Lecture-06 Analysis and Design of Slab Systems

Lecture-06 Analysis and Design of Slab Systems By: Prof Dr. Qaisar Ali Civil Engineering Department UET Peshawar drqaisarali@uetpeshawar.edu.pk www.drqaisarali.com 1 Topics Addressed Organization of the lecture Analysis & design of one way slab system Analysis & design of one way joist system Analysis & design of two way slab system with beams Analysis & design of two way slab system without beams (flat plate & flat slabs) Analysis & design of two way joist system 2 1

Organization of the Lecture Analysis & design of one way slab system The ACI code approximate method of analysis called as strip method is used for the analysis of one way slabs. This topic has already been covered in BSc lectures 3 & 4, Please download the same from the website. Analysis & design of one way joist system This topic will be covered in this lecture. 3 Organization of the Lecture Analysis & design of two way slab system with beams The ACI code approximate method of analysis called as moment coefficient method is generally used for the analysis of two way slabs supported on stiff beams or walls. This topic has been covered in BSc lecture 4, please download the same from the website. 4 2

Organization of the Lecture Analysis & design of two way slab system without beams (flat plate & flat slabs) The ACI code approximate method of analysis called as Direct Design Method (DDM) is used for the analysis of flat plates and flat slabs. This topic has been covered in BSc lecture 5. The students can download the same from the website. However this topic will also be quickly discussed in this lecture. Please note that DDM can also be used for the analysis of two way slabs with beams. However as the application of this method to such systems is relative difficult therefore these systems are generally analyzed using moment coefficient method instead of DDM, provided that beams are relatively stiff. 5 Organization of the Lecture Analysis & design of two way joist system This topic will be discussed in this lecture. 6 3

Organization of the Lecture Therefore in this lecture, only the following topics will be discussed. Analysis & design of one way joist system Analysis & design of two way slab system without beams (flat plate & flat slabs) Analysis & design of two way joist system 7 Analysis & design of one way joist system 8 4

Contents General Characteristics Basic Steps Serviceability Requirements 9 General Joist: T-beams called joists are formed by creating void spaces in what otherwise would be a solid slab. Joist Construction: Joist construction consists of a monolithic combination of regularly spaced ribs and a top slab (T beam or Joist) arranged to span in one direction or two orthogonal directions. Rib 10 5

General ACI code 8.11 contains provisions for joist construction aimed at facilitating the analysis and design of joist system as compared to regular slab beam system. A structural system will be however called as joist system if the pan width (clear spacing between ribs) is less than or equal to 30 inches (ACI 8.11.3). 11 General If the system does not fulfill the requirements of joist system then it shall be designed as regular slab beam system which means that slab shall be designed for flexure and beam shall be designed for flexure and shear. 12 6

Characteristics Suitable for long spans Economical range: 30 ft 50 ft 13 Characteristics Pan voids reduce dead load. Electrical/mechanical equipment can be placed between joists. Clear, unobstructed tenant space. Less effect of vibrations due to stiffer system. Standard forms for the void spaces between ribs are either 20 or 30 inches wide, and 8, 10, 12, 16, or 20 inches deep. 14 7

Characteristics Economical for buildings such as apartment houses, hotels, and hospitals, where the live loads are fairly small and the spans comparatively long. Forms are tapered in cross section generally at a slope of 1 to 12 to facilitate removal. 15 Basic Steps for Structural Design Step No. 01 (Sizes): Sizes of all structural and non structural elements are decided. Step No. 02 (Loads): Loads on structure are determined based on occupational characteristics and functionality. Step No. 03 (Analysis): Effect of loads are calculated on all structural elements. Step No. 04 (Design): Structural elements are designed for the respective load effects following code provisions. 16 8

Basic Steps for Structural Design Sizes: Depth of joist: Same table as used for one way slabs can be used to find depth of joist. l = Span length (same as for solid one way slab) The depth of the joist includes the slab depth. 17 Basic Steps for Structural Design Sizes: Width of joist: Minimum width of rib = Depth of rib/3.5 but not less than 4. Depth of slab Minimum slab thickness = clear distance between ribs/12 but not less than 2. 18 9

Basic Steps for Structural Design Loads: One way Joist systems are usually designed for gravity loading (U = 1.2D + 1.6L). The joist is analyzed for load (per running foot) over a width equal to the center to center spacing between the joists as shown: b b 19 Basic Steps for Structural Design Analysis: For the purpose of analysis, a single T shaped joist is considered. 20 10

Basic Steps for Structural Design Analysis: ACI approximate analysis is applicable. Uniformly distributed load (L/D 3) 1.2l n l n Prismatic members Two or more spans 21 Basic Steps for Structural Design Analysis: w u Integral with support l n l n Simple support l n Spandrel support Column support 1/14 1/16 1/11 1/24 1/10* 1/11 1/11 1/10* 0 1/16 *1/9 (2 spans) * 1/12 (for all spans with l n < 10 ft) Note: For simply supported slab, M = w u l 2 /8, where l = span length (ACI 8.7). Positive Moment x w u l n 2 Negative Moment x w u l n 2 22 11

Basic Steps for Structural Design Design: Design of joist for flexure: Design of one way joist is just like design of a beam. 23 Basic Steps for Structural Design Design: Flexural Reinforcement Placement in joist Reinforcement for the joists usually consists of two bars in the positive bending region, with one bar discontinued where no longer needed or bend up to provide a part of negative steel requirement over the supporting girder. One way joists are generally proportioned with the concrete providing all of the shear strength, with no shear reinforcement used. 24 12

Basic Steps for Structural Design Design: Flexural Reinforcement Placement in joist Slab Main Reinforcement (located at mid depth): As required by flexural demand but ACI 7.12 (shrinkage reinf.) For Joist negative bending: As required by flexural demand ACI 10.5.1 (Minimum reinforcement for flexural members) For Joist positive bending: As required by flexural demand but ACI 10.5.1 (Minimum reinforcement for flexural members) 25 Basic Steps for Structural Design Design: Flexural Reinforcement Placement in joist Skin reinforcement ACI 10.6.7 states that if the effective depth d of a beam or joist exceeds 36 inches, longitudinal skin reinforcement shall be provided as per ACI section 10.6.7. 26 13

