### Steel_Ch5 -Beam-Column 1 - An

```62323: Architectural Structures II
Design of Beam-Columns
Monther Dwaikat
Assistant Professor
Department of Building Engineering
An-Najah National University
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Slide # 1
Beam-Column - Outline

Beam-Columns

Moment Amplification Analysis

Braced and Unbraced Frames

Analysis/Design of Braced Frames

Design of Base Plates
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Slide # 2
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 # 3
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 # 4
Beam-Columns

Structural elements subjected to combined flexural moments and axial

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
 Q
i
i
Rn

 1.0
Thus: a column beam interaction can be written as
 M ux
M uy 
Pu


  1.0
c Pn b M nx b M ny 

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.
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Slide # 5
Beam-Columns

AISC code makes a distinct difference between lightly and heavily axial
P
for u  0.2
c Pn
M uy 
Pu
8  M ux
 

  1.0
c Pn 9 b M nx b M ny 
AISC Equation
P
for u  0.2
c Pn
 M ux
M uy 
Pu


  1.0
2c Pn  b M nx b M ny 
AISC Equation
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Slide # 6
Beam-Columns

Definitions
Pu = factored axial compression load
Pn = nominal compressive strength
Mux = factored bending moment in the x-axis, including second-order effects
Mnx = nominal moment strength in the x-axis
Muy = same as Mux except for the y-axis
Mny = same as Mnx 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 # 7
Beam-Columns

The increase in slope for lightly axial-loaded columns represents the less
Pu/cPn
Unsafe Element
Safe Element
0.2
Mu/bMn
These are design charts that are a bit conservative than behaviour envelopes
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Slide # 8
Moment Amplification

When a large axial load exists, the axial load produces moments due to
any element deformation.
x
d
P
P
d
M

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.
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Slide # 9
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
Unbraced Frames
Member deflection
Member sidesway
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Slide # 10
Unbraced and Braced Frames

In braced frames amplification moments can only happens due to
member deflection
Braced Frames
Sidesway bracing system
Member deflection
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Slide # 11
Unbraced and Braced Frames

Braced frames are those frames prevented from sidesway.

In this case the moment amplification equation can be simplified to:
M ux  B1x M ntx
 EAg
2
Pe 
Cm
B1 
 Pu
1  
 Pe
KL / r 



1
M uy  B1 y M nty
AISC Equation
2

KL/r for the axis of bending considered

K ≤ 1.0
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Slide # 12
Unbraced and Braced Frames

The coefficient Cm is used to represent the effect of end moments on the
maximum deflection along the element (only for braced frames)
Cm
 M1 
 0.6  0.4 
M 

2 

M1
 ve
M2
M1
  ve
M2

the beam either of the following
case applies
Conservatively Cm  1.00
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Slide # 13
Ex. 5.1- Beam-Columns in Braced Frames
A 3.6-m W12x96 is subjected to bending and
compressive loads in a braced frame. It is bent in
single curvature with equal and opposite end
50 steel. Is the section satisfactory if Pu = 3200 kN
and first-order moment Mntx = 240 kN.m
Step I: From Section Property Table
W12x96 (A = 18190 mm2, Ix = 347x106 mm4, Lp = 3.33 m, Lr
= 14.25 m, Zx = 2409 mm3, Sx = 2147 mm3)
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Slide # 14
Ex. 5.1- Beam-Columns in Braced Frames
Step II: Compute amplified moment
- For a braced frame let K = 1.0
KxLx = KyLy = (1.0)(3.6) = 3.6 m
- From Column Chapter: cPn = 4831 kN
Pu/cPn = 3200/4831 = 0.662 > 0.2  Use eqn.
- There is no lateral translation of the frame: Mlt = 0
 Mux = B1Mntx
Cm = 0.6 – 0.4(M1/M2) = 0.6 – 0.4(-240/240) = 1.0
Pe1 = 2EIx/(KxLx)2 = 2(200)(347x106)/(3600)2 = 52851 kN
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Slide # 15
Ex. 5.1- Beam-Columns in Braced Frames
B1 
Cm
1.0

 1.073  1.0
P
3200
1 u 1
52851
Pe1
(OK )
Mux = (1.073)(240) = 257.5 kN.m
Step III: Compute moment capacity
Since Lb = 3.6 m
Lp < Lb< Lr
b M n  739 kN.m
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Slide # 16
Ex. 5.1- Beam-Columns in Braced
Frames
Step IV: Check combined effect
M uy
Pu
8  M ux



c Pn 9  b M nx b M ny
 3200 8  257.5


 
 0   0.972  1.0
 4831 9  739


 Section is satisfactory
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Slide # 17
Ex. 5.2- Analysis of Beam-Column

Check the adequacy of an ASTM A992 W14x90 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:
•
•
ky = 1.0
Lateral unbraced length of the compression flange is 4.2 m.
