Closed and Open Tubes, 531 



A Tuhe whose boundaries are similar Ellipses. 



If we apply equation (34) to the tube whose section is 



bounded by the simils 



ir ellipses 



and J + ^ = l + 2* (39) 



A- being very small compared with a or &, we find, neglecting 

 all powers of k above the first, that 



U = d0(a 2 sin 2 + Z> 2 cos 2 0)§ . . . . (40 ) 

 t=kab(a 2 sm 2 + b 2 cos?0)-*. . . (41) 



6 being the eccentric angle. 

 Hence 



ds _ Tr(a 2 + b 2 ) 



" t kab 



(42) 



Now equation (34) gives 



r\ k°b i r in- 



7r{a 2 -tb 2 ) K ' 



4z7r?ikra z b' d 



+ // 



(43) 



The precise value of the torque can be obtained by taking 

 the difference of the torques on the complete ellipses. This 

 method of differences is justified because the shear lines in 

 an elliptic section are similar ellipses. Thus, since the torque 

 on an elliptic section is 



Q = 7rnT ., . /0 , (44) 



a~ -j- It- 

 it follows that the torque in the tube is 



W \ (aP + VXl + k)* -tf + b*j 



=™-^$jJ( 1 + *) 4 - 1 }> («) 



which becomes, when powers of h above the first are 

 neglected, 



r\ ±<r ] b 3 k 



®=™ r tf + V> (4 "> 



exactly as by the approximate equation (34). 



