68402: Structural Design of Buildings II
61420: Design of Steel Structures
62323: Architectural Structures II
Design of Compression Members
Monther Dwaikat
Assistant Professor
Department of Building Engineering
An-Najah National University
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Slide # 1
Design of Compression
Members

Short and long columns

Buckling load and buckling failure modes

Elastic and Inelastic buckling

Local buckling

Design of Compression Members

Effective Length for Rigid Frames

Torsional and Flexural-Torsional Buckling

Design of Singly Symmetric Cross Sections
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Slide # 2
Axially Loaded Compression Members
Columns
 Struts
 Top chords of trusses
 Diagonal members of trusses

Column and Compression member are often used
interchangeably
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Slide # 3
Axially Loaded Compression Members

Commonly Used Sections:
• W/H shapes
• Square and Rectangular or round HSS
• Tees and Double Tees
• Angles and double angles
• Channel sections
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Slide # 4
Columns

Failure modes (limit states):
•
•
•
Crushing (for short column)
Flexural or Euler Buckling (unstable under bending)
Local Buckling (thin local cross section)
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Slide # 5
Short Columns


Compression Members: Structural elements that are
subjected to axial compressive forces only are called
columns. Columns are subjected to axial loads thru the
centroid.
Stress: The stress in the column cross-section can be
calculated as
f 
P
A
f - assumed to be uniform over the entire cross-section.
Short columns - crushing
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Slide # 6
Long Columns

This ideal state is never reached. The stress-state will be
non-uniform due to:
•
•
•



Accidental eccentricity of loading with respect to the centroid
Member out-of –straightness (crookedness), or
Residual stresses in the member cross-section due to fabrication
processes.
Accidental eccentricity and member out-of-straightness
can cause bending moments in the member. However,
these are secondary and are usually ignored.
Bending moments cannot be neglected if they are acting
on the member. Members with axial compression and
bending moment are called beam-columns.
“Long” columns
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Slide # 7
Long Columns


The larger the slenderness ratio (L/r), the greater the
tendency to buckle under smaller load
Factors affecting tendency to buckle:
•
•
•
•
•
•
•
end conditions
unknown eccentricity (concentric & eccentric loads)
imperfections in material
initial crookedness
out of plumbness
residual stress
buckling can be on one or both axes (major or minor axis)
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Slide # 8
Column Buckling

Consider a long slender
compression member. If
an axial load P is applied
and increased slowly, it
will ultimately reach a
value Pcr that will cause
buckling of the column.
Pcr is called the critical
buckling load of the
column.
P
(a)
P
Pcr
(b)
Pcr
Figure 1. Buckling of axially loaded compression members
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Slide # 9
Buckling Load
•
Now assume we have a pin connected column. If we apply a similar
concept to that before here we find
Pcr
Pcr • The internal resisting moment M in the column
is
M   P
•

We can write the relationship between the
deflected shape and the Moment M
d 2

2
dx
M
M
 P cr 

EI
EI
d 2
Pcr

0
2
dx
EI
Pcr 
 2 EI
L2
Pcr - The load at which bucking starts to happen


x
(Critical buckling load)
P
P
x
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Slide # 10
Column Buckling

•
•
What is buckling?
Buckling occurs when a straight column subjected
to axial compression suddenly undergoes bending
as shown in the Figure 1(b). Buckling is identified as
a failure limit-state for columns.
The critical buckling load Pcr for columns is theoretically given by
Equation (3.1):
 2E I
Pcr 
[3.1]
K L 2
I - moment of inertia about axis of buckling.
K - effective length factor based on end boundary conditions.
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Slide # 11
Effective Length
KL- Distance between inflection points in column.
K- Effective length factor
L- Column unsupported length
KL=0.7L
KL=L
K = 1.0
KL=0.5L L
K = 0.5
K = 0.7
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Slide # 12
L
Column Buckling
Table C-C2.2
Approximate Values of Effective Length Factor, K
Boundary
conditions
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Slide # 13
Ex. 3.1- Buckling Loads

