Structural Concrete - Hassoun

20.5 Special Requirements in Design of Structures Subjected to Earthquake Loads 827
Solution
1. Seismic design category, base shear, lateral seismic force, and seismic shear are determined in
Example 20.3.
2. Load combinations are given as follows:
1.4D (I)
1.2D + 1.6L (II)
1.2D + 1.0E + f 1 Lf 1 = 0.5 according to Section 20.4 (III)
0.9D + 1.0E (IV)
Redundancy factor, ρ, cannot be taken less than 1.0. For seismic design category D to F, use
ρ = 1.3. Under special cases, ρ can be taken as 1.0 (ASCE 7-10, Section 12.3.4.2). Use ρ = 1.0
assuming one of the conditions is met.
Seismic load effect, E, can be determined using Eqs. 20.30 and 20.31:
Replacing the E in Eq. III gives:
E = ρQ E + 0.2S DS D = Q E + 0.2(1.0)D = Q E + 0.2D
E = ρQ E − 0.2S DS D = Q E − 0.2(1.0)D = Q E − 0.2D
1.4D + 0.5L + Q E
Replacing the E in Eq. IV gives:
D + 0.5L + Q E
1.1D + Q E
0.7D + Q E
The member forces for the beam AB on the second floor (Fig. 20.17) are calculated using
the software for load analysis, and the values of required flexural strengths are determined using
different load combinations, as shown in Table 20.10.
From the previous table 20.10, the most critical loads are chosen and summarized in
Table 20.11. Longitudinal reinforcement for the beam is also determined in Table 20.12.
Table 20.12 summarizes the reinforcement used for the beam.
3. General requirements for flexural members of special moment frame are checked as follows:
a. Clear span ≥ 4 × (effective depth)
28 ft ≥ 4 21.5
12
= 7.2ft (OK)
b. Width-to-depth ratio ≥ 0.3
20
= 0.83 > 0.3 (OK)
24
c. Width = 20 in. ≥ 10 in. (OK)
d. Width ≤ width of supporting member + distance on each side of the supporting member not
exceeding smaller of C 2 or 1.5C 1
20in. ≤ C 2 + 2C 2 = 3 × 24 = 72 in.
(OK)
20in. ≤ C 2 + 1.5C 1 = 24 +(1.5 × 26) =63 in.
(OK)