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COLD FORMED STEEL STRUCTURES

LECTURES PREPARED BY:PROF. DR. Abdelrahim Khalil Dessouki

Lecture 8

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COMPRESSION MEMBERSCold-formed sections are made of thin material, and in many cases the shear center does not coincide with the centroid of the section. Therefore in the design of such compression members, consideration should be given to the following limit states depending on the configuration of the section, thicknessof material, and column length used:

1. Yielding2. Overall column buckling

a. Flexural buckling: bending about a principal axisb. Torsional buckling: twisting about shear centerc. Torsionalflexural buckling: bending and twisting

simultaneously3. Local buckling of individual elements

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YIELDINGIt is well known that a very short, compact column under axial load may fail by yielding. For this case, the yield load is simply

Py = AFy (5.1)where A = full cross-sectional area of column

Fy = yield point of steel FLEXURAL COLUMN BUCKLING

Elastic Buckling

A slender axially loaded column may fail by overall flexural buckling if the cross section of the column is a doubly symmetric shape (I-section), closed shape (square or rectangular tube), cylindrical shape, or point-symmetricshape (Z-shape or cruciform). For singly symmetric shapes, flexural buckling is one of the possible failure modes.

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The elastic critical buckling load for a long column can be determined by the Euler formula:

where KL/r is the effective slenderness ratio and r is the least radius of gyration.

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Inelastic BucklingFor the analysis of flexural column buckling in the inelastic range, two concepts have been used in the past. They are the tangent modulus method and the reduced modulus method.The tangent modulus method was proposed by Engesser in 1889. Based on this method, the tangent modulus load is :

and the critical buckling stress is :

Where : Et is the tangent modulus.The tangent modulus concept did not include the effect of elasticunloading. Engesser then corrected his theory And developed the reduced modulus or double modulus concept, as follows :

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Practical Equations :

In the 1996 edition of the AISI Specification, the design equations for calculating the nominal inelastic and elastic flexural buckling stresses were changed to those used in the AISC LRFD Specification as follows:

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where : (Fn)I is the nominal inelastic buckling stress, (Fn)e is the nominal elastic buckling stress, lc is the column slenderness parameter Fy/se,

in which se is the theoretical elastic flexural buckling stress of the column determined by Eq. (5.3).

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TORSIONAL BUCKLING ANDTORSIONAL FLEXURAL BUCKLING

Usually, closed sections will not buckle torsionally because of their large torsional rigidity. For open thin-walled sections, however, three modes of failure are considered in the analysis of overall instability (flexural buckling, torsional buckling, and torsionalflexural buckling) as previously mentioned. Distortional buckling has been considered in some design standards but not in the 1996 edition of the AISI Specification.

When an open section column buckles in the torsionalflexural mode, bending and twisting of the section occur simultaneously.As shown in Fig. next slide, the section translates u and v in the x and y directions and rotates an angle F about the shear center.

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Displacement of a nonysmmetric section

during torsionalflexural buckling.

The equilibrium of a column subjected to an axial load P leads to the following differential equations:

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where : Ix = moment of inertia about x-axis, in.4Iy = moment of inertia about y-axis, in.4u = lateral displacement in x direction, in.v = lateral displacement in y direction, in.F = angle of rotation, rad.X0= x-coordinate of shear center, in.y0 = y-coordinate of shear center, in.E = modulus of elasticity, = 29.5 X 103 ksi (203 GPa)G = shear modulus,= 11.3 X103 ksi (78 GPa)J = St. Venant torsion constant of cross section, in.4, =1/3 S liti3Cw = warping constant of torsion of cross section, in.6 (Appendix B)ECw = warping rigidityGJ = torsional rigidityr0 = polar radius of gyration of cross section about shear center,

= rx 2 + ry2 + x02 + y02rx, ry = radius of gyration of cross section about x- and y-axis, in.

All derivatives are with respect to z, the direction along the axis of the member.

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Boundary ConditionsConsidering the boundary conditions for a member with completely fixed ends, that is, at z = 0, L,

and for a member with hinged ends, that is, at z = 0, L,

Equations (5.8) to (5.10) result in the following characteristic equation:

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The buckling mode of the column can be determined by Eq.(5.13). The critical buckling load is the smallest value of the threeroots of Pcr. The following discussion is intended to indicate thepossible buckling mode for various types of cross section.

