Simplified Design for Beams and One-Way Slabs - Simplified Design of Reinforced Concrete Buildings - page 91

(d) Bottom bars in interior spans:

A s = 72.9/(4 18.0) = 1.01 in. 2

From Table 3-5: Use 2 No. 7 (A s = 1.20 in. 2 )

(e) Top bars at interior beams:

A s = [106/(4 18) = 1.47 in. 2 ]/3 ft = 0.50 in. 2 /ft

From Table 3-6, with s max = 11.6 in.

Use No. 6 @ 10 in. (A s = 0.59 in. 2 /ft)

(f) Slab reinforcement normal to joists:

Use M u = w u ˜ n 2 /12 (see Fig. 2-5)

= 0.20(6.25) 2 /12 = 0.65 ft-kips

where w u = 1.2(56 + 30) + 1.6(60) = 200 psf = 0.20 kips/ft

Place bars on slab centerline: d = 4.5/2 = 2.25 in.

A s = 0.65/4(2.25) = 0.07 in. 2 /ft
(but not less than required temperature reinforcement).

From Table 3-4, for a 4 1 ⁄ 2 in. slab: Use No. 3 @ 13 in. (A s = 0.10 in. 2 /ft)

Check for fire resistance: from Table 10-3, for restrained members, required cover for
fire resistance rating of 2 hours = 3 ⁄ 4 in. O.K.

(6) Reinforcement details shown in Fig. 3-20 are determined directly from Fig. 8-3(a).*

For structural integrity (ACI 7.13.2.3), one of the No. 7 and No. 8 bars at the bottom must be
spliced over the support with a Class B tension splice; the No. 8 bar must be terminated with
a standard hook at the non-continuous supports.

(7) Design of Shear Reinforcement

(a) End spans:

V u at face of interior beam = 25.3 kips
V u at distance d from support face = 25.3 - 1.6(1.50) = 22.9 kips

* The bar cut-off points shown in Fig. 8-3(a) are recommended for beams without closed stirrups. The reader may consider deter-
mining actual bar lengths using the provisions in ACI 12.10.