Determine the buckling strength
of a W 12 x 50 column. Its
length is 6 m. For minor axis
buckling, it is pinned at both
ends. For major buckling, is it
pinned at one end and fixed at
the other end.
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x
y
Slide # 14
Ex. 3.1- Buckling Loads
Step I. Visualize the problem
• For the W12 x 50 (or any wide flange section), x is the major
axis and y is the minor axis. Major axis means axis about which
it has greater moment of inertia (Ix > Iy).
Step II. Determine the effective lengths
• According to Table C-C2.2:
•
•
•
For pin-pin end conditions about the minor axis
Ky = 1.0 (theoretical value); and Ky = 1.0 (recommended design
value)
• For pin-fix end conditions about the major axis
• Kx = 0.7 (theoretical value); and Kx = 0.8 (recommended design
value).
According to the problem statement, the unsupported length for
buckling about the major (x) axis = Lx = 6 m.
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Slide # 15
Ex. 3.1- Buckling Loads
•
•
•
The unsupported length for buckling about the minor (y) axis = Ly =
6 m.
Effective length for major (x) axis buckling = Kx Lx = 0.8 x 6 = 4.8 m.
Effective length for minor (y) axis buckling = Ky Ly = 1.0 x 6 = 6 m.
Step III. Determine the relevant section properties
• Elastic modulus of elasticity = E = 200 GPa (constant for all steels)
• For W12 x 50: Ix = 163x106 mm4. Iy = 23x106 mm4
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Slide # 16
Ex. 3.1- Buckling Loads
Step IV. Calculate the buckling strength
•
Critical load for buckling about x - axis = Pcr-x =
Pcr-x =
 2  200  163 106
 4800 
2
  200  23  10
Pcr-y =
 6000 
2
K x Lx 2
= 13965 kN.
Critical load for buckling about y-axis = Pcr-y =
2
 2E I x
6
2 E I y
K y L y 2
= 1261 kN.
Buckling strength of the column = smaller (Pcr-x, Pcr-y) = Pcr = 1261 kN.
Minor (y) axis buckling governs.
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Slide # 17
Ex. 3.1- Buckling Loads
Notes:
• Minor axis buckling usually governs for all doubly
•
symmetric cross-sections. However, for some cases,
major (x) axis buckling can govern.
Note that the steel yield stress was irrelevant for
calculating this buckling strength.
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Slide # 18
Inelastic Column Buckling





Let us consider the previous example. According to our
calculations Pcr = 1261 kN. This Pcr will cause a uniform
stress f = Pcr/A in the cross-section.
For W12 x 50, A = 9420 mm2. Therefore, for Pcr = 1261
kN; f = 133.9 MPa.
The calculated value of f is within the elastic range for a
344 MPa yield stress material.
However, if the unsupported length was only 3 m, Pcr
=would be calculated as 5044 kN, and f = 535.5 MPa.
This value of f is ridiculous because the material will
yield at 344 MPa and never develop f = 535.5 kN. The
member would yield before buckling.
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Slide # 19
Inelastic Column Buckling



Eq. (3.1) is valid only when the material everywhere in the crosssection is in the elastic region. If the material goes inelastic then
Eq. (3.1) becomes useless and cannot be used.
What happens in the inelastic range?
Several other problems appear in the inelastic range.
•
•
•
•

The member out-of-straightness has a significant influence on the buckling
strength in the inelastic region. It must be accounted for.
The residual stresses in the member due to the fabrication process
causes yielding in the cross-section much before the uniform stress f
reaches the yield stress Fy.
The shape of the cross-section (W, C, etc.) also influences the buckling
strength.
In the inelastic range, the steel material can undergo strain hardening.
All of these are very advanced concepts and beyond the scope of this
course.
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Slide # 20
AISC Specifications for Column Strength





