11-168

Beam Design for Use with the 2007 North American Cold-Formed Steel Soe

Example 11-7: Tubular Section - Round

Given:

1.

Steel: Fy = 42 ksi

2.

Section: Shown in sketch above

Required:

. 1.
2.

Determine the ASD allowable flexural strength, M;

/b .

Determine the LRFD design flexural strength, $bMn .

Solution:
1.

Nominal Flexural Strength (Section C3.1.3):

Ratio of outside diameter to wall thickness,

D/t = 8.0010.100 = 80.0
Check limit
D/t

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

Full Section Properties

Sf

t-

_ [Outsde Diameter
[Inside Diameter
-n~------~--~~----~----~
32 (Outside Diameter)

(8.00t -(7.80t
(32)(8.00)

= n -'---~.,.....,.-'---:~

=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 + 0.D20(~;: )]F

Fe = [0.970 + 0.020( 29~~~b42)]42 = 48.1 ksi

ign for Use with the 2007 North American Cold-Formed Steel Specification

(Eq. C3.1.3-1)

- in = FeSE
=

(48.1)(4.84)

11-169

= 233 kip-in.

allowable flexural strength

- in _ 233 -140 ki p-m.

.
----o, 1.67
deslgn flexural strength

M, = (0.95)(233) = 221 kip-in.

11-170

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

Example 11-8: Tubular Section - Rectangular

y
t

1\

'

R = 0.0494 inl

I
6.00 in.]

-If------l------

f--

----+x

I,
-1t- t = 0.125

in.

I
~:

:1

6.00 in.

Given:
1.

Steel: Fy

2.

Section:

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

3.

Calculated
A
1
S

=
=
=

46 ksi

gross properties

using nominal

dimensions

above

2.91 in.2
16.7 in.s
5.56 in.3

3.

Simple span length

4.

Laterally braced at both ends

10.0 ft

Required:

Mn/nb

1.

Determine

the ASD flexural allowable

strength,

2.

Determine

the LRFD design flexural strength,

3.

Compare the calculated available flexural strengths to those calculated using the AIS
cedures. The inside radius given above is selected to give the same flat width, w, f,
flanges and webs used in the AISC calculations.

cj>bMn .

Solution:
Compute the nominal flexural strength,
C3.1.2.2 for closed box members.
1.

Mnt according

to Specification Sections C3.1.1(a) ~-

Nominal Flexural Strength, Mn, based on initiation of yielding (Section C3.1.1):

Check the compression flange as a uniformly compressed compression element in a

dance with Section B2.1. By calculations not shown, the effective width of the compressiar
flange width is found to be 4.608 in.
Check webs in accordance
calculations not shown.
The effective section properties
I,

15.5 in.s

with Section B2.3. The webs are found to be fully effective.E

can then be calculated

as:

11-171

:: "or Use with the 2007 North American Cold-Formed Steel Specification

flexural strength is calculated as:

" ~__ o....",1

(Eq. C3.1.1-1)

"Fy
= =95)(46) = 228 kip-in. or 19.0 kip-ft
".Ia.1 Flexural Strength, Mn, based on lateral-torsional

buckling (Section C3.1.2.2):

"''''-'--="''" _ Specification Eq. C3.1.2.2-1,the minimum unbraced length of a closed box member
ateral-torsional buck1ing, Lw is calculated as:

_ = 0.36Cb7t ~EGJly

(Eq. C3.1.2.2-1)

FySf

1.0 (assumed)

:Ull section)

16.7 n.s

nill section) = 5.56 in."

2( ab)2

(Eq. C-C3.1.2.2-1)

- e
a=b=D-t=6.00-0.125=5.875
tI

= t2 = 0.125 in.
2(5.875)4

(5.875/0.125)

in.

. 4
= 25.3m.

= ~.3~1.0)) J(29500)(11300)(25.3)(16.7) = 1660 in.

46 5.56

(Eq. C-C3.1.2.2-1)

138 ft

(Eq. C3.1.2.2-1)

> ..,0.0 ft, the member is not subject to lateral-torsional buckling and the flexural

~ oased on Section C3.1.1(a). That is,

- = 19.0 kip-ft
owable flexural strength

~ - 19.0 -11 .4 k'lp- ft

----1.67
deslgn flexural strength
_.~ = (0.95)(19.0)

= 18.1kip-ft.

c:::.mparison of AISI and AISC available flexural strengths

- available flexural strengths calculated above are compared with those calculated in
- ordance with the 2005 AISC Specification in the table below. AISC strengths are
- cenfrom Table 3-13 of the 13th Edition of the AISC Steel Construction Manual.

