Cold Form Steel

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    Yu, W.W.Cold-Formed Steel StructuresStructural EngineeringHandbookEd. Chen Wai-Fah

    Boca Raton: CRC Press LLC, 1999

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    C o l d -Fo r m ed S t e el S t r u c t u r es

    Wei -Wen Yu

    Depar tm ent of Civ il Engineer ing,

    U niver sity of Missour i- Rolla,

    R o l l a , M O

    7.1 Introduction7.2 Design Standards7.3 Design Bases

    Allowable StressDesign (ASD) Limit StatesDesign or Loadand ResistanceFactor Design (LRFD)

    7.4 Materialsand Mechanical Properti es

    Yield Point , TensileStrength, andStress-Strain Relationship Strength Increase from Cold Work of Forming ModulusofElasticity,Tangent Modulus, andShear Modulus Ductility

    7.5 Element StrengthMaximum Flat-Width-to-Thickness Ratios Stiffened Ele-mentsunder Uniform Compression Sti ffened ElementswithStressGradient Unstiffened Elementsunder Uniform Com-pression Uniformly CompressedElementswithanEdgeSti ff-ener Uniformly Compressed Elementswith IntermediateStiffeners

    7.6 Member DesignSectional Properties Linear Methodfor ComputingSectionalProperties Tension Members Flexural Members Concen-tr icallyLoadedCompression MembersCombinedAxialLoadand Bending Cylindrical Tubular Members

    7.7 Connectionsand JointsWelded Connections Bolted Connections Screw Connec-tions

    7.8 Structural Systemsand AssembliesMetal Buildings Shear Diaphragms Shell Roof StructuresWall Stud Assemblies Residential Construction CompositeConstruction

    7.9 Defining TermsReferencesFurther Reading

    7.1 Introduction

    Cold-formedsteel membersasshown in Figure7.1 arewidely used in buildingconstruction, bridgeconstruction, storage racks, highway products, drainage facil it ies, grain bins, transmission towers,car bodies, railway coaches, and varioustypesof equipment. Thesesectionsare cold-formed fromcarbon or low alloy steel sheet, str ip, plate, or flat bar in cold-rolling machinesor by pressbrakeorbending brakeoperations. The thicknesses of such members usually range from 0.0149 in. (0.378mm) to about 1/4 in. (6.35mm) even thoughsteel platesandbarsasthick as1 in. (25.4mm) can becold-formed into structural shapes.

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    FIGURE7.1: Variousshapesof cold-formed steel sections. (From Yu, W.W. 1991. Cold-Formed SteelDesign, John Wiley & Sons, New York. With permission.)

    Theuseof cold-formed steel membersin buildingconstruction began in the1850sin both theU.S.and Great Britain. However, such steel memberswerenot widely used in buildingsin theU.S. untilthe1940s. At thepresent time, cold-formed steel membersarewidely used asconstruction materialsworldwide.

    Comparedwith other materialssuchast imber and concrete,cold-formed steel memberscan offerthe followingadvantages: (1) lightness, (2) highstrength and sti ffness, (3) easeof prefabrication andmassproduction, (4) fast and easy erection and installation, and (5) economy in transportation andhandling, just to namea few.

    From the structural design point of view, cold-formed steel members can be classified into twomajor types: (1) individual structural framing members (Figure 7.2) and (2) panels and decks(Figure7.3).

    In view of the fact that the major function of the individual framing members is to carry load,structural strength and sti ffness are the main considerations in design. The sections shown in

    Figure7.2 can beused asprimary framing members in buildingsup to four or fivestoriesin height.In tall multistorybuildings, themain framingistypicallyofheavyhot-rolledshapesandthesecondaryelementssuch aswall studs, joists, decks, or panels may be of cold-formed steel members. In thiscase, the heavy hot-rolled steel shapesand the cold-formed steel sectionssupplement eachother.

    Thecold-formed steel sectionsshown in Figure7.3are generally used for roof decks, floor decks,wall panels, and sidingmaterial in buildings. Steel decksnot onlyprovidestructural strength to carryloads, but they also providea surfaceon which flooring, roofing, or concrete fill can beapplied asshown in Figure7.4. They can alsoprovidespacefor electrical conduits. Thecellsof cellular panels

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    FIGURE7.2: Cold-formed steel sectionsused for structural framing. (From Yu, W.W. 1991. Cold-Formed Steel Design, John Wiley & Sons, New York. With permission.)

    FIGURE7.3: Decks, panels,and corrugatedsheets. (From Yu, W.W. 1991. Cold-Formed Steel Design,John Wiley & Sons, New York. With permission.)

    can also beused asducts for heatingand air conditioning. For composite slabs, steel decksareusednot only asformwork during construction, but also asreinforcement of thecompositesystem aftertheconcrete hardens. In addition, load-carrying panelsand decksnot only withstand loadsnormalto their surface, but they can alsoact asshear diaphragmsto resist forcesin their own planesif theyareadequately interconnected to each other and to supportingmembers.

    During recent years,cold-formed steel sectionshavebeen widely used in residential constructionand pre-engineered metal buildingsfor industr ial, commercial, and agricultural applications. Metalbuilding systemsare also used for community facilit ies such as recreation buildings, schools, andchurches. For addit ional information on cold-formed steel structures, seeYu [49], Rhodes[ 36], andHancock [28].

    7.2 DesignStandards

    DesignstandardsandrecommendationsarenowavailableinAustralia[39],Austria[31],Canada[19],Czechoslovakia [21], Finland [26] , France[20], Germany [23] , India[30], Japan [14], TheNether-lands[27],NewZealand[40] ,ThePeoplesRepublicofChina[ 34] ,TheRepublicof SouthAfrica[38],Sweden [44], Romania[37],U.K.[17], U.S. [7] , USSR[41], and elsewhere. Since1975, theEuropeanConvention for Constructional Steelwork [24] hasprepared several documents for the design and

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    FIGURE 7.4: Cellular floor decks. (From Yu, W.W. 1991. Cold-Formed Steel Design, John Wiley &Sons, New York. With permission.)

    testingof cold-formedsheet steel used in buildings. In 1989,Eurocode3provideddesign informationfor cold-formed steel members.

    Thischapter presentsdiscussionson thedesign of cold-formed steel structural membersfor useinbuildings. It ismainly basedonthecurrentAISI combinedspecification[7] for allowablestressdesign(ASD) and load and resistancefactor design (LRFD). It should benoted that in addition to theAISIspecification, in the U.S., many tradeassociationsand professional organizationshaveissued special

    designrequirementsfor usingcold-formedsteel membersasfloor androof decks[42], roof trusses[6],open web steel joists [43], transmission poles [10], storage racks [35], shear diaphragms [7, 32],compositeslabs[11], metal buildings[33], light framingsystems[15] ,guardrails,structural supportsfor highway signs, luminaries, and traffic signals [4], automotive structural components [5], andothers. For the design of cold-formed stainless steel structural members, see ASCE Standard 8-90[12] .

    7.3 DesignBases

    For cold-formedsteel design, two designapproachesarebeingused. Theyare: (1) ASD and(2) LRFD.Both methodsarebriefly discussed in thissection.

    7.3.1 AllowableStressDesign(ASD)

    In theASD approach, the required strengths(moments, axial forces, and shear forces) in structuralmembers are computed by accepted methods of structural analysis for the specified nominal orworking loadsfor all applicable load combinationsl isted below [7] .

    1. D

    2. D + L + (Lr or Sor Rr )

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    3. D + (W or E)4. D + L + (Lr or Sor Rr ) + (W or E)

    where

    D = dead loadE = earthquakeloadL = liveload dueto intended useand occupancyLr = roof liveloadRr = rain load, except for pondingS = snow loadW = wind load

    In addition, dueconsiderationshould also begiven to theloadsdueto (1) fluidswith well-definedpressureand maximum heights, (2) weight and lateral pressureof soil and water in soil, (3) ponding,and (4) contraction or expansion resulting from temperature, shrinkage, moisture changes, creep incomponent materials,movement dueto different sett lement, or combinationsthereof.

    Therequiredstrengthsshould not exceedtheallowabledesignstrengthspermi ttedbytheapplicabledesign standard. Theallowabledesign strength isdetermined bydividing thenominal strength bya

    safety factor asfollows:Ra = Rn/ (7.1)

    whereRa = allowabledesign strengthRn = nominal strength = safety factor

    For thedesign of cold-formed steel structural membersusing theAISI ASD method [7], thesafetyfactorsaregiven in Table7.1.

    Whenwind or earthquakeloadsact in combinationwithdeadand/or liveloads, it hasbeenageneralpracticeto permit theallowabledesign strength to beincreased byafactor of one-third becausetheaction of wind or earthquake on a structure is highly localized and of very short duration. Thiscan also beaccomplished by permitt ing a 25% reduction in thecombined load effects without theincreaseof the allowabledesign strength.

    7.3.2 Limit StatesDesignor LoadandResistanceFactor Design (LRFD)

    Two types of limit states are considered in the LRFD method. They are: (1) the limit state ofstrength required to resist theextremeloadsduringthelifeof thestructureand (2) thel imit stateofserviceability for astructureto perform itsintended function.

    For thelimit stateof strength, thegeneral format of theLRFD methodisexpressed bythefollowingequation:

    i Qi Rn (7.2)wherei Qi = requiredstrengthRn

    =design strength

    i = load factorsQi = load effects =

    resistancefactorRn = nominal strength

    Theload factorsand load combinationsarespecified in variousstandards. According to theAISISpecification [7], the following load factorsand load combinations are used for cold-formed steeldesign:

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    TABLE7.1 Safety Factors, , and ResistanceFactors, , used in theAISI

    Specification [7]ASD LRFD

    safety resistanceTypeof strength factor, factor,

    (a) StiffenersTransversestiffeners 2.00 0.85Shear stiffenersa 1.67 0.90

    (b) Tension members(seealsoboltedconnections) 1.67 0.95(c) Flexural members

    Bending strengthFor sectionswith sti ffened or part ially sti ffened

    compression flanges 1.67 0.95For sectionswithunsti ffenedcompression flanges 1.67 0.90

    Laterally unbraced beams 1.67 0.90Beamshaving onefl angethrough-fastened to deck or

    sheathing (C- or Z-sections) 1.67 0.90Beamshavingoneflangefastened to astandingseam roof system 1.67 0.90Web design

    Shear strengtha 1.67 0.90Web crippling

    For singleunreinforced webs 1.85 0.75For I-sections 2.00 0.80

    For two nested Z-sections 1.80 0.85(d) Concentrical ly loadedcompressionmembers 1.80 0.85(e) Combinedaxial load andbending

    For tension 1.67 0.95For compression 1.80 0.85For bending 1.67 0.90-0.95

    (f) Cylindrical tubular membersBending strength 1.67 0.95Axial compression 1.80 0.85

    (g) Wall studsandwall assembliesWall studsin compression 1.80 0.85Wall studsin bending 1.67 0.90-0.95

    (h) Diaphragm construction 2.00-3.00 0.50-0.65(i) Weldedconnections

    GrooveweldsTension or compression 2.50 0.90Shear (welds) 2.50 0.80Shear (basemetal) 2.50 0.90

    Arc spot weldsWelds 2.50 0.60Connected part 2.50 0.50-0.60

    Minimum edgedistance 2.00-2.22 0.60-0.70Tension 2.50 0.60

    Arc seam weldsWelds 2.50 0.60Connected part 2.50 0.60

    Fillet weldsLongi tudinal l oading (connected par t) 2.50 0.55-0.60Transverseloading (connected part) 2.50 0.60Welds 2.50 0.60

    FlaregrooveweldsTransverseloading (connected part) 2.50 0.55Longitudinal loading (connected part) 2.50 0.55Welds 2.50 0.60

    ResistanceWelds 2.50 0.65(j) Bolted connections

    Minimum spacingandedgedistance 2.00-2.22 0.60-0.70Tension strength on net section

    With washersDoubleshear connection 2.00 0.65Singleshear connection 2.22 0.55

    Without washers 2.22 0.65

    Bearingstrength 2.22 0.55-0.70Shear strength of bolts 2.40 0.65Tensilestrength of bolts 2.00-2.25 0.75

    (k) Screw connections 3.00 0.50(l) Shear rupture 2.00 0.75(m) Connect ions to other mater ial s(Bear ing) 2.50 0.60

    a When h/t 0.96Ekv /Fy , = 1.50, = 1.0

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    1. 1.4D + L2. 1.2D + 1.6L + 0.5(Lr or Sor Rr )3. 1.2D + 1.6(Lr or Sor Rr ) + (0.5L or 0.8W )4. 1.2D

    +1.3W

    +0.5L

    +0.5(Lr or Sor Rr )

    5. 1.2D + 1.5E + 0.5L + 0.2S6. 0.9D (1.3W or 1.5E)All symbolsweredefined previously.

