Sway Column Example. PCA Notes on ACI 318

Transcription

1 Sway Column Example PCA Notes on ACI 318 ASDIP Concrete is available for purchase online at

2 Example 11.2 Slenderness Effects for Columns in a Sway Frame Design columns C1 and C2 in the first story of the 12-story office building shown below. The clear height of the first story is 13 ft-4 in., and is 10 ft-4 in. for all of the other stories. Assume that the lateral load effects on the building are caused by wind, and that the dead loads are the only sustained loads. Other pertinent design data for the building are as follows: Material properties: Concrete: 6000 psi for columns in the bottom two stories (w c 150 pcf) 4000 psi elsewhere (w c 150 pcf) Reinforcement: f y 60 ksi Beams: in. Exterior columns: in. Interior columns: in. Superimposed dead load 30 psf Roof live load 30 psf Floor live load 50 psf Wind loads computed according to ASCE '-0" 24'-0" 24'-0" A B C D E F 24'-0" 24'-0" 24'-0" 24'-0" 24'-0" 7" slab N 4 R '-0" 132'-0" '-0" 2 G 24'-0" 120'-0" 11-31

3 1. Factored axial loads and bending moments for columns C1 and C2 in the first story Since this is a symmetrical frame, the gravity loads will not cause appreciable sidesway. Column C1 Bending moment Load Case Axial Load (ft-kips) (kips) Top Bottom Dead (D) Live (L)* Roof live load (L r ) Wind (W) (N-S) Wind (W) (S-N) No. Load Combination M 1 M 2 M 1ns M 2ns M 1s M 2s D D + 1.6L + 0.5L r D + 0.5L + 1.6L r D + 1.6L r + 0.8W D + 1.6L r - 0.8W D + 0.5L + 0.5L r + 1.6W D + 0.5L + 0.5L r - 1.6W D + 1.6W D - 1.6W *includes live load reduction per ASCE 7 Column C2 Bending moment Load Case Axial Load (ft-kips) (kips) Top Bottom Dead (D) 1, Live (L)* Roof live load (L) r Wind (W) (N-S) Wind (W) (S-N) No. Load Combination M 1 M 2 M 1ns M 2ns M 1s M 2s D 1, D + 1.6L + 0.5L r 1, D + 0.5L + 1.6L r 1, D + 1.6L r + 0.8W 1, D + 1.6L r - 0.8W 1, D + 0.5L + 0.5L r + 1.6W 1, D + 0.5L + 0.5L r - 1.6W 1, D + 1.6W D - 1.6W *includes live load reduction per ASCE 7 2. Determine if the frame at the first story is nonsway or sway The results from an elastic first-order analysis using the section properties prescribed in are as follows: ΣP u total vertical load in the first story corresponding to the lateral loading case for which ΣP u is greatest The total building loads are: D 17,895 kips, L 1991 kips, Lr 270 kips. The maximum ΣP u is from Eq. (9-4): ΣP u ( ,895) + ( ) + ( ) ,605 kips 11-32

4 V us factored story shear in the first story corresponding to the wind loads kips Eq. (9-4), (9-6) Δ o first-order relative deflection between the top and bottom of the first story due to V u 1.6 (0.28 0) 0.45 in. Stability index Q ΣP u Δ o V us l c 22, > ( 15 12) ( 20 / 2) Eq. (10-10) Since Q > 0.05, the frame at the first story level is considered sway Design of column C1 a. Determine if slenderness effects must be considered. Determine k from alignment chart in R I col ,665 in E c 57, For the column below level 2: 4,415 ksi E c I l c 4,415 13, ( ) in.-kips For the column above level 2: E c I l c 4,415 13, in.-kips I beam ,600 in For the beam: E c I l c ψ A ΣE c I/l c ΣE c I/l 57 4,000 5, in.-kips Assume ψ Β 1.0 (column essentially fixed at base) 11-33

