Simplified Design for Two-Way Slabs - Simplified Design of Reinforced Concrete Buildings - page 132

Direct shear from gravity loads:

q u = 1.20(136) + 1.60(29.5*) = 210.4 psf
V u = 0.210(24 x 20 - 1.94 2 ) = 100.2 kips

1.94 '

where d = 8.50 - 1.25 = 7.25 in.

and b 1 = b 2 = (16 + 7.25) /12 = 1.94 ft

Critical
section

Gravity + wind load:

24'-0"

Two load combinations have to be considered ACI Eq. (9-3)
and Eq. (9-4).

ACI Eq. (9-3) [1.2D + 0.8W or 1.2D + 0.5L]

V u = [1.2(136) + 0.5(29.5)] x [24 x 20 - 1.94 2 ] = 84.7 kips
M u = 0.8 111.56 = 89.25 ft-kips

ACI Eq. (9-4) [1.2D + 0.5L + 1.6W]

V u = [1.2(136) + 0.5(29.5)] [24 20 - 1.94 2 ] = 84.75 kips

M u = 1.6 111.56 = 178.5 ft-kips

From Table 4-7, for 8 1 ⁄ 2 in. slab with 16 x 16 in. columns:

A c = 674.3 in. 2
J/c = 5352 in. 3

Shear stress at critical transfer section:

v u = V u /A c + γ v M u c/J

= (84,700/674.3) + (0.4 x 178.5 12,000/5352)

φ 4 f c ʹ

= 125.6 + 160.1 = 285.7 psi > = 215 psi

The 8 1 ⁄ 2 in. slab is not adequate for gravity plus wind load transfer at the interior columns.

Increase shear strength by providing drop panels at interior columns. Minimum slab thickness at drop
panel = 1.25(5.5) = 10.63 in. (see Fig. 4-2). Dimension drop to actual lumber dimensions for economy of
formwork. Try 2 1 ⁄ 4 in. drop (see Table 9-1).

h = 8.5 + 2.25 = 10.75 in. > 10.63 in.

d = 7.25 + 2.25 = 9.5 in.

* Live load reduction: A I (4 panels) = 24 x 20 x 4 = 1920 sq ft
L = 50(0.25 + 15/

1920

) = 29.5 psf