EQUILIBRIUM OF AN ISOTROPIC ELASTIC ROD OF CIRCULAR SECTION. 941 

 and, from 30 (5), 



df3 



: = 8 727r-?COsN7r(N7r)l 



(4) 



Thus for a unit element of transverse traction at {a, w', z') (Art. 7), 

 cosNtj-X ) 



^ 



III 



V 



1 



N^ 



TT- 2N27r2 



1 



'<'+^U/.-' 



( 



_N7r 

 a 

 m 



PP = 2mN7rri-A 



sin + 6,1 — 



e! = mN7r(^- 



ap a' 



NV 

 a2 



.„/2p 1 



/* 



Va'' ap 



cos 



sm 



in < — - + 0r,' >e « ^ '??i((o-oO, (5) 



I a a J cos 



e-<^"+*"^^(-)sinm( 



(- 



) sin 7n{(i} - 



-0)'), 



(- 



) sin m{(ti - 



- 0)'), 





cos 7n{o> - 



-co'), 



(- 



) sin m{<j} - 



-0,'). 



(U — w ), 



cos Ntt' 



(6) 



cosNt 

 ■ /xN-^tt^ 



(7) 



For a unit element of normal and longitudinal traction change into . m{ui - w'), 



cos sin 



in 



(5), (6), (7), and multiply by^^^^^, - ™ respectively. 



(8) 



For a unit force at an internal point (jo', w', 2') multiply the results for the 

 corresponding surface force by the coefficients 



■ (9) 



(O or Z force), cos NttI^^) 'cos^^' + Sq^^) • 



It may be remarked that 



(i) the displacement u^ and the stresses pw, zw predominate, so that this is 



approximately a purely torsional mode ; 

 (ii) the coefficient with which the mode occurs in the solution for a unit force is 



of highest order in N when the force is a transverse one ; 

 (iii) for ?n = 0, the mode is exactly torsional, and occurs only for transverse force. 



32. The same for a complex zero of D,,^. z>2'. 



By 30 (2), (10), for a complex zero of D,„, with positive imaginary part, 



^-S^ = a = N7r + ^+i-ilog4N7r, 



^_1 /2 



"a'~nV ^ 



M-t-(^-^°). 



Aj = 4N37r2, Bj = - 4«NV3, Ci = - 4:mW7r\ ] 



A2 = 4-imN27r, E2 = imW7r\ Cg = - iim^K^Tr, l 

 A3 = - 4zN*7r3, B3 = - 4 N^ttS C3 = iimN^TT^, J 



(1) 



(2) 



(3) 



