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

Answer:

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

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

$\therefore q_n = 0.00256 k_z k_{zt} k_d v^2$

$q_n = 0.00256 \times k_z (1)(1)(120)^2$

    $= 36.86 k_z$

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

$\therefore q_n = 36.86 \times 0.98$

        = 36.16

Now, $\frac{L}{B}= \frac{200}{200} =1$ ,   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