BARS 



135 



the bending moment being now proportional to the increase of 

 curvature. 



Resolving along and perpendicular to the radius vector the 

 forces on a mass-element pwaSO, we have (see Fig. 47) 

 P a>aS0. u = SP- QSe, pvaW.v = BQ + PS0 ; 

 and, taking moments about a normal to the plane of the ring, 



the rotational inertia being neglected as in the case of a straight 

 bar ( 45). Thus 



8 2 w 8P n 



a* = ;S|-C. P^ a ^ = 



dt* 80 



.(4) 



and 



80 



= Pa. 



-(5) 



P+bP 



These, together with (3), are 

 the equations of our problem. 

 It is easily seen that they cannot 

 be satisfied on the assumption 

 that the tension Q vanishes, and 

 that accordingly some degree of 

 extension is involved in any 

 mode of vibration. This is 

 readily accounted for, a stress 

 of this kind being necessarily 

 called into play by the inertia of 

 the different portions swinging 

 in opposite directions. It may be shewn however that in the 

 " flexural " modes to be referred to presently the corresponding 

 strains are small compared with those involved in the change of 

 curvature. 



Eliminating P, Q, M between (3), (4), and (5), we find 



E ( dv 



(6) 



Fig. 47. 



, 



** + 



_ L _ u 



pa 2 + ~ + = 



To ascertain the normal modes we assume that u and v vary 

 as cos (nt + e). Again, the ring being complete, u and v are 



