Structural Concrete - Hassoun

3.15 Analysis of T- and I-Sections 135
4. Check ε t : a = 1.24 in., c = 1.24/0.85 = 1.46 in., d t = d = 16 in.
5. Calculate:
ε t = 0.003(d t − c)
c
=
0.003(16 − 1.46)
1.46
= 0.0299 > 0.005,φ= 0.9
φM n = φA s f y (d − a∕2) =0.9(2.37)(60)(16 − 1.24∕2)
= 1968 K ⋅ in. = 164 K ⋅ ft.
6. You may check that A s used is less than or equal to Max A s (Eq. 3.72), which is not needed when
a < t:
Max A s = 0.0425[(45 − 10)+0.31 × 10 × 16] =8.11 in. 2 ;
A s = 2.37 in. 2 < Max A s
Example 3.12
Calculate the design moment strength of the T-section shown in Fig. 3.34 using f c ′ = 3.5ksi and
f y = 60 ksi.
Figure 3.34 Example 3.12.
Solution
1. Given b = b e = 36 in., b w = 10 in., d = 17 in., and A s = 6.0 in. 2 , check if a ≤ t:
a =
A s f y
0.85f c ′ b = 6 × 60
= 3.36 in.
0.85 × 3.5 × 36
Since a > t, it is a T-section analysis.
2. Find:
A sf = 0.85f c ′ t(b − b w ) 0.85 × 3.5 × 3(36 − 10)
= = 3.87in. 2 (A
f y 60
s − A sf )=A s1 (web)
= 6 − 3.87 = 2.13 in. 2
3. Check ε t : a (web) = A s1 f y ∕(0.85f c ′ b w )=2.13 × 60/(0.85 × 3.5 × 10) = 4.3 in. c = 4.3/0.85 =
5.06 in., d t = 20.5 −2.5 = 18 in., and c/d t = 0.281 < 0.375. Or ε t = 0.003(d t − c)/c = 0.0077 >
0.005, then φ = 0.9