lecture 09 analysis and design of flat plate slabs

1

Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan

Lecture-09

Analysis and Design of Two-way Slab System without Beams

(Flat Plate and Flat Slabs)By: Prof Dr. Qaisar Ali

Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 1

Civil Engineering Department

UET [email protected]

[email protected]


Two Way Slabs


2


Topics Addressed

Two Way Slabs

Behavior

Types

Analysis and Design Considerations



Topics Addressed

Direct Design Method

Introduction

Limitations

Frame Analysis Steps for Flat Plates and Flat Slabs

Frame marking

Column and middle strips marking

Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures

Column and middle strips marking

Static moment calculation

Longitudinal distribution of static moment

Lateral distribution of longitudinal moment

4

3


Two Way Slabs

Behavior

A slab having bending in both directions is called two-way slab (Long span/short span < 2).

25′ 25′ 25′ 25′

20′


20′

20′


Two Way Slabs

Behavior

Short direction moments in two-way slab.

Short

25′

25′

25′

25′

20′

20′

20′


Short Direction

4


Two Way Slabs

Behavior

Long direction moments in two-way slab.

25′

25′

25′

25′

20′

20′

20′


Long Direction


Two Way Slabs

Behavior: More Demand (Moment) in short directiondue to size of slab

∆central Strip = (5/384)wl4/EI

As these imaginary strips are part of monolithic slab, the deflection at anypoint, of the two orthogonal slab strips must be same:

∆a = ∆b

(5/384)w l 4/EI = (5/384)w l 4/EI


(5/384)wala4/EI = (5/384)wblb4/EI

wa/wb = lb4/la4 wa = wb (lb4/la4)

Thus, larger share of load (demand) is taken by the shorter direction.

8

5


Two Way Slabs

Types

Wall Supported

Beam supported

Flat Plate

Flat slab


Waffle Slab

9


Two Way Slabs

Analysis

Unlike beams and columns, slabs are two dimensionalmembers. Therefore their analysis except one-way slabsystems is relatively difficult.

Design

Once the analysis is done the design is carried out in the


Once the analysis is done, the design is carried out in theusual manner. So no problem in design, problem is only inanalysis of slabs.

10

6


Two Way Slabs

Approximate Analysis Methods of ACI

Slab System Applicable Analysis MethodsOne-Way Slab Strip Method for one-way slabs

Two-way slabs supported on stiff beams and walls

Moment Coefficient Method,

Direct Design Method,

Equivalent Frame Analysis Method


q y

Two-way slabs with shallow beams or without beams

Direct Design Method,

Equivalent Frame Analysis Method




7



IntroductionIn DDM, frames rather than panels are analyzed as is done inanalysis of two way slabs with beams using ACI momentcoefficients.

Interior Frame

Exterior Frame

25′

20′

25′ 25′ 25′


Interior Frame

Interior Frame

Exterior Frame

20′

20′



Introduction

25′

20′

25′ 25′ 25′

For complete analysis of slab system, frames areanalyzed in E-W and N-S directions.


20′

20′

14

E-W FramesN-S Frames

8



Introduction

Though DDM is useful for analysis of slabs, speciallywithout beams, the method is applicable with somelimitations as discussed next.




Limitations (ACI 13.6.1)Uniformly distributed loading (L/D ≤ 2)Uniformly distributed loading (L/D ≤ 2)

ll11 ll11≥2≥2ll1 1 /3 /3 Three or more Three or more spans spans


ll22 Column offsetColumn offset≤ ≤ ll2 2 /10/10

Rectangular slab Rectangular slab panels (2 or less:1)panels (2 or less:1)

16

9



Limitations (ACI 13.6.1): Example

1515′′ 1515′′If ≥10If ≥10′′DDM APPLICABLE as 2/3 (15) = 10DDM APPLICABLE as 2/3 (15) = 10′′


1515′′ 1515′′If <10If <10′′DDM NOT APPLICABLEDDM NOT APPLICABLE

17



Frame Analysis

Step No. 01: Slab is divided in frames (for E-W analysis,slab is divided into E-W frames and vice versa for N-Sanalysis).25′

20′

25′ 25′ 25′

Interior Frame

Exterior Frame


20′

20′

Interior Frame

Interior Frame

Exterior Frame

10



Frame Analysis

Step No. 01 (continued):

An interior frame

Interior Frame l

25′

20′

25′ 25′ 25′


Interior Frame

l1

l2

20′

20′



Frame Analysis

25′

20′

25′ 25′ 25′


Interior Frame lHalf width of panel

Marking an E-W Interior Frame

Panel Centerline


20′

20′

20

Interior Frame

l1

l2 on one side

Half width of panel on other side

Col Centerline

Panel Centerline

11



Frame Analysis

25′

20′

25′ 25′ 25′


Exterior Frame

l

Marking an E-W Exterior Frame

Note: For exterior framesl2 = Panel width/2 +h2/2

l2Half width of panel on one side

h2/2


20′

20′

21

l1



Frame Analysis

Step No. 02: A frame is divided further into strips known ascolumn and middle strips (Defined in ACI 13.2).