    Exceptions:

    1. Theload factor for E in combinations(5) and (6) should beequal to 1.0when theseismicload model specified by theapplicablecodeor specification is limit statebased.

    2. Theload factor for L in combinations(3), (4), and (5) should beequal to 1.0 for garages,areas occupied as places of public assembly, and all areas where the live load is greaterthan 100 psf.

    3. For wind load on individual purlins, girts, wall panels, and roof decks, multiply theload

    factor for W by0.9.4. Theload factor for Lr in combination (3) should beequal to 1.4 in lieu of 1.6 when the

    roof liveload isdueto thepresenceof workmen and materialsduringrepair operations.

    In addit ion, the following LRFD criteria apply to roof and floor composite construction usingcold-formed steel:

    1.2Ds + 1.6Cw + 1.4C

    where

    Ds = weight of steel deckCw = weight of wet concrete during constructionC = construction load, including equipment, workmen, and formwork, but excluding the

    weight of thewet concrete.

    Table 7.1 lists the factors, which are used for the AISI LRFD method for the design of cold-formed steel members and connections [7]. It should be noted that different load factors andresistance factors may be used in dif ferent standards. These factors are selected for the specificnominal strength equationsadopted by thegiven standard or specification.

    7.4 MaterialsandMechanical Properties

    In theAISI Specification [7], 14different steelsarepresently listed for thedesign of cold-formed steelmembers. Table7.2 lists steel designations, ASTM designations, yield points, tensilestrengths, andelongationsfor thesesteels.

    From astructural standpoint, themost important propert iesof steel areasfollows:

    1. Yield point or yield strength, Fy2. Tensilestrength, Fu

    3. Stress-strain relationship

    4. Modulusof elasticity, tangent modulus, and shear modulus

    5. Ductility

    6. Weldability

    7. Fatiguestrength

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    TABLE7.2 Mechanical Propertiesof Steels Referred to in the AISI 1996 SpecificationElongation (%)

    Yield Tensile In 2-in. In 8-in.ASTM point, Fy strength, Fu gage gage

    Steel designation designation (ksi) (ksi) length length

    Structural steel A36 36 58-80 23 High-strengthlow-alloy A242(3/4in.structural steel and under) 50 70 18

    (3/4in. to1-1/2 in.) 46 67 21 18

    Low and intermediate A283 Gr. A 24 45-60 30 27tensilestrength B 27 50-65 28 25carbon plates, shapes C 30 55-75 25 22and bars D 33 60-80 23 20

    Cold-formed welded A500andseamlesscarbon Round tubingsteel structural tubing A 33 45 25 in roundsand shapes B 42 58 23

    C 46 62 21 D 36 58 23

    Shaped tubingA 39 45 25 B 46 58 23 C 50 62 21 D 36 58 23

    Structural steel with 42 ksi A529 Gr. 42 42 60-85 19minimum yield point 50 50 70-100 18

    Hot-rolled carbon steel A570 Gr. 30 30 49 21-25 sheetsand stripsof 33 33 52 18-23 structural quality 36 36 53 17-22

    40 40 55 15-21 45 45 60 13-19 50 50 65 11-17

    High-strength low-alloy A572 Gr. 42 42 60 24 20columbium-vanadium 50 50 65 21 18steelsof structural 60 60 75 18 16quality 65 65 80 17 15

    High-strength low-alloy A588 50 70 21 18structural steel with50ksi minimum yield point

    Hot-rol ledandcold-rol led A606high-strength low-alloy Hot-rolled assteel sheet and strip with rolled coils; 45 65 22 improved corrosion resistance annealed, or

    normalized; andcold-rolled

    Hot-rolled asrolledcutlengths 50 70 22

    Hot-rol ledandcold- rol led A607Gr. 45 45 60(55) Hot-rol led23-25high-strength low-alloy Cold-rolled 22columbium and/or vanadium 50 50 65 (60) Hot-rolled 20-22 steel sheet and strip Cold-rolled 20

    55 55 70 (65) Hot-rolled 18-20 Cold-rol led18

    60 60 75 (70) Hot-rolled 16-18 Cold-rol led16

    65 65 80 (75) Hot-rolled 14-16 Cold-rol led15

    70 70 85 (80) Hot-rolled 12-14 Cold-rol led14

    Cold-rolled carbon A611 Gr. A 25 42 26 structural steel sheet B 30 45 24 C 33 48 22 D 40 52 20 E 80 82

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    TABLE7.2 Mechanical Propertiesof SteelsReferred to in the AISI 1996

    Specification (continued)Elongation (%)

    Yi eld Tensi le I n 2-i n. In 8-in.ASTM point, Fy strength, Fu gage gage

    Steel designat ion designat ion ( ksi ) ( ksi ) l ength length

    Zinc-coated steel sheets A653 SQ Gr. 33 33 45 20 of structural quality 37 37 52 18

    40 40 55 16 50 (class1) 50 65 12 50 (class3) 50 70 12

    80 80 82 HSLA Gr. 50 50 60 20

    60 60 70 16 70 70 80 12(14) 80 80 90 10(12)

    Hot-rolled high-strength A715 Gr. 50 50 60 22-24 low-alloy steel sheets 60 60 70 20-22 and strip with improved 70 70 80 18 formability 80 80 90 14

    Aluminum-zinc A792 Gr. 33 33 45 20 alloy-coated by the 37 37 52 18 hot-dip process 40 40 55 16 general requirements 50 50 65 12

    80 80 82

    Notes:

    1. The tabulatedvaluesarebasedon ASTM Standards.2. 1in. = 25.4mm; 1 ksi = 6.9MPa.3. A653 Structural Quali ty Grade80, GradeE of A611, and Structural Quali ty Grade80 of A792 are

    allowed in the AISI Specification under special condit ions. For thesegrades, Fy = 80ksi, Fu = 82ksi,elongationsareunspecified. SeeAI SI Specification for reduction of yield point andtensilestrength.

    4. For A653 steel, HSLA Grades70 and 80, the elongation in 2-in. gagelength given in theparenthesis isfor TypeII . Theother value is for TypeI.

    5. For A607steel, the tensilestrength given in theparenthesis is for Class2. Theother value is for Class1.

    In addition, formabil ity, durabili ty, and toughnessarealso important propert iesfor cold-formedsteel.

    7.4.1 YieldPoint, TensileStrength,andStress-Strain Relationship

    Aslisted in Table7.2, theyield pointsor yield strengthsof all 14different steelsrangefrom 24to 80ksi(166to 552MPa). Thetensilestrengthsof thesamesteelsrangefrom 42to 100ksi (290to 690MPa).The ratios of the tensile strength-to-yield point vary from 1.12 to 2.22. As far as the stress-strainrelationship is concerned, thestress-strain curvecan either bethesharp-yielding type(Figure7.5a)or thegradual-yielding type(Figure7.5b).

    7.4.2 StrengthIncreasefrom ColdWork of Forming

    Themechanical propert ies(yield point, tensilestrength, and ducti li ty) of cold-formed steel sections,

    part icularly at the corners, are sometimes substantially different from those of the flat steel sheet,str ip, plate, or bar before forming. This is because the cold-forming operation increases the yieldpoint and tensilestrength and at thesametimedecreasesthe ducti li ty. Theeffects of cold-work onthe mechanical propert iesof corners usually depend on several parameters. The ratios of tensilestrength-to-yield point, Fu/Fy , and insidebend radius-to-thickness, R/t, are considered to be themost important factors to affect the changein mechanical propert iesof cold-formed steel sections.Design equationsare given in theAISI Specification [7] for computing the tensileyield strength ofcornersand theaveragefull-section tensileyield strength for design purposes.

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    FIGURE7.5: Stress-strain curvesof steel sheet or str ip. (a) Sharp-yielding. (b) Gradual-yielding.(From Yu, W.W. 1991. Cold-Formed Steel Design, John Wiley & Sons, New York. With permission.)

    7.4.3 ModulusofElasticity, TangentModulus, andShear Modulus

    Thestrength of cold-formed steel members that are governed by buckling dependsnot only on theyield point but also on themodulusof elasticity, E, and the tangent modulus, Et. A valueofE =29,500 ksi (203GPa) isused in theAISI Specification for the design of cold-formed steel structuralmembers. ThisE value is slightly larger than thevalueof 29,000ksi (200 GPa), which isbeing usedin theAISC Specification for thedesign of hot-rolledshapes. Thetangent modulusisdefinedby theslope of thestress-strain curveat any given stress level asshown in Figure7.5b. For sharp-yieldingsteels, Et = E up to theyield, but with gradual-yieldingsteels, Et = E only up to theproportionallimit, fpr (Figure7.5b). Once the stress exceeds the proportional limit, the tangent modulus Etbecomesprogressively smaller than the initial modulus of elasticity. For cold-formed steel design,theshear modulusis taken asG = 11,300 ksi (77.9 GPa) according to the AISI Specification.

    7.4.4 Ductility

    According to the AISI Specification, the ratio of Fu/Fy for the steels used for structural framingmembers should not be less than 1.08, and the total elongation should not be less than 10% for a

    2-in. (50.8 mm) gage length. If these requirements cannot be met, an exception can be made forpurlinsand girtsfor which thefollowingl imitationsshould besatisfied when such amaterial isused:(1) local elongation in a 1/2-in. (12.7 mm) gage length across the fractureshould not be less than20% and (2) uniform elongation outsidethefractureshould not belessthan 3%. It should benotedthat the required ducti li ty for cold-formed steel structural members depends mainly on the typeof application and the suitability of thematerial. Thesameamount of ducti li ty that is considerednecessary for individual framing members may not be needed for roof panels, siding, and similarapplications. For thisreason,even thoughStructural Grade80of ASTM A653steel, GradeEof A611

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    steel, and Grade80 of A792 steel do not meet the AISI requirements of the Fu/Fy ratio and theelongation, thesesteelscan beused for roofing,siding, and simi lar applicationsprovided that (1) theyield strength, Fy , used for design is taken as75% of the specified minimum yield point or 60 ksi(414 MPa), whichever is less, and (2) thetensilestrength, Fu, used for design is taken as75% of the

    specified minimum tensile stressor 62ksi (427 MPa), whichever isless.

