Structural Concrete - Hassoun

20.5 Special Requirements in Design of Structures Subjected to Earthquake Loads 837
Solution
1. The load combinations gave the following results:
2.
P u = 1022 K
P u = 935 K
(maximum forceatthe first floor)
(maximum forceatthe second floor)
P u = 1022 K > A g f c
′ (24 × 24)
= = 230 K
10 10
The member is subjected to bending and axial loads. General requirements should be checked as
follows:
a. Shortest cross-section dimension = 24 in. ≥ 12 in., which is OK.
b. The ratio of shortest cross-sectional dimension to the perpendicular dimension, 24
24 = 1 ≥ 0.4,
which is OK.
3. Longitudinal reinforcement for the column with P u = 1022 K is eight no. 8 bars.
The reinforcement ratio is ρ g = 0.011 < 0.06, which is OK, and > 0.01, which is also OK.
∑
Mnc ≥ 6 ∑
Mnb
5
For P u = 1022 K, M n = 580 kip ⋅ ft. For P u = 935 K, M n = 528 kip ⋅ ft. A minimum nominal flexural
strength of the beam at the joint including the slab reinforcement is M n = 723 kip ⋅ ft.
∑
Mnc = 580 + 528 = 1108 kip ⋅ ft
∑
Mnb = 723 kip ⋅ ft
∑
Mnc = 1108 kip ⋅ ft ≥ 6 ∑
Mnb = 6 723 = 868 kip ⋅ ft
5 5 (OK)
4. Length l 0 is determined as follows:
Choose l 0 = 24 in.
⎧depth of the member = 24 in.
⎪ 1
l 0 ≥ ⎨ clear height = 1 (12 × 12) = 24 in.
6 6
⎪
⎩18 in.
⎧h
⎪
⎪ 4 = 24 4 = 6in.
From Eq.20.46, spacings ≤ ⎨6 × Smallest longitudinal diameter bar = 6 × 1.0 = 6in.
( )
⎪ 14 − 11
⎪s 0 = 4 +
= 5in.
⎩
3
Therefore, s = 5in.
Required cross-section area of reinforcement is
⎧ (
⎪ sbc f ′ )( ) Ag
( )( )
c
5 × 20.5 × 4 576
0.3
− 1 = 0.3
⎪ f
A sh ≥
yt A ch 60 441 − 1 = 0.63 in. 2
⎨
⎪0.09 sb c f c
′ = 0.09 5 × 20.5 × 4 = 0.62 in. 2
⎪ f
⎩ yt 60
Choose no. 4 hoops and no. 5 crossties:
A sh = 2 × 0.2 + 0.31 = 0.71 in. 2 > 0.63 in. 2
Detailing of the reinforcement can be found in Fig. 20.22.