# Steel Design

Urban and Civil

Nov 29, 2013 (4 years and 5 months ago)

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LRFD
-
Steel Design

Chapter
6

6
.
1

INTRODUCTION

Most

beams

and

columns

are

subjected

to

some

degree

of

both

bending

and

axial

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

.

Such

members

are

called

beam
-
column
.

Consider

the

rigid

frame

shown

in

the

Figure
:

For

the

given

condition,

The

horizontal

member

AB

must

not

only

support

the

vertical

uniform

but

must

also

assist

the

vertical

members

in

resisting

the

concentrated

P
1
.

Member

CD

is

a

more

critical

case,

because

it

must

resist

the

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

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
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:

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

Solution:

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
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
-
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

)
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

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.

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
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

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
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:

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

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