1. 62323: Architectural Structures II
Design of Beam-Columns
Monther Dwaikat
Assistant Professor
Department of Building Engineering
An-Najah National University
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2. 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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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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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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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.
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6. Beam-Columns
AISC code makes a distinct difference between lightly and heavily axial loaded columns
AISC Equation
AISC Equation
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7. 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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8. Beam-Columns Unsafe Element Safe Element 0.2
The increase in slope for lightly axial-loaded columns represents the less effect of axial load compared to the heavily axial-loaded columns
Unsafe Element
Pu/fcPn
Safe Element
0.2
Mu/fbMn
These are design charts that are a bit conservative than behaviour envelopes
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9. Moment Amplification P d M
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
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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
Unbraced Frames
Member deflection
Member sidesway
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11. 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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12. Unbraced and Braced Frames
Braced frames are those frames prevented from sidesway.
In this case the moment amplification equation can be simplified to:
AISC Equation
KL/r for the axis of bending considered
K ≤ 1.0
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13. 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)
When there is transverse loading on the beam either of the following case applies
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14. 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 = mm2, Ix = 347x106 mm4, Lp = 3.33 m, Lr = m, Zx = 2409 mm3, Sx = 2147 mm3)
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15. 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 = > 0.2  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 = kN
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16. Ex. 5.1- Beam-Columns in Braced Frames
Mux = (1.073)(240) = kN.m
Step III: Compute moment capacity
Since Lb = 3.6 m Lp < Lb< Lr
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17. Ex. 5.1- Beam-Columns in Braced Frames
Step IV: Check combined effect
 Section is satisfactory
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18. 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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19. Ex. 5.2- Analysis of Beam-Column
Step I: Compute the capacities of the beam-column
cPn = 4577 kN Mnx = 790 kN.m
Mny = 380 kN.m
Step II: Check combined effect OK
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20. Design of Beam-Columns
Trial-and-error procedure
Select trial section
Check appropriate interaction formula.
Repeat until section is satisfactory
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21. 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.
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
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22. 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) = kN.m.
B1 = 1.0  Mux = B1Mntx = 1.0(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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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 Pn = 2247 kN and Mnx = 386 kN.m
 Use a W12 x 58 section OK
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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
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25. 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 B
where:
If plate covers the area of the footing
0.95d N
If plate covers part of the area of the footing
The dimensions of the plate are computed such that m and n are approximately equal.
A1 = area of base plate
A2 = area of footing
f’c = compressive strength of concrete used for footing
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26. Design of Base Plates Thickness of plate Nominal bearing strength
However  may be conservatively taken as 1
Nominal bearing strength
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27. 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 kN. Assume 3 m x 3 m concrete footing with concrete strength of 20 MPa.
W14x211
0.95d N
Steel base plates for steel columns can be welded or bolted to columns to transfer load from column to concrete or masonry foundation.
0.8bf B
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28. Ex. 5.4 - Design of Base Plate
Step I: Plate dimensions
Assume thus:
Assume m = n
N = mm say N = 730 mm
B = mm say B = 680 mm
To determine plate thickness, moments are taken in two directions as though the plate is cantilevered out by the dimension m and n
tp = max(m,n, or n’) (2Pu/0.9FyBN)1/2
fp = Pu/BN
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29. Ex. 5.4 - Design of Base Plate
Step II: Plate thickness
To determine plate thickness, moments are taken in two directions as though the plate is cantilevered out by the dimension m and n
tp = max(m,n, or n’) (2Pu/0.9FyBN)1/2
fp = Pu/BN
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30. Ex. 5.4 - Design of Base Plate
Selecting the largest cantilever length
use 730 mm x 670 mm x 80 mm Plate
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