THE SPRING CLUTCH , 729 



fibers above the neutral axis will be compressed while those below 

 the neutral axis will be stretched. Let y be the distance of any given 

 fiber from the neutral axis. The length of the undistorted fiber will 

 be L = {Ri + 3')^! while after bending this same fiber will have the 

 length L' — {Ri + y)Qi- The strain or the change in length per unit 

 length is 



L-V ^ _ {R2 + y)d, 



L (R, + y)e, ^""^ 



Since along the neutral axis there is no change in length, 9i = L0/R2 

 and 61 — Lq/Ri. Substituting these values in equation (9) gives 



Strain = {y/R2)(R2 - Ri)/{Ri + 3')- (10) 



The potential energy per unit volume in a material strained in tension 

 or compression is 



W/V = (E/2)(Strain)2. (11) 



Substituting the value of strain from equation (10) in equation (11) 

 the energy density at any point of the deflected wire will be 



W E 



y{R2 - Ri) 



V 2 I R^iRi + y) J 



(12) 



Let b represent the width of the wire at the point y. Then for a wire 

 symmetrical about the neutral axis the strain energy per unit length 

 of the wire becomes 



r^l^ET y(Ro - RQ 



'^bdy, (13) 



where the factor (Ri + y)/Ri represents the ratio of the length of the 

 fiber at the point y to the length along the neutral axis. If all values of 

 y, and hence also h/2, are small compared to Ri there will be little error 

 made in neglecting the y, which carries both positive and negative 

 values, in the expression Ri + y. Hence 



W C''^E( R^- R, \\ ,^ ,.,. 



T = l^,,2[-R^)'y''y- ^''^ 



The integral of hyHy is the area moment / of the section and therefore 



T=l-(^')- 



This must be equal to the work done per unit length of the neutral 



