Two-Way Flat Plate Concrete Floor System Analysis and Design Version: Aug-10-017
Two-Way Flat Plate Concrete Floor System Analysis and Design The concrete floor slab system shown below is for an intermediate floor to be designed considering partition weight = 0 psf, and unfactored live load = 40 psf. Flat plate concrete floor system does not use beams between columns or drop panels and it is usually suited for lightly loaded floors with short spans typically for residential and hotel buildings. The lateral loads are independently resisted by shear walls. The two design procedures shown in ACI 318-14: Direct Design Method (DDM) and the Equivalent Frame Method (EFM) are illustrated in detail in this example. The hand solution from EFM is also used for a detailed comparison with the analysis and design results of the engineering software program spslab. Figure 1 - Two-Way Flat Concrete Floor System Version: Aug-10-017
Contents 1. Preliminary Member Sizing... 1. Two-Way Slab Analysis and Design... 3.1. Direct Design Method (DDM)... 4.1.1. Direct design method limitations... 4.1.. Design moments... 4.1.3. Flexural reinforcement requirements... 5.1.4. Factored moments in columns... 7.. Equivalent Frame Method (EFM)... 8..1. Equivalent frame method limitations... 9... Frame members of equivalent frame... 9..3. Equivalent frame analysis... 11..4. Design moments... 1..5. Distribution of design moments... 13..6. Flexural reinforcement requirements... 14..7. Column design moments... 16 3. Design of Interior, Edge, and Corner Columns... 18 3.1. Determination of factored loads... 19 3.. Column Capacity Diagram (Axial-Moment Interaction Diagram)... 0 4. Two-Way Slab Shear Strength... 4 4.1. One-Way (Beam action) Shear Strength... 4 4.. Two-Way (Punching) Shear Strength... 5 5. Two-Way Slab Deflection Control (Serviceability Requirements)... 8 5.1. Immediate (Instantaneous) Deflections... 8 5.. Time-Dependent (Long-Term) Deflections (Δ lt)... 37 6. Computer Program Solution... 38 7. Summary and Comparison of Two-Way Slab Design Results... 58 8. Comparison of Two-Way Slab Analysis and Design Methods... 60 Version: Aug-10-017
Code Building Code Requirements for Structural Concrete (ACI 318-14) and Commentary (ACI 318R-14) Minimum Design Loads for Buildings and Other Structures (ASCE/SEI 7-10) International Code Council, 01 International Building Code, Washington, D.C., 01 Reference Notes on ACI 318-11 Building Code Requirements for Structural Concrete, Twelfth Edition, 013 Portland Cement Association, Example 0.1 Concrete Floor Systems (Guide to Estimating and Economizing), Second Edition, 00 David A. Fanella Simplified Design of Reinforced Concrete Buildings, Fourth Edition, 011 Mahmoud E. Kamara and Lawrence C. Novak Design Data Floor-to-Floor Height = 9 ft (provided by architectural drawings) Superimposed Dead Load, SDL =0 psf for framed partitions, wood studs plaster sides ASCE/SEI 7-10 (Table C3-1) Live Load, LL = 40 psf for Residential floors ASCE/SEI 7-10 (Table 4-1) f c = 4000 psi (for slabs) f c = 6000 psi (for columns) f y = 60,000 psi Required fire resistance rating = hours Solution 1. Preliminary Member Sizing a. Slab minimum thickness - Deflection ACI 318-14 (8.3.1.1) In this example deflection will be calculated and checked to satisfy project deflection limits. Minimum member thickness and depths from ACI 318-14 will be used for preliminary sizing. Using ACI 318-14 minimum slab thickness for two-way construction without interior beams in Table 8.3.1.1. l n 00 Exterior Panels: h 6.67 30 30 ACI 318-14 (Table 8.3.1.1) But not less than 5 ACI 318-14 (8.3.1.1(a)) l n 00 Interior Panels: h 6.06 33 33 ACI 318-14 (Table 8.3.1.1) But not less than 5 ACI 318-14 (8.3.1.1(a)) 1
Where l n = length of clear span in the long direction = 16 16 = 00 Try 7 slab for all panels (self-weight = 87.5 psf) b. Slab shear strength one way shear Evaluate the average effective depth (Figure ): db 0.5 dl tslab cclear db 7 0.75 0.5 5.50 db 0.5 dt tslab cclear 7 0.75 6.00 d d l t 5.50 6.00 d 5.75 avg Where: c clear = 3/4 for # 4 steel bar ACI 318-14 (Table 0.6.1.3.1) d b = 0.5 for # 4 steel bar Figure - Two-Way Flat Concrete Floor System Factored dead load, qdu 1. (87.5 0) 19 psf Factored live load, qlu 1.6 40 64 psf ACI 318-14 (5.3.1) Total factored load qu 193 psf Check the adequacy of slab thickness for beam action (one-way shear) ACI 318-14 (.5) at an interior column: Consider a 1- wide strip. The critical section for one-way shear is located at a distance d, from the face of support (see Figure 3) 18 16 5.75 1 Tributary area for one-way shear is A [( ) ( ) ( )] ( ) 7.854 ft Tributary 1 1 1 V u qu A 0.193 7.854 1.5 kips Tributary Vc fc ' bwd ACI 318-14 (Eq..5.5.1) where 1for normal weight concrete
5.75 V c 0.75 1.0 4000 1 6.6 kips 1000 Slab thickness of 7 is adequate for one-way shear. c. Slab shear strength two-way shear V u Check the adequacy of slab thickness for punching shear (two-way shear) at an interior column (Figure 4): 16 5.75 Tributary area for two-way shear is A (18 14) ( ) 48.7 ft Tributary 1 V u qu A 0.193 48.7 48.0 kips Tributary Vc 4 fc ' bo d (For square interior column) ACI 318-14 (Table.6.5.(a)) 5.75 V c 4 4000 (41.75) 16.6 kips 1000 V 0.7516.6 95.0kips V c Slab thickness of 7 is adequate for two-way shear. u Figure 3 Critical Section for One-Way Figure 4 Critical Section for Two-Way d. Column dimensions - axial load Check the adequacy of column dimensions for axial load: Tributary area for interior column is ATributary P u qu ATributary 0.193 5 48.6 kips (18 14) 5 ft P 0.80P 0.80 (0.85 f '( A A ) f A ) (For square interior column) ACI 318-14 (.4.) n o c g st y st P 0.80 (0.85(6000)(16 16 0) 0) 1,044,480 lb 1,044 kips n P 0.65 1, 044 679 kips P 48.6 kips n u Column dimensions of 16 x 16 are adequate for axial load.. Two-Way Slab Analysis and Design ACI 318 states that a slab system shall be designed by any procedure satisfying equilibrium and geometric compatibility, provided that strength and serviceability criteria are satisfied. Distinction of two-systems from one-way systems is given by ACI 318-14 (R8.10..3 & R8.3.1.). 3
