Slenderness Effects for Concrete Columns in Sway Frame - Moment Magnification Method (CSA A )
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1 Slenderness Effects for Concrete Columns in Sway Frame - Moment Magnification Method (CSA A )
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3 Slender Concrete Column Design in Sway Frame Buildings Evaluate slenderness effect for columns in a sway frame multistory reinforced concrete building by designing the first story exterior column. The clear height of the first story is 4.75 m, and is 2.75 m for all of the other stories. Lateral load effects on the building are governed by wind forces. Compare the calculated results with the values presented in the Reference and with exact values from spcolumn engineering software program from StructurePoint. Figure 1 Slender Reinforced Concrete Column Cross-Section Version: June
4 Contents 1. Factored Axial Loads and Bending Moments Service loads Load Combinations Factored Loads Slenderness Effects and Sway or Nonsway Frame Designation Determine Slenderness Effects Moment Magnification at Ends of Compression Member Gravity Load Combination #2 (Gravity Loads Only) Lateral Load Combination #5 (Gravity Plus Wind Loads) Moment Magnification along Length of Compression Member Gravity Load Combination #2 (Gravity Loads Only) Load Combination #5 (Gravity Plus Wind Loads) Column Design c, a, and strains in the reinforcement Forces in the concrete and steel P r and M r Column Interaction Diagram - spcolumn Software Summary and Comparison of Design Results Conclusions & Observations Version: June
5 Code Design of Concrete Structures (CSA A ) Explanatory Notes on CSA Standard A Reference Reinforced Concrete Mechanics and Design, First Canadian Edition, 2000, James MacGregor and Michael Bartlett, Prentice Hall, Example 12-3, 4 and 5. Design Data f c = 25 MPa for columns f y = 400 MPa Slab thickness = 180 mm Exterior Columns = 500 mm 500 mm Interior Columns = 500 mm 500 mm Interior Beams = 450 mm 750 mm 9 m Exterior Beams = 450 mm 750 mm 9.5 m Total building loads in the first story from the reference: Table 1 Total building factored loads CSA A Reference No. Load Combination P f, kn D 59, D + 1.5L 77, D + 1.5W 59, D - 1.5W 59, D L W 72, D L W 72, D + 1.5W 40, D - 1.5W 40,460 1
6 1. Factored Axial Loads and Bending Moments 1.1. Service loads Table 2 - Exterior column service loads Load Case Axial Load, Bending Moment, kn.m kn Top Bottom Dead, D 1, Live, L Wind, W Load Combinations Factored Loads CSA A (8.3) CSA A Reference Table 3 - Exterior column factored loads Axial Bending Moment, MTop,ns MBottom,ns MTop,s No. Load Combination Load, kn.m kn.m kn.m kn.m kn Top Bottom D 2, D + 1.5L 2, MBottom,s kn.m D + 1.5W 2, D - 1.5W 2, D L W 2, D L W 2, D + 1.5W 1, D 1.5W 1,
7 2. Slenderness Effects and Sway or Nonsway Frame Designation Columns and stories in structures are considered as nonsway frames if the stability index for the story (Q) does not exceed CSA A ( ) P f is the total factored vertical load in the first story corresponding to the lateral loading case for which P f is greatest (without the wind loads, which would cause compression in some columns and tension in others and thus would cancel out). CSA A ( ) V fs is the total factored shear in the first story corresponding to the wind loads, and Δ o is the first-order relative deflection between the top and bottom of the first story due to V f. CSA A ( ) From Table 1, load combination (1.25D + 1.5L) provides the greatest value of P f. Pf 1.25 D 1.5 L 77,500 kn CSA A (8.3.2) Since there is no lateral load in this load combination, the reference applied an arbitrary lateral load as 1.05W representing (V f) since the deflection calculated for this loading and calculated the resulting story lateral deflection (Δ o). V 1,105 kn (given) f 7.58 mm (given) o Pf o 77, Q V l 1,105 5, 500 f c CSA A (Eq ) Thus, the frame at the first story level is considered sway. 3
