Circular Slab

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1. Data

Diameter of circular slabl= 3.00 m

Grade of concrete M 20

Permissible stress in concrete fc= 70.00 kg/sq.cm

Grade of steel Fe 415

Permissible stress in steel fst= 1900.00 kg/sq.cm Thickness of circular wall 0.60 m

Thickness of slab 0.75 m Poissons ratio 0.2 E= 2550000 t/sq.m D= 93383.79 t-m2 2. Support conditions: Freely supported Partially Fixed Rigidly fixed

3. Calculation of moment of resistence

m= 13.333 j= 0.890

k= 0.329 Q= 10.263 bd2 kg-cm

4. Calculation of Loads

a) Load due to concentric ring

thickness of ring 0.3 Height of ring 2

Weight/m say

b) Self weight of concrete slab= LL

Weight of eqipment= Total

Other loads impact factor 15%

Net downward load 6.756 t/sq.m say 7.00 t/sq.m

5. B.M.calculation b/a 0.305 a= 4.00 b= 1.220 a/b= 3.27869 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 1.50 t/sq.m 1.5 t/sq.m

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c 1

q= 1.50 t/sq.m

Radial moment Mr at centre 0.22848 t-m

Deflection= 5.3E-06 m

0.005 mm Simply supported plate with UDL inside concentric circle:

2.982 P P q = UDL= 7.00 Due to UDL r r/a= a/r Mr Mt SF 1.00 1.22 0.31 3.28 0.00 35.18 0.00 2.00 1.78 0.44 2.25 8.08 26.14 3.28 3.00 2.05 0.51 1.95 9.16 23.96 4.65 4.00 2.33 0.58 1.72 9.28 22.36 5.93 5.00 2.61 0.65 1.53 8.74 21.08 7.14 6.00 2.89 0.72 1.39 7.70 19.98 8.30 7.00 3.17 0.79 1.26 6.26 18.95 9.44 8.00 3.44 0.86 1.16 4.47 17.96 10.54 9.00 3.72 0.93 1.07 2.38 16.98 11.63 10.00 4.00 1.00 1.00 0.00 15.97 12.70 Max. 9.28 35.18 12.70

With simply supported edge

b/a 0.305 5.40 -0.12179 ) )( 3 ( 16 2 2 r a q Mr       D qa w ) 1 ( 64 4 5  2b 2a 2c                                                      b a b a b r a r a b a a r qb qr Mr 1 1 ln 4 3 ln 4 16 3 2 2 2 2 2 2 2 2 2 2      4 1 p   a b b a b ln 2 2 2

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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 3.28 0.00 -11.83 2 1.78 0.69 0.44 2.25 -1.71 -7.99 3 2.05 0.59 0.51 1.95 -1.76 -6.90 4 2.33 0.52 0.58 1.72 -1.64 -6.07 5 2.61 0.47 0.65 1.53 -1.42 -5.41 6 2.89 0.42 0.72 1.39 -1.15 -4.85 7 3.17 0.39 0.79 1.26 -0.87 -4.38 8 3.44 0.35 0.86 1.16 -0.58 -3.97 9 3.72 0.33 0.93 1.07 -0.29 -3.60 10 4.00 0.31 1.00 1.00 0.00 -3.28 -1.76 -11.83 Total radial Moment due to UDL & concentric load 11.042 t-m Total tangential moment due to UDL & concentric 47.006 t-m

max Shear= 12.698 t

6. Design of slab

Effective depth of slab due to Mr= 32.80 cm

Effective depth of slab due to Mt= 67.67542 cm

Cover to reinforcem 5 cm

Provide 20 mm bars, over all depth= D= 72.18 cm

say 75.00 cm

7. Area of reinforcement /width for Mr and M

Dia of the bar 20 mm

Area= ast= 3.14 sq.cm

Effective depth of slab for bottom layer 69 cm Effective depth of slab for top layer 67 cm

Mx Q b 105

                 ln 2 2 2 ln 1 22 4 1 r a a b b a b a r p Mr   Mx Q b 105

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r Mr Mt M Ast Ast

adopted dia of bar spacing

Spacing adopted 1.22 0.00 47.01 47.01 41.48 41.4802 20 7.57 75 1.78 9.79 34.12 34.12 30.11 30.1133 20 10.43 100 2.05 10.92 30.86 30.86 27.23 27.2307 20 11.54 115 2.33 10.92 28.43 28.43 25.09 25.0917 20 12.52 125 2.61 10.16 26.49 26.49 23.38 23.3760 20 13.44 130 2.89 8.86 24.83 24.83 21.91 21.9090 20 14.34 140 3.17 7.13 23.33 23.33 20.59 20.5888 16 9.77 95 3.44 5.05 21.93 21.93 19.35 19.3527 16 10.39 100 3.72 2.66 20.58 20.58 18.16 18.1602 16 11.07 110 4.00 0.00 19.25 19.25 16.98 16.9845 16 11.84 115

Min. area of steel= 11.25 sq.cm (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

say 165 mm c/c 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

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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 3. Point of inflection+12 107.96 cm Maximum= 155.46 cm Say 155 cm 10. Distribution steel

Dia of the bar 16 mm

ast= 2.01 sq.cm

Area of steel= @.12% of Ac 9 sq.cm

Spacing 22.3 cm c/c

Say 220 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 126.98

Shear stress N/mm2 0.18

% of steel= 1.219 %

correction factor= k 1.3

Permissible shear stress= N/mm2 0.44 Permissible Shear stress kN/mm2

0.57

Figure

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