/        r0        (r_x) squared + R squared \
t_x = T ( 1 - ---------- . ------------- )     (2)
\     R0 + r0         (r_x) squared   /

The greatest value t_x = t0 corresponds to the surface of the bore and must be t0 = -T, therefore

r0 squared + R squared
--------------- = 2
r0 (R + r0)

whence P0 = T sqrt(2) = 1.41 T.

From the whole of the preceding, it follows that in a homogeneous cylinder under fire we can only attain simultaneous expansion of all the layers when certain relations between the radii obtain, and on the assumption that the maximum pressure admissible in the bore does not exceed 1.41 U.

Equation (2) may be written thus—­

R       r_x — Rr
t_x = T -------- . ----------        (3)
R + r0     (r_x) squared

Substituting successively r_x = r0 and r_x = R, we obtain expressions for the stresses on the external and internal radii—­

R — r0                     R   R — r0
t_R  = T --------  and  t_r0  = -T ---- --------
R + r0                    r0   R + r0

Therefore, in a homogeneous hollow cylinder, in which the internal stresses are theoretically most advantageous, the layer situated next to the bore must be in a state of compression, and the amount of compression relative to the tension in the external layer is measured by the inverse ratio of the radii of these layers.  It is further evident that the internal stresses will obey a definite but very simple law, namely, there will be in the hollow cylinder a layer whose radius is sqrt(R r0), in which the stress is nil; from this layer the stresses increase toward the external and the internal radii of the cylinder, where they attain a maximum, being in compression in the internal layers and in tension in the external ones.

The internal pressures corresponding to these stresses may be found by means of very simple calculations.  The expression for this purpose, reduced to its most convenient form, is as follows: 

R      /  R     \   /      r0 \
p_x = T -------- (  --- - 1 ) ( 1 - ----- )     (4)
R + r0   \ r_x    /   \     r_x /

In order to represent more clearly the distribution of stresses and pressures in the metal of a homogeneous ideally perfect hollow cylinder, let us take, as an example, the barrel of a 6 in. gun—­153 mm.  Let us suppose T = 3,000 atmospheres; therefore, under the most favorable conditions, P0 = 1.41 T, or 4,230 atmospheres.  From Equation (1) we determine R = 184.36 mm.  With these data were calculated the internal stresses and the pressures from which the curve represented in Fig. 1 is constructed.  The stresses developed under fire with a pressure in the bore of 4,230 atmospheres are represented by a line parallel to the axis of the abscissae, since their value is the same throughout all the layers of metal and equal to the elastic limit, 3,000 atmospheres.  If, previous to firing, the metal of the tube were free from any internal stresses, then the resistance of the tube would be