arriostre de vigas
DESCRIPTION
Paper de J. Yura sobre arriostre de vigas.TRANSCRIPT
, FUNDAMENT AlS OF BEAM BRACING ~""") 1
Revised 5195Joseph A. Yuca
University of Texas at Austin
Introduction
The purpose of this paper is to provide a fairly comprehensive view of the subject of beamstability bracing. Factors that affect bracing requirements will be discussed and design methods proposedwhich are illustrated by design examples. The design examples emphasize simplicity. Before going intospecific topics related to beam bracing, some important concepts developed for column bracing by Winter(1960) will be presented because these concepts will be extended 10 beams later.
For a perfectly straight column with a p:midheight brace stiffnes~ .BL' the ~elati~nship e "" -/between P cr and .BL IS shown In FIg. 1 full bracing
t (fimoshenko, 1961). The column buckles pI between brace points at full or ideal bracing; in P cr ~t this case the ideal brace stiffness .Bi = 2 P el Lb lt¡
where P e = ~ EI/Lb 2. Any brace with astiffness up to the ideal value will significantly p: p = ~ ti Lincrease the column buckling loado Winter t e L 2(1960) showed that effective braces require not b tonly adequate stiffness but also sufficient 2 4strengtb. The strength requirement is directly O PL Lb / ~related to the magnitude of the initial out-of- e
straightness of the member to be braced. F. 1 Effi f B S .ffni Ig. ect o cace ti e..~sI
¡ For a column with an initial out-of-I straightness (half sine curve) with a displacement .6.0 at midheight and a midheight brace stiffness equalf. to the ideal value, the heavy solid line in Fig. 2(a) shows fue relationship between tJ.r and P. For P =l:- O, tJ.r= .6.0. When P increases and approaches the buckling load, r2 EI/Lb 2, fue total deflection tJ.rt, becomes very large. For example, when the applied load is within 5% ofthe buckling load, tJ.r= 20.6.0.t If abrace stiffness twice fue value of the ideal stiffness is used, much smaller deflections occur. When
the load just reaches the buckling load, the tJ.r= 2~. For .BL = 3.Bi and P = P e , tJ.r= 1.5~. Thebrace force, Fbr , is equal to (tJ.r - .6.0 ).BL and is directly related to fue magnitude of the initialimperfection. If a member is fairly straight, the brace force will be small. Conversely, members with
r~ large initial out-of-straightness will require larger braces. Ifthe brace stiffness is equal to the ideal value,: then the brace force gets very large as the buckling load is approached because tJ.r gets very large as
1 1
30.8 0.8 7P ti¡.¡.p ~
P O6 ~ Pe 0.6 ~Ie . Lb
04 ~ 0.40.2 L 0.2 40 = 0.002~
Ao = 0.002L b t p /0.8 % p.
00 4 8 12 16 20 00 1% 2% 3% 4'1
!J. / Fbr(%ot P)(a) T !J.o (b)
Fig. 2 Braced Column with lnitial Out-of-Straightness
~ ""e- -
2! shown in Fig. 2(a). For example, at P = 0.95P cr and .1.0= Lb / 500, the brace force is 7.6% of P e whichi is off the scale of the graph. Theoretically the brace force will be infinity when the buckling load isI reached if the ideal brace stiffness is used. Thus. a brace s~stem will not be satisfactorv if the theoretical
- ideal stiffness is ¡!rovided because the brace forces get toa large. If the brace stiffness is overdesigned,as represented by the.BL = 2.B¡ and 3.B¡ curves in Fig. 2(b), then the brace forces will be more reason-able. For abrace stiffness twice the ideal value and a .1.0= Lb / 500, the brace force is onIy 0.8 %P e atP = P e ' not infinity as in the ideal brace stiffness case. For abrace stiffness ten times the ideal value,the brace force will reduce even further to 0.44%. The brace force cannot be less than 0.4%P corres-ponding to .1. = O (an infinitely stiff brace) for .1.0= Lb / 500. For design Fbr = 1 %P is recommendedbased on abrace stiffness of twice the ideal value and an initial out-of-straightness of Lb / 500 becausethe Winter model gives slightly unconservative results for the midspan brace problem (Plaut, 1994 ).
Published bracing requirements for beams usually onIy consider the effect of brace stiffnessbecause perfectly straight beams are considered. Such solutions should not be used directly in designoSimilarly, design roles based on strength considerations onIy, such as a 2 % role, can result in inadequatebracing systems. Both strength and stiffness of the brace system must be checked.
Beam Bracing Systems LATERAL BRACING
. . h Beam "b" has lower load so it canBeam bracmg IS a much more com- brace the top flange of girder "a",
plicated topic compared to column bracing.,r This is due mainIy to the fact that most a b
b kl d h k f '
1 b kl . . 1 . . 1 b d . - uc e s ape -wea rammg
! co umn uc mg mvo ves pnmarl y en mg '.'---::: !; whereas beam buckling involves both flexure ~-:-~~A::~ buckled shape - strong framing
,i an~ torsion. An eff~tive beam brace res~sts ~, "'" "'...~.~:- '
:¡ twlst of the cross sectlon. In general bracmg Glrder Top Flange Frammg;¡ mar be divided into two main categories, PLAN VIEW'¡ lateral and torsional bracing as illustrated in 11MtFfMt11Mt1tM'1;.t; Fig. 3. Lateral bracing restrains lateral dis- l,+-ij~.u+~.u+~~+l-r Metal Deck Forms
1 placement as its name implies. The effective-1r ness of a lateral brace is related to the degreeI~ that twist of the cross section is restrained. TORSIONAL BRACING1 F . 1 ed b b. ed Through'1 or a slmp y support eam su ~ect to , GirdersI1 uniform moment, the center of twist is located f=5[llaPhragms Cross ~. .d th . fl th ~rames
at a pomt Outsl e e tenslon ange; e top :
1: flange moves laterally much more than thebottom flange. Therefore, a lateral brace /restricts twist best when it is located at the PLAN VIEW floor beamtop flange. Lateral bracing attached at the /eckingbottom flange of a simply supported beam is 11===::II:~ almost totally ineffective. A torsional bracecan be differentiated from a lateral brace in A A B~kl d Sh
. f th ... ed e apethat tWlSt O e cross sectlon IS restramdirectl y, as in the case of twin beams with a SECT A - Across frame or diaphragm between the. .members. The cross frame location, while Flg. 3 Types of Beam Bracmg
able to displace laterally, is still considered abrace point because twist is prevented' Some systems such as concrete slabs can act both as lateral andtorsional braces. Bracing that controls both lateral movement and twist is more effective than lateral ortorsional braces acting alone (Tong and Chen, 1988; Yura, 1992). However, since bracing requirementsare so minimal, it is more practical to develop separate design recommendations for these two types of
systems.
Lateral bracing can be divided into four categories: relative, discrete, continuous and lean-on.A relative brace system controls the relative lateral movement between two points along the span of the
1 2 3 :v- ten. flg. restraint 3
1 ~ ..A is best
a) brace location ~ Af .tive - center o twist
c e ~ ~:~A centroid brace
~ ~ R- relatively i.neffectiveange A . center of twistrder b) top flange loadlng
brac~ ~A]0- brace location
~, ,\ comp. flg. can\~ ..-J A ~) move laterally
c) cross section distortion A Asect -
Fig. 4 Relative Bracing Fig. 5 Factors That Affect Brace Effectiveness
girder. The top flange horizontal truss system shown in Fig. 4 is an example of a relative brace system.The system relies on fue fact that if fue girders buck1e laterally, points a and b would move differentamounts. Since fue diagonal brace prevents points a and b from moving different amounts, lateralbuck1ing cannot occur except between fue brace points. Typically, if a perpendicular cut anywhere alongfue span length passes through one of fue bracing members, fue brace system is a relative type. Discretesystems can be represented by individual lateral springs along fue span length. Temporary guy cablesattached to fue top flange of a girder during erection would be a discrete bracing system. A lean-onsystem relies on fue lateral buck1ing strength oflightly loaded adjacent girders to laterally support a moreheavily loaded girder when all fue girders are horizontally tied together. In a lean-on system all girdersmust buck1e simultaneously. In continuous bracing systems, there is no "unbraced" length. In this paperonly relative and discrete systems that provide full bracing will be considered. Design recornmendationsfor lean-on systems and continuous lateral bracing are given elsewhere (Yura, 1992,1993). Torsionalbrace systems can be discrete or continuous as shown in Fig. 3. Both types are considered herein.