The AISC specifications for column design are based on
several years of research.
These specifications account for the elastic and inelastic
buckling of columns including all issues (member
crookedness, residual stresses, accidental eccentricity
etc.) mentioned above.
The specification presented here will work for all doubly
symmetric cross-sections.
The design strength of columns for the flexural buckling
limit state is equal to cPn
Where, c = 0.9
(Resistance factor for
compression members)
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Slide # 21
Inelastic Buckling of Columns
•
Elastic buckling assumes the material to follow Hooke’s law and thus
assumes stresses below elastic (proportional) limit.
•
If the stress in the column reaches the proportional limit then Euler’s
assumptions are violated.
Stress “F”
Proportional
limit
Elastic Buckling
(Long Columns)
F  FP
Euler assumptions
 2E
Fcr 
( L / r )2
Inelastic Buckling
(Short columns)
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L/r
Slide # 22
AISC Specifications for Column
Strength
Pu   Pn
Pn = Ag Fcr
[E3-1]
 2E
[E3-4]
Fe 
KL r 2
F y Fe 

0
.
658
 Fy


Fcr  
0.877Fe

 KL r  4.71 E Fy
Inelastic [E3-2]
 KL r  4.71 E Fy
Elastic [E3-3]
Fe- Elastic critical Euler buckling load
Ag - gross member area
K - effective length factor
L - unbraced length of the member
r - governing radius of gyration
The 0.877 factor in Eq (E3-3) tries to account for initial crookedness.
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Slide # 23
AISC Specifications For Column
Strength
1.0


Fy Fe 


FFcrcr = 00.658
.658 Fy  Fy


Fcr/Fy
2
c
Elastic Buckling
(Long columns)
 0.877 
Fye
FF
=  02.877
 F
crcr
  c 
0.39
Inelastic Buckling
(Short columns)
K L Fy4.711.5E
c =
Fy
r E
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KL
r
Slide # 24
AISC Specifications For Column
Strength

For a given column section:
•
•
•
•
•

Calculate I, Ag, r
Determine effective length K L based on end boundary conditions.
Calculate KL/r
E
If KL/r is greater than 4.71
, elastic buckling occurs and use
Fy
Equation (E3.4)
E
If KL/r is less than or equal to 4.71
, inelastic buckling
Fy
occurs and use Eq. (E3.3)
Note that the column can develop its yield strength Fy as
KL/r approaches zero.
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Slide # 25
Ex. 3.2 - Column Strength

Calculate the design strength of W14 x 74 with length of 6
m and pinned ends. A36 steel is used.
•
•
Step I. Calculate the effective length and slenderness ratio for the
problem
Kx = Ky = 1.0
Lx = Ly = 6 m
Major axis slenderness ratio = KxLx/rx = 6000/153.4 = 39.1
Minor axis slenderness ratio = KyLy/ry = 6000/63 = 95.2
Step II. Calculate the buckling strength for governing slenderness
ratio
The governing slenderness ratio is the larger of (KxLx/rx, KyLy/ry)
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Slide # 26
Ex. 3.2 - Column Strength
•
KyLy/ry is larger and the governing slenderness ratio;
E
200000
 4.71
 133.7
Fy
248
K y Ly / ry  4.71
•
Fe 
 2E
K L
y
•
•
•
Therefore,
y
/ ry 
2
3.14162  200000

 217.8 MPa
2
95.2

Fcr  0.658
Fy Fe
F
y
 154
MPa
Design column strength = cPn = 0.9 (Ag Fcr) = 0.9 (14060x154)/1000=
1948.7 kN.
Design strength of column = 1948.7 kN.
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Slide # 27
Local Buckling Limit State

The AISC
specifications for
column strength
assume that column
buckling is the
governing limit state.
However, if the
column section is
made of thin (slender)
plate elements, then
failure can occur due
to local buckling of
the flanges or the
webs.
Figure 4. Local buckling of columns
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Slide # 28
Local Buckling Limit State
•
•
Local buckling is another limitation
that represents the instability of the
cross section itself.
If local buckling occurs, the full
strength of the cross section can
not be developed.
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Slide # 29
Local Buckling Limit State





