Lecture 09 Analysis and Design of Flat Plate Slabs_2011

8/12/2019 Lecture 09 Analysis and Design of Flat Plate Slabs_2011

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

Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Fall 2011

Lecture-09

Analysis and Design of Two-way Slab

System without Beams

(Flat Plate and Flat Slabs)

By: Prof Dr. Qaisar Ali

Civil Engineering Department

UET Peshawar

[email protected]

1



Two Way Slabs

2


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Topics Addressed

Two Way Slabs

Behavior

Types

Analysis and Design Considerations

3



Topics Addressed

Direct Design Method

Introduction

Limitations

Frame Analysis Steps for Flat Plates and Flat Slabs

Frame marking

Column and middle strips marking

Static moment calculation

Longitudinal distribution of static moment

Lateral distribution of longitudinal moment

4


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Two Way Slabs

Behavior

A slab having bending in both directions is called two-way

slab (Long span/short span < 2).

5

25 25 25 25

20

20

20



Two Way Slabs

Behavior

Short direction moments in two-way slab.

6

Short

Direction

25

25

25

25

20

20

20


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Two Way Slabs

Behavior

Long direction moments in two-way slab.

7

Long

Direction

25

25

25

25

20

20

20



Two Way Slabs

Behavior:More Demand (Moment) in short direction

due to size of slab

central Strip= (5/384)wl4/EI

As these imaginary strips are part of monolithic slab, the deflection at any

point, of the two orthogonal slab strips must be same:

a = b(5/384)wala

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

wa/wb= lb4/la

4 wa= wb(lb4/la

4)

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

8


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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 dimensional

members. Therefore their analysis except one-way slab

systems is relatively difficult.

Design

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

usual manner. So no problem in design, problem is only in

analysis of slabs.

10


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Two Way Slabs

Approximate Analysis Methods of ACI

11

Slab System Applicable Analysis Methods

One-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

Two-way slabs with shallow

beams or without beams

Direct Design Method,

Equivalent Frame Analysis Method




12


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Introduction

In DDM, frames rather than panels are analyzed as is done in

analysis of two way slabs with beams using ACI moment

coefficients.

13

Interior Frame

Exterior Frame

Interior Frame

Exterior Frame

25

20

25 25 25

20

20



25

20

25 25 25

20

20


Introduction

For complete analysis of slab system, frames are

analyzed in E-W and N-S directions.

14

E-W FramesN-S Frames


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Introduction

Though DDM is useful for analysis of slabs, specially

without beams, the method is applicable with some

limitations as discussed next.

15




Limitations (ACI 13.6.1)

16

Uniformly distributed loading (L/D 2)

2

1 121/3Three or more spans

Column offset 2/10

Rectangular slab

panels (2 or less:1)


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Limitations (ACI 13.6.1): Example

17

15 15If 10

DDM APPLICABLE as 2/3 (15) = 10

15 15If


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Frame Analysis

Step No. 01 (continued):

An interior frame

19

Interior Frame

l1

l2

25

20

25 25 25

20

20



25

20

25 25 25

20

20


Frame Analysis


20

Interior Frame

l1

l2Half width of panel

on one side

Half width of panel

on other side

Marking an E-W Interior Frame

Col Centerline

Panel Centerline

Panel Centerline


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25

20

25 25 25

20

20


Frame Analysis


21

Exterior Frame

l1

Marking an E-W Exterior Frame

Note: For exterior frames

l2= Panel width/2 +h2/2


on one side

h2/2




Frame Analysis

Step No. 02: A frame is divided further into strips known as

column and middle strips (Defined in ACI 13.2).

Column Strip:A column strip is a design strip with a width on each

side of a column centerline equal to 25 percent of l1or l2, whichever

is less.

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

strips.

22

25

20

25 25 25

20

20

l2Column strip

Full Middle strip

Half Middle strip

l2


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Frame Analysis

Step No. 02 (continued): Why a frame is divided into column

and middle strips?

Because the slab portion on the column centerline will offer more

resistance than the rest of the slab.

23

25

20

25 25 25

20

20




Frame Analysis


24

25

20

25 25 25

20

20

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

CS/2

CS/2C.S

M.S/2

M.S/2

l2

l1

ln

Half Column strip

a) Marking Column Strip

b) Middle Strip


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


25

-Middle strip

-Middle strip

n

Column strip

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

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

(l2)A

(l2)B



25

20

25 25 25

20

20


Frame Analysis

Step No. 02 (continued): For l1 = 25 and l2 = 20, CS and MS

widths are given as follows.

