=Paper=
{{Paper
|id=Vol-1853/p02
|storemode=property
|title=Analysis of the transversely isotropy, due to the
production process, of carbon black filled SBR
|pdfUrl=https://ceur-ws.org/Vol-1853/p02.pdf
|volume=Vol-1853
|authors=Gianfilippo Dottore
|dblpUrl=https://dblp.org/rec/conf/system/Dottore17
}}
==Analysis of the transversely isotropy, due to the
production process, of carbon black filled SBR==
https://ceur-ws.org/Vol-1853/p02.pdf
Analysis of the transversely isotropy, due to the
production process, of carbon black filled SBR
Gianfilippo Dottore
Department of Civil Engineering and Architecture
University of Catania (Italy)
email: gianfilippo.dottore@gmail.com
Abstract—Through tensile tests of uniaxial and planar type, Transversely isotropic behavior, caused by the rubber cal-
was proceeded to the mechanical characterization of SBR rubber endering process [20], has made more difficult the analysis.
sheets filled with particles of carbon black. The analysis has The phenomenon is well known in the literature, so that were
allowed defining the elastic features, the storage modulus, the
transversal contraction coefficient (both obtained from uniaxial already made attempts to characterize elastomers which have
testing) and the shear modulus (from the planar/pure shear transverse isotropy [21], [22], [23].
tests). Due to the calendering process by which was obtained, the Therefore, tests were performed at different loading di-
elastomer sheet it’s shown transversely isotropic, with different rections, in order to assess the behavioral differences of the
mechanical behavior in the direction of processing with respect to material between the direction of calendering and the direction
the transverse direction. Therefore, it was essential to evaluate
two different values for each storage modulus, depending that orthogonal to it.
the sample was loaded in the direction of calendering or in that
orthogonal to it. II. C ONSTITUTIVE EQUATIONS
Index Terms—Transversal Isotropy, Hyperelasticity, SBR, Car- Most of engineering materials belongs to the category of
bon Black, Calendering. simple materials, for which the stress tensor T is a function
of the history of the position gradient F = ∇ x [2]:
I. I NTRODUCTION
The Styrene Butadiene Rubber (SBR) arises from copoly- T (x, t) = T (F (x, τ ), x0 , t) τ ∈ [0, t] (1)
merization of Butadiene and Styrene. The latter, present to the
extent of 20-25% of the copolymer, increases both strength and Among the simple materials, those elastic exhibit a stress
stiffness from SBR with respect to the butadiene rubber (BR) state as a function of the single value of the deformation
[1], [2]. Often, it is filled with carbon black, carbon charges gradient:
of nanometric particles obtained from the combustion of
hydrocarbons, the presence of which improves the mechanical T (x, t) = T (F (x, τ ), x0 , t) (2)
properties and slows the aging. The SBR is the most usage According to this constitutive relationship, stress does not
rubber [3], [4], [5], [6], [7], [8], [9], being able to provide depend on the deformation path (Cauchy elasticity), but it
mechanical properties similar to those of natural rubber at a is not the same for the deformation work that it performs.
lower cost. The elastomer is used mainly where there is a need The hyperelastic materials are, instead, conservative elastic
for high-friction surfaces, so that the 70% of the production materials (Green elasticity), i.e. for which, there being an
is for tires tread [10], [11], [12], [13]. elastic potential function ϕ = ϕ(F), which can be derived
The mechanical characterization of a material consists in from stress, the differential form of the deformation work is
determining the current constitutive law between stress and exact
strain [3], [14], [15], [16].
t
The characterization process is particularly critical for elas-
∂ϕ
tomers, due the large deformations that, for the same elastomer = TI (3)
∂F
definition [17], [18], can tolerate without reaching the failure.
The aim of this experimental analysis is reached the where TI := JTF−t is the first tensor of Piola-Kirchhoff,
mechanical characterization of styrene-butadiene rubber, im- with J = det F.
proved in its quality by the introduction of fillers of carbon The simple assumption of linear elastic material is correct if
black (25%) in the blend [4], [19]. you can overlook the time-dependent effects (such as sliding
The analysis developed is focused on tracking of math- and relaxation) [2], [24], [25] and for small deformations.
ematical functions, which correlated stress and strain, and Circumstance, the latter, in which it is allowed the stress-
on evaluation, starting with these functions, of the storage strain relationship linearization. The linear stress-strain law
modules assumed at the beginning of the rubber test. is represented in classic form, tensorial, from equations Tij =
P 3 4
h,k=1 Cijhk Ehk , which contain the 3 = 81 constants Cijhk
Copyright c 2017 held by the authors. of the elastic tensor di of order 4. Since the stress and strain
7
symmetry of the tensors requires six independent components and
of the stress tensor as a function of the six independent 1 v v
− Exyy − Exyy 0 0 0
components of the strain tensor, only 36 independent constants Ex
v v
− yx 1
are needed to determine the linear elastic relationship. Ex Ey − Eyzy 0 0 0
− vyx v
− Eyzy 1
0 0 0
C−1 =
Ex Ey
0 0 0 1/Gxy 0 0
σx σ(1) ε̇x ε̇(1) 0 0 0 0 1/Gxy 0
σy σ(2) ε̇y ε̇(2) 2(1+vyz )
0 0 0 0 0 Ey
σz σ(3) ε̇z ε̇(3)
σ := =
; ε̇ := =
τxy σ(4) γ̇xy ε̇(4) where the 6 modules present are defined as follows
τxz σ(5) γ̇xz ε̇(5)
τyz σ(6) γ̇yz ε̇(6) σi
Ei = σ(i) 6= 0, σ(k) = 0 ∀k 6= i
εi
We can express the linear relationship in a matrix simplified
using independent components vectors εj
vji = σ(i) 6= 0, σ(k) = 0 ∀k 6= i
εi
σx
c11 c12 c13 c14 c15 c16
εx τij
Gij =
σy c21 c22 c23 c24 c25 c26 εy γij
σz c31 c32 c33 c34 c35 c36
εz
= The elements on the main diagonal are the modules of
τxy c41 c42 c43 c44 c45 c46
γxy
τxz c51 c52 c53 c54 c55 c56 γxz normal deformability 1/Ei and tangential 1/Gij ; those that
τyz c61 c62 c63 c64 c65 c66 γyz appear outside of the same diagonal such as the modules
of transverse deformability −vji /Ei . The symmetry of the
previous matrix adds the condition vyx /Ex = vxy /Ey , which
The linear elastic relationship is defined by 6 × 6 = 36 reduces to 5 the independent components. In the case of
components of the C tensor of order 2 of elastic constants. transversely isotropic material plates, which present the x or
Assuming the occurrence of a potential elastic (hyperelastic y axis of transverse symmetry, assuming the status of plane
material), according to the Schwartz theorem, the elasticity stress σz = τxz = τyz = 0 and writing separately the
tensors are symmetrical and the constants are reduced to 21 equations:
[3].
They are called transversely isotropic those materials, which vyx 1
have an isotropic response in the plane orthogonal to an axis, εz = − σx + (σz − vyz σy )
Ex Ey
said axis of transverse symmetry. By choosing a base having
the x-axis parallel to symmetry transverse
γxz = 0
σx
c11 c12 c13 0 0 0
εx
γyz = 0
σy c12 c22 c23 0 0 0 εy
the constitutive relation can be narrowed to only three
σz c13 c23 c33 0 0 0 εz
=
stress and three strain components in the x and y directions,
τxy 0 0 0 c44 0 0 γxy
becoming [27], [3]:
τxz 0 0 0 0 c44 0 γxz
c22 −c23
τyz 0 0 0 0 0 2 γyz
Ex vxy Ex
0
σx 1−vxy vyx 1−vxy vyx εx
In this case, the elastic response is described by 5 indepen- σy = vyx Ey Ey
0 = εy
dent parameters [26].
1−v xy vyx 1−vxy vyx
τxy 0 0 Gxy γxy
The constitutive equation can be formulated in terms of
engineering constants [3], inverting the stress-strain relation
1 v
− Exyy 0
εx Ex σx
vyx 1
εx
σx
εy =
Ex Ey 0
= σy
γxy 1 τxy
εy σy 0 0 Gxy
εz
= C−1 σz
γxz
τxy
where there is a symmetry condition vyx /Ex = vxy /Ey ,
γxz τxz whose effect is that only 4 of the 5 elastic constants appearing
γyz τyz in the above matrix are independent.
8
III. P URE SHEAR IN PLANAR TEST
A valid constitutive model must properly describe the
material behavior for each stress state. A single test as a
simple uniaxial tensile test does not allow the construction
of a reliable model in every situation. It then requires the
programming of tests set that provide different stress mode.
The uniaxial tensile test is then flanked by planar tensile and
equibiaxial tests [4], [19].
The planar tensile test replaces the pure shear test (torsion of
cylindrical sample), not applicable on samples cut from rubber
sheets. The central portion of the specimen is subject to a pure
shear strain and stress [4], [24]. To prove this assumption, we
consider an orthogonal axes system, formed by the loading Fig. 1. Reference system, x-y plane
direction, from the direction perpendicular to this plane in the
specimen and the direction normal to the sample itself. Noted
these directions, respectively, with the letters x, y, z (Figg. 1,
2):
The test consists to creating a stretching in the x-direction.
For implementation of the test (high ratio width/length of the
sample), it is considered negligible the strain along the y-
direction
λy = 1 ⇒ ε̇y = 0
Consequently, the incompressibility condition J = det F =
1 ⇒ ε̇V = ε̇x + ε̇y + ε̇z = 0 for planar test is written [2]:
1
J = λx λz ⇔= 1 ⇒ λz =
λx
ε̇V = ε̇x + ε̇z ⇔ ε̇z = ε̇x
Therefore, assuming the incompressibility of the mate-
rial, the coordinates of tensors F and V=1/2(grad v +
grad vt )(strain rate) respect to the system of axes x, y, z
are., Fig. 2. Reference system, x-z plane
λx 0 0
F xyz = FD xyz = 0 1 0
1 σ˙x 0 0
0 0 λx
dT σ˙x
dt xyz
= 0
2 0
0 0 0
ε˙x 0 0
V xyz = 0 0 0 Unlike from strain, a condition of pure shear not realized
0 0 −ε˙x for stress. However, the deviator stress tensor T0 still takes a
pure shear connotation (see Fig. 4):
being FD the right tensor of the deformation gradient. As σ˙x
can be seen from the following image (Fig. 3), relative to the h 0i 2 0 0
circumferences of Mohr for the strain rate V, at each instant dT =0 0 0
dt xyz
of the test, the strain occurs in pure shear mode. 0 0 − σ2˙x
The deformation in the z-direction is free, therefore in the
n and t are the orthogonal directions to the y-axis and
absence of external loads in this direction is not generated
forming angles of 45◦ with the x- and z-axis (Fig. 2). Since
stress (σz = 0). Moreover, from the generalized Hooke’s law
τ̇nt = σ̇x /2, and being γ̇nt = 2ε̇x , the calculation of the shear
ε̇y = σ̇y − ν(σ̇x + σ̇z ), being for incompressible materials
modulus G is carried out using the formula
ν = 1/2 [3], for the y-direction stress is σ̇y = ν σ̇x = σ̇x /2. In
conclusion, during the planar test, the stress tensor T follows dτnt 1 dσx
the variation equation G= =
dγnt 4 dεx
9
Fig. 3. Circumferences of Mohr of the strain rate tensor.
Fig. 5. Grid for the optical relief of the strain.
20 mm wide and 180 mm long, with a gauge length of 120
mm.
In the planar tests, was adopted a rectangular specimen, 150
mm wide and 70 mm long, having a gauge length of 50 mm.
For the measurement of deformations of the specimen was
adopted an optical method. In effect, the large deformations
do not allow easy application of the classic method of the
strain gages. The used solution is based on tracing a grid on
the specimen surface, which allows to highlight the location
of certain specific points. By known distances between various
points of the grid in the absence of load, just evaluate their
new position for load applied to achieve the deformation.
For image acquisition, carried out at a constant rate, it used
a digital camera Basler acA1300-30gm. It is equipped with a
Sony CCD sensor ICX445 that provides 8-bit greyscale images
Fig. 4. Circumferences of Mohr of the stress rate deviator tensor. at a maximum frame rate of 30 per second, with a resolution
of 1280x960 px.
To provide an accurate strain measurement, the Vision
IV. E XPERIMENTAL TESTS
Builder AI software of National Instruments acquires and
The tested material is SBR filled with 25% of carbon black. processes images through the Vision Assistant command. At
The tests were performed on specimens cut from a sheet first a convolution filter is applied to emphasize the grid.
having a 3 mm thickness. These were carried out with a test In fact, the alteration of the grid, caused by the specimen
machine Zwick/Roel, model Z100, equipped with 1 kN load stretching, reduces the brightness and thus the contrast with the
cell, to imposed deformation, while maintaining constant the background, making difficult the identification of the points by
speed of the movable crossbar. the software [33]. Finally, a median filter removes the image
The uniaxial tests were performed on rectangular specimens, residual noise.
10
Fig. 8. Localization of points on the processed image.
Fig. 6. Uniaxial Test Setup.
Fig. 9. Uniaxial Test: Normal Stress vs Normal Strain.
Fig. 7. Not processed image.
to the growing of the test speed. A decrease of the failure
strain is associated. It also occurs a different location of the
V. R ESULTS sample crack, depending on the execution test rate. At the
By the stress and strain values, the relationships of vari- ends, near the gripping point, for tests with crossbar that moves
ous functional dependencies between the physical quantities 5 mm/min; at the center of the specimen for the tests with
involved is obtained. The storage modules were determined speed of 60 mm/min.
from the mathematical derivation of these functions. In all performed tests, the value of Young’s modulus E at the
The uniaxial testing allows to express both the normal stress O(0, 0) ranged between 6.45 and 7.70 MPa, comparable with
as a function of the normal strain in the vertical direction values reported in the literature for PDMS filled by carbon
(loading direction), and the normal strain in the horizontal black (25%) [34]. The stretching at failure is much lower than
direction (orthogonal to the loading direction) as a function the average obtained in the styrene-butadiene rubber [34].
of vertical strain. From the first of these applications the In addition, at the O(0, 0), the Poisson’s ratio ν is greater
normal storage modulus E = dσ/dεν [24] is obtained, from for the high-speed tests (0.460 and 0.490) compared to those
the second application the transverse contraction coefficient at low speed (0.360 and 0.387). For the same rate of the
ν = −dεo /dεν . crossbar, instead, the tested elastomer shows a greater Young’s
As already verified during a previous study on a silicone, modulus and Poisson’s ratio in the transverse test, but a lesser
developed at the University of Catania [25], a small strain failure strain. It is interesting to observe that for faster tests
rate, lower than ASTM standards, could rise viscoelastic the coefficient ν, evaluated by means of real strains, assumes
relaxation phenomena in elastomers (decreasing of stress level constant value and close to the theoretical limit value of 0.5,
compared to the strain achieved). Increasing the execution maximum value for isotropic materials [26], which character-
test rate, this aspect can be kept under control, drastically izes the incompressibility, during the entire test. This statement
mitigating the effects. As a consequence of this, the stress- is true only for the Hencky’s tensor. Adopting instead other
strain curve is higher for the higher test rate, then the rubber tensors of deformation, it decreases during the experimental
manifests an increase of stiffness and mechanical resistance test, being equal to about 0.5 only at the beginning of the test
11
account the highly hyperelastic material characteristics. Then,
they applied optical methods for overcoming this limit to
characterize the elastomer.
The collected data were analyzed to determine the main
elastomer storage modules. Different responses in relation
to the different direction of load application are obtained,
studying anisotropy expressed by the material due to the
production process of calendering elastomeric sheet. Given
the high variability of the materials properties, the results
were compared with those indicated in the literature for the
styrene-butadiene rubber filled with carbon black. The normal
storage modulus (stiffness) and the transverse contraction
coefficient (Poisson’s ratio) of the SBR as a function of
the filler percentage have appeared, in accordance with the
Fig. 10. Uniaxial Test: Horizontal Strain vs Vertical Strain. data available, lower for the tests with load applied in the
calendaring direction.
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