Biaxial Bending Short Columns

Reinforced Concrete II Hashemite University

Dr. Hazim Dwairi 1

The Hashemite University

Department of Civil Engineering

Lecture 6 Lecture 6 –– Biaxial Bending Biaxial Bending of Short Columnsof Short Columns

Dr Hazim DwairiDr Hazim Dwairi

Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University

Dr. Hazim DwairiDr. Hazim Dwairi

BiaxiallyBiaxially Loaded ColumnLoaded Column



Dr. Hazim Dwairi 2

Interaction DiagramInteraction Diagram

Uniaxial Bending about y-axis


Uniaxial Bending about x-axis

Approximation of Section Approximation of Section Through Intersection SurfaceThrough Intersection Surface



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NotationNotation

•• PPuu = factored axial load, positive in compression= factored axial load, positive in compression•• ee = eccentricity measured parallel to the x= eccentricity measured parallel to the x axis positive toaxis positive to•• eexx = eccentricity measured parallel to the x= eccentricity measured parallel to the x--axis, positive to axis, positive to

the right.the right.•• eeyy = eccentricity measured parallel to y= eccentricity measured parallel to y--axis, positive axis, positive

upward.upward.•• MMuxux = factored moment about x= factored moment about x--axis, positive when causing axis, positive when causing

compression in fibers in the +compression in fibers in the +veve yy--direction = direction = PPuu.e.eyy

Reinforced Concrete IIReinforced Concrete II

•• MMuyuy = factored moment about y= factored moment about y--axis, positive when causing axis, positive when causing compression in fibers in the +compression in fibers in the +veve xx--direction = direction = PPuu.e.exx

Dr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University

Analysis and DesignAnalysis and Design

•• Method I:Method I: Strain Compatibility MethodStrain Compatibility MethodThi i th t l th ti ll t th dThi i th t l th ti ll t th dThis is the most nearly theoretically correct method This is the most nearly theoretically correct method of solving of solving biaxiallybiaxially--loadedloaded--column (see Macgregor column (see Macgregor example 11example 11--5)5)•• Method II:Method II: Equivalent Eccentricity MethodEquivalent Eccentricity MethodAn approximate method. Limited to columns that An approximate method. Limited to columns that

t i l b t t ith ti f idt i l b t t ith ti f id


are symmetrical about two axes with a ratio of side are symmetrical about two axes with a ratio of side lengths llengths lxx//llyy between 0.5 and 2.0 (see Macgregor between 0.5 and 2.0 (see Macgregor example 11example 11--6)6)



Dr. Hazim Dwairi 4

Strain Compatibility MethodStrain Compatibility Method


Equivalent Eccentricity MethodEquivalent Eccentricity Method•• Replace the biaxial eccentricities eReplace the biaxial eccentricities exx & & eeyy by an by an

equivalent eccentricity eequivalent eccentricity e0x0xe

ePM and Pfor column design then

0

0xu0yu

+=

=≥

y

xyxx

y

y

x

x

lle

ee

le

leif

α


5.0696

27631 0.6

696276

5.0

4.0 4.0

'

''

≥+

⎟⎟⎠

⎞⎜⎜⎝

⎛−=≥

+⎟⎟⎠

⎞⎜⎜⎝

⎛+=

>≤

y'cg

uy

cg

u

cgucgu

ffA

P.f

fAP

fAPforfAPfor

αα


Dr. Hazim Dwairi 5

Analysis and DesignAnalysis and Design

•• Method III:Method III: 4545oo Slice through Interaction SurfaceSlice through Interaction Surface( M 524)( M 524)(see Macgregor page 524)(see Macgregor page 524)•• Method IV:Method IV: BreslerBresler Reciprocal Load MethodReciprocal Load MethodACI commentary sections 10.3.6 and 10.3.7 give ACI commentary sections 10.3.6 and 10.3.7 give the following equation, originally presented by the following equation, originally presented by BreslerBresler for calculating the capacity under biaxial for calculating the capacity under biaxial b dib di


bendingbending..

•• Method VMethod V:: BreslerBresler Contour Contour Load MethodLoad MethodDr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University

n0nynxu

1111PPPP φφφ

−+≅

BreslerBresler Reciprocal Reciprocal Load MethodLoad Method

1. Use Reciprocal 1. Use Reciprocal Failure surface SFailure surface SFailure surface SFailure surface S22(1/(1/PPnn,e,exx,e,eyy))2. The ordinate 1/2. The ordinate 1/PPnn on on the surface Sthe surface S22 is is approximated by approximated by ordinate 1/ordinate 1/PP on theon the


ordinate 1/ordinate 1/PPnn on the on the plane S’plane S’22 (1/(1/PPnn eexx,e,eyy))3. Plane S3. Plane S22 is defined is defined by points A,B, and C.by points A,B, and C.