26


l2/4 = 20/4 = 5

55

10

5

5

l2

l1

ln

Half Column strip


b) Middle Strip


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Frame Analysis

Step No. 03: Calculate Static Moment (Mo) for interior span of

frame.

27

25

20

25 25 25

20

20

MoMo=wu 2 n

2

8l2

ln

Span of frame




Frame Analysis

Step No. 04: Longitudinal Distribution of Static Moment (Mo).

28

M+

M M

M = 0.65Mo

M + = 0.35Mo

25

20

25 25 25

20

20


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Frame Analysis

Step No. 05: Lateral Distribution to column and middle strips.

29

M = 0.65Mo

M + = 0.35Mo

0.60M +

0.75M 0.75M

25

20

25 25 25

20

20




Frame Analysis

Step No. 03: Calculate Static Moment (Mo) for exterior span of

frame.

30

25

20

25 25 25

20

20

MoMo=wu 2 n

2

8l2

ln


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Frame Analysis

Step No. 04: Longitudinal distribution of static moment (Mo).

31

Mext = 0.26Mo

M ext+ = 0.52Mo

Mint- = 0.70Mo

Mext+

Mext Mint

25

20

25 25 25

20

20




Frame Analysis


32

M ext+ = 0.52Mo

Mint- = 0.70Mo

Mext = 0.26Mo0.60Mext+

1.00Mext 0.75Mint

25

20

25 25 25

20

20


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Frame Analysis


33

M ext+ = 0.52Mo

Mint- = 0.70Mo

M - = 0.65Mo

M + = 0.35Mo

Mext = 0.26Mo

25

20

25 25 25

20

20

0.60Mext+

1.00Mext 0.75Mint 0.60M+

0.75M 0.75M




Frame Analysis

Example 1:Analyze the flat slab shown below using DDM. The slab

supports a live load of 144 psf. All columns are 14square. Take fc=

4 ksi and fy= 60 ksi.

34

25

20

25 25 25

20

20


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Step A: Sizes

ACI table 9.5 (c) is used for finding flat plate and flat slab

thickness.

hmin= 5 inches (slabs without drop panels)

hmin= 4 inches (slabs with drop panels)

35




Step A: Sizes

Exterior panel governs. Therefore,

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

requirement)

Take hf= 10

36


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Step B: Loads

Service Dead Load (D.L) = slabhf

= 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 Analysis

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

38

Interior Frame

l1


on one side

Half width of panel

on other side

Col Centerline

Panel Centerline

Panel Centerline

25

20

25 25 25

20

20


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Frame Analysis

Step No. 02 : Marking column and middle strips.

39


l2/4 = 20/4 = 5

55

10

5

5

l2

l1

ln

Half Column strip


b) Middle Strip

25

20

25 25 25

20

20



25

20

25 25 25

20

20


Frame Analysis

Step No. 03: Static Moment (Mo) calculation.

40

Mo = wul2ln2/8

= 540 ft-kip

l2

l1

ln =23.83

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


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

41

Mext = 0.26Mo = 140

Mext+ = 0.52Mo = 281

Mint = 0.70Mo = 378

M = 0.65Mo = 351M+ = 0.35Mo = 189

25

20

25 25 25

20

20

Mext+

Mext Mint-

M+

M M




Frame Analysis

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

42

Mext = 0.26Mo = 140

Mext+ = 0.52Mo = 281

Mint = 0.70Mo = 378

M = 0.65Mo = 351M+ = 0.35Mo = 189

25

20

25 25 25

20

20

281

140 378

189

351 351


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25

20

25 25 25

20

20


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

43

Mext = 0.26Mo = 140

Mext+ = 0.52Mo = 281

Mint = 0.70Mo = 378

M = 0.65Mo = 351M+ = 0.35Mo = 189

0.60M+

0.75M 0.75M

0.60Mext+

1.00Mext 0.75Mint



0.60M+


Frame Analysis


44

0.75M 0.75M

0.60Mext+

140 0.75Mint

Mext = 0.26Mo = 140

Mext+ = 0.52Mo = 281

Mint = 0.70Mo = 378

M = 0.65Mo = 351M+ = 0.35Mo = 189

100 % of Mext-goes to

column strip and

remaining to middle strip

25

20

25 25 25

20

20


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25

20

25 25 25

20

20


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

45

113

0.75M 0.75M

168

140 0.75Mint

Mext = 0.26Mo = 140

Mext+ = 0.52Mo = 281

Mint = 0.70Mo = 378

M = 0.65Mo = 351M+ = 0.35Mo = 189

60 % of Mext+& M+goes

to column strip and




25

20

25 25 25

20

20


Frame Analysis

Step No. 05: Lateral distribution to column and middle strips.

