Two-Way Flat Slab (Concrete Floor with Drop Panels) System Analysis and Design (CSA A )

Transcription

1 Two-Way Flat Slab (Concrete Floor with Drop Panels) System Analysis and Design (CSA A3.3-14)

2 Two-Way Flat Slab (Concrete Floor with Drop Panels) System Analysis and Design (CSA A3.3-14) Design the concrete loor slab system shown below or an intermediate loor considering partition weight = 1 kn/m, and unactored live load = 3 kn/m. The lateral loads are independently resisted by shear walls. The use o lat plate system will be checked. I the use o lat plate is not adequate, the use o lat slab system with drop panels will be investigated. Flat slab concrete loor system is similar to the lat plate system. The only exception is that the lat slab uses drop panels (thickened portions around the columns) to increase the nominal shear strength o the concrete at the critical section around the columns. The Elastic Frame Method (EFM) shown in CSA A is used in this example. The hand solution rom EFM is also used or a detailed comparison with the model results o spslab engineering sotware program. Figure 1 - Two-Way Flat Concrete Floor System Version: Oct

3 Contents 1. Preliminary member sizing Flexural Analysis and Design Elastic Frame Method (EFM) Limitations or use o elastic rame method Frame members o elastic rame Elastic rame analysis Factored moments used or Design Distribution o design moments Flexural reinorcement requirements Factored moments in columns Design o Columns by spcolumn Determination o actored loads Moment Interaction Diagram Shear Strength One-Way (Beam action) Shear Strength At distance d v rom the supporting column At the ace o the drop panel Two-Way (Punching) Shear Strength Around the columns aces Around drop panels Serviceability Requirements (Delection Check) Immediate (Instantaneous) Delections Time-Dependent (Long-Term) Delections (Δ lt) spslab Sotware Program Model Solution Summary and Comparison o Design Results Conclusions & Observations One-Way Shear Distribution to Slab Strips Two-Way Concrete Slab Analysis Methods Version: Oct

4 Code Design o Concrete Structures (CSA A3.3-14) and Explanatory Notes on CSA Group standard A Design o Concrete Structures Reerence CAC Concrete Design Handbook, 4 th Edition, Cement Association o Canada Design Data Story Height = 4 m (provided by architectural drawings) Superimposed Dead Load, SDL =1 kn/m Live Load, LL = 3 kn/m c = 35 MPa (or slab) c = 4 MPa (or columns) y = 400 MPa Solution 1. Preliminary member sizing For slabs without Drop Panels a. Slab minimum thickness Delection CSA A (13..3) In lieu o detailed calculation or delections, CSA A3.3 Code gives minimum slab thickness or two-way construction without interior beams in Clause For this lat plate slab systems the minimum slab thicknesses per CSA A are: Exterior Panels: h s,min ln 0.6 y /1, mm CSA A (13..3) 30 But not less than 10 mm. CSA A (13..1) Exterior Panels: h s,min ln 0.6 y /1, mm CSA A (13..3) 30 But not less than 10 mm. CSA A (13..1) Where l n = length o clear span in the long direction = 9, = 8,500 mm Try 300 mm. slab or all panels (sel-weight = 4 kn/m m = 7. kn/m ) b. Slab shear strength one way shear At a preliminary check level, the use o average eective depth would be suicient. However, ater determining the inal depth o the slab, the exact eective depth will be used in lexural, shear and delection calculations. Evaluate the average eective depth (Figure ): 1

5 db 16 dl hs cclear db mm db 16 dt hs cclear mm d avg dl dt mm Where: c clear = 0 mm CSA A (Annex A. Table 17) d b = 16 mm or 15M steel bar Figure - Two-Way Flat Concrete Floor System Factored dead load, 1.5 (7. 1) 10.5 kn/m wd CSA A (Annex C. Table C.1 a) Factored live load, w kn/m Total actored load, w kn/m l Check the adequacy o slab thickness or beam action (one-way shear) CSA A (13.3.6) At an interior column: The critical section or one-way shear is extending in a plane across the entire width and located at a distance, d v rom the ace o support or concentrated load (see Figure 3). CSA A ( ) Consider a 1 m wide strip Tributary are or one-way shear is A Tributary V w ATributary kn ' Vc c c bwd v 9, , 000 1, m CSA A (Eq.11.6)

6 Where: 1 For normal weight concrete CSA A (8.6.5) 0.1 For slabs with overall thickness not greater than 350 mm CSA A ( ) d Max (0.9 d, 0.75 h) =37.6 mm CSA A (3.) v ' 5.9 MPa 8 MPa CSA A (11.3.4) c 37.6 Vc , kn 1,000 V Slab thickness o 300 mm is adequate or one-way shear. c. Slab shear strength two-way shear Check the adequacy o slab thickness or punching shear (two-way shear) at an interior column (Figure 4): Tributary area or two-way shear is A Tributary (9 9) 80.4 m 1,000 V w A ,186.1 kn Tributary v V 1, MPa bd 3, o v v 0.38 CSA A ( ) ' r c c c v MPa > v c Slab thickness o 300 mm is not adequate or two-way shear. It is good to mention that the actored shear (V ) used in the preliminary check does not include the eect o the unbalanced moment at supports. Including this eect will lead to an increase o V value as shown later in section 4.. 3

7 Figure 3 Critical Section or One-Way Shear Figure 4 Critical Section or Two-Way Shear In this case, our options could be used: 1) to increase the slab thickness, ) to increase columns cross sectional dimensions or cut the spacing between columns (reducing span lengths), however, this option is assumed to be not permissible in this example due to architectural limitations, 3) to use headed shear reinorcement, or 4) to use drop panels. In this example, the latter option will be used to achieve better understanding or the design o two-way slab with drop panels oten called lat slab. 4

8 Check the drop panel dimensional limitations as ollows: 1) The additional thickness o the drop panel below the soit o the slab (Δ h) shall not be taken larger than h s. CSA A (13..4) Since the slab thickness (h s) is 60 mm (see page 7), the thickness o the drop panel should be at less than 60 mm. Drop panel dimensions are also controlled by ormwork considerations. The ollowing Figure shows the standard lumber dimensions that are used when orming drop panels. Using other depths will unnecessarily increase ormwork costs. The Δ h dimension will be taken as the lumber dimension plus the thickness o one sheet o plywood (19 mm). For nominal lumber size: hdp = = 57 mm or h dp = = 108 mm Try h dp = 108 mm < 60 mm The total thickness including the slab and the drop panel (h) = h s + h dp = = 368 mm Nominal Lumber Size, mm Actual Lumber Size, mm Plyorm Thickness, mm h dp, mm x x Figure 5 Drop Panel Formwork Details 5

