Yu W., LaBoube R.A. Cold-Formed Steel Design

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NORTH AMERICAN DESIGN FORMULAS FOR CONCENTRICALLY LOADED COMPRESSION MEMBERS 205

4. 45

◦

≤ θ ≤ 90

◦

5. 2 ≤ h

≤ 8, and

6. 0.04 ≤ D sin θ/b

≤ 0.5.

where h

= out-to-out web depth as deﬁned in Fig. 3.30

= out-to-out ﬂange width as deﬁned in Fig. 3.48

D = out-to-out lip dimension as deﬁned in Fig. 3.48

t = base steel thickness

θ = lip angle as deﬁned in Fig. 3.48

The distortional buckling stress, F

, shall be calculated in

accordance with Eq. (5.65):

= αk

12(1 − μ

)





(5.65)

where α is a value that accounts for the beneﬁt of an unbraced

length, L

, shorter than L

but can be conservatively taken as

α =

⎧

⎪

⎨

⎪

⎩

1.0forL

≥ L





ln(L

)

for L

(5.66)

where L

is the distance between discrete restraints that restrict

distortional buckling [for continuously restrained members L

= L

, but the restraint can be included as a rotational spring,

, in accordance with the provi si ons in C4.2(b) or (c)],

= 1.2h



D sin θ



0.6

≤ 10h

(5.67)

= 0.05 ≤ 0.1



D sin θ



1.4

≤ 8.0 (5.68)

E is the modulus of elasticity of steel, and μ is Poisson’s ratio.

(b) For C- and Z-Sections or Hat Sections or Any Open Section

with Stiffened Flanges of Equal Dimension Where the Stiffener

is Either a Simple Lip or a Complex Edge Stiffener. The

provisions of this section of the speciﬁcation shall apply to

any open section with stiffened ﬂanges of equal dimension,

including those meeting the geometric limits of C4.2(a):

φfe

+ k

φwe

+ k

φfg

φwg

(5.69)

where k

φfe

is the elastic rotational stiffness provided by the

ﬂange to the ﬂange/web juncture, in accordance with Eq.

(4.116); the elastic rotational stiffness provided by the web

to the ﬂange/web juncture is given as

φwe

(1 − μ

)

(5.70)

is the rotational stiffness provided by restraining elements

(brace, panel, sheathing) to the ﬂange/web juncture of a

member (zero if the ﬂange is unrestrained) (if rotational stiff-

ness provided to the two ﬂanges is dissimilar, the smaller

rotational stiffness is used);

φfg

is the geometric rotational

stiffness (divided by the stress F

) demanded by the ﬂange

from the ﬂange/web juncture, in accordance with Eq. (4.118);

and the geometric rotational stiffness (divided by the stress

) d emanded by t he web from the ﬂange/web juncture is

given as

φwg





(5.71)

where L is the minimum of L

and L



6π



1 − μ





(

− h

)

+ C

−

xyf

(

− h

)



(5.72)

and L

is the distance between discrete restraints that restrict

distortional buckling (for continuously restrained members L

= L

See Section C3.1.4(b) in Section 4.2.4.2 of this volume for

deﬁnition of variables in Eq. (5.72).

ling analysis that considers distortional buckling shall be

permitted to be used in lieu of the expressions given in Section

C4.2(a) or (b). The safety and resistance factors in Section C4.2

shall apply.

