TECHNICAL NOTE On Cold-Formed Steel Construction Ligt Gauge Steel Engineers Association Wasington, D.C. Toll-Free: 1 (866) 465-4732 www.lgsea.com $5.00 DESIGN CONSIDERATIONS FOR FLEXURAL AND LATERAL-TORSIONAL BRACING SUMMARY: Load bearing cold-formed/ligt gauge steel (CFS/LGS) framed walls are typically designed for a combination of axial and lateral out-of-plane (flexural) loading. Under tis loading condition, common C-section studs may be susceptible to local, torsional, flexural, torsional-flexural, lateral-torsional or distortional buckling. Te response performance of te stud depends on a number of parameters most notably ow it is supported along its lengt (including its ends), te relative magnitudes of te applied loads and te distribution of tese loads. Tis tecnical note discusses te beavior of te typical wall stud and provides some practical considerations for design of torsional-flexural and lateral-torsional bracing. Recommendations and considerations suggested in tis tecnical note are done in accordance wit acceptable practices and existing design documents. Introduction: Te response of any structural member depends on, but is not limited to, te loads applied to or induced in te member, te member support conditions (along te lengt of te member), te cross-section configuration and te members material properties. In cold-formed/ligt gauge steel design, two of te most often realized loading conditions for framing members involve a combination of axial and bending (flexural) loads, or bending only. Current design guidelines for members under tese loading conditions require, initially, separate analysis for eac type of loading (axial and bending) separately. If overall buckling of te member is inibited, failure of te section will be governed by te local and possibly distortional buckling beavior of te crosssectional elements. Assuming, neiter local or distortional buckling occurs, Figure 1 illustrates te teoretical overall beavior of te common C- sections under concentric axial compression and under bending. Te more likely response of te loaded C-section will involve a coupling of te local, distortional and overall buckling modes. Tus, it can be appreciated tat te resulting response of te C-section member becomes relatively complex. Tis complex beavior is usually treated using some form of an interaction equation. Overall failure modes for CFS/LGS C-section members in axial compression or bending Figure 1 Ligt Gauge Steel Engineers Association 1 TECH NOTE (559) 6/01
Te 1996 AISI Specification for te Design of Cold-Formed Steel Structural Members, Section D4 (and associated Sections C3, C4 and C5), provides a basis for evaluating te strengt of wall studs subjected to axial and bending, and combined axial and bending loads. Te design equations in Section D4 for estimating te stud capacity are based on te assumption tat seating is attaced to bot flanges of te stud. Section D4, owever, also states tat te effect of seating attaced to te wall stud may be neglected in te calculation of te stud capacity. Tis allowance implies tat te stud capacity may be computed on te basis of te requirements in Section C, wit due consideration for te use of mecanical bracing. Te nominal capacities of a member will depend on te unbraced lengt of tat member and te distribution of load (axial, sear and moment) in tat unbraced lengt. Te unbraced lengt is usually taken as te member lengt between brace points, per Section D3 of te AISI Specification. Te effective unbraced lengt depends on te degree of rotational and lateral restraint (simple, partial or full) provided by te brace point support. Te common bracing scemes currently used in residential and commercial construction are illustrated in Figure 2. As sown in tis figure, mecanical bracing may be located at Alternative bracing scemes for walls studs Figure 2 discrete points (strap and blocking or bridging) or it can be more uniformly distributed as is te case wit seating attaced to te framing members wit fasteners. Wen seating is used as bracing wall members it is important tat due consideration be given to te designed function of tat wall. Specifically, if te load-bearing wall also