C8: DESIGN OF SUNSHADE AS PER IS CODE

SUNSHADE DESIGN

LONGER SPAN:

Length of sunshade       = 5.248 m

Width of sunshade        = 450 mm = 0.45 m

Thickness @ fixed end = 150 mm = 0.15 m

Thickness @ free end   = 60 mm = 0.06 m

f_{ck =} 20 N/mm² & f_y = 415 N/mm²

Clear cover                   = 25 mm

           SUse 8 mm dia bar

EFFECTIVE DEPTH:

d = D – cc – (dia/2) – dia

   = 150 – 25 – (8/2) – 8

d = 113 mm

LOAD CALCULATION:

Self weight of sunshade = L x b x d x γ

                                       = 5.248 x 0.45 x [(0.15+0.06) / 2] x 25

                                       = 6.1992 KN

                Imposed load = L x b x d x 15

                                       = 5.248 x 0.45 x 0.150 x 15

                                       = 5.3136 KN

                    Total load   = 6.1992 + 5.3136 = 11.5128 KN

                 Design load   = 1.5 x 11.5128 = 17.2692 KN

BENDING MOMENT:

Max BM, Mu = Wl / 8

                       = (17.2692 x 5.248) / 8

                       = 11.32 KNm.

EFFECTIVE DEPTH REQUIRED:

d² = (Mu) / (0.138 x fck x b)

     = (11.32 x 10⁶) / (0.138 x  20 x 450)

d = 95.46 mm ≃ 96 mm

∴ 113 mm > 96 mm

               Hence safe.

AREA OF STEEL:

   Xu (max) = 0.48 x d = 0.48 x 113 = 54.24 mm

             Mu = 0.87 x fy x Ast x [ d – (0.42 x Xu max)]

11.32 x 10⁶ = 0.87 x 415 x Ast x [ 113 – ( 0.42 x 54.24 )]

            Ast = 347.5 mm² ≃ 350 mm²

     Ast min = 0.12 % of Ag

                  = ( 0.12 / 100 ) x 450 x 150

                  = 81 mm²

Provide 8 mm dia bars.

SPACING:

Spacing = ( ast / Ast ) x 1000

              = ( 81 / 350) x 1000

              = 231.4 mm ≃ 200 mm spacing.

SPACING LIMIT:

i ) Max spacing = 3d = 3 x 113 = 339 mm

ii ) 300 mm

Provide 8 mm dia bars @ 200 mm c/c spacing.

DISTRIBUTION BARS:

       Ast = 81 mm²

Provide 6 mm dia bars

∴ ast = ( π / 4) x d²

          = ( π / 4 ) x 6²

          = 28.27 mm²

                    Spacing = ( ast / Ast ) x 1000

          = ( 28.27 / 81 ) x 1000

          = 349 mm ≃ 350 mm.

SPACING LIMIT:

i ) Max spacing = 5d = 5 x 113 = 565 mm

ii ) 300 mm

Provide 6 mm dia bars @ 350 mm c/c spacing.

RESULT:

Overall depth,    D = 150 mm

Effective  depth, d = 113 mm

Provide 8 mm dia of main bars @ 200 mm c/c spacing.

Provide 6 mm dia of distribution bars @ 250 mm c/c spacing.

SHORTER SPAN:

Length of sunshade       = 3.280 m

Width of sunshade        = 450 mm = 0.45 m

Thickness @ fixed end = 150 mm = 0.15 m

Thickness @ free end   = 60 mm = 0.06 m

fck = 20 N/mm² &  fy  = 415 N/mm²

Clear cover                   = 25 mm

          Use 8 mm dia bar

EFFECTIVE DEPTH:

d = D – cc – (dia/2)

   = 150 – 25 – (8/2)

d = 121 mm

LOAD CALCULATION:

Self weight of sunshade = L x b x d x γ

                                       = 3.280 x 0.45 x [(0.15+0.06) / 2] x 25

                                       = 3.8745 KN

                Imposed load = L x b x d x 15

                                       = 3.280 x 0.45 x 0.150 x 15

                                       = 3.321 KN

                      Total load = 3.8745 + 3.321 = 7.1955 KN

                   Design load = 1.5 x 7.1955 = 10.79325 KN

BENDING MOMENT:

Max BM, Mu = Wl / 8

                       = (10.8 x 3.280) / 8

                       = 4.428 KNm.

EFFECTIVE DEPTH REQUIRED:

d² = (Mu) / (0.138 x fck x b)

     = (4.428 x 10⁶) / (0.138 x  20 x 450)

  d = 59.70 mm ≃ 60 mm

∴ 121 mm > 60 mm

               Hence safe.

AREA OF STEEL:

  Xu (max) = 0.48 x d = 0.48 x 121 = 58.08 mm

             Mu = 0.87 x fy x Ast x [ d – (0.42 x Xu max)]

4.428 x 10⁶ = 0.87 x 415 x Ast x [ 121 – ( 0.42 x 58.08 )]

             Ast = 126 mm²

      Ast min = 0.12 % of Ag

                    = ( 0.12 / 100 ) x 450 x 150

                    = 81 mm2

    Provide 8 mm dia bar

SPACING:

Spacing = ( ast / Ast ) x 1000

              = ( 81 / 126 ) x 1000

              = 642.8 mm ≃ 640 mm spacing.

SPACING LIMIT:

i ) Max spacing = 3d = 3 x 121 = 363 mm

ii ) 300 mm

Provide 8 mm dia bars @ 300 mm c/c spacing.

DISTRIBUTION BARS:

        Ast = 81 mm2

  Provide 6 mm dia bars

      ∴ ast = ( π / 4) x d2

              = ( π / 4 ) x 62

              = 28.27 mm2

Spacing = ( ast / Ast ) x 1000

              = ( 28.27 / 81 ) x 1000

              = 349 mm ≃ 350 mm.

SPACING LIMIT:

i ) Max spacing = 5d = 5 x 121 = 605 mm

ii ) 300 mm

Provide 6 mm dia bars @ 350 mm c/c spacing

RESULT:

Overall depth,    D = 150 mm

Effective  depth, d = 121 mm

Provide 8 mm dia of main bars @ 300 mm c/c spacing.

                        Provide 6 mm dia of distribution bars @ 350 mm c/c spacing