ACI 318 permits the use of Direct Design Method (DDM) and Equivalent Frame Method (EFM) for the gravity load analysis of orthogonal frames and is applicable to flat plates, flat slabs, and slabs with beams. The following sections outline the solution per DDM, EFM, and spslab software respectively..1. Direct Design Method (DDM) Two-way slabs satisfying the limits in ACI 318-14 (8.10.) are permitted to be designed in accordance with the DDM..1.1. Direct design method limitations There is a minimum of three continuous spans in each direction ACI 318-14 (8.10..1) Successive span lengths are equal ACI 318-14 (8.10..) Long-to-short span ratio is 1.9 < ACI 318-14 (8.10..3) Columns are not offset ACI 318-14 (8.10..4) Loads are uniformly distributed over the entire panel ACI 318-14 (8.10..5) Service live-to-dead load ratio of 0.37 <.0 ACI 318-14 (8.10..6) Slab system is without beams and this requirement is not applicable ACI 318-14 (8.10..7) Since all the criteria are met, Direct Design Method can be utilized..1.. Design moments a. Calculate the total factored static moment: M o qu n 0.19314 16.67 93.6 ft-kips ACI 318-14 (8.10.3.) 8 8 b. Distribute the total factored moment, M o, in an interior and end span: ACI 318-14 (8.10.4) Exterior Span Table 1 - Distribution of Mo along the span Location Total Design Strip Moment, MDS (ft-kips) Exterior Negative 0.6 x M o = 4.3 Positive 0.5 x M o = 48.7 Interior Negative 0.70 x M o = 65.5 Interior Span Positive 0.35 x M o = 3.8 c. Calculate the column strip moments. ACI 318-14 (8.10.5) That portion of negative and positive total design strip moments not resisted by column strips shall be proportionally assigned to corresponding two half-middle strips. ACI 318-14 (8.10.6.1) 4
Table - Lateral Distribution of the Total Design Strip Moment, MDS Location Total Design Strip Moment, MDS (ft-kips) Column Strip Moment, (ft-kips) Moment in Two Half Middle Strips, (ft-kips) Exterior Negative * 4.3 1.00 x M DS = 4.3 0.00 x M DS = 0.0 Exterior Span Positive 48.7 0.60 x M DS = 9. 0.40 x M DS = 19.5 Interior Negative * 65.5 0.75 x M DS = 49.1 0.5 x M DS = 16.4 Interior Span Positive 3.8 0.60 x M DS = 19.7 0.40 x M DS = 13.1 * All negative moments are at face of support..1.3. Flexural reinforcement requirements a. Determine flexural reinforcement required for column and middle strips at all critical sections The following calculation is for the exterior span exterior negative location of the column strip. M 4.3 u ft-kips Use average d avg = 5.75 To determine the area of steel, assumptions have to be made whether the section is tension or compression controlled, and regarding the distance between the resultant compression and tension forces along the slab section (jd). In this example, tension-controlled section will be assumed so the reduction factor is equal to 0.9, and jd will be taken equal to 0.95d. The assumptions will be verified once the area of steel in finalized. Assume jd 0.95 d 5.46 Column strip width, b 14 1 / 84 Middle strip width, b 141 84 84 A M u 4.31000 s f jd 0.99 in y 0.9 60000 0.95 5.75 Recalculate a for the actual A s = 0.99 : Af s y 0.99 60000 a 0.08 in 0.85 f ' b 0.854000 84 c a 0.08 c 0.44 in 0.85 1 0.003 0.003 t ( ) dt 0.003 ( ) 5.75 0.003 0.0676 0.005 c 0.44 5
Therefore, the assumption that section is tension-controlled is valid. A M u 4.31000 s f 0.96 ( d a in y / ) 0.9 60000 (5.75 0.08 / ) Min As 0.018 847 1.06 in 0.96 in ACI 318-14 (4.4.3.) Maximum spacing smax h 7 14 in 18 in ACI 318-14 (8.7..) Provide 6 - #4 bars with A 1.0 s in and s 84 / 6 14 in s max Based on the procedure outlined above, values for all span locations are given in Table 3. Column Strip Middle Strip Column Strip Middle Strip Span Location Table 3 - Required Slab Reinforcement for Flexure (DDM) Mu (ft-kips) b () d () End Span As Req d for flexure (in ) Min As (in ) Reinforcement Provided As Prov. for flexure (in ) Exterior Negative 4.3 84 5.75 0.96 1.06 6-#4 1. Positive 9 84 5.75 1.15 1.06 6-#4 1. Interior Negative 49.6 84 5.75 1.99 1.06 10-#4 Exterior Negative 0 84 5.75 0 1.06 6-#4 1. Positive 19.7 84 5.75 0.77 1.06 6-#4 1. Interior Negative 15.9 84 5.75 0.6 1.06 6-#4 1. Interior Span Positive 19.7 84 5.75 0.77 1.06 6-#4 1. Positive 13.1 84 5.75 0.51 1.06 6-#4 1. b. Calculate additional slab reinforcement at columns for moment transfer between slab and column The factored slab moment resisted by the column ( f Mu ) shall be assumed to be transferred by flexure. Concentration of reinforcement over the column by closer spacing or additional reinforcement shall be used to resist this moment. The fraction of slab moment not calculated to be resisted by flexure shall be assumed to be resisted by eccentricity of shear. ACI 318-14 (8.4..3) Portion of the unbalanced moment transferred by flexure is f Mu ACI 318-14 (8.4..3.1) Where 1 f ACI 318-14 (8.4..3.) 1 ( / 3) b / b 1 b1 Dimension of the critical section determined in ACI 318, Chapter 8 (see Figure 5). b o measured in the direction of the span for which moments are 6
b Dimension of the critical section bo measured in the direction perpendicular to b1 in ACI 318, Chapter 8 (see Figure 5). b b = Effective slab width = c 3h ACI 318-14 (8.4..3.3) Figure 5 Critical Shear Perimeters for Columns Column Strip Table 4 - Additional Slab Reinforcement required for moment transfer between slab and column (DDM) Span Location M u * (ft-kips) γ f γ f M u (ft-kips) End Span Effective slab width, b b () d () A s req d within b b (in ) A s prov. For flexure within b b (in ) Exterior Negative 4.3 0.6 15.1 37 5.75 0.6 0.53 1-#4 Interior Negative 0.0 0.60 0.0 37 5.75 0.0 0.97 - *M u is taken at the centerline of the support in Equivalent Frame Method solution. Add l Reinf..1.4. Factored moments in columns a. Interior columns: M 0.07 [( q 0.5 q ) ' ' ' ] Du Lu n q n ACI 318-14 (8.10.7.) u Du 16 16 M u 0.07 (19 0.5 64) 1418 1918 14 17.75 ft-kips 1 1 7
