262
XV
STRESSES IN A RIGHT CIRCULAR CYLINDER
The integrations extend all round the circumference of the
cylinder.
Proof. We shall first show that the expression (1)
represents a possible system of stress. Let x and y be
rectangular coordinates in a plane perpendicular to the axis
of the cylinder; then the pressures X, Y,, Yr, which are
independent of the third coordinate, form a possible system if
they satisfy these differential equations
ǝy, ǝyǝX, Ə²Y.
ΟΥ
+
0:
=
ΟΧ ƏY
+
дх dy
ay x
x
0
=
›
y
"
+
მე by dy2 dx²
I
Ə²Y
= 2
дудх
(2).
Let P, o be polar coordinates, p cos w = x, p sin w = y, and
in the stress-components P., P, let a denote the direction
perpendicular to w; then the system of stress
Pp
=
COS @
P=0, = 0·
(3)
P
satisfies the equations (2). This is proved by calculating the
values of X, Y, Y, which follow from (3), and substituting
them in (1). The three equations (3) may be replaced by the
one equation
Mn
M₁ =
cos o cos (n,p) cos (m,p) cos (x,p) cos (n,p) cos (m,p)
P
P
A sum of such M, with different poles and multiplied by
arbitrary constants will represent a system which satisfies the
differential equations (2). Now the integral which occurs in
(1) is such a sum, and as the expression in front of the
integral merely represents a constant pressure p uniformly
distributed through the cylinder, it follows that the system
expressed by the equation (1) is a possible one.
n
We shall prove secondly that when we approach infinitely
near to the curved surface and make the direction n coincide
with that of the radius p, then M, coincides with the
component of F in the direction m; so that M₁ = F cos (m,n).
For this purpose we separate the integral into two parts, one
relating to the portion of the boundary infinitely near the
element considered, the other to the remaining more distant
portion. For this latter, and for it alone, we have