Structural Concrete - Hassoun

164 Chapter 4 Flexural Design of Reinforced Concrete Beams
For ε t = 0.005,
c =
( 3
8)
d t and a = β 1 c
b. Calculate the compression force in the concrete:
C 1 = 0.85f ′ cab = T 1 = A s1 f y
Determine A s1 . Calculate M u 1 = φ A s1 f y (d − a/2); ρ 1 = A s1 /bd, φ = 0.9.
c. Calculate M u2 = M u − M u1 ; assume d ′ = 2.5 in.
d. Calculate A s2 : M u2 = φ A s2 f y (d − d ′ ), f ′ c = f y ,φ= 0.9. Total A s = A s1 + A s2 .
e. Check if compression steel yields similar to step 6 in Section 4.4.1.
Example 4.5
A beam section is limited to a width b = 10 in. and a total depth h = 22 in. and has to resist a factored
moment of 226.5 K ⋅ ft. Calculate the required reinforcement. Given: f c ′ = 3ksi and f y = 50 ksi.
Solution
1. Determine the design moment strength that is allowed for the section as singly reinforced based
on tension-controlled conditions. This is done by starting with ρ max .Forf c ′ = 3ksi and f y = 50 ksi
and from Eqs. 3.18, 3.22, and 3.31,
ρ b = 0.0275 ρ max = 0.01624 R u = 614 psi
M u = R u bd 2 b = 10 in. d = 22 − 3.5 = 18.5in.
M u = 226.5 × 12 = 2718K ⋅ in.
(This calculation assumes two rows of steel, to be checked later.) Assume M u1 = 0.614 × 10 ×
(18.5) 2 = 2101 K⋅in. = max φM n , as singly reinforced. Design M u = 2718 K⋅in. > 2101 K⋅in.
Therefore, compression steel is needed to carry the difference.
2. Compute A s1 , M u1 ,andM u2 :
A s1 = ρ max bd = 0.01624 × 10 × 18.5 = 3.0in. 2
M u1 = 2101K ⋅ in.
M u2 = M u − M u1 = 2718 − 2102 = 617K ⋅ in.
3. Calculate A s2 and A ′ s , the additional tension and compression steel due to M u2 . Assume d′ = 2.5 in.;
M u2 = φ A s 2 f y (d − d ′ ):
A s2 =
Total tension steel is equal to A s :
M u2
φf y (d − d ′ ) = 617
= 0.86 in.2
0.9 × 50(18.5 − 2.5)
A s = A s1 + A s2 = 3.0 + 0.86 = 3.86 in. 2
The compression steel has A ′ s = 0.86 in.2 (in A ′ s yields).
4. Check if compression steel yields:
ε y =
f y
29, 000 = 50
29, 000 = 0.00172