286 
DR. G. B. JEFFERY ON PLANE STRESS AND 
Hence if the centre distance is less than half the internal radius the compression in 
the inner surface decreases steadily from a maximum at the thinnest part of the 
cylinder to a minimum at the thickest part; otherwise there is a minimum at each of 
the points and maxima at the points corresponding to cos /3 = — rj2d. Similarly if the 
centre distance is less than half the external radius the compression in the outer 
surface decreases steadily from a maximum at the thinnest part of the cylinder to a 
minimum at the thickest part ; if the centre distance exceeds this value the 
compression is a maximum at each of these points and minima at the points corre¬ 
sponding to cos (3 = rJZd. 
If in these results we put d = 0, we have, for a concentric tube under internal 
pressure, tensions at the inner and outer surfaces which are respectively 
rj + r 
r 2 2 —r 
, 2 
i_ 
. 2 
1 
Pi, 
2 Vy 
r 2 2 -rj 
while for a tube under external pressure the compressions at the inner and outer 
surfaces are respectively 
r, 
Ozv» 2 
2 p 
2 „ 2 X 25 
1 ' 1 
r_l + rl 
ro-rd 
P 2 , 
These are the well-known formulae for thick tubes. 
6. A Semi-infinite Plate with a Circular Hole Subject to a Uniform 
Normal Pressure. 
If in the results of the last section we put a 2 = 0 and P 2 = 0, we have the solution 
for a semi-infinite plate containing a circular hole, which is subject to a uniform 
normal pressure, and bounded by a straight edge which is free from stress. 
We have on the boundary of the hole 
/3/3 = — Px + 2 P l cosech 2 <x x (cosh 2 ol x — cos 2 /3) 
and on the straight edge 
/3/3 = — 2Pi cosech 2 a. x (l — cos f3) cos f3. 
If r is the radius of the hole, d the perpendicular distance of its centre from the 
straight edge, and x the distance measured along the straight edge from the foot of 
the perpendicular, 
d = a coth a, r = a cosech a u d 2 —r 2 = a\ 
and 
x — a sin /3/(l — cos /3). 
We have therefore on the straight edge 
~4Pj 
r 2 (x 2 —d 2 + r s ) 
(x 2 + d 2 —r 2 ) 2 
(61) 
