IN THE NAME OF ALLAH, THE MOST BENEFICENT,

THE MOST MERCIFUL

DESIGN OF FLAT PLATE

DESIGN OF FLAT PLATE

A five storey building has a line plan as shown below.

The floor consist of reinforced concrete flat plate with no edge

beam and has a ceiling height of 10 ft. The building is

subjected to gravity loads only. The dead load consist of 2 ½”

F.F, ½” ceiling plaster, 10 psf for mechanical fixtures and 25

psf for partition load. The live load = 60 psf. The external wall

weighs 350 Ib/ft. f’c = 4 ksi and fy = 60 ksi. Design the end

panel Q of the floor system. Check the conditions of DDM.Q

S

W

N

E

16'

16'

16'

18' 18' 18'

P

RR S

Q

16“ x 12"

16“ x 16“ COL

16“ x 12" COL

LINE PLAN

SOLUTIONSlab Thickness Refer to table 9.5 (c) of ACI Code.

h = ln/30 = 200/30 = 6.66" say 7.0"ln= 18x12 – 16 = 200"

Check for Geometry and Loading Condition of DDMACI 13.6.1 Refers

Three or more spans is each direction Panels are rectangular and 18/16 = 1.125 <2.0 Successive span don't differ No column offset Loads are due to gravity only

wd= 7x12.5+30+6+10+25= 158.5 psfwl= 60 psf2wd > wl ok

No beam present

Check for Shear LOADS

wu = 1.4(158.5) + 1.7x60 = 324 psf

Assuming ¾" clear cover and # 4 bar being used.

d= 7- 0.75 - 0.5/2 = 6"

Interior Column Critical section for punching shear is at a

distance d/2 from face of support.

Vu = [18x16 – (22/12)2]x324

= 92220 Ib

bo = 22x4 = 88"

According to ACI 11.12.2.1, Vc is smallest of the following

16"

22"

22" 16"

16"

22"

22" 16’

18’

Assumed Loaded Area for Interior Column

16"

• Vc= (2+4/βc)x √fc' bod βc = 1.0

=(2+4/1.0) √4000x88x6 =200362 lb

• Vc = (αsd/bo+2)√fc'bod

=(40x6/88+2) √4000x88x6 =157860 Ib.

αs = 40 for interior column

• Vc = 4√fc' bod = 4x√4000x88x6 = 133574 Ib.

Vc is the lowest of above three values i.e. 133574

Ib.

ΦVc = 0.85x133574= 113538 Ib.

ΦVc > Vu Safe

Exterior Column

bo = 15x2 + 22 = 52"

22"

15"

15”

22”

18’

8.5’

Assumed loaded area for exterior column

Shear is caused by floor load and weight of exterior wall.

Vu= [18x (8+0.5) – 22x15/144] 324 +[(18-16/12) 350 x 1.4]

= 57000 Ib.

Vc is smallest of the following

Vc = (2+4/βc)√ fc‘ bod = (2+4/1.33)√4000x52x6 = 98678 Ib

βc= 16/12 = 1.33

Vc = (αsd/bo+2) √fc' bod αs = 30 for exterior column

=(30x6/52+2) √4000x52x6 =107770 lb

Vc =4 √fc’ bo d =4x√4000x52x6 = 78930 lb

ΦVc = 0.85x78930 = 67090 Ibs.

ΦVc > Vu Safe

Total Factored Static Moment in E-W Dir and its Distr

Equivalent Rigid Frame on Inner Column Line

Mo = wul2ln2 /8 = 0.324x16(16.67)2/8 = 180.07 kft

ln = 18 - 16/12 = 16.67 ft

D.F ACI 13.6.3.2

- ve moment = 0.65Mo= 117.05 k'

+ ve moment = 0.35Mo= 63.02 k'

Moment in Column Strip ACI 13.6.4

l2/l1 = 16/18= 0.89, αl2/l1 = 0

- ve moment in C.S = 75 %

+ ve moment in C.S = 60 %

0.65 0.35 0.65

S

W

N

E

18' 18' 18'

P

RR S

Q

16“ x 12“ COL

16“ x 16“ COL

16“ x 12" COL

LINE PLAN

16’

16'

16’

