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

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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.

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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 = 0.01TL = 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 WeightWeightUnitAreaRoof:

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 7.813rd floor 77.48 20 1549.6 0.211 5.212nd 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.

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Determine the Design Shear Force for the Shearwall on Grid A and the 2nd Floor

Story shear from structural analysis is 11.03 kips

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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

LRFD5: 1.2(0) + 1.0(11.03) + (0) + 0.2(0) = 11.03 k7: 0.9(0) + 1.0(11.03) = 11.03 k

ASD5: (0) + 0.7(11.03) = 7.72 k6: (0) + 0.75(0.7(11.03)) + 0.75(0) + 0.75(0) = 5.79

k8: 0.6(0) + 0.7(11.03) = 7.72 k

See ASCE 7-05 2.3 & 2.4

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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