Introduction
This page looks at the strain states present in a typical tire. Understanding
such deformation modes is relevant to knowing how to best perform material
tests in the lab to characterize rubber behavior in support of tire design and
development. For example, it can be important to know whether a portion
of a tire is deforming primarily in uniaxial tension or in shear so that
lab tests can be developed to characterize the rubber's strength, stiffness,
damping, etc, in the appropriate deformation mode.
The page contains notes I compiled many years ago. So enjoy the old graphics!
Deformation States
Any strain state can be transformed to its principal orientation, allowing it to
be described by only the three principal strain values, along with its
orientation vector,
QQ.
Look at the three deformation states below that are described in terms of
the principal strains.
In each case, the cube is being stretched the most in the
1 direction, and
shrinking the most in the
3 direction. But it is the
2
direction that dictates the deformation mode. In the tension case at the far left,
ϵ2=ϵ3, both of which are negative.
In the center case, the second principal strain is zero. This is shear in which
γMax=ϵ1−ϵ3, while the remaining
strain,
ϵ2, is zero.
In the far right case, the second principal strain is the maximum possible,
which is equal to
ϵ1. This is called
equibiaxial tension,
although it is in fact the same as compression.
This key here is that it is the middle principal strain, which is bounded between
the max and min principals, which is the key to identifying the deformation state.
Deformation States and Strain Transformations
If the material is rubber, which is incompressible, then only two of the
principal strains are independent. So it is useful to transform the
problem from the principal strains to a set of three new parameters
such that the third, dependent one, is zero, or at least very close.
One way of doing this is through invariants.
But invariants are practically impossible to intuitively relate directly
to a deformed state (such as uniaxial tension or shear). So an alternative
is desirable.
One such alternative is as follows:
[ϵ1000ϵ2000ϵ3]=ϵHyd[100010001]+(ϵ1−ϵ3)[1/20000000−1/2]+(ϵ2−ϵHyd)[−1/20001000−1/2]
where
ϵ1>ϵ2>ϵ3.
This can be checked by multiplying each term of the matrix out.
For example, looking at the
11 slot gives
ϵ1=ϵHyd+12(ϵ1−ϵ3)−12(ϵ2−ϵHyd)=ϵ1+ϵ2+ϵ33+12(ϵ1−ϵ3)−12(ϵ2−ϵ1+ϵ2+ϵ33)=ϵ1
Doing the same steps for the
22 and
33 components will confirm the identities.
The relationship can be simplified somewhat by noting that
γMax=ϵ1−ϵ3 because
ϵ1 and
ϵ3 are the max and min principal strains.
Likewise, (
ϵ2−ϵHyd) can be named
γSecondary because its matrix is traceless, so
it is a shear condition.
So the equation can be written a little more concisely as
[ϵ1000ϵ2000ϵ3]=ϵHyd[100010001]+γMax[1/20000000−1/2]+γSec[−1/20001000−1/2]
For uniaxial tension,
ϵ2=ϵ3, so
ϵHyd=ϵ1+2ϵ33
and
γSec=ϵ2−ϵHyd=ϵ3−ϵ1+2ϵ33=ϵ3−ϵ13
This is exactly -1/3 of
γMax, so
Uniaxial Tension:γSecγMax=−13
At the other extreme is the case of equibiaxial tension. In this case,
ϵ2=ϵ1, so
ϵHyd=2ϵ1+ϵ33
and
γSec=ϵ2−ϵHyd=ϵ1−2ϵ1+ϵ33=ϵ1−ϵ33
This is exactly 1/3 of
γMax, so
Equibiaxial Tension:γSecγMax=13
The shear state actually depends on which strain tensor is being used. If it is true strain,
then
γSec/γMax=0 is exactly pure shear. For Green
strain, the ratio for pure shear drifts slightly negative at large strains. It equals
−γMax/12.
Nevertheless, the ratio
γSec/γMax varies from -1/3 to +1/3
and completely characterizes the deformation state.
Deformation States in a Tire
So now let's take a look at the conditions in a tire. It is a 205/70R14 XW4
(from long ago). And element #128 is in the shoulder of the tire. It is rolling
straight ahead at nominal load and pressure. (Everything below is based on Green strains.)
The strain components, in cylindrical components, in the shoulder (element #128)
are shown below as a function of angle around the tire.
It is hard to tell in the above graphs just what the deformation state(s) is.
But this can be made clearer by transforming the strains into principals and
calculating
γMax and
γSec.
The graphs show that this portion of the shoulder actually spends the majority
of its time in shear as it passes through the contact patch because
γSec does not approach
±γMax/3.
In fact, the next graph shows that the majority of the tire is in shear
in the contact patch.
Here is the tread at the center of the contact patch. It has the highest fraction of material undergoing
uniaxial tension.
And here are the belts. Back to shear. Probably coming from intercable shear and belt edge shear.
And here is the bead. Definitely shear.
And finally the entire cross-section opposite the contact patch.
These examples have shown that rubber in the XW4 passenger tire deforms
predominantly in shear.