    7.5 ElementStrength

    For cold-formed steel members, the width-to-thickness ratios of individual elements are usuallylarge. Thesethin elementsmaybucklelocally at astresslevel lower than theyield point of steel whenthey are subject to compression in flexural bending and axial compression asshown in Figure 7.6.Therefore, for the design of such thin-walled sections, local buckling and postbuckling strength ofthin elementshaveoften been themajor design considerations. In addition, shear buckling and webcripplingshould also beconsidered in thedesign of beams.

    FIGURE 7.6: Local buckling of compression elements. (a) Beams. (b) Columns. (From Yu, W.W.1991. Cold-Formed Steel Design, John Wiley & Sons, New York. With permission.)

    7.5.1 MaximumFlat-Width-to-ThicknessRatios

    In cold-formed steel design, the maximum flat-width-to-thicknessratio, w/t, for flangesis limitedto thefollowingvaluesin theAISI Specification:

    1. Sti ffenedcompression element havingonelongitudinal edgeconnected to awebor flangeelement, theother stiffened by

    Simple lip . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

    Anyother kind of st iffener . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

    2. Sti ffened compression element with both longitudinal edgesconnected to other sti ffenedelement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 500

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    3. Unsti ffened compression element and elementswith an inadequateedgestiffener . . . 60

    For thedesign of beams, themaximum depth-to-thicknessratio, h/t, for websare:

    1. For unreinforced webs: (h/t)max=

    200

    2. For websthat areprovided with transversestif feners:

    Using bearingsti ffenersonly: (h/t)max = 260Using bearingsti ffenersand intermediatesti ffeners: (h/t)max = 300

    7.5.2 StiffenedElementsunder Uniform Compression

    The strength of a sti ffened compression element such as the compression flangeof a hat section isgoverned byyielding if itsw/tratio isrelatively small. It maybegovernedbylocal bucklingasshownin Figure7.7at astresslevel lessthan theyield point if its w/t ratio is relatively large.

    FIGURE7.7: Local buckling of sti ffened compression flangeof hat-shaped beam.

    The elastic local buckling stress, fcr , of simply supported square plates and long platescan bedetermined asfollows:

    fcr =k 2E

    12(1 2)(w/t)2 (7.3)

    wherek = local buckling coefficientE = modulus of elasticity of steel = 29.5 103 ksi (203GPa)w = width of theplatet = thicknessof theplate = Poissonsratio

    It iswell knownthatsti ffenedcompression elementswi ll not collapsewhen thelocal bucklingstressisreached. Anadditional loadcanbecarriedbytheelement after bucklingbymeansofaredistr ibution

    of stress. Thisphenomenon isknownaspostbucklingstrength and ismost pronounced for elementswith largew/t ratios.Themechanism of thepostbuckling action can beeasily visualized from asquare platemodel as

    shown in Figure7.8 [48]. I t representstheportion abcdof thecompression flangeof thehat sectionil lustrated in Figure7.7. Assoon astheplate starts to buckle, thehorizontal bars in thegrid of themodel wil l act astierodsto counteract the increasing deflection of the longitudinal struts.

    In the plate, the stress distribution is uniform prior to its buckling. After buckling, a portionof the prebuckling load of the center str ip transfers to the edge portion of theplate. Asa result, a

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    FIGURE7.8: Postbuckling strength model. (From Yu, W.W. 1991. Cold-Formed Steel Design, John

    Wiley & Sons, New York. With permission.)

    nonuniform stress distr ibution is developed, asshown in Figure 7.9. The redistr ibution of stresscontinuesunti l thestressat theedgereachestheyield point of steel and then theplatebeginsto fail.

    FIGURE7.9: Stressdistr ibution in sti ffened compression elements.

    For cold-formed steel members,aconcept of effectivewidth hasbeen used for practical design.In thisapproach, instead of considering thenonuniform distribution of stressover theentirewidthof the plate, w, it isassumed that thetotal load iscarried byafictitiouseffectivewidth, b, subjectedto auniformly distr ibutedstressequal to theedgestress, fmax, asshown in Figure7.9. Thewidth, b,isselected so that theareaunder thecurveof theactual nonuniform stressdistr ibution isequal to thesum of thetwo partsof theequivalent rectangular shaded areawith atotal width, b, and an intensity

    of stressequal to the edgestress, fmax. Based on the research findingsof von Karman, Sechler, andDonnell [45] , andWinter [47], thefollowingequationshavebeen developed in theAISI Specificationfor computing theeffectivedesign width, b, for sti ffened elementsunder uniform compression [7]:

    (a) Strength Determination

    1. When 0.673, b = w (7.4)2. When > 0.673, b = w (7.5)

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    whereb = effective design width of uniformly compressed element for strength determination (Fig-

    ure7.10)w = flat width of compression element = reduction factor determined from Equation 7.6:

    = (1 0.22/)/ 1 (7.6)where = plateslendernessfactor determined from Equation 7.7:

    = (1.052/

    k)(w/t)(

    f/E) (7.7)

    wherek = platebucklingcoefficient =4.0for sti ffenedelementssupportedbyaweboneachlongitudinal

    edgeasshown in Figure7.10t = thicknessof compression elementE = modulusof elasticityf = maximum compressiveedgestressin the element without considering the safety factor

    FIGURE7.10: Effectivedesign width of sti ffened compression elements.

    (b) Deflection DeterminationFordeflectiondetermination,Equations7.4through7.7canalsobeusedfor computingtheeffectivedesign width of compression elements, except that the compressivestress should be computed onthebasisof theeffectivesection at theload for which deflection iscalculated.

    Therelationship between and according to Equation 7.6 isshown in Figure7.11.

    EXAMPLE7.1:

    Calculate theeffectivewidth of thecompression flangeof thebox section (Figure7.12) to beusedasabeam bending about thex-axis. UseFy = 33ksi. Assumethat thebeam websarefully effectiveand that thebendingmoment isbased on initiation of yielding.

    Solution Becausethe compression flangeof thegiven section is auniformly compressedsti ffened element, which issupported byaweb on each longitudinal edge, theeffectivewidth of theflangefor strength determination can becomputed byusing Equations7.4through 7.7with k = 4.0.

    Assume that the bending strength of thesection isbased on Init iation of Yielding, y 2.50 in.Therefore, the slendernessfactor for f = Fy can becomputed from Equation 7.7, i.e.,

    k = 4.0w = 6.50 2(R + t) = 6.192 in.

    w/t = 103.2

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    FIGURE7.11: Reduction factor, , vs. slenderness factor, . (From Yu, W.W. 1991. Cold-FormedSteel Design, John Wiley & Sons, New York. With permission.)

    FIGURE7.12: Example 7.1. (From Yu, W.W. 1991. Cold-Formed Steel Design, John Wiley & Sons,New York. With permission.)

    f = 33 ksi = (1.052/

    k)(w/t)

    f/E

    =(1.052/

    4.0)(103.2)33/29,500 = 1.816

    Since > 0.673, useEquations7.5and 7.6 to computetheeffectivewidth, b, asfollows:

    b = w = [(1 0.22/)/]w= [(1 0.22/1.816)/1.816](6.192) = 3.00 in.

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    7.5.3 StiffenedElementswith StressGradient

    When a flexural member is subject to bending moment, the beam web isunder thestressgradientcondition (Figure7.13), in which thecompression portion of theweb may buckledueto thecom-pressivestresscaused bybending. Theeffectivewidth of thebeam web can bedetermined from thefollowingAISI provisions:

    FIGURE7.13: Sti ffened elementswith stressgradient.

    (a) Strength Determination

    Theeffectivewidths,b1 and b2, asshown in Figure7.13, should bedetermined from thefollowingequations:

    b1 = be/(3 ) (7.8)

    For 0.236 b2 = be/2 (7.9)b1 + b2 should not exceed the compression port ion of the web calculated on the basis of effectivesection.

    For > 0.236b2 = be b1 (7.10)

    wherebe = effectivewidth b determined byEquation 7.4or Equation 7.5with f1 substi tuted for fand with k determined asfollows:

    k = 4 + 2(1 )3 + 2(1 ) (7.11) = f2/f1 (7.12)

    f1, f2 = stressesshown in Figure7.13 calculated on thebasisof effectivesection. f1 iscompression(+) and f2 can be either tension () or compression. In case f1 and f2 are both compression,f1 f2

    (b) Deflection Determination

    The effectivewidthsused in computing deflectionsshould bedetermined as above, except thatfd1 and fd2 aresubsti tuted for f1 and f2, wherefd1 and fd2 are thecomputed stressesf1 and f2 asshown in Figure7.13 based on theeffectivesection at the load for which deflection is determined.

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    7.5.4 UnstiffenedElementsunder UniformCompression

    Theeffectivewidth of unsti ffened elementsunder uniform compression asshown in Figure7.14canalso becomputed by using Equations7.4 through 7.7, except that thevalue ofk should be taken as0.43 and theflat width w ismeasured asshown in Figure7.14.

    FIGURE7.14: Effectivedesign width of unsti ffened compression elements.

    7.5.5 Uniformly CompressedElementswith an EdgeStiffener

    Thefollowing equationscan beused to determine the effectivewidth of theuniformly compressedelementswith an edgesti ffener asshown in Figure7.15.

    FIGURE7.15: Compression elements with an edgesti ffener.

    CaseI: For w/t S/3

    Ia = 0 (no edgesti ffener needed)

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    b = wds = ds for simple lip stiffener

    As = As for other stiffener shapes (7.13)

    CaseII : For S/3 < w/t < S

    Ia /t4 = 399{[(w/t)/S]

    ku/4}3

    n = 1/2C2 = Is /Ia 1C1 = 2 C2 (7.14)

    b should becalculated according to Equations7.4 through 7.7, where

    k = Cn2

    (ka ku) + ku

    ku =0

    .43

    For simple lip stiffener with 140 40 and D/w 0.8 where isasshown in Figure7.15:

    ka = 5.25 5(D/w) 4.0ds = C2ds

    For asti ffener shapeother than simplelip:

    ka = 4.0As = C2As

    CaseII I: For w/t

    S

    Ia /t4 = [115(w/t)/S] + 5 (7.15)C1, C2, b , k , d s , and As arecalculated per CaseII with n = 1/3where

    S = 1.28E/fk = bucklingcoefficientd , w , D = dimensionsshown in Figure7.15ds = reduced effectivewidth of thesti ffenerds = effectivewidth of thesti ffener calculated asunstiffened element under uniform com-

    pressionC1, C2 = coefficientsshown in Figure7.15As = reduced areaof thesti ffenerIa

    =adequate moment of inertia of the sti ffener, so that each component element wil lbehaveasasti ffened element

    Is , As = moment of inertia of the full section of the sti ffener about its own centroidal axis

    parallel to theelement to besti ffened,andtheeffectiveareaof thesti ffener,respectivelyFor thesti ffener shown in Figure7.15,

    Is = (d3tsin2 )/12As = ds t

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    7.5.6 Uniformly CompressedElementswith IntermediateStiffeners

    Theeffectivewidth of uniformly compressed elementswith intermediatesti ffenerscan also bedeter-mined from the AISI Specification, which includesseparate design rules for compression elementswith only one intermediate sti ffener and compression elements with more than one intermediatestiffener.

