Ch. 10 Design o Short Columns Subjet to Axial Load and Bending Axial Loading and Bending Development o Interation Diagram Column Design Using P-M Interation Diagram Shear in Columns Biaxial Bending Examples
Axial Load and Bending Equivalent
Development o Interation Diagrams
b u =0.003 0.85 M P h Y X As3, d3 As2, d2 As4, d4 s2 s3 s4 a s3 s2 s4 P F s4 C F s3 F s2 P n As1, d1 s1 Setion Strain Stress Equilibrium C F si 0.85 ab A si si where h a h M n C Fsi di 2 2 2 P-M interation urve an be onstruted by plotting (M n,p n ) or various s1 or. s1 si y F s1
Example 10.2: olumn setion 6 #29 bars 356 mm 64 mm 482 mm 64 mm 610 mm
Example 10.2: Strain Proile s 0.002 180 244 0.001475 a 0.85366 311. 1mm 1 0.003 s 302 366 0.002475 y 64 mm 180 mm 302 mm 64 mm 244 mm =366 mm 610 mm
Example 10.2: Fore Equilibrium s 0.001475 200000 295 MPa T s A s s 1935 29510 571kN 3 571 kn 2598 kn 801 kn 92 mm C 0.85 ab 0.85 27.6311.135610 2598 kn C s A 801kN 155 mm s s 1935 41410 3 3 P n M 149 mm 64 mm 241 mm 241 mm 64 mm C n C C h 2 s T s 305 mm 305 mm 2598 801 571 2828kN a F 2 si h d 2 si 311.1 2598305 571 241 801 241 10 2 718.9 kn m 3
Example 10.2: Tension Failure New Strain Proile s > y ( y ) 64 mm 610 mm
Example 10.2: P-M Interation Curve 6595 kn 2828 kn 718.9 kn-m 2245 kn 759 kn-m 403 kn-m 1602 kn
P-M Interation Diagram
P-M Interation Depending on Steel
P-M Curve with Strength Redution Fator (M n, P n ) (M n, P n )
st y st g n st y st g n A A A P A A A P 0.85 0.85 0.85 0.80 P-M Curve or Design
Modiiation o Strength Redution Fator
Column Design Using P-M Interation Diagrams Column design harts are nothing but P-M interation urves arranged in one o various ways The primary purpose o design harts is to make olumn design quik and easy without onstruting the P-M urve or every speii olumn It is important to keep the bar arrangement as lose as possible between the hart and reality
Estimation o Column Size Least dimension o retangular setion 200 mm Minimum diameter o irular setion 300 mm For tied olumns For spiral olumns A g 0. 45 Pu y A g 0. 55 Pu y
Example 10.3 The short 1420 in (356508 mm) tied olumn is to be used to support P D =125 k (556 kn), P L =140 k (623 kn), M D =75 k-t (102 kn-m) and M L =90 k-t (122 kn-m). I =4 ksi (27.6 MPa) and y =60 ksi (414 MPa), selet reinoring bars to be plaed in its end aes only using appropriate ACI olumn interation diagrams. (64 mm) Sine olumn design harts in SI units are not available, US ustomary units are used. (356 mm) (381) M u 1.2M D (508 mm) M u M n (64 mm) P u P e 1.2P n M P 1.6P 15 0.75 20 1.6M 234 0.65 36012 575.4 1.2125 1.6140 374 kips Pu 374 575.4 kips ( t y assumed ) 0.65 n n D L L 1.2 75 1.690 234 k t 360 k t 7.51in
Example 10.3 The short 1420 in (356508 mm) tied olumn is to be used to support P D =125 k (556 kn), P L =140 k (623 kn), M D =75 k-t (102 kn-m) and M L =90 k-t (122 kn-m). I =4 ksi (27.6 MPa) and y =60 ksi (414 MPa), selet reinoring bars to be plaed in its end aes only using appropriate ACI olumn interation diagrams. (356 mm) (64 mm) Pn 575.4 0.513 Ag 414 20 Pn e 575.47.51 0.193 Ag h 414 20 20 0.022 ( 0.7, Graph A3) 0.0185 ( 0.8, Graph A4) 0.0202 ( 0.75) (381) (508 mm) Ast bh 0.0202356508 3653mm 2 Use 6 #29( Ast, provided 3870 mm ) s Sine 0.55 1.0 t 0.002 y (64 mm) 0.65 O.K. 2
0.513 0.022 0.193
0.513 0.0185 0.193
