building systems snow and wind load

7/27/2019 Building Systems snow and wind load

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PCI 6th Edition

Building Systems

(Snow + Wind)


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Presentation Outline

Building System Loads Snow

Uniform loading and drifting

Example Wind

Main lateral wind resisting system

Component example


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Structural Systems

Gravity Load Systems Beams

Columns

Floor Member Double Tees, Hollow Core

Spandrels


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Structural Systems

Lateral Load Systems Shear Walls

Moment Resisting Frames

Cantilever Columns Braced Frames K Frames


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System Loads

Dead Loads

Live Loads

Snow Loads Roof / Ground

Drifting

Wind Loads Earthquake

Loads

This session willplace emphasison these forces.


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Snow Loads

Based on ASCE 7

Different than other live loads due to

transient nature


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Load Combinations

U = 1.2D +1.6(LrorS or R) + (1.0L or 0.8W)

U = 1.2D + 1.6W + 1.0L + 0.5(LrorS or R)

U = 1.2D + 1.0E + f1L + 0.2S


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Snow Loads

Roof Snow Load, pf is now based on theGround Snow Load, pg

pf= 0.7CeCtpg

Where

pf= flat roof snow load (psf)

Ce = exposure factor

Ct = thermal factor = importance factor

pg = ground snow load (psf)


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Snow Loads

Limited by

pf pg where pg 20 psf

pf 20 where pg > 20 psf


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Exposure Factor, CePage 3-106

The terrain category and roof exposure condition chosenshall be representative of the anticipated conditions duringthe life of the structure. An exposure factor shall bedetermined for each roof of a structure.


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Thermal Factor, CtPage 3-106

These conditions shall be representative of theanticipated conditions during winters for the life of thestructure.


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Drifting Loads

Consideration for Windward and Leeward


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Snow Drift Configuration

If hc/hb 0.2, drift loads need not be applied


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Otherwise (Example)

Given: A flat roofed office building 450 ft long has a 50 ft

long, 8 ft high penthouse centered along the

length. The building is located in downtown

Milwaukee, Wisconsin.


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Snow Load Example

Problem: Determine the following

Roof Snow Load. pf

Drift Load and Location


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Solution Steps

Step 1 Calculate the Ground Snow Load

Step 2 Calculate Drift Requirements

Step 3 Calculate Balanced Snow HeightStep 4 Determine if drifting is considered

Step 5 Determine Drift Height

Step 6 Drift Force


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Step 1Roof Snow Load, pf

Recall - pf= 0.7Ce Ct I

Wherepf= flat roof snow load (psf)Ce = exposure factorCt = thermal factor = importance factor

pg = ground snow load (psf)


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Step 1Terrain Category

Assumption

Within the life of the structure, taller

buildings may be built around it Exposure B

Ce =1.2


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Step 1Thermal Condition

Office - Heated Structure

Ct = 1.0


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Step 1Importance factor

For office buildings

= 1.0

From Figure 3.10.1 pg 3-103

Building Category II


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Step 1Determine Ground Snow Load

Downtown Milwaukee, Wisconsin(pg 3-105 figure 3.10.2)

pg = 30 psf


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Step 1 - Alternate

Special Case

StudyRegion

CS Site Specific Case Studies

are required to establish theground snow load


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Step 1Roof Snow Load, pf

pf= 0.7Ce Ct I pg

= 0.7(1.2)(1.0)(1.0)(30) = 25.2 psf


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Step 2Calculate Drift Requirements

Balanced snow load height - hb


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Step 3Balanced Snow Height, hb

hb = Roof Snow Load / Unit Weight

= pf/

Unit Weight of Snow,

= 0.13pg + 14 30 pcf

= 0.13(30 pcf ) + 14 pcf = 17.9 pcf

17.9 30 pcf

hb = pf/ = 25.2 psf /17.9 pcf = 1.40 ft

OK


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Step 4Determine if Drifting is Considered

hc penthouse height in this case = 8.0 ft

hc/hb = (8.0 -1.4)/1.4 = 4.7

4.7 > 0.2

Drifting must be considered!


