#### Transcript Sample Problem NEHRP Seismic Provisions

### 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

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 3

### Determine the Seismic Design Category

The building is in Occupancy Category II Get S S and S 1 from the maps or online Using USGS software with a 99801 zip code: S S = 61.2%; S 1 = 28.9% The building Site Class is D From Tables F a = 1.311; F v = 1.822

Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05 4

### Seismic Design Category

continued….

Determine S MS and S M1 S MS = F a S S = 1.311(0.612) = 0.802

S M1 = F v S 1 = 1.822(0.289) = .526

Determine S DS and S D1 S DS = (2/3) S MS = 2(0.802)/3 = 0.535

S D1 = (2/3) S M1 = 2(0.526)/3 = 0.351

Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05 5

### Seismic Design Category

continued….

S D1 = 0.351

S DS = 0.535

Use Seismic Design Category D Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05 6

### 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 7

### Determine the Analysis Method

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

### 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 9

### Determine I and T

a From Table 11.5-1, I = 1.0

Determine the approximate fundamental period for the building (Section 12.8.2.1) T a = 0.020(40’) 3/4 = .318 sec.

Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05 10

### Determine C

s From section 12.8.1.1: C s = S DS /(R/I) = .535/(6.5/1) = 0.0823

lower limit = 0.01

T L = 12 (Figure 22-17) Upper limit = S D1 /(T(R/I)) = .351/(.318*6.5/1) Upper limit = 0.169

**USE C S = 0.0823**

Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05 11

### Determine Building Weight

Roof: Roof Ext. Walls Snow/4 Area ft^2 2040 1120 2040 Unit psf 15 10 12.5

Weight lb 30600 11200 25500 67300 Typ. Floor Roof Ext. Walls Partitions Total Building Area ft^2 2040 2240 2040 Level Roof 4th flr 3rd flr 2nd flr Unit psf 12 10 15 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 12

### Compute the Base Shear, V

V = C s W = 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 13

### Compute the Vertical Distribution

Base Shear, V = Level Roof 4th floor 3rd floor 2nd floor Sum: w x (k) 67.3

77.48

77.48

77.48

299.74

h x (ft) 40 30 20 10 24.67 kips w x h x k (ft-k) 2692 2324.4

1549.6

774.8

7340.8

C vx 0.367

0.317

0.211

0.106

1.000

k = 1 F x (k) 9.05

7.81

5.21

2.60

24.67

Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05 14

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 Seismic Provisions Example – A Beginner’s Guide to ASCE 7-05 16

### Compute E

There is no Dead Load story shear so E = D Q E = 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).

Q E = 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) + 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

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 19