DESIGN OF FIXED ENDED ARCHES BY THE ELASTIC THEORY 147 



to satisfy that result. Hence the true frame must be so formed as 

 to make 2^,M,y=0. 



Since — is constant 2_^,Mjy is also equal to the total horizontal 

 EI ^ 



deflection of the true frame between the limits V andV. The same 



is true for the rib for the expression '2^,'K^^t/. See equations (9) 



and (10), If other limits than V and V be chosen, 2Mjy and 2M„y 

 then represent the total horizontal deflections for the true frame and 

 the rib respectively between those limits. Neither expression need 

 be zero. But they must be equal. 

 Consider the equations: — 



M=^Fx, (19) 



D,=^2My. (20) 



When forces are laid off on a load line and a pole is taken and 

 a funicular polygon is drawn in the space diagram, by graphics, it 

 can be shown that the vertical interscepts in the funicular polygon 

 are proportional to its moments. Hence, by analogy, see equations 

 (19) and (20), if a series of moments be laid off to scale upon a line 

 and be treated like forces, the interscepts in the corresponding frame 

 will be proportional to deflections. Frames so drawn are termed 

 deflection polygons. 



Column 13 is found by taking the algebraic sums of the figures 

 in column 12 in the following way: (1+18), (l-fl8)+(2+17), etc. 

 Column 14 is obtained in like manner from column 11. These are 

 the moment quantities M of equation (20) to be laid off on the so- 

 called load lines of the deflection polygons. 



Let rv be the load line for the deflection polygon of the trial 

 polygon and rw for the rib. Select a common pole O. Lay off the 

 moments of column 14, like forces, upon rv and those from column 

 13 upon rw. Draw rays from O. The lines of action of the 

 moments (treated as forces) are respectively along the chords of the 



