Biaxial Bending Short Columns

Reinforced Concrete II

Hashemite University

T h e H a s h e m i t e U n i ve ve r s i t y D e p a r t m e n t o f C i vi vi l E n g i n e e r i n g

L e ct c t u r e 6 – B iiaa x iiaa l B e n d i n g o f S h o r t Co C o lu lu m n s .

Dr. D r. Hazim Hazim Dwairi Dwairi

The The Hashemite Hashemite University University


B i a x iiaa l lly y L o a d e d Co C o lu lu m n

Dr. Hazim Dwairi

Dr Hazim Dwairi

The Hashemite University


1



I n t e r a c tio n D ia g r a m

Uniaxial Bending about y-axis

Uniaxial Bending about x-axis

Dr. Hazim Dwairi



Ap p r o x im a t i o n o f S e ct io n Th r o u gh I n t e r s e ct io n S u r fa c e

Dr. Hazim Dwairi

Dr Hazim Dwairi



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Notation • Pu = factored axial load, positive in compression • x , the right. • ey = eccentricity measured parallel to yy--axis, axis, positive positive upward. • Mux = factored moment about x x--axis, positive positive when when causing compression in fibers in the +ve + ve y-direction = P u.ey • Muy = factored moment about yy -axis, positive when causing compression in fibers in the +ve + ve x-direction = P u.ex

Dr. Hazim Dwairi



An a l ys i s a n d D e s i gn • Method I: Strain Compatibility Method s s e mos near y eore ca y correc me o of solving biaxially biaxially--loadedloaded-column (see Macgregor example 1111-5) • Method II: Equivalent Eccentricity Method An approximate method. Limited to columns that are symmetrical about two axes with a ratio of side lengths lx / /lly between 0.5 and 2.0 (see Macgregor example 1111-6) Dr. Hazim Dwairi

Dr Hazim Dwairi



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S tr a i n Co m p a t ib i lit y M e t h o d

Dr. Hazim Dwairi



E q u iva l e n t E cc e n t r ic it y M e t h o d • Replace the biaxial eccentricities e x & ey by an equivalent eccentricity e 0x if

e x l x

≥

y

l y

then design column for Pu and M 0y e0 x

for Pu Ag f c'

⎛ ⎜ ⎝

α = ⎜ 0.5 + Dr. Hazim Dwairi

Dr Hazim Dwairi

= e x +

α e y l x l y

≤ 0.4

for Pu Ag f c'

⎞ f y + 276 ⎟ ≥ 0.6 ' Ag f c ⎠⎟ 696 Pu

= Pu e0x

⎛ ⎜ ⎝

α = ⎜1.3 −


> 0.4

⎞ f y + 276 ⎟ ≥ 0.5 ' Ag f c ⎠⎟ 696 Pu


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An a l ys i s a n d D e s i gn • Method III: 45 o Slice through Interaction Surface see acgregor page Bresler Reciprocal Load Method • Method IV: Bresler ACI commentary sections 10.3.6 and 10.3.7 give the following equation, originally presented by Bresler for calculating the capacity under under biaxial bending.. bending 1 1 1 1

≅

Pu

φ Pnx

+

φ Pny

−

φ Pn0

V: Bresler Bresler Contour Contour Load Method • Method V: Dr. Hazim Dwairi



B r e s llee r R e cip r o ccaa l Lo a d M e t h o d Br 1. Use Reciprocal 2

(1/ Pn,ex,ey) (1/P 2. The ordinate 1/P 1/ Pn on the surface S2 is approximated by n

plane S’2 (1/ (1/P Pn ex,ey) 3. Plane S2 is defined by points A,B, and C. Dr. Hazim Dwairi

Dr Hazim Dwairi



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B r e s llee r R e cip r o ccaa l Lo a d M e t h o d Br P0 = Axial Load Strength under pure axial Mnx = Mny = 0 P0x = Axial Load Strength under uniaxial eccentricity, ey (corresponds to point B ) Mnx = Pn ey P0y = Axial Load Strength under uniaxial eccentricity, ex (corresponds to point A ) Mny = Pn ex Dr. Hazim Dwairi



