Steel_Ch5 -Beam-Column 1 - An-Najah Staff

reelingripebeltΠολεοδομικά Έργα

15 Νοε 2013 (πριν από 3 χρόνια και 10 μήνες)

163 εμφανίσεις

68402

Slide #
1

Design of

Beam
-
Columns

Monther Dwaikat

Assistant Professor

Department of Building Engineering

An
-
Najah National University



62323:
Architectural Structures II

68402

Slide #
2


Beam
-
Columns


Moment Amplification Analysis


Braced and Unbraced Frames


Analysis/Design of Braced Frames


Design of Base Plates


Beam
-
Column
-

Outline

68402

Slide #
3

Design for Flexure


LRFD Spec.


Commonly Used Sections:


I


shaped members (singly
-

and doubly
-
symmetric)


Square and Rectangular or round HSS


68402

Slide #
4

Beam
-
Columns

Likely failure modes due to combined bending and axial forces:


Bending and Tension: usually fail by yielding


Bending (uniaxial) and compression: Failure by buckling in the
plane of bending, without torsion


Bending (strong axis) and compression: Failure by LTB


Bending (biaxial) and compression (torsionally stiff section):
Failure by buckling in one of the principal directions.


Bending (biaxial) and compression (thin
-
walled section): failure by
combined twisting and bending


Bending (biaxial) + torsion + compression: failure by combined
twisting and bending

68402

Slide #
5

Beam
-
Columns


Structural elements subjected to combined flexural moments and axial
loads are called
beam
-
columns


The case of beam
-
columns usually appears in structural frames


The code requires that the sum of the load effects be smaller than the
resistance of the elements




Thus: a column beam interaction can be written as





This means that a column subjected to axial load and moment will be
able to carry less axial load than if no moment would exist.

0
.
1











ny
b
uy
nx
b
ux
n
c
u
M
M
M
M
P
P



0
.
1


n
i
i
R
Q


68402

Slide #
6

Beam
-
Columns


AISC code makes a distinct difference between lightly and heavily axial
loaded columns


0
.
1
9
8











ny
b
uy
nx
b
ux
n
c
u
M
M
M
M
P
P



0
.
1
2











ny
b
uy
nx
b
ux
n
c
u
M
M
M
M
P
P



2
.
0

n
c
u
P
P
for

2
.
0

n
c
u
P
P
for

AISC Equation

AISC Equation

68402

Slide #
7

Beam
-
Columns


Definitions


P
u

= factored axial compression load


P
n

= nominal compressive strength


M
ux

= factored bending moment in the x
-
axis, including second
-
order effects


M
nx
= nominal moment strength in the x
-
axis


M
uy

= same as M
ux

except for the y
-
axis


M
ny

= same as M
nx

except for the y
-
axis



c

= Strength reduction factor for compression members = 0.90




b

= Strength reduction factor for flexural members = 0.90

68402

Slide #
8


The increase in slope for lightly axial
-
loaded columns represents the less
effect of axial load compared to the heavily axial
-
loaded columns

P
u
/

c
P
n

Safe Element

0.2

M
u
/

b
M
n

Beam
-
Columns

Unsafe Element

These are design charts that are a bit conservative than behaviour envelopes

68402

Slide #
9

Moment Amplification


When a large axial load exists, the axial load produces moments due to
any element deformation.







The final moment “M” is the sum of the original moment and the
moment due to the axial load. The moment is therefore said to be
amplified.



As the moment depends on the load and the original moment, the
problem is nonlinear and thus it is called second
-
order problem.



P

M

d

x

d

P

68402

Slide #
10

Braced and Unbraced Frames


Two components of amplification moments can be observed in unbraced
frames:


Moment due to member deflection (similar to braced frames)


Moment due to sidesway of the structure

Member deflection

Member sidesway

Unbraced Frames

68402

Slide #
11


In braced frames amplification moments can only happens due to
member deflection

Member deflection

Braced Frames

Sidesway bracing system

Unbraced and Braced Frames

68402

Slide #
12


Braced frames are those frames prevented from sidesway.


In this case the moment amplification equation can be simplified to:

ntx
x
ux
M
B
M
1

1
1
1











e
u
m
P
P
C
B
AISC Equation



2
2
/
r
KL
EA
P
g
e



KL/r for the axis of bending considered


K


1.0

nty
y
uy
M
B
M
1

Unbraced and Braced Frames

68402

Slide #
13


The coefficient C
m

is used to represent the effect of end moments on the
maximum deflection along the element (only for braced frames)












2
1
4
.
0
6
.
0
M
M
C
m
ve
M
M


2
1
ve
M
M


2
1

When there is transverse loading on
the beam either of the following
case applies


00
.
1
vely
Conservati

m
C
Unbraced and Braced Frames

68402

Slide #
14

Ex. 5.1
-

Beam
-
Columns in Braced Frames


A

3
.
6
-
m

W
12
x
96

is

subjected

to

bending

and

compressive

loads

in

a

braced

frame
.

