Rotatory Inertia on the Vibrations of Bars. 45 



By means of (9) the latter equation reduces. to 



7 2 i'= 7l d'". 

 Thus 



2 



/ = ^(coS7 1 f+ ^cosh 72 ^+cf(sm 7l f+psinh 72 f). (28) 

 1 2 /i 



Again, fj _n ) 



j 7 , ,., , when £ = —-—i. 



and d/ ^/_n I * 



,\ d( — cos^Z' + COSI17/) + d' (— siny^' + ~, sinh y 2 Z') = 

 and ^ s | (29) 



<Z(— sin 7/— "^ sinh 7/) -M'(cos yj'— cosh y 2 V) =0 j 



Thus, eliminating d and d', 



y G — 7/ 

 1— cos 7]_Z' cosh y 2 Z' + t^ — 3 — -r sin 7i^ smn 72^ = 0. 



Equations (9) make 



fc?T=^+^), • • • • (30) 



.'. 1- cos 7/ cosh 7/+ — (3+A 4 ) sin y Y V sinh 7 2 Z'=0 (E) 



is the period equation for a free-free bar. 

 Using (29) in (28) we find 



f= D[ (7 2 2 cos 7^ + 7 X 2 cosh 72?) (71 3 sin 7^ — 7/ sinh 7^) 

 — 7i 2 72 2 (71 SU1 7i£ + 72 sinh 72?) (cos yjf — cosh 7 2 Z')] . (F) 



2. Correction for Rotatory Inertia in a Free-free Bar. 

 The period equation omitting A, 4 and substituting 



71 = X(l + iX 2 ) 



7 2 = X(l-iX 2 ) 

 becomes 



cos n (1 + 5X 2 ) cosh ?* (1 — ^X 2 ) — 1 - |X. 2 sin n sinh w = 0. 



cos n cosh ft — j A, 2 (sin n cosh >i + cos n sinh w) 



— 1 — 1\ 3 sin ?i sinh n = } 



where n = \— . 



