Given A oneway slab has a simple span of 15 ft which carries

Given: A one-way slab has a simple span of 15 ft which carries a service live load of 100 psf in addition to its self weight. Normal weight concrete of fe4000 psi is specified for use with steel of yield stress equal to 60,000 psi. Required: Design the one-way slab for flexure following the current ACI 318 Code.

Solution

Let us assume a 5\" thick slab

self weight of slab = (5/12)*150=62.5 psf

Factored load on slab per load combination 1.2D+1.6L = (1.2*62.5)+(1.6*100)=235 psf

Span of slab=15 ft

Let us consider unit width of slab.It is like a simply supported beam with factored load 235 plf and span 15 ft

Moment in beam = 0.235*15*15/8=6.61 kip-ft=79320 lb-in

Let us consider effective depth of beam = 4\"

Let area of steel required be A_st

Let depth of compression zone be a

0.85*4000*12*a=0.9*60000*A_st

a=1.32A_st

moment capapcity = 0.9*60000*A_st*(4-1.32A_st/2)=54000*A_st(4-0.66A_st)=79320

Or, 216000A_st-35640A_st²=79320

A_st = 0.393 in²

Provide 2-#4

therefore, in the slab provide 2-#4 bar in every 12\" width of slab. This means provide #4@6\" bottom rebar