Transcript Steel_Ch5 -Beam-Column 1 - An

68402: Structural Design of Buildings II
61420: Design of Steel Structures
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

Second Order Analysis

Compact Sections for Beam-Columns

Braced and Unbraced Frames

Analysis/Design of Braced Frames

Analysis/Design of Unbraced Frames

Design of Bracing Elements
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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
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Slide # 4
Beam-Columns
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Slide # 5
Beam-Columns
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Slide # 6
Beam-Columns
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Slide # 7
Beam-Columns
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Slide # 8
Beam-Columns
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Slide # 9
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 # 10
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 # 11
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
2c Pn  b M nx b M ny 
AISC Equation
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Slide # 12
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 # 13
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 # 14
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 # 15
Moment Amplification
Second-order Moments, Puδ and Pu
Moment amplification in column
braced against sidesway
Moment amplification in
unbraced column
Mu = Mnt + Puδ
Mu = Mlt + Pu
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Slide # 16
Moment Amplification

Using first principles we can prove that the final moment Mmax is
amplified from M0 as
M max



1
 M 0 B  M 0 
P
1   u
P

 e







The amplification factor B can be


1
B

 Pu
1  
P

 e









Where
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M max  bending moment
Slide # 17
Second Order Analysis
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Slide # 18
Second Order Analysis
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Slide # 19
Second Order Analysis
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Slide # 20
Second Order Analysis
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Slide # 21
Second Order Analysis
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Slide # 22
Second Order Analysis
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Slide # 23
Second Order Analysis
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Slide # 24
Compact Sections for BeamColumns

The axial load affects the ratio for compactness. When the check for
compactness for the web is performed while the web is subjected to axial
bf
load the following ratios shall be
for
for
Pu
 0.125
b Py
Pu
 0.125
b Py
for all
Pu
b Py
Flange limit is
similar to beams
2.75Pu
E 
 p  3.76
1


Fy 
b Py



tf
h
tw
E
 p  1.12
Fy

Pu 
E
2
.
33


1
.
49



P
Fy

b y 


E
r  5.70
Fy

0.74Pu
1


b Py

 p  0.38
E
Fy
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r  0.83



E
Fy  68 .9
Slide # 25
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 # 26
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 # 27
Unbraced and Braced Frames


The AISC code approximate the effect by using two amplification
factors B1 and B2
M u  B1 M nt  B2 M lt
AISC Equation
Pr  Pnt  B2 Plt
AISC Equation
Where

B1 amplification factor for the moment occurring in braced member

B2 amplification factor for the moment occurring from sidesway

Mnt and Pnt is the maximum moment and axial force assuming no sidesway

Mlt and Plt is the maximum moment and axial force due to sidesway

Pr is the required axial strength
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Slide # 28
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 # 29
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 # 30
Unbraced and Braced Frames

AISC requires stability bracing to have

Specific strength to resist the lateral load

Specific axial stiffness to limit the lateral deformation.
Pbr  0.004 Pu
bbr 
bbr 
3 Pu
2 Pu
L
Braced
Frames
Unbraced
Frames
L

Where Pu is the sum of factored axial load in the braced story

Pbr is bracing strength and bbr is braced or unbraced frame stiffness (
= 0.75)
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Slide # 31
Unbraced and Braced Frames

Unbraced frames can observe loading + sidesway

In this case the moment amplification equation can be simplified to:
BMD
M u  B1 M nt  B2 M lt
1.0
B1 
 Pu
1  
 Pe



1
AISC Equation
B2 
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1.0
 Pu   oh 
1
RM  H  L 
Slide # 32
Unbraced and Braced Frames

