Structural Concrete - Hassoun

120 Chapter 3 Flexural Analysis of Reinforced Concrete Beams
Table 3.4
Values of K for Different f ′ c and f y
f ′ c (ksi) f y (ksi) K K (for d′ = 2.5 in.)
3 40 0.1003d ′ /d 0.251/d
3 60 0.1164d ′ /d 0.291/d
4 60 0.1552d ′ /d 0.388/d
5 60 0.1826d ′ /d 0.456/d
Therefore, Eq. 3.37 becomes (A s − A ′ s)f y = 0.85f ′ cab:
Also,
Therefore,
(ρ − ρ ′ )bdf y = 0.85f cab
′
( f
ρ − ρ ′ ′
) (a )
= 0.85 c
f y d
a = β 1 c = β 1
(
87
87 − f y
)
d ′
( f
ρ − ρ ′ ′
)( )( )
= 0.85β c d
′ 87
1 = K (3.49)
f y d 87 − f y
The quantity ρ − ρ ′ is the steel ratio, or (A s − A ′ s)∕bd = A s1 ∕bd = ρ 1 for the singly reinforced basic
section.
If ρ − ρ ′ is greater than the value of the right-hand side in Eq. 3.49, then compression steel will
also yield. In Fig. 3.25 we can see that if A s1 is increased, T 1 and, consequently, C 1 will be greater
and the neutral axis will shift downward, increasing the strain in the compression steel and ensuring
its yield condition. If the tension steel used (A s1 ) is less than the right-hand side of Eq. 3.49, then
T 1 and C 1 will consequently be smaller, and the strain in compression steel, ε ′ s, will be less than
or equal to ε y , because the neutral axis will shift upward, as shown in Fig. 3.25c, and compression
steel will not yield.
Therefore, Eq. 3.49 can be written
ρ − ρ ′ ≥ 0.85β 1
f ′ c
f y
× d′
d × 87
87 − f y
= K (3.49a)
where f y is in ksi, and this is the condition for compression steel to yield.
For example, the values of K for different values of f ′ c and f y are as shown in Table 3.4.
Example 3.9
A rectangular beam has a width of 12 in. and an effective depth of d = 22.5 in. to the centroid of tension
steel bars. Tension reinforcement consists of six no. 9 bars in two rows; compression reinforcement
consists of two no. 7 bars placed as shown in Fig. 3.26. Calculate the design moment strength of the
beam if f c ′ = 4ksiandf y = 60 ksi.