    UniformlyCompressedElementswith OneIntermediateStiffener

    Thefollowing equation can be used to determine the effectivewidth of the uniformly com-pressed elementswith oneintermediatesti ffener asshown in Figure7.16.

    FIGURE7.16: Compression elementswith oneintermediatesti ffener.

    CaseI: For b0/t S

    Ia

    =0 (no intermediatesti ffener needed)

    b = wAs = As

    CaseII : For S < b0/t < 3S

    Ia /t4 = [50(b0/t)/S] 50

    b and As arecalculated according to Equations7.4 through 7.7, where

    k = 3(Is /Ia )1/2 + 1 4As = As (Is /Ia ) As

    CaseII I: For b0/t 3SIa /t

    4

    = [128(b0/t)/S] 285b and As arecalculated according to Equations7.4 through 7.7, where

    k = 3(Is /Ia )1/3 + 1 4As = As (Is /Ia ) As

    In theaboveequations, all symbolsweredefinedpreviously.

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    Uniformly CompressedElementswithMoreThanOneIntermediate

    Stiffener

    For the determination of the effectivewidth of sub-elements, thesti ffeners of a sti ffened ele-ment with more than onesti ffener should bedisregarded unlesseach intermediatesti ffener hasthe

    minimum Is asfollows:

    Imin/t4 = 3.66

    (w/t)2 (0.136E)/Fy 18.4

    where

    w/t = width-thicknessratio of the larger sti ffened sub-elementIs = moment of inertiaof thefull sti ffener about its own centroidal axisparallel to theelement

    to besti ffened

    For additional requirements, seethe AISI Specification.

    7.6 MemberDesign

    Thischapter dealswith thedesign of thefollowingcold-formed steel structural members: (a) tensionmembers,(b) flexural members,(c) concentrically loadedcompression members,(d) combinedaxialload and bending, and (e) cylindrical tubular members. Thenominal strength equationswith safetyfactors() and resistancefactors() areprovided in theSpecification [7] for thegiven limit states.

    7.6.1 Sectional Properties

    Thesectional propert iesof amember such asarea, moment of inertia, section modulus, and radiusof gyration are calculated by using the conventional methodsof structural design. Thesepropert iesarebased on either ful l cross-section dimensions, effectivewidths,or net section, asapplicable.

    For thedesignof tensionmembers,thenominal tensilestrengthispresentlybasedonthenet section.However, for flexural members and axially loaded compression members, the ful l dimensions are

    used when calculating the criti cal moment or load, while the effective dimensions, evaluated at thestresscorresponding to thecritical moment or load, areused to calculate thenominal strength.

    7.6.2 Linear Methodfor ComputingSectional Properties

    Becausethethicknessof cold-formed steel membersisusuallyuniform, thecomputationof sectionalpropertiescan besimplified byusinga linear or midline method. In thismethod,thematerial ofeach element is considered to beconcentrated along the centerline or midline of the steel sheet andthe areaelementsare replaced by straight or curved l ineelements. The thicknessdimension, t, isintroduced after the linear computationshavebeen completed. Thus, the total area is A = Lt, andthemoment of inertiaof thesection is I = It, whereL is the total length of all lineelementsand Iis the moment of inertiaof thecenterline of thesteel sheet. Themoments of inert ia of straight lineelementsand circular lineelementsareshown in Figure7.17.

    7.6.3 TensionMembers

    Thenominal tensilestrength of axially loaded cold-formed steel tension members is determined bythefollowingequation:

    Tn = AnFy (7.16)

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    FIGURE7.17: Propert iesof lineelements.

    whereTn = nominal tensilestrengthAn = net areaof thecross-sectionFy = design yield stress

    When tension members usebolted connections or circular holes, the nominal tensile strength isalso limited bythetensilecapacity of connected partstreated separately bytheAISI Specification [7]under thesection ti tleof BoltedConnections.

    7.6.4 Flexural Members

    For thedesignof flexural members,considerationshould begiven to severaldesign features: (a) bend-ing strength and deflection, (b) shear strength of websand combined bending and shear, (c) webcrippling strength and combined bending and web crippling, and (d) bracing requirements. Forsomecases, special consideration should also begiven to shear lagand flangecurling dueto theuseof thin materials.

    BendingStrength

    Bending strengths of flexural members are differentiated according to whether or not themember is laterally braced. If such membersare laterally supported, theyaredesigned according tothe nominal section strength. Otherwise, if they are laterally unbraced, then the bending strengthmay be governed by the lateral buckling strength. For channels or Z-sections with tension flangeattached to deck or sheathingand with compression flangelaterally unbraced, and for such members

    having oneflangefastened to astanding seam roof system, thenominal bending strength should bereduced according to the AISI Specification.

    Nominal Section Strength

    Twodesignproceduresarenowused in theAISI Specificationfor determiningthenominal bendingstrength. Theyare: (I) Initiation of Yieldingand (II) Inelastic Reserve Capacit y.

    According to Procedure I on the basis of ini tiation of yielding, the nominal moment, Mn, of thecross-section is theeffectiveyield moment, My , determined for theeffectiveareasof flangesand the

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    beam web. Theeffective width of the compression flangeand the effectivedepth of the web can becomputed from thedesign equationsgiven in Section 7.5. Theyield moment of a cold-formed steelflexural member isdefined asthemoment at which an outer fiber (tension, compression, or both)first attains the yield point of the steel. Figure7.18 showsthreetypesof stressdistr ibution for yield

    FIGURE7.18: Stressdistr ibution for yield moment (based on ini tiation of yielding).

    moment based on different locationsof the neutral axis. Accordingly, the nominal section strengthfor initiation of yielding can becomputed asfollows:

    Mn = My = SeFy (7.17)

    whereSe = elasticsection modulusof theeffectivesection calculated with theextremecompression or

    tension fiber at FyFy = design yield stress

    For cold-formed steel design, Se isusually computed byusing oneof thefollowing two cases:

    1. If the neutral axis is closer to the tension than to the compression flange (Case c), themaximum stress occurs in the compression flange, and therefore the plate slendernessratio (Equation 7.7) and theeffectivewidth of thecompression flangearedetermined

    by thew/t ratio and f = Fy . Thisprocedureisalso applicableto thosebeamsfor whichtheneutral axis is located at themid-depth of thesection (Casea).

    2. If the neutral axis is closer to the compression than to the tension flange (Case b), themaximum stressofFy occursin the tension flange. Thestressin thecompression flangedependson thelocation of the neutral axis, which is determined by the effectiveareaofthesection. Thelatter cannot bedetermined unlessthecompressivestressisknown. Theclosed-form solution of this type of design is possiblebut would bea very tedious andcomplex procedure. It is, therefore, customary to determine the sectional propertiesofthe section by successiveapproximation.

    SeeExamples7.2and 7.3 for thecalculation of nominal bending strengths.

    EXAMPLE7.2:

    Usethe ASD and LRFD methods to check the adequacyof the I-section with unsti ffened flangesasshown in Figure7.19. Thenominal moment isbased on the ini tiation of yielding using Fy = 50ksi. Assume that lateral bracing isadequately provided. Thedead load moment MD = 30 in.-kipsand theliveload moment ML = 150 in.-kips.

    Solution

    (A) ASD Method

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    FIGURE7.19: Example 7.2. (From Yu, W.W. 1991. Cold-Formed Steel Design, John Wiley & Sons,New York. With permission.)

    1. Location of Neutral Axis. For R = 3/16 in. and t = 0.135 in., thesectional propert iesofthe corner element areasfollows:

    Ix = Iy = 0.0003889 in.4

    A = 0.05407 in.2x = y = 0.1564 in.

    For theunsti ffened compression flange,

    w = 1.6775 in.,w/t = 12.426

    Using k = 0.43 and f = Fy = 50ksi,

    = (1.052/

    k)(w/t)

    f/E = 0.821 > 0.673b = [(1 0.22/0.821)/0.821](1.6775) = 1.496 in.

    Assuming theweb is fullyeffective, theneutral axisis located at ycg = 4.063in. asshownin Figure7.20. Sinceycg > d/2, ini tial yield occursin thecompression flange.Therefore, f = Fy .

    2. Check theweb for full effectivenessasfollows(Figure7.20):

    f1 = 46.03 ksi (compression)f2 = 44.48 ksi (tension) = f2/f1 = 0.966.

    UsingEquation 7.11,

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    FIGURE7.20: Stressdistr ibution in webs. (From Yu, W.W. 1991. Cold-Formed Steel Design, JohnWiley & Sons, New York. With permission.)

    k = 4 + 2(1 )3 + 2(1 ) = 23.13h = 7.355 in.

    h/t = 54.48 = (1.052/

    k)(54.48)

    46.03/29,500

    = 0.471 < 0.673be = h = 7.355 in.b1 = be/(3 ) = 1.855 in.b2 = be/2 = 3.6775 in.

    Sinceb1 + b2 = 5.5325in. > 3.7405in., theweb is fully effective.3. Themoment of inertia Ix is

    Ix = (Ay2) + 2Iweb (A)(ycg)2= 25.382 in.4

    Thesection modulusfor thetop fiber is

    Se = Ix /ycg = 6.247 in.3

    4. Based on initiation of yielding, thenominal moment for section strength is

    Mn = SeFy = 312.35 in.-kips

    5. Theallowablemoment or design moment is

    Ma = Mn/ = 312.35/1.67 = 187.04 in.-kipsBased on thegiven data, therequired moment is

    M = MD + ML = 30 + 150 = 180 in.-kipsSinceM < Ma, the I-section isadequate for theASD method.

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    (B) LRFD Method

    1. Based on thenominal moment Mn computed above, thedesign moment is

    b

    Mn =

    0.90(312.35)=

    281.12 in.-kips

    2. Therequiredmoment for combined dead and livemoments is

    Mu = 1.2MD + 1.6ML= (1.2 30) + (1.6 150)= 276.00 in.-kips

    SincebMn > Mu, theI-section isadequatefor bending strength according to theLRFDapproach.

    EXAMPLE7.3:

    Determine the nominal moment about thex-axis for the hat section with sti ffened compressionflangeasshown in Figure7.21. Assumethat theyield point of steel is50ksi. Usethe linear method.Thenominal moment isdetermined byinitiation of yielding.

    FIGURE7.21: Example 7.3. (From Yu, W.W. 1991. Cold-Formed Steel Design, John Wiley & Sons,New York. With permission.)

    Solution

    1. CalculationofSectional Propert ies. Inorder tousethelinear method,midlinedimensionsareshown in Figure7.22.

    A. Corner element (Figures7.17 and 7.22)

    R = R + t/2 = 0.240 in.