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Doubly Symmetric Shapes

For a doubly symmetric shape, such as an I-section or a cruciform, the shear center coincides with the centroid of the section (Fig. 5.5), that is,

xo = yo = 0 (5.17)For this case, the characteristic equation, Eq. (5.13), becomes

(Pcr - Px )(Pcr - Py )(Pcr - Pz ) = 0 (5.18)

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The critical buckling load is the lowest value of the following three solutions:

(Pcr )1 = Px (5.19) (Pcr )2 = Py (5.20) (Pcr )3 = Pz (5.21)

An inspection of the above possible buckling loads indicates that for doubly symmetric sections, the column fails either in pure bending or in pure torsion, depending on the column length and the shape of the section. Usually compression members are so proportioned that they are not subject to torsional buckling. However, if the designer wishes to evaluate the torsional buckling stress st, the following formula based on Eq. (5.16) can be used:

All notations are as explained before.

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EFFECT OF LOCAL BUCKLING ON COLUMN STRENGTHCold-formed steel compression members may be so proportioned that local buckling of individual component plates occurs before the applied load reaches the overall collapse load of the column. The interaction effect of thelocal and overall column buckling may result in a reduction of the overall column strength. In general, the influence of local buckling on column strength depends onthe following factors:

1. The shape of the cross section2. The slenderness ratio of the column3. The type of governing overall column buckling (flexural buckling,

torsional buckling, or torsionalflexural buckling)4. The type of steel used and its mechanical properties5. Influence of cold work6. Effect of imperfection7. Effect of welding8. Effect of residual stress9. Interaction between plane components10. Effect of perforations

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If an axially loaded short column has a compactcross section, it will fail in yielding rather than buckling, and the maximum load P can be determined as follows:

P = AFy (5.44) where : A = full cross-sectional area

Fy = yield point of steelHowever, for the same length of the column if the w/t ratios of compression elements are relatively large, the member may fail through local buckling at a stress less than Fy. Assuming the reduced stress is QFy instead of Fy, then:

P = A(QFy ) (5.45) where Q is a form factor, which is less than unity, representing the weakening influence due to local buckling.

From the above two equations it can be seen that the effect of local bucklingon column strength can be considered for columns that will fail in localbuckling by merely replacing Fy by QFy. The same is true for columns havingmoderate KL/r ratios.

Q-Factor Method

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The value of form factor Q depends on the form or shape of the section. It can be computed as follows for various types of sections:

1. Members Composed Entirely of Stiffened ElementsWhen a short compression member is composed entirely of Stiffened elements such as the tubular member shown in Fig. 5.1a, it will fail when the edge stress of the stiffened elements reaches the yield point. Using the effective width concept, the column will fail under a load :

P = Aeff Fy (5.46)Where: Aeff is the sum of the effective areas of all stiffened elements.

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Comparing the above equation with Eq. (5.45), it is obvious thatQa = Aeff/A (5.47)

where : Qa is the area factor.

2. Members Composed Entirely of Unstiffened ElementsIf a short compression member is composed entirely of unstiffened elements, it will buckle locally under a load

P = A scr (5.48)Where : scr is the critical local buckling stress of the unstiffenedelement.Comparing the above equation again with Eq. (5.45), it is found that :

Qs = scr /Fy = Fc/F (5.49) where : Qs = stress factor

Fc = allowable compressive stressF = basic design stress (0.60Fy) for the ASD method

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3. Members Composed of Both Stiffened and Unstiffened Elements:

If a short compression member is composed of both stiffened and unstiffened elements as shown in Fig. 5.1c, the useful limit of the member may be assumed to be reached when the stress in the weakest unstiffened element reaches the critical local buckling stress cr. In this case, the effective area Aeffwill consist of the full area of all unstiffened elements and the sum of the reduced or effective areas of all stiffened elements. That is,

P = Aeff scr (5.50)Comparing the above equation with Eq. (5.45),

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North American Specifications DESIGN FORMULAS FOR CONCENTRICALLY LOADED COMPRESSION MEMBERS

Appropriate design provisions are included in the North American Specifications for the design of axially loaded compression members. The following excerpts are adopted from Sec. C4 of the 2007 edition of the North American Specification for the ASD and LRFD methods :

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(a) The nominal axial strength, Pn, shall be calculated as follows:Pn = Ae Fn (5.53) Wc = 1.80 (ASD)Fc = 0.85 (LRFD)

where : Ae = Effective area at the stress Fn. For sections with circular holes, Ae shall be determined according to Section B2.2a, subject to the limitations of that section. If the number of holes in the effective length region times the hole diameterdivided by the effective length does not exceed 0.015, Ae canbe determined ignoring the holes.

