of varying Elasticity . 263 



Also from (12) and (14), 



A ' = -B(<r 1 -a)n 1 a 1 'b\a:^a 1 2 )A 2 , 1 



A/= B{a l -a)na; 2 a 2 {a 1 2 -b 2 )^. S K } 



If F denote the total amount of traction, or negative pressure, 

 applied over the end z= I of the bar, we can determine B from 

 the condition 



F = 7r{(a 1 2 -6 2 )R + (a 2 -a 1 2 )R 1 }. 



n 

 m 



Referring to (4) and (6), using (15), and denoting — (3m — n) 



by M, and — ^dm l —n l ) by M 1? we find very simply 



F = ttB[M(« 1 2 -6 2 ) +M 1 (a 2 -a 1 2 ) 



+ 4( - 1 -cr) 2 ^^ 1 a 1 2 (a 2 -a 1 2 )(a 1 2 -5 2 )A 2 ] ; (17) 



which determines B in terms of given quantities. 



It should be noticed that, strictly, F should be distributed 

 so that there should be at every point of the terminal section 

 in the one material the traction R, and in the other the trac- 

 tion Rj. In any practical case this is not likely actually to 

 occur ; but if the traction be applied symmetrically so that its 

 resultant acts along the axis of the beam, and if the radius of 

 the cross section be small compared to the length of the beam, 

 the above solution may be regarded as sufficiently correct, 

 except perhaps for points close to the terminal section. 



From (15), (16), and (17) all the constants of the solu- 

 tions (1) and (2) are expressed directly in terms of F and 

 other given quantities ; thus these solutions are complete. 



If, as in the case of ordinary wire, the cylinder be solid to 

 the centre, we have 6 = 0, and so 



— - = na 2 a, 2 , 

 and 



— =na L 2 [> + ^ {rn^-a 2 ) H-mc^ 2 }]. 



Thus the expressions (15) are comparatively short ; while 

 from (16) A x = 0, and A/ is a simple expression. 



In the event of a 1 and a being equal a great simplification 

 occurs. From (16) we see that A' and A/ identically vanish, 

 while from (15), 



A=-B<7, A^-Bo-^A. 



Also from (17), 



F = 7rB[M(a 1 2 -6 2 ) + M 1 (a 2 -a 1 2 )]. . . (18) 



