWS 0 1.1, Type B)
12.7 kips per anchor
1-..=
Steel headed stud anchors contbnn1ng to AWS 01 .1 Type B quality as a ducuJe steel element. and thus
the strength reductton factor 1)1s 0.70 per Secuon 17.3.3. In addition. four anchors are present to evenly
provide shear strength:
.~-..=0.70(4)(12 . 7
klps) = 35.6 ktps
From Figure5- 12,lhe design shear force ts 7.42 ktps.
•
~-..= 35.6
ktps > 7.42 kips .. . OK
Cont rttt brt-akoul s trtngtb in sbu r V~
AC I §1 7.5.2
As previously mentioned. it is asswned in lhls example that the embed plate is not located near an edge of
the panel. In this S1tuauon. rrtt wdl not govern (ACI Sec.tion Rl 7.5.2.1 ). Where purl in embeds are located
near panel JOints. J·eb must be evaluated.
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O.sigl Example 5 • Til-Up Butilg
ConcrttC' pryout stnnglb in shear V..,
For a group of cast-In-place anchors, the nominal pryout suength 1n shear J ~ IS a fWlCLion of the concre-te
breakout strength N~., determined earlier.
Eq 17.5.3.1b
v"" = kq/'1""
k~ = 2.0
for anchor embedment~ h4 ;:: 1.5 1n
N.,.=N,..=33. 1 ktps
V"" = 2(33. I)= 66.2 kips
The strength reduction factor 4) is 0.70 per SectJon 17.3.3 because no supple:menm.ry re-inforcing is
provided, an<l thus
$~:,.,=0.70(66.2
ktps)=46.J ktps> 7.42 ktps ... OK
GO\'trning shtar strengtb
For thes gravity-only load combu1.auon. the steel suength at 35.6 kips tbr lhe four anchoi'S 1s less lhan t11e
other various concrete shear failure modes checked, and this strenglh IS greater lhan the design shear load
of7.42 ktps. Ji0\1/e\'er, acceptability is pending the ronbi1led tens1le and shear tnterac.tJon check presented
be-low.
l nttrattion oftus ilt and sbtar rortt:S
ACI § 17.6
An interactton equatiOn check (ACI Equation 17.6.3) is required unles..~ either the shear r:w or tenston N...a
does not exceed 20 percent ofthe1r des1gn strength.~ Qf.>~ or 9-N.. respecuvely. The following shear 3Jld
1ens1on checks are made for lhis load combu'lation:
7.42 ktps/35.6 ktps = 0.2 I > 0.20 .. . Not satisfied
Check tf N,j~N. ,;0.20
6. 19 ktps/ 19.0 ktps=O.JJ > 0.20 .. . Not satisfied
Thl.t.\ mteractton ACI Equauon 17.6.3 1s required to be chec.ked.
For the four-headed stud anchorage conflgurauon
O.J3+0.2 1=0.54SI.2 . .. OK
In summary~ the headed sruds under the gravity load c.ombina.tion (3) are acceptable.
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Design Example 5 • Tlr-Up Building
l.Md Combi.aalion Equa tion 16-5 Analysis
Thts load combtnation contau\S contrtbuuon.o; from eanhquake loadtng £. Because the design strengths
wdl be compared wath destg.n load.~ that anclude earthquake forces wtthm Se!SmtC Desagn C.ruegoty Cor a
hagher SlfUCture. the provl.~tons of ACI Section 172.3 are applicable.
The seismic anchorage provistons ofACI 3 18 Section 17.2.3 aam 10 prevent sudden brmle fatlures when
subjected to htgh seasmac loads because design earthquake forces are normally less than those thought to
po<enLially occur during !he design hfe of !he structure. More specifically, ACI 318 Sec.uons 17.2.3.4.3 (for
tenston) and 17.2.3.5.3 (f« shear) requlre ducule anchor behavior or alternatively a seismac anchot design
force wath ove-rstreng.th n However. fOC' anchorage coonections wuh less than 20 pe.rcenr of the factored
tmaJ desagn force coming ti"om a seismic component. the ductility andlor oversttength requirement~ ru-e
ellmu\ated(ACI 318 Secuoos 17.2.3.4.1 and 17.2.3.5.1). Observing Figure 5- 13, more than 20 percent
of the anchor's tension force 1S from £ mggering the d:uctJJt[y/ovetstrength requtrements for the tensaon
d~ign, but less than 20 percent of its shear force is from £, allowtng those requiremenrs to be agnored for
shear design.
Ovemding this diSCussion is 20 18lBC Section 1905.1.8, \\ihich modifies the requtrementsof ACI Section
17.2.3.4.2 (<ens ion design). In response LO brittle connection failures observed afiet lhe 1994 Notthndge
earthquake, the wall-anchorage loads from ASCE 7 Equauon 12. I 1-1 already approximate ma.'<imum
expecred force levels and anclude an embedded overstrength factor to elimutue the need f« a oonnector's
ductilny in tenston (SEAOC Blu~ Book, 1999). Wtth this in mtnd, the IBC eliminates ACI's reqUirement for
anchorage ductiluy or addittonal oversu-ength f« tension desagn when forces are detennined from Equatiott
12.1 1-1. Thu.o; in thjs example, the connecuon des1gn \\111no1 c-onsider ductility or addttlonal overstrength
requtrements fOr etther tension or shear.
An addJttonaJ requtrement to conside-r is ACI 318 Secttoo 172.3.4.4, which employs a 0.75 factor 0 11
the coocrete strength Iinln states based Otl a cracked ooM.rete assumption, unless n can be demonsLroted
thal the coocre.te will remain LU'l(;J"OCk:ed at lhe.ru'lChor. ln this exarnple,lhe bending mome.nt at the wallanc.horage loc:aucm ts mintnlal due to the very shon parapet, and cracklng due to shnnkage as unlikely due
to the mm1mal tn-plane panel restratnt at the roofhne; thus, at can be demonstrated that the modulus of
rupture wtll not be e.'<ceeded.
ACI §1 7.4. 1
Stttl Slrt ng th in tu.sion N ..
As was determmed for the prevaous load combinaliott, N., = 12.74 kaps pe.r anchor and
<>N» =0.75(12.74 kips)= 9.56 kips
The most highly stressed anchors tn thts group are the bottom two headed stud anchors. The design tension
force tor an indav1dual illlchor an thts group as N-.J = 8508/2 = 4254 lb.
t-¥~
= 9.56 k1ps > 4.25 kipS ... OK
Conrntt brtakout s tnngtb in ttnsion N.-..
ACI §1 7.4.2
With all four anchors tn tenston. the projected failure area ha." ancreased. TI'II! ~-Inch-diameter headed studs
have an after-weld length of 5 mches, and consade-rtng lhear ¥.vinch-th1ck heads, they have an effect1ve
embedment of h4 = 4.688 tnches. The-plate's thtckness m.ay be added to h,1 because the frulure surface
e.xtends out pasLthe outhne of the embed plaie. ThiS results ln h4 = 4.688 + 0.375 = 5.06 inches... rounded to
h,1 = 5 tnChes. For the different possible failure modes being evalumed. ACJ Sectton 17.2.1.1 is consulted to
detennine whether group action is occurring. For concrete breakout 1n tension. the criucal spacing is 3h4 =
15 anches, whach IS greater than the-actual spacang and therefore group actton IS occurrang.
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O.sigl Example 5 • Til-Up Butilg
I'
anchor rod
" "_,
/
/
~r
/
--<(
I
Jr - - ~
T
'
--
/
I "''r
<
'
~
,__
"
~
1' -()"
)
;;
Figure 5-16. ProJtcledfailure ~11
ACI Equation 17A2.1 b:
N
riJ
= A,, ., (o
A.v..,
~..v
'f -.~,.,. Xo t:pv'f
""'.N)NII
-r
A,,,= [2(7.5 1n) + 6 .nj [2(7.5 .n) + 12 in]= 567 •n'
A,,.t u=9h1q =9(5f=-225 m2
F 5- 16
Eq 17.4.2.1c
Per ACI Section 17.4.2. 1. AM shall nm e.xceed nA,, ..,..
The modification tbr any posstble loading ecceJUricJty is per Equauon I7.4.2:4:
where~~ ts tl'!e ecce.ntricny of the resultant te.ns1le fO!ce from lhe-oenuoid of the anchor group acting 1n
tensJon. lJstng stattcs. e~. ts computed tbr thlS load combinatjorl (see Figure 5- 13):
t', = 6 .n _ 6 •n(50681b) _ 0. in
16
·'
2
13,576 lb
I
§17.4.2.5
§17.4.2.6
§17.4.2.7
Eq 17.4.2.2a
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Design Example 5 • Tlr-Up Building
The strenglh reduction factor I) is 0.70 per Se<:t:ion 17.3.3 because no supplementary reinforc1ng tS prov1ded
and thus
<I>N"-=0.70(42.1 ktps) = 29.5 kips
In conduioos where lhe concrete 1n lhe \'IC'inny of the aoch<>r could crack. ACI 3 1S Se<:tion 17.2.3.4.4
requires lhe concrete design strength to be taken as 0.75qN,... However. tn the case of this tJit-up wall panel,
the conc.rete is nor expected to crac-k as mentioned previously.
The total te.ns1on force on the embed per Figure 5-1 3 IS
N.=50681b<+85081bs=l3,576 1bs < 29.5kips .. . OK
ACI §1 7.4.3
Pullout stnngth in ttnsion
As was determined for ll1e previou.'i load combinanon. /\',. = 19.8 l::i.ps per anchor. The most highly stressed
anchors tn this group are the bonom two headed studs.
,tN,.. =2(19.8) = 39.6 ktps
The strength reduction fac.tor 4'1s 0.70 per S«:non 17.3.3 because no supple..nentary reinforc1ng ts
provided. In addlliOJ\, as was menuoned previously. the concrete remauts uncmc:ked~ thus,
.,V,...=0.70(39.6ktps) :27.7ktps>8.511aps . .. OK
Conrntt sidt-ratt blo""-out .strtag«b in ttnsioa
ACI §17.4.4
As stated prev1ously, it will be a.:;:sumed tn th1s example that the concrete anchorage ts not located near an
edge; thus, N., = 1.0.
Stttl !Urtngtb i.11 she;ar V_..
