68402
Slide #
1
Design of Compression Members
Monther Dwaikat
Assistant Professor
Department of Building Engineering
An

Najah National University
68402:
Structural Design of Buildings II
61420:
Design of Steel Structures
62323:
Architectural Structures II
68402
Slide #
2
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
Design of Compression Members
68402
Slide #
3
Axially Loaded Compression Members
Columns
Struts
Top chords of trusses
Diagonal members of trusses
Column
and
Compression member
are often used
interchangeably
68402
Slide #
4
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
68402
Slide #
5
Columns
Failure modes (limit states):
•
Crushing
(for short column)
•
Flexural
or
Euler Buckling
(unstable under bending)
•
Local Buckling
(thin local cross section)
68402
Slide #
6
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

assumed
to
be
uniform
over
the
entire
cross

section
.
Short
columns

crushing
A
P
f
68402
Slide #
7
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
68402
Slide #
8
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)
68402
Slide #
9
Column Buckling
Consider
a
long
slender
compression
member
.
If
an
axial
load
P
is
applied
and
increased
slowly,
it
will
ultimately
reach
a
value
P
cr
that
will
cause
buckling
of
the
column
.
P
cr
is
called
the
critical
buckling
load
of
the
column
.
P
cr
P
cr
P
P
(a)
(b)
P
cr
P
cr
P
P
P
P
(a)
(b)
Figure 1.
Buckling of axially loaded compression members
68402
Slide #
10
Buckling Load
P
M
•
Now assume we have a pin connected column. If we apply a similar
concept to that before here we find
P
cr
P
•
The internal resisting moment M in the column
is
P
P
cr
M
•
We can write the relationship between the
deflected shape and the Moment M
EI
P
EI
M
dx
d
cr
2
2
0
2
2
EI
P
dx
d
cr
x
x
cr
P

The load at which bucking starts to happen
(Critical buckling load)
2
2
L
EI
P
cr
68402
Slide #
11
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
P
cr
for
columns
is
theoretically
given
by
Equation
(
3
.
1
)
:
•
[
3
.
1
]
I

moment of inertia about axis of buckling.
K

effective length factor based on end boundary conditions.
2
2
L
K
I
E
P
cr
68402
Slide #
12
Effective Length
KL

Distance between inflection points in column.
K

Effective length factor
L

Column unsupported length
KL=L
KL=0.5L L
KL=0.7L L
K = 1.0
K = 0.5
K = 0.7
68402
Slide #
13
Column Buckling
Boundary
conditions
Table C

C2.2
Approximate Values of Effective Length Factor, K
68402
Slide #
14
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
.
x
y
68402
Slide #
15
Ex. 3.1

Buckling Loads
Step
I
.
Visualize
the
problem
•
For
the
W
12
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
(I
x
>
I
y
)
.
Step
II
.
Determine
the
effective
lengths
•
According
to
Table
C

C
2
.
2
:
•
For pin

pin end conditions about the minor axis
•
K
y
= 1.0 (theoretical value); and K
y
= 1.0 (recommended design
value)
•
For pin

fix end conditions about the major axis
•
K
x
= 0.7 (theoretical value); and K
x
= 0.8 (recommended design
value).
•
According to the problem statement, the unsupported length for
buckling about the major (x) axis = L
x
= 6 m.
68402
Slide #
16
Ex. 3.1

Buckling Loads
•
The unsupported length for buckling about the minor (y) axis = L
y
=
6 m.
•
Effective length for major (x) axis buckling = K
x
L
x
= 0.8 x 6 = 4.8 m.
•
Effective length for minor (y) axis buckling = K
y
L
y
= 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:
I
x
= 163x10
6
mm
4
.
I
y
= 23
x10
6
mm
4
68402
Slide #
17
Ex. 3.1

