MODULE C: COMPRESSION MEMBERS

MODULE C: COMPRESSION MEMBERS This module of CIE 428 covers the following subjects Column theory Column design per AISC Effective length Torsional and flexural-torsional buckling Built-up members READING: Chapter 4 of Segui AISC LRFD Manual of Steel Construction, 3rd Ed. INTRODUCTION Compression members are structural elements that are subjected only to compression forces, that is, loads are applied along a longitudinal axis through the centroid of the cross-section. In this idealized case, the axial stress f is calculated as f = P A Note that the ideal state is never realized in practice and some eccentricity of load is inevitable. Unless the moment is negligible, the member should be termed a beam-column and not a column. Beam columns are addressed in Module E of CIE 428. 8/21/2002 8:47 AM 1

COLUMN THEORY Consider the long compression member shown in the figure below from Segui. If the axial load P is applied slowly, it will ultimately become large enough to cause the member to become unstable and assume the shape shown by the dashed line. The member has then buckled and the corresponding load is termed the critical buckling load (also termed the Euler buckling load after the mathematician Euler who formulated the relationship in 1759). If the member is slender such that the axial stress at the onset of buckling is below the yield stress and there are no lateral loads on the member, the critical buckling load is given by P cr 2 π EI = 2 L where E is Young s modulus, I is the moment of inertia (second moment of area) about the minor principal axis, and L is the length of the member between supports. For the above to apply: Member must be elastic Hinged or pinned ended 8/21/2002 8:47 AM 2

The derivation of the above equation is well established and is presented below to illustrate the importance of end or boundary conditions. The differential equation giving the deflected shape of an elastic member subject to bending is 2 d y 2 dx = where x is a location along the longitudinal axis of the member, y is the deflection of the axis at that point, M is the bending moment at that point, and other terms have been defined previously. If the analysis is undertaken at the onset of buckling, and in the absence of lateral loads 1, M EI M = Py P y + ( ) y = 0 EI 1 If transverse loads are applied, which acting alone would produce a moment m, the total bending moment would be x M = Py + m x and the above differential equation becomes P m x y + ( ) y = EI EI The solution of this differential equation will be presented later in the semester. 8/21/2002 8:47 AM 3

The latter equation is a linear, second-order ordinary differential equation with the solution y = Acos( cx) + Bsin( cx) where A and B are constants and c= P/ EI. The constants are evaluated by applying the boundary conditions y(0) = 0 and y(l) = 0. This yields A = 0 [BC 1] and 0= Bsin( cl) [BC 2]. For a nontrivial solution (the trivial solution is B = 0), sin( cl ) = 0, or and cl = 0, π,2 π,4 π,... = nπ 2 2 P, n π cl = L = nπ P = P = EI EI cr 2 L Different values of n correspond to different buckling modes. A value of n = 0 gives the trivial case of no load; n = 1 represents the first mode, n = 2 represents the second mode, etc. These buckling modes are shown on the following page (from Popov) together with the corresponding values of the buckling load. 8/21/2002 8:47 AM 4

The solution to the differential equation is y = nπ x Bsin( ) L where the coefficient B is indeterminate. For the case of n = 1, the lowest non-trivial value of the buckling load is P cr 2 π EI = 2 L Noting that the radius of gyration can be written as r 2 I Then the critical buckling load can be re-written as = A P cr 2 π EA = L 2 ( ) r where L/r is the slenderness ratio. The above equation does not give reliable results for stocky (L/r < 40 say) columns for which the critical buckling stress exceeds the proportional limit. Why? Because the relationship between stress and strain is not linear. Consider a material with a stress-strain relationship similar to that shown by the figure below. In this figure E is not constant for stresses greater than the proportional limit. For stresses between 8/21/2002 8:47 AM 5

the proportional limit and the yield stress, a tangent modulus is used, which is defined as the slope of the stress strain curve for values of f between these two limits. Such a curve is seen from tests of stocky columns and is due primarily to residual stresses. In the transition region be written as F < f F, the critical buckling load can pl y P cr 2 π EA t = L 2 ( ) r but this is not particularly useful because the tangent modulus is strain dependent. Accordingly, most design specifications contain empirical formulae for inelastic columns. The critical buckling stress is often plotted as a function of slenderness as shown in the figure at the top of the following page. This curve is called a column strength curve. From this figure it 8/21/2002 8:47 AM 6