9 Figure 6 Drop Panels Dimensions 6

10 For Flat Slab (with Drop Panels) For slabs with changes in thickness and subjected to bending in two directions, it is necessary to check shear at multiple sections as deined in the CSA A The critical sections or two-way action shall be located with respect to: 1) Perimeter o the concentrated load or reaction area. CSA A ( ) ) Changes in slab thickness, such as edges o drop panels. CSA A ( ) a. Slab minimum thickness Delection In lieu o detailed calculation or delections, CSA A3.3 Code gives minimum slab thickness or two-way construction with drop panel in Clause For this lat plate slab systems the minimum slab thicknesses per CSA A are: The value o x d/l n is not known at this point. The upper limit is 1/4. A reasonable preliminary estimate is 1/6. ln 0.6 y /1,000 x d Exterior Panels: hs,min mm 30 l n CSA A (13..4) But not less than 10 mm. CSA A (13..1) Interior Panels: h s,min ln 0.6 y /1,000 xd 47.3 mm CSA A (13..4) 30 l n But not less than 10 mm. CSA A (13..1) Where l n = length o clear span in the long direction = 9, = 8,500 mm Try 60 mm slab or all panels Sel-weight or slab section without drop panel = 4 kn/m m = 6.4 kn/m Sel-weight or slab section with drop panel = 4 kn/m m = 8.83 kn/m b. Slab shear strength one way shear For critical section at distance d rom the edge o the column (slab section with drop panel): Evaluate the average eective depth: db 16 dl hs cclear db mm db 16 dt hs cclear mm d avg dl dt mm 7

11 Where: c clear = 0 mm CSA A (Annex A. Table 17) d b = 16 mm or 15M steel bar Factored dead load 1.5 (8.83 1) 1.9 kn/m w d CSA A (Annex C. Table C.1 a) Factored live load kn/m w l Total actored load kn/m w Check the adequacy o slab thickness or beam action (one-way shear) rom the edge o the interior column CSA A (13.3.6) Consider a 1 m wide strip. The critical section or one-way shear is located at a distance d v, rom the edge o the column (see Figure 7) Tributary area or one-way shear is A Tributary V w ATributary kn 9, , 000 1, m ' Vc c c bwd v CSA A (Eq. 11.6) Where 1 or normal weight concrete This slab contains no transverse reinorcement and it is assumed the speciied nominal maximum size o coarse aggregate is not less than 0 mm, β shall be taken as: CSA A ( ) (1, 000 ) (1, ) d v d Max (0.9 d, 0.75 h) =99 mm CSA A (3.) v 99 Vc , kn 1,000 Slab thickness o 368 mm is adequate or one-way shear or the irst critical section (rom the edge o the column). V u For critical section at the edge o the drop panel (slab section without drop panel): Evaluate the average eective depth: db 16 dl hs cclear db mm 8

12 db 16 dt hs cclear mm d avg dl dt mm Where: c clear = 0 mm CSA A (Annex A. Table 17) d b = 16 mm or 15M steel bar Factored dead load 1.5 (6.4 1) 9.05 kn/m w d CSA A (Annex C. Table C.1 a) Factored live load kn/m w l Total actored load kn/m w Check the adequacy o slab thickness or beam action (one-way shear) rom the edge o the interior drop panel. CSA A (13.3.6) Consider a 1 m wide strip. The critical section or one-way shear is located at a distance, d v rom the ace o support (see Figure 7) Tributary area or one-way shear is A Tributary V w ATributary kn ' Vc c c bwd v 9,000 3, , 000 1, m CSA A (Eq. 11.6) Where 1 or normal weight concrete 0.1 or slabs with overall thickness not greater than 350 mm CSA A ( ) d Max (0.9 d, 0.75 h) =0 mm CSA A (3.) v ' 5.9 MPa 8 MPa CSA A (11.3.4) c 0 Vc , kn V 1,000 Slab thickness o 60 mm is adequate or one-way shear or the second critical section (rom the edge o the drop panel). 9

13 Critical Section rom the Edge o the Column Figure 7 Critical Sections or One-Way Shear Critical Section rom the Edge o the Drop Panel c. Slab shear strength two-way shear For critical section at distance d/ rom the edge o the column (slab section with drop panel): Check the adequacy o slab thickness or punching shear (two-way shear) at an interior column (Figure 8): Tributary area or two-way shear is A (9 9) m Tributary V w A , kn Tributary v V 1, MPa bd 3, 3833 o v v 0.38 CSA A ( ) ' r c c c v MPa > v c Tributary area or two-way shear is A (9 9) m Tributary V w A kn Tributary v V MPa bd 1,896 4 o v v 0.38 CSA A ( ) ' r c c c v MPa > v c 10

14 Slab thickness o 60 mm is adequate or two-way shear or the irst critical section (rom the edge o the column). For critical section at the edge o the drop panel (slab section without drop panel): Check the adequacy o slab thickness or punching shear (two-way shear) at an interior drop panel (Figure 8): Tributary area or two-way shear is A (9 9) m Tributary V w A kn Tributary v V MPa bd 1,896 4 o v v 0.38 CSA A ( ) ' r c c c v MPa > v c Slab thickness o 60 mm is adequate or two-way shear or the second critical section (rom the edge o the drop panel). Critical Section rom the Edge o the Column Figure 8 Critical Sections or Two-Way Shear Critical Section rom the Edge o the Drop Panel 11

15 d. Column dimensions - axial load Check the adequacy o column dimensions or axial load: Tributary area or interior column or live load, superimposed dead load, and sel-weight o the slab is A m Tributary Tributary area or interior column or sel-weight o additional slab thickness due to the presence o the drop panel is A 33 9 m Tributary Assuming ive story building P nw A , kn Tributary Assume 500 mm square column with 1 30M vertical bars with design axial strength, P r,max o P,max ( h) P 0.80P (For tied column along ull length) CSA A (Eq. 10.9) r ro ro P A A A A F A A CSA A (Eq ) ' ( ) A ro 1 c c g st t p s y st y t pr p P ( ) (1 700) 0 7, 39.4 kn ro Pr,max ( ) 7, , 39.4 P r,max 5,791.5 kn P Where: ' c CSA A (Eq. 10.1) Column dimensions o 500 mm 500 mm are adequate or axial load.. Flexural Analysis and Design CSA A3.3 states that a slab system shall be designed by any procedure satisying equilibrium and geometric compatibility, provided that strength and serviceability criteria are satisied. Distinction o two-systems rom oneway systems is given by CSA A (3..) CSA A permits the use o Direct Design Method (DDM) and Elastic Frame Method (EFM) or the gravity load analysis o orthogonal rames and is applicable to lat plates, lat slabs, and slabs with beams. The ollowing sections outline the solution per EFM and spslab sotware. For the solution per DDM, check the lat plate example..1. Elastic Frame Method (EFM) EFM (as known as Equivalent Frame Method in the ACI 318) is the most comprehensive and detailed procedure provided by the CSA A3.3 or the analysis and design o two-way slab systems where these systems may, or purposes o analysis, be considered a series o plane rames acting longitudinally and transversely through the 1