Column Strip: A column strip is a design strip with a width on eachside of a column centerline equal to 25 percent of l1 or l2, whicheveris less.

Middle Strip: Middle strips are design strips bounded by two column


Middle Strip: Middle strips are design strips bounded by two columnstrips.

22

25′

20′

25′ 25′ 25′

20′

20′

l2Column stripFull Middle strip

Half Middle strip

l2

12



Frame Analysis

Step No. 02 (continued): Why a frame is divided into columnand middle strips?

Because the slab portion on the column centerline will offer moreresistance than the rest of the slab.

25′ 25′ 25′ 25′


20′

20′

20′



Frame Analysis


25′

20′

25′ 25′ 25′

CS/2C S

M.S/2l2Half Column strip

a) Marking Column Strip

b) Middle Strip


20′

20′

CS/2 = Least of l1/4 or l2/4

CS/2 C.S

M.S/2

l1

ln

p

13


Step No. 02 (continued): Frame and strips in 3D.


½ l t i idth l /4 (l ) /4 hi h i i i

(l2)B

½-Middle strip

½-Middle strip

Column strip

½ column strip width: l1/4 or (l2)B/4, whichever is minimum(l2)A


½ Middle strip

ln

½ column strip width: l1/4 or (l2)A/4, whichever is minimum

25



Frame Analysis

25′

20′

25′ 25′ 25′

Step No. 02 (continued): For l1 = 25′ and l2 = 20′, CS and MSwidths are given as follows.

5′10′

5′l2Half Column strip


b) Middle Strip


20′

20′

26


l2/4 = 20/4 = 5′

5′10′

5′

l1

ln

p

14



Frame AnalysisStep No. 03: Calculate Static Moment (Mo) for interior span ofp ( o) pframe. 25′

20′

25′ 25′ 25′

MoMo =wu l2 ln

2

8l2

ln

Span of frame


20′

20′

27



Frame AnalysisStep No. 04: Longitudinal Distribution of Static Moment (Mo).p g ( o)

M+

M − M −M − = 0.65Mo

M + = 0.35Mo

25′

20′

25′ 25′ 25′


+ 0.35 o 20′

20′

28

15



Frame AnalysisStep No. 05: Lateral Distribution to column and middle strips.p p

M − = 0.65Mo

M + = 0.35Mo

0.60M +

0.75M − 0.75M −

25′

20′

25′ 25′ 25′


+ 0.35 o 20′

20′

29



Frame AnalysisStep No. 03: Calculate Static Moment (Mo) for exterior span ofp ( o) pframe. 25′

20′

25′ 25′ 25′

MoMo =wu l2 ln

2

8l2

ln


20′

20′

30

16



Frame AnalysisStep No. 04: Longitudinal distribution of static moment (Mo).p g ( o)

Mext − = 0.26Mo

M ext+ = 0.52Mo

Mext+

Mext− Mint−

25′

20′

25′ 25′ 25′


ext+ 0.5 o

Mint- = 0.70Mo

20′

20′

31




M ext+ = 0.52Mo

Mext − = 0.26Mo0.60Mext+

1.00Mext− 0.75Mint−

25′

20′

25′ 25′ 25′


ext+ 0.5 o

Mint- = 0.70Mo

20′

20′

32

17




M ext+ = 0.52Mo

Mint- = 0.70Mo

Mext − = 0.26Mo

25′

20′

25′ 25′ 25′

0.60Mext+

1.00Mext− 0.75Mint− 0.60M+

0.75M− 0.75M−


M - = 0.65Mo

M + = 0.35Mo

20′

20′

33



Frame AnalysisExample 1: Analyze the flat slab shown below using DDM The slabExample 1: Analyze the flat slab shown below using DDM. The slabsupports a live load of 144 psf. All columns are 14″ square. Take fc′ =4 ksi and fy = 60 ksi.