Example 10.4 Design a short olumn or P u =600 kips (2670 kn), M u =125 k-t (170 kn-m), =4 ksi (27.6 MPa), y =60 ksi (414 MPa). Plae bars uniormly around all our aes. Determination o olumn setion, assuming 0.02 A g 0.45 Use 8#32( A u 0.454000 0.02 60000 1616in (256in 16 21.6 23/ 8 1 0.69 16 0.65assumed Pn A M A g n g Pu A 0.04 ( 0.6, 0.036 ( 0.7, ) 0 0.65, Graph A6) 0 0.65, Graph A7) 0.036 0.04 0.04 0.09 0.0364 ( 0.69) 0.1 2 2 A bh 0.0364 256 25.4 6012 mm st P g M u h A h g y st, provided 2 6552 mm 6001000 600 0.901 0.65 4 256 12512 0.141 0.65 4 25616 s s y y 2 ) 256in 2 64 279 mm 406 mm 64 64 mm 279 64 mm 406
0.901 0.04 0.141
0.901 0.036 0.141
Example 10.5 Selet reinoring bars or the short round spiral olumn i =4 ksi (27.6 MPa), y =60 ksi (414 MPa), P u =500 kips (2225 kn), M u =200 k-t (271 kn-m). 202580 mm 2 64 mm 381 mm 64 mm 508 mm Pu A M A 0.0175( 0.75 A u st g g Use 6#29( A Sine 500 0.531 0.75 4 314 200 12 0.127 h 0.75 4 314 20 15 0.75 20 0.0195( 0.7, 0.0155( 0.8, A s g y by interpolation) 0.0175 202580 3545mm st, provided 3870 mm 1.0,, 0.7 t s s y y y 0.45, Graph A11) 0.55, Graph A12) 2 ) 2
0.024 0.531 0.127
0.531 0.02 0.127
Example 10.7 Using the appropriate interation urves, determine the value o P n or the short tied olumn i e x =10 in (254 mm). Assume =4 ksi (26.7 MPa) and y =60 ksi (414 MPa). (76 mm) 3#32(2445 mm 2 ) 3#32(2445 mm 2 ) (305 mm) From Fig.10.15, Pn e A h g 0.24 is read P n 0.24 (356) (508) e (76 mm) A g h e 10 0.5 h 20 23.79 0.0316 12 20 14 0.7 20 0.24 412 20 20 10 460.8 kips
0.0316
Shear in Columns mm., units in positive in ompression and all load, atored axial where 6 1 0 3 1 tension, with axial ii) when aounting or lexure noted above or 8 4 (approximate) 6 1 14 1 ompression, with axial ii) 1 where (aounting or lexure) 0 3 7 120 (approximate) 6 1 load, without axial i) N N d b A N. V M d h N M M d b A N V M d V d b. d b M d V ρ V d b V u w g u u u u m w g u u u w w u u w w
Biaxial Bending
1 P P P P P ni where ni nx ny o 1 P Biaxial Bending Bresler Formula nx 1 P ny 1 P at an eentriity e at an eentriity e o the nominal axial load apaity o the setion when the load is plaed at a given eentriity along both axes the nominal axial load apaity o the setion when the load i splaed x the nominal axial load apaity o the setion when the load is plaed y the nominal axial load apaity o the setion when the load is plaed with a zero eentriity. It is usually taken as 0.85 A g A st y
Example 10.8 Determine the design apaity P ni o the short tied olumn subjeted to biaxial bending. Use =4 ksi (27.6 MPa), y =60 ksi (414 MPa), e x =16 in (406 mm) and e y =8 in (203 mm). (64 mm) Bending about X axis (8 #29) (64 mm) 254 (381 mm) (508) (635) (64 mm) 20 0.8 25 8 0.0213 15 25 e 16 0.64 h 25 Pn e 0.185 (Graph A8) A h P nx g 0.185 415 25 25 16 434 kips
e/h=0.64 0.0213 0.185
Example 10.8 Determine the design apaity P ni o the short tied olumn subjeted to biaxial bending. Use =4 ksi (27.6 MPa), y =60 ksi (414 MPa), e x =16 in (406 mm) and e y =8 in (203 mm). (64 mm) (8 #29) (64 mm) 254 (381 mm) (508) (635) (64 mm) Bending about Y axis 10 0.67 15 8 0.0213 15 25 e 8 0.53 h 15 Pn e 0.161 (Graph A6 and A7) Ag h 0.161 415 2515 Pny 452.3kips 8
e/h=0.53 0.0213 0.159
e/h=0.53 0.0213 0.177
(8 #29) (64 mm) Example 10.8 Determine the design apaity P ni o the short tied olumn subjeted to biaxial bending. Use =4 ksi (27.6 MPa), y =60 ksi (414 MPa), e x =16 in (406 mm) and e y =8 in (203 mm). (64 mm) 254 (381 mm) (508) (635) (64 mm) Axial loading apaity or M P 0.85 o Ag y Ast 0.85 415 25 608 1755 kips Using Bresler's ormula 1 1 1 1 Pni Pnx Pny Po 1 1 1 434 452.3 1755 P 253.5kips ni 0