8 3. Determine Slenderness Effects c Icolumn mm CSA A (10.14) E 1.5 ' c c 3,300 fc 6,900 2,300 CSA A (Eq. 8-6) E c 2,400 3, , , MPa 2,300 For the column below level 2: Ec Icolumn 24, l 5,500 c N.mm For the column above level 2: 9 Ec Icolumn 24, l 3,500 c N.mm For beams framing into the columns: 9 Eb Ibeam 24, l 9,500 b N.mm Where: E 1.5 ' c c 3,300 fc 6,900 2,300 CSA A (Eq. 8-6) E c 2,400 3, , , MPa 2, I bh beam mm CSA A (10.14) EI lc columns A 2.92 EI 1.45 l beams CSA A (Figure N ) 1.0 (Column considered fixed at the base) CSA A (Figure N ) B Using Figure N from CSA A k = 1.51 as shown in the figure below for the exterior column. 4
9 Figure 2 Effective Length Factor (k) for Exterior Column (Sway Frame) Note: CSA A (Cl ) allows to neglect the slenderness in a non-sway frame. However, there is no such clause in for sway frames. The CSA A committee intended that all columns in sway frames should be designed for slenderness. 4. Moment Magnification at Ends of Compression Member A detailed calculation for load combinations 2 and 5 is shown below to illustrate the slender column moment magnification procedure. Table 4 summarizes the magnified moment computations for the exterior columns Gravity Load Combination #2 (Gravity Loads Only) M M M CSA A (Eq ) 2 2ns s 2s Where: M _ M _ M2_ 0kN.m Top s Bottom s s M M 2 2ns M nd M _2 Top, ns 235 kn.m Top M nd M _2 Bottom, ns 257 kn.m Bottom 5
10 M max M, M M 257 kn.m M M 257 kn.m nd nd nd nd st st 2 _ 2 Top _ 2 Bottom _ 2 Bottom _ 2 2 _1 Bottom _1 M min M, M M 235 kn.m M M 235 kn.m nd nd nd nd st st 1_ 2 Top _ 2 Bottom _ 2 Top _ 2 1_1 Top _1 P f = 2,563 kn 4.2. Lateral Load Combination #5 (Gravity Plus Wind Loads) M M M CSA A (Eq ) 2 2ns s 2s Where: (1) Second-order analysis 1 (2) s moment magnifier Pf 1 mp c 1 (3), if Q < 1/3 1 Q CSA A ( ) There are three options for calculating δ s. CSA A ( ) will be used since it does not require a detailed structural analysis model results to proceed and is also used by the solver engine in spcolumn. P f is the summation of all the factored vertical loads in the first story, and P c is the summation of the critical buckling load for all sway-resisting columns in the first story. 2 EI Pc CSA A (Eq ) 2 kl Where: u 0.2E I E I (a) EI 1 d (b) 0.25 EI c g c g s st CSA A ( ) There are two options for calculating the flexural stiffness of slender concrete columns EI. The first equation provides accurate representation of the reinforcement in the section and will be used in this example and is also used by the solver in spcolumn. Further comparison of the available options is provided in Effective Flexural Stiffness for Critical Buckling Load of Concrete Columns technical note. c Icolumn mm CSA A (10.14) E 1.5 ' c c 3,300 fc 6,900 2,300 CSA A (Eq. 8-6) 6
11 E c 2,400 3, , , MPa 2, β d in sway frames, is the ratio of maximum factored sustained shear within a story to the maximum factored shear in that story associated with the same load combination. The maximum factored sustained shear in this example is equal to zero leading to β d = 0. CSA A (10.0) For exterior columns with one beam framing into them in the direction of analysis (14 columns): With 12 25M reinforcement equally distributed on all sides I st = mm 4 0.2E I E I EI 1 d c g s st CSA A (Eq ) , ( ) 200, 000 ( ) EI N.mm 1 0 k = 1.51 (calculated previously) , P c N 11, kn 13 2 For exterior columns with two beams framing into them in the direction of analysis (4 columns): EI lc columns A 1.42 EI l beams CSA A (Figure N ) 1(Column considered fixed at the base) CSA A (Figure N ) B Using Figure N from CSA A k = 1.38 as shown in the figure below for the exterior columns with two beams framing into them in the directions of analysis. 7
12 Figure 3 Effective Length Factor (k) for Exterior Columns with Two Beams Framing into them in the Direction of Analysis , P c 2 2 For interior columns (10 columns): N 13, kn EI lc columns A 1.42 EI l beams CSA A (Figure N ) 1.0 (Column essentially fixed at base) CSA A (Figure N ) B Using Figure N from CSA A k = 1.38 as shown in the figure below for the interior columns. 8
13 Figure 4 Effective Length Factor (k) Calculations for Interior Columns With 12 25M reinforcement equally distributed on all sides I st = mm 4 0.2E I E I EI 1 d c g s st CSA A (Eq ) , ( ) 200, 000 ( ) EI N.mm , P c N 13, kn P n P n P n P c 1 c1 2 c2 3 c3 1013, , , ,960 kn P c 72,100 kn (Table 1) P f 1 s Pf 1 P m c CSA A (Eq ) 1 s = , ,960 9
