The Effective Length Concept — AISC 360-22 Commentary C-A-7

The effective length KL of a column is the length of an equivalent pin-ended column that would have the same Euler buckling load as the actual column with its actual end restraints. The K factor relates the effective length to the unbraced length L:

Pcr = pi^2 x E x I / (K x L)^2

K depends on:

  1. The rotational restraint at each end of the column (beam stiffnesses framing in)
  2. Whether lateral translation of the joint is prevented (sidesway inhibited) or not (sidesway uninhibited)
  3. The axial load distribution in the column

Sidesway Inhibited vs. Uninhibited — Chapter C

AISC 360-22 makes a fundamental distinction between two frame behaviours:

Sidesway inhibited (braced frame): Lateral stability is provided by a bracing system (X-bracing, shear walls, concrete cores). Columns buckle in single-curvature mode between floors. End moments can reverse curvature but the ends do not translate laterally relative to one another.

For sidesway-inhibited frames: K is bounded between 0.5 and 1.0.

Sidesway uninhibited (moment frame / sway frame): The frame resists lateral loads through flexure in the columns and beams. Columns can buckle in a sway mode where the ends translate laterally relative to one another.

For sidesway-uninhibited frames: K >= 1.0 and can exceed 100 for very flexible frames.


G Factors and Alignment Charts — Commentary Figures C-A-7.1 and C-A-7.2

The Jackson-Moreland alignment charts (nomographs) provide K values based on the relative rotational stiffness at the top (GA) and bottom (GB) of each column:

GA = sum(Ic/Lc) / sum(Ib/Lb) at joint A (top) GB = sum(Ic/Lc) / sum(Ib/Lb) at joint B (bottom)

Where Ic/Lc is the column stiffness ratio and Ib/Lb is the beam stiffness ratio, summed over all members rigidly connected at the joint.

End condition adjustments:

To use the chart: Draw a straight line between GA (left scale) and GB (right scale). Read K at the intersection with the centre scale. For sidesway-inhibited frames, use Figure C-A-7.1. For sidesway-uninhibited frames, use Figure C-A-7.2.

For sidesway-uninhibited frames with column inelasticity, the alignment chart is unconservative. AISC recommends applying the stiffness reduction factor tau_b = 0.8 (or the inelastic K formula from the Commentary).


K Factor Quick Reference Table

End Condition (Sidesway Inhibited) Theoretical K Recommended Design K
Both ends fixed 0.50 0.65
One end fixed, one end pinned 0.70 0.80
Both ends pinned 1.00 1.00
One end fixed, one end free (sway) 2.00 2.10
One end fixed, one end guided (sway) 2.00 2.00

For sidesway-uninhibited frames, the AISC commentary recommends K = 1.0 as a minimum.


Direct Analysis Method — AISC 360-22 Chapter C

AISC 360-22 Appendix 1 (now integrated into Chapter C) endorses the Direct Analysis Method (DAM) as the preferred stability analysis approach for all steel frames. Under DAM:

  1. A second-order analysis (P-Delta and P-delta effects) is mandatory.
  2. Notional loads of 0.002 x Yi (gravity load at each level) are applied to model initial out-of-plumbness.
  3. Stiffness reduction: EI* = 0.8 x tau_b x EI for all members contributing to lateral stability, where tau_b = 1.0 when alpha x Pr/Py <= 0.5.
  4. K = 1.0 for all columns when using DAM — the second-order effects and stiffness reduction replace the need for K > 1.0.

This eliminates the alignment chart uncertainties and provides a consistent, rational treatment of frame stability. DAM is mandatory for frames with delta_2nd / delta_1st > 1.5.


Worked Example — Braced Frame Column

A W12x65 column (A992 steel, Ix = 533 in^4) in a braced frame has unbraced length Lc = 14 ft. Beams framing into the top: W18x35, Lb = 30 ft, Ib = 510 in^4. Bottom: essentially fixed (GB = 1.0).

G factors: Ic/Lc = 533 / (14 x 12) = 3.17 in^3 At top: two W18x35 beams restrain the column. For sidesway-inhibited, multiply beam stiffness by 1.5: Ib/Lb = 510 / (30 x 12) = 1.42 in^3 GA = 3.17 / (2 x 1.5 x 1.42) = 0.74

From Figure C-A-7.1 (sidesway inhibited): GA = 0.74, GB = 1.0, K = 0.79

Effective length: KL = 0.79 x 14 = 11.1 ft. The elastic buckling load about the strong axis: Pcr = pi^2 x 29000 x 533 / (0.79 x 14 x 12)^2 = 4,820 kips

For design, use the column curve per Chapter E to compute phi_Pn from KL/r and Fy = 50 ksi.


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