Some of fue factors that affect brace design are shown in Fig. 5. A lateral brace should beattached where it best offsets fue twist. For a cantilever beam in (a), fue best location is fue top tensiontlange , not fue compression flange. Top tlange loading reduces fue effectiveness of a top flange bracebecause such loading causes fue center of twist to shift toward fue top flange as shown in (b). Largerlateral braces are required for top flange loading. If cross members provide bracing above fue top flange,case (c),the compression tlange can still deflect laterally if cross-section is not prevented by stiffeners.In fue following sections fue effect of loading conditions, load location. brace location and cross-sectiondistortion on brace requirements will be presented. All fue cases considered were solved using fue elasticfinite element program BASP (Akay, 1977; Chao, 1987) which considers local and lateral-torsionalbuck1ing including cross-section distortion. The BASP program will handle many types of restraintsincluding lateral and torsional braces at any nade point along fue span along with transverse andlongitudinal stiffeners. The solutions and fue design recornmendations presented are consistent with thework of others: Kirby and Nethercot (1979), Linder and Schmidt (1982), Medland (1980), Milner (1977),Nakamura (1981, 1988), Nethercot (1989), Taylor and Ojalvo (1966), Tong and Chen (1988), Trahairand Nethercot (1982), Wakabayashi (1983), and Wang and Nethercot (1989).
Lateral Bracing of Beams
Behavior. The uniform moment condition is the basic case for lateral buckling of beams. If alateral brace is placed at the midspan of such a beam, the effect of different brace sizes (stiffness) isillustrated by the BASP solutions for a W16x26 section 20 ft long in Fig. 6. Fo. abrace attached to thetop (compression) flange, the beam buckling capacity initially increases almost linearly as the bracestiffness increases. If the brace stiffness is less than 1.6 k/in., the beam buckles in a shape resembling
.'
i ' 4
a half sine curve. Even though there is .lateral rnovement at fue brace point fue load Top Flange Brace
, rincrease can be more than three times fue 3unbraced case. The ideal brace stiffness M required to force fue beam to buckJe between -:!:- ñQ-stItTéñér
lateral supports is 1.6 k/in. in this example. M no br 2 ,,--,,-- Centroid BraceAny brace stiffness greater than this value ~
does not increase fue beam buckJing capacity 1 ,. No Brace 'C~ W16x26W Mand fue buckJed shape is a full sine curve. ~~~ cr
When fue brace is attached at fue top flange, 8 j:;ids~an bracEj there is no cross section distortion. No 00 . 8 12 16
stiffener is required at fue brace point. LATERAL BRACE STIFFNESS (k/in)
! A lateral brace placed at fue centroid Fig. 6 Effect of Lateral Brace Locationf . of fue cross section requires an ideal stiffness
; of 11.4 kIffi. if a 4 x 1/4 stiffener is attached; I at rnidspan and 53.7 k/in. (off scale) if no 5
T FI B.a . ed S b .all b . op ange racestlllener IS us . U Stantl y more racmg ~ /is required for fue no stiffener case because of 4
b d'. th b . Th ' ideal brace. we Istortlon at e race pomt. e , ,. Brace at Centroid
centroid bracing system is less efficient than Pfue top flange brace because fue centroid ~brace force causes fue center of twist to no br 2 P at centroidrnove ~ fue bottom flange and closer to ~W1~~;;
, fue brace point which is undesirable for 1 ~1~f~¡ lateral bracing. 8 j~mid~pan brace
Oj 'f For fue case of a beam with a O 30 60 90 120 150
.~ .con~entrated centro id loa~ at midspan, shown LATERAL BRACE STIFFNESS (k/in)
1 m Flg. 7, fue rnornent varles along fue length.The ideal centroid brace (110 k/in.) is 44 Fig. 7 Midspan Load at Centro id
( times larger than fue ideal top flange brace:'t (2.5 k/in.). For both brace locations cross d C o. ... oa at entroldlo! sectlon dl~tortlon had a rnmor effect « 3 %). 50 \
---~-- ""'~-'~..c~'~ ;~ "" ~--~ ~
cross section distonion since a stiffener was used at fue brace point. The top flange loading causes fue 5
center of twist at buckJing to shift to a position clase to mid-depth for most practical unbraced lengths,as shown in Fig. 5. Since there is vinually no lateral displacement near fue centroid for top flangeloading, a lateral brace at fue centro id will not brace fue bearn. Because of cross-section distonion andtop flange loading effects, lateral braces at fue centro id are not recornmended. Lateral braces must beplaced near fue top flange of simply supponed and overhanging spans. Design recornmendations willbe developed only for fue top flange lateral bracing situation. Torsional bracing near fue centro id or evenfue bottom flange can be effective as discussed later.
The load position effect discussed aboveassumes that the load remains vertical duringbuckJing and passes through fue plane of fue web.In fue laboratory, a top flange loading conditionis achieved by loading through a knife edge at fuemiddle of fue flange. In structures fue load isapplied to fue bearns through secondary membersor fue slab itself. Loading through fue deck can Restoring Torque Cross Sectlon Distortionprovide a beneficial "tipping" effect illustrated in ( a) ( b )Fig. 9. As fue bearn tries to buckJe, fue contact .. . Epoint shifts from mid-flange to fue flange tip Flg. 9 Tlppmg ffect
resulting in a restoring torque which increases fuebuckling capacity. Unfonunately, cross- section distonion severely limits fue beneficial effects oftipping.Linder (1982, in German) has developed a solution for fue tipping effect which considers fue flange-webdistonion. The test data (Linder,1982; Raju, 1992)indicates that a cross member merely resting (notpositively attached) on fue top flange can significantly increase fue lateral buckling capacity. The tippingsolution is sensitive to fue initial shape of fue cross section and location of fue load point on fue flange.Because of fuese difficulties, it is recornmended that fue tipping effect not be considered in designo
When a bearn is bent in 4000 ~ ~
double curvature fue compression flange 4switches from fue top flange to fue bottom ""flange at fue inflection point. Bearns with M 3000 --
compression in both fue top and bottom ( inC!k ) t~-~Muflanges along fue span have more severe 2000 /IJ~ "'ItIlIID"
bracing requirements than bearns with rcompression on just one sirle as illustrated by 1000 ífue comparison of fue cases given in Fig. 10. ~1~~~The salid lines are BASP solutions for a 20 ft ""Tnldspan flg.bracelong W16x26 beam subjected to equal but 00 5 10 15 20 25opp~site end ~omen~ ando with. lateral BRACE STIFFNESS ( k/in )bracmg at fue mldspan mflectlon pomt. Forno bracing fue buckJing moment is 1350 in-k. F. 10 B .th I fl . P .
