Steel_Ch5 -Beam-Column 1 - An
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Transcript 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
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
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
loaded columns
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
2c 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
effect of axial load compared to the heavily axial-loaded columns
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
When there is transverse loading on
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
moments and is not loaded transversely. Use Grade
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
subjected to the service-load
axial force and bending moments
shown (the end shears are not
shown). Bending is about the
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
for light, heavy loaded, small and large concentrically loaded base plates
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 cA1
If plate covers part of the area of the footing
PP 0.85 f cA1
A2
1.7 f cA1
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 maxn
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 cA1 Pu
0.6 1.7 20 A1 10000103
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
10000103
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