Additional Comments on North American Design

Formulas

In addition to the discussions of Section 5.2–5.6, the

following comments are related to some of the North Amer-

ican design provisions for concentrically loaded compres-

sion members:

1. Safety Factor. In the 1986 and earlier editions of the

AISI Speciﬁcation, the allowable axial load for the ASD

method was determined by either a uniform safety factor

of 1.92 or a variable safety factor ranging from 1.67 to 1.92

for wall thickness not less than 0.09 in. (2.3 mm) and F

/2. The use of the smaller safety factors for the type of

relatively stocky columns was occasioned by their lesser

sensitivity to accidental eccentricities and the difference

in structural behavior between the c ompression members

having different compactness. The latter fact is illustrated

by the stress–strain curves of the more compact and less

compact stub-column specimens, as s hown in Fig. 5.18. In

the experimental and analytical investigations conducted by

Karren, Uribe, and Winter,

2.14,2.17

both types of specimens

shown in Fig. 5.18 were so dimensioned that local buckling

would not occur at stresses below the yield stress of the

material. Test data did indicate that the less compact stub

column (curve A for cold-reduced killed press braked hat

section) reached the ultimate compressive load at strains in

the range of 3 × 10

−3

–5 × 10

–3

in./in., after which the

load dropped off suddenly because yielding was followed

by local crippling. However, the more compact stub column

206 5 COMPRESSION MEMBERS

Figure 5.18 Stress–strain curves for more compact and less

compact stub columns.

2.17

Figure 5.19 Comparison between the design axial strengths, P

for the ASD method.

(curve B for hot-rolled semi killed roll-formed channel

section) showed a long stable plateau and, after some strain

hardening, reached the ultimate load at much higher values

of strain in the range of 16 × 10

−3

–27 × 10

–3

in./in. For

this reason, the use of a smaller safety factor for more

compact sections is justiﬁed and appropriate as far as the

overall safety of the compression member is concerned.

As discussed in Section 5.3.2, the AISI design equations

were changed in 1996 on the basis of a different strength

model. These equations enable the use of a single safety

factor of 1.80 for all λ

values. Figure 5.19 shows a

comparison of design axial strengths permitted by the 1986

and 1996 AISI Speciﬁcations and the 2007 North American

Speciﬁcation. It can be seen that the design capacity is

increased by using the 1996 and 2007 Speciﬁcations for thin

columns with low slenderness parameters and decreased

for high slenderness parameters. However, the difference

would be less than 10%.

Figure 5.20 Comparison between the nominal axial strengths,

, for the LRFD method.

For the LRFD method, the differences between the

nominal axial strengths used for the 1991, 1996, and 2007

LRFD design provisions are shown in Fig. 5.20. The resis-

tance factor of φ

= 0.85 is the same as the 1999 AISC

Speciﬁcation

3.150

and the 1991 edition of the AISI LRFD

Speciﬁcation.

3.152

In the 2007 edition of the North American Speciﬁcation,

the safety factor for the ASD method and the resistance

factor for the LRFD method are the same as that used in

the 1996 AISI Speciﬁcation.

2. Maximum Slenderness Ratio. The maximum allow-

able slenderness ratio KL/r of compression members has

been preferably limited to 200, except that during construc-

tion the maximum KL/ r ratio should not exceed 300.

1.346

This limitation on the slenderness ratio is the same as that

used by the AISC for the design of hot-rolled heavy-steel

compression members. Even though the design formulas

are equally applicable to columns having s lenderness ratios

larger than 200, any use of the unusually long columns will

result in an uneconomical design due to the small capacity

to resist buckling. In 1999, the AISI Committee on Speciﬁ-

cations moved the KL/r limit from the Speciﬁcation to the

Commentary.

1.333

3. Flexural–Torsiona Buckling. The simpliﬁed equation

for ﬂexural–torsional buckling [Eq. (5.58)] is based on the

following formula given by Pekoz and Winter in Ref. 5.66:

TFO

(5.73)

TFO

(5.74)

EFFECTIVE LENGTH FACTOR K 207

For singly symmetric angle sections with unstiffened

legs, a study conducted at the University of Sydney

by Popovic, Hancock, and Rasmussen

5.100

indicated that

the relatively compact angle section will not fail in a

ﬂexural–torsional buckling mode if the effective area (A

)

under yield stress F

is equal to the full, unreduced cross-

sectional area (A). For this case, the concentrically loaded

angle section can be designed for ﬂexural buckling alone

in accordance with Section C4.1.1 of the Speciﬁcation.

