EQUILIBRIUM OF AN ISOTROPIC ELASTIC ROD OF CIRCULAR SECTION. 901 



The equations for A, B, C are 



- Po = A(v - 2/32 J + 2yS3J') + B2^2j" 



-0„ = A(-m/32J) 



«2 



+ C(2?n/3J'-2mJ), 



+ B( - 2m/3J' + 2?«J) + C( - yS^J" + f3J' - m'^J), 



H- 



Z(, = A( - 2y82j + vySJ' + ??i2J) + B( - 2;8J') 



+ C( - mJ), 



(4) 



where the argument of the function J and of its first and second derived functions J' 

 and J" (written for brevity instead of J,„, J J, J,„"), is /8. 



The determinant of the equations (4) will be written throughout the paper in 

 the form 



a 





(5) 

 (6) 



^3, mi "3,7;i) ''S.i/i 



The first minors of D,„ will be written Ai ,„, Bi ,„, Ci,,„ ; A._, „, , etc. ; 



but tti, Ai, etc., will often be written instead of c<i ,„, Aj „j, etc. 



We may also note here that the functions of m and /3 obtained when J, J', J" are 

 replaced by G, G', G'' * in 



^2, 7H ) ^2, til ) ^2. r. 



*3, »K ) "3, 7(1 ) ^3, III ^ 



*l,m ) /-'l.m ) yi.i 



will be denoted by a.^,,,,, /3.i,i«> T-',) 



*3, 711 ) Ps, m ) 73, > 



(7) 



The solution of equations (4) is 



A = - ~(PoA,,,„ + 0,,A2,„, + Z,A3.7„), ^ 

 /*-L)„i 



B = *- J-(P„B,,„, + n„B,,,„ + Z„B3,,„), 



/A 



D,. 



a2 1 



C = — =— (PoCl, ,„ + i^oCs, 7,1 + ZuCg,,,,). 



(8) 



M-L'm 



The problem (1) is thus solved, unless (3 is a zero of D^^. To get a solution for this 

 exceptional case, let A', B', C' be written for the coefficients of ^=— on the right of 

 (8), and suppose to begin with that /3 is not a zero of D^. Then 



\\l; j = i C' 7'"^' " sin '"(" - "^ ) Sive, at p = a, 

 /pp p5,\ ^ 1^ p, ,J3Z, ,\ -f cos ^^^ _ 



Differentiate as to /8, and then let /3 become a zero of D,^. 



Thus 



M, ^,\ 3 / /A', B',\. /?P J-^ ) cos , ,, . , \ 



rVrS^iv C' 7"'«' "fsin'"^^-'") give,atp = a, 



(9) 



PP^P-A = ^dBJdd^o '/^o A,-f cos _ ^, 

 , pa) J a^ ' '^V Oq / sm ^ ^ 



(10) 



* Art. 15, footnote. 