A b hed fl .. && . Ig. earns WI n ectlon omts
cace attac to one ange lS mellectlvefor reverse curvature because twist at midspan is not prevented. If lateral bracing is attached to bothflanges, fue buckling moment increases nonlinearly as fue brace stiffness increases to 24 k/in, fue idealvalue shown by fue black doto Greater brace stiffness has no effect because buckJing occurs between fuebrace points. The ideal brace stiffness for a bearn with a concentrated midspan load is 2.6 k/in at Mcr= 2920 in-k as shown by fue dashed lines. For fue two load cases fue moment diagrams between bracepoints are similar, maximum moment at one end and zero moment at fue other end. In design a Cb =1.75 is used for fuese cases which corresponds to an expected maximum moment of 2810 in-k. Thedouble curvature case reached a maximum moment 25% higher because of warping restraint at midspanprovided by fue adjacent tension flange. In fue concentrated load case no such restraint is available sincefue compression flanges of both unbraced segments are adjacent to each other. On fue other hand, fuebrace stiffness at each flange must be 9.2 times fue ideal value of fue concentrated load case to achievefue 25 % increase. Since warping restraint is usually ignored in design Mcr = 2810 in-k is fue maximum
..; .
-~::;~I-=-:~'~¿:
6design momento At this moment level the double curvature case requires abrace stiffness of 5.6 k/inwhich is about twice that required for the concentrated load case. The results in Fig. 10 show that notonly is it incorrect to assume that an inflection point is abrace point but also that bracing requirementsfor beams with inflection points are greater than cases of single curvature. For other cases of doublecurvature such as uniform1y loaded beams with end restraint (moments), fue observations are similar.
20'Up to this point only beams with a 1600 ~f=~~ "" ?0_00_0_00_00_._0 single midspan lateral brace have been 1 brace .,.0--' ~ ...m ¿¡¡y
discussed. The bracing effect of a beam with o/
multiple braces is shown in Fig. 11. The M 1200
response of a beam with three equally spaced (in:k )braces is shown by fue solid line. When fue 800
lateral brace stiffness, .BL' is less than 0.14 ~ W16~ o;;¡k/in., fue beam will buckle in a single wave. 40 -.:!!1~ ~rIn this region a small increase in brace 8--¡:7--}~stiffness greatly increases fue buckling loado O 3 braces 1
For 0.14 < .BL < 1.14, fue buckled shape O 1 2 3 o 4switches to two waves and fue relative BRACE STIFFNESS @ EACH LOCATION (k/ In)
: effectiveness of fue lateral brace is reduced.i For 1.4 < .BL < 2.75, fue bucked shape is Fig. 11 Multiple Lateral Bracing
three waves. The ideal brace stiffness is 2.75i¡:, k/in. at which fue unbraced length can be considered 10 ft. For fue 20 ft span with a single brace at
l' midspan discussed previously which is shown by the dashed line , abrace stiffness ofonly 1.6 k/in. was
i, required to reduce fue unbraced length to 10 ft. Thus fue number of lateral braces along fue span affectsJ fue brace requirements. A similar behavior has been derived for columns (fimoshenko and Gere, 1961):¡ where changing from one brace to three braces required an increase in ideal column brace stiffness of1ft 1.71, which is fue same as that shown in Fig. 9 for beams, 2.75/1.6 = 1.72.
, :; Yuca and Phillips (1992) report fue - 7 9'1 results of a test program on fue lateral and ~ 911' torsional bracing of bearns for comparison ;g, 6 \ ~ '---f with the theoretical studies presented above. ~ 5 BASP
1\ Some typical test results show good O Ea. (1)1( correlation with fue BASP theory in Fig. 12. ~ 4
i: Since the theoretical results were found to be ~ 3 Top Flange Load¡, reliabl~, signific~t variables from the theory ~ l..,{: were mcluded m the development of the g A .~ -. -~
design recornmendations given in the m 1 ~ W12x14 - 24 n.. .1
following section. In surnmary, moment Ogradient brace location load location brace O 0.5 1 1.5 2 205, , ,stiffness and number of braces affect the LATERAL BRACE STIFFNESS (k/in)buckling strength of laterally braced bearns.The effect of cross section distortion can be Fig. 12 Lateral Bracing Testseffectively eliminated by placing the lateralbrace near the top flangeo
Lateral Brace Design. In the previous section it was shown that fue buckling load increases asthe brace stiffness increases until full bracing causes the beam to buckle between braces. In manyinstances the relationship between bracing stiffness and buckling load is nonlinear as evidenced by theresponse shown in Fig. 11 for multiple braces. A general design equation has been developed for bracedbeams which is gives good correlation with exact solutions for the entire range of lero bracing to fullbracing (Yura, 1992b). That braced beam equation is applicable to both continuous and discrete bracingsystems, but it is fairly complicated. In most design situations full bracing is assumed or desired, thatis, buckling between the brace points is assumed. For full bracing a simpler design alternative basedon Winter's approach wa..~ developed (Yura, 1992b) and is presented below.
~
For elastic beams under uniforrn moment fue Winter idealT bl 1 B C ffi . t 7
1 al b .ffn . ed ~ b kI. b th b a e. cace oe Iclenater cace stl ess requlr to .orce uc mg etween e facesiS.6i = #P{ / Lb where p{ = ~ EIyc / L2b' lyc is the out-Qf-planemoment of inertia of the compression flange which is Iy/2 for doubly Number Bracesyrnmetric cross sections, and # is a coefficient depending on the of Braces Coef.number of braces n within the span, as given in Table 1 (Winter, 1 21960) or approximated by # = 4 - ( 2/n). The Cb factor given in 2 3design specifications for nonuniforrn moment diagrams can be used to 3 341estímate the increased brace requirements for other loading cases. For .example, for a simply supported beam with a load and brace at 4 3.63midspan shown in Fig. 7, fue full bracing stiffness required is 1.56 Many 4.0times greater than the uniform moment case. The Cb = 1.75 for thisloading case provides a conservative estímate of the increase. Anadditional modifying factor Cd = 1 + (Ms / MJ2 is required when there are inflection points along thespan (double curvature), where Ms and ML are the maximum moments causing compression in the topand bottom flanges as shown in Fig. 13. The moment ratio must be equal to or less than one, so Cdvaríes between 1 and 2. In double curvature cases lateral braces must be attached to both flanges. Top ..
¡
flange loading increases the brace requirements even when bracing is provided at the load point. Themagnitud e of the increase is affected by the number of braces alongfue span as given by the modifying factor CL = 1 + ( 1.2/n). For MSone brace CL = 2.2; for many braces top flange loading has noeffect on brace requirements, i.e. Cd = 1.0.
In summary, a modified Winter's ideal bracing stiffness can M MLdefined as follows,
. # CbPj.6i = --~ CLCd (1)
MSFor the W12x14 beams laterally braced at midspan shown in Fig. Fig. 13 Double Curvature12, Lb = 144 in., # = 2, Cb = 1.75, CL = 1 + 1.2/1 = 2.2, andp{ = ~ (29000) (2.32/2)/(144)2 = 16.01 kips, .6i. = 0.856 k/in.which is shown by the * in Fig. 12. Equation (1) compares very favorably with the test results and withthe theoretical BASP results. For design the ideal stiffness given by Eq. (1) must be doubled for beamswith initial out-Qf-straightness so brace forces can be maintained at reasonable levels as discussed earlier.The brace force requirement for beams follows directly from the column Fbr = O.OIP for discrete bracesgiven earlier. The column load P is replaced with the equivalent compressive beam flange force, either(Cb p{) or M{ /h, where M{ is the maximum beam moment and h is the distance between flange centroids.The M{ /h estímate of the flange force is applicable for both fue elastic and inelastic regions. For relativebracing the force requirement is one half the discrete value. The lateral brace design recommendationswhich follow are based on an initial out-Qf-straightness ofadjacent brace points ofLb/500. The combined
LATERAL BRACING DESIGN REQUlREMENTS
Stirrn~s: .6i =2 # (CbPpCLCd/4 or 2#(Mj/h)CLCD/4 (2)"
where # = 4 - (2/n) or the coefficient in Table 1 for discrete bracing; = 1.0 for relative bracinCbP { = Cb ~ E Iyc / Lb 2 ; or = (M{ / h) where M{ is the maximum beam moment
CL = l + ( 1.2/n ) for top flange loading; = 1.0 for other loadingCd = 1 + (Ms / MJ2 for double curvature; = 1.0 for single curvaturen = number of braces
Strength: Discrete bracing: Fbr = 0.01 CL Cd Mf /h (3)Relative bracing: Fbr = 0.004 CL Cd Mj /h (4)
. C;.-.n I e rv q,; \'1~ ~ \~ lt ,~ E't('¡) .O;L C 10 (M+/~) C<::t / lb ~r- Q~y Y\
,
, ","'*". .