ACI §17.5. 1
As determlned in the pre\•ious load combmat.ion analysts, the four headed stud group has a design sue-ngth
of
~V.=0. 70(4XI2 . 7
k1ps) = 35.6 kips
From Ftgure 5- 13. the des:1gn shear force is 2.69 kips.
~~--.. =
35.6 ktps > 2.69 kips ... OK
Coucntt bn:a.kout stn.ngtb i.n shu r Va
ACI §1 7.5.2
As Pte\'IOusly mentioned. it is assumed in th1s example that the embed plate is not located near an edge of
the panel. In this suuaUOt\ 11d will not go\•em (ACI Sectioo R17.5.2.1 ).
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O.sigl Example 5 • Til-Up Butilg
ConcrttC' pryout stnnglb in shear V..,
For easHn-place rutchors. the nominal pryout strength tn shear. VtAr ts a funcuon of the OOJ'lcrete breakout
strength Nt.bf determtned earlier.
V~= kq.Ntllf
krp = 2.0 for anchor embedments where ht/~ 2.5 1n
ACI Eq 17.5.J . Ib
N.,.. = N... =42. 1 k1ps
v..,.= 2{42.1)= 84.2 kip<
ll\e strength reduction factor 9 is 0.70 per Secdon 17.3.3 because oo supple.memary re-tnforcing is
provided. In addttion. as was mentior~ed pre\•tously. the concrete remains uocracked~ thtls
~l-...=0.70(84.2k•ps)=58. 9k•ps>2.69kips
... OK
G <wtruing shtar s-trt ng tb
For this load c.ombmatio~ the steel shear sttength at 35.6 ktps f(l( the tbur aochots is less than the other
vattous concrete shear failure nlodes checked. ruld th1s design sue.ngth lS greater than lhe design shear load
of2.69 kips.
lnttrat tion oftus ilt and sbtar rortt:S
ACI § 17.6
An interactton equatiOn check (ACI Equation 17.6.3) is required unles..~ either the shear r:w or tenston N...a
does not exceed 20 percent ofthetr destgn strength.~ Qf.>~ or 9-N.. respecuvely. The following shear 3Jld
tenston checks are made for lhis load combtnation.
Check •fl
;.;w. ~ 0.20
2.69 k•ps/35.6 k1ps = 0.08 < 0.20 .. . Satisfied
Thll\ the anteraclion Equauon 17.6.3 need nOl be chec.ked, and lhe fu ll strength in tenston ts pennined as
has been already checked above.ln summary. the steel headed stud anchors under thts load combanauon are
acceptable.
Load Combiruuion Equalion 16-7 A nai)'Sis
Simalar to load combtnation Equation 16-5, this loadtng also conuuns contrabutions from eanhquake
loading£, and IBC Section 1905.1.8 agam is applicable by modifymg the requ.rements ofACI Section
17.2.3.4.2. As smted previously. IBC SecttOEt 1905.1 .8 c.onta1ns explicit e.xc.eptiol'IS to the ACt ductility and
overstrength requtreme.nts fOI' wall-anchorage design tfASCE 7 Equation 12. 11-1 or I2.14-10 is utilized.
Sttthtnngtb i.o ttn.sion Nu
AC I §17.4. 1
As was deteJ'm.ined for the prevtous load combination, Ar.,= 12.74 kips per anchor and
~"'~
=0.75(12.74 kips)= 9.56 kips
The most htghly Sltes..<;ed anchotS in thts group are the bemoan two headed studs. The destgn tension force
fo( an t.ndwtdual anchor tn lhis group ts N_ , = 962812 =48 14 lb.
$N~=9.56kips>4 . 81
kips .. . OK
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Design Example 5 • Tlr-Up Building
ACI §17.4.2
Contrttt brt-akoul s trtngtb in ltnsion N,.,.
The projected failure area im•olves all tOur anchors, \\<hich are in tension. stmilar to the prevtous Jood
combination checked.
ACJ Eqwuion 11.4.2.1b:
A,
A·t <.,N.,...'t tJ,)oxe>t,....'t
NU/6=
tp.,,,.)N.
Ne.
[2(7.5 in)+6tn]l2(7.5 in)+ 12 tnj = 567 •n'
A.,,=
tf:::. 9(Si~
2
A,,'e. = 9h
= 225 tn?
ACI Eq 17.4.2. 1<
Per ACI Section 17.4.2.1 . A,,'t shall not exceed n4,, .tll).
'
'
llA,,.e. :::. 4(9h ,.1):::. 900 tn > A,v,. •.• OK
The modification for any poss1ble loading ecce..ttrictty is per Equation 17.42.4:
v~..- -(
1
2e'
)<
I.Omax
I +~
3h"
where e~. is the eccenttictl)' of the resultant tensile force fi'om the centroid of the anch« group ac.ung tn
ten.o:;aon. Usmg smttcs, t,~ is computed for thts load combU'Iauon (see F1gure 5-14):
t'
•
= 6 Ill- 6 U1(39481b) -1.26 Ill
2
13,5761b
I
v~..-· - (I+ 2(1.26))- o.86
3(5.0)
"'"'·'= 1.0 (no adjacent edge efl'e<:IS)
§17.4.2.5
v,,,= 1.25 (uncraeked section dLe to <h<XI parapet)
Vfcp..\' ::
§17.4.2.6
1.0 {ca"'>Hn-place anchor)
N• = k,>../iJr'.J = 24(I.O)JJ000(5)u = 14.7kips
§17.4.2.7
Eq 17.4.2.23
N,._ = ~~~ (0.86XI.OX I.25)( 1.0)14.7 = 39.8 kips
The strength reduction factor 4' IS 0.70 per Secuon 17.3.3 because no supplementary reinforcang
provided. In addtlton, as ·was menuoned prevaous:ly, the concrete remains UJ'Ic:.racked; thus
t."'"" =0.70(39.8 ktps) =27.9 ktps
The total te.nsaon force on the embed per Figure 5- 14 IS
Nw
330
=39481b + 9628 lb =13,5761b < 27.9 ktps
... OK
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tS
O.sigl Example 5 • Til-Up Butilg
PuUout Slrtngrb in ttnsio n
ACI §17.4-3
As was determmed for the prevaous load combut.a[Jon~ N,.:: 19.8 ktps per anchor. The most highly stressed
anchors in tht~ group are lhe bottom two headed sruds.
nN,. = 2( 19.8) = 39.6 k1ps
The strengrh reducuon factor q ts 0.70 pe.r Secuon 17.3.3 because no supplememary retnforcang is
provided. In addttion. as was men6onecl pre"'aously,lhe concrete remains uncracked; thus
~N,..=0.70(39.6k1ps)=27. 7kips>9.63kips
. .. OK
Cont ntt sidt-ra.u hlowour s rrragth lu ttnsion
As stated previously, 11 will be assumed
edge~
ACI §17.4.4
an lhis example that the concrf'le anchorage IS not located neat M
lhus,Nu:: 1.0.
Stttl strtngth in sbttr
v_..
AC I §17.5.1
As determtned tn the previous load comhtn.ation analysi~ the four
or
~~ -.. =
headed stud group has a design strength
0.70(4)(12.7 kips)= 35.6 kops
From Ftgure 5- 14. the destgn shear fOI'ce is 1.34 ktps.
~~-..=
35.6 k1ps > 1.34 k1ps . . . OK
Contrttt brtakour strtngtb in shur Vtt
As prevaously menuoned.. at
the panel. In thts Situation.
1S
ACI §17.5.2
asswned 1n lhts example that the embed plate tS nOl located near an edge of
not govern (ACt Section R 17.5.2.1 ).
Vl 6 Will
Concntt pryout s tnngth in sbu r Vw
AC I §17.5.3
For easHn-place rutehors. the nominal pryout strength •n shear VlJII is a funchon of the concrete breakout
strength Nt.bf detemuned earlier.
v"" = k,/'1qy
AC I Eq 17.5.J. Ib
k..,; 2.0 fOr anchor embedment.<.; where h;f~ 2.5 1n
N.,..=N,..=39.8k1ps
v..,. = 2(39.8) = 79.6 kips
The strength reduction factor Q l<.; 0.70 per Secuon 17.3.3 because no supplementary retnfbrcing is
provided. In addttion. as \'laS mentioned previously. the coocrete remau\S uncracked; thus
$~:,.,=0.70(79.6klps)=55. 7k1pS>
1. 34klps ... OK
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331
Design Example 5 • Tlr-Up Building
GM1trning sbttr s-lrtnglh
For this load oombinatton.the sreel shear strength at 35.6 ktps for the four anchots is less than the other
\1ariOlL~ concrete shear failure tnodes checked. and lhts design strength is gre.ater lhan the destgn shear load
of 1.34 k1ps.
lnltratHon or ltns-ilt and s bur rortts
ACI §17.6
An tnteracuon equauon check (ACI EquatJon. 17.6.3) is required un.les:s e1ther the shear VMl or renston N..
does not exceed 20 percent oflhetr desagn strenglhs qJ··., or filA~. respecuvely. The following shear and
tenston checks are made for this Jood combinatJOO.
Check 1f V./9V. ~ 0.20
1.34 k•ps/35.6 k•ps = 0.04 < 0 .20 ... Sausfied
Thus, the mreracuon. Equatjon 17.6.3 need not be chec.ked. and lhe full strength in tenston is perrniued and
has already been checked above. In summary, the sreel headed stud anchors under this load cotnb1natton are
acceptable.
Cbttk nquirtmtnts lo pruludt s pliulng failurt
ACI §17.7
For the cast-to-place headed sruds, the followutg hmtts are c.)1e('.ked:
Minimum cen.ter-to-cemer spacing= 4 diamerers = 2 lnc-hes < 6 tnc:hes
Mmimum edges dismnce =Concrete cover per ACI Secuon 20.6 .1 ... OK
In summary,the four ~tnch-dtameter x 5-tnch headed sruds are acceptable.
9.3 CHECK SHELF ANGLE AT TYPICAL WALL·ROOF TIE
ASCE7
In this example-. the Sleel JOist purhn s.its on a steel shelf angle (LS x 5 x ~ ln x I ft). Without addttion.al
tnfOrmauon, u ts assumed lhe load act" at the tip oflhe leg. The honzontalleg 1s subJect to bend1ng and
se-tSrntc tension suesses. Evaluanng the array of load cotnbananons from IBC Section 1605.2 and ASCE 7
Section 12.4.2 tbr strength design forces, combu>auons (3) and (5) pmem1ally SO''em.
