Analysis and Design of Waffle Slabs

Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Lecture-11 Analysis and Design of Two-way Slab Systems (Two-way Slab with Beams & Two Way joist Slabs) B Prof By: P f Dr. D Qaisar Q i Ali Civil Engineering Department NWFP UET Peshawar [email protected] Prof. Dr. Qaisar Ali

CE 5115 Advance Design of Reinforced Concrete Structures

1


Topics Addressed y Moment Coefficient Method for Two way slab with beams y Introduction y Cases y Moment Coefficient Tables y Reinforcement R i f t Requirements R i t y Steps y Example

Prof. Dr. Qaisar Ali

2

1


Topics Addressed y Two-way Joist Slab y Introduction y Behavior y Characteristics y Basic Steps for Structural Design y Example

3



Two Way Slabs y Moment Coefficient Method (Introduction) z

The Moment Coefficient Method included for the first time in 1963 ACI Code is applicable to two-way slabs supported on four sides of each slab panel by walls, steel beams relatively deep, stiff, edge beams (h = 3hf).

z

Although, not included in 1977 and later versions of ACI code, its continued use is permissible under the ACI 318-08 code provision (13.5.1). Visit ACI 13.5.1.


4

2


Two Way Slabs Ma,neg

la

y Moment Coefficient Method y Moments: Ma, neg =

Mb,neg

Mb,pos

Ca, negwula2

Mb, neg = Cb, negwulb2

lb

Ma,pos Mb,neg Ma,neg

Ma, pos, (dl + ll) = M a, pos, dl + M a, pos, ll = Ca, pos, dl × wu, dl × la2 + Ca, pos, ll × wu, ll × la2 Mb, pos, (dl + ll) = Mb, pos, dl + Mb, pos, ll = Cb, pos, dl × wu, dl × lb2 + Cb, pos, ll × wu, ll × lb2

y Where Ca, Cb = Tabulated moment coefficients wu = Ultimate uniform load, psf la, lb = length of clear spans in short and long directions respectively.

5



Two Way Slabs y Moment Coefficient Method: Cases y Depending on the support conditions, several cases are possible:


6

3



7





8

4



9



Two Way Slabs y Moment Coefficient Tables:


10

5



11





12

6



13





14

7



15



Two Way Slabs y Load Coefficient Table:


16

8


Two Way Slabs y Maximum spacing and minimum reinforcement requirement: z

Maximum spacing (ACI 13.3.2): smax = 2 hf in each direction.

z

Minimum Reinforcement (ACI 7.12.2.1): Asmin = 0.0018 b hf for grade 60. Asmin = 0.002 b hf for grade 40 and 50.

17



Two Way Slabs y Special Reinforcement at exterior corner of Slab z

The reinforcement at exterior ends of the slab shall be provided as per ACI 13.3.6 in top and bottom layers as shown.

z

The positive and negative reinforcement in any case, should be of a size and spacing equivalent to that required for the maximum positive moment (per foot of width) in the panel.


18

9


Two Way Slabs y Moment Coefficient Method y Steps y

Find hmin = perimeter/ 180 = 2(la + lb)/180

y

Calculate loads on slab (force / area)

y

Calculate m = la/ lb

y

Decide about case of slab,

y

Use table to pick moment coefficients,

y

Calculate moments and then design.

y

Apply reinforcement requirements (smax = 2hf, ACI 13.3.2)

19



Two Way Slabs y Moment o e t Coe Coefficient c e t Method: et od Example a pe y A 100′ × 60′, 3-storey commercial building is to be designed. The grids of column plan are fixed by the architect.


20

10


Two Way Slabs y Moment o e t Coe Coefficient c e t Method: et od Example a pe y Complete analysis of the slab is done by analyzing four panels

Panel I

Panel III

Panel III

Panel I

Panel II

Panel IV

Panel IV

Panel II

Panel I

Panel III

Panel III

Panel I 21



Two Way Slabs y Moment Coefficient Method: Example p y A 100′ × 60′, 3-storey commercial building: Sizes and Loads. y Sizes: y

Minimum slab thickness = perimeter/180 = 2 (20+25)/180 = 6″ However, for the purpose of comparison, take hf = 7″

y

Columns = 14″ × 14″ (assumed)

y

Beams = 14″ × 20″ (assumed)

y Loads:


y

S.D.L = Nil ; Self Weight = 0.15 x (7/12) = 0.0875 ksf

y

L.L = 144 psf ; wu = 0.336 ksf 22

11


Two Way Slabs y Moment Coefficient Method: Example y Panels are analyzed using Moment Coefficient Method Case = 4 m = la/lb = 0.8 Ma,neg Mb,neg

