11-174 Beam Design for Use with the 2007 North American Cold-Formed Steel Specificatio'"

At center, away from holes

PL (0.900)(8.0)M = - = = 1.80 kip-ft = 21.6 kip-in.4 4

At edge of hole closest to center

M = V[L/2 -(12.0 - 2.25)J = (0.450)[96.0/2 -9.75] = 17.2 kip-in.

LRFD Required Strength

P, = 1.2Po + 1.6PL = (1.2)(0.150)+ (1.6)(0.750) = 1.38kips

V, = Pu/2 = 1.38/2 = 0.690 kips

At center, away from holes

P L (1.38)(8.0)M, = _u_ = = 2.76 kip-ft = 33.1 kip-in,4 4

At edge of hole closest to center

M, = Vu [L/2 - (12.0 - 2.25)J = (0.690)[ (96.0/2) -9.75 J = 26.4kip-in.

b) Flexural Strength without Holes

The member is not subject to lateral-torsional buckling, so compute strength using SectionC3.1.1 with effective section modulus, Se,at f = Fy.

It can be shown that, in the area without holes, the section is eligible for strength increaseusing the cold work of forming provisions of Section A7.2.

F, = Fya= 56.6 ksi (calculations not shown)Se = 0.670 in.' (calculations not shown)Mn = SeFy (Eq. C3.1.1-:

= (0.670)(56.6) = 37.9 kip-in.

c) Nominal Flexural Strength with Holes

The member is not subject to lateral-torsional buckling, so compute strength using SectionC3.1.1 with effective section modulus, Se,at f = Fy.

Check web using Section B2.4 - "C-Section Webs with Holes under Stress Gradient".dh = 1.5 in.Lh = 4.5 in.

h = 4.00 - 2(0.1070 + 0.0713) = 3.643 in.

Check limi tsdh/h = 1.5/3.643 = 0.412 < 0.7 OK

hit = 3.643/0.0713 = 51.1 < 200 OK

Holes are centered at mid-depth of web OKClear distance between holes = 24.0 - 4.5 = 19.5 in. > 18.0 in. OK

Comer radii = 0.25 in. > (2)(0.0713) = 0.143 in. OK

dh < 2.5 in. OKLi, = 4.5 in. OK

-- Design for Use with the 2007 North American Cold-Formed Steel Specification 11-175

dh> 9/16 in. OK

Since dh/h > 0.38, treat compression portíon of web as a uniformly compressed unstiffenedelement as follows:

w = (h-dh)/2 =(3.643-1.50)/2 = 1.072 in.

k = 0.43

:aIculate first estímate of fl at the top of the flat width using similar triangles with gross:operties.

f = fl

= 50(4.00/2 -0.0713 - 0.1070J = 45.5ksi4.00/2

F" = k 12(:2~ ~2) (: r (Eq. B2.1-S)

= 0.43 11:2(29500) (0.0713)2 =50.7ksi

12 (1- 0.32) 1.072

A = J~r (Eq. B2.1-4)

= ~45.5 = 0.947> 0.673 :. web is subject to local buckling50.7

p = (1-0.22/A)/A

= (1- 0.22/0.947)/0.947 = 0.811

b =pw

= (0.811)(1.072) =0.869 in.

(Eq. B2.1-3)

(Eq. B2.1-2)

Since the web is not fully effective, the cross section is not eligible for design using the coldwork of forming provision in this area.

Check Flange and Lip

It can be shown that the flange and lip are fully effectíve at this stress level (calculations notshown).

Recompute Section Properties

Calculate the effective section modulus, Se, deducting both the 1.50 inch hole and the inef-fective portion of the compression area of the web. Using the methods illustrated in the ex-amples in Part 1, the effective flexural properties can be computed as:

ye = 2.03 in. (from top fiber)

Ixe = 1.32 in.s

Sxe = 0.648 in.>

Further Iterations

The shift in the centroid causes a very slight change to the stress distribution and conse-quently a very small change in the value of h at the top of the flat width of the web, but notenough to change the values calculated above.

2. ShearStrength

a) Required Strength

ASD Required StrengthV = 0.450 kips (from above)

LRFD Required StrengthVu = 0.690 kips (from above)

b) Shear Strength without Holes - Section C3.2.1hit = 51.1 (computed above)

~Ekv/Fy = ~(29500)(5.34)/50 = 56.1

Since hit < ~Ekv/Fy ,

P, = 0.60Fy

= (0.60)(50) = 30 ksi(Eq. C3.2.1-

11-176 Beam Design for Use with the 2007 North American Cold-Formed Steel Specifica .~-

Nominal Flexural StrengthMn -s,r, (Eq. C3....:

= (0.648)(50) = 32.4 kip-in.

Alternatively, M¿ can be taken from Table 11-2. For a 400S162-68 with Fy= 50 ksi,M¿ = 32.4 kip-in.

d) Available Strength

ASD Allowable StrengthQb =1.67

At center, away from holes

Mn = 37.9 = 22.7 kip-in. > 21.6 kip-in. OKn, 1.67

At holes nearest center

Mn = 32.4 = 19.4 kip-in. > 17.2 kip-in. OKo, 1.67

LRFD Design StrengthQ>b =0.95

At center, away from holesQ>bMn = (0.95)(37.9) = 36.0 kip-in. > 33.1 kip-in. OK

At holes nearest centerQ>bMn = (0.95)(32.4) = 30.8 kip-in. > 26.4 kip-in. OK

(Eq. C3.2.1-1= (3.643)(0.0713)(30) = 7.79 kips

e) Shear Strength with Holes - Section C3.2.2

Limits same as those checked above OK

Design for Use with the 2007 North American Cold-Formed Steel Specification 11-177

e = h/2-dh/2

= 3.643/2 -1.50/2 = 1.07 in.

e/ t = 1.07/0.0713 = 15.0

- ee 5 < e/ t < 54,

qs = ej(54t)

= 1.07/[(54)(0.0713)J = 0.278

v: = qsVn =(0.278)(7.79) = 2.17kips

(Eq. C3.2.2-3)

(Eq. C3.2.2-1)

_ ternatively, V, ean be taken from Table U-2. For a 4005162-68 with F, = 50 ksi,Vn = 2.17 kips

Available 5trength

__Allowable 5trength=1.60

~ = 2.17 = 1.36 kips > 0.450 kips. OK1.60

Design 5trength=0.95

- = (0.95)(2.17) = 2.06 kips > 0.690 kips. OK

mbined Bending and Shear Strength

_-ear the eenter of the beam (no holes)

