Structural Concrete - Hassoun

250 Chapter 6 Deflection and Control of Cracking
2. Fig. 6.9, section b:
a. Calculations of spacing of bars are similar to those in section a.
b. For this section, d c = 2.5 in., and the steel bars are placed in two layers. The centroid
of the steel bars is 3.5 in. from the bottom fibers. The effective tension concrete area is
A = 12(2 × 3.5)/6 = 14 in. 2 Then
which is adequate.
W = 0.076 × 1.2 × 36,000 3 √
14 × 2.5 × 10 −6 = 0.0107 in.
Discussion
It can be seen that the spacing, s, in Eq. 6.17 is a function of the stress in the tension bars or, indirectly,
is a function of the strain in the tension steel, f s = E s × ε s ,andE s for steel is equal to 29,000 ksi. The
spacing also depends on the concrete cover, C c . An increase in the concrete cover will reduce the limited
spacing s, which is independent on the bar size used in the section.
In this example, the expected crack width was calculated by Eq. 6.17 to give the student or the
engineer a physical feeling for the crack width and crack control requirement. The crack width is usually
measured in beams when tested in the laboratory or else in actual structures under loading when serious
cracks develop in beams or slabs and testing is needed. If the crack width measured before and after
loading is greater than the yield strain of steel, then the main reinforcement is in the plastic range and
ineffective. Sheets with lines of different thickness representing crack widths are available in the market
for easy comparisons with actual crack widths. In addition to the steel stress and the concrete cover, W
depends on a third factor, A, representing the tension area of concrete surrounding one bar in tension.
Example 6.7
Design a simply supported beam with a span of 24 ft to carry a uniform dead load of 1.5 K/ft and a live
load of 1.18 K/ft. Choose adequate bars; then check their spacing arrangement to satisfy the ACI Code.
Given: b = 16 in., f c ′ = 4ksi,f y = 60 ksi, a steel percentage = 0.8%, and a clear concrete cover of 2 in.
Solution
1. For a steel percentage of 0.8%, R u = 400 psi = 0.4 ksi (φ = 0.9). The external factored moment is
M u = w u × L 2 /8, and
w u = 1.2(1.5)+1.6(1.18) =3.69 K∕ft
M u = 3.69(24)2 = 265.68 K ⋅ ft = 3188.2K⋅ in.
8
M u = R u bd 2 d = 22.32 A s = 0.008 × 16 × 22.32 = 2.86 in. 2
Figure 6.10 Example 6.7.