Question

Wind blows on the side of a fully enclosed hospital located on open flat terrain where V= 120 mi/h. Determine the external pressure acting on the leeward wall, if the length and width of the building are 200 ft and the height is 30 ft. Take Ke 1.0.

Answer 1

The external pressure is p = -21.9 psf or p = -8.85 psf

Step-by-step explanation:

Given :

Velocity of wind, v = 120 mi / hr

`$k_d = k_c =1 $` (wind direction factor)

`$k _(zt) = 1 $` = topographical factor (for flat terrain)

`$ q_n$` = velocity pressure at height h

`$ \therefore q_n = 0.00256 k_z k_(zt) k_d v^2 $`

`$ q_n = 0.00256 * k_z (1)(1)(120)^2 $`

`$ = 36.86 k_z$`

But for height h = 30 ft,
`$ k_z$` = 0.98 (from table)

`$ \therefore q_n = 36.86 * 0.98 $`

= 36.16

Now,
`$ (L)/(B)= (200)/(200) =1$` , so
`$C_p=-0.5 $` (from table)

`$p = q(G)(C_p)-q_n(GC_(pi))$`

where, p = external pressure

G = 0.85 = gust factor (for typical rigid building)

`$GC_(pi) = \pm 0.18 $` (internal pressure co efficient)

Therefore putting the values,

`$p = (36.13)(0.85)(-0.5)-(36.13)(\pm 0.18)$`

p = -21.9 psf or p = -8.85 psf