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    FIGURE7.22: Lineelements. (From Yu, W.W. 1991. Cold-Formed Steel Design, John Wiley & Sons,New York. With permission.)

    Arc length

    L = 1.57R = 0.3768 in.c = 0.637R = 0.1529 in.

    B. Location of neutral axis

    a. First approximation. For the compression flange,

    w = 15 2(R + t) = 14.415 in.w/t = 137.29

    UsingEquations7.4 through 7.7 and assuming f = Fy = 50ksi,

    =1.052

    4(137.29) 50

    29500 =2.973 > 0.673

    =

    1 0.222.973

    /2.973 = 0.311

    b = w = 0.311(14.415) = 4.483 in.

    By using the effectivewidth of thecompression flangeand assuming the webis fully effective, theneutral axiscan belocated asfollows:

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    DistancefromEffectivelength L top fiber y Ly

    Element (in.) (in.) (in.2)

    1 2 1.0475 = 2.0950 9.9475 20.84002 2

    0.3768

    =0.7536 9.8604 7.4308

    3 2 9.4150 = 18.8300 5.0000 94.15004 2 0.3768 = 0.7536 0.1396 0.10525 4.4830 0.0525 0.2354

    Total 26.9152 122.7614

    ycg = (Ly)L =122.761426.9152

    = 4.561i n.

    Becausethedistanceycg is less than thehalf-depth of 5.0 in., theneutral axisis closer to the compression flangeand, therefore, the maximum stressoccursin the tension flange. The maximum compressive stresscan be computed asfollows:

    f = 50

    4.561

    10 4.561

    = 41.93 ksi

    Sincetheabovecomputedstressislessthan theassumed value,another trial isrequired.

    b. Second approximation. Assuming that

    f = 40.70 ksi = 2.682 > 0.673b = 4.934 in.

    Distancefrom

    Effectivelength L top fiber y Ly Ly2

    Element (in.) (in.) (in.2) (in.3)

    1 2.0950 9.9475 20.8400 207.30592 0.7536 9.8604 7.4308 73.27073 18.8300 5.0000 94.1500 470.75004 0.7536 0.1396 0.1052 0.01475 4.9340 0.0525 0.2590 0.0136

    Total 27.3662 122.7850 751.3549

    ycg = 122.785027.3662 = 4.487in.

    f =

    4.487

    10 4.487

    = 40.69 ksi

    Sincetheabovecomputed stressis closeto theassumed value, it isO.K.

    C. Check the effectiveness of the web. Use the AISI Specification to check the effec-tivenessof the web element. From Figure7.23,

    f1 =50(4.1945/5.513)

    =38.04 ksi (compression)

    f2 = 50(5.2205/5.513) = 47.35 ksi (tension) = f2/f1 = 1.245. UsingEquation 7.11,k = 4 + 2(1 )3 + 2(1 )

    = 4 + 2(2.245)3 + 2(2.245) = 31.12

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    h/t = 9.415/0.105 = 89.67 < 200 O.K.

    = 1.05231

    .12

    (89.67)38.04

    29,500= 0.607 < 0.673

    be = h = 9.415 in.b1 = be/(3 ) = 2.218 in.

    Since < 0.236,

    b2 = be/2 = 4.7075 in.b1 + b2 = 6.9255 in.

    Becausethe computed valueof(b1 + b2) is greater than the compression portionof theweb (4.1945 in.), theweb element is fully effective.

    FIGURE7.23: Effectivelengths and stressdistr ibution using ful ly effective webs. (From Yu, W.W.1991. Cold-Formed Steel Design, John Wiley & Sons, New York. With permission.)

    D. Moment of inert ia and section modulus. The moment of inert ia based on lineelementsis

    2I3

    = 2

    1

    12(9.415)3 = 139.0944

    (Ly2) = 751.3549Iz = 2I3 + (Ly2) = 890.4493 in.3

    (L)(ycg )2 = 27.3662(4.487)2 = 550.9683 in.3

    Ix = Iz (L)(ycg )2 = 339.4810 in.3

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    Theactual moment of inertiais

    Ix = Ix t = (339.4810)(0.105) = 35.646 in.4

    Thesection modulusrelativeto theextremetension fiber is

    Sx = 35.646/5.513 = 6.466 in.3

    2. Nominal Moments. Thenominal moment for section strength is

    Mn = SeFy = Sx Fy = (6.466)(50) = 323.30 in.-kips

    Once the nominal moment is computed, the design moments for the ASD and LRFDmethodscan bedetermined asil lustrated in Example7.2.

    According to Procedure II of theAISI Specification, the nominal moment, Mn, is themaximumbending capacity of thebeam byconsideringthe inelastic reservestrength through part ial plastifica-tionof thecross-sectionasshown in Figure7.24. Theinelasticstressdistr ibution in thecross-section

    FIGURE 7.24: Stress distr ibution for maximum moment (inelastic reserve strength). (From Yu,W.W. 1991. Cold-Formed Steel Design, John Wiley & Sons, New York. With permission.)

    depends on the maximum strain in the compression flange, which is limited by the Specificationfor the given width-to-thickness ratio of the compression flange. On the basis of the maximumcompression strain allowed in theSpecification, theneutral axiscan belocated byEquation 7.18andthenominal moment, Mn, can bedetermined byusing Equation 7.19:

    dA = 0 (7.18)ydA = M (7.19)

    where is thestressin thecross-section. For additional information, seeYu [49].

    Lateral Buckli ng Str ength

    Thenominal lateral bucklingstrength of unbraced segmentsof singly-, doubly-, and point- sym-metr ic sectionssubjected to lateral buckling, Mn, can bedetermined asfollows:

    Mn = ScMc

    Sf(7.20)

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    whereSf = elastic section modulusof the full unreduced section for theextremecompression fiberSc = elastic section modulusof theeffectivesection calculated at astressMc/Sf in theextreme

    compression fiber

    Mc = critical moment for singly-, doubly-, and point-symmetr ic sectionscalculated asfollows:1. For Me 2.78My :

    Mc = My (7.21)2. For 2.78My > Me > 0.56My :

    Mc =10

    9My

    1 10My

    36Me

    (7.22)

    3. For Me 0.56My :Mc = Me (7.23)

    whereMy = moment causing ini tial yield at theextremecompression fiber of thefull section

    = SfFyMe = elastic critical moment calculated according to (a) or (b) below:

    (a) For singly-, doubly-, and point-symmetr ic sections:Me = Cbr0Aey t for bendingabout thesymmetryaxis. For singly-symmetr ic sections, x-axis

    is the axis of symmetry oriented such that the shear center has a negativex-coordinate.For point-symmetr ic sections, use 0.5 Me. Alternatively, Me can be calculated using theequation for doubly-symmetr ic I-sectionsor point-symmetr ic sectionsgiven in (b)

    Me = Cs Aex [j+ Cs

    j2 + r 20

    (t/ex )]/CT F for bendingabout thecentroidal axisperpendic-ular to thesymmetry axis for singly-symmetr ic sectionsonly

    Cs = +1 for moment causingcompression on the shear center side of thecentroidCs = 1 for moment causing tension on theshear center sideof thecentroidex = 2E/(Kx Lx /rx )2

    ey = 2

    E/(Ky Ly /ry )2

    t = [GJ + 2ECw/(KtLt)2]/Ar20A = full cross-sectional area

    Cb = 12.5Mmax/(2.5Mmax + 3MA + 4MB + 3MC ) (7.24)In Equation 7.24,Mmax = absolute valueof maximum moment in theunbraced segmentMA = absolutevalueof moment at quarter point of unbraced segmentMB = absolutevalue of moment at centerlineof unbraced segmentMC = absolutevalueof moment at three-quarter point of unbraced segment

    Cb ispermitted to beconservatively taken asunity for all cases. For canti leversor overhangswherethefreeend isunbraced, Cb shall be taken asunity. For memberssubject to combined axial loadand

    bending moment, Cb shall betaken asunity.E = modulusof elasticityCT F = 0.6 0.4(M1/M2)where

    M1 is the smaller and M2 the larger bending moment at the endsof the unbraced length in theplaneofbending,andwhereM1/M2, theratioof endmoments, ispositivewhen M1 and M2 havethesamesign (reversecurvature bending) and negativewhen they areof oppositesign (singlecurvaturebending). When the bending moment at any point within an unbraced length is larger than that at

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    both endsof thislength, and for memberssubject to combined compressiveaxial load and bendingmoment, CT F shall betaken asunity.

    r0 = Polar radiusof gyration of thecross-section about theshear center

    = r2x + r2y + x20rx , ry = radii of gyration of thecross-section about thecentroidal principal axesG = shear modulusKx , Ky , Kt = effectivelength factorsfor bending about thex- andy-axes, and for twistingLx , Ly , Lt = unbraced length of compression member for bendingabout thex- andy-axes, and

    for twistingx0 = distancefrom theshear center to thecentroid along theprincipal x-axis, taken as

    negativeJ = St. Venant torsion constant of thecross-sectionCw = torsional warpingconstant of thecross-sectionj = [

    A

    x3dA +

    Axy 2dA]/(2Iy ) x0

    (b) For I - or Z-sectionsbent about the centroidal axisperpendicular to theweb (x-axis):

    In lieu of (a), thefollowingequationsmay beused to evaluate Me:

    Me = 2ECbdIyc /L2 for doubly-symmetr ic I-sections (7.25)Me = 2ECbdIyc /(2L2) for point-symmetr ic Z-sections (7.26)

    In Equations7.25and 7.26,

    d = depth of sectionE = modulusof elasticityIyc = moment of inert ia of the compression portion of a section about the gravity axis of the

    entiresection parallel to theweb,using the full unreduced sectionL = unbraced length of themember

    EXAMPLE7.4:

    Determinethenominal moment for lateral bucklingstrength for theI -beam used in Example7.2.Assumethat thebeam isbraced laterally at both endsand midspan. UseFy = 50ksi.

    Solution

    1. Calculation of Sectional Propert ies

    Based on thedimensionsgiven in Example7.2 (Figures7.19 and 7.20), themoment of inertia, Ix ,and thesection modulus, Sf, of thefull section can becomputed asshown in thefollowing table.

    Distancefrom

    Area A mid-depth y Ay2

    Element (in.2) (in.) (in.4)

    Flanges 4(1.6775)(0.135) = 0.9059 3.9325 14.0093Corners 4(0.05407) = 0.2163 3.8436 3.1955Webs 2(7.355)(0.135) = 1.9859 0 0Total 3.1081 17.2048

    2 Iweb = 2(1/12)(0.135)(7.355)3 = 8.9522Ix = 26.1570in.4

    Sf = Ix /(8/2) = 6.54in.3

    ThevalueofIyc can becomputed asshown below.

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    Distance

    AreaA from y axis, x Ax2

    Element (in.2) (in.) (in.4)

    Flanges 4(1.6775)(0.135) = 0.9059 1.1613 1.2217Corners 4(0.05407)

    =0.2163 0.1564 0.0053

    Webs 2(7.355)(0.135) = 1.9859 0.0675 0.0090Total 3.1081 1.2360

    Iflanges = 4(1/12)0.135(1.6775) 3 = 0.2124Iy = 1.4484 in.4

    Iyc = Iy/2 = 0.724in.4

    Considering the lateral supportsat both endsand midspan, and the moment diagram shown inFigure7.25, thevalueofCb for the segment AB or BC is 1.30according to Equation 7.24. Using

    FIGURE7.25: Example 7.4. (From Yu, W.W. 1991. Cold-Formed Steel Design, John Wiley & Sons,New York. With permission.)

