The interesting special case at this value of ql is a pin-ended beam 

 vibrating with G = , v^ = , and F = 6 /a, p. If G 7^ 0, then 

 Vq = - O.9I+6 G/(EIq^). 



(f) Second built-in frequency: As q£ increases toward 1.5'^, c is 

 negative but decreases numerically until cC = -1. Because of the huge size 

 of S and C, c must be very small; cC = -1 at q«. = 1.1+95^^ = ^+.695, with 



c = -O.OIT, S = C=5i+.7 to three figures, and 



^11 ^ ^^'^ ' ^12 " " ^'^'^ ' ^22 " ^^-^ ' ^ " ° 



whereas b , b p, and b „ are all infinite. As in Case (b), it follows 



2 

 from a,,a^„ - a,. = that, in general, 6 /v is fixed at the value 

 11 22 12 > & '00 



^12^^11' °^ ^ere at - 1.017 q. If, however, either F = G = or F/G = 



-a, _/a, ^ = + 1.017 q, then v = G =0 as for a built-in beam. 

 12 11 ^' o o 



(g) At ql = 1.5-n = 4.712: cC = and a = - a = a = 55.6 

 (to 3 figures), and A = 1. 



(h) First free-free frequency: At (very nearly) qS- = 1.505^^ = U.730, 



cC = 1, as for free-free vibration. Now, finally, it is the turn of a and 



2 

 A to become infinite. Also, ^11^22 - ^-lo ~ '^ ' hence b ,/b,p = b p/bpp, 



so that the right-hand members of Equations [l5a,b], if not zero, are in 



the ratio b /b , and, if G ^ 0: 



F l^n 



- = — ^^ = (cot qJi + coth q£) q = 0.982 q 



G bi2 



a finite quantity. Thus, when a beam is forced so that F = 0.982qG, the 



beajn amplitude remains finite! An attempt to force with F/G in a 



15 



