Closed and Open Tubes. 529 



The Torque in terms of the Twist. 



Let us take the .r-axis parallel to the tangent at K to the 

 middle line of a thin tube, as indicated in fig. 5. The stress 



Fig. 5. 



at K is 



whence 



Let us now writers for the element of length KK', which 

 is approximately the same as d.v. Then 



~div c 

 n^— - = b + iiry. 

 OS 



This equation is true at any point of the shear line, but 

 — y must be regarded as the perpendicular from on to the 

 line of the shear stress S. Writing — p for y, 



and now we no longer need the axes. 



Now w must return to the same value if we travel once 

 round the shear line. Then, integrating once round the 

 shear line from K back again to K, we get 



— = S— nrp, (33) 



that 



is. 



n 1 ds= 1 Sds — ut 1 pds, 



o= 2iiT-" r (^- 



But pds is twice the area of the triangle OKK'. Writing 

 dA for this, w T e get 



= % (' h -2nrA. 



