OHAP. VII.j SIMPLE GIRDERS. 87 



CHAPTER VII. 



SIMPLE GIRDERS. 



69. Action of Concentrated Loads Invariable in Posi- 

 tion. By "simple girder" we understand a girder resting 

 upon two supports only, in opposition to a continuous girder 

 which rests upon more than two. 



Suppose a number of forces P t . . . P 5 acting at various points. 

 [Fig. 41, PI. 13.] We form the force polygon by laying off the 

 forces to scale one after another ; then choose a pole O, and 

 draw O 0, 1, 2, etc., to the points of division. Parallel to these 

 lines we draw the lines of the equilibrium polygon between the 

 corresponding force lines prolonged. If now we close the poly- 

 gon thus formed by the line A B, and draw through O the 

 parallel O L to A B, the segments L and L 5 of the force 

 line give the reactions V t and V 2 . Further, the shearing force 

 between A and P x is S x = V t L ; between P! and P 2 , S 2 = 

 V t P! ; at P 3 , S 8 = V t P! P 2 , etc. That is, the shearing forces 

 are the distances of the points of the force polygon from I*. It 

 is easy, then, to construct them, as shown in the lower shaded 

 area of the Fig. (See also Art. 46.) 



If in the equilibrium polygon we let fall at any p'int a ver- 

 tical as I K, and from K draw K L perpendicular to A B, and 

 indicate by H the horizontal pull, by L the strain in A. B, and 

 by M the sum of the moments of all forces left of I K, then, 

 for equilibrium about K, we have M ^LxKL^LxIK cos 

 I K L, or, since the angle IKL^LOHin force polygon, 

 L x cos I K L = H, and hence M = H x I K, or representing 

 the variable ordinate I K by y : 



m = Hy. 



But H is the distance of the pole O from the force line ; 

 the moment at any point is therefore proportional to the verti- 

 cal height of the equilibrium polygon. (See also Art. 38.) If 

 we take H equal to the unit of force, we have 