16 building. Each rame shall be composed o equivalent line members intersecting at member centerlines, shall ollow a column line, and shall include the portion o slab bounded laterally by the centerline o the panel on each side. CSA A ( ) Probably the most requently used method to determine design moments in regular two-way slab systems is to consider the slab as a series o two-dimensional rames that are analyzed elastically. When using this analogy, it is essential that stiness properties o the elements o the rame be selected to properly represent the behavior o the three-dimensional slab system Limitations or use o elastic rame method In EFM, live load shall be arranged in accordance with which requires slab systems to be analyzed and designed or the most demanding set o orces established by investigating the eects o live load placed in various critical patterns. CSA A (13.8.4) Complete analysis must include representative interior and exterior equivalent elastic rames in both the longitudinal and transverse directions o the loor. CSA A ( ) Panels shall be rectangular, with a ratio o longer to shorter panel dimensions, measured center-to-center o supports, not to exceed. CSA A (13..) For slab systems with beams between supports, the relative eective stiness o beams in the two directions is not less than 0. or greater than. CSA A (13..) Column osets are not greater than 0% o the span (in the direction o oset) rom either axis between centerlines o successive columns. CSA A (13..) The reinorcement is placed in an orthogonal grid. CSA A (13..).1.. Frame members o elastic rame Determine moment distribution actors and ixed-end moments or the equivalent rame members. The moment distribution procedure will be used to analyze the equivalent rame. Stiness actors k, carry over actors COF, and ixed-end moment actors FEM or the slab-beams and column members are determined using the design aids tables at Appendix 0A o PCA Notes on ACI These calculations are shown below. a. Flexural stiness o slab-beams at both ends, K sb. c 500 c , ,000 9,000 N1 N 1 For c c, stiness actors, k k 5.55 PCA Notes on ACI (Table A&3) F1 N1 NF FN Ecs Is Ecs Is Thus, K k 5.55 PCA Notes on ACI (Table A&3) sb NF

17 K sb , , N.m 3 3 sh 9, 000 (60) Where, Is mm E 1.5 ' c cs (3, 300 c 6, 900) E cs,300,447 (3, , 900) 9, 00 MPa, CSA A3.3-14(8.6..) Carry-over actor COF = PCA Notes on ACI (Table A&3) Fixed-end moment, FEM = m w l PCA Notes on ACI (Table A&3) n i1 NFi i 1 Uniorm load ixed end moment coeicient, m NF1 = Fixed end moment coeicient or (b-a) = 0. when a = 0, m NF = Fixed end moment coeicient or (b-a) = 0. when a = 0.8, m NF3 = The coeicient o ixed end moment or (b-a) = 0. when a = 0.8 is taken as in order to be conservative and take the upper bound and be conservative. b. Flexural stiness o column members at both ends, K c. Reerring to Table A7, Appendix 0A, For the Bottom Column (Below): t a 60 / mm, t 60 / 130 mm b t t a b H 4 m 4,000mm, H c 4,000mm 60 mm 108 mm 3,63 mm H 4, H 3,63 c Thus, k 5.31 and C 0.55 by interpolation. c, bottom AB AB 5.31E cc I K c PCA Notes on ACI (Table A7) c K c, bottom ,000 1, , N.m 4 4 c (500) Where Ic mm

18 E 1.5 ' c cc (3, 300 c 6, 900),300 CSA A3.3-14(8.6..) E cc,447 (3, , 900) 31, 047 MPa,300 l c = 4 m = 4,000 mm 1.5 For the Top Column (Above): t t b a H 4, H 3,63 c Thus, k 4.88 and C 0.6 by interpolation. BA BA 4.88E cc I K c PCA Notes on ACI (Table A7) c K c, top c ,000 1, , N.m c. Torsional stiness o torsional members, K t. K t 9E C cs c t 1 t 3 CSA A3.3-14( ) K t 9 9 9, ,000 (1 500 / 9,000) 1, N.m Where 3 x x y C y 3 CSA A3.3-14( ) C mm c 500 mm, 9 m = 9,000 mm Equivalent column stiness K ec. K ec K ec K K c c Kt K t ( )( 1.53) 10 [( ) (1.53)] K ec N.m

19 Where K t is or two torsional members one on each side o the column, and K c is or the upper and lower columns at the slab-beam joint o an intermediate loor. Figure 9 Torsional Member Figure 10 Column and Edge o Slab d. Slab-beam joint distribution actors, DF. At exterior joint,.36 DF 0.57 ( ) At interior joint,.36 DF 0.36 ( ) COF or slab-beam =0.576 Figure 11 Slab and Column Stiness 16

20 .1.3. Elastic rame analysis Determine negative and positive moments or the slab-beams using the moment distribution method. Since the unactored live load does not exceed three-quarters o the unactored dead load, design moments are assumed to occur at all critical sections with ull actored live on all spans. CSA A ( ) L D (6.4 1) 4 a. Factored load and Fixed-End Moments (FEM s). For slab: Factored dead load w 1.5 (6.4 1) 9.05 kn/m Factored live load w kn/m l d Factored load w wd wl kn/m For drop panels: Factored dead load w 1.5 ( ) 3.4 kn/m Factored live load w kn/m l d Factored load w wd wl 3.4 kn/m n i1 NFi i 1 Fixed-end moment, FEM = m w l PCA Notes on ACI (Table A1) FEM (9 / 6) (9 / 3) 9 FEM kn.m b. Moment distribution. Computations are shown in Table 1. Counterclockwise rotational moments acting on the member ends are taken as positive. Positive span moments are determined rom the ollowing equation: M, midspan ( MuL MuR ) Mo Where M o is the moment at the midspan or a simple beam. When the end moments are not equal, the maximum moment in the span does not occur at the midspan, but its value is close to that midspan or this example (6% dierence compared with the exact value rom spslab). Positive moment in span 1-: M u 9 (3.4 9 / 6) 9 / 6 ( ) ( ) 9 / / 8 9 M kn.m 17

21 Table 1 - Moment Distribution or Equivalent Frame Joint Member DF COF FEM Dist CO Dist CO Dist CO Dist CO Dist CO Dist CO Dist CO Dist CO Dist CO Dist M, kn.m Midspan M, kn.m Factored moments used or Design Positive and negative actored moments or the slab system in the direction o analysis are plotted in Figure 1. The negative moments used or design are taken at the aces o supports (rectangle section or equivalent rectangle or circular or polygon sections) but not at distances greater than l 1 rom the centers o supports. CSA A ( ) 500 mm , 000 1,575 mm (use ace o supporting location) 18