ShnpUOtd load combin.a tion IBC Equ.a lion 16-J
1.20 + 1.6L,
Joist reaction= 1.2( 1920 Ib)+ 1.6(3200 lb) = 7424 lb
Mome.n.t arm to criltcal section= leg - k dtmension = 5 - 1.25 = 3.75 tn
M,= 74241b(3.75 in)= 27,840 u>-lb
Plastte secuon modulus:
z = 12 111(0.75)'
- 1.69 in'
4
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O.sigl Example 5 • Til-Up Butilg
Per AISC Section Fl J.l.the nomtnal flexural suength, M,.. may be chec.ked as follows:
AISC Eq Fil-l
M. =M,=Fj!S1.6.\fy
M, = 36,000 k.<t( 1.69 •n' ) = 60,840 1n-lb
2
•12 m(0.75)
1.6M, = 1.6F, S = 1.6(36,000)l
6
)
= 64,800 1n-lb
Thus, M. = 60,840 in-lb.
The design fte.xural sttength is c-hecked as follows:
~,M.
= 0.90(60,840) = 54,756 1n-lb <: 27,840 1n-lb ... OK
Simplifitd load comblnatJon IBC Equation 16-S
1.4D+Q,
A combmauoo of gravny forc.es wnh hottzontal Lie forces wtll be evalualed.
Joist gravity reaction= 1.4( 1920 lb) = 2688 lb (dead load).
Moment arm to cm.ical section= leg - k dtmension = 5 - 1.25 = 3.75 u\.
M, =26881b(3.75 in)= 10,080 in-lb
1
Z= 12ul(0.75) - 1.
,
6910
4
= =
M. M, F,:Z s 1.6,\fy
M, = 36,000 ks1( 1.69 m') = 60,840 in-lb
AISC Eq Fil- l
•
.12 in(0.75)2 )
= 64,800 10-lb
1.6M, = 1.6F,.S = 1.6(36,000)l
6
Thus, M. = 60,840 •n- Ib.
The des1gn ftexural strengtb 1s cheeked as follows:
~~~~.
=0.90(60,840)= 54,756•n-lb<: 10,080 1n-lb ... OK
JOist honzontalue force= 13,5761b (from Pan 9.1 ).
Per ASCE 7 Secuon 12.11.2.22. steeJ ele.menlS of the sLnJctural wall anchorage system (SOC C and abtwe)
are designed for strength forces v.'ith an additional 1.4 multiplier. This material-specific muh..iplter is based
on the obsenred poor performance of steel straps during the 1994 Nonhrtdge earthquake. It was derermtned
that an inadequate O"\<erstrength range extsted in vartous steel elements to acc.ommodare the maximum
expected rooftop accelerattons. Thas 1.4 force muJtjplier ts appl1ed to all steel element'i resisung the
wall-anchorage forces ofSec.uon 12.11.2.1 (SOC C and above). tncludmg wa11connecto~ subdiaphtagm
sttapptng. continuous cross-ties, and their connections. Reanforeing steel. &'IChor rods and steel headed
sruds, wood boiHng. and nailing are not subject to tJus force multtplier.
Required tie force P,= 1.4(13,5761b) = 19,0061b
ASCE 7 §12.11.2.2.2
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Design Example 5 • Tlr-Up Building
The tie force is applied across a Wh1tmore secuon wad th at the end of the joiSt seaL Assumtng a 5-mch jo1st
bearing seat width and a 2-•nc.h weld length (SJI mtnimurn).lhe Whiunore sectton width is as follows:
W= 5 an+ 2(2an tan 30°)= 7.3tn
Tensile area A,= 7.3 ln x0.75 in= 5.48 ln2
Oetermll'le the destgn tensale sttength for chedang oomblned forces per AISC Sectton H 1.2:
P, = oil/'. = oi>,Fy'f, = 0.90(36,000)5.48 = 177,552 lb
P, =
P,
AISCEqD2-I
19 006
•
=0. 11 < 0.2
177,552
The.rtfore, AISC Equation H 1-1 b is applicable fbr checking the combined forces of tension ru'ld be.ndtng
flexure.
Therefote1 the she.!f ang.le suppon as adequate.
9.4 CHECK THE SHELF-ANGLE WELD TO THE EMBED PLATE
AISC 360
Check the use of a ~- tnch fille t \\'eld all around the shelfangle"s peruneter. Per AISC Table J2.4, the ~ - inch
fillet weld meeLo~t the minimum we.ld stze hmttattons tbr the thinner plate jomed (V*-tnch embed plate). and
per AISC Secuon J2.2b, tJlf' ~-tnch fillet weld meelS lhe maxtmum weld s.tze hmnauons fOr the ~-tnch
edge lhickness of the shelf angle.
Si.m.tlar to the process in Part 9.2. the fOfce disrrtbution to the shelf angle's upper and I0\1/er we.lds ts shown
an Figure 5-17 tor the various poten[ially governing load combinations.
5
,..~·!;,,.,.
5
74241b.
~ lb
..
74241b.
-1~
10, 1~~ i
j
load tO<TI>o (3)
..
lo~d combo {5)
Rgure 5-17. Fac1orrd foods 011 siH!Iftmgl~s
334
t344i
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"'
lead oombo (7)
'; 5161b
O.sigl Example 5 • Til-Up Butilg
Bocause load combiJtataons of IBC Equations 16-5 and 16-7 mvolve seismically tnduced wall-anchorage
force to the weld. they are subject to the 1.4 force muh1plier ofASCE 7 Section 12.1 I.2.2.2 (SOC C and
above). The followmg IistS the etl"'ecuve resullS oflhe worst vertical and horizontal force vectors acting on
the fille1 weld:
2
P,=h424 +3712' =83001b
IBC Eq 16-3 Combinmion
~=
J(l .4 x8852)' + 1344' = 12,4651b
IBC Eq 16-5 Combinalion
f',. = J(l .4 X 10,196)2 +672' = 14,290 lb
IBC Eq 16-7 Combonalion
In this e.xample~ the load combination of Equatton 16-7 governs at 14,290 pounds. Where smaller wa.JI.
at\('hofage loads occur, ofte.n the gravity load combination ofEquatiOil 16-3 will govern.
Check1ng 1he strength of the Y.- io\Ch x 12- onch-long fillel 1wld (AISC Section 12.2 and Table 12.5) gives
fiR, =.F. A.
= 0 .75(0.6 X 70 k.<i>( O.htn X 12 tn) = 66.8 kops > 14.3 kips ... OK
11teref0re, the shelf-angle weld to the embed plate is adequate.
9.5 CHECK JOIST-SEAT WELD AT TYPICAL WALL-ROOF TIE
The connection oflhe joist to the embed's shelf angle is lhrough a fille.t weld. Gwen its orientation, the steel
shelf angle fLS x 5 x ~ m x I ft) has a flat run-out dtstance of 3~ tnches swtable fOr jotst seat be.anng..
Per the Steel Joisl Institute ·s Standard SfNcificationl, SJJ 100-2015. the nunimum weld at the joist seat
anac.hlnems varies from Y.~~o-tnch x 2~- ioc.h- long fillet on each side of the seat up to Y~tnch x 4-tneh-long
fillet (SJI, 2015 Sectton 5.7. LH-series joists), depenchng on how heavy the steel jotst configuration is.
Ofte.n the mtnimum sae weld w~U be suffic1etlL In thts e.xample. the mtntmum Y..-tnch x 21h-inch-long fille-t
on each stde wtll be checked and upsized tf necessruy.
Chedang lhe stte.ngth of the tw·o rows of Y.e.-tnc.h x
follows:
2~ tnch-long
fillet welds per AISC Secuon J2.4 is as
Requited tie force P, = 1.4{ 13,576 Ib)= 19,0061b < 20,900 lb . .. Montnum weld os adequate.
TherefOre, thts new JOISt seat weld to lhe shelf angle support is adequate. Because larger JOtsts may requite
a larger Y.'eld to comply with the SJJ Standard SpecifteatiOnl, thJs weld should be Hsted as a mimmum, w1th
a note that lhe jo1·st designer may requtre a larger weld fbr Installation.
9.6 DESIGN STEEL JOIST FOR TYPICAL WALL-ROOF ANCHORAGE FORCES
ASCE7
Whelher comb1ned \\1th a panellzed WSP-shealhed roof or a metal deck roof. steel ~or jol'iiS are lhe
most common roof framing system now 1n uh-up buildings. Spec:iahy engineers an assoclauon with joist
manufucturers typtc.aHy design the stei'l joist members.. As requared by IBC Secti-On 2207.2, lhe bmldang·s
destgn engmeer is respons1ble for providing axlal wall. tie and c:onunUJty-tle torces lOthe manufacturer
along with informatton staung wbkh load facto(s. if any. have already been appJied.
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Design Example 5 • Tlr-Up Building
In this example. the destgner should repon to the JOtSl manufacture that the un(actored \vall ue a.x&:al
foi'Ce (tension and compres.sjon) acung through the joist top chord is F,. = I 3,5761b increased by the sreel
mru:enaJ oversttength factor 1.4 per ASCE 7 Sectjon 12.11.2.2.2, resulung tn FP = 13,576 x 1.4 = 19,006
lb. It is necessary to indicate to the josst ananutilcturer that this Lie fOrce is fi'om setsmic efli!-crs so that the
JOtst·s specialty engineer as able to apply the prope,r JBC load combtna.taons with Section 12.4.2.
Though nOl shO\Ir'fl. m this e.xample, the top chord axial effects of wand IV must also be considered Lf it could
lead 10 a g<:weming destgn. of the joist. Because the lood combinations ofiBC Soct.iott 1605.2 (strength
destgn) and Section 1605.3 (allowable stress desJgn) oontatn yery dtft"'ereru formulas w~n eonstdering
se!Sintc E and wind W, the destgn engineer cannot s1mply compare E and W to determu'le wtuch govems.
In condnjons where axial loads are transfe-rred through the jotst seat at e.nher the wall ue or at interior
spliees at jotst girders, it must be made dear to the manufacturer so that the seat strength wtll be checked
also. 11tere are limils to the amount of load that manufacturers can transfer through these JOist seats, so the
designer must check with the manufacture-r·s specialty engmeer.
Occastonally, the tad end of a collector hne terminates at the wall and provides out-of-pla.'le wall
anchorage, similar to line 3 at Iine E {see Fagute 5-1). ln these situations, the building's design e,nganeer
mUSt specifY a highe-.r collector Joad as well as an E waJI.ue.lood tOr lhe joast. The joist manutOCturefs
spedahy enganeer \VIII have to check both the baste IBC load ootnbmanons wnh SecLion 12.4.2 tOtE as
well as the basic load combtnations with the overstrength factor of Section 12.4.3 for £.,. AJtematively,
some dtaphragms and collectors are des1gned to Sec.tion 12.10.3 whe-re a I.5 multtplier as used U\Stead of
the overstrength fuctor E., from Secuon 12.102.
For this example, the following is the type of infoonatiott to be placed on the drowings for the steel joist
manufacturer to properly design the joists fot Lateral loading.. The wall anchorage. force E shown should
already 1nclude the 1.4 multtplier fOr steel elemet\ts.
Jmst Axjal Forces
E = I9.0 ktps (wlfactored)
E. =0.0 kips (unfactored per ASCE 7 Secnoo 12.10.2)Applicable only at collectors.
IV= 7.0 kips (unfactore<l)
Forces shall be checked 1n both te.nsioo and oompressJon.
Axial fOrce shaJI be transferred through the joi.(!t seats where ooted in the detatls.
9.7 CHECK JOIST·TO..JOlST SPUCE AT THE GIRDER UNES
ASCE7
The interronnectton of elementS wathm the butld1ng IS required per ASCE 7 Sections 12.1.3 and 12.1.4.
In addtuon, the JOist a.xlal load t!om the waJI anchorage must be dJSlributed across the building's matn
dtaphragm from cltord to chord per Secuon 12.11.2.2.1 u~tng conunoou.'(! cross-ues {SOC C and above).
Seismac Jooding in the nonh-south di.J'e(;uon utdtres the steel jotsts as the c:onunuous Lies. and thus the joist
ax1aJ load must be spliced across the Interior g.1rder lines. See F1gure 5- 18 for a typical OOttnectJott. In Pan
9.6, the '"all anchorage force and thus continuous cross-tie force for the steel jotsts tS P, = 1.4{ 13,576 Ib)=
19,006 lb.
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O.sigl Example 5 • Til-Up Butilg
Per Secuon 12.1.3. the minimum interconnection force lS0.133SusJV=O. 1 33W~ bot not less than O.OSJV,
where W is the dead load of the smaller portion of the building being connected together. Unltke the
\\'aU-anchorage fOI'ce, W m thJS case mcludes the d1aphragm wetght and thtJS could govern at the mtenor of
buildtngs.. In this case, the diaphragm v.-eight has i.nc.reased from 12 psf to 14 psf to account for the glrder
self-weight. The WOI'Sl- ca<ie value for W is at gnd line C with the fbllowtng result
P, (m1n) = 0_ 133( 14 psf)(8 ft)(40 ft + 40 I\)+ 0 .133( 116 psf)(S ft)(32)(3212)128 = 34491b
Per Secuon 12.1.-1, the mimmum support connection tbrce ts5 percent oftl'lf' dead and hve load reactions.
P, (m1n)= 0.05( 14 psf + 20 psf)(8 f\)(40 fi/2)=2721b
Thus. the wall-andlotage conunoous ue force P, = 1.4{ 13,.576 Ib)= 19,006lb gove.rns.
llte sphce c,an be attomphshed ''~th a welded c.over plate from jotst top chord to jolst top chord (see Ftgure
5- 18). Chec.k the use of a~ x 3-lnch-wtde cover plate with Yio-1nch fillel welds.
Check the design tensile sueng.th per AISC Section 02:
~/'. = ~,F,A
1 = 0 .90(36,000)(0.25)(3) = 24,300 lb
AISC EqD2- I
ReqUired lie force P,= 19,006 1b <24,300 lb ... OK
3:tnJJiet mile<:! fQf rl.l$h f.C
O'l<lr splice pliJte. Too "<lll
to rtUSs 11af~&r eaeh et~d
1 lt.:."x3" .x0'·8"~k»~
FtguJ? j - /8. Joutto girder detail
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Design Example 5 • Tlr-Up Building
Uslng two lanes of%. x 2~utCh-long fille-t \\'tlds. check the des1gn weld strength per AI SC Section n .4:
~R. =+F.A, = 0.75(0.6x 70 kso>( 0· 1
j:5'" )2
x2.5
= 20.9 kops
Requored ue force P, = 19.0 ktps < 20.9 ktps . . . OK
The-refore, the steel JOISt splice across the interior gtrde.rs is adequate.
Commentary
h L~ possable to sphce the JOist ax1al loads across the 1nteraor girders through thetr joist seaL-e; a<t is done
atlhe wall-anchorage JOISt end However. thJs requ1res added jotst seat c-OStS and requtres the JOISt gtrder
double-angle top chords to be jou'led together for thiS pe.rpendicular force. If this is the destgn enganeer"s
totem. tt must be made clear to the joist Jnanu10Cturer that the joist seat~ and joist-girder top chords are to
be destgned tor these forces. iocluding the 1.4 ovetstrength factor.
10. Wall Anchorage at Subpurllns (East-West Loading)
ASCE7
On the east and west \\-all elevations. wall-roof lies are used to trnns1tr out-of-plane seism1c forces on
the ult-up wall panels to lhe subd1aphtagms. Applicable requtre.ments tor connection of out-of-plane wall
anchorages to ftextble dl3jlltrogms are specofied tn ASCE 7 Secuon 12.1 1.2. 1.
10.1 SEISMIC FORCE ON WALL-ROOF TIE
Sei.sm1c forces aredetermtned usang Equ;u,on 12.11 -1 wtth SJJ$= 1.0 and 1, = 1.0. Because the wan
thickness and height are the same. these are the same fbrc:es as those determined 1n Pan 9 .I for the north
and south waHs..
F,= 1697 plfalongt he roofline
10.2 DESIGN TYPICAL WALL ·ROOF TIE
Try ties at 4-foor spactng~ and detemune F/1:
F,= 4 t\ X 1697 p lf= 6788 1b
Commentary
Whe.n ue spactng.exceeds 4 feet. ASCE 7 Section 12.11 .2. 1 requJJesthm walls be desJgned to res1st
bendtng between 31\('.hors..
Try prefubncated mood hold-downs w1th two JA-tnch bolts toto a 3x subpurhn and two *-tnch and\Or
rods connecting. t11e hold-<loo•ns to the wall panel. Thjs cooneclion. IllUstrated in F1gure 5- 19, IS des1gned
to take both ten.~ion and compression as recommended by the SEAOSC/COLA Nonhndge Ea.nhquake
Tilt-up Buildtng Task Force (SEAOSC/COLA 1994) and the 2008 Blue Book "Tilt-Up Ilutldings'' anicle
(SEAOC Seismology Committee 2008). Design oflhe hold--dO\\'n hard,~;are is not s.hoYm. Consult ICC-E.S
Evaluaoon Repon.s for the aJiowab1e load capacity of premanuf'hctured hold-dov..ns. Note that tf a onesided hohk lown is used• .stresses 1n the subpurlln due to eccentr1c load1ng should be considered per Section
12.11 .2.2.6. Generally, one-sided wall-roof anchorage IS not recommended 1n SOC C and above.
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O.sigl Example 5 • Til-Up Butilg
Precas.t Wall Panel
Wooo Structural
Panel Sheathing
513''
Aochor Bolt
(2 to!al)
Hold·dovm each
side of subpur1in
w/2-3/d" M.B.
3x Subp\.lrlin
Figure 5-I 9. Wa/1-tmchorage deuul
Check the capacity of the two lA-inch bolts tn lhe Douglas Ftr-Larch l x subpurlin ustng2018 A\VC NOS
Table 12G, where C0 = 1.6 and C1 = 0.97 and assuming ~- inch steel s1de members.
(2630)(2 bohs)(l.6)(0.97) = 8164 1b > 6788(0.7)= 4752lb ... OK
Mmunwn requ1red end dtStaJ'ICe = 1D= 7(0.75)= 5.25 tn
NOS T 12.5.1A
A distance of6 tnches fi'om dle through-- bolt tn the hold-down 10 the ledger will be u<;ed. Often, there IS a
gap of~ inch or more between the end of the subpurlin and the side of the ledger caused by panelized roof
erectiOn methods. and lhe use of a 6-inch edge distance will ensw-ecompliance with the 7D requiremenL A
larger dtsta.nce can be used to ensure that bah row· tear-ow per NOS Appendix E does not occur in the 3x
subpurl an, prov1ded that the I imtts for compression buckltng are checked.
Check the te~JOI'l: capac.Jl)' of two %-1nch ASTM F 1554 (Grade 36) allchor rods us1ng LRFO:
F,= 0.75F,= 0.75(58) =43.5 ksi
$,11, =$/',A,= 0.75(43.5)(2)(0.307) = 20.0 kips . .. OK
AISC 360 T. J3.2
AISC360 Eq Jl- 1
R. =F,=67881b < 20.0klps ... OK
Note: The 1.4 factor nomlally apphed to steel elemems of the wall anchorage system JS not apphed to
an<hor rods per ASCE 7 Section 12.11.2.2.2.
Check !he compression capacity oftwo%-mch ASTM F l554 Grade 36 anchor rods using LRFD:
P,=F,A,
AISC 360 Eq El- l
A1 =Ab = 0.307 in~
Rad1us of gyrauon of%-in rod= (0.625 Jn/4) = 0. 1563
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Design Example 5 • Tlr-Up Building
Assume L = 4~ inches and K
=- 1.0.
L, = KL = 1.0(4.5) = _
28 8
r
r
0.1563
29 000
4.71 II= 4.71,/ •
= 133.7> L,
VF,:
' 36
r
Thus, AJSC Equation EJ-2 ts apphcable, so
F~ =[0.658t]r;_
where
x'E
· l:r
F = - - , =- 345 kst
F~ J0.65S~]F, J0.65S~L6,000
AISC 360 Eq EJ-4
= 34,462 pst
P,= 34,462(0.307X2 rods)= 21,160 lb
+,1',=0.90(21,160)= 19,044Jb<:678S Jb ... OK
Check lhe tension capacity or anchor rods 1n the wall panel considering cotlC:rete anchorage.
The tjlt-up panels ate exteJ'tor waU elements, but the requarements of ASCE 7 Sections 13.3 and 13.5.3
do nm apply. This ts because the oh-up panels are strucu.nl \valls tnstead of nonstructural arcl\l[ecruraJ
cladding. The requtrementsofSecuon 12.l I are Lhe appfOpriate design rules in thissituatton. Section
12. I 1.2.2.5 requires that \\>all anchorage using straps be anached or hooked so as to uansfer the fOrces
to the retnforc1ng steel. In this case, ca'\1-tn-place boiL'l instead of straps are used, and the bolts are not
requtred to be ..hooke<f' around lhe waJI reanforcement.
Recall lhm for wall anchomge, F, =6788 lb. Try a * -•nth-diameter ASTM Fl554 Gro<le 36 hex-headed
bolt embedded 1n the ooncrete panel with 5 tnches of embedmem (htf=- 5 inches). Because the anchorage
spacmg is only 48 tnches, 11 ts qutte posstble that lhe bolt embedment may be near an edge, unless the
panel joints are specifically located to mtss anchorage locauons. Normally, spec tal considerations are
necessary for eondttions where anc.horage occurs near a panel joint. For Lh1s example, assume that the bolt
embedment is nm near a concrete edge and that the venicaJ shear load is neglig.tble.
The wall"s concrete anchorage needs to be checked usjngstrengthdestgn under ACI 318-1 4 Chap[er 17.
The verucaJ shear load on the anc.hor 1S ve-ry low because of the smaJI subpurhn tnbutary roof lood. ACJ
Section 17.6.1 aJIO\vs the full tenstOn sttenglh to be used witltout reduction when the tactored shear load is
less than 20 percent of the design shear capacity of the anchorage. as m lh.is case.
ACI Table 17.3. 1.1 requtres tlte anchorage tensale destgn strength $N,. > N~ for the vattous fajlure modes.
ts detemuned by checkang the stee-l strength in tenston NMJ (ACI Secuon 17.4.1 ). the OOI'lcrete breakout
strength tn tensiott N~ (ACI Sectton 17.42). the pullout strength tn ten.o;ion N,.. (ACI Section 17.4.3), and
the concrete side-face blowout suength in tension N1• (ACI Secuon 17.4.4 ). For anchorage forc.es lO\'Oiving
setSmic loads for structures in SOC C and above. ACI Secuon 17.2.3.4.4 requjres the tension design
strengths QN, be multtphed by 0.75 unless it can be demonstrated that lhe concrete rematns uncracked. 1lle
~v,
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O.sigl Example 5 • Til-Up Butilg
duculity and overstrength requirements of ACI Section 17.2.3.4.2 are not modified based on IBC Sec-tion
1905.1.8's exception for when ASCE 7 Equation 11.1 1-1 ts used, as is the case here.
Stttl Strtngrh in Ttnsion N~.
ACI §1'-4.1
ll\e nominal steel strength tOr %-tnc-h-diame.ter ASTM F I554 Gtade 36 headed anchor rods ts as follows.:
Eq 17.4.1.2
AISC Sleet Couslructitm Manual T 7- 17
Nsa=A.JfAff,.v = 0.226 in~ (net tensile area)
f-=58 ksi
AISC Stu/ Ctmstruction Mamw/T 2-6
The specified tensile strength ofllte steel anchor/., shall be not greater than 1.9F,.. = 1.9(36) = 68 kso, per
ACI Section 17.4. 1.2. In thts case,[... = 58 ksi governs.
Thus, N~ = (0.226)58= 13.1 kips per anchor.
Threaded boiL~ oonfomung [OASTM F15S4 Grade 36 qualtf)' as a ductile steel eletnem aJ'l.d thus'= 0.75
for steel per ACI Sectton 17.3.3. Addttionally, conside-ring that two bolts ate being provided. the desagn
sue.ngt.h ts
ojoN.=2(0.75)(13.1 kips)= 19.7kops>6.79kops .. . OK
Contrttt Brt-akoul Slrtaglb in Tu.sion N,.,
ACI §17.4.2
For the difttret\t possible ti:ulure mode.~ being evaluated, AO Section 17.2.1.1 IS COI'I$ulted [0 detennine
whethe-r group ac[ton tS occurnng. For conc.rete bteakout in [ens ton, the crttteal spactng is 31r4 ; 15 inches.
which is greater than the acrual spactng oflhe two embedded anchors (one on eac.h side of the subpurhn)~
therefore, they are spaced close enough to be coosidered acung as a group~ as shown in Figure 5-20.
" "_,
anchor head
">
6i
/
"
/
/
/
/
~
...~"'
/
/
'"r
' ' __,,_
it>
...:
'~
7.5"
r'
7.5"
~-
Figw·~ 5-20. Projtct~d tJ!Ya dtugram
N
rill
=
A,\"t c~
A,\'..,
N:,.\ '
~ .-.~,.v Xq ~..\' ~
ti4\' ) N•
A,,., = 2(7.5) X (7.5 + 7.0 + 7.5) = 330 1n2
ANt.o) =9h2t/=9(5f::225tn2
Eq 17.4.2.1 b
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Design Example 5 • Tlr-Up Building
Per ACI SecLion l7.4.2. I• .4,\·c- shall not exceed nA,\'tJO
'""·•·~=2(9h .}=4501n >A,, ... OK
2
2
v.,._,,= I.O(ooece<ntricloodmg)
'V..o(.v = 1.0 (no adjacent e.dge effeclS)
'Vc~v = 1.25 (uncracked secuon due to shoo parapet)
'Vc-"'.\' = 1.0 (cast-t~place anchor)
N• = k,1..fj:h'.j = 24(J.O)J3000(5)1~ = 14.7 kips
§17.4.2.4
§17.4.2.5
§17.4.2.6
§17.4.2.7
Eq 17.4.2.23
N,_., = ~~ ( I.OXJ.OXJ.25XJ.O)I4.7= 27.0 kips
The strength-reduction factor 41 is 0.70 per ACt Section I7.3.3 because no suppleme11tary reinfO!clng IS
provided, and thus
<>N,_.,=0.70(27.0 k1ps) = 18.9 k1ps> 6.79 k1ps ... OK
Pulloul Stnnglb in Ttnsion
ACI §17.4.3
NI" = Vf.,/",
I.4 (assume uncracked section due 10 shon parapet he•g.ht)
Eq 17.4.3.1
N,= &A,..f; (where headed studs or bolts are used)
Eq 17.4.3.4
Vft.J> =
§17.4.3.6
A.., =hear1ngarea of boll head= (hex head area) - (shank area)
A..,= ~E- (shank area)= 0.761-0.307 = 0.454 in'
N,_ = 1.4(8X0.454X3000 psiX2 boles)= 30.5 k1ps
The strength reduction factor 4l1s 0.70 per ACI Secuon 17.3.3 because no supplementary retnf«clng JS
provided., and thus
t-'-',..=0.70(30.5kips)=21.41ops>6.79klps ... OK
Conrntt Sidt-Fact Blo"'~oul Strtugtb in Ttnsion
ACI §17.4.4
Because itlS assumed that this concrete anchor is not located near an edge, N4 will not afltc.t the design.
Cbtck Rutuinmtnts to r ·ndudt Splitting Failurt
ACI §17.7
For lhe cast-in-place headed studs, the followtng ltmits are c.hed:ed:
Minunum centeHo-center spacing= 4 diamere.r s = 2.5 inches< 6 inches
Min unum edges dJstance = Conc.rete C<Wer per ACI Secuon 20.6.1 ... OK
In summal)•. all applicable eoncrete anchOt'age fadure modes have been checked and found satisfactory;
thus. the two %-inch-diameter x 5-i.nc.h embedded he.x bohs are acceptable.
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O.sigl Example 5 • Til-Up Butilg
Comprt:Ssion Distussion
Wall anchorage forces act t.n oompress1on as well as tenston. Panellzed wood roof systems by thetr vel)•
nature are not erected tight against the perimeter wallledger.leavanga small gap to pmenually close during
seismlC compression forces. Strap-type: \vaJI anc.hofs that may have yielded and suetched under tenstle
fOfces are vulnerable to buckhng and low-cycle faugue as lhe gaps close. Cast-in-place anchor rods used
in connectors Call be checked fOr compression, but it is imponant to provide an addttlooal nut agatnstlhe
imen« wall surface to prevent the anchor puoch1.ng through the wall. A cornmon wall-roofue connectton
shown in Figure 5-21 does nm offer the same c.ompres.<iion reststance a'i the anchor rod scheme presented in
thtS e.xample. Although there ha\<e been no fiulures of waU panels collapstng tnto the bu.ild1ng. consideration
of compressive forces will maintain the integrity oflhe wall anchorage ue and prmect the diaphragm edge
nailing unde-r the reverstble setsmtc forces.
Premanufactured 12-gage strap
Diaphragm sheallling
116 bar
3x Subp<lrlin
Ledger
Ftg1n 5-2 I. Uh/1-to-roofstrop tmdu>r
.~ncb()ngt
Ddormation Discussion
No prescnp1ive defotmallOn l1mns of the wall ue system have been anuoduced amo the IBC or ASCE 7~
howe"\·er. the eornpaubtlity of the 3.1\('homge system ·s flexabtl1ty and the diaphragm shear nadmg along
the ledger should be oonsjdered. Wall anchorage systems with too much fte.xabdity wall inadvene..ttly load
the \VSP edge natllng and either pull the-nails through the sheathing_ edge or place tlte wood ledgers tn
cross-yatn bend1ng « te-ns1on. Premanufactured strap-type walJ ties are destgned 10 l1m1t the maxamum
deformation toY. inch m thear rated allowable load, and premanufacrured hold-down devaces ustng anchor
rods also have tnhetMt deformation. The designer should contact the device manufacturer for addmonaJ
deforrn.ation utformation. Thls reponed hold-d0\"\1'1 device fte.xtbiluy is solely v.'ithin the steel compone.nt
itself ruld IS additive 10 othe-r sources of defoonation. AdditaonaJ deformation can be coaur1buted by other
anchorage components (e.g.• bolts and nails) and ansaallauon practices (e.g.• O\•ersaed holes).
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Design Example 5 • Tlr-Up Building
10.3 DESIGN CONNECTION TO TRANSFER SEISMIC FORCE ACROSS FIRST ROOF TRUSS
PURLIN
Under ASCE 7 Section 12. I 1.2.2.1 for SOC C and higher, conlinutty Lies are provtded in diaphragms and
subdaaphtagms to dtSU'tbute wall anchorage loads. Consequently, the forces used to design the walt.rooflJes
must also be used to design the conunuity ties w1thtn the s:ubdiaphragm. From Pan 10.2:
F,= wall-roof lie load =6788 1b
For thjs example, a subdiaphragm depth of 32 tbe1 wtll be assumed to start wnh; and with steel jotst purlans
spaced at8 feet. the connection at the fii'Sl pttrhn from the wall must carry dtree-quane:rs of the wall·
roof ue force. Some engineers use the full. unreduced force for all the cross--purhn connections, bot this
simpltficauon is not required by rational analysis.
(32 -8)
- - - xF, =%x6788•5091 1b
32
At the second and third purhns frorn the watt, the force to be transferred JS one-half and one-founh~
respecti\·ely, of the wall-roof tie fOrce:
16x6788=33941b
V.x6788= 16971b
Try 12-gage metal strap w11h IOd common natls. Consult ICC-ES Evaluation Reports for aliO\\-able toad
capacity of premanufactured straps aoo ues.
The tbliO\\~ng calculauon shows the determmauon of the number of IOd ootnmon nads lntO Douglas Firlarch required at the first connection using alloY..'able stress design:
0 7 5091
· (
) - 17.5 < 18 nads
1.6(127 lb)
NOS T 12P, T 11.3.1, and T 2.3.2
Therefore, use a 12-gage metal strap wuh 18-IOd common nails (or equivalent)on eac-h s1de.
The design of the 12-gage metal strap is not presented here, but thedestgn lS ba~ ott forces tncreased
b)' 1.4 ttmes the forces otherwise requ1red under Secuoo 12. I I. Thts requirement ofSecuon 12.11.2.2.2
ts a result of the unexpected suap failures observed tn the 1994 Nonhridge earthquake.. It was fOund that
many steel componems lac.ked suffictem ducllhry and oversuength to adequately accommodate sets:mic
overloads. It is the tntent of the 1.4 steel-material multiplier to pc'Ovide suffictent oversueng.th to resist
maximum anticipated wall anchorage forces wtthout relying on duculity. Doth the gross and net secuons
shall be c.hecked.
Where premanutactured and pre-engineered straps and ties are utilized using capactty values published
tn ICC-ES Evaluatton ReportS. the eng_~.neer should compare the published capacity with the 1.4 steeltncreased force unless sufficie.nt tnformauon is available to deternune steel material \•alues tndepe.ndently of
other compone 1\tS..
Both subpurli.ns in Figure 5-22 hkely \\'Ould be Jx me-mbers because of the heavy strap nathng..
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O.sigl Example 5 • Til-Up Butilg
12-gage6l1'8p wilh 18-10d l'lOmmon
- nail& ea.eh s\de of root pUtlitl
Subi)Uflin
Open WGb $l~el joist
p!Mlin wi!h WQCXf n;aler
Figure j -21. Scrapdeuu/
Destgns of the se-cond and thtrd connecltons from the wall are stmilar [0 lhat shown above-.
Nme-: Addtuonal requirements fOr eccentric wall anchorage and walls wtlh pilasters ate provided tn
Sections 12.11.2.2.6 and 12.11.2.2.7.
11. Subdlaphragm Design (East-West Loading)
Subdiaphragms transfer forces from the Lndtvtdu.al \\'all anchorage ttes to lhe ma1t1 dtaphragm ·s conunuous
cross-ues. To transfer setSrruc forces from the heavy peri.t'l'k!ter walls toto lhe main roof dtaphragm.
continuous c:ross-ues are necessary to dmg Lhe load unifonnly across the diaphragm depth. Instead of
creaung a conunoous cross-tte at every wall anchorage locmioo. main conunuous c.ross-ties can be placed
at wider spacings usmg subdiaphragms. Subdiaphragms are poniol'IS of lhe main diaphragm lhat span
between the oonunuous cro~ties and garher the Ylall-anchorage loads and ttai'ISfer these loads to the crossties. Once the load is collected tnto the continuous cros:s-ue~ at is distributed across the maln diaphragm for
funher distnbut.ion to the butld1ng•s shear \'t>aJis and frames.
Subdiaphragrns are provided fOr wtder ASCE 7 Secuon I2. I I.22.1 as an anal)1ical tool to prov1de a
rauonaJ load path for wall anchorage. ConsequentJy. sulxhaphragms are considered patt of the \\'all3.1\('homge system and are S:ubJ~l to loads per Section 12.11. Us1ng this tecluuque, the subdiaphraga.n shears
are not combined wtlh the global diaphragm shears. For SOC C and above. the subdjaphragm aspect ranos
are limited to 2~ to I, and n is asswned that this provtdes sufficie.nt suH'iless such that the Independent
deflection be-tween the subchaphrag.m and the mam dtaphragm may be 1goored.
11.1 CHECK SUB OIAPHRAGM ASPECT RATIO
The maximum aJiowable subdiaphragm rntto IS 2.5 to I.
ASCE 7 § 12.11.2.2.1
From Ftgures 5-1 and 5-23. the ma.x1mum north-south subchaphragm span= 40 feeL
The minimum subdiaphragm de.pth
=40 r..,t/2.5 =16 feet.
With the cypic.aJ roof puthn spaclng gtven as 8 feet. 1t IS desttable for the engmeet to keep the subd1aph.ragm
depth dlmensioo at incre.menlS of this 8-foot spacing. In this case, the 16-foot mimmum subdiaphragm
depth rematns., which IS less than the initial 32-foot depth asswned. This is acceprable.
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Design Example 5 • Tlr-Up Building
11.2 FORCES ON SUBDIAPHRAGM
Because subdiaphragms are part oflhe out-of-plane wall anchorage system.. they are also designed under
the requirementS ofSectton 12.11.2.1. Seismic forces on a typical east-west subdiaphragm are determmed
from Equa<ion 12.11-1 wnh SDS= 1.0 Md I,= 1.0.
Asde<ermined previoosly 1n Pan IO.I,F, = 1697 plfalong !he wall.
11.3 CHECK SUBDIAPHRAGM SHEAR
Assunle a 32-foot-dee-p subdiaphragm. as shown 1n Figure 5-23. n us 1s done for two reasons. Fiest. the
steel jo1·st purl in along line 8 can be used as a sutxJiaphragm ch«d (see Figure 5-23). SecOild, the deeperthan-required subdiaphragm depth (32 fee.t vs. 16 feet) reduces the subdiaphragm shear to manageable
le-v-els.
Shear reaction lOconunuily cross-ties along hnes C and 0:
R - 1697 plf(40 ft) - 33,940 lb
2
Max1mum shear =33,940/32 = 1061 plf
ApplyiJtg the ASD load combmation:
ASD shear:0.7( 106 1 plf) = 742 plf
1--
J
40' -0'
Joist~ Cootlnuity Tle ~
,.-
"k---I C
1>-
.....,
~
Jo1St Chord
..,....,
X
~
£..:>
~·
1>1>-
~ 1>-
I=
1-
,..
I
Jolst-Gude' Conlfnu.ly l ie
1>-
Q
b
•
~
lo-•
~
-'k--
J
Figure j -23. Subtltaphragm
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-{0
O.sigl Example 5 • Til-Up Butilg
Assume the matn diaphragm lS to be constructed Wtth 1 ~-mch SU\Ictural-1 OSB sheaxhing (WSP) wnh all
edges supported (blocked), a< ts typical in a panelized hybnd roof system. Refer 10 !he 2015 SDPWS Table
4.2A fOr n.aJhng requarements of d1fferent assemblies. 1lte sheathJng arrangement (shown tn Ftgure 5- I)
for east-west seismtc forces is case I wtlh the long panel dltecuon of the sheatJung parallel to suppons.
The sheathing edges are supponed only by 2x subpurlin frnmmg, and the strength ts lherefore ltmited by
the nad spactngs associated with 2-ulc.h nominal framing \\~dth unles:s upgraded to Jx framing. Vanous
natl spacittgs a1 she-athing panel edges and thelr respecuve se-tsmic shear capacities ate giVen in SDPWS
Table 4.2A at nomtnal strength levels and need to be divtded by 2 tOr allowable suess design (SDPWS
Section 42.3). Considenng the shear demand of742 plf(ASD), detennine the assembly acceptable tn the
subdiaphmgm areas. Nomtnal capaeny for u<e tn SDPWS Table 4 .2A = 2(742 pll) = 1484 plf. Per Table
4.2A. 3x frnrntng at adjointng. paneJ edges and boundaries is required based on:
1n SuucturaJ.J OSB sheathing wnh the tOIIowing c.o.umon na.hng:
IOd x J~ m penemmon in framtngat 2 in o.c. boundaries and conunuous edges
IOd xI~ tn penerration in framtngat 3 tn o.c. other edges
IOd x 10.: in pe11ettauon in ti'anungat 12 in o.c. intennedtate area.;;
1~
Nomtnal capaeny = 1640 plf> 1484 plf ... OK
11.4 CHECK STEEL JOIST AS SUBDIAPHRAGM CHORD
The steel joists 32 feet from the perimeter waH and lhe continuous honzonml reinfOlcement m the
penmeter walls aloog h.nes I and 9 act as chords for the subdtaphtagms.
Check to see tfthe steel joist can carry addittonal seismtc fOI'ce.
4
Chord li>rre = 1697 piQ 0)' -10 6061b
8(32)
'
Because the subdiaplvagm chord IS a steel element of the wall anchorage system, lll'i .subject to a 1.4 force
increase pe.rsection 12. 11.22.2.
Chord force (steel)= 10,606( 1.4)= 14,8481b
The< cl10rd force ( 14.8 ktps) L< less than the wall-anchorage joe<! force ( 19.0 kips) found in P'olrt 9 .6, and
thus does not govern.
Commentary
In reahty,lhe steel JO ISL'I acting as chOlds may tlot act tn tenston ao; a subcltaphragm chord a.;; shown above.
They will be loaded in tens ton only when oompres.:stve waJI.anchorage fOrces act on the dtaphragm.