4. Point-Symmetric Sections. As discussed in Section

5.4.3, point-symmetric sections may fail e ither in ﬂexural

buckling about the minor principal axis or twisting. Section

C4.1.3 was added in the North American Speciﬁcation

in 2001 for the design of concentrically loaded point-

symmetric reactions.

5. Design Tables. Part III of the AISI Design Manual

1.349

contains a number of design tables for column properties

and nominal axial strengths of C-sections with and without

lips including SSMA studs and tracks. The tables are

prepared for F

= 33, 50, and 55 ksi (228, 345, and

379 MPa, or 2320, 3515, and 3867 kg/cm

). In addition,

Tables III-4 to III-6 of the Design Manual provide computed

distortional buckling properties and strengths under axial

load for the representative C-shapes, SSMA studs, and Z-

shapes with lips, respectively. The values in these tables

have been calculated for use with Section C4.2(b).

5.9 EFFECTIVE LENGTH FACTOR K

In steel design, lateral bracing is generally used to resist

lateral loads, such as wind or earthquake loads, or to

increase the strength of members by preventing them from

deforming in their weakest direction.

4.111

The use of such

bracing may affect the design of compression members.

In Sections 5.3–5.8, the effective length KL of the c olumn

was used to determine buckling stresses. The factor K (a

ratio of the effective column length to the actual unbraced

length) represents the inﬂuence of restraint against rotation

and/or translation at both ends of the column.

The theoretical K values and the design values recom-

mended by the Structural Stability Research Council are

tabulated in Table 5.1. In practice, the value of K = 1 can

be used for columns or studs with X-bracing, diaphragm

bracing, shear-wall construction, or any other means that

will prevent relative horizontal displacements between both

ends.

1.161

If translation is prevented and restraint against

rotation is provided at one or both ends of the member, a

value of less than 1 may be used for the effective length

factor.

In the design of trusses, it is realized that considerable

rotational restraint could be provided by continuity of the

compression chord as long as the tension members do not

yield. In view of the fact that for the ASD method tension

Table 5.1 Effective Length Factor K for Axially Loaded Columns with Various End Conditions

Source: Reproduced from Guide t o Stability Design Criteria for Metal Structures, 4th ed., 1988. (Courtesy of John Wiley and Sons, Inc.)

208 5 COMPRESSION MEMBERS

Figure 5.21 Laterally unbraced portal frame.

1.161

members are designed with a safety factor of 1.67 and

compression members are designed with a relatively large

safety factor of 1.80, it is likely that the tension members

will begin to yield before the buckling of compression

members. Therefore, the rotational restraint provided by

tension members may not be utilized in design as discussed

by Bleich.

3.3

For this reason, compression members in

trusses can be designed for K = 1.

1.161

However, when

sheathing is attached directly to the top ﬂange of a contin-

uous chord, additional research has shown that the K value

may be taken as 0.75,

5.107

as discussed in Section 13.2.2.7.

Figure 5.22 Effective length factor K in laterally unbraced

portal frames.

1.161

For unbraced frames, the structure depends on its own

bending stiffness for lateral stability. If a portal frame is not

externally braced in its own plane to prevent sidesway, the

effective length KL is larger than the actual unbraced length,

as shown in Fig. 5.21, that is, K > 1. This will result in

a reduction of the load-carrying capacity of columns when

the sidesway is not prevented.

For unbraced portal frames, the effective column length

can be determined from Fig. 5.22 for the speciﬁc ratio of

(I /L)

beam

/(I /L)

col

and the condition of the foundation. If

the actual footing provides a rotational restraint between

hinged and ﬁxed bases, the K value can be obtained by

interpolation.

Figure 5.23 Alignment charts developed by L.S. Lawrence for effective length of column in

continuous frames. (Courtesy of Jackson & Moreland Division of United Engineers & Constructors,

Inc.)

