AXIAL STRENGTH AISC 360-05

1 Index

Contenido1 Index......................................................................................................................................................1

2 AISC 360-05............................................................................................................................................1

2.1 Record Section properties:.............................................................................................................1

2.2 Determine K (Effective Length factor) for the X-axis and Y-axis.....................................................1

2.2.1 Columns.................................................................................................................................2

2.2.2 Frames....................................................................................................................................2

3 Methods:................................................................................................................................................3

3.1 Tables.............................................................................................................................................3

3.2 Manual Calculation........................................................................................................................8

3.2.1 Check Local stability...............................................................................................................8

3.2.2 Determine Fcr.........................................................................................................................9

3.2.3 Nominal strength..................................................................................................................10

4 Columns in Frames (BEAM COLUMN)..................................................................................................11

4.1 Shapes different to I-shapes.........................................................................................................14

2 STRUCTURAL CONCEPTS

The design compressive strength φPn and the allowable compressive strength Pn/Ω are determined using φc = 0.90 (LRFD) and Ωc = 1.67 (ASD) for all cases. The nominal compressive strength Pn is determined as the least value calculated for the limit states of

- flexural buckling, - torsional buckling, and - flexural-torsional buckling, as applicable.

Flexural buckling is applicable for doubly symmetric and singly symmetric members.

For singly symmetric and unsymmetric members, and certain doubly symmetric members, such as cruciform or built-up columns, the limit states of torsional or flexural torsional buckling are also applicable.

An important factor in the design of compression members is the slenderness ratio, KL/r, where L = laterally unbraced length of member, in (mm), r = governing radius of gyration, in (mm), andK = effective length factor.

Although the AISC Specification imposes no maximum slenderness limit, it is sometimes considered good practice to limit KL/r to 200 for members designed on the basis of compression.

The buckling coefficient K is the ratio of the effective column length to the unbraced length L.

(Also see Art. 5.2.) Values of K depend on the support conditions of the column to be designed. The AISC Specification indicates that K should be taken as unity for columns in braced frames unless analysis indicates that a smaller value is justified. Analysis is required for determination of K for unbraced frames, but K should not be less than unity. Design values for K recommended by the Structural Stability Research Council for use with six idealized conditions of rotation and translationat column supports are illustrated in Fig. 5.1.The following articles give the compressive strength of members without slender elements.Provisions for the latter may be found in Art. E7 of the AISC Specificatio

2.1 EFFECTIVE WIDTH CONCEPT (COLD FORMED STEEL)The design of cold-formed steel differs from heavier construction in that elements of members typically have large width-to-thickness (w/t) ratios and are thus subject to local buckling.

Figure 9.2 illustrates local buckling in beams and columns. Flat elements in compression that have both edges parallel to the direction of stress stiffened by a web, flange, lip, or stiffener are referred to as stiffened elements.

Examples in Fig. 9.2 include the top flange of the channel and the flanges of the I-cross-section column.

To account for the effect of local buckling in design, the concept of effective width is employed for elements in compression. The background for this concept can be explained as follows.

FIGURE 9.2 Local buckling of compression elements. (a) In beams. (b) In columns. (Source: Commentary on the North American Specification for the Design of Cold-Formed Steel Structural Members, American Iron and Steel Institute, Washington, D.C., 2001, with permission.)

Unlike a column, a plate does not usually attain its maximum load-carrying capacity at the buckling load, but usually shows significant postbuckling strength.

This behavior is illustrated in Fig. 9.3, where longitudinal and transverse bars represent a plate that is simply supported along all edges. As the uniformly distributed end load is gradually increased, the longitudinal bars are equally stressed and reach their buckling load simultaneously. However, as the longitudinal bars buckle, the transverse bars develop tension in restraining the lateral deflection of the longitudinal bars. Thus, the longitudinal bars do not collapse when they reach their buckling load but are able to carry additional load because of the transverse restraint. The longitudinal bars nearest the center can deflect more than the bars near the edge, and therefore, the edge bars carry higher loads after buckling than do the center bars.

The postbuckling behavior of a simply supported plate is similar to that of the grid model. However, the ability of a plate to resist shear strains that develop during buckling also contributes to its postbuckling strength.

Although the grid shown in Fig. 9.3a buckled into only one longitudinal half-wave, a longer plate may buckle into several waves, as illustrated in Figs. 9.2 and 9.3b.

