36 MOMENT OF ROTATION. PARALLEL FORCES. [CHAP. V. 







CHAPTER Y. 



MOMENT OF ROTATION. PARALLEL FORCES. 



3. Equilibrium Polygon. Since the forces acting upon 

 structures are generally due to the action of gravity, these 

 forces may be considered as parallel and vertical, and in all 

 practical cases therefore, we have to do with a system of paral- 

 lel forces. 



Given any number of parallel forces P^, PI. 6, Fig. 20 ; 

 required to find the direction, intensity and position of the 

 resultant, and the moment of rotation at any point. 



1st. Draw the force polygon, In this case it is, of course, a 

 straight lino. 



2d. Choose a pole C, and draw the lines S , Sj, S 2 . etc. 



3d. Draw the string or equilibrium polygon a, T> c d e f. 

 Considering this polygon as a system of strings, the forces will 

 be held in equilibrium if we join any two points, as a and </, 

 by a strut or compression piece, and apply at a and g the up- 

 ward forces Y! and V 2 . 



4th. Prolong a b and fg to their intersections. Through 

 this point the resultant must pass. It is of course parallel and 

 equal to the sum of the forces. 



Now, if a g is assumed horizontal, the perpendicular H to 

 the force line, or the "pole distance" divides the resultant 5 

 into the two reactions YX and V 2 (Art. 22). 



All the forces in the equilibrium polygon have the same 

 horizontal projection H, in the force polygon. 



Let a g represent a beam resting upon supports at a and g. 

 We have then at once the vertical reactions V x and V 2 or k 

 and 5 k, which, in order to cause equilibrium, must act up- 

 wards. 



For the moment at any point, as 0, due to V l5 we have, by 

 Culmann's principle, m o multiplied by H. The triangle formed 

 by a b, a g, and P 15 gives then the moment of rupture at any 

 point of the beam as far as Pj. For a point o, beyond P x , the 