Basic Steps for Structural Design Design: Maximum Spacing Requirement for flexural reinforcement in slab If spacing is obtained by flexural demand than compare spacing with: Least of 3h or 18 (ACI 7.6.5) If shrinkage reinforcement is governing than compare spacing with: Least of 5h or 18 (ACI 7.12.2.2) 27 Basic Steps for Structural Design Design: Design for Shear If V u ΦV c, then shear capacity shall be increased either by increasing rib depth or width or by providing single legged shear reinforcement. If V u < ΦV c the section is safe against shear. Even minimum reinforcement as required for the beams with [ΦV c /2 < V u < ΦV c ] is not required as per ACI 11.5.5.1. 28 14

Basic Steps for Structural Design Design: Design for Shear ACI 11.5.5.1: Minimum shear reinforcement A minimum area of shear reinforcement shall be provided in all reinforced concrete flexural members (prestressed and nonprestressed) where factored shear force V u exceeds one-half the shear strength provided by concreteφv c, except: (a) Slabs and footings; (b) Concrete joist construction defined by 8.11; (c) Beams with total depth not greater than 10 inches., 29 Basic Steps for Structural Design Design: Design for Shear For joist construction, contribution of concrete to shear strength V c shall be permitted to be 10 percent more than that specified in Chapter 11. Critical shear demand section shall be at a distance d from the face of the support. 30 15

Determine the required depth and reinforcement for the oneway joist system shown next. The joists are 6 inches wide and are spaced 36 inches c/c. The slab is 3.5 inches thick. f c = 4000 psi ; f y = 60,000 psi Service Live Load (LL) = 60 psf Superimposed Dead Load (SDL) = 64 psf Width of spandrel beams = 20 inches ; Width of interior beams = 36 inches Columns: interior = 18 18 inches ; exterior = 16 16 inches Story height (typical) = 13 ft 31 Plan View Section A-A 32 16

Solution: Step No 01: Sizes l/18.5 = 30 12/18.5 = 19.5 ; Use 19.5 deep joist (16 + 3.5 ) d = 19.5 0.75 clear cover ½ bar dia. (Assuming #8 bar) = 18.25 Width of the joist = 6 > (rib depth/ 3.5 = 16/3.5 = 4.57 ), OK. 33 Solution: Step No 02: Loads Area of rib (A) = 6 16 = 96 in 2 Dead Load of rib = Aγ c = (96/144) 0.15 = 0.10 kip/ft b =36 6 16 3.5 19.5 Per foot dead load of slab = 0.15 (3.5 36)/144 = 0.131 kip/ft Per foot superimposed dead load on slab = 0.064 36/12 = 0.192 kip/ft Per foot live load on 36 width = 0.060 36/12 = 0.18 kip/ft w u = 1.2DL+1.6LL=1.2 (0.10+0.131+0.192)+1.6 0.18 = 0.795 kip/ft 34 17

Solution: Step No 03: Analysis w u Integral with support l n l n l n Spandrel support 1/14 1/16 1/16 1/24 1/10 1/11 1/11 1/11 1/11 Positive Moment x w u l n 2 Negative Moment x w u l n 2 35 Solution: Step No 03: Analysis 0.795 kip/ft 10.93 27.5 ' 27 ' 12.34 V u,ext = 9.72 12.34 V u,int = 11.18 SFD (kip) 42.94 12.57 36.22 12.34 BMD (k-ft) 25.05 60.12 52.68 52.68 52.68 36 18

Solution: Step No 04: Design Design of joist for shear: V u, ext = 9.72 kips ; V u, int = 11.18 kips ΦV c = 1.10Φ2 (f c )bd = 1.10 0.75 2 (4000) 6 18.25/1000 = 11.42 kips ΦV c > V u, ext and V u, int,ok 37 Solution: Step No 04: Design Design of joist for flexure: Design for positive moment (42.94 kip-ft) b eff = 36 (c/c spacing between joists) Check if joist is to be designed as rectangular beam or T-beam. Assume a = h f = 3.5 A s = M u / {Φf y (d a/2)}=(42.94 12)/ {0.9 60 (18.25 3.5/2)}= 0.57 in 2 Re-calculate a : a = A s f y / (0.85f c b eff ) = 0.57 60/ (0.85 4 36) = 0.27 < h f Therefore design joist as rectangular beam. After trials, A s = 0.53 in 2 (2 #5 bars, 0.6 in 2 ) 38 19

Solution: Step No 04: Design Design of joist for flexure: Design for exterior negative moment (25.05 kip-ft) b = 6 (bottom width of joist) ; h = 19.5 ; d = 18.25 After trials, A s = 0.31 in 2 A smin = (3 (f c )/f y )bd (200/f y )bd = 0.365 in 2, governs Therefore, A s = 0.365 in 2 Distribute bars uniformly in top slab: A s = 0.365/ 3 = 0.121 in 2 /ft (#3 @ 10 c/c) 39 Solution: Step No 04: Design Design of joist for flexure: Design for interior negative moment (60.12 kip-ft) b = 6 (bottom width of joist) ; h = 19.5 ; d = 18.25 After trials, A s = 0.78 in 2 A smin = (3 (f c )/f y )bd (200/f y )bd = 0.365 in 2 Therefore, A s = 0.78 in 2 Distribute bars uniformly in top slab: A s = 0.78/ 3 = 0.26 in 2 /ft (#3 @ 5 c/c) 40 20