B r e s llee r Lo a d Co n t o u r M e t h o d Br • In this method, the surface S 3 is approximated values of Pn. These curves may be regarded as “load contours.” where Mnx and Mny are the nominal biaxial moment strengths in the direction of the xand y-axes, respectively. equivalent of the nominal uniaxial moment M n . The moment M n0x 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

Dr Hazim Dwairi



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B r e s llee r Lo a d Co n t o u r M e t h o d Br • The general expression for the contour curves can be approximated as: β

α M ⎞ ⎛ M nx ⎞ ⎛ ⎜⎜ ⎟⎟ + ⎜ ny ⎟ = 1.0 ⎝ M n 0 x ⎠ ⎜⎝ M n 0 y ⎠⎟

• The values of the exponents α and β are a , of reinforcement, the dimensions of the column, and the strength and elastic properties of the steel and concrete. Bresler indicates that it is reasonably accurate to assume that α = β Dr. Hazim Dwairi



B r e s llee r Lo a d Co n t o u r M e t h o d Br • Bresler indicated that, typically, α varied from 1.15 . , . accurate for most square and rectangular sections having uniformly distributed reinforcement. A value of α = 1.0 will yield a safe design.

⎞ ⎛ M nx ⎞ ⎛ ⎜ M ny ⎟ = ⎝ M n 0 x ⎠ ⎝ M n 0 y ⎠

.

• Only applicable if: Pn Dr. Hazim Dwairi

Dr Hazim Dwairi

< 0.1 f c' Ag The Hashemite University


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Bia x ia l Co lu m n E xa m p l e 66mm

The section of a short tied and is reinforced with 6 φ32 bars as shown. Determine the allowable ultimate load on the section φPn if its acts at e = 200mm. and e = 300mm. Use fc’ = 35 MPa and fy = 420 MPa MPa..

234mm m m 0 0 6

234mm

400mm 66mm

Dr. Hazim Dwairi



Bia x ia l Co lu m n E xa m p l e • Compute P0 load, pure axial load Ast = 6 × 804 = 4824mm Ag

2

= 400 × 600 = 240000mm 2

= 0.85 f c' ( Ag − Ast ) + Ast f y P = 0.85 × 35 × 240000 − 4824 + 4824 × 420 P0 = 9023kN Pn 0 = 0.8 × 9023 = 7218kN Pn 0 = 7218kN P0

Dr. Hazim Dwairi

Dr Hazim Dwairi



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Bia x ia l Co lu m n E xa m p l e • Compute Pnx, by starting with ey term and . e y

= 300mm < 2 / 3d = 2 / 3(534) = 356mm

OK!

• Compute the nominal load, Pnx and assume second compression steel does not contribute .

Pn

= C c + C s1 + C s 2 − T

Dr. Hazim Dwairi



Bia x ia l Co lu m n E xa m p l e • Brake equilibrium equation into its components: c = . c c = .

= (1608)(420 − 0.85 × 35) = 627715 N 534 − c 534 − c = (1608)( )(600) = 964800( )

C s1 T s

c c • Com ute the moment about tension steel:

β c ⎞ = C c ⎛ ⎜ d − 1 ⎟ + C s1 (d − d ' ) 2 ⎠ ⎝ Pn (300 + 234) = 9639c(534 − 0.405c ) + (627715)(534 − 66) Pn .e '

Dr. Hazim Dwairi

Dr Hazim Dwairi



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Bia x ia l Co lu m n E xa m p l e • The resulting equation is:

= 9,639c − 7.311c 2 + 550,132

Pn

• Recall equilibrium equation: Pn

= 9,639c + 627715 − 1608 f s

• Set the two equation equal to one another and solve for f :

f s

= 0.0046c 2 + 390.4

Dr. Hazim Dwairi



Bia x ia l Co lu m n E xa m p l e • Recall fs definition: f s

534 − c

= 600⎜ ⎝

c

⎟ ⎠

• Combine both equations:

⎛ 534 − c ⎞ ⎟ c ⎝ ⎠ 0.0046c 3 + 990.4c − 320400 = 0 0.0046c 2 + 390.4 = 600⎜

• Solve cubic equation by trial and error  c = 323 mm Dr. Hazim Dwairi

Dr Hazim Dwairi



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Bia x ia l Co lu m n E xa m p l e • Check the assumption that f s2 = 0.0 323 − 300 = 600⎜ ⎟ = 42.72 MPa 323 ⎝ ⎠ C s 2 = 68.7 kN TOO SMALL f s 2

• Calculate Pnx n

=

−

,

Pnx Dr. Hazim Dwairi

.

