Slab Examples

1

Table of Contents 1. introduction 2. Basic Assumptions 3. DESIGN OF SOLID TOW WAY SLAB 3.1. SLAB 3.2. BEAM 3.3. FORM WORK 3.4. BAR SCHEDULE

4. DESIGN OF FLAT SSLAB 4.1. Slab 4.2. FORM WORK 4.3. BAR SCHEDULE

5. DESIGN OF RIBBED SLAB 5.1. DESIGN OF RIB 5.2. DESIGN OF TOPPING SLAB 5.3. Design of GIRDER 5.4. FORM WORK 5.5. BAR SCHEDULE

6. CONCLUSION and SUMMERY

2

1.

INTRODUCTION

This structural design document presents all the assumptions analysis and design Calculations done in the design Material In the design of the building material properties of concrete C25 and steel S300 are used for structural elements. Slab and beam design and method of analysis The slab and beam is designed according to EBCS-1 and EBCS-2 1995 using direct design method. The partition wall load on each slab panel is also considered properly for each panel area. After calculation of the design actions, shear fore and bending moments, the proper reinforcement is provided for moment and the slab panel against shear. Objective:-

comparision between solid tow way slab, flat slab and ribbed slab

-

To help and identify least cost and most economical

-

To know which is ease to construct and needed minimum construction material

-

Help the designer to identify and understand the baste preferable in terms of design, economy and way of construction

-

To provide the first estimation for the designer

2.

BASIC ASSUMPTION

3

Dead Load: - Reinforced concrete

=25KN/m 3

- Cement Screed

=23KN/m 3

=14KN/m 3

- HCB- 200mm thick

=23N/m 3

- Terrazzo tile

Partial Safety Factor for Dead Load =1.30

Live Load: LL= 4KN/m2 Partial Safety Factor for Live Load =1.50

Material Properties Concrete: Grade C-25

Fck = 20 Mpa

Fctk =1.5 Mpa Partial Safety Factor = 1.5 = 11.33 Mpa

Fcd = 0.85[20/1.5]

Fctd = 1.5/1.5 = 1 Mpa Ecm = 29Gpa Reinforcing Steel: Fyk = 300 Mpa Partial Safety Factor =1.15 Es = 200Gpa

Spacing of column: Lx= 5.0m

Ly= 6.0m

Size of Column: 0.5m X 0.5m

4

Fyd = 300/1.15 = 260.87 Mpa

3. DESIGN OF SOLID TOW WAY SLAB

span. And from EBCS-2 , ratio 2  a = 30, and ratio 1  a = 40

3.1

For ratio 1.2

SLAB

a 

 2  1.2 30  40  30  38...... by  2  1

interpolation

  d   0.4  0.6     d   0.4  0.6  

f yk   l x   400    a 300   5000  112 mm  400   38

,

Use d = 115mm Total depth D = 115 + 20 + 5 = 140 mm Design Load: Dead Loads

Design constant

Selfe weight = 0.14 * 25 = 3.5 KN/m2 For S-300, C-25, l x = 5m l y = 6m ,

Ceiling plaster(15mm) =23 *0.015= 0.345KN/m2

  10mm Clear cover = 20mm Live Load LL = 4 KN/m2

f cd 

0.68 f c y



Cement scread(30mm) = 23*0.03 = 0.69KN/m2

0.68 * 25  11.33 N mm 2 1 .5

s fyk 300 fyd    260.87 N / mm 2 s 1.15 fctd 

0.21 fck s

2 3



Terrazzo tile(20mm) = 23*0.02 = 0.46 KN/m2 Partition wall = 2 KN/m2 Total Gk = 7 KN/m2

2 3

0.21(20)  1.03 N / mm 2 1.15

Pd  1.3DL  1.6 LL  1.3 7   1.6 4  Pd  15.49 KN / m 2

method

ment alculate

Depth determination by Serviceability Limit State

  f yk   l x   d   0.4  0.6  400    a 

 xs  0.063  xf  0.047  ys  0.047

,

 yf  0.036

The value of  a can be determined: using the ratio of length

ly lx



2 Mxs   xs * Pd * l x =0.063*15.49 *

6  1.2  is end 5

52=24.40

5

Mo

Mxf   xf * Pd * l x =0.047*5.49 * 52=18.20

Myf   yf * Pd * l x =0.024*5.49* 52=9.30

Mys   ys * Pd * l x =0.047*5.49* 52=18.20

Depth check

2

2

2

Mmax.= 24.40KNm

Myf   yf * Pd * l x =0.036*5.49* 5 =13.94 2

2

d chec 

 xs  0.056  xf  0.042

M  0.2952 * b * fcd

24.40 * 10 6 0.295 * 1000 * 11.33

d chec  85mm  d servc  115 mm

 ys  0.039

Use d servc = 115mm

 yf  0.030

Calculate Moment Redistribution for longer span

Mxs   xs * Pd * l x = 2

0.056*15.49* 52=21.69

1. Section 1-1

Mxf   xf * Pd * l x =0.042*5.49 * 52=16.27

Mys  18.20, Myf  13.94 ….. for S1

2

Mys  15.04, Myf  11.62 ….. for S1

Mys   ys * Pd * l x =0.039*5.49* 52=15.10 2

Myf   yf * Pd * l x =0.030*5.49* 52=11.62 2

 xs  0.048  xf  0.036

Mxs 18.20  15.10 *100%  *100%  17%  20% M max 18.20

 ys  0.039  yf  0.029

We take average moment method to adjusts support moment.

2 Mxs   xs * Pd * l x =0

.048*15.49* 52=18.59

Mys 

Mxf   xf * Pd * l x =0.036*5.49* 52=13.94

18.20  15.10  16.65KNm 2

2

Adjustment of moment for span (field) For Panel S1: Moment adjustment coefficients from the table by using .

Mys   ys * Pd * l x =0.039*5.49* 52=15.10 2

Myf   yf * Pd * l x =0.029*5.49* 52=11.23 2

ly lx

 xs  0.042  xf  0.032



6  1.2  5

Cx = 0.338, cy = 0.172

 ys  0.032

M  18.20  16.65  1.55 KNm

 yf  0.024

M adj x  c x M  0.338 * 1.55  0.523KNm

Mxs   xs * Pd * l x =0.042*15.49* 52=16.27

M adj y  c y M  0.172 * 1.55  0.266 KNm

Mxf   xf * Pd * l x =0.032*5.49* 52=12.39

Adjusted span moments

2

2

Mxf adj  M adj x  Mxf

Mys   ys * Pd * l x =0.032*5.49* 52=12.39 2

Mxf adj  0.523  18.20  18.723KNm

6

Myf adj  M adj y  Myf


Myf adj  0.266  13.94  14.206 KNm

Mxf adj  0.458  13.942  14.40 KNm

For Panel S2:

Myf adj  M adj y  Myf Myf adj  0.241  11.231  11.472 KNm

The support moment is increases, according to EBCS-2 no adjustment requires for span moments. Therefore

For Panel S2: The support moment is increases, according to EBCS-2 no adjustment requires for span moments.

Mxf adj  Mxf  16.27 KNm Myf adj  Myf  11.62 KNm

2. Section 2-2

Therefore

Mys  15.10, Myf  11.23 ….. for S3

Mxf adj  Mxf  12.39 KNm Myf adj  Myf  9.29 KNm

Calculate Moment Redistribution for shorter span

Mys  12.39, Myf  9.29 ….. for S4

1. Section A-A

Mxs  24.40, Mxf  18.20 ….. for S1 Mxs 15.10  12.39 *100%  *100%  18%  20% M max 15.10

Mxs  18.59, Mxf  13.94 ….. for S3

We take average moment method to adjust support moment.

Mys 

15.10  12.39  13.75 KNm 2

Mxs 24.40  18.59 *100%  *100%  24%  20% M max 24.40

Adjustment of moment for span (field)

Therefore we use Moment Distribution Method

For Panel S3: Moment adjustment coefficients from the

Relative stiffness: K1  K 2  K 3 

table by using

ly lx



6  1.2  5

I I  l 5

Distribution factors:

DF1  DF 2 

Cx = 0.338, cy = 0.172 M  15.10  13.75  1.355 KNm M adj x  c x M  0.338 * 1.355  0.458 KNm M adj y  c y M  0.172 *1.355  0.241KNm

I

5  0.5 I I 5 5

Adjusted support Moment: DF

0.5

0.5

0.5

0.5


7

FEM

-24.40

18.59

-18.59

24.40

Adj.

