166 TIME-DEPENDENT AND NON-LINEAR PROBLEMS
Poisson ratio is constant. Let us use the second assumption with
v = 0. This gives λ = 0 and reduces μ to E/2. Hence equation (6.68)
becomes,
*ij(t) = ß7v +
ß*7ü
(6.69)
where y
tj
= (u
h t
+ u
u}
) = y
jt
.
In order to represent the time-dependent behaviour of the material
McHenry
17
proposed the following equation to approximate the
relaxation modulus:
μ(ί) =
$
0
+Σ
S^l-e-"")
(6.70)
/ = i
This equation can be fitted to experimental data by reducing it to a
set of linear equations in S
k
and giving y
l
n values of time over the
transition range of the experimental data.
Assuming y° = 0, equation (6.69) can be reduced to the following
by taking the Laplace transform,
L(a
l}
) =
K
^)L(y%) (6.71)
where from (6.70)
LOi(ie)) = - s
0
+ X (-^— ) 1 (6.72)
The stress-strain relationship of the viscoelastic material has thus
been reduced to the normal static form (<r
i;
= μγ^) in the transform
plane (equation (6.71)). That is, we use a 'quasi-static' analysis.
To obtain a quasi-static analysis in boundary element terms we can
set up an integral equation relationship for the displacements similar
to equation (5.23), but using the constitutive relations (6.71) to
construct the fundamental traction tensor components Pfj. κ will
appear as a parameter of our problem and we must solve for a
sufficient number of values in order that the inversion of our Laplace
transforms can be accurately performed.
For the fully transient viscoelastic problem it is possible to follow
Cruse
1
but instead of the constitutive equations (6.42) we have to use
(6.68),
which becomes equivalent to the normal static form under the
Laplace transformation. The fundamental influence tensor U* would
remain the same but the traction influence tensor P£ would become
altered by the new constitutive relations. The normal integral