# Chapter (8) BC-1-10 - KFUPM

Urban and Civil

Nov 15, 2013 (4 years and 6 months ago)

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BC
-
1

Members subjected to combined axial loads and bending moments are called

“Beam
-
Columns”. Examples of this, are floor or roof beams resisting later wind

Interaction Formula:

1
BC

0.9)
(

1.0

M
M
P
P
or
1.0

resistance
effects

Σ
b
c
n
b
u
n
c
u

where:

P
u

= factored axial compression.

M
u

= factored bending moment.

P
n

= nominal axial strength.

M
n

= nominal bending strength.

w w w w w w w

Top chord with bending

BC
-
2

Equation

(BC
-
1
)

is

the

basic

of

AISC

design

criteria

as

stated

in

(chapter

H)

of

AISC

LRFD

specs
:
-

1a)
H1-

Equation

(AISC

1.0

M
M
M
M
P
2
Pu
0.2,
P
Pu

For
1a)
H1-

Equation

(AISC

1.0

M
M
M
M
9
8
P
Pu
0.2,
P
Pu

For
ny
b
uy
nx
b
ux
n
c
n
c
ny
b
uy
nx
b
ux
n
c
n
c

BC
-
3

Example BC
-
1

The

beam
-
column

shown

in

Figure

below

is

pinned

at

both

ends

and

is

subjected

to

the

factored

shown
.

Bending

is

the

strong

axis
.

Determine

weather

this

member

satisfy

the

appropriate

AISC

Specification

interaction

equation
.

Solution

From the column load tables (Table 4.1) the
axial

compressive design strength of W8x58 with
F
y
=50 ksi and K
y
L
y
=17 ft

c
P
n

= 286 kips

BC
-
4

From the beam design charts (Table 3
-
10 page 3

125) for

un braced length of L
b
=17, and C
b
=1.0

b
M
n

= 202 k.ft.

b
=1.32 (table 3.1)

b
M
n

= 1.32 x 202 = 267 k.ft.

b
M
p

= 224 k.ft. (Table 3.2 page 3
-
18).

b
M
n

= 224 k.ft.

b
M
p

OK

1

0.889
224
93.5
9
8
386
200
M
M
9
8
P
P
1.1a

H

AISC
Use

0.2

0.518
386
200
P
P
ft.
k

93.5
4
17

22

M

:
moment

maximum

factored

The
nx
b
ux
n
c
u
n
c
u
u

This member satisfies the AISC specifications.

BC
-
5

Moments caused by eccentricity of axial load cannot be ignored for beam
-
columns.

P.
δ

8
wL

Moment

Maximum
2

The

value

of

(P∙

)

is

called

“Moment

magnification”

due

to

initial

beam

column

initial

crookedness

or

from

bending

due

to

transverse

(

)
.

It can be proven that a beam column with initial

crookedness (e) and initial moment (M
o

= P
u
∙e),

that the total moment becomes:

M = P
u

( e + y
max
)

e
u
o
P
P
1
1
M

M

BC
-
6

where:
-

M = Magnified moment.

M
o
= Initial moment (due to initial crookedness or more often due to transverse loads).

1.15
1567
200
1
1
P
P
1-
1

Factor

ion
Amplificat
kips.

1567

17.1

x
(55.9)
x29000
π
P
55.9
3.65
12

x

17

x

1.0
r
L
K
Bending

of

Axis
r
L
K
r
KL
re
whe

.Ag
r
KL
E
π
P
e
u
2
2
e
x
x
x
x
x
x
2
2
e

So M = 1.15 M
o

= 1.15

93.5 = 107.5 k∙ft.

buckling

Euler
L
EI
π
Pe
2

Example BC
-
2

Compute the amplification factor for example (BC
-
1)

BC
-
7

Moment amplification is covered in chapter C of the AISC code.

Two amplification factors are used in LRFD:
-

A)
* One to account for amplification due to deflection.

B)
* One to account for amplification due to frame

sideway to lateral forces in unbraced frames.

LRFD account for both effects:

M
u

= M
r

= B
1

M
nt

+ B
2

M
lt

AISC C2
-
1a

Where:

M
r

= M
u

= factored load combination as affected by amplification.

M
nt

= Maximum moment assuming no sidesway (no translation)

M
lt

= Maximum moment caused by sidesway (lateral translation).

(M
lt

= 0 for braced frames)

B
1

= amplification factor for braced frames.

B
2

= amplification factor for unbraced frames.

BC
-
8

The

maximum

moment

in

a

beam
-
column

depend

on

the

end

bending

moments

in

a

braced

frame,

the

various

cases

are

accounted

for

by

a

factor

(Cm)

as

follows
:

1.0
P
P
α
1
Cm
B
e1
r
1

(AISC C2
-
2)

BC
-
9

Where:

Cm = Coefficient whose value taken as follows:

1: If there are no transverse loads acting on the member,

M
1
/
M
2

is

a

ratio

of

the

bending

moments

at

the

ends

of

the

member
.

M
1

is

the

end

moment

that

is

smaller

in

absolute

value,

M
2

is

the

larger,

and

the

ratio

is

positive

for

moment

bent

in

reverse

curvature

and

negative

for

single
-
curvature

bending
.

Reverse

curvature

(a

positive

ratio)

occurs

when

M
1

and

M
2

are

both

clockwise

or

both

counterclockwise
.

2
1
m
M
M
0.4
0.6
C
(AISC Equation C2

4)

BC
-
10

2
.

For

transversely

members,

C
m

can

be

taken

as

0
.
85

if

the

ends

are

restrained

against

rotation

and

1
.
0

if

the

ends

are

unrestrained

against

rotation

(pinned)
.

End

restraint

will

usually

result

from

the

stiffness

of

members

connected

to

the

beam
-
column
.

The

pinned

end

condition

is

the

one

used

in

the

derivation

of

the

amplification

factor
;

hence

there

is

no

reduction

for

this

case,

which

corresponds

to

C
m

=

1
.
0
.

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
.

Evaluation of C
m

Factor

2
2
e1
u
r
KL
EI
π
P
P
P
1.0
α

(AISC C2

5)