With the same column size and length above and below the slab, 17.75 M 8.87 ft-kips column b. Exterior Columns: Total exterior negative moment from slab must be transferred directly to the column: ftkips. With the same column size and length above and below the slab, M 4.3 u 4.3 M column 1.15 ft-kips The moments determined above are combined with the factored axial loads (for each story) for design of column sections as shown later in this example... Equivalent Frame Method (EFM) EFM is the most comprehensive and detailed procedure provided by the ACI 318 for the analysis and design of two-way slab systems where the structure is modeled by a series of equivalent frames (interior and exterior) on column lines taken longitudinally and transversely through the building. The equivalent frame consists of three parts: 1) Horizontal slab-beam strip, including any beams spanning in the direction of the frame. Different values of moment of inertia along the axis of slab-beams should be taken into account where the gross moment of inertia at any cross section outside of joints or column capitals shall be taken, and the moment of inertia of the slab-beam at the face of the column, bracket or capital divide by the quantity (1-c /l ) shall be assumed for the calculation of the moment of inertia of slab-beams from the center of the column to the face of the column, bracket or capital. ACI 318-14 (8.11.3) ) Columns or other vertical supporting members, extending above and below the slab. Different values of moment of inertia along the axis of columns should be taken into account where the moment of inertia of columns from top and bottom of the slab-beam at a joint shall be assumed to be infinite, and the gross cross section of the concrete is permitted to be used to determine the moment of inertia of columns at any cross section outside of joints or column capitals. ACI 318-14 (8.11.4) 3) Elements of the structure (Torsional members) that provide moment transfer between the horizontal and vertical members. These elements shall be assumed to have a constant cross section throughout their length consisting of the greatest of the following: (1) portion of slab having a width equal to that of the column, bracket, or capital in the direction of the span for which moments are being determined, () portion of slab specified in (1) plus that part of the transverse beam above and below the slab for monolithic or fully composite construction, (3) the transverse beam includes that portion of slab on each side of the beam extending a distance equal to the projection of the beam above or below the slab, whichever is greater, but not greater than four times the slab thickness. ACI 318-14 (8.11.5) 8
..1. Equivalent frame method limitations In EFM, live load shall be arranged in accordance with 6.4.3 which requires slab systems to be analyzed and designed for the most demanding set of forces established by investigating the effects of live load placed in various critical patterns. ACI 318-14 (8.11.1. & 6.4.3) Complete analysis must include representative interior and exterior equivalent frames in both the longitudinal and transverse directions of the floor ACI 318-14 (8.11..1) Panels shall be rectangular, with a ratio of longer to shorter panel dimensions, measured center-to-center of supports, not to exceed. ACI 318-14 (8.10..3)... Frame members of equivalent frame Determine moment distribution factors and fixed-end moments for the equivalent frame members. The moment distribution procedure will be used to analyze the equivalent frame. Stiffness factors k, carry over factors COF, and fixed-end moment factors FEM for the slab-beams and column members are determined using the design aids tables at Appendix 0A of PCA Notes on ACI 318-11. These calculations are shown below. a. Flexural stiffness of slab-beams at both ends, K sb. c N 1 16 0.07 (18 1) 1, c N 16 0.1 (14 1) For cf1 cf, stiffness factors, knf kfn 4.13 PCA Notes on ACI 318-11 (Table A1) E I E I Thus, K k cs s 4.13 cs s NF PCA Notes on ACI 318-11 (Table A1) sb 1 1 3 480 K 4.13 3834 10 3510 6 -lb sb 16 where, 3 168 (7) 3 s h I s 480 in 4 1 1 E w 1.5 33 f 150 1.5 33 4000 3834 10 3 cs c c psi ACI 318-14 (19...1.a) Carry-over factor COF 0.509 PCA Notes on ACI 318-11 (Table A1) Fixed-end moment FEM 0.0843w u PCA Notes on ACI 318-11 (Table A1) 1 b. Flexural stiffness of column members at both ends, K c. Referring to Table A7, Appendix 0A, ta 3.5, tb 3.5, ta H H 9 ft 108, Hc 101, 1, 1.07 tb Hc 9
Thus, k k 4.74 by interpolation. AB BA 4.74E cc I K c PCA Notes on ACI 318-11 (Table A7) c c 3 5461 K 4.74 4696 10 115.5 10 6 c -lb 108 Where 4 4 c (16) Ic 5461 4 1 1 E w 1.5 33 f 150 1.5 33 6000 4696 10 3 cc c c psi ACI 318-14 (19...1.a) c 9 ft 108 c. Torsional stiffness of torsional members, K t. K t 9EcsC c 3 [ (1 ) ] ACI 318-14 (R.8.11.5) 9383410 3 135 K 367 10 6 t -lb 168(0.905) 3 Where 3 x x y C (1 0.63 )( ) ACI 318-14 (Eq. 8.10.5.b) y 3 7 3 16 C (1 0.63 )(7 ) 135 in 4. 16 3 c 16, and 14 ft 168 d. Equivalent column stiffness Kc K K t ec Kc Kt K ec. K ( 115.5)( 367) 10 6 ec [(115.5) (367)] Figure 6 - Torsional Member 10
K 553.7 10 6 ec -lb Where K t column, and is for two torsional members one on each side of the K c is for the upper and lower columns at the slabbeam joint of an intermediate floor. e. Slab-beam joint distribution factors, DF. At exterior joint, 35 DF 0.389 (35 553.7) Figure 7 Column and Edge of Slab At interior joint, 35 DF 0.80 (35 35 553.7) COF for slab-beam 0.509..3. Equivalent frame analysis Determine negative and positive moments for the slab-beams using the moment distribution method. Since the unfactored live load does not exceed three-quarters of the unfactored dead load, design moments are assumed to occur at all critical sections with full factored live on all spans. ACI 318-14 (6.4.3.) L 40 3 0.37 D (87.5 0) 4 a. Factored load and Fixed-End Moments (FEM s). Factored dead load qdu 1.(87.5 0) 19 psf Factored live load qlu 1.6(40) 64 psf Factored load q q q 193 psf u Du Lu Figure 8 Slab and Column Stiffness FEM s for slab-beams m q PCA Notes on ACI 318-11 (Table A1) NF u 1 0.0841 (0.193 14) 18 73.8 ft-kips 11
