CHAP. I.] COMMON POINT OF APPLICATION. 9 



P n =P sin. iu*.i-i 

 P h =P sin. i L84cos - i 

 P V =P cot. isin. i 1 - 84008 - 1 



where P is the intensity of the wind pressure in Ibs. per sq. ft. 

 upon a surface perpendicular to its direction, i is the inclination 

 of any plane surface to this direction ; P n is the normal pres- 

 sure, P h the horizontal component of this normal pressure, and 

 P y its vertical component. 



That is, if the wind blows horizontally, P h is the horizontal 

 and P v the vertical component of the pressure on the roof. If 

 we take P=40 Ibs., which probably allows sufficient margin 

 for the heaviest gales, we have the following values of the nor- 

 mal pressure and its components for various inclinations of 

 roof surface : 



Vf . Lbs. per square foot of surface. 



Pn Pv Ph 



5 5.0 4.9 0.4 



10 9.7 9.6 1.7 



20 18.1 17.0 6.2 



30 26.4 22.8 13.2 



40 33.3 25.5 21.4 



50 38.1 24.5 29.2 



60 40.0 20.0... 34.0 



70 41.0 14.0 38.5 



80 40.4 7.0 39.8 



90" 40. 0.0 40.0 



The load at each joint may be taken as equal to the pressure 

 of the wind striking a surface whose area is equal to that por- 

 tion of the roof supported by one bay of the rafter, and inclined 

 at the same angle as the tangent to the rib at- the joint. Thus 

 we can calculate P 1? P 2 , P 3 , P 4 , (Fig. 6), resolve these forces into 

 their horizontal and vertical components, and find the reactions 

 at the supports as well as the horizontal force at the left abut- 

 ment, which in our construction is supposed to be fixed. Should 

 the wind be supposed to blow from the right side, the strains 

 would be entirely different, and it would be necessary to form 

 a second diagram. Each piece must be proportioned to resist 

 the strains arising in either case. The forces P w and their 

 horizontal and vertical components, as also the reactions, being 

 known, we can now form the force polygon. 



Thus in Fig. 6 (), we lay off the forces P^, make a c equal 



