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Example 1: Design of one-way slab

Design a one-way slab for the inside of a building using the span, loads, and other data given in Figure 1. Normal-weight aggregate concrete is specified with a density of 145 pcf. (assuming cover 3/4 in)

Figure 1

The minimum thickness for one-way slab simply supported=L/20 using table 1(ACI

Table 1

now we will calculate d=(6-3/4(cover)-1/4(estimated as half diameter of reinforcement)
d=5 in 

now we will calculate dead load
concrete density=145pcf, Usually 5 pcf is added to account for the weight of reinforcement, so 150
pcf is used in calculating the weight of a normal-weight concrete member.
dd=(6in*ft/12)*1*150/=75 lb/ft
Wu=1.2*75+1.6*200=410 lb/ft

maximum moment for simply supported span =(Wu*L^2)/8

now we can calculate ρ



checking for ρmin
ρmin=(3√fc'/fy) and not less than 200/fy
ρmin=3*√4000/6000 and not less than 200/60000
ρ>ρmin   ok

As=0.00393*12*6=0.282 in2/ft
Use #4 @ 10 in. from Table A.6 (As = 0.24 in2/ft)
Spacing < maximum of 18 in. as per ACI 7.6.5

checking for ϕ=0.9??

β1=0.85 for fc'=4000 psi

εs=0.033>0.005 so the section is tension control..............ok

Transverse direction-shrinkage and creep

steel G60
then As=0.0018*b*d=0.0018*12*6=0.1296in2

Use #3 @ 10 in. (0.13 in2/ft)

Use #3 @ 10 in. (0.13 in2/ft)  ok

The bar #4 is placed below bar #3 because the effective depth is important for main reinforcement calculation (flexural calculation)


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