    Equation 7.25,

    Me = 2ECbdI

    ycL2

    = 2(29,500)(1.30) (8)(0.724)(5 12)2 = 608.96 in.-kips

    My = SfFy = (6.54)(50) = 327.0 in.-kips0.56My = 183.12 in.-kips2.78My = 909.06 in.-kips

    Since2.78My > Me > 0.56My , from Equation 7.22,

    Mc =10

    9My

    1 10My

    36Me

    = 109

    (327.0)

    1 10(327.0)36(608.96)

    = 309.14 in.-kips

    Based on Equation 7.20, thenominal moment for lateral bucklingstrength is

    Mn = ScMc

    Sf

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    in which Sc istheelasticsection modulusof theeffectivesection calculated at acompressivestressoff = Mc/Sf = 309.14/6.54 = 47.27 ksi. By using the sameprocedure il lustrated in Example7.2,Sc = 6.295 in.3. Therefore, thenominal moment for lateral bucklingstrength is

    Mn = (6.295)

    309.146.54

    = 297.6 in.-kipsFor channelsor Z-sectionshavingthetension flangethrough-fastenedtodeckor sheathingwith the

    compression flangelaterally unbraced and loaded in aplaneparallel to theweb, thenominal flexuralstrength isdetermined by Mn = RSeFy , where R is a reduction factor [ 7]. A similar approach isused for beamshavingoneflangefastened to astanding seam roof system.

    Unusuall y Wide Beam Flanges and Short Span Beams

    When beam flangesareunusuallywide, special considerationshould begiven to thepossibleeffectsof shear lagandflangecurling. Shear lagdependson thetypeof loadingand thespan-to-width ratioand isindependent of thethickness. Flangecurlingisindependent of span length but dependson thethicknessand width of theflange, thedepth of thesection, and thebendingstressesin both tension

    and compression flanges.In order to consider the shear lag effects, the effectivewidths of both tension and compressionflangesshould beused according to theAISI Specification.

    When astraight beam with unusually wide and thin flangesis subject to bending, the portion ofthe flange most remote from the web tends to deflect toward the neutral axis dueto the effect oflongitudinal curvatureof thebeam and theapplied bendingstressesin both flanges. For thepurposeof controlling the excessiveflangecurling, theAISI Specification providesan equation to l imit theflangewidth.

    ShearStrength

    Theshear strength of beam websisgoverned byeither yieldingor bucklingof theweb element,depending on the depth-to-thickness ratio, h/t, and the mechanical propert iesof steel. For beamwebshaving small h/t ratios, the nominal shear strength isgoverned by shear yielding. When the

    h/t ratio is large, thenominal shear strength is controlled byelastic shear buckling. For beam webshavingmoderate h/t ratios, the shear strength isbased on inelastic shear buckling.

    For the design of beam webs, the AISI Specification provides the following equations for deter-mining the nominal shear strength:

    For h/t 0.96

    Ekv/Fy :

    Vn = 0.60Fy ht (7.27)For 0.96

    Ekv/Fy < h/t 1.415

    Ekv/Fy :

    Vn = 0.64t2

    kv Fy E (7.28)

    For h/t > 1.415Ekv /Fy :Vn = 2Ekvt3/[12(1 2)h] = 0.905Ekvt3/ h (7.29)

    whereVn = nominal shear strength of beamh = depth of theflat portion of theweb measured alongtheplaneof thewebt = web thicknesskv = shear bucklingcoefficient determined asfollows:

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    1. For unreinforced webs, kv = 5.342. For beam webswith transversestif fenerssatisfying the AISI requirements

    when a/ h 1.0:

    kv = 4.00 +5.34

    (a/h)2

    when a/h > 1.0:

    kv = 5.34 +4.00

    (a/h)2

    where

    a = the shear panel length for unreinforced web element= the clear distancebetween transversesti ffeners for reinforced web elements

    For a web consistingof two or moresheets,eachsheet should beconsideredasaseparateelementcarrying itsshareof theshear force.

    CombinedBendingandShear

    For continuous beamsand cantilever beams, high bending stressesoften combine with highshear stresses at the supports. Such beam webs must be safeguarded against buckling due to thecombination of bending and shear stresses. Based on theAISI Specification, the moment and shearshould satisfy theinteraction equationslisted in Table7.3.

    TABLE7.3 Interaction EquationsUsed for CombinedBending and ShearASD LRFD

    Beamswith unreinforcedwebs

    M

    Maxo

    2 + VVa

    2 1.0 Mub Mnxo

    2 + Vuv Vn

    2 1.0(7.30) (7.32)

    Beams with transverseweb sti ffeners

    M Ma and V Va Mu b Mn and Vu v Vn

    0.6 MMaxo

    + VVa 1.3 0.6Mu

    bMnxo+

    Vu

    v Vn 1.3

    (7.31) (7.33)

    M = bending momentMa = allowablemoment when bending aloneexistsMaxo = allowablemoment about thecentroidal x-axisdetermined in accordancewith thespec-

    ification excluding theconsideration of lateral bucklingV = unfactored shear forceVa = allowableshear forcewhen shear aloneexistsb = resistancefactor for bendingv = resistancefactor for shearMn = nominal flexural strength when bending aloneexistsMnxo = nominal flexural strength about the centroidal x-axis determined in accordance with

    the specification excluding theconsideration of lateral bucklingMu = required flexural strengthVn = nominal shear strength whenshear alone existsVu = requiredshear strength

    WebCrippling

    For cold-formed steel beams, transversesti ffenersarenot frequently used for beam webs. Thewebsmay cripple due to the high local intensity of the load or reaction as shown in Figure 7.26.Becausethe theoretical analysis of web crippling is rather complex dueto the involvement of manyfactors, the present AISI design equationsare based on extensiveexperimental investigationsunderfour loading conditions: (1) end one-flange (EOF) loading, (2) interior one-flange (IOF) loading,

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    FIGURE7.26: Web crippling of cold-formed steel beams.

    (3) endtwo-flange(ETF) loading, and (4) interior two-flange(ITF) loading[ 29, 46, 50]. Theloadingconditionsused for thetestsare illustrated in Figure7.27.

    FIGURE7.27: Loadingconditionsfor webcrippling tests. (a) EOFloading. (b) IOFloading. (c) ETFloading. (d) ITFloading. (From Yu, W.W. 1991. Cold-Formed Steel Design, John Wiley & Sons, NewYork. With permission.)

    The nominal web crippling strength for a given loading condition can bedetermined from theAISI equations[7] on thebasisof thethicknessof web element, design yield stress, thebend radius-to-thicknessratio, thedepth-to-thicknessratio, thebearing length-to-thicknessratio, and theanglebetween theplane of theweb and the plane of thebearing surface. Tables 7.4a and Table 7.4b list

    the equationsfor determining the nominal web crippling strengths of one- and two-flange loadingconditions, respectively.

    CombinedBendingandWebCrippling

    For combined bending and web crippling, the design of beam websshould be based on theinteraction equations provided in the AISI Specification [ 7] . These equations are presented inTable7.5.

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    Table7.4a Nominal Web CripplingStrength for One-FlangeLoading, per Web,Pn

    Shapes having Shapes having I-Sections orsinglewebs singlewebs similar sections

    Stiffened or Stiffened,partially partially stiffened,stiffened Unstiffened and unstiffenedflanges flanges flanges

    Endreaction t2kC3C4C[331 0.61(h/t)] t2kC3C4C[217 0.28(h/t)] t2Fy C6(10.0 + 1.25

    N/t)

    opposing loads [1 + 0.01(N/t)]C9 a [1 + 0.01(N/t)]C9 a,bspaced > 1.5h

    Interior reactions t2kC1C2C[538 0.74(h/t)] t2kC1C2C[538 0.74(h/t)] t2Fy C5(0.88 + 0.12m)opposing loads [1 + 0.007(N/t)]C9 c [1 + 0.007(N/t)]C9 c (15.0 + 3.25

    N/t)

    spaced > 1.5h

    a When Fy 66.5 ksi (459 MPa), thevalueofKC3 shall betaken as1.34.b When N/t > 60, thefactor [1 + 0.01(N/t)] may beincreased to [0.71 + 0.015(N/t)]c When N/t > 60, thefactor [1 + 0.007(N/t)] may beincreased to [0.75 + 0.011(N/t)]

    C1 = (1.22 0.22k)C2

    =(1.06

    0.06R/t)

    1.0

    C3 = (1.33 0.33k)C4 = (1.15 0.15R/t) 1.0 but not lessthan 0.50C5 = (1.49 0.53k) 0.6C6 = 1 + (h/t)/750 when h/t 150

    = 1.20, when h/t > 150C9 = 1.0 for U.S. customary units, kipsand in.

    = 6.9 for SI units,N and mmC = 0.7 + 0.3(/90)2Fy = design yield stressof the webh = depth of theflat portion of the web measured alongtheplaneof thewebk = 894Fy /Em = t/0.075, when t isinin.m = t/1.91, when t isinmmt = web thicknessN = actual length of bearingR = insidebend radius = anglebetween the planeof the web andthe planeof the bearing surface 45, but not morethan 90

    Table7.4b Nominal Web CripplingStrength for Two-FlangeLoading, per Web, Pn

    Shapes havi ng Shapes havi ng I -Secti ons orsinglewebs single webs similar sections

    Stiffened or Stiffened,partially partially stiffened,stiffened Unstiffened and unstiffenedflanges flanges flanges

    Endreaction t2kC3C4C[244 0.57(h/t)] t2kC3C4C[244 0.57(h/t)] t2Fy C8(0.64 + 0.31m)opposing loads [1 + 0.01(N/t)]C9 a [1 + 0.01(N/t)]C9 a (10.0 + 1.25

    N/t)

    spaced 1.5hInterior reaction t2kC1C2C[771 2.26(h/t)] t2kC1C2C[771 2.26(h/t)] t2Fy C7(0.82 + 0.15m)opposing loads [1 + 0.0013(N/t)]C9 [1 + 0.0013(N/t)]C9 (15.0 + 3.25

    N/t)

    spaced 1.5ha When Fy 66.5 ksi (459 MPa), thevalueofKC3 shall betaken as1.34.C7 = 1/k , when h/t 66.5

    = [1.10 (h/t)/665]/k , when h/t > 66.5C8 = [0.98 (h/t)/865]/kC1, C2, C3, C4, C9, C, Fy , h , k , m , t , N , R, and aredefined in Table7.4a.

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    TABLE7.5 Interaction Equationsfor Combined Bendingand Web CripplingASD LRFD

    Shapes having single unrein-forced webs

    1.2

    PPa

    +

    MMaxo

    1.5 1.07

    Pu

    w Pn

    +

    MubMnxo

    1.42

    (7.34) (7.37)

    Shapeshavingmultipleunrein-forced webssuch

    1.1

    PPa

    +

    MMaxo

    1.5 0.82

    Pu

    w Pn

    +

    MubMnxo

    1.32

    asI-sections (7.35) (7.38)

    Support point of two nestedZ-shapes

    MMno

    + PPn

    1.0 MuMno

    + PuPn

    1.68

    (7.36) (7.39)

    Note: The AISI Specification includessomeexception clauses, under which the effect of combined bendingand web crippli ngneed not bechecked.