Solution: Step No 04: Design Design of slab for flexure: l n = 2.5 The slab reinforcement normal to the ribs is often located at mid-depth of the slab to resist both positive and negative moments. w u = 1.2 (0.044+0.064)+1.6*0.06 =0.23 ksf Over the support, negative moment coefficient of 1/12 will be used. M u = w u l n2 /12 = 0.23 2.5 2 /12 = 0.12 ft-kip (1.44 in-kip) After trials, As = 0.015 in 2 /ft A smin = 0.0018 12 3.5 = 0.08 in 2 /ft, governs (#3 @ 16.5 c/c) 41 Solution: Step No 04: Design Design of slab for flexure: Maximum spacing allowed for temperature steel reinforcement : 5h f = 5 3.5 = 17.5 18 Finally provide #3 @ 16 c/c. 42 21

Solution: Step No 05: Drafting A #3 @ 5 c/c #3 @ 10 c/c l n /4 = 6-9 l n /3 = 9-3 l n /3 = 9-0 #3 @ 16 c/c 3.5 Beam A l n = 27.5 l n = 27 2 #5 bars in joist 43 Solution: Step No 05: Drafting #3 @ 16 c/c, at mid depth (shrinkage & temperature) #3 @ 10 c/c (Main ve) Section A-A 2 #5 bars (Main +ve) 44 22

Analysis and Design of Two-way Slab System without Beams (Flat Plates and Flat Slabs) 45 There are two sections under this topic Section I: Flexural Analysis of Two-Way Slab System without Beams (Direct Design Method) Section II: Shear Design for Two-Way Slab System without Beams (Flat Plates and Flat Slabs) 46 23

Contents of Section-1 Introduction to Direct Design Method (DDM) Steps in Direct Design Method (DDM). Detailing of Flexural Reinforcement Summary of Direct Design Method (DDM) 47 Introduction to DDM General: In the moment coefficient method of analysis used for slabs with beams, the system is analyzed panel by panel. In DDM, frames rather than panels are analyzed. Exterior Frame Interior Frame Interior Frame Exterior Frame 48 24

Introduction to DDM General: For complete analysis of slab system frames, are analyzed in E- W and N-S directions. E-W N-S Frames 49 Introduction to DDM Limitations (ACI 13.6.1): Though DDM is useful for analysis of slabs, specially without beams, the method is applicable with some limitations as discussed next. 50 25

Introduction to DDM Limitations (ACI 13.6.1): Uniformly distributed loading (L/D 2) l 1 2l 1 /3 l 1 Three or more spans l 2 Rectangular slab panels (2 or less:1) Column offset l 2 /10 51 Step 01: Sizes Steps in DDM ACI table 9.5 (c) are used for finding the slab thickness. 52 26

Step 02: Loads Steps in DDM The slab load is calculated in usual manner. 53 Step 03: Analysis Steps in DDM Step I: Marking E-W Frame (Interior Frame) Panel Centerline Col Centerline Panel Centerline Interior Frame l 1 l 2 Half width of panel on one side Half width of panel on other side According to ACI 13.6.2.3: Where the transverse span of panels on either side of the centerline of supports varies, l 2 shall be taken as the average of adjacent transverse spans. 54 27

Step 03: Analysis Steps in DDM Step I: Marking E-W Frame (Exterior Frame) Exterior Frame l 2 h 2 /2 Half width of panel on one side l 1 l 2 = Panel width/2 +h 2 /2 55 Step 03: Analysis Steps in DDM Step II: a) Marking Column Strip (For Interior Frame) b) Marking Middle Strip (For Interior Frame) Half Column strip CS/2 CS/2 l 2 M.S/2 C.S M.S/2 CS/2 = Least of l 1 /4 or l 2 /4 l n l 1 56 28

Steps in DDM Step 03: Analysis Step II: a) Marking Column Strip (For Interior Frame) b) Marking Middle Strip (For Interior Frame) Half Column Strip 5 5 l 2 5 10 For l 1 = 25 and l 2 = 20, CS 5 and MS widths are given as follows: l n CS/2 = Least of l 1 /4 or l 2 /4 l 2 /4 = 20/4 = 5 l 1 57 Steps in DDM Step 03: Analysis Step II: a) Marking Column Strip (For Exterior Frame) b) Marking Middle Strip (For Exterior Frame) h/2 CS Min. Panel Width/4 MS l 2 CS = Min. Panel Width/4 + ½ (Col. Size) MS = l 2 - CS For Given Frame: CS and MS widths are given as follows: l 2 = ½ (20) + ½ (14/12) = 10.58 CS = (20/4) + ½ (14/12) = 5.58 MS = 10.58 5.58 = 5 l n l 1 58 29

Step 03: Analysis Steps in DDM Step III: Calculate Static Moment (M o ) for interior span of frame. M o = w u l 2 l n 2 8 M o Span of frame l n l 2 59 Step 03: Analysis Steps in DDM Step IV: Longitudinal Distribution of Static Moment (M o ). M = 0.65M o M + M + = 0.35M o M M 60 30

Step 03: Analysis Steps in DDM Step V: Lateral Distribution to column and middle strips. M = 0.65M o M + = 0.35M o 0.60M + 0.75M 0.75M 61 Step 03: Analysis Steps in DDM Step VI: Calculate Static Moment (M o ) for exterior span of frame. M o = w u l 2 l n 2 8 M o Span of frame l n l 2 62 31

Step 03: Analysis Steps in DDM Step VII: Longitudinal distribution of static moment (M o ). M ext = 0.26M o M ext+ = 0.52M o M ext + M ext M int M int- = 0.70M o 63 Step 03: Analysis Steps in DDM Step VIII: Lateral Distribution to column and middle strips. M ext- = 0.26M o 1.00M ext 0.75M int M ext+ = 0.52M o 0.60M ext + M int- = 0.70M o 64 32

