Answer
This depends on what is meant by the fracture stress. The yield stress must be larger
than the overall fracture propagation stress, since otherwise the specimen put under
stress will undergo yielding rather than fracture. But the yield stress must be smaller
than the local (or ‘‘true’’) fracture stress, since otherwise blunting of the crack tip will
not occur and pure elastic fracture results.
17.1.3 Fracture in Elongational Flow
Consider a long and thin cylinder of a viscoelastic material upon which an
extensional stress is acting, which stress is larger than the yield stress.
Elongational flow will then occur. This happens during spinning of, say, a
highly concentrated protein solution into a thread. In such a configuration,
somewhere in the thread a slightly thinner spot or ‘‘neck’’ will frequently
form by chance, and this may lead to breaking of the thread. The forceF
acting on and in the direction of the thread will be everywhere the same.
SinceF¼sA, the stress will be higher for a smaller value of the cross-
sectional areaA, if the value of the Young modulusEuis constant. This
means that the thread will be extended faster at the neck than at other
places, hence the thin part will become ever thinner, and the thread will
break. Nevertheless, several materials can be spun. How is this possible?
It will only succeed if the reaction force (stress times area) in the thin
part can be larger than that in the thick part of the cylinder. Let the cross-
sectional area at the neck beAþdA, where dAis small and negative. The
condition then becomes
ðAþdAÞ?sðAþdAÞ>A?sðAÞ
which can be reduced to
s?dAþA?ds> 0
Dividing byA?dsresults in
dlnAþdlns> 0
Sinceeu¼lnðL=L 0 Þ, whereLis specimen length and the subscript 0 denotes
the original value, andL?Ais constant, we have deu¼dlnA. Inserting
this we obtain the relation
deuþdlns> 0