Structural Concrete - Hassoun

20.3 Analysis Procedures 817
9. Determine the seismic lateral story drift using Eq. 20.24:
Δ 1 = 0.01h 1 = 0.01 × 15 = 0.15 ft = 1.8in. (first floor)
Δ 2 = 0.01h 2 = 0.01 × 12 = 0.12 ft = 1.44 in. (second floor)
Risk category I
Check for allowable drift using Table 20.9:
Δ a = 0.020h sx
where h sx is the story height below level x and
= 0.020h 1 = 0.020 × 15 = 0.3ft= 3.6in.>1.8in.(OK) (first floor)
Δ a1
Δ a2
= 0.020h 2 = 0.020 × 12 = 0.24 ft = 2.88 in. >1.44 in. (OK) (second floor)
Example 20.5 Torsional Effects
Determine the shear forces V 1 and V 2 acting on the shear wall 1 and 2 of the building with the floor plan
shown in Fig. 20.9. Assume that the value of story shear, V y , is 15 K. Consider torsional effect.
y
1
25 ft
V = 15 K
10 ft
35 ft
25 ft
C m
2
10 ft
90 ft
30 ft 30 ft
x
Figure 20.9
Example 20.5: Floor plan.
Solution
Center of mass is in the centroid of the rigid diaphragm. The center of rigidity x can be determined as
follows (Fig. 20.10):
2 × 25 × 30 + 2 × 10 × 120
x = = 55.7ft
2 × 25 + 2 × 20
e x = 150∕2 − 55.7 = 19.3ft
For the story shear force V y = 15 K and eccentricity of 19.3 ft, the torsional moment is
T = 15 × 19.3 = 289.5kip⋅ ft
The shear force acting on the wall is the sum of the shear force due to story shear, V x , and shear
force due to torsional moment, T y .
For wall 1, shear force V 1 is
25
V 1 =
25 + 25 + 10 + 10 (15) =5.4K (due to V y )
V 1 = 2 × 25 × 25.7 2 + 2 × 10 × 64.3 2 ∕289.5(25 × 25.7) =1.6kip⋅ ft (due to T)