﻿104 Mr. A. E. Young on the Form of a Suspended 



The point of contrary flexure can be found by solving for x 

 or x 1 from the equation 



al cosh a I s — x' j 

 d*n \ 2 I 



2 sinh 

 hence 



log £ | 2-30Iog 10 | Z 

 #' — — = — =3*6 inches in the example. 



In this short distance the effect of the stiffness on the slope 

 has decreased from 9° 9' to about 0''*5, which shows how 

 rapidly this effect dies away. 



If we make a'= ~ in the expressions for y we obtain the 



deflexion of the middle point of the tape below the ends, and 

 if we make T = in these resulting expressions w r e should 

 obtain the deflexion of the middle point of a beam loaded 

 uniformly and supported at the ends either free or clamped. 

 In the first case we have 



wx 2 w cosh ax— 1 wl 2 w /' , al 



2I 



cosh-^ 



wl" w / _. , al\ 



= 8f -^l 1 - se %J 



ioP . w /' , , d ! P 5a 4 / 4 , . . , . „ A 



SlT ^\* ~ * + T ~ "SS* + terms mvolving To ) 



5 wV 

 384 EI 



when T =0. 



In the second case, we have 



us rZ 2 Z /aZ a 3 Z 3 \1 



'■" r F I 8 ~~ 9a V 4 ~ f9"9 + termS involvin & a * and higher P owers ;J 



when T =0. 



384EI ' 



These results agree with those given in books on Applied 

 Mechanics. 