QbM)2+(nvV)2 ::;;1.0- lnxo Vn

(Eq. C3.3.1-1)

1.67)(21.6))2 +((1.60)(0.450))2 = 0.956 < 1.0 OK37.9 . 7.79

_-_edge of the hole closest to the eenter

1.67)(17.2))2 + ((1.60)(0.450))2 = 0.947 < 1.0 OK32.4 2.17

(Eq. C3.3.1-1)

- tematively, this ease ean be eheeked with Table Il-Tl a. For a 4005162-68 with Fy = 50 ksi,. g a required allowable moment, M, of 17.2 kip-in., eonservatively interpolate the maxi-

cmm permitted shear, V.or M = 16.8 kip-in., V::;;0.678 kipsor M = 18.7 kip-in., V::;;0.351 kipsor M = 17.2 kip-in., interpolating,

V::;;0.351+(18.7-17.2)(0.678_0.351) =0.609kips>0.450kips OK18.7 -16.8

11-178 Beam Design for Use with the 2007 North American Cold-Formed Steel Speciflcati

LRFD

Near the center of the beam (no holes)

( - J2 ( - J2M V+ -- <10~bMnxo ~v Vn _.

(Eq. 0--

M =Mu

V =V,

( J2 ( J233.1 + 0.690 = O924 < 1 O OK

(0.95)(37.9) (0.95)(7.79) . .(Eq. C3-

At edge of hole closest to the center

( J2 ( J226.4 + 0.690 = 0.921 < 1.0 OK

(0.95)(32.4) (0.95)(2.17)(Eq. O:

Alternatively, this case can be checked with Table Il-llb. For a 400S162-68with Fy = 50using a required moment, Me. of 26.4 kip-in., conservatively interpolate the maximum pe:-mitted factored shear, Vu.

for M, = 21.8 kip-in., Vu:S;1.46 kipsfor M, = 26.7 kip-in., Vu:S;1.03 kipsfor M, = 26.4 kip-in., interpolating,

Vu:S;1.03+(26.7-26.4)(1.46_1.03) = 1.06kips > 0.690kips OK26.7 -21.8

4. Web CripplingStrength

a) Required Strength

ASD Required Strength

End ConditionP = V = 0.450kips

Interior ConditionP = PD + PL= 0.150 + 0.750= 0.900kips

LRFD Required Strength

End ConditionP, = Vu= 0.690kips

Interior Condition

Pu = t.zr, +1.6PL=(1.2)(0.150)+(1.6)(0.750)=1.38kips

b) Web Crippling Strength without Holes - Section C3.4.1e = 90 degreesR = 0.1070in.t = 0.0713in.

- Design for Use with the 2007 North American Cold-Formed Steel Specification 11-179

h = 3.643 in.

End ConditionN = 3.0 in.

From Table C3.4.1-2

Check limitshjt = 51.1 < 200 OK (computed above)Njt = 3.0/0.0713 = 42.1 < 210 OK

Njh = 3.0/3.643 = 0.823 < 2.0 OK

For conditions of Fastened to SupportjStiffened or Partially Stiffened FlangesjOne FlangeLoadingjEnd Condition:

C =4CR = 0.14CN = 0.35

Ch = 0.02Ow = 1.75~w = 0.85R/t = 0.1070/0.0713 = 1.50 < 9 OK

Po = Ct2FySme(1-CR~)(l+CN~)( l-Ch~) (Eq. C3.4.1-1)

= (4)(0.0713)2 (50)sin(90)[1- 0.14 0.1070 )[1 + 0.35 /3.0)[1-0.020.0713 ~o.om

=2.36 kips

Alternatively, Pn can be conservatively interpolated from Table Il-14. For a 400S162-68 withFy = 50 ksi, fastened to support, case A:

for N = 2 in., Pn = 2.06 kipsfor N = 4 in., P, = 2.61 kipsfor N = 3 in., interpolating, Pn = 0.5(2.06 + 2.61) = 2.34 kips

3.643 )0.0713

Interior ConditionN = 5.0 in.

From Table C3.4.1-2

Check limits (other limits checked above)N j t = 5.0/0.0713 = 70.1 < 210 OK

Njh = 5.0/3.643 = 1.37 < 2.0 OK

For conditions of Fastened to SupportjStiffened or Partially Stiffened FlangesjOne FlangeLoadingjInterior Condition

C =13CR = 0.23

CN = 0.14Ch = 0.01!1w = 1.65~w = 0.90R/t = 1.50<5.0 OK

Po ~ Ct2FySll+-CR~J(1+CN~J(l-ChJfJ= (13)(0.0713)2 (50)Sin(90)(1-0.23 0.1070 J(l +0.14 /5.OJ(1-0.01

0.0713 ~ü.07i3=4.79 kips

Alternatively, Pn can be conservatively interpolated from Table II-14. For a 4005162-68 -,F, = 50 ksi, fastened to support, case B:

for N = 4 in., P,= 4.51 kipsfor N = 6 in., Pn = 5.03 kipsfor N = 5 in., interpolating, P, = 0.5( 4.51 + 5.03) = 4.77 kips

(Eq. c:-

11-180 Beam Designfor Usewith the 2007 North American Cold-FormedSteel Sp~-~-

3.643 J0.0713

e) Web Crippling 5trength with Holes - Section C3.4.2

Limits same as those checked above OK

End Conditionx = 12.0 - 4.50/2 - 3.0/2 = 8.25 in. (distance between web hole and edge of bearing)

Re = 1.01-0.325dh/h + 0.083x/h ~ 1.0 (Eq. C3.4. .:

= 1.01- (0.325)(1.50)/3.643 + (0.083)(8.25)/3.643 = 1.06 > 1 Use 1.0

Pn = RcPn = (1.0)(2.36) = 2.36 kips

Alternatively, Re can be extrapolated from Table II-16b. For stud depth = 4 in., x »5 in.,Re = 1.00

Interior Conditionx = 12.0 - 4.50/2 - 5.0/2 = 7.25 in. (distance between web hole and edge of bearing)

Re = 0.90 - 0.047dh/h + 0.053x/h S; 1.0 (Eq. C3.4.2-:

= 0.90 - (0.047)(1.50)/3.643 + (0.053)(7.25)/3.643 = 0.986 < 1.0 OK

P, = RcPn = (0.986)( 4.79) = 4.72 kips

Alternatively, Re can be conservatively interpolated from Table II-16a. For depth = 4 in.,for x = 4 in., Re = 0.94for x = 8 in., Re = 0.99

for x = 7.25 in., interpolating, Re = 0.94+(7.25-4)(0.99_0.94) = 0.988-4

~=am Design for Use with the 2007 North American Cold-Formed Steel Specification 11-181