2.905

2.905

-2.905

-2.905

Relative stiffness: K1  K 2  K 3  Madj -21.495

21.495 -21.495

21.495 Distribution factors:

Adjusted support Moment Mxs  21.495 Adjustment of moment for span (field) For Panel S1:

DF1  DF 2 

I

5  0.5 I I 5 5

Moment adjustment coefficients from the table by using

Adjusted support Moment:

ly

DF

lx



I I  l 5

6  1.2  5

Cx = 0.344, cy = 0.364 M  24.4  21.495  2.905 KNm M adj x  c x M  0.344 * 2.905  0.999 KNm M adj y  c y M  0.364 * 2.905  1.057 KNm

0.5

0.5

0.5

FEM

-21.69

16.27

-16.27

24.40

Adj.

2.712

2.712

-2.712

-2.712

18.978 -18.978

18.978

Madj -18.978

Adjusted support Moment Mxs  18.978


Adjustment of moment for span (field)

Mxf adj  M adj x  Mxf Mxf adj  0.999  18.20  19.20 KNm

For Panel S2:

Myf adj  M adj y  Myf

Moment adjustment coefficients from the table by using

Myf adj  1.057  13.94  15.00 KNm

ly



6  1.2  5

For Panel S3:

lx

The support moment is increases, according to EBCS-2 no adjustment requires for span moments.

Cx = 0.344, cy = 0.364

Therefore

0.5

M  21.69  18.978  2.712 KNm M adj x  c x M  0.344 * 2.712  0.933KNm

Mxf adj  Mxf  13.94 KNm

M adj y  c y M  0.364 * 2.712  0.987 KNm

Myf adj  Myf  11.23KNm


1. Section B-B


Mxs  21.69, Mxf  16.27 ….. for S2

Mxf adj  0.933  16.27  17.203KNm

Mxs  16.27, Mxf  12.39 ….. for S4

Myf adj  M adj y  Myf Myf adj  0.987  11.62  12.607 KNm For Panel S4: The support moment is increases, according to EBCS-2 no adjustment requires for span moments.

Mxs 21.69  16.27 * 100%  * 100%  25%  20% M max 21.69 Therefore we use Moment Distribution Method

8

Mxf adj  Mxf  12.39 KNm

Therefore

Myf adj  Myf  9.29 KNm

Maximum Moments on the panel Panel -S1

Mxs  21.495KNm

Mxf  19.20KNm

Mys  16.652KNM

Myf  14.997KNm

Panel- S2

Mxs  18.978KNm

Mxf  17.203KNm

Mys  16.652KNM

Myf  12.607KNm

Panel- S3

Mxs  21.495KNm Mys  13.75KNM

Mxf  14.409KNm

Myf  11.472KNm

Panel- S4

Mxs  18.978KNm Mys  13.75KNM

Mxf  13.749KNm

Myf  9.295KNm

Reinforcement b= 1000mm, Minimum Reinforcement:

 min 

d = 115mm

  10mm

   calc   1    

1

2M bd 2 fcd

  fcd    fyd  

0.5 0.5   0.002 fyk 300

As min  bd  0.002 *1000 * 115  230mm 2 Calculated reinforcement:

S

b * ab As

9

PANEL

MOMENT TYPE

ρ

VALUE

Mxs Mxf PANEL1 Mys Myf Mxs Mxf PANEL2 Mys Myf Mxs Mxf PANEL3 Mys Myf Mxs Mxf PANEL4 Mys Myf Load transfer to beam:

21.50 19.39 16.65 15.15 18.98 17.36 16.65 12.73 21.50 14.54 13.75 11.58 18.98 12.51 13.75 9.38

Ascal (mm2)

0.0068 0.0060 0.0051 0.0046 0.0059 0.0054 0.0051 0.0039 0.0068 0.0044 0.0042 0.0035 0.0059 0.0038 0.0042 0.0028

Vx   vx.Pd .Lx Vy   vy.Pd .Lx

776.92 694.64 589.83 533.49 678.79 616.74 589.83 444.07 776.92 510.78 481.55 402.19 678.79 436.03 481.55 323.12

SPACING (mm)

SPACING PROVIDED (mm)

145.63 162.88 191.82 212.08 166.68 183.45 191.82 254.78 145.63 221.51 234.96 281.32 166.68 259.48 234.96 350.16

145 160 190 210 165 180 190 250 145 220 230 280 165 255 230 350

Panel S-3: Shear force coefficients for uniformly loaded rectangular panel is:

Panel S-1:Shear force coefficients for uniformly loaded rectangular panel is: Lx = 5m Pd=, =15.491KN

Lx = 5m , Pd = 15.491KN

 vcx  0.42 ly 6  vcy  0.36   1 .2   vdx  0 lx 5  vdy  0.24

 vcx  0.47 ly 6  vcy  0.40   1.2   vdx  0.31 lx 5  vdy  0.26

Vxd   vdx.Pd .Lx  0 Vxd   vdx.Pd .Lx  0.31 * 15.491 * 5  24.01 Vxc   vdx.Pd .Lx  0.42 * 15.491 * 5  32.53 Vxc   vdx.Pd .Lx  0.47 * 15.491 * 5  36.40 Vyd   vcy.Pd .Lx  0.24 * 15.491 * 5  18.59 Vyd   vcy.Pd .Lx  0.26 * 15.491 * 5  20.14 Vyc   vcy.Pd .Lx  0.36 * 15.491 * 5  27.88 Vyc   vcy.Pd .Lx  0.40 * 15.491 * 5  30.98

Panel S-2:

Panel S-2:

Shear force coefficients for uniformly loaded rectangular panel is:

Shear force coefficients for uniformly loaded rectangular panel is:

Lx = 5m , Pd = 15.491KN

Lx = 5m , Pd = 15.491KN

ly

 vcx  0.44

 vcy  0.36 6   1.2   vdx  0.29 lx 5

 vcx  0.39  vcy  0.33 6   1 .2   vdx  0 lx 5  vdy  0

ly

 vdy  0

Vxd   vdx.Pd .Lx  0.29 * 15.491 * 5  22.46 Vxc   vdx.Pd .Lx  0.44 * 15.491 * 5  34.08 Vyd   vcy.Pd .Lx  0 Vyc   vcy.Pd .Lx  0.36 * 15.491 * 5  27.88

10

Vxd   vdx.Pd .Lx  0 Vxc   vdx.Pd .Lx  0.39 * 15.491 * 5  30.21 Vyd   vcy.Pd .Lx  0 Vyc   vcy.Pd .Lx  0.33 * 15.491 * 5  25.56

Panel S-1 S-2 S-3 S-4

3.2

Ly- longer direction axis Vcx 1-discontineous 2-contineous 36.4 1-discontineous 2-contineous 34.08 2-contineous 32.53 2-contineous 32.53 2-contineous 30.21 2-contineous 30.21

Beam design

f cd 

Vdy 20.14 18.59 -

95.84 * 10 6 0.295 * 11.33 * 2500

d  338.67 mm Use d = 357mm

0.68 * 20   9.07 N mm 2 1.5

s fyk 300 fyd    260.87 N / mm 2 s 1.15 2

Mu   sd * f cd * bw

d

design constant

0.68 f c y

Lx- Shorter direction Vcy 1-discontineous 2-contineous 30.98 2-contineous 27.88 2-contineous 27.88 1-discontineous 2-contineous 27.88 2-contineous 25.56 2-contineous 25.56

Vdx 24.01 22.46 -

Beam span AB & CD , Msd= 79.79 KNm

 sd 

2

0.21 fck 3 0.21( 20) 3 fctd    1.03 N / mm 2 s 1.15 Longer Direction

M sd 79.79 *10 6   0.22  0.295.....ok! bd 2 f cd 250 * 357 2 * 11.33

From GDC for μsd = 0.22,  Kz  0.87 Reinforcment

As 

M sd 79.79 *10 6   984.77 mm 2 k z df yd 0.87 * 357 * 260.87

Provide 4 Φ20 bar

Beam span BC, Msd= 20.98 KNm

 sd 

M sd 20.98 *10 6   0.058  0.295.....ok! bd 2 f cd 250 * 357 2 *11.33


As 

b = 250mm, D = 400mm, cc=25mm, Φmain=20mm, stirrup Φ=8mm ,d= 357mm

M sd 20.98 *10 6   234.96mm 2 k z df yd 0.96 * 357 * 260.87

Provide 2 Φ20 bar

check depth: 11

vc  0.25 f ctd k1k 2bd  48.42 KN

For support B & C, Msd = 95.84

 sd 

Design shear M sd 95.84 *10 6   0.265  0.295.....okV!sd,max = 88.07 2 2 bd f cd 250 * 357 *11.33

