The effective length factor K was introduced in

The effective length factor (K) was introduced in page (C-7) for six ideal conditions, these are not encountered in practical field conditions. LRFD commentary provides both real conditions and standard ideal conditions (C-C 2. 2) (page 16. 1 -239 to 242) Braced Frames: Unbraced Frames: No lateral movement is allowed (0. 5 < K < 1. 0) (sideway prevented) Lateral movement possible (1. 0 < K < 20. 0) (sideway allowed) a) Diagonal bracing b) Shear Walls (masonry, reinforcement concrete or steel plate) C-23

where A is top of column where B is bottom of column * For fixed footing G = 1. 0 * For pinned support G = 10. 0 C-24

Example C – 8 : - W 12 x 96 In the rigid frame shown below, Determine Kx for columns (AB) & (BC). Knowing that all columns webs are in the plane. Solution: Column (AB): W 12 x 120 (A): W 24 x 55 W 12 x 120 Joint W 24 x 55 12' W 24 x 68 A 12' W 24 x 68 B 15' C 20' 18' C-25

For joint B, : - From the alignment chart for sideways uninhibited, with GA = 0. 94 and GB = 0. 95, Kx = 1. 3 for column AB. Column (BC): For joint B, as before, G = 0. 95 For joint C, at a pin connection the situation is analogous to that of a very stiff column attached to infinitely flexible girders – that is, girders of zero stiffness. The ratio of column stiffness to girder stiffness would therefore be infinite for a perfectly frictionless hinge. This end condition is only be approximated in practice, so the discussion accompanying the alignment chart recommends that G be taken as 10. 0. From the alignment chart with GA = 0. 95 and GB = 10. 0, Kx = 1. 85 for column BC. C-26