11-172

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

Available
Strength

AISI
AISC

AISI
kip-ft

AISC
kip-ft

ASD: Mn/nb

11.4

10.8

1.06

LRFD: <PbMn

18.1

16.2

1.12

--

The differences between the AISI and AISC strengths shown in the table above are d
following factors:

~-

.~

a.

The AISI calculations were performed using the nominal wall thickness of the HS5
the AISC calculations were performed using 93% of the nominal wall thickness as
by AISC.

b.

The calculations of the effective width of the compression flange are slightly diff~~
two specifications, including slightly different values of E.

c.

For the LRFD method, the resistance factor used in the AISI Specification is <Pb = O_~
the resistance factor used in the AISC Specification is <Pb = 0.90.

d.

The differences shown in the table above are representative of HSS sections ha" .
elements, but actual ratios vary somewhat based on section dimensions and yield _

:2.-.--........

tor Use with the 2007 North American

))-173

CoJd-Formed SteeJ Specification

C-Section with Openings

~625 in.

:;::::::::::=:::::::)j 10.500

<i.

in.

, PD: 0.150kips
PL: 0.750 kips

: 0.1070in.
1.5x 4.5 in. punchouts

5 in.

3i~n.~~~======~====================~========~==~
= 0.0713in.

12 in.
0

---.--.

--48i-n.

24 in.

I
--.~_.--:!
12 in.

'

Fy = 50 ksi, Fu = 65 ksi
400S162-68 as shown above

'E~n;'

-3e::..:'n is simply supported, fully braced against lateral-torsional and distortional buckling,
.s fastened to support.
y 4.5 in. web punchouts with 0.25 inch comer radii located as shown above. Note
_ e location of punchouts is often not known with this precision.

- adequacy of the section considering:

-='exure
Shear
eb Crippling
.7

Combined Bending and Shear

Combined Bending and Web Crippling

_. D - ASCE/SEI 7-05 ASD load combination D + L

:RFD - ASCE/SEI 7-05 LRFD load combination l.2D + 1.6L
_ ect self weight of beam

_ - ed Strength
nired Strength

= PD + PL = 0.150 + 0.750 = 0.900 kips

-

P/2

= 0.900/2 = 0.450 kips

Beam Design for Use with the 2007 North American Cold-Formed Stee' ~

11-174

At center, away from holes

M =

PL
4

(0.900)(8.0)
A

= 1.80 kip-ft = 21.6 kip-in.

At edge of hole closest to center

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

LRFD Required Strength

P,

1.2PD + 1.6PL

Vu

P /2

= (1.2)(0.150)+

= 1.38/2 = 0.690

(1.6)(0.750)

= 1.38kips

kips

At center, away from holes

M,

P L
_u_

(1.38)(8.0)

2.76 kip-ft

33.1 kip-in.

At edge of hole closest to center

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

M,

b)

Flexural Strength without Holes

The member is not subject to lateral-torsional buckling, so compute strength usinz
C3.1.1 with effective section modulus, Se, at f = Fy.

~~"Ir

It can be shown that, in the area without holes, the section is eligible for strength in.. ~
using the cold work of forming provisions of Section A7.2.
Fy = Fya= 56.6 ksi (calculations not shown)

e)

Se

0.670 in.' (calculations not shown)

M,

SeFy

(0.670)(56.6)

= 37.9kip-in.

Nominal Flexural Strength with Holes

The member is not subject to lateral-torsional buckling, so compute strength using
C3.1.1 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.

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

Check limits
dh/h = 1.5/3.643 = 0.412 < 0.7 OK
h/ t

= 3.643/0.0713 = 51.1 < 200 OK

Holes are centered at mid-depth of web OK

Clear distance between holes = 24.0 - 4.5 = 19.5 in. > 18.0 in. OK
Comer radii

di, < 2.5 in. OK

Li,

4.5 in. OK

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

gn for Use with the 2007 North American Cold-Formed Steel Specification

11-175

h> 9/16 in. OK

.::rxe dh/h > 0.38, treat compression portion of web as a uniform1y compressed unstiffened
ent as follows:
: = (h-dh)/2=(3.643-1.50)/2=1.072in.
= 0.43
- ate first estmate of I at the top of the flat width using similar triangles with gross
es.
- f1 -- 50(4.00/2-0.0713-0.1070)
4.00/2

71?E
zr

= 45.5 k SI.

