Page 223 - T. Anderson-Fracture Mechanics - Fundamentals and Applns.-CRC (2005)

1656_C004.fm Page 203 Thursday, April 21, 2005 5:38 PM

Dynamic and Time-Dependent Fracture 203

material in the failure zone may be severely damaged and contain voids and other discontinuities,
it is assumed that the surrounding material can be treated as a continuum. If σ does not vary with
m
x, applying Equation (3.44) gives
J = σδ e (4.78)
m
v
e
where δ is the pseudo-crack-tip-opening displacement, which is related to the actual CTOD through
a hereditary integral of the form of Equation (4.77). Thus the CTOD is given by

J
δ ( / σ = {Dd m )} (4.79)
v
Although σ was assumed to be independent of x at time t, Equation (4.79) permits σ to vary
m
m
with time. The CTOD can be utilized as a local failure criterion: If crack initiation occurs at δ , i
the J at initiation can be inferred from Equation (4.79). If δ is assumed to be constant, the critical
i
v
J would, in general, depend on the strain rate. A more general version of Equation (4.79) can be
v
derived by allowing σ to vary with x.
m
An alternative local failure criterion is the fracture work w . Equating the work input to the
f
crack tip to the energy required to advance the crack tip by da results in the following energy
balance at initiation:
∫ 0 δ i σ m δ d w = 2 f (4.80)

assuming unit thickness and Mode I loading. This energy balance can also be written in terms of
a time integral:

∫ 0 τ ι σ m δ ∂ t ∂ dt = 2 w f (4.81)

Inserting Equation (4.79) into Equation (4.81) gives

∫ 0 τ ι σ m ∂{ Dd J t ∂ v σ / m )} dt = 2 w f (4.82)
(

If σ is independent of time, it cancels out of Equation (4.82), which then simplifies to
m
J ∂
E R ∫ 0 τ ι D t i t ( − τ i ∂τ v d , ) τ w = 2 f (4.83)

−1
For an elastic material, D = E and J = 2w . If the failure zone is viscoelastic and the surrounding
v
f
R
continuum is elastic, J may vary with time. If the surrounding continuum is viscous, D t = − ( t )/( v E τ R , )
v
where t is a constant with units of time. Inserting this latter result into Equation (4.83) and integrating
v
by parts gives
t v −1 ∫ t i J dt = 2 w f (4.84)
v
0