It

is

bent

in

single

curvature

with

equal

and

opposite

end

moments

and

is

not

loaded

transversely
.

Use

Grade

50

steel
.

Is

the

section

satisfactory

if

P
u

=

3200

kN

and

first
-
order

moment

M
ntx

=

240

kN
.
m


Step I:

From Section Property Table


W12x96 (
A
= 18190 mm
2
,
I
x
= 347x10
6

mm
4
,
L
p

= 3.33 m,
L
r

= 14.25 m, Z
x

= 2409 mm
3
,
S
x

= 2147 mm
3
)

68402

Slide #
15

Step II:
Compute amplified moment

-

For a braced frame let K = 1.0


K
x
L
x

= K
y
L
y

= (1.0)(3.6) = 3.6 m

-

From Column Chapter:

c
P
n

= 4831 kN


P
u
/

c
P
n

= 3200/4831 = 0.662
> 0.2


Use eqn.

-

There is no lateral translation of the frame:
M
lt

=
0




M
ux
= B
1
M
ntx




C
m

=
0.6


0.4(
M
1
/M
2
) =
0.6


0.4(
-
240/240)

=
1.0


P
e1

=

2
EI
x
/(K
x
L
x
)
2

=

2
(200)(347x10
6
)/(3600)
2

= 52851 kN


Ex. 5.1
-

Beam
-
Columns in Braced Frames

68402

Slide #
16

Ex. 5.1
-

Beam
-
Columns in Braced Frames

)
(
0
.
1
073
.
1
52851
3200
1
0
.
1
1
1
1
OK
P
P
C
B
e
u
m







M
ux

= (1.073)(240) = 257.5 kN.m


Step III:
Compute moment capacity


Since

L
b

= 3.6 m


L
p

< L
b
< L
r






m
kN
M
n
b
.
739


68402

Slide #
17

Ex. 5.1
-

Beam
-
Columns in Braced
Frames

0
.
1
972
.
0
0
739
5
.
257
9
8
4831
3200
9
8





















ny
b
uy
nx
b
ux
n
c
u
M
M
M
M
P
P





Section is satisfactory

Step IV:
Check combined effect

68402

Slide #
18

Ex. 5.2
-

Analysis of Beam
-
Column


Check

the

adequacy

of

an

ASTM

A
992

W
14
x
90

column

subjected

to

an

axial

force

of

2200

kN

and

a

second

order

bending

moment

of

400

kN
.
m
.

The

column

is

4
.
2

m

long,

is

bending

about

the

strong

axis
.

Assume
:


k
y

=

1
.
0


Lateral

unbraced

length

of

the

compression

flange

is

4
.
2

m
.

68402

Slide #
19

Ex. 5.2
-

Analysis of Beam
-
Column


Step

I
:

Compute

the

capacities

of

the

beam
-
column



c
P
n

= 4577 kN



M
nx

= 790 kN.m




M
ny

= 380 kN.m



Step

II
:

Check

combined

effect


2
.
0
481
.
0
4577
2200



n
c
u
P
P

OK

0
.
1
931
.
0
0
790
400
9
8
4577
2200
9
8





















ny
b
uy
nx
b
ux
n
c
u
M
M
M
M
P
P



68402

Slide #
20

Design of Beam
-
Columns


Trial
-
and
-
error procedure


Select trial section


Check appropriate interaction formula.


Repeat until section is satisfactory


68402

Slide #
21

Ex. 5.3


Design
-
Beam Column


Select

a

W

shape

of

A
992

steel

for

the

beam
-
column

of

the

following

figure
.

This

member

is

part

of

a

braced

frame

and

is

subjected

to

the

service
-
load

axial

force

and

bending

moments

shown

(the

end

shears

are

not

shown)
.

Bending

is

about

the

strong

axis,

and

K
x

=

K
y

=

1
.
0
.

Lateral

support

is

provided

only

at

the

ends
.

Assume

that

B
1

=

1
.
0
.


P
D

= 240 kN

P
L

= 650 kN

M
D

= 24.4 kN.m

M
L

= 66.4 kN.m

4.8 m

M
D

= 24.4 kN.m

M
L

= 66.4 kN.m

68402

Slide #
22

Ex. 5.3


Design
-
Beam Column



Step I:

Compute the factored axial load and bending moments



P
u

= 1.2P
D

+ 1.6P
L
= 1.2(240)+ 1.6(650) = 1328 kN.



