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

68402 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

68402 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≤

++

nyb

uy

nxb

ux

nc

u

M

M

M

M

P

P

φφφ

0.1≤∑n

ii

R

Q

φ

γ

68402 Slide # 6

Beam-Columns

� AISC code makes a distinct difference between lightly and heavily axial

loaded columns

0.19

8≤

++

nyb

uy

nxb

ux

nc

u

M

M

M

M

P

P

φφφ

0.12

≤

++

nyb

uy

nxb

ux

nc

u

M

M

M

M

P

P

φφφ2.0≤

nc

u

P

Pfor

φ

2.0≥nc

u

P

Pfor

φ

AISC Equation

AISC Equation

68402 Slide # 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

φc = Strength reduction factor for compression members = 0.90

φb = Strength reduction factor for flexural members = 0.90

68402 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

Pu/φφφφcPn

Safe Element

0.2

Mu/φφφφbMn

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

δδδδ

x

δδδδ 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

68402 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

68402 Slide # 12

� Braced frames are those frames prevented from sidesway.

� In this case the moment amplification equation can be simplified to:

ntxxux MBM 1=

1

1

1 ≥

−

=

e

u

m

P

P

CB AISC Equation

( )22

/ rKL

EAP

g

e

π=

� KL/r for the axis of bending considered

� K ≤ 1.0

ntyyuy MBM 1=

Unbraced and Braced Frames

68402 Slide # 13

� The coefficient Cm is used to represent the effect of end moments on the

maximum deflection along the element (only for braced frames)

−=

2

14.06.0M

MC m

veM

M−=

2

1ve

M

M+=

2

1

� When there is transverse loading on

the beam either of the following

case applies

00.1vely Conservati =mC

Unbraced and Braced Frames

68402 Slide # 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 = 18190 mm2, Ix = 347x106 mm4, Lp = 3.33 m, Lr

= 14.25 m, Zx = 2409 mm3, Sx = 2147 mm

3)

68402 Slide # 15

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: φcPn = 4831 kN Pu/φcPn = 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 = π2EIx/(KxLx)2 = π2(200)(347x106)/(3600)2 = 52851 kN

Ex. 5.1- Beam-Columns in Braced Frames

68402 Slide # 16

Ex. 5.1- Beam-Columns in Braced Frames

)(0.1073.1

52851

32001

0.1

11

1 OK

P

P

CB

e

u

m >=−

=−

=

Mux = (1.073)(240) = 257.5 kN.m

Step III: Compute moment capacity

Since Lb = 3.6 m Lp < Lb< Lr

mkNMnb

.739=φ

68402 Slide # 17

Ex. 5.1- Beam-Columns in Braced

Frames

0.1972.00739

5.257

9

8

4831

3200

9

8<=

++=

++

nyb

uy

nxb

ux

nc

u

M

M

M

M

P

P

φφφ

∴ Section is satisfactory

Step IV: Check combined effect

68402 Slide # 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.

68402 Slide # 19

Ex. 5.2- Analysis of Beam-Column

� Step I: Compute the capacities of the beam-column

φcPn = 4577 kN φMnx = 790 kN.m

φMny = 380 kN.m

� Step II: Check combined effect

2.0481.04577

2200>==

nc

u

P

P

φ

OK

0.1931.00790

400

9

8

4577

2200

9

8<=

++=

++

nyb

uy

nxb

ux

nc

u

M

M

M

M

P

P

φφφ

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 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

68402 Slide # 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) = 135.5 kN.m.

B1 = 1.0 � Mux = B1Mntx = 1.0(135.5) = 135.5 kN.m

� Step II: compute φMnx, φPn • 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 φPn = 2369 kN and φMnx = 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 φPn = 2247 kN and φMnx = 386 kN.m

� Use a W12 x 58 section

2.056.02369

1328>==

nc

u

P

P

φ

OK 0.191.00344

5.135

9

8

2369

1328

9

8<=

++=

++

nyb

uy

nxb

ux

nc

u

M

M

M

M

P

P

φφφ

0.190.00386

5.135

9

8

2247

1328

9

8<=

++=

++

nyb

uy

nxb

ux

nc

u

M

M

M

M

P

P

φφφ

2.059.02247

1328>==

nc

u

P

P

φ

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

up PP <φ

185.0 AfP cP′=

� The design approach presented here combines three design approaches

for light, heavy loaded, small and large concentrically loaded base plates

B 0.8 bf

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

21

7.185.0 AfA

AAfP

ccP′≤′=

6.0=φIf plate covers the area of the footing

If plate covers part of the area of the footing

A1 = area of base plate

A2 = area of footing

f’c = compressive strength of concrete used

for footing

68402 Slide # 26

Design of Base Plates

2

95.0 dNm

−=

='

max

n

n

m

l

λ

2

8.0 fbBn

−=

pc

u

f

f

P

P

bd

dbX

φ

+=

2)(

4 X

X

dbn f

−−=

=

11

2

4

1'

λ

λλ

6.0=cφ

=pP Nominal bearing strength

y

u

y

upl

NFB

Pl5.1

FNB9.0

P2lt ≈=

Thickness of plate

However λ may be conservatively taken as 1

68402 Slide # 27

B

N

0.8bf

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

21

2 >A

A

23

1

3

1

1

102.490

1010000207.16.0

7.1

mmA

A

PAfP ucp

×=

×=×××

=′=φ

( )( ) mmmmmNBA

mmmbB

mmmdN

f

4.175102.49023212379

232124018.028.0

2379239995.0295.0

3

1=⇒×=++==

+=+×=+=

+=+×=+=

228.41

2 >=A

A

68402 Slide # 29

� Step II: Plate thickness

y

pp

F

f)'n or,n,m(5.1t =

mmdbn

mmbBn

mmdNm

f

f

1004

1'

5.1792/)8.0(

5.1752/)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

mmt

MPaf

req

p

7.76248

14.20)5.179(5.1

14.20730680

1010000 3

==

=×

×=

Ex. 5.4 - Design of Base Plate