5 From the alignment chart (Fig. R (b)), k 1.9. kl u r > Thus, slenderness effects must be considered. b. Determine total moment M 2 (including slenderness effects) and the design load combinations, using the approximate analysis of The following table summarizes magnified moment computations for column C1 for all load combinations, followed by detailed calculations for combinations no. 4 and 5 to illustrate the procedure. No. Load Combination ΣP u Δ o V us Q δ s M 2ns M 2s M 2 (kips) (in.) (kips) (ft-kips) (ft-kips) (ft-kips) 1 1.4D 25, D+1.6L+0.5L r 24, D+0.5L+1.6L r 22, D+1.6L r +0.8W 21, D+1.6L r -0.8W 21, D+0.5L+0.5L r +1.6W 22, D+0.5L+0.5L r -1.6W 22, D+1.6W 16, D-1.6W 16, M 2 M 2ns + δ s M 2s Eq. (10-18) δ s M 2s M 2s I Q M 2s Eq. (10-20) For load combinations no. 4 and 5: U 1.2D + 1.6L r ± 0.8W ΣP u (1.2 17,895) + ( ) ± 0 21,906 kips Δ ο 0.8 (0.28-0) 0.22 in. V us kips l c (15 12) (20/2) 170 in. Q ΣP u Δ o V us l c 21,

6 δ s 1 1 Q For sidesway from north to south (load combination no. 4): δ s M 2s ft-kips M 2 M 2ns + δ s M 2s ft-kips P u kips For sidesway from south to north (load combination no. 5): M 2s ft-kips M 2su ft-kips δ s M 2s 1.14 (-110.4) ft-kips M ft-kips P u kips c. For comparison purposes, recompute δ s M 2s using the magnified moment method outlined in M δ s M 2s 2S M ΣΡ 2s 1 u Eq. (10-21) 0.75ΣΡc The critical load P c is calculated from Eq. (10-13) using k from and EI from Eq. (10-14) or (10-15). Since the reinforcement is not known as of yet, use Eq. (10-15) to determine EI. For each of the 12 exterior columns along column lines 1 and 4 (i.e., the columns with one beam framing into them in the direction of analysis), k was determined in part 3(a) above to be 1.9. EI 0.4E c I 1+β dns ( ) in. 2 -kips Eq. (10-15) β ds P c π 2 EI ( ) 2 kl u π ( ) 2 3,686 kips Eq. (10-13) 11-35

7 For each of the exterior columns A2, A3, F2, and F3, (i.e., the columns with two beams framing into them in the direction of analysis): ψ A 2 70 ψ n From the alignment chart, k P c π ( ) 2 4,345 kips Eq. (10-13) For each of the 8 interior columns: I col ,354 in For the column below level 2: E c I l c 4,415 19,354 ( 15 12) in.-kips For the column above level 2: E c I l c 4,415 19, in.-kips ψ A 2 70 ψ A From the alignment chart, k EI 0.4 4, in.-kips P c π 2 EI ( ) 2 π ( ) kl u ,683 kips Eq. (10-13) 11-36

8 Therefore, ΣP c 12( 3,686) + 4( 4,345) + 8( 5,683) 107,076 kips The following table summarizes magnified moment computations for column C1 using for all load conditions. The table is followed by detailed calculations for combinations no. 4 and 5 to illustrate the procedure. No. Load Combination ΣP u δ s M 2ns M 2s M 2 (kips) (in.) (ft-kips) (ft-kips) (ft-kips) 1 1.4D 25, D + 1.6L + 1.6L r 24, D + 0.5L + 1.6L r 22, D + 1.6L r + 0.8W 21, D + 1.6L r - 0.8W 21, D + 0.5L + 0.5L r + 1.6W 22, D + 0.5L + 0.5L r - 1.6W 22, D + 1.6W 16, D - 1.6W 16, For load combinations No. 4 and 5: U 1.2D + 1.6L r ± 0.8W 1 δ s ΣP 1 u 0.75ΣP c 1 21, , For sidesway from north to south (load combination no. 4): δ s M 2s ft-kips M ft-kips P u kips For sidesway from south to north (load combination no. 5): δ s M 2s 1.38 (-110.4) ft-kips M ft-kips P u kips 11-37

9 A summary of the magnified moments for column C1 for all load combinations is provided in the following table. P u No. Load Combination (kips) δ s M 2 δ s M 2 (ft-kips) (ft-kips) 1 1.4D D + 1.6L + 0.5L r D + 0.5L + 1.6L r D + 1.6L r + 0.8W D + 1.6L r - 0.8W D + 0.5L + 0.5L r + 1.6W D + 0.5L + 0.5L r - 1.6W D + 1.6W D - 1.6W d. Determine required reinforcement. For the in. column, try 8-No. 8 bars. Determine maximum allowable axial compressive force, φp n,max : ( ) + f y A st φp n,max 0.80φ 0.85 f c A g A st Eq. (10-2) ( )[(0.85 6) ( ) + ( )] 1,464.0 kips > maximum P u kips O.K. The following table contains results from a strain compatibility analysis, where compressive strains are taken as positive (see Parts 6 and 7). Use M u M 2 from the approximate method in " 11" 22" 2.375" 22" 11-38

