Answer :
Answer:
The external pressure is p = -21.9 psf or p = -8.85 psf
Explanation:
Given :
Velocity of wind, v = 120 mi / hr
[tex]$k_d = k_c =1 $[/tex] (wind direction factor)
[tex]$k _{zt} = 1 $[/tex] = topographical factor (for flat terrain)
[tex]$ q_n$[/tex] = velocity pressure at height h
[tex]$ \therefore q_n = 0.00256 k_z k_{zt} k_d v^2 $[/tex]
[tex]$ q_n = 0.00256 \times k_z (1)(1)(120)^2 $[/tex]
[tex]$ = 36.86 k_z$[/tex]
But for height h = 30 ft, [tex]$ k_z$[/tex] = 0.98 (from table)
[tex]$ \therefore q_n = 36.86 \times 0.98 $[/tex]
= 36.16
Now, [tex]$ \frac{L}{B}= \frac{200}{200} =1$[/tex] , so [tex]$C_p=-0.5 $[/tex] (from table)
[tex]$p = q(G)(C_p)-q_n(GC_{pi})$[/tex]
where, p = external pressure
G = 0.85 = gust factor (for typical rigid building)
[tex]$GC_{pi} = \pm 0.18 $[/tex] (internal pressure co efficient)
Therefore putting the values,
[tex]$p = (36.13)(0.85)(-0.5)-(36.13)(\pm 0.18)$[/tex]
p = -21.9 psf or p = -8.85 psf