Structural Concrete - Hassoun

158 Chapter 4 Flexural Design of Reinforced Concrete Beams
Solution
For f c ′ = 3ksi, f y = 60 ksi, and β 1 = 0.85, ρ max for a tension-controlled section is calculated as follows
(φ = 0.9):
( f
′
)[ ]
ρ b =(0.85)β c 87
1
f y 87 + f y
( )( )
ρ =(0.85) 2 3 87
= 0.0214
60 147
( )
0.003 + fy ∕E s
ρ max = ρ b = 0.63375ρ
0.008
b = 0.01356 (Table 4.1)
R u,max = φρ max f y
(
1 − ρ max f y
1.7f ′ c
= 0.9 × 0.01356 × 60 ×
)
(
1 −
(Or, use the tables in Appendix A or Table 4.1.)
Since M u = R u bd 2 ,
bd 2 = M u
R u
=
)
0.01356 × 60
= 0.615 ksi
1.7 × 3
( ) 361 × 12
= 4332 = 7043 in.3
0.615 0.615
Thus, for the following assumed b, calculate d and A s = ρbd:
√
ρ = 0.85f c
′ ⎡
⎢⎢⎣ 1 − 1 − 4M ⎤
u ⎥⎥⎦
f y
1.7f c ′ bd 2
b = 10 in. d = 26.5in. A s = 3.59 in. 2
b = 12 in. d = 24.2in. A s = 3.94 in. 2 six no. 8bars(A s = 4.71 in. 2 )
b = 14 in. d = 22.4in. A s = 4.95 in. 2 five no. 9 bars (A s = 5.0in. 2 )
b = 16 in. d = 21.0in. A s = 4.55 in. 2
The choice of the effective depth d depends on three factors:
1. Width b Required. A small width will result in a deep beam that decreases the headroom available.
Furthermore, a deep narrow beam may lower the design moment strength of the structural member
due to possible lateral deformation.
2. Amount and Distribution of Reinforcing Steel. A narrow beam may need more than one row of
steel bars, thus increasing the total depth of the section.
3. Wall Thickness. If cement block walls are used, the width b is chosen to be equal to the wall thickness.
Exterior walls in buildings in most cases are thicker than interior walls. The architectural
plan of the structure will show the different thicknesses.
A reasonable choice of d/b varies between 1 and 3, with practical value about 2. It can be seen from
the previous calculations that the deeper the section, the more economical it is, as far as the quantity of
concrete used, expressed by the area bd of a 1-ft length of the beam. Alternatively, calculate bd 2 = M u /R u
and then choose adequate b and d.
The area of the steel reinforcement, A s ,isequaltoρbd. The area of steel needed for the different
choices of b and d for this example was shown earlier. Because the steel percentage required is constant
(ρ max = 0.01356), A s is proportional to bd. For a choice of a 12 ×24.2-in. section, the required A s is
4.65 in. 2 Choose six no. 8 bars in two rows (actual A s = 4.71 in. 2 ). The minimum b required for three