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dISEÑO DE DOMOS CIRCULARES
India Pavilion at World Expo 2010, Shangai, China
Top Circular Slab Design
Slab Design of Partially Fixed Circular Slab INPUT DATA :Diameter of circular slab
=
7350 mm
Grade of concrete
=
25 N/mm2
Grade of steel
=
415 N/mm2
Basic L/d ratio
20
Percentage of steel i.e pt
0.5 %
Modification Factor
1.5
Thickness of slab
=
I.S 456:2000 Cl 23.2.1
I.S 456:2000 Fig 4
245 mm
Provided depth of slab:--
250 .mm
Cover
=
25 mm
Assume the diameter of bar
=
12 mm
Effective depth of slab (d)
=
219 mm
Live load
=
5 kN/m2
Design Load
=
16.6875 kN/m2
Max. moment at center i.e Mr =
=
28.17 kN.m
Max. Shear
=
30.66328 kN/m
Effective depth of slab
=
90.36462 mm
Shear i.e v
=
0.140015 N/mm2
Minimum shear i.e Tc for M25
=
0.37 N/mm2
Calculation of Moments and Shear
Design of depth of slab 219 O.K.
0.140015 O.K.
Variation of Moments along the radius Radius at where the radial moment become 3000.625 mm Distance of Point of Contraflexu
=
674.3751 mm
Steel Calculation 1) Rectangular mesh for positive B.M ( MΦ & Mr ) in the central portion at the bottom Note : Mesh(Radial) reinforcement are provided in both ways at the bottom of slab.
(M/b*d*d)
=
0.587392
1
From IS:456 :2000 Table 19
India Pavilion at World Expo 2010, Shangai, China
Top Circular Slab Design
Percentage of steel i.e pt
=
0.2 %
Area of steel i.e Ast
=
438 mm2
Diameter of reinforcement
=
10 mm
Number of steel reinforcement
=
5.579618 Nos
Spacing of reinforcement
=
179.2237 mm
Spacing provided:-Provide
10
.mm dia at
SP:16 ( Depending of fck and fy)
Maximum limit of spacing is (3*d).
100 .mm 100 c/c spacing
2) Radial steel for negative B.M.(Mr ) at the ends placed on top of the slab (Provide in perpendicular dir.) Radius of the slab
=
Radial Moment (-ve) at the end
= -140.8594 kN.m
(M/b*d*d)
=
Percentage of steel i.e pt
=
0.94 %
Area of steel i.e Ast
=
2058.6 mm2
Diameter of reinforcement
=
12 mm
Number of steel reinforcement
=
18.21125 Nos
Spacing of reinforcement
=
54.9111 mm
Spacing provided:--
7.35 M
2.936958
Maximum limit of spacing is (3*d) .
100 .mm
Curtailment of reinforcement beyond point o 818.3751 mm
Provide
SP:16 ( Depending of fck and fy)
12
mmdia at
100
I.S 456:2000 Cl.26.2.3.1
mm c/c spacing
3) Circumferential steel for negative B.M.(Mθ ) at the ends placed on the bottom of the slab(Provide in perpendicular dir.) Note : Provide circumferential steel for development length at the ends on the bottom of the slab. Diameter of reinforcement
=
10 mm
Development length ( Ld )
=
818.3751 mm
SP:16 Table 65
Spacing required
=
224.0297 mm
Maximum limit of spacing is (3*d) .
Spacing provided:--
=
100 .mm
Number of rings
=
4.463694 Nos
Provide
10
mmdia at
100
mm c/c spacing ( 6 Nos.)
4) Circumferential steel for negative B.M.Mθ ) at the ends placed on the top of the slab(Provide in perpendicular dir.) Note : These circumferential steel are provided same as given in the above (3rd case) calculation.
