ASCE705.Seismic.sampleProb

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

Quimby & Associates www.bgstructuralengineering.com

Seismic Provisions Example – Example – A  A Beginner’s Guide to ASCE 7-05

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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

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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

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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

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Seismic Design Category continued…. 

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

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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

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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

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Determine the Analysis Method



Use ELF Method Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05

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Determine R, I, and T a 

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

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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.

Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05

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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

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Determine Building Weight Roof:

Unit psf 

Roof  Ext. Walls Snow /4

 Area ft^2 2040 1120 2040

Unit psf 

Roof  Ext. Walls Partitions

 Area ft^2 2040 2240 2040

Typ. Floor 

Total Building

Level Roof  4th flr  3rd f lr  2nd flr 

15 10 12.5

12 10 15

Weight lb 30600 11200 25500 67300 Weight lb 24480 22400 30600 77480

Weight k 67.3 77.48 77.48 77.48 299.74 Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05

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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

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Compute the Vertical Distribution Base Shear, V =

24.67 kips

k= 1

wx

hx

wxhxk

(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

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

Level

Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05

Cvx

Fx

(k)

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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

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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

Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05

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Compute E 

There is no Dead Load story shear so



E = DQE = 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).



D



QE = 11.03 k

Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05

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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) + 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

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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

Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05

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