sample problem asce 7-05 seismic provisions a beginner’s guide to asce 7-05 dr. t. bart quimby,...
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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
1Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05
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.
2Seismic 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
3Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05
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
4Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05
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
5Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05
Seismic Design Categorycontinued….
SD1 = 0.351
SDS = 0.535
Use Seismic Design Category D
6Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05
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
7Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05
Determine the Analysis Method
Use ELF Method
8Seismic 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.
9Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05
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.
10Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05
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
11Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05
Determine Building Weight
WeightUnitAreaRoof:lbpsfft^230600152040Roof11200101120Ext. Walls2550012.52040Snow/467300
WeightUnitAreaTyp. Floorlbpsfft^224480122040Roof22400102240Ext. Walls30600152040Partitions77480
WeightLevelTotal Buildingk67.3Roof
77.484th flr77.483rd flr77.482nd flr
299.74
12Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05
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.
Compute the Vertical Distribution
Base Shear, V = 24.67 kips k = 1
Level wx hx wxhxk Cvx Fx
(k) (ft) (ft-k) (k)
Roof 67.3 40 2692 0.367 9.05
4th floor 77.48 30 2324.4 0.317 7.81
3rd floor 77.48 20 1549.6 0.211 5.21
2nd floor 77.48 10 774.8 0.106 2.60
Sum: 299.74 7340.8 1.000 24.67
14Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05
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.
15Seismic 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 11.03 kips
16Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05
Compute E
There is no Dead Load story shear so E = QE = 1.0 (11.03 k ) = 11.03 k
= 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
17Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05
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
18Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05
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
19Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05