b. Moment distribution. Computations are shown in Table 5. Counterclockwise rotational moments acting on the member ends are taken as positive. Positive span moments are determined from the following equation: ( MuL MuR ) (midspan) M o M u Where M o is the moment at the midspan for 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 for this example. Positive moment in span 1-: 18 (46.6 84.0) M u (0.19314) 44.1 ft-kips 8 Positive moment span -3: 18 (76. 76.) M u (0.19314) 33. ft-kips 8 Table 5 Moment Distribution for Equivalent Frame Joint 1 3 4 Member 1- -1-3 3-3-4 4-3 DF 0.389 0.80 0.80 0.80 0.80 0.389 COF 0.509 0.509 0.509 0.509 0.509 0.509 FEM +73.8-73.8 +73.8-73.8 +73.8-73.8 Dist CO Dist CO Dist CO Dist CO Dist -8.7 0.0 0.0.1-0.8 0.3-0.1 0.1 0.0 0.0-14.6 4.1 0.0 0.6-0.4 0. -0.1 0.0 0.0 0.0 4.1 -.1 0.6-0.3 0. -0.1 0.0 0.0 0.0-4.1.1-0.6 0.3-0. 0.1 0.0 0.0 14.6-4.1 0.0-0.6 0.4 -.0 0.1 0.0 8.7 0.0 0.0 -.1 0.8-0.3 0.1-0.1 0.0 Neg. M 46.6-84.0 76. -76. 84.0-46.6 M at midspan 44.1 33. 44.1..4. Design moments Positive and negative factored moments for the slab system in the direction of analysis are plotted in Figure 9. The negative moments used for design are taken at the faces of supports (rectangle section or equivalent rectangle for circular or polygon sections) but not at distances greater than 0.175 1 from the centers of supports. ACI 318-14 (8.11.6.1) 16 0.67 ft 0.175 18 3. ft (use face of support location) 1 1
Figure 9 - Positive and Negative Design Moments for Slab-Beam (All Spans Loaded with Full Factored Live Load)..5. Distribution of design moments a. Check whether the moments calculated above can take advantage of the reduction permitted by ACI 318-14 (8.11.6.5): If the slab system analyzed using EFM within the limitations of ACI 318-14 (8.10.), it is permitted by the ACI code to reduce the calculated moments obtained from EFM in such proportion that the absolute sum of the positive and average negative design moments need not exceed the value obtained from the following equation: qu n (16.67) M o 0.19314 93.9 ft-kips ACI 318-14 (Eq. 8.10.3.) 8 8 13
End spans: (3.3 67.0) 44.1 93.8 ft-kips Interior span: (60.8 60.8) 33. 94 ft-kips The total design moments from the Equivalent Frame Method yield a static moment equal to that given by the Direct Design Method and no appreciable reduction can be realized. b. Distribute factored moments to column and middle strips: After the negative and positive moments have been determined for the slab-beam strip, the ACI code permits the distribution of the moments at critical sections to the column strips, beams (if any), and middle strips in accordance with the DDM. ACI 318-14 (8.11.6.6) Distribution of factored moments at critical sections is summarized in Table 6. End Span Interior Span Table 6 - Distribution of factored moments Slab-beam Strip Column Strip Middle Strip Moment (ft-kips) Percent Moment (ft-kips) Percent Moment (ft-kips) Exterior Negative 3.3 100 3.3 0 0 Positive 44.1 60 6.5 40 17.7 Interior Negative 67 75 50.3 5 16.7 Negative 60.8 75 45.6 5 15. Positive 33. 60 19.9 40 13...6. Flexural reinforcement requirements a. Determine flexural reinforcement required for strip moments The flexural reinforcement calculation for the column strip of end span exterior negative location is provided below. M u 3.3 ft-kips Use average d avg = 5.75 To determine the area of steel, assumptions have to be made whether the section is tension or compression controlled, and regarding the distance between the resultant compression and tension forces along the slab section (jd). In this example, tension-controlled section will be assumed so the reduction factor is equal to 0.9, and jd will be taken equal to 0.95d. The assumptions will be verified once the area of steel in finalized. Assume jd 0.95 d 5.46 14
Column strip width, b 14 1 / 84 Middle strip width, b 141 84 84 Mu 3.31000 As 1.31 f jd 0.9 60000 0.95 5.75 y Recalculate a for the actual A s = 1.31 : Af s y 1.31 60000 a 0.76 0.85 f ' b 0.854000 84 c a 0.76 c 0.35 0.85 1 0.003 0.003 t ( ) dt 0.003 ( ) 5.75 0.003 0.0501 0.005 c 0.35 Therefore, the assumption that section is tension-controlled is valid. Mu 3.31000 As 1.8 f ( d a / ) 0.9 60000 (5.75 0.76 / ) y Min As 0.018 847 1.06 in 1.8 ACI 318-14 (4.4.3.) Maximum spacing smax h 7 14 18 ACI 318-14 (8.7..) Provide 7 - #4 bars with A s = 1.40 and s = 84/7 = 1 s max Based on the procedure outlined above, values for all span locations are given in Table 7. Column Strip Middle Strip Column Strip Middle Strip Span Location Table 7 - Required Slab Reinforcement for Flexure [Equivalent Frame Method (EFM)] Mu (ft-kips) b * () d ** () As Req d for flexure (in ) End Span Min As (in ) Reinforcement Provided As Prov. for flexure (in ) Exterior Negative 3.3 84 5.75 1.8 1.06 7-#4 1.4 Positive 6.5 84 5.75 1.04 1.06 6-#4 1. Interior Negative 50.3 84 5.75.0 1.06 11-#4. Exterior Negative 0 84 5.75 0 1.06 6-#4 1. Positive 17.7 84 5.75 0.69 1.06 6-#4 1. Interior Negative 16.7 84 5.75 0.65 1.06 6-#4 1. Interior Span Positive 19.9 84 5.75 0.78 1.06 6-#4 1. Positive 13. 84 5.75 0.51 1.06 6-#4 1. b. Calculate additional slab reinforcement at columns for moment transfer between slab and column by flexure 15
The factored slab moment resisted by the column ( M f u ) shall be assumed to be transferred by flexure. Concentration of reinforcement over the column by closer spacing or additional reinforcement shall be used to resist this moment. The fraction of slab moment not calculated to be resisted by flexure shall be assumed to be resisted by eccentricity of shear. ACI 318-14 (8.4..3) Portion of the unbalanced moment transferred by flexure is f Mu ACI 318-14 (8.4..3.1) Where 1 f ACI 318-14 (8.4..3.) 1 ( / 3) b / b 1 b1 Dimension of the critical section determined in ACI 318, Chapter 8 (see Figure 5). b o measured in the direction of the span for which moments are b Dimension of the critical section 8 (see Figure 5). b o measured in the direction perpendicular to b 1 in ACI 318, Chapter b b = Effective slab width = c 3 h ACI 318-14 (8.4..3.3) Column Strip Table 8 - Additional Slab Reinforcement required for moment transfer between slab and column (EFM) Span Location M u * (ft-kips) γ f γ f M u (ft-kips) End Span Effective slab width, b b () d () A s req d within b b (in ) A s prov. For flexure within b b (in ) Exterior Negative 46.6 0.60 8.9 37 5.75 1.17 0.6 3-#4 Interior Negative 7.8 0.60 4.7 37 5.75 0.18 0.97 - *M u is taken at the centerline of the support in Equivalent Frame Method solution. Add l Reinf...7. Column design moments The unbalanced moment from the slab-beams at the supports of the equivalent frame are distributed to the support columns above and below the slab-beam in proportion to the relative stiffness of the support columns. Referring to Figure 9, the unbalanced moment at joints 1 and are: Joint 1= +46.6 ft-kips Joint = -84.0 + 76. = -7.8 ft-kips The stiffness and carry-over factors of the actual columns and the distribution of the unbalanced slab moments (M sc) to the exterior and interior columns are shown in Figure 10a. 16
