1. Sample Problem ASCE 7-05 Seismic Provisions
A Beginner’s Guide to ASCE 7-05
Dr. T. Bart Quimby, P.E.
Quimby & Associates
Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05

2. The Problem Definition
The wood framed office building shown here is to be constructed in a “suburban” area in Juneau, Alaska out near the airport. The site conditions consist of deep alluvial deposits with a high water table.
Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05

3. Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05
Other Given Data
Roof DL = 15 psf
Typical Floor DL = 12 psf
Partition Load = 15 psf
Snow Load = 30 psf
Exterior Wall DL = 10 psf
Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05

4. Determine the Seismic Design Category
The building is in Occupancy Category II
Get SS and S1 from the maps or online
Using USGS software with a zip code:
SS = 61.2%; S1 = 28.9%
The building Site Class is D
From Tables
Fa = 1.311; Fv = 1.822
Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05

5. Seismic Design Category continued….
Determine SMS and SM1
SMS = FaSS = 1.311(0.612) = 0.802
SM1 = FvS1 = (0.289) = .526
Determine SDS and SD1
SDS = (2/3) SMS = 2(0.802)/3 = 0.535
SD1 = (2/3) SM1 = 2(0.526)/3 = 0.351
Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05

6. Seismic Design Category continued….
SD1 = 0.351
SDS = 0.535
Use Seismic Design Category D
Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05

7. Categorize the Plan Irregularities
The building has re-entrant corners (type 2) since the projection is more than 15% of dimension
0.15(40’) = 6’ < 10’ and 0.15(60’) = 9’ < 30’
No Vertical Irregularities
Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05

8. Determine the Analysis Method
Use ELF Method
Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05

9. Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05
Determine R, I, and Ta
From Table 5.2.2, R = 6.5 for bearing wall systems consisting of light framed walls with shear panels.
Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05

10. Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05
Determine I and Ta
From Table , I = 1.0
Determine the approximate fundamental period for the building (Section )
Ta = 0.020(40’)3/4 = .318 sec.
Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05

11. Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05
Determine Cs
From section :
Cs = SDS/(R/I) = .535/(6.5/1) =
lower limit = 0.01
TL = 12 (Figure 22-17)
Upper limit = SD1/(T(R/I)) = .351/(.318*6.5/1)
Upper limit = 0.169
USE CS =
Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05

12. Determine Building Weight
Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05

13. Compute the Base Shear, V
V = CsW = ( k) = k
This is the total lateral force on the structure.
Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05

14. Compute the Vertical Distribution
Base Shear, V =
24.67
kips
k = 1
Level wx hx
wxhxk
Cvx Fx
(k)
(ft)
(ft-k)
Roof
67.3 40
2692
0.367
9.05
4th floor
77.48 30
2324.4
0.317
7.81
3rd floor 20
1549.6
0.211
5.21
2nd floor 10
774.8
0.106
2.60
Sum:
299.74
7340.8
1.000
Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05

15. Typical Level Horizontal Distribution
Load is distributed according to mass distribution.
Since the loading is symmetrical, each of the two supporting shear walls receives half the story shear.
Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05

16. Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05
Determine the Design Shear Force for the Shearwall on Grid A and the 2nd Floor
Story shear from structural analysis is kips
Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05

17. Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05
Compute E
There is no Dead Load story shear so
E = DQE = 1.0 (11.03 k ) = k
D = 1.0 since the stories resisting more than 35% of the base shear conform to the requirements of Table (other).
QE = k
Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05

18. ASCE 7 Load Combinations
See ASCE & 2.4
LRFD
5: 1.2(0) + 1.0(11.03) + (0) + 0.2(0) = k
7: 0.9(0) + 1.0(11.03) = k
ASD
5: (0) + 0.7(11.03) = 7.72 k
6: (0) (0.7(11.03)) (0) (0) = 5.79 k
8: 0.6(0) + 0.7(11.03) = 7.72 k
Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05

19. ASCE 7-05 Load Combinations
Combinations 3 & 4 have E in them.
For the wall shear:
D = L = 0
E = k
Design Wall Shear = k
Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05