﻿524 Mr. S. Lees on a Simple Model 



F — x relation will always be given by a portion of the line 

 KL of fig. 4. Similarly when slip occurs with diminution 

 of F, the F — x relation is given by a portion of the line MJ. 

 The actual loop for unsymmetrical stress limits will therefore 

 be obtained by choosing the points L x and K x on these lines 

 respectively, so that the stress at Lj is F + Fi and at J 1 is 

 F — F l9 where Fq + Fj are the unsymmetrical stress limits. 



It will readily be perceived from the diagram that the 

 area of the loop will only depend on the range of variation 

 of stress, i. e. on Fj ; and for a given value of ¥ 1 the diagram 

 is got by taking F o =0 (i.e. for the cyclic condition), and 

 displacing it parallel to the line KL of fig. 4 until the stress 

 limits come right. 



In particular, the results of § § 4 and 5 for the area of the 

 loop in terms of F x will hold, provided we interpret F t as 

 being one-half the greatest variation in stress. 



We may conveniently term the state of affairs here referred 

 to the asymmetric cyclic state. 



For such a state we may briefly indicate the results 

 involved. 



From equation (29), provided our assumptions hold, we 

 see that for no hysteresis to occur, F must always lie 

 between 



Fo± ^%+^), (31) 



where, of course, F may have any fixed value. 



When this condition (31) is not satisfied, the area of the 

 loop is given by (30), where ¥ 1 has the meaning of half 

 the greatest variation of stress. 



§ 8. Particular Case of F = F x . 

 If the lower limit of stress for the asymmetric cyclic con- 

 dition be zero, we get an interesting case. We have here 

 to take F = F l5 and hence Fmaj^^F^ From (31) we see 

 that the greatest value of F for no hysteresis to occur will 

 accordingly be 



2/*N(A 1 + 2\ 2 )/A 2 . ..... (32) 



This is exactly double the maximum stress for the cyclic 

 condition, which just fails to produce hysteresis. 



If the stress varies between and F, and hysteresis does 

 occur, equation (30) shows that the area of the steady loop 

 will be given by 



VNr F 2 / tN(2X, + X 1 ) n _ _ _ (33) 



It is thus proportional to the excess of F over the value (32). 