( t J2

(Eq. B2.1-5)

k 12(1-.t2)

2
0.43 rc (29500) (0.0713J2 = 50.7ksi
12(1-0.32)
1.072

=)t

(Eq. B2.1-4)

= ~45.5 = 0.947> 0.673 .'. web is subject to local buckling

50.7
=

(1-0.22/A)/A

(1- 0.22/0.947)/0.947

(Eq. B2.1-3)
= 0.811

=pw

(Eq. B2.1-2)

(0.811)(1.072) = 0.869 in.

- - e web is not fully effective, the cross section is not eligible for design using the cold
t: orming provision in this area.
- ange and Lip
hown that the flange and lip are fully effective at this stress level (calculations not
-,,--.---_"'te Section Properties
_

...__

the effectve secton modulus, Se, deducting both the 1.50 inch hole and the inef"::'rtion of the compression area of the web. Using the methods illustrated in the ex~ Part 1,the effectve flexural properties can be computed as:
u_

= _03 in. (from top fiber)

=

:.32 in.'

= .648 in.>
-'

the centroid causes a very slight change to the stress distribution and conse'ery small change in the value of I at the top of the flat width of the web, but not
_c:::::: - change the values calculated above.

Beam Design for Use with the 2007 North American Cold-Formed Stee

11-176

Nominal Flexural Strength

Mn =SeFy
= (0.648)(50) = 32.4 kip-in.
Alternatively, M, can be taken from Table II-2. For a 400S162-68 with Fy = 50
M, = 32.4 kip-in.

d)

Available Strength

ASD Allowable Strength

o, =1.67
At center, away from holes
Mn

o,

37.9 = 22.7 kip-in. > 21.6 kip-in. OK

1.67

At holes nearest center

Mn

= 32.4
1.67

o,

= 19.4 kip-in,

> 17.2 kip-in. OK

LRFD Design Strength

</>b =0.95
At center, away from holes
</>bMn = (0.95)(37.9)

= 36.0 kip-in. > 33.1 kip-in. OK

At holes nearest center

</>bMn

(0.95)(32.4) = 30.8 kip-in. > 26.4 kp-in. OK

2.

Shear Strength

a)

Required Strength

ASD Required Strength

V = 0.450 kips (from above)
LRFD Required Strength
V, = 0.690 kips (from above)
b)

Shear Strength without Holes - Section C3.2.1

hit = 51.1 (computed above)
~Ekv/Fy

~(29500)(5.34)/50

= 56.1

Since hit < ~Ekv/Fy ,

Fv = 0.60Fy
= (0.60)(50)=30
Vn

ksi

AwFv

= (3.643)(0.0713)(30) = 7.79 kips

c)

Shear Strength with Holes - Section C3.2.2

Limits same as those checked above OK

:-

ign for Use with the 2007 North American Cold-Formed Steel Specification

e] t

h/2-dh/2

3.643/2 -1.50/2 = 1.07 in.

1.07/0.0713 = 15.0

11-177

(Eq. C3.2.2-3)

.5< c/t < 54,

s

'n

cj(54t)

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

qsVn = (0.278)(7.79) = 2.17 kips

(Eq. C3.2.2-1)

- :ematively, Vn can be taken from Table Il-2. For a 4005162-68 with Fy = 50 ksi,

-n
-

2.17 kips

. able 5trength

- owable 5trength
=1.60
-

2.17
1.60

= 1.36 kips

> 0.450 kips. OK

.gn 5trength
=0.95
=

(0.95)(2.17)

= 2.06 kips

> 0.690 kips. OK

ined Bending and Shear Strength

- .,..the center of the beam (no holes)

(Eq. C3.3.1-1)

=--

)(21.6)J2 [(1.60)(0.450)J
+
37.9
7.79

0.956 < 1.0 OK

ge of the hole closest to the center

2
2
r )(17.2)J
[(1.60)(0.450)J
+
= 0.947 < 1.0 OK
32.4
2.17

(Eq. C3.3.1-1)

atively, this case can be checked with Table Il-l1a. For a 4005162-68 with Fy = 50 ksi,
a required allowable moment, M, of 17.2 kip-in., conservatively interpolate the maxipermitted shear, V.
~ T

