Structural Concrete - Hassoun

6.7 ACI Code Requirements 249
Checking the spacing S by Eq. 6.18 and assuming f s = 2/3f y = 2/3 × 60 = 40 ksi, and C c = 2 in.,
then S = 15(40/40) − 2.5 × 2 = 10 in., which is less than 12(40/40) = 12 in. The spacing used is adequate.
Note that C c = 1.5 in. may be used for the skin reinforcement concrete cover.
It is recommended to use smaller spacing to control the propagation of tensile cracks along
the side of the tension zone with the first side bar to be placed at 4 to 6 in. from the main tensile steel.
Example 6.6
The sections of a simply supported beam are shown in Fig. 6.9.
a. Check if the bar arrangement satisfies the ACI Code requirements.
b. Determine the expected crack width.
c. Check the Z-factor based on Eq. 6.19.
Given: f c ′ = 4ksi,f y = 60 ksi, and no. 3 stirrups.
Solution
1. Fig. 6.9, section a:
a. For three no. 8 bars, A s = 2.35 in. 2 , clear cover, C c = 2.5 − 8/16 = 2.0 in. Assume f s =
2
f 3 y = 2∕3 × 60 = 40 ksi. Maximum spacing s = 600/40–2.5 × 2 = 10 in., which is less than
12(40/40) = 12 in. Spacing provided equals 0.5(12–2.5–2.5) = 3.5 in., center to center of bars,
which is less than 10 in.
b. For this section, d c = 2.5 in. The effective tension area of concrete for one bar is
12(2 × 2.5)
A = = 20 in. 2
3
Estimated crack width using Eq. 6.16 is
W = 0.076(1.2)(36,000) 3 √
20 × 2.5 × 10 −6 = 0.0121 in.
This is assuming β = 1.2 for beams and f s = 36 ksi. The crack width above is less than
0.016 in. and 0.013 in. for interior and exterior members.
Figure 6.9 Two sections for Example 6.6.