55 and the bending moment at B' of the deflected ring is: M=Mo+VoA'C-Mpo
(2.69)
where:
MPO
=
bending moment (per unit length) due to the external pressure Po at any section of ring. bending moment about O.
Mo =
From Fig. 2.9(b), the vertical and horizontal components of force po are given by: cosa
V-/pods
(2.70)
H - / pods sin c~
(2.71 )
Referring to Fig. 2.9 (b), the bending moment due to pressure po, i.e., Mpo, at any point on the arc A ' B ' can be expressed as: Mpo-
/podscosc~(A'C-z)+/podssina(B'C-y)
A'C -
-
ax=0
[B'C p~ (A'C - x) d , + , y = o
Po
p~ (B'C - y) dy
(2.72)
2 (A'B')2
Substituting Eqs. 2.67 and 2.72 in Eq. 2.69 and applying the laws of cosines one obtains: M - Mo - po --f (
2 _ -X70 2 )
(.2.7a)
However, substituting O B ' = r* + u, and A'O = r" - Uo into Eq. 2.73 and then neglecting the squares of small quantities u and Uo, the bending moment becomes: M=Mo-por*(uo-u)
(2.74)
Finally, substituting Eq. 2.74 into Eq. 2.66, the final expression of the differential equation for the deflected ring becomes: