Chapter 2 Shear angle design

7/30/2019 Chapter 2 Shear angle design

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Chapter - ShearTh eAISC Specification evotes neentire entenceo shear !) in chapterc, paragraph .4 ,whileth econ.nentarys aboutonepage ong. The Specification ives he ,rpr"rrion tit sirea.s asilnplematter. ut afterreadinghe comrrentary ou will relizehat hre s more o it than eetsth eeye. At thebeginning f thisdesignmanual here s a questior.rboutchecking heshearn asingleangle. Yo u might hink hat he shear tresss 4,,,,, ivicle y theareaof te I.g puruir.l othedirectio'of load. If yoLr toppedhere. ou wourdhaveconsideredlexLrrarhear,"bithis sonly partof theshear brceequatiorr.Everyangle hat s usedasa treamn a conventionalor erration.(such sa relieving ngre t the headof a window),unless ateray braced gainstwisr,is sirbjectedo a tor.siotaltomentwhich prodLrcesorsional hearTh e strategy doptedn thisdesignmanual s to conrbinehe lexuraland orsional hear tresses,an dcomparehat value o the imit given n th eAISC Specificatiou.2.1 Flexun ShearFlexural hear s h e sirearha tmostengineersorrnalryrrinkabout. The commentaryo theAISCSpecification rovideshe ollowingequation.or lexuralshear tress,,:

, I 5V,.D t Eqn. C-C4- )

This.q,utlon applieso equal eg anglesoaded longone ofthe prrncipal xes. when wa s he asttime ,ou oaded n arrgle longon eofthe principal xes?probablrnever.Do you evenknowho wto.deterninewhere heprincipar xe sare?(After using his designmanuai ou wi .) The usualoriertationhas he angle oaded rong rr eof thegeomtric xes, n whichcase accordi'g o thlComnrentarv)heshear tress eco.es I .35Zl(t), or equal eg angles.So ar,sogood."Herecotnes he tricky part. The tenn v1,s theconrponen f tlle sheai 'oice parallel o tl.r-"eg *ith "rossection ength and hickness, so t's no t he otalshear, utrather t i i the porrionofre shearneach eg' The comnrentaryoes ro texprain ow to detefminerrisproportion.F*rh"r;;.", il ; "corstarts 5 and 1.35 n these quations re or equaregangles.For uneqLraregungr"s, i"ualreis soe"vheren between ndcouldbe carcLrratedsing rrgit. (Assuming, f coirrselha tyo uknow row o derermine ,). t rcalry sn't practicaror an ngineern a cre-signffice o spd timedohg ltt,o/lt calctLlationsr trying o frgureout how muclrsrlears in each Jg. A ,nu.n simpre,approachs needed.Assu,nehat heshear s resisted y the eg paraler o trredirectionof trre oacr.For hecase farelieving ngleabovean opening n a bric k wa , this s thevertical eg . Thecapacity rth" ungi" .thendetelmined rsing ISC Specification quationG2_ :

I,, 0.6F,,A,,C,.In this equaton ", s the areaofthe reg esisti 'g ir eshear ndc" is theshear-bucklingeductioniactorwhich_is ivenby parag.aph 4 in the Specification s r.0 in trriscase. nexpliJabry,he.\lSC Specificatiorrlso ellsus ' paragraph 4 that ' = I .2,but,t, doesnot enter nto equationCl- I so his sof no Lrse.

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UsingC"= 1.0,wecan earrangehis equationo determinehe shear tress,jfi:= 0.6FrC,

There is still the matterofthe LRFD and ASD reduction actors,but first we need o determine hetorsional shearstress. Then we can combine he two shears nd apply the reduction actors.|=r

2,2 Tonional ShearWhen an angle s loadedconventionally,as n the caseofa relievingangle,a torsionalmoment s produced, eeFigure 2. I . The torsionalmomentM7 is the result of theeccentricity between he applied oadP and the shearcenter.

M r = P eRecall the conceptofthe shearcenter rom elementarymechanics.This is the point on the crosssection hroughwhich applied oadsmustact n order o produce endingwithout wisting. Forboth equal eg andunequal egangles, he shearcenter s assumed o be locatedat theintersection fthe legs.Technicallyhis isn't heexactlocation, but the eror resulting rom this assumption sinsignificant compared o the level of effort required odetermine he true shearcenter. If you really want toknow where he shear enter s, consult bookon thetheorv of elasticiw.