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Step 5Determine Drift Height

Leeward lu = 50fthd ~ 2.5ft

Windward lu = 200ft

hd = 75% (Graph value)

hd ~ 0.75 ( 4.8) = 3.6ft


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Step 6Drift Force

Drift Width = w = 4hd4(3.6ft)=14.4ft

Force = hdw

(17.9pcf)(3.6)(14.4) =464 plf

Location - acts 1/3w frompenthouse wall1/3(14.4) = 4.8ft


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Wind Load

Method Presented

ASCE 7 02

Method 1 Simplified Procedure


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Wind Load

Limitations of Simplified Procedure Height 60 ft or least lateral dimension

Enclosed building (includes parking structures)

Regular shaped No expansion joints

Fundamental frequency 1 Hz.

Flat or shallow pitched roof

No unusual topography around the building


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Wind Load Procedure

Determine Basic wind speed

Directionality Factor

Exposure

Pressure Zone

Load per unit area

Importance Factor


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Determine the Basic Wind Speed

Basic Wind Speed Chart (pg 3-108, 3-109)


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Determine the Directionality Factor

The directionality factor is 0.85 for buildings

For additional FactorsMinimum Design Loads for Buildings and Other Structures,Revision of ASCE 7-98 (SEI/ASCE 7-02), American Societyof Civil Engineers, Reston, VA, 2003 (Co-sponsored by theStructural Engineering Institute). includes more detaileddescriptions)


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Determine Exposure Category

Applies to upwind direction

Exposure B:

Urban and suburban areas, wooded areas Exposure D:

Flat, unobstructed areas outside hurricane-proneregions

Exposure C:All others

i


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Determine the Pressure Zones

P L l S


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Pressure on Lateral System

where:

ps = combined windward and leeward net pressures

30 represents average 30 ft building height

The pressure on Main Wind Force Resisting System (MWFRS)

ps = ps30

P L l S


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where:

l = building height and exposure coefficientfrom

Figure 3.10.6(c)

pg 3-110


ps = ps30

P L t l S t


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where:

= importancefactor for wind

Figure 3.10.1


ps = ps30

P L t l S t


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where:ps30 = simplified

design windpressure from

Figure 3.10.6(a)pg 3-110


ps = ps30

Wi d L d F


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Wind Load Force

The force on the MWFRS is then determined bymultiplying the values of ps30 by their respective zoneareas

Zone A can be at both ends of the structure


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Cl ddi Wi d L d E l


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Cladding Wind Load Example

Given: A 114 ft wide by 226 ft long by 54 ft

tall hospital building in Memphis, TN.

Cladding panels are 7 ft tall by 28 ftlong. A 6 ft high window is attachedto the top of the panel, and an 8 ft

high window is attached to thebottom.

Cl ddi Wi d L d E l


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Cladding Wind Load Example

Problem: Part A

Determine the design wind load on the MWFRS

Part B

Determine the design wind load on the claddingpanels.

Solution Method:

As this is an enclosed building under 60 ft high,Method 1 may be used.

Suburban Area - Exposure Category B

P t A S l ti St (MWFRS)


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Part ASolution Steps (MWFRS)

Step 1MWFRS Determine Wind Speed

Step 2MWFRS Determine Zone Coefficients

Step 3MWFRS Calculate Zone Pressure Step 4MWFRS Calculate Zone Area

Step 5MWFRS Calculate Zone Force

Step 6MWFRS Calculate Force Location

St 1 D t i Wi d S d


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Step 1MWFRSDetermine Wind Speed

Figure 3.10.5 (page 3-109)

Memphis, TN

90 Mph

St 2 Z C ffi i t


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Step 2MWFRSZone Coefficient, l

Height / Exposure Coefficient Building Height 54ft, Exposure - B

Figure 3.10.6(c) (pg 3-110)

