Page 146 - Engineered Interfaces in Fiber Reinforced Composites

Chapter 4. Micromechanics of stress transfer 129
1992b). The external stress, 0, is represented by 00, o:, 0; and ufr for (frictionless)
initial debond stress, partial debond stress, maximum debond stress and initial
frictional pull-out stress after complete debonding, respectively, at different stages of
the fiber pull-out process. A typical fiber pull-out stress-displacement ((r-6) diagram
along with these characteristic external stresses is schematically shown in Fig. 3.7.
For the cylindrical coordinates (Y, 0,z) in the fiber pull-out test, the basic
governing equations and the mechanical equilibrium conditions between the
composite constituents are essentially the same as those given in Section 4.2.3, i.e.
Eqs. (4.8)-(4.18). The only exception is the equilibrium condition between the
external stress and the axial stresses in the fiber and the matrix given by Eq. (4.1 I),
which has to be modified to
1
cr = $(z) + dm (z) . (4.87)
I
Therefore, in a procedure similar to that used in the fiber fragmentation test,
combining Eqs. (4. lo), (4.17), (4.1 8), and (4.87) yields a second-order differential
equation for the FAS

(4.88)

where the coefficients A, and A2 are given in Eqs. (4.20) and (4.21). The solution of
the FAS is subjected to the following boundary conditions:

of([) = (TI = cr - w(i7 - cr)[exp(At) - 11, $(z) = o . (4.89)

FS~ is the crack tip debond stress at the boundary between the bonded and debonded
regions at z = e, as defined in Section 4.2.3. It should be noted, however, that the
actual values of (rt are different for different specimen geometry even for an identical
debond length, C. Therefore, the solutions of FAS, and the corresponding MAS,
MSS and IFSS are obtained for the bonded region (CGzGL):

[i 2 atop] sinh [ fi(L -z)] - 2 cr sinh [~(Gz)]
y
4(z) = sinh[fi(L - e)] -"%a. (4.90)

b??(r+ (rp] sinh[fi(L -z)] - y2asinh[fi(e -z)]
d,(Z) = -1'
sinh [a(L - e)]
2A2 (4.91)
+?/ -0,
[ y 2 cr + CT~] cosh [ fi(L - z)] - y 2 CJ cash [a([
A1
z)]
-
zm (r, 2) = y
sinh [a(L - C)]
(4.92)