22 Figure 1 - Positive and Negative Design Moments or Slab-Beam (All Spans Loaded with Full Factored Live Load) 19

23 .1.5. Distribution o design moments Check Applicability o Direct Design Method: 1. There is a minimum o three continuous spans in each direction. CSA A ( ). Successive span lengths are equal. CSA A ( ) 3. Loads are uniormly distributed over the entire panel CSA A ( ) 4. Factored live-to-dead load ratio o 0.5 <.0 CSA A ( ) (Note: The sel-weight o the drop panels is not uniormly distributed entirely along the span. However, the variation in load magnitude is small). Ater the negative and positive moments have been determined or the slab-beam strip, the ACI code permits the distribution o the moments at critical sections to the column strips, beams (i any), and middle strips in accordance with the DDM. CSA A ( ) Distribution o actored moments at critical sections is summarized in Table. End Span Interior Span Table - Distribution o actored moments Slab-beam Strip Column Strip Middle Strip Moment (kn.m) Percent Moment (kn.m) Percent Moment (kn.m) Exterior Negative Positive Interior Negative Negative Positive Flexural reinorcement requirements a. Determine lexural reinorcement required or strip moments The lexural reinorcement calculation or the column strip o end span exterior negative location is provided below. M kn.m Use d avg = 33 mm In this example, jd is assumed equal to 0.98d. The assumption will be veriied once the area o steel in inalized. Assume jd 0.98 d 35.4 mm Column strip width, b 9, 000 / 4,500 mm Middle strip width, b 9, 000 4,500 4,500 mm 6 M As,803 mm jd s y 0

24 CSA A (10.1.7) ' c A s s y Recalculate ' a' or the actual As 803 mm a mm c 1 ' cb , 500 jd d a 0.98d Thereore, the assumption that jd equals to 0.98d is valid. The slab have two thicknesses in the column strip (368 mm or the slab with the drop panel and 60 mm or the slab without the drop panel). The weighted slab thickness: h w 9 / 3 9 / 9 / / / 9 / 3 33 mm As,min mm > 814 mm CSA A (7.8.1) Maximum spacing: CSA A ( ) - Negative reinorcement in the band deined by b b: 1.5hw 498 mm 50 mm - Remaining negative moment reinorcement: 3hw 996 mm 500 mm For the negative reinorcements at the exterior span the maximum spacing is 50 mm. To distribute the bars uniormly, the same minimum spacing is applied to the remaining negative moment reinorcements within the column. Reinorcement or the total actored negative moment transerred to the exterior columns shall be placed within a band width b b. Temperature and shrinkage reinorcement determined as speciied in Clause shall be provided in that section o the slab outside i the band region deined by b b. CSA A ( ) b 500 ( ) 1,604 mm b Provide 10 15M bars with A s = 3,000 mm within b b width and 9 15M bars with A s = 1,800 mm out o the band. Based on the procedure outlined above, values or all span locations are given in Table 3. 1

25 Column Strip Middle Strip Span Location Table 3 - Required Slab Reinorcement or Flexure [Equivalent Frame Method (EFM)] Mu (kn.m) b (mm) d (mm) End Span As Req d or lexure (mm ) Min As (mm ) Reinorceme nt Provided As Prov. or lexure (mm ) Exterior Negative ,500 33,803, M * 4,800 Positive , ,933, M 4,00 Interior Negative , ,04, M 7,400 Exterior Negative 0.0 4, , M *,400 Positive ,500 4,579, M,800 Interior Negative ,500 4,07, M *,400 Interior Span Column Strip Positive , M *,400 Middle Strip Positive , M *,400 * Design governed by minimum reinorcement b. Calculate additional slab reinorcement at columns or moment transer between slab and column by lexure The actored slab moment resisted by the column ( M ) shall be assumed to be transerred by lexure. When gravity load, wind, earthquake, or other lateral orces cause transer o moment between slab and column, a raction o unbalanced moment given by shall be transerred by lexural reinorcement placed within a width b b. CSA A (13.10.) Portion o the unbalanced moment transerred by lexure is M r 1 CSA A (13.10.) 1 ( / 3) b / b 1 Where b 1 = Width o the critical section or shear measured in the direction o the span or which moments are determined according to CSA A3.3-14, clause 13 (see the ollowing igure). b = Width o the critical section or shear measured in the direction perpendicular to b1 according to CSA A3.3-14, clause 13 (see the ollowing igure). b b = Eective slab width = c 3 hs CSA A (3.)

26 Figure 13 Critical Shear Perimeters or Columns For exterior support: d = 33 mm b1 c 1 + d/ = / = 666 mm b c + d = = 83 mm b ,604 mm b ( / 3) 666 / kn.m Msc Using the same procedure in.1.6.a, the required area o steel: A,41 mm s The area o steel provided to resist the lexural moment within the eective slab width b b: As, provided 3,000 mm Then, there is no need to add additional reinorcement or the unbalanced moment. Based on the procedure outlined above, values or all supports are given in Table 4. 3

27 Column Strip Table 4 - Additional Slab Reinorcement required or moment transer between slab and column (EFM) Span Location M sc * (kn.m) γ γ M sc (kn.m) Eective slab width, b b (mm) End Span d (mm) A s req d within b b (mm ) A s prov. For lexure within b b (mm ) Exterior Negative ,604 33,41 3,000 - Interior Negative , ,000 - *M sc is taken at the centerline o the support in Equivalent Frame Method solution. Add l Rein Factored moments in columns The unbalanced moment rom the slab-beams at the supports o the equivalent rame are distributed to the support columns above and below the slab-beam in proportion to the relative stiness o the support columns. Reerring to Figure 1, the unbalanced moment at the exterior and interior joints are: Exterior Joint = kn.m Joint = -1, = kn.m The stiness and carry-over actors o the actual columns and the distribution o the unbalanced slab moments (M sc) to the exterior and interior columns are shown in the ollowing igure. 4

28 Figure 14 - Column Moments (Unbalanced Moments rom Slab-Beam) In summary: For Top column (Above): For Bottom column (Below): M col,exterior= kn.m M col,exterior= 0.73 kn.m M col,interior = kn.m M col,interior = 53.8 kn.m The moments determined above are combined with the actored axial loads (or each story) and actored moments in the transverse direction or design o column sections. The moment values at the ace o interior, exterior, and corner columns rom the unbalanced moment values are shown in the ollowing table. 5

29 Mu kn.m Table 5 Factored Moments in Columns Column Location Interior Exterior Corner Mux Muy Design o Columns by spcolumn This section includes the design o interior, edge, and corner columns using spcolumn sotware. The preliminary dimensions or these columns were calculated previously in section one Determination o actored loads Interior Column: Assume 5 story building Tributary area or interior column or live load, superimposed dead load, and sel-weight o the slab is A m Tributary Tributary area or interior column or sel-weight o additional slab thickness due to the presence o the drop panel is A 33 9 m Tributary Assuming ive story building P nw ATributary , kn M,x = kn.m (see the previous Table) M,y = kn.m (see the previous Table) Edge (Exterior) Column: Tributary area or edge column or live load, superimposed dead load, and sel-weight o the slab is A Tributary m Tributary area or edge column or sel-weight o additional slab thickness due to the presence o the drop panel is A Tributary m P nw ATributary ,981.4 kn M,x = 05.5 kn.m (see the previous Table) M,y = kn.m (see the previous Table) 6

