LRFD

Steel Design
Chapter
6
6
.
1
INTRODUCTION
Most
beams
and
columns
are
subjected
to
some
degree
of
both
bending
and
axial
load
especially
in
statically
indeterminate
structures
.
Many
columns
can
be
treated
as
pure
compression
members
with
negligible
error
.
For
many
structural
members,
there
will
be
a
significant
amount
of
both
bending
moment
and
axial
load
.
Such
members
are
called
beam

column
.
Consider
the
rigid
frame
shown
in
the
Figure
:
For
the
given
loading
condition,
The
horizontal
member
AB
must
not
only
support
the
vertical
uniform
load
but
must
also
assist
the
vertical
members
in
resisting
the
concentrated
load
P
1
.
Member
CD
is
a
more
critical
case,
because
it
must
resist
the
load
P
1
+
P
2
without
any
assistance
from
the
vertical
members
.
The
reason
is
that
the
bracing
members,
prevents
sidesway
in
the
lower
story
.
(in
the
direction
of
P,
ED
will
be
in
tension
and
CF
will
be
slack)
Member CD must transmit the load P
1
+ P
2
from C to D.
The
vertical
members
of
this
frame
must
also
be
treated
as
beam

column
.
In
addition,
at
A
and
B,
B
.
M
.
are
transmitted
from
the
horizontal
member
through
the
rigid
joints
.
This
is
also
occur
at
C
and
D
and
is
true
in
any
rigid
frame
.
Most
columns
in
rigid
frame
are
actually
beam

columns,
and
the
effects
of
bending
should
not
be
ignored
.
Another
example
of
beam

column
can
sometimes
be
found
in
roof
trusses
if
purlins
are
placed
between
the
joints
of
the
top
chord
.
6.2 INTRODUCTION FORMULAS
The inequality Equation may be written in the following form:
If both bending and axial compression are acting the interaction
formula would be
Where
P
u
is the factored axial compressive load.
Ф
c
P
n
is the compressive design strength.
M
u
is the factored bending moment.
Ф
c
M
n
is the flexural design strength.
1.0
resistance
effects
load
φR
Q
γ
n
i
i
1.0
M
φ
M
P
φ
p
n
b
u
n
c
u
For biaxial bending, there will be two bending ratios:
Two formulas are given in the specification:
One for small axial load and one for large axial load.
1.0
M
φ
M
M
φ
M
P
φ
p
ny
b
uy
b
ux
n
c
u
nx
1.0
M
φ
M
M
φ
M
P
2
φ
p
P
φ
1.0
M
φ
M
M
φ
M
P
φ
p
P
φ
ny
b
uy
b
ux
n
c
u
n
c
ny
b
uy
b
ux
n
c
u
n
c
nx
u
nx
u
P
For
P
For
2
.
0
9
8
2
.
0
Example 6.1
Determine whether the member shown in the Figure satisfies the
appropriate AISC Specification interaction equation if the bending
is about the strong axis.
Solution:
From the column load tables:
Ф
c
P
n
= 382 kips.
Since bending is about the strong axis,
Ф
b
M
n
for C
b
can be obtained from the beam
Design chart in Part 5 of the Manual.
For L
b
= 17 ft,
Ф
b
M
n
= 200
ft.kips
. For the end condition and
loading of this problem, C
b
= 1.32.
Ф
b
M
n
= C
b
*
200
=
1.32
*
200
=
264.0
ft

kips.
This moment is larger than
Ф
b
M
p
=
227
ft

kips (also from Manual)
So the design moment must be limited to
Ф
b
M
p.
Therefore,
Ф
b
M
n
=
227
ft

kips.
Max. B.M. occurs at midheight, so
M
u
=
25
*
17
/
4
=
106.3
ft

kips.
Determine which interaction equation controls:
This member satisfies the AISC Specification.
1.0
0
106.3
382
200
M
φ
M
M
φ
M
P
φ
p
P
φ
ny
b
uy
b
ux
n
c
u
n
c
94
.
0
227
9
8
9
8
2
.
0
05236
382
200
nx
u
P
6.3
MOMENT AMPLIFICATION
The presence of the axial load produces secondary moment.
The total moment =
P
8
2
wL
The second term may be neglected if P is small.
Because the total deflection cannot be found
directly, this problem is nonlinear, and without
knowing the deflection, we cannot compute the
moment.
Ordinary structural analysis methods that do not take the displaced into
account are referred to as first

order methods.
Iterative numerical techniques, called second

order methods, can be
used to find the deflection and secondary moments.
These method are usually implemented with a computer program.
Most current design codes permit the use of either a second

order
analysis or the moment
amplification method.
This method entails computing the maximum B.M. resulting from
flexural loading by a first