A minimum lateral load in each combination shall be added so that the
shear in each story is given by:
Hu  0.0042 Pu
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Slide # 33
Analysis of Unbraced Frames
 Pu
is the sum of factored axial loads on all columns in floor
oh
is the drift due to the unfactored horizontal forces
L
is the story height
H
 oh 
 L 


story shear produced by unfactored horizontal forces
is the drift index (is generally between 1/500 to 1/200)
Pe
is the sum of Euler buckling loads of all columns in floor
Pu
is the factored axial load in the column
RM
can be conservatively taken as 0.85
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Slide # 34
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 # 35
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 # 36
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 # 37
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 # 38
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 # 39
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 # 40
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 # 41
Design of Unbraced Frames

Design can
procedure:
 Use
be
performed
using
the
following
a procedure similar to that of braced frames
 To
start the design assume B1 = 1.0 and compute B2
by assuming the ratio
oh
L

1
1
to
500
200
 Compute
Mu and perform same procedure used for
braced frames
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Slide # 42
Ex. 5.3- Analysis-External Column
• Check the exterior column of an unbraced frame shown in the
figure for the following load combination. All columns are 3.8 m
long and all beams are 9 m long. Assume A992 steel.
LoadCase  1.2 D  0.5L  1.6W
Pnt  2250kN Plt  0
W24x76
M nt  128kN - m
M lt  210 kN - m
W14x90
For this frame
 H  660 kN
W24x76
 Pu  10800kN
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Slide # 43
Ex. 5.3- Analysis-External Column
Step I: Calculate Kx and Ky
 416 
2

3.8 

GT  GB 
 2.25
 874 


 9 
K x  1.66
W14x90
Effective length, Ky ,
assumed braced frame
W24x76
W24x76
K y  1.0
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Slide # 44
Ex. 5.3- Analysis-External Column
Step II: Calculate Pn and p
K x Lx 1.66(3800)

 40.4
rx
156
K y Ly
ry
1 3800

 40.4
94
Pn  4700kN
 M nx  797 kN.m
 M ny  380 kN.m
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Slide # 45
Ex. 5.3- Analysis-External Column
Step III: Determine second-order moments-No translation, Mnt
Due to lack of information, assume Cm = 1.0
Pe1 
 2 (200)(416106 )
1.66 3800
B1 
2
 20637kN
1
 1.12
2250
1
20637
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Slide # 46
Ex. 5.3- Analysis-External Column
Step IV: Determine second-order moments - Translation, Mlt.
Don’t know all columns in story, thus assume the frame will have a
deflection limit
 oh  L 400
For this frame
 H  660 kN
 Pu  10800kN
Thus,
B2 
1
 1.04
10800 1 
1


660  400 
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Slide # 47
Ex. 5.3- Analysis-External Column
•
Step V: Second-order moment
M u  1.12(128)  1.04(210)  361.7 kN - m

Step VI: Check combined effect
Pu
2250

 0.479  0.2
c Pn 4700
M uy
Pu
8  M ux



c Pn 9  b M nx b M ny
•
•
 2250 8  361.7


 
 0   0.882  1.0
 4700 9  797


OK
Thus, the W14x90, Fy = 344 MPa will work for this loading case.
Now it should be checked for any other load case, such as
1.2D+1.6L
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Slide # 48
Ex. 5.4 – 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 # 49
Ex. 5.4 – 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 # 50
Ex. 5.4 – 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 # 51
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 # 52
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 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 # 53
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 # 54
Ex. 5.5 – 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 # 55
N
Ex. 4.7- 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 # 56
Ex. 4.7- 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 # 57
Ex. 4.7- 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 # 58
Eccentrically Loaded Columns



For eccentrically loaded columns
Compute dimensions such that stress (q) is less than concrete
compressive strength.
Compute thickness so that the ultimate moment on the plate equals
the full plastic moment multiplied by , where  = 0.9.

6e 
1 
  f c
 N or B 
qmax 
Pu
BN
qmin 
Pu 
6e 
1 
  0
BN  N or B 
t p  2. 1
Mu
Fy
no tension
e = eccentricity
Mu = ultimate moment per (mm) width on the plate
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Slide # 59