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ASCE 7-05, 10 SEISMIC LOAD CALCULATION

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Important Note on seismic load effects on ASCE 7-05,10 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 12.3.4.1 and 12.3.4.2),  horizontal seismic load effect shall include redundancy factor, r (= 1.3)

Eh = r 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 + r QE + L* + 0.2S

(0.9- 0.2SDS) D + r QE

4. For allowable stress design, vertical seismic load effect shall be include in load combination as

(1.0 + 0.14SDS)D + 0.7 r QE

(1.0 + 0.105SDS)D + 0.525 r QE + 0.75 L + 0.75 (Lr or S)

(0.6 - 0.14SDS)D + 0.7 r QE

ASCE 7-05, 10 Equivalent lateral force procedure

USGS website for seimic parameters: http://geohazards.usgs.gov/designmaps/us/application.php

IBC website for free online code information:

http://publicecodes.cyberregs.com/icod/index.htm

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 hnx

where

hn is the height of building above base.

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 steel eccentrically braced frame and buckling-restrained braced 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  £ T

Csmax= SD1 / [T2(R/I)] for T  > T

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

Csmin³ 0.01

      If S1 ³ 0.6g, Csmin³ 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