25′

20′

25′ 25′ 25′


20′

20′

34

18



Step A: Sizes

ACI table 9 5 (c) is used for finding flat plate and flat slabACI table 9.5 (c) is used for finding flat plate and flat slabthickness.


hmin = 5 inches (slabs without drop panels)

hmin = 4 inches (slabs with drop panels)

35



Step A: Sizes

Exterior panel governs ThereforeExterior panel governs. Therefore,

hf = ln/30 = [{25 – (2 × 14/2)/12}/30] × 12 = 9.53″ (ACI minimum requirement)

Take hf = 10″


19



Step B: Loads

Service Dead Load (D L) = γ l bhfService Dead Load (D.L) γslabhf

= 0.15 × (10/12) = 0.125 ksf

Superimposed Dead Load (SDL) = Nil

Service Live Load (L.L) = 144 psf or 0.144 ksf

Factored Load (wu) = 1.2D.L + 1.6L.L


= 1.2 × 0.125 + 1.6 × 0.144 = 0.3804 ksf

37



Frame AnalysisStep No. 01 : Marking E-W Interior Frame.

Interior Frame lHalf width of panel

Panel Centerline

25′

20′

25′ 25′ 25′


Interior Frame

l1

l2 on one side

Half width of panel on other side

Col Centerline

Panel Centerline 20′

20′

20



Frame AnalysisStep No. 02 : Marking column and middle strips.

5′ 10′

5′l2Half Column strip


b) Middle Strip

25′

20′

25′ 25′ 25′



l2/4 = 20/4 = 5′

5′10′

5′

l1

ln

p

20′

20′



Frame AnalysisStep No. 03: Static Moment (Mo) calculation.

25′

20′

25′ 25′ 25′

20′

Mo = wul2ln2/8

= 540 ft-kip

l2

Mo = 540 ft-k Mo = 540 ft-k


20′

20′l1

ln =23.83′

40

21



Frame AnalysisStep No. 04: Longitudinal distribution of Static Moment (Mo).

Mext − = 0.26Mo = 140 Mext+ = 0.52Mo = 281 Mint − = 0.70Mo = 378

M 0 65M 351

25′

20′

25′ 25′ 25′

20′

Mext+

Mext− Mint-

M+

M − M −


M − = 0.65Mo = 351M + = 0.35Mo = 189

20′

20′

41



Frame AnalysisStep No. 04: Longitudinal distribution of Static Moment (Mo).


M 0 65M 351

25′

20′

25′ 25′ 25′

20′

281

140 378

189

351 351


M − = 0.65Mo = 351M + = 0.35Mo = 189

20′

20′

42

22



Frame AnalysisStep No. 05: Lateral Distribution to column and middle strips.

25′

20′

25′ 25′ 25′

20′


M 0 65M 351

0.60M+

0.75M − 0.75M −0.60Mext+

1.00Mext − 0.75Mint−


20′

20′

M − = 0.65Mo = 351M + = 0.35Mo = 189

43




0.60M+

0.75M − 0.75M −0.60Mext+

140 0.75Mint−


M 0 65M 351

25′

20′

25′ 25′ 25′

20′


M − = 0.65Mo = 351M + = 0.35Mo = 189

100 % of M ext- goes to column strip and remaining to middle strip

20′

20′

44

23




25′

20′

25′ 25′ 25′

20′

113

0.75M − 0.75M −168

140 0.75Mint −


M 0 65M 351


20′

20′

M − = 0.65Mo = 351M + = 0.35Mo = 189

60 % of Mext+ & M + goes to column strip and remaining to middle strip

45



Frame AnalysisStep No. 05: Lateral distribution to column and middle strips.

25′

20′

25′ 25′ 25′

20′

113

263 263168

140 283


M 0 65M 351


20′

20′

M − = 0.65Mo = 351M + = 0.35Mo = 189

75 % of Mint- goes to column strip and remaining to middle strip

46

24



Frame Analysis (E-W Interior Frame)Step No. 05: Lateral distribution to column and middle strips.

25′

20′

25′ 25′ 25′

20′

113

263 263168

140 283


M 0 65M 351

112/2 94/2 88/2 76/20 88/2

112/2 94/2 88/2 76/20 88/2


20′

20′

M − = 0.65Mo = 351M + = 0.35Mo = 189

112/2 94/2 88/2 76/20 88/2

47



Frame Analysis (E-W Interior Frame)Step No. 05: Lateral distribution to column and middle strips.

25′

20′

25′ 25′ 25′

20′

263 263168

140

112/2 94/2 76/20 88/2

113

283

88/211.2 9.4 8.8 7.6

14.016.8

28.3 26.3

11.3

26.3

8.8 5′ half middle strip

5′ half middle strip

10′ column strip

Mu (per foot width)= M / strip width


20′

20′

5 half middle strip

48

25



Frame Analysis (E-W Exterior Frame)Step No. 05: Lateral distribution to column and middle strips.

