From AISC Eq. E3.2 or Table 4-22 (pg. 4-323), obtain
φ
c
F
cr
= 29.8 ksi, based on KL/r = 75.
3.Compute required ?𝑔=𝑃?𝜑𝑐?𝑐?=27429.8= 9.2 sq. in.
4.
Select section based on buckling in weak direction, (
KL/r
)
y
.
From Column Load Tables in Part 4 of AISC manual, select a W 10×33, the lightest section
that has the required area.
5.
Check the W 10 × 33 section:
A
=9.71 sq in.,
r
x
/
r
y
=2.16,
r
y
=1.94,
KL/r
y
= 99,
φ
c
F
cr
= 22 ksi
(Note: (
KL/r
x
) = (
KL/r
y
)(
r
x
/
r
y
)= 99/2.16 =45.8 (not govern).)
φ
c
P
n
=
φ
c
F
cr
A
g
= 22(9.71) = 213.62 kips
(<
P
u
= 274) NG
Note that Table 4-22 provides
φ
c
F
cr
for given values of
KL/r
.
Other possible sections are:
Section
Area (sq.in.)
KL/r
y
φ
c
F
cr
(ksi)
φ
c
P
n
(kips)
W 8 × 40
11.7
94.1
23.60
276 (lightest section)
W 10 × 45
13.3
95.5
22.75
307
W 12 × 45
13.1
98.5
22.15
290

88
P
P
P
P
15 ft
15 ft
15 ft
15 ft
Potential buckled
shape
Example
:
Select the lightest Grade 50 W-section for a pin-end column of a braced structure
main column length 30 feet
100 kips dead and 200 kips live compressive loads
weak direction support at mid-height
Solution
:
1.
Factored load:
P
u
=1.2
D
+1.6
L
=1.2(100) + 1.6(200) = 440 kips
2.
For hinged ends,
K
=1 and effective length are
(
KL
)
x
= 30 feet and (
KL
)
y
= 15 feet
3.
AISC Column Load Tables are computed assuming
(
KL/r
)
y
controls.
Enter these tables with the effective length (
KL
)
y
= 15 ft and
P
u
= 440 kips.
4.
Starting with the shallow W10 sections and working towards deeper sections
Section
φ
c
P
n
(kips)
r
x
/r
y
A
g
(in
2
)
W10 × 49
449
1.71
14.4
W12 × 53
478
2.11
15.6
5. Try
2
12
53:
15.6 in
;
2.48 in;
2.11 (> 2.0) ;
5.23 in
x
g
y
x
y
r
W
A
r
r
r
Since (
KL
)
x
= 2(
KL
)
y
, if
r
x
/r
y
≥ 2, weak axis (i.e. bending about y
-y axis) controls and tabular
loads give the correct answer. That is,
30(12)
15(12)
68.8
72.6
5.23
2.48
x
y
KL
KL
r
r
Thus W 12 × 53 is acceptable.
(
φ
c
P
n
= 478 kips
(as tabulated)
Try
2
10
49:
14.4 in
;
2.56 in;
1.71 (<2.0) ;
4.34 in
x
g
y
x
y
r
W
A
r
r
r
with strong axis
bending controls. That is,
30(12)
15(12)
82.95
70.32
4.34
2.56
x
y
KL
KL
r
r
.
(a)
Using Table 4-22,
c
82.95,
27.2 ksi
cr
KL
F
r
;
φ
c
P
n
=
φ
c
F
cr
A
g
= 27.2(14.4) = 392 kips
(b)
Using Column Table, compute equivalent (
KL
)
y
corresponding to (
KL
)
x
= 30 ft by setting
(??)
?
?
?
=
(??)
?,𝑒??𝑖?𝑎𝑙𝑒𝑛?
?
?
⇒
Equivalent (
KL
)
y
= (KL)
x
/(
r
x
/
r
y
)= 30/1.71 =
17.54ft
For (
KL
)
y
= 17.54 ft, W10×49's design strength is 393 kips (<
P
u
= 440) (NG)
5.
Check section W 12 × 53: (
KL/r
)
y
= 15(12)/2.48 = 72.5,
φ
c
F
cr
= 30.65 ksi.
φ
c
P
n
=
φ
c
F
cr
A
g
=
30.65(15.6) = 478.1 kips (>
P
u
= 440)
Use section W 12 × 53. ((
KL
)
y,equivalent
= 30/2.11
≤ 15ft, strong axis ok.)

89
More on Effective Length Factor K
Basic Cases:
L
-unsupported length,
KL
-effective length
In order to take into account of the different boundary conditions, the unsupported length
L
is
replaced by the effective length
KL
in the calculation of buckling stresses and the slenderness
ratio
L/r
is replaced by the effective slenderness ratio
KL/r
.

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- Spring '14
- GregoryG.Deierlein
- Column, Buckling, KL, AISC, effective length