serves as a sear wall, te seating sould not be relied upon to provide bracing for members under axial load. Tis recommendation is based on te presumption tat at te design lateral load attaced seating will be damaged in acieving te desired lateral response. In its damaged state, te amount of bracing resistance at te fastener connections would be compromised wic in turn can compromise te integrity of te gravity load resisting system. For members in bending (out-of-teplane of a wall, for example) and axial load only, seating alone may be satisfactory as bracing for bot bending and axial load. Given tis design pilosopy, te design process ten involves two steps: (1) coice of a bracing system and (2) determination of te strengt and stiffness demands for bracing. Lateral Bracing Design Requirements: Once te system tat will be used for bracing is cosen (example: see Figure 2 details), te next step is to determine te required strengt and stiffness of tat system. For flexure of studs, were te studs are attaced to te top and bottom tracks (plates) wit fasteners, te stud may be treated as laterally and torsionally supported at tese points of attacment. Were conventional seating (plywood, oriented strand board, or gypsum wallboard/seating) is attaced to te wall stud (on bot flanges), it is generally assumed tat te seating provides lateral support for flexure via te fasteners. If seating is attaced to only one flange of te stud, te stud is considered flexurally unbraced except at its ends (stud-track connection). In tis case, lateral-torsional support/bracing, if needed, may be developed per te requirements of Section D3.2.2 of te AISI Specification. Using Section D3.2.2 of te AISI Specification, te brace lateral force demand at eac flange, P brace, may be determined as: m a l a r P brace = 1.5 w ( + ) Eq. 1 d 2 2 were w is te uniform load on te stud (plf) a 1 is te distance between braces to te left of te brace under consideration, a r is te distance between braces to te rigt of te brace under consideration, d is te dept of te stud and m is te distance from te sear center of te studs to te mid-plane of te web. Te TECH NOTE (559) 6/01 2 Ligt Gauge Steel Engineers Association
distance m may be computed as: w f dt 4D 2 m = [w f d + 2D (d - )] Eq. 2 4I x 3d were w f is te orizontal projection of te flange from te inside face of te web, t is te stud tickness, I x is te moment of inertia of te gross section about te x-axis ( to te web) and D is te overall dept of te lip. In addition to providing te required brace strengt, a minimum brace stiffness must be imposed to limit te secondary stress in te stud due to deformation of te brace. Assuming a maximum overstrees of 15% is permitted and tat tis overstress is related to a 0.026 radian rotation of te section (Yu, 1991), te axial stiffness of te brace,, may be computed as: 2P brace = Eq. 3 0.026d Tus, equations 1 and 3 above provide a straigtforward metod of determining te brace requirements for mecanically braced studs in flexure. Similar expressions are now needed for members in axial compression. Apart from studs seated bot sides (Section D4.1), te AISI Specification provides no guidance on te bracing requirements for te wall studs in compression. However, following te recommendations in te Guide to Stability Design Criteria for Metal Structures (Galambos 1998), te brace demand at eac flange (P brace and ) for discrete bracing systems may be taken as: ( ) 2 P P brace = 0.004 4 - Eq. 4 n 2 and k 2 2 P brace = ( 4 - ) ( ) Eq. 5 n L 2 were P is te axial load in te stud, L is te unbraced lengt (assuming equaly unbraced lengts), and n is te number of braces (not including te member end braces). Note tat as n goes to infinity, 2/n goes to zero. Tus, te maximum value of te (4-2/n) term is 4.0. Tis value may be conservatively used to estimate te brace requirements (strengt and stiffness). Te above equations are written in general terms and need to be modified for ASD and LRFD design (i.e. use appropriate safety and resistance factors). In addition, tese equations apply to a single stud. In te actual bearing wall conditions multiple studs will be connected and loaded. As a result, te brace force will teoretically be cumulative and te required stiffness will be te larger of te stiffness defined above. As a practical matter owever, because an entire wall acts as a system tere will be some built-in redundancy. Terefore, it can be expected te teoretical cumulative brace force would be reduced. Te amount of tis reduction depends on a number of factors: tributary wall lengt for te brace, ratio of required stud strengt to stud capacity, and te weter or not te wall also acts as a sear wall. Te following example illustrates te design of lateral bracing for a 14-ft. long, 8 ft.