476 REPORTS ON THE STATE OF SCIENCE. — 1919. ; 



2. Framework of thin rods. '~ 



In the case of thin rods forming part of a plane framework under stress 

 in its own plane, we have, if T be the total tension and M the bending 

 moment at any point of a rod, and T', M' the corresponding quantities in 

 the model, 



T = EAs ; 



T'= E'A's'. 



when s, s' are the longitudinal strains of the rod in the full size and model, 

 respectively. 



E, E' are the Young's Moduli, A, A' the cross-sections of the rods. 



Now s = s' by geometrical similarity. 



Hence T : T'=EA : E'A' (1). 

 Further 



where I is the moment of inertia of the cross-section of the rod about its 



neutral axis and R is the radius of curvature of the rod. 



Similarly 



where clearly R' = IcR by geometrical similarity. Also, let f be the 

 ratio of forces {not stresses) in the model and full size. 

 Then M' ^ Mph, i.e., 



E'I'/R' = j>k (EI/R) 



and ET/EI = fh"^. (2) 



But from (1) since T' = pT 



E'A' 



EA=^^- (3) 



From (2) and (3) by division 



V|^' 



I/A 



=P (4) 



Let K be the swing radius of the cross-section of any rod in the full size, 

 and K' in the model. 

 Then 1 = K^A 



r = (K')2A' 



(4) then gives (K'f/K- = F- 

 or K' = Ids. (5) 



That is, although the cross-sections need not be geometrically similar 

 (the rods being thin), the radii of gyration of the cross-sections must be 

 in the ratio of geometrical similarity h. 



