282 Cilley — Fundamental Propositions in the 



pressure. Such a tube would evidently support much larger 

 pressures without overstrain than would a tube not so prepared. 

 As another illustration let us consider a rectangular beam 

 and let us suppose that on a cross section under a given bend- 

 ing moment the longitudinal liber stresses do not vary simply 

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

 of beams, but that they vary according to a law which makes 

 them at first increase very rapidly as we leave the neutral axis, 

 hut thereafter less and less rapidly, so that at last they are not 

 increasing at all as the outer fibers are reached."* It will at 

 once be seen that such a distribution would utilize the material 

 of the beam in a particularly effective manner. Not only the 

 outer fibers but the other fibers for some distance in would be 

 acting with nearly their full effective strength, and so, with the 

 same maximum fiber stress the moment resisted would be 

 largely increased, or the same moment would be resisted with 

 a considerably smaller maximum fiber stress. An increased 

 resistance of from twenty to twenty -five per cent could be 

 obtained from such a stress distribution, of course with a pro- 

 portionally larger deflection. Such a stress distribution is 

 similar to that which would result from systematically over- 

 straining a beam. It would perhaps pay to roll beams curved, 



* Let the beam be of depth 2a and of width 2b and resist at the section under 

 consideration the moment M. Let the fiber stress vary with the departure from 

 the neutral axis (central) according to the law 



R = Bo — ( 3 I where R = R at x = +a 



2a \ a 1 i 



&R i / 3^2 \ 



This makes - — = R - — ( 3 — ) or zero at %= ± a, that is it makes R 



dx 2a V a 2 / 



constant as the outer fibers are reached. 



We have M = / " ~R ~ ( 3 - %-) bx dx = %R~ tfb 



J - a 2a V a 1 J 



— 5Jf — 5Mx / x 1 \ 

 whence R = -^ or R = ^- (3 - ^) 



The stress by the ordinary solution involving no primary stress would be 



x /*^~ a x 2 3 M 33fx 



R = R — whence M= / R — bxdx = — i? a 2 & and R = r 57 and R= -—^r 



a J — a a 



Thus the primary or self balancing stress which would remain if Mwere removed 

 Mx 



has the same sign as R and without has the opposite sign. We note that 

 b (R) dx = and / b{R)xdx = as they should. Under the distribution 



a J -a 



of stress here considered the same moment is supported with a maximum fiber 

 stress but five-sixths that for the beam without primary stress supporting the 

 same moment. 



is (R) = R — R= - n —ls — j which is zero at x = ± a V f , within this value 