55 50

1.19 1.16

54 50

l 1.16

l 1.18

Step 2 Zone Coefficient p


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Step 2MWFRSZone Coefficient,ps 30

Pressure Coefficient From Table 3.10.6(a) (pg 3-110)

Zone A ps 30 = 12.8 psf

Zone C ps 30 = 8.5 psf

Step 2 Zone Coefficient I


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Step 2MWFRSZone Coefficient, I

Importance Factor From Table 3.10.1

(page3-103)

= 1.15

Step 3 Calculate Zone Pressures


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Step 3MWFRSCalculate Zone Pressures

ps zone A = ps 30 zone A

1.18(1.15)(12.8) = 17.4 psf

ps zone C = ps 30 zone C

1.18(1.15)(8.5) = 11.5 psf

Step 4 Calculate Zone A Dimensions


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Step 4MWFRSCalculate Zone A Dimensions

Length of building 226 ft Lesser of

0.2(114) = 22.8

Or0.8(54) = 43.2

A226 = 22.8 ft

Step 4 Calculate Zone C Dimensions


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Step 4MWFRSCalculate Zone C Dimensions

Length of building 226 ft C226 = 226 22.8 = 203.2 ft

Step 5 MWFRS Zone Forces


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Step 5MWFRSMWFRS Zone Forces

F1

= A226

h ps Zone A

= 22.8(54)(17.4)/1000 = 21.4 kips

F2 = C226 h ps Zone C

= 203.2(54)(11.5)/1000 = 126.2 kips

Total force

= 21.4 + 126.2

= 147.6 kips

Step 6 MWFRS Forces Location


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Step 6MWFRSMWFRS Forces Location

F1 = 21.4 kips, F2 = 126.2 kips Resultant Location, eleft

eleft

F1

A

226

2

F2

C

226

2

A226

F

1 F

2

eleft

21.4 22.8

2

126.2 203.22

22.8

21.4 126.2e

left 107.8ft 108ft

or about 5 ft to the left of center

Part B Solution Steps (Cladding)


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Part BSolution Steps (Cladding)

Step 1Clad Determine Wind Speed

Step 2Clad Determine Zone Coefficients

Step 3Clad

Calculate Tributary Area

Step 4Clad Calculate Zone Pressure

Step 5Clad Calculate Panel Force

Step 6Clad Calculate Window Force

From Previous Solution

Step 3 Calculate the Cladding Tributary Area


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Step 3CladCalculate the Cladding Tributary Area

Tributary area per panel =one-half of upper window

+

panel

+

one-half of lower window times the width

(6/2 + 7 + 8/2)(28) = 392 ft2

Step 4 Cladding p Zone Pressure


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Step 4CladCladding pnet 30 Zone Pressure

Table 3.10.6(b) (page 3-110) Interpolating panel area between

100 and 500 ft2:

392 100

500 100 0.73



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Inward pressure Zone A12.4 0.73(12.4 10.9) = 11.3 psf

pnet30 = 1.18(1.15)(11.3) = 15.3 psf



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Outward pressure Zone A15.1 0.73(15.1 12.1) = 12.9 psf

pnet30 = 1.18(1.15)(12.9) = 17.5 psf

Step 4Clad

Cladding pnet 30 Wind


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Force

Panel Size Height 7 ft

Length 28 ft

Force on panel:Inward: 7.0(28)(15.3) = 2999 lb

Outward: 7.0(28)(17.5) = 3430 lb

Step 5Cl d Window Forces


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Step 5Clad Window Forces

Force on panel from upper window:

Inward: (6.0/2)(28)(15.3) = 1285.2 lb

Outward: (6.0/2)(28)(17.5) = 1470 lb

Step 5Cl d Window Forces


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Step 5Clad Window Forces

Force on panel from lower window:

Inward: (8.0/2)(28)(15.3) = 1713.6 lb

Outward: (8.0/2)(28)(17.5)= 1960 lb

Step 6Cl d Resultant Cladding FDB


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Step 6Clad Resultant Cladding FDB


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building systems snow and wind load

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