Reinforced Concrete II Hashemite University The Hashem ite University Departm ent of Civil Engineering Lecture 6 – Biaxial Bending of Short Colum ns Dr. Hazim Dwairi Dr Hazim Dwairi Dr. Hazim Dwairi The Hashemite University Reinforced Concrete II Biaxially Loaded Colum n Dr. Hazim Dwairi The Hashemite University Reinforced Concrete II Dr. Hazim Dwairi 1
Reinforced Concrete II Hashemite University Interaction Diagram Uniaxial Bending about y-axis Uniaxial Bending about x-axis Dr. Hazim Dwairi The Hashemite University Reinforced Concrete II Approxim ation of Section Through Intersection Surface Dr. Hazim Dwairi The Hashemite University Reinforced Concrete II Dr. Hazim Dwairi 2
Reinforced Concrete II Hashemite University Notation • P u = factored axial load, positive in compression • e = eccentricity measured parallel to the x axis positive to • e x = eccentricity measured parallel to the x-axis, positive to the right. • e y = eccentricity measured parallel to y-axis, positive upward. • M ux = factored moment about x-axis, positive when causing compression in fibers in the +ve y-direction = P u .e y • M uy = factored moment about y-axis, positive when causing compression in fibers in the +ve x-direction = P u .e x Dr. Hazim Dwairi The Hashemite University Reinforced Concrete II Analysis and Design • Method I: Strain Compatibility Method This is the most nearly theoretically correct method Thi i th t l th ti ll t th d of solving biaxially-loaded-column (see Macgregor example 11-5) • Method II: Equivalent Eccentricity Method An approximate method. Limited to columns that are symmetrical about two axes with a ratio of side t i l b t t ith ti f id lengths l x /l y between 0.5 and 2.0 (see Macgregor example 11-6) Dr. Hazim Dwairi The Hashemite University Reinforced Concrete II Dr. Hazim Dwairi 3
Reinforced Concrete II Hashemite University Strain Com patibility Method Dr. Hazim Dwairi The Hashemite University Reinforced Concrete II Equivalent Eccentricity Method • Replace the biaxial eccentricities e x & e y by an equivalent eccentricity e 0x e e e ≥ = y x if then design column for P and M P e u 0y u 0x l l x y α e l = + y x e e 0 x x l y ≤ > ' ' for P A f 0 . 4 for P A f 0 . 4 u g c u g c ⎛ ⎞ ⎛ ⎞ + + f 276 f 276 P P ⎜ ⎟ ⎜ ⎟ α = + ≥ α = − ≥ y y u u 0 . 5 0.6 1 . 3 0 . 5 ⎜ ⎟ ⎜ ⎟ ' ' A f 696 A f 696 ⎝ ⎠ ⎝ ⎠ g c g c Dr. Hazim Dwairi The Hashemite University Reinforced Concrete II Dr. Hazim Dwairi 4
Reinforced Concrete II Hashemite University Analysis and Design • Method III: 45 o Slice through Interaction Surface ( (see Macgregor page 524) M 524) • Method IV: Bresler Reciprocal Load Method ACI commentary sections 10.3.6 and 10.3.7 give the following equation, originally presented by Bresler for calculating the capacity under biaxial bending. b di 1 1 1 1 ≅ + − φ φ φ P P P P u nx ny n0 • Method V: Bresler Contour Load Method Dr. Hazim Dwairi The Hashemite University Reinforced Concrete II Bresler Reciprocal Load Method 1. Use Reciprocal Failure surface S Failure surface S 2 (1/P n ,e x ,e y ) 2. The ordinate 1/P n on the surface S 2 is approximated by ordinate 1/P on the ordinate 1/P n on the plane S’ 2 (1/P n e x ,e y ) 3. Plane S 2 is defined by points A,B, and C. Dr. Hazim Dwairi The Hashemite University Reinforced Concrete II Dr. Hazim Dwairi 5
Reinforced Concrete II Hashemite University Bresler Reciprocal Load Method P 0 = Axial Load Strength under pure axial compression (corresponds to point C ) compression (corresponds to point C ) M nx = M ny = 0 P 0x = Axial Load Strength under uniaxial eccentricity, e y (corresponds to point B ) M nx = P n e y P 0y = Axial Load Strength under uniaxial eccentricity, e x (corresponds to point A ) M ny = P n e x Dr. Hazim Dwairi The Hashemite University Reinforced Concrete II Bresler Load Contour Method • In this method, the surface S 3 is approximated by a family of curves corresponding to constant by a family of curves corresponding to constant values of P n . These curves may be regarded as “load contours.” where M nx and M ny 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 vectorial equivalent of the nominal uniaxial moment M n . The moment M n0x is the nominal uniaxial moment strength about the x-axis, and M n0y is the nominal uniaxial moment strength about the y-axis. Dr. Hazim Dwairi The Hashemite University Reinforced Concrete II Dr. Hazim Dwairi 6