8 values of # and CL vary between 4.0 and 4.8 for all values of n so Eq. (2) can be conservativelysimplified for all situations tO.8L. = 10 Mf / hLb for single curvature and .8L. = 20 Mf / hLb for double
curvature.
Some adjustments to the design requirements are necessary to account for the different design codemethodologies, i.e. allowable stress design, load factor designo etc.. In AASHTO-LFD and AISC-LRFD,Mf is the factored moment; in Allowable Stress Design, Mf is based on service loads. The CbPf formof Eq (2) can be used directly for all specifications because it is based on geometric properties of thebeam, i.e., .8L ~ .8L. where .8L is the brace stiffness provided. The brace strength requirements, Eqs.(3) and (4), can also be used directly since the design strengths or resistances given in each code areconsistent with the appropriate factored or service loads. 'Only the Mf I h form of Eq. (2) which relieson the applied load level used in the structural analysis must be altered as follows:
AISC-LRFD: .8L ~ .8L. / 4> where 4> = 0.75 is suggestedAISC-ASD: .8L ~ 2 .8L. where 2 is a safety factor.AASHTO-LFD: .8L ~ .8L no change
The discrete and relative lateral bracing requirements are illustrated in the following two design examples.
.:. Lateral Brace Design Examples. Two different lateral bracing systems are used to stabilize five~t composite steel plate girders during bridge construction; a discrete system in Example 1 and a relativej". bracing in Example 2. The AASHTO- Load Factor Design Specification is used. Each brace shown,~ dashed in Example 1 control s the lateral movement of one point along the span, whereas the diagonals.,1 in the top flange truss system shown in Example 2 control the relative lateral displacement oftwo adjacent
points. Relative systems require 1/2 the brace force and from 1/2 to 1/4 of the stiffness for discrete'í systems. In both examples, a tension type structural system was used but the bracing formulas are also
applicable to compression systems such as K-braces. In Example 1 the full bracing requirements forstrength and stiffness given by Eqs. (2) and (3) are based on each brace stabilizing five girders. Since
; the moment diagram gives compression in one flange, Cd for double curvature is not considered.
!: In both examples, stiffness control s the brace area, not the strength requirement. In Example 1 thestiffness criterio n required abrace area 3.7 times greater than the strength formula. Even if the bracewas designed for 2 % of the compression flange force (a cornmonly used bracing rule), the brace systemwould be inadequate. It is important to recognize that bQ1h stiffness and strength must be adequate for
a satisfactory bracing system.
¡¡ Torsional Bracing of Beams 4 Tong & Chen "
d I b BASP\, " I ea race,11. " / /---:\ Examples of torsional bracing systems 3 4x1/4 sti~
were shown in Fig. 3. Twist can be Mcrprevented by attaching a deck to the top ~ bno r2flange of a simply supported beam, by floor "
beams attached near the bottom tension flange no stlffener
of through girders or by diaphragms located 1 no brace 111 {Hr W~:~6 W MT . -2~ crnear the centroid of the stringer. Wlst can 8 @ _,~_8also be restrained by cross frames that O midspan braceprevent the relative movement of the top and O 1000 2000 3000 4000
bottom flanges. The effectiveness oftorsional TORSIONAL BRACE STIFFNESS (in-k/rad)braces attached at different locations on thecross section will be presented. Fig. 14 Torsional Brace at Midspan
Behavior. The BASP solution for a simply supported beam with a top flange torsional braceattached at midspan is shown in Fig. 14. The buckling strength - brace stiffness relationships are non-linear and quite different from the top flange lateral bracing linear response given in Fig. 6 for the samebeam and loading. For top flange lateral bracing a stiffener has no effect. A torsional brace can onlyincrease the buckling capacity about fifty percent above the unbraced case if no stiffener is used. Localcross-section distortion at midspan reduces the brace effectiveness. If a web stiffener is used with fue
--
9torsional brace attached to the compression flange, then the buckling strength will increase until bucklingoccurs between the braces at 3.3 times the no-brace case. The ideal or full bracing requires a stiffnessof 1580 in-k/radian for a 4 x 1/4 stiffener and 3700 in-k/radian for a 2.67 x 1/4 stiffener. Tong andChen (1988) developed a closed form solution for ideal torsional brace stiffness neglecting cross-sectiondistortion that is given by the sol id dot at 1450 in-k/radian in Fig. 14. The difference between the Tongsolution and the BASP results is due to web distortion. Their solution would require a 6 x 3/8 stiffenerto reach the maximum buckling loado If the Tong ideal stiffness (1450 in-k/radian) is used with a 2.67x 1/4 stiffener, the buckling load is reduced by 14%; no stiffener gives a 51% reduction.
LATERAL BRACING - DESIGN EXAMPLE 1
... .. Span = 80 ft.; 10 in. concrete slab
, \, \\\ 5 girders @ 8 ft spacing, A36 steel
-r'- ,~? Girder Properties \\ D . I I b . t t t b.l'th~,'~ \\ eslgn a atera raclng sys em o s a Ilze e
~.' 14 x 8 \\\ girders during the deck pour. Use the external. \\ tension system shown. The form supports.
~ ~ = 561 i - transmit some load to the botlom flange so~ x 48 t .d I d.~= 32.0 i assume cen rol oa Ing.
J = 12.9 i 1/4 x 15 Use Load Factor Design for the constructioncondition - L.F. = 1.3
Loads: Steel girder: A = 48.75 in.2. wt = 165 Ib/ftCanco slab: . 8' x~ x 150 Ib/ft 3 = 1000 Ib/ft
w = 1165Ib/ft=1.165k/ft
1 2 1 2M = 8 w l x L.F. = 8 (1.165 ) (80) 1.3 = 1211 k-ft
My= 36 (561)/12 = 1682 k-ft > 1211 k-ft = 1.0
Try 4 lateral braces @ 16-ft spacing
Check lateral buckling - center 16-ft is most critical (AASHTO 10-102c)
6 32.0 _1 12.9 / ~n \2( 50 ) 2 M = 91x10 (1.0)16x12~ 0.772 ~ + 9.87l16"~1
= 15020000 lb-in = 1251 k-ft> 1211 k-ft 4 braces required
Brace Design: Use the full bracing formula - discrete system -See Eq 2 & 3
7t2 (29000) (32.0) - '. - 2 - . - . -~ = (16 x 12)2 - 248 klpS, # - 4 - 4" - 3.5, Cb - 1.0. ~ - 1.0
. 3.5 (248)(1.0)(1.0) 9 04 k/. f . d .~L = 2 16 x 12 =. In. or ea. glr er = 45.2 k/ln. for 5 girders = F/~
2 ~AE) Ab (29000) ABrace stiffness = cos e T = 2 = 45.2 k/in. /~-).. Fb <--r5 ) 335 ~Ab
I ~ = 2.61 in~ I .(-- CONTROLS
Brace Strength: F b = 0.01 (5) (1211 x 12 149.0) = 14.83 k(A36 steel) r ~ve girders
A F = 14.831 cose; ~ = 14.3863...[i; = 0.92 in~bY
.
