﻿520 Mr. S. Lees on a Simple Model 



For the point M, the values of x and F are clearly the nega- 

 tives of (23) and (24) respectively. 



The value of the maximum slip (measured from the 

 neutral state) can easily be obtained, for this must vary 

 between (say) y x and — y lm The range of slip is accordingly 

 2y 1? and this range of slip is incurred during the description 

 of both the lines KL and MJ. Thus 2y x must equal the 

 difference between the values of x at L and K, i. e. 



__ F 1 -(X 1 + A 1i > 1 _ (X, +2\ 2 > 1 -F 1 



4/i ~*i-- -^- ----- x 2 • ■ V*) 



This result can also be obtained from equations (12). 



Yet another expression can be obtained by remembering 

 that at the point L, where the slip is y lt , we must have 

 from (12) 



x 2 Oi— yd = ^N. 



Substituting the value of y given by (25), we get * 



F 1 -# 1 X 1 =2/*N. ..... (26) 



§ 5. Area of Hysteresis Loop* 



The area of the hysteresis loop can now be obtained. It is 

 clearly twice the area of the triangle JKL, the coordinates 

 (x, F) of the vertices of which are respectively (— x 1} — Fj), 



(Fj — X x + X 2 . x x \i 4- X 2 . F-, — X 1 .h 1 + 2X 2 . %i 

 X 2 X 2 



and "(#!, Fi). The area f is therefore 



— X X y — F 2 , 1 



F 1 -(X 1 +X 2 )^ 1 F x (Xt +X 2 )-X 1 (X 1 + 2\ 2 )x 1 ' 



)• 



X 2 Xj 



x ± , F 2 ,1 



= 2(F 1 -^ 1 X 1 ) [(Xi + 2X2>i _ Fi] = Sfl xy u . (27) 

 x 2 



on using (25) and (26). 



• This result might also have been obtained by remarking 

 that the work lost in a complete cycle can be accounted for 

 as due to a force of friction 2/jlN overcome twice (during 



* When hysteresis exists, the locus of points L is therefore a straight 

 line. 



t The negative sign is used so as to make the area positive. 



