Sample Problem ASCE 7 05 Seismic Provisions A

Sample Problem ASCE 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 1

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 2

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

Determine the Seismic Design Category n The building is in Occupancy Category II n Get SS and S 1 from the maps or online n Using USGS software with a 99801 zip code: SS = 61. 2%; S 1 = 28. 9% n The building Site Class is D n From Tables Fa = 1. 311; Fv = 1. 822 Seismic Provisions Example – A Beginner’s Guide to ASCE 7 -05 4

Seismic Design Category continued…. n Determine SMS and SM 1 SMS = Fa. SS = 1. 311(0. 612) = 0. 802 SM 1 = Fv. S 1 = 1. 822(0. 289) =. 526 n Determine SDS and SD 1 SDS = (2/3) SMS = 2(0. 802)/3 = 0. 535 SD 1 = (2/3) SM 1 = 2(0. 526)/3 = 0. 351 Seismic Provisions Example – A Beginner’s Guide to ASCE 7 -05 5

Seismic Design Category continued…. n SD 1 = 0. 351 n SDS = 0. 535 n Use Seismic Design Category D Seismic Provisions Example – A Beginner’s Guide to ASCE 7 -05 6

Categorize the Plan Irregularities n 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’ n No Vertical Irregularities Seismic Provisions Example – A Beginner’s Guide to ASCE 7 -05 7

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

Determine R, I, and Ta n 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 9

Determine I and Ta n From Table 11. 5 -1, I = 1. 0 n 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 10

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 = SD 1/(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 11

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

Compute the Base Shear, V V = Cs. W = 0. 0823(299. 74 k) = 24. 67 k n This is the total lateral force on the structure. Seismic Provisions Example – A Beginner’s Guide to ASCE 7 -05 13

Compute the Vertical Distribution Base Shear, V = 24. 67 kips k = 1 Level wx hx k Cvx Fx (k) (ft-k) (k) Roof 67. 3 40 2692 0. 367 9. 05 4 th floor 77. 48 30 2324. 4 0. 317 7. 81 3 rd floor 77. 48 20 1549. 6 0. 211 5. 21 2 nd floor 77. 48 10 774. 8 0. 106 2. 60 Sum: 299. 74 7340. 8 1. 000 24. 67 Seismic Provisions Example – A Beginner’s Guide to ASCE 7 -05 14

Typical Level Horizontal Distribution n Load is distributed according to mass distribution. n 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 15

Determine the Design Shear Force for the Shearwall on Grid A and the 2 nd Floor n Story shear from structural analysis is 11. 03 kips Seismic Provisions Example – A Beginner’s Guide to ASCE 7 -05 16

Compute E n There is no Dead Load story shear so n E = DQE = 1. 0 (11. 03 k ) = 11. 03 k n D = 1. 0 since the stories resisting more than 35% of the base shear conform to the requirements of Table 12. 3 -3 (other). n QE = 11. 03 k Seismic Provisions Example – A Beginner’s Guide to ASCE 7 -05 17

ASCE 7 Load Combinations See ASCE 7 -05 2. 3 & 2. 4 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) = 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 18

ASCE 7 -05 Load Combinations n Combinations 3 & 4 have E in them. n For the wall shear: n D = L = 0 n E = 11. 23 k n Design Wall Shear = 11. 23 k Seismic Provisions Example – A Beginner’s Guide to ASCE 7 -05 19