Wind Load Analysis Chai2.docx

Wind Load Analysis

Mean Roof Height =

R HE + R E 2

=

8.5+6 2

= 7.25 m

2.5 4 θ = 32.005o

tan θ =

 P.1-6; Table 103-1 Occupancy Category: IV – Standard Occupancy

 P.2-38; Figure 207A.5-1A to 5-10 V = 250 km/hr V = 69.44 m/s

 P.2-42 Kd = 0.85

 P.2-47 Kzt = 1

 P.2-67; Section 207B.32 qz = 0.613 Kz Kzt Kd V2 qz = 0.613 (Kz) (1) (0.85) (69.44)2 qz = 2512.4581 Kz

 P.2-68; Section 207B4.1 P = q GCp – qi GCpi; Pa (N/m2)

 P.2-55; Section 207A.9.1 G = 0.85 Use qi = qh Use q = 2512.4581 Kz

 P.2-42; Section 207A.7.3 Exposure Category = B  qi = 2512.4581 Kz

 P.2-68; Table 207B.3-1 Kz = use hmean h 6 7.25 7.5

6−7.25 6−7.5

Kz 0.62 Kz 0.66

=

0.62−K z 0.62−0.66

Kz = 0.653 qi = 2512.4581 Kz qi = 2512.4581 (0.653) qi = 1640.635 Pa  P = q (0.85) (Cp) – (1640.635) ( ± 0.18 ) P = 0.85 q Cp ± 295.314

 Solving P for each face of the building

i.

Windward Roof; use q = qh = qi (Cp is in P.2-74) @ θ = 32.005o h 7.25 = L 8

= 0.906

h/L

Cp 32.005o X1 X2 Cp1 Cp2 Y1 Y2

30o -0.2 0.2 A1 A2 -0.3 0.2

0.5 0.906 ≥10

(For 30o – Vertical Direction) 0.5−0.906 0.5−10

=

−0.2− A 1 −0.2−(−0.3)

0.5−0.906 0.5−10

=

0.2− A 2 0.2−0.2

≈ A1 = -0.204

≈ A2 = 0.2

(For 35o – Vertical Direction) 0.5−0.906 0.5−10

=

−0.2−B 1 −0.2−(−0.2)

0.5−0.906 0.5−10

=

0.3−B 2 0.3−0.2

≈ B1 = -0.2

≈ B2 = 0.279

(For 0.5 – Horizontal Direction) 0

30 −32.005 0 o 30 −35

o

30 0−32.005o 30 0−35o

=

−0.2−X 1 −0.2−(−0.2)

=

0.2−X 2 0.2−0.3

≈ X1 = -0.2

≈ X2 = 0.2401

35o -0.2 0.3 B1 B2 -0.2 0.2

(For ≥10 – Horizontal Direction) 30 0−32.005o 30 0−35o

=

−0.3−Y 1 −0.3−(−0.2)

30 0−32.005o 0 o 30 −35

=

0.2−Y 2 0.2−0.2

≈ Y1 = -0.2599

≈ Y2 = 0.2

(For 0.906 – Horizontal Direction) 30 0−32.005o 0 o 30 −35 0

30 −32.005 0 o 30 −35

=

−0.204−Cp 1 −0.204−(−0.2)

=

0.2−Cp2 0.2−0.279

o

≈ Cp2 = 0.232

 Using +GCpi; Using Cp = -0.202 P = 0.85 q Cp + 295.314 P = 0.85 (1640.635) (-0.202) +¿ 295.314 P = 13.617 Pa  Using -GCpi; Using Cp = -0.202 P = 0.85 q Cp - 295.314 P = 0.85 (1640.635) (-0.202) P = -577.011 Pa

– 295.314

 Using +GCpi; Using Cp = 0.232 P = 0.85 q Cp + 295.314 P = 0.85 (1640.635) (0.232) +¿ 295.314 P = 618.847 Pa = 2.919 kN/m  Using -GCpi; Using Cp = 0.232 P = 0.85 q Cp - 295.314 P = 0.85 (1640.635) (0.232) P = 28.219 Pa ii.

– 295.314

Leeward Roof; use q = qh = qi h/L

Cp ≥20o

≈ Cp1 = -0.202

0.5 0.906 ≥10 0.5−0.906 0.5−10

-0.6 Cp -0.6 =

−0.6−Cp −0.6−(−0.6)

≈ Cp = -0.6

 Using +GCpi; Using Cp = -0.6 P = 0.85 q Cp + 295.314 P = 0.85 (1640.635) (-0.6) +¿ 295.314 P = -541.41 Pa = -2.554 kN/m  Using -GCpi; Using Cp = -0.6 P = 0.85 q Cp – 295.314 P = 0.85 (1640.635) (-0.6) P = -1132.038 Pa iii.

– 295.314

Windward Wall Cp is in P.2-74 Cp = 0.8 L=8m B=7m

L/B = 1.143

Use q = qz = 2512.4581 Kz (Use all values) hmean = 7.25 m (Use only up to “6”)  Using +GCpi; Using Kz = 0.57 for h = 0 – 4.5 P = 0.85 q Cp + 295.314 P = 0.85 (2512.4581) (0.57) (0.8) + 295.314 P = 1269.143 Pa = 1.269 kN/m  Using +GCpi; Using Kz = 0.62 for h = 6

P = 0.85 q Cp + 295.314 P = 0.85 (2512.4581) (0.62) (0.8) + 295.314 P = 1354.566 Pa = 8.127 kN/m  Using -GCpi; Using Kz = 0.57 for h = 0 – 4.5 P = 0.85 q Cp – 295.314 P = 0.85 (2512.4581) (0.57) (0.8) – 295.314 P = 678.515 Pa  Using -GCpi; Using Kz = 0.62 for h = 6 P = 0.85 q Cp – 295.314 P = 0.85 (2512.4581) (0.62) (0.8) – 295.314 P = 763.938 Pa iv.

Leeward Wall Cp is in P.2-74 L=8m B=7m

L/B = 1.143

L/B 0-1 1.143 2 1−1.143 1−2

Cp -0.5 Cp -0.3

=

−0.5−Cp −0.5−(−0.3)

≈ Cp = -0.4714

Use q = qh = qi qi = 1640.635 Pa  Using +GCpi P = 0.85 q Cp + 295.314 P = 0.85 (1640.635) (-0.4714) + 295.314 P = -362.072 Pa  Using -GCpi P = 0.85 q Cp – 295.314 P = 0.85 (1640.635) (-0.4714) – 295.314 P = -952.700 Pa