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**Contents:****Important notes on sisimic load effect on ASCE 7-05, 10, 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__

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 = E_{h} ± E_{v}
(AISC 7-05 Eq. 12.4.1 & 12.4.2)

where E_{h} is
effect of horizontal seismic force, E_{v} 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)

E_{h} = r Q_{E }(AISC
7-05 Eq. 12.4.3.)

where Q_{E} is
effect of horizontal seismic shear force calculated based on seismic
base shear V, or F_{p} from
equation for components, non-structural elements, etc.

3. For strength design, vertical seismic load effect shall be calculated as

E_{v} = 0.2 S_{DS}D
(AISC 7-05 Eq. 12.4.4)

where S_{DS} is
seismic parameter, D is effect of dead load.

The load combinations become

(1.2+0.2S_{DS}) D + r Q_{E} +
L^{* }+ 0.2S

(0.9- 0.2S_{DS})
D + r Q_{E}

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

(1.0 + 0.14S_{DS})D + 0.7 r Q_{E}

(1.0 + 0.105S_{DS})D + 0.525 r Q_{E }+
0.75 L + 0.75 (L_{r }or
S)

(0.6 - 0.14S_{DS})D + 0.7 r Q_{E}

IBC website for free online code information:

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

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 T_{s} 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 T_{s}.

5. Irregular structure in Seismic design category D, E, and F with
vertical irregularities Type 4, 5a and 5b and T < 3.5 T_{s} .

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 T_{s} in
seismic design caterogy D, E, and F.

where T_{s} = S_{D1}/S_{DS}.

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, F_{a}, from Table 11.4.1.

5. Determine site coefficient, F_{v}, 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.

S_{MS} =
F_{a} S_{s} (Eq.
11.4-1)

S_{M1 }=
F_{v} S_{1} (Eq.
11.4.2)

7. Determine deign spectral response acceleration parameters for short period, SDS

and at 1 second period, SD1.

S_{DS} =(2/3)
S_{MS} (Eq.
11.4.3)

S_{D1} =(2/3)
S_{M1} (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

C_{s}= S_{DS} /
(R/I)

12. Determine approximate fundamental period from Eq. 12.8-7

T = C_{t} h_{n}^{x}

where

h_{n} is
the height of building above base.

C_{t} =
0.028, x = 0.8 for steel moment resisting frame,

C_{t} =
0.016, x = 0.9 for concrete moment resisting frame,

C_{t} =
0.03, x = 0.75 for steel eccentrically braced frame and
buckling-restrained braced frame.

C_{t} =
0.02, x = 0.75 for all other structural frame

13. Check if T < 3.5 S_{D1}/S_{DS}.

14. Determine Maximum seismic response coefficient Eq. 12.8-3 and 12.8-4

C_{smax}= S_{D1} /
[ T(R/I)] for T £ T_{L }

C_{smax}= S_{D1} /
[T^{2}(R/I)] for T > T_{L }

where T_{L} is
long-period transition period in Figure 22.15 to 22.20.

15. Minimum seismic response coefficient, Eq 12.8-5 & 12.8.6

C_{smin³} 0.01

If S_{1} ³ 0.6g,
C_{smin}³ 0.5
S1 /
(R/I)

16. Determine Seismic response for strength design from 12.8-1

V = C_{s} W

__Given:__

Code: ASCE 7-05 Equivalent lateral force procedure

Design information:

Weight of building, W = 500 kips

0.2 second response spectral acceleration, S_{s }=
0.25

1 second response spectral acceleration, S_{1} =
0.1

Long-period transition period, T_{L} =
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, F_{a} =
2.5

Site coefficient, F_{v} =
3.5

Design spectral response acceleration parameters

S_{MS} =
F_{a} S_{s} =
0.625 (Eq.
11.4.1)

S_{M1} =
F_{v} S_{1}=
0.35 (Eq.
11.4.2)

S_{DS} =(2/3)
S_{MS} =0.417 (Eq.11.4.3)

S_{D1} =(2/3)
S_{M1} =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)

C_{s}= S_{DS} /
(R/I) = 0.128

Fundamental period (Eq. 12.8-7), C_{T} =
0.02, *x* =
0.75

T = C_{T} h_{n}^{0.75}^{ }=
0.256 sec

< (3.5)(S_{DS}/S_{D1}) = (3.5)(0.417/0.233)=1.96
O.K.

Maximum seismic response coefficient (Eq. 12.8-3)

T < T_{L} = 12
sec

C_{smax}= S_{D1} /
[(R/I) T] = 0.192

Since S_{1} = 0.1
< 0.6, Minimum seismic response coefficient (Eq. 12.8-5)

C_{smin}= 0.01

Seismic base shear for strength design

V = C_{s} W
= (0.128)(500) = 64 kips

__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, F_{a} =
1.2

Site coefficient, F_{v} =
1.65

Design spectral response acceleration parameters

S_{MS} =
F_{a} S_{s} =
0.6 (Eq.
11.4.1)

S_{M1} =
F_{v} S_{1}=
0.247 (Eq.
11.4.2)

S_{DS} =(2/3)
S_{MS} =0.4 (Eq.
11.4.3)

S_{D1} =(2/3)
S_{M1} =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)

C_{s}= S_{DS} /
(R/I) = 0.08

Fundamental period (Eq. 12.8-7), C_{T} =
0.02, *x* =
0.75

T = C_{T} h_{n}^{0.75}^{ }=
0.318 sec

< (3.5)(S_{DS}/S_{D1}) = (3.5)(0.4/0.165)=1.44 sec
O.K.

Maximum seismic response coefficient (Eq. 12.8-3)

T < T_{L} = 12
sec

C_{smax}= S_{D1} /
[(R/I) T] = 0.104

Since S_{1} =
0.15 < 0.6, Minimum seismic response coefficient (Eq. 12.8-5)

C_{smin}= 0.01

Seismic base shear for strength design

V = C_{s} W
= (0.08)(1000) = 80 kips