COMPLEX STRESS DISTRIBUTIONS IN ENGINEERING MATERIALS. 347 
the pipe, and by F the shearing force per unit length of the circumference, the 
equivalent pressure tending to increase the radius will be 
oF 
Na a > 
being the width of the ring, and therefore 
R2 2k é 
“= 4 (P-p +=) ; i ¥ ? a 
The value of P depends on the amount of ‘ shrinkage’ allowed in fitting the rings, 
and on the radial displacement z). Let s be the shrinkage, ¢.e. the amount by which 
the radius of the ring is less than that of the pipe when both are in the unstressed 
state. Then the actual stress in the ring is 
E(s + 29) 
R 
and _ Els ms an)ih i 
nae 4 2 : : : 
Again, applying the method described in the previous paper to the determination 
of the shearing force F, we have 
Em? 8B dz 
B m2—1 12 dx 
which from equation (2) leads to 
Em fe 
m—1_3 \ Et 
iF= “s) oe amet : : =) (dD) 
Substituting the values of p and F given by equations (4) and (5) in equation (3), we 
obtain . 
2 2 3 
PRP ppae a 
cm? —1- 3 
Em? # : 
m—1 3 Hr 
o= 
B(tth) +R. 
_ PR? — cEsh + 1-55 PR? RH 
cE(t+h)+1-55 Eit/ik.H 
The radial displacement z at any point is then obtained by substitution in equa- 
tion (2), and the circumferential stress 
ote 
eS 
In all cases of any practical importance q will be greatest at the point midway between 
consecutive rings, and an expression for its value at this point may be more con- 
_ yeniently obtained by determining the constants A, B, C, D in equation (1), taking 
_ the middle point as origin, so that the boundary conditions are 
l 
z—2,and = 0 forx = + 3 
We then obtain, for the radial displacement at the centre, where x == G, 
PR? PR? ) 
At a —*) fu 
7 
sinh sa cos - -+- cosh S sin 5 
where M = 
MN ieceere Tob ni. nl 
cosh 2 sinh ) + cos 3 sins 