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Slide # 18
Ex. 5.2- Analysis of Beam-Column

Step I: Compute the capacities of the beam-column
cPn = 4577 kN
Mny = 380 kN.m

Mnx = 790 kN.m
Step II: Check combined effect
Pu
2200

 0.481 0.2
c Pn 4577
M uy
Pu
8  M ux



c Pn 9  b M nx b M ny
 2200 8  400 


 0  0.931 1.0
 4577 9  790 

OK
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Slide # 19
Design of Beam-Columns

Trial-and-error procedure
• Select trial section
• Check appropriate interaction formula.
• Repeat until section is satisfactory
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Slide # 20
Ex. 5.3 – Design-Beam Column

Select a W shape of A992 steel
for the beam-column of the
following figure. This member is
part of a braced frame and is
axial force and bending moments
shown (the end shears are not
strong axis, and Kx = Ky = 1.0.
Lateral support is provided only at
the ends. Assume that B1 = 1.0.
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PD = 240 kN
PL = 650 kN
MD = 24.4 kN.m
ML = 66.4 kN.m
4.8 m
MD = 24.4 kN.m
ML = 66.4 kN.m
Slide # 21
Ex. 5.3 – Design-Beam Column

Step I: Compute the factored axial load and bending moments
Pu = 1.2PD + 1.6PL = 1.2(240)+ 1.6(650) = 1328 kN.
Mntx = 1.2MD + 1.6ML = 1.2(24.4)+ 1.6(66.4) = 135.5 kN.m.
B1 = 1.0  Mux = B1Mntx = 1.0(135.5) = 135.5 kN.m

Step II: compute Mnx, Pn
•
•
•
The effective length for compression and the unbraced length for
bending are the same = KL = Lb = 4.8 m.
The bending is uniform over the unbraced length , so Cb=1.0
Try a W10X60 with Pn = 2369 kN and Mnx = 344 kN.m
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Slide # 22
Ex. 5.3 – Design-Beam Column

Step III: Check interaction equation
Pu
1328

 0.56  0.2
c Pn 2369
M uy
Pu
8  M ux


c Pn 9  b M nx b M ny

 1328 8  135.5


 
 0   0.91  1.0
 2369 9  344


OK
Step IV: Make sure that this is the lightest possible section.
 Try W12x58 with Pn = 2247 kN and Mnx = 386 kN.m
Pu
1328

 0.59  0.2
c Pn 2247
M uy
Pu
8  M ux



c Pn 9  b M nx b M ny
 1328 8  135.5


 
 0   0.90  1.0
 2247 9  386


 Use a W12 x 58 section
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Slide # 23
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
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Slide # 24
Design of Base Plates

The design approach presented here combines three design approaches
Area of Plate is computed such that
n
m
0.8 bf
0.95d
N

The dimensions of the plate
are computed such that m and
n are approximately equal.
B
Pp  Pu
where:
  0.6
If plate covers the area of the footing
PP  0.85 f cA1
If plate covers part of the area of the footing
PP  0.85 f cA1
A2
 1.7 f cA1
A1
A1 = area of base plate
A2 = area of footing
f’c = compressive strength of concrete used
for footing
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Slide # 25
Design of Base Plates
Thickness of plate
t pl  l
m

l  maxn
 n '

2 Pu
Pu
 1.5 l
0.9 B N Fy
B NFy
N  0.95 d
m
2
n
B  0.8 b f
1
dbf 
4
2 X

1 1 X
n ' 
2
 4dbf  Pu
X 
2
(
d

b
)

 c Pp
f
However

may
be
conservatively taken as 1
c  0.6
Pp  Nominal bearing strength
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Slide # 26
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.
0.95d
W14x211
0.8bf
B
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Slide # 27
N
Ex. 5.4 - Design of Base Plate

Step I: Plate dimensions
•
Assume
A2
2
A1
thus:
 Pp  1.7 f cA1  Pu
0.6 1.7  20 A1  10000103
A1  490.2 103 m m2
•
A2
 4.28  2
A1
Assume m = n
N  0.95d  2m  0.95 399 2m  379 2m
B  0.8b f  2m  0.8  401 2m  321 2m
•
A1  NB  379 2m 321 2m   490.2  103  m  175.4 m m
N = 729.8 mm say N = 730 mm
B = 671.8 mm say B = 680 mm
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Slide # 28
Ex. 5.4 - Design of Base Plate

Step II: Plate thickness
t p  1.5( m ,n ,or n' )
fp
Fy
m  ( N  0.95d ) / 2  175.5 m m
n  ( B  0.8b f ) / 2  179.5 m m
1
n' 
dbf  100m m
4
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Slide # 29
Ex. 5.4 - Design of Base Plate

Selecting the largest cantilever length
10000103
fp 
 20.14 MPa
680 730
20.14
t req  1.5(179.5)
 76.7 m m
248

use 730 mm x 670 mm x 80 mm Plate
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Slide # 30
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