1.158

BRACING OF AXIALLY LOADED COMPRESSION MEMBERS 209

The K values to be used for the design of unbraced

multistory or multibay frames can be obtained from the

alignment chart in Fig. 5.23.

1.158

In the chart, G is

deﬁned as

G =



(

)



(

)

(5.75)

in which I

is the moment of inertia and L

the unbraced

length of the column and I

is the moment of inertia and

the unbraced length of the beam.

In practical design, when a column base is supported by

but not rigidly connected to a footing or foundation, G is

theoretically inﬁnity, but unless actually designed as a true

friction-free pin, it may be taken as 10. If the column end

is rigidly attached to a properly designed footing, G may

be taken as 1.0.

1.158,5.67

In the use of the chart, the beam stiffness I

should

be multiplied by a factor as follows when the c onditions at

the far end of the beam are known:

1. Sidesway is prevented:

1.5 for far end of beam hinged

2.0 for far end of beam ﬁxed

2. Sidesway is not prevented:

0.5 for far end of beam hinged

0.67 for far end of beam ﬁxed

After determining G

and G

for joints A and B at two

ends of the column section, the K value can be obtained

from the alignment chart of Fig. 5.23 by constructing a

straight line between the appropriate points on the scales

for G

and G

5.10 BUILT-UP COMPRESSION MEMBERS

For built-up compression members composed of two

sections in contact, the available axial strength (factored

axial resistance) shall be determined in accordance with

Section C4.1(a) of the North American Speciﬁcation

(Section 5.8 in this volume) subjected to modiﬁcation as

necessary. Based on Section D1.2 of the 2007 edition of

the North American speciﬁcation, if the buckling mode

involves relative deformations that produce shear forces in

the connections between individual shapes, the effective

slenderness ratio KL/r is replaced by the modiﬁed effective

slenderness ratio (KL/r )

calculated by Eq. (5.76):















(5.76)

where (KL/r)

= overall slenderness ratio of entire section

about built-up member axis

a = intermediate fastener or spot weld

spacing

= minimum radius of gyration of full

unreduced cross-sectional area of an

individual shape in built-up member

See Section 5.8 for deﬁnition of other symbols. In addition,

the fastener strength (resistance) and spacing shall satisfy

the following:

1. The intermediate fastener or spot weld spacing a is

limited such that a/r

does not exceed one-half the

governing slenderness ratio of the built-up member.

2. The ends of a built-up compression member are

connected by a weld having a length not less than

the maximum width of the member or by connectors

spaced longitudinally not more than four diameters

apart for a distance equal to 1.5 times the maximum

width of the member.

3. The intermediate fastener(s) or weld(s) at any longi-

tudinal member tie location are capable of transmit-

ting a force in any direction of 2.5% of the nominal

axial strength (compressive resistance) of the built-up

member.

In the above design criteria, Eq.(5.76) was added to the

North American Speciﬁcation since 2001 on the basis of the

1999 AISC Speciﬁcation and the 1994 CSA Standard.

1.117

The overall slenderness ratio, (KL/r)

, is computed about

the same axis as the modiﬁed slenderness ratio, (KL/r)

The (KL/r)

ratio replaces KL/r for both ﬂexural and

ﬂexural–torsional buckling.

Section D1.2 of the North American Speciﬁcation

includes the above three requirements concerning interme-

diate fastener spacing a, end connection of the built-up

member, and the applied force for the design of inter-

mediate fastener(s). The intermediate fastener spacing

requirement (1) for [a/r

≤ 0.5(KL/r)] is to prevent

ﬂexural buckling of individual shapes between interme-

diate connectors to account for any one of the connectors

becoming loose or ineffective. Requirements 2 and 3 are

related to connection design as discussed in Section 8.8

and illustrated in Example 8.6.