Steps in DDM Step 03: Analysis (Summary) Distribution factors for longitudinal & Lateral Distribution to column and middle strips. M ext- = 0.26M o M ext+ = 0.52M o M int- = 0.70M o 0.60M ext + 0.75M 0.75M 1.00M ext 0.75M int 0.60M + M - = 0.65M o M + = 0.35M o 65 Detailing of flexural reinforcement Maximum spacing and minimum reinforcement requirement for slab systems without beams: Maximum spacing (ACI 13.3.2): s max = 2 h f in each direction. Minimum Reinforcement (ACI 7.12.2.1): A smin = 0.0018 bh f for grade 60. A smin = 0.002 bh f for grade 40 and 50. 66 33

Detailing of flexural reinforcement Reinforcement placement: In case of two way slabs supported on beams, short-direction bars are normally placed closer to the top or bottom surface of the slab, with the larger effective depth because of greater moment in short direction. 67 Detailing of flexural reinforcement Reinforcement placement: However in the case of flat plates/slabs, the long-direction negative and positive bars, in both middle and column strips, are placed closer to the top or bottom surface of the slab, respectively, with the larger effective depth because of greater moment in long direction. 68 34

Detailing of flexural reinforcement Splicing: ACI 13.3.8.5 requires that all bottom bars within the column strip in each direction be continuous or spliced with length equal to 1.0 l d. Splicing shall be provided in location where yielding is not expected. 69 Detailing of flexural reinforcement Continuity of bars: At least two of the column strip bars in each direction must pass within the column core and must be anchored at exterior supports (ACI 13.3.8.5). 70 35

Detailing of flexural reinforcement Standard Bar Cut off Points (Practical Recommendation) for column and middle strips both. 71 Detailing of flexural reinforcement Reinforcement at Exterior Corners: Reinforcement should be provided at exterior corners in both the bottom and top of the slab, for a distance in each direction from the corner equal to one-fifth the longer span of the corner panel as shown in figure. The positive and negative reinforcement should be of size and spacing equivalent to that required for maximum positive moments (per foot of width) in the panel (ACI 13.3.6) l /5 l /5 l = longer clear span 72 36

Summary of Direct Design Method Decide about sizes of slab and columns. The slab depth can be calculated from ACI table 9.5 (c). Find Load on slab (w u = 1.2DL + 1.6LL) On given column plan of building, decide about location and dimensions of all frames (exterior and interior) For a particular span of frame, find static moment (M o = w u l 2 l n2 /8). 73 Summary of Direct Design Method Find longitudinal distribution of static moment: Exterior span (M ext - = 0.26M o ; M ext + = 0.52M o ; M int - = 0.70M o ) Interior span (M int - = 0.65M o ; M int + = 0.35M o ) Find lateral Distribution of each longitudinal moment: 100 % of M ext goes to column strip 60 % of M ext + and M int+ goes to column strip 75 % of M int goes to column strip The remaining moments goes to middle strips Design and apply reinforcement requirements (s max = 2h f ) 74 37

Analyze the flat plate shown below using DDM. The slab supports a live load of 144 psf. All columns are 14 square. Take f c = 4 ksi and f y = 60 ksi. See BSc Lecture 5 for solution of this example. 25 25 25 25 20 20 20 75 Contents of Section-II General Various Design Options 76 38

General Punching Shear in Flat Plates Punching shear occurs at column support points in flat plates and flat slabs. Shear crack Punch Out 77 General Critical Section for Punching Shear 78 39

General Critical Section for Shear Design In shear design of flat plates, the critical section is an area taken at a distance d/2 from all face of the support. Column Critical perimeter Slab Tributary Area, A t d/2 d/2 Slab thickness (h) d = h cover 79 General Punching Shear: Critical Perimeter, b o c 1,S d/2 c 1,L d/2 b o = 2(c 1,S + d) + 2(c 1,L + d) 80 40

General Punching Shear: Critical Perimeter, b o c 1,S d/2 d/2 c 1,L d/2 b o = 2(c 1,S + d/2) + (c 1,L + d) 81 General Punching Shear: Critical Perimeter, b o c 1,L d/2 c 1,S d/2 b o = (c 1,S + d/2) + (c 1,L + d/2) 82 41

General Punching Shear Demand (V u ): For Square Column Critical Perimeter, b o : b o = 4(c + d) Area Contributing to Load (Excluding Area of b o ), A t : l 2 A t = (l 1 l 2 ) (c + d) 2 / 144 [l 1 & l 2 are in ft. units and c & d are in inches] Punching Shear Demand: l 1 V u = W u A t 83 General Capacity of Slab in Punching Shear: ΦV n = ΦV c + ΦV s ΦV c is least of: Φ4 (f c )b o d Φ(2 + 4/β c ) (f c )b o d Φ{(α s d/b o +2} (f c )b o d β c = longer side of column/shorter side of column α s = 40 for interior column, 30 for edge column, 20 for corner columns 84 42

Various Design Options When ΦV c V u (Φ = 0.75): Nothing is required. When ΦV c < V u, then we need to increase the punching shear capacity of the slab. Punching shear capacity of the flat plates can be increased by either of the following ways: i. Increasing d,depth of slab: This can be done by increasing the slab depth as a whole or in the vicinity of column (Drop Panel) ii. iii. Increasing b o, critical shear perimeter: This can be done by increasing column size as a whole or by increasing size of column head (Column capital) Increasing f c (high Strength Concrete) 85 Various Design Options And/ or provide shear reinforcement (ΦV s ) in the form of: Integral beams Bent Bars Shear heads Shear studs 86 43

Various Design Options Drop Panels (ACI 9.5.3.2 and 13.3.7.1): 87 Various Design Options Column Capital: ACI 13.1.2 requires the column capital should be oriented no greater than 45 0 to the axis of the column. ACI 6.4.6 requires that the capital concrete be placed at the same time as the slab concrete. As a result, the floor forming becomes considerably more complicated and expensive. The increased perimeter can be computed by equating V u to ΦV c and simplifying the resulting equation for b 0 88 44