10
LATERAL BRACING - DESIGN EXAMPLE 2
; , 'v' 'v' Same as Example 1 except the bracing system is a relative system -¡ '
1 lA, R ti I~' - a top flange horizontal truss. Each truss stabilizes 2-1/2 girders.'v' 'v' A-. The unbraced length of the girder flange is 16 ft which was checked
, ~ A 16ft . Ex I 1i 1 " ", t In amp e .i , 1 " -
" \1 B st.ftlA, ,~, Top flg race I ness: Pf = 248 kips, V !girder 2 P C C" '1 f b L C _ 10" " R * b - .
A ~ P =" l' L1 , , , Lb C = 1.0" " L': 't 2(248) 1.0 (1.0),", ,", = 16 x 12 = 2.58 k/in for ea. girder
PLAN VIEW = 6.45 k/in for 2 1/2 girders
2~) ( 1 ) .Ao (29000) 2cos26 AE = - ,_w = 6.45 k/in; A b = 0.239 in +-L b -.J5"" (8 x 12 AV 5)
(1211x12) .Brace strength: Fbr = 0.004 (2 1/2) 49 = 2.97 klps
2.97...[5 2AbFy = 2.97-f5 ; Ab= 36 = ~~Stiftness controls the brace size; 9/16 41 OK A = 0.248 in2
Figure 15 shows that torsional bracing 3000 .on the tension flange (dashed line) is just as no distortion ! ngeseffective as compression flange bracing (solid "---1 edline), even with no stiffener. If the beam has 2000 !
.&& l . . b . all !! 11 bracedno Stllleners, sp Ittmg racmg equ y ~between the two flanges gives a greater (in-k)capacity than placing all the bracing on just 1000 9 braced
one flange. The dot-dash curve is thesolution if web distortion is prevented by )Mcrtransverse stiffeners. The distortion does not Ohave to be gross to affect strength, as shown 1 10 10 10 10 101in Fig. 16 for a total torsional brace stiffness TOTAL TORSIONAL BRACE STIFFNESS (in-k I rad)of 3000 in-k/radian. If the W16x26 sectionhas transverse stiffeners, the buckled cross Fig. 15 Effect of Torsional Brace Locationsection at midspan has no distortion as shownby the heavy solid lines and Mcr= 1582 in-k.If no stiffeners are used, the buckling load
. k 2801 d th Total top flg bracedrops to 1133 m- , a 70 ecrease, yet ere 3000 in-k/radianis only slight distortion as shown b.y the 1- - - _/- ' I ."'..
dashed shape. The overall angle of tWISt for ~'e e e e e ~jMcr ¡¡o
the braced beam is much smaller than the L- 30' -..1 .I¡
twist in the unbraced case (dot-dash curve). W16x26 / stiffened/ -- Mcr= 1582 in-k
¡~The effect of load position on -', ¡¡
torsionally braced beams is not very no bracing -.'..tiff.&& M M 283. k no s eners: significant, as shown in Fig. 17. The d111er- 0= cF In- McF 1133 in-k
! ence in load between the two curves for topflange and centroid loading for braced beamsis almost equal to the difference in strength Fig. 16 Effect of Cross Section Distortion
,
",,--11
for fue unbraced beams (zero brace stiffness). 50 --The ideal brace stiffness for top flange loading is 18% greater than for centro id P 40 ~"", ---Top Flange Load
cr .- loading. This behavior is different from that ( kips ~ """ Load at Centroid
shown in Fig. 8 for lateral bracing where fue O ""
top flange loading ideal brace is 2.5 times ,,'that for centro id loading. 2 "" rfr t P
10 " W ,Ir, W16x26 M
Figure 18 surnmarizes fue behavior of mid;;;~~ top flga 4Q-ft span with three equal torsional braces o
ed 10ft art Th b t '"-' ed o 1000 2000 3000 4000 5000spac - ap. e eam was s 111~n . .at each brace point to control fue distortion. TORSIONAL BRACE STIFFNESS (In- k I radian
The response is non-linear and follows the. . .pattern discussed earlier for a single braceo Flg. 17 Effect of Load POSltlOn
For brace stiffness less than 1400 in- .k/radian, fue stringer buckled into a single 1600 ~f~~~=:;~~=ft" --_o ~/-~
wave. Only in fue stiffness range of 1400- 1b "" ~ ~. race "" ,.' ..rrIrrtt wJJY
1600 in-k/radian did multl-wave buckled 1200 ,/' ~shapes appear. The ideal brace stiffness at ~r ,/// "-each location was slightly greater than 1600 (in-k) """", ~in-k/radian. This behavior is very different 800 /,," ~Hr W19x26 W ~from fue multiple lateral bracing case for fue /" H~8~gsame beam shown in Fig. 11. For multiple 400 101 al b . th b b kled . 3 bracesater racrng e eam uc rnto two
waves when fue moment reached 600 in-k Oand then into three waves at M = 1280 in- O 400 800 1200 1600 2000
crk. For torsional bracing, fue single wave TORSIONAL STIFFNESS @ EACH BRACE (in-k/rad)controlled up to Mcr = 1520 in-k. Since fuemaximum moment of 1600 corresponds to Fig. 18 Multiple Torsional Bracesbuckling between fue braces, it can be assumed, for design purposes, that torsionally braced beams bucklein a single wave until fue brace stiffness is sufficient to force buckling between fue braces. The figurealso shows that a single torsional brace at midspan of a 2Q-ft span (unbraced length = 10 ft) requiresabout fue same ideal brace stiffness as three braces spaced at 10 ft. In fue lateral brace case fue threebrace system requires 1.7 times fue ideal stiffness of fue single brace system, as shown in Fig. 11.
Tests have been conducted on torsionally braced beams with various stiffener details which arepresented elsewhere (Yura, 1992). The tests show good agreement with fue Basp solutions.
Buckling Strength 01 Torsionally Braced Beams. Taylor and Ojalvo (1973) give fue followingexact equation for fue critical moment of a doubly syrnmetric beam under uniform moment withcontinuous torsional bracing
6 Topflg brace ~ M .~Q~V M~ + P' b El y (5) ., - ~ ~ ~ ~ ~ ~- ~ cr ",0(\ ,.,M = M + P' E 1 ~vW16x26v *J o\s\<?""g\"s
cr o b y 5 t-::.: l ~ ~""3\\ \e{\.,.'" l = 30'.'
. .. 4 ",""" = 283 in-k)where Mo lS fue bucklrng capaclty of fue ~r .,." o
- 3" 20'unbraced beam and :B b = attached torsional ~ .".,.' (504 in-k)
brace stiffness (k-in/rad per in. length).Equation (5), which assumes no cross s~cti~n 1 (161~~~k)distortion, is shown by fue dot-dash Ime rnFig. 19. The salid lines are BASP results for 00 10 20 30 40
a W16x26 section with no stiffeners and spans ~EI 1M2of 10ft, 20 ft, and 30 ft under uniform y o
moment with braces attached to fuecompression flange. Cross-section distortion Fig. 19 Approximate Buckling Formula
::~"tt;~"",~;:";~:t~j - 'c,
12 causes the poor correlation between Eqo (5) and the Diaphragms Thro gh GOrders
BASP results. Milner (1977) showed that cross- or DecksJ&~::J[U I
section distortion could be handled by using an ~ ~ Ieffective brace stiffness, .aor, which has been b
expanded (Yura, 1992) to include the effect of ---5--'"
stiffeners and other factors as follows,"\~=~'~ ;¡- ~ r
-.!- c -.!- + -..!.- + -.!- (6) -[~~~ JL et3T t3b t3aec t3g
6 E lb 2 E lb~- ~-where t3b is the stiffness of the attached brace, t3aec is b S b S
the cross-section web stiffness and t3g is the girder . o ..system stiffness. The effective brace stiffness is less Flg. 20 Torslonal Bracmg StlffnesS
than the smallest of t3b, t3aec or t3g.