Figure 10a - Column Moments (Unbalanced Moments from Slab-Beam) In summary: M col,exterior=.08 ft-kips M col,interior = 3.66 ft-kips The moments determined above are combined with the factored axial loads (for each story) and factored moments in the transverse direction for design of column sections. Figure 10b shows the moment diagrams in the longitudinal and transverse direction for the interior and exterior equivalent frames. Following the previous procedure, the moment values at the face of interior, exterior, and corner columns from the unbalanced moment values can be obtained. These values are shown in the following table. 17
Figure 10b Moment Diagrams (kips-ft) Mu kips-ft Column number (See Figure 10b) 1 3 4 Mux 3.66.08.04 1.39 Muy.3 1.8 1.49 6.79 3. Design of Interior, Edge, and Corner Columns This section includes the design of interior, edge, and corner columns using spcolumn software. The preliminary dimensions for these columns were calculated previously in section one. The reduction of live load per ASCE 7-10 will be ignored in this example. However, the detailed procedure to calculate the reduced live loads is explained in the wide-module Joist System example. 18
3.1. Determination of factored loads Interior Column (Column #1): Assume 4 story building Tributary area for interior column is ATributary P 4q A 40.193 5 194.5 kips u u Tributary M u,x = 3.66 ft-kips (see the previous Table) M u,y =.3 ft-kips (see the previous Table) (18 14) 5 ft Edge (Exterior) Column (Column #): Tributary area for interior column is ATributary P 4q A 40.193 16 97.3 kips u u Tributary M u,x =.08 ft-kips (see the previous Table) M u,y = 1.8 ft-kips (see the previous Table) (18 / 14) 16 ft Edge (Exterior) Column (Column #3): Tributary area for interior column is ATributary P 4q A 40.193 16 97.3 kips u u Tributary M u,x =.04 ft-kips (see the previous Table) M u,y = 1.49 ft-kips (see the previous Table) (18 14 / ) 16 ft Corner Column (Column #4): Tributary area for interior column is ATributary P 4q A 40.193 63 48.6 kips u u Tributary M u,x = 1.39 ft-kips (see the previous Table) M u,y = 6.79 ft-kips (see the previous Table) (18 / 14 / ) 63 ft The factored loads are then input into spcolumn to construct the axial load moment interaction diagram. 19
3.. Column Capacity Diagram (Axial-Moment Interaction Diagram) Interior Column (Column #1): 0
Edge Column (Column #): 1
Edge Column (Column #3):
Corner Column (Column #4): 3
4. Two-Way Slab Shear Strength Shear strength of the slab in the vicinity of columns/supports includes an evaluation of one-way shear (beam action) and two-way shear (punching) in accordance with ACI 318 Chapter. 4.1. One-Way (Beam action) Shear Strength ACI 318-14 (.5) One-way shear is critical at a distance d from the face of the column as shown in Figure 3. Figure 11 shows the factored shear forces (V u) at the critical sections around each column. In members without shear reinforcement, the design shear capacity of the section equals to the design shear capacity of the concrete: Vn Vc Vs Vc, Where: ( V s 0) ACI 318-14 (Eq..5.1.1) V f ' c cbwd ACI 318-14 (Eq..5.5.1) 1for normal weight concrete 4000 V c 0.75 1.0 (14 1) 5.75 91.64 kips 1000 Because Vc Vu at all the critical sections, the slab has adequate one-way shear strength. Figure 11 One-way shear at critical sections (at distance d from the face of the supporting column) 4
4.. Two-Way (Punching) Shear Strength ACI 318-14 (.6) Two-way shear is critical on a rectangular section located at d/ away from the face of the column as shown in Figure 5. a. Exterior column: The factored shear force (V u) in the critical section is computed as the reaction at the centroid of the critical section minus the self-weight and any superimposed surface dead and live load acting within the critical section (d/ away from column face). 1.7 18.88 Vu. 0.193 1.65 144 kips The factored unbalanced moment used for shear transfer, M unb, is computed as the sum of the joint moments to the left and right. Moment of the vertical reaction with respect to the centroid of the critical section is also taken into account. 18.88 5.99 16 / M 46.6 1.65( ) 37.8 kips-ft unb 1 For the exterior column in Figure 5, the location of the centroidal axis z-z is: moment of area of the sides about AB (18.8 5.7518.8 / ) cab 5.99 area of the sides 18.8 5.75 1.75 5.75 The polar moment J c of the shear perimeter is: b d 3 db 3 J 1 1 b 1 b d 1 c b dc c 1 1 AB AB 18.885.75 3 5.7518.88 3 18.88 J 18.88 5.75 5.99 1.75 5.75 5.99 14109 c 4 1 1 γ 1 γ 10.617 0.383 ACI 318-14 (Eq. 8.4.4..) v f The length of the critical perimeter for the exterior column: bo (16 5.75/ ) (16 5.75) 59.5 The two-way shear stress (v u) can then be calculated as: V γ M c v u v unb u AB b d J o c ACI 318-14 (R.8.4.4..3) 5
1.651000 0.383 (37.81 1000) 5.99 vu 137.0 59.5 5.75 14109 psi ' 4 v min 4, ', s ' c fc fc d f b c o ACI 318-14 (Table.6.5.) 4 405.75 vc min 41 4000, 1 4000, 1 4000 1 59.5 vc min 53, 79.5, 371 psi vc 0.75 53 189.7 psi 53 psi Since v c v u b. Interior column: at the critical section, the slab has adequate two-way shear strength at this joint. 1.7 1.7 Vu 4.3 6.4 0.193 50.07 144 kips M 84 76. 50.07(0) 7.8 unb kips-ft For the interior column in Figure 5, the location of the centroidal axis z-z is: b 1.75 c 1 AB 10.88 The polar moment J c of the shear perimeter is: b d 3 db 3 J 1 1 b 1 b d c b dc c 1 1 1 AB AB 1.755.75 3 5.751.75 3 J 1.75 5.750 1.75 5.755.99 40131 c 4 1 1 γ 1 γ 10.600 0.400 ACI 318-14 (Eq. 8.4.4..) v f The length of the critical perimeter for the interior column: bo (16 5.75) (16 5.75) 87 V γ M c v u v unb u AB b d J o c ACI 318-14 (R.8.4.4..3) 50.07 1000 0.400 7.811000 10.88 vu 110. psi 875.75 40131 6