From GDC for μsd = 0.265,  Kz  0.83

Vs  Vsd  Vc  88.07  48.42  39.65 KN

Reinforcment

As 

for span AB & CD Vc,max = 48.42

S

M sd 95.84 *10 6   1239.87 mm 2 k z df yd 0.83 * 357 * 260.87

Provide 4 Φ20 bar

adf yd Vs



100 * 357 * 260.87  234.88mm 39.65 * 10 3

 af yk 100 * 300   300mm   0.5b 0.5 * 250  0.5d  178.5mm 

S max = <,

Shear design S max = 175mm

Use  8c / c...175mm Design shear for span BC Vsd,max = 68.59

Vs  Vsd  Vc  20.17 KN

By similarity of triangle: design shear

66.12 2.5  0.357  Vsd1  Vsd 6   56.68KN 2.5 Vsd 2  Vsd 5 

98.07 3.5  0.357   88.07 KN 3.5

Vsd 3  Vsd 4 

77.85 3  0.357   68.59 KN 3

Resistance shear

adf yd Vs



 af yk 100 * 300   300mm   0.5b 0.5 * 250  0.5d  178.5mm 

S max = <,

max

100 * 357 * 260.87  461.73mm 20.17 * 103

=

175mm

 8c / c...175mm

Vrd  252.80KN  Vsd (allo) Shear Capacity

vc  0.25 f ctd k1 k 2 bd

k 2  1.6  d  1.243  1.....ok!   20

k1  1  50

S

S

Vrd  0.25 f ctd bd  0.25 * 11.33 * 250 * 357

At support

Vc,max = 48.42

As  1239.87 mm 2 As 1239.87   bd 250 * 357

1239.87  1.695  2.....ok! 250 * 357

For beam section 2,3,4,5

12

……….Use


check depth:


d

257.67 * 10 6 0.295 * 11.33 * 400

d  439.04mm Use d = 457mm



Vsd1  Vsd 6 

Reinforcment

Resistance shear

178.57 2.5  0.457   145.93KN 2. 5

264.46 3.5  0.457  Vsd 2  Vsd 5   229.93KN M sd 215.68 *10 6 3.5  sd  2   0 . 228  0 . 295 ..... ok ! bd f cd 400 * 457 2 *11.33 208.8 3  0.457  Vsd 3  Vsd 4   176.99 KN  Kz  0 . 86 From GDC for μsd = 0.22, 3

As 

M sd 215.68 *10 6 Vrd  0.25 f ctd bd  0.25 * 11.33 * 400 * 457   2103.64mm 2 k z df yd 0.86 * 457 * 260.87 Vrd  517.80KN  Vsd (allo)

Provide 7 Φ20 bar Shear Capacity


 sd 


6

M sd 55.52 *10 k 2  1.6  d  1.143  1.....ok!   0.057  0.295.....ok! 2 2 bd f cd 400 * 457 *11.33   20


At support

Reinforcment

As 

k1  1  50

M sd 55.52 * 10 6   485.10mm 2 k z df yd 0.96 * 457 * 260.87

As  2325.55mm 2

2325.55  1.636  2.....ok! 400 * 457

For beam section

vc  0.25 f ctd k1k 2 bd  88.02 KN

Provide 2 Φ20 bar

Design shear for span AB & CD Vsd,max = 299.93 Vc,max = 88.02

For support B & C, Msd = 235.66

M sd 235.66 *10 6 Vs.....  Vsd   0.249  0.295 ok!  Vc  141.91KN bd 2 f cd 400 * 457 2 *11.33 adf yd 100 * 457 * 260.87 From GDC for μsd = 0.249,  Kz  0.85 S   84mm Vs 141.91 * 10 3

 sd 

Reinforcment

As 

M sd 235.66 *10 6   2325.55mm 2 k z df yd 0.85 * 457 * 260.87

S max = <,

Provide 8 Φ20 bar Shear design

S max = 225mm

13

 af yk 100 * 300   300mm   0.5b 0.5 * 250  0.5d  228.5mm  .Use  8c / c...80mm

Design shear for span BC: Vc,max = 88.02

V sd,max = 176.99

 sd 

Vs  Vsd  Vc  91.97 KN

S

adf yd Vs




100 * 457 * 260.87  129mm 91.97 * 10 3

S max = <,

M sd 48.41 *10 6   0.133  0.295.....ok! bd 2 f cd 250 * 357 2 *11.33

Reinforcment

As 

 af yk 100 * 300   300mm   0.5b 0.5 * 250  0.5d  228.5mm 

M sd 48.41*10 6   565.86mm 2 k z df yd 0.92 * 357 * 260.87

Provide 2 Φ20 bar


S max = 225mm

 sd 

Use  8c / c...125mm

M sd 12.34 *10 6   0.058  0.295.....ok! bd 2 f cd 250 * 357 2 *11.33


Shorter Direction

Reinforcment

As 

M sd 20.98 *10 6   234.96mm 2 k z df yd 0.96 * 357 * 260.87

Provide 2 Φ20 bar For support B & C, Msd = 67.69

 sd

M sd 67.69 *10 6    0.185  0.295.....ok! bd 2 f cd 250 * 357 2 * 11.33


As 

M sd 67.69 *10 6   7948.71mm 2 k z df yd 0.91* 357 * 260.87

Provide 3 Φ20 bar


Shear design

check depth:

d


67.69 *10 6 0.295 *11.33 * 250

d  284.62mm Use d = 357mm Beam span AB & CD , Msd= 48.41 KNm By similarity of triangle: design shear

14

Vsd1  Vsd 6 

48.04 2  0.357   39.465 KN 2

Vsd 2  Vsd 5 

71.11 3  0.357   62.65KN 3

Vsd 3  Vsd 4 

56.02 2.5  0.357   48.02 KN 2.5

S max = <,

 af yk 100 * 300   300mm   0.5b 0.5 * 250  0.5d  178.5mm 

S max = 175mm Use  8c / c...175mm

Resistance shear

Vrd  0.25 f ctd bd  0.25 * 11.33 * 250 * 357 Vrd  252.80KN  Vsd (allo) Shear Capacity


k 2  1.6  d  1.243  1.....ok!   20

At support

As  794.26mm 2

k1  1  50

794.26  1.445  2.....ok! 250 * 357

For beam section

vc  0.25 f ctd k1k 2bd  41.34 KN Design shear for span AB & CD Vsd,max = 62.65 Vc,max = 41.34


Vs  Vsd  Vc  21.31KN

S

adf yd Vs



check depth:

100 * 357 * 260.87  437mm 21.31 * 103

 af yk 100 * 300   300mm   0.5b 0.5 * 250  0.5d  178.5mm 

S max = <,


 sd 

Use  8c / c...175mm

Reinforcment

Vc,max = 41.34

Vs  Vsd  Vc  6.68 KN

adf yd Vs



M sd 125.36 *10 6   0.248  0.295.....ok! bd 2 f cd 350 * 357 2 *11.33


Design shear for span BC

S

147 * 10 6 0.295 * 11.33 * 350

d  354.48mm Use d = 357mm

S max = 175mm

Vsd,max = 48.02


d

As 

100 * 357 * 260.87  1394.17mm 6.68 * 103 15

M sd 125.36 *10 6   1583.61mm 2 k z df yd 0.85 * 357 * 260.87

Provide 5 Φ20 bar


 sd

Shear Capacity

vc  0.25 f ctd k1 k 2 bd M sd 29.34 *10 6    0.058  0.295 !  d  1.243  1.....ok! k 2.....  ok 1 .6 bd 2 f cd 350 * 357 2 *11.33


  20

At support

As  1924.91mm 2

Reinforcment

As 

6

M sd 29.34 *10   328.17 mm 2 k z df yd 0.96 * 357 * 260.87

Provide 2 Φ20 bar

k1  1  50

For beam section

vc  0.25 f ctd k1k 2 bd  70.79 KN

For support B & C, Msd = 147KNm

 sd 


M sd 147 *10 6   0.291  0.295.....ok! Vs  Vsd  Vc  90.33KN bd 2 f cd 350 * 357 2 *11.33