25′

20′

25′ 25′ 25′

20′

Mext- = 0.26Mo = 74 Mext+ = 0.52Mo = 148 Mint- = 0.70Mo = 200

89Mo = 285.68 ft-kip

l2 =10.58′

60

0

74 150 140 140

59 50 46 40 46


20′

20′

M - = 0.65Mo = 186M + = 0.35Mo = 100

49



Frame Analysis (E-W Exterior Frame)Step No. 05: Lateral distribution to column and middle strips.

15.94l2 =10.58′

10.75

0

13.26 26.8 25.1 25.1

11.87 10 9.2 8 9.2


Mo = 285.68 ft-kip

25′

20′

25′ 25′ 25′

20′


M - = 0.65Mo = 186M + = 0.35Mo = 100

20′

20′

50

26



Frame Analysis (N-S Interior Frame)Step No. 05: Lateral distribution to column and middle strips.


Mo = 421.5 ft-kip131 88/2

1100 0

22174/2

206 69/2

25′

20′

25′ 25′ 25′

20′


M - = 0.65Mo = 274M + = 0.35Mo = 148

l2 =25′

88.859/2

206 69/2

20′

20′

51



Frame Analysis (N-S Interior Frame)Step No. 05: Lateral distribution to column and middle strips.


Mo = 421.5 ft-kip13.1

11.00 0

22.14.9

20.6

8.8

4.6

25′

20′

25′ 25′ 25′

20′


M - = 0.65Mo = 274M + = 0.35Mo = 148

l2 =25′

8.883.9

20.6 4.6

20′

20′

52

27



Frame Analysis (N-S Exterior Frame)Step No. 05: Lateral distribution to column and middle strips.

Mext − = 0.26Mo = 58Mext+ = 0.52Mo = 114 Mint − = 0.70Mo = 154

Mo = 220.5 ft-kip69

58

115.5

107.3

0

45

38.5

35.75

25′

20′

25′ 25′ 25′

20′


M − = 0.65Mo = 143M + = 0.35Mo = 77

l2 =13.08′

46.2

107.3

30.8

35.75

20′

20′

53



Frame Analysis (N-S Exterior Frame)Step No. 05: Lateral distribution to column and middle strips.

Mext − = 0.26Mo = 58Mext+ = 0.50Mo = 110 Mint − = 0.70Mo = 154

Mo = 220.5 ft-kip12.32

10.4

20.69

19.2

0

6

5.13

4.76

25′

20′

25′ 25′ 25′

20′


M − = 0.65Mo = 143M + = 0.35Mo = 77

l2 =13.08′

8.27 4.12

19.2 4.76

20′

20′

54

28



Frame Analysis (E-W Direction Moments)

25′ 25′ 25′ 25′

8.8

11.3

26.3 26.316.8

14.0 28.3

9.4 7.60

11.2 8.80

15.94 10.75

0

13.26 26.8 25.1 25.1

11.87 10 9.2 8 9.2

25

20′

25 25 25

20′


20′

55



Frame Analysis (N-S Direction moments)

25′ 25′ 25′ 25′

8.88

13.1

3.9

11.00 0

22.14.9

20.6

8.8

4.6

8.27

12.32

10.4

20.69

19.2

0

6

4.12

5.13

4.76

0

8.8

4.6

25

20′

25 25 25

20′


20.6 4.619.2 4.76 4.6

20′

56

29


Comparison with SAPEW direction moments from SAP

14.9 (15.94)

24 (13)

24 (26.8)

10.0 (10.75)

24 (25.1)

24 (25.1)

11.0(11.87)

16.0(16.8)

0(0)

20(14)

8(10)

28(28.3)

8(8)

10.5(11.3)

8(9.2)

28(26.3)

6.8(9.2)

28(26.3)

16.0(16.8)

20(14)

28(28)

10.5(11.3)

28(26.3)

28(26.3)

12.5(11.2)

0(0)

9(9.4)

8(7.6)

9(8.8)

7.7(8.8)



Comparison with SAPNS direction moments from SAP

4.5(0)

24 (10.4)

25(11)

4.5 (0)

25(11)

7(5.8)

2.6(5.13)

10(11.9)

20(20.69)

12(12.7)

22(22.1)

7(5.6)

1.8(5.13)

1.5(4.76)

20(19.2)

22(20.6)

1(4.6)

4.5(4.12)

9(8.27)

9(8.88)

3.8(4.12)

9(12.7)

22(22.1)

22(20.6)

9(8.88)


30


Direct Design MethodExample 2

Analysis results of the slab shown below using DDM are presentednext. The slab supports a live load of 60 psf. Superimposed deadload is equal to 40 psf. All columns are 14″ square. Take fc′ = 3 ksiand fy = 40 ksi.