Buckling Loads
Step IV.
Calculate the buckling strength
•
Critical load for buckling about x

axis = P
cr

x
=
P
cr

x
=
=
13965
kN
.
Critical load for buckling about y

axis = P
cr

y
=
P
cr

y
=
= 1261 kN.
Buckling strength of the column = smaller (P
cr

x
, P
cr

y
) =
P
cr
= 1261 kN.
Minor (y) axis buckling governs.
2 6
2
200 163 10
4800
2
2
y
y
y
L
K
I
E
2 6
2
200 23 10
6000
2
2
x
x
x
L
K
I
E
68402
Slide #
18
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.
68402
Slide #
19
Let
us
consider
the
previous
example
.
According
to
our
calculations
P
cr
=
1261
kN
.
This
P
cr
will
cause
a
uniform
stress
f
=
P
cr
/A
in
the
cross

section
.
For
W
12
x
50
,
A
=
9420
mm
2
.
Therefore,
for
P
cr
=
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,
P
cr
=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
.
Inelastic Column Buckling
68402
Slide #
20
Inelastic Column Buckling
Eq
.
(
3
.
1
)
is
valid
only
when
the
material
everywhere
in
the
cross

section
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
F
y
.
•
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
.
68402
Slide #
21
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
c
P
n
Where,
c
= 0.9
(Resistance factor for
compression members)
68402
Slide #
22
Inelastic Buckling of Columns
P
F
F
•
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
.
L/r
Stress “F”
Euler assumptions
2
2
)
/
(
r
L
E
F
cr
Elastic Buckling
(Long Columns)
Inelastic Buckling
(Short columns)
Proportional
limit
68402
Slide #
23
AISC Specifications for Column
Strength
P
u
P
n
P
n
=
A
g
F
cr
[E
3

1
]
[E
3

4
]
F
e

Elastic
critical
Euler
buckling
load
A
g

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.
2
2
r
KL
E
F
e
y
e
y
y
e
F
y
F
cr
F
E
r
KL
F
F
E
r
KL
F
F
71
.
4
877
.
0
71
.
4
658
.
0
Inelastic [E3

2]
Elastic [E3

3]
68402
Slide #
24
AISC Specifications For Column
Strength
c
=
E
F
r
L
K
y
F
cr
/
F
y
1.0
1.5
0.39
F
cr
=
F
y
2
c
877
.
0
F
cr
=
F
y
2
c
658
.
0
c
=
E
F
r
L
K
y
c
=
E
F
r
L
K
y
F
cr
/
F
y
1.0
1.5
0.39
F
cr
=
F
y
2
c
877
.
0
F
cr
=
F
y
2
c
877
.
0
F
cr
=
F
y
2
c
658
.
0
F
cr
=
F
y
2
c
658
.
0
r
KL
y
F
E
71
.
4
y
e
F
y
F
cr
F
F
658
.
0
e
cr
F
F
877
.
0
Inelastic Buckling
(Short columns)
Elastic Buckling
(Long columns)
68402
Slide #
25
AISC Specifications For Column
Strength
For
a
given
column
section
:
•
Calculate
I,
A
g
,
r
•
Determine
effective
length
K
L
based
on
end
boundary
conditions
.
•
Calculate
KL/r
•
If
KL/r
is
greater
than
,
elastic
buckling
occurs
and
use
Equation
(E
3
.
4
)
•
If
KL/r
is
less
than
or
equal
to
,
inelastic
buckling
occurs
and
use
Eq
.
(E
3
.
3
)
Note
that
the
column
can
develop
its
yield
strength
F
y
as
KL/r
approaches
zero
.
y
F
E
71
.
4
y
F
E
71
.
4
68402
Slide #
26
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
K
x
= K
y
= 1.0
L
x
= L
y
= 6 m
Major axis slenderness ratio = K
x
L
x
/r
x
= 6000/153.4 = 39.1
Minor axis slenderness ratio = K
y
L
y
/r
y
= 6000/63 = 95.2
•
Step II.
Calculate the buckling strength for governing slenderness
ratio
The governing slenderness ratio is the larger of (K
x
L
x
/r
x
, K
y
L
y
/r
y
)
68402
Slide #
27
Ex. 3.2