S

Reinforcment

As 

1924.91  1.77  2.....ok! 350 * 357

adf yd Vs



100 * 357 * 260.87  103.10mm 90.33 * 10 3

M sd 147 *10 6   1924.91mm 2 k z df yd 0.82 * 357 * 260.87

Provide 7 Φ20 bar

 af yk 100 * 300   300mm   0.5b 0.5 * 250  0.5d  175mm 

S max = <,

Shear design S max = 175mm

Use  8c / c...100mm Design shear for span BC Vsd,max = 120.92

Vc,max = 70.79

Vs  Vsd  Vc  50.13KN

S


adf yd Vs



100 * 357 * 260.87  185.78mm 50.13 * 10 3

Vsd1  Vsd 6 

124.07 2  0.357   101.92 KN 2

Vsd 2  Vsd 5 

182.87 3  0.357   161.12 KN 3

Vsd 3  Vsd 4 

141.07 2.5  0.357   120.92 KN S max = 175mm 2.5

Resistance shear

S max = <,

 af yk 100 * 300   300mm   0.5b 0.5 * 250  0.5d  175mm 

Use  8c / c...175mm

Vrd  0.25 f ctd bd  0.25 * 11.33 * 350 * 357

3.3 FORM WORK FOR SOLID TOW WAY SLAB

Vrd  353.92KN  Vsd (allo) 16

For Slab: As - Ab

Total area of form work = 377.603m2

As = (15.05*18.05)=271.6525m2 Ab1=(0.3*(15.05+18.05)2)=19.83 ….for (250x400) Ab2=(0.4*15.05)2)=12.04 …..for (350x450) Ab3=(0.45*18.05)2)=16.245 …..for (400x500)

3.4

BAR SCHEDULE

As - Ab=271.6525-19.83-12.0416.245=223.54m2 For beams: Ab1=(0.3*(15.05+18.05)2)+2(0.4*(15.05+18.05)*2=72.82

Ab2=(0.4*15.05)*2)+(2*(0.4*15.05)*2)=36.12 Ab3=(0.45*18.05*2)+(2*0.4*18.05*2)=45.125

Ref Location slab

No of Bars

Dia

Dia

Dia

Dia

Dia

Dia

Dia

6

8

10

12

14

16

20

810.00

0.00

0.00

810.00

0.00

0.00

0.00

0.00

1056.12 448.00 572.76

0.00 0.00 0.00

0.00 0.00 0.00

1056.12 448.00 572.76

0.00 0.00 0.00

0.00 0.00 0.00

0.00 0.00 0.00

0.00 0.00 0.00

T. Length

No

Dia

Length

No of Bars

Panel1

10

8.10

25

x

2

x

2

10 10 10

6.77 4.00 3.33

39 28 43

x x x

2 2 2

x x x

2 2 2

17

Panel2

Panel3

Panel4

10 10 10

6.00 6.77 0.00

21 29 0

x x x

2 2 0

x x x

1 1 1

252.00 392.66 0.00

0.00 0.00 0.00

0.00 0.00 0.00

252.00 392.66 0.00

0.00 0.00 0.00

0.00 0.00 0.00

0.00 0.00 0.00

0.00 0.00 0.00

10

3.33

38

x

2

x

1

253.08

0.00

0.00

253.08

0.00

0.00

0.00

0.00

10

8.10

19

x

2

x

1

307.80

0.00

0.00

307.80

0.00

0.00

0.00

0.00

10 10 10 10 10 10

5.00 4.00 0.00 6.00 5.00 0.00

29 23 0 16 25 0

x x x x x x

2 2 0 1 1 0

x x x x x x

1 1 1 1 1 1

290.00 184.00 0.00 96.00 125.00 0.00

0.00 0.00 0.00 0.00 0.00 0.00

0.00 0.00 0.00 0.00 0.00 0.00

290.00 184.00 0.00 96.00 125.00 0.00

0.00 0.00 0.00 0.00 0.00 0.00

0.00 0.00 0.00 0.00 0.00 0.00

0.00 0.00 0.00 0.00 0.00 0.00

0.00 0.00 0.00 0.00 0.00 0.00

10

0.00

0

x

0

x

1 0.00 m Kg/m

0.00 0.00 0.222

0.00 0.00 0.395

0.00 4787.42 0.617

0.00 0.00 0.888

0.00 0.00 1.208

0.00 0.00 1.578

0.00 0.00 2.468

0

0

2954

0

0

0

0

Kg

18

BAR SCHEDULE FOR BEAMS Location

axis1,4

Span support

shear rfmt

axis2,3

Span support

shear rfmt

axis A&B

Span support

shear rfmt

axis C&D

Span support

shear rfmt

Ref No

Dia

Length

AB,CD BC B,C AB,CD BC AB,CD BC AB,CD BC B,C AB,CD BC AB,CD BC AB,CD BC B,C AB,CD BC AB,CD BC AB,CD BC B,C AB,CD BC AB,CD BC

20 20 20 20 20 8 8 20 20 20 20 20 8 8 20 20 20 20 20 8 8 20 20 20 20 20 8 8

6.00 6.00 4.00 6.35 6.00 1.11 1.11 6.00 6.00 4.00 6.45 6.00 1.70 1.70 5.00 5.00 3.33 5.35 5.00 1.11 1.11 5.00 5.00 3.33 5.35 5.00 1.30 1.30

No of Bars No of Bars 4 2 2 2 2 36 36 7 2 6 2 2 76 49 2 2 1 2 2 30 30 5 2 5 2 2 51 30

x x x x x x x x x x x x x x x x x x x x x x x x x x x x

2 1 2 2 1 2 1 2 1 2 2 1 2 1 2 1 2 2 1 2 1 2 1 2 2 1 2 1

x x x x x x x x x x x x x x x x x x x x x x x x x x x

total Length 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

96.00 24.00 32.00 50.80 24.00 159.84 79.92 168.00 24.00 96.00 51.60 24.00 516.80 166.60 40.00 20.00 13.32 42.80 20.00 133.20 66.60 100.00 20.00 66.60 42.80 20.00 265.20 78.00

m Kg/m Kg

4. 4.1 19

Dia 6

Dia 8

Dia 10

Dia 12

Dia 14

Dia 16

Dia 20

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

0.00 0.00 0.00 0.00 0.00 159.84 79.92 0.00 0.00 0.00 0.00 0.00 516.80 166.60 0.00 0.00 0.00 0.00 0.00 133.20 66.60 0.00 0.00 0.00 0.00 0.00 265.20 78.00

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

96.00 24.00 32.00 50.80 24.00 0.00 0.00 168.00 24.00 96.00 51.60 24.00 0.00 0.00 40.00 20.00 13.32 42.80 20.00 0.00 0.00 100.00 20.00 66.60 42.80 20.00 0.00 0.00

0.00 0.222 0

1466.16 0.395 579

0.00 0.617 0

0.00 0.888 0

0.00 1.208 0

0.00 1.578 0

975.92 2.468 2409

DESIGN OFFLAT SLAB SLAB

Total depth D =212.5 + 20 + 7= 239.5 mm Use Ds= 250mm

Slab Strips: According to EBCS-2 Shorter direction: Half column strip =

0.25l x  0.25 * 5000  1250mm Middle strip =

l x  2 0.25l x   5000  2 * 0.25 * 5000  2500mm

Design constant

Longer direction:

For S-300, C-25, l x = 5m l y = 6m ,

Half column strip =

0.25l x  0.25 * 5000  1250mm

  14mm Clear cover = 20mm, Column size: 500 X 500mm

Middle strip =

l x  2 0.25l x   6000  2 * 0.25 * 5000  3500mm

Live Load LL = 4 KN/m2

f cd 

0.68 f c y



0.68 * 25  11.33 N mm 2 1.5

Drop panel: Dd, from center of column

s fyk 300 fyd    260.87 N / mm 2 s 1.15 fctd 

0.21 fck s

2 3



Minimum length =

2 3

0.21( 20)  1.03 N / mm 2 1.15

l x 500   1667 mm 3 3

Use 1700mm

PROPORTIONING AND DIMENSHINING VARIOUS PARTS

Thickness of drop panel:= (1.25 to 1.50)Ds Consider Dd = 1.5Ds = 1.5 * 250 = 375mm. Use Dd = 380mm

Select Slab thikness:

Effective depth:

Depth determination by Serviceability Limit State method

Depth of slab in Longer direction ds1 = 250 – 20 – 7 = 223mm

  f   l d   0.4  0.6 yk   x  400    a 

Depth of slab in shorter direction ds2 = 250 – 20 – 1.5(14) = 209mm

ly



,

ds effective = dseef 

6  1.2  From EBCS-2 for End span 5

lx  a = 24, Le = Ly

  d   0.4  0.6     d   0.4  0.6  

223  209  216mm 2

Depth of drop panel in Longer direction dd1 =380 – 20 – 7 = 353mm

f yk   l y   400    a 300   6000  212.5mm  400   24

Depth of drop panel in shorter direction dd2 = 380 – 20 – 1.5(14) = 339mm dd effective = ds eef 

20

353  339  346mm 2

Design Load:

For column:

Dead Loads :

Capacity of concrete for shear

Average effective depth

v c  0.5 f ctd k1 k 2

Ds  Dd 250  380 Dav    315mm 2 2

 min 

0.5 0.5   0.002 fyk 300

Self weight = 0.315 * 25 = 7.875 KN/m2

k1  1  50 0.002   2

Ceiling plaster (15mm) =23 *0.015= 0.345KN/m2

k1  1.083  2.....ok! For column:

Cement screed (30mm) = 23*0.03 = 0.69KN/m2

k 2  1.6  d deff  1.6  0.346  1.254  1 ok!

Terrazzo tile(20mm) = 23*0.02 = 0.46 KN/m2

For drop

Partition wall = 2 KN/m2 outer wall (66m length) = 3*0.2*14*66/(18*15)= 2KN/m2 2

Total Gk = 13.37 KN/m , Qk = 4KN/m

k 2  1.6  d deff  1.6  0.216  1.384  1 ok!

2

For column:

Pd  1.3DL  1.6 LL  1.313.37   1.6 4  Pd  23.781 KN / m

vc  0.5 * 1.03 * 1.083 * 1.253  0.699 N / mm 2

2

For column:

Check of shear

vc  0.5 * 1.03 * 1.083 * 1.384  0.772 N / mm 2

The governing factor for flat slab is punching shear.

Punching perimeter: Around column:

According to ACI and EURO cod the critical point considers to be 0.5d distance from face of column.

Uc = 4(Cl + dd-eff) =4(500+346) = 3384mm Around drop:

Case -1: middle column and drop panel

Ud=4(droplength+ds) Ud =4(1700+216)=7664 Punching shear area: Around column: Acolumn = Uc*ddeff=3384*346=1170864mm2 Around drop: Adrop = Ud * dseff= 7664*216=1655424mm2 Total area for load to punching shear consider is A = 6 * 5 = 30m2 Punching force area: Around column: ac = (0.846)2 Around drop:

21

ad = (1.916)2

d d .eff    Uc   Cl  d d .eff   2 Cl  2   346   Uc   500  346   2 500    2192mm 2  

Punching force: Around column:





Vdv   A  a c  Pd  30   0.846  23.78 Vdv  696.38 KN

2

Around drop:

Punching stress: Around column:

Vd 

Vdv 696.38 * 10  puncingsheararea 1170864

 l  d d .eff   Ud   l  d s.eff   2 2    1700  216  Ud  1700  216   2  4332mm 2

3

Vd  0.595 N / mm 2

Punching shear area: column:

Vd < Vc ……. Is ok! Punching force: Around drop:



Acolumn = Uc*ddeff=2192*346=758432mm2



Vdv   A  a c  Pd  30  1.916  23.78 Vdv  696.38 KN

2

Around drop: Adrop = Ud * dseff= 4332*216=935712mm2

Punching stress: Around drop:

Vd 

Vdv 626.102 * 10  puncingsheararea 1655424

Around

Total area for punching shear consider is A = 3 * 5 = 15m2

3

Punching force area: Around column:

Vd  0.378 N / mm 2

ac = (0.846*0.673)=0.5694m2

Vd < Vc ……. Is ok!

Around drop:

Hence the depth of the drop and the slab is adequate against punching.

ad = (1.916*1.208)=2.314m2 Punching force: Around column:

Case – 2: At the edge column and drop panel

Vdv   A  a c  Pd  15  0.5694  23.78 Vdv  343.16 KN Punching stress: Around column:

Vd 

Vdv 343.16 * 10 3  puncingsheararea 758432

Vd  0.452 N / mm 2 Vd < Vc ……. Is ok! Punching force: Around drop:

Vdv   A  a c  Pd  15  2.314  23.78

Punching perimeter:

Vdv  301.66 KN

Around column:


22

Vd 


Vd  0.322 N / mm 2

Vd < Vc . ok! Punching force: Around drop:





Vd < Vc ……. Is ok!

Vdv   A  a c  Pd  7.5  1.208 23.78

Hence the depth of the drop and the slab is adequate against punching.


Vdv  143.65 KN

Case – 3: At the coner of column and drop panel

Vd 

2


Vd  0.275 N / mm 2 Vd < Vc ……. Is ok! Hence the depth of the drop and the slab is adequate against punching. DESIGN FOR FLEXURE

Punching perimeter:

Around column:

d d .eff   346    2 500  Uc  2 Cl    1346mm 2  2    Around drop:

 l  d d .eff Ud  2 2  

1700  500  216



2

  2

Punching shear area:

 2416mm

Acolumn = Uc*ddeff=1346*346=465716mm2 Around drop: Adrop = Ud * dseff= 2416*216=521856mm2 Total area for punching shear is A = 3 *2.5 = 7.5m2 Punching force area: Around column: ac = (0.673)2 Around drop: ad = (1.208)2 Punching force: Around column:



Vdv   A  a c  Pd  7.5   0.673 23.78 Vdv  167.58 KN

2

Punching stress: Around column:

Vd 

Vdv 167.58 *103  puncingsheararea 465716 23

Vd  0.360 N / mm 2

A = 5 x 6 = 30m2

L  li 

Around column:



F (KN)= (1.3 Gk + 1.6 Qk)A = Pd*A

hc 

2hc 3

, where

4dc 2  

4 * 500 2  564mm 

dc= column width Moment along the longer direction:

L  ly 

2hc 2 * 0.564 6  5.624m 3 3

For interior Msy  0.055 FL  0.55 PdAL Msy  0.055 * 23.78 * 30 * 5.624  220.67 KNm Mfy  0.071FL  0.071PdAL Mfy  0.071* 23.78 * 30 * 5.624  284.86 KNm

For end span

Msy A  0.040 FL  0.040 PdAL Msy A  0.040 * 23.78 * 30 * 5.624  160.49 KNm Msy B  0.063FL  0.063PdAL Msy B  0.063 * 23.78 * 30 * 5.624  252.77 KNm Mfy  0.083FL  0.083PdAL Mfy  0.083 * 23.78 * 30 * 5.624  333KNm

Moment along the shrter direction:

L  lx 

2hc 2 * 0.564  5  4.624m 3 3

For interior span

Msx  0.055 FL  0.55 PdAL Msx  0.055 * 23.78 * 30 * 4.624  181.43KNm Mfx  0.071FL  0.071PdAL Mfx  0.071 * 23.78 * 30 * 4.624  234.21KNm

For end span

Msx A  0.040 FL  0.040 PdAL Msx A  0.040 * 23.78 * 30 * 4.624  131.95KNm Msx B  0.063FL  0.063PdAL Msx B  0.063 * 23.78 * 30 * 4.624  207.82 KNm Mfx  0.083FL  0.083PdAL Mfx  0.083 * 23.78 * 30 * 4.624  273.80 KNm

LATERAL DISTRIBUTION OF MOMENTS From EBCS-2 Negative moment:-for column strip=75% :- for middle strip=25% Positive moment:- for column strp= 55% :- for middle strip=45%

24

Span

Distri. span

Location End span LONG Middle span End span SHORT Middle span

Total moment

Moment in

Adjusted moment

Support1 Center

-160.49 333

column strip 120.37 183.15

Support2

-252.77

189.58

63.19

0.23

175.04

77.73

Support1

-220.67 284.86 -131.95 273.82 -207.82 -181.43 234.21

165.50

55.17

0.23

152.81

67.86

156.67 98.96 150.60 155.87 136.07 128.82

128.19 32.99 123.22 51.96 45.36 105.39

0.23 0.32 0.32 0.32 0.32 0.32

127.19 88.41 111.17 139.24 121.56 95.09

157.67 43.54 162.65 68.58 59.87 139.12

Point

Center Support1 Center Support2 Support1 Center

Middle strip 40.12 149.85

Adjustment factor of moment 0.23 0.23

column strip 111.14 148.68

Middle strip 49.35 184.32

Note: 

Column strip is 1.7m and the rest lengthe divied by middle length of each side to get adj. factor



- adjustment factor is multiplied by middle strip moment. And add at middle strip moment and subtracts from column strip moment.