30 Corner Column: Tributary area or corner column or live load, superimposed dead load, and sel-weight o the slab is A Tributary m Tributary area or corner column or sel-weight o additional slab thickness due to the presence o the drop panel is A Tributary m P nw ATributary ,578. kn M u,x = 117. kn.m (see the previous Table) M u,y = 117. kn.m (see the previous Table) 7

31 3.. Moment Interaction Diagram Interior Column: 8

32 Edge Column: 9

33 Corner Column: 30

34 4. Shear Strength Shear strength o the slab in the vicinity o columns/supports includes an evaluation o one-way shear (beam action) and two-way shear (punching) in accordance with CSA A Chapter One-Way (Beam action) Shear Strength CSA A (13.3.6) One-way shear is critical at a distance d rom the ace o the column as shown in Figure 3. Figures 15 and 16 show the actored shear orces (V ) at the critical sections around each column and each drop panel, respectively. In members without shear reinorcement, the design shear capacity o the section equals to the design shear capacity o the concrete: V V V V V, ( V V 0) CSA A (Eq. 11.4) r c s p c s p Where: V b d CSA A (Eq. 11.5) ' c c c w v Note: The calculations below ollow one o two possible approaches or checking one-way shear. Reer to the conclusions section or a comparison with the other approach At distance dv rom the supporting column h weighted 3689 / / 3 96 mm 9 d / 60 mm w dv Max (0.9 d, 0.7 h) Max (0.9 60, ) 34 mm CSA A (3.) 1 or normal weight concrete (1,000 ) (1,000 34) d v CSA A ( ) 34 Vc ,000 1,506.4 kn > V 1,000 Because V V at all the critical sections, the slab has adequate one-way shear strength. r 31

35 Figure 15 One-way shear at critical sections (at distance d rom the ace o the supporting column) At the ace o the drop panel h 60 mm d / 4 mm dv Max (0.9 d, 0.7 h) Max (0.9 4, ) 0 mm CSA A (3.) 1 or normal weight concrete 0.1 or slabs with overall thickness not greater than 350 mm CSA A ( ) 0 Vc ,000 1,465. kn > V 1,000 Because V V at all the critical sections, the slab has adequate one-way shear strength. r Figure 16 One-way shear at critical sections (at the ace o the drop panel) 3

36 4.. Two-Way (Punching) Shear Strength CSA A (13.3.) Around the columns aces Two-way shear is critical on a rectangular section located at d/ away rom the ace o the column as shown in Figure 13. a. Exterior column: The actored shear orce (V ) in the critical section is computed as the reaction at the centroid o the critical section minus the sel-weight and any superimposed surace dead and live load acting within the critical section (d/ away rom column ace) V V w b1 b kn 6 10 The actored unbalanced moment used or shear transer, M unb, is computed as the sum o the joint moments to the let and right. Moment o the vertical reaction with respect to the centroid o the critical section is also taken into account / Munb M V b1 cab c1 / kn.m 3 10 For the exterior column in Figure 13, the location o the centroidal axis z-z is: c AB moment o area o the sides about AB (666 (83 666) 666 / ) 05 mm area o the sides The polar moment J c o the shear perimeter is: 3 3 b1 d db1 b1 J b1d c AB bdc J J mm v AB γ 1 γ CSA A (Eq. 13.8) The length o the critical perimeter or the exterior column: b ,164 mm o The two-way shear stress (v ) can then be calculated as: 33

37 v V vmunbc b d J o AB CSA A (Eq.13.9) v , ( ) MPa, a) b) c) The actored resisting shear stress, V r shall be the smallest o: CSA A ( ) v v ' r c c c MPa c 1 sd ' 333 vr vc 0.19c c MPa bo 164 v v ' r c 0.38c c MPa I the eective depth. D, used in two-way shear calculations exceeds 300 mm, the value o v c obtained shall be multiplied by 1,300/ (1,000+d). CSA A ( ) v r 1,300 vc MPa (1, ) Since v c v at the critical section, the slab has adequate two-way shear strength around this drop panel. b. Interior column: 8383 V V w b1 b ,159.8 kn M M V b c c / ,159.8(0) kn unb 1 AB For the interior column in Figure 13, the location o the centroidal axis z-z is: b 1 83 cab 416mm The polar moment J c o the shear perimeter is: 3 3 b d db b J b d c b 1 dc 1 1 AB J J mm AB 34

38 γ v 1 γ CSA A (Eq. 13.8) The length o the critical perimeter or the interior column: b ,38 mm o The two-way shear stress (v ) can then be calculated as: v V vmunbc b d J o AB CSA A (Eq.13.9) v 6 1, , ( ) MPa 3, The actored resisting shear stress, V r shall be the smallest o: CSA A ( ) a) b) v v ' r c c c MPa c 1 sd ' 433 vr vc 0.19c c MPa bo 338 ' c) v v MPa r c c c I the eective depth. D, used in two-way shear calculations exceeds 300 mm, the value o v c obtained shall be multiplied by 1,300/ (1,000+d). CSA A ( ) 1,300 vr vc Mpa (1, ) Since v c v c. Corner column: at the critical section, the slab has adequate two-way shear strength around this drop panel. In this example, interior equivalent rame strip was selected where it only have exterior and interior supports (no corner supports are included in this strip). However, the two-way shear strength o corner supports usually governs. Thus, the two-way shear strength or the corner column in this example will be checked or educational purposes. Same procedure is used to ind the reaction and actored unbalanced moment used or shear transer at the centroid o the critical section or the corner support or the exterior equivalent rame strip V V w b1 b kn / Munb M V b1 cab c1 / kn.m 3 10 For the interior column in Figure 13, the location o the centroidal axis z-z is: 35

39 c AB moment o area o the sides about AB ( / ) mm area o the sides The polar moment J c o the shear perimeter is: 3 3 b1 d db1 b1 J b1d cab bdc 1 1 J c AB J.5 10 mm c 10 4 γ v 1 γ CSA A (Eq. 13.8) The length o the critical perimeter or the corner column: b ,33 mm o The two-way shear stress (v u) can then be calculated as: v V vmunbc b d J o AB CSA A (Eq.13.9) v , ( ) MPa 1, The actored resisting shear stress, V r shall be the smallest o: CSA A ( ) a) b) c) v v ' r c c c MPa c 1 sd ' 33 vr vc 0.19c c MPa bo 1,33 v v ' r c 0.38c c MPa I the eective depth. D, used in two-way shear calculations exceeds 300 mm, the value o v c obtained shall be multiplied by 1300/ (1000+d). CSA A ( ) 1,300 vr vc MPa (1, ) Since v c v at the critical section, the slab has adequate two-way shear strength around this drop panel. 36

