Structural Concrete - Hassoun

7.8 Splices of Reinforcement 279
Solution
a. For no. 8 bars, d b = 1.0 in., and ψ t = ψ e = λ = 1.0: Check first for √ 5000 = 70.7psi < 100psi, and
then calculate l d from Eq. 7.8 or Table 7.1, l d = 42.5d b , conditions for clear spacings and cover are
met. l d = 42.5(1.0) = 42.5 in., or 43 in. For (A s provided)/(A s required) > 2, class A splice applies,
l st = 1.0l d = 43 in. > 12 in. (minimum). Bars spliced are less than half the total number.
b. Let l d = 43 in., as calculated before. Because (A s provided)/(A s required) is less than 2, class B
splice applies, l st = 1.3l d = 1.3(42.5) = 55.25 in., say, 56 in., which is greater than 12 in. and more
than half the total number of bar spliced.
c. Class B splice applies and l st = 56 in. > 12 in.
Example 7.7
Calculate the lap-splice length for a tied column. The column has eight no. 10 longitudinal bars and no.
3 ties. Given f ′ c = 5ksi, solve for (a) f y = 60 ksi and (b) f y = 80 ksi.
Solution
Tie spacing s is the smaller of 16 × 1.128 = 18 in., 48 × 3/8 = 18 in. or 20 in. therefore s = 18 in.
8 no. 10 bars
20
20
a. Determine lap-splice length for f y = 60,000 psi
l sc = 0.0005f y d b > 12in.
= 0.0005 × 60,000 × 1.27
= 38.1in. ≃ 39in. >12in. (Eq. 7.16)
Determine column tie requirements to allow 0.83 reduce lap-splice length according to ACI
Code, Section 10.7.5.2.1.
Effective area of ties ≥0.0015 hs
2 × 0.11 ≥ 0.0015 × 20 × 18
0.22 < 0.54
Modifier 0.83 will not apply. Lap-splice length = 39 in.
b. Determine lap-splice length for f y > 60,000 psi
l sc =(0.0009f y − 24)d b
=(0.0009 × 80000 − 24)×1.27
Modifier 0.83 will not apply as previously calculated.
= 60.96in ≃ 61in. (Eq. 7.17)