Mb,pos

Mb,neg

Ma,pos Ma,neg

23



Two Way Slabs y Moment Coefficient Method: Example y Panels are analyzed using Moment Coefficient Method Case = 4 m = la/lb = 0.8 Ca,neg = 0.071 Cb,neg = 0.029 Ca,posLL = 0.048 Cb,posLL = 0.020 Ca,posDL = 0.039 0 039 Cb,posDL = 0.016


24

12



25





26

13



27





28

14



29



Two Way Slabs y Moment Coefficient Method: Example y Panels are analyzed using Moment Coefficient Method Case = 4 m = la/lb = 0.8 Ca,neg = 0.071 Cb,neg = 0.029 Ca,posLL = 0.048 Cb,posLL = 0.020 Ca,posDL = 0.039 0 039 Cb,posDL = 0.016 Ma,neg = 9.5 k-ft Ma,pos = 6.1 k-ft Mb,neg = 6.1 k-ft Mb,pos = 3.9 k-ft Prof. Dr. Qaisar Ali

Panel I Mb,neg

Ma,neg Mb,pos

Mb,neg

Ma,pos Ma,neg

For slab supported on Spandrals, Mneg,ext = 1/3Mpos

30

15



Panel II

Ma,neg Mb,neg

Mb,pos

Mb,neg

Ma,pos Ma,neg

Ma,neg = 10.1 k-ft Ma,pos = 5.1 k-ft Mb,neg = 3.6 k-ft Mb,pos = 3.1 k-ft

31




Ma,neg

Panel III Mb,neg

Mb,pos

Mb,neg

Ma,pos Ma,neg

Ma,neg = 7.4 k-ft Ma,pos = 5.4 k-ft Mb,neg = 8.6 k-ft Mb,pos = 3.7 k-ft Prof. Dr. Qaisar Ali

32

16


Two Way Slabs y Moment Coefficient Method: Example y Panels are analyzed using Moment Coefficient Method Case = 2 m = la/lb = 0.8 Panel IV

Ca,neg = 0.065 Cb,neg = 0.027 Ca,posLL = 0.041 Cb,posLL = 0.017 Ca,posDL = 0.026 0 026 Cb,posDL = 0.011

Ma,neg Mb,neg

Mb,pos

Mb,neg

Ma,pos Ma,neg

Ma,neg = 8.7 k-ft Ma,pos = 4.9 k-ft Mb,neg = 5.7 k-ft Mb,pos = 3.2 k-ft

33



Two Way Slabs y Moment Coefficient Method: Example y Slab analysis summary 9.5

7.4

3.9 6.1

3.7 6.1

6.1 8.6

5.4

9.5 10.1 3.6

3.2 5.1 10.1


8.6

7.4 8.7 3.6 5.7

3.2 4.9

5.7

8.7

34

17


Two Way Slabs y Moment Coefficient Method: Example y Slab Reinforcement Details C

C

A B

A B

C

B

C A

A B C

C

C B

A B

B

A= #4 @ 12″ B = #4 @ 6″ C = #4 @ 4″

C

35


Two-Way Joist Slab

36

18

Two-Way Joist y Introduction z

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

37


Like one-way joist system, a two way system will be qualified to be said as two-way joist system if clear spacing between ribs (dome width) does not exceed 30 in.

38

19

Two-Way Joist y Introduction

39



The joists are commonly formed by using Standard Square “dome” forms and the domes are omitted around the columns to form the solid heads.


40

20



Standard Dome Data z

Generally the dome for waffle slab can be of any size. However the commonly used standard domes are discussed as follows: z

30-in × 30-in 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.

z

19 i h × 19-inch 19-inch 19 i h square domes d with ith 2 ½-inch ½ i h flanges, fl f from which hi h 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.

41





Standard Dome Data

42

21


Two-Way Joist y Behavior z

The behavior of two-way joist slab is similar to a two way flat Slab system.



Two-Way Joist y Characteristics z

Dome voids reduce dead load

z

Attractive ceiling (waffle like appearance)

z

Electrical fixtures can be placed in the voids

z

Particularly advantageous where the use of longer spans and/or heavier loads are desired without the use of deepened drop panels or supported beams.


