CHAP. Vm.] COOTINTJOTJS GIEDEBS. 137 



drawn, and the moments at the supports found. ' We conceive, 

 therefore, the moment area as the difference of the trapezoid 

 A' A" B" B' [PL 15, Fig. 53] and the parabolic area A" C" B", 

 or equal to A" C" B" -minus triang. A' A" B' minus triang. 

 B / B // A //^ The area A ,/ c // B // we ca n the gfayfo OT parabolic 



moment area. 



If we indicate the moments at the supports A' A" and B' B" 

 by M' and M", then, for a given span I, 



A' A" B' = M' Z and 

 A'' B' B" = M" I. 



If we indicate further the height of a rectangle of base I and 

 area A" C" B", that is, the mean value of the moments of the 

 corresponding simple girder by 2ft, we have 



area A" C" "B" = 3ft Z. 



The verticals through the centres of gravity of the triangles 

 divide the span into three equal parts. We call these the third 

 verticals. The load 9ft I acts at the centre of gravity of the 

 parabolic area A" B" C". 



The four-sided equilibrium polygon A U S V B corresponding 

 to these forces we call the second equilibrium polygon. The 



pole distance must be (Art. 81) -El. Instead of this, we take, 



Vlt 



which amounts to the same thing, the forces G F = - M' , 



A A> 



E H = - M"- and F E = 2ft -, and the pole distance I = 



A A. A< 



E I 



-, where X is any assumed length. For X, we may take the 



/ fo 



arithmetical mean of all the spans, or, as we have seen, one of 

 the actual spans. If the outer spans are both equal to 1 6 and the 

 other spans equal to ^, we should naturally choose X = ? t , since 



then the forces would be M', - M" and 3ft. 



2i A 



If the position of the tangents at the supports were known 

 or found, the equilibrium polygon could be easily drawn as 

 follows : Upon the two verticals distant each side of the centre 



E I 



S [Fig. 53] by the pole distance b = - lay off the distance 



n A. 