For long plates, the half-wavelength approaches the width b.

After a simply supported plate buckles, the compressive stress will vary from a maximum near the supported edges to a minimum at the mid-width of the plate, as shown by line 1 of Fig. 9.3c. As the load is increased the edge stresses will increase, but the stress in the mid-width of the plate may decrease slightly. The maximum load is reached and collapse is initiated when the edge stress reaches the yield stress—a condition indicated by line 2 of Fig. 9.3c.

The postbuckling strength of a plate element can be considered by assuming that, after buckling, the total load is carried by strips adjacent to the supported edges which are at a uniform stress equal to the actual maximum edge stress. These strips are indicated by the dashed lines in Fig. 9.3c. The total width of the strips, which represents the effective width of the element b, is defined so that the product of b and the maximum edge stress equals the actual stresses integrated over the entire width.

FIGURE 9.3 Effective width concept. (a) Buckling of grid model. (b) Buckling of plate. (c) Stress distributions.

The effective width decreases as the applied stress f increases. At maximum load, the stress on the effective width is the yield stress fy.

Thus, an element with a small enough w/t will be able to reach the yield point and will be fully effective. Elements with larger ratios will have an effective width that is less than the full width, and that reduced width will be used in section property calculations.

The behaviour of elements with other edge-support conditions is generally similar to that discussedabove. However, an element supported along only one edge will develop only one effective strip.Equations for calculating effective widths of elements are given in subsequent articles based onthe AISI NAS. These equations are based on theoretical elastic buckling theory but modified to reflectthe results of extensive physical testing.9.7 MAXIMUM WIDTH-TO-THICKNESS RATIOSThe AISI NAS gives certain maximum width-to-thickness ratios that must be adhered to.

For flange elements, such as in flexural members or columns, the maximum flat width-to thickness ratio, w/t, disregarding any intermediate stiffeners, is as follows:

Stiffened compression element having one longitudinal edge connected to a web or flange element, the other stiffened by

• A simple lip, 60• Other stiffener with IS Ia, 90• Other stiffener with IS ≥Ia, 90

Stiffened compression element with both longitudinal edges connected to other stiffened elements, 500

Unstiffened compression element, 60

In the above,- IS is the moment of inertia of the stiffener about its centroidal axis, parallel to the element to be stiffened, and - Ia is the moment of inertia of a stiffener adequate for the element to behave as a stiffened element.

Note that, although greater ratios are permitted, stiffened compression elements

2.2 Equations for Axially Loaded Metal Columns

3 AISC 360-05

Brace and unbraced frames. Buckling resistance and torsional buckling according to AISC 13 th edition . Personal development. 1 hour. Chart to obtain K. Rotational stiffness.

If AMERICAN SECTIONS, check table 22 to determine Available Critical Stress

3.1 Record Section properties: Ag , rx , ry , b , be, k , tf, t , kdes

where kdes is the design value of k. (Different manufacturers will produce this shape with different values of k. The design value is the smallest of these values. The detailing value is the largest.)

Determine Lb

Determine if braced or unbraced, if there is translation/sway or non-sway/non-translation

3.2 Determine K (Effective Length factor) for the X-axis and Y-axis Select case

3.2.1 ColumnsTable C-C2.2

For isolated columns that are not part of a continuous frame, Table C-C2.2 in the Commentary to the Specification will usually suffice.

3.3 Determine KL/r

3.3.1 Columns FY (N/mm2)

355 113.2275 128.6

3.3.2 FramesBraced frames Unbraced frames

Kx·L/rx

Ky·L/ry

If there is sway then K·Lrxry

Take the larger of them ,

Condition: KL/r < 200

wherePr = required axial strengthPc = available axial strength

3.3.3 Notes On effective length:

3.4 Compressive Strength for Flexural BucklingThe nominal compressive strength Pn for the limit state of flexural buckling isPn = AgFcr where Fcr is the buckling stress, determined according to Eqs. (5.20) and (5.21).

When (KL/r) ≤ 4.71 √E/Fy or Fe ≥ 0.44 Fy,

When (KL/r) > 4.71√E/Fy or Fe < 0.44Fy,

Strength of section. Manual Calculation Notes on this issue. Exercises and learn on design

where Fe = elastic buckling stress determined according to Eq. (5.22) or from the stability analysis:

FY (N/mm2)

355 113.2275 128.6

If KL/r > 200 then Not appropriate

If KL/r < or Fe ≥ 0.44 Fy then Eq. E3-2

If KL/r > then Eq. E3-3

3.4.1 Check Local stability. Determine compactness or slenderness of section The strength corresponding to any buckling mode cannot be developed, however, if the elements of the cross section are so thin that local buckling occurs. This type of instability is a localized buckling or wrinkling at an isolated location.

