Post on 21-Apr-2015
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
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Dr. Hazim DwairiDr. Hazim Dwairi
BiaxiallyBiaxially Loaded ColumnLoaded Column
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Dr. Hazim Dwairi 2
Interaction DiagramInteraction Diagram
Uniaxial Bending about y-axis
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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
Reinforced Concrete IIReinforced Concrete II
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 DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University
Reinforced Concrete II Hashemite University
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Strain Compatibility MethodStrain Compatibility Method
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University
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
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University
5.0696
27631 0.6
696276
5.0
4.0 4.0
'
''
≥+
⎟⎟⎠
⎞⎜⎜⎝
⎛−=≥
+⎟⎟⎠
⎞⎜⎜⎝
⎛+=
>≤
y'cg
uy
cg
u
cgucgu
ffA
P.f
fAP
fAPforfAPfor
αα
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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
Reinforced Concrete IIReinforced Concrete II
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
Reinforced Concrete IIReinforced Concrete II
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 DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University
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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
Reinforced Concrete IIReinforced Concrete II
PP0y0y = Axial Load Strength under = Axial Load Strength under uniaxialuniaxialeccentricity, eeccentricity, exx (corresponds to point A ) (corresponds to point A )
MMnyny = = PPnn eexx
Dr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University
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
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University
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.
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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
Reinforced Concrete IIReinforced Concrete II
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
Dr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University
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
Reinforced Concrete IIReinforced Concrete II
•• Only applicable if:Only applicable if:
Dr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University
0.100⎟⎟⎠
⎜⎜⎝
+⎟⎟⎠
⎜⎜⎝ ynxn MM
gcn AfP '1.0<
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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
Reinforced Concrete IIReinforced Concrete II
xx yy= 300mm. Use = 300mm. Use ffcc’ = 35 ’ = 35 MPaMPa and and ffyy = 420 = 420 MPaMPa..
Dr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University
400mm66mm
Biaxial Column ExampleBiaxial Column Example
•• 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
×+−××=
+−=
=×=
=×=
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University
( )
kNPkNP
n 721890238.09023
0
0
0
=×==
kNPn 72180 =
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Biaxial Column ExampleBiaxial Column Example
•• 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
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University
TCCCP sscn −++= 21
Assume = 0.0
Biaxial Column ExampleBiaxial Column Example
•• 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:
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University
Compute the moment about tension steel:Compute the moment about tension steel:
( )( ) )66534)(627715(405.05349639)234300(
2. '
11'
−+−=+
−+⎟⎠⎞
⎜⎝⎛ −=
ccP
ddCcdCeP
n
scnβ
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Biaxial Column ExampleBiaxial Column Example
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::
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University
solve for solve for ffss::
4.3900046.0 2 += cfs
Biaxial Column ExampleBiaxial Column Example
⎞⎛
•• Recall Recall ffss definition:definition:
⎟⎠⎞
⎜⎝⎛ −
=c
cfs534600
•• Combine both equations:Combine both equations:5346004.3900046.0 2 ⎟
⎠⎞
⎜⎝⎛ −
=+c
cc
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University
03204004.9900046.0 3 =−+ cc•• Solve cubic equation by trial and errorSolve cubic equation by trial and error
c = 323 mmc = 323 mm
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Biaxial Column ExampleBiaxial Column Example
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
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University
132,550)323(311.7)323(639,9 +−=nP
kNPnx 2900=
Biaxial Column ExampleBiaxial Column Example
•• 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
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University
TCCP scn −+= 1
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Biaxial Column ExampleBiaxial Column Example
•• 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:
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University
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
Biaxial Column ExampleBiaxial Column Example
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::
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University
solve for solve for ffss::
12.770073.0 2 += cfs
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Biaxial Column ExampleBiaxial Column Example
⎞⎛
•• Recall Recall ffss definition:definition:
⎟⎠⎞
⎜⎝⎛ −
=c
cfs334600
•• Combine both equations:Combine both equations:33460012.770073.0 2 ⎟
⎠⎞
⎜⎝⎛ −
=+c
cc
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University
020040012.6770073.0 3 =−+ cc•• Solve cubic equation by trial and errorSolve cubic equation by trial and error
c = 295 mmc = 295 mm
Biaxial Column ExampleBiaxial Column Example
•• Calculate Calculate PPnyny2
kNPny 3498=
281,755)295(50.17)295(5.458,14
281,75550.175.458,142
2
+−=
+−=
n
n
P
ccP
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Biaxial Column ExampleBiaxial Column Example
•• Calculate Nominal Biaxial Load Calculate Nominal Biaxial Load PPnn
1111
72181
34981
290011
1111
0
−+=
−+=
n
nnynxn
P
PPPP
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University
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
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University
lx
ly
xu
uy
x
x
yx
stt
lPM
le
llA
=
=ρ
yu
ux
y
y
lPM
le
=
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Design of Biaxial ColumnDesign of Biaxial Column
φPnx
φPny
φPn0
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University
Design of Biaxial ColumnDesign of Biaxial Column
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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