Elastomere Friction - The Best Friend international

36 H.Lorenz,J.Meier,andM.Klüppel
Because the relative stress is always positive, it turns out that the clusters
give a positive stress contribution in the up cycle and a negative one in the
down cycle.
The local relative deformation κ1(ε) is calculated from the global deformation
and the strain amplification factor X. Bythisway,σR,μ and the cluster
deformation depend on X which we calculate in the following section.
3.2 Hydrodynamic Strain Amplification
Huber and Vilgis calculated an amplification factor X for overlapping CB aggregates
[19] assuming a fractal geometry. They found that X is proportional
to powers of filler volume fraction and relative aggregate diameter. Because
undamaged clusters are also stiff and show a fractal geometry, we can use
the same result. All of the stiff clusters contribute to X, according to their
diameter. Damage of stiff clusters causes stress softening by decreasing X,
which is expressed as an integral over the “surviving”, hard, section of the
cluster size distribution, while broken clusters are treated as having a relative
size of x = 1. Essentially, the strain amplification factor is a function of the
maximum deformation the material has been subjected to during its entire
deformation history [10]:
Xmax = X(εmin,εmax)
=1+ c
3 Φ2/(3−df )
eff
3�
�� xµ,min
μ=1
0
x dw−df
� ∞
φ(x)dx +
xµ,min
�
φ(x)dx (13) ,
where the constant c is taken to be ≈ 2.5, the Einstein coefficient for spherical
inclusions. The exponent dw stands for the anomalous diffusion exponent
which amounts to ≈ 3.1 [14], and df is the fractal dimension of the filler
clusters (≈ 1.8 for cluster-cluster aggregation [15]).
To solve the integrals analytically, we make use of an approximation for
the exponent: dw − df = 1. This results in:
Xmax =1+ c
3 Φ2/(3−df )
eff
3�
μ=1
�
1+
� xµ,min
0
�
(x − 1)φ(x)dx . (14)
The solution of this expression in the case of uniaxial loading with λ ≡ λ1
and λ2 = λ3 = λ−1/2 is given as follows:
Xmax =1+ c
3 Φ2/(3−df � �
)
eff 3x0 − exp −2 x1,min
�
x0
� �
× (x0 − 1) 1+ 2x1,min
�
+
x0
2x2 �
1,min
x0