104 



THE ATlCn. 



ivoly to OB, 13 A and tlio dotted lino throiigli A; tlion G x, 

 Cj/, and C 2 will represent tho tlirnsts along 015,1! A, and 

 the dotted lino through A. 



Thus we have in this triangular fi'aine ^ m lines repre- 

 senting all the thrusts in tho entire structure, those thrusts 

 in the direction of the arch rods A B, BO, D, D E, E F, 

 and FGbyCy, O.t, OD, 0/, 0/,-, and 0/, the horizontal 

 thrust by Gh and tho vortical loads AT, B S, ('N, O D, 

 P E, Q F, and E, G by zy, y x, ;» D, D i, i k, k I, and I m re- 

 spectively. The line 2; m represents therefore the weight 

 of tho whole bridge. 



From what has been proved by means of this diagram, 

 it may also be shown that if we wish to sketch an arch of 

 the correct form to suit a given span under a given road line, 

 wo can, by an inversion of the process just described, obtain 

 tho true form of the equilibrated curve, by construct.ion on 

 a sheet of paper, in the following manner : 



For instance, lot our object bo to construct an equili- 

 brated arch to carry a level roadway T R over an opening 

 30 feet wide between the abutments A and G in Diagram 2. 

 Let us begin by dividing the roadway Til into six equal 

 lengths of five foot each at T, R, N, 0, V, Q, and R, and 

 through all these points draw vertical lines. AVe next as- 

 sume a crown thickness D in the middle division of tlie 

 roadway Tli,, lotus say equal to two feet, being one foot 

 for the thickness of tho arch and another for tho super- 

 structure or metalling, etc. We then take D i equal to 

 DO, and from h, the middle of D i, we draw the horizontal 

 line h 0, meeting N in C, then the lino joining D will be 

 the first rod, and D E, parallel to i, will be the correspond- 

 ing rod on the other side of the centre D, meeting P E in E. 

 We then take i k equal to N or P E and join /r, and draw 

 E F parallel to k, then talving k I equal to Q F we join I 