If it occurs, the cross section is no longer fully effective, and the member has failed. I- and H-shaped cross sections with thin flanges or webs are susceptible to this phenomenon, and their use should be avoided whenever possible.

Otherwise, the compressive strength given by:- AISC Equations E3-2 and E3-3 must be reduced.

The measure of this susceptibility is the width–thickness ratio of each cross-sectional element. Two types of elements must be considered: 1. unstiffened elements, which are unsupported along one edge parallel to the direction of load, and 2. stiffened elements, which are supported along both edges.

Limiting values of width–thickness ratios are given in AISC B4, “Classification of Sections for Local Buckling,” where cross-sectional shapes are classified as - compact, - noncompact, or - slender,

according to the values of the ratios.

For uniformly compressed elements, as in an axially loaded compression member, the strength must be reduced if the shape has any slender elements.

The width–thickness ratio is given the generic name of λ. Depending on the particular cross-sectional element, λ for I- and H-shapes is either the ratio b/t or h/tw, both of which are defined presently.

- If λ is greater than the specified limit, denoted λr , the shape is slender, and the potential for local buckling must be accounted for.

For I- and H-shapes, the projecting flange is considered to be an unstiffened element, and its width can be taken as half the full nominal width.

Using AISC notation givesCheck limiting width-thickness ratio for compression elements

(We postpone a discussion of the compactand noncompact categories until Chapter 5, “Beams.”)

3.4.1.1 Limits on flange/web ratios on compression member See 7.4 COMPRESSION MEMBERS from Structural Steel Designers Handbook (3rd Edition) Brockenbrough & Merritt- bookmarked.pdfFlange: bf/2tf // Allowable: AISC 9th edition: 171/sqr(Fy) / AISC 13th edition: 253.55/sqr(Fy) (S.I) Web: D/tw // Allowable: AISC 9th edition: 1683/sqr(Fy) / AISC 13th edition: 674.76/sqr(Fy) (S.I)

Values of λr for I-beams / Channels:FY (N/mm2)355 13.5 35.8275 15.3 40.7

If slender: Stiffener or use factor Q (foot note C in AISC manual section tables)

KL/r , an upper limit of 200 is recommended

3.4.1.2 Compact (no instability)Calculate Euler load Fe: (critical buckling load according to the Euler equation)

Determine Fcr

Fe = elastic critical buckling stress, calculated using Equations E3-4 and E4-4 for doubly symmetric members,

CHANNELS & ANGLESEquations E3-4 and E4-5 for singly symmetric members, and

OTHER MEMBERSEquation E4-6 for unsymmetric members, except for single angles where Fe is calculated using Equation E3-4.

3.4.1.3 Slender For uniformly compressed elements, as in an axially loaded compression member, the strength mustbe reduced if the shape has any slender elements.(TAKE INTO ACCOUNT IF SLENDER STIFFENED OR SLENDER UNSTIFFENED)SEE E7. MEMBERS WITH SLENDER ELEMENTS AISC 2005 - Specification for Structural Steel Buildings

The nominal strength is Pn = Fcr·Ag (AISC Equation E7-1)

Fe = elastic critical buckling stress, calculated using Equations E3-4 and E4-4 for doubly symmetric members, Equations E3-4 and E4-5 for singly symmetric members, and Equation E4-6 for unsymmetric members, except for single angles where Fe is calculated using Equation E3-4.

3.4.2 Nominal strength Pn

3.4.2.1 For LRFD,Pu ≤ Φc ·Pn Φc = 0.90Pu =Factored load sum of the factored loads

3.4.2.2 For ASDPa = sum of the service loads and being Ωc= 1.67 (compression factor)

fa computed axial compressive stress Pa/Ag

Fa allowable axial compressive stress

3.4.3 Check width-thickness ratiosFrom the dimensions and properties table in the Manual, the width-thickness ratio for the larger overall dimension is h/t and the ratio for the smaller dimension is b/t

From AISC Table B4.1, Determine case XX , and upper and lower limit CHECK COMPRESSION

the upper limit for nonslender elements is

The reduction factor Q is the product of two factors—Qs for unstiffened elements and Qa for stiffened elements.- Q = 1.0 for members with compact and noncompact sections, as defined in Section B4, - If the shape has no slender unstiffened elements, Qs = 1.0. - If the shape has no slender stiffened elements, Qa = 1.0.- for uniformly compressed elements = QsQa for members with slender-element sections, as defined in Section B4, for uniformly compressed elements.

