7/24/2019 Built-Up Cold-Formed Steel Compression Member Design

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aids for the structuralengineer’s toolbox

ENGINEER’S NOTEBOOK

Roger LaBoube, Ph.D, P.E.( laboube@mst.edu ), is Curator’sTeaching Professor Emeritus ofCivil Engineering and Directorof the Wei-Wen Yu Center forCold-Formed Steel Structuresat the Missouri University ofTechnology. Roger is active inseveral professional organizations

and societies, including the American Iron and Steel Institute’sCommittee on Specifications andCommittee on Framing StandardsHe also serves on STRUCTURE’sEditorial Board.

By Roger LaBoube, Ph.D., P.E.

Built-Up Cold-FormedSteel CompressionMember Design

Built-up cold-formed steel compressionmembers are commonly used as shear wallchord members, and at openings of doorsand windows (stud packs) to resist the

additional load transferred from an opening header.Te provisions in North American Specification forthe Design of Cold-Formed Steel Structural Members , AISI S100 Section D1.2 are limited to concentricallyloaded compression members composed of two

shapes joined together at discrete points along theaxis of the member. Tus, the AISI S100 provisionsare limited to either an I-shaped cross section or abox-shaped cross section.oday there are various assumptions employed

when designing the stud packs. An often employedassumption by inexperienced cold-formed steeldesign engineers is that each stud in a stud packhas the same tributary area as a typical wall stud.

What this assumption consists of is adding studsto the stud pack equal to the number of studsdisplaced by an opening. Tus, the stud pack isnot engineered but, in fact, is simply assembled to

provide an equal number of studs as if the openingdid not occur. Tis assumption can be both uneco-nomical and can result in poor framing designs asillustrated by Figure 1.

Another questionable assumption made byinexperienced cold-formed steel design engi-neers is that the axial load is shared equallyto each individual member of the stud pack,and each member’s strength is based on thebehavior as a discrete member. Making thisassumption can lead to a suspect load path oran uneconomical, design as any synergy of theindividual stud pack members is not accountedfor in the design.Te following discussion introduces design con-

cepts and practical considerations for built-up

member design, for which AISI S100 and AISIframing standards, AISI S211 North AmericanStandard for Cold-Formed Steel Framing – WallStud Design have specific design provisions.

Design Methodology

Built-up compression members interconnectedat discrete points have a reduced shear rigidity

which reduces the buckling stress of the member.o reflect the reduced shear rigidity, AISI S100Section D1.2 requires the use of a modified slen-derness ratio, (KL/r)m as follows:

where

(KL/r)o = Overall slenderness ratio of entiresection about built-up member axis

a = Intermediate fastener or spot weldspacing

ri = Minimum radius of gyration of full

unreduced cross-sectional area of anindividual shape (single C-section) ina built-up member

Note, the modified slenderness ratio is onlyapplied to the buckling axis that requires theinterconnecting fasteners to transfer shear. Foran I-shaped section, this means the (KL/r)y axis isthe slenderness ratio to be modified. Te (KL/r)x axis is not modified. When applying the modified slenderness ratio,the following additional the fastener strength [resis-tance ] and spacing shall be satisfied:

1) Te intermediate fastener

or spot weld spacing,a, is limited such thata/ri does not exceedone-half the governingslenderness ratio of thebuilt-up member.

2) Te ends of a built-upcompression memberare connected by a weld having a lengthnot less than the maximum width ofthe member, or by connectors spacedlongitudinally not more than 4 diametersapart for a distance equal to 1.5 times the

maximum width of the member.3) Te intermediate fastener(s) or weld(s) atany longitudinal member tie location arecapable of transmitting required strength [factored forces] in any direction of 2.5percent of the available axial strength [ factored resistance ] of the built-up member.

AISI S211 requires that if the above criteria arenot met, the design strength of the built-upmember shall be taken as the sum of the indi-vidual members of the built-up section. AISI S100 Section D1.2 imposes stringent connec-tion requirements for the ends of built-up members

(requirement 2 above). However, based on research,the following provision has been adopted for thenext edition of the AISI framing standards which will combine the current framing standards intoone document, North American Standard for Cold-Formed Steel Structural Framing AISI S240:

Exception: Where a built-up axial load bearingsection comprised of two studs oriented back-to-back forming an I-shaped cross-section isproperly seated in a track in accordance withthe requirements of Section C3.4.3, and thetop and bottom end bearing detail of the studsconsists of full support by steel or concrete

Figure 1.

KLr

KLr

a ri

( )m = √( )

2

o + ( )

2

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components with adequate strength andstiffness to preclude differential end slipof the built-up studs, the compliance withthe end connection provisions of AISIS100 Section D1.2(b) is not required.

Te current framing standards are a freedownload from www.aisistandards.org.

Example Problem

A typical 9-foot jamb stud as shown consistingof two 600S162-54 (50 ksi) sections intercon-nected by two self-drilling screws 24 incheson center in the web (Figure 2 ). Te tracksection is not considered to be a structural

member to resist axial loads, but is needed tocreate a closure for the opening at the dooror window. Weak axis bracing (in the planeof the wall) is provided at 4-foot intervals.

