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Sample Problem ASCE 7-05 Seismic Provisions

Learn how to determine seismic design category, analyze irregularities, compute base shear, and more for a wood-framed office building in Juneau, Alaska. This beginner’s guide follows ASCE 7-05 provisions.

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Sample Problem ASCE 7-05 Seismic Provisions

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  1. Sample ProblemASCE 7-05 Seismic Provisions A Beginner’s Guide to ASCE 7-05 Dr. T. Bart Quimby, P.E. Quimby & Associates www.bgstructuralengineering.com 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. 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 99801 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 Categorycontinued…. • Determine SMS and SM1 SMS = FaSS = 1.311(0.612) = 0.802 SM1 = FvS1 = 1.822(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 Categorycontinued…. • 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 • 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. 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. Determine I and Ta • From Table 11.5-1, I = 1.0 • Determine the approximate fundamental period for the building (Section 12.8.2.1) Ta = 0.020(40’)3/4 = .318 sec. Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05

  11. Determine Cs From section 12.8.1.1: Cs = SDS/(R/I) = .535/(6.5/1) = 0.0823 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 = 0.0823 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 = 0.0823(299.74 k) = 24.67 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 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. Determine the Design Shear Force for the Shearwall on Grid A and the 2nd Floor • Story shear from structural analysis is 11.03 kips Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05

  17. Compute E • There is no Dead Load story shear so • E = DQE = 1.0 (11.03 k ) = 11.03 k • D = 1.0 since the stories resisting more than 35% of the base shear conform to the requirements of Table 12.3-3 (other). • QE = 11.03 k Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05

  18. ASCE 7 Load Combinations LRFD 5: 1.2(0) + 1.0(11.03) + (0) + 0.2(0) = 11.03 k 7: 0.9(0) + 1.0(11.03) = 11.03 k ASD 5: (0) + 0.7(11.03) = 7.72 k 6: (0) + 0.75(0.7(11.03)) + 0.75(0) + 0.75(0) = 5.79 k 8: 0.6(0) + 0.7(11.03) = 7.72 k See ASCE 7-05 2.3 & 2.4 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 = 11.23 k • Design Wall Shear = 11.23 k Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05

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