User Note: - For cross sections composed of only unstiffened slender elements, Q = Qs (Qa = 1.0).- For cross sections composed of only stiffened slender elements, Q = Qa (Qs = 1.0). - For cross sections composed of both stiffened and unstiffened slender elements, Q = Qs·Qa.

Many of the shapes commonly used as columns are not slender, and the reduction will not be needed. This includes most (but not all) W-shapes. However, a large number of hollow structural shapes (HSS), double angles, and tees have slender elements.

3.4.3.1 AISC E.7.1. Slender Unstiffened Elements, Qs

AISC Specification Section E7.1 gives the procedure for calculating Qs for slender unstiffened elements: The procedure is straightforward, and involves comparing the width-thickness ratio with a limiting value and then computing Qs from an expression that is a function of the width-thickness ratio, Fy, and E.

The reduction factor Qs for slender unstiffened elements is defined as follows:- (a) For flanges, angles, and plates projecting from rolled columns or other compression

members:

- (b) For flanges, angles, and plates projecting from built-up columns or other compression members:

- (c) For single angles:

(d) For stems of tees:

3.4.3.2 AISC E.7.2. Slender Stiffened Elements, Qa

The computation of Qa for slender stiffened elements is given in AISC E7.2 and is slightly more complicated than the procedure for unstiffened elements. The general procedure is as follows.

- Compute an effective area of the cross section. This requires a knowledge of the stress in the effective area, so iteration is required. The Specification allows a simplifying assumption, however, so iteration can be avoided.

- Compute Qa = Aeff/A, where Aeff is the effective area and A is the actual area:

Aeff is the reduced effective area. The Specification user note for square and rectangular sections permits a value of f = Fy to be used in lieu of determining f by iteration.

wheref = Pn/Aeff

User Note: In lieu of calculating f = Pn/Aeff, which requires iteration, f may be taken equal to Fy . This will result in a slightly conservative estimate of column capacity.

be = effective width of the slender element

3.5 Compressive Strength for Torsional and Flexural-Torsional BucklingThis article applies to singly symmetric and unsymmetric members, and certain doubly symmetricmembers, such as cruciform or built-up columns. The elements of the members must have width–thickness ratios for axially compressed elements such that the sections are classified as compact ornoncompact (see Art. 5.1.5). For single-angle members, see Art. 5.4.3.The nominal compressive strength Pn is calculated from Eq. (5.19) based on the limit states of flexural-torsional buckling and torsional buckling. Fcr is determined as follows:

where The following definitions apply:Kz = effective length factor for torsional bucklingG = shear modulus of elasticity of steel = 11,200 ksi (77,200 MPa)Cw = warping constant, in6 (mm6)J = torsional constant, in4 (mm4)Ix, Iy = moment of inertia about the principal axes, in4 (mm4)xo, yo = coordinates of shear center with respect to the centroid, in (mm)ṝo= Polar radius of gyration about shear center, in (mm)

For doubly symmetric I-shaped sections, Cw may be taken conservatively as Iyd2/4. For tees and double angles, take Cw and xo as 0.

3.5.1 For double-angle and tee-shaped compression members,

where Fcry is determined according to Eq. (5.20) or (5.21) for flexural buckling about the y axis of symmetry with (KL/r) (KL/ry), and

For all other cases, Fcr is determined from Eq. (5.20) or (5.21), but with Fe determined as follows:

3.5.2 For doubly symmetric members,

3.5.3 For singly symmetric members where y is the axis of symmetry

3.5.4 For unsymmetric members, Fe is the lowest root of the cubic equation

3.6 Compressive Strength of Single-Angle MembersThe compressive strength of single-angle members can be determined from Eqs. (5.19)–(5.21), usingKL/r as given in this article and neglecting eccentricity, provided the angles are (1) loaded at the endsin compression through the same one leg, (2) attached by welding or by two-bolt-minimum connections,and (3) subjected to no intermediate transverse loads. For other conditions, see the AISCSpecification. The modified slenderness ratios are intended to account indirectly for bending due toeccentricity of loading and end restraint from truss chords.For equal-leg angles, or unequal-leg angles connected through the longer leg, which are individualmembers or are web members of planar trusses with adjacent web members attached to the sameside of the gusset plate or chord:When 0 ≤ (L/rx) ≤ 80,