Reinforced Concrete II Hashemite University Bresler Load Contour Method • The general expression for the contour curves can be approximated as: β α ⎛ ⎞ ⎛ ⎞ M M ⎜ ⎟ ⎜ ⎟ + = ny nx 1 . 0 ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ M M ⎝ ⎠ n 0 x n 0 y • The values of the exponents α and β are a function of the amount, distribution and location function of the amount distribution and location 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 The Hashemite University Reinforced Concrete II Bresler Load Contour Method • Bresler indicated that, typically, α varied from 1.15 to 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 having uniformly distributed reinforcement. A value of α = 1.0 will yield a safe design. ⎛ ⎞ ⎛ ⎞ M M ⎜ ⎜ ⎟ ⎟ ⎜ ⎜ ⎟ ⎟ + + = ny nx 1 1 . 0 0 ⎜ ⎜ ⎟ ⎟ ⎜ ⎜ ⎟ ⎟ ⎝ ⎠ M ⎝ M ⎠ n 0 x n 0 y • Only applicable if: < ' P 0 . 1 f A n c g Dr. Hazim Dwairi The Hashemite University Reinforced Concrete II Dr. Hazim Dwairi 7
Reinforced Concrete II Hashemite University Biaxial Colum n Exam ple 66mm The section of a short tied column is 400 x 600 mm column is 400 x 600 mm and is reinforced with 6 φ 32 234mm bars as shown. Determine 600mm the allowable ultimate load on the section φ P n if its 234mm acts at e x = 200mm. and e y x y = 300mm. Use f c ’ = 35 MPa and f y = 420 MPa. 400mm 66mm Dr. Hazim Dwairi The Hashemite University Reinforced Concrete II Biaxial Colum n Exam ple • Compute P 0 load, pure axial load = × = 2 A 6 804 4824 mm st = × = 2 400 600 240000 A mm g ( ) = − + ' P 0 . 85 f A A A f 0 c g st st y ( ( ) ) = × × − + × P 0 . 85 35 240000 4824 4824 420 0 0 = P 9023 kN 0 0 = = × = P 7218 kN P 0 . 8 9023 7218 kN n 0 n Dr. Hazim Dwairi The Hashemite University Reinforced Concrete II Dr. Hazim Dwairi 8
Reinforced Concrete II Hashemite University Biaxial Colum n Exam ple • Compute P nx , by starting with e y term and assume that compression controls. Check by: assume that compression controls Check by: = < = = e y 300 mm 2 / 3 d 2 / 3 ( 534 ) 356 mm OK! • Compute the nominal load, P nx and assume second compression steel does not contribute Assume = 0 0 Assume = 0.0 = + + − P C C C T n c s 1 s 2 Dr. Hazim Dwairi The Hashemite University Reinforced Concrete II Biaxial Colum n Exam ple • Brake equilibrium equation into its components: = = C C 0 0 . 85 85 ( ( 35 35 )( )( 0 0 . 81 81 c )( )( 400 400 ) ) 9639 9639 c c = − × = C ( 1608 )( 420 0 . 85 35 ) 627715 N s 1 − − 534 c 534 c = = T ( 1608 )( )( 600 ) 964800 ( ) s c c • Compute the moment about tension steel: Compute the moment about tension steel: β ⎛ ⎞ ( ) c = − + − ⎜ ⎟ ' ' 1 . P e C d C d d n c s 1 ⎝ ⎠ 2 ( ) + = − + − P ( 300 234 ) 9639 c 534 0 . 405 c ( 627715 )( 534 66 ) n Dr. Hazim Dwairi The Hashemite University Reinforced Concrete II Dr. Hazim Dwairi 9
Reinforced Concrete II Hashemite University Biaxial Colum n Exam ple • The resulting equation is: 2 + = − 2 P 9 , 639 c 7 . 311 c 550 , 132 n • Recall equilibrium equation: = + − P 9 , 639 c 627715 1608 f n s • Set the two equation equal to one another and solve for f s : solve for f s : 2 + = 0 . 0046 390 . 4 f s c Dr. Hazim Dwairi The Hashemite University Reinforced Concrete II Biaxial Colum n Exam ple • Recall f s definition: − ⎛ ⎛ ⎞ ⎞ 534 c = ⎜ ⎟ f s 600 ⎝ ⎠ c • Combine both equations: − ⎛ ⎞ 534 c + = ⎜ ⎟ 2 0 . 0046 c 390 . 4 600 ⎝ ⎠ c + − = 3 0 . 0046 c 990 . 4 c 320400 0 • Solve cubic equation by trial and error � c = 323 mm Dr. Hazim Dwairi The Hashemite University Reinforced Concrete II Dr. Hazim Dwairi 10
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