16.8 kip-in., V s 0.678 kips

:\1 = 18.7 kip-in., V s 0.351 kips

- T

17.2 kip-in., interpolating,

V s 0.351+(18.7 -17.2)(0.678_0.351)
~18.7 -16.8

= 0.609 kips > 0.450 kips OK

11-178

Beam Design for Use with the 2007 North American Cold-Formed Steel Sp""''''='--

LRFD
Near the center of the beam (no holes)

Ir

J2 + ( -- VV J2 <10

M
~bMnxo

~v

_.

M =M,
V

-v;
2

33.1
[

(0.95)(37.9))

0.690

2_

[(0.95)(7.79)) - 0.924 < 1.0 OK

At edge of hole closest to the center

26.4
[

(0.95)(32.4)

0.690

+ (0.95)(2.17)) = 0.921 < 1.0 OK

Alternatively, this case can be checked with Table I1-11b. Por a 400S162-68 with Fy =
using a required moment, M of 26.4 kip-in., conservatively interpolate the maximum mitted factored shear, Vu.
for M,

21.8 kip-in., Vu ~ 1.46 kips

for M,

26.7 kip-in., Vu ~ 1.03 kips

for M,

26.4 kip-in., interpolating,

v: ~ 1.03+(26.7

-26.4)(1.46_1.03)
26.7 -21.8

4.

Web CripplingStrength

a)

Required Strength

1.06 kips > 0.690 kips OK

ASD Required Strength

End Condition
P

=V =

0.450 kips

Interior Condition
P

= Po + PL = 0.150 + 0.750 = 0.900 kips

LRFD Required Strength

End Condition
Pu

Vu = 0.690 kips

Interior Condition
Pu
b)

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

Web Crippling Strength without Holes - Section C3.4.1

90 degrees

0.1070 in.

0.0713 in.

__esign for Use with the 2007 North American Cold-Formed Steel Specification

11-179

= 3.643 in.
-

- Condition
_-

3.0 in.

Table C3.4.1-2
-

limits

_/t

= 51.1 < 200 OK (computed above)

_-/t

= 3.0/0.0713 = 42.1 < 210 OK

_-/h = 3.0/3.643 = 0.823 < 2.0 OK

onditions of Fastened to SupportjStiffened
........
=LULL',gjEnd
Condition:

or Partially Stiffened FlangesjOne

Flange

=4

Lo

= 0.14

. = 0.35
= 0.02
= 1.75
= 0.85
~ t = 0.1070/0.0713 = 1.50 < 9 OK

sme[1- e,~)[ 1+ eN~)[

~ Ct Fy

= 4)(0.0713)2 (50)Sin(90)[I-

1-

e, Jf)

(Eq. C3.4.1-1)

~o.om

0.14 0.1070 )[1 + 0.35 3.())[1-0.02

0.0713

3.643 )
0.0713

=_36 kips
-_-=:...LLatively,
Pn can be conservatively interpolated from Table I1-14. For a 400S162-68 with
- ksi, fastened to support, case A:
= 2 in., P, = 2.06 kips
= 4 in., Pn = 2.61 kips
~ = 3 in., interpolating,

Pn = 0.5(2.06 + 2.61) = 2.34 kips

or Condition
_-

5.0 in.

Table C3.4.1-2
limits (other limits checked above)
_- t = 5.0/0.0713 = 70.1 < 210 OK
- h = 5.0/3.643 = 1.37 < 2.0 OK
ditions of Fastened to SupportjStiffened
___- -..
,gj Interior Condition
-

=13

--,. = 0.23

or Partially Stiffened FlangesjOne

Flange

11-180

Beam Design for Use with the 2007 North American Cold-FormedSteel __

CN = 0.14
0.01

Ch

nw

= 1.65

<pw = 0.90
R/t = 1.50 < 5.0 OK

Ct

r, ~

2Fysine (

l-CR~J( l+C fIJ( l-Ch~J

J(l
N

= (13)(0.0713)2 (50)Sin(90)(1-

0.23 0.1070
0.0713

+ 0.14

[5JJJ(l-

VM713

0.01 3.643
0.0713

=4.79 kips
Alternatively, Pn can be conservatively interpolated from Table II-14. For a 400S16::Fy = 50 ksi, fastened to support, case B:
for N = 4 in., P n = 4.51 kips
for N = 6 in., Pn = 5.03 kips
for N = 5 in., interpolating,
e)