25′

20′

25′ 25′ 25′


20′

20′

59



Example 2Calculation summaryCalculation summary

Slab thickness hf = 10″

Factored load (wu) = 0.294 ksf

Column strip width = 5′


31



Example 2E-W Direction Moments (units: kip-ft)

33.9

88

204 204130

108.6 219

36.5 29.20

43.4 33.90

68 46.4

0

57 116 107 107

23 19.3 18 15.5 18

25′

20′

25′ 25′ 25′

20′


20

20′

61



Example 2N-S Direction moments (units: kip-ft)

3 9

0 0

1714.9

33.9

26.5

35 8

53.2

44.3

89.5

83.1

0

17.7

11 9

14.9

13.9

0

28.5

22.8

25′

20′

25′ 25′ 25′

20′68 4

101

84.7

159


3.9

26.5

35.8 11.9

83.1 13.9

22.8 20

20′

68.4

159

62

32


Direct Design MethodExample 3

Analysis results of the slab shown below using DDM are presentednext. The slab supports a live load of 60 psf. Superimposed deadload is equal to 40 psf. All columns are 12″ square. Take fc′ = 3 ksiand fy = 40 ksi.

20′

15′

20′ 20′ 20′


15′

15′

63



Example 3Calculation summaryCalculation summary

Slab thickness hf = 8″

Factored load (wu) = 0.264 ksf

Column strip width = 3.75′


33



Example 3E-W Direction Moments (units: kip-ft)

14.5

37.5

87.1 87.155.8

46.5 93.8

15.6 12.50

18.6 14.50

29.7 20

0

24.8 50 46.5 46.5

9.9 8.3 7.7 6.7 7.7

20′

15′

20′ 20′ 20′

15′


15

15′

65



Example 3N-S Direction moments (units: kip-ft)

3 9

0 0

67.94.9

13.5

10.5

14 3

21.2

17.7

35.7

33.1

0

7.1

4 8

5.9

5.5

0

11.3

9.1

20′

15′

20′ 20′ 20′

15′27 2

40.4

33.6

63.1


3.9

10.5

14.3 4.8

33.1 5.5

9.1 15

15′

27.2

63.1

66

34


Two Way Slabs (Requirements of ACI Code)

Maximum spacing and minimum reinforcementMaximum spacing and minimum reinforcement requirement

Maximum spacing (ACI 13.3.2):

smax = 2 hf in each direction.

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.

67



Detailing of flexural reinforcement for columnDetailing of flexural reinforcement for columnsupported two-way slabs

At least 3/4” cover for fire or corrosion protection.

Slab


3/4″

Support

68

35




In case of two way slabs supported on beams, short-direction barsare normally placed closer to the top or bottom surface of the slab,with the larger effective depth because of greater moment in shortdirection.





However in the case of flat plates/slabs, the long-direction negativeand positive bars, in both middle and column strips, are placedcloser to the top or bottom surface of the slab, respectively, with thelarger effective depth because of greater moment in long direction.


36




ACI 13.3.8.5 requires that all bottom bars within the column strip ineach direction be continuous or spliced with length equal to 1.0 ld , ormechanical or welded splices.





At least two of the column strip bars in each direction mustpass within the column core and must be anchored at exteriorsupports (ACI 13.3.8.5).


37



Detailing of flexural reinforcement for columnsupported two-way slabs




Standard Bar Cut off Points (Practical (Recommendation):

For column and middle strips both


38



Summary

Decide about sizes of slab and columns The slab depth canDecide about sizes of slab and columns. The slab depth canbe calculated from ACI table 9.5 (c).

Find Load on slab (wu = 1.2DL + 1.6LL)

On given column plan of building, decide about location anddimensions of all frames (exterior and interior)


For a particular span of frame, find static moment (Mo =wul2ln2/8).

75



Summary

Find longitudinal distribution of static moment:Find longitudinal distribution of static moment:

Exterior span (Mext - = 0.26Mo; M ext + = 0.52Mo; Mint - = 0.70Mo)

Interior span (Mint - = 0.65Mo; M int + = 0.35Mo)

Find lateral Distribution of each longitudinal moment:

100 % of Mext – goes to column strip


60 % of Mext + and Mint+ goes to column strip

75 % of Mint – goes to column strip

The remaining moments goes to middle strips

Design and apply reinforcement requirements (smax = 2hf)

76

39


Design of Two Way Slab Systems for Shear

(Flat Plate and Flat Slabs)



TopicsShear in Slabs Without Beams

Two-way shear (punch out shear)o ay s ea (pu c out s ea )

Shear strength of slab in punching shear

Various Design Options for Shear

Example


40


Two Way Slabs (General)Shear in slab without beams

Two way shear (Punch out shear)

In addition to flexure, flat plates shall also be designed for two way shear (punch out shear) stresses.




