20
A. M. Antonov et al.
∂σ xx
∂ x
+
∂σ yx
∂ y
= ρ
∂
2 u
∂t 2 ,
∂σ xy
∂ x
+
∂σ yy
∂ y
= ρ
∂
2 v
∂t 2 .
(2.4)
Here, u—longitudinal and ν—transversal components of displacements vector.
We assume that at the boundary y = 0 there are no stresses and moment stresses,
i.e., boundary conditions for the system of equations (2.4) have the form (Sabodash
and Filippov 1971):
μ y = 0, σ yy y=0 = 0, σ yx y=0 = 0, σ yz y=0 = 0,
(2.5)
Moreover, the third condition is identically satisfied by virtue of the assumption
that deformations are independent of the variable z.
The components of the stress tensor included in (2.4), (2.5) are related to
displacements u, the following relations:
σ xx = λ
∂u
∂ x
+
∂v
∂ y
+ 2μ
∂u
∂ x
, σ yy = λ
∂u
∂ x
+
∂v
∂ y
+ 2μ
∂v
∂ y
,
σ xy = μ
∂v
∂ x
+
∂u
∂ y
− L
2
∂
2
∂ x 2 +
∂
2
∂ y 2
∂v
∂ x
−
∂u
∂ y
,
σ yx = μ
∂v
∂ x
+
∂u
∂ y
+ L
2
∂
2
∂ x 2 +
∂
2
∂ y 2
∂v
∂ x
−
∂u
∂ y
.
(2.6)
It can be easily seen that σ xy = σ yx .
Moment stresses μ x and μ y can be expressed through u and v:
μ x = 2μL
2 ∂
∂ x
∂v
∂ x
−
∂u
∂ y
, μ y = 2μL
2 ∂
∂ y
∂v
∂ x
−
∂u
∂ y
.
(2.7)
We introduce scalar ϕ and vector ψ potentials so that the displacement vector u
can be written as (Erofeyev 2003):
u = ∇ϕ + ∇ ∗ ψ,
(2.8)
Since the displacement does not depend on the coordinate z, the only nonzero
component of vector potential is the component along the z axis, and we denote this
component by ψ.
With the help of (2.8), the system (2.4) is reduced to the equations
ϕ −
1
c
2
1
∂
2
ϕ
∂t 2 = 0, = 0,
(2.9)
1 − L
2
ψ −
1
c
2
2
∂
2
ψ
∂t 2 = 0,
(2.10)