- Available Strength

- 3D Allowable StrengthEnd Condition

0w = 1.75

Pn = 2.36 = 1.35 kips > 0.450 kips OK0w 1.75

Interior Condition0w = 1.65

Pn = 4.72 = 2.86 kips > 0.900 kips OK0w 1.65

:..RFDDesign StrengthEnd Condition

<I>w =0.85

<pwPn= (0.85)(2.36)=2.01kips> 0.690kips OK

Interior Condition<pw =0.90<l>wPn= (0.90)(4.72) = 4.25kips > 1.38 kips OK

5. Combined Bending and Web Crippling

Concentrated load at center of beam controlsASD

0.91(~) + (~):s:: 1.33Pn Mnxo °

0.91(0.900)+(21.6):s:: 1.334.72 37.9 1.70

0.743 < 0.782 OKLRFD

0.91(P]+( M ]:S::1.33<PPn Mnxo

(Eq. C3.5.1-1)

(Eq. C3.5.2-1)

-P =P,

M =M,

0.91(1.38) + (33.1):s:: 1.33(0.90)4.72 37.9

1.14 < 1.20 OK

11-182 Beam Design for Use with the 2007 North American Cold-Formed Steel Specifica --

Example 11-10: C-Section with Combined Bending and Torsional Loading

w

t •..C' = 0.773 inTI . t

R = 0.1875 In. ,

I,I+H_X"'-D-+I

X_---

B' = 2.500 in.Given:

. 1

s.c. A' = 9.000 in.

1. Steel: Fy = 55 ksi2. Section: 9CS2.5x0593. Gross Section Properties (from Example 1-1or Table 1-1)

t, = 10.3 in." s, = 2.29 ín.' J = 0.00102in.4 Cw = 11.9 in.6

xo = -1.66 in. m = 1.05 in. x = 0.641in.

4. Effective Section Properties (from Example 1-8or Table II-1)

Ixe= 9.18 in." Sxe= 1.89 in.' y = 4.859 in.

5. The member is a simply supported beam spanning 25 feet supporting a uniformly distrir-uted loado

6. The load is applied vertically in the plane of the web.7. The beam has torsional braces at both ends of the member and at the brace points specífíe;

below.

Required:

Determine the nominal flexural strength, M11! based on initiation of yielding of the effectisection considering the effects of torsion. Consider aIternate conditions of:

1. A single brace at mid-span2. Two braces, each at the one-third points of the span

Assump tions:

1. Rotation is completely restrained at the member ends and at the braces.2. The member is free to warp at both ends.

~m Design for Use with the 2007 North American Cold-Formed Steel Specification 11-183

- uiion:

A torsional reduction factor, R, is calculated using Section C3.6 and applied to the nominalstrength calculated using Section C3.1.1(a). Note that this reduction factor is not applied toother limit states, such as lateral-torsional buckling or distortional buckling.

This solution is based on the method described in the AISC Steel Design Cuide Series 9:"Torsional Analysis of Structural Steel Members"7 (DC 9). The actualloading is modeledby superimposing the three conditions as shown in the figure below.

w t,

d along Load through Distributed Brace at Brace atveb shear center torque mid-span 1/3 points

Simple DG 9 Case 4 DG 9 Case 3 DG 9 Case 3bending tr=wxo a. = 0.5 a. = 0.33 and 0.67

Ti T2 = T2

Loading = #1 #2 #3 #3A

Torsional warping stresses are calculated using the second derivative of the angle of rota-tion, S,with respect to the position, z, along the length of the member.

The sign convention for use with all torsionexpressions are shown in the figure to theright. Note that calculated values for 8 andS" may be either positive or negative. Theproper sign for these calculated values mustbe used for torsional stress calculations. Cal-culated positive values are in the directionsshown.

~ +... !

\~, ,, ,, ,'; i

Compression~1',

Point A

,-r' Z-direction

\\\\\\\

_?.), Compression

Positive rotation and warping stresses

burg, P_Aand Carter,C], "TorsionalAnalysisofStructuralSteelMembers- SteelDesignGuide Se-- 9", AmericanInstitute ofSteelConstruction,Chicago,IL,1997

11-184 Beam Design for Use with the 2007 North American Cold-Formed Steel Specifica . -

For the síngly-symmetric channel, only the compression side need be checked for combinedbending and warping.

Normal Stresses Due to Warping~s = EWns8" (AISe Design Guide 9, Eq. 4.3a

where Wns are normalized warping functions (section properties) of the cross-section ateach point of consideration given by:

Point A, at tip of flange stiffener

WA

= a:(~-b) -c(m+b)

Point B, at junction of the flange and stiffener

a:(m-b)W

B= ---'----'-

2

Point e, at junction of the flange and webarnv.=:2

where,a: = centerline web height = 8.941 in.

b = centerline flange width = 2.441 in.e = centerline lip length = 0.744in.m = distance from shear center to web centerline = 1.05 in.

The torsional warping properties for this section are:(8.941)(1.05-2.441) .

WA = (0.744)(1.05+2.441) =-8.82 m.22

(8.941)(1.05 - 2.441)WB = = -6.22 in.?

2(8.941)(1.05)

Wc = =4.69 in.?2

Formulas for rotation due to a number of torsionalloadings are given in Appendix C.4 of DG 9.Summarized below are those used in subsequent calculations.

For Loading #2 above, use DG 9 Case 4 - Uniforrnly distributed torque on member with pinnedends.

tra2[r:(z Z2J (z) (L). (z) 18t =-- -- --- +cosh - -tanh - sinh - -1.0

GJ 2a2 L L2 a 2a a

where

a=JE~Jw

Differentiating twice with respect to z yields

= '=' Design for Use with the 2007 North American Cold-Formed Steel Specification 11-185

-= _ Loadings #3 and #3A above, use DG 9 Case 3 - Concentrated torque at aL.

for O::O;z::O;aL

and

8~ =~[[Sjnh( ~) -COSh(aL)lSinh(~)laGJ tanh(~) a a

ote that the reduction factor, R, defined in Eq. C3.6-1, is a ratio of calculated stresses.These calculated stresses are directly proportional to the value of the applied uniform loadoThus a load of any magnitude can be used to calculate R. In this example, a load of w = 10pounds/ foot is used.