Various Design Options Integral Beam and Bent Bars: In case of integral beam or bent bar reinforcement following must be satisfied. ACI 11.12.3 requires the slab effective depth d to be at least 6 in., but not less than 16 times the diameter of the shear reinforcement. When bent bars and integral beams are to be used, ACI 11.12.3 reduces ΦV c by 2 89 Various Design Options Integral Beams Integral Beams require the design of two main components: i. Vertical stirrups ii. Horizontal bars radiating outward from column faces. Vertical Stirrups Horizontal Bars 90 45

Various Design Options Integral Beams (Vertical Stirrups) Vertical stirrups are used in conjunction with supplementary horizontal bars radiating outward in two perpendicular directions from the support to form what are termed integral beams contained entirely within the slab thickness. In such a way, critical perimeter is increased Increased Critical Perimeter, b o Vertical stirrups For 4 sides, total stirrup area is 4 times individual 2 legged stirrup area 91 Various Design Options Integral Beams (Horizontal Bars) How much should be the length of the horizontal bars, l v Critical Perimeter, b o c ¾ (l v c/2) l v can be determined using the critical perimeter b o Distance from the face of column to the boundary of critical perimeter = ¾ (l v c/2) X = ¾ (l v c/2) R l v X = ¾ (l v c/2) b o = 4R + 4c 92 46

Various Design Options Integral Beams For Square Column of Size c : b o = 4R + 4c... (1) Critical Perimeter, b o ¾ (l v c/2) R = (X 2 + X 2 ) = (2) X Eq (1) => b o = 4 (2) X + 4c c putting value of l v : b o = 4 (2){(3/4)(l v c 1 /2)} + 4c R after simplification, we get: b o = 4.24 l v + 1.88c The above equation can be used for determining the X = ¾ (l v c/2) length up to which the horizontal bars should be extended beyond the face of column. X = ¾ (l v c/2) 93 Design the flat plate as shown below for punching shear. The slab supports a live load of 144 psf. All columns are 14 square. Take f c = 4 ksi and f y = 60 ksi. See BSc Lecture 5 for solution of this example. 25 25 25 25 20 20 20 94 47

Two-Way Joist System 95 Contents General Behavior Characteristics Basic Steps for Structural Design Some Important Points 96 48

General A two-way joist system, or waffle slab, comprises evenly spaced concrete joists spanning in both directions and a reinforced concrete slab cast integrally with the joists. Joist 97 General Like one-way joist system, a two way system will be called as two-way joist system if clear spacing between ribs (dome width) does not exceed 30 inches. 98 49

General 99 General The joists are commonly formed by using standard square dome forms and the domes are omitted around the columns to form the solid heads. Solid Head 100 50

General Standard Dome Data: The dome for waffle slab can be of any size. However the commonly used standard domes are discussed as follows: 30-inch 30-inch square domes with 3-inch flanges; from which 6-inch wide joist ribs at 36-inch centers are formed: these are available in standard depths of 8, 10, 12, 14, 16 and 20 inches. 19-inch 19-inch square domes with 2 ½-inch flanges, from which 5-inch wide joist ribs at 24-inch centers are formed. These are available in standard depths of 8, 10, 12, 14 and 16 inches. 101 General Standard Dome Data 102 51

Behavior The behavior of two-way joist slab is similar to a two way flat Slab system. 103 Characteristics Dome voids reduce dead load. Attractive ceiling (waffle like appearance). Electrical fixtures can be placed in the voids. Particularly advantageous where the use of longer spans and/or heavier loads are desired without the use of deepened drop panels or supported beams. 104 52

Basic Steps for Structural Design Step No. 01 (Sizes): Sizes of all structural and non structural elements are decided. Step No. 02 (Loads): Loads on structure are determined based on occupational characteristics and functionality. Step No. 03 (Analysis): Effect of loads are calculated on all structural elements. Step No. 04 (Design): Structural elements are designed for the respective load effects following code provisions. 105 Basic Steps for Structural Design Sizes Minimum Joist Depth For Joist depth determination, waffle slabs are considered as flat slab (ACI 13.1.3, 13.1.4 & 9.5.3). The thickness of equivalent flat slab is taken from table 9.5 (c). The thickness of slab and depth of rib of waffle slab can be then computed by equalizing the moment of inertia of equivalent flat slab to that of waffle slab. However since this practice is time consuming, tables have been developed to determine the size of waffle slab from equivalent flat slab thickness. 106 53

Basic Steps for Structural Design Sizes Minimum Joist Depth Equivalent Flat Slab Thickness ACI 318-05 Sect. 9.5.3 107 Basic Steps for Structural Design Sizes Minimum Joist Depth Slab and rib depth from equivalent flat slab thickness Table 01: Waffle flat slabs (19" 19" voids at 2'-0")-Equivalent thickness Rib + Slab Depths (in.) Equivalent Thickness t e (in.) 8 + 3 8.89 8 + 4 ½ 10.11 10 + 3 10.51 10 + 4 ½ 11.75 12 + 3 12.12 12 + 4 ½ 13.38 14 + 3 13.72 14 + 4 ½ 15.02 16 + 3 15.31 16 + 4 ½ 16.64 Reference: Table 11-2 of CRSI Design Handbook 2002. Note: Only first two columns of the table are reproduced here. 54

Basic Steps for Structural Design Sizes Minimum Joist Depth Slab and rib depth from equivalent flat slab thickness Table 02: Waffle flat slabs (30" 30" voids at 3'-0")-Equivalent thickness Rib + Slab Depths (in.) Equivalent Thickness t e (in.) 8 + 3 8.61 8 + 4 ½ 9.79 10 + 3 10.18 10 + 4 ½ 11.37 12 + 3 11.74 12 + 4 ½ 12.95 14 + 3 13.3 14 + 4 ½ 14.54 16 + 3 14.85 16 + 4 ½ 16.12 20 + 3 17.92 20 + 4 ½ 19.26 Reference: Table 11-2 of CRSI Design Handbook 2002. Note: Only first two columns of the table are reproduced here. Basic Steps for Structural Design Sizes Minimum Width of Rib ACI 8.11.2 states that ribs shall be not less than 4 inches in width. Maximum Depth of Rib ACI 8.11.2 also states that ribs shall have a depth of not more than 3 ½ times the minimum width of rib. Minimum Slab Thickness ACI 8.11.6.1 states that slab thickness shall be not less than one-twelfth the clear distance between ribs, nor less than 2 inch. 55