Column Strength
•
K
y
L
y
/r
y
is larger and the governing slenderness ratio;
•
MPa
•
Therefore,
MPa
•
Design column strength =
c
P
n
= 0.9 (A
g
F
cr
) = 0.9 (14060x154)/1000=
1948.7 kN.
•
Design strength of column = 1948.7 kN.
200000
/4.71 4.71 133.7
248
y y y
y
E
K L r
F
0.658 154
y e
F F
cr y
F F
2 2
2
2
3.1416 200000
217.8
95.2
/
e
y y y
E
F
K L r
68402
Slide #
28
Local Buckling Limit State
Figure 4.
Local buckling of columns
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.
68402
Slide #
29
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.
68402
Slide #
30
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

to

thickness
b/t
ratio)
of
the
plate
element
and
the
yield
stress
(F
y
)
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
.
Local Buckling Limit State
68402
Slide #
31
Local Buckling Limit State
•
Local buckling can be prevented by limiting the width to thickness ratio
known as “
” to an upper limit
r
y
f
F
E
t
b
56
.
0
y
w
F
E
t
h
49
.
1
b
h
w
t
f
t
b
f
t
68402
Slide #
32
The
AISC
specification
provides
two
slenderness
limits
(
p
and
r
)
for
the
local
buckling
of
plate
elements
.
Local Buckling Limit State
Compact
Non

Compact
Slender
Compact
Non

Compact
Slender
b
t
F
Axial shortening,
Axial Force, F
F
y
Compact
Non

Compact
Slender
Compact
Non

Compact
Slender
b
t
F
Axial shortening,
Axial Force, F
F
y
68402
Slide #
33
•
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
F
y
•
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
F
y
•
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
F
y
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
.
Local Buckling Limit State
68402
Slide #
34
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
F
cr
shall
be
reduced
•
If
then
the
section
is
slender,
a
reduction
factor
for
capacity
shall
be
computed
from
•
It
is
not
recommended
to
use
slender
sections
for
columns
.
r
AISC Manual for Steel
Design
68402
Slide #
35
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
Local Buckling Limit State
68402
Slide #
36
Local Buckling Limit State
68402
Slide #
37
Ex. 3.3
–
Local Buckling
200000
1.49 1.49 42.3
248
r
y
E
F
Determine
the
local
buckling
slenderness
limits
and
evaluate
the
W
14
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
For the webs of I

shapes section in pure compression
Use
E = 200000 MPa
200000
0.56 0.56 15.9
248
r
y
E
F
68402
Slide #
38
Ex. 3.3
–
Local Buckling
•
Step
II
.
Calculate
the
slenderness
ratios
for
the
flanges
and
webs
of
W
14
x
74
•
For the flanges of I

shape member, b = b
f
/2 = flange width / 2
Therefore, b/t = b
f
/2t
f
.
For W 14 x 74, b
f
/2t
f
= 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/t
w
= 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/t
w
<
r
. Therefore the web is non

compact
Therefore, the section is non

compact
Therefore,
local buckling will not limit the column strength
.
68402
Slide #
39
Design of Compression Members
•
Steps for design of compression members
•
Calculate the factored loads P
u
•
Assume (a cross section) or (
KL/r ratio between 50 to 90)
•
Calculate the slenderness ratio
KL/r and the ratio F
e
•
Calculate
c
F
cr
based on value of
F
e
•
Calculate the Area required A
g
•
Choose a cross section and get (
K
x
L/r
x
)and (
K
y
L/r
y
) (
KL/r
)
max
•
Recalculate
c
F
cr
and thus check
•
Check local buckling requirements
cr
c
u
required
F
P
A
u
cr
c
g
n
c
P
F
A
P
2
2
r
KL
E
F
e
68402
Slide #
40
Ex. 3.4
–
Design Strength
Determine
the
design
strength
of
an
ASTM
A
992
W
14
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:
r
x
= 159.5 mm;
r
y
= 95.5 mm;
A
g
=25030 mm
2
K
x
= 1.0
and
K
y
= 1.0
L
x
= 9 m
and
L
y
= 4.5 m
K
x
L
x
= 9 m and
K
y
L
y
= 4.5 m
68402
Slide #
41
Ex. 3.4
–
Design Strength
344 620.5
0.658 344 272.8
cr
F
•
Step
II
.
Determine
the
governing
slenderness
ratio
K
x
L
x
/r
x
=
9000
/
159
.
5
=
56
.
4
K
y
L
y
/r
y
=
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
2 2
2
2
200000
620.5
56.4
e
x x x
E
F
K L r
MPa
200000
/4.71 4.71 113.6
344
x x x
y
E
K L r
F
MPa
68402
Slide #
42
Ex. 3.4
–
Design Strength
5
.
13
56
.
0
y
r
F
E
•
Step IV.
Check the local buckling limits
For the flanges, b
f
/2t
f
= 7.15 <
For the web, h/t
w
= 17.7<
Therefore, the section is non

