94
L. Igumnov et al.
Fig. 6.1 Cylinder cross
section
fixed, while the outer surface is exposed to an axisymmetric radial load P(t) evenly
distributed along the cylinder’s element since the moment t = 0.
In a polar coordinate system R, θ with the origin at the center of the cross section,
let us introduce dimensionless values
r = R
R 1 , r 0 = R 0
R 1 , τ = t
t 0 ,u(r, τ ) = u R (R, t)/R 1 , σ 11 (r, τ ) = σ R R (R, t)/2μ 0
σ 22 (r, τ ) = σ θθ (R, t)/2μ 0 , P 0 f (τ ) = P(t)/2μ 0 , γ(τ) = t 0 T (t)
where t 0 = R 1
c; R 0 , R 1 are interior and outer radiuses of the cylinder, c =
√
(λ 0 + 2μ 0 )/ρ is the longitudinal elastic wave velocity, u R is the radial displacement, σ R R , σ θθ are stresses, P 0 is the dimensionless constant, T (t) is the kernel of
relaxation (Poisson’s ratio ν 0 is constant). The mathematical statement of the problem
consists of equations (prime marks and dots denote derivatives with respect to r and
τ correspondingly):
(1 − ˆ
γ )[u
(r, τ ) + u(r, τ )/r ]
= ¨
u(r, τ ),
(6.23)
σ 11 (r, τ ) = (1 − ˆ
γ )[wu
(r, τ ) + (w − 1)u(r, τ )/r ],
(6.24)
and boundary and initial conditions
u(r 0 , τ ) = 0, σ 11 (1, τ ) = −P 0 f (τ ); u(r, 0) = 0, ˙
u (r, 0) = 0
(6.25)
where w = (1 − ν 0 )/(1 − 2ν 0 ).
Let us assume that γ (τ ) = ae
−bτ , 0 < a < b and f (τ ) = h(τ ) is Heaviside
function. Then, after Laplace transform and inversion operation, the solution will
take the form