BULETINUL INSTITUTULUI POLITEHNIC DIN IAŞI

20 Ion Crăciun
Ψ r
Having assumed that in (30) only does not vanishes, we obtain the following
representation for a regular solution in B of the governing Eqs. (30) of the
second axially-symmetric problem
0
⎧ uθ = −2 α
3
∂zΨr,
⎪
0 0 0 2 −1
⎨φr
= {
2 3− [( β+ γ−ε) 2−4 α ] ∂
r( r +∂r)
} Ψ
r,
⎪
2 −1
⎪⎩ φz
= − {[( β+ γ−ε) 2
−4 α ] ∂
z( r +∂r)
} Ψ
r,
where the stress function
Ψ z
satisfies the differential equation
(45)
0 0
3
Ω Ψ + Y = 0.
(46)
r
r
For Y = Y = 0, the relationship between the stress functions Ψ and Ψ takes
r
z
0
the form
3
( ∂
rΨz +∂
zΨr)
= 0.
Finally, let us consider that the body loadings are
X= (0, X ,0), Y = (0, Y ,0).
(47)
The body force
Y θ
θ
θ
is connected to the displacements and rotation fields of the
first axially-symmetric problem. Assuming that only Φ
θ
does not vanishes
identically in Eqs. (36), we obtain the following representation for u θ
, ϕ
r
, and
ϕ
z
in B :
⎧
0
uθ
=
4Φθ,
⎪
⎨φr = 2 α∂zΦθ
,
(48)
⎪
−1
⎪⎩ φz = − 2 α( r +∂r)
Φθ
,
where Φ
θ
is a solution in B of the equation
0
Ω + =
Φ r
X θ
7. Concentrated Loadings
Supposing that Yr
= X θ
= 0, the corresponding solutions of the second
axially-symmetric problem can be obtained by means of differential equation
(43) which can be write in the form
( μ + α)( β+ 2 γ)( ∇ + k )( ∇ + k )( ∇ + σˆ
) Ψ = Y .
(49)
2 2 2 2 2 2
1 2 3
Having applied Fourier transformations, direct and inverse,
⎧
1 +∞
f%
(, r η) = f(,)exp(i r z ηz)d,
z
⎪
2π
∫−∞
⎨
1 +∞
⎪ f(,) r z = f%
(, r η)exp(i − ηz)dη
⎪⎩
2π
∫−∞
over z and Hankel transformations
z
0.
z
r
z
(50)