Figure 2.1- Eccentricityromlheshear enter.

Theresultingorsionalmoments resisted y two shearmechanisms: t.Venant's pure)shear ndwarpingshear.To make t evenworse,hereare wokindsof warping hear.However,lrecommentaryo the Specificationxplainshat hesewo warping hears ccountor lesshan20oloofthe St.Venant's pure)shear nd huscanbe gnoredf it is assumedhat heentire orsionalmoment s resisted y St.Venant's pure)shear.The orsional hear tresssgivenby Equation(C-G4-2) n theCommentaryo theAISC Specification:r -M7. t_3Mr"JA t

Notice hat hese quationsre hesame,akea few minutes ndprove t toyourself.One s interms fl the orsionalonstant,hiletheothers n erms f the otalarea,4. he ermrepresentshe hicknessftheangle. f J isn'tavailablen a table.henuse hesecond quationwhichhas rea nd hicknessn hedenominator.heAISCManualists hevalue fJforallstandardingle ngles.

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23 Total SheaThe total shearstesss the surnofthe flexuralshearand he torsionalshear:

- l/_ M-tt =---!-+------!-I11w ,t )Using paragraph GI of the AISC Specification with C, : 1.0, the limits for the shearstressof asingle anglewith ayield sfength of 3 ksiare:LRFDQ"v,=o.sr,=0.s(0.6Fy)=o.s+4= l9.4ksi. t, : : j .

Pdge2-3

ASD:v/o, =' 1.u, =o'u",' /.u7 =0.36y=1 .e si .

u;' * - .

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EXAMPLE 2.7Determine heshear tressn a 5x5x5/16 elieving ngle, pporting 5' tall brickwall. Assumethe ollowing:Fy 36 ksi ; span ength6'-0"; simplysupported; nbraced gainst ater.alisplacerrent;oadactsdown,any archingaction n the masonrywill be gnored.Step l Load DeterminationAssume hata brick wall weighs40 lbs/ft2,hus

LRFD:P,,:40x5'x1.4 280 bs/f t .ASD: P,, 40x5' :200 bs/R. . -Step2. TorsionDeterminehe maximum orsion n the angle. Git enth egeometryn FigureE2.landusing3.63" for ihewidth ofa standard rick, he eccentricity. , can bedetermined s:

5+ 0.5 (3.6312)(0.312s12)3.53".l5 ]125Theunit orsionn his ase ill be:

'l:

i . ' r

((((((

((I(I(({(I(; . ((III((III(((I(((

T = p oLRFD= 280x 3.53= 988 nch. bs./foot.ASD= 200x 3.53= 706 nch. bs./foot.' )c6 '2 ' 'Th e maximum orsion or a relievinganglewill be :

LRFD= 988x6/2 2964 nch. bs .ASD= 106x6/2 21 7 inch. bs.Step3. VerticalShearDetemine heverticalshear t lie endsupport.

v,,: I4/L/2LRFD= 280x6/2840 bs.ASD 200x6/2600 bs.

M, = TL/2 a ',,,a\tr.- i .: ' ' - . . ) t 1 .\ : , . .

Figure E2.ll 1

. / |i - , : i ' : i i t o . l2 g 4 . f l a ' r ' : ' 7 \ Y \ \ ) -l ' F


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Step4. CombineheTorsional hear nd heVertical hearTheequationor hemaximumhear tresss

" l/ . M, tbt .I

840 2964 0 3125LRFD , 5x 0 .3 25 =518+8576=91 l4ps i .0. 08ASD / , =- - i90 +2 ] ,17 '0 ' 3 t25-384+6126=6510s i .5v 0 .1 25 0 .108

Comparehese tresseso he imts:LRFD Q, ,, = 19,400pstASD:rt /n = 2.900psi/ " t 'In both caseshe actual lrear tresss ess han he allowable, o he design s acceptable.Ofcourse. eflectiolthould lsobechecked. ote hatasignificantortion fthe otal hear tresssrepresentedy the orsional ompolent.Clearly, his contributiono the otalshear annotbeisnoted.

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Chapter 2 Shear angle design