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
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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
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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
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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
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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
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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
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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
)
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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
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