46

113

263 263

168

140 283

Mext = 0.26Mo = 140

Mext+ = 0.52Mo = 281

Mint = 0.70Mo = 378

M = 0.65Mo = 351M+ = 0.35Mo = 189

75 % of Mint-goes to

column strip and



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25

20

25 25 25

20

20


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

47

113

263 263

168

140 283

Mext = 0.26Mo = 140

Mext+ = 0.52Mo = 281

Mint = 0.70Mo = 378

M = 0.65Mo = 351M+ = 0.35Mo = 189

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

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



25

20

25 25 25

20

20


Frame Analysis (E-W Interior Frame)


48

263 263

168

140

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

113

283

88/211.2 9.4 8.8 7.6

14.0

16.8

28.3 26.3

11.3

26.3

8.8 5 half middle strip

5 half middle strip10 column strip

Mu(per foot width)

= M / strip width


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25

20

25 25 25

20

20


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

49

Mext- = 0.26Mo = 74

Mext+ = 0.52Mo = 148

Mint- = 0.70Mo = 200

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

89Mo = 285.68 ft-kip

l2 =10.58

60

0

74 150 140 140

59 50 46 40 46




Frame Analysis (E-W Exterior Frame)


50

15.94

l2 =10.5810.75

0

13.26 26.8 25.1 25.1

11.87 10 9.2 8 9.2

Mext- = 0.26Mo = 74

Mext+ = 0.52Mo = 148

Mint- = 0.70Mo = 200

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

Mo = 285.68 ft-kip

25

20

25 25 25

20

20


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

51

Mext- = 0.26Mo = 110

Mext+ = 0.52Mo = 219

Mint- = 0.70Mo = 295

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

Mo = 421.5 ft-kip

l2 =25

88.8

131 88/2

59/2

1100 0

22174/2

206 69/2

206 69/2

25

20

25 25 25

20

20




Frame Analysis (N-S Interior Frame)


52

Mext- = 0.26Mo = 110

Mext+ = 0.52Mo = 219

Mint- = 0.70Mo = 295

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

Mo = 421.5 ft-kip

l2 =25

8.88

13.1

3.9

11.00 0

22.14.9

20.6

20.6

8.8

4.6

4.6

25

20

25 25 25

20

20


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

53

Mext = 0.26Mo = 58

Mext+ = 0.52Mo = 114

Mint = 0.70Mo = 154

M = 0.65Mo = 143M+ = 0.35Mo = 77

Mo = 220.5 ft-kip

l2 =13.08

46.2

69

58

115.5

107.3

107.3

0

45

30.8

38.5

35.75

35.75

25

20

25 25 25

20

20




Frame Analysis (N-S Exterior Frame)


54

Mext = 0.26Mo = 58

Mext+ = 0.50Mo = 110

Mint = 0.70Mo = 154

M = 0.65Mo = 143M+ = 0.35Mo = 77

Mo = 220.5 ft-kip

l2 =13.08

8.27

12.32

10.4

20.69

19.2

0

6

4.12

5.13

4.76

19.2 4.76

25

20

25 25 25

20

20


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Frame Analysis (E-W Direction Moments)

55

8.8

11.3

26.3 26.3

16.8

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




Frame Analysis (N-S Direction moments)

56

8.88

13.1

3.9

11.00 0

22.14.9

20.6

20.6

8.8

4.6

4.6

8.27

12.32

10.4

20.69

19.2

0

6

4.12

5.13

4.76

19.2 4.76

0

8.8

4.6

4.6

25

20

25 25 25

20

20


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Comparison with SAP

EW direction moments from SAP

57

11.0

(11.87)

14.9

(15.94)

16.0

(16.8)

0

(0)

24

(13)

20

(14)

8

(10)

24

(26.8)

28

(28.3)

8

(8)

10.0

(10.75)

