The Tovf^ion Problem for Bodies of Bevohdion. 
Taking the stream -line i/'- = — 2 as one of the boundary lines of the 
body of revolution in a meridian plane we get a body such as shown on 
Plate XV. 
If ^ -> ± oc, the stress-lines become parallel to the axis of the body, 
and if 4^- = j^. be a definite stress-line, then its distance from the axis 
at = + oc is given by 
- _ ^ _vjr_^_^ ^^^^ 
i. e. ri = 
Again, for z = — oc we have 
45;- + 16a A 
V, 
= 4«A, 
where ro is the distance of the stress-line 4/- = constant from the axis at 
2; =r — oc. This u'ives 
4,^5 — 16aX 
V Vo 
In the special case, where we took v,, = 64, X = 1 , we have 
For 4^- — - 2, = 707, r.-, = 0, which corresponds with the value got 
from Plate XIV. 
Further, we note that if a sufficiently great negative value is taken for 
then the stream-line approximates to a straight line parallel to the .^-axis, 
and hence we get the case of a circular cylinder with a hole bored in 
at the one end, the equation of the hole in a meridian plane being given 
by ^5 = - 2. 
If = — 62, i\ — 1*414 and v., = 1'392, the difference between the 
radii then being "022. 
In such a body, then, the stress-components are given by 
_ _ _ f 1 1 '[ 
' - M { + «) 2 + r ^ }f {{z _ a) ^ + r }t ) ■ 
The displacement ve is given by 
lie = (<A + c). 
12 
