Page 76 - Mechanics Analysis Composite Materials
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Chapter 3. Mechanics ?fa unidirectional ply 61
The answer for the second question (why the fibers are stronger than the
corresponding bulk materials) was in fact given by Griffith (1920) whose results
have formed the basis of Fracture Mechanics.
Consider a fiber loaded in tension and having a thin circumferential crack as
shown in Fig. 3.10. The crack length, f, is much less than the fiber diameter, A.
For a linear elastic fiber, CT =EE,and the elastic potential in Eq. (2.51) can be
presented as
When the crack appears, the strain energy is released in a material volume adjacent
to the crack. Assume that this volume is comprised by conical ring whose generating
lines are shown in Fig. 3.10 by broken lines and heights are proportional to the
crack length, 1. Then, the total released energy, Eq. (2.52), is
1 02
W = -kn-lI"d (3.10)
2 E ~
where k is some constant coefficient of proportionality. On the other hand,
formation of new surfaces consumes the energy
S=2n~ld, (3.1 1)
where y is the surface energy, Eq. (3.9). Now assume that the crack length is
increased by an infinitesimal increment, dl. Then, if for some value of acting stress, CT
0
ttttttt
111
0
Fig. 3.10. A fiber with a crack.
