CHAP. Vm.] CONTINUOUS GrRDEKS. 135 



struction. "We see at once that we shall thus obtain in each end 

 span two points, in the middle span four points, which last, be- 

 ing joined by lines crossing each other, give in the middle span 

 two sides in proper position. It is also evident how the poly 

 gon may then be completed. 



7. Construction of the If o in cuts at the Supports. Thus 

 we are able to construct the equilibrium polygon, or rather the 

 extreme tangents to the elastic line for each span. We have 

 now to determine the two moments over the supports. This is 

 very simple. The first moment to the left is cut off by the 

 fourth side, the second by the fifth side of the polygon, from 

 the verticals through the supports. "We have therefore only to 

 prolong these two sides, take off the segments in the dividers, 

 and lay them off in Fig. 52 (a) in A A' and B B'. "We have, 

 then, in PL 14, Fig. 52 (a) the moments for the given case and 

 loading at any point, as shown by the shaded area. 



The proof is simple. The two lines N M and N L [Fig. 52, 

 5] evidently cut off upon the vertical at the support the moment 

 of the force acting at N. This force is the area of the triangle 



A A' B' equal to A A' 4. This reduced to the basis - ^ gives 



A A'. If we multiply this by the lever arm of the force, we 

 have its moment. This moment is, however, equal to the seg- 

 ment A A' multiplied by the pole distance, and since this pole 



distance is itself l, the segment itself to the assumed pole 



distance gives us the moment. 



We see that it is not necessary to draw the line N L as it 

 passes through the support. We have simply to prolong the 

 side M N to intersection with the vertical through the support. 

 It is to be observed that the moments at the supports are cut off 

 at the supports only by those lines which pertain to the " stand- 

 ard " span, or that span from which we take our reduction basis 

 and pole distance. For lines in the other spans the above does 

 not hold good without modification. It is, however, always 

 possible, at least for from two to five symmetrical spans, to 

 observe the above conditions. In those cases where this is not 

 possible, an easy graphical multiplication of the segments by 

 the square of the ratio of the spans will give the moments. 

 We see also the reason why, for four symmetrical spans, the 

 second and not the first must be taken as the standard span. 