Mid-Span Bracing .

rnid-span bracing, the stresses are maximum at mid-span. Combine Loadings #1, #2 and

ding #1 - Simple bending through the shear center

e _Mylb -

I

_ wL2 _ 10(25)2 (12) = ki _.M - - ( ) 9.38 P m.8 8 1000

Stresses at top flange points A, B and C are all compression stresses.

_ 9.38(4.859-0.773)_ k.fbA - - - -4.18 SI

9.189.38( 4.859) .

fbB = fbe = - = -4.97ksl9.18

ding #2 - Uniformly distributed torque - use DG 9 Case 4.

t a2

[ e (z z2) (z) ( L ) (z) 18 =_r_ - --- +cosh - -tanh - sinh - -1.0t GJ 2a 2 L L2 a 2a a

where

10(1.05) . . .t = ( ) = 0.000875 kip-in.Zin.r 12 1000

a = 29500(11.9) = 175 in.11300(0.00102)

L= 25(12) = 300in. L/a = 300/175= 1.71

z = 150in. z/L = 150/300 = 0.500 z/a = 150/175 = 0.857

(1.71)2( 2) )= 0.000875(175)2 2 0.500-(0.500) + cosh(0.857

. St --,--''----'-:-1

11300(0.00102) -tanhC·~l )sinh(0.857) -1.0

11-186 Beam Design for Use with the 2007 North American Cold-Formed Steel Spf>ri=·_~:T

= 0.199radians

Sil = 0.000875 [-1.0 + cosh(0.857) -tanh(1.71)Sinh(0.857)]t 11300(0.00102) 2

= 21.2 X 10-6

Loading #3 - Brace at Mid-Span - use De 9 Case 3 with a = 0.5.

for O~z~aL

r[

'nhaL ] 1SI -TL z a a aL. zST =- (1.0-0,)-+- -cosh- sinh-eJ L L tanh L a a

a

and

S~=~r[Sinh~ +cosh aL]Sinh~laeJ tanh L a a

a

Set T = 1.0 to find the rotation per kip-in.

- 1.0(300) [ 1 (Sinh(0.5(l.71)) J . -eT - ( ) (1-0.5)0.5+- () -cosh(0.5(l.71)) sinh(0.85/11300 0.00102 1.71 tanh 1.71

= 1.21radians

S" = 1.0 [(Sinh(0.5(l.71)) - h(O 5(1 71))J inh(O 857)]T 175(11300)(0.00102) tanh(l.71) cos .. s .

= -172 X 10-6

Calculate the required value of torque provided by mid-span brace to prevent rotatíon at mid-span.

S = et+TIeT=0.199+TI(1.21)=0

TI = -0.164 kip-in.

Using this brace force, combine the calculated values for en from each loading to obtain e" forthe mid-span braced condition.

Sil = e;+ TIe; = -21.2x10-6 -0.164( -172 X 10-6 ) = 7.01X 10-6

The torsional warping stresses are:

f.v = EWne"= 29500Wn(7.01xl0-6) = 0.207Wn

-- Design for Use with the 2007 North American Cold-Formed Steel Specification 11-187

f.vA=0.207(-8.82)=-1.83ksi

f.vB= 0.207 (-6.22) = -1.29 ksi

f.vc = 0.207( 4.69) = 0.971 ksi

::EIIDine the location oí the maximum combined flexural and warping stress.

fA = fbA+ f.vA= -4.18 -1.83 = -6.01 ksi

fB= fbB+ f.vB= -4.97 -1.29 = -6.26 ksi CONTROLS

fe = fbC+ f.vc = -4.97 + 0.971 = -4.00 ksi

- enlate the reduction factor.

Rfbending

(Eq. C3.6-1)fbending+ ftorsion

-4.97 = 0.794-4.97 -1.29

~-ote that this value occurs at the intersection of the flange and stiffener; therefore, no in-crease is permitted.

- - zulate the nominal yielding strength.

Mn = RSeFy

= (0.794)(1.89)(55) = 82.5 kip-in,

applicable limit states should al so be evaluated (not shown) .

. d-Point Bracing

_ this condition, stresses are calculated at both the third-points and at mid-span, since it is notious by inspection which location will govem. Superimpose the stresses from Loadings #1,

: - d #3A. Use DG 9 Case 3 to calculate e and e" at these points due to the torsional restraintrided by the braces. The value of the torque at the brace points is calculated by requiring

=- fue value of e be zero at these two points. Note by symmetry, the torques at the braces are

~ . g #1 - Simple bending through the shear center

Flexural stresses mid-span are the same as previously calculated. Those at the third-pointsare:

wL2 10(25)212 ..M = -9-= 9(1000) =8.33 kip-in.

fbA = - 8.33 (4.859 - 0.773) = -3.71 ksi9.18

fbB = fbC= - 8.33 (4.859) = -4.41 ksi9.18

ading #2 - Uniformly Distributed Torque - Use DG 9 Case 4

alues at míd-span are aspreviously calculated. Those at third-points are:

z = Lj3 = 100 in. zjL = 0.333 z/a = 0.571

11-188 Beam Design for Use with the 2007 North American Cold-Formed Steel Specifica '_

(1.71)2( 2)= 0.000875(175)2 2 0.333-(0.333) +cosh(0.571)

11300(0.00102) (1.71). ( )-tanh - xsinh 0.571 -1.02

= 0.173rads

By symmetry, rotation at the 2/3 point is equal to the rotation at the 1/3 point: 8tl/3= 8e:

8;1/3 = 0'70875 )[-1.0+COSh(0.571)-tanh(1.71)Sinh(0.571)]11300 0.00102 2

= -19.0 X 10-6

By syrnmetry, 8" at the 2/3 point is equal to 8"at the 1/3 point: 8;2/3= 8;1/3

Loading #3A - Braces at third-points - Use DG 9 Case 3 with a = 0.667

Apply the brace torque at 2/3 point and calculate 8T and 8~ at z = L/3, z = L/2 and z =

2L/3.