can be seen that the tangent modulus curve is tangent to the Euler curve at the point corresponding to the proportional limit. The above equations for the critical buckling load (Euler buckling load) were derived assuming A perfectly straight column Axial load with no eccentricity Column pinned at both ends If the column is not straight, that is, the column is initially crooked, bending moments will develop in the column. Similarly, if the axial load is applied eccentric to the centroid, bending moments will develop. Often these moments can be ignored. The third assumption is a serious limitation and other boundary conditions will give rise to different critical loads. As noted earlier, the bending moment will generally be a function of x (and not y alone), resulting in a non-homogeneous differential equation. The solution procedure proceeds similarly to before except that the 8/21/2002 8:47 AM 7

problem is formulated as a fourth-order differential equation. Consider the column shown below (part c) that is pinned at one end ( y(0) = y (0) = 0) and fixed against translation and rotation at the other end ( y(0) = y (0) = 0). The critical buckling load is: P cr 2 2 2.05π EI π EI = = L (0.7 L) 2 2 For a column that is fixed at one end and free to translate and rotate at the other (part a), the critical buckling load is: P cr 2 2 π EI π EI = = 2 2 4 L (2 L) It is clear that boundary conditions play a key role in determining the critical buckling load. For convenience, the equation for the critical buckling load can be written as 8/21/2002 8:47 AM 8

P cr = 2 π EI ( KL) 2 where KL is the effective length and K is the effective length factor. For the above two cases, the effective length factor is equal to 0.7 (fixed-pinned) and 2.0 (fixed-free), respectively. The AISC LRFD specification presents a table to aid in the calculation of effective length. Theoretical and design values are recommended. The conservative design values should generally be used unless the proposed end conditions truly match the theoretical conditions. 8/21/2002 8:47 AM 9

COLUMN DESIGN PER AISC Introduction The basic requirements for compression members are covered in Chapter E of the AISC LRFD specification. The basic form of the relationship is that presented in Module A, namely, P φ P = φ ( A F ) u c n c g cr where φ is the resistance factor for compression members (=0.85) c and F is the critical buckling stress (inelastic or elastic). The cr LRFD specification introduces a dimensionless slenderness parameter λ = c KL rπ F y E For elastic columns, the critical buckling stress can be written as F cr 2 π E = = ( KL / r) 1 2 2 λ c F y which is reduced to the following equation to account for initial crookedness (L/1500 per Galambos). F cr 0.877 = F 2 λ c y For inelastic columns, the critical buckling stress is given by 8/21/2002 8:47 AM 10

F cr λ 2 c = (0.658 ) F y where this equation also accounts for initial crookedness. The use of this equation avoids the trial and error approach needed with the tangent modulus equation. The AISC LRFD specification sets the boundary between the inelastic and elastic buckling ranges at λ = 1.5. c The above equations for the critical buckling stress are given in Section E.2 of the specification. The figure below, from Segui, illustrates the above equations and the transition point. AISC specifies a maximum slenderness ratio, KL/r, of 200 for compression members. This is a practical upper limit because more slender columns will not be economical. Below is a design example from Segui that applies the design equations to a W14x74 column with a length of 20 feet and pinned ends. A36 steel is used. What is K? 8/21/2002 8:47 AM 11

Why use r y and not r x? Column sections versus beam sections? Note that the mode of failure considered to date is flexural buckling. With some cross sections, the member will fail by twisting (torsional buckling) or a combination of twisting and bending (flexural-torsional buckling). These (somewhat rare) cases will be considered later in this module. Flange and web compactness For the strength associated with a buckling mode to develop, local buckling of elements of the cross section must be prevented. If local buckling (flange or web) occurs The cross-section is no longer fully effective Compressive strengths given by F must be reduced cr 8/21/2002 8:47 AM 12

Section B5 of the LRFD specification provides limiting values of width-thickness ratios (denoted λ ) where shapes are classified as Compact Noncompact Slender AISC writes that if λ exceeds a threshold value λ r, the shape is considered slender and the potential for local buckling must be addressed. Two types of elements must be considered Unstiffened elements Unsupported along one edge parallel to the direction of load Stiffened elements Supported along both edges parallel to the load The figure on the following page, from Segui, presents compression member limits ( λ r ) for different cross-section shapes that have traditionally been used for design. Note that these limits are in U.S. units. 8/21/2002 8:47 AM 13

