Structural Concrete - Hassoun

7.5 Summary for Computation of I d in Tension 267
d. Bars are confined with no. 3 stirrup. The conditions are met. Then
l d
d b
= ψ t ψ e f y
20λ √ f ′ c
(for bars > no. 7) (Eq. 7.8)
2. Determine the multiplication factors: ψ t = 1.0 (bottom bars), ψ e = 1.0 (no coating), and λ = 1.0
(normal-weight concrete). Also check that √ f ′ c = 54.8psi < 100psi.
l d
d b
=
60,000
20 × 1 × √ = 54.8 (Eq. 7.7)
3000
So, l d = 54.8(1.0) = 54.8 in., say, 55 in. These values can be obtained directly from Tables 7.1
and 7.2.
If we calculate l d from Eq. 7.7,
l ( ) ( )(
)
d 3 f y ψt ψ
=
e ψ s
√ ( )
d b 40 f
′ cb c
+ K tr ∕db
= ψ t = ψ e = ψ s = λ = 1.0
Also c b = smaller of distance from center of bar to the nearest concrete surface c b1 or one-half
the center-to-center bar spacing c b2 :
( )( )
1 12 − 5
c b1 = 2.5in. c b2 =
= 1.17in. (controls)
2 3
k tr = 40A tr
sn
A tr = 2 ×(0.11) =0.22in. 2
s = 6in.
n = 2
40 × 0.22
k tr = = 0.73
2 × 6
C b + k tr 1.17 + 0.73
= = 1.9in.