14 sm, kn.m Top s M M M CSA A ( ) _2,, kn.m Top nd Top ns s Top s smbottom, s kn.m sm, kn.m Bottom s M M M CSA A ( ) _2,, kn.m Bottom nd Bottom ns s Bottom s M max M, M M kn.m M M nd nd nd nd st st 2 _ 2 Top _ 2 Bottom _ 2 Bottom _ 2 2 _1 Bottom _1 M max M, M M kn.m M M kn.m nd nd nd nd st st 2 _ 2 Top _ 2 Bottom _ 2 Bottom _ 2 2 _1 Bottom _1 M min M, M M kn.m M M kn.m nd nd nd nd st st 1_ 2 Top _ 2 Bottom _ 2 Top _ 2 1_1 Top _1 P f = 2,400 kn A summary of the moment magnification factors and magnified moments for the exterior column for all load combinations using both equation options CSA A23.3 (Eq ) and (Eq ) to calculate δ s is provided in the table below for illustration and comparison purposes. Note: The designation of M 1 and M 2 is made based on the second-order (magnified) moments and not based on the first-order (unmagnified) moments. No. Table 4 - Factored Axial loads and Magnified Moments at the Ends of Exterior Column Axial Using CSA Eq Using CSA Eq Load Load Combination M kn δ 1, M 2, M s δ 1, M 2, kn.m kn.m s kn.m kn.m D 2,019 * * * D + 1.5L 2, D + 1.5W 2,019 * * * D - 1.5W 2,019 * * * D L W 2, D L W 2,400 * * * D + 1.5W 1,373 * * * D - 1.5W 1,373 * * * * Not covered by the reference 10
15 5. Moment Magnification along Length of Compression Member In sway frames, if an individual compression member has: lu 35 CSA A (Eq ) r P f A ' f / ( c g ) It shall be designed for the factored axial load, P f and moment, M c, computed using Clause (Nonsway frame procedure), in which M 1 and M 2 are computed in accordance with Clause CSA A ( ) CmM2 Mc M Pf 1 P Where: m c 2 CSA A ( ) C m M CSA A ( ) M 2 M 2 = the second-order factored moment (magnified sway moment) And, the member resistance factor would be m 0.75 CSA A ( ) 2 EI Pc CSA A (Eq ) 2 kl Where: u 0.2E I E I (a) EI 1 d (b) 0.25 EI c g c g s st CSA A ( ) There are two options for calculating the effective flexural stiffness of slender concrete columns EI. The first equation provides accurate representation of the reinforcement in the section and will be used in this example and is also used by the solver in spcolumn. Further comparison of the available options is provided in Effective Flexural Stiffness for Critical Buckling Load of Concrete Columns technical note Gravity Load Combination #2 (Gravity Loads Only) 4 I g 500 /12 r mm CSA A ( ) 2 A 500 l u g r P ,564 1, ' f / ( fc Ag ) CSA A (Eq ) 11
16 Since < 54.63, calculating the moments along the column length is not required. Check minimum moment: CSA A ( ) CSA A does not require to design columns in sway frames for a minimum moment. However, the reference decided conservatively to design the column for the larger of computed moments and the minimum value of C mm 2. CmM 2 P ( h) min f CM m 2 2,563 ( ) / 1, kn.m min 5.2. Load Combination #5 (Gravity Plus Wind Loads) P ,400 1, ' f / ( fc Ag ) CSA A (Eq ) Since < 56.48, calculating the moments along the column length is not required. Check minimum moment: CSA A ( ) CmM 2 P ( h) min f CM m 2 2, 400 ( ) / 1, kn.m min M c1 and M c2 will be considered separately to ensure proper comparison of resulting magnified moments against negative and positive moment capacities of unsymmetrical sections as can be seen in the following figure. Figure 5 Column Interaction Diagram for Unsymmetrical Section A summary of the moment magnification factors and magnified moments for the exterior column for all load combinations using both equation options CSA A23.3 (Eq ) and (Eq ) to calculate δ s is provided in the table below for illustration and comparison purposes. 12
17 No. Table 5 - Factored Axial loads and Magnified Moments along Exterior Column Length Load Combination Axial Load, kn Using CSA Eq Using CSA Eq δ M c1, kn.m M c2, kn.m δ M c1, kn.m M c2, kn.m D 2,019 * * * D + 1.5L 2, D + 1.5W 2,019 * * * D - 1.5W 2,019 * * * D L W 2, D L W 2,400 * * * D + 1.5W 1,373 * * * D 1.5W 1,373 * * * * Not covered by the reference For column design, CSA A23.3 requires that δ s to be computed from Clause using P f and P c corresponding to 1.25 dead load and 1.5 live load shall be positive and shall not exceed 2.5. β d shall be taken as the ratio of the factored sustained axial dead load to the total axial load. For values of δ s above the limit, the frame would be very susceptible to variations in EI, foundation rotations and the like. If this value is exceeded, the frame must be stiffened to reduce δ s. CSA A ( & N ) Total factored sustained axial load d CSA A ( ) Total factored axial load 59,500 d , π EI Pc CSA A (Eq ) 2 (kl ) Where: u 0.2E I E I EI 1 c g s st d , ( ) 200, 000 ( ) EI N.mm For exterior columns with two beams framing into them in the direction of analysis: CSA A (Eq ) P c π (1.51 4,750) , kn For interior columns and exterior columns with two beams framing into them in the direction of analysis: P c π (1.38 4,750) , kn (10 4) 7, , ,142 kn P c 13