Dr. Hazim Dwairi 6

BreslerBresler Reciprocal Reciprocal Load MethodLoad Method

PP00 = Axial Load Strength under pure axial = Axial Load Strength under pure axial compression (corresponds to point Ccompression (corresponds to point C ))compression (corresponds to point C compression (corresponds to point C ))

MMnxnx = = MMnyny = 0= 0PP0x0x = Axial Load Strength under = Axial Load Strength under uniaxialuniaxialeccentricity, eccentricity, eeyy (corresponds to point B ) (corresponds to point B )

MMnxnx = = PPnn eeyy


PP0y0y = Axial Load Strength under = Axial Load Strength under uniaxialuniaxialeccentricity, eeccentricity, exx (corresponds to point A ) (corresponds to point A )

MMnyny = = PPnn eexx


BreslerBresler Load Contour MethodLoad Contour Method

•• In this method, the surface SIn this method, the surface S33 is approximated is approximated by a family of curves corresponding to constantby a family of curves corresponding to constantby a family of curves corresponding to constant by a family of curves corresponding to constant values of values of PPnn. These curves may be regarded as . These curves may be regarded as “load contours.”“load contours.”

where Mnx and Mny are the nominal biaxial moment strengths in the direction of the x-and y-axes, respectively.Note that these moments are the vectorial


Note that these moments are the vectorialequivalent of the nominal uniaxial moment Mn. The moment Mn0x is the nominal uniaxial moment strength about the x-axis, and Mn0y is the nominal uniaxial moment strength about the y-axis.


Dr. Hazim Dwairi 7

BreslerBresler Load Contour MethodLoad Contour Method•• The general expression for the contour curves The general expression for the contour curves

can be approximated as:can be approximated as:

•• The values of the exponents The values of the exponents αα and and ββ are a are a function of the amount distribution and locationfunction of the amount distribution and location

0.100

=⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛βα

yn

ny

xn

nx

MM

MM


function of the amount, distribution and location function of the amount, distribution and location of reinforcement, the dimensions of the column, of reinforcement, the dimensions of the column, and the strength and elastic properties of the and the strength and elastic properties of the steel and concrete. steel and concrete. BreslerBresler indicates that it is indicates that it is reasonably accurate to assume that reasonably accurate to assume that αα = = ββ


BreslerBresler Load Contour MethodLoad Contour Method

•• BreslerBresler indicated that, typically, indicated that, typically, αα varied from 1.15 varied from 1.15 to 1 55 with a value of 1 5 being reasonablyto 1 55 with a value of 1 5 being reasonablyto 1.55, with a value of 1.5 being reasonably to 1.55, with a value of 1.5 being reasonably accurate for most square and rectangular sections accurate for most square and rectangular sections having uniformly distributed reinforcement. A having uniformly distributed reinforcement. A value of value of αα = 1.0 will yield a safe design.= 1.0 will yield a safe design.

01=⎟⎟⎞

⎜⎜⎛

+⎟⎟⎞

⎜⎜⎛ nynx MM


•• Only applicable if:Only applicable if:


0.100⎟⎟⎠

⎜⎜⎝

+⎟⎟⎠

⎜⎜⎝ ynxn MM

gcn AfP '1.0<


Dr. Hazim Dwairi 8

Biaxial Column ExampleBiaxial Column Example

The section of a short tied The section of a short tied column is 400 x 600 mmcolumn is 400 x 600 mm

66mm

column is 400 x 600 mm column is 400 x 600 mm and is reinforced with 6and is reinforced with 6φφ32 32 bars as shown. Determine bars as shown. Determine the allowable ultimate load the allowable ultimate load on the section on the section φφPPnn if its if its acts at eacts at exx = 200mm. and = 200mm. and eeyy

600m

m

234mm

234mm


xx yy= 300mm. Use = 300mm. Use ffcc’ = 35 ’ = 35 MPaMPa and and ffyy = 420 = 420 MPaMPa..


400mm66mm


•• Compute PCompute P00 load, pure axial loadload, pure axial load

( )( )P

fAAAfP

mmA

mmA

yststgc

g

st

420482448242400003585.0

85.0

240000600400

48248046

0

'0

2

2

×+−××=

+−=

=×=

=×=


( )

kNPkNP

n 721890238.09023

0

0

0

=×==

kNPn 72180 =


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•• Compute Compute PPnxnx, by starting with , by starting with eeyy term and term and assume that compression controls Check by:assume that compression controls Check by:

OK! 356)534(3/23/2300 mmdmmey ==<=

assume that compression controls. Check by:assume that compression controls. Check by:

•• Compute the nominal load, Compute the nominal load, PPnxnx and assume and assume second compression steel does not contributesecond compression steel does not contribute

Assume = 0 0


TCCCP sscn −++= 21

Assume = 0.0


•• Brake equilibrium equation into its components:Brake equilibrium equation into its components:9639)400)(810)(35(850C

)534(964800)600)(534)(1608(

627715)3585.0420)(1608(9639)400)(81.0)(35(85.0

1

cc

ccT

NCccC

s

s

c

−=

−=

=×−===

•• Compute the moment about tension steel:Compute the moment about tension steel:


Compute the moment about tension steel:Compute the moment about tension steel:

( )( ) )66534)(627715(405.05349639)234300(

2. '

11'

−+−=+

−+⎟⎠⎞

⎜⎝⎛ −=

ccP

ddCcdCeP

n

scnβ


Dr. Hazim Dwairi 10


2

•• The resulting equation is:The resulting equation is:132,550311.7639,9 2 +−= ccPn

•• Recall equilibrium equation:Recall equilibrium equation:

sn fcP 1608627715639,9 −+=•• Set the two equation equal to one another and Set the two equation equal to one another and

solve forsolve for ffss::


solve for solve for ffss::

4.3900046.0 2 += cfs


⎞⎛

•• Recall Recall ffss definition:definition:

⎟⎠⎞

⎜⎝⎛ −

=c

cfs534600

•• Combine both equations:Combine both equations:5346004.3900046.0 2 ⎟

⎠⎞

⎜⎝⎛ −

=+c

cc


03204004.9900046.0 3 =−+ cc•• Solve cubic equation by trial and errorSolve cubic equation by trial and error

c = 323 mmc = 323 mm


Dr. Hazim Dwairi 11


300323 ⎞⎛•• Check the assumption that fCheck the assumption that fs2s2 = 0.0= 0.0

SMALL TOO 7.68

72.42323

300323600

2

2

kNC

MPaf

s

s

=

=⎟⎠⎞

⎜⎝⎛ −

=

•• Calculate Calculate PPnxnx

132550)323(3117)323(6399 2 +P


132,550)323(311.7)323(639,9 +−=nP

kNPnx 2900=


•• Compute Compute PPnyny, by starting with , by starting with eexx term and term and assume that compression controls Check by:assume that compression controls Check by:

OK! 223)334(3/23/2200 mmdmmex ==<=

assume that compression controls. Check by:assume that compression controls. Check by:

•• Compute the nominal load, Compute the nominal load, PPnyny


TCCP scn −+= 1


Dr. Hazim Dwairi 12


•• Brake equilibrium equation into its components:Brake equilibrium equation into its components:514458)600)(810)(35(850C

)334(1447200)600)(334)(2412(

941283)3585.0420)(2412(5.14458)600)(81.0)(35(85.0

1

cc

ccT

NCccC

s

s

c

−=

−=

=×−===

•• Compute the moment about tension steel:Compute the moment about tension steel:


Compute the moment about tension steel:Compute the moment about tension steel:

( )( ) )66334)(941283(405.03345.14458)134200(2

. '1

1'

−+−=+

−+⎟⎠⎞

⎜⎝⎛ −=

ccP

ddCcdCeP

n

scnβ


2

•• The resulting equation is:The resulting equation is:281,75550.175.458,14 2 +−= ccPn

•• Recall equilibrium equation:Recall equilibrium equation:

sn fcP 412,2283,9415.458,14 −+=•• Set the two equation equal to one another and Set the two equation equal to one another and

solve forsolve for ffss::


solve for solve for ffss::

12.770073.0 2 += cfs


Dr. Hazim Dwairi 13


⎞⎛

•• Recall Recall ffss definition:definition:

⎟⎠⎞

⎜⎝⎛ −

=c

cfs334600

•• Combine both equations:Combine both equations:33460012.770073.0 2 ⎟

⎠⎞

⎜⎝⎛ −

=+c

cc


020040012.6770073.0 3 =−+ cc•• Solve cubic equation by trial and errorSolve cubic equation by trial and error

c = 295 mmc = 295 mm


•• Calculate Calculate PPnyny2

kNPny 3498=

281,755)295(50.17)295(5.458,14

281,75550.175.458,142

2

+−=

+−=

n

n

P

ccP



Dr. Hazim Dwairi 14


•• Calculate Nominal Biaxial Load Calculate Nominal Biaxial Load PPnn

1111

72181

34981

290011

1111

0

−+=

−+=

n

nnynxn

P

PPPP


kNPn 2032=

kNPP nu 1321)2032)(65.0( ===φ

Design of Biaxial ColumnDesign of Biaxial Column

1)1) Select trial sectionSelect trial sectionP

2)2) Compute Compute γγ3)3) Compute Compute φφPPnxnx, , φφPPnyny, , φφPPn0n0

( ) 0015.0 use ;40.0 ')( =

+≥ t

ytc

utrialg ff

PA ρρ

ly

γlx

stA


lx

ly

xu

uy

x

x

yx

stt

lPM

le

llA

=

=ρ

yu

ux

y

y

lPM

le

=


Dr. Hazim Dwairi 15


φPnx

φPny

φPn0



4)4) Solve for Solve for φφPPnn

5)5) If If φφPPnn < < PPuu then design is inadequate, increase then design is inadequate, increase either area of steel or column dimensions either area of steel or column dimensions

0

1111

nnynxn PPPP φφφφ−+=


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Biaxial Bending Short Columns