The t3b of so me cornmon torsional brace systems are given in Figs. 20 and 21. Systems comprised:, of diaphragms, slabs, and floor systems for through girders in Fig. 20 assume that the connection,;! between the girder and the brace can support a bracing moment Mbr If partially restrained connections
are used, their flexibility should also be included in Eq. (5). Elastic truss analyses were used to derivethe stiffness of the cross trame systems shown in Fig. 21. If the diagonals of a X-system are designedfor tension only, then horizontal members are required in the system. In the K-brace system a tophorizontal is not required.
~ ~ATENSION SYSTEM I/..~~ /f/ e- 6 + 6 h . M = F h
F ~~/ .If hb' b
~ ~ A h ~b = M I eb 2 2
F: - ES hb .F Pb - 2C 53 T enslon System-
2Fh S 2Fh ~A + _A horizontals are required=...:..:p =...:..:p c h
S S
COMPRESSION SYSTEMF ~F 2 2
1><1 ~b = ~ ES hb Tension - Compression System
and L~ horizontals not requiredF: F
K BRACEF O~ F 22
~ 2E5 hb = -~-7 K Brace System -.::;=..s. + A diagonals designed for
F: F A c h tension and compression
At¡ = Brea of horizontal members Lc = length of diagonal membersAc = area of diagonal members S = spacing of girdersE = modulus of elasticity ~ = height of the cross trame
Figo 21 Stiffness Formulas for Cross Frames
In crossframes and diaphragms the brace Beam Loadl 13
moments Mbr are reacted by vertical forces on the ..main girders as shown in Fig. 22. These forces m' .",' mincrease some main girder moments and decreases ~others. The effect is greater for the twin girder M r Brace Load 5 r
system B compared to the interconnected system 5A. The vertical couple causes a differentialdisplacement in adjacent girders which reduces ~;f-,.//f",~;r fi:\the torsional stiffness of the cross trame system. .J¿"""'~/I'~~~ ~For abrace only at midspan in a twin girder t S S S I ~rsystem the contribution of the inplane girder "" 35flexibility to the brace system stiffness is
2 ~;r-~;r @- 12 S EI;¡; (7) .l/"~~ -.J¿"I' ~
.Bg - L3 t ~ t ~ ~r
where.lx is the stro~g axis moment of inertia of Fi . 22 Beam Load from Bracesone glrder and L lS the span length. As the g
number of girders increase, the effect of girder stiffness will be less significant. In multi-girder systems,the factor 12 in Eq. 7 can be conservatively changed to 24 (ng - 1)2/ng where n~ is the number of girders.For example, in a six-girder system, the factor becomes 100 or more than elght times the twin girdervalue of 12. Helwig (1993) has shown that for twin girders the strong axis stiffness factor .Bg issignificant and Eq. (7) can be used even when there is more than one brace along the span.
Cross-section distortion can be approximated byconsidering the flexibility of the web, including ful1 depthstiffeners if any, as follows: tiffener
[ 3 3 ] h - E ( N + 1.5h ) t w t s b s (8).B - 3.3 - + - tleC h 12 12
Torsional Bracewhere fw = thickness of web, h = depth of web, t. = Nthickness of stiffener, b. = width of stiffener, and N = rcontact length of the torsional brace as shown in Fig. 23. For Icontinuous bracing use an effective net width of 1 in. instead hof (N + 1.5h) in .B1eC and lib in place of.Bb to get liT, The L (N + 1.5 h)
dashed liDes in Fig. 19 based on Eqs. (5) and (6) show goodagreement with the BASP theoretical solutions. For the 10 Fig. 23 Effective Web Widthft and 20 ft spans, BASP and Eq. (6) are almost identical.Other cases with discrete braces and different size stiffeners also show good agreement.
In general, stiffeners or connection details such as clip angles, can be used to control distortion.For decks and through girders, the stiffener must be attached to the flange that is braced. Diaphragmsare usually W shapes or channel sections connected to the web of the stringer or girders through clipangles, shear tabs or stiffeners. When full depth stiffeners or connection details are used to controldistortion,the stiffener size to give the desired stiffness can be determined from Eq. (8). For partial depthstiffening illustrated in Fig. 24, the stiffness of the various sections of the web can be evaluatedseparately, then combined as follows:
3 3~ 3.3E ( h )2 ( (N+1.5h¡)tw tsbs
) 9.B - -+- ()I h¡ h. 12 12
I
;
--
14 where h¡ = hc' h.. or ht andhc
-Y- = ~ + ~ + ~ (10) --.l
, .8sec .8c .8s {3,I h ha
The portion of fue web within hb can be consideredinfinite1y stiff. For rolled sections, if fue diaphragrnconnection extends over at least one-half fue bearn ht ~depth' then cross-section distortion will not be signif- ht
icant because the webs are fairly stoCky compared to Fig. 24 Partially Stiffened Websbuilt-up sections. The depth of fue diaphragm, h.,can be less than one-half fue girder depth as long as it provides the necessary stiffness to reach fue
I required momento Cross trames without web stiffeners should have a depth h. of at least 3/4 of fue bearn
depth to minimize distortion. The location of a diaphragm or cross trame on the cross section is not very
l . important; it does not have to be located close to the compression flange. The stiffeners or connectionangles do not have to be welded to the flanges when diaphragrns are used. For cross trames, (3., should
. be taken as infinity; only ht and hc will affect distortion. If stiffeners are required for flange connectedtorsional braces on rolled beams, they should extend at least 3/4 depth to be fully effective.
Equation (5) was developed for doubly-syrnrnetric sections. The torsional bracing effect forsingly-syrnrnetric sections can be approximated by replacing !y in Eqs. (5) with leff defined as follows:
1eff = 1yc + .: 1y, (11)c
where lyc and ~ are the lateral moment of inertia of the compression flange and tension flangerespectively, and c and t are the distances from fue neutral bending axis to fue centro id of fuecompression and tension flanges respectively, as shown in Fig. 25(a). For a doubly syrninetric sectionc = t and Eq. (11) reduces to !y. A comparison between BASP solutions and Eqs. (5) and (11) for threedifferent girders with torsional braces is shown in Fig. 25(b). The curves for a W16x26 show verygood agreement. In the other two cases, one of the flanges of the W16x26 section was increased to10xl/2. In one case fue small flange is in tension and in the other case, fue compression flange is fuesmallest. In all cases Eq. (11) is in good agreement with the theoretical buckling load given by BASP.
Equation (5) shows that the buckling load increases without limit as fue continuous torsional bracestiffness increases. When enough bracing is provided, yielding will control fue bearn strength so Mcr cannot exceed My, fue yield or plastic strength of the section. It was found that Eq. (5) for continuousbracing could be adapted for discrete torsional braces by surnrning the stiffness of each brace along thespan and dividing by fue bearn length to get an equivalent continuous brace stiffness. In this case Mcr
~b M5000 ~6I@ eee~*' cr Flg.10x1/2
comp. flg. - 1..;5.5x1/4 stiffel}-,r ~.E 4000 40' l.
I~-
c ~ 3000 BASP .lE o x ~ 2000 ---
Y t ~ 1:"§ 1000 W16x26
. Utenslon flg. OO 2 4 6 8 10
( a ) ( b ) P b Brace Stiffness (k-in/rad/in. length)
Fig. 25 Singly Syrnrnetric Girders
15will be limited to M., fue moment corresponding to buckJing between fue brace points. By adjusting Eq.