For the research work on built-up compression

members, recent studies have been made by Yang and

Hancock,

5.188,5.191

Brueggen and Ramseyer,

5.189

Stone and

LaBoube,

5.190

Young and Chen,

5.192

and others.

5.11 BRACING OF AXIALLY LOADED

COMPRESSION MEMBERS

In the 2007 edition of the North American Speciﬁcation,

new design provisions are included for bracing of axially

loaded compression members. According to Section D3.3

210 5 COMPRESSION MEMBERS

of the Speciﬁcation, the required brace strength (resistance)

and brace stiffness to restrain lateral translation at a brace

point for an individual compression member shall be calcu-

lated by Eqs. (5.77) and (5.78), respectively, as follows:

br,1

= 0.01P

(5.77)

for required brace strength,

br,1

2[4 − (2/n)]P

(5.78)

for required brace stiffness

where P

br,1

= required nominal brace strength (resistance)

for a single compression member

= required axial compression strength

(resistance) for a single compression

member

br,1

= required brace stiffness for a single

compression member

n = number of equally spaced intermediate

brace locations

= distance between braces on one

compression member

The above design requirements were developed from a

study conducted by Green, Sputo, and Urala in 2004.

5.193

These provisions for lateral translation assume that the

braces are perpendicular to the compression member being

braced and located in the plane of buckling. The stiff-

ness requirements include the contributions of the bracing

members, connections, and anchorage details.

1.346

It should

be noted that these requirements are only for lateral trans-

lation of the compression member.They do not account

for torsion or ﬂexural–torsional buckling, which may be

designed through rational analysis or other methods. Addi-

tional design information on bracing can be found from

the publications of Helwig and Yura,

5.194

Green, Sputo,

and Urala,

5.195

Sputo and Turner,

5.196

Sputo and Beery,

5.197

Tovar, Helwig, and Sputo,

5.207

and others.

5.12 DESIGN EXAMPLES

Example 5.1 Determine the allowable axial load for

the square tubular column shown in Fig. 5.24. Assume

that F

= 40 ksi, K

= K

= 10 ft, and the dead

load–live load ratio is

.UsetheASDandLRFD

methods.

SOLUTION

A. ASD Method

Since the square tube is a doubly symmetric closed section,

it will not be subject to torsional or ﬂexural–torsional

buckling. It can be designed for ﬂexural buckling.

Figure 5.24 Example 5.1.

1. Sectional Properties of Full Section

w = 8.00 − 2(R + t) = 7.415 in.

A = 4(7.415 × 0.105 + 0.0396) = 3.273 in.

= I

= 2(0.105)



(7.415)

+ 7.415



4 −

× 0.105





+ 4(0.0396)(4.0 −0.1373)

= 33.763 in.

= r



33.763

3.273

= 3.212 in.

2. Nominal Buckling Stress F

. According to Eq. (5.56),

the elastic ﬂexural buckling stress F

is computed as

follows:

10 × 12

3.212

= 37.36 < 200 OK

(KL/r)

(

29,500

)

(

37.36

)

= 208.597 ksi





208.597

= 0.438 < 1.5



0.658





0.658

0.438



= 36.914 ksi

3. Effective Area A

. Because the given square tube

is composed of four stiffened elements, the effec-

tive width of stiffened elements subjected to

uniform compression can be computed from Eqs.

(3.41)–(3.44):

7.415

0.105

= 70.619

λ =

1.052

√







λ =



1.052

√



(

70.619

)



36.914

29,500

= 1.314

DESIGN EXAMPLES 211

Since λ>0.673, from Eq. (3.42),

b = ρw

where

ρ =

1 − 0.22/λ

1 − 0.22/1.314

1.314

= 0.634

Therefore,

b = (0.634)(7.415) = 4.701 in.

Theeffectiveareais

= 3.273 − 4(7.415 − 4.701)(0.105)

= 2.133 in.

