﻿to Illustrate Elastic Hysteresis. 533 



the states (F x , .<•), ( — F l9 — x { ) respectively ; whilst the cor- 

 responding slips are y { and —y x respectively. We shall 

 suppose that y l is less than /*/2a, so that the cyclic condition 

 is possible. By (42), we shall have 



g, Ql 2 -2aN ) (Xa.fx-zxNo) _ 



^ 55* + 2^3 * 2a/3 - U ' (45) 



Fj is then given by 



F 1 =\ 1 x 1 + 2\ 2 {x 1 -y l ) = (\ 1 + 2\ 2 )x l -2\ 2 y l . (46) 



Now imagine F diminished, corresponding to the path LM. 

 At first, we shall have 



g=(\ 1 + 2\ s ), (20) 



and this will go on till slip occurs at M. If the value of x 

 at M. be # M , the value of T 2 there will be 



(T 2 ) M =(K -(3y 1 *)(fi-2ay 1 )-\ 2 (x 1 -x K ), . (47) 



which must be a compression sufficient to cause slip. Hence 

 also 



-(T 2 ) M =(N -/3 i/l 2 )( / , + 2ay 1 ). • • • (48) 



From (47) and (48) we get 



X 2 (« 1 -^m)=/*(N -.% 1 2 ). . . . (49) 



For a given a?i, (45) determines y h and then (46) gives F x . 

 From (49) we can find x^i, and the appropriate value of F at 

 M is easily obtained, since the point L and the slope of LM 

 are both known. The points J and K are got by symmetry 

 from L and M respectively. 



To get the shape of the curve KL (and therefore by sym- 

 metry JM), let at any point of the curve the slip be y. Then 

 (42) holds and gives y in terms of x, say y=f 1 (x). Also 



F = \ s x + 2T a = \ 1 x + 2($ -/3y*XfjL-2ay) ; . (50) 



hence on eliminating x between f\{x) and F, we get the 

 required F — x relation for the curve KL. Without going 

 into the algebra, it is quite clear physically that the shape 

 must be something like that shown. 



The area of the complete loop is most easily obtained by 

 observing that energy is lost on the whole, only during the 