' 4 v min 4, ', s ' c fc fc d f b c o ACI 318-14 (Table.6.5.) 4 405.75 vc min 41 4000, 1 4000, 1 4000 1 87 vc min 53, 79.5, 93.7 psi vc 0.75 53 189.7 psi 53 psi Since vc v at the critical section, the slab has adequate two-way shear strength at this joint. u c. Corner column: In this example, interior equivalent frame 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 of corner supports usually governs. Thus, the two-way shear strength for the corner column in this example will be checked for educational purposes. Same procedure is used to find the reaction and factored unbalanced moment used for shear transfer at the centroid of the critical section for the corner support for the exterior equivalent frame strip. 18.8818.88 Vu 13.18 0.193 1.70 144 kips 18.88 4.7 16 / M 5.75 1.70( ) 19.3 kips-ft unb 1 For the corner column in Figure 5, the location of the centroidal axis z-z is: moment of area of the sides about AB (18.8 5.7518.8 / ) cab 4.7 area of the sides 18.85.75 The polar moment J c of the shear perimeter is: b d 3 db 3 J 1 1 b 1 b d 1 c b dc c 1 1 AB AB 18.885.75 3 5.7518.88 3 18.88 J 18.88 5.75 5.99 18.88 5.75 4.7 8354 c 4 1 1 γ 1 γ 10.600 0.400 ACI 318-14 (Eq. 8.4.4..) v f Where: 1 f 1 ( / 3) b / 1 b ACI 318-14 (8.4..3.) 7
1 0.600 f 1 ( / 3) 18.88 /18.88 The length of the critical perimeter for the exterior column: b o (16 5.75 / ) (16 5.75 / ) 37.75 The two-way shear stress (v u) can then be calculated as: V γ M c v u v unb u AB b d J o c ACI 318-14 (R.8.4.4..3) 1.70 1000 0.400 (19.3 11000) 4.7 vu 110.6 37.75 5.75 8354 psi ' 4 v min 4, ', s ' c fc fc d f b c o ACI 318-14 (Table.6.5.) 4 05.75 vc min 41 4000, 1 4000, 1 4000 1 37.75 vc min 53, 79.5, 319. psi = 53 psi vc 0.75 53 189.7 psi Since vc v at the critical section, the slab has adequate two-way shear strength at this joint. u 5. Two-Way Slab Deflection Control (Serviceability Requirements) Since the slab thickness was selected based on the minimum slab thickness tables in ACI 318-14, the deflection calculations are not required. However, the calculations of immediate and time-dependent deflections are covered in this section for illustration and comparison with spslab model results. 5.1. Immediate (Instantaneous) Deflections The calculation of deflections for two-way slabs is challenging even if linear elastic behavior can be assumed. Elastic analysis for three service load levels (D, D + L sustained, D+L Full) is used to obtain immediate deflections of the two-way slab in this example. However, other procedures may be used if they result in predictions of deflection in reasonable agreement with the results of comprehensive tests. ACI 318-14 (4..3) The effective moment of inertia (I e) is used to account for the cracking effect on the flexural stiffness of the slab. I e for uncracked section (M cr > M a) is equal to I g. When the section is cracked (M cr < M a), then the following equation should be used: 3 3 Mcr Mcr I I 1 I I Ma M a e g cr g ACI 318-14 (Eq. 4..3.5a) 8
Where: M a = Maximum moment in member due to service loads at stage deflection is calculated. The values of the maximum moments for the three service load levels are calculated from structural analysis as shown previously in this document. These moments are shown in Figure 1. M cr = cracking moment. Figure 1 Maximum Moments for the Three Service Load Levels M cr fi r g 474.34 480 1 54.3 ft-kip y 3.5 11000 t f r = Modulus of rapture of concrete. ACI 318-14 (Eq. 4..3.5b) f r 7.5 f ' 7.51.0 4000 474.34 psi ACI 318-14 (Eq. 19..3.1) c I g = Moment of inertia of the gross uncracked concrete section I g 3 lh 480 1 1 3 14 1 7 h 7 yt 3.5 I cr = moment of inertia of the cracked section transformed to concrete. PCA Notes on ACI 318-11 (9.5..) The calculations shown below are for the design strip (frame strip). The values of these parameters for column and middle strips are shown in Table 9. As calculated previously, the exterior span frame strip near the interior support is reinforced with 17 #4 bars located at 1.5 along the section from the top of the slab. Figure 13 shows all the parameters needed to calculate the moment of inertia of the cracked section transformed to concrete. 9
E cs = Modulus of elasticity of slab concrete. Figure 13 Cracked Transformed Section E w 33 f 150 33 4000 3834 10 psi ACI 318-14 (19...1.a) 1.5 1.5 3 cs c c Es 9000000 n 7.56 PCA Notes on ACI 318-11 (Table 10-) E 3834000 cs b 141 B 6.54 na 7.56 170. s 1 PCA Notes on ACI 318-11 (Table 10-) db 11 5.75 6.54 11 kd 1.18 PCA Notes on ACI 318-11 (Table 10-) B 6.54 b( kd) I na d kd 3 3 cr s ( ) PCA Notes on ACI 318-11 (Table 10-) I cr 141 (1.18) 3 7.56 17 0. 5.75 1.18 69 3 4 The effective moment of inertia procedure described in the Code is considered sufficiently accurate to estimate deflections. The effective moment of inertia, I e, was developed to provide a transition between the upper and lower bounds of I g and I cr as a function of the ratio M cr/m a. For conventionally reinforced (nonprestressed) members, the effective moment of inertia, Ie, shall be calculated by Eq. (4..3.5a) unless obtained by a more comprehensive analysis. Ie shall be permitted to be taken as the value obtained from Eq. (4..3.5a) at midspan for simple and continuous spans, and at the support for cantilevers. ACI 318-14 (4..3.7) For continuous one-way slabs and beams. I e shall be permitted to be taken as the average of values obtained from Eq. (4..3.5a) for the critical positive and negative moment sections. ACI 318-14 (4..3.6) For the exterior span (span with one end continuous) with service load level (D+LL full): 3 3 M M cr cr I I 1 I, since M cr 54.3 ft-kips < M a = 64.17 ft-kips e M g M cr a a ACI 318-14 (4..3.5a) - Where I e is the effective moment of inertia for the critical negative moment section (near the support). 30
3 3 54.3 54.3 I e 480 1 69 3148 64.17 64.17 4 I I M M e g 4 480, since cr 54.3 ft-kips > a =34.5 ft-kips Where I e + is the effective moment of inertia for the critical positive moment section (midspan). Since midspan stiffness (including the effect of cracking) has a dominant effect on deflections, midspan section is heavily represented in calculation of I e and this is considered satisfactory in approximate deflection calculations. The averaged effective moment of inertia (I e,avg) is given by: I e, avg 0.85 I 0.15 I for end span PCA Notes on ACI 318-11 (9.5..4(1)) e I, 0.85 480 0.15 3148 4554 e avg Where: I e I e e 4 = The effective moment of inertia for the critical negative moment section near the support. = The effective moment of inertia for the critical positive moment section (midspan). For the interior span (span with both ends continuous) with service load level (D+LL full): 3 3 M M cr cr I I 1 I, since M cr 54.3 ft-kips < M a = 58.7 ft-kips e M g M cr a a ACI 318-14 (4..3.5a) 3 3 54.3 54.3 I e 480 1 69 3993 58.7 58.7 4 I I M M e g 4 480, since cr 54.3 ft-kips > a = 5.34 ft-kips The averaged effective moment of inertia (I e,avg) is given by: e, avg e e, l e, r I 0.70 I 0.15 I I for interior span PCA Notes on ACI 318-11 (9.5..4()) Ie, avg 0.70 480 0.15 3993 3993 4559 Where: 4 I el, = The effective moment of inertia for the critical negative moment section near the left support. I er, = The effective moment of inertia for the critical negative moment section near the right support. 31
