C12: DESIGN OF SLAB AS PER ACI METHOD

DESIGN OF SLAB

DATA:

f_y = 60000psi

                        f_c’ = 3000psi

                        l_x = 15’; l_y = 20’

                    l_y/l_x = 20/15 = 1.33

            Slab should be designed as a two way slab.

MINIMUM THICKNESS:

                       h = 2(15+20) x (12/180)

                         = 4.667”

        Assume h = 8”

LOAD CALCULATION:

            Dead load  = ((8/12) x 120) + 20

                              = 120 psf

            Live load  = 40 psf

            Factored load  = (1.2 x 120) + (1.6 x 40)

                                    = 208 psf

                                W = 0.208 klf

MOMENT CALCULATION:

            Middle, Mu = wl²/8 = 0.208 x 15²/8

                                             = 5.85 kft

    End, Mu = 0.1 x 0.208 x 15²

                                                             = 4.68 kft

REINFORCEMENT:

            Effective depth = 8”- 2” = 6”

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                        Mu/φbd²  = (5.85 x 1000 x 12) / (0.9 x 12 x 6²)

                                         = 180.55

                        Use ρ_mini   = 0.0033

                                  As  = ρbd

                                        = 0.0033 x 12 x 6

                                        = 0.2376 in² / ft

            Provide #5 bars @ 12” c/c.

                        Mu/φbd² = (4.68 x 1000 x 12) / (0.9 x 12 x 6²)

                                        = 144.44

                        Use ρ_mini = 0.0033

                                  As = ρbd

                                        = 0.0033 x 12 x 6

                                        = 0.2376 in² / ft

            Provide #5 bars @ 12” c/c.