    P = concentrated loador reaction in the presenceof bending momentPa = allowableconcentrated loador reaction in the absenceof bending momentPn = nominal web crippli ng strength for concentrated load or reaction in the absenceof bending

    moment ( for Equations7.34, 7.35,7.37,and 7.38)Pn = nominal web crippli ngstrength assuming single web interior one-flangeloading for thenested

    Z-sections,i.e., sum of the two websevaluated individually (for Equations7.36and 7.39)Pu = required strength for theconcentrated loador reaction in the presenceof bending momentM = applied bending moment at, or immediately adjacent to,thepoint of application of theconcen-

    trated loador reactionMaxo = allowablemoment about thecentroidal x-axisdetermined in accordancewith the specificationexcluding the consideration of lateral buckling

    Mno = nominal fl exural strength for the nested Z-sections, i .e., thesum of the two sections evaluatedindividually,excluding lateral buckling

    Mnxo = nominal flexural strength about the centroidal x-axisdetermined in accordancewi th thespeci-fication excluding theconsideration of lateral buckling

    Mu = required flexural strength at,or immediately adjacent to,thepoint ofapplication of theconcen-trated loador reaction Pu

    = resistancefactor = 0.9b = resistancefactor for bendingw = resistancefactor for web crippling

    BracingRequirements

    In cold-formed steel design, bracesshould bedesigned to restrain lateral bending or twistingof a loaded beam and to avoid local crippling at the points of attachment. When channels andZ-shaped sectionsare used asbeamsand loaded in theplaneof theweb, the AISI Specification [7]providesdesign requirements to restrain twisting of thebeam under the following two conditions:(1) the top flange is connected to deck or sheathing material in such a manner as to effectivelyrestrain lateral deflection of the connected flange, and (2) neither flange isconnected to sheathing.In general, bracesshould bedesigned to satisfy thestrength and stiffnessrequirements. For beamsusing symmetr ical crosssections, such asI-beams, theAISI Specification doesnot providespecificrequirements for braces. However, the braces may bedesigned for a capacity of 2% of the forceresisted by the compression portion of the beam. This is a frequently used rule of thumb but is aconservativeapproach,asproven byarigorousanalysis.

    7.6.5 Concentrically LoadedCompression Members

    Axially loaded cold-formed steel compression members should be designed for the followinglimit states: (1) yielding, (2) overall column buckling (flexural buckling, torsional buckling, ortorsional-flexural buckling), and (3) local buckling of individual elements. The governing failuremodedependson theconfiguration of thecross-section, thicknessof material, unbraced length, andend restraint.

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    Yielding

    A very short, compact column under axial load may fail byyielding. For thiscase, thenominalaxial strength is theyield load, i.e.,

    Pn = Py = AFy (7.40)whereA is the full cross-sectional areaof thecolumn and Fy is theyield point of steel.

    Overall ColumnBuckling

    Overall column buckling may beoneof thefollowingthreetypes:1. Flexural buckling bending about aprincipal axis. Theelastic flexural buckling stressis

    Fe = 2E

    (KL/r)2(7.41)

    whereE

    =modulusof elasticity

    K = effectivelength factor for flexural buckling(Figure7.28)L = unbraced length of member for flexural bucklingr = radiusof gyration of thefull section

    FIGURE7.28: Effective length factor K for concentrically loaded compression members.

    2. Torsional buckling twisting about shear center. Theelastic torsional bucklingstressis

    Fe =1

    Ar20

    GJ +

    2ECw

    (KtLt)2

    (7.42)

    whereA = full cross-sectional area

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    Cw = torsional warping constant of thecross-sectionG = shear modulusJ = St. Venant torsion constant of thecross-sectionKt

    =effectivelength factor for twisting

    Lt = unbraced length of member for twistingr0 = polar radiusof gyration of thecross-section about shear center3. Torsional-flexural buckling bendingand twistingsimultaneously. Theelastictorsional-flexuralbucklingstressis

    Fe = [(ex + t)

    (ex + t)2 4ex t]/2 (7.43)where = 1(x0/r0)2ex = 2E/(Kx Lx /rx )2t = thesameasEquation 7.42x0 = distancefrom shear center to thecentroid along theprincipal x-axis

    For doubly-symmetr ic and point-symmetr ic shapes (Figure 7.29), the overall column bucklingcan beeither flexural typeor torsional type. However, for singly-symmetric shapes(Figure7.30),theoverall column bucklingcan beeither flexural bucklingor torsional-flexural buckling.

    FIGURE7.29: Doubly-symmetric shapes.

    FIGURE 7.30: Singly-symmetric shapes. (From Yu, W.W. 1991. Cold-Formed Steel Design, JohnWiley & Sons, New York. With permission.)

    For overall column buckling, thenominal axial strength is determined byEquation 7.44:

    Pn = AeFn (7.44)

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    where

    Ae = effectiveareadetermined for thestressFnFn = nominal buckling stressdetermined asfollows:

    For c 1.5:Fn = (0.658

    2c )Fy (7.45)

    For c > 1.5:

    Fn =

    0.877

    2c

    Fy (7.46)

    Theuse of the effectivearea Ae in Equation 7.44 is to reflect the effect of local buckling on thereduction of column strength. In Equations7.45 and 7.46,

    c =

    Fy /Fe

    in which Fe is the least of elastic flexural buckling stress (Equation 7.41), torsional buckling stress(Equation 7.42), and torsional-flexural bucklingstress(Equation 7.43), whichever is applicable.

    For the design of compression members, the slendernessratio should not exceed 200, except thatduringconstruction, KL/r preferably should not exceed 300.

    For nonsymmetricshapeswhosecross-sectionsdo not haveanysymmetry, either about an axisorabout apoint, theelastic torsional-flexural bucklingstressshould bedetermined byrational analysisor by tests. SeeAISI Design Manual [8].

    In addition to the above design provisions for the design of axially loaded columns, the AISISpecification also provides design cri teria for compression members having one flange through-fastened to deck or sheathing.

    EXAMPLE7.5:

    Determinethe allowableaxial load for the square tubular column shown in Figure7.31. Assumethat Fy = 40 ksi, Kx Lx = Ky Ly = 10ft, and the dead-to-live load ratio is1/5. UsetheASD andLRFD methods.

    FIGURE7.31: Example 7.5. (From Yu, W.W. 1991. Cold-Formed Steel Design, John Wiley & Sons,New York. With permission.)

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    Solution

    (A) ASD MethodSince the square tube is a doubly-symmetr ic closed section, it will not be subject to torsional-

    flexural buckling. It can bedesigned for flexural buckling.

    1. Sectional Propert iesof Full Section

    w = 8.00 2(R + t) = 7.415 in.A = 4(7.415 0.105 + 0.0396) = 3.273 in.2Ix = Iy = 2(0.105)[(1/12)(7.415)3 + 7.415(4 0.105/2)2] + 4(0.0396)(4.0 0.1373)2

    = 33.763 in.4rx = ry =

    Ix /A =

    33.763/3.273 = 3.212 in.

    2. Nominal Buckling Stress, Fn. According to Equation 7.41, the elastic flexural bucklingstress, Fe, is computed asfollows:

    KL

    r= 10 12

    3.212= 37.36 < 200 O.K.

    Fe = 2E

    (KL/r)2=

    2(29500)

    (37.36)2= 208.597 ksi

    c =

    Fy

    Fe=

    40

    208.597= 0.438 < 1.5

    Fn = (0.6582c )Fy = (0.6580.438

    2

    )40 = 36.914 ksi

    3. Effective Area, Ae. Because the given square tube is composed of four sti ffened ele-ments, theeffectivewidth of sti ffened elementssubjected to uniform compression can be

    computed from Equations7.4through 7.7byusing k = 4.0:

    w/t = 7.415/0.105 = 70.619

    = 1.052k

    wt

    FnE

    = 1.052/

    4(70.619)

    36.914/29,500 = 1.314

    Since > 0.673, from Equation 7.5,

    b

    =w

    where

    = (1 0.22/)/ = (1 0.22/1.314)/1.314 = 0.634Therefore, b = (0.634)(7.415) = 4.701 in.Theeffectiveareais

    Ae = 3.273 4(7.415 4.701)(0.105) = 2.133 in.2

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    4. Nominal and AllowableLoads. UsingEquation 7.44, thenominal load is

    Pn = AeFn = (2.133)(36.914) = 78.738 kipsTheallowableload is

    Pa = Pn/ c = 78.738/1.80 = 43.74 kips(B) LRFD MethodIn Item (A) above, the nominal axial load, Pn, wascomputed to be78.738kips. The design axial

    load for theLRFD method is

    cPn = 0.85(78.738) = 66.93 kips.Based on theload combination of dead and liveloads, therequired axial load is

    Pu = 1.2PD + 1.6PL = 1.2PD + 1.6(5PD) = 9.2PDwherePD

    =axial load dueto dead load

    PL = axial load dueto liveloadBy using Pu = cPn, thevaluesofPD and PL arecomputed asfollows:

    PD = 66.93/9.2 = 7.28 kipsPL = 5PD = 36.40 kips

    Therefore, theallowableaxial load is

    Pa = PD + PL = 43.68 kipsIt can be seen that the allowable axial loads determined by the ASD and LRFD methods are

    practically the same.

    7.6.6 CombinedAxial LoadandBending

    TheAISI Specification providesinteraction equationsfor combined axial load and bending.

    CombinedTensileAxial LoadandBending

    For combinedtensileaxial load and bending, therequired strengthsshould satisfy theinterac-tion equationspresented in Table 7.6. Theseequationsare to prevent yielding of thetension flangeand to prevent failureof the compression flangeof the member.

    CombinedCompressiveAxial LoadandBending

    Cold-formed steel members under combined compressive axial load and bending are usuallyreferred to as beam-columns. Such members are often found in framed structures, trusses, andexterior wall studs. For thedesign of thesemembers, the required strengthsshould satisfy the AISI

    interaction equationspresented in Table7.7.

    7.6.7 Cylindrical Tubular Members

    Thin-walled cylindrical tubular members are economical sections for compression and torsionalmembers because of their large ratio of radius of gyration to area, the same radiusof gyration inall directions, and the largetorsional rigidity. TheAISI design provisionsare limited to theratio ofoutsidediameter-to-wall thickness, D/t, not greater than 0.441E/Fy .