4. Nominal and Allowable Loads. Using Eq. (5.53), the

nominal load is

= A

= (2.133)(36.914) = 78.738 kips

Theallowableloadis



78.738

1.80

= 43.74 kips

B. LRFD Method

In item A above, the nominal axial load P

was computed

to be 78.738 kips. The design axial load for the LRFD

method is

= 0.85(78.738) = 66.93 kips

Based on the load combination of dead and live loads, the

required load is

= 1.2P

+ 1.6P

= 1.2P

+ 1.6(5P

) = 9.2P

where P

= axial load due to dead load

= axial load due to live load

By using P

= φ

, the values of P

and P

are

computed as follows:

66.93

9.2

= 7.28 kips

= 5P

= 36.40 kips

Therefore, the allowable axial load is

= P

+ P

= 43.68 kips

It can be seen that the allowable axial loads determined

by the ASD and LRFD methods are practically the same.

Example 5.2 Use the ASD and LRFD methods to deter-

mine the available strengths of the I-section (Fig. 5.25) to be

used as a compression member. Assume that the effective

length factor K is 1.0 for the x and y axes and that the

Figure 5.25 Example 5.2.

unbraced lengths for the x and y axes are 12 and 6 ft,

respectively. Also assume that K

= 6ft.UseF

= 33 ksi.

The intermediate fastener spacing is assumed to be 12 in.

SOLUTION

A. ASD Method

1. Properties of Full Section. Based on the method used

in Chapter 4, the following full section properties can be

computed:

A = 2.24 in.

= 22.1in.

= 4.20 in.

= 3.15 in. r

= 1.37 in.

2. Nominal Buckling Stress, F

. Since the given I-section

is a doubly symmetric section, the nominal buckling stress

will be governed by either ﬂexural buckling or torsional

buckling as discussed in Section 5.4.1.

a. Elastic Flexural Buckling. By using Eq. (5.56), the

elastic ﬂexural buckling stress can be computed as

follows:

(1)(12 × 12)

3.15

= 45.714

(1)(6 × 12)

1.37

= 52.555

Use

= 52.555

Since the slenderness ratio KL/r is governed by the

column buckling about the y axis of the I-section, which

involves relative deformations that produce shear forces

in the connections between individual channels, the

modiﬁed slenderness ratio (KL/r )

should be used to

compute the elastic ﬂexural buckling s tress F

. Based

on Eq. (5.76),















212 5 COMPRESSION MEMBERS

where (KL/r)

= 52.555

a = intermediate fastener spacing,

=12 in.

= radius of gyration of a channel

section about its y axis, =1.08 in.

Therefore,







(52.555)



1.08



= 53.717

Since a/r

< 0.5(KL/r)

, requirement 1 of Section 5.10

is satisﬁed,

(KL/r)

(29,500)

(53.717)

= 100.902

b. Elastic Torsional Buckling. Using Eq. (5.22) of Section

5.4.1, the torsional buckling stress is

= σ



GJ +

(

)



where A = 2.24 in.



+ r



(

3.15

)

(

1.37

)

= 3.435 in.

G = 11,300 ksi

J = 0.00418 in.

= 70.70 in.

E = 29,500 ksi

= 6ft

Therefore

= σ

(

2.24

)(

3.435

)



(

11,300

)(

0.00418

)

(

29,500

)(

70.70

)

(

6 × 12

)



= 152.02 ksi

The nominal buckling stress F

is determined by using

the smaller value of the elastic ﬂexural buckling stress

and torsional buckling stress, that is,

= 100.902 ksi





100.902

= 0.572 < 1.5

From Eq. (5.54),

= (0.658

0.572

)(33) = 28.777 ksi

3. Effective Area A

at Stress, F

= 0.7 − (R + t) = 0.5313 in.