(5) for top flange loading and other loading conditions, fue following general formula can be used for fuebuckJing strength of torsionally braced beams :
C2 p ElM = C2 M2 + bb T ~ ~ M o, M (12)cr bll o C 'y.r
T
where Cbu and Cbb are fue two 1imiting Cb factors corresponding to an unbraced beam (very weak braces)and an effectively braced beam (buckJing between fue braces); Cr is a top flange loading modificationfactor; Cr = 1.2 for top flange loading and Cr = 1.0 for centro id 10ading; and 13 T is fue equivalent
effective continuous torsional brace (in-k/radian/in. 1ength) from Eq.(6). The following two casesillustrate fue accuracy of Eq. (12). 7
4x1/4 stiffener
For fue case of a single torsional brace - 6 BASP ~at midspan shown in Fig. 26, Cbu= 1.35 for .9. 5 " E 12a concentrated 10ad at fue midspan of an ~ 4 q ~unbraced beam (G al ambos , 1988). Usually ~designers conservatively use Cb = 1.0 for this = 3 no stiffenercase. For the beam assumed braced at mid- ~ 2 Top Flange Load Cr= 1.2span, Cbb = 1.75 for a straight 1ine moment "8 ~.~~~~ . ~ -. -,h/, ~ = 1.35diagram with zero moment at one end afilie 1 ~W12x14-24ft. .1 Cbb= 1.75
unbraced 1ength. These two values of Cb are 00 100 200 300 400 500 600used with any value of brace torsional . . .stiffness in Eq. (12). For accuracy at small Torslonal Brace Stlffness (k-ln/rad)
values of brace stiffness fue unbraced buck-1ing capacity ChuMo should algO consider top Fig. 26 Effect of Stiffener
flange 10ading effects. Equation (12) shows exce1lent agreement with fue BASP theory. With nostiffener, .Bacc from Eq. (8) is 114 in-k/radian, so fue effective brace stiffness ~ from Eq. (6) cannot begreater than 114 regard1ess of fue brace stiffness magnitude at midspan. Equations (6), (8) and (12)predict fue buck1ing very accurate1y for al1 values of attached bracing, even at very 10w values of bracingstiffness. A 4 x 1/4 stiffener increased .Bacc from 114 to 11000 in-k/radian. This makes fue effectivebrace stiffness very c10se to fue app1ied stiffness, .Bb' With a 4 x 1/4 stiffener, fue effective stiffness is138 in-k/radian if fue attached brace stiffness is 140 in-k/radian. The bracing equations can be used todetermine fue required stiffener size to reduce fue effect of distortion to some to1erance 1evel, say 5 %.
Figure 27 shows fue correlation between fue approximate buckJing strength, Eq. (12) and fue exactBASP solution for fue case of a concentrated midspan load at fue centroid with three equally spacedbraces along fue span. Stiffeners at fue three brace points prevent cross-section distortion so PT =3.Bb/288 in.. Two horizontal cutoffs for Eq. (12) corresponding to fue theoretical moment at buckJingbetween fue braces are shown. The K = 1.0 25
limit assumes that fue critical unbraced íi)' Eq 121ength, which is adjacent to fue midspan load, .9- 20 K = .88is not restrained by fue more lightly loaded ::, "U
end spans. To account for fue effect of fue ~ 15end span restraint, an effective length factor -.J K = 1.0
K = 0.88 was calculated using fue procedure ~ 10given in fue SSRC Guide (Galambos, 1988). ~ - -~.! ce~tr~ LoadFigure 27 shows that it is impractical to rely U 5 Cbb= 1'~~~ñr--~on side span end restraint in determining fue Cbu- 1.35 ~W'2X1~ - 24 ft~
buckJing load between braces. An infinitely 00 500 1000 1500 2000 2500 3000stiff brace is required to reach a moment . .corresponding to K = 0.88. If a K factor of Brace Stlffness per Brace (k-ln/rad)
I is used in fue buckling strength formula, fueFig. 27 Multiple Discrete Braces
16 comparison between Eq. (12) and the BASP solution is good. Equation (12) should not be used with K
factors less than 1.0; the results will be unconservative at moments approaching the full bracing case.Similar results were obtained for laterally braced beams (Yura, 1992).
Torsiona/ Brace Design. There are two basic torsional bracing systems shown in Figs. 20 and 21:bending members represented by diaphragms, decks or floor beams; and trusses for the cross trames.The two systems can be correlated by noting that Mbr = Fbr hb' where hb is the depth of the cross trame.The term "brace forces" used hereinafter refers to both Mbr and Fbr" Equation (12) gives the relationshipbetween brace stiffness and Mcr for an ideally straight beam. For beams with an initial twist, 80, it isassumed that the brace design requirements are affected in a similar manner as that developed for lateral-.bracing of beams with initial out-of-straightness . The required brace stiffness ~ T' which must be at least
twice the ideal stiffness to keep brace forces small, can be obtained by rearranging Eq. (12)
'X7 * = 2 (M2 - C2 M2 ) :~ (13)#.IT cr bu o 2Cbb E Ieff
For discrete braces ~ * = p;L/n. The brace force Mbr = ~ * 80. An initial twist 80= 10 ( 0.0175
radians) is recornmended. For a 14-in deep section this assumed initial twist corresponds to a 0.25 in.relative displacement between the top and bottom flanges. Equation (13) can be conservatively simplifiedby neglecting the ChuMo term which will be small compared to Mcr at full bracing and by taking themaximum ~, which is 1.2 for top flange loading. The simplified stiffness and brace force requirementsare given in fue following surnmary.
TORSIONAL BRACING DESIGN REQillREMENTS
. * * 2 2 (14)Stlrrn~s: .8T =:BT L In = 2.4LM¡ I (nEleffCbb)
Strength: M = F h = O 04LM2 I ( nEI C2 ) (15)br br b . 1 eff bb
where Mf = maximum beam momentIcff = Iyc + (t Ic) ~; = 1 for doubly syrnmetric sections (see Fig. 25 )Cbb = moment diagram modification factor for the full bracing conditionL = span lengthn = number of braces along the span
The available effective stiffness of the brace system ~ is calculated as follows:
~ = ~ + ~ + ..!. + ~ + ~ (16).BT .8c .8. .8t .8b .8g
h3 3
= 3.3E ( h )2 ( (N+1.5h¡)lw Isbs).8 .8 .B - - + - (17)c' s' I h. 12 12
I h.I
h t where h¡ = hc, h.. or ht ; N = bearing length ( see Fig. 23 )
.8b = stiffness of attached brace (see Figs. 20 and 21);
.8g = 24( ng- 1 )2 52 Elxl (L 3 ng) (18)
where n is the number of interconnected girders (see Fig. 22)g
I I
The torsional brace stiffness requirement, Eq. (14), must be adjusted for the different designspecifi-cations as discussed earlier for the lateral brace requirements:
AISC-LRFD: Pr ~ Pro / <t> where <t> = 0.75 is suggested
AISC-ASD: Pr ~ 3 Pro , ~'d ~ 1.5x (Eqn. 15) w/ service moment
AASHTO-LFD: Pr ~ Pro no change
Torsional Brace Design Examples. In Example 3 a diaphragm torsional bracing system isdesigned by fue AASHTO-LFD specification to stabilize fue five steel girders during construction asdescribed in Examples 1 and 2 for lateral bracing. The strength criterion, Eq. 15, is initially assumed tocontrol the size of the diaphragm. A C 1 Ox 15.3 is sufficient to brace the girders. Both yielding andbuckling of fue diaphragrn are checked. The stiffness of the C 1 Ox 15.3 section, 195,500 in-k/radian, ismuch greater than required but the connection to fue web afilie girder and the in-plane girder flexibilitya1so affect the stiffness. In this example, the in-plane girder stiffness is very large and its affect on tbe bracesystem stiffness is only 2%. In most practical designs, except for twin girders, this effect can be ignored.If a full depth connection stiffener is used, a 3/8 x 3-1/2 in. section is required. The weld design betweenthe channel and the stiffener, which is not shown, must transmit the bracing moment of293 in-k.
