CHAP. Vm.] CONTINUOUS GIRDERS. 131 



other as ^ to 1 , or inversely as the forces. Since the entire dis- 

 tance is - 4 + o k-, the distance of the resultant or of the point 

 o o 



of intersection M from L is ^ ^ ; from N it is ^ 1 . The point 



o o 



M therefore, must lie somewhere in the vertical at ^ \ from L, 



o 



the point of application of the triangular area force for the 

 span 1 . The verticals through the centres of the parabolic 

 areas we call the parabolic or middle verticals ; those through 

 the centres of gravity of the triangular areas, the third verti- 

 cals / those through a point as M, the point of application of 

 the resultant of two contiguous triangular area forces, the lim- 

 ited third verticals. Upon these verticals two sides must always 

 intersect. 



5. Polygon for the Poitive Moment Areas. It will be 



found best to take as the reduction base for areas / 15 i.e., half 



2 



the second span, and for pole distance ^. Reducing the 



o 



areas of the parabolas to this basis, and considering the heights 

 thus obtained as forces, we can foraa a force polygon with pole 



distance - ^. It is not necessary to draw this polygon ; our ob- 



o 



ject is to find the corresponding equilibrium polygon. This 

 last, since we consider the entire parabolic area as a force act- 

 ing at its centre of gravity, consists of two lines which inter- 

 sect in the vertical through the middle of the span. We pro- 

 long each of these lines and obtain two lines as shown in PI. 14, 

 Fig. 52 (c). The segments cut off by these lines from the verti- 

 cals through the supports are the moments of the parabolic 

 areas with respect to the supports. These moments we can 

 easily find. 



Thus, let the deflection of the parabola in the second span at 



o 



the centre be/*, then its area is - f \. This reduced to the 



o 



1 4: 



basis - li gives f as the force. The moment of this force 

 2i o 



4. i 1 



with reference to the supports is -/" x -Ii = 2fxI 1 . This 



o 2 3 



moment is equal to the segment sought multiplied by the pole 



