334 TIIE INVERTED ARCH. [CHAP. XVI. 



the corresponding equilibrium polygon A 3, 4, 5, 6, 7, B. Pro- 

 lone: then the outer side B 7 to intersection with P', and draw 

 the closing line A P'. A parallel through O to this line cute 

 off from the force line P the reaction at A and the cable reac- 

 tion P'. 



Now P' being thus found and the form of cable given, we 

 have only to lay off P' vertically, draw from its extremities 

 lines parallel to the two outer sides of the given cable arc, and 

 from the pole thus determined, lines parallel to the other sides 

 will give us the forces 1', 2', 3', etc. These when thus found 

 we lay off on our force line for the pole O, as shown in the 

 Fig., and then construct the corresponding equilibrium poly- 

 gon A 1', 2' 9', 10', B. 



Thus the vertical ordinates between A P', P' B and this 

 polygon give us the moment at any point for a truss acted 

 upon by the forces I/. 2', 3', etc., alone. The ordinates between 

 A P', P' B and the polygon A 3, 4, 5, 6, 7, give, in like manner, 

 the moments for a truss acted upon by the forces 3, 4, 5, 6, 7, 

 whose reactions are A and P'. The ordinates, then, included 

 between both polygons give us the moment at any point of the 

 stiffening truss. Thus the ordinate y, multiplied by the pole 

 distance, gives us the moment in the truss at the point o. If 

 we had taken the pole distance O equal to the height of the 

 truss, then these ordinates would give us at once the strain in 

 the flanges. We can thus easilv find the strains in the stiffen- 



O / 



ing truss for any weight or system of weights in any position. 



1S6. most unfavorable method of Loading. Let us in- 

 vestigate the action of a single weight P at any point. In PI. 

 27, Fig. 109, we have a single weight P acting between A and 

 P'. 



The Fig. is nothing more than a repetition of Fig. 108, only 

 we have a single load P instead of a system of four loads, and 

 therefore the polygon for P consists only of two straight lines 

 instead of having as many angles 3, 4, 5, etc., as there are apex 

 loads in the first case. All lines have the same position as in 

 Fig. 108, and hence the construction needs no further explana- 

 tion. 



We see at once from the Fig. that any load between A and 

 P' increases the moment at every point of the span A B, and 

 therefore at the point of rupture or of maximum moment also. 

 So als"o for the shearing force. When, therefore, the moment 