40 4... Around drop panels Two-way shear is critical on a rectangular section located at d/ away rom the ace o the drop panel. Note: The two-way shear stress calculations around drop panels do not have the term or unbalanced moment since drop panels are a thickened portion o the slab and are not considered as a support. a. Exterior drop panel: 1,86 3, 4 V V w A kn 6 10 The length o the critical perimeter or the exterior drop panel: b 1,86 3, 4 6,948 mm o The two-way shear stress (v ) can then be calculated as: v V b d o CSA A (N ) v , MPa 6, The actored resisting shear stress, V r shall be the smallest o: CSA A ( ) a) b) c) v v ' r c c c MPa c 1 sd ' 34 vr vc 0.19c c MPa bo 6,948 v v ' r c 0.38c c MPa Since v c v at the critical section, the slab has adequate two-way shear strength around this drop panel. b. Interior drop panel: V V w A V 3,4 3, , kn 6 10 The length o the critical perimeter or the interior drop panel: b 3, 4 3, 4 1,896 mm o The two-way shear stress (v ) can then be calculated as: 37

41 v V b d o CSA A (N ) v 1, , MPa 1,896 4 The actored resisting shear stress, V r shall be the smallest o: CSA A ( ) a) b) c) v v ' r c c c MPa c 1 sd ' 44 vr vc 0.19c c MPa bo 1,896 v v ' r c 0.38c c MPa Since v c v at the critical section, the slab has adequate two-way shear strength around this drop panel. c. Corner drop panel: V V w A V 1,86 1, kn 6 10 The length o the critical perimeter or the corner drop panel: b 1,86 1, mm o The two-way shear stress (v ) can then be calculated as: v V b d o CSA A (N ) v , MPa 3, 74 4 The actored resisting shear stress, V r shall be the smallest o: CSA A ( ) a) b) c) v v ' r c c c MPa c 1 sd ' 4 vr vc 0.19c c MPa bo 3,74 v v ' r c 0.38c c MPa Since v c v at the critical section, the slab has adequate two-way shear strength around this drop panel. 38

42 5. Serviceability Requirements (Delection Check) Since the slab thickness was selected below the minimum slab thickness equations in CSA A3.3-14, the delection calculations o immediate and time-dependent delections are required and shown below including a comparison with spslab model results Immediate (Instantaneous) Delections When delections are to be computed, delections that occur immediately on application o load shall be computed by methods or ormulas or elastic delections, taking into consideration the eects o cracking and reinorcement on member stiness. Unless delections are determined by a more comprehensive analysis, immediate delection shall be computed using elastic delection equations. CSA A (9.8.. & ) The calculation o delections or two-way slabs is challenging even i linear elastic behavior can be assumed. Elastic analysis or three service load levels (D, D + L sustained, D+L Full) is used to obtain immediate delections o the two-way slab in this example. However, other procedures may be used i they result in predictions o delection in reasonable agreement with the results o comprehensive tests. The eective moment o inertia (I e) is used to account or the cracking eect on the lexural stiness o the slab. I e or uncracked section (M cr > M a) is equal to I g. When the section is cracked (M cr < M a), then the ollowing equation should be used: M I I I I I cr e cr g cr g M a 3 CSA A (Eq.9.1) Where: M a = Maximum moment in member due to service loads at stage delection is calculated. The values o the maximum moments or the three service load levels are calculated rom structural analysis as shown previously in this document. These moments are shown in Figure

43 Figure 17 Maximum Moments or the Three Service Load Levels (No live load is sustained in this example) For positive moment (midspan) section: M Cracking moment. cr M cr I / ( ) r g kn.m CSA A (Eq.9.) Y t 130 r = Modulus o rapture o concrete. r I g 0.6 ' MPa CSA A (Eq.8.3) c = Moment o inertia o the gross uncracked concrete section. 40

44 I g 3 3 lh 9, mm y t = Distance rom centroidal axis o gross section, neglecting reinorcement, to tension ace, mm. h 60 yt 130 mm I cr = Moment o inertia o the cracked section transormed to concrete. CAC Concrete Design Handbook 4 th Edition (Table 6.(a)) As calculated previously, the positive reinorcement or the end span rame strip is 35-15M bars are located along the section rom the bottom o the slab. Two o these bars are not continuous and will be conservatively excluded rom the calculation o I cr since they might not be adequately developed or tied (33 bars are used). Figure 18 shows all the parameters needed to calculate the moment o inertia o the cracked section transormed to concrete at midspan. Figure 18 Cracked Transormed Section (positive moment section) E cs = Modulus o elasticity o slab concrete. E cs ' c, 447 (3, 300 c 6, 900) (3, , 900) 9, 00 MPa,300,300 CSA A3.3-14(8.6..) Es 00, 000 n 6.9 CAC Concrete Design Handbook 4 th Edition (Table 6.a) E 9, 00 cs b 9,000 B 0. mm na s 1 CAC Concrete Design Handbook 4 th Edition (Table 6.a) db kd 4.81 mm B 0. CAC Concrete Design Handbook 4 th Edition (Table 6.a) b( kd) I na d kd 3 3 cr s ( ) CAC Concrete Design Handbook 4 th Edition (Table 6.a) I cr 3 9, 000 (4.81) mm

45 For negative moment section (near the interior support o the end span): The negative reinorcement or the end span rame strip near the interior support is 36-15M bars along the section rom the top o the slab. M cr I / ( ) r g kn.m CSA A (Eq.9.) Y t 15.4 r 0.6 ' MPa CSA A (Eq.8.3) c I mm g 10 4 y 15.4 mm t Figure 19 I g calculations or slab section near support E 1.5 ' c cs (3, 300 c 6, 900),300 CSA A3.3-14(8.6..) E cs,447 (3, , 900) 9, 00 MPa, Es 00, 000 n 6.9 CAC Concrete Design Handbook 4 th Edition (Table 6.a) E 9, 00 cs b 3,000 B 0.06 mm na s 1 CAC Concrete Design Handbook 4 th Edition (Table 6.a) db kd 89.6 mm B 0.06 CAC Concrete Design Handbook 4 th Edition (Table 6.a) b( kd) I na d kd 3 3 cr s ( ) CAC Concrete Design Handbook 4 th Edition (Table 6.a) I cr 3 3, 000 (89.58) mm

46 Figure 0 Cracked Transormed Section (negative moment section) The eective moment o inertia procedure described in the Code is considered suiciently accurate to estimate delections. The eective moment o inertia, I e, was developed to provide a transition between the upper and lower bounds o I g and I cr as a unction o the ratio M cr/m a. For conventionally reinorced (nonprestressed) members, the eective moment o inertia, I e, shall be calculated by Eq. (9.1) unless obtained by a more comprehensive analysis. For continuous prismatic members, the eective moment o inertia may be taken as the weighted average o the values obtained rom Equation 9.1 or the critical positive and negative moment sections CSA A3.3-14(9.8..4) For the middle span (span with two ends continuous) with service load level (D+LL ull): 3 M cr Ie Icr Ig Icr, since M cr 69.4 kn.m < M a =739 kn.m M a CSA A3.3-14(Eq. 9.1) Where I e - is the eective moment o inertia or the critical negative moment section (near the support) I e mm 739 For the middle span (span with two ends continuous) with service load level (D+LL ull): 3 M cr Ie Icr Ig Icr, since M cr kn.m < M a =0.9 kn.m M a CSA A3.3-14(Eq. 9.1) I e mm 0.9 Where I e + is the eective moment o inertia or the critical positive moment section (midspan). Since midspan stiness (including the eect o cracking) has a dominant eect on delections, midspan section is heavily represented in calculation o I e and this is considered satisactory in approximate delection calculations. Both the midspan stiness (I + e ) and averaged span stiness (I e,avg) can be used in the calculation o immediate (instantaneous) delection. 43

