Specified Wind Load
Figure 4.1.7.6.-A (NBC2020)
v1.0.1
Thinking...
Load Case A: Winds generally perpendicular to ridge
Load Case B: Winds generally parallel to ridge
Factors
Location: 100 Mile House, British Columbia
q50: 0.39kPa
Importance Factor, ULS: Iw = 1.0 / SLS: Iw = 0.75
Roof slope = 0 degrees
Ce = (h/10)0.2 = 0.9
q50: 0.39kPa
Importance Factor, ULS: Iw = 1.0 / SLS: Iw = 0.75
Roof slope = 0 degrees
Ce = (h/10)0.2 = 0.9
External Wind Pressure
p = Iw*q*Ce*Cg*CpLoad Case A: Winds generally perpendicular to ridge
Load Case B: Winds generally parallel to ridge
Side
1
1E
2
2E
3
3E
4
4E
5
5E
6
6E
Load Case A
CpCg
0.75
1.15
-1.3
-2
-0.7
-1
-0.55
-0.8
n/a
n/a
n/a
n/a
ULS
p
(kPa)
0.26
0.4
-0.46
-0.7
-0.25
-0.35
-0.19
-0.28
n/a
n/a
n/a
n/a
SLS
p
(kPa)
0.2
0.3
-0.34
-0.53
-0.18
-0.26
-0.14
-0.21
n/a
n/a
n/a
n/a
Load Case B
CpCg
-0.85
-0.9
-1.3
-2.0
-0.7
-1.0
-0.85
-0.9
0.75
1.15
-0.55
-0.8
ULS
p
(kPa)
-0.3
-0.32
-0.46
-0.7
-0.25
-0.35
-0.3
-0.32
0.26
0.4
-0.19
-0.28
SLS
p
(kPa)
-0.22
-0.24
-0.34
-0.53
-0.18
-0.26
-0.22
-0.24
0.2
0.3
-0.14
-0.21
Internal Wind Pressure
Cgi = 2
Cpi = -0.15 to 0
pi = Iw*q*Ce*Cgi*Cpi