compact.
OK
.
5
.
13
56
.
0
y
r
F
E
0.9 25030 272.8 1000 6145
n
P kN
68402
Slide #
43
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
A
992
steel
and
select
a
W
shape
.
•
Step I.
Calculate the factored design load P
u
P
u
= 1.2 P
D
+ 1.6 P
L
= 1.2 x 700 + 1.6 x 2400 = 4680 kN.
•
Step II.
Calculate F
cr
by assuming KL/r = 80
2 2
2
2
200000
308.4
80
e
E
F
KL r
MPa
344 308.4
0.658 344 215.3
cr
F
200000
/4.71 4.71 113.6
344
y
E
KL r
F
MPa
68402
Slide #
44
Ex. 3.5
–
Column Design
•
Step
III
.
Calculate
the
required
area
of
steel
A = 4680*1000/(0.9*215.3) = 24156 mm
2
•
Step IV.
Select a W shape from the Section Property Tables
Select W14 x 132. It has A = 25030 mm
2
OR W12 x 136 A = 25740 mm
2
Select W14 x 132 because it has lower weight.
K
y
L
y
/r
y
=7800/95.5 = 81.7
F
e
= 295.7
F
cr
= 211.4
P
n
= 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
W
12
or
W
14
.
Generally
sections
bigger
than
W
14
are
not
used
as
columns
.
68402
Slide #
45
Effective Length
Specific Values of
K
shall be known
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
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
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)
68402
Slide #
46
•
If we assume all connections are pinned
then:
K
x
L
= 3 m and
K
y
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
•
Determine a G factor
3
m
3
m
'
10
L
K
x
'
20
L
K
y
g
g
g
c
c
c
L
I
E
L
I
E
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
g
g
c
c
L
I
L
I
G
/
/
AISC Manual
–
Figure C

C2.2
K Factor for Rigid Frames
3 m
6 m
68402
Slide #
47
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

C
2
.
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
.
68402
Slide #
48
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
.
It
must
be
calculated
for
both
ends
of
the
column
b
b
c
c
L
I
E
L
I
E
G
c: for columns
b: for beams
68402
Slide #
49
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
.
,
G
A
and
G
B
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
.
68402
Slide #
50
Effective Length
Monograph or
Jackson and Moreland
Alignment Chart
for Unbraced Frame
68402
Slide #
51
Effective Length
Monograph or
Jackson and Moreland
Alignment Chart
for braced Frame
68402
Slide #
52
Ex. 3.6
–
Effective Length Factor
Calculate
the
effective
length
factor
for
the
W
12
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
out

of

plane
buckling
.
Assume
that
the
same
column
section
is
used
for
the
stories
above
and
below
.
10 ft.
10 ft.
12 ft.
15 ft.
20 ft.
18 ft.
18 ft.
W14 x 68
W14 x 68
W14 x 68
B
A
W12 x 79
W12 x 79
W12 x 79
10 ft.
10 ft.
12 ft.
15 ft.
20 ft.
18 ft.
18 ft.
W14 x 68
W14 x 68
W14 x 68
B
A
W12 x 79
W12 x 79
W12 x 79
5.4 m
5.4 m
6 m
4.5 m
3.6 m
3.0 m
3.0 m
68402
Slide #
53
Ex. 3.6
–
Effective Length Factor
•
Step
I
.
Identify
the
frame
type
and
calculate
L
x
,
L
y
,
K
x
,
and
K
y
if
possible
.
It
is
an
unbraced
(sidesway
uninhibited)
frame
.
L
x
=
L
y
=
3
.
6
m
K
y
=
1
.
0
K
x
depends
on
boundary
conditions,
which
involve
restraints
due
to
beams
and
columns
connected
to
the
ends
of
column
AB
.
Need
to
calculate
K
x
using
alignment
charts
.
•
Step II.
Calculate K
x
I
xx
of W 12 x 53 = 425 in
4
I
xx
of W14x68 = 753 in
4
021
.
1
360
.
6
493
.
6
12
20
723
12
18
723
12
12
425
12
10
425
L
I
L
I
G
b
b
c
c
A
68402
Slide #
54
Ex. 3.6
–
Effective Length Factor
Using G
A
and G
B
: K
x
= 1.3