2

,

= 2900kN The Hashemite University


Bia x ia l Co lu m n E xa m p l e • Compute Pny, by starting with ex term and .

e x

= 200mm < 2 / 3d = 2 / 3(334) = 223mm

OK!

• Compute the nominal load, Pny

Pn Dr. Hazim Dwairi

Dr Hazim Dwairi

= C c + C s1 − T The Hashemite University


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Bia x ia l Co lu m n E xa m p l e • Brake equilibrium equation into its components: = . c . c c = .

= (2412)(420 − 0.85 × 35) = 941283 N 334 − c 334 − c = (2412)( )(600) = 1447200( )

C s1 T s

c c • Com ute the moment about tension steel:

β c ⎞ = C c ⎛ ⎜ d − 1 ⎟ + C s1 (d − d ' ) 2 ⎠ ⎝ Pn (200 + 134) = 14458.5c (334 − 0.405c ) + (941283)(334 − 66) Pn .e '

Dr. Hazim Dwairi



Bia x ia l Co lu m n E xa m p l e • The resulting equation is: Pn

= 14,458.5c − 17.50c + 755,281

• Recall equilibrium equation: Pn

= 14,458.5c + 941,283 − 2,412 f s

• Set the two equation equal to one another and solve for f :

f s Dr. Hazim Dwairi

Dr Hazim Dwairi

= 0.0073c 2 + 77.12 The Hashemite University


12



Bia x ia l Co lu m n E xa m p l e • Recall fs definition: f s

334 − c

= 600⎜ ⎝

c

⎟ ⎠

• Combine both equations:

⎛ 334 − c ⎞ ⎟ ⎝ c ⎠ 0.0073c 3 + 677.12c − 200400 = 0 0.0073c 2 + 77.12 = 600⎜

• Solve cubic equation by trial and error  c = 295 mm Dr. Hazim Dwairi



Bia x ia l Co lu m n E xa m p l e • Calculate Pny

= 14,458.5c − 17.50c + 755,281 Pn = 14,458.5( 295) − 17.50( 295) 2 + 755,281 Pn

Pny

Dr. Hazim Dwairi

Dr Hazim Dwairi

= 3498kN



13



Bia x ia l Co lu m n E xa m p l e • Calculate Nominal Biaxial Load Pn Pn

1 Pn

= =

1 2900

Pn Pu

+

Pnx

Pny

+

−

Pn 0

1 3498

−

1 7218

= 2032kN

= φ Pn = (0.65)(2032) = 1321kN

Dr. Hazim Dwairi



D e s i gn o f Bi a x ia l Co l u m n 1) Select trial section Ag ( trial )

≥

0.40( f

u

' c

+ ρ t f y )

; use ρ t = 0.0015 γlx

2) Compute 3) Compute φPnx, φPny, φPn0 ρ t = e x l x Dr. Hazim Dwairi

Dr Hazim Dwairi

=

As

ly

l x l y

M uy

e y

Pu l x

l y

=

M ux Pu l y The Hashemite University

lx Reinforced Concrete II

14



D e s i gn o f Bi a x ia l Co l u m n

Pn0

Pnx Pn

Dr. Hazim Dwairi



D e s i gn o f Bi a x ia l Co l u m n 4) Solve for φPn

1

φ Pn

=

1

φ Pnx

+

1

φ Pny

−

1

φ Pn 0

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

Dr. Hazim Dwairi

Dr Hazim Dwairi



15

Biaxial Bending Short Columns

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