Basic Steps for Structural Design Sizes Solid Head Dimension of solid head on either side of column centerline is equal to l/6. The depth of the solid head is equal to the depth of the combined depth of ribs and top slab. Basic Steps for Structural Design Loads Floor dead load for two-way joist with certain dome size, dome depth can be calculated from the table shown for two options of slab thicknesses (3 inches and 4 ½ inches). Dome Size Table 03: Standard Dome Dimensions and other Data Dome Depth (inches) Volume of Void (ft 3 ) Floor Dead Load (psf) per slab thickness 3 inches 4 ½ inches 8 3.98 71 90 10 4.92 80 99 30 inches 12 5.84 90 109 14 6.74 100 119 16 7.61 111 129 20 9.3 132 151 8 1.56 79 98 10 1.91 91 110 19 inches 12 2.25 103 122 14 2.58 116 134 16 2.9 129 148 Reference: Table 11-1, CRSI Design Handbook 2002 56

Basic Steps for Structural Design Loads Floor dead load (w dj ) for two-way joist can also be calculated as follows: Volume of solid: V solid = (36 36 11)/1728 = 8.24 ft 3 3 36 Volume of void: V void = (30 30 8)/1728 = 4.166 ft 3 8 Total Load of joists per dome: w dj = (V solid V void ) γ conc = ( 8.24 4.166) 0.15 = 0.61 kips/ dome 30 Total Load of joists per sq. ft: w dj / (dome area) = 0.61/ (3 3) = 0.0679 ksf = 68 psf 71 psf (from table 03) The difference is because sloped ribs are not considered. Plan Loads Basic Steps for Structural Design Loads in an interior span of a two way joist system can be calculated as follows: a a w dj w sj w sj l 2 b a a l n l n l 1 a = b/2 c/2 half joist width W sh = W sj + W dj W sj = W sh - W dj W sh = Dead load of solid head W dj = Dead load of joist W sj = Dead load of solid head excluding joist 57

Basic Steps for Structural Design Loads Factored loads can be calculated as: w uj = 1.2 w dj + 1.6w L If w L = live load (load/area) 1.2 w sj 1.2 w sj w dj = dead load of joists, then Factored load due to joists (w j ) a l n a w uj = 1.2 w dj + 1.6w L Factored load due to w sj l n w usj = 1.2 w sj Where, w sj = (w dsh w dj ) l 2 b Basic Steps for Structural Design Analysis ACI code allows use of DDM for analysis of waffle slabs (ACI R13.1). In such a case, waffle slabs are considered as flat slabs, with the solid head acting as drop panels (ACI 13.1.3). 116 58

Basic Steps for Structural Design Analysis Static moment calculation for DDM analysis: w uj w usj w usj l n a l n a M oj l n M osj l 2 b M oj = w uj l 2 l n2 /8 M osj = w usj ba 2 /2 M o = M oj + M osj 117 Basic Steps for Structural Design Design Design for Flexure The design of waffle slab for flexure is done like solid slab design. 118 59

Basic Steps for Structural Design Design Placement of Flexural Reinforcement Slab Reinforcement (located at mid depth): As required by flexural demand but ACI 7.12 (shrinkage reinf.), with max spacing least of 5h or 18 Each joist rib contains two bottom bars. Straight bars are supplied over the column centerlines for negative factored moment. For Main Negative Reinforcement: As required by flexural demand but ACI 7.12 (shrinkage reinf.), with max spacing least of 2h or 18 For Main Positive Reinforcement: As required by flexural demand but ACI 7.12 (shrinkage reinf.) 119 Basic Steps for Structural Design Design Design for punching shear The solid head shall be checked against punching shear. The critical section for punching shear is taken at a section d/2 from face of the column, where d is the effective depth at solid head. 120 60

Two-Way Joist Design Design for Beam Shear Beam shear is not usually a problem in slabs including waffle slabs. However for completion of design, beam shear may also be checked. Beam shear can cause problem in case where larger spans and heavier loads with relatively shallow waffle slabs are used. The critical section for beam shear is taken at a section d from face of the column, where d is the effective depth at solid head. For joist construction, contribution of concrete to shear strength V c shall be permitted to be 10 percent more than that specified in Chapter 11. If required, one or two single legged stirrups are provided in the rib to increase the shear capacity of waffle slab. 121 Some Important Points For layouts that do not meet the standard 2-feet and 3-feet modules, it is preferable that the required additional width be obtained by increasing the width of the ribs framing into the solid column head. 122 61

Some Important Points The designer should sketch out the spacing for a typical panel and correlate with the column spacing as a part of the early planning. 123 Design the slab system of hall shown in figure as waffle slab, according to ACI 318. Use Direct Design Method for slab analysis. f c = 4 ksi f y = 60 ksi Live load = 100 psf 124 62

Solution: A 108 144 building, divided into twelve (12) panels, supported at their ends on columns. Each panel is 36 36. The given slab system satisfies all the necessary limitations for Direct Design Method to be applicable. 125 Step No 01: Sizes Columns Slab Let all columns be 18 18. Adopt 30 30 standard dome. Minimum equivalent flat slab thickness (h f ) can be found using ACI Table 9.5 (c): Exterior panel governs. Therefore, h f = l n /33 l n = 36 (2 18/2)/12 = 34.5 h f = (34.5/33) 12 = 12.45 126 63