Pn = 0.5( 4.51 + 5.03) = 4.77 kips

Web Crippling Strength with Holes - Section C3.4.2

Limits same as those checked above OK
End Condition
x

= 12.0 - 4.50/2 - 3.0/2 = 8.25 in. (distance between web hole and edge of bearr

Re = 1.01- 0.325dh /h + 0.083x/h :-:;1.0

= 1.01-(0.325)(1.50)/3.643

+ (0.083)(8.25)/3.643 = 1.06> 1 Use 1.0

P, = RePn = (1.0)(2.36) = 2.36 kips

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

= 12.0 - 4.50/2 - 5.0/2 = 7.25 in. (distance between web hole and edge of bearinz

Re = 0.90 - 0.047dh/h + 0.053x/h:-:;1.0'

= 0.90 - (0.047)(1.50)/3.643 + (0.053)(7.25)/3.643 = 0.986 < 1.0 OK
Pn = RePn =(0.986)(4.79) =4.72 kips
Alternatively, Re can be conservatively interpolated from Table II-16a. For depth = 4 .
for x = 4 in., Re = 0.94
for x = 8 in., Re= 0.99
for x = 7.25 in., interpolating,

Re= 0.94+(7.25-4)(0.99-0.94)
8-4

= 0.98

-z tor Use with the 2007 North American Cold-Formed Steel Specification

11-181

- ble Strength
.able Strength
- Condition
T

= 1.75
-

= 2.36 = 1.35 kips > 0.450 kips OK

1.75

- ior Condition
__ = 1.65
-

= 4.72 = 2.86 kips > 0.900 kips OK

1.65

=0.85
Po = (0.85)(2.36) = 2.01kips > 0.690 kips OK
-or Condition
=0.90
P; = (0.90)(4.72) = 4.25kips> 1.38kips OK
ined Bending and Web Crippling
1""'

---::rated
-SI)

load at center of beam controls

0.91(R.)
+ (~)
r;

~ 1.33

Mnxo

0.91(0.900)+(21.6)~
4.72
37.9

(Eq. C3.5.1-1)

1.33
1.70

0.743 < 0.782 OK

~

0.91(

P )+(
Pn

M )~1.33<l>
Mnxo

=P,
=M,

0.91(1.38) + (33.1) s 1.33(0.90)

4.72
37.9
1.14 < 1.20 OK

(Eq. C3.5.2-1)

11-182

Beam Design for Use with the 2007 North American Cold-Formed Stee

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

w

l--....

m
S.C.I

I A' = 9.000 in.

B' = 2.500 in.

Given:

1. Steel: F, = 55 ksi
2. Section: 9CS2.5x059
3. Gross Section Properties (from Example 1-1or Table 1-1)
I, = 10.3 in."

Sx = 2.29 in.'

J = 0.00102in."

xo = -1.66 in.

m = 1.05 in.

x = 0.641in.

Cw = 11.9 in."

4. Effective Section Properties (from Example 1-8or Table I1-1)

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 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 s;
below.
Required:

Determine the nominal flexural strength, Mn, based on initiation of yielding of the
section considering the effects of torsion. Consider alternate conditions of:
1. A single brace at mid-span
2. Two braces, each at the one-third points of the span
Assumptions:

1. Rotation is completely restrained at the member ends and at the braces.

2. The member is free to warp at both ends.

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

11-183

tion:
A torsional reduction factor, R is calculated using Section C3.6 and applied to the nominal
strength calculated using Section C3.1.1(a). Note that this reduction factor is not applied to
other limit states, such as lateral-torsional buckling or distortional buckling.
This solution is based on the method described in the AISC Steel Design Guide Series 9:
"Torsional Analysis of Structural Steel Members"7 (DG 9). The actualloading is modeled
y superimposing the three conditions as shown in the figure below.

