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## Q. 2.2

Determine the external wind pressure on the roof of the rigid gabled frame of a nonessential industrial building shown in $Fig. 2.6(a)$. The structure is located in a suburb of Boston, Massachusetts, where the terrain is representative of exposure B. The wind direction is normal to the ridge of the frame as shown.

## Verified Solution

Roof Slope and Mean Roof Height From $Fig. 2.6(a)$, we obtain

$\tan \theta =\frac{16.83}{20}=0.842$  ;   or   $\theta = 40.1°$

$h = 11.58 + \frac{16.83}{2}=20.0$

$\frac{h}{L}=\frac{20}{40}=0.5$

Velocity Pressure at $z = h = 20^′$ From $Fig. 2.4$ , we obtain the basic wind speed for Boston as

$V = 110 mph$

From Table 2.3, we can see that the importance factor for wind loads for nonessential buildings (category II) is

 TABLE 2.3 CLASSIFICATION OF BUILDINGS FOR ENVIRONMENTAL LOADS Occupancy or use Importance Factor, I Category Wind loads Snow loads Earthquake loads Buildings representing low hazard to human life in the case of failure, such as agricultural and minor storage facilities I 0.87 for V ≤100 mph 0.77 for V > 100 mph 0.8 1.00 All buildings other than those listed in Categories I, III, and IV II 1.00 1.0 1.00 Buildings representing a substantial hazard to human life in the case of failure, such as: those where more than 300 people congregate in one area; day-care facilities with capacity greater than 150; schools with capacity greater than 250; colleges with capacity greater than 500; hospitals without emergency treatment or surgery facilities but with patient capacity greater than 50; jails; power stations and utilities not essential in an emergency; and buildings containing hazardous and explosive materials III 1.15 1.1 1.25 Essential facilities, including hospitals, fire and police stations, national defense facilities and emergency shelters, communication centers, power stations, and utilities required in an emergency IV 1.15 1.2 1.5 Source: Adapted with permission from ASCE 7-02, Minimum Design Loads for Buildings and Other Structures. This information is extracted from ASCE 7-02; for further information, the complete text of the manual should be referenced (http://www.pubs.asce.org/ASCE7.html?9991330).
$I = 1.0$

and from Table 2.4, for the exposure category B, we obtain the following values of the constants:

 TABLE 2.4 EXPOSURE CATEGORIES FOR BUILDINGS FOR WIND LOADS Exposure Constants Category $z_g$ ft(m) $\alpha$ Urban and suburban areas with closely spaced obstructions of the size of single family houses or larger. This terrain must prevail in the upwind direction for a distance of at least 2,630 ft (800 m) or 10 times the building height, whichever is greater B 1,200(366) 7.0 Applies to all buildings to which exposures B or D do not apply C 900(274) 9.5 Flat, unobstructed areas and water surfaces outside hurricane-prone regions. This terrain must prevail in the upwind direction for a distance of at least 5,000 ft (1,524 m) or 10 times the building height, whichever is greater D 700(213) 11.5 Source: Adapted with permission from ASCE 7-02, Minimum Design Loads for Buildings and Other Structures. This information is extracted from ASCE 7-02; for further information, the complete text of the manual should be referenced (http://www.pubs.asce.org/ASCE7.html?9991330).

$z_g = 1,200 ft$    and     $\alpha = 7.0$

By using Eq. $K_z=\begin{cases} 2.01(z/z_g)^{2/\alpha } for 15 ft (4.6 m )\leq z\leq z_g\\ 2.01(15/z_g)^{2/\alpha } for z\lt 15 ft (4.6 m) \end{cases}$, we determine the velocity pressure exposure coefficient:

$K_h = 2.01(\frac{h}{z_g})^{2/\alpha}= 2.01(\frac{20}{1,200})^{2/7}=0.62$

Using $K_{zt} = 1$ and $K_d = 1$, we apply Eq. ($q_z = 0.00256K_zK_{zt}K_dV^2I$) to obtain the velocity pressure at height $h$ as

$q_h = 0.00256K_hK_{zt}K_dV^2I$

$= 0.00256(0.62)(1)(1)(110)^2(1.0)$

$= 19.2 psf$

External Wind Pressure on Roof For rigid structures, the gust effect factor is

$G = 0.85$

For $\theta \approx 40°$ and $h=L = 0.5$, the values of the external pressure coefficients are (Fig. 2.5):

For windward side: $C_p = 0.35$  and  $-0.1$

For leeward side: $C_p = -0.6$

Finally, by substituting the values of $q_h, G$, and $C_p$ into Eq. ($p_z = q_zGC_p$ for windward wall

$p_h = q_hGC_p$ for leeward wall; sidewalls; and roof), we obtain the following wind pressures: for the windward side,

$p_h = q_h G C_p = (19.2)(0.85)(0.35) = 5.71 psf$          Ans.

and

$p_h = q_h G C_p = (19.2)(0.85)(-0.1) = -1.63 psf$         Ans.

and for the leeward side

$p_h = q_h G C_p = (19.2)(0.85)(-0.6) = -9.79 psf$         Ans.

These wind pressures are applied to the roof of the frame, as shown in Fig. 2.6(b). The two wind pressures (positive and negative) on the windward side are treated as separate loading conditions, and the structure is designed for both conditions.