Structural Concrete - Hassoun

20.5 Special Requirements in Design of Structures Subjected to Earthquake Loads 829
Table 20.13
Longitudinal Reinforcement Requirements According to the Limits of the Reinforcement
Reinforcement
Limits
A s
(in. 2 ) Min A s
(in. 2 ) Max A s
(in. 2 ) Provided A s
(in. 2 ) φM n
(kip ⋅ ft)
Support 5.53 5.53 –474
(joint face)
(7 no. 8 at the top)
3.6 3.6 358
(6 no. 7 at the bottom) 1.43 10.75
Midspan 1.58 1.58 –148
(2 no. 8 at the top)
1.2 a 1.8 a 168
(2 no. 7 at the bottom) (3 no. 7)
a Since 1.2 in. 2 < min A s
= 1.43 in. 2 , use three no. 7 bars at the bottom. (A s
= 1.8 in. 2 )
4. Special requirements for longitudinal reinforcement are
a. Use Eq. 20.3.4
A − s or A+ s
⎧3 √ f ⎪⎪⎨⎪⎪⎩
c ′ b w d
= 3√ 4000 × 20 × 21.5
= 1.36 in. 2
≥ f y 60000
200b w d 200 × 20 × 21.5
= = 1.43 in. 2
f y 60000
max A s = 0.025b w d = 0.025 × 20 × 21.5 = 10.75 in. 2
Check the reinforcement limits against the required reinforcement, as shown in Table 20.13.
b. Positive moment strength at joint face ≥ 1 negative moment strength at that face of the joint:
2
M n + = 358 kip ⋅ ft ≥ 1 2 M− n = 1 474 = 237 kip ⋅ ft
2 (OK)
c. (Mn − or M+ n ) at any section ≥ 1 (max M 4 n at either joint) (ACI Code, Section 18.6.3.2):
M n = 148 K-ft > 1 (474) =119 kip ⋅ ft
4 (OK)
Anchorage of flexural reinforcement in exterior column is determined as follows using
Eq. 20.37:
For no. 8 bars,
⎧ 60,000 × 1.0
⎪
⎪ 65 √ = 14.6 ′′
4000
l dh = ⎨
⎪
8 × 1.0 = 8in.
⎪6in.
⎩
Therefore, l dh = 14.6 in. For no. 7 bars,
⎧60,000 × 0.875
⎪
⎪ 65 √ = 12.8in.
4000
l dh = ⎨
⎪
8 × 0.875 = 7in.
⎪6in.
⎩
Therefore, l dh = 12.8 in.