If local buckling of the individual plate elements occurs, then
the column may not be able to develop its buckling strength.
Therefore, the local buckling limit state must be prevented
from controlling the column strength.
Local buckling depends on the slenderness (width-tothickness b/t ratio) of the plate element and the yield stress
(Fy) of the material.
Each plate element must be stocky enough, i.e., have a b/t
ratio that prevents local buckling from governing the column
strength.
The AISC specification provides the slenderness (b/t) limits
that the individual plate elements must satisfy so that local
buckling does not control.
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Slide # 30
Local Buckling Limit State
•
Local buckling can be prevented by limiting the width to thickness ratio
known as “” to an upper limit r
b

b
E
 0.56
tf
Fy
b
tf
tf
h
tw

h
E
 1.49
tw
Fy
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Slide # 31
Local Buckling Limit State
The AISC specification provides two slenderness limits (p
and r) for the local buckling of plate elements.
Compact
Fy
Axial Force, F

Non-Compact
Slender
b
F
t
Axial shortening, 
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Slide # 32
Local Buckling Limit State
•
•
•

If the slenderness ratio (b/t) of the plate element is greater than r
then it is slender. It will locally buckle in the elastic range before
reaching Fy
If the slenderness ratio (b/t) of the plate element is less than r but
greater than p, then it is non-compact. It will locally buckle
immediately after reaching Fy
If the slenderness ratio (b/t) of the plate element is less than p,
then the element is compact. It will locally buckle much after
reaching Fy
If all the plate elements of a cross-section are compact,
then the section is compact.
•
•
If any one plate element is non-compact, then the cross-section is
non-compact
If any one plate element is slender, then the cross-section is
slender.
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Slide # 33
Local Buckling Limit State
•
•
•
Cross section can be classified as “compact”, “non compact” or
“slender” sections based on their width to thickness ratios
If the cross-section does not satisfy local buckling requirements its
critical buckling stress Fcr shall be reduced
If   r then the section is slender, a reduction factor for capacity
shall be computed from
AISC Manual for Steel
Design
•
It is not recommended to use slender sections for columns.
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Slide # 34
Local Buckling Limit State


The slenderness limits p and r for various plate elements with
different boundary conditions are given in the AISC Manual.
Note that the slenderness limits (p and r) and the definition of plate
slenderness (b/t) ratio depend upon the boundary conditions for the
plate.
•
•


If the plate is supported along two edges parallel to the direction of
compression force, then it is a stiffened element. For example, the webs of
W shapes
If the plate is supported along only one edge parallel to the direction of the
compression force, then it is an unstiffened element. Ex., the flanges of W
shapes.
The local buckling limit state can be prevented from controlling the
column strength by using sections that are compact and non-compact.
Avoid slender sections
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Slide # 35
Local Buckling Limit State
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Slide # 36
Ex. 3.3 – Local Buckling

Determine the local buckling slenderness limits and
evaluate the W14 x 74 section used in Example 3.2. Does
local buckling limit the column strength?
•
Step I. Calculate the slenderness limits
See Tables in previous slide.
For the flanges of I-shape sections in pure compression
r  0.56 
E
200000
 0.56 
 15.9
Fy
248
For the webs of I-shapes section in pure compression
r  1.49 
Use
E
200000
 1.49 
 42.3
Fy
248
E = 200000 MPa
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Slide # 37
Ex. 3.3 – Local Buckling
•
Step II. Calculate the slenderness ratios for the flanges and webs
of W14 x 74
•
•
•
For the flanges of I-shape member, b = bf/2 = flange width / 2
Therefore, b/t = bf/2tf.
For W 14 x 74, bf/2tf = 6.41
(See Section Property Table)
For the webs of I shaped member, b = h
h is the clear distance between flanges less the fillet / corner radius of
each flange
For W14 x 74, h/tw = 25.4
(See Section Property Table).
Step III. Make the comparisons and comment
For the flanges, b/t < r. Therefore, the flange is non-compact
For the webs, h/tw < r. Therefore the web is non-compact
Therefore, the section is non-compact
Therefore, local buckling will not limit the column strength.
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Slide # 38
Design of Compression Members
• Steps for design of compression members
•
•
•
•
Calculate the factored loads Pu
Assume (a cross section) or (KL/r ratio between 50 to 90)
Calculate the slenderness ratio KL/r and the ratio Fe F   2 E
e