Two way shear (Punch out shear): Critical section

In shear design of beams, the critical section is taken at adistance “d” from the face of the support.

Beam

Shear crack


d

Beam

80

41




In shear design of flat plates, the critical section is an areataken at a distance “d/2” from all face of the support.

Critical perimeterColumn


Slab thickness (h)d/2

d/2d = h − cover

Tributary Area, At

Slab

81





42



Two way shear (Punch out shear): Critical perimeter, bo


bo = 2(c1+d)+2(c2+d)

83



Two way shear (Punch out shear): Critical perimeter, bo


bo = 2(c1+d/2)+ (c2+d)

84

43



Shear Strength of Slab in punching shear:

ΦVn = ΦVc + ΦVs

ΦVc is least of:

Φ4√ (fc′)bod

Φ(2 + 4/βc) √ (fc′)bod

Φ{(αsd/bo +2} √ (fc′)bod


Φ{(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

85



Shear Strength of Slab:

When ΦVc ≥ Vu (Φ = 0.75) O.K, Nothing required.

When ΦVc < Vu, then either increase ΦVc = Φ4√ (fc′)bod by:

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)


Increasing bo, critical shear perimeter: This can be done by increasing column size as a whole or by increasing size of column head (Column capital)

Increasing fc′ (high Strength Concrete)

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44



Shear Strength of Slab:Shear Strength of Slab:

And/ or provide shear reinforcement (ΦVs) in the form of:

Integral beams

Bent Bars

Shear heads

Shear studs


Shear studs

87



Drop Panels (ACI 9.5.3.2 and 13.3.7.1):


45



Column Capital:Column Capital:




Minimum depth of slab in case of shear reinforcement to beprovided as integral beams or bent bars:

ACI 11.12.3 requires the slab effective depth d to be at least6 in., but not less than 16 times the diameter of the shearreinforcement.


When bent bars and integral beams are to be used, ACI11.12.3.1 reduces ΦVc by 2

90

46


Example: Calculate the shear capacity of slab at 14″ column C1of the 10″ flat plate shown.

Two Way Slabs (General)

p

Calculation of Punching shear demand (Vu):

Critical perimeter:d = h – 1 = 9″bo = 4(c+d)

= 4(14+9) = 92″

Tributary area (excluding area


y ( gof bo):At = (25×20) – (14+9)2/144

= 496.3 ft2

wu = 0.3804 kip/ft2

Vu = wu × At = 189 kip

91


Example: Calculate the shear capacity of slab at 14″ column C1of the 10″ flat plate shown.

Two Way Slabs (General)

p

Calculation of Punching shear capacity (ΦVc):

√ (fc′)bod=√(4000)×92×9/1000=52 k

ΦVc is least of:

Φ4√ (fc′)bod = 156 k

Φ (2 + 4/β ) √ (f ′)b d = 234 k


Φ (2 + 4/βc) √ (fc )bod = 234 k

Φ{(αsd/bo +2} √ (fc′)bod = 230 k

Therefore,

ΦVc = 156 k < Vu (190 k) , N.G

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47


Two Way Slabs (General)Example: Calculate the shear capacity of slab at 14″ column C1of the 10″ flat plate shown.p

Design for shear (option 01): Drop panels

In drop panels, the slab thickness in the vicinity of the columns is increasedto increase the shear capacity (ΦVc = Φ4√ (fc′)bod) of concrete.

The increased thickness can be computed by equating Vu to ΦVc andsimplifying the resulting equation for “d” to calculate required “h”.




Design for shear (option 01): Drop panels

25/6 = 4.25′

20/6 = 3.5′

Equate Vu to ΦVc:Vu = ΦVc

189 = 0.75 × 4 √ (fc′) × 92 × dd = 10.82″Therefore, h = d+1≈ 12″


This gives 2″ drop panel. According to ACI, minimum thickness of drop panel = h/4 = 10/4 = 2.5″, which governs.Drop Panel dimensions:25/6 ≈ 4.25′; 20/6 ≈ 3.5′

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Design for shear (option 02): Column Capitals

Occasionally, the top of the columns will be flared outward, as shown infigure. This is known as column capital.

This is done to provide a larger shear perimeter at the column and toreduce the clear span, ln, used in computing moments.


