ASCE 7-05 SEISMIC LOAD CALCULATION
Contents: Important notes on sisimic load effect on ASCE 7-05, Sec 12.4 ASCE 7-05 Equivalent lateral force procedure Example 1: Building frame systems with ordinary steel concentric braced frame Example 2: Building frame systems with ordinary reinforced concrete shear wall
Important Note on seismic load effects on ASCE 7-05, Sec 12.4 1. Seismic design shall include both effective from horizontal seismic and vertical seismic force. The seismic force, E, appears in load combinations 1.2 D 1.0 E + L* + 0.2S 0.9 D + 1.0 E shall include both horizontal and vertical effects, E = Eh Ev
(AISC 7-05 Eq. 12.4.1 & 12.4.2)
where Eh is effect of horizontal seismic force, Ev is effect of vertical seismic force. 2. In seismic design category, D, E, & F (except special condition in section section 12.3.4.1 and 12.3.4.2), horizontal seismic load effect shall include redundancy factor, 1.3) Eh = QE (AISC 7-05 Eq. 12.4.3.) where QE is effect of horizontal seismic shear force calculated based on seismic base shear V, or Fp from equation for components, non-structural elements, etc.
3. For strength design, vertical seismic load effect shall be calculated as Ev = 0.2 SDSD (AISC 7-05 Eq. 12.4.4) where SDS is seismic parameter, D is effect of dead load. The load combinations become *
(1.2+0.2SDS) D QE + L + 0.2S (0.9- 0.2SDS) D + QE 4. For allowable stress design, vertical seismic load effect shall be include in load combination as (1.0 + 0.14SDS)D + 0.7 QE (1.0 + 0.105SDS)D + 0.525 QE + 0.75 L + 0.75 (Lr or S) (0.6 - 0.14SDS)D + 0.7 QE
ASCE 7-05 Equivalent lateral force procedure Applicability Equivalent lateral force procedure is limited to be used in 1. Seismic design category A. 2. Seismic design category B and C, (except light framed construction, see 6) 3. Regular structure with T < 3.5 Ts in Seismic design category D, E, and F 4. Irregular structure in Seismic design category D, E, and F with horizontal irregularities Type, 2, 3, 4, and 5 and T < 3.5 Ts.
5. Irregular structure in Seismic design category D, E, and F with vertical irregularities Type 4, 5a and 5b and T < 3.5 Ts . 6. Light framed construction in occupancy I & II, more than 3 stories high, or occupancy III & IV, more than 2 stories high, and regular light frame structures with T < 3.5 Ts in seismic design caterogy D, E, and F. where Ts = SD1 /SDS.
Procedure 1. Determine weight of building, W. 2. Determine 0.2 second and 1 second response spectral acceleration, Ss and S1 from Figure 22-1 to 22-14 3. Determine Site class from Chapter 20 or from soil report. 4. Determine site coefficient, Fa, from Table 11.4.1. 5. Determine site coefficient, Fv, from Table 11.4.2 6. Determine adjusted maximum considered earthquake spectral response acceleration parameters for short period, SMS and at 1 second period, SM1. SMS = Fa Ss (Eq. 11.4-1) SM1 = Fv S1 (Eq. 11.4.2) 7. Determine deign spectral response acceleration parameters for short period, SDS and at 1 second period, SD1. SDS =(2/3) SMS (Eq. 11.4.3) SD1 =(2/3) SM1 (Eq. 11.4.4) 8. Determine Important factor, I, from Table 11.5.1 9. Determine Seismic design category from Table 11.6-1, & 11.6-2 10. Determine Response modification factor, R. from Table 12.2-1 check building height limitation 11. Determine seismic response coefficient from Eq. 12.8-1 Cs= SDS / (R/I) 12. Determine approximate fundamental period from Eq. 12.8-7 T = Ct hn where hn is the height of building above base. x