10.5

(11.3)

8

(9.2)

24

(25.1)

28

(26.3)

6.8

(9.2)

24

(25.1)

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 SAP

NS direction moments from SAP

58

7

(5.8)

4.5

(0)

2.6

(5.13)

10

(11.9)

24

(10.4)

20

(20.69)

12

(12.7)

25

(11)

22

(22.1)

7

(5.6)

4.5

(0)

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)

25

(11)

22

(22.1)

22

(20.6)

9

(8.88)


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Example 2

Analysis results of the slab shown below using DDM are presented

next. The slab supports a live load of 60 psf. Superimposed dead

load is equal to 40 psf. All columns are 14square. Take fc= 3 ksi

and fy= 40 ksi.

59

25

20

25 25 25

20

20




Example 2

Calculation summary

Slab thickness hf= 10

Factored load (wu) = 0.294 ksf

Column strip width = 5

60


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Example 2

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

61

33.9

88

204 204

130

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




Example 2

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

62

3.9

0 0

1714.9

33.9

26.5

26.5

35.8

53.2

44.3

89.5

83.1

0

17.7

11.9

14.9

13.9

83.1 13.9

0

28.5

22.8

25

20

25 25 25

20

20

68.4

101

84.7

159

159


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Example 3

Analysis results of the slab shown below using DDM are presented

next. The slab supports a live load of 60 psf. Superimposed dead

load is equal to 40 psf. All columns are 12square. Take fc= 3 ksi

and fy= 40 ksi.

63

20

15

20 20 20

15

15




Example 3

Calculation summary

Slab thickness hf= 8

Factored load (wu) = 0.264 ksf

Column strip width = 3.75

64


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Example 3

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

65

14.5

37.5

87.1 87.1

55.8

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




Example 3

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

66

3.9

0 0

67.94.9

13.5

10.5

10.5

14.3

21.2

17.7

35.7

33.1

0

7.1

4.8

5.9

5.5

33.1 5.5

0

11.3

9.1

20

15

20 20 20

15

15

27.2

40.4

33.6

63.1

63.1


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Two Way Slabs

(Requirements of ACI Code)

67

Maximum 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 hffor grade 60.

Asmin = 0.002 b hf for grade 40 and 50.



Two Way Slabs


Detailing of flexural reinforcement for column

supported two-way slabs

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

68

3/4

Slab

Support


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Two Way Slabs

(Requirements of ACI Code) Detailing of flexural reinforcement for column


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.

69



Two Way Slabs




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.

70


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Two Way Slabs



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 ld, or

mechanical or welded splices.

71

ld

Slab

Support



Two Way Slabs




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).

72


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Two Way Slabs



73



Two Way Slabs

(Requirements of ACI Code) Standard Bar Cut off Points (Practical

Recommendation):

For column and middle strips both

74


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Summary

Decide about sizes of slab and columns. The slab depth can

be 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 and

dimensions of all frames (exterior and interior)

For a particular span of frame, find static moment (Mo =

wul2ln

2/8).

75




Summary

Find longitudinal distribution of static moment:

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

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

Find lateral Distribution of each longitudinal moment:

100 % of Mextgoes to column strip

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

75 % of Mintgoes to column strip

The remaining moments goes to middle strips

Design and apply reinforcement requirements (smax= 2hf)

76


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Design of Two Way Slab Systems for

Shear

(Flat Plate and Flat Slabs)

77



Topics

Shear in Slabs Without Beams

Two-way shear (punch out shear)

Shear strength of slab in punching shear

Various Design Options for Shear

Example

78


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Two Way Slabs (General)

79

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.




80


Two way shear (Punch out shear): Critical section

In shear design of beams, the critical section is taken at a

distance dfrom the face of the support.

d

Beam

Shear

crack


41/68




81



In shear design of flat plates, the critical section is an area

taken at a distance d/2from all face of the support.

Slab thickness (h)

Critical perimeter

d/2d/2

d = h cover

Tributary Area, At

Column

Slab




82




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85


Shear Strength of Slab in punching shear:

Vn= Vc+ Vs

Vcis least of:

4 (fc)bod

(2 + 4/c) (fc)bod

{(sd/bo+2} (fc)bod

c= longer side of column/shorter side of column

s= 40 for interior column, 30 for edge column, 20 for corner columns




86


Shear Strength of Slab:

When Vc Vu( = 0.75) O.K, Nothing required.

