Structural Concrete - Hassoun

4.5 Design of T-Sections 173
A s
Figure 4.10 Example 4.9.
Let M u = φ A sft f y (d − t/2) and calculate d:
812 × 12 = 0.9(8.16)(60)
(
d − 4 )
2
d = 24.1in.
Now, if an effective d = 24.1 in. is chosen, then A s = A sft = 8.16 in. 2
2. If a depth d > 24.1 in. is chosen, say 26.5 in., then a < t and it is a rectangular analysis. The steel
ratio can be calculated from Eq. 4.2 with ρ = 0.00574 and A s = ρbd = 0.00574 × 48 × 26.5 =
7.3 in. 2 (six no. 10 bars in two rows, A s = 7.62 in. 2 ).
3. If a depth d < 24.1 in. is chosen, say, 23.5 in., then a > t, and the section behaves as a T-section.
Calculate
A sf = 0.85f c ′ t(b − b w ) 0.85(3)(4)(48 − 16)
= = 5.44 in. 2
f y 60
M u2 = φA sf f y
(
d − 1 2 t )
= 0.9(5.44)(60)
M u1 = 812 × 12 − 6316 = 3428K ⋅ in.
(
23.5 − 4 )
= 6316K ⋅ in.
2
4. For the basic singly reinforced section, b w = 16 in., d = 23.5 in., and M u1 = 3428 K ⋅ in.,
R u = 387 psi. Calculate ρ 1 fromEq.4.2togetρ 1 = 0.0079:
A s1 = ρ 1 b w d = 0.0079(16)(23.5) =2.97 in. 2
Total A s = A sf + A s1 = 5.44 + 2.97 = 8.41 in. 2
A s = 8.89 in. 2 )
(seven no.10 bars in two rows,
5. Check ε t : a = 2.97 × 60/(0.85 × 3 × 16) = 4.368 in., c = a/0.85 = 5.14 in., d t = 24.5 in., and
ε t = 0.003 (d t − c)/c = 0.0113 > 0.005, a tension-controlled section.
6. Calculate the total max A s that can be used for the T-section by Eq. 3:
MaxA s =
= 0.0425[(b − bw)t + 0.319bwd] =10.54 in. 2
A s (used) ≤ max A s
7. Note: If there are no restrictions on the total depth of the beam, it is a common practice to adopt
the case when a ≤ t (step 2). This is because an increase in d produces a small increase in concrete
in the web only while decreasing the quantity of A s required.