For z = L/3 = 100and a = 0.667

8Tl/3= 1.0/300) )l(1-0.667)0.333+_1_[SW;~~;;~~1)) ]SiOO(0.571)]11300 0.00102 1.71 ( )

-cosh 0.667(1.71)

= 0.826radians

8" - 1.0 [(SiOO(0.667(1.71)) ( )J 1Tl/3 - 175(11300)(0.00102) taOO(1.71) cosh 0.667(1.71) sinh(0.571)

= -67.1 x 10-6

For z = L/2 = 150

8Tl/2 = 113~~0(~~~l02)l(1-0.667)0.500+ 1.~1[SW;~7~;;~~1)) )]Sinh(0.857)]-cosh 0.667(1.71)

= 1.03radians

"- 1.0 [(SiOO(0.667(1.71)) J . 18Tl/2 - ( )( ) () -cosh(0.667(1.71)) sIOO(0.857)175 11300 0.00102 tanh 1.71

= -108 X 10-6

For z = 2L/3 = 200

8T2/3= 113~~0(~~~l02)l(1-0.667)0.667 + 1.~1[SW;~~;;~~1)) ]Sinh(1.14)]-cosh( 0.667(1.71))

= 0.982radians

_ ígn for Use with the 2007 North American Cold-Formed Steel Specification 11-189

• - 1.0 [(Sinh(0.667(1.71)) J . 18T2/3- ( )( ) () -cosh(0.667(1.71)) sinh(1.14)175 11300 0.00102 tanh 1.71

= -156 X 10-6

"te the value of the torques at third-points required to prevent rotation at those brace

81/3 = Stl/3 + T2STl/3+ T2ST2/3= 0.173 + 0.826T2+ 0.982T2 = O

T2 = -0.0957 kip-in.

ate torsional warping stresses at the 1/3 and 2/3 points.

==L/38" - 8" T S" T 8"1/3 - tl/3 + 2 Tl/3 + 2 T2/3

= -19.0xl0-6 + (-0.0957)( -67.1xl0-6) + (-0.0957)( -156xl0-6)

=2.35xl0-6

~ = 29500Wn (2.35xl0-6) = 0.0693Wn

~A = 0.0693( -8.82) = -0.611 ksi

~B = 0.0693( -6.22) = -0.431 ksi

~c = 0.0693( 4.69) = 0.325 ksi

_':ermine the location of the maximum combined flexural and warping stress.fA=fbA+~A =-3.71-0.611=-4.32ksi

fB= fbB+ ~B = -4.41- 0.431 = -4.84 ksi CONTROLS

fe = fbC+ ~c = -4.41 +0.325 = -4.09 ksi

culate the reduction factor at the 1/3 and 2/3 points.

R = -4.41 = 0.911-4.41- 0.431

ate the nominal yielding strength at the 1/3 and 2/3 points.

Mn = RSeFy

= (0.911)(1.89)(55) = 94.7kip-in.

(Eq. C3.6-1)

-""xulate torsional warping stress es at mid-span.

-z = L/2

Si/2 = 8;1/2+ 2T2Si1/2= -21.2xl0-6 + 2( -0.0957)( -108xl0-6)

= -0.529 X 10-6

~ = 29500Wn (-0.529x10-6) = -0.0156Wn

~A = -0.0156( -8.82) = 0.138 ksi

~B = -0.0156(-6.22) = 0.0970ksi

11-190 Beam Design for Use with the 2007 North American Cold-Formed Steel Specifica :

~c = -0.0156( 4.69) = -0.0732ksi

Determine the location of the maximum combined flexural and warping stress.fA= fbA+ fwA= -4.18 + 0.138 = -4.04 ksi

fB= fbB+ ~B = -4.97 + 0.0970 = -4.87 ksi

fe = fbC+ ~c = -4.97 - 0.0732 = -5.04 ksi CONTROLS

Calculate the reduction factor at mid-span.

R = (1.15) -4.97 = 1.13 > 1.0-4.97 - 0.0732

(Eq. C3.6-_

Since R exceeds 1.0, take R as 1.0 at midspan. The 15% increase is permitted since the maximcombined stress occurs at the junction of the flange and web.

Calculate the nominal yielding strength at mid-span.Mn = RSeFy

= (1.0)(1.89)(55) = 104 kip-in.

ample 11-11: Web Crippling

3eam Design for Use with the 2007 North American Cold-Formed Steel Specification 11-191

3.0 in.

12.0 in.

O--+lf-l-

3.0 in.O--+lf-j-

3.0 in.O--+lf-l-

3.0 in.

V/////l. 4.0 in. I

- ~en:

Flexural member: SSMAStud 1200S200-68(50ksi)

Bearing stiffener: SSMAStud 362S162-33(33ksi)

- uired:

Calculate the available bearing strength of the joist section with the C-section bearing stiff-ener using both ASD and LRFD

:: tution:

~culate the available ASD and LRFD strength using Section C3.7.

~-e Section C3.7.1if the wIt, limits for the stiffener are not exceeded.

Check Applicability Limits for Section C3. 7.1

eck web of stiffener:D -2(R +ts)w/ts =

3.625-2(0.0765+0.0346) =98.30.0346

Limit= 1.28M

= 1.28~29500/33 = 38.3< 98.3 NG; therefore, try Section C3.7.2

2. Check Applicability Limits for Section C3. 7.2

1) The stiffener has full bearing; therefore, use 100%of the calculated capacity. OK

_) The stiffener is a C-section with a web depth of 3.625 in. > 3.5 in. minimum. The stiffenerhas a thickness of 0.0346in. > 0.0329in. minimum. OK

3) The stiffener is attached to the flexural member with three screws. OK

~) The distance from the flexural member flanges to the first fastener is d/4 > d/8 minimum.OK

5) The length of the stiffener is equal to the depth of the flexural member. OK

6) The bearing width is greater than 1112in. OK

2. Calculate nominal strength, Pn, using Section C3.7.2

Calculate the nominal bearing strength.

Pn = 0.7(Pwc+AeFy)2Pwc (Eq. C3.7_-:

11-192 Beam Design for Use with the 2007 North American Cold-Formed Steel Specifican

From Table II-14 for a 1200S200-68 (50 ksi), Fastened to the support, Case e N = 4 in.Pwc= 1.26 kips (flexural member)

From Table I1I-2for a 362S162-33 (33 ksi)Pn = 5.72 kips (stiffener) = AeFy

Nominal StrengthPn = 0.7(1.26+5.72) 21.26kips

= 4.89 kips > 1.26 kips; therefore, use 4.89 kips

(Eq. C3.7_-_

3. Available Strength

ASD - Allowable strength

Pn = 4.89 = 2.88 kipsn 1.70

(Eq. A4.1.1-:

LRFD - Design strength<!>Pn=0.90(4.89)=4.40kips (Eq. A5.1.:-:

3eam Design for Use with the 2007 North American Cold-Formed Steel Specification 11-193

ample 11-1.2: Web-Stiffened C-Section by the Direct Strength Method - Flexure

8.00 in.

roen:

IO.875 in.