Distribution of Moment

Location Total C.S Moment (k') M.S.Moment (k')

E-W Dir 117.05 117.05x0.75 = 87.78 29.26

- ve moment

E-W Dir 63.02 63.02x0.6 = 37.81 25.21+ ve moment

Equivalent Rigid Frame on Outer Column Line

Mo= 0.324(8+0.5)x (16.67)2/8+ 0.35 (16.67)2/8 x 1.4 = 112.68 kftD.F. For interior span

- ve moment = 0.65Mo = 0.65x112.68 = 73.24 kft

+ ve moment = 0.35Mo = 0.35x112.68 = 39.44 kft Percentage moment in C.S.=Same as for inner column line.

Distribution of momentsLocation Total Moment C.S.(kft) M.S .(kft)E-W Dir 73.24 73.24x0.75= 54.93 18.31- ve momentE-W Dir 39.44 39.44x0.60 = 23.66 15.78+ ve moment

0.65 0.35 0.65

Total Factored Static Moment in N-S Dir and its Distr

Mo= wu l2 ln2/8 = 0.324x18(14.83)2/8 = 160.40 kft

ln = 16 - (6+8)/12 = 14.83 ftD.F. ACI 13.6.3.3

Ext –ve moment = 0.26Mo= 41.70 k'

+ve moment = 0.52Mo= 83.41 k'

Int –ve moment = 0.70Mo= 112.28 k’Percentage Moment in C.S. ACI 13.6.4

l2/l1 = 18/16, = 1.13

α l2/l1 = 0 βt = 0Ext –ve moment in C.S = 100 %

+ve moment in C.S = 60 % Int –ve moment in C.S = 75 %

0.26

0.52

0.70

S

W

N

E

18' 18' 18'

P

RR S

Q

16“ x 12"

16“ x 16"

16“ x 12" Col

LINE PLAN

16’

16'

16’

Distribution of Moments.

Location Total moment C.S kft MS kft

N-S Dir 41.7 41.7 0.0

Ext -ve

N-S Dir 83.41 0.6x83.41=50.05 33.36

+ ve moment

N-S Dir 112.28 0.75x112.28=84.2 28.07

Int -ve moment

Design of Slab Reinforcement Panel QStrip Loc Muk’ b ft Mu/ft

kft

d” ρ As

in2

No of

bars

Remarks

E-W Dir

2x1/2 C.S

-ve

+ve

87.78

37.81

8

8

10.97

4.72

6

6

0.006

0.00258

3.46

1.49

18

8

E-W Dir

2x1/2 M.S

-ve

+ve

29.26

25.21

8

8

3.66

3.15

6

6

0.00208

0.00208

1.2

1.2

7

7

Use ρmin

Use ρmin

E-W Dir

1/2 C.S

-ve

+ve

54.93

23.66

4.5

4.5

12.2

5.26

6

6

0.00669

0.00288

2.17

0.93

12

5 Use ρmin

E-W Dir

1/2 M.S

-ve

+ve

18.31

15.78

4

4

4.58

3.95

6

6

0.0025

0.00216

0.72

0.63

4

4

N-S Dir

2x1/2 C.S

Ext-ve

+ve

Int-ve

41.7

50.05

84.21

8

8

8

5.21

6.26

10.53

5.5

5.5

5.5

0.0034

0.00408

0.00686

1.8

2.15

3.62

10

11

19

N-S Dir

2x1/2 M.S

Ext-ve

+ve

Int-ve

0

33.36

28.07

10

10

10

0

3.34

2.81

5.5

5.5

5.5

0.00227

0.00227

0.00227

1.5

1.5

1.5

8/9

8/9

8/9

Use ρmin

Use ρmin

Use ρmin

Asmin = 0.0018 bxh = 0.0018x12x7 = 0.15 in2

ρmin in E-W direction = 0.15/(12x6) = 0.00208

ρmin in N-S direction = 0.15/(12x5.5)= 0.00227

Area of steel can be calculated from flexural formula.

Mu = ɸρbd2fy(1-.59ρfy/fc’)

C.S8' - 0”

M.S10' - 0”

C.S8' - 0”

C.S4' - 6”

M.S8' - 0”

C.S8' - 0”

REINFORCEMENT PLAN

12#4T

10#4T

8#4T

11#4B

18#4T

19#4T

12#4T

10#4T

8#4T

11#4B

18#4T

19#4T

5#4B9#4T

8#4B9#4B

8#4B

9#4T

DESIGN THE INTERIOR PANEL OF THE ABOVE FLOOR SYSTEM

Solution

1. Slab Thickness Same as for exterior panel i.e. 7"

3. Total Factored Static Moment in E-W Dir and its Distribution Same as for exterior panel on interior column line

4. Total Factored Static Moment in N-S Dir and its Distr

Mo = wul2ln2/8 = 0.324 x 18 (16-16/12)2/8 = 156.82 k’

0.65

0.35

0.65D.F

S

ACI Code 13.6.3.2

S

W

N

E

18' 18' 18'

P

RR S

Q

16“ x 12" Col

LINE PLAN

16’

16'

16’

- ve Moment = 0.65 Mo = 101.93 k'

+ve Moment = 0.35 Mo = 54.89 k'

Percentage Moment in C.S.

l2/l1 = 18/16 = 1.13 αl2 / l1 = 0

+ve moment in C.S = 60%

- ve moment in C.S = 75%

Distribution of Moment

Location Total Moment C.S moment M.S moment

N-S Dir 0.75x101.93=

-ve moment 101.93 76.45 25.48

N-S Dir 0.6x54.89=

+ve moment 54.89 32.93 21.96

5.Design of Slab Reinforcement - Panel

Strip Loc M kft b ft Mu/ft

k'

d

in

ρ As

in2

No of

#4 Bar

Remark

E-W Dir

2x1/2 C.S

-ve

+ve

87.78

37.81

8

8

10.97

4.72

6

6

0.006

0.0025

8

3.46

1.49

18

8

E-W Dir

2x1/2 M.S

-ve

+ve

29.26

25.21

8

8

3.66

3.15

6

6

0.0020

8

0.0020

8

1.2

1.2

7

7

ρmin

“

N-S Dir

2x1/2 C.S

- ve

+ve

76.45

32.93

8

8

9.56

4.12

5.5

5.5

0.0061

9

0.0026

7

3.27

1.41

17

8

N-S Dir

2x1/2 M.S

- ve

+ve

25.48

21.96

10

10

2.55

2.20

5.5

5.5

0.0022

7

0.0022

7

1.5

1.5

8/9

8/9

ρmin

“

S

Asmin =0.0018 bxh= 0.15 in2

ρmin in E-W direction = 0.15/(12x6) = 0.00208

ρmin in N-S direction =0.15/(12x5.5) = 0.00227

Area of steel is calculated using flexural formula.

Mu = ɸρbd2fy(1- 0.59ρfy/fc’)

For example for moment of 10.97 kft, As is calculated as fol

12x10.97 = 0.9 ρ 12 (6)2 60 (1- 0.59 ρ 60/4)

8.85ρ2 – ρ + Mu/1944 = 0

ρ = 0.0060

As = 0.006x8x12x6 = 3.46 in2

C.S8' - 0”

M.S10' - 0”

C.S8' - 0”

C.S8' - 0”

M.S8' - 0”

C.S8' - 0”

Reinforcement Plan

6 Sketch

18#4T

17#4T

7#4T

8#4B

18#4T

17#4T

8#4B

9#4T

7#4B

9#4B

8#4B

9#4T

18#4T

17#4T

7#4T

8#4B

18#4T

17#4T

6#4T

6#4T

C.S M.S C.S M.S C.S

8#4T

7#4B

C.S

M.S

C.S

M.S

C.S

10#4T

12#4T

7#4T

5#4B

10#4T

11#4T

8#4B

9#4T

10#4B

10#4B

12#4B

9#4T

8#4B

9#4B

12#4B

9#4T

12#4T

10#4T

8#4T

11#4B

19#4T

19#4T

7#4T

8#4B

19#4T

17#4T

5#4B

9#4T

8#4B

9#4B

8#4B

9#4T

7#4B

9#4B

8#4B

9#4T

12#4T

10#4T

8#4T

11#4B

18#4T

19#4T

7#4T

8#4B

18#4T

17#4T

P

R S

Q

ANY QUESTION ?

ThanksThanks