ACI 6.4.6 requires that the capital concrete be placed at the same time asthe slab concrete. As a result, the floor forming becomes considerablymore complicated and expensive.

95



Design for shear (option 02): Column CapitalsEquate Vu to ΦVc:Vu = ΦVc

190 = 0.75 × 4 √ (fc′) × bo × 9bo = 111.26″


Now,bo = 4 (c + d) 111.26 = 4(c + 9)Simplification gives,c = 18.8 ≈ 19″

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49



Design for shear (option 02): Column Capitals

According to ACI code, θ < 45o

y = 2.5/ tanθ

c = 19″

2.5″

yθ


Let θ = 30o, then y ≈ 4.35″

For θ = 20o, y ≈ 7″ 14″capital

column

y

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Design for shear (option 03): Integral Beams

Vertical stirrups are used inconjunction with supplementaryhorizontal bars radiating outwardin two perpendicular directionsfrom the support to form what are

Vertical stirrupsFor 4 sides, total stirrup area is 4 times individual 2 legged stirrup area

lv


pptermed integral beams containedentirely within the slab thickness.

In such a way, critical perimeter isincreased

Horizontal barsIncreasedcriticalperimeter

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50





bo = 4R + 4c

99




ΦVc = 156 kips

When integral beams are to be used, ACI 11.12.3 reduces ΦVc by 2.Therefore ΦVc = 156/2 = 78 kips

Using 3/8″ Φ, 2 legged (0.22 in2), 4 (side) = 4 × 0.22 = 0.88 in2


Spacing (s) = ΦAvfyd/ (Vu – ΦVc)

s = 0.75 × 0.88 × 60 × 9/ (190 – 78) = 3.18 ≈ 3″

Maximum spacing allowed d/2 = 6/2 = 3″ controls.

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Design for shear (option 03): Integral BeamsFour #5 bars are to be provided in each direction to hold the stirrups. We know minimum bo =111.26″

bo = 4R + 4c1 ........ (1)

R = √ (x2 + x2)

From figure, x = (3/4)(lv – c1/2), therefore,


R = √ (2) x, and eqn. (1) becomes,

bo = 4√ (2) x + 4c1

bo = 4√ (2){(3/4)(lv – c1/2)} + 4c1

Or bo = 4.24lv – 2.12c1 + 4c1= 4.24lv + 1.88c1

Therefore lv ≈ 20″

101



Design for shear (option 03): Integral Beams details.

lv = 20″ ≈ 24″ or 2′


52


Additional Requirements for Slab with Beams


DDM Limitations:

For slabs with beams between supports on all sides (ACI 13.6.1.6):

Where,

0.2 ≤ α1l22/α2l1

2 ≤ 5.0

α = EcbIb / EcsIs


Ecb = Modulus of elasticity of beam concrete

Ecs = Modulus of elasticity of slab concrete

Ib = Moment of inertia of beam section

Is = Moment of inertia of slab section

103




DDM Limitations:

Explanation of Ib and Is:

α = EcbIb / EcsIs


53




Example on α calculation

hf = 7″, hw = 18″, bw = 12″

Effective flange width

bw + 2hw = 48″, bw + 8hf = 68, 48″ governs

Ib = 33060 in4 OR,

20′20′

25′


IT-section ≈ 2Irectangle section & IL-section ≈ 1.5Irectangle section

Ib = 2 × 12 × 243/12 = 27648 in4

Is = (10 + 10) × 12 × 73/12 = 6860 in4

α = Ib/Is = 33060/ 6860 = 4.82

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Longitudinal Distribution of Static Moments


54




Longitudinal Distribution of Static Moments





Lateral Distribution of Longitudinal Moments

Column Strip Moments

ACI tables 13.6.4.1, 13.6.4.2 and 13.6.4.4 of the ACI are used toassign moments to column strip.


55















56





Middle Strip Moments

The remaining moments are assigned to middle strip in accordancewith ACI 13.6.6.

Beams between supports shall be proportioned to resist 85 percent ofcolumn strip moments if α1l2/l1 {Where l2 shall be taken as full span


column strip moments if α1l2/l1 {Where l2 shall be taken as full spanlength irrespective of frame location (exterior or interior)} is equal to orgreater than 1.0 (ACI 13.6.5.1).

111





Graph A4

Lateral distribution of longitudinal moments can also be done usingGraph A.4 (Design of Concrete Structures, Nilson 13th Ed)


57


Additional


Requirements for Slab with Beams


In graph A.4, l2 shall be


taken as full spanlength irrespective offrame location (exterioror interior).

113




Example on graph A4:

Find the lateral distribution to column strip of positive andinterior negative moments using graph A4. Take

l2/l1 = 1.3

αl /l > 1


αl2/l1 > 1.

