Updegraff — Flexure of Telescopes. 

 Differentiating Eq. (25), 



255 



dx 



'© 



aW A (w) — 



x 2 J x (u) 



3 



Differentiating (26) and reducing, 



(26) 



(27) 



Combining Eqs. (25) and (27) to eliminate the constant of 

 integration C we get, 



— £ = — axy — bx 2 , 



dx i 



dy 



which shows that Eq. (25) is correct. 



From the conditions of the problem we know that ^- = 



for x = I, the half length of the tube, and substituting I for x 

 in Eq. (26) we have, 



ty ° 



(I) 



I 2 J JL (u 1 ) — a^lJ^Uj) 



3 3 



= 0, (28) 



2 A i 



in which u L = «7 2 a , 



Eqs. (25) and (28) then give on eliminating C, 



V = 



x 2 J- L {u) 



I 2 J ± (u 1 ) — a 2 / e / i (M 1 ) 



(29) 



which is the approximate equation of the neutral axis of the 

 upper half of the tube. From Eq. (20) we deduce in the 

 same manner, as the equation of the elastic curve of the lower 

 half of the tube, 



