95 



(61) 



Solid Cylinders of Elliptic Section. 

 and we obtain the simple result 



If </>i> 02 be the eccentric angles of the points of section in 

 the positive quadrant, and 1? 6 2 be the corresponding vec- 

 torial angles, then 



& + & = 7r/2, (62) 



1 + <9 2 = tan- 1 (??)- | =2<H)say. ... (63) 



Thus the eccentric angles are always complementary, and the 

 arithmetic mean of the vectorial angles is independent of the 

 shape of the disk. 



§ 28. When the two points of section coincide and the 

 curve of no radial strain touches the rim, we at once deduce 



• • (64) 

 . . (65) 

 . . (66) 



The angle 20, it may be noticed, is the same as the inclination 

 O to the major axis of the tangent at the centre to the curves 

 of no radial displacement and strain in the limiting form of 

 the disk for which b/a = 0. 



When b/a exceeds b 3 /a, the area of radial compression ex- 

 tends uninterruptedly round the rim of the disk. The values 

 of bja and of the corresponding vectorial angle © — answering 

 to the point of contact of the curve of no radial displacement 

 and the rim in the positive quadrant — are given approximately 

 in Table X. 



</>=t/4, 



^=©=|tan- 1 (^)-^, . . 



b/a= \/l +77 — */r) = b B /a say 





Table X 



. — Values 



of bz/a and ©. 





r) = 



•1. 



•2. 



•25. 



•3. 



•5. 



bja= 



e = 



■733 

 36° 13|' 



•648 



32° 57' 



•618 



31° 43' 



•592 

 30° 39' 



•518 



27° 22' 



When b/a is less than b$a the curve of no radial strain 

 crosses the rim in the two points in the positive quadrant 

 whose vectorial angles 1? 6 2 are approximately as follows : — 



