Circular Slab

DOCUMENDATE 03-05-2007

PROJECT:

TITLE:

DESIGNED CHECKED SHEET

CIRCULAR SLAB

1. Data Diameter of circular slabl= Grade of concrete Permissible stress in concrete fc= Grade of steel Permissible stress in steel fst= Thickness of circular wall Thickness of slab Poissons ratio E= D= 2. Support conditions:

3.00 M 70.00 Fe 1900.00 0.60 0.75 0.2

m 20 kg/sq.cm 415 kg/sq.cm m m

2550000 t/sq.m 93383.79 t-m2

Freely supported Partially Fixed Rigidly fixed 3. Calculation of moment of resistence

a)

b)

m= 13.333 j= 0.890 k= 0.329 Q= 10.263 bd2 kg-cm 4. Calculation of Loads Load due to concentric ring thickness of ring 0.3 Height of ring 2 Weight/m 1.50 t/sq.m 1.5 t/sq.m say

5. B.M.calculation 4.00 a=

1

1.88 t/sq.m 3.00 t/sq.m 1.00 t/sq.m 5.88 t/sq.m 0.88 t/sq.m 7.00 t/sq.m

Self weight of concrete slab= LL Weight of eqipment= Total Other loads 15% impact factor Net downward load 6.756 t/sq.m say b/a

b= 1.220

a/b=

0.305 3.27869

DOCUMENDATE 03-05-2007

PROJECT:

TITLE:

c q=

DESIGNED CHECKED SHEET

CIRCULAR SLAB

1 1.50

t/sq.m

Radial moment Mr at centre w

Deflection=

Mr 

q (3   )( a 2  r 2 ) 16

0.22848 t-m

5qa 4  64(1   ) D

5.3E-06 m 0.005 mm

Simply supported plate with UDL inside concentric circle: Mr 

2  b 2  3 2 qb 2   r  3  a2 a2   a  ln    1  2  2       1 2 qr  2    16 4   a  4  b r    r   a  b

 a  ln    b

2.982

2c 2b

P

P q

=

2a

UDL=

7.00

Due to UDL 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 Max.

r 1.22 1.78 2.05 2.33 2.61 2.89 3.17 3.44 3.72 4.00

r/a= 0.31 0.44 0.51 0.58 0.65 0.72 0.79 0.86 0.93 1.00

a/r 3.28 2.25 1.95 1.72 1.53 1.39 1.26 1.16 1.07 1.00

Mr 0.00 8.08 9.16 9.28 8.74 7.70 6.26 4.47 2.38 0.00 9.28

Mt 35.18 26.14 23.96 22.36 21.08 19.98 18.95 17.96 16.98 15.97 35.18

With simply supported edge

b/a

2

0.305 p1     4

5.40

b2 b ln -0.12179  a  b2 a 2

SF 0.00 3.28 4.65 5.93 7.14 8.30 9.44 10.54 11.63 12.70 12.70

DOCUMENDATE 03-05-2007

PROJECT:

TITLE:

Mr 

DESIGNED CHECKED SHEET

CIRCULAR SLAB

p1     r b2 b a 2    ln 1  ln  2 4  a a  b 2 a  r 2 

P=total circular load= 37.699 t Due to circular conc.load at inner edge r b/r r/a a/r Mr Mt 1 1.22 1.00 0.31 0.00 -11.83 3.28 0.69 0.44 -1.71 -7.99 2 1.78 2.25 0.59 0.51 -1.76 -6.90 3 2.05 1.95 0.52 0.58 -1.64 -6.07 4 2.33 1.72 0.47 0.65 -1.42 -5.41 5 2.61 1.53 0.42 0.72 -1.15 -4.85 6 2.89 1.39 0.39 0.79 -0.87 -4.38 7 3.17 1.26 0.35 0.86 -0.58 -3.97 8 3.44 1.16 0.33 0.93 -0.29 -3.60 9 3.72 1.07 0.31 1.00 0.00 -3.28 10 4.00 1.00 -1.76 -11.83 Total radial Moment due to UDL & concentric load Total tangential moment due to UDL & concentric max Shear= 6. Design of slab Effective depth of slab due to Mr=

Effective depth of slab due to Mt=

Provide

11.042 t-m 47.006 t-m 12.698 t

Mx105 Qb

32.80 cm

Mx105 Qb

67.67542 cm

Cover to reinforcem 5 cm 20 mm bars, over all depth= D= say

7. Area of reinforcement /width for Mr and M Dia of the bar Area= ast= Effective depth of slab for bottom layer Effective depth of slab for top layer

3

20 3.14 69 67

mm sq.cm cm cm

72.18 cm 75.00 cm

DOCUMENDATE 03-05-2007

PROJECT:

TITLE:

r

Mr 1.22 1.78 2.05 2.33 2.61 2.89 3.17 3.44 3.72 4.00

0.00 9.79 10.92 10.92 10.16 8.86 7.13 5.05 2.66 0.00

Min. area of steel=

CIRCULAR SLAB

M

Mt 47.01 34.12 30.86 28.43 26.49 24.83 23.33 21.93 20.58 19.25

47.01 34.12 30.86 28.43 26.49 24.83 23.33 21.93 20.58 19.25

11.25 sq.cm

Ast 41.48 30.11 27.23 25.09 23.38 21.91 20.59 19.35 18.16 16.98

DESIGNED CHECKED SHEET

Ast Spacing dia of bar spacing adopted adopted 20 41.4802 7.57 75 20 30.1133 10.43 100 20 27.2307 11.54 115 20 25.0917 12.52 125 20 23.3760 13.44 130 20 21.9090 14.34 140 16 20.5888 9.77 95 16 19.3527 10.39 100 16 18.1602 11.07 110 16 16.9845 11.84 115

(0.15%)

Provide the steel as mesh on bothways

8. Area of reinforcement /width for -ve B.M. Mr at edges Moment=M= 11.042 t-m Dia of bar 16 mm Area of each bar= 2.01 sq.cm effective depth of slab for bottom layer 71.7 cm Ast=M/tjd= 9.10 sq.cm Provide bars in the form of mesh, Min. area of steel= 11.25 sq.cm Spacing of steel 17.87 cm c/c 165 mm c/c say The above steel will be provided in the form of rings, The above steel is given for a length of= 1.550m from the support 10 No.bars=

4

DOCUMENDATE 03-05-2007

PROJECT:

TITLE:

DESIGNED CHECKED SHEET

CIRCULAR SLAB

10nos of16dia. Bars @165c/c at to 9. Distance from face to of the support upto which radial reinforcement is to be provided Point of inflection= 1.21m from centre = 0.89m from edge This will be greater of the following Point of inflection= 1. Ld=bd = 84.82 cm 2. Point of inflection+d 155.46 cm 107.96 cm 3. Point of inflection+12 Maximum= 155.46 cm Say 155 cm

10. Distribution steel Dia of the bar ast= Area of steel= Spacing

16 2.01 9 22.3 220

@.12% of Ac Say

mm sq.cm sq.cm cm c/c mm c/c

At the edge of slab, the mesh bars are free and are not capable of taking full tension. 11. Check for shear shear force=wr/2 kN Shear stress N/mm2 % of steel= correction factor= k Permissible shear stress= N/mm2 2 Permissible Shear stress kN/mm Balance shear

5

N/mm2

126.98 0.18 1.219 % 1.3 0.44 0.57 -

Hence okay