Step No 01: Sizes Slab The closest depth of doom that will fulfill the requirement of equivalent thickness of flat slab equal to 12.45 is 12 in. with a slab thickness of 4 ½ in. for a dome size of 30-in. Table: Waffle flat slabs (30" 30" voids at 3'-0")-Equivalent thickness Rib + Slab Depths (in.) Equivalent Thickness t e (in.) 8 + 3 8.61 8 + 4 ½ 9.79 10 + 3 10.18 10 + 4 ½ 11.37 12 + 3 11.74 12 + 4 ½ 12.95 14 + 3 13.3 14 + 4 ½ 14.54 16 + 3 14.85 16 + 4 ½ 16.12 20 + 3 17.92 20 + 4 ½ 19.26 127 Step No 01: Sizes Planning of Joist layout l = 36-0 = 432 Standard module = 36 36 No. of modules in 36-0 : n = 432/36 = 12 Joist width = 6 Planning: First joist is placed on interior column centerline with progressive placing of other joists towards exterior ends of panel. To flush the last joist with external column, the width of exterior joist comes out to be 15 (6 +Column size /2) as shown in plan view. 128 64

Step No 01: Sizes Solid Head Solid head dimension from column centerline = l/6 = 36/6 = 6 Total required length of solid head= 2 6 = 12 As 3 3 module is selected, therefore 4 voids including joist width will make an interior solid head of 12.5 12.5. (Length of solid head = c/c distance between rib + rib width ) Depth of the solid head = Depth of standard module = 12 + 4.5 = 16.5 c/c between ribs= 4x3 = 12 Rib width= 3+3 inches Total = 12 +0.5 =12.5 129 Step No 02: Loads Floor (joist) dead load (w dj ) = 109 psf = 0.109 ksf Dome Size Table: Standard Dome Dimensions and other Data Dome Depth (in.) Volume of Void (ft3) Floor Dead Load (psf) per slab thickness 3 inches 4 ½ inches 30-in 19-in 8 3.98 71 90 10 4.92 80 99 12 5.84 90 109 14 6.74 100 119 16 7.61 111 129 20 9.3 132 151 8 1.56 79 98 10 1.91 91 110 12 2.25 103 122 14 2.58 116 134 16 2.9 129 148 Reference: Table 11-1, CRSI Design Handbook 2002 130 65

Step No 02: Loads Floor (joist) dead load (w dj ) = 109 psf = 0.109 ksf Solid head dead load w sh = γ c h sh = 0.15 {(12 + 4.5)/12 = 0.20625 Solid Head dead load excluding joist (w sj ) = w sh w dj = 0.20625 0.109 = 0.097 ksf w sj w dj w sj a a l n Step No 02: Loads w L = 100 psf = 0.100 ksf Load due to joists plus LL (w uj ) w uj = 1.2 w dj + 1.6w L w uj = 1.2 w dj + 1.6w L 1.2 w sj 1.2 w sj = 1.2 0.109 + 1.6 0.100 = 0.291 ksf a l n a Load due to solid head (w ush ) w ush = 1.2w sj = 1.2 0.097 = 0.1164 ksf 66

Step No 03: Frame Analysis (E-W Interior Frame) Marking E-W Interior Frame: l 2 = 36 133 Step No 03: Frame Analysis (E-W Interior Frame) Marking of Column and Middle Strips: MS/2 = 9 CS/2 = Least of l 1 /4 or l 2 /4 l 2 /4 = 36/4 = 9 MS/2 = 9 CS/2 = 9 CS/2 = 9 l 2 = 36 134 67

Step No 03: Frame Analysis (E-W Interior Frame) Static Moment Calculation w uj w usj w usj Solid head a l n a l n a M oj M osj b l 2 M oj = w uj l 2 l n2 /8 M osj = w usj ba 2 /2 b a M o = M oj + M osj a a = b/2 c/2 half joist width a = 12.5/2-1.5/2-0.25 = 5.25 ft b l n l 1 b = 12.5 ft 135 Step No 03: Frame Analysis (E-W Interior Frame) Static Moment Calculation M oj (due to joists) = w oj l 2 l n2 /8 = 0.291 36 34.5 2 /8 = 1557.56 ft-kip M osj (moment due to solid head excluding joists) = w usj ba 2 /2 = 0.1164 12.5 5.25 2 /2 = 20 ft-kip M o (total static moment) = M oj + M osj = 1557.56 + 20 = 1577.56 ft-kip Note: Since normally M osj is much smaller than M oj, the former can be conveniently ignored in design calculations. 136 68

Step No 03: Frame Analysis (E-W Interior Frame) Static Moment Calculation. l 2 = 36 137 Step No 03: Frame Analysis (E-W Interior Frame) Longitudinal distribution of Total static moment (M o ). 0.52 0.35 0.52 0.26 0.70 0.65 0.65 0.70 0.26 l 2 = 36 138 69

Step No 03: Frame Analysis (E-W Interior Frame) Longitudinal distribution of Total static moment (M o ). M L = M o (D.F) L 820 552 820 410 1104 1025 1025 1104 410 l 2 = 36 Units: ft-kip 139 Step No 03: Frame Analysis (E-W Interior Frame) Lateral Distribution of Longitudinal moment (L.M). 0.60 0.60 0.60 1.00 0.75 0.75 0.75 0.75 1.00 l 2 = 36 140 70

Step No 03: Frame Analysis (E-W Interior Frame) Lateral Distribution of Longitudinal moment (L.M). M L,ext- = 410 kip-ft M L,ext+ = 820 kip-ft M L,int- = 1104 kip-ft M L,- = 1025 kip-ft M L,+ = 552 kip-ft M Lat = M L (D.F) Lat 0 328/2 276/2 256/2 221/2 256/2 276/2 328/2 0 492 331 492 410 828 769 769 828 410 0 328/2 276/2 256/2 221/2 256/2 276/2 328/2 0 l 2 = 36 141 Step No 03: Frame Analysis (E-W Interior Frame) Lateral Distribution of Longitudinal moment (L.M). M L,ext- = 410 kip-ft M L,ext+ = 820 kip-ft M L,int- = 1104 kip-ft M L,- = 1025 kip-ft M L,+ = 552 kip-ft M Lat per foot = M lat /strip width 0 18.22 15.33 14.22 12.3 14.22 15.33 18.22 0 27.3 18.38 27.3 22.8 46 42.7 42.7 46 0 18.22 15.3 3 14.22 12.3 14.22 15.33 18.22 22.8 0 l 2 = 36 142 71