22


Two-Way Joist y Basic Steps for Structural Design z

Step No. No 01 (Sizes): Sizes of all structural and non structural elements are decided.

z

Step No. 02 (Loads): Loads on structure are determined based on occupational characteristics and functionality (refer Appendix C of class notes).

z

Step No. No 03 (Analysis): Effect of loads are calculated on all structural elements.

z

Step No. 04 (Design): Structural elements are designed for the respective load effects following code provisions. 45



Two-Way Joist y Sizes z

Minimum Joist Depth z

For Joist depth determination, waffle slabs are considered as flat slab (ACI 13.1.3, 13.1.4 & 9.5.3).

z

The thickness of equivalent flat slab is taken from table 9.5 (c).

z

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. slab

z

However since this practice is time consuming, tables have been developed to determine the size of waffle slab from equivalent flat slab thickness.


46

23


Two-Way Joist y Sizes z

Minimum Joist Depth z

Equivalent Flat Slab Thickness

z

ACI 318-05 – Sect. 9.5.3 z

Minimum thickness = ln/33

47


Two-Way Joist y Sizes y Minimum Joist Depth y

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 te (in.)

8+3 8.89 8+4½ 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.

48

24

Two-Way Joist y Sizes y Minimum Joist Depth y

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.)


8+3 8.61 8+4½ 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.

49

Two-Way Joist y Sizes y Minimum Width of Rib y

ACI 8.11.2 states that ribs shall be not less than 4 inch in width.

y Maximum Depth of Rib y

A rib shall have a depth of not more than 3 ½ times the minimum width of rib.

y Minimum Slab Thickness y

ACI 8.11.6.1 states that slab thickness shall be not less than onetwelfth the clear distance between ribs, nor less than 2 in.

50

25

Two-Way Joist y Loads y

Floor dead load for two-way y jjoist with certain dome size,, dome depth p can be calculated from the table shown for two options of slab thicknesses (3 inches and 4 ½ inches). Table 03: Standard Dome Dimensions and other Data Dome Size

Dome Depth (in.)

Volume of Void (ft3)

8 10 12 14 16 20 8 10 12 14 16

3.98 4 92 4.92 5.84 6.74 7.61 9.3 1.56 1.91 2.25 2.58 2.9

30-in

19-in

Floor Dead Load (psf) per slab thickness 3 inches

4 ½ inches

71 80 90 100 111 132 79 91 103 116 129

90 99 109 119 129 151 98 110 122 134 148

Reference: Table 11-1, CRSI Design Handbook 2002

51

Two-Way Joist y Loads y Floor dead load (wdj) for two-way two way joist can also be calculated as follows: Volume of solid: Vsolid = (36 × 36 × 11)/1728 = 8.24 ft3

3″

36″

8″

Volume of void: Vvoid = (30 × 30 × 8)/1728 = 4.166 ft3

30″

Total Load of joists per dome: wdj = (Vsolid – Vvoid) × γconc = ( 8.24 – 4.166) × 0.15 = 0.61 kip Total Load of joists per sq. ft: wdj/ (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. 52

26

Two-Way Joist y Loads y At locations where solid head is present, present the floor dead load can be calculated as follows: y

If, wdj = dead load in joist area

y

Wsh = dead load in solid head area

Wdj+sh

Wdj+sh

a

= hsolid × γconc y

wdj

a ln

Wdj+sh = {{wshb + wdj((l2-b)}/l )} 2

b

l2

a

a 53

Two-Way Joist y Loads y Factored loads can be calculated as:

Wish

wosh

Wish

y If wL = live load (load/area), then y

a

Load out of solid head region

ln

wosh = 1.2 wdj + 1.6wL y

wish

Load in solid head region wish = 1.2w 1 2 dj+sh+1.6w 16 L

a

l2

b

wish wosh

a

a

54

27


Two-Way Joist y Analysis ACI code allows use of DDM for analysis of waffle slabs (ACI

z

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).

55



Two-Way Joist y Analysis Static moment calculation for DDM analysis:

z

Wish

wosh

wosh

a

ln Mosh

ln

Wish

a

ln

Mish b a

Mosh = woshl2ln2/8

Mish = (wish-wosh)ba2/2

a

l2

Mo = Mosh + Mish Prof. Dr. Qaisar Ali

56

28


Two-Way Joist y Design Design of slab for punching shear

z

z

The solid head shall be checked against punching shear.

z

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.

