102 



THE AECH. 



between the soflfit of llio arch and tho road lino is throngl;> 

 out of tho same specific gravity. 



Now it will bo i-ocollected that in tho case of Diagram 1 , 

 in order to find all tho tliruHtH acting upon the three rods, 

 we drew tho parallelograms B 7i and Gg, and proved that 

 tho thrusts along A B, B C, and C D were represented re- 

 spectively by tho lines gli, H C, and C h, that tho vortical 

 loads on the angles B and C were repi-esented by the verti- 

 cals B e and C/, and that tho horizontal thrust in every case 

 was represented by B h. 



Following the same rule of construction in tho proaont 

 case, but without drawing the ontii-o paxallelograms, which 

 are not now necessary and would entirely confuse the 

 figure ; let us begin at the central point D. First draw a 

 vertical lino zm through D, and Gi parallel to D E, meeting 

 zm in i; wo now have a triangle of forces, C D i, in which, 

 as proved in the former case, each side will roprosont tho 

 thrust in its own direction — Ci tho thrust along D E, DC 

 that in its own direction, and D i tho vortical load on tlio 

 angle D, which vortical load is also represented by D, 

 therefore D'i is equal to DO, and, farther, tho horizontal 

 line C h drawn from C to the line z m will represent the 

 liorizontal thrust throughout tho arch. To prooood now to 

 the next angle E, C i is already drawn parallel to D E, lot 

 us draw Gk parallel to E F"', mooting zm in k; thon it fol- 

 lows, as hoforo, that C k will I'oprosont the thrust along E .F, 

 and ik (equal to P E) tho vortical load on E. In tho same 

 manner draw Gl parallel to F Gr and Cm parnllol to tho 

 dotted line throngh G, then CL and CM will roprosont the 

 thrusts along F(t and tho dotted line through G respect- 

 ively, and kl and Im (equal to QF and E, G) the vei'tical 

 loads on F and G. 



On tho other side draw C ,r, C y, and C z parallel respect- 



