1848.] Quantity of Iron necessary in a Tension Chain Bridge. 



RC 



Hence, tension of BC = * tension of BA 



BD 



% tension of BA, 



BC 



since the triangle BhC is similar to the triangle BCD. 



But the transverse section of iron is to be proportional to the ten- 

 sion. Hence 



Tth 



Section of BC = % section of B A. 



BC 



.'. Quantity of iron in BC = BC * section of BC. 



= Bb X, section of BA. 



Hence the quantity of iron in AB and BC together = AB % section 

 of AB -f Bb % section of AB =zAby> section of AB = quantity in a 

 bar of length Ab, and thickness at A. — Q. E. D. 



We shall now proceed to give, first a Geometrical, and then an 

 Analytical demonstration of the Fundamental Proposition which is the 

 subject of this communication. 



1. Geometrical Demonstration. 



Let fig. 2 represent the bridge, the dark lines representing the iron 

 work. The lower parts EB, BC of the rods in fig. 1 are removed, 

 and replaced by EF, FC, and EG, GC, on both sides the bridge : the 

 rod FC is necessary to counteract the horizontal strain of FE, and the 

 rod GC is necessary to hold down EG, EG in position. 



We have to show, that if these four new rods are proportional, in 

 transverse section, to their strains, the quantity of iron in them is the 

 same as in those which they replace, viz. in EB, BC. 



Draw Ch perpendicular to EF produced, and Cg perpendicular to 

 EG produced. Then, by the property already proved in case of fig. 

 1, the quantity of iron in EF and FC = quantity in a length Eh of 

 same section as EF, and the quantity of iron in EG and \ GC* = the 

 quantity in a length Eg of the same section as EG. Now the tensions 

 of EA, EF, and EG acting at E are in equilibrium. Draw the 

 parallelogram JH. Hence the sides of the triangle BIIE (as also of 

 EJB), being parallel to the directions of these three forces, are pro- 

 portional also to them in magnitude. 



* The other half of GC's substance belongs to the other half of the bridge, 



R 2 



