Clase 29 - Columnas Biaxiales

Biaxial Bending and AxialBiaxial Bending and Axial Load LoadUnaxial bending about y-axisBiaxial Bending and AxialBiaxial Bending and Axial Load LoadThe biaxial bending momentsMx=P*eyMy=P*exApproximate AnalysisApproximate Analysis Methods MethodsUse Reciprocal Failure surface S2(1/Pn,ex,ey)The ordinate 1/Pnon the surface S2is approximated by ordinate 1/Pnon the plane S2(1/Pnex,ey)Plane S2is defined by points A,B, and C.Approximate AnalysisApproximate Analysis Methods MethodsP0=Axial Load Strength under pure axial compression(corresponds to point C ) Mnx=Mny=0P0x=Axial Load Strength under uniaxial eccentricity, ey(corresponds to point B ) Mnx=PneyP0y=Axial Load Strength under uniaxial eccentricity, ex(corresponds to point A ) Mny=PnexApproximate AnalysisApproximate Analysis Methods MethodsDesign: PuMuy, Mux Pu, Puex, PueyApproximate AnalysisApproximate Analysis Methods MethodsPn=Nominal axial load strength at eccentricities, ex& eyLimited to cases when0 0y 0xn0 0y 0x n n1 1 111 1 1 1 1P P PPP P P P P +~ + ='~g c n1 . 0 A f P >Biaxial Bending in ShortBiaxial Bending in Short Columns Columns1) Calculate P02) Calculate P0y( Pnfor e =ex, ey=0 )3)Calculate P0x( Pnfor ex=0, e =ey)4) Calculate Pn(from Breslers Formula )Analysis Procedure: Reciprocal Load Method Breslers Formula:Steps:0 0y 0x n1 1 1 1P P P P + ~Biaxial Bending in ShortBiaxial Bending in Short Columns Columnswhere, f =0. 65 n u P P sBiaxial Column Example Biaxial Column ExampleThe section of a short tied column is 16 x 24 in.and is reinforced with 8 #10 bars as shown.Determine the allowable ultimate load on the section fPnif its acts at ex=8 in. and ey=12 in.Use fc=5 ksi and fy=60 ksi.Biaxial Column Example Biaxial Column ExampleCompute the P0load, compression with no moments( )( )( ) ( )( )( )( )( )( )2 2st0 c g st st y22n0 08 1.27 in 10.16 in0.850.85 5 ksi 24.0 in. 24.0 in. 10.16 in10.16 in 60 ksi2198.4 k0.8 2198.4 k 1758.7 kAP f A A AfP rP= == += +== = =Biaxial Column Example Biaxial Column ExampleCompute Pnx, by starting with eyterm and assume that compression controls.Check byCompute the nominal load, Pnx and assume second compression steel does not contributeassume small( )y2 212 in.21.5 in. 14.33 in.3 3e d = s = =n c s1 s2P C C C T = + + Biaxial Column Example Biaxial Column ExampleThe components of the equilibrium equation are:Use similar triangles to find the stress in the steel, fs( )( )( )( ) ( )( )( )( ) ( )c2s12ss0.85 5 ksi 16 in. 0.8 54.4 3.81 in 60 ksi 0.85 5 ksi 212.4 kips3.81 in21.5 in.1 29000 ksi 0.003 1 87 ksiC c cCT fdfc c= == ==| | | |= = ||\ . \ .Biaxial Column Example Biaxial Column ExampleCompute the moment about the tension steel:whereThe resulting equation is:( )1n c s12cP e C d C d d| |' '= + |\ .( ) ( )( )n2n9.5 in. 12 in. 21.5 in.21.5 in. 54.4 21.5 in. 0.4212.4 k 21.5 in. 2.5 in.54.4 1.01 187.7eP c cP c c' = + == + = +Biaxial Column Example Biaxial Column ExampleCombine the two equations and solve for Pnusing an iterative solutionSet the two equation equal to one another and sole for fsand the definition:n s2n54.4 212.4 3.8154.4 1.01 187.7P c fP c c= + = +2ss0.265 6.48321.5 in.87 1cf cf= +| |= |\ .Biaxial Column Example Biaxial Column ExampleCombine the two equations and solve for c using an iterative techniqueYou are solving a cubic equation221.5 in.87 1 0.265 6.483cc| | = + |\ .c (i n.) fs (ksi ) RHS15 37.7 66.1281910 100.05 32.9919413 56.8846251.2831513.3 53.6391 53.3747113.315 53.4806653.48054Biaxial Column Example Biaxial Column ExampleCheck the assumption that Cs2is close to zero( ) ( )( ) ( )s22s212 in. 12 in.1 87 ksi 1 87 ksi13.315 in.8.59 ksi2.54 in 8.59 ksi 0.85 5 ksi11.0 kipsfcC| | | |= = ||\ . \ .== =This value is small relative to the othersBiaxial Column Example Biaxial Column ExampleThisCs2=11 kips relatively small verses the overall load, which isSo Pnx=733.0 kips( ) ( )n s54.4 212.4 3.8154.4 13.315 in. 212.4 k 3.81 53.48 ksi733.0 kP c f = + = + =Biaxial Column Example Biaxial Column ExampleStart with exterm and assume that compression controls.Compute the nominal load, Pny and assume second compression steel does not contributeassume small( )x2 25.5 in.13.5 in. 9 in.3 3e d = s = =n c s1 s2P C C C T = + + Biaxial Column Example Biaxial Column ExampleThe components of the equilibrium equation are:( ) ( )( )( ) ( )( )( )( ) ( )c2s12ss0.85 5 ksi 24 in. 0.8 81.6 3.81 in 60 ksi 0.85 5 ksi 212.4 kips3.81 in13.5 in.1 29000 ksi 0.003 1 87 ksiC c cCT fdfc c= == ==| | | |= = ||\ . \ .Biaxial Column Example Biaxial Column ExampleCompute the moment about the tension steel:whereThe resulting equation is:( )1n c s12cP e C d C d d| |' '= + |\ .( ) ( )( )n2n5.5 in. 8 in. 13.5 in.13.5 in. 81.6 13.5 in. 0.4 212.4 k 13.5 in. 2.5 in.81.6 2.42 173.07eP c cP c c' = + == + = +Biaxial Column Example Biaxial Column ExampleCombine the two equations and solve for Pnusing an iterative solutionSet the two equation equal to one another and sole for fsand the definition:n s2n81.6 212.4 3.8181.6 2.42 173.07P c fP c c= + = +2ss0.634 10.32413.5 in.87 1cf cf= +| |= |\ .Biaxial Column Example Biaxial Column ExampleCombine the two equations and solve for cusing an iterative techniqueYou are solving a cubic equation213.5 in.87 1 0.634 10.324cc| | = + |\ .c (i n.) fs (ksi ) RHS10 30.45 73.763718 59.8125 50.925318.5 51.1764756.159118.3 54.5060254.027538.31735 54.2108454.21043Biaxial Column Example Biaxial Column ExampleCheck the assumption that Cs2is close to zero( )( ) ( )s22s28 in.1 87 ksi8.317 in.3.32 ksi2.54 in 3.32 ksi 0.85 5 ksi2.10 kipsfC| |= |\ .== = This value is negative so it does not contributeBiaxial Column Example Biaxial Column ExampleThis Cs2=- 2.1 kips relatively small verses the overall load, which isSo Pnx=684.6 kips( ) ( )n s81.6 212.4 3.8181.6 8.317 in. 212.4 k 3.81 54.21 ksi684.6 kP c f = + = + =Biaxial Column Example Biaxial Column ExampleCompute the nominal load( )n nx ny n0n u n1 1 1 11 1 1733.0 k 684.6 k 1758.7 k443.2 k 0.65 443.2 k 288.1 kP P P PP P P = + = + = = = =