Table 9 provides a summary of the required parameters and calculated values needed for deflections for exterior and interior equivalent frame. It also provides a summary of the same values for column strip and middle strip to facilitate calculation of panel deflection. Table 9 Averaged Effective Moment of Inertia Calculations Span Ext Int zone Left Ig, 4 Icr, 4 D Ma, ft-kip D + LLSus For Frame Strip D + Lfull 499-6.10-6.10-35.78 Mcr, k-ft D Ie, 4 Ie,avg, 4 D + LLSus Right 69-46.76-46.76-64.17 480 480 3148 480 54.3 Left 69-4.47-4.47-58.7 480 480 3993 D + Lfull 480 480 480 Midspan 465 4.95 4.95 34.5 480 480 480 Mid 465 18.47 18.47 5.34 480 480 480 Right 69-4.47-4.47-58.7 480 480 3993 D D + LLSus D + Lfull 480 480 4554 480 480 4559 Deflections in two-way slab systems shall be calculated taking into account size and shape of the panel, conditions of support, and nature of restraints at the panel edges. For immediate deflections two-way slab systems the midpanel deflection is computed as the sum of deflection at midspan of the column strip or column line in one direction (Δ cx or Δ cy) and deflection at midspan of the middle strip in the orthogonal direction (Δ mx or Δ my). Figure 14 shows the deflection computation for a rectangular panel. The average Δ for panels that have different properties in the two direction is calculated as follows: ( ) ( ) cx my cy mx PCA Notes on ACI 318-11 (9.5.3.4 Eq. 8) Figure 14 Deflection Computation for a rectangular Panel 3
To calculate each term of the previous equation, the following procedure should be used. Figure 15 shows the procedure of calculating the term Δ cx. same procedure can be used to find the other terms. For exterior span - service dead load case: Figure 15 Δ cx calculation procedure 4 wl frame, fixed PCA Notes on ACI 318-11 (9.5.3.4 Eq. 10) 384EI Where: c frame, averaged frame, fixed = Deflection of column strip assuing fixed end condition. w (0 150 7 / 1)(14) 1505 lb/ft E w 33 f 150 33 4000 3834 10 psi ACI 318-14 (19...1.a) 1.5 1.5 3 c c c I frame,averaged = The averaged effective moment of inertia (I e,avg) for the frame strip for service dead load case from Table 9 = 480 4 4 3 (1505)(18) (1) frame, fixed 0.039 3 384(383410 )(480) I frame, averaged c, fixed LDFc frame, fixed PCA Notes on ACI 318-11 (9.5.3.4 Eq. 11) Icg, Where LDF c is the load distribution factor for the column strip. The load distribution factor for the column strip can be found from the following equation: 33
LDF LDF c LDF l LDF R And the load distribution factor for the middle strip can be found from the following equation: LDF m 1 LDF c For the end span, LDF for exterior negative region (LDF L ), interior negative region (LDF R ), and positive region (LDF + L ) are 1.00, 0.75, and 0.60, respectively (From Table 6 of this document). Thus, the load distribution factor for the column strip for the end span is given by: 1.0 0.75 0.6 LDF c 0.738 I c,g = The gross moment of inertia (I g) for the column strip for service dead load = 401 4 480 c, fixed 0.738 0.039 0.057 401 ( M ) net, L frame cl, PCA Notes on ACI 318-11 (9.5.3.4 Eq. 1) K Where: ec cl, = Rotation of the span left support. ( M ) 6.3 ft-kips = Net frame strip negative moment of the left support. net, L frame K ec = effective column stiffness = 553.7 x 10 6 -lb (calculated previously). 6.31 1000 0.0006 rad 553.7 10 cl, 6 I l PCA Notes on ACI 318-11 (9.5.3.4 Eq. 14) g c, L c, L 8 I e frame Where: cl, = Midspan deflection due to rotation of left support. 34
I I g e frame = Gross-to-effective moment of inertia ratio for frame strip. 181 480 cl, 0.0006 0.0161 8 480 ( M ) (46.76 4.47) 11000 0.00010 rad net, R frame cr, 6 Kec 553.7 10 Where cr, = rotation of the span right support. ( M ) = Net frame strip negative moment of the right support. net, R frame I l g 181 480 cr, cr, 0.00010 0.007 8 I 8 480 e frame Where: cr, = Midspan delfection due to rotation of right support. cx fixed PCA Notes on ACI 318-11 (9.5.3.4 Eq. 9) cx, cx, R cx, L 0.057 0.007 0.0161 0.076 cx Following the same procedure, Δ mx can be calculated for the middle strip. This procedure is repeated for the equivalent frame in the orthogonal direction to obtain Δ cy, and Δ my for the end and middle spans for the other load levels (D+LL sus and D+LL full). Assuming square panel, Δ cx = Δ cy= 0.076 and Δ mx = Δ my= 0.039 The average Δ for the corner panel is calculated as follows: ( ) ( ) cx my cy mx cx my cy mx 0.076 0.039 0.115 35
Table 10 Immediate (Instantaneous) Deflections in the x-direction Column Strip Middle Strip Span LDF Δ frame-fixed, Δ c-fixed, θ c1, rad D θ c, rad Δθ c1, Δθ c, Δ cx, LDF Δ frame-fixed, Ext 0.738 0.0386 0.0570 0.00059 0.00010 0.0161 0.007 0.076 0.6 0.0386 0.00 0.00059 0.00010 0.0161 0.007 0.039 Int 0.675 0.0386 0.051-0.00010-0.00010-0.007-0.007 0.047 0.35 0.0386 0.051-0.00010-0.00010-0.007-0.007 0.00 Δ m-fixed, θ m1, rad D θ m, rad Δθ m1, Δθ m, Δ mx, Span LDF Δ frame-fixed, Δ c-fixed, θ c1, rad D+LL sus θ c, rad Δθ c1, Δθ c, Δ cx, LDF Δ frame-fixed, Ext 0.738 0.0386 0.0570 0.00059 0.00010 0.0161 0.007 0.076 0.6 0.0386 0.00 0.00059 0.00010 0.0161 0.007 0.039 Int 0.675 0.0386 0.051-0.00010-0.00010-0.007-0.007 0.047 0.35 0.0386 0.051-0.00010-0.00010-0.007-0.007 0.00 Δ m-fixed, θ m1, rad D+LL sus θ m, rad Δθ m1, Δθ m, Δ mx, Span LDF Δ frame-fixed, Δ c-fixed, θ c1, rad D+LL full θ c, rad Δθ c1, Δθ c, Δ cx, LDF Δ frame-fixed, Ext 0.738 0.0559 0.085 0.0008 0.00014 0.0333 0.00388 0.110 0.6 0.0559 0.093 0.0008 0.00014 0.0333 0.00388 0.017 Int 0.675 0.0558 0.0753-0.00014-0.00014-0.00387-0.00387 0.068 0.35 0.0558 0.0363-0.00014-0.00014-0.00387-0.00387 0.009 Δ m-fixed, θ m1, rad D+LL full θ m, rad Δθ m1, Δθ m, Δ mx, Span LDF LL Δ cx, Ext 0.738 0.034 0.6 0.017 Int 0.675 0.01 0.35 0.009 LDF LL Δ mx, 36
