A slab spans 25 and has a minimum thickness of 15 Design the

A slab spans 25\' and has a minimum thickness of 15\". Design the slab to support a live load of 65 PSF and a dead load of 30 PSF plus the weight of the slab.

F\'c = 4000 PSI

Fy = 60 KSI

Solution

Solution:- Design of one way slab:-

Given

D = 15 inch

Span = 25 feet

Let Width = 16 feet

Live load = 65 psf

Live load = 65 *16 = 1040 pound/feet

Weight of slab = 15/12 *16*150 =3000 pound/feet

Density of concrete = 150 pcf

Total dead load = 30*16 + 3000 = 3480 pound/feet

Total load = live load + total dead load

Total load = 1040 + 3480 = 4520 pound/feet

Factored load (w_u) = 1.5 *4520 = 6780 pound/feet

Let effective cover of slab = 1 inch

Effective depth(d) = 15 – 1 = 14 inch

Effective span(l) = 25 +14/12 = 26.16 feet

B.M = w_u*l²/8

B.M. = 6780*(26.16)²/8

B.M. = 579.98 kips –feet

We know that

B.M. = 0.87f_yA_st*(d – f_y*A_st/(f_cb))

579.98 *10³*12 * (14 – 60000*A_st/(4000 *16*12))

Area of steel(A_st) = 10.09 inch²   

Provide 0.7 inch diameter bars @ 6 inch spacing C/C

Shrinkage and temperature reinforcement = 0.15% area of concrete

Shrinkage and temperature reinforcement = 0.0015 *15*(16*12)

= 4.32 inch²

Provide 0.5 inch diameter bars @ 4 inch spacing C/C.