KL r 2
Calculate c Fcr based on value of Fe
Pu

c Fcr
•
Calculate the Area required Ag
•
Choose a cross section and get (KxL/rx )and (KyL/ry)
•
•
Arequired
Recalculate c Fcr and thus check
(KL/r) max
c Pn  Ag c Fcr  Pu
Check local buckling requirements
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Slide # 39
Ex. 3.4 – Design Strength

Determine the design strength of an ASTM A992 W14 x
132 that is part of a braced frame. Assume that the
physical length L = 9 m, the ends are pinned and the
column is braced at the ends only for the X-X axis and
braced at the ends and mid-height for the Y-Y axis.
•
Step I. Calculate the effective lengths.
From Section Property Table
For W14 x 132: rx = 159.5 mm; ry = 95.5 mm;
Kx = 1.0 and
Ky = 1.0
Lx = 9 m and
Ly = 4.5 m
Ag =25030 mm2
KxLx = 9 m and KyLy = 4.5 m
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Slide # 40
Ex. 3.4 – Design Strength
•
•
Step II. Determine the governing slenderness ratio
KxLx/rx = 9000/159.5 = 56.4
KyLy/ry = 4500/95.5 = 47.1
The larger slenderness ratio, therefore, buckling about the major
axis will govern the column strength.
Step III. Calculate the column strength
Fe 
 2E
 K x Lx
rx 
2
K x Lx / rx  4.71

 2  200000
56.4
2
 620.5 MPa
E
200000
 4.71
 113.6
Fy
344
Fcr   0.658344 620.5   344  272.8 MPa
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Slide # 41
Ex. 3.4 – Design Strength
Pn  0.9  25030  272.8 1000  6145 kN
•
Step IV. Check the local buckling limits
For the flanges, bf/2tf = 7.15 < r  0.56
For the web, h/tw = 17.7< r  0.56
E
 13.5
Fy
E
 13.5
Fy
Therefore, the section is non-compact. OK.
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Slide # 42
Ex. 3.5 – Column Design

A compression member is subjected to service loads of 700 kN DL and 2400
kN of LL. The member is 7.8 m long & pinned at each end. Use A992 steel and
select a W shape.
•
•
Step I. Calculate the factored design load Pu
Pu = 1.2 PD + 1.6 PL = 1.2 x 700 + 1.6 x 2400 = 4680 kN.
Step II. Calculate Fcr by assuming KL/r = 80
Fe 
 2E
 KL r 
2

KL / r  4.71
 2  200000
80
2
 308.4 MPa
E
200000
 4.71
 113.6
Fy
344
Fcr   0.658344 308.4   344  215.3 MPa
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Slide # 43
Ex. 3.5 – Column Design
•
•


Step III. Calculate the required area of steel
A = 4680*1000/(0.9*215.3) = 24156 mm2
Step IV. Select a W shape from the Section Property Tables
Select W14 x 132. It has A = 25030 mm2 OR W12 x 136 A = 25740 mm2
Select W14 x 132 because it has lower weight.
KyLy/ry =7800/95.5 = 81.7
Fe = 295.7
Fcr = 211.4
Pn = 4897.3 kN
OK
W14 x 145 is the lightest.
Section is non-compact but students have to check for that
Note that column sections are usually W12 or W14. Generally sections bigger
than W14 are not used as columns.
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Slide # 44
Effective Length