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    TABLE7.6 Interaction Equationsfor CombinedTensileAxial Load and BendingASD LRFD

    Check tension flangeb MxMnxt

    + b MyMnyt

    + tTTn

    1.0 Muxb Mnxt

    + Muyb Mnyt

    + TutTn

    1.0(7.47) (7.49)

    Check compression flange b MxMnx + b MyMny + tTTn 1.0 Muxb Mnx + Muyb Mny TutTn 1.0(7.48) (7.50)

    Mnx , Mny = nominal flexural strengths about the centroidal x- and y-axesMnxt, Mnyt = Sf tFyMux , Muy = required flexural strengthswi th respect to the centroidal axesMx , My = required moments with respect to thecentroidal axesof t hesectionSf t = section modulusof the full section for theextremetension fiber about theappropriateaxisT = required tensileaxial loadTn = nominal tensileaxial strengthTu = requiredaxial strengthb = resistancefactor for bendingt = resistancefactor for tensionb = safety factor for bendingt = safety factor for tension

    TABLE7.7 Interaction Equationsfor CombinedCompressiveAxial

    Load and BendingASD LRFD

    when cP /Pn 0.15, when Pu/cPn 0.15,

    c PPn

    + b MxMnx

    + b MyMny

    1.0 Puc Pn

    + Muxb Mnx

    + Muyb Mny

    1.0(7.51) (7.54)

    when cP /Pn > 0.15, when Pu/cPn > 0.15,c P

    Pn+ b Cmx Mx

    Mnx x+ b CmyMy

    Mnyy 1.0 Pu

    c Pn+ Cmx Mux

    b Mnx x+ CmyMuy

    b Mnyy 1.0

    (7.52) (7.55)c PPno

    + b MxMnx

    + b MyMny

    1.0 Puc Pno

    + Muxb Mnx

    + Muyb Mny

    1.0(7.53) (7.56)

    Mx , My = required moments with respect to thecentroidal axesof theeffectivesection determined for therequired axial strength alone

    Mnx , Mny = nominal flexural strengthsabout thecentroidal axesMux , Muy = required flexural strengths with respect to the centroidal axesof the

    effectivesection determinedfor the required axial strength aloneP = required axial loadPn = nominal axial strength determinedin accordancewith Equation 7.44Pno = nominalaxial strengthdeterminedin accordancewith Equation7.44,

    for Fn = FyPu = required axial strengthx = 1 cP /PEX (for Equation 7.52)y = 1 cP /PEY (for Equation 7.52)x = 1 Pu/PEX (for Equation 7.55)y = 1 Pu/PEY (for Equation 7.55)PEX = 2EIx /(Kx Lx )2PEY = 2EIy /(Ky Ly )2b = safety factor for bendingc = safety factor for concentr ically loaded compressionCmx , Cmy = coefficientswhosevalueshall betaken asfollows:1. For compression membersin framessubject to joint translation (sidesway) Cm = 0.852. For restrained compression members in f ramesbraced against joint t ranslation and

    not subject to transverseloading between their supports in theplaneof bending Cm =0.6

    0.4 (M1/M2), where M1/M2 is theratio of thesmaller to thelarger moment at

    the endsof that port ion of the member under consideration which is unbraced in theplaneof bending. M1/M2 is posit ive when the member is bent in reversecurvatureand negativewhen it isbent in singlecurvature

    3. For compression members in frames braced against joint translation i n the planeofloadingand subject to transverseloadingbetween their supports, thevalueofCm maybe determined by rational analysis. However, i n l ieu of such analysis, the followingvaluesmay be used: (a) for members whose endsare restrained, Cm = 0.85; (b) formemberswhoseendsareunrestrained, Cm = 1.0

    Ix , Iy , Lx , Ly , Kx , and Ky were defined previously.

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    BendingStrength

    For cylindrical tubular members subjected to bending, the nominal flexural strengths are asfollowsaccordingto theD/t ratio:

    1. For D/t

    0.070E/Fy :

    Mn = 1.25Fy Sf (7.57)2. For 0.070E/Fy < D/t 0.319E/Fy :

    Mn = [0.970 + 0.020(E/Fy )/(D/t)]Fy Sf (7.58)

    3. For 0.319E/Fy < D/t 0.441E/Fy :

    Mn = [0.328E/(D/t)]Sf (7.59)

    whereSf = elastic section modulusof the full, unreduced cross-section

    Other symbolsweredefined previously.

    CompressiveStrength

    When cylindrical tubesareused asconcentrically loaded compression members, the nominalaxial strength is determined by Equation 7.44, except that (1) the elastic buckling stress, Fe, isdetermined for flexural bucklingby using Equation 7.41 and (2) theeffectivearea, Ae, is calculatedbyEquation 7.60.

    Ae = [1 (1 R2)(1 A0/A)]A (7.60)whereR =

    Fy /2Fe

    A0 = {0.037/[(DFy )/(tE)] + 0.667}A AA = areaof theunreduced cross-section

    In theaboveequations, thevalueA0 is thereduced areadueto theeffect of local buckling [8, 49].

    7.7 ConnectionsandJoints

    Welds,bolts,screws,rivets,andother specialdevicessuchasmetalstitchingandadhesivesaregenerallyused for cold-formed steel connections. TheAISI Specification contains only the design provisionsfor welded connections, bolted connections, and screw connections. These design equations arebased primari ly on theexperimental data obtained from extensivetest programs.

    7.7.1 WeldedConnections

    Welds used for cold-formed steel construction may beclassified asarc welds (or fusion welds) and

    resistancewelds. Arcweldingisusuallyused for connectingcold-formed steel membersto eachotheraswell asconnecting such thin members to heavy, hot-rolled steel framing members. I t isused forgroovewelds, arc spot welds, arc seam welds, fil let welds, and flare groove welds. The AISI designprovisions for welded connectionsareapplicableonly for cold-formed steel structural members, inwhich the thicknessof the thinnest connected part is 0.18 in. (4.57 mm) or less. Otherwise, whenthe thicknessof connected parts is thicker than 0.18 in. (4.57 mm), thewelded connection shouldbedesigned accordingto theAISCSpecifications[1, 2]. Additional design informationonstructuralweldingof sheet steelscan also befound in theAWSCode[ 16].

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    Arc Welds

    According to the AISI Specification, the nominal strengths of arc welds can be determinedfrom theequationsgiven in Table7.8. Thedesign strengthscan then becomputed byusing thesafetyfactor or resistancefactor provided in Table 7.1.

    TABLE7.8 Nominal Strength Equationsfor Arc WeldsTypeof weld Type of strength Nominal strength Pn (kips)

    Groove Tension or compression LteFywelds Shear strength of weld Lte(0.6Fxx )

    (Figure7.32) Shear strength of connected part Lte(Fy /

    3)

    Arc spot Shear strengthwelds Strength of weld 0.589 d2e Fxx

    (Figure7.33) Strength of connected part1. da /t 0.815

    E/Fu 2.20tda Fu

    2. 0.815

    E/Fu < (da /t) < 1.397

    E/Fu 0.28[1 + (5.59

    E/Fu)/(da /t )]tda Fu3. da /t

    1.397

    E/Fu 1.40tda Fu

    Shear strength of connected partbased on end distance eFutTensile strengthStrength of weld 0.785d2e FxxStrength of connected part1.Fu/E < 0.00187 [6.59 3150(Fu/E)](tda Fu)

    1.46tda Fu2. Fu/E 0.00187 0.70tda Fu

    Arc seam Shear strength [d2e /4 + Lde]0.75Fxxwelds Strengthof connectedpar t 2.5tFu(0.25L + 0.96da )

    (Figure7.34)

    Fillet welds Shear strength of weld 0.75tw LFxx(Figure7.35) (for t > 0.15in.)

    Strength of connected part1. Longitudinal loading

    L/t < 25: [1(0.01L/t)]tLFuL/t 25: 0.75tLFu

    2. Transverseloading tLFuFlaregroovewelds

    Shear strength of weld 0.75t

    wLF

    xx(Figure7.36) (for t > 0.15in.) Strength of connected part1. Transverse loading 0.833tLFu2. Longitudinal loadingFor t tw < 2t or if lip height < L 0.75tLFuFor tw 2 t and lip height L 1.50tLFu

    d = visiblediameter of outer surfaceof arc spot weldda = averagediameter of thearcspot weld at mid-thicknessoftda = (d t) for singlesheetda = (d2 t) for multiplesheetsde = effectivedi ameter of fusedareaat planeof maximum shear transferde = 0.7d 1.5 t 0.55de = distancemeasured in thelineof forcefromthecenterlineof aweld to thenearest edgeof an adjacent

    weld or to theendof theconnected part toward whichtheforceis directedFu = tensilestrength of the connected partFy = yield point of steelFxx = fil ler metal strength designation in AWSelectrodeclassificationL = length of weldPn = nominal strength of weldt = thi cknessof connectedsheettw = effectivethroat dimension for grooveweld, seeAI SI specificationtw = effectivethroat = 0.707w1 or 0.707 w2, whichever is smallerw1 = legof weldw2 = legof weld

    SeeAI SI Specification for additi onal design information.

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    ResistanceWelds

    The nominal shear strengths of resistance welds are provided in the AISI Specification [7]according to the thicknessof the thinnest outside sheet. Theyareapplicable for all structural gradesof low-carbon steel, uncoated or galvanized with 0.9 oz/ft2 of sheet or less, and medium carbon and

    low-alloysteels.

    7.7.2 BoltedConnections

    Due to the thinnessof the connected parts, the design of bolted connections in cold-formed steelconstruction is somewhat different from that in hot-rolled heavy construction. The AISI designprovisionsareapplicableonly to cold-formed membersor elements lessthan 3/16in. (4.76 mm) inthickness. For materialsnot lessthan3/16in. (4.76mm), thebolted connection should bedesignedin accordancewith theAISC Specifications[ 1, 2].

    In the AISI Specification, five types of bolts (A307, A325, A354, A449, and A490) are used forconnectionsin cold-formed steel construction, in which A449 and A354 boltsshould beused asanequivalent of A325 and A490 bolts, respectively, whenever boltswith smaller than 1/2-in. diameters

    arerequired.On the basis of the failure modes occurring in the tests of bolted connections, the AISI criteriadeal wi th threemajor design considerations for the connected parts: (1) longitudinal shear failure,(2) tensilefailure,and (3) bearing failure. Thenominal strength equationsaregiven in Table7.9.

    TABLE7.9 Nominal Strength Equationsfor Bolted ConnectionsType of strength Nominal strength, Pn

    Shear strength based on spacingandedge teFudistance

    Tensile strength1. Withwashersunder bolt head andnuts (1 0.9 r + 3rd/s)FuAn FuAn2. Nowashersor onlyonewasher under (1r+ 2.5 rd/s)FuAn FuAn

    bolt head and nutsNote: Thetensilestrength computedabove

    should not exceedAnFy .Bearingstrength

    1. With washers under bolt head andnut Insidesheet of double shear connection

    Fu/Fy 1.08 3.33FudtFu/Fy < 1.08 3.00Fudt

    Singleshear andoutsidesheetsof doubleshear 3.00Fudtconnection

    2. Without washers under bolt headand nutor with only onewasher

    Insidesheet of doubleshear connection 3.00Fudt Singleshear andoutsidesheetsof doubleshear 2.22Fudt

    connection

    An = net areaof t heconnected partd = diameter of bolte = distancemeasured in the line of force from thecenter of bolt to the nearest edgeof

    an adjacent holeor to theend of the connected partFu = tensilestrength of the connected partFy = specified yield point of steelr = forcetransmittedby thebolt or boltsat thesectionconsidered,divided bythetension

    forcein themember at that section. I fr is lessthan 0.2, it may betaken equal tozero

    s = spacing of boltsperpendicular to lineof forcet = thi cknessof thinnest connectedpart

    In addition, design strength equationsare provided for shear and tension in bolts. Accordi