= 3.0 − 2(R + t) = 2.6625 in

= 8.0 − 2(R + t) = 7.6625 in.

a. Effective Width of the Compression Flanges (Section

3.5.3.1). From Eq. (3.80),

S = 1.28



= 1.28



29,500

28.777

= 40.982

0.328S = 13.442

2.6625

0.075

= 35.50

Since w

/t > 0.328S , use Eq. (3.81) to compute the

required moment of inertia of the e dge stiffener I

follows:

= 399t



− 0.328



= 399(0.075)



35.50

40.982

− 0.328



= 0.002 in.

The above computed value should not exceed the

following value:

= t



115(w

/t)



+ 5



= (0.075)



115(35.50)

40.982



+ 5



= 0.0033 in.

Therefore, use I

= 0.002 in.

For the simple lip edge

stiffener,

D = 0.7in. d = 0.5313 in.

0.5313

0.075

= 7.08

By using Eq. (3.83), the moment of inertia of the full

edge stiffener is

)

(0.5313)

(0.075)

= 0.000937 in.

From Eq. (3.82),

0.000937

0.0020

= 0.469 < 1.0OK

The effective width b of the compression ﬂange can be

calculated as follows:

0.7

2.6625

= 0.263

From Eq. (3.84),

n = 0.582 −

= 0.582 −

35.50

4 × 40.982

= 0.365 >

DESIGN EXAMPLES 213

Use n = 0.365. Since 0.25 < D/ w

< 0.8 and θ = 90

◦

k =



4.82 −



)

+ 0.43

= [4.82 − 5(0.263)](0.469)

0.365

+ 0.43 = 3.09 < 4.0

Use k = 3.09 to compute the effective width of the

compression ﬂange.

From Eqs. (3.41)–(3.44),

λ =

1.052

√







1.052

√

3.09

(35.50)



28.777

29,500

= 0.664 < 0.673

ρ = 1.0

b = w

= 2.6625 in.

b. Effective Width of Edge Stiffeners

0.5313

0.075

= 7.084

λ =

1.052

√







1.052

√

0.43

(7.084)



28.777

29,500

= 0.355 < 0.673



= w

= 0.5313 in.

Based on Eq. (3.79), the reduced effective width of the

edge stiffener is

= R



= (0.469)(0.5313) = 0.249 in. <d



c. Effective Width of Web Elements

7.6625

0.075

= 102.167 < 500 OK

λ =

1.052

√







1.052

√

(102.167)



28.777

29,500

= 1.678 > 0.673

ρ = 1 −

0.22/λ

= 1 −

0.22/1.678

1.678

= 0.518

b = ρw

= (0.518)(7.6625) = 3.969 in.

d. Effective Area A

= 2.24 − [4(0.5313 − 0.249)

+ 2(7.6625 − 3.969)](0.075)

= 2.24 − 0.639 = 1.601 in.

4. Nominal Axial Load for Flexural Buckling. The

nominal load is

= A

= (1.601)(28.777) = 46.07 kips

5. Distortional Buckling Strength. Because edge-

stiffened ﬂanges are used for the I-section, the nominal

axial load for distortional buckling should be checked

according to Section C4.2 of the North American

Speciﬁcation as provided here in Section 5.8.

I. Application of Section C4.2(a) of the Speciﬁcation. At

the present time (2009), the AISI Committee intends

to move Section C4.2(a) of the Speciﬁcation to the

Commentary because it is just for preliminary design

and has been speciﬁcally derived to always be more

conservative than Section C4.2(b). However, the

following calculations are included for the purpose of

comparison:

a. Yield load P

: Based on Eq. (5.63),

= A

= (2.24)(33) = 73.92 kips

b. Distortional Buckling Load P

crd

: From Eq. (5.64),

crd

= A

in which, the elastic distortional buckling stress F

is calculated in accordance with Section C4.2(a) of

the North American Speciﬁcation. In order to use

Eq. (5.65) to compute F

, the following geometric

limits should be satisﬁed:

i. 50 <(h

/t = 106.67)<200 OK

ii. 25 <(b

/t = 40)<100 OK

iii. 6.25 <(D/t= 9.33)<50 OK

iv. 45

◦

<(θ= 90

◦

) = 90

◦

v. 2 <(h

= 2.67)<8OK

vi. 0.04 <(Dsin θ/b

= 0.23)<0.5OK

From Eq. (5.65), the elastic distortional buckling stress

is calculated as follows:

