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Structural Concrete - Hassoun

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816 Chapter 20 Seismic Design of Reinforced <strong>Concrete</strong> Structures<br />

w 2 = 40 K<br />

12'<br />

w 1 = 35 K<br />

15'<br />

Figure 20.7<br />

Example 20.4: Building elevation.<br />

F 2 = 14.9 K<br />

2<br />

14.9 K<br />

0<br />

F 1 = 13 K 1<br />

27.8 K<br />

178.8<br />

0<br />

27.8 K<br />

597.3<br />

V = 27.8 K<br />

V x<br />

M x<br />

Figure 20.8 Example 20.4: Distribution of lateral seismic force, F x<br />

, base shear, V x<br />

,and<br />

overturning moment, M x<br />

.<br />

4. Calculate the total gravity load (Fig. 20.7): W = w 1 + w 2 = 35 + 40 = 75 Kips.<br />

5. For two-story building, F = 1.1 as described in Section 20.3.3:<br />

V = FS DS<br />

R W = 1.1(3.5) × 75 = 27.8 K (Eq. 20.19)<br />

4<br />

6. Calculate the seismic lateral forces acting at the first and second floors using Eq. 20.20 (Fig. 20.8):<br />

F 1 = 1.1S DS<br />

R w 1 = 1.1(1.35) × 35 = 13 K<br />

4<br />

(first floor)<br />

F 2 = 1.1S DS<br />

R w 2 = 1.1(1.35) × 40 = 14.9K<br />

4<br />

(second floor)<br />

7. Calculate the story shear force using Eq. 20.23:<br />

V 2 = 14.9K (second floor)<br />

V 1 = 27.8K (first floor)<br />

8. Calculate the overturning moment using Eq. 20.24:<br />

M 2 = 0 (second floor)<br />

M 1 = 014.9 × 12 = 178.8kip⋅ ft (first floor)<br />

M 0 = 14 ⋅ 9(12 + 15)+13 × 15 = 597.3kip⋅ ft (at the base of the structure)

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