18 Where the member resistance factor is m 0.75 CSA A ( ) 1 s Pf 1 P m c CSA A (Eq ) 1 s = 2.13 < , ,142 Thus, the frame is stable. 6. Column Design Based on the factored axial loads and magnified moments considering slenderness effects, the capacity of the assumed column section (500 mm 500 mm with 12 25M bars distributed all sides equal) will be checked and confirmed to finalize the design. A column interaction diagram will be generated using strain compatibility analysis, the detailed procedure to develop column interaction diagram can be found in Interaction Diagram - Tied Reinforced Concrete Column example. The factored axial load resistance P r for all load combinations will be set equals to P f, then the factored moment resistance M r associated to P r will be compared with the magnified applied moment M f. The design check for load combination #5 is shown below for illustration. The rest of the checks for the other load combinations are shown in the following Table. Figure 6 Strains, Forces, and Moment Arms (Load Combination 5) The following procedure is used to determine the nominal moment capacity by setting the factored axial load resistance, P r, equal to the factored axial load, P f and iterating on the location of the neutral axis c, a, and strains in the reinforcement Try c 335 mm Where c is the distance from extreme compression fiber to the neutral axis. CSA A (10.0) a 1 c mm CSA A (10.1.7a) Where: 14
19 CSA A (Eq. 10-2) ' f c cu CSA A (10.1.3) f y 400 y E 200, 000 s s ( d1 c) ( ) (Tension) < y c 335 tension reinforcement has not yielded 0.60 c 0.85 s CSA A (8.4.2) CSA A (8.4.3) ' s4 ( c d4) (335 54) (Compression) > y c 335 ' s3 c d3 ( ) (Compression) < y c 335 ' s2 c d2 ( ) (Compression) < y c Forces in the concrete and steel C f ab CSA A (10.1.7a) ' rc 1 c c ,852.6 kn Where: CSA A (Eq. 10-1) f ' f c E , MPa s s s T f A kn rs s s s1 Since > compression reinforcement has yielded ' s4 y ' fs4 f 400 MPa y Since < compression reinforcement has not yielded ' s3 y ' ' fs3 s3 Es , MPa Since < compression reinforcement has not yielded ' s2 y 15
20 f 2 2 E , MPa ' ' s s s The area of the reinforcement in third and fourth layers has been included in the area (ab) used to compute C rc. As a result, it is necessary to subtract α 1 f c from f s before computing C rs: ' ' ' f f A C / 1, kn rs4 s s4 1 c c s4 ' ' ' f f A C / 1, kn rs3 s s3 1 c c s3 ' ' f A C / 1, kn rs2 s s2 s P r and M r P C C 2 C 3 C 4 T 1, , kn r rc rs rs rs rs P 2, kn 2,400 kn = P r The assumed value of c = 335 mm is correct. f h a h h h h M r Crc Crs 4 d4 Crs3 d3 Crs 2 d2 Trs d M r M r , ,541 N.m = kn.m M kn.m f Table 6 Exterior Column Axial and Moment Capacities No. P f, kn M u = M 2(2nd), kn.m c, mm ε t = ε s P r, kn M r, kn.m 1 2, , , , , , , , , , , , , , , , Since M r > M f for all P r = P f, use mm column with 12 25M bars. 16
21 7. Column Interaction Diagram - spcolumn Software spcolumn program performs the analysis of the reinforced concrete section conforming to the provisions of the Strength Design Method and Unified Design Provisions with all conditions of strength satisfying the applicable conditions of equilibrium and strain compatibility and includes slenderness effects using moment magnification method for sway and nonsway frames. For this column section, we ran in investigation mode with control points using the CSA A In lieu of using program shortcuts, spsection (Figure 7) was used to place the reinforcement and define the cover to illustrate handling of irregular shapes and unusual bar arrangement. Figure 7 spcolumn Model Editor (spsection) 17
22 Figure 8 spcolumn Model Input Wizard Windows 18
23 Figure 5 Column Section Interaction Diagram about X-Axis Design Check for Load Combination 5 (spcolumn) 19