Under this load1ng,lhe seismic forces probably do not follow lhe subdiaphragm path shown above bur are
uanstniued through the wood framing to other pans oflhe diaphragm. Even if subdiaphragm beha\'tOr
does occur,lhe subdiaphragtn may effectively be tnuch deeper than shown. However, because It is
necessary to demonstrate that there IS a system to res1stlhe out-of. plane tOn~es on the dtaphragm edge,lhe
subdiaphtagtn system shown abcwe is provided.
11.5 DETERMINE MINIMUM CHORD REINFORCEMENT AT EXTERIOR CONCRETE WALLS
Th1s destgn example assumes that lhe.re lS conunuous hortzootal reutforce.ment ln lhe walls at lhe roof le\·el
tl'!at acts as a chord for both the matn diaphragm and lhe sutxhaphragms.
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Design Example 5 • Tlr-Up Building
Subdiap~ragm chord force= P. = 10,6061b
J:, = I0,606 - () 20 1n
' =·r,
0.9(60,000)
.
2
A
Thls ts a relauvely small amount of retnfbrcemem. Generally, the mrun diaphragm c.hord retnforcement
e.xeeeds this amount. In present practtce. the subdtaphragm chord steel requiremem tS not added to the
chord steel requtremMt for the matn diaphragm.
12. Continuity Ties across the Main Diaphragm (East-West Loading)
In a uh-up building. continuoll~ ues have tv:o func.tions. lbe first is to transmttthe heavy out-of-p1ane
waJI loods tnto the mam dtaphrogm. The second funcuon •s that of'"'tytng" the tnten or pontons of the roof
together. In this example. the conununy ties on lines C and D will be destgned.
12.1 SEISMIC FORCES ON CONTINUITY CROSS-TIES ALONG LINES C AND 0
A mimmal tn[e,rconneclton of elements '"'ithm the buddtng ts requtred per ASCE 7 Secuons 12.1.3 and
12.1.4. Addtttonally. conbnuous cross-ttes are requtred per Sectton 12.11.2.2.l (SOC C and abo\re) to
ttansfer setsmtc fOI'Ces from the heavy peruneter walls tnto the main dtaphragm. In the east-west load
dtrecuon. the subclt.aphragm load is collected tnto the continuous cross-ties and then distributed across the
main diaphragm for further distrlbullon to the building·s shear walls atld frames.
The comtnuous cross. tie axtal force at the line 8 connection is the sum of both subd~aphragm reaction..:;..
Because the c:onunoous cross-ues are considered pan of the wall-anchotage system. the1r design (()(ce is
subject to the steel matertal ove.rsue.ngth multtplier 1.4 per Section 12.11.2.2.2.
P = 1697 pU(40 ft) (2)1.4 = 95,0321b
'
2
Per Section 12.1.3. the mtntmum t.ntercoMecuon force is0.133SDSW= 0.133JY. but not less lhan O.OSW.
where JV as the dead load of the smaller pon1ons of the build1ng being connected together. Unltke the
waJI-anchomge force, W m thiS ca~ includes the diaphragm wetght and thus could govern at the intenor
ofbuddlngs. In th1s case. the diaphragm weightlS 14 psflo include the g1tder self-we1ght. lne worst-case
value for Wfor the conttnuous cross-ue lS near the c.ente:r of the buildmg m gnd line 5 (Figure 5-1) wtth the
following result:
P,(m1n) = 0.133(14 pst)(40 ft)(4)(40 ft) + 0.1 33( 116 psf)(40 1\)(32 ft)(32 ft/2)/28
=23,2011b
Per Section 12.1.4.the muumum suppon connection force tS 5 percent of the dead and ltve load reaction.
P, (m1n) = 0.05(14 psf + 16 psf)(40 ft)(40 fli2) = 1200 lb
Thus. the wall anchorage conunoous cross-ue force ts governed by the subd.iaphragm de~ilgn.
P, = 93,032 lb
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O.sigl Example 5 • Til-Up Butilg
Note: The conunuous uesaloogltnesC and 0 ate not collectore.lementsand lhusare nm subject to the
speclal overstreng.th load combtnations of Section 12.10.2.1 or lhe 1.5 mulupher of Section 12.10.3.4. The
gtrder hne along line B funcuons bodl as a coounuous cross-tie and as a collector~ therefore.. both baste and
htgh.er collector load eombtnations must be considered. "The collector destgn may go\rem at some locatjons,
while cross-tie forces may govern at others.
12.2 DESIGN OF JOIST GIRDERS AS CONTINUITY TIES ALONG UNES C AND D
Whelher cotnbu-.ed walh a WSP-sheathed hybrid roof or a metal deck roof.. open-we-b steel jotst gtrders are
common roofgtrders an Ult-up bUIIding.'t. Specially enguteers tn associatton wath lhe joist manufacturer
l)'p!cally des1gn !he steel joist-g~rder members. As reqUJred by IBC Sec.tion 2207.2, the bulld•ng's design
engmeer ts responstble for pmvHhng axtal conunutty cross-ue fOrces to lhe manufb.cturer along with
informarjon staung whtch lood facuxs. tf any, have already been applied.
In this example. the designer should repon to the joist manufacturer that the unfactored wall anchorage
axlal fOrce (tension and c.o.npl'ession) acung oo the joist-girder top chord is P,= 93.032 lb. h is necessary
to uxhcate to the JOist manufacturer that thLs a'<tal force is t'i'om seismic effects so that the JOtst-girder"s
specialty engtneer ts able to apply the proper IBC lood combination<; with ASCE 7 Section 12.4.2.
Though not shown an Lhlse.xample. the top chord axtal eflects ofw1.nd Wmust also be constdered if the
effects could lead to a governing des.g.n oi the joist g_~rder. Becau.~e the load c.o.nbinauons of IBC Secuon
1605.2 (strength destgn) and Section 1605.3 (allowable suess des1gn) contatn very different formulas whe-n
c.onsidertng seiSJ11.ic E and wl.fld W~ the desagn engineer cannot simply compare£ and W to de.termine
wh1c.h governs. For lhjs reason. n lS recommended that unfactored loads be reported.
Wnh ltne B acung as a collector (Fagure 5-1), any JOISt g.~rders occurnng lhere requtre an addttJonal check
of !he higher collector loads of ASCE 7 Sec!lon 12.10.2 or 12.10.3.4 (SOC C and higher). In lh1s sm.auon,
the srrucrure's engtneer must speed)' the highe-r collector lood as well as an E conunuous cross-ue load. The
JOlst manufacturer's specialty enguteer will have to check both the basic IBC load combi.nauons walh the
c.ross-tie load£ as well as the basic load combinatjons wuh lhe higher collector load.
The tbllowang tS an example oflhe tnfonnat.ion to be placed on the drawmg.c; for the stee.l joist manufacturer
to properly destgn lhe joiSt girders fbr lateralload1ngs at hnes C and D. The wall-anchorage force£ shown
should already include the 1.4 mulupher for stee.l elements.
JoJst-Gtrder A'<ial Forces
£
= 93.0 k1ps (unfuctored)
£. = 0.0 k1ps (uniactored per ASCE 7 Section 12.102) Applicable only at collectors.
W = 18.4 k1ps (unfi>ctored)
Forces shall be checked in both tension and compression.
12.3 DESIGN OF JOIST-GIRDER SPLICES ALONG LINES C AND D
Splicing large axlaJ loads between jolst-gtrder top chords ts best done watha knife plate betw·een the lOp
chords at the JOist seat (f1gure 5-24). Top c.hords have a 1-lnch gap between them~ and the joist-gtrder
manufucturer \ViU keep thts space cle.ar if it is known in adYaJlce that a kmte plate ·w1ll be installed heJe. To
factlttate in<;taJiauon. the knife plate should be no thmner than% 1nch. The hetght of the kmte plate is lhat
which is necessa.r')' to obtain splice welding access. and often the stze of the kntfe plate ts excessive JUSt to
accommodate installation.
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Design Example 5 • Tlr-Up Building
7t0~x8·h'2"'1!10f!
ll1 '·4'~klllfe
""'
'JJI' 081-'.A &Ubirut·
plete ~I! lo:c!un"4.
Pl~~bl:~"~
clle«<~t:411'SWII'IC\II
........
Check the '%-tnch x 6~~tnch splic.e plate's design tens1Je sttength per AISC Secuon 02:
9,P, = 9,f';.A1
= 0.90(36,000X0.875X6.5) = 184,275 lb
Requ<red tie force P, = 93,032 lb < 184,275 lb ... OK
Ustng two lines of'l'artneh x 7-inch-long fillet welds., check the design we-ld strength per AJSC §12.4:
Requ<red tie forceP,= 93.0 k1ps < 97.4 k1ps . .. OK
Therefore. the JO tst~g.srder sphce acmss the columns 1s adequate.
12.4 COMMENTS ON METAL DECK DIAPHRAGMS
Although less common in the southwes[eJn United Stales than WSP sheathing. ftextble metal dec:k
dtaphtagms (walhout concre-le 611) are su11 somewhat common tn tdt-up conSttuctJon tn se.is:micaUy active
areas. When designed ptoper1y. metal dec.ktng c-an assist 1n prov1d10g wan anchorage and elimlnate the
need fOr subdiaphragms by acung ttself as the contlnuous cross-tte. However. tmportant detailing tssues
must be carefully considered.
Metal decks can only provtde coounuous cross-ues parallel to the deck span duectton. ASCE 7 Sectton
12.11.2.2.4 specifically pfOhtbits use of a metal deck perpe.ndtcular to the direction of span for conunuity
because the deck flutes will sttetch out and fl311en. Where the steel decking is spliced at the ends.. a common
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O.sigl Example 5 • Til-Up Butilg
structural member is needed to receive the anachmem from both deck panels. In common steel joist
systems w1lh double-top chords. it is necessary that both deck panels be attached to the same tndividual
to~hord half; othem·ise, cross-ue loads wdl be tnadvenently transferred through the steel JOist tO)H:I'lord
separation plate or web Y.'elding. dependtng on the configuration. Another concern at the memJ deck panel
splice and direct ledger attachment is the weJd tear-out through the metal deck edge. Proper deck gauge and
puddle-weld edge diSlaJ\('.e must be maintained for adeqwue wall-anchorage strength.
