486 13 Stability of Elastic Systems
Thus, differential equation (13.13a) becomes
d^2 w
d^2
C
1 C
qR^3
EI
wD
k'
EI
R^2
sin ̨
sin: (13.15)
Denote
n^2 D 1 C
qR^3
EI
: (13.15a)
CD
k' R^2
EIsin ̨
: (13.15b)
Pay attention thatCis unknown, since the angles of rotation'of the supports are
unknown. Differential equation (13.15) may be rewritten as follows
d^2 w
d^2
Cn^2 wDCsin: (13.16)
Solution of this equation is
wDAcosnCBsinnCw: (13.17)
The partial solutionwshould be presented in the form of the right part of (13.16),
mainlywDC 0 sin,whereC 0 is a new unknown coefficient. Substituting of this
expression into (13.16) leads to equation
C 0 sinCn^2 C 0 sinDCsin;
so
C 0 D
C
n^2 1
:
Thus the solution of equation (13.15) becomes
wDAcosnCBsinnC
C
n^2 1
sin: (13.17a)
Unknown coefficientsA, B,andCmay be obtained from the following boundary
conditions:
1.For point of the arch on the axis of symmetry.D0/, the radial displacement
wD 0 (because the antisymmetrical form ofthe loss of stability); this condition
leads toAD 0 ;
2.For point of the arch at the support.D ̨/the radial displacement iswD 0 ,so
Bsinn ̨C
C
n^2 1
sin ̨D0: (13.18)