10 No. P u M u c ε t φ φp n φm n (kips) (ft-kips) (in.) (kips) (ft-kips) Therefore, since φm n > M u for all φp n P u, use a in. column with 8-No. 8 bars (rg 1.3%). The same reinforcement is also adequate for the load combinations from the magnified moment method of Design of column C2 a. Determine if slenderness effects must be considered. In part 3(c), k was determined to be 1.82 for the interior columns. Therefore, kl u r x Slenderness effects must be considered > b. Determine total moment M 2 (including slenderness effects) and the design load combinations, using the approximate analysis of The following table summarizes magnified moment computation for column C2 for all load combinations, followed by detailed calculations for combinations no. 4 and 5 to illustrate the procedure. No. Load Combination ΣP u Δ o V us Q δ s M 2ns M 2s M 2 (kips) (in.) (kips) (ft-kips) (ft-kips) (ft-kips) 1 1.4D 25, D+1.6L+0.5L r 24, D+0.5L+1.6L r 22, D+1.6L r +0.8W 21, D+1.6L r -0.8W 21, D+0.5L+0.5L r +1.6W 22, D+0.5L+0.5L r -1.6W 22, D+1.6W 16, D-1.6W 16, M 2 M 2ns + M 2s δ s M 2s M 2s I Q M 2s Eq. (10-19) Eq. (10-20) 11-39

11 For load combinations no. 4 and 5: U 1.2D + 1.6L r ± 0.8W From part 3(b), δ s was determined to be For sidesway from north to south (load combination no. 4): M 2s ft-kips M 2ns 1.2(-1.0) ft-kips δ s M 2s ft-kips M 2 M 2ns + δ s M 2s ft-kips P u 1,332.6 kips For sidesway from south to north (load combination no. 5): δ s M 2s 1.14 (-164) ft-kips M ft-kips P u 1,333.0 kips c. For comparison purposes, recompute using the magnified moment method outlined in Use the values of computed in part 3(c). No. Load Combination ΣP u δ s M 2ns M 2s M 2 (kips) (in.) (ft-kips) (ft-kips) (ft-kips) 1 1.4D 25, D + 1.6L + 0.5L r 24, D + 0.5L + 1.6L r 22, D + 1.6L r + 0.8W 21, D + 1.6L r - 0.8W 21, D + 0.5L + 0.5L r + 1.6W 22, D + 0.5L + 0.5L r - 1.6W 22, D + 1.6W 16, D - 1.6W 16, U 1.2D + 1.6L r ± 0.8W δ s 1.38 from part 3(c) For sidesway from north to south (load combination no. 4): 11-40

12 δ s M 2s ft-kips M ft-kips P u 1,332.6 kips For sidesway from south to north (load combination no. 5): δ s M 2s 1.38 (-164.0) ft-kips M ft-kips P u 1,333.0 kips A summary of the magnified moments for column C2 under all load combinations is provided in the following table. P u No. Load Combination (kips) δ s M 2 δ s M 2 (ft-kips) (ft-kips) 1 1.4D 1, D + 1.6L + 0.5L r 1, D + 0.5L + 1.6L r 1, D + 1.6L r + 0.8W 1, D + 1.6L r - 0.8W 1, D + 0.5L + 0.5L r + 1.6W 1, D + 0.5L + 0.5L r - 1.6W 1, D + 1.6W D - 1.6W d. Determine required reinforcement. For the in. column, try 8-No. 8 bars. Determine maximum allowable axial compressive force, φp n,max : ( ) + f y A st φp n,max 0.80φ 0.85 f c A g A st ( )[(0.85 6) ( ) + ( )] 1,708 kips > maximum P u 1,529.0 kips O.K. Eq. (10-2) The following table contains results from a strain compatibility analysis, where compressive strains are taken as positive (see Parts 6 and 7). Use M u M 2 from the approximate method in

13 21.625" 12" 24" 2.375" 24" No. P u M u c ε t φ φp n φm n (kips) (ft-kips) (in.) (kips) (ft-kips) 1 1, , , , , , , , , , , , , , Therefore, since φm n > M u for all φp n P u, use a in. column with 8-No. 8 bars (ρ g 1.1%)

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