Specific Values of K shall be known
End conditions
K
Pin-Pin
1.0
Pin-Fixed
0.8
Fixed-Fixed
0.65
Fixed-Free
2.1
Recommended
design values (not
theoretical values)
Values for K for different end conditions range from 0.5 for
theoretically fixed ends to 1.0 for pinned ends and are given by:
Table C-C2.2 AISC Manual

For compression elements connected as rigid frames the effective length
is a function of the relative stiffness of the element compared to the
overall stiffness of the joint. This will be discussed later in this chapter
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Slide # 45
K Factor for Rigid Frames
•
•
•
•
•
If we assume all connections are pinned
then: Kx L = 3 m and Ky L = 6 m
However the rigidity of the beams affect
the rotation of the columns. Thus in rigid
frames the K factor can be determined
from the relative rigidity of the columns
Kx L  3
10m
'
K y L 620
m'
3m
3m
Determine a G factor
Ec I c / Lc

G  E I /L
 gg g
I c / Lc

G  I /L
g g
Where “c” represents column and “g” represents girder
The G value is computed at each end of the member and K is
computed factor from the monograms in
AISC Manual – Figure C-C2.2
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Slide # 46
Effective Length of Columns in Frames



So far, we have looked at the buckling strength of
individual columns. These columns had various boundary
conditions at the ends, but they were not connected to
other members with moment (fix) connections.
The effective length factor K for the buckling of an
individual column can be obtained for the appropriate end
conditions from Table C-C2.2 of the AISC Manual .
However, when these individual columns are part of a
frame, their ends are connected to other members (beams
etc.).
•
•
Their effective length factor K will depend on the restraint offered by
the other members connected at the ends.
Therefore, the effective length factor K will depend on the relative
rigidity (stiffness) of the members connected at the ends.
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Slide # 47
Effective Length of Columns in Frames

The effective length factor for columns in frames must be
calculated as follows:
•
First, you have to determine whether the column is part of a braced
frame or an unbraced (moment resisting) frame.
•
•
•
If the column is part of a braced frame then its effective length factor 0
<K≤1
If the column is part of an unbraced frame then 1 < K ≤ ∞
Then, you have to determine the relative rigidity factor G for both
ends of the column
•
G is defined as the ratio of the summation of the rigidity (EI/L) of all
columns coming together at an end to the summation of the rigidity
(EI/L) of all beams coming together at the same end.
E Ic
c: for columns
Lc
It must be calculated for both ends of the column
G
E Ib
b: for beams
Lb


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Slide # 48
Effective Length of Columns in Frames
• Then, you can determine the effective length factor K for the
column using the calculated value of G at both ends, i.e., GA and
GB and the appropriate alignment chart
•
There are two alignment charts provided by the AISC manual,
•
•
•
One is for columns in braced (sidesway inhibited) frames. 0 < K ≤ 1
The second is for columns in unbraced (sidesway uninhibited) frames.
1<K≤∞
The procedure for calculating G is the same for both cases.
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Slide # 49
Effective Length
Monograph or
Jackson and Moreland
Alignment Chart
for Unbraced Frame
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Slide # 50
Effective Length
Monograph or
Jackson and Moreland
Alignment Chart
for braced Frame
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Slide # 51
Ex. 3.6 – Effective Length Factor
3.0ft. m
10
W14 x 68
10 ft.m
3.0
A
W14 x 68
12 ft.m
3.6
W14 x 68
B
18 ft.m
5.4
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W12 x 79
W12 x 79
Calculate the effective
length factor for the
W12 x 53 column AB of
the
frame
shown.
Assume that the column
is oriented in such a
way that major axis
bending occurs in the
plane of the frame.
Assume
that
the
columns are braced at
each story level for outof-plane
buckling.
Assume that the same
column section is used
for the stories above
and below.
W12 x 79