order analysis, then multiplying it by a
moment amplification factor to account for the secondary moment.
Where, M
0
is the unamplified maximum moment.
P
e
is the Euler buckling load = and
P
u
is factored load.
)
e
u
0
max
/P
(P
1
1
M
M
2
2
(kL)
EI
π
As we describe later, the exact form of the AISC moment
amplification factor can be slightly different.
Example 6.2.
Compute the amplification factor for the beam

column of example 6.1.
This represents a 12 % increase in B.M.
M
max
= 1.12 * 106.3 = 119 ft

kips
kips
1874
12)/4.35)
*
(17
*
(1.0
14.4
*
29000
*
π
P
(kL/r)
EA
π
(kL)
EI
π
P
2
2
e
2
2
2
2
e
1.12
(200/1874)
1
1
factor
Amp.
)
/P
(P
1
1
factor
ion
Amplificat
e
u
6.4 WEB LOCAL BUCKLING IN BEAM

COLUMNS
The determination of the design moment requires that the cross
section be checked for compactness.
The web is compact for all tabulated shapes if there is no axial
load.
If λ ≤
λ
p
, the shape is compact
If
λ
p
≤ λ ≤
λ
r
, the shape is noncompact; and
If
λ
r
≤ λ, the shape is slender
AISC B5, Table 5.1, prescribes the following limits:
.
y
b
u
y
p
y
b
u
P
φ
2.75P
1
F
E
3.76
λ
0.125,
P
φ
P
For
y
y
b
u
y
p
y
b
u
F
P
φ
P
2
F
E
.
λ
0.125,
P
φ
P
For
E
49
.
1
33
.
12
1
For any value of
Where
P
y
= A
g
F
y
Because
P
y
is variable, compactness of the web cannot be
checked and tabulated.
Some rolled shapes satisfy the worst case limit of
Shapes listed in the column load tables in Part
4
of the Manual
that do not satisfy this criterion are marked, and these shapes
need to be checked for compactness of the web.
Shapes whose flanges are not compact are also marked
y
b
u
y
r
y
b
u
P
φ
P
F
E
λ
,
P
φ
P
74
.
0
0
.
1
70
.
5
y
F
E
/
49
.
1
Example
6.3
A W
12
x
58
of A
992
steel is subjected to a bending moment
and a factored axial load of
300
kips. Check the web for
compactness.
The upper limit is
From the dimension and properties tables
λ
= h/
t
w
=
27.0
<
λ
p
The web is therefore compact
0.125
A
φ
P
P
φ
P
b
u
y
b
u
3922
.
0
50
*
0
.
17
*
90
.
0
300
y
g
F
27
.
52
3922
.
33
.
12
1
33
.
12
1
0
2
50
29000
.
P
φ
P
2
F
E
.
λ
y
b
u
y
p
27
.
52
27
.
52
88
.
35
29000
49
.
1
49
.
1
p
E
50
F
y
6.5
BRACED VERSUS UNBRACED FRAMES
Two amplification factors are used in LRFD
One to account for amplification resulting from the member
deflection and the other to account for the effect of sway when the
member is part of unbraced frame. The following Figure illustrates
these two components.
.
In Figure a, the member is restrained against
sidesway, and the max. secondary moment is P
δ
.
If the frame is unbraced, there is an additional
component of the secondary moment, shown in
Figure b, and the max value of it is P
Δ
, which
represents an amplification of the end moment.
The amplified moment to be used in design is:
M
u
= B
1
M
nt
+ B
2
M
lt
Where
M
nt
= maximum moment assuming that no sidesway occurs.
M
lt
= maximum moment caused by sidesway, =
0.0
in the
actuall
braced frame.
B
1
= amplification factor for the moment occurring in the member
when it is braced against sidesway.
B
2
= amplification factor for the moment resulting from sidesway.
The following sections explain the evaluation of B
1
and B
2
.
6.6 MEMBERS IN BRACED FRAMES
The amplification factor given in section 6.3 was derived for a
member braced against sidesway.
The following Figure shows a member of this type
Maximum moment amplification occurs at the
center, where the deflection is largest.
For equal end moment, the moment is constant
throughout the length of the member,
So the maximum primary moment also occurs at the center.
If the end moments are not equal, the maximum primary and
secondary moments will occur near each other.
If the end moments produce reverse

curvature bending as shown.
Here the max. primary moment is at one of the ends, and the max.
amplification occurs between the ends.
Therefore, the max. moment in a beam

