1\-166 Beam Design for Use with the 2007 North American Cold-Formed Steel Specifica:

Example 11-6: Fully Braced Hat Section

I~~I 45~~Oi" lf----------'--R = 0.1875 in. 1

I... -e--t = 0.135 in. L'- -------- -------

6.00 ft

---- ... y

al rr 1]m 1 I !------:>fl'

3.000 in.

670 in.~I l. .1

1.670 in.

l. teel: Fy= 50 ksi

Section: 3HU4.5x135 shown in sketch above

o. Top flange is in compression and is fully braced.

Re ired:

1. Check the flexural adequacy of a 6 foot long simple span beam with:Dead Load, Wdead = 120 plfLi e Load, wu-, = 700 plf

Do not consider inelastic reserve,

3. Check using both ASD and LRFD.

ution:

1. Calculate Nominal Strength, Mn (Section C3.1.1)

inee the member is fully braced and not subject to lateral-torsional or distortional buckling,calculate the nominal strength using Section C3.1.1.

From Example 1-13or Table II-6, Se= 1.52 in."Mn = SeFy (Eq. C3.1.1-_

= (1.52)(50) = 76.0kip-in.

2. Calculate Required and Available Strength

Bending Moments

M = WdeadL2_ (0.120)(6)2D 8 - 8

ML = Wlivee (0.700)(6)28 8

= 0.540 kip-ft = 6.48 kip-in.

= 3.15 kip-ft = 37.8 kip-in.

- m Design for Use with the 2007 North American Cold-Formed Steel Specification 11-167

ASDRequired Allowable Strength

M = Mo +ML= 6.48 + 37.8 = 44.3 kip-in.Allowable Strength

M 44.3 kip-in. OKn, 1.67LRFD

Required StrengthM, = l.2Mo + 1.6 ML

= (1.2)(6.48) + (1.6)(37.8)

= 68.3 kip-in.Design Strength

Mu < 68.3 kip-in. OK

The resistance factor of 0.95 is permitted per Section C3.1.1 because the compressionflange is stiffened.

(Eq. A4.1.1-1)

(Eq. A5.1.1-1)

:1.. Nominal Flexural Strength (Section C3.:1..3):

Ratio of outside diameter to wall thickness,D/t = 8.00/0.100 = 80.0

ChecklimitD/t < 0.441E/Fy = 0.441(29500/42) = 310 OK

Full Section Properties

(Outside Diameter t -[Insde Diameter tSf =n~------~--~--~----~----~

32(Outside Diameter)

(8.00t -(7.80t= n~-'-"':'---'--------''-(32)(8.00)

=4.84 in.>Determine the governing equation

0.0714E/Fy= 0.0714(29500/42) = 50.2

0.318E/Fy= 0.318(29500/42) = 223

Since 0.0714E/Fy< D/t < 0.318E/Fy

F, = [0970 +0020(b;: J]F,Fe = [0.970+ 0.020( 29~~~b42JJ42 = 48.1 ksi

(Eq. C3.1.3-

11-168 Beam Design for Use with the 2007 North American Cold-Formed Steel Specificc::

Example 11-7: Tubular Section - Round

T78.00 in.

~---......--Given:

1. Steel: F, = 42 ksi

2. Section: Shown in sketch above

Required:

. 1. Determine the ASD allowable flexural strength, Mn/nb .

2. Determine the LRFD design flexural strength,

3eam Design for Use with the 2007 North American Cold-Formed Steel Specification 11-169

(Eq. C3.1.3-1)

M, = (48.1)(4.84) = 233 kip-in.