(5.33)When (L/rx) > 80,

(5.34)For unequal-leg angles with leg length ratios less than 1.7 connected through the shorter leg, theKL/r from Eqs. (5.33) and (5.34) should be increased by adding 4[(bl/bs)2 − 1], but KL/r should not be less than 0.95L/rz.For equal-leg angles or unequal-leg angles connected through the longer leg that are web membersof box or space trusses with adjacent web members attached to the same side of the gusset plate or chord:When 0 ≤ (L/rx) ≤ 75,

(5.35)When (L/rx) > 75,

(5.36)For unequal-leg angles with leg length ratios less than 1.7, connected through the shorter leg, theKL/r from Eqs. (5.35) and (5.36) should be increased by adding 6[(bl/bs)2 − 1], but KL/r should notbe less than 0.82L/rz.The following definitions apply:L = length of member between work points at truss chord centerlines, in (mm)rz = radius of gyration about minor axis, in (mm)rx = radius of gyration about axis parallel to connected leg, in (mm)bl = longer leg of angle, in (mm)bs = shorter leg of angle, in (mm)

3.7 Compressive Strength of Built-up Members

The compressive strength of built-up members comprised of two or more shapes interconnected by stitch bolts or welds should be determined from Eqs. (5.19) to (5.22), but with KL/r replaced by a modified column slenderness ratio (KL/r)m determined as follows.For intermediate connectors that are snug-tight bolted,

For intermediate connectors that are welded or fully tensioned bolted,

where (KL/r)o slenderness ratio of built-up member acting as a unit in the buckling directionbeing considereda distance between connectors, in (mm)ri minimum radius of gyration of individual component, in (mm)rib radius of gyration of individual component relative to its centroidal axis parallelto member axis of buckling, in (mm)separation ratio h/2rib

h distance between centroids of individual components perpendicular to the memberaxis of buckling, in (mm) The following dimensional requirements apply for built-up members. Individual components ofcompression members composed of two or more shapes should be connected to one another at intervals

a such that the effective slenderness ratio Ka/ri of each of the component shapes, between theconnectors, does not exceed three-fourths of the governing slenderness ratio of the built-up member.Use the least radius of gyration ri to compute the slenderness ratio of each component part. The endconnection must be welded or fully tensioned bolted with Class A or B faying surfaces. The end connectionmay be designed for the full compressive load in a bearing-type bolted connection, but thebolts must be fully tightened (see Art. 5.9.5).At the ends of built-up compression members bearing on base plates or milled surfaces, all componentsin contact with one another must be connected by a weld having a length not less than themaximum width of the member, or by bolts spaced longitudinally not more than four diameters apartfor a distance equal to 11/2 times the maximum width of the member.Along the length of built-up compression members between the end connections, longitudinalspacing for intermittent welds or bolts should be adequate to transfer required forces. Where a componentof a built-up compression member consists of an outside plate, and intermittent welds are providedalong the edges of the components or bolts are provided on all gage lines at each section, themaximum spacing should not exceed the thickness of the thinner outside plate times 0.75 √E/Fy ,nor12 in (305 mm). When fasteners are staggered, the maximum spacing on each gage line should notexceed the thickness of the thinner outside plate times 1.12 √E/Fy nor 18 in (460 mm).Open sides of compression members built up from plates or shapes should be provided withcontinuous cover plates perforated with a succession of access holes. According to the AISCSpecification, the unsupported width of such plates at access holes contributes to the design strengthprovided the following requirements are met: (1) the width–thickness ratio conforms to the limitationsof Art. 5.1.5, (2) the ratio of hole length in the direction of stress to hole width of hole does notexceed 2, (3) the clear distance between holes in the direction of stress is not less than the transversedistance between the nearest lines of connecting fasteners or welds, and (4) the periphery of the holeshas a radius no less than of 11/2 in (38 mm).As an alternative to perforated cover plates, lacing can be used with tie plates at each end and atpoints where the lacing is interrupted. In members providing design strength, the end tie platesshould have a length of not less than the distance between the lines of fasteners or welds connectingthem to the components of the member. Intermediate tie plates should have a length not less thanone-half this distance. The thickness of tie plates should be not less than 1/50th the distance betweenlines of welds or fasteners connecting them to the segments of the members. In welded construction,the welding on each line connecting a tie plate should total at least one-third the length of the plate.In bolted construction, the spacing in the direction of stress in tie plates should be not more than sixdiameters and the tie plates should be connected to each segment by at least three fasteners.Lacing, including flat bars, angles, channels, or other shapes employed as lacing, must be sospaced that L/r of the flange included between their connections does not exceed three-fourths of thegoverning slenderness ratio for the member as a whole. Lacing must be proportioned to provide ashearing strength normal to the axis of the member equal to 2% of the compressive design strengthof the member. The L/r ratio for lacing bars must not exceed 140 for single lacing systems, or 200for double lacing systems. Join double lacing bars where they intersect. For single lacing bars incompression, take L as the unsupported length of the lacing bar between welds or fasteners connectingit to the components of the built-up member. For double lacing, take L as 70% of that distance. It is considered good practice to keep the inclination of lacing bars to the axis of the memberto not less than 60for single lacing or 45for double lacing. Also, use double lacing or lacingmade up of angles where distance between the lines of bolts or welds exceeds 15 in (380 mm).For limitations on the longitudinal spacing of connectors between elements in continuous contactconsisting of a plate and a shape or two plates, and other limitations, see Art. 5.9.7.