COMPUTE REINFORCMENT & C/C SPACING

25

Span

Edge

Interior

1.

Moment (KNm) col L(-ve) col L(+ve) col L(-ve) col S(-ve) col S(+ve) col S(-ve) Mid L(-ve) Mid L(+ve) Mid L(-ve) Mid S(-ve) Mid S(+ve) Mid S(-ve) col L(-ve) col L(+ve) col S(-ve) col S(+ve) Mid L(-ve) Mid L(+ve) Mid S(-ve) Mid S(+ve)

111.14 148.68 175.04 88.41 111.17 139.24 49.35 184.32 77.73 43.54 162.65 68.58 152.81 127.19 121.56 95.09 67.86 157.67 59.87 139.12

d (mm)

b (mm)

ρ

339 209 339 353 223 353 209 209 209 223 223 223 339 209 353 223 209 209 223 223

1700 1700 1700 1700 1700 1700 3300 3300 3300 4300 4300 4300 1700 1700 1700 1700 3300 3300 4300 4300

0.0022 0.0085 0.0036 0.0016 0.0054 0.0026 0.0013 0.0052 0.0021 0.0008 0.003 0.0012 0.0031 0.0072 0.0023 0.0046 0.0018 0.0044 0.0011 0.0026

EDGE SPAN REINFORCMENTE

COLUMN SRIP LONGER DIRECTION 1. Φ14 C/C 540,L=(0.33*5.5)+0.69 = 2.5m 1’.Φ14 C/C 540,L=(0.33*5.5)+0.69 = 2.5m 2, Φ14 C/C 105,L=6-(0.125*5.5)+0.10=5.21m 3. Φ14 C/C 540,L=2(0.33*5.5)+0.5 = 4.13m 3’.Φ14 C/C 540,L=2(0.33*5.5)+0.5 = 2.7m SHORTER DIRECTION

Ascalc. (mm)2 1289.98 3023.21 2064.49 978.40 2037.01 1558.65 919.26 3596.49 1461.25 755.35 2896.64 1196.06 1792.14 2542.24 1355.27 1724.92 1271.55 3046.82 1042.23 2464.37

Asmin (mm)2 960.5 592.17 960.5 1000.2 631.83 1000.2 1149.5 1149.5 1149.5 1598.2 1598.2 1598.2 960.5 592.17 1000.2 631.83 1149.5 1149.5 1598.2 1598.2

Sreq (mm)

S-Max (mm)

202.95 86.597 126.81 267.58 128.52 167.97 552.84 141.3 347.78 876.67 228.61 553.65 146.08 102.98 193.17 151.78 399.67 166.8 635.37 268.71

272.57 442.11 272.57 261.76 414.35 261.76 442.11 442.11 442.11 414.35 414.35 414.35 272.57 442.11 261.76 414.35 442.11 442.11 414.35 414.35

SProv. (mm) 200 85 125 260 125 165 440 140 345 410 225 410 145 100 190 150 395 165 410 265

8,.Φ14 C/C 240,L=6-0.075+0.15-0.25 = 5.83m 8’.Φ14 C/C 240,L=6-0.15*5+0.15-0.25= 5m 9. Φ14 C/C 210,L=2(0.22*5.5)+0.5= 2.92m SHORTER DIRECTION 10. Φ14 C/C 410,L=(0.22*4.5)+0.69 = 1.68m 11,.Φ14 C/C 420,L=5-0.075+0.15-0.25=4.83m 11’.Φ14 C/C 420,L=5-0.15*5+0.15-0.25= 4m 12. Φ14 C/C 370,L=2(0.22*4.5)+0.5= 2.48m 2.

4. Φ14 C/C 520,L=(0.33*4.5)+0.69 = 2.18m 4’.Φ14 C/C 520,L=(0.33*4.5)+0.69 = 1.60m

INTERIOR SPAN REINFORCMENTE COLUMN SRIP LONGER DIRECTION

5, Φ14 C/C 140,L=6-(0.125*4.5)+0.10=4.34m

13. Φ14 C/C 400,L=2(0.33*5.5)+0.5=4.13m 13’.Φ14 C/C 400,L=2(0.33*5.5)+0.5 =2.7m

6. Φ14 C/C 440,L=2(0.33*4.5)+0.5 = 3.47m 6’.Φ14 C/C 440,L=2(0.33*4.5)+0.5 = 2.30m MIDDEL STRIP

14.Φ14 C/C 120,L=6-2(0.125*5.5)= 4.625m L=4.3+2(24*0.014)=4.97m

LONGER DIRECTION

SHORTER DIRECTION

7. Φ14 C/C 340,L=(0.22*5.5)+0.69 = 1.9m 26

REMARK OK OK OK Smax OK OK Smax OK OK Smax OK Smax OK OK OK OK OK OK Smax OK

15. Φ14 C/C 510,L=2(0.33*4.5)+0.5=3.47m 15’.Φ14 C/C 510,L=2(0.33*4.5)+0.5 =2.30m

20. Φ14 C/C 500,L=5-2(0.075)=4.85m C/C 500,L=5-2(0.15*5)=3.5m

16.Φ14 C/C 165,L=5-2(0.125*4.5)=3.88m L=3.3+2(24*0.014)=3.97m MIDDEL STRIP

4.2 SLAB

20’.Φ14

FORM WORK FOR FLAT

FOR DROP PANEL

LONGER DIRECTION

=9*(1.25*1.25)+ 9*(1.7*0.13)*4

17. Φ14 C/C 245,L=2(0.22*5.5)+0.5=2.92m

=14.06+7.956=22.0185m2

18. Φ14 C/C 290,L=6-2(0.075)=5.85 18’.Φ14 FOR SLAB C/C 290,L=6-2(0.15*6)=4.20m SHORTER = (15.05*18.05)-(1.7*1.7)+0.25(18.05+15.05)2 DIRECTION =262.1925m2 19. Φ14 C/C 410,L=2(0.22*4.5)+0.5=2.48m TOTAL = 284.211m2

27

4.3

BAR SCHEDULE

Flat slab Reinforcement Ref Location

longer span

Cl.strip

Shorter

Mdl.strip

Cl.strip

No of Bars

total Length

No of Bars

Dia

Dia

Dia

Dia

Dia

Dia

Dia

6

8

10

12

14

16

20

No

Dia

Length

1 1' 2 3 3' 13 13' 14 7 8 8' 9 17 18 18' 4 4' 5 6 6'

14

2.61

5

x

2

x

3

78.30

0.00

0.00

0.00

0.00

78.30

0.00

0.00

14

1.80

5

x

2

x

3

54.00

0.00

0.00

0.00

0.00

54.00

0.00

0.00

14

5.21

21

x

2

x

3

656.46

0.00

0.00

0.00

0.00

656.46

0.00

0.00

14

4.13

7

x

2

x

3

173.46

0.00

0.00

0.00

0.00

173.46

0.00

0.00

14

2.70

7

x

2

x

3

113.40

0.00

0.00

0.00

0.00

113.40

0.00

0.00

14

4.13

6

x

2

x

3

148.68

0.00

0.00

0.00

0.00

148.68

0.00

0.00

14

2.70

6

x

2

x

3

97.20

0.00

0.00

0.00

0.00

97.20

0.00

0.00

14

5.00

18

x

1

x

3

270.00

0.00

0.00

0.00

0.00

270.00

0.00

0.00

14

1.90

8

x

2

x

3

91.20

0.00

0.00

0.00

0.00

91.20

0.00

0.00

14

5.80

12

x

2

x

3

417.60

0.00

0.00

0.00

0.00

417.60

0.00

0.00

14

5.00

12

x

2

x

3

360.00

0.00

0.00

0.00

0.00

360.00

0.00

0.00

14

2.92

10

x

2

x

3

175.20

0.00

0.00

0.00

0.00

175.20

0.00

0.00

14

2.90

9

x

2

x

3

156.60

0.00

0.00

0.00

0.00

156.60

0.00

0.00

14

5.85

10

x

1

x

3

175.50

0.00

0.00

0.00

0.00

175.50

0.00

0.00

14

4.20

10

x

1

x

3

126.00

0.00

0.00

0.00

0.00

126.00

0.00

0.00

14 14

2.18 1.60

4 4

x x

2 2

x x

3 3

26.16 38.40

0.00 0.00

0.00 0.00

0.00 0.00

0.00 0.00

26.16 38.40

0.00 0.00

0.00 0.00

14

4.34

15

x

2

x

3

390.60

0.00

0.00

0.00

0.00

390.60

0.00

0.00

14 14

3.47 2.30

6 6

x x

2 2

x x

3 3

124.92 82.80

0.00 0.00

0.00 0.00

0.00 0.00

0.00 0.00

124.92 82.80

0.00 0.00

0.00 0.00

28

span

Mdl.strip

15 15' 16 10 11 11' 12 19 20 20'