When Vc< Vu, then either increase Vc= 4 (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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87


Shear Strength of Slab:

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

Integral beams

Bent Bars

Shear heads

Shear studs




88


Drop Panels (ACI 9.5.3.2 and 13.3.7.1):


45/68




89


Column Capital:




90


Minimum depth of slab in case of shear reinforcement to be

provided as integral beams or bent bars:

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

11.12.3.1 reduces Vcby 2


46/68



Example:Calculate the shear capacity of slab at 14column C1

of the 10flat plate shown.

Calculation of Punching shear demand (Vu):


91

Critical perimeter:

d = h1 = 9bo= 4(c+d)

= 4(14+9) = 92

Tributary area (excluding area

of bo):

At= (2520)(14+9)2/144

= 496.3 ft2

wu= 0.3804 kip/ft2

Vu= wuAt= 189 kip



Example: Calculate the shear capacity of slab at 14column C1


Calculation of Punching shear capacity (Vc):


92

(fc)bod=(4000)929/1000=52 k

Vcis least of:

4 (fc)bod = 156 k

(2 + 4/c) (fc)bod= 312 k

{(sd/bo+2} (fc)bod= 307 k

Therefore,

Vc= 156 k < Vu(190 k) , N.G


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93



Design for shear (option 01): Drop panels

In drop panels, the slab thickness in the vicinity of the columns is increased

to increase the shear capacity (Vc= 4(fc)bod) of concrete.

The increased thickness can be computed by equating Vu to Vc and

simplifying the resulting equation for dto calculate required h.




94



Design for shear (option 01): Drop panels

25/6 = 4.25

20/6 = 3.5

Equate Vuto Vc:

Vu= Vc

189 = 0.75 4 (fc) 92 d

d = 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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95



Design for shear (option 02): Column Capitals

Occasionally, the top of the columns will be flared outward, as shown in

figure. This is known as column capital.

This is done to provide a larger shear perimeter at the column and to

reduce the clear span,ln, used in computing moments.

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.




96




Equate Vuto Vc:

Vu= Vc

190 = 0.75 4 (fc) bo9

bo= 111.26

Now,

bo= 4 (c + d)

111.26 = 4(c + 9)

Simplification gives,

c = 18.8 19


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97




According to ACI code, < 45o

y = 2.5/ tan

Let = 30o, then y 4.35

For = 20o, y 7

c = 19

14

2.5

capital

column

y




98



Design for shear (option 03): Integral Beams

Vertical stirrups are used in

conjunction with supplementary

horizontal bars radiating outward

in two perpendicular directionsfrom the support to form what are

termed integral beams contained

entirely within the slab thickness.

In such a way, critical perimeter is

increased

Vertical stirrups

For 4 sides, total

stirrup area is 4

times individual 2

legged stirrup area

Horizontal bars

lv

Increased

critical

perimeter


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99




bo= 4R + 4c




100




Vc= 156 kips

When integral beams are to be used, ACI 11.12.3 reduces Vc by 2.

Therefore Vc = 156/2 = 78 kips

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

Spacing (s) = Avfyd/ (VuVc)

s = 0.75 0.88 60 9/ (19078) = 3.18 3

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


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101




Four #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)(lvc1/2), therefore,

R = (2) x, and eqn. (1) becomes,

bo= 4(2) x + 4c1

bo= 4(2){(3/4)(lvc1/2)} + 4c1

Or bo= 4.24lv2.12c1+ 4c1= 4.24lv+ 1.88c1

Therefore lv20




102



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

lv= 2024or 2


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Additional Requirements for Slab with Beams

DDM Limitations:

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

Where,

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

0.2 a1l22/a2l1

25.0

a= EcbIb / EcsIs




DDM Limitations:

Explanation of Iband Is:


104

= EcbIb/ EcsIs


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Example on calculation

hf= 7, hw= 18, bw= 12

Effective flange width

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

Ib= 33060 in4 OR,

IT-section2Irectangle section& IL-section1.5Irectangle section

Ib= 2

12

243

/12 = 27648 in4

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

= Ib/Is= 33060/ 6860 = 4.82


105

20

2025




Longitudinal Distribution of Static Moments


106


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Longitudinal Distribution of Static Moments


107




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 to

assign moments to column strip.


