FIGURE 61. Ribbed floor.Try the cross section shown in Fig. 61, which has an average weight of 54 lb/ft^2 (2.59
kPa). Although the forms are tapered to facilitate removal, assume for design purposes
that the joist has a constant width of 5 in (127 mm). The design of a ribbed floor is gov-
erned by the ACI Code. The ultimate-strength design of reinforced-concrete members is
covered in Sec. 1.
Referring to the ACI Code, compute the ultimate load carried by the joist. Or, wu =
2.08[1.5(54 + 20 +10 + 15) + 1.8(8O)] = 608 Ib/lin ft (8.9 kN/m).
- Determine whether the joist is adequate with respect to shear
Since the joist is too narrow to permit the use of stirrups, the shearing stress must be lim-
ited to the value given in the ACI Code. Or, vc = 1.1(2<£)(/C')(2<£)^05 = 1.1(2)(0.85)
(3000)°^5 = 102 lb/in^2 (703.2 kPa).
Assume that the reinforcement will consist of no. 4 bars. With % in (19.1 mm) for fire-
proofing, as required by the ACI Code, d = 8 + 2.5 - 1.0 = 9.5 in (241.3 mm). The vertical
shear at a distance d from the face of the support is Vu = (8.5 - 0.79)608 = 4690 Ib (20.9
kN).
The critical shearing stress computed as required by the ACI Code is vu = VJ(bd) =
4690/[5(9.5)] = 99 lb/in^2 (682.6 kPa) < vc, which is satisfactory. - Compute the ultimate moments to be resisted by the joist
Do this by applying the moment equations given in the ACI Code. Or, MMpos =
(>/i6)608(17)^212 = 132,000 in-lb (14.9 kN-m); MMneg = (^1 Xii)608( 17)212 = 192,000 in-lb
(21.7 kN-m).
Where the bending moment is positive, the fibers above the neutral axis are in com-
pression, and the joist and tributary slab function in combination to form a T beam.
Where the bending moment is negative, the joist functions alone. - Determine whether the joist is capable of resisting the negative
moment
Use the equation #max = 0.6375^87,000/(87,0OO + fy), or qmax = 0.6375(0.85)
87,000/127,000 - 0.371. By Eq. 6 of Sec. 1, M 14 =b(Pfc'q(l - 0.590), or Mu =
0.90(5)9.5^2 x (3000)0.371(0.781) = 353,000 in-lb (39.9 kN-m), which is satisfactory.
5. Compute the area of negative reinforcement
Use Eq. 7 of Sec. 1. Or,/c = 0.85(3) = 2.55 kips/in^2 (17.6 MPa); bdfc = 5(9.5)2.55 = 121.1;
2bfcMu/= 2(5)2.55(192)/0.90 = 5440; A 5 = [121.1 - (121.1^2 - 5440)^05 ]/40 = 0.63 in^2
(4.06 cm^2 ).
6. Compute the area of positive reinforcement
Since the stress block lies wholly within the flange, apply Eq. 7 of Sec. 1, with b = 25 in
(635 mm). Or, bdfc = 605.6; 2bfcMul(f> = 18,700; A 3 = [605.6 - (605.6^2 - 18,700)°^5 ]/40 =
0.39 in^2 (2.52 cm^2 ).
7. Select the reinforcing bars and locate the bend points
For positive reinforcement, use two no. 4 bars, one straight and one trussed, to obtain As =