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E-18
Example E.4a W-Shape Compress ion Member (Moment Frame)
This example is primarily intended toillustrate the use of the
alignment chart for
sidesway uninhibited columns.
Given:
The member sizes shown for the moment
frame illustrated here (sideswayuninhibited in the plane of the
frame) have
been determined to be adequate for lateralloads. The
material for both the column
and the girders is ASTM A992 grade 50.
The loads shown at each level are theaccumulated dead loads and
live loads at
that story. The column is fixed at the base
about the x- x axis of the column.
Determine if the column is adequate to
support the gravity loads shown. Assumethe column is
continuously supported in
the transverse direction (the y- y axis of
thecolumn).
Material Properties:
ASTM A992 F y = 50 ksi
F u = 65 ksi ManualTable 2-1
Geometric Properties:
W18×50 I x = 800 in.4 W24×55
I x = 1350 in.4
W14×82 Ag = 24.0 in.2
I x = 881 in.4
Manual
Table 1-1
Calculate the required strength for the column between the roof
and floor
LRFD ASD
( ) ( )1.2 41.5 kips 1.6 125 kips 250 kipsuP = +
=
= + =41.5 125 167 kipsa
P
Calculate the effective length factor, K
LRFD ASD
2
250 kips10.4 ksi < 18 ksi
24.0 in.
u
g
P
A
= =
1.00τ =
2
167 kips= 6.96 ksi
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E-19
Determine Gtop and Gbottom
Gtop =( / )
( / )
c c
g g
I L
I Lτ
∑∑
=
4
4
881 in.
14.0 ft(1.00)
800 in.
2 35.0 ft
⎛ ⎞⎜ ⎟⎝ ⎠⎛ ⎞⎜ ⎟⎝ ⎠
= 1.38
Gbottom =( / )
( / )
c c
g g
I L
I Lτ
∑∑
=
4
4
881 in.2
14.0 ft(1.00)
1,350 in.2
35.0 ft
⎛ ⎞⎜ ⎟⎝ ⎠
⎛ ⎞⎜ ⎟⎝ ⎠
= 1.63
From the alignment chart, K is slightly less than
1.5. Because the column available strengthtables are based on the
KL about the y-y axis, the equivalent effective column length
of the
upper segment for use in the table is:
( )
= = =⎛ ⎞⎜ ⎟⎝ ⎠
1.5 14.0 ft( )
8.61 ft2.44
x
x
y
KL
KL r
r
Commentary
C2.2
Commentary
C2.2 &Fig. C-C2.4
Take the available strength of the W14x82 from Manual Table
4.1
At KL = 9 ft, the available strength in axial compression
is:LRFD ASD
942 kips > 250 kipsc nPφ = o.k. / 627 kips >
167 kipsn cP Ω = o.k.
ManualTable 4-1
Calculate the required strength for the column segment between
the floor and the foundation
LRFD ASD
Pu = 1.2(100 kips) + 1.6 (300 kips)= 600 kips
Pa = 100 kips + 300 kips= 400 kips
Calculate the effective length factor, K
LRFD ASD
2
600 kips25.0 ksi
24.0 in.
u
g
P
A= =
0.890τ =
( )
4
4
881 in.
2 14.0 ft0.890 1.45
1350 in.2
35.0 ft
c
top
g
I
LG
I
L
τ
⎛ ⎞⎛ ⎞
Σ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠= = =⎛ ⎞ ⎛ ⎞Σ⎜ ⎟ ⎜ ⎟⎝ ⎠
⎝ ⎠
2
400 kips16.7 ksi
24.0 in.
a
g
P
A= =
τ = 0.890
( )τ
⎛ ⎞⎛ ⎞
Σ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠= = =⎛ ⎞ ⎛ ⎞
Σ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
4
4
881 in.
2 14.0 ft0.890 1.45
1350 in.2
35.0 ft
c
top
g
I
LG
I
L
Manual
Table 4-21
Commentary
Section
C2.2b
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E-20
( )1 fixed bottomG =
From the alignment chart, K is approximately 1.42.
Because the column available strengths
are based on the KL about the y-y axis, the effective
column length of the lower segment foruse in the table is:
( )1.42 14.0 ft( )8.15 ft
2.44
x
x
y
KLKL
r
r
= = =⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠
Take the available strength of the W14x82 from Manual Table
4-1
at L = 9 ft, (conservative) the available strength in axial
compression is:LRFD ASD
942 kips > 600 kipsc nPφ = o.k. / 627 kips > 400
kipsn cP Ω = o.k. ManualTable 4-1
A more accurate strength could be determined by interpolation
from Manual Table 4-1
