Cavitation DeformationDeformation

UNIVERSAL DEFORMATIONS OF NEMATIC ELASTOMERS

Family 2: Straightening, stretching, and shearing of a sector of a tube

3.7 Cavitation DeformationDeformation

(a) (b)

Figure 3.9: Cylindrical balloon results: (a) Results for cylindrical balloon expansion at varying anisotropy parameter. (b) Progression of the cylindrical balloon solution through individual case numbers.

the surrounding polymer network deforms accordingly, with a stretch along that preferred direction and contraction in the transverse axes. Macroscopically, this means that the cylindrical balloon experiences finite stretch with zero stress.

Figure 3.9b shows the solution’s progression through Cases 0 through 5 throughout the deformation. For 𝑟 = 1 (rubber), the balloon is in Case 0 (corresponding to region𝐿) at𝜆_𝑜 =1 because there is no deformation, but after that point the balloon is entirely in Case 5 (region 𝑆), corresponding to a solid-like response without microstructure formation. For𝑟 > 1, the balloons begin in Case 0 (entirely in region 𝐿). Then the balloons will develop an inner portion that lies in region𝑀 (Case 1), corresponding to fine-scale microstructure formation in the inner part of the balloon, due to the boundary condition that there is internal pressure at the inner radius of the balloon. Then the balloon will then become entirely in region𝑀(Case 2) before the inner portion of the balloon will develop a solid-like response because it will be in region𝑆(Case 3). Finally, the balloon will become entirely in region𝑆, and the balloon will have a purely rubber response (Case 5). Note that we do not encounter the case in which 𝐿, 𝑀, and 𝑆 are all present within the balloon (Case 4) in these calculations.

3.7 Cavitation

ruptures form and grow in radius [30]. A schematic for the deformation of cavitation is shown in Figure 3.10.

Figure 3.10: Schematic showing the cross-section of a disk of nematic elastomer bonded to parallel plates, which are stretched in uniaxial tension.

This experiment can be modeled as a spherical void inside of an infinite medium subjected to a state of triaxial extension, as seen in Figure 3.11. We do not consider nucleation (we assume that the spherical void already exists), and we assume that the cavity remains spherical throughout the deformation, which is consistent with the experiments. We are interested in the behavior as the spherical void becomes

Figure 3.11: Nematic elastomeric sphere with radius 𝑅_𝑜 with spherical void of radius𝑅_𝑖, subjected to external pressure 𝑝_outer.

infinitesimally small (the limit as 𝑅_𝑖 → 0) or, equivalently, when the sphere of nematic elastomer is infinitely large (the limit as 𝑅_𝑜 → ∞). The deformation mapping for the growth of such a spherical void is as follows:

















𝜌 =𝜌(𝑅) 𝜃 = Θ 𝜙 = Φ

. (3.183)

The deformation gradient for this specific mapping then is F= 𝜕 𝜌

𝜕 𝑅

eh𝜌i ⊗Eh𝑅i + 𝜕 𝜃

𝜕Θ

𝜌sin𝜙

𝑅sinΦeh𝜃i ⊗EhΘi + 𝜕 𝜙

𝜕Φ 𝜌 𝑅

eh