from
Alignment Chart on Page
16.1

242
•
Step III.
Design strength of the column
K
y
L
y
= 1.0 x 12 = 12 ft.
K
x
L
x
= 1.3 x 12 = 15.6 ft.
r
x
/ r
y
for W12x53 = 2.11
(KL)
eq
= 15.6 / 2.11 = 7.4 ft.
K
y
L
y
> (KL)
eq
Therefore, y

axis buckling governs. Therefore
c
P
n
= 547 kips
835
.
0
360
.
6
3125
.
5
12
20
723
12
18
723
12
15
425
12
12
425
L
I
L
I
G
b
b
c
c
B
68402
Slide #
55
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
A
992
steel
.
68402
Slide #
56
Ex. 3.8
–
Column Design
10 ft.
10 ft.
12 ft.
15 ft.
20 ft.
18 ft.
18 ft.
W14 x 68
W14 x 68
W14 x 68
B
A
W12 x 79
W12 x 79
W12 x 79
10 ft.
10 ft.
12 ft.
15 ft.
20 ft.
18 ft.
18 ft.
W14 x 68
W14 x 68
W14 x 68
B
A
W12 x 79
W12 x 79
W12 x 79
5.4 m
5.4 m
6 m
4.5 m
3.6 m
3.0 m
3.0 m
68402
Slide #
57
Ex. 3.8
–
Column Design
•
Step I

Determine the design load and assume the steel material.
Design Load = P
u
= 2300 kN.
Steel yield stress = 344 MPa (A992 material).
•
Step II.
Identify the frame type and calculate L
x
, L
y
, K
x
, and K
y
if
possible.
It is an unbraced (sidesway uninhibited) frame.
L
x
= L
y
= 3.6 m
K
y
= 1.0
K
x
depends on boundary conditions, which involve restraints due
to beams and columns connected to the ends of column AB.
Need to calculate K
x
using alignment charts.
Need to select a section to calculate K
x
68402
Slide #
58
Ex. 3.8
–
Column Design
•
Step III

Select a column section
Assume minor axis buckling governs.
K
y
L
y
= 3.6 m
Select section W12x53
K
y
L
y
/r
y
= 57.2
F
e
= 604.4
F
cr
= 271.1
c
P
n
for y

axis buckling = 2455.4 kN
•
Step IV

Calculate K
x
I
xx
of W 12 x 53 = 177x10
6
mm
4
I
xx
of W14x68 = 301x10
6
mm
4
68402
Slide #
59
Ex. 3.8
–
Column Design
Using GA and GB
: K
x
= 1.3

from Alignment Chart
177 177
3 3.6
1.02
301 301
5.4 6
c
c
A
b
b
I
L
G
I
L
177 177
3.6 4.5
0.836
301 301
5.4 6
c
c
B
b
b
I
L
G
I
L
68402
Slide #
60
Ex. 3.8
–
Column Design
Step V

Check the selected section for X

axis buckling
K
x
L
x
= 1.3 x 3.6 = 4.68 m
K
x
L
x
/r
x
= 35.2
F
e
= 1590.4
F
cr
= 314.2
For this column,
c
P
n
for X

axis buckling = 2846.3
Step VI

Check the local buckling limits
For the flanges, b
f
/2t
f
= 8.69 <
For the web, h/t
w
= 28.1
<
Therefore, the section is non

compact. OK, local buckling is not a
problem
5
.
13
56
.
0
y
r
F
E
9
.
35
49
.
1
y
r
F
E
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