Ct = 0.028, x = 0.8 for steel moment resisting frame, Ct = 0.016, x = 0.9 for concrete moment resisting frame, Ct = 0.03, x = 0.75 for eccentrically braced steel frame, Ct = 0.02, x = 0.75 for all other structural frame 13. Check if T < 3.5 SD1 /SDS. 14. Determine Maximum seismic response coefficient Eq. 12.8-3 and 12.8-4 Csmax= SD1 / [ T(R/I)] for T TL 2 Csmax= SD1 / [T (R/I)] for T TL where TL is long-period transition period in Figure 22.15 to 22.20. 15. Minimum seismic response coefficient, Eq 12.8-5 & 12.8.6 Csmin 0.01 If S1 0.6g, Csmin 0.5 S1 / (R/I) 16. Determine Seismic response for strength design from 12.8-1 V = Cs W
Example 1: Building frame systems with ordinary steel concentric braced frame Given: Code: ASCE 7-05 Equivalent lateral force procedure Design information: Weight of building, W = 500 kips 0.2 second response spectral acceleration, Ss = 0.25 1 second response spectral acceleration, S1 = 0.1 Long-period transition period, TL = 12 sec Building frame systems with ordinary steel concentric braced frame Building category II Building height: 30 ft Soil Profile: E Requirement: Determine seismic base shear
Solution: Site coefficient, Fa = 2.5 Site coefficient, Fv = 3.5 Design spectral response acceleration parameters SMS = Fa Ss = 0.625 (Eq. 11.4.1) SM1 = Fv S1= 0.35 (Eq. 11.4.2) SDS =(2/3) SMS =0.417 (Eq.11.4.3) SD1 =(2/3) SM1 =0.233 (Eq. 11.4.4) Seismic design category B from Table 11.6.1, category C based on Table 11.6.2. Use category D. From Table 12.2.1, No limit for building frame system with ordinary steel concentric braced frame Response modification factor, R = 3.25 Important factor, I = 1 (Table 11.5.1) Seismic response coefficient (Eq. 12.8-1) Cs= SDS / (R/I) = 0.128 Fundamental period (Eq. 12.8-7), CT = 0.02, x = 0.75 T = CT hn0.75 = 0.256 sec < (3.5)(SDS /SD1) = (3.5)(0.417/0.233)=1.96 O.K. Maximum seismic response coefficient (Eq. 12.8-3) T < TL = 12 sec Csmax= SD1 / [(R/I) T] = 0.192 Since S1 = 0.1 < 0.6, Minimum seismic response coefficient (Eq. 12.8-5) Csmin= 0.01 Seismic base shear for strength design V = Cs W = (0.128)(500) = 64 kips
Example 2: Building frame systems with ordinary reinforced concrete shear wall Given: Code: ASCE 7-05 Equivalent lateral force procedure Design information: Weight of building, W = 1000 kips
0.2 second response spectral acceleration, Ss = 0.5 1 second response spectral acceleration, S1 = 0.15 Soil profile class: C Building frame systems with ordinary reinforced concrete shear wall Building category II Building height: 40 ft Requirement: Determine seismic base shear for strength design Solution: Site coefficient, Fa = 1.2 Site coefficient, Fv = 1.65 Design spectral response acceleration parameters SMS = Fa Ss = 0.6 (Eq. 11.4.1) SM1 = Fv S1= 0.247 (Eq. 11.4.2) SDS =(2/3) SMS =0.4 (Eq. 11.4.3) SD1 =(2/3) SM1 =0.165 (Eq. 11.4.4) Seismic design category C from Table 11.6.1, category C based on Table 11.6.2. Use category C. From Table 12.2.1, no limit for building frame system with ordinary shear wall for category C. Response modification factor, R = 5 Important factor, I = 1 (Table 11.5.1) Seismic response coefficient (Eq. 12.8-1) Cs= SDS / (R/I) = 0.08 Fundamental period (Eq. 12.8-7), CT = 0.02, x = 0.75 T = CT hn0.75 = 0.318 sec < (3.5)(SDS /SD1) = (3.5)(0.4/0.165)=1.44 sec O.K. Maximum seismic response coefficient (Eq. 12.8-3) T < TL = 12 sec Csmax= SD1 / [(R/I) T] = 0.104 Since S1 = 0.15 < 0.6, Minimum seismic response coefficient (Eq. 12.8-5) Csmin= 0.01 Seismic base shear for strength design V = Cs W = (0.08)(1000) = 80 kips
