13-Effective Length of Columns (Steel Structural Design & Prof. Shehab Mourad)

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This document discusses the effective length of columns, which depends on: 1) The end conditions of the column (fixed, hinged, or free) represented by the factor K. 2) Whether the frame is braced or unbraced. 3) The relative stiffnesses of the column to beams at each end, represented by factors G1 and G2. Formulas and charts are provided to calculate the buckling length factor K based on these parameters. An example problem demonstrates calculating K for three columns in a two-story frame.

1 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
Effective Length of Columns
The effective column length (the equivalent half wave of buckling
length), depends on;
1) The fixed end conditions (fixed, hinged or free ends), of the column
(K factor).
2) The framing condition, whether it is braced or unbraced
L L
Braced Frame
(Sway prevented) Unbraced Frame
(Sway Permitted)
P P P P
0.5 < K< 1.0 1.0 < K 5.0
K≤ 1 K≥ 2
Ib

2 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
3) The column - beam connections, and the relative stiffnesses of columns
to beams at each end of the column
S (I/L) columns
G =
S (I/L) beams
Note : G = 10 for hinged (pinned) support
G = 1.0 for fixed support
For braced Frame
Km = 3 Gi Gj + 1.4 (Gi + Gj) + 0.64
3 Gi Gj + 2 (Gi+Gj) + 1.28
For unbraced Frame
Km = 1.6 Gi Gj + 4 (Gi+Gj) + 7.5
Gi + G j + 7.5
Or by using charts:

3 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
Example
For the shown two storey frame with the indicated relative
Inertias, computed the Buckling length factor (K) for columns 1, 2 and 3.
Solution:
G1= 10 for hinged support
G3 = G6 = 1.0 for fixed support
G2 = (I/5) = 0.60
(2I /6)
G4 = ( 2I/5 + I/3 ) = 1.257
(2I/6 + 2I/8)
G5 = (I/3) = 1.332
(2I/8)
For Column 1, (unbraced frame) Values from charts
K1 = 1.6 x 10 x 0.6 + 4 (10 +0.6) + 7.5 = 1.81 1.80
10 + 0.60 + 7.5
For Column 2, (unbraced frame)
K2 = 1.6 x 1 x 1.257 + 4 (1 + 1.257) + 7.5 = 1.378 1.375
1 + 1.257 + 7.5
For Column 3, (braced Frame)
K3 = ( 3 x 1.257x1.332) + 1.4 (1.257 + 1.332) + 0.64 = 0.81 0.80
3 x 1.257 x 1.332 + 2 ( 1.257 + 1.332) + 1.28
5 m
3m
6m 8m
1
2
3
4
5
6
7
8
I 2I
2I 2I
2I
I I
I1 2
3

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