(losed and Open Tubes. 535 



function satisfying this is the log function. That is, 



w + iyjr = t'H log, {re id ) + M + t'N 



= /Hlog e ?'-H0 + M + tN, 

 whence 



^=Hlog e r + N. 

 Therefore 



f=filog e r-L Tr 2 + N. 



Now £ has the same constant value over each boundary 

 circle. Thus 



H log (a + K) ~ ir(a + Mf + N = C, 



Hlog(a-iO-iT(«-i0 2 + N = 0, 



whence, by subtraction, 



H loo- — —rat 



°a — ±t 



or H = 



Oo 



Since we are only aiming to get an approximate solution 

 we may as well neglect powers of t beyond the first at once. 



Then, expanding the logarithm in powers of -, we get 



Tj rat 2 



a 

 Therefore 



% = ra 2 \og e r- |rf*+N (51) 



The resultant shear stress, which acts perpendicular to r, is 



<=n(-^+Tr) (52) 



If we write (a + x) for r in this, # being thus measured 

 radially from the mean circle of the section, we get 



S = nr | — a(l+-j + a + #"l 



= nx{2,-J + ....}, 



which becomes, on neglecting powers of a; beyond the first, 



S = 2nrx, (5; , >) 



exactly as along a straight strip. 