57



Two-Way Joist y Design z

Design

of

slab

for

punching shear z

Load on tributary area will

l1

cause punch out shear. z

Within tributary area, two types of loads are acting:

z

z

Solid head load

z

Joist load

Both

types

l2 shall

d/2

be

considered while calculating punching shear demand Prof. Dr. Qaisar Ali

58

29



Design of slab for punching shear z

Total area = l1 × l2

z

Solid area = Asolid

z

Joist part area (Aj) = (l1×l2) -Asolid

z

Critical perimeter area = Acp

z

Vu =Aj×wosh+ (Asolid – Acp) × wish

z

Where,

l1

l2

d/2

wosh = joist part load wish = load inside solid head 59




Shear Strength of Slab in punching shear: z

ΦVn = ΦVc + ΦVs

z

ΦVc is least of: z

Φ4√ (fc′)bod

z

(2 + 4/βc) √ (fc′)bod

z

{(αsd/bo +2} √ (fc′)bod

βc = longer side of column/shorter side of column αs = 40 for interior column, 30 for edge column, 20 for corner columns


60

30



Design of Joist for Beam Shear: z

Beam shear Demand z

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.

z

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.

61




Design g of Joist for Beam Shear: z

Beam shear capacity of concrete joist z

ΦVn = ΦVc + ΦVs

z

ΦVc is least of: z

z z

Stirrup

Φ2√ (fc′)bribd

ΦVs = ΦAvfy/bribs

If required, one or two single legged stirrups are provided in the rib to increase the shear capacity of waffle slab.


62

31



Design for Flexure z

The design of waffle slab is done by usual procedures.

z

However, certain reinforcement requirements apply discussed next.

63



Two-Way Joist y ACI recommendations on reinforcement requirement of waffle slab: z

ACI 10.6.7 states that if the effective depth d of a beam or joist exceeds 36 in., longitudinal skin reinforcement shall be provided as per ACI section 10.6.7.

z

According to ACI 13.3.2, for cellular or ribbed construction reinforcement shall not be less than the requirements of ACI 7.12.

z

As per ACI 7.12, Spacing of top bars cannot exceed 5h or 18 inches.


64

32


Two-Way Joist y ACI recommendations on reinforcement requirement of waffle slab:

65



Two-Way Joist y Other important points: z

The amount of reinforcement and, and if necessary, necessary the top slab thickness can be changed to vary the load capacities for different spans, areas, or floors of a structure.

z

Each joist rib contains two bottom bars. Straight bars are supplied over the column centerlines for negative factored moment.

Bottom bar


66

33


Two-Way Joist y Other important points: z

For layouts that do not meet the standard 2-feet 2 feet and 3-feet 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.

z

The designer should sketch out the spacing for a typical panel p g as a p part of the earlyy and correlate with the column spacing planning.

67



Two-Way Joist y Example: Design the slab system of hall shown in figure as waffle slab, according to ACI 318. Use Direct Design Method for slab analysis. z

fc′ = 4 ksi

z

fy = 60 ksi

z

Live load = 100 psf


68

34


Two-Way Joist y Solution: z

A 108 108′ × 144 144′ building, building divided into twelve (12) panels, panels supported at their ends on columns. Each panel is 36′ × 36′.

z

The given slab system satisfies all the necessary limitations for Direct Design Method to be applicable.

69



Two-Way Joist y Step No 01: Sizes z

Columns z

z

Let all columns be 18″ × 18″.

Slab z

Adopt 30″ × 30″ standard dome.

z

Minimum equivalent flat slab thickness (hf) can be found using ACI Table 9 5 (c): 9.5

z

Exterior panel governs. Therefore, hf = ln/33 = [{36 – (2 × 18/2)/12}/33] × 12 = 12.45″


70

35



Slab z

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.)



8+3 8.61 8+4½ 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.

71



Planning of Joist layout

l = 36′-0″ = 432″ Standard module = 36″ × 36″ No. of modules in 36′-0″: n = 432/36 = 12 Planning: First module is placed on interior column centerline and provided towards exterior ends of panel. In this way, width of exterior joist comes out to be 15″.