M
ntx

= 1.2M
D

+ 1.6M
L
= 1.2(24.4)+ 1.6(66.4) = 135.5 kN.m.



B
1

= 1.0


M
ux

= B
1
M
ntx

= 1.0(135.5) = 135.5 kN.m



Step II:

compute

M
nx
,

P
n


The effective length for compression and the unbraced length for
bending are the same = KL = L
b

= 4.8 m.


The bending is uniform over the unbraced length , so C
b
=1.0


Try a W10X60 with

P
n

= 2369 kN and

M
nx

= 344 kN.m




68402

Slide #
23

Ex. 5.3


Design
-
Beam Column



Step III:

Check interaction equation










Step IV:

Make sure that this is the lightest possible section.




Try W12x58 with

P
n

= 2247 kN and

M
nx

= 386 kN.m











Use a W12 x 58 section

2
.
0
56
.
0
2369
1328



n
c
u
P
P

OK

0
.
1
91
.
0
0
344
5
.
135
9
8
2369
1328
9
8





















ny
b
uy
nx
b
ux
n
c
u
M
M
M
M
P
P



0
.
1
90
.
0
0
386
5
.
135
9
8
2247
1328
9
8





















ny
b
uy
nx
b
ux
n
c
u
M
M
M
M
P
P



2
.
0
59
.
0
2247
1328



n
c
u
P
P

68402

Slide #
24

Design of Base Plates


We are looking for design of concentrically loaded columns. These base
plates are connected using anchor bolts to concrete or masonry footings



The column load shall spread over a large area of the bearing surface
underneath the base plate

AISC Manual Part 16, J8

68402

Slide #
25

Design of Base Plates

u
p
P
P


1
85
.
0
A
f
P
c
P



The design approach presented here combines three design approaches
for light, heavy loaded, small and large concentrically loaded base plates


B

0.8 b
f

0.95d

N


The

dimensions

of

the

plate

are

computed

such

that

m

and

n

are

approximately

equal
.


m

n

Area of Plate is computed such that

where:

1
1
2
1
7
.
1
85
.
0
A
f
A
A
A
f
P
c
c
P




6
.
0


If plate covers the area of the footing

If plate covers part of the area of the footing

A
1

= area of base plate

A
2

= area of footing

f’
c

= compressive strength of concrete used
for footing

68402

Slide #
26

Design of Base Plates

2
95
.
0
d
N
m








'
max
n
n
m
l

2
8
.
0
f
b
B
n


p
c
u
f
f
P
P
b
d
db
X











2
)
(
4
X
X
db
n
f




1
1
2
4
1
'



6
.
0

c


p
P
Nominal bearing strength

y
u
y
u
pl
NF
B
P
l
5
.
1
F
N
B
9
.
0
P
2
l
t


Thickness of plate

However



may

be

conservatively

taken

as

1

68402

Slide #
27



B

N

0.8b
f

0.95d

Ex. 5.4


Design of Base Plate


For the column base shown
in the figure, design a base
plate if the factored load on
the column is 10000 kN.
Assume 3 m x 3 m concrete
footing with concrete
strength of 20 MPa.


W14x211

68402

Slide #
28

Ex. 5.4
-

Design of Base Plate


Step I:

Plate dimensions


Assume



thus:





Assume
m

=
n






N

= 729.8 mm say N = 730 mm


B = 671.8 mm say B = 680 mm



2
1
2

A
A
2
3
1
3
1
1
10
2
.
490
10
10000
20
7
.
1
6
.
0
7
.
1
mm
A
A
P
A
f
P
u
c
p















mm
m
m
m
NB
A
m
m
m
b
B
m
m
m
d
N
f
4
.
175
10
2
.
490
2
321
2
379
2
321
2
401
8
.
0
2
8
.
0
2
379
2
399
95
.
0
2
95
.
0
3
1






















2
28
.
4
1
2


A
A
68402

Slide #
29


Step II:

Plate thickness

y
p
p
F
f
)
'
n

or
,
n
,
m
(
5
.
1
t

mm
db
n
mm
b
B
n
mm
d
N
m
f
f
100
4
1
'
5
.
179
2
/
)
8
.
0
(
5
.
175
2
/
)
95
.
0
(








Ex. 5.4
-

Design of Base Plate

68402

Slide #
30


Selecting the largest cantilever length






use 730 mm x 670 mm x 80 mm Plate

mm
t
MPa
f
req
p
7
.
76
248
14
.
20
)
5
.
179
(
5
.
1
14
.
20
730
680
10
10000
3






Ex. 5.4
-

Design of Base Plate