47 The averaged eective moment o inertia (I e,avg) is given by: e, avg e e, l e, r I 0.70 I 0.15 I I or two ends continuous CSA A (Eq.9.3) I, 0.85 I 0.15 I or one end continuous CSA A (Eq.9.4) e avg e e However, these expressions lead to improved results only or continuous prismatic members. The drop panels in this example result in non-prismatic members and the ollowing expressions should be used according to ACI : e, avg e e, l e, r I 0.50 I 0.5 I I or interior span ACI 435R-95 (.14) For the middle span (span with two ends continuous) with service load level (D l): I, mm e avg I, 0.50 I 0.50 I or end span ACI 435R-95 (.14) e avg e e For the end span (span with one end continuous) with service load level (D l): I, mm e avg Where: I el, I el, = The eective moment o inertia or the critical negative moment section near the let support. = The eective moment o inertia or the critical negative moment section near the right support. I e = The eective moment o inertia or the critical positive moment section (midspan). Table 6 provides a summary o the required parameters and calculated values needed or delections or exterior and interior spans. 44

48 Table 6 Averaged Eective Moment o Inertia Calculations For Frame Strip Span zone I g, mm 4 ( 10 9 ) I cr, mm 4 ( 10 9 ) D M a, kn.m D + LL Sus D + L ull M cr, kn.m D I e, mm 4 ( 10 9 ) I e,avg, mm 4 ( 10 9 ) D + LL Sus D + L ull D D + LL Sus D + L ull Ext Int Let Midspan Right Let Midspan Right Delections in two-way slab systems shall be calculated taking into account size and shape o the panel, conditions o support, and nature o restraints at the panel edges. For immediate delections in two-way slab systems, the midpanel delection is computed as the sum o delection at midspan o the column strip or column line in one direction (Δ cx or Δ cy) and delection at midspan o the middle strip in the orthogonal direction (Δ mx or Δ my). Figure 1 shows the delection computation or a rectangular panel. The average Δ or panels that have dierent properties in the two direction is calculated as ollows: ( ) ( ) cx my cy mx PCA Notes on ACI ( Eq. 8) Figure 1 Delection Computation or a rectangular Panel 45

49 To calculate each term o the previous equation, the ollowing procedure should be used. Figure shows the procedure o calculating the term Δ cx. Same procedure can be used to ind the other terms. Figure Δ cx calculation procedure For end span - service dead load case: 4 wl rame, ixed PCA Notes on ACI ( Eq. 10) 384EI c rame, averaged Where: rame, ixed = Delection o column strip assuming ixed end condition. w ( )(9) kn/m E 1.5 ' c cs (3, 300 c 6, 900),300 CSA A3.3-14(8.6..) E cs,447 (3, , 900) 9, 000 MPa, I rame,averaged = The averaged eective moment o inertia (I e,avg) or the rame strip or service dead load case rom Table 6 = 5.75 x 10 9 mm (65.16)(9 10 ) rame, ixed 5.31 mm 9 384(9, 000)( ) I rame, averaged c, ixed LDFc rame, ixed PCA Notes on ACI ( Eq. 11) Icg, 46

50 For this example and like in the spslab program, the eective moment o inertia at midspan will be used. LDF c is the load distribution actor or the column strip. The load distribution actor or the column strip can be ound rom the ollowing equation: LDF LDF c LDF l LDF R And the load distribution actor or the middle strip can be ound rom the ollowing equation: LDF m 1 LDF c For the end span, LDF or exterior negative region (LDF L ), interior negative region (LDF R ), and positive region (LDF + ) are 1.00, 0.85, and 0.60, respectively (From Table o this document). Thus, the load distribution actor or the column strip or the end span is given by: LDF c I c,g = The gross moment o inertia (I g) or the column strip or service dead load = 6.59 x 10 9 mm c, ixed mm ( Mnet, L ) rame cl, PCA Notes on ACI ( Eq. 1) K ec Where: cl, = Rotation o the span let support. ( M ) 31 kn-m = Net rame strip negative moment o the let support. net, L rame K ec = eective column stiness = 1.76 x 10 5 kn-m/rad (calculated previously). 31 cl, rad l I g c, L c, L 8 Ie rame PCA Notes on ACI ( Eq. 14) Where: cl, = Midspan delection due to rotation o let support. 47

51 I I g e rame = Gross-to-eective moment o inertia ratio or rame strip cl, mm M 7 net, R rad rame cr, 11 Kec Where cr, = Rotation o the end span right support. ( M ) Net rame strip negative moment o the right support. net, R rame l I mm 9 c, R g c, R 8 Ie rame Where: cr, = Midspan delection due to rotation o right support. PCA Notes on ACI ( Eq. 9) cx cx, ixed cx, R cx, L mm cx Following the same procedure, Δ mx can be calculated or the middle strip. This procedure is repeated or the equivalent rame in the orthogonal direction to obtain Δ cy, and Δ my or the end and middle spans or the other load levels (D+LL sus and D+LL ull). Since in this example the panel is squared, Δ cx = Δ cy= 1.07 mm and Δ mx = Δ my= 6.63 mm The average Δ or the corner panel is calculated as ollows: cx my cy mx cx my cy mx mm 48

52 Table 7 Immediate (Instantaneous) Delections in the x-direction Column Strip Middle Strip Span LDF Δ rame-ixed, mm Δ c-ixed, mm θ c1, rad D θ c, rad Δθ c1, mm Δθ c, mm Δ cx, mm LDF Δ rame-ixed, mm Ext Int Δ m-ixed, mm θ m1, mm D θ m, mm Δθ m1, mm Δθ m, mm Δ mx, mm Span LDF Δ rame-ixed, mm Δ c-ixed, mm θ c1, rad D+LL sus θ c, rad Δθ c1, mm Δθ c, mm Δ cx, mm LDF Δ rame-ixed, mm Ext Int Δ m-ixed, mm θ m1, mm D+LL sus θ m, mm Δθ m1, mm Δθ m, mm Δ mx, mm Span LDF Δ rame-ixed, mm Δ c-ixed, mm θ c1, rad D+LL ull θ c, rad Δθ c1, mm Δθ c, mm Δ cx, mm LDF Δ rame-ixed, mm Ext Int Δ m-ixed, mm θ m1, mm D+LL ull θ m, mm Δθ m1, mm Δθ m, mm Δ mx, mm Span LDF LL Δ cx, mm Ext Int LDF LL Δ mx, mm 49