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58


Additional



Example on graph A4

l2/l1 = 1.3

αl /l > 1


αl2/l1 > 1

65 % of positive longitudinal moment will go to column strip

65 % of interior negative longitudinal moment will go to column strip

115





Torsional Stiffness Factor (βt)

In the presence of an exterior beam, all of the exterior negativefactored moment goes to the column strip, and none to the middlestrip, unless the beam torsional stiffness is high relative to the flexuralstiffness of the supported slab.


Torsional stiffness factor βt is the parameter accounting for this effect.βt reflects the relative restraint provided by the torsional resistance ofthe effective transverse edge beam.

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59






For a considered frame, the transverse edge beam


provides restraint through its torsional resistance.

117







60





Determination of βt:

Where walls are used as supports along column lines, they can beregarded as very stiff beams with an α1l2/l1 value greater than one.

Where the exterior support consists of a wall perpendicular to thedirection in which moments are being determined βt may be taken as


direction in which moments are being determined, βt may be taken aszero if the wall is of masonry without torsional resistance.

βt may be taken as 2.5 for a concrete wall with great torsionalresistance that is monolithic with the slab.

119






βt can be calculated using the following formula:


61






Where,

Ecb = Modulus of elasticity of beam concrete;

Ecs = Modulus of elasticity of slab concrete


C = cross-sectional constant to define torsional properties

x = shorter overall dimension of rectangular part of cross section, in.

y = longer overall dimension of rectangular part of cross section, in.

Is = Moment of inertia of slab section spanning in direction l1 and having width bounded by panel centerlines in l2 direction.

121






C for βt determination can be calculated using the following formula.

y2y2


x2x1

y1

x2

x1

y1122

62





Determination of βt (Example): For determination of E-W frame exteriornegative moment distribution to column strip, find βt for beam marked. Takeslab depth = 7″ and Ecb = Ecs.


Exterior edge beam(12″ × 24″)

123





Determination of βt (Example):

βt = EcbC/(2EcsIs) = C/ (2Is)

Calculation of C:

7″

hw ≤ 4hf = 17″


12″

24″

124

63







Calculation of C:

C = {1 – 0.63×12/24}{123 ×24/3} + {1 – 0.63×7/17}{73 ×17/3} = 10909 in4

Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structuresx1 =12″

y1= 24″x2 = 7″

y2 = 17″

12

125







Calculation of C:

C = {1 – 0.63×12/17}{123 ×17/3} + {1 – 0.63×7/29}{73 ×29/3} = 8249 in4

Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structuresx1 =12″

y1= 17″

x2 = 7″

y2 = 17″ + 12″ = 29″

1

2

126

64







Calculation of C:

Therefore, C = 10909 in4






Determination of βt (Example): βt = EcbC/(2EcsIs) = C/ (2Is)

Calculation of Is:

Is = bhf3/12 = (20 × 12) × 73/12 = 6860 in4


b

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65






βt = C/ (2Is)

= 10909/ (2 × 6860) = 0.80



Additional



Lateral Distribution ofLongitudinal Moments

Once βt is known,

βt = 0.8 90 % of exterior negative moment goes to column strip


exterior negative moment in column strip can be found. For,

l2/l1 = 1.3

αl2/l1 > 1 and βt = 0.8

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66


Two Way Slabs (General)Minimum thickness for two way slab:

For 0 2 ≤ α ≤ 2:For 0.2 ≤ αm ≤ 2:

But not less than 5 in. fy in psi.

For αm > 2:

( )2.0536200,000

8.0

m

yn

−+

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=αβ

fl

h


But not less than 3.5 in. fy in psi.β936

200,0008.0 y

n

+

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

fl

h

131


Two Way Slabs (General)Minimum thickness for two way Slab:

h = Minimum slab thickness without interior beams.

ln = length of clear span in direction that moments are beingdetermined, measured face-to-face of supports.

β = ratio of clear spans in long to short direction of two-wayslabs.


αm = average value of α for all beams on edges of a panel.

For αm < 0.2, use the ACI table 9.5 (c).

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Special Reinforcement at exterior corner of SlabThe reinforcement at exterior ends of the slab shall be provided as per ACIThe reinforcement at exterior ends of the slab shall be provided as per ACI13.3.6 in top and bottom layers as shown.

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



References

ACI 318-02

Design of Concrete Structures (Chapter 13) 13th Ed byDesign of Concrete Structures (Chapter 13), 13 Ed. by Nilson, Darwin and Dolan.


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The End


lecture 09 analysis and design of flat plate slabs

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