building systems snow and wind load
TRANSCRIPT
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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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Pressure on Lateral System
where:
l = building height and exposure coefficientfrom
Figure 3.10.6(c)
pg 3-110
The pressure on Main Wind Force Resisting System (MWFRS)
ps = ps30
P L t l S t
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Pressure on Lateral System
where:
= importancefactor for wind
Figure 3.10.1
The pressure on Main Wind Force Resisting System (MWFRS)
ps = ps30
P L t l S t
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Pressure on Lateral System
where:ps30 = simplified
design windpressure from
Figure 3.10.6(a)pg 3-110
The pressure on Main Wind Force Resisting System (MWFRS)
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
Step 4 Cladding p Zone Pressure
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Step 4CladCladding pnet 30 Zone Pressure
Inward pressure Zone A12.4 0.73(12.4 10.9) = 11.3 psf
pnet30 = 1.18(1.15)(11.3) = 15.3 psf
Step 4 Cladding p Zone Pressure
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Step 4CladCladding pnet 30 Zone Pressure
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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Questions?