53 5.. Time-Dependent (Long-Term) Delections (Δlt) The additional time-dependent (long-term) delection resulting rom creep and shrinkage (Δ cs) may be estimated as ollows: ( ) PCA Notes on ACI ( Eq. 4) cs sust Inst The total time-dependent (long-term) delection is calculated as: ( ) ( ) (1 ) [( ) ( ) ] CSA A (N9.8..5) total lt sust Inst total Inst sust Inst Where: ( ) Immediate (instantaneous) delection due to sustained load, in. sust Inst 1 50 ' ACI ( ) ( ) Time-dependent (long-term) total delection, in. total lt ( ) Total immediate (instantaneous) delection, in. total Inst For the exterior span =, consider the sustained load duration to be 60 months or more. ACI (Table ) ' = 0, conservatively mm cs mm total lt Table 8 shows long-term delections or the exterior and interior spans or the analysis in the x-direction, or column and middle strips. Table 8 - Long-Term Delections Column Strip Span (Δsust)Inst, mm λδ Δcs, mm (Δtotal)Inst, mm (Δtotal)lt, mm Exterior Interior Middle Strip Exterior Interior

54 6. spslab Sotware Program Model Solution spslab program utilizes the Elastic Frame Method described and illustrated in details here or modeling, analysis and design o two-way concrete loor slab systems. spslab uses the exact geometry and boundary conditions provided as input to perorm an elastic stiness (matrix) analysis o the equivalent rame taking into account the torsional stiness o the slabs raming into the column. It also takes into account the complications introduced by a large number o parameters such as vertical and torsional stiness o transverse beams, the stiening eect o drop panels, column capitals, and eective contribution o columns above and below the loor slab using the o equivalent column concept (CSA A ( )). spslab Program models the elastic rame as a design strip. The design strip is, then, separated by spslab into column and middle strips. The program calculates the internal orces (Shear Force & Bending Moment), moment and shear capacity vs. demand diagrams or column and middle strips, instantaneous and long-term delection results, and required lexural reinorcement or column and middle strips. The graphical and text results will be provided rom the spslab model in a uture revision to this document. 51

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75 7. Summary and Comparison o Design Results Table 9 - Comparison o Moments obtained rom Hand (EFM) and spslab Solution (kn.m) Hand (EFM) spslab Exterior Span Exterior Negative * Column Strip Positive Interior Negative * Exterior Negative * Middle Strip Positive Interior Negative * Interior Span Column Strip Interior Negative * Positive Middle Strip Interior Negative * Positive * negative moments are taken at the aces o supports Table 10 - Comparison o Reinorcement Results Span Location Reinorcement Provided or Flexure Additional Reinorcement Provided or Unbalanced Moment Transer* Total Reinorcement Provided Column Strip Middle Strip Column Strip Middle Strip Exterior Negative Hand spslab Hand spslab Hand spslab Exterior Span 4-15M 4-15M n/a n/a 4-15M 4-15M Positive 1-15M 1-15M n/a n/a 1-15M 1-15M Interior Negative Exterior Negative 37-15M 37-15M M 37-15M 1-15M 1-15M n/a n/a 1-15M 1-15M Positive 14-15M 14-15M n/a n/a 14-15M 14-15M Interior Negative 1-15M 1-15M n/a n/a 1-15M 1-15M Interior Span Positive 1-15M 1-15M n/a n/a 1-15M 1-15M Positive 1-15M 1-15M n/a n/a 1-15M 1-15M 7

76 Table 11 - Comparison o One-Way (Beam Action) Shear Check Results Span dv, kn drop panel, kn dv, kn drop panel, kn Hand spslab Hand spslab Hand spslab Hand spslab Exterior , ,587. 1,465. 1,465. Interior , ,587. 1,465. 1,465. * x u calculated rom the centerline o the let column or each span Table 1 - Comparison o Two-Way (Punching) Shear Check Results (around Columns Faces) Support b1, mm b, mm bo, mm V, kn cab, mm Hand spslab Hand spslab Hand spslab Hand spslab Hand spslab Exterior ,164, Interior ,38 3,38 1, , Corner ,33 1, Support Jc, mm 4 γv Munb, kn.m v, MPa vc, MPa Hand spslab Hand spslab Hand spslab Hand spslab Hand spslab Exterior Interior Corner Table 13 - Comparison o Two-Way (Punching) Shear Check Results (around Drop Panels) Support b1, mm b, mm bo, mm V, kn cab, mm Hand spslab Hand spslab Hand spslab Hand spslab Hand spslab Exterior 1,86 1,86 3,4 3,4 6,948 6, Interior 3,4 3,4 3,4 3,4 1,896 1,896 1, Corner 1,86 1,86 1,86 1,86 3,74 3, Support Jc, mm 4 γv Munb, kn.m v, MPa vc, MPa Hand spslab Hand spslab Hand spslab Hand spslab Hand spslab Exterior N.A. N.A. N.A. N.A Interior N.A. N.A. N.A. N.A Corner N.A. N.A. N.A. N.A Note: Shear stresses rom spslab are higher than hand calculations since it considers the load eects beyond the column centerline known in the model as right/let cantilevers. This small increase is oten neglected in simpliied hand calculations like the one used here. 73

77 Span Table 14 - Comparison o Immediate Delection Results (mm) Column Strip D D+LLsus D+LLull LL Hand spslab Hand spslab Hand spslab Hand spslab Exterior Interior Span Middle Strip D D+LLsus D+LLull LL Hand spslab Hand spslab Hand spslab Hand spslab Exterior Interior Table 15 - Comparison o Time-Dependent Delection Results Column Strip Span λδ Δcs, mm Δtotal, mm Hand spslab Hand spslab Hand spslab Exterior Interior Middle Strip Span λδ Δcs, mm Δtotal, mm Hand spslab Hand spslab Hand spslab Exterior Interior In all o the hand calculations illustrated above, the results are in close or exact agreement with the automated analysis and design results obtained rom the spslab model. 74

78 8. Conclusions & Observations 8.1. One-Way Shear Distribution to Slab Strips In one-way shear checks above, shear is distributed uniormly along the width o the design strip (9 m). StructurePoint inds it necessary sometimes to allocate the one-way shears with the same proportion moments are distributed to column and middle strips. spslab allows the one-way shear check using two approaches: 1) calculating the one-way shear capacity using the average slab thickness and comparing it with the total actored one-shear load as shown in the hand calculations above; ) distributing the actored one-way shear orces to the column and middle strips and comparing it with the shear capacity o each strip as illustrated in the ollowing igures. An engineering judgment is needed to decide which approach to be used. Figure 3a Distributing Shear to Column and Middle Strips (spslab Input) 75