72

36



Solid Head z

Solid head dimension from column centerline = l/6 = 36/6 = 6′

z

Total length of solid head= 2 × 6 = 12′

z

As 3′ × 3′ module is selected, therefore 4 voids will make an interior solid head of 12.5′ × 12.5′.

z

Depth of the solid head = Depth of standard module = 12 + 4.5 = 16.5′

73



Two-Way Joist y Step No 02: Loads z

Floor (joist) dead load (wdj) = 109 psf = 0.109 0 109 ksf

Dome Size

30-in

19-in

Table: Standard Dome Dimensions and other Data Floor Dead Load (psf) per slab thickness Volume of Void Dome Depth (in.) (ft3) 3 inches 4 ½ inches 8 10 12 14 16 20 8 10 12 14 16

3.98 4.92 5.84 6.74 7.61 61 9.3 1.56 1.91 2.25 2.58 2.9

71 80 90 100 111 132 79 91 103 116 129

90 99 109 119 129 151 98 110 122 134 148

Reference: Table 11-1, CRSI Design Handbook 2002


74

37

Two-Way Joist y Step No 02: Loads y Floor (joist) dead load (wdj) = 109 psf = 0.109 0 109 ksf y Solid Head dead load (wsh) = {(12 + 4.5)/12} × 0.15 = 0.206 ksf y Wdj+sh = {wshb + wdj(l2-b)}/l2

Wdj+sh

Wdj+sh

wdj

= {0.206×12.5 + 0.109 (36 – 12.5)}/36 a

= 0.143 ksf

a ln

l2

b = 12.5′ a = 5.25′

a 75

Two-Way Joist y Step No 02: Loads y wL = 100 psf = 0.100 0 100 ksf y Load out of solid head region wosh = 1.2 wdj + 1.6wL = 1.2×0.109 + 1.6×0.100 = 0.291 ksf y Load in solid head region

wish l2

b

wish wosh

a

a

wish = 1.2wdj+sh+1.6wL = 1.2 × 0.143 + 1.6 × 0.100 = 0.33 ksf 76

38


Two Way Joist Two-Way y Step No 03: Frame Analysis (E-W Interior Frame) z

Step 1: Marking E-W Interior Frame:

l1 = 36′-0″ 36′ 0″ ln = 34′-6″


l2 = 36′-0″

77



Step 01: Marking E-W Interior Frame z

Design Span of frame (c/c) = l1 = 36′ Design Length of frame = ln = 36 – (2 × 18/2)/12 = 34.5′ Width of frame = l2 = 36′ Half column strip width = (Shorter span)/ 4 = 36/4 = 9′


78



Step 2: Marking Column and Middle Strips

MS/2 = 9′-0″ a = 55′-3″ -3 b= 12′-6″

CS/2 = 9′-0″ CS/2 = 9′-0″ MS/2 = 9′-0″

CS/2 = Least of l1/4 or l2/4 l2/4 = 36/4 = 9′ 9


79



Step 03: Static Moment Calculation z

Mosh (outside head) = woshl2ln2/8 = 0.291 × 36 × 34.52/8 = 1557.56 ft-k Mish (solid head) = (wish – wosh) ba2/2 = (0.33–0.291)×12.5×5.252/2 = 6.70 ft-k Mo (total static moment) = Mosh + Mish = 1557.56 + 6.70 = 1564.26 ft-k

Note: Since normally, Mish is much smaller than Mosh the former can be conveniently ignored in design calculations


80




Step 04: Longitudinal distribution of Total static moment (Mo).

81



Step 05: Lateral Distribution of Longitudinal moment (L.M). α INT36 =0 {no interior beams} l2/l1 = 36/36 = 1 α INT36l2/l1 = 0


82


Two Way Joist Two-Way y Step No 03: Frame Analysis (E-W Exterior Frame) z

Step 01: Marking E-W exterior Frame

l1 = 36′-0″ 36′ 0″ ln = 34′-6″

l2 = 18′-0″ + ((9/12)) = 18.75′


83



Step 01: Marking E-W exterior Frame z

Design Span of frame (c/c) = l1 = 36′ Design Length of frame = ln = 36 – (2 × 18/2)/12 = 34.5′ Width of frame = l2 = 9′ + 9′ + (9/12)″ = 18.75′ Half column strip width = (Shorter span)/ 4 = 36/4 = 9′


84



Step 02: Marking Column and Middle Strips

l1 = 36′-0″ 36′ 0″ ln = 34′-6″

CS/2 = Least of l1/4 or l2/4 l2/4 = 36/4 = 9′ 9


MS/2 = 9′-0″ a = 55′-33″ b= 7′-0″

CS/2 = 9′-0″ 85



Step 03: Static Moment Calculation z

Mosh (outside head) = woshl2ln2/8 = 0.291 × 18.75 × 34.52/8 = 811.78 ft-k Mish (solid head) = (wish – wosh) ba2/2 = (0.33–0.291)×7×5.252/2 = 3.76 ft-k Mo (total static moment) = Mosh + Mish = 811.78 + 3.76 = 815.54 ft-k

Note: Since normally, Mish is much smaller than Mosh the former can be conveniently ignored in design calculations


86


Two Way Joist Two-Way y Step No 03: Analysis z


Step 04: Longitudinal distribution of Total static moment (Mo).