_ along
eb

Load through
shear center
Simple
bending

Loading = #1

t,

Distributed
torque
DG 9 Case 4
t,=wxo

#2

Brace at
mid-span
DG 9 Case 3
u= 0.5
Ti

Brace at
1/3 points
DG 9 Case 3
u = 0.33 and 0.67
T2 = T2

#3

#3A

Torsional warping stresses are calculated using the second derivative of the angle of rotaion, 8, with respect to the postion, 2, along the length oEthe member.
The sign convention for use with all torsion
ressions are shown in the figure to the
rizht. Note that calculated values for 8 and
may be either positive or negative. The
::-Topersign for these calculated values must
_ used for torsional stress calculations. Cal:ulated positive values are in the directions
: own.

"

;~

,
,

,
,

Point Brc---:>~,:--ro

Compression
/-,

_.~ Z-<lirection

PointA

Positive rotation and warping stresses

~ P.A and Carter, Cl, "Torsional Analysis of Structural Steel Members - Steel Design Guide Se- ,American Institute of Steel Construction, Chicago, IL, 1997

Beam Design for Use with the 2007 North American Cold-Formed

11-184

,,-=..

For the singly-syrnmetric channel, only the compression side need be checked for f"" c;'
bending and warping.

d~

Normal Stresses Due to Warping

= EWns8"

~s

(AISC Design G_--

where Wns are normalized warping functions (section properties) of the cross-seczx __ ;
each point of consideration given by:
Point A, at tip of flange stiffener

a(n:-b) -c(m+b)

WA =

Point B, at junction of the flange and stiffener

a(m-b)

WB = ---'-_--...

Point C, at junction of the flange and web

W. _ am

c-

where,

a=

centerline web height = 8.941 in.

b = centerline flange width

2.441 in.

= centerline lip length = 0.744 in.

m = distance from shear center to web centerline

1.05 in.

The torsional warping properties for this section are:

WA = (8.941)(1.05-2.441) -(0.744)(1.05+2.441)=-8.82in.2
2
WB =

(8.941)(1.05 - 2.441)
= -6.22 in.?
2

Wc =

(8.941)(1.05)
=4.69 in.?
2

Formulas for rotation due to a number of torsionalloadings are given in Appendix C.4 o: _
Surnmarized below are those used in subsequent ca1culations.
For Loading #2 above, use DG 9 Case 4 - Uniformly distributed torque on member with _ends.
2
- -tanh (L).
- sinh (z)
- -1.0
[ -L (z--- Z2J +cosh (z)
Q ~2 L ~
a
~
a
a2

8t =-tr
where
a=

JECGJ

Differentiating twice with respect to z yields

WIIII

:o

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 s z:S;aL

and

e~ =~[[Sinh(~)
aGJ

tanh(~ J

-COSh(aLJlSinh(~JI
a
a

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

mid-span bracing, the stresses are maximum at mid-span. Combine Loadings #1, #2 and
_ .......
,g #1 - Simple bending through the shear center
f _My
b I
2

= _wL_

= 10(25/ (12) = 9.38 kip-in.

8(1000)

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

_ 9.38(4.859-0.773) _
k .
- -4.18 SI
9.18

fbA - -

fbB= fbC= -

9.38(4.859)
.
= -4.97ksl
9.18

___iU.LL,g
#2 - Uniformly distributed torque - use DG 9 Case 4.
2

=--tra [L2(Z
---- Z2) +cosh (zJ
- -tanh (LJ'
- snh (zJ
- -1.0
GJ 2a2 L L2
a
2a
a

-.rhere

tr

a=

10(1.05)
. . .
(
) = 0.000875kp-n.Zin.
12 1000
29500(11.9) = 175 in.
11300(0.00102)

11-186

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

L= 25(12) = 300 in.

L/a =300/175 =1.71

z = 150 in.

z/L = 150/300 = 0.500

(1.71)2(

0.000875(175)2

=
t

0.500-(0.500)

z/a = 150/175 = 0.857

2)

+cosh(0.857)
2
11300(0.00102) _tanh(_L~_l)sinh(0.857) -1.0
'---'-:-1

= 0.199 radians
e" =
t

0.000875 [-1.0 + cosh(0.857) - tanh(1.71)Sinh(0.857)]