2. ASOallowable flexural strength

Mn _ 233 -140ki .----- p-m.o, 1.67

3. LRFOdeslgn flexural strength

11-170 Beam Deslgn for Use with the 2007 North American Cold-Formed Steel Specification

Example 11-8: Tubular Section - Rectangular

6.00 in. - ------l------ I----+x

I,

I-+ .- t = 0.125 in. ,

I

Given:

ytI

~- 'R= 0.0494 inl

I

1. Steel: F, = 46 ksi

2. Section: HSS 6x6x1fs(from Table 1-12, AISC Steel Construction Manual, 2005)

3. Calculated gross properties using nominal dimensions aboveA = 2.91 in.2I = 16.7 in.sS = 5.56 in.'

3. Simple span length = 10.0 ft

4. Laterally braced at both ends

Required:

1. Determine the ASD flexural allowable strength, M; /0.b .2. Determine the LRFD design flexural strength,

~oam Design for Use with the 2007 North American Cold-Formed Steel Specification 11-171

Se= 4.95 in.'nominal flexural strength is calculated as:Mn= SeFy

= (4.95)(46) = 228kip-in. or 19.0 kip-ft(Eq. C3.1.1-1)

Nominal Flexural Strength, Mn, based on lateral-torsional buckling (Section C3.1.2.2):

.r: ccording to Specification Eq. C3.1.2.2-1,the minimum unbraced length of a closed box member:- - iect to lateral-torsional buckling, L is calculated as:

L = 0.36Cbn JEGJI (E C )u F S y q. 3.1.2.2-1y f

whereCs = 1.0 (assumed)Iy (full section) = 16.7 in.sSf(full section) = 5.56 in.>

2( ab)2J = ...,........,..---7---:-,...--:-(ajt1) + (b/t2)

(Eq. C-C3.1.2.2-1)

wherea = b =D- t = 6.00- 0.125= 5.875 in.tI = t2 = 0.125 in.

2(5.875t 4J = 2 (5.875jO.125) = 25.3in. (Eq. C-C3.1.2.2-1)

--erefore,

L = ~.3~1.0)) J(29500)(11300)(25.3)(16.7) = 1660 in. = 138 ft46 5.56

(Eq. C3.1.2.2-1)

- ceL, > 10.0 ft, the member is not subject to lateral-torsional buckling and the flexuralsrrength is based on Section C3.1.1(a). That is,

M, = 19.0 kip-ft

ASO allowable flexural strength

Mn _ 19.0 -114 k' f----- . lp- to, 1.67

LRFO deslgn flexural strength

11-172 Beam Design for Use with the 2007 North American Cold-Forrned Steel Specifica . -

Available AISI AISC AISI--Strength kip-ft kip-ft AISC

ASD: Mn/nb 11.4 10.8 1.06

LRFD: (hMn 18.1 16.2 1.12

The differences between the AISI and AISC strengths shown in the table above are due to thefollowing factors:

a. The AISI calculations were performed using the nominal wall thickness of the HSS, whilethe AISC calculations were performed using 93% of the nominal wall thickness as requirec.by AISC.

b. The calculations of the effective width of the compression flange are slightly different intwo specifications, including slightly different values of E.

c. For the LRFD method, the resistance factor used in the AISI Specification is

_ m Design for Use with the 2007 North American Cold-Formed Steel Specification 11-173

Example 11-9: C-Section with Openingsy

t

, R = 0.1070 in.

I,I- ----x, 3i~n'~~~==~~~~~~~~~~~~~~~~~~==T====1

it 00713 in t...:_-=1:=.2.:c..inc.....- ..~I+~-----=:4...:...8:c..:i~n'--. ---_~ I_~_....c1=2...:...in.:.:...- ..~

ti

! Po = 0.150 kipsPL= 0.750 kips

-.000 in.

1.5 x 4.5 in. punchouts5 in.

-. en:

Steel: F, = 50 ksi, Fu= 65 ksiSection: 400S162-68as shown above

Section is simply supported, fully braced against lateral-torsional and distortional buckling,and is fastened to support.

1.5 in. by 4.5 in. web punchouts with 0.25 inch comer radii located as shown above. Notethat the location of punchouts is often not known with this precision.

. uired:

eck the adequacy of the section considering:a. Flexureb. Shearc. Web Cripplingd. Combined Bending and Sheare. Combined Bending and Web Crippling

1. ASD - ASCE/SEr 7-05ASD load combination D + L2. LRFD- ASCE/SEI 7-05LRFD load combination l.2D + 1.6LNeglect self weight of beam

- utian:

Flexural Strength

Required Strength

_.:8DRequired StrengthP = Po + PL= 0.150+ 0.750= 0.900kipsV = P/2 = 0.900/2 = 0.450kips