284 
DR, G. B. JEFFERY ON PLANE STRESS AND 
In order to investigate these results further we will consider separately the cases 
when the cylinder is subject to either internal or external pressures. There is no 
greater difficulty in the consideration of the general case, should the necessity arise, 
except that the formulae are correspondingly longer. 
A Cylinder under Internal Pressure. 
If we put P 2 = 0, we have on the external surface 
fi/3 =— 4Pj M (cosh a 2 —cos /3){sinh (a l — a 2 ) cos 6 — sinh a 2 cosh (e^ — a 2 )}, 
if d u d 2 denote the distances of the circles a 1? a 2 from the origin, r u r 2 their radii and 
d the distance apart of their centres, so that d = d 2 —d 1 we may show from (2) that 
and 
d 1 = a coth a l5 
r x = a cosech a x , 
do = a coth a 2 
r 2 = a cosech a 2 
d\ = (r 2 —r 2 —d 2 )/2d, 
a 2 - Wi 2 — (n + c2) a } {r 2 
d 2 = (r 2 —r 2 + d 2 )/2d 
-(n-d) 2 }/4d 2 . 
By means of these relations we can reduce the expression for fi/3 to the form 
Zq = 2P x r 2 {r 2 2 (r 2 -2d cos (3) 2 -{r 2 -d 2 ) 2 } 
(r 2j rr 2 ){r 2 — (r 1 + d) 2 } {r 2 2 —(r 1 — d) 2 } 
From this and the obvious inequality d <r 2 — r x we easily see that— 
(l) The numerically greatest stress is when (3 = x, i.e., on the line of centres at the 
thinnest part of the cylinder. This is always a tension if Pj is positive and is 
given by 
2Pin 2 {r 2 2 + r 1 2 + 2r 2 d-d 2 ) , , 
(r 1 2 + r 2 2 ) (r 2 2 —r 2 —2r 2 d~ed 2 ) 
(2) If the centre distance is greater than half the external radius there is minimum 
stress at the points corresponding to cos /3 = r 2 /2c?. This is always negative 
when Pj is positive and we have maximum compressions equal to 
_2P ,r 2 {r 2 —d 2 ) 2 _ 
(n 2 + r 2 ) {r 2 - (r x + d) 2 } {r 2 2 - (r x -d) 2 } 
(53) 
This is always numerically less than the maximum tension. There is a 
secondary maximum at /3 = 0, i.e., on the line of centres at the thickest part of 
the cylinder, which is equal to 
2? 1 {r 2 + r 2 -2r 2 d-d 2 ) 
( r 2 + r 2 ) (r 2 — r 2 + 2 r 2 d + d 2 ) 
(54) 