Fn is determined as follows:

Where : And, Fe = the least of the elastic flexural, torsional, and torsionalflexural buckling stress determined accordingto Sections C4.1.1 through C4.1.5.

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(b) concentrically loaded angle sections shall be designed for an additional bending moment as specified in the definitions of Mx, My (ASD), or Mux, Muy, (LRFD) in Section C5.2.

Sections Not Subject to Torsional or TorsionalFlexural Buckling

For doubly symmetric sections, closed cross sections and any other sections which can be shown not to be subject to torsionalor torsionalflexural buckling, the elastic flexural buckling stress, Fe, shall be determined as follows:

Where : E = modulus of elasticityK = effective length factorL = laterally unbraced length of memberr = radius of gyration of the full, unreduced

cross section about axis of buckling.

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EFFECTIVE LENGTH FACTOR K

The factor K (a ratio of the effective column length to the actual unbraced length) represents the influence of restraint against rotation and translation at both ends of the column.

The theoretical K values and the design values recommended bythe Structural Stability Research Council are tabulated in Tablenext slide. 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 relativehorizontal displacements between both ends. 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.

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

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The K values to be used for the design of unbraced multistory orMultibay frames can be obtained from the alignment chart in Fig. on next slide. In the chart, G is defined as :

in which Ic is the moment of inertia and Lc is the unbraced lengthof the column, and Ib is the moment of inertia and Lb is 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-fee 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

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Example: Determine the allowable axial load for the square tubular column shown in Fig. below. Assume that Fy = 40 ksi, KxLx = KyLy = 10 ft, and the dead-to-live load ratio is 1/5. Use the ASD and LRFD methods.

Solution:A. ASD Method

Since the square tube is a doubly symmetric closed section, it will not be subject to torsionalflexural buckling. It can be designed for flexural buckling.

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1. Sectional Properties of Full Section:w = 8.00 - 2(R + t) = 7.415 in.A = 4(7.415 X 0.105 + 0.0396) = 3.273 in.2Ix =Iy = 2(0.105)[(1/12)(7.415)3 + 7.415(4 - 0.105/2)2]

+ 4(0.0396)(4.0 - 0.1373)2 = 33.763 in.4rx=ry= Ix/A = 33.763/3.273 = 3.212 in.

2. Nominal Buckling Stress, Fn According to Eq. (5.56), the elastic flexural buckling stress, Fe, is computed as follows:

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3. Effective Area, Ae. Because the given square tube is composed of four stiffened elements, the effective width of stiffened elements subjected to uniform compression can be computed from Eqs. (3.41) through (3.44).

w/t = 7.415/0.105 = 70.619l=(1.052/ K)(w/t)( Fn/E )

= (1.052/ 4)(70.619) 36.914/29,500 =1.314 Since l> 0.673, from Eq. (3.42),

b =r wWhere :

r=(1 - 0.22/l)/ l = (1 - 0.22/1.314)/1.314 = 0.634Therefore, b = (0.634)(7.415) = 4.701 in.The effective area is:

Ae = 3.273 - 4(7.415 - 4.701)(0.105) = 2.133 in.24. Nominal and Allowable Loads. Using Eq. (5.53), the nominal load is:

Pn = Ae Fn = (2.133)(36.914) = 78.738 kips The allowable load is :

Pa = Pn / Wc = 78.738/1.80 = 43.74 kips

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B. LRFD MethodIn Item (A) above, the nominal axial load, Pn, was computed to be = 78.738 kips. The design axial load for the LRFD method is:

Fc Pn = 0.85(78.738) = 66.93 kips Based on the load combination of dead and live loads, the required load is :

Pu = 1.2PD + 1.6PL = 1.2PD + 1.6(5PD ) = 9.2 PDwhere

PD axial load due to dead loadPL axial load due to live load

By using Pu = FcPn , the values of PD and PL are computed as follows:

PD = 66.93/9.2 = 7.28 kips PL = 5PD = 36.40 kips

Therefore, the allowable axial load is:Pa = PD + PL = 43.68 kips compared with 43.74 (ASD)

COMPRESSION MEMBERS