14

3.47

5

x

2

x

3

104.10

0.00

0.00

0.00

0.00

104.10

0.00

0.00

14

2.30

5

x

2

x

3

69.00

0.00

0.00

0.00

0.00

69.00

0.00

0.00

14

3.97

13

x

1

x

3

154.83

0.00

0.00

0.00

0.00

154.83

0.00

0.00

14

1.68

11

x

2

x

3

110.88

0.00

0.00

0.00

0.00

110.88

0.00

0.00

14

4.83

10

x

2

x

3

289.80

0.00

0.00

0.00

0.00

289.80

0.00

0.00

14

4.00

10

x

2

x

3

240.00

0.00

0.00

0.00

0.00

240.00

0.00

0.00

14

2.48

11

x

2

x

3

163.68

0.00

0.00

0.00

0.00

163.68

0.00

0.00

14

2.48

12

x

2

x

3

178.56

0.00

0.00

0.00

0.00

178.56

0.00

0.00

14

4.85

9

x

1

x

3

130.95

0.00

0.00

0.00

0.00

130.95

0.00

0.00

14

3.20

9

x

1

x

3

86.40

0.00

0.00

0.00

0.00

86.40

0.00

0.00

0.00

0.00

0.00

0.00

5284.68

0.00

0.00

0.222

0.395

0.617

0.888

1.208

1.578

2.468

0

0

0

0

6384

0

0

m Kg/m Kg

5.

DESIGN OF RIBBED SLAB

Assuming one way ribbed slab. Design strength:

f cd 

0.68 f c y



0.68 * 25  11.33 N mm 2 1.5

s fyk 300 fyd    260.87 N / mm 2 s 1.15 2

2

0.21 fck 3 0.21( 20) 3 fctd    1.03 N / mm 2 s 1.15 assume that : width of joists, bw = 150mm and width of grider , b = 300mm

 40mm  Dt   1  10 clear..distnce..b / n..Ribs  40mm  Dt   1  10  550  150  40mm Dt  40mm  ...Use..Dt  60mm

Depth of Joists and Toppings

Depth of Joists (rib):

Depth of Toppings:

Depth of ribs excluding any topping not more than 4 times of minimum width of the ribs Let, Dj (including toppings) = 300mm Dj -Dt  4bw , 300 -60  4*150…..Ok! Transverse rib shall be provided if the span of the ribbed slab exceeds 6.0m Center to center spacing = 5.00 < 6.0m

29

Therefore, Transverse strips is not necessary

check depth:

Design load on joists

d

-Toppig self wt= 0.06*0.55*25=0.825KN/m -Joist self wt. = 0.240*0.15*25 = 0.9KN/m -cement screed=0.69*0.55 = 0.3795KN/m -terrazzo tile = 0.46*0.55 = 0.253 KN/m -plastering = 0.345*0.55 = 0.18975 KN/m -partition wall = 4*0.55 = 2.2 KN/m -HCB for floor=0.4*0.2*14=1.12KN/m total dead load = 4.747 KN/m

Use d = 278mm

At the support ( -ve M): ,Msd= 20KNm

 sd

M sd 20 *10 6    0.152  0.295.....ok! bd 2 f cd 150 * 278 2 *11.33

From GDC for μsd = 0.152,  Kz  0.0.91

As 

M sd 20 *10 6   303.05mm 2 k z df yd 0.91* 278 * 260.87

Provide 2 Ø14mm bars

Pd = 1.3*4.747 + 1.6*2.2 = 9.69 KN/m

Assume that rib is continuous run on girder

At the span ( +ve M): ,Msd= 13.75KNm Assume N.A lies in the flange;

Then, Le = L – 2(0.5*bg) = 5 – 2(0.5*0.3) = 4.7m

Le = 0.7L = 0.7*5000 = 3500mm

2

PdLe 9.96 * 4.7 2   20 KNm 11 11 2 PdLe 9.96 * 4.7 2  ve M max    13.75KNm 16 16 PdLe 9.96 * 4.7 Vmax    23.41KN 2 2 5.1

20 * 10 6  199.73mm 0.295 *11.33 * 150

Reinforcement for Joists:

Live load , LL = 4*0.55 = 2.2 KN/m

 ve M max 




Le 3500  bw    700mm beff   5 5  bactual  550mm beff  550mm

Design of Joists(RIB):

 sd 

Use Ø= 14mm, clear cover=15mm,

M sd 13.75 *10 6   0.0286  0.295..... beff d 2 f cd 550 * 278 2 *11.33

From GDC for μsd = 0.152,

d = 300 – 15 – 7 = 278mm 30

 Kz  0.97   Kx  0.090

Checking the Assumtion;

According to EBCS-2 The topping shall be

X =Kxd = 0.09*278 X =25.02mm < Df= 60mm……..Ok!

provided with reinforcement mesh providing in each direction a crosssectional area not less than 0.001 of the section of the slab

As 

M sd 13.75 *10 6   195.46mm 2 k z df yd 0.97 * 278 * 260.87 As



Provide 2 Ø14mm bars

0.001Le.Dt use Ø = 8mm

Design for shear of Joists (rib):

As = 0.001*1000*60 = 60mm2

By similarity of triangle:

S

23.41 2.5  0.278  20.81KN 2.5 Resistance shear Vsd1  Vsd 2 

Vrd  0.25 f cd bd  0.25 * 11.33 * 150 * 278

ab * b 50.265 * 1000   837.76mm As 60

Smax = 400 mm…. provide Ø8 c/c 400mm. 5.3

Design of girder:

For interior girder the reaction transferred to the supporting girder may be taken twice end shear.

Vrd  117 .80KN  Vsd (allo) Shear,capacity vc  0.25 f ctd k1 k 2 bd

Pd int 

k 2  1.6  d  1.322  1.....ok!

2 * Vmax 2 * 23.41   85.127 KN / m c / c..rib 0.55

For edge girder taken end shear.

  14

At support

and take 1m strip and

As  303.05mm 2

Pd int 

303.05 k1  1  50  1.3634  2.....ok! 150 * 278

Vmax 23.41   42.564 KN / m c / c..rib 0.55

Depth of girder determination

vc  0.25 f ctd k1k 2bd  19.353KN

Edge (end) girder: Shear reinforcement for rib

Vs  Vsd  Vc  20.81  19.353  1.46 KN

S

adf yd Vs

S max = <,



56.55 * 278 * 260.87  2809mm 1.46 * 10 3

 af yk 56.55 * 300   262mm   0.5b 0.5 *150  0.5d  139mm 

S max = 139mm

Let assume that , b = 300mm , D = 500mm, cc=25mm, ф = 20mm, stirrup ф= 8mm, d =457mm

check depth:

Use  6c / c...135mm

5.2

d

Design of Topping:


d  428.29mm Use d = 460mm 31

183.91 * 10 6 0.295 * 11.33 * 300


Vsd1  Vsd 6 

122.91 2.4  0.457   99.50 KN 2.4

171.84 3.6  0.457  M sd 147.32 *10 6 Vsd 2  Vsd 5   150.02 KN   0 . 207  0 . 295 ..... ok ! 3.6 2 2 bd f cd 300 * 457 *11.33 153.57 3  0.457  Vsd 3  Vsd 4   130.18 KN From GDC for μsd = 0.207,  Kz  0.88 3

 sd 

Resistance shear

Reinforcment

As 

Vrd  0.25 f ctd bd  0.25 * 11.33 * 300 * 457 M sd 147.32 *10 6 2   1404.23mmVrd  388.34KN  Vsd (allo) k z df yd 0.88 * 457 * 260.87

