Column Interaction Diagram Lecture21

Lecture 21 Columns July 25, 2003 CVEN 444 Lecture Goals Columns Interaction Diagrams Using Interaction Diagrams Example: Axial Load vs. Moment Interaction Diagram Consider an square column (20 in x 20 in.) with 8 #10 ( = 0.0254) and fc = 4 ksi and fy = 60 ksi.Draw the interaction diagram. Example: Axial Load vs. Moment Interaction Diagram Given 8 # 10 (1.27 in2) and fc = 4 ksi and fy = 60 ksi ( )( )2 2st22g2st2g8 1.27 in 10.16 in20 in. 400 in10.16 in0.0254400 inAAAA= == == = =Example: Axial Load vs. Moment Interaction Diagram Given 8 # 10 (1.27 in2) and fc = 4 ksi and fy = 60 ksi ( )( )( )( )( )0 c g st y st2 220.850.85 4 ksi 400 in 10.16 in 60 ksi 10.16 in1935 kP f A A f A = += +=( )n 00.8 1935 k 1548 kP rP == =[ Point 1 ] Example: Axial Load vs. Moment Interaction Diagram Determine where the balance point, cb. Example: Axial Load vs. Moment Interaction Diagram Determine where the balance point, cb.Using similar triangles, where d = 20 in. 2.5 in. = 17.5 in., one can find cb bbb17.5 in.0.003 0.003 0.002070.00317.5 in.0.003 0.0020710.36 in.ccc=+| | =|+\ .=Example: Axial Load vs. Moment Interaction Diagram Determine the strain of the steel ( )( )bs1 cubbs2 cub2.5 in. 10.36 in. 2.5 in.0.00310.36 in.0.0022810 in. 10.36 in. 10 in.0.00310.36 in.0.000104ccccc cc c| | | |= = | |\ .\ .=| | | |= = | |\ .\ .=Example: Axial Load vs. Moment Interaction Diagram Determine the stress in the steel ( )( )s1 s s1s2 s s129000 ksi 0.0022866 ksi60 ksicompression29000 ksi 0.0001043.02 ksicompressionf Ef Ecc= == = ==Example: Axial Load vs. Moment Interaction Diagram Compute the forces in the column ( )( ) ( ) ( )( )( )( ) ( )( )( ) ( )c c 1s1 s1 s1 c22s20.850.85 4 ksi 20 in. 0.85 10.36 in.598.8 k0.853 1.27 in 60 ksi 0.85 4 ksi215.6 k 2 1.27 in 3.02 ksi 0.85 4 ksi0.97 kneglectC f b cC A f fC| ==== = == = :Example: Axial Load vs. Moment Interaction Diagram Compute the forces in the column ( )( )2s s sn c s1 s2 s3 1.27 in 60 ksi228.6 k 599.8 k 215.6 k 228.6 k 585.8 kT A fP C C C T= === + + = + =Example: Axial Load vs. Moment Interaction Diagram Compute the moment about the center ( )c s1 1 s 32 2 2 20.85 10.85 in.20 in.599.8 k2 220 in.215.6 k 2.5 in.220 in.228.6 k 17.5 in.26682.2 k-in556.9 k-fth a h hM C C d T d| | | | | |= + + |||\ . \ . \ .| |= |\ .| |+ |\ .| |+ |\ .= Example: Axial Load vs. Moment Interaction Diagram A single point from interaction diagram, (585.6 k, 556.9 k-ft).The eccentricity of the point is defined as 6682.2 k-in11.41 in.585.8 kMeP= = =[ Point 2 ] Example: Axial Load vs. Moment Interaction Diagram Now select a series of additional points by selecting values of c.Select c = 17.5 in. Determine the strain of the steel. (c is at the location of the tension steel) ( )( )s1 cus1s2 cus22.5 in. 17.5 in. 2.5 in.0.00317.5 in.0.00257 74.5 ksi 60 ksi (compression)10 in. 17.5 in. 10 in.0.00317.5 in.0.00129 37.3 ksi (compression)ccfccfc cc c | | | |= = ||\ . \ .= = | | | |= = ||\ . \ .= =Example: Axial Load vs. Moment Interaction Diagram Compute the forces in the column ( )( )( )( )( )( )( ) ( )( )( ) ( )c c 12s1 s1 s1 c2s20.85 0.85 4 ksi 20 in. 0.85 17.5 in.1012 k0.85 3 1.27 in 60 ksi 0.85 4 ksi216 k 2 1.27 in 37.3 ksi 0.85 4 ksi86 k C f b cC A f fC| = === = == =Example: Axial Load vs. Moment Interaction Diagram Compute the forces in the column ( )( )2s s sn3 1.27 in 0 ksi0 k 1012 k 216 k 86 k 1314 kT A fP= === + +=Example: Axial Load vs. Moment Interaction Diagram Compute the moment about the center ( )c s1 12 2 20.85 17.5 in.20 in.1012 k2 220 in.216 k 2.5 in.24213 k-in351.1 k-fth a hM C C d| | | |= + ||\ . \ .| |= |\ .| |+ |\ .= Example: Axial Load vs. Moment Interaction Diagram A single point from interaction diagram, (1314 k, 351.1 k-ft).The eccentricity of the point is defined as 4213 k-in3.2 in.1314 kMeP= = =[ Point 3 ] Example: Axial Load vs. Moment Interaction Diagram Select c = 6 in.

Determine the strain of the steel, c =6 in. ( )( )s1 cus1s2 cus2s3 cu2.5 in. 6 in. 2.5 in.0.0036 in.0.00175 50.75 ksi(compression)10 in. 6 in. 10 in.0.0036 in.0.002 58 ksi (tension)17.5 in. 6 in.ccfccfccc cc cc c | | | |= = ||\ . \ .= = | | | |= = ||\ . \ .= = | |= = |\ .( )s317.5 in.0.0036 in.0.00575 60 ksi (tension) f| | |\ .= =Example: Axial Load vs. Moment Interaction Diagram Compute the forces in the column ( )( )( )( )( )( )( ) ( )( )( )( )( )c c 1s1 s1 s1 c22s20.850.85 4 ksi 20 in. 0.85 6 in.346.8 k0.853 1.27 in 50.75 ksi 0.85 4 ksi180.4 kC2 1.27 in 58 ksi147.3 kTC f b cC A f fC| ==== = ===Example: Axial Load vs. Moment Interaction Diagram Compute the forces in the column ( )( )2s s sn3 1.27 in 60 ksi228.6 k 346.8 k 180.4 k 147.3 k228.6 k 151.3 kT A fP= === + =Example: Axial Load vs. Moment Interaction Diagram Compute the moment about the center ( )( )( )c s1 1 s 32 2 2 20.85 6 in.346.8 k 10 in.2180.4 k 10 in. 2.5 in.228.6 k 17.5 in. 10 in.5651 k-in470.9 k-fth a h hM C C d T d| | | | | |= + + |||\ . \ . \ .| |= |\ .+ + = Example: Axial Load Vs. Moment Interaction Diagram A single point from interaction diagram, (151 k, 471 k-ft).The eccentricity of the point is defined as 5651.2 k-in37.35 in.151.3 kMeP= = =[ Point 4 ] Example: Axial Load vs. Moment Interaction Diagram Select point of straight tension.

The maximum tension in the column is ( )( )2n s y8 1.27 in 60 ksi610 kP A f = ==[ Point 5 ] Example: Axial Load vs. Moment Interaction Diagram Pointc (in) PnMne 1- 1548 k0 0 2201515 k253 k-ft 2 in 317.51314 k351 k-ft 3.2 in 412.5 841 k500 k-ft 7.13 in 510.36 585 k556 k-ft11.42 in 6 8.0393 k531 k-ft16.20 in 7 6.0151 k471 k-ft37.35 in 8~4.50 k395 k-ftinfinity 90-610 k0 k-ftExample: Axial Load vs. Moment Interaction Diagram Column Analysis-1000-50005001000150020000 100 200 300 400 500 600M (k-ft)P (k)Use a series of c values to obtain the Pn verses Mn. Example: Axial Load vs. Moment Interaction Diagram Column Analysis-800-600-400-2000200400600800100012000 100 200 300 400 500|Mn (k-ft)fPn (k)Max. compression Max. tension Cb Location of the linearly varying |. Behavior under Combined Bending and Axial Loads Interaction Diagram Between Axial Load and Moment ( Failure Envelope ) Concrete crushes before steel yields Steel yields before concrete crushes Note: Any combination of P and M outside the envelope will cause failure. Design for Combined Bending and Axial Load (short column) Column Types Tied Column - Bars in 2 faces (furthest from axis of bending. - Most efficient when e/h > 0.2 - rectangular shape increases efficiency 3) Design for Combined Bending and Axial Load (short column) Spices Typically longitudinal bars spliced just above each floor. (non-seismic) Type of lap splice depends on state of stress (ACI 12.17) Design for Combined Bending and Axial Load (short column) Spices All bars in compressionUse compression lap splice (ACI 12.16) 15 . 12 ACIsplice lap tensionB Classspliced) bars 1/2 (B Class) splice bars 2 / 1 (lap A tensionClass5 . 0face on tension 5 . 0 0y sy s)`>s sf ff fDesign for Combined Bending and Axial Load (short column) Column Shear ( ) 4 - 11 ACI20001 2w cgucd b fANV|||.|

\|+ =Recall ( Axial Compression ) 5 . 0 Ifc u > V V | Ties must satisfy ACI 11 and ACI Sec. 7.10.5 Design for Combined Bending and Axial Load (short column) Additional Note on Reinforcement Ratio ( ) 10.9.1 ACI 0.08 0.01Recall s s For cross-section larger than required for loading: Min. reinforcement may be computed for reduced effective area, Ag, ( 1/2 Ag (total) ) Provided strength from reduced area and resulting Ast must be adequate for loading. >(ACI 10.8.4 ) -Non-dimensional Interaction Diagrams See Figures B-12 to B-26 or ACI Common 340 Design Handbook Vol 2 Columns (ACI 340.2R-91) n nc g c gversusP Mf A f A hn nn nc g c geversusRP PKf A f A h= =or Non-dimensional Interaction Diagrams Design using Non-dimensional Interaction diagrams Calculate factored loads (Pu , Mu ) and e for relevant load combinations Select potentially governing case(s) Use estimate h to calculate h, e/h for governing case(s) 1.) 2.) 3.) Design using Non-dimensional Interaction diagrams Use appropriate chart (App. A) target g (for each governing case) Select 4.) 5.) nc g Pf Au cgnc g P fAPf A|=| | | |\ .Read Calculate required h b A b * h&g = Design using Non-dimensional Interaction diagrams If dimensions are significantly different from estimated (step 3), recalculate ( e / h ) and redo steps 4 & 5. Revise Ag if necessary. Select steel 6.) 7.) g st A A = Design using non-dimensional interaction diagrams Using actual dimensions & bar sizes to check all load combinations ( use charts or exact: interaction diagram). Design lateral reinforcement.8.) 9.) Example: Column design usingInteraction Diagrams Determine the tension and compression reinforcement for a16 in x 24 in. rectangular tied column to support Pu= 840 k and Mu = 420 k-ft.Use fc = 4 ksi and fy = 60 ksi.Using the interaction diagram. Example: Interaction Diagrams Compute the initial componentsun840 kips1292 k0.65PP|= = =unu12 in.420 k-ft fte 6.0 in.840 kMP| | |\ .= = =Example: Interaction Diagrams Compute the initial components24 in. 5.0 in. 19.0 in. h = =19.0 in.0.7924 in.= =Example: Interaction Diagrams Compute the coefficients of the column( )( )( )nng c1292 k16 in. 24 in. 4 ksi0.84PKAf= ==( )( )( )( )( )( )nng c1292 k 6 in.e16 in. 24 in. 4 ksi 24 in.0.21PRAf h= ==Example: Interaction Diagrams Using an interaction diagram, B-13( ) ( )n nc y, 0.21, 0.840.74 ksi60 ksi0.042R Kf f=== ==Example: Interaction Diagrams Using an interaction diagram, B-14( ) ( )n nc y, 0.21, 0.840.94 ksi60 ksi0.034R Kf f=== ==Example: Interaction Diagrams Using linear interpolation to find the of the column( )( )( )( )( )0.9 0.70.70.70.9 0.70.034 0.0420.042 0.79 0.70.9 0.70.0384 = + = =Example: Interaction Diagrams Determine the amount of steel required Select the steel for the column, using #11 bars( )( )( )st g20.0384 16 in. 24 in.14.75 inA A = ==2st2b14.75 in9.45 bars 10 bars1.56 inAA= = Example: Interaction Diagrams The areas of the steel: The loading on the column2st2 2s1 t15.6 in7.8 in ,7.8 inAA A== =Example: Interaction Diagrams The compression components are ( )( ) ( )( )( )( )2s1 s1 y cc c0.85 7.8 in 60 ksi 0.85 4 ksi441.5 k0.85 0.85 4 ksi 16 in. 0.8546.24C A f fC f ba cc= = == ==Example: Interaction Diagrams The tensioncomponent is ( )2s1 s ss s cu7.8 in21.5 in.29000 ksi 0.00321.5 in.87 ksiT Af fd c cf Ec cccc= = | | | |= = ||\ . \ .| |= |\ .Example: Interaction Diagrams Take the moment about the tension steel ( ) ( )n s1 ce2aP C d d C d| |' '= + |\ .e 6 in.9.5 in. 15.5 in.' = +=Example: Interaction Diagrams The first equation related to Pn ( ) ( )n22n15.5 in. 441.5 k 21.5 in. 2.5 in.0.8546.24 21.5 in.28388.5 k-in. 994.2 19.65541.2 k 64.14 1.27Pccc cP c c= | |+ |\ .= + = + Example: Interaction Diagrams The second equation comes from the equilibrium equation and substitute in for Pn