From the analysis in the transverse direction the deflection values below are obtained: For DL loading case: my cy For DL+LL sust loading case: my cy For DL+LL full loading case: my cy These values for the x-direction are shown in Table 10. Then, the total midpanel deflection is calculated by combining the contributions of the column and middle strip deflections from the X and Y directions: cx my cy mx PCA Notes on ACI 318-11 (9.5.3.4 Eq. 8) 5.. Time-Dependent (Long-Term) Deflections (Δlt) The additional time-dependent (long-term) deflection resulting from creep and shrinkage (Δ cs) may be estimated as follows: ( ) PCA Notes on ACI 318-11 (9.5..5 Eq. 4) cs sust Inst The total time-dependent (long-term) deflection is calculated as: ( ) ( ) (1 ) [( ) ( ) ] CSA A3.3-04 (N9.8..5) total lt sust Inst total Inst sust Inst Where: ( ) Immediate (instantaneous) deflection due to sustained load, sust Inst 1 50 ' ACI 318-14 (4..4.1.1) 37
( ) Time-dependent (long-term) total delfection, total lt ( ) Total immediate (instantaneous) deflection, total Inst For the exterior span =, consider the sustained load duration to be 60 months or more. ACI 318-14 (Table 4..4.1.3) ' = 0, conservatively. 1500 0.076 0.15 cs 0.076 1 0.110 0.076 0.6 total lt Table 11 shows long-term deflections for the exterior and interior spans for the analysis in the x-direction, for column and middle strips. Table 11 - Long-Term Deflections Column Strip Span (Δ sust) Inst, in λδ Δ cs, in (Δ total) Inst, in (Δ total) lt, in Exterior 0.076.000 0.15 0.110 0.61 Interior 0.047.000 0.095 0.068 0.16 Middle Strip Exterior 0.039.000 0.078 0.056 0.134 Interior 0.00.000 0.041 0.09 0.069 6. Computer Program Solution spslab program utilizes the Equivalent Frame Method described and illustrated in details here for modeling, analysis and design of two-way concrete floor slab systems. spslab uses the exact geometry and boundary conditions provided as input to perform an elastic stiffness (matrix) analysis of the equivalent frame taking into account the torsional stiffness of the slabs framing into the column. It also takes into account the complications introduced by a large number of parameters such as vertical and torsional stiffness of transverse beams, the stiffening effect of drop panels, column capitals, and effective contribution of columns above and below the floor slab using the of equivalent column concept (ACI 318-14 (R8.11.4)). spslab Program models the equivalent frame as a design strip. The design strip is, then, separated by spslab into column and middle strips. The program calculates the internal forces (Shear Force & Bending Moment), moment 38
and shear capacity vs. demand diagrams for column and middle strips, instantaneous and long-term deflection results, and required flexural reinforcement for column and middle strips. The graphical and text results are provided below for both input and output of the spslab model. 39
40
41
4
43
44
45
46
47
48
49
50
51
5
53
54
55
56
57
7. Summary and Comparison of Two-Way Slab Design Results Table 1 - Comparison of Moments obtained from Hand (EFM) and spslab Solution Hand (EFM) spslab Exterior Span Exterior Negative * 3.0 3.66 Column Strip Positive 6.50 6.87 Interior Negative * 50.30 50.1 Exterior Negative * 0.00 0.0 Middle Strip Positive 17.70 17.91 Interior Negative * 16.70 16.74 Interior Span Column Strip Interior Negative * 45.60 45.47 Positive 19.90 19.90 Middle Strip Interior Negative * 15.0 15.16 Positive 13.0 13.7 * negative moments are taken at the faces of supports Table 13 - Comparison of Reinforcement Results with Hand and spslab Solution Span Location Reinforcement Provided for Flexure Additional Reinforcement Provided for Unbalanced Moment Transfer* Total Reinforcement Provided Column Strip Middle Strip Column Strip Middle Strip Exterior Negative Hand spslab Hand spslab Hand spslab Exterior Span 7-#4 7-#4 3-#4 3-#4 10-#4 10-#4 Positive 6-#4 6-#4 n/a n/a 6-#4 6-#4 Interior Negative Exterior Negative 11-#4 11-#4 --- --- 11-#4 11-#4 6-#4 6-#4 n/a n/a 6-#4 6-#4 Positive 6-#4 6-#4 n/a n/a 6-#4 6-#4 Interior Negative 6-#4 6-#4 n/a n/a 6-#4 6-#4 Interior Span Positive 6-#4 6-#4 n/a n/a 6-#4 6-#4 Positive 6-#4 6-#4 n/a n/a 6-#4 6-#4 *In the EFM, the unbalanced moment (M sc, M unb) at the support centerline is used to determine the value of the additional reinforcement as compared with DDM using the moments at the face of support. 58
Table 14 - Comparison of One-Way (Beam Action) Shear Check Results Using Hand and spslab Solution Span Vu, kips xu *, φvc, kips Hand spslab Hand spslab Hand spslab Exterior 3.3 3.3 1.15 1.15 91.64 91.64 Interior 1. 1. 16.85 16.85 91.64 91.64 * x u calculated from the centerline of the left column for each span Table 15 - Comparison of Two-Way (Punching) Shear Check Results Using Hand and spslab Solution b 1, b, b o, A c, V u, kips v u, psi Support Hand spslab Hand spslab Hand spslab Hand spslab Hand spslab Hand spslab Exterior 18.88 18.88 1.75 1.75 59.5 59.5 34.00 34.13 1.65.79 63.3 66.6 Interior 1.75 1.75 1.75 1.75 87 87 500.00 500.5 50.07 50.07 100.1 100.1 c AB, J c, 4 γ v M unb, kips-ft v u, psi φv c, psi Support Hand spslab Hand spslab Hand spslab Hand spslab Hand spslab Hand spslab Exterior 5.99 5.99 14109 14110 0.383 0.383 37.8 37. 137.0 139. 189.7 189.7 Interior 10.88 10.88 40131 40131 0.400 0.400-7.8-7.7 110. 110.1 189.7 189.7 Table 16 - Comparison of Immediate Deflection Results Using Hand and spslab Solution () Column Strip Span D D+LLsus D+LLfull LL Hand spslab Hand spslab Hand spslab Hand spslab Exterior 0.076 0.07 0.076 0.07 0.110 0.103 0.034 0.031 Interior 0.047 0.045 0.047 0.045 0.068 0.064 0.01 0.019 Middle Strip Span D D+LLsus D+LLfull LL Hand spslab Hand spslab Hand spslab Hand spslab Exterior 0.039 0.038 0.039 0.038 0.056 0.054 0.017 0.016 Interior 0.00 0.019 0.00 0.019 0.09 0.07 0.009 0.008 Table 17 - Comparison of Time-Dependent Deflection Results Using Hand and spslab Solution Column Strip Span λδ Δcs, Δtotal, Hand spslab Hand spslab Hand spslab Exterior.0.0 0.15 0.145 0.61 0.48 Interior.0.0 0.095 0.089 0.16 0.153 Middle Strip Span λδ Δcs, Δtotal, Hand spslab Hand spslab Hand spslab Exterior.0.0 0.078 0.076 0.134 0.19 Interior.0.0 0.041 0.038 0.069 0.065 59