-1 in. tall wall wit studs at 2 ft. on center. Te studs are SSMA designated 350S162-43 sections wit a specified minimum yield strengt of 33 ksi. Lateral bracing will be provided at te stud mid-eigt and it is required to develop te axial compressive strengt of te studs wen an out-of-plane factored lateral load of 15 psf is acting. Seating is attaced to one side of te wall and is assumed to provide no lateral resistance under flexural loading. Te lateral brace configuration will comprise a 33 mil (20 gauge, or 0.0346 in.) flat strap attaced to 33 mil solid Figure 3 DESIGN EXAMPLE track-section blocking, as illustrated in Figure 2. Te solid blocking is assumed to provide te basic mecanism for ancoring te strap load. Te entire end bay wic includes te block, studs, upper and lower tracks (and teir attacments), and attaced seating or coverings, collectively take te sear in te blocking to diapragm or foundation levels above and below te wall. Oter engineered systems may be used to ancor te strap. Determine te design loads and brace requirements: Maximum stud factored moment: wl 2 M Max = 8 (were w = 30 plf and l = 8.083 ft.) = 245 lb. - ft. Using a commercially available software package, te axial compressive strengt of te 350S162-43 stud wit mid-eigt bracing was computed as = 3320 lb. wen te out-of-plane load is acting. Applying te equations above for te brace force and te stiffness, and noting tat te brace force for axial compression and bending are cumulative but stiffness is not, te required brace force per stud is: Ligt Gauge Steel Engineers Association 3 TECH NOTE (559) 6/01
P brace (total) = P brace (bending) + P brace (axial compression) P brace = 41 + 13 lb. (See appendix for sample calculation) Tus, and te required brace stiffness is: = max [ (bending), (axial compression) ] = max [ 902, 137] lb. / in. (See appendix for sample calculation) = P brace (total) = 54 lb. 902 lb. / in In tis example, te sear strengt of te solid block (in te end bays) can be directly computed. Alternatively, te designer can determine te size of solid blocking required on te basis of a practical coice for spacing of te blocking. Te strap and spacing of te blocking must be suc tat te cumulative force transferred to te blocking does not exceed te blocking capacity. Table 1 Te sear strengt of te 33 mil track blocking was computed as 1585 lb. (see appendix for sample calculation). Tus, if te total required brace force per stud is 54 lbs., te solid block can support 1585/54 = 23 studs. Terefore, te blocking can be placed a maximum of 46 ft. on center. Since te wall in tis example is 14 ft. long, solid blocking will be required in te two end bays only (i.e. 12 ft on center). Under tis condition, te maximum force, P strap, to be developed in te flat strap will be: P strap = 4 * 54 lb P strap = 216 lb Te tensile brace strengt may be computed as P strengt = 0.95 A n F y, per te AISI Specification were A n is te net crosssectional area of te strap. Assuming No. 10 screws (nominal diameter = 0.183 in.) are used for te framing connections, te following table gives te strap strengt and stiffness for tree practical 33 mil (33 ksi) strap widts (Table 1): Te strap tensile stiffness in te above table was computed from te expression: A g E k strap = Eq. 6 L Strap Dimensions Strap Nominal Strengt Strap Stiffness (P strengt ) lb. (k strap ) lb / in. 1 x 0.0346 885 10632 1-1/2 x 0.0346 1429 15948 2 x 0.0346 1971 21265 were A g is te gross cross-sectional area of te strap and L is te lengt of te strap in tension. In tis example, te strap supports 4 studs, terefore te strap lengt L in equation 6 will be 8 ft. (96 in.). Tus, on te basis of tese calculations, 33 mil solid blocking on te end bay wit a 1-in. 33 mil (33 ksi) strap will be sufficient to develop te prescribed stud loads. References 1. AISI, 1996. Specification for te Design of Cold-Formed Steel Structural Members. American Iron and Steel Institute, Wasington, DC, October. 