Theory of Elasticity. 283 



systematically overstrain them to make them straight, then use 

 them, having a care to keep the right side up. 



A very similar but simpler case is that of a circular shaft 

 subject to a twisting moment, in which the stress does not vary 

 simply as the distance from the axis, as in the ordinary theory 

 of torsion without primary stress, but in such a manner that, 

 under a given twisting moment, the stress increases at first 

 very rapidly as we leave the axis, but thereafter less and less 

 rapidly, so that at last it is not increasing at all as the outer 

 surface is reached.* Here again the distribution of stress. 

 imagined would considerably increase the resistance of the 

 shaft, say ten to fifteen per cent. Such a distribution would 

 result from systematically overstraining a shaft and suggests 

 the preparation of shafts to be driven in one direction only, by 

 such preliminary overstraining. 



The three foregoing illustrations have not only shown us 

 some solutions of the equations of equilibrium other than those 

 ordinarily recognized, and their bearing on actual cases, but 

 have further shown how it may be advantageous to introduce 

 definite primary stresses in certain cases, and how these stresses 

 may be produced. This has long been appreciated in gun con- 

 struction but apparently not elsewhere. The illustrations 

 explain to us, moreover, why, in part at least, the elastic limit 



*Let the shaft be of radius r Q and subject to the twisting moment M. Let the 

 shear vary with the departure from the center according to the law 



— — r I r 2 \ — 



S= S — - 1 3 : where S is the shear at r = r 



2r V r 2 / 



This makes = S — -i 3 s- ) or zero at r = r , that is, makes & constant 



dr 2r \ r 2 / 



as the outer surface is reached. We have 



p+r r / r 2 \ 7 — „ 



whence S~ = - 1 ^, S~= ^r(s - —, A We may take M constant and F— 



Wo 3 777To 4 V ?0 ' 



~Q = lR '= T = ~U=z 0. The ordinary solution has S = S — 



1 1M IMr 



M= I " rS —2 nrdr = —rrS r 5 S = — -„ S = — ■ 



2 TTTV 7Tr 



-r*k 



The primary or self balancing shear which would remain if the moment M were 



removed is (S) = S— S = , ( 4-6 -=■ ) which is zero at r = r V $, within 



v ' 77rr 4 V To/ 



this has the same sign as ~S and without has opposite sign. We note that 

 / ° r(S)2Tir dr = as it should. Under the distribution of stress here con- 

 sidered the same moment is supported with a maximum shear but six-sevenths 

 that for a shaft without primary stress supporting the same moment. 