Step No 03: Frame Analysis (E-W Exterior Frame) Static Moment Calculation M oj (due to joists) = w oj l 2 l n2 /8 l 2 = 18 + c/2 = 18 + (18/12)/2 = 18.75 c = column dimension = 0.291 18.75 34.5 2 /8 = 811.78 ft-kip M osj (moment due to solid head excluding joists) = w usj ba 2 /2 = 0.1164 7 5.25 2 /2 = 12.83 ft-kip M o (total static moment) = M oj + M osj = 811.78 + 12.83 = 825 ft-kip 143 Step No 03: Frame Analysis (E-W Exterior Frame) Static Moment Calculation. l 2 = 18.75 144 72

Step No 03: Frame Analysis (E-W Exterior Frame) Longitudinal distribution of Total static moment (M o ). 0.52 0.35 0.52 0.26 0.70 0.65 0.65 0.70 0.26 145 Step No 03: Frame Analysis (E-W Exterior Frame) Longitudinal distribution of Total static moment (M o ). M L = M o (D.F) L 429 289 429 215 578 536 536 578 215 Units: ft-kip 146 73

Step No 03: Frame Analysis (E-W Exterior Frame) Lateral Distribution of Longitudinal moment (L.M). 0.60 0.60 0.60 1.00 0.75 0.75 0.75 0.75 1.00 147 Step No 03: Frame Analysis (E-W Exterior Frame) Lateral Distribution of Longitudinal moment (L.M). M L,ext- = 215 kip-ft M L,ext+ = 429 kip-ft M L,int- = 578 kip-ft M L,- = 536 kip-ft M L,+ = 289 kip-ft M Lat = M L (D.F) Lat 215 257 434 402 173 769 402 257 215 0 172 144 134 116 134 144 172 0 148 74

Step No 03: Frame Analysis (E-W Exterior Frame) Lateral Distribution of Longitudinal moment (L.M). M L,ext- = 215 kip-ft M L,ext+ = 429 kip-ft M L,int- = 578 kip-ft M L,- = 536 kip-ft M L,+ = 289 kip-ft M Lat per foot = M lat /strip width 23.8 28.6 48.2 44.7 19.2 44.7 48.2 28.6 23.8 0 17.64 14.76 13.74 11.89 13.74 14.76 17.64 0 149 Step No 03: Frame Analysis Analysis of N-S Interior and Exterior Frame will be same as E-W respective frames due to square panels. 150 75

Step No 04: Design E-W Interior Slab Strip: 0 18.22 15.33 14.22 12.3 14.22 15.33 18.22 0 27.3 18.38 27.3 22.8 46 42.7 42.7 46 22.8 l 2 = 36 0 18.22 15.33 14.22 12.3 14.22 15.33 18.22 0 151 Step No 04: Design E-W Exterior Frame. 23.8 28.6 48.2 19.2 48.2 28.6 23.8 0 17.64 14.76 11.89 14.76 17.64 0 152 76

Step No 04: Design Design of N-S Interior and Exterior Frame will be same as E-W respective frames due to square panels and also for the reason that d avg is used in design. d avg = 16.5 (0.75 inch (cover) + ¾ inch (Assumed bar diameter) = 15 inch This will be used for both directions positive as well as negative reinforcement. 153 Step No 05: Detailing (E-W Frames) Negative Reinforcement M = 46 ft-k =46 x12 = 552 in-k d = 15 f y = 60 ksi A s = 0.7 in 2 S= 0.44/0.7 x 12 = 7.2 inch Finally using #6 @ 6 c/c & 12 c/c in the column strips at all interior & exterior supports respectively. in the middle strip use #6 @ 18 c/c. #6 @ 12 #6 @ 6 #6 @ 6 #6 @ 12 #6 @ 18 #6 @ 18 #6 @ 18 #6 @ 18 #6 @ 12 #6 @ 6 #6 @ 6 #6 @ 12 154 77

Step No 05: Detailing (E-W Frames) Positive reinforcement For M = 27 ft-k =27x12 = 324in-k d = 15 f y = 60 ksi A s = 0.373 in 2 This is per foot reinforcement. For 18 feet col strip, this will be equal to 0.373 x 18 = 6.714 in 2 There are 6 joists in 18 feet with. Therefore per rib reinforcement = 1.12 Using # 7 bars, 2 bars per joist rib will be provided in the column as well as middle strips. 155 Step No 05: Detailing (E-W Interior Frame) 18-0 #6 @ 6 c/c 2 #7 Bars Column Strip at interior support #6 @ 12 c/c 2 #7 Bars Column Strip at exterior support 156 78

Step No 05: Detailing (E-W Interior Frame) 18-0 #6 @ 18 c/c 2 #7 Bars Middle Strip at interior support #6 @ 18 c/c 2 #7 Bars Middle Strip at Exterior support 157 Step No 05: Detailing (E-W Exterior Frame) 9-0 #6 @ 6 c/c 2 #7 Bars Column Strip at interior support #6 @ 12 c/c 2 #7 Bars Column Strip at exterior support 158 79

Step No 05: Detailing (E-W Exterior Frame) 9-0 #6 @ 18 c/c 2 #7 Bars Middle Strip at interior support #6 @ 18 c/c 2 #7 Bars Middle Strip at exterior support 159 Step No 04: Design Note: For the completion of design problem, the waffle slab should also be checked for beam shear and punching shear. 160 80

References ACI 318-02 CRSI Design Handbook Design of Concrete Structures by Nilson, Darwin and Dolan [13 th Ed] 161 The End 162 81