132 CONTINUOUS GIRDERS. [CHAP. VI IT. 



distance. This last is - ^. The segment, therefore, is 2y. We 

 o 



do not need, therefore, to draw the force polygon, but have sim- 

 ply to take off with the dividers the middle ordinate of the para- 



bola f = -f- , and lay it off twice on the verticals right and left 

 8 



through the supports, and join the four points thus obtained by 

 lines crossing each other under the centre of the span. The 

 equilibrium polygon for the positive parabolic area is then 

 ready for the middle span. [We advise the reader to construct 

 it for himself.] 



For the two side spans the construction is different. Here 



2 4 In 



the area of the parabola is -f IQ, the reduced area -f -^ and 



3 3 ^ 



the moment 'x = 2'A. 



Dividing by - Zj, we have for the segment required 

 8 



Therefore, in the end spans, or generally in any span not equal 

 to the standard span, or that which furnishes the constants - \ 



and - ^, we must multiply the middle ordinate of the parabola 



o 



by the square of the ratio of the " standard " span to the span 

 in question, and then lay the product off twice upon the verti- 

 cals through the supports. This multiplication is easily per- 

 formed graphically. If from the middle of span 1 we lay off 

 I? horizontally and join the end with the end of f', then lay off 

 7 2 in same direction and draw a parallel to the first line, the 



72 



segment ony will bey'-y- 2 . For we shall have 



n 



tf :/'.-:#: or *=f$- 



Since these cross-lines depend upon given quantities, they 

 can be constructed for every span, and thus we have Fig. 52, c. 

 They give us not only the moments over the supports, but also 

 the moments for any point of the parabolic area forces. We 

 shall hereafter make use of them. 



6. Construction of the Fixed Points, and of the Equi- 

 librium Polygon. We have thus all the given and known 



