20 = sum of perimeters of reinforcing bars, in (mm). Then the ultimate shear flow at any
plane between the neutral axis and the reinforcing steel is hu = VJ(d-al2).
In conformity with the notational system of the working-stress method, the distance d- all is designated asyW. Dividing the shear flow by the area of contact in a unit length
and introducing the capacity-reduction factor yield
«•-£&
(18)
A section of the ACI Code sets <f> = 0.85 with respect to bond, andy is usually assigned the
approximate value of 0.875 when this equation is used.- Calculate the bond stress
Thus, 20 = 11.0 in (279.4 mm), from the ACI Handbook. Then uu = 72,000/[0.85(11.0)
(0.875) x (15)] = 587 lb/in^2 (4047.4 kPa).
The allowable stress is given in the Code as
W _ 9.5(/;r
M,allow~ j^ \^)but not above 800 lb/in
2
(5516 kPa). Thus, wMallow = 9.5(4,000)° V0.875 - 687 lb/in
2
(4736.9 kPa).DESIGN OF INTERIOR SPAN OF A
ONE-WAYSLAB
A floor slab that is continuous over several spans carries a live load of 120 Ib/ft^2 (5745
N/m^2 ) and a dead load of 40 Ib/ft^2 (1915 N/m^2 ), exclusive of its own weight. The clear
spans are 16 ft (4.9 m). Design the interior span, using/; = 3000 lb/in^2 (20,685 kPa) and
fy = 50,000 lb/in^2 (344,750 kPa).
Calculation Procedure:- Find the minimum thickness of the slab as governed by
the Code
Refer to Fig. 8. The maximum potential positive or negative moment may be found by ap-
plying the type of loading that will induce the critical moment and then evaluating this
moment. However, such an analysis is time-consuming. Hence, it is wise to apply the
moment equations recommended in the ACI Code whenever the span and loading condi-
tions satisfy the requirements given there. The slab is designed by considering a 12-in
(304.8-mm) strip as an individual beam, making b = 12 in (304.8 mm).
Assuming that L— 17 ft (5.2 m), we know the minimum thickness of the slab is tmin =
L/35 = 17(12)735 = 5.8 in (147.32 mm). - Assuming a slab thickness, compute the ultimate load
on the member
Tentatively assume t = 6 in (152.4 mm). Then the beam weight = (6/12)(150 Ib/ft^3 = 75
Ib/lin ft (1094.5 N/m). Also, wtt - 1.5(40 + 75) + 1.8(120) = 390 Ib/lin ft (5691.6 N/m).