11300(0.00102)
2

= 21.2 X 10-6
Loading #3 - Brace at Mid-Span - use DG 9 Case 3 with a = 0.5.
for O~z~aL
TL
z
eT =(1.0-a)-+GJ
L

a
L

SI 'nhaLa

[ tanh L

-cosh-

J snh >z 1

aL.
a

and
T
eT =_
aGJ
N

r[

sinh a
aL
z
aL
L - coshsinhtanha
a

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

eT

_
-

ei =

1.0(300)
[
1 (Sinh( 0.5(1.71))
) .
(
) (1-0.5)0.5+()
-cosh(0.5(l.71)) sinh
11300 0.00102
1.71
tanh 1.71
1.21 radians
1.0
[(Sinh(0.5(1.71))
)
175(11300)(0.00102)
tanh (1.71) - cosh (0.5(1.71)) sinh (0.857)

-172 X 10-6

Calculate the required value of torque provided by md-span brace to prevent rotation a

span.
e

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

TI = -0.164 kip-in.
Using this brace force, combine the calculated values for e" from each loading to obtain e'
the mid-span braced condition.

eN = e; + T1ei = -21.2x10-6 -0.164( -172x10-6) = 7.01x10-6

The torsional warping stresses are:
~

= EWne"= 29500Wn(7.01x10-6) = 0.207Wn

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

f.vA = 0.207 (-8.82)

= -1.83

ksi

f.vB= 0.207 (-6.22)

= -1.29

ksi

f.vc = 0.207( 4.69)

= 0.971 ksi

:e::a:rrti'TI1ethe location of the maximum

fA = fbA+ f.vA = -4.18 -1.83
fB = fbB+.f.vB= -4.97 - 1.29
fe

11-187

combined

flexural and warping

stress.

= -6.01 ksi
= -6.26 ksi

CONTROLS

= fbe + f.vc = -4.97 + 0.971 = -4.00 ksi

ate the reduction

factor.

fbending

(Eq. C3.6-1)

fbendmg+ ~orsion
-4.97
-4.97 -1.29

= 0.794

_.ote that this value occurs at the intersection

ease is permitted.
nlate the nominal
M,

yielding

RSeFy

(0.794)(1.89)(55)

applicable

of the flange and stiffener; therefore,

no in-

strength.

= 82.5 kip-in.

lirnit states should also be evaluated

(not shown) .

. d-Point Bracing

=-.

condition, stresses are calculated at both the third-points and at mid-span, since it is not
inspection which location will govem.
Superimpose the stresses from Loadings #1,
#3A. Use DG 9 Case 3 to calculate
and 9" at these points due to the torsional restraint
-=-"_'1'ied by the braces. The value of the torque at the brace points is calculated by requiring
:. e value of be zero at these two points. Note by syrnrnetry, the torques at the braces are
~-:"'"11.1S by

-":;"'-~l.g #1 - Simple bending

:?:exural stresses mid-span
are:
_ wL2 _ 10(25)212
M - -9-9(1000)

through

the shear center

are the sarne as previously

calculated.

Those at the third-points

_
..
-8.33 kip-in,

4A

= -

8.33 (4.859 - 0.773)

9.18

= -3.71

4B

= fbC = - 8.33 (4.859) = -4.41 ksi

ksi

9.18

o-

#2 - Uniformly

--a!ues at mid-span
z

= L/3 = 100in.

Distributed

Torque - Use DG 9 Case 4

are as previously
z/L

= 0.333

calculated.
z/a

Those at third-points

= 0.571

are:

Beam Design for Use with the 2007 North American Cold-Formed Steel S--

11-188

0.000875(175)2

8t1/3

11300(0.00102)

(1.71)2 (
2)
2
0.333-(0.333) +cosh(0.571)
-tanh

(1.71)

xsinh(0.571)-1.0

0.173 rads

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

0.000S75 [-1.0 + cosh(0.571) - tanh( 1.71) sinh (0.571)]

11300(0.00102)
2

-19.0

u/s

10-Q

By symmetry, 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 ex = 0.667
Apply the brace torque at 2/3 point and calculate 8T and 8~ at z
2L/3.

L/3, z

L/2 anc -::.

