XV
263
STRESSES IN A RIGHT CIRCULAR CYLINDER
r = 2R cos (p,r),
cos (p,r)
cos (n,r)
=
r
=
1
2R
so that this part of the integral
=
=
1
Rπ
1
2Rπ
F cos (fr) cos (m,r)ds
{F cos (f,m)+F cos [(f,p)-(m,p)]}ds,
-
since (f,p)+(m,r)=(f,m); (f,r) — (m,r) = (ƒ,p) — (m,p).
Now since the forces F must not produce a displacement
of the cylinder in the direction m, nor a rotation round the
axis,
F cos (f,m)ds = 0, F sin (f,p)ds = 0.
[F
Hence the part of the integral examined is equal to
1
2RT
cos (m,p) F cos (ƒ,p)ds,
(m,p) [1
and thus cancels the first term in M,; and M, reduces to
that part of the integral which is due to the part of the
curved surface near the element considered. Here we have
rd(p,r) = ds cos (p,r). Hence
cos (n,r)
ds
cos (p,") ds = d(p,r),
7°
r
and as F, ƒ may be regarded as constant over the smal-
portion of surface considered, we have
Mn
=
2
+
ㅠ
​F cos (fr) cos (m,r)d(p,r)
=
-
211
π
+
Fcos [(r,p)-(ƒ,p)] cos [(r,p) − (m,p)]d(p,r)