3.8 Strength From Tables 1. (FROM TABLES): Table 4-22 Available Critical Stress for Compression MembersFrom tables obtain for LFRD: Fcr · Φc From tables obtain for ASD: Fcr/Ωc

Being Ωc= 1.67 and Φc = 0.90

These expressions can be written in the general form

4 FRAMES (BEAM COLUMN)

A more rational procedure, however, will account for the degree of restraint provided by connecting members

The rotational restraint provided by the beams, or girders, at the end of a column is a function of the rotational stiffnesses of the members intersecting at the joint. The rotational stiffness of a member is proportional to EI/L, where I is the moment of inertia of the cross section with respect to the axis of bending. Gaylord, Gaylord, and Stallmeyer (1992) show that the effective length factor K depends on the ratio of column stiffness to girder stiffness at each end of the member, which can be expressed as

4.1 G, Ratio of column stiffness to girder:

Assumptions• All columns under consideration reach

buckling Simultaneously• All joints are rigid• Consider members lying in the plane of

buckling• All members have constant A

Use alignment charts (Structural Stability Research Council SSRC) LRFD Commentary Figure C-C2.2 p 16.1-241,242

When Connections to foundations(a) Hinge: G is infinite - Use G=10(b) Fixed: G=0 - Use G=1.0

Values of G: - for a pinned support, G should be taken as 10.0;- for a fixed support, G should be taken as 1.0. (The latter support condition corresponds to an infinitely stiff

girder and a flexible column, corresponding to a theoretical value of G = 0).

the discussion accompanying the alignment chart in the Commentary recommends a value of G = 1.0 because true fixity will rarely be achieved.

The relationship between G and K has been quantified in the Jackson–Mooreland Alignment Charts (Johnston, 1976), which are reproduced in Figures C-C2.3 and C-C2.4 in the Commentary. To obtain a value of K from one of these nomograms, first calculate the value of G at each end of the column, letting one value be GA and the other be GB. Connect GA and GB with a straight line, and read the value of K on the middle scale. The effective length factor obtained in this manner is with respect to the axis of bending, which is the axis perpendicular to the plane of the frame.

A separate analysis must be made for buckling about the other axis. Normally the beam-to-column connections in this direction will not transmit moment; sidesway is prevented by bracing; and K can be taken as 1.0.

Unbraced frames are able to support lateral loads because of their moment resisting joints. Often the frame is augmented by a bracing system of some sort; such frames are called braced frames. The additional resistance to lateral loads can take the form of diagonal bracing or rigid shear walls, as illustrated in Figure 4.16.

Unbraced frames are able to support lateral loads because of their moment resisting joints

for a fixed support, G should be taken as 1.0. The latter support condition corresponds to an infinitely stiff girder and a flexible column, corresponding to a theoretical valueof G = 0. The discussion accompanying the alignment chart in the Commentary recommendsa value of G = 1.0 because true fixity will rarely be achieved.Unbraced frames are able to support lateral loads because of their moment resisting joints. Often the frame is augmented by a bracing system of some sort; such frames are called braced frames.