Table B5.1 of the 3rd Edition of the LRFD specification presents limiting width-thickness ratios for compression elements. Note that the values in this table now account for Flange yield stresses ( F ( F ) y yf ) larger than the nominal value Residual stresses ( F r ) Flange-web interaction (k) for built-up members Figures on the following pages reproduce Table B5.1 from the LRFD specification and a figure from the commentary to the specification. Note that S.I. units are used in the 3rd edition. 8/21/2002 8:47 AM 14

For unstiffened elements 8/21/2002 8:47 AM 15

For stiffened elements 8/21/2002 8:47 AM 16

A summary of the key limits for standard shapes is shown below in S.I. units. 8/21/2002 8:47 AM 17

If λ > λ in an element of a member, the design strength of that r member must be reduced because of local buckling. The general procedure for this case is as follows: 1. Compute a reduction factor Q per Appendix B5.3a (unstiffened compression elements) or B5.3b (stiffened compression elements). 2. Compute KL F y λ c = rπ E as before 3. If 4. If Qλ c cr y λ Q 1.5, F = Q(0.658 ) F from Appendix B5.3d λ 0.877 Q > 1.5, F = [ ] F from Appendix B5.3d c cr 2 y λ c 2 c 5. Calculate the design strength as φ P = 0.85AF Tables for design of compression members c n g cr Tables 4.2 through 4.17 in Part 4 of the LRFD specification present design strengths in axial compression for columns with specific yield strengths, for example, 50 ksi for W shapes. Data are provided for slenderness ratios of up to 200. Sample data are provided on the following page for some W14 shapes. How can these tables be used if slenderness is controlled by the major principal axis? 8/21/2002 8:47 AM 18

Example 1 A compression member is subjected to service loads of D = 200 k and L = 600 k. The member is 30 feet long and pinned at each end. Select a Grade 50, W14 shape from the above table. 8/21/2002 8:47 AM 19

1. Calculate the factored load and thus the required design strength. φ P P = 1.2 D + 1.6 L = n u 2. For KL = 30 feet, a W14x*** has a design strength of **** kips. 3. Select a W14x*** Example 2: Example 4.1 from the LRFD specification 8/21/2002 8:47 AM 20

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Example 3: Example 4.7 from Segui 8/21/2002 8:47 AM 22

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Effective length of columns The LRFD table presented on page 19 presents values for the design load based on a slenderness ratio calculated using the minimum radius of gyration, r y. Consider now the figure below. In such a case, slenderness about the minor axis may not control because the effective length for minor axis buckling is half that for major axis buckling. In this case, the effective slenderness ratio must be checked about each axis. The tables in Part 4 of the LRFD specification can still be used but one must now check for the following two slenderness ratios: K L x r x x ; K L y r y y 8/21/2002 8:47 AM 24

For columns in moment-resisting frames, the tabulated values of K presented on page 9 will not suffice for design. Consider the moment-frame below that is permitted to sway. Columns neither pinned not fixed. Columns permitted to sway. Columns restrained by members framing into the joint at each end of the column The rotational restraint provided by the beams is a function of the rotational stiffness of the members intersecting at the joint. Rotational stiffness is proportional to 8/21/2002 8:47 AM 25

Consider column AB above and assume that the frame is not free to sway. If the beams above and below the column are 1000 times stiffer than the column, what is the degree of fixity at the joint? If the beams above and below the column are 1000 times stiffer than the column, what is the degree of fixity at the joint? The effective length factor for a column along a selected axis can be calculated using simple formulae and a nomograph. The procedure is as follows: 1. Compute a value of G, defined below, for each end of the column, and denote the values as G and G, respectively. A B G EI / L I / L = = E I / L I / L c c c c c g g g g g where the subscripts c and g refer to column and girder, respectively. 2. Use the nomograph provided by AISC (and reproduced on the following pages). Interpolate between the calculated values of G and G to determine K. A B 8/21/2002 8:47 AM 26