108


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109









110


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Middle Strip Moments

The remaining moments are assigned to middle strip in accordance

with ACI 13.6.6.

Beams between supports shall be proportioned to resist 85 percent of

column strip moments if 1l2/l1 {Where l2 shall be taken as full span

length irrespective of frame location (exterior or interior)} is equal to or

greater than 1.0 (ACI 13.6.5.1).


111





Graph A4

Lateral distribution of longitudinal moments can also be done using

Graph A.4 (Design of Concrete Structures, Nilson 13thEd)


112


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Additional

Requirements for

Slab with Beams

Lateral Distribution of

Longitudinal Moments

In graph A.4, l2shall be

taken as full span

length irrespective of

frame location (exterior

or interior).


113




Example on graph A4:

Find the lateral distribution to column strip of positive and

interior negative moments using graph A4. Take

l2/l1= 1.3

l2/l1> 1.


114


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Additional

Requirements for

Slab with Beams

Example on graph A4

l2/l1= 1.3

l2/l1> 1


115

65 % of positivelongitudinal moment

will go to column

strip

65 % of interiornegative longitudinal

moment will go to

column strip





Torsional Stiffness Factor(t)

In the presence of an exterior beam, all of the exterior negative

factored moment goes to the column strip, and none to the middle

strip, unless the beam torsional stiffness is high relative to the flexural

stiffness of the supported slab.

Torsional stiffness factor tis the parameter accounting for this effect.

t reflects the relative restraint provided by the torsional resistance of

the effective transverse edge beam.


116


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Torsional Stiffness Factor (t)


117

For a considered frame, the

transverse edge beam

provides restraint through its

torsional resistance.





Torsional Stiffness Factor (t)


118


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Determination of t:

Where walls are used as supports along column lines, they can be

regarded as very stiff beams with an 1l2/l1value greater than one.

Where the exterior support consists of a wall perpendicular to the

direction in which moments are being determined, t may be taken as

zero if the wall is of masonry without torsional resistance.

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


119





Determination of t:

tcan be calculated using the following formula:


120


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Determination of t:

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 l1and having

width bounded by panel centerlines in l2direction.


121





Determination of t:

C for tdetermination can be calculated using the following formula.


122

x2x1

y1

y2

x2

x1

y1

y2


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Determination of t (Example): For determination of E-W frame exterior

negative moment distribution to column strip, find tfor beam marked. Take

slab depth = 7and Ecb= Ecs.


123

Exterior edge beam

(12 24)





Determination of t(Example):

t= EcbC/(2EcsIs) = C/ (2Is)

Calculation of C:


124

12

24

7

hw 4hf= 17


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Calculation of C:

C = {10.6312/24}{12324/3} + {10.637/17}{7317/3} = 10909 in4


125x1=12

y1= 24x2= 7

y2= 17

12







Calculation of C:

C = {10.6312/17}{12317/3} + {10.637/29}{7329/3} = 8249 in4


126x1=12

y1= 17

x2= 7

y2= 17 + 12 = 29

1

2


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Calculation of C:

Therefore, C = 10909 in4


127





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

Calculation of Is:

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


128

b


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t= C/ (2Is)

= 10909/ (2 6860) = 0.80


129



Additional

Requirements for

Slab with Beams

Lateral Distribution of

Longitudinal Moments

Once tis known,

exterior negative

moment in column strip

can be found. For,

l2/l1= 1.3

l2/l1> 1 and t= 0.8


130

t= 0.890 % of exterior negative moment goes to column strip


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131

Minimum thickness for two way slab:

For 0.2 m 2:

But not less than 5 in. fyin psi.

For m> 2:

But not less than 3.5 in. fyin psi.

2.0536200,000

8.0

m

y

n

a

fl

h

936200,000

8.0 y

n

fl

h




132

Minimum thickness for two way Slab:

h = Minimum slab thickness without interior beams.

ln = length of clear span in direction that moments are being

determined, measured face-to-face of supports.

= ratio of clear spans in long to short direction of two-way

slabs.

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

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


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Special Reinforcement at exterior corner of Slab

The reinforcement at exterior ends of the slab shall be provided as per ACI

13.3.6 in top and bottom layers as shown.

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.

133



References

ACI 318-02

Design of Concrete Structures (Chapter 13), 13thEd. by

Nilson, Darwin and Dolan.

134


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Lecture 09 Analysis and Design of Flat Plate Slabs_2011

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