2 Design a oneway floor slab for an industrial building The

2. Design a one-way floor slab for an industrial building. The slab will be indoors (not exposed to weather) and will have a simple span of 8\'-0\". The floor must support a superimposed dead load of 25 psf (in addition to the slab selfweight) and a live load of 150 psf. Use a slab thickness of 5\" and use #4 reinforcing (choose most efficient spacing allowed by code). Select temperature and shrinkage steel and check minimum reinforcing requirements. Concrete compressive strength, fe 4000 psi, and the steel is grade 60 Wuve 9-0

Solution

Solution:-

Design one way slab

Depth = 5 inch

Dead load = 25 psf

Live load = 150 psf

Total load = 175 psf

Factored load = 1.5 * 175 = 262.5 psf

Factored B.M. = w_ul²/8

Assume steel consist of #4 bars , 0.5 inch diameter bars are used with 0.59 inch clear cover

Effective depth (d) = 5 – 0.59 = 4.41 inch

Effective span of slab = 8 + 4.41/12 = 8.36 feet

Bending moment = 262.5 * 8.36²/8

                              = 2293.25 pound – feet

Area of steel

M_u = 0.87f_yA_st*( d – f_yA_st/(f_ckb))

2293.25 *12 = 0.87* 60000 * A_st*(4.41 – 60000*A_st/(4000*39.36))

A_st = 11.47 inch²                         

Use #4 bars @ 6 inch spacing

Temperature reinforcement equal to 0.15 % of the gross concrete area will be provided in the longitudinal direction

   = 0.0015*39.36*4.41 =0.26 inch²

Use #4 bars @ 5 inch spacing

Minimum area of reinforcement = 0.12% of total cross sectional area

                                                           = 0.12/100 * 39.36*5 = 0.23 inch²  

So provided reinforcement is safe