W0.500 in.

2.25 in.

0.625 in.-I----""

2.25 in. 0.0451 in.

l· ~I2.50 in.

Steel: Fy = 50 ksi, Fu = 65 ksi

Sigma section (C-section with web stiffener) as shown above

J. The member is a simply supported flexural member fully braced against lateral-torsionalbuckling.

equired:

Calculate the ASD and LRFD available flexural strengths using the Direct Strength procedure~ om Specification Appendix 1

~olution:

Although the Direct Strength method may be used for any cross-section, it is particularly wellsuited to this example, since the cross-section is somewhat complex and the Specification has no

rovisions for the complex edge stiffeners on the flanges.

Perform a finite strip analysis

A finite strip analysis of the cross-section is performed using a program such as CUFSM8. Aure flexural stress distribution is assumed with the extreme fibers at Fy. Results from the

analysis include the bending moment under the assumed stress distribution, My, and a graph ofthe section buckling strength versus unbraced length, shown below.

From the analysis:

ield momentM, = 86.4 kip-in,

< Schafer, B.W., Ádány, S. "Buckling analysis of cold-formed steel members using CUFSM:conventionaland constrained finite strip methods." Eighteenth International Specialty Conference on Cold-FormedSteel Structures, Orlando, FL. October 2006. Available at www.ce.jhu.edujbschaferjcufsm

11-194 Beam Design for Use with the 2007 North American Cold-Formed Steel Speciñcet -

Critical elastic local buckling momentMere =0.96My =(0.96)(86.4) =82.9 kip-in.

--, ..

Critical elastic distortional buckling momentMerd =1.16My =(1.16)(86.4)=100 kip-in.

-'-"'--'--''''''--'--'---"-'' ---,--,--"--"--------~, -

¡,,- ",,,,...t:=::J '

r¡"t:lt:.Io«..,_ .... _... _... __ .....~..... _. ..... _.~'.~

tFSMI~C1~.jiQl'I

--·-0;;;;;·_·---·--1;;...•.•• . ,1d

2. Calculate the nominal flexural strength

Per section 1.2.2 of Appendix 1, take M, as the lowest of the nominal strengths for lateral-torsional buckling, Mne,local buckling, Mne and distortional buckling, Mnd.

1) Lateral-torsional buckling: In this case, since the member is fully braced against lateral-torsional buckling,Mne = M, = 86.4 kip-in. (Eq. 1.2.2-3

2) Local buckling:

Al = ~Mne/Mcre (Eq.1.2.2-

= ~86.4/82.9 = 1.02Since x, >0.776,

Mnt = [1-015(~: rJ(~:r Mne (Eq.1.2.2-6)

= (1_0.15(82.9)OAJ(82.9)OA 86.4 = 72.4 kip-in.86.4 86.4

3) Distortional buckling:

Ad ~My/Merd

~86.4/100 = 0.93

(Eq.1.2.2-10

Design for Use with the 2007 North American Cold-Formed Steel Specification 11-195

inee x, >0.673,

~M = [1-022[~d rl~7rM, (Eq. 1.2.2-8)

[1- 0.22( 100 )0.5)( 100 )0.586.4 = 71.0 kip-in,

86.4 86.4

The nominal flexural strength is therefore 71.0 kip-in, governed by distortional buekling.

Calculate the available strengths

the limitations for prequalified beams in Table 1.1.1-2 to determine the appropriategth reduetion faetors. Sinee there is no prequalified eategory for C-sections with web stiff--and eomplex lips, use the strength reduetion faetors from Seetion Al.2(b)

- - - Allowable strength

M, _71.0 -355 ki .----- . p-m.n 2.00(Eq. A4.1.1-1)

-ill -Design strength

~Mn = 0.80(71.0) = 56.8 kip-in. (Eq. A5.1.1-1)

- mn Design for Use with the 2007 North American Cold-Formed Steel Specification 111-1

TABLE OF CONTENTS

PART 11ICOLUMN DESIGN

For Use With the2007 Edition of the

North American Specification for the Designof Cold-Formed Steel Structural Members

CTION 1- CONCENTRICALLY LOADED COLUMNS 3

.1 Notes On The Tables 31.2 Nominal Axial Strength Tables - Braced Columns 3

Table 11I-1 Braced Column Properties - C-Sections With Lips 4Table 11I-2 Braced Column Properties - SSMA Studs - C-Sections With Lips 6Table 11I-3 Braced Column Praperties - SSMA Tracks - C-Sections Without Lips 8

.3 Distortional Buckling Axial Strength Tables 10Table 11I-4 Distortional Buckling Properties - Axial Strength - C-Sections With Lips 11Table 11I-5 Distortional Buckling Properties - SSMA Studs - Axial Strength

- C-Sections With Lips 13Table 11I-6 Distortional Buckling Praperties - Axial Strength - Z-Sections With Lips 16

.4 Nominal Axial Strength Tables - Unbraced Columns 18Table 11I-7 Nominal Axial Strength, Pntkips - C-Sections With Lips 18Table 11I-8 Nominal Axial Strength, Pntkips - SSMA Studs C-Sections With Lips 24Table 11I-9 Nominal Axial Strength, Pn, kips - SSMA Tracks C-Sections Without Lips ..35

CTION 2 - EXAMPLE PROBLEMS 45

Braced C-Section With Lips - Bending And Compression 46C-Section With Lips With Holes - Compression 50C-Section Subject to Distortional Buckling - Compression 55Unbraced Equal Leg Angle With Lips - Compression 63Tubular Section - Round - Bending and Compression 67Stiffened Z-Section With One Flange Through-Fastened To Deck OrSheathing - Compression 71

Example 11I-7 Stiffened Z-Section With One Flange Fastened To a Standing SeamRoof - Compression 74

Example 11I-8 Hat Sec.tion- Bending and Compression 76Example 11I-9 1Section - Built-Up fram Channels 80Example 11I-10 Square HSS Section - Bending and Compression 86Example 11I-11 Frame Design by Second Order Analysis : 92Example 11I-12 Web-Stiffened C- Section by the Direct Strength Method-

Compression 97

Example 11I-1Example 11I-2Example 11I-3Example 11I-4Example 11I-5Example 11I-6

111-2 Column Design for Use with the 2007 North American Cold-Formed Steel Speci -

PART 11I- COLUMN DESIGN

The design of cold-forrned steel columns requires the consideration of the limit states of:L Combined overall member buckling and local buckling, and2. Distortional buckling

Specification Section C4 includes provisions for the evaluation of these limit states. For colthat are parts of certain structural systems, Section D6 includes provisions that supersedeof the general provisions of Section C4 .