= αk

12(1 − μ

)





From Eq. (5.67), the distortional buckling half-

wavelength L

= 12h



D sin θ



0.6

≤ 10h

= 12(8.00)



3.00 × 0.7 × sin 90

◦

8.00 × 0.075



0.6

≤ 10(8.00)

= 20.36 in. ≤ 80.00 in.

Use L

= 20.36 in. Since (L

= L

= 72 in.) > L

, α

= 1.0.

214 5 COMPRESSION MEMBERS

Based on Eq. (5.68), the plate buckling coefﬁcient k

for distortional buckling is

= 0.1



D sin θ



1.4

= 0.1



3.00 × 0.7 × sin 90

◦

8.00 × 0.075



1.4

= 0.578

Since 0.05 <k

< 8.00, use k

= 0.578. The elastic

distortional buckling stress F

= (1)(0.578)

(29,500)

12(1 − 0.3

)



0.075

3.00



= 9.63 ksi

Therefore,

crd

= A

= 2.24(9.63) = 21.57 kips

Based on Eq. (5.62),



crd



73.92

21.57

= 1.85 > 0.561

From Eq. (5.61), the nominal axial load for distortional

buckling based on Section C4.2(a) is



1 − 0.25



crd



0.6





crd



0.6



1 − 0.25



21.57

73.92



0.6





21.57

73.92



0.6

(73.92)

= 31.09 kips

II. Application of Section C4.2(b) of the Speciﬁcation.

Based on the equations listed in Table 4.4, the geometric

ﬂange properties for the C-section used for the given

I-section can be computed as follows. The reason

for the following calculations is that the mechanical

model for prediction of distortional buckling strength

considers the ﬂange itself as a “column” which may

undergo restrained ﬂexural–torsional buckling, and the

restraint comes from the web and additional attach-

ments. It should be noted that in the equations below the

symbols x

and y

are presented as x

and y

, respec-

tively, to prevent confusion with other sections of the

speciﬁcation.

h = h

− t = 8.000 − 0.075 = 7.925 in.

b = b − t = 3.000 − 0.075 = 2.925 in.

d = D −

t = 0.700 −

× 0.075 = 0.6625 in.

= (b + d)t = (2.925 +0.6625)(0.075)

= 0.269 in.

= t[t

+ 4bd

+ t

bd + d

]/12(b + d)

= (0.075)[(0.075)

(2.925)

+ 4(2.925)(0.6625)

+ (0.075)

(2.925)(0.6625)

+ (0.6625)

]/12(2.925 + 0.6625)

= 0.00637 in.

= t[b

+ 4db

]/12(b + d)

= (0.075)[(2.925)

+ 4(0.6625)(2.925)

12(2.925 + 0.6625)

= 0.243 in.

xyf

= tb

/4(b + d)

= (0.075)(2.925)

(0.6625)

/4(2.925 + 0.6625)

= 0.0196 in.

= b

/2(b + d) = (2.925)

/2(2.925 + 0.6625)

= 1.192 in.

=−d

/2(b + d)

=−(0.6625)

/2(2.925 + 0.6625)

=−0.0612 in.

=−[b

+ 2db]/2(b + d)

=−[(2.925)

+ 2(0.6625)(2.925)]/

2(2.925 + 0.6625)

=−1.733 in.

= [bt

+ dt

]/3

= [(2.925)(0.075)

+ (0.6625)(0.075)

]/3

= 0.000504 in.

= 0.0in.