n s1 c2s2s2s541.2 k 64.14 1.27 441.5 k 46.24 7.87.8 1.27 17.9 99.70.1628 2.282 12.782P C C Tc c c ff c cf c c= + + = + = = Example: Interaction Diagrams Substitute the relationship of c for the stress in the steel. The problem is now a cubic solution cfs RHS 15in.37.7-10.38 19in.11.45 2.6419.5 in.8.924.6320.0 in.6.526.7019.98 in.6.626.62 221.5 in.87 0.1628 2.282 12.782cc cc| | = |\ .Example: Interaction Diagrams Compute Pn

Compute Mn about the center ( ) ( )2n541.2 k 64.14 19.98 in. 1.27 19.98 in.1313.7 k1292 kP= + = >n s1 c2 2 2 2h h a hM C d C T d| | | | | |'= + + |||\ . \ . \ .Example: Interaction Diagrams Compute Mn about the center ( )( )( )( )( )( )n2441.5 k 12 in. 2.5 in.0.85 19.98 in.46.24 19.98 in. 12 in.27.8 in 6.62 ksi 21.5 in. 12 in.4194.25 k-in. 3241.4 k-in. 490.54 k-in.7926.2 k-in. 660.5 k-ft.M= | |+ |\ .+ = + += Example: Interaction Diagrams Check that Mn is greater than the required Mu

Check the Pn is greater than the required Pu ( )n0.65 660.5 k-ft.429.33 k-ft.420 k-ft.M | == >( )n0.65 1313.7 k853.9 k 840 kP | == >Example: Interaction Diagrams Determine the tie spacing using #4 bars ( )( )bstirrup16spacing smallest 48smallest dimension16 1.41 in. 22.56 in.48 0.5 in. 24 in.16 in.dd== = =Use 16 in.