2. SSMA, 2000. Product Tecnical Information: ICBO ER-4943P. Steel Stud Manufacturers Association, Cicago, IL April. 3. Yu, W-W, 1991. Cold-Formed Steel Design: 2 nd edition. Jon Wiley and Sons, Inc. 4. Galambos, T. V., 1998. Guide to Stability Design Criteria for Metal Structures: 5 t edition. Jon Wiley and Sons, Inc. Primary autor of tis Tec Note: Reynaud Serrette, P.D. Tis Tecnical Note on Cold-Formed Steel Construction is publised by te Ligt Gauge Steel Engineers Association. Te information provided in tis publication sall not constitute any representation or warranty, express or implied, on te part of LGSEA or any individual tat te information is suitable for any general or specific purpose, and sould not be used witout consulting wit a qualified engineer, arcitect, or building designer. ANY INDIVIDUAL OR ENTITY MAKING USE OF THE INFORMATION PROVIDED IN THIS PUBLICATION ASSUMES ALL RISKS AND LIABILITIES ARISING OR RESULT- ING FROM SUCH USE. LGSEA believes tat te information contained witin tis publication is in conformance wit prevailing engineering standards of practice. However, none of te information provided in tis publication is intended to represent any official position of te LGSEA or to exclude te use and implementation of any oter design or construction tecnique. Copyrigt 2001 Ligt Gauge Steel Engineers Association Wasington, D.C. Toll-Free: 1 (866) 465-4732 www.lgsea.com TECH NOTE (559) 6/01 4 Ligt Gauge Steel Engineers Association
TECHNICAL NOTE On Cold-Formed Steel Construction Ligt Gauge Steel Engineers Association DESIGN CONSIDERATIONS FOR LATERAL-TORSIONAL BRACING APPENDIX (Sample Calculations) Computation of Brace Force and Stiffness Bracing Requirements for Flexure (AISI Specification, Section D3.2.2): Input -- Applied uniform load (plf): w := 30 Dist. between braces to te left of tis brace (in.): a 1 := 48.5 Dist. between braces to te rigt of tis brace (in.): a r := 48.5 Stud web dept --out-to-out (in.): d := 3.5 Horizontal proj. of stud flange from web inside face (in.): w f := 1.625 -.0451 -.0712 Stud tickness (in.): t := 0.0451 Stud gross moment of intertia (in. 4 ): Ix := 0.6546 Overall--out-to-out--dept of stud lip (in.): D := 0.5 Computed Values -- w f * d * t 4 * D 2 Dist. from stud sear center to web mid-plane (in.): m := *[(w f ) * d + 2 * D * ( d - )] 4 * I x 3 * d m w a l a r 2 * P brace_bending P brace_bending := 1.5 * * * ( + ) _bending := d 12 2 2 0.026 * d P brace_bending = 41.048 lb. _bending = 902.164 lb./in. Bracing Requirements for Axial Load (Stability Design Guide, Section 12.5): Input -- Applied factored axial load (lb.): P = 3320 Stud unbraced lengt (in): L := 48.5 Number of braces (installed mecanical braces): n := 1 Computed Values -- 2 P P a_brace := 0.004 * (4 - ) * n 2 ( 2 2 P k a_brace := (4 - ) * n L 2 ) P a_brace = 13.28 lb. k a_brace = 136.907 lb./in. Ligt Gauge Steel Engineers Association 5 TECH NOTE (559) 6/01
Capacity of Sear Blocking Design Data: F y := 33 ksi b flange := 1.625 r := 0.0764 in. (inside corner radius) t := 0.0346 in. (33 mil / 20 gauge) d stud := 3.500 in. s := 24.0 E := 29500 ksi Determine te Sear Buckling Factor: a := s - b flange a = 22.375 in. := d stud - 2 * r - 2 * t = 3.278 in. a 4.00 5.34 k v := if > 1.0, 5.34 +, 4.00 + 2 2 a a ( ) () k v = 5.426 Compute te Sear Strengt of te Block Based in te AISI Equations and te Following Assumptions: web_slenderness := web_slenderness = 94.74 t E * k S l := v S l = 69.645 F y S 2 := 1.1415 * S 1 S 2 = 98.547 V 1n := 0.577 * F y * * t V 2n := 0.9 * 0.64 * t 2 * k v * F y * E V 3n := 0.9 * 0.905 * E * K v * V 1n = 2.16 kips V 2n = 1.585 kips V 3n = 1.647 kips t 3 φv n := if (web_slenderness < S l, V ln, if (web_slenderness > S 2, V 3n, V 2n )) THE CAPACITY OF THE SOLID BLOCK IS: φv n = 1.585 kips TECH NOTE (559) 6/01 6 Ligt Gauge Steel Engineers Association