Lx = 9 ft, Ly = Lt = 4 ft

Properties:Single 600S162-54, ry = 0.5699 inch

Double 600S162-54, rx = 2.2677 inches,ry = 0.7042, a = 24 inches (center to centerspacing of web fasteners)

KL/r for the y-axis,

= [(48/0.7042)2 + (24/0.5699)2 ]0.5 = 80.12

KL/r for the x-axis,

(KL/r)x = (9 x 12)/2.2677 = 47.63

Te y-axis slenderness ratio controls theaxial capacity.

Fe = = 45.35 ksi

λ c =

√

= 1.05

For λ c ≤ 1.5

Fn (0.658 λ 2c )Fy = 31.52 ksi

Te effective area, A e, is computed at f = Fn, A e = 0.7010 square inches

Pn = Fn A e = 31.52 ksi x 0.7010 in2 =24.93 kips

Available Strength, Pa = Pn/Ω = 24.93 kips/1.80 = 13.85 kips

A design consideration is the spacing of the web connectors which will influence bothload capacity and economics as summarized:

“a” “Pa”(inches) (kips) 24 13.85 18 14.50 12 14.99

Te design engineer should carefully consider ifthe increased load will provide the most economi-cal design solution because of the added laborexpense of providing screws at a closer spacing. When creating built-up sections, orienta-tion of the individual members should beconsidered. For example, if two 600S162-54 (50 ksi) sections were oriented in a boxconfiguration (Figure 3), with the toe-to-toe welds spaced 24 inches on center, the availablestrength, Pa , is 16.92 kips vs 13.85 kips forthe I-section configuration. Furthermore, theI-section configuration requires the additional

track section for the jamb closure.

Check the following additional the fastenerand spacing requirements:

1) Te intermediate fastener or spot weld spacing, a, is limited such that

a/ri does not exceed one-half thegoverning slenderness ratio of thebuilt-up member.a/ri = 24/0.5699 = 42.110.5(KL/r) = 0.5 (80.12) = 40.06

Although the a/ri is 5% larger than one-half the governing slenderness ratio ofthe built-up member, the 24-inch spac-ing is deemed to be acceptable. Tiscriteria is to ensure that the individualmember will not buckle prior to overallbuckling of the built-up member. Teadditional track section, as well assheathing attached to both flanges ofthe individual member, will enhance itsbuckling strength.

2) Te ends of a built-up compressionmember are connected by a weldhaving a length not less than themaximum width of the member, orby connectors spaced longitudinallynot more than 4 diameters apartfor a distance equal to 1.5 times themaximum width of the member.Te built-up member will beproperly seated in track sections top

and bottom, thus by utilizing theprovision of AISI S240, additionalfasteners are not required.

3) Te intermediate fastener(s) or weld(s)at any longitudinal member tielocation are capable of transmittingrequired strength [factored forces]in any direction of 2.5 percent ofthe available axial strength [ factored

resistance ] of the built-up member.Using No. 12 self-drilling screws, thenominal shear capacity of a screw is1.29 kips. Te available strength of ascrew is 1.29 kips/3.0 = 0.86 kips perscrew. Where 3.0 is the safety factor.2.5% x 13.85 kips = 0.35 kips <

0.86 kips, Okay!

Practical Considerations

Specific design methodology for two membersinterconnected is presented here, but in many

cases more than two members are used tocreate a stud pack. Te following are designthoughts offered by several experienced cold-formed steel design engineers:

• Te design varies from job-to-jobbased on the contractor preferenceand politics (e.g. on some union jobs,the cold-formed steel contractor can’tinstall HSS thicker than ⅛-inch – iron

worker vs carpenter unions).• Where we see a pair of 97-mil,

S200 or bigger studs, our firm startsthinking HSS. Te cost of buying

two (2) heavy studs and then weldingthem together seldom makes sense when compared to buying a tube.

• Our firm is not a big fan of triple ormore built-ups both due to cost and, inthe case of jamb stud packs, the notionthat equal load-sharing is questionablefor the studs away from the opening.

• Our firm limits studs to a maximumof three (3) in a stud pack because ofconcern that equal load sharing doesnot occur as well as the economics offabricating the stud pack. Our firmuses a light-weight welded I-section

whenever three (3) studs are notadequate. Also, we do not changethe thickness of the stud in the panelto create a stud pack. If the panelconsisted of 6-inch 54 mil studs, welimited the stud pack to three 6-inch54 mil studs; we did not up thethickness of the column studs.

Giving consideration to load path and fabrica-tion costs can help cold-formed steel designengineers to develop the most efficient cold-formed steel wall framing assembly.

Figure 2.

Figure 3.

KLr

KLr

a ri( )m

= √( )2

o + ( )

2

Studs – back to back

Reinforcing track (no shearconnection top and bottom)

π2E(KL/r)2

Fy

Fe

Studs – toe to toe