87



Step 05: Lateral Distribution of Longitudinal moment (L.M) [Refer to ACI 13.6.4 to ACI 13.6.6].


z

α EXT36 =0 {no exterior beams}

z

l2/l1 = 36/36 = 1

z

α EXT36l2/l1 = 0

88



Analysis of N-S Interior and Exterior Frame will be same as E-W respective frames due to square panels.

N-S Exterior Frame l2 = 18′-9″ N-S Interior Frame l2 = 36′-0″


89


Two Way Joist Two-Way y Step No 04: Design z

For E-W Interior slab strip: z

davgg = 12 + 4.5 – 1″ (concrete cover) – 0.75 (avg. bar dia) = 14.75″

z

Asmin = 0.0018bte (Where te = equivalent flat slab thickness) Asmin = 0.0018 × 12 × 12.95 = 0.279 in2

z

Now, Equation used to calculate (ρ) in table below is as follows: Mu = Φfyρbdavg2{1– 0.59ρfy/fc′} = 0.9×60×ρ×12×14.752×{1– 0.59×ρ×60/4}

z

After solving the above equation for ρ, we get: ρ = [140980.5 ±√{(140980.5)2 – (4 × 1247677 × Mu′ × 12)}]/2(1247677)….(A)


90




For E-W Interior slab strip:

91



For E-W exterior slab strip: z

davgg = 12 + 4.5 – 1– 0.75 = 14.75″

z

Asmin = 0.0018bte (Where te = equivalent flat slab thickness) Asmin = 0.0018 × 12 × 12.95 = 0.279 in2

z

Now, Equation used to calculate (ρ) in table below is as follows: Mu = Φfyρbdavg2{1– 0.59ρfy/fc′} = 0.9×60×ρ×12×14.752×{1– 0.59×ρ×60/4}

z

After solving the above equation for ρ, we get: ρ = [140980.5 ±√{(140980.5)2 – (4 × 1247677 × Mu′ × 12)}]/2(1247677)….(A)


92




For E-W exterior slab strip:

93



Design of N-S Interior and Exterior Frame will be same as EW respective frames due to square panels and also for the reason that davg is used in design.


94



Note: For the completion of design problem, the waffle slab should also be checked for beam shear and punching shear.


95


Two Way Joist Two-Way y Step No 05: Detailing (E-W Frames)


#6 @ 12″

#6 @ 6″ #6 @ 6″

#6 @ 12″

#6 @ 18″

#6 @ 18″

#6 @ 18″

#6 @ 18″

#6 @ 12″

#6 @ 6″

#6 @ 6″

#6 @ 12″

96


Two Way Joist Two-Way y Step No 05: Detailing (N-S Frames) #6 @ 12″ #6 @ 18″


#6 @ 12″

#6 @ 6″

#6 @ 18″

#6 @ 6″

#6 @ 6″

#6 @ 18″

#6 @ 6″

97


Two Way Joist Two-Way y Step No 05: Detailing (E-W Interior Frame) 18′-0″ #6 @ 6″ c/c

Column Strip (Interior Frame); section taken over support 2 #7 Bars #6 @ 12 12″ c/c

2 #7 Bars


Column Strip (Exterior Frame); section taken over support

98


Two Way Joist Two-Way y Step No 05: Detailing (E-W Interior Frame) 18′-0″ #6 @ 18″ c/c

Middle Strip (Interior Frame); Section taken over column line 2 #7 Bars #6 @ 18 18″ c/c

2 #7 Bars Prof. Dr. Qaisar Ali

Middle Strip (Exterior Frame); Section taken over column line 99


Two Way Joist Two-Way y Step No 05: Detailing (E-W Exterior Frame) 9′-0″ #6 @ 6″ c/c

2 #7 Bars

Column Strip (Interior Frame); section over support

#6 @ 12″ c/c


Column Strip (Exterior Frame); section over support 100


Two Way Joist Two-Way y Step No 05: Detailing (E-W Exterior Frame) 9′-0″ #6 @ 18″ c/c

2 #7 Bars

Middle Strip (Interior Frame) ; section over support

#6 @ 18″ c/c


Middle Strip (Exterior Frame); section over support 101


The End


102

Analysis and Design of Waffle Slabs

Recommend Documents