Is the limit of stress intensity factors "1" when the crack length is close to "0"?

Dear Zhanqi:

I'm not familiar with fracture in functionally graded materials (it
shouldn't really be that different from classical fracture mechanics as
a functionally graded material is a material in which mechanical
properties are position dependent) but I would say when the crack
length goes to zero (in 2D) stress intensity factor (SIF) should go to
zero as well. For example, when you have a crack of length 2a in an
infinite body, SIF is \sigma\sqrt{\pi a} (\sigma is the far-filed
stress) and clearly goes to zero when "a" goes to zero.

Perhaps, you should post a couple of those papers where SIF is assumed to be "1" for small cracks.

You can also think of a crack as a special "defect" in a solid. SIF is
related to the so-called "material" or "configurational" force (energy
release rate in this case). A configurational force is, by definition,
the thermodynamic force that drives propagation of the crack. Now, when
crack length goes to zero, in the limit there is no defect and hence no
configurational force.

If you want to use SIF as a strength criterion, there are problems with
small cracks. Assuming that a crack propagates when SIF reaches a
critical value, let's say K_c, in principle, you can calculate the
corresponding critical stress or "strength". Assuming that K_c is
independent of crack length, you will end up having infinite strength
in the limit of a very small crack. Of course, this is not physically
meaningful because for very small cracks you need to consider things
like surface effects, etc. and this would mean that K_c is explicitly a
function of crack length. This (and similar problems) has been a
motivation for the so-called nonlocal failure criteria, e.g. the ones
proposed by Neuber and Novozhilov.

I hope this helps.

Regards,
Arash