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

1.0 300

8T1/3=

/))

11300 0.00102 [

(1-0.667)0.333+-

Sinh(0.667(1.71))
tanh(l.71)
sinh(0.571)
1.71 [
(
)
-cosh 0.667(1.71)
1

0.S26radians

8"
=
1.0
[(Sinh(0.667(1.71))
Tl/3 175(11300)(0.00102)
tanh (1.71)
- cosht 0.667(1.71)) sinh( 0.571
-67.1

For z = L/2

10-Q

150

8
=
1.0(300)
[
1 [Sinh(0.667(1.71))
Tl/2 11300(0.00102) (1-0.667)0.500+ 1.71
tanh(1.71)
Sinh(0.S57)J
=coshf 0.667(1.71))
=

1.03 radians

8"
1.0
[(Sinh(0.667(171))
Tl/2 - 175(11300)(0.00102)
tanh(l.71)
=

-lOS

J-

cosh(0.667(1.71)) sinh(0.S57)

10-6

For z = 2L/3 = 200

8
=
1.0(300)
[
1 [Sinh(0.667(1.71))
T2/3 11300(0.00102) (1-0.667)0.667 + 1.71
tanh(l.71)
sinh(1.14)
-cosh( 0.667(1.71))
=

0.982 radians

::

tgn for Use wth the 2007 North American Cold-Formed Steel Specification

"1.0
[[Sinh(0.667(1.71))
BT2/3(
)(
)
()
175 11300 0.00102
tanh 1.71
= -156

-cosh(0.667(1.71))

11-189

J slnh(1.14)
.
1

10-6

....-.----.:::te the value of the torques at third-points required to prevent rotation at those brace
1/3

= 9tl/3 + T2BTl/3 + T29T2/3= 0.173 + 0.826T2 + 0.982T2 = O

= -0.0957 kip-in.

______ = ~

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

= -19.0xl0-6

+ (-0.0957)( -67.1xl0-6)

+ (-0.0957)( -156xl0-6)

= 2.35xl0-6

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

= 0.0693( -8.82) = -0.611ksi
- - = 0.0693(-6.22) = -0.431 ksi
-

= 0.0693( 4.69) = 0.325 ksi

e location of the maximum combined flexural and warping stress.

-_ =,A +lA

=-3.71-0.611=-4.32ksi

= ,B+ lB = -4.41- 0.431 = -4.84 ksi CONTROLS

:: = 4c + le = -4.41 + 0.325 = -4.09 ksi
~-=:.. e reduction factor at the 1/3 and 2/3 points.
-4.41
= 0.911
-4.41- 0.431

(Eq. C3.6-1)

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

= 0.911)(1.89)(55) = 94.7kip-in.
ional warping stresses at mid-span.

_ = 9;1/2+ 2T29i1/2 = -21.2x10-6 + 2( -0.0957)( -108xl0-6)

= -0.529

10-6

500Wn (-O.529x10-6)

= -0.0156Wn

=-0.0156 (-8.82) = 0.138 ksi

.0156(-6.22) = 0.0970 ksi

Beam Design for Use with the 2007 North American Cold-Formed Stee. 5::.....-

11-190

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

Determine the location of the maximum combined flexural and warping stress.
fA=fbA+~A =-4.18+0.138=-4.04ksi
fB= fbB+ ~B = -4.97 + 0.0970= -4.87 ksi
fe = fbe + ~e = -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

Since R exceeds 1.0, take R as 1.0 at midspan. The 15% increase is permitted since the
combined 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.

.LL""-~

sign for Use with the 2007 North American Cold-Formed Steel Specification

11-191

le 11-11: Web Crippling

3.0 in.
O--+ll-f-

3.0 in.
O-+~-f-

12.0 in.

3.0 in.
O-+ll-f-

3.0 in.

v//////
4.0 in.

- exural member: SSMA Stud 1200S200-68(50 ksi)

--,aring stiffener: SSMA Stud 362S162-33(33 ksi)
_ ed:
Calculate the available bearing strength of the joist section with the C-section bearing stiffer using both ASD and LRFD
um:

ate the available ASD and LRFD strength using Section C3.7.
Section C3.7.1 if the w I t, limits for the stiffener are not exceeded.
eck Applicability

Limits for Section C3. 7.1

- web of stiffener:

3.625- 2(0.0765 + 0.0346) = 98.3

0.0346
Limit= 1.28M
= 1.28~29500/33 = 38.3< 98.3 NG; therefore, try Section C3.7.2
eck Applicability

Limits for Section C3.7.2

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 stiffener
has a thickness of 0.0346 in. > 0.0329in. minimum. OK
The stiffener is attached to the flexural member with three screws. OK
_ The distance from the flexural member flanges to the first fastener is di 4 > di 8 minimum.
OK
-

The length of the stiffener is equal to the depth of the flexural member. OK
The bearing width is greater than 11/2 in. OK