TORSIONAL OSCILLATIONS OF WIRES. 437 



Consider unit length of the wire. Let £ be the relative linear displacement per 

 unit length at which a particular group breaks down, and let vd£ be the number of such 

 groups which break in the increment of displacement d£. Then, in the element of 

 volume Zirrdr, the number 2-n-vrdrd^ break down in the increment dg. Let be the 

 angular distortion per unit length of the wire. Then rO is the shear in the element of 

 volume under consideration. Let 



m to - 1 



where m is a whole number. If we assume that a group which breaks at the shear £ is, 

 on the average, formed again into a group which also breaks at the shear £, those 

 groups which break at £ and if will also break at rd. Now take 



r-'+*?-e-<i+i£i)'. 



where p is a proper fraction. 



A group which breaks at £", has had, when the total shear is rd, m — 1 breaks, its 

 last being at (m— l)£" = (m — l+p)%. The shear to which it is subjected, when the 

 total shear is rd, is therefore 



(m-l)(r-0 = (l-^- 



Hence, if we divide the shear £' — £ into an infinite number of equal parts dg, the 

 average value of p is ^, so that the average value of the stretch to which the group 

 which breaks at %" is subjected, when the total shear is rO, is rBj2m. 



Now the number 2Trvrdrd%, when summed over the range corresponding to two con- 

 secutive values of m, becomes 



2rrvrdr 



m(m - 1) 



rO. 



So, if the stress to which a group is subjected when it sustains a shear x is, on the 

 average, hx, the total stress for the above number of groups is 



Trkvr^O-dr 

 m 2 (m - 1) 



And the total stress due to groups which break at shears lying between and rO is 



ehrkvt 2 1 /"W? r = L fogvf -*— , (1) 



•2W 2 (m-iy 4 om 2 (m-l) 



2 (m-l)J 4 2 m 2 (m-l) 



where a is the radius of the wire, and v and k are assumed to be constants. 



If N be the total number of groups per unit volume, the number of unbroken groups 

 is, in the volume 2-irrdr, 



(N-jvdA 



rB 

 li)2irrdr; 