•Overall and Local Buckling: The strength of all columns is limited by the combined limit stateglobal and local buckling, which is evaluated using Section C4.1. Although the specifics v~somewhat for different cross-section shapes, the general procedure involves 1) determinationthe elastic column buckling stress, 2) transformation of the elastic buckling stress to a criticalbuckling stress, taking into account the effects of inelasticity and 3) determination of the effec-tive area with the section at the critical buckling stress. See Manual Section 3.6 and Examples -and 1-10 through 1-13 for further information on the calculation of effective area of compre __members.

The elastic buckling stress is taken as the lowest of the applicable buckling stresses for flexura(Euler) buckling, torsional buckling and flexural-torsional buckling. All cross-sections are sur-ject to flexural buckling about their principal axes, per Section C4.1.1.

All doubly-symmetric sections and most singly-symmetric sections, such as C-shapes, are su -ject to flexural-torsional buckling per Section C4.1.2. Unlipped singly-symmetric angles havi _fully effective areas, Ae, at a stress of Fy are exempt from the flexural-torsional provisions andesigned based on flexural buckling about the principal axis. Point-symmetric sections, such zsZ-shapes, are subject torsional and flexural buckling per Section C4.1.3.

Section D6 of the Specification provides specialized provisions for the flexural-torsional bucklir ;of compression members that are elements of metal roof and wall systems, including through-fastened purlins and girts and standing seam roofs.

Distortíonal Buckling: The distortional buckling limit state involves the cross-sectional deforma-tions of two or more elements acting as a group, e.g., the rotation of the flange and lip of a C-shape about the web-to-flange junction. The Specification provides three levels of provisions f _this limit state in Section C4.2. Section C4.2(a) requires a simple calculation using basic cros -section dimensions and produces a conservative, and sometimes very conservative, result. 1: -approach can sometimes be used to quickly establish that distortional buckling is not a control-ling limit state. For those cases where the extra work is justified, Section C4.2(b) can be used,which requires considerably more complex calculations, but produces accurate results. SectiC4.2(c) provides a framework for the use of computerized numerical methods to evaluate dis-tortional buckling. This approach requires fewer calculations than Section C4.2(b) and is espe-cially useful for cross-sections that do not meet the limits of applicability of the other two ap-proaches. For all three approaches, the general procedure involves 1) determination of the elas-tic distortional buckling stress, 2) determination of the corresponding elastic buckling force us-ing the gross area of the cross-section and 3) transformation of the elastic buckling force to anominal axial strength, taking into account the effects of inelasticity and post-buckling streng

For members whose required strengths are determined by first-order analysis, combined flex-ure and axial force must be checked using Section CS. Altematively, Appendix 2 permits theuse of second-order analysis for the determination of required strengths. In this case, SectionCS is still used to evaluate members subject to combined flexure and axial force, but the mo-ment modifiers and effective length factors used in Section CS are set to unity.

_ n Design for Use with the 2007 North American Cold-Formed Steel Specification 111-3

ON 1 - CONCENTRICALLY LOADED COLUMNS

1..1 Notes On The Tables

a) With the exception of the SSMA studs and tracks, the sections listed in these tables arenot necessarily stock sections. They are included primarily as a guide in the design ofcold-formed steel structural members.

b) The section designations listed in these tables correspond to those for which dimen-sions and properties are given in Tables 1-1,1-2and 1-3.

e) Tabulated properties and capacities are shown to three significant figures.d) Where they apply, the algebraic formulae presented in Section 3 of Part 1formed the

basis of the calculations for these tables.e) The strengths listed in Tables III-1 to III-9 inclusive were computed using the yield

stress listed in the tables. Cold work of forming increases were not included.f) Tables III-1, III-2 and III-3 give the nominal axial strength, Pn, for fu11ybraced C-

sections at the yield stress listed in the respective tables. Distortional buckling is notconsidered.

a) The values labeIed Pweb,Pflangeand Plipin Tables III-1, III-2 and I11-3are the highestnominal forces at which the web, flange and lip (if applicable) respectiveIy are fu11yef-fective. These values are only meaningful where they do not exceed Pnofor the sectionand yield stress in question. A vaIue of 0.00 for Pwebin Table III-2 indicates that a re-duction in web area is required at any stress level when standard punchouts are used.

) Tables III-4, III-5 and III-6 give tabuIated criticaI buckling lengths, stiffness coeffi-cients, elastic buckling stresses and nominal axiaI strengths for the limit state of distor-tional buckling for use with Section C4.2(b). Rotational restraint from sheathing ordiscrete bracing is not considered in the values given for the stiffness coefficients, elas-tic buckling stresses and nominal flexural strengths. To incorporate the strength in-creases resulting from significant continuous rotational bracing or discrete distortionalbracing spaced at Iess than Ler,use the provisions of Section C4.2(b)or C4.2(c).

;) Tables III-7, III-8and III-9give the nominal axial strength, Pn, for C-sections with vary-ing x- and y-axis unbraced Iengths. In a11cases, the torsional unbraced length is as-sumed to equal the y-axis unbraced Iength and K, = K, = 1.0. Lengths are arbitrarilycut off at a KL/rx ratio of approximately 100.

-) The calculated values in Tables III-1 through III-9 are nominal strengths. These valuesmust be modified by a safety factor, Qc, for ASO or a resistance factor <Pc, for LRFD.See the appropriate Specification section for more information.

,} The effects of standard factory punchouts in SSMAstuds have been included in TablesIII-2 and III-8. These punchouts are considered in SSMA studs with flange widths of1.625in. or less. Standard punchout sizes are 1.5 in. by 4.5 in. for sections with depthsof 3.5 inches or more and 0.75 in. by 4.5 in. for sections with shallower depths.Dashes in the place of data vaIues in the P, columns of Tables III-2, III-3 and III-5 indi-cate that the section is not available in the listed grade of steel. Blank data values inTabIes III-7, III-8 and III-9 indicate that the section is not avaiIable in the listed gradeof steel or that KL/ry exceeds 200.