column depends on the
distribution of bending moment within the member.
This distribution is accounted for by a factor, C
m
, applied to the
amplification factor given in section
6.3
.
The amplification factor given in section
6.3
was derived for the
worst case, so C
m
will never be greater than
1.0
.
The final form of the
amplification factor is
2
2
1
)
/
(
1
)
r
KL
EA
where
g
e1
e1
u
m
P
/P
(P
1
C
B
Note:
When computing p
e
1
, use KL/r for the axis of bending and an effective length
factor K less than or equal to
1.0
(braced condition)
Evaluation of C
m
The factor Cm applies only to the braced condition.
There are two categories of members:
1.
Members with transverse loads applied between it’s ends.
2.
Members with no transverse loads.
If there are no transverse loads acting on the member,
M
1
/M
2
is the ratio of the bending moments at the ends of the
member. (
M
1
is the smallest value and M
2
is the largest one).
The ratio is positive for member bent in reverse curvature and
negative for single curvature bending as shown in the Figure.
2
1
4
.
0
6
.
0
M
M
C
m
For transversely loaded members, C
m
can be taken as 0.85 if the
ends are restrained against rotation (fixed) and 1.0 if the ends are
unrestrained against rotation (pinned).
End restrained will usually result from the stiffness of members
connected to the beam

column.
Although the actual end condition may lie between full fixity and a
frictionless pin, use of one of the two values given here will give
satisfactory results.
Example 6.5
The horizontal beam

column shows in the Figure is subjected to
the service live loads shown. This member is laterally braced at its
ends, and bending is about the x

axis. Check for compliance with
the AISC Specification.
Solution
The factored load =
P
u
=1.6*28=44.8 kips,
w
u
=12*0035=0.042 kips
The maximum moment is
The member is braced against end translation, so
M
lt
=0.0
kips
ft
112.5
8
(10)
*
0042
4
10
*
44.8
M
2
nt
Compute the moment amplification factor
The member braced against sidesway, transversely loaded, and
with unrestrained ends, C
m
can be taken as 1.0.
The amplification factor is:
For the axis of bending
kips
2522
(34.19)
10.3
*
29000
*
π
(Kl/r)
EA
π
P
2
2
2
g
2
e1
1
1.018
)
(44.8/2522

1
1
)
/P
(P

1
C
B
e1
u
m
1
kips

ft
114.5
0
112.5
*
1.018
M
B
M
B
M
Lt
2
nt
1
u
compact
is
shape
the
,
p
λ
λ
since
9.152
50
29000
0.38
y
F
E
0.38
p
λ
8.10,
f
2t
f
b
λ
flange
the
of
s
compactnes
for
check
first
strength,
flexural
the
For
kips
339.1
38.73
*
10.3
*
0.85
cr
F
g
A
c
φ
n
P
c
φ
ksi
38.73
(50)
2
(0.7813)
(0.658)
y
)F
2
c
λ
(0.658
cr
F
inelastic
1.5
c
λ
0.7813
29000
50
π
29.11
E
y
F
r
π
KL
c
λ
59.11
2.03
12)
*
1.0(10
y
r
L
y
K
r
KL
Maximum
r
L
b
L
p
L
Since
ft
in
r
L
y
F
X
y
F
X
y
r
r
L
ft
in
y
F
E
y
r
p
L
ft
09
.
24
1
.
289
2
)
10
50
)(
6
10
*
763
(
1
1
)
10
50
(
3610
*
03
.
2
2
)
10
(
2
1
1
)
10
(
1
17
.
7
04
.
86
50
29000
03
.
2
*
76
.
1
76
.
1
10
b
L
buckling.
torsional
Lateral
kips
ft
122.8
1473
*
0.9
M
φ
kips
in
1473
10
24.09
7.17
10
1248)
(1735
1735
M
kips
in
1248
31.2
*
10)
(50
10)S
(F
M
kips
in
1735
34.7
*
50
Z
F
M
)
M
(M
M
M
n
b
n
y
r
y
p
r
p
p
n
p
L
r
L
p
L
b
L
Because the beam weight is very small in relation to the
concentrated live load, C
b
may be taken from Figure
5.15
c as
1.32
. This value results in a design moment of
kips
162.1ft
122.8
*
1.32
M
φ
n
b
This moment is greater than the plastic moment =
0.90
*
1735
/
12
=
130.1
ft

kips, so the design strength must be limited to this value.
Check the interaction formula:
adequate
is
35
x
W8
a
So
(OK)
1.0
0.943
0
130.1
114.1
2(339.1)
44.8
M
φ
M
M
φ
M
P
2
φ
P
0.20
0.1321
339.1
44.8
P
φ
P
ny
c
uy
nx
c
ux
n
c
u
n
c
u
Please read the remaining examples,
6.4
and
6.6
from the text
book.
6.7
MEMBERS IN UNBRACED FRAMES
The amplification factor given in section
6.3
was derived for a
member braced against sidesway.
The following Figure shows a member of this type
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