Wires of varying Elasticity. 83 



cylinder will remain at rest if originally so. Differentiate (1) 

 with respect to r, and (3) with respect to z ; then using the 

 above simplifications we get 



dr 2 r dr dz 2 ^ ' 



Substituting in ( 1 ) the value of % from (6), we find 

 d 2 u d,, dw _ m + n dS 



dz 2 dr dz n dr 



Substituting for -=- from (4), this becomes 



d?u 1 du u d 2 u __ m dS 



dr 2 r dr r 2 dz 2 ~ n dr ' ' ^ ' 



By a similar method we obtain from (3), 



d 2 iv 1 dw d 2 w _ m dS 



dr 2 r dr dz 2 n dz' ' ' ' ^ ' 



If A, B, C denote constants, obvious solutions of (12), (13), 

 (14) agreeing with the identity (4) are 

 8 =2A + B,a 



u—Ar, V (15) 



iv=Bz + C.) 



Referring to (8) and (9), we see that the corresponding- 

 stresses are 



P = 2mA + (m-?i)B, 



R = 2(m-n)A + (m + n)B, J* ... 



Q = S = T = U = 0. 



If the cylindrical surface be free from forces, we see from 

 (11) that P is zero, and therefore from (16), 



A=-^B, (17) 



"} 



and therefore 



n 

 m 



R=-{'3m-n)B (18) 



If, then, — (3m— n) = M, mid —^ — —a, and if F denote the 



traction, or negative pressure, exerted over the end section of 

 the cylinder per square unit of surface, corresponding to the 

 above solution, we see from (10) that 



F = MB, 



G2 