The 40-in. deep cross frarne design in Example 4 required a brace force of7.13 kips from Eq. (15).The factored girder moment of 1211 k-ft. gives an approximate compression force in the girder of 12 11 x12/49 = 296 kips. Thus, the brace force is 2.5% of the equivalent girder force in this case. The framing
details provide sufficient stiffi1ess. The 3-in. unstiffened web at the top and bottom flanges was smallenough to keep P- well above the required value. For illustration purposes, a 30-in. deep cross frarneattached near fue compression flange is also considered. In this case, the cross frame itself provides a largestiffness, but fue 14-in. unstiffened web is too flexjble. Cross-section distortion reduces fue system stiffnessto 16,900 in.-k/radian, which is less than the required value. If this same cross frame was placed at thegirder midheight, fue two 7-in. unstiffened web zones top and bottom would be stiff enough to satisfy thebrace requirements. For a fixed depth of cross frarne, attachment at the mid-<lepth provides more effectivebrace stiffness than attachment close to either flange.
Closing Remarks and Limitations
Two general structural systerns are available for bracing bearns, lateral systems and torsionalsystems. Torsional bracing is less sensitive than lateral bracing to conditions such as top flange loading,brace location, and number of braces, but more affected by cross-section distortion. The bracingrecommendations can be used in the inelastic buckling range up to ~ ifthe Mf fonn ofthe lateral bracestiffness equation is used (Ales, 1993).
The recommendations do not address the bracing requirements for moment redistribution orductility in seismic designo The bracing fonnulations will be accurate for design situations in which thebuckling strength does not reir on effective lengths less than one. Lateral restraint provided by ligbtlyloaded sirle spans should, in general, not be considered because the brace requirements would be muchlarger than fue recornrnendations herein. AIso, laboratory observations in fue author's experience ( usuallyunplanned failures of test setups ) show that brace forces can be very large when local flange or webbuckling occurs prior to lateral instability. After local buckling the cross section is unsymrnetric andverticalloads develop very significant out of plane load components. The bracing recommendations donot address such situations.
ReferencesAkay, H.U., Johnson, C.P., and Will, K.M., 1977, "Lateral and Local Buckling ofBeams and Frames,"
Journal o/ Ihe Structural Division, ASCE, ST9, September, pp. 1821-1832.
Ales, J. M. and Yum, J. A., 1993, "Bracing Design for lnelastic Structures", Structural Stability ResearchCouncil Conference-ls Your Structure Suitably Braced?, April 6-7, Milwaukee, WI..
r 18"1
TORSIONAL BRACING - DESIGN EXAMPLE 3~~:~1f~~~~~~: Same as Example 1 except use the diaphragmsystem shown. In Ex. 1, tour lateral braces wererequired which gave Mr = 1251 k-ft, just 3%
96 greater than the required momo Since torsional, Girder Properties braces will impose some additional vertical forces¡ . on the girders as shown in Fig. 22, probably five
h = 49.0 ,c = 30.85 ,t = 18.15 In braces should be used. However, for comparisonI = 17500 . I = 32.0 , L ~ = 352 i~ with Examples 1 & 2, a four-brace system will bex yc -yt: designed. Mreq'd = 1211 k-ft (see Ex. 1)
E ( 18,15 5 239 . 4! q. 11): I eff= 32 + 3Q85 3 2 = In
,! 2¡ Eq. (15) ~r = 0.0.4.~~~~_~2!_(!~~~.~ 12)- = 293 in-k A36 Steel:
4 (29000) 239 (1.0) Req'd S x = 293/36 =~~ ~ 3
. Try C10 x 15.39 - 4 . 3 . . . 4; Ix = 67.4 in. Sx = 13.5 In. t f = 0.436 In , b = 2.60 In , J = 0.21 In
j" Check lateral b~Ckling of the diaphragm T ~ \ T ={ Iyc = (2.60) (0.436)/12 = 0.639 in .!-' ~Cb 2.3
;1 M = 91 x 106(2.3) ~-- I .772 (0.21) + /7t 10 \2(.,!E.1Q..) (AASHTO 10-102c)1 r 96 \J 0.639 \ -gs- J
'1 = 837000 Ib-in = 837 in-k> M y = 13.5(36) = 486> 293 in-k; OK
~ Check stiftness: Eq. (14); P ,= 2.4(80 x 12) (1211 x 12) = 17550 in-k/radianf~ "'" =;'. T req d 4(29000) 239 (1.0) 2~ ~ 1-(- bsJ¡ I The stiffness of the diaphragms on the exterior girders is 6EI br ISo
~ < 110 Since there are diaphragms on both sides of each interior girder, the) "1 stiffness is 2 x 6EI br ISo The average stiffness available to each girder.. I -+. I is (2 x 6 + 3 x 12)/5 = 9.6 El br ISo, .J¡ 19.5:; P b = 9.6 (29000) 67.4/96 = 195500 in-k/radian
2 2Girder: Eq. (18) P = 24 (5-1) 29000 (96) 17500 = 406000 in-k/radian
g 5 (80x12) 3
Distortion: From Fig 23 and Eq. (17) dete~ine the required stiffener size . t = 3/8 ins
2 3 3R = 3.3(29000)(~' ( 1.5X19.5X.5 + ~~ ) (1)Ps 19.5 19.5'} 12 12 .
From Eq (16) 1 = 1 + 1 + -..?:-.- '~ s= 40500 in-k/radian (2). 17550 195500 406000 ~s'
Equating (1) and (2) gives b = 3.17 in, - Use 3/8 x 3-1/2 stiffeners
'-
19
TORSIONAL BRACING - DESIGN EXAMPLE 4
Same as Example 3, but use cross trames. Make allmember sizes the same. A K-frame system will beconsidered using double angle members welded toconnection gusset plates. Member lengths are shown ininches. Use tour crossframes. See Examples 1 and 3 forsection properties. Use A36 steel.
Assume brace strength criterion controls - Eq. (15)
0.04 (80 x 12) (1211 X 12)2~r (40) = ,2 = 293 in-k; Fb = 7.31 kips
4 (29000) 239 (1.0) r
2Fbr Lc 2(7.31) 62.5From Fig. 21 : Max force = diagonal force = = = 9.52 kips - compS 96
The AASHTO Load Factor method does not give a strength formula for compression membersso the formulation in Allowable Stress Design will be used. Convert to ASO by dividing themember force by the 1.3 load factor to get an equivalent service load force.
Diagonal Force (ASO) = 9.52/1.3 = 7.3 kips
Try 2L - 21/2 x 21/2 x 1/4 rx = .769 in. ,A = 2.38 in.2
t/ r = 62.5/.769 = 81.2 ; ~ = 16980 - .53 (81.2f = 13490 psi = 13.5 ksi
Pallow = 13.49 (2.38) = 32.1 kips > 7.3 kips OK
Check brace stiffness:Eq. ( 14); ~ = 17550 in-k/radian - see Example 3
T req'd
Fig. 21 : ~ = 2(29000) (96)2 (40t (2.38) = 717000 in-k/radianb 8 (62.5) 3 + (96) 3
Girder : ~ = 406000 in-k/radian - see Example 3g
~c = ~t = 3.3 ~~OOO) ( -* )2(.~~~_1~~) = 399000 in-k/rad
.1- 1 1 2. - .Eq. (16). ~ - 717000 + 406000 + 399000 ' ~T - 113000 > 17550 In-k/rad OK
T
Evaluate the cross trame shown below
R 2(29000) (96) 2(30) 2(2.38)ti b = 3 3 = 490000 in-k/radian8 (56.6) + (96)
2 3~ = 3.3 (29000) (~ )(1.5 (14.0) (.5)' ) - 18300 . k/ dt 14.0 14.0 12 - In- ra
1 t = "4'§~555'" + 4aiI5OD + ~ ; ~T = 16900 < 17550 in-k/rad NG
T
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