18 ft.m
5.4
20
ft.
6m
Slide # 52
15 ft. m
4.5
Ex. 3.6 – Effective Length Factor
•
•
Step I. Identify the frame type and calculate Lx, Ly, Kx, and Ky if
possible.
It is an unbraced (sidesway uninhibited) frame.
Lx = Ly = 3.6 m
Ky = 1.0
Kx depends on boundary conditions, which involve restraints due to
beams and columns connected to the ends of column AB.
Need to calculate Kx using alignment charts.
Step II. Calculate Kx
Ixx of W 12 x 53 = 425 in4
Ixx of W14x68 = 753 in4
Ic
425
425


L c 10  12 12  12 6.493
GA 


 1.021
Ib
723
723
6.360


L b 18  12 20  12
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Slide # 53
Ex. 3.6 – Effective Length Factor
Ic
425
425

L c 12  12 15  12 5.3125
GB 


 0.835
Ib
723
723
6.360


L b 18  12 20  12

Using GA and GB: Kx = 1.3
16.1-242
•
- from Alignment Chart on Page
Step III. Design strength of the column
KyLy = 1.0 x 12 = 12 ft.
Kx Lx = 1.3 x 12 = 15.6 ft.
rx / ry for W12x53 = 2.11
(KL)eq = 15.6 / 2.11 = 7.4 ft.
KyLy > (KL)eq
Therefore, y-axis buckling governs. Therefore cPn = 547 kips
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Slide # 54
Ex. 3.8 – Column Design





Design Column AB of the frame shown below for a design
load of 2300 kN.
Assume that the column is oriented in such a way that
major axis bending occurs in the plane of the frame.
Assume that the columns are braced at each story level
for out-of-plane buckling.
Assume that the same column section is used for the
stories above and below.
Use A992 steel.
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Slide # 55
Ex. 3.8 – Column Design
3.0ft. m
10
W14 x 68
10 ft.m
3.0
A
W14 x 68
12 ft.m
3.6
W14 x 68
18 ft.m
5.4
W12 x 79
W12 x 79
W12 x 79
B
18 ft.m
5.4
15 ft. m
4.5
20
ft.
6m
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Slide # 56
Ex. 3.8 – Column Design
•
•
Step I - Determine the design load and assume the steel material.
Design Load = Pu = 2300 kN.
Steel yield stress = 344 MPa (A992 material).
Step II. Identify the frame type and calculate Lx, Ly, Kx, and Ky if
possible.
It is an unbraced (sidesway uninhibited) frame.
Lx = Ly = 3.6 m
Ky = 1.0
Kx depends on boundary conditions, which involve restraints due
to beams and columns connected to the ends of column AB.
Need to calculate Kx using alignment charts.
Need to select a section to calculate Kx
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Slide # 57
Ex. 3.8 – Column Design
• Step III - Select a column section
Assume minor axis buckling governs.
Ky Ly = 3.6 m
Select section W12x53
KyLy/ry = 57.2 Fe = 604.4 Fcr = 271.1
cPn for y-axis buckling = 2455.4 kN
• Step IV - Calculate Kx
Ixx of W 12 x 53 = 177x106 mm4
Ixx of W14x68 = 301x106 mm4
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Slide # 58
Ex. 3.8 – Column Design
I c  177 177 
 L  3  3.6 
c
GA 

 1.02
Ib
301 301
 L 5.4  6
b
I c  177 177 
 L  3.6  4.5 
c
GB 

 0.836
Ib
301 301
 L 5.4  6
b
Using GA and GB: Kx = 1.3 - from Alignment Chart
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Slide # 59
Ex. 3.8 – Column Design

Step V - Check the selected section for X-axis buckling
Kx Lx = 1.3 x 3.6 = 4.68 m
Kx Lx/rx = 35.2
Fe = 1590.4
Fcr = 314.2
For this column, cPn for X-axis buckling = 2846.3

Step VI - Check the local buckling limits
For the flanges, bf/2tf = 8.69 <
For the web, h/tw = 28.1
<
r  0.56
r  1.49
E
 13.5
Fy
E
 35.9
Fy
Therefore, the section is non-compact. OK, local buckling is not a
problem
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Slide # 60