If the metal deckang is expected to carry waJI-anchorage forces. 11 must be uwesugated for te-nsion
and compression IDUaJ loads m conjunction wuh acung gravny loads. The axtal compression loads are
associated wnh inward wall force.~ and reqUJre a ~tal axiallbend1.ng a.natysls of the decklng. The No1'1lr
American Sumdardfor Cold-Formed Ste~l Framing (AJSJ, 20 16) provides destgn crite.rw tOr the decking.
and the Structural Steel Educauon Councd {Mayo. 200 1) illusttmes one approac.h for th1s style of wall
anchorage. A more robu.<;t approach to metal deck wall anchorage tS to use small sreeJ angles or tubes that
provide tension and compress1on wall support and dtSlflbute the load imo a memJ dec.k subdiaphragm.
Another challe.nge with metal deck diaphragms is the need for theJ'mal expansion jo1nt~. Me.tal deck roof
dtaphmgms are much more vulnerable to temperalUre swtngs than wood diaphragm sysretns~ and wtth
the trend tOYo'ilrd larget roof dtmenstons, thermal expanst<>n jotnts become very imponanL However, these
expansion jotnts uuerrupt the corumuous cross-Lies of the waU-.anchorage system and thus create several
tndependent buiJdmgs to be anaJyozed separately. llte \\'all-anchorage forces must be fuJiy deveh)ped tnto
the main dlaphragm and transferred to the applicab1e shear walls before reaching the e.~pansion joinL This
can resuh m larger diaphragm shears.
12-5 DESIGN GIRDER (CONTINUITY TIE) CONNECTION TO WALL PANEL
Jn llus example. walls are beanng walls and pllao;ters are not used to suppott the JOISt gtrder vert.tcaJiy.
Consequently. the kind of detatl shown tn Ftgure S-25 must be used. Thls detail pro\•tdes both vert.ical
support for the girde.r and the necessary waJI.andlomge c.apacity. The oe force IS the same as that fOr the
waJI.roofue of Part 10.2 (F,=67881b)~ but not less than 5 percemof the dead·lood plus live-Joad reaction
per ASCE 7 Sec:uon 12.1.4. Thedetatl has the capacuy to take both tel'L~ion and shear forces. DetaiJs of the
des 1g.n are no[ giVen.
Embcxt l)bto ""h
he;ti:SN ~1-1 st~
, . ·~ .
..
, •, L
Ftgurt 5-25. lkuul ofgu-tler to wal/ panel
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Design Example 5 • Tlr-Up Building
13. Shear Wall Design Loads
After the wall-anchorage forces are dtstribured into the main diaphragm, the waJI \\'"etght and diaphragm
self-we.ight are then ttanSferred to the coocre-le shear walls for a load path t0\\'3td the ground. Tdt-up
concrete butldmgs utiltze their tnherem concrete panels as a laternl-force-resJsting system. In SOC C and
htgher. only tntermediate precast shear walls and special reinfa«ed c.oncrete shear \li<l.lls are permitted.
\Vhde spectaJ reinfOrced conc.rete shear walls benefit from htgher respo1lse modification coeffietents.
addtuonal detadtng and analysts provtstons are applicable.
In one- and two-story tilt-up buildings. iauermediate precast shear wall systeJ"ns are more c.ommon. This
lateral-force-reststtng system has a heiglu limttatton of 40 feet, e.xc.epl in single-story storage ware.houses
where the limitation extends to 45 feet. These hetght limitations were selected to coorduwe wilh similar
hetght restricuons fOund in the fire-protectton destgn nonnally found m these h1gh-pale Slorage warehouses.
The use of the intermediate precast shear wall system will be illustrated tbr lhis building example.
13.1 DESIGN SPECTRAl RESPONSE ACCELERATIONS Sos AND S01
The sne coeffictent.o; F. and Fr are used to modify the mapped spectral acceferottons Ssand S1• Ustng
the given specual accele:ratiolls Ss= 1.5 and S 1 ; 0.6 and Site Clao;s 0, the folfowtng site coefficients are
deternuned accurately from the ooltne ASCE 7-16 earthquake hazard maps based on informaLion from the
United Stares Geologica] Survey (www.usgs.gov); however, JBC Ftgure 1613.2.1 may also be consulted.
F, = I (shoo penod)
F,= 1.1 (1-second period)
Using these site coefficients.. the site-adjusted specual accelerations are determined:
s.vs = F.Ss = 1.0(1.5) = 1.5 (shOJt period)
S.vo = F,S0 = 1.7(0.6): 1.02 (l ·se<Ond period)
IBC Eq 16-36
IBC Eq 16-37
The des1g.n specual response accelerauons are obmmed as tbllows:
s.,. = 2/3Sw =
IBC Eq 16-38
IBC Eq 16-39
1.0 (short penod)
S01 =213S,.,. =0.68 (1-second period)
Uslng the design specttal response acc.eletations ruld the rtsk care-gory from rae Table 1604.5, the Jlext ste.p
is tode~eronone theappropnate "'""nic design categoty from IBC Tables 1613.2.5(1) and (2). Woth S, less
than 0.75, both the short-penod and 1-second. peraod desagn categoraes are level 0~ thus, SOC 0 govems.
Short-period categoty = D
1-second-penod category = D
Governtng SOC; 0
352
IBCT 1613.2.5(1)
IBC T 1613.2.5(2)
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O.sigl Example 5 • Til-Up Butilg
Use ASCE 7 Sec.tion 12.6 1n conjunction wnh Table 12.6- I to oblatn the appropriate analysis procedure..
Use the equivalent lateral-force procedure ofSe(:tion 12.8 to detennlne lhe seismtc base shear coefficient.
For a concrete shear \\'311 build1ng. lhe appro:umate fundamental perl(xl T tS oblamed tL<;tng ASCE 7
Equation 12.8-7 (or 12.8-9) wuh a C1 = 0.020 andx:: 0.75. For thtsbuddmg, use an average roof height h.
= 28 feet.
r., = C,h."' = (0.02)28"·" = 0.24 see
ASCE 7 Eq 12.8-7
IfthlS example invoh·ed a regular structure five stories or fewer tn heig,ht ha\1tng a period Tless than 0.5
seconds, the OOse shear coefficient C1 could have been ba~ on an SM equal to I.0 (Secnon 12.8.1.3). But
m this example.lhe structure has a re-entrant comer trregularity per ASCE 7 Table 12.3-1. Type 2, and thus
the destgn spectral response accelerauons and SOC remain as ol'igu\aJiy calculated.
Sos.w..,. = 1.0 (shon period)
S016-=0.68 (! -second penod)
13.2 BASE SHEAR USING THE EQUIVALENT LATERAL-FORCE PROCEDURE
ASCE7
ASCE 7 Secuon 12.8.1 defines the seismic ba<e shear as
V= C,IV
Eq 12.8-1
s
c, =..!!!..
R
Eq 12.8-2
where
I,
Because these ttlt-up concrete walls will be considered load-be.artng walls and lntermedJate prec.ast shear
walls~
then
R = 4 (response modjficatioo factor)
In addition. the imponance factor descrtbed in ASCE 7 Sectton I 1.5 ts obtained from Table 1.5-2 based on
the bullding·s given Risk Categocy II:
1,= 1.0
T 1.5-2
Therelore, per Equation 12.8-2:
c'
5
= ""
R
'~
=~=0.25
4
1.0
Chedang the nla:.amurn l imn for
·-=;mR
Tl-
C
Sm
i,
c$ With Equatton 12.8·3 where TS: TL
0 ·6
062 - 0 2·
=~4 =. »
. > ...
OK
0.24 l 1.0
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20!8/BC SEAOC Structural/Seismic Design Manual. \tbl. 2
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353
Design Example 5 • Tlr-Up Building
Checkmg the minimum allowed value for C.., ASCE 7 Equruions 12.8-5 and 12.8-6 ate applicable. In tJus
e.\:ample. S1 JS equal to 0.6g; therefore, Equation 12.8-6 1S \1alid to check tl'le manimum allowed Cr
c,_ =0.044S,..,I, =0.044(1.0XI.O) =0.044 <025
... OK
C,-=0.0 1 <0.25 ... OK
Eq 12.8-5
Eq 12.8-5
-- 0.07)' < 0 .2)' .. . (lK
C1 . . ,-- O.SS,
R -- 0.5(0.6)
4
Eq 12.8-6
-I,
-1.0
The calculated value. for C1 = 0.25 ts be.ru-ee1l the maxunwn and mimmwn allov.'ed \'alues.
Subsutuung into EquaLion 12.8- 1:
,. = c,w = 0.251V
13.3 BASE SHEAR USING THE SIMPLIFIED ALTERNATIVE STRUCTURAL DESIGN
CRITERIA
Instead of the lengthy setsmtc analysis shown abcwe. stmple buddmgs that meet the twelve lurutauons
of ASCE 7 Secuon 12.14.1.1 may use the Slmphfied arutl)'SIS procedure u'l Secuon 12.14. Usu'lgSecuon
12.1.1,lhe simplified analysis procedure of Section 12.14 is allo\\'ed a~ an alte-rnatjve nlethod tOr
determming this example"s seismic forces.
13.4 SHEAR WALL DESIGN LOADS
Wnh the base shear coeffic1em de-tennJned, the-dimibwon of the lateral loads to the participating shear
walls may be determu1ed next. Because thts ts a fle.xtble diaphragm, nts assumed the loads are dtstrtbtned
on a tributary basis wath the diaphragm modeled as a seftes of simply supponed beams v.ith consideJ"auon
for the relauve dtstrlbutton of mass. Example 2 tn this volume ptOVJdes an IIIU.'ru'au,•e des:ign e.xample.
The seismtc load Jnto the walls 1ncludes the lood from the djaphtagm and from the \Valls· self-\vetglus.
In lhese heavy-walled bujldmgs., the wall self-weight is a s1gnificant source oflhe \~,tails· shear loads.
Frequently, numerous ·wall peneuations oomplicate the shear \\'311 design. resulting in some port1ons
resembling frame-type column or pae-r elemenlS. ACI Sect1on 18.10.8 provtdes design provisions associated
wnh walls containing w-all-pier elements and frame-l1ke column elements. The design of remforced
concrete shear \\'ails m general is presented m the 1018/BC Scn tcturaiJSeismlc DeJig" Afanual. Volume 3.
14. References
See the refe-re-nce hstmg at the front of t.hJs design volume.
354
2018/BC SEAOC Structural/Seismic Desi gn Manual, \til. 2
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