Structural Concrete - Hassoun

6.7 ACI Code Requirements 251
Choose three no. 9 bars (area = 3.0 in. 2 ) in one row, and a total depth of h = 25.0 in. Actual
d = 25–2–9/16 = 22.44 in. (Fig. 6.10).
2. Check spacing of bars using Eq. 6.18. Calculate the service load and moment, w = 1.5 +
1.18 = 2.68 K/ft:
M = 2.68(24)2 = 193 K ⋅ ft = 2315 K ⋅ in.
8
3. Calculate the neutral axis depth kd and the moment arm jd (Eq. 6.12):
1
2 b(kd)2 − nA s (d − kd) =0 n = 8 A s = 3.0 d = 22.44 in.
kd = 6.85 in. jd = d − kd 20.16
= 20.16 in. j =
3 22.44 = 0.898
Note that an approximate value of j = 0.87 may be used if kd is not calculated.
4. Calculate the stress f s :
5. Calculate the spacing s by Eq. 6.18:
M = A s f s jd 2315 = 3(f s )(20.16) f s = 38.3ksi
s = 600∕38.3 − 2.5 × 2 = 10.7in.
(controls)
which is less than 12(40/40) = 12.0 in. Spacing provided = 0.5 (16–2.56–2.56) = 5.44 in., which
is less than 10.7 in.
Example 6.8
Design a simply supported beam of 7.2-m span to carry a uniform dead load of 22.2 kN/m and a live
load of 17 kN/m. Choose adequate bars, and check their spacing arrangement to satisfy the ACI Code.
Given: b = 400 mm, f c ′ = 30 MPa, f y = 400 MPa, a steel percentage of 0.8%, and a clear concrete
cover of 50 mm.
Solution
1. For a steel percentage of 0.008 and from Eq. 3.22, R u = 2.7 MPa. Factored load w u = 1.2(22.2) +
1.6(17) = 53.8 kN/m. Then M u = w u L 2 /8 = 53.8(7.2) 2 /8 = 348.6 kN ⋅ m; M u = R u ⋅bd 2 , or
348.6 × 10 6 = 2.7 × 400d 2 , d = 568 mm, A s = ρbd = 0.008 × 400 × 568 = 1818 mm 2 . Choose four
bars, 25 mm (no. 25 M), A s = 2040 mm 2 ,inonerow(b min = 220 mm). Let h = 650 mm, the actual
d = 650–50–25/2 = 587.5 mm, say 585 mm. Final section: b = 400 mm, h = 650 mm, with four
no. 25 mm bars (Fig. 6.11).
Figure 6.11 Example 6.8.