Nominal Axial Strength Tables - Braced Columns

111-4 Column Design for Use with the 2007 North American Cold-Formed Steel Speciñca;

Table 11I-1 eBraced Column Properties 3 nc = 1.80

C-Sections With Lips ~c = 0.85

Pn at r-r, Pn at r-r,kips 1 Maximum kips 1 Maximum

r, Effective Force, kips 2 Fy Effective Force, kip¿:-Section 33 55

Pweb Pflange PlipSection 33 55

Pweb Pflange p-ksi ksi ksi ksi

12CS4xl05 47.6 64A 8.99 45.8 33.8 8CS2xl05 36.2 54.9 13.2 108 10312CS4x085 33.1 45.6 4.74 26.0 20.2 8CS2x085 27.0 41.2 6.91 62.1 59._

12CS4x070 24.3 33.3 2.63 15.6 12.9 8CS2x070 20.7 31.8 3.83 37.7 37.-

12CS3.5xl05 46.2 63.5 8.56 54.0 41.2 8CS2x065 18.7 28.8 3.06 30.9 31..':

12CS3.5x085 33.0 44:9 4.51 30.8 24.5 8CS2x059 16.5 25.1 2.28 23.3 25.=12CS3.5x070 24.0 33.0 2.51 18.5 15.6 7CS4xl05 45.7 62.5 21.7 44.0 32.=12CS2.5xl05 40.9 62.0 7.70 82.2 70.1 7CS4x085 32.2 44.6 l1A 25.1 19_12CS2.5x085 30.5 44.8 4.06 47.0 41.2 7CS4x070 23.8 32.7 6.28 15.0 12.~12CS2.5x070 23A 31.8 2.25 28.3 25.9 7CS4x065 21.2 29.1 5.02 12.4 ro.s10CS4xl05 47.1 63.9 12.0 45.3 33.3 7CS4x059 18.2 24.9 3.74 9.63 8':-=

10CS4x085 32.9 45.3 6.29 25.8 20.0 7CS2.5xl05 39.1 60.1 17.6 80A 68-::

10CS4x070 24.2 33.1 3A9 15A 12.7 7CS2.5x085 29.5 43.9 9.22 46.0 4010CS4x065 21.5 29A 2.79 12.7 10.8 7CS2.5x070 22.9 31.3 5.10 27.8 25.-10CS3.5xl05 45.7 63.0 11.3 53.5 40.7 7CS2.5x065 20.8 27.7 4.07 23.0 21....!.

10CS3.5x085 32.7 44.7 5.96 30.5 24.2 7CS2.5x059 18.0 24.1 3.03 18.0 17_

10CS3.5x070 23.8 32.9 3.31 18A 15A 6CS4xl05 45.0 61.7 28.5 43.2 31..2

10CS3.5x065 21.3 29.3 2.64 15.2 13.0 6CS4x085 31.8 44.2 14.9 24.7 18.

10CS2.5xl05 40.4 61.5 10.1 81.7 69.6 6CS4x070 23.5 32.5 8.23 14.8 12.':·10CS2.5x085 30.2 44.6 5.29 46.7 40.9 6CS4x065 21.0 28.9 6.57 12.2 10.~10CS2.5x070 23.3 31.7 2.94 28.2 25.8 6CS4x059 18.1 24.8 4.89 9.50 8.3..:

10CS2.5x065 21.1 28.0 2.35 23.3 21.7 6CS2.5xl05 38.3 59A 22.8 79.6 67.::10CS2xl05 36.9 55.7 9A3 109 104 6CS2.5x085 29.1 43.5 11.9 45.6 39.10CS2x085 27A 41.6 4.96 62.5 60.3 6CS2.5x070 22.6 31.1 6.58 27.6 2510CS2x070 20.9 32.0 2.75 38.0 37.7 6CS2.5x065 20.6 27.6 5.25 22.8 21..2

10CS2x065 18.9 29.0 2.20 31.1 31.6 6CS2.5x059 17.8 24.0 3.91 17.8 17.'::

9CS2.5xl05 40.1 61.1 11.8 81.4 69.3 4CS4xl05 41.1 58.8 51.6 39.3 27.~9CS2.5x085 30.0 44A 6.20 46.5 40.8 4CS4x085 30.2 42.7 27.8 22.7 16.-

9CS2.5x070 23.2 31.6 3A4 28.1 25.7 4CS4x070 22.7 31.6 15.8 13.8 11..

9CS2.5x065 21.0 28.0 2.75 23.2 21.7 4CS4x065 20.3 28.2 12.8 11A 9 -~9CS2.5x059 18.2 24.3 2.05 18.1 17.4 4CS4x059 17.6 24.3 9.70 8.92 tss8CS4xl05 46.3 63.1 17.2 44.6 32.5 4CS2.5xl05 34A 56.4 47.7 76.7 64.:::

8CS4x085 32.5 44.9 9.05 25A 19.5 4CS2.5x085 27.6 41.9 24.7 44.0 38.-::

8CS4x070 23.9 32.9 5.01 15.2 12.5 4CS2.5x070 21.8 30.2 13.6 26.7 24.~8CS4x065 21.3 29.2 4.00 12.5 10.6 4CS2.5x065 19.9 26.9 10.8 22.1 20.8CS4x059 18.3 25.0 2.98 9.73 8.56 4CS2.5x059 17.3 23.5 8.02 17.3 16.c8CS3.5xl05 44.9 62.2 16.2 52.7 27.3 4CS2xl05 31.6 51.7 43.8 103 1048CS3.5x085 32.3 44.3 8.51 30.1 16.7 4CS2x085 25.3 39.8 22.7 58A 59.=8CS3.5x070 23.6 32.7 4.72 18.2 11.0 4CS2x070 19.9 31.3 12.5 33.8 35.8CS3.5x065 21.1 29.1 3.77 15.0 9.38 4CS2x065 18.2 28A 9.92 27A 28.8CS3.5x059 18.2 25.0 2.81 11.7 7.68 4CS2x059 16.1 24.5 7.37 20.7 21.88CS2.5xl05 39.6 60.7 14.2 80.9 64.58CS2.5x085 29.8 44.2 7A4 46.3 38.38CS2.5x070 23.0 31.4 4.13 28.0 24.38CS2.5x065 20.9 27.9 3.29 23.1 20.68CS2.5x059 18.1 24.2 2A6 18.1 16.6

Oesign tor Use with the 2007 North American Cold-Formed Steel Specitication 111-5

_ "al strengths given are nominal strengths. To obtain the available strength, thesevalues must be modified by safety factors (ASD)or resistance factors (LRFD).

reb, Pflange and Plip are the highest nominal axial compression forces at which the-"eb, flange and lip, respectively, are fully effective.

The distortional buckling limit state is not considered in this table, Distortional_ ckling strengths are provided in Table III-4.