Sample%20Problem%20ASCE%207-05%20Seismic%20Provisions

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Sample ProblemASCE 7-05 Seismic Provisions

• A Beginners Guide to ASCE 7-05
• Dr. T. Bart Quimby, P.E.
• Quimby Associates
• www.bgstructuralengineering.com

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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.
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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

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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

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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

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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 lt 10 and 0.15(60) 9 lt 30
• No Vertical Irregularities

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Determine the Analysis Method

• Use ELF Method

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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.

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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/(.3186.5/1)
• Upper limit 0.169
• USE CS 0.0823

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Determine Building Weight
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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.

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Compute the Vertical Distribution
Base Shear, V 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
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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.

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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 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

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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

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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