Returning to the example above and column AB, what is the value of G if the beams above and below the column are 10 and 100 times stiffer than the column, and the column is continuous through the two joints? What are the values of K? What is the value of G if the beams above and below the column are 10 and 100 times more flexible than the column, and the column is continuous through the two joints? What are the values of K? 8/21/2002 8:47 AM 27

What is the value of G if the beams at A are 10 times stiffer than the column and the beams at B are 10 times more flexible than the column, and the column is continuous through the two joints? What are the values of K? Repeat the three exercises from above. What are the values of K? AISC writes that G = 10 for a pinned support. Why? G = 1.0 for a fixed support. Why? 8/21/2002 8:47 AM 28

The distinction between braced (sidesway inhibited) and unbraced (sidesway inhibited) frames is important, as evinced by difference between the values of K calculated above. What are bracing elements? The figure below from Segui provides sample information. The question remains, how much lateral bracing is necessary for sidesway to be considered inhibited? The presentation above assumed that all behavior in the frame was elastic. If the column buckles inelastically ( λ < 1.5), then the c effective length factor calculated from the alignment chart will be conservative. One simple strategy for accommodating this observation is to adjust each value of G using a stiffness reduction factor (SRF). 8/21/2002 8:47 AM 29

One strategy for calculating an updated value of G is as follows: G inelastic EI / L E t c c t = = EI / L E g g G elastic but this requires knowledge of the tangent modulus. AISC uses an approximation for the inelastic portion of the column strength curve, as follows. First, remember that F cr P P / φ cr u c = A A then F cr(inelastic) F cr(elastic) = P u f ( ) φ A c For example, if P /( φ A) = 26 ksi and u c F = 36 ksi, then y F 2 2 λ c c λ 2 F λ cr(inelastic) y c = 26 = 0.658 = 0.658 36; = 0.778 F cr(elastic) 0.877 0.877 = F = y 36 = 40.7 λ 0.778 2 c and the stiffness reduction factor is equal to F cr(inelastic) 26 SRF = = = 0.639 F 40.7 cr(elastic) 8/21/2002 8:47 AM 30

Table 4.1 of the LRFD specification, shown below, presents values for the SRF (or τ ) for various values of F and P / A. y u g 8/21/2002 8:47 AM 31

The use of the SFR is shown below using an example from Segui. Consider a four-story, four-bay frame. 8/21/2002 8:47 AM 32

Note that if the end of the column is fixed (G = 10) or pinned (G = 0), the value of G should not be multiplied by the SRF. TORSIONAL AND FLEXURAL-TORSIONAL BUCKLING An axially loaded compression member can buckle in one of three ways, as shown in the figure below from Segui. 1. Flexural buckling: The type of buckling considered to date. Caused by bending or flexure about the axis with the largest slenderness ratio. Any compression member can fail in this manner. 2. Torsional buckling: Twisting about the longitudinal axis of the member. See part b of the figure below. Can only occur with doubly symmetric cross sections with very slender elements (e.g., built-up thin-plate elements). Standard hotrolled shapes are not susceptible to this type of buckling. 3. Flexural-torsional buckling: A combination of flexural and torsional buckling. See part c of the figure below. Can only occur with cross-sections that are either symmetric about one axis only or those with no axis of symmetry. 8/21/2002 8:47 AM 33

Section 4.6 of Segui provides an excellent presentation of this subject. Additional information can be found in Part E3 (and commentary) of the LRFD specification. Because standard rolled shapes are used for most engineered steel construction and because such shapes are not susceptible to either torsional or flexuraltorsional buckling, this subject is not discussed further in CIE 428. BUILT-UP MEMBERS Built-up members from rolled shapes are not widely used for columns in buildings but are sometimes used for bracing elements. For built-up column elements such as that shown in the figure below from Segui, the procedure to calculate the capacity is straightforward, namely, 8/21/2002 8:47 AM 34

1. Calculate the area, centroid ( y ), the moment of inertia about the y-axis, and the radius of gyration ( r y ). 2. Calculate λ = c KL rπ F y E 3. Calculate F cr and φ P = 0.85AF c n g cr The most common built-up shape is that composed of angle sections, back to back. Such a cross-section is often used for bracing members. The figure below from Segui shows a compression member connected to gusset plates at each end. 8/21/2002 8:47 AM 35