Structural Concrete - Hassoun

814 Chapter 20 Seismic Design of Reinforced Concrete Structures
2. According to Table 20.7, for special moment-resisting frame, select R = 8.
3. Equation 20.16 is not applicable since h x > 10 ft. The period of vibration of the structure is calculated
according to Eq. 20.15 as follows:
T a = C T h 3∕4
n = 0.016 ×(6 × 12) 0.9 = 0.75 s
4. Calculate seismic response coefficient as follows, and check for the limits:
C s = S DS
= 1.0
R∕I e 8∕1.25 = 0.156
C s,max =
S DI
(R∕I e )T = 0.4
(8∕1.25)0.74 = 0.083
Since S 1 = 0.6 g, Eq. should be used to calculate C s,min :
C s,min = 0.5S 1 0.5 ×(0.6)
= = 0.047
(R∕I e ) 8∕1.25
Since C s > C s,max , C s = 0.083.
5. The total seismic weight is calculated as follows:
W = w 1 + w 2 + w 3 + w 4 + w 5 + w 6 = 6 ×(1700) =10,200 K
6. Calculate the seismic base using Eq. 20.10:
V = C s W = 0.083 × 10200 = 846.6K≈ 847 K
7. Calculation of F x , V x ,andM x (Fig. 20.6). Use Eq. 20.17 to calculate the seismic lateral force, F x ,
as shown in the following table. The table also calculates the shear force for each floor level and
the overturning moments as described in Eqs. 20.21 and 20.23.
Floor
Level
Weight
W i
(K)
Height
h i
(ft)
W i
h k*
j Lateral
(K ⋅ ft) C vx
Force, F x
(K)
Shear
Force, V x
(K)
Overturning
Moment M x
(K ⋅ ft)
6 1,700 72 208,905 0.30 254 254 0
5 1,700 60 170,165 0.24 207 461 3,052
4 1,700 48 132,387 0.19 161 622 8,559
3 1,700 36 95,783 0.14 117 739 16,061
2 1,700 24 60,700 0.09 74 813 24,931
1 1,700 12 27,831 0.04 34 847 34,689
0 695,770 847 44,853
*To calculate K, use Section 20.3.1. For T = 0.75 s, using linear interpolation k = 1.125.
8. Calculation of drift: According to Table 20.7 for special moment-resisting frame C d = 5.5,
I e = 1.25 (Table 20.1)
h sx = 12 ft = 144 in.
Floor Level δ xe
(in.) δ x
(in.) Δ (in.) P x
(K) V x
(K) θ
6 1.26 5.54 1.23 1,700 254 0.013
5 0.98 4.31 1.19 3,400 461 0.014
4 0.71 3.12 1.10 5,100 622 0.014
3 0.46 2.02 0.96 6,800 739 0.014
2 0.24 1.06 0.80 8,500 813 0.013
1 0.06 0.26 0.26 10,200 847 0.005