Page 219 - T. Anderson-Fracture Mechanics - Fundamentals and Applns.-CRC (2005)

1656_C004.fm Page 199 Thursday, April 21, 2005 5:38 PM

Dynamic and Time-Dependent Fracture 199

By performing a Laplace transform on Equation (4.59) and Equation (4.60), it can be shown that
the creep compliance and the relaxation modulus are related as follows:

∫ t Et −τ ) dD( −ττ o ) d τ H = ( − t τ o ) (4.61)
(
τ ο dτ

For deformation in three dimensions, the generalized hereditary integral for strain is given by

kl
ε ij ∫ t D () = ijkl    t − τ t d στ ()    τ d (4.62)
0  τ d 
but symmetry considerations reduce the number of independent creep compliance constants. In
the case of a linear viscoelastic isotropic material, there are two independent constants, and the
mechanical behavior can be described by E(t) or D(t), which are uniquely related, plus ν (t), the
c
Poisson’s ratio for creep.
Following an approach developed by Schapery [59], it is possible to define a pseudo-elastic
strain, which for uniaxial conditions is given by
(
e
ε t() = σ t) (4.63)
E R
where E is a reference modulus. Substituting Equation (4.63) into Equation (4.59) gives
R
e
ε E = R ∫ t D ( − t τ() t ) d ετ() τ d (4.64)
0 τ d

The pseudo-strains in three dimensions are related to the stress tensor through Hooke’s law,
assuming isotropic material behavior:

ε ij e −1 ν = σ E [( i j − ν 1 σ + kk ij ] (4.65)
δ )
R
where δ is the Kronecker delta, and the standard convention of summation on repeated indices is
ij
followed. If ν = ν = constant with time, it can be shown that the three-dimensional generalization
c
of Equation (4.64) is given by
e
ij
ε ij E () = R ∫ t D ( − t τ t ) d ετ() τ d (4.66)
0 τ d
and the inverse of Equation (4.66) is as follows:

ij
ε ij e E () = R −1 ∫ t E ( − t τ t ) d ετ() τ d (4.67)
0 τ d
The advantage of introducing pseudo-strains is that they can be related to stresses through
Hooke’s law. Thus, if a linear elastic solution is known for a particular geometry, it is possible to
determine the corresponding linear viscoelastic solution through a hereditary integral. Given two
identical configurations, one made from a linear elastic material and the other made from a linear
viscoelastic material, the stresses in both bodies must be identical, and the strains are related through