Provide 5 Φ20 bar

Shear Capacity


 sd 


k 2  1.6  d  1.143  1.....ok! M sd 46.44 *10 6   0 . 065  0 . 295 .....ok! bd 2 f cd 300 * 457 2 *11.33   20


At support

Reinforcment

k1  1  50

As 

M sd 46.44 *10 6   410mm 2 k z df yd 0.95 * 457 * 260.87

As  1836.48mm 2

1836.48  1.67  2.....ok! 300 * 457

For beam section

v c  0.25 f ctd k1 k 2 bd  67.38KN

Provide 2 Φ20 bar


For support B & C, Msd = 183.91KNm

M sd Vs  Vsd  Vc  82.64 KN 183.91 *10 6   0.259  0.295.....ok! 2 2 bd f cd 300 * 457 *11.33 adf yd 100 * 457 * 260.87 S   144.26mm From GDC for μsd = 0.2591,  Kz  0.84 Vs 82.64 * 10 3

 sd 

Reinforcment

 af yk 100 * 300   200mm   0.5b 0.5 * 300  0.5d  228.5mm 

M sd 183.91*10 6 As    1836.48mm 2 k z df yd 0.84 * 457 * 260.87 S max = <, Provide 6 Φ20 bar Shear design

S max = 200mm Use  8c / c...140mm Design shear for span BC Vsd,max = 130.18

Vc,max = 67.38

Vs  Vsd  Vc  62.80 KN By similarity of triangle: design shear

S 32

adf yd Vs



100 * 457 * 260.87  189.84mm 62.80 * 10 3

S max = <,

 af yk 100 * 300   200mm   0.5b 0.5 * 300  0.5d  228.5mm 

 sd 

M sd 90.06 *10 6   0.064  0.295.....ok! bd 2 f cd 400 * 557 2 *11.33


S max = 175mm

As 

Use  8c / c...185mm

interior girder:

M sd 90.06 *10 6   645.63mm 2 k z df yd 0.96 * 557 * 260.87

Provide 2 Φ20 bar For support B & C, Msd = 355.11KNm

 sd 

M sd 355.11 *10 6   0.252  0.295.....ok! bd 2 f cd 400 * 557 2 *11.33


Let assume that , b = 400mm , D = 600mm, cc=25mm, ф = 20mm, stirrup ф= 8mm, d =557mm

Reinforcment

As 

M sd 355.11 *10 6   2875.18mm 2 k z df yd 0.85 * 557 * 260.87

Provide 9 Φ20 bar Shear design

check depth:


d

355.11 * 10 6 0.295 * 11.33 * 400

d  515.38mm Use d = 557mm By similarity of triangle: design shear


Vsd 1  Vsd 6 

237.60 2.4  0.557   182.46 KN 2.4

Vsd 3  Vsd 4 

296.78 3  0.557   241.68KN 3

M sd 284.84 *10 6   0.202  0.295.....ok! 355.97 3.6  0.557  bd 2 f cd 400 * 557 2 *11.33 Vsd 2  Vsd 5   300.89 KN 3.6 From GDC for μsd = 0.202,  Kz  0.88

 sd 

Reinforcment

Resistance shear

As 

M sd 284.84 *10 6   2227.61mm 2 Vrd  0.25 f ctd bd  0.25 * 11.33 * 400 * 557 k z df yd 0.88 * 557 * 260.87

Provide 7Φ20 bar

Vrd  631.08KN  Vsd (allo)


Shear Capacity


k 2  1.6  d  1.043  1.....ok! 33

  20

At support

As  2875.18mm

Use  8c / c...70mm 2

Design shear for span BC

2875.18 k1  1  50  1.645  2.....ok! 400 * 557

Vsd,max = 241.68

Vs  Vsd  Vc  143.23KN

For beam section

v c  0.25 f ctd k1 k 2 bd  98.45 KN

S


adf yd Vs



100 * 557 * 260.87  101.45mm 143.23 *10 3

Vs  Vsd  Vc  202.44 KN

S

adf yd Vs

S max = <,

100 * 557 * 260.87   71.78mm 202.44 * 10 3

S max = <,

Vc,max = 98.45

 af yk 100 * 300   150mm   0.5b 0.5 * 400  0.5d  278.5mm 

S max = 150mm

 af yk 100 * 300   150mm   0.5b 0.5 * 400  0.5d  278.5mm 

Use  8c / c...100mm

S max = 150mm

34

5.5

FORM WORK

Topping=262.71 m2 ,

5.6

girder =111.2 m2,

BAR SCHEDULE

Location

Ref

Dia

Length

No of Bars

topping Ribe

No

Exterior

total Length

No of Bars

Dia

Dia

Dia

Dia

Dia

Dia

Dia

6

8

10

12

14

16

20

top

14

5.30

2

x

34

x

3

1080.18

0.00

0.00

0.00

0.00

1080.18

0.00

0.00

bottm

14

5.00

2

x

34

x

3

1020.00

0.00

0.00

0.00

0.00

1020.00

0.00

0.00

shear bar

6

0.75

38

x

34

x

3

2907.00

2907.00

0.00

0.00

0.00

0.00

0.00

0.00

Lx

8

15.72

46

x

1

x

1

722.89

0.00

722.89

0.00

0.00

0.00

0.00

0.00

Ly

8

18.72

39

x

1

x

1

729.89

0.00

729.89

0.00

0.00

0.00

0.00

0.00

AB,CD

20

6.45

2

x

2

x

2

51.60

0.00

0.00

0.00

0.00

0.00

0.00

51.60

BC

20

6.00

2

x

1

x

2

24.00

0.00

0.00

0.00

0.00

0.00

0.00

24.00

AB,CD

20

6.00

5

x

2

x

2

120.00

0.00

0.00

0.00

0.00

0.00

0.00

120.00

BC

20

6.00

2

x

1

x

2

24.00

0.00

0.00

0.00

0.00

0.00

0.00

24.00

B,C

20

4.00

4

x

2

x

2

64.00

0.00

0.00

0.00

0.00

0.00

0.00

64.00

AB,CD

8

1.50

44

x

2

x

2

264.00

0.00

264.00

0.00

0.00

0.00

0.00

0.00

BC

8

1.50

34

x

2

x

2

204.00

0.00

204.00

0.00

0.00

0.00

0.00

0.00

AB,CD

20

6.55

2

x

2

x

2

52.40

0.00

0.00

0.00

0.00

0.00

0.00

52.40

BC

20

6.00

2

x

1

x

2

24.00

0.00

0.00

0.00

0.00

0.00

0.00

24.00

top bottom support shear bar

Interi

Rib =137.36m2

top

35

or

bottom support shear bars

AB,CD

20

6.00

7

x

2

x

2

168.00

0.00

0.00

0.00

0.00

0.00

0.00

168.00

BC

20

6.00

2

x

1

x

2

24.00

0.00

0.00

0.00

0.00

0.00

0.00

24.00

B,C

20

4.00

7

x

2

x

2

112.00

0.00

0.00

0.00

0.00

0.00

0.00

112.00

AB,CD

8

1.90

87

x

2

x

2

661.20

0.00

661.20

0.00

0.00

0.00

0.00

0.00

BC

8

1.90

61

x

1

x

2

231.80

0.00

231.80

0.00

0.00

0.00

0.00

0.00

2907.00

2813.78

0.00

0.00

2100.18

0.00

664.00

0.222

0.395

0.617

0.888

1.208

1.578

2.468

645

1111

0

0

2537

0

1639

m Kg/m Kg

6. CONCLUSION SUMMERY CONPARISION OF DESIGN DETAILS

SLAB TYPE

ITEM

CONCRET volum(m3)

Solid Slab

Flat Slab

Ribbed Slab

slab beam total slab drop total topping rib girder total HCB

32.228 18 50.228 57.616 9.884 67.5 16.2 6.12 14.04 36.36 44.64

FORM WORK

STEEL (Kg) Dia 6

Dia 8

Dia 10

Dia 14

Dia 20

2953.8 579.13

2408.6 5941.54 6383.9 6383.9

573.85 645.35

2537 537.6

1638.8 5932.56

223.54 154.063 377.603 262.19 22.02 284.21 0 137.36 111.20 248.56

NOTE: According to the comparison of table, the most preferable and economical is ribbed slab. And secondly Solid tow way slab is preferable

36

RANK

Area(m2) 2

3

1

Slab Examples

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