889 MR- L. X. G. FILOX OX THE RESISTANCE TO 



This will always be the case, provided /8 be greater than the root of tin- equation 



2/3 = cos ft. 



IT 



This root is 53 31' nearly. 



Hence, for values of ft > 53 31' the maximum stress always occurs at the thinnest 

 point of the section. This is the case for the sections of fig. 4 with broad keyways, 

 for which ft = GO . 



lleturuing to the case of sections for which ft < Tr/4 we have OQ given by the 

 equation 



, = scch 



tan 2/3= (cosh 2a +cos2/3) 2 i sech -- + ~- *' 2 2/9 



"- o 



Putting ft = 7T/6 in this we have for the transcendental equation giving in this 

 case 



! v [""I* 16 >'= sech [27 



~- = (cosh 2 + *) [2^ aeoh [2 + 1] 3o + F ^ 



Now, for fairly large values of a,,, we may neglect all terms in this summation 

 except the first. We then have 



TT 23 2 cosh 2 + 1 23 , 



" = (sech a + 2 sech 3ao) 



s 



whence I find a = 1'225. 



Hence for values of a greater than this the maximum occurs at certain points on 

 the contour, given by previously obtained equation in 



15. Calculation of the Position of Fail-Points and the Magnitude of the Maximum 

 Stress for the Sections ft = ir/6, a. = Tr/2 and ft = ir/6, a. = 2ir/3. 



We have, therefore, if we wish to find the maximum stress for the sections a = Tr/2 

 and * = 27T/3 when ft = TT/(J, to solve this transcendental equation : 



TV/JJ siuh 2g (2 cosh 2g + 1) y 2 + 1 aiuh(6n 



6 2 cosh 2 + 1 l ^ i (g + 1) (6n + 5) cosh (6 + 3) 



"=* fsmh(6re + 5) g sinh (6?t + l)f] 1 



,,to I 6 + 5 6;t + 1 J cosh (6re + :!) a 



which may be put into the slightly simpler form 



^v/3~ _ 12 cosh f '= ___ (2<t + 1) siuh (6?t + 3) g 

 6 (cosh 2 + i) ~ sinh 3f B = (6w + 1) (6;i + 5) cosh (671 + 3) 



4- - 1 - "tT I sinh ( " + "^ ^ siuh(6n+l)f _ 1 



cosh f sinh 3g n=0 t(<m + 5)cosh(0/i + 3) ^ " i/./i + 1) cosh (8+3)j ' 



