Structural Concrete - Hassoun

4.5 Design of T-Sections 171
The design moment that the concrete flange can resist is greater than the factored applied
moment. Therefore, the section behaves as a rectangular section.
2. Determine the area of tension steel, considering a rectangular section, b = 40 in.:
R u = φM n 3, 350, 000
=
bd2 40 × 14.5 = 398psi
2
From Eq. 4.2 or from tables in Appendix A, for R u = 398 psi and ρ = 0.00817,
A s = ρbd = 0.00817 × 40 × 14.5 = 4.74in. 2
Use six no. 8 bars, A s = 4.74 in. 2 (in two rows).
3. Check that ρ w = A s /b w d ≥ ρ min ; ρ w = 4.74/(10 ×14.5) = 0.0327 >ρ min = 0.00333. Note that the A s
used is less than A s,max of 7.06 in. 2 calculated by Eq. 3.72.
Also, a = 2.788 in., c = 3.28 in., d t = 14.5 in.,
and ε t = 0.003(d t − c)/c = 0.01 > 0.005, which is OK.
Example 4.8
The floor system shown in Fig. 4.8 consists of 3-in. slabs supported by 14-ft-span beams spaced 10 ft
on center. The beams have a web width, b w , of 14 in. and an effective depth, d, of 18.5 in. Calculate the
necessary reinforcement for a typical interior beam if the factored applied moment is 5080 K ⋅ in. Use
f c ′ = 3ksi and f y = 60 ksi.
Solution
1. Find the beam flange width: The flange width is the smallest of
b = 16t + b w = 3 × 16 + 12 = 60in.
and
b = span 14 × 12
= = 42in.
4 4
Center-to-center spacing of adjacent slabs is 10 ×12 = 120 in. Use b = 42 in.
2. Check the position of the neutral axis assuming a = t:
(
φM n (based on flange) =φ × 0.85f c ′ bt d − 1 )
2 t
= 0.9 × 0.85 × 3 × 42 × 3(18.5 − 1.5) =4916K ⋅ in.
The applied moment is M u = 5080 K ⋅ in. > 4916 K ⋅ in.; the beam acts as a T-section, so a > t.
Figure 4.8
Example 4.8: Effective flange width.