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30 8. Summary and Comparison of Design Results Analysis and design results from the hand calculations above are compared for the one load combination used in the reference (Example 12-3,4 and 5) and exact values obtained from spcolumn model. Table 7 Parameters for Moment Magnification at Column Ends Q k EI, N.mm 2 P c, kn M 1(2nd), kn.m M 2(2nd), kn.m Hand * , Reference , spcolumn * , * From nomographs (CSA A23.3 charts) Conservatively estimated not using exact formulae without major impact on the final results in this special case Exact formulated answer In this table, a detailed comparison for all considered load combinations are presented for comparison. Table 8 - Factored Axial loads and Magnified Moments at Column Ends No. P f, kn δ s M 1(2nd), kn.m M 2(2nd), kn.m Hand spcolumn Hand spcolumn Hand spcolumn Hand spcolumn 1 2,019 2,019.0 N/A N/A ,563 2,563.3 N/A N/A ,019 2, ,019 2, ,400 2, ,400 2, ,373 1, ,373 1,
31 Table 9 - Design Parameters Comparison No. c, mm ε t = ε s P f, kn M r, kn.m Hand spcolumn Hand spcolumn Hand spcolumn Hand spcolumn , , , , , , , , , , , , , , , , All the results of the hand calculations illustrated above are in precise agreement with the automated exact results obtained from the spcolumn program. 27
32 9. Conclusions & Observations The analysis of the reinforced concrete section performed by spcolumn conforms to the provisions of the Strength Design Method and Unified Design Provisions with all conditions of strength satisfying the applicable conditions of equilibrium and strain compatibility and includes slenderness effects using moment magnification method for sway and nonsway frames. CSA A23-3 provides multiple options for calculating values of EI and δ s leading to variability in the determination of the adequacy of a column section. Engineers must exercise judgment in selecting suitable options to match their design condition as is the case in the reference where the author conservatively made assumptions to simplify and speed the calculation effort. The spcolumn program utilizes the exact methods whenever possible and allows user to override the calculated values with direct input based on their engineering judgment wherever it is permissible. It was concluded in the CSA A that the probability of stability failure increases rapidly when the stability index Q exceeds 0.2 and a more rigid structure may be required to provide stability. CSA A ( ) If a frame undergoes appreciable lateral deflections under gravity loads, serious consideration should be given to rearranging the frame to make it more symmetrical because with time, creep will amplify these deflections leading to both serviceability and strength problems. One of these limitations is to limit the second-order lateral deflections to first-order lateral deflections to 2.5 (the ratio should not exceed 2.5) for loads applied to the structure with 1.25 dead load and 1.5 live load plus a lateral load applied to each story equal to multiplied by factored gravity load in that story. CSA A ( & N ) The limitation on δ s is intended to prevent instability under gravity loads alone. For values of δ s above the limit, the frame would be very susceptible to variations in EI, foundation rotations and the like. If δ s exceeds 2.5 the frame must be stiffened to reduce δ s. CSA A (N ) Exploring the impact of other code permissible equation options provides the engineer added flexibility in decision making regarding design. In some cases resolving the stability concern may be viable through a frame analysis providing values for V f and Δ o to calculate magnification factor δ s. Creating a complete model with detailed lateral loads and load combinations to account for second order effects may not be warranted for all cases of slender column design nor is it disadvantageous to have a higher margin of safety when it comes to column slenderness and frame stability considerations. 28
ε t increases from the compressioncontrolled Figure 9.15: Adjusted interaction diagram
CHAPTER NINE COLUMNS 4 b. The modified axial strength in compression is reduced to account for accidental eccentricity. The magnitude of axial force evaluated in step (a) is multiplied by 0.80 in case
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