.

Braced frames Often the frame is augmented by a bracing system of some sort. The additional resistance to lateral loads can take the form of diagonal bracing or rigid shear walls. The additional resistance to lateral loads can take the form of diagonal bracing or rigid shear walls, as illustrated in Figure 4.16

A frame must resist not only the tendency to sway under the action of lateral loads but also the tendency to buckle, or become unstable, under the action of vertical loads. Bracing to stabilize a structure against vertical loading is called stability bracing. Appendix 6 of the AISC Specification, “Stability Bracing for Columns and Beams,” covers this type of bracing.

Two categories are covered: relative and nodal. 1. With relative bracing, a brace point is restrained relative to adjacent brace points. A relative brace is

connected not only to the member to be braced but also to other members, as with diagonal bracing. With relative bracing, both the brace and other members contribute to stabilizing the member to be braced.

2. Nodal bracing provides isolated support at specific locations on the member and is not relative to other brace points or other members.

The provisions of AISC Appendix 6 give equations for the required strength and stiffness (resistance to deformation)of stability bracing. The provisions for columns are from the Guide to StabilityDesign Criteria (Galambos, 1998). The required strength and stiffness forstability can be added directly to the requirements for bracing to resist lateral loading.Stability bracing is discussed further in Chapter 5, “Beams,” and Chapter 6,“Beam–Columns.”

Columns that are members of braced rigid frames are prevented from sideswayand have some degree of rotational restraint at their ends. Thus they are in a categorythat lies somewhere between cases (a) and (d) in Table C-C2.2 of the Commentary,and K is between 0.5 and 1.0. A value of 1.0 is therefore always conservative for membersof braced frames and is the value prescribed by AISC C1.3a unless an analysis ismade. Such an analysis can be made with the alignment chart for braced frames. Useof this nomogram would result in an effective length factor somewhat less than 1.0,and some savings could be realized.*As with any design aid, the alignment charts should be used only under the conditionsfor which they were derived. These conditions are discussed in Section C2 of theCommentary to the Specification and are not enumerated here. Most of the conditionswill usually be approximately satisfied; if they are not, the deviation will be on the conservative

side. One condition that usually is not satisfied is the requirement that all behaviorbe elastic. If the slenderness ratio KLr is less than 4.71 √E/Fy the column will buckle inelastically, and the effective length factor obtained from the alignment chart will be overly conservative. A large number of columns are in this category.A convenient procedure for determining K for inelastic columns allows the alignment charts to be used (Yura, 1971; Disque, 1973). To demonstrate the procedure we begin with the critical buckling load for an inelastic column given by Equation4.6b. Dividing it by the cross-sectional area gives the buckling stress:The rotational stiffness of a column in this state would be proportional to EtIc_Lc, andthe appropriate value of G for use in the alignment chart isBecause Et is less than E, Ginelastic is less than Gelastic, and the effective length factor Kwill be reduced, resulting in a more economical design. To evaluate Et_E, called thestiffness reduction factor (denoted by ta), consider the following relationship for a columnwith pinned ends:(4.13)AISC uses an approximation for the inelastic portion of the column strength curve, soEquation 4.13 is an approximation when AISC Equations E3-2 and E3-3 are used for Fcr.We can approximate Fcr by the compressive strength:Then in the elastic range, Fcr (inelastic) is approximatelyandWe can solve for Fe, then computeFcr(elastic) 0.877Fe

The stiffness reduction factor ta can then be computed

4.2 K-VALUESThese charts, originally developed by Julian and Lawrence, present a practical method for estimating K-values. They were developed from a slope-deflection analysis of the frames that includes the effects of column loads.

One chart was developed for columns braced against sidesway and one for columns subject to sidesway

The stiffness reduction can only be applied if the buckling is inelastic. Take into account that KL/ry can be larger than KL/rx. Use larger to determine buckling elastic or inelastic

Chart for effective length (for columns braced against sidesway and one for columns subject to sidesway)

4.3 STIFFENSS REDUCTION FACTOR, τNote: It only applies for INELASTIC BUCKLINGBecause Et is less than E, Ginelastic is less than Gelastic, and the effective length factor K will be reduced, resulting in a more economical design.

To evaluate Et/E, called the stiffness reduction factor (denoted by τa), consider the following relationship for a column with pinned ends

Fcr(elastic) 0.877·FeAlternatively from tables

Axial capacity Pc: ASD LFRD

Ωc= safety factor for compression = 1. 67Pn /Ωc= allowable compressive strength

Being,φc= resistance factor for compression = 0 . 90φcPn= design compressive strength

4.4 Shapes different to I-shapesA ) Compute buckling strength as normal (min KL/rx or KL/ry)

B) Compute flexural torsional buckling

Min strength of a and b governs

Compression AISC 9TH EIDTION

5 EXAMPLES

5.1 Example . Investigate Column with pinned ends.A W14×74 of A992 steel has a length of 20 feet and pinned ends. Compute the design compressive strength for LRFD and the allowable compressive strength for ASD. Investigate Local instability

5.1.1 Nominal StrengthL = 20 ftry=2.48 inK = 1 (both ends pinned)

Determine Slenderness:λ=K L/ry = 1 20 12/2.48 = 96.77∙ ∙ ∙

Check λ < 200 =>OK

For S355 , Cc = 113

Depending of λ and Cc use

If KL/r < then Eq. E3-2

If KL/r > then Eq. E3-3

Being

Fe = π 2∙ Eλ2 = 30.56 ksi

E3-2 : Fcr = 0.658^(Fy/Fe) Fy = 25.21 ksi∙

Nominal Strength :

LFRD Øc = 0.9

ASD Ωc=1.67 Pa=

5.1.2 Local instabilityI-beam

For I-beams and channels:FY (N/mm2)355 13.5 35.8275 15.3 40.7

5.2 Example 2. Frame. Determine Effective length factor, without consider the stiffness reduction factor.

4.7-8 The frame shown in Figure P4.7-8 is unbraced, and bending is about the x-axis of the members. All beams are W18 × 35, and all columns are W10 × 54. a. Determine the effective length factor Kx for column AB. Do not consider the stiffness reduction factor.b. Determine the effective length factor Kx for column BC. Do not consider the stiffness reduction factor.c. If Fy = 50 ksi, is the stiffness reduction factor applicable to these columns?

a) For column AB

Ratio of column stiffness to girder:

For column AB

b) For column BC

c) For column AB

Since Kx for column BC is smaller, Kx·L/rx, is smaller, so column BC is also inelastic.

5.3 Example 3. Frame axial strength4.7-9 The given frame is unbraced, and bending is about the x axis of each member. The axial dead load supported by column AB is 204 kips, and the axial live load is 408 kips. Fy = 50 ksi. Determine Kx for member AB. Use the stiffness reduction factor if possible.a. Use LRFD.b. Use ASD

Ratio of column stiffness to girder:Lc = 13’ , Lg = 25’GA pinned , thus GA = 10From section tables:

- Column W18x97 => Ic=2460 - Girder / beam W18 × 50 : Ig=800

5.3.1 LFRD. Factored ultimate load

GB’= GB·τa=

To calculate the section strength:Determine Fe: See if inelastic or elastic buckling.

Slenderness K·L/rmin

Compare to cc

Determine Fcr:

Pn = Fcr · Ag

Strength of the section= Φc·Pn

5.3.2 ASD

GA= 1

To calculate the section strength:Determine Fe: See if inelastic or elastic buckling.

Slenderness K·L/rmin

Compare to cc

Determine Fcr:

Pn = Fcr · Ag

Strength of the section= Pn/Ωc

5.4 Example 4. Multi Frame The frame shown in Figure P4.7-13 is unbraced against sidesway. The columns are HSS 6 × 6 × 5⁄8, and the beams are W12 × 22. ASTM A500 grade B steel (Fy = 46 ksi) is used for the columns, and Fy = 50 ksi for the beams. The beams are oriented so that bending is about the x-axis. Assume that Ky = 1.0.a. Use the alignment chart to determine Kx for column AB. Use the stiffness reduction factor if applicable. For column AB, the service dead load is 17 kips and the service live load is 50 kips.b. Compute the nominal compressive strength of column AB.

5.4.1 Column ABLc = 13’ , Lg = 20’GA pinned , thus GA = 10From section tables:

- HSS 6 × 6 × 5⁄8 => Ic=55.2 - Girder / beam W12 × 22 : Ig=156

5.4.2 Nominal compressive strength

Elastic buckling: