EQUILIBRIUM OF AN ISOTROPIC ELASTIC ROD OF CIRCULAR SECTION. 971 

 SO that their general solution is 



dz 



V_i = - a-rj — - — -2 + V_.i + |l2_j , 

 rlz 



>(iU_„ dY_„ , ^Tfj 

 dz dz 



(21) 



Here, as throughout the investigation, the symbols U, V, W, Q. with numerical 

 suffixes denote functions of z alone. 



The equations for Uq, Vq, Wq are of the same form as those for u_2, v_2, w~2 and 

 u_i, v_i, ^t;_i. If we write X', Y', Z' and X'^, Y'^, Z/ for the applied forces which arise 

 from the part of the solution already determined, then 



3 dw_, 3%. 



dm_. 





M- 



and 



„., .J .d2U_o dW. dW_\ 



dz^ 



dz"^ 



dz^ ' 



^d^V_ 



""■'''^-^^'-".H' 



dW_n dW 

 ■ + 



- z;= 



Mi'+v'') 



dz 



dz ^ dz 



d^W_, ^dU. dY_, 



From these 



2X' = 0, 2Y' = 0, 2Z' = 7rE' 



2(^Y'-^X') = !^^2. 

 2 dz^ 



d^W_ 



dz^ 



Hence the equations for n^, v^, iv^ are soluble only if 2X = and 2Y = 0. 

 These relations not holding, the given forces cannot be taken in at this point. 

 Hence 



^^^..0, and 'i!S--0, 



dz^ 



dz^ 



and the equations for Wo, v^, w^ are 





and 







D3K) = 0, 83K) + ;.(^^l+,,^l) = 0. 



The solutions of these are 



,d^ 

 1^ 



dz^ 



"o = ^<^{e - v')^^i^ + ^^V-^^ - ^t-^-' + Uo - '?"o > 





— crt; 



dz 



+ Vo + efio>^ 



(22) 



(23) 



(24) 



(25) 



(26) 



(27) 



