350 -MK. L. N. G. FILON <>.\ THK KKS1STANCK TO 



Hence we have, roughly, 



If the reduction oi torsional rigidity were simply proportional to the deptli of the 

 key way, the last two results ought to be 26 '4 per cent, and 5 '3 per cent, respectively. 



It may be objected to these deductions that in the above sections, the fact that the 

 stress at the vertex of the key way is infinite, violates the physical conditions assumed 

 by the theory of elasticity and renders our results untrustworthy. 



My answer to this is that these cases should really be considered as limiting cases. 

 If, instead of considering a section where the keyway is actually a straight line, we 

 consider a hyperbola with a very sharp bend, we can easily ensure, provided the 

 angle of torsion be not too great, that the physical conditions shall not be violated, 

 and, on the other hand, the values of the torsional rigidity (since they tend to a finite 

 limit) will differ but little from the values obtained above. The very fact that the 

 torsioual rigidity tends to a finite limit shows that, even in the extreme case, the area 

 where the physical considerations are violated is infinitesimal. 



Finally, in drawing conclusions from such results we shall only be following the 

 example of SAINT- VENANT and of THOMSON and TAIT, who have not hesitated to use 

 results found in a precisely similar way for such keyways cut into a circle. 



23. Values of the Stresses. 



I have also determined the stresses at three points on the boundary of the section, 

 in order to find where the stress was greatest. The points selected are denoted in 

 figs. 6, 7, and 8 by the letters A, B, C. 



A is the vertex of the hyperbola, B is the point opposite to A, and C is the point 

 corresponding to 17 = tr/2, g = a. 



A is given by = 0, 17 = ft, and B by = , 17 = 0. 

 If S A , SB, S c denote the corresponding stresses, I find 

 g 

 = cos ft (1 sech 2) 



(J.TC 



( 1 \ foriV, 



8a (cosh 2 - cos 20) ; ( 



2 



11 = 



2 



= tan 2ft sin ft 



8/3 (cosh 2 - cos 2/3) "=* 

 sin/3 n f 



sech 



'2n+ 



. - (34), 



