33C THE INVERTED AROII. [OHAP. XVI 



let us take the structure represented in PI. 28, Fig. 111. Span, 

 60 ft. ; depth of truss and panel length, 5 ft. Scale, 10 ft. to an 

 inch. We suppose the live load to be 2 tons per ft., giving 

 thus 10 tons for each lower apex, and take the scale of force 50 

 tons per inch. 



On the left we have laid off the force lines for the loads 2, 3, 

 4, 5, 6 and 7 to 11, and have taken the poles for each, so that 

 the first lines are all parallel to each other and to the first link 

 of the cable ; the common pole distance being 2 times the 

 height of truss, or 1.5 inches. The moment scale is then 

 1.5 x 10 x 50 = 750 ft. tons per inch. Since the full load is 

 entirely supported by the cable, we have only to investigate the 

 effect of the live load upon the truss. 



Precisely as in Fig. 108, we construct the polygons for forces 

 2-11, 3-11, 4-11, etc., and draw the closing lines as indicated 

 by the broken lines radiating from the centre O. Parallels to 

 these from the poles cut off from the force lines the end and 

 chain reactions. The upper portions are the chain reactions, 

 the lower the reactions at the right end for the loads 2-11, 

 3-11, etc. 



Now we ha.ve to divide these chain reactions into as many 

 parts as there are load apices by lines parallel to the sides of 

 the chain. This we have done by drawing two lines parallel to 

 the two chain ends, inserting the chain reactions between these 

 lines, and then drawing parallels to the chain sides. If, as in 

 this case, the curve of the chain is a parabola, these reactions 

 are divided into 11 equal parts. If the chain has any other 

 form, the parallels to the chain sides determine the relative 

 lengths of these portions. 



It will only be found necessary to construct the moment 

 polygons for 4, 5 and 6-11 ; the other polygons already drawn 

 are necessary for the determination of the shearing forces only. 



Thus, on the force line for loads 4 to 11 we can now lay off 

 the 11 equal parts just found, into which the chain reaction is 

 divided. So for 5-11 and 6-11. These portions we have indi- 

 cated by Roman numerals. We can now draw the correspond- 

 ing polygons precisely as in Fig. 108, which are indicated also 

 by Roman numerals. 



It is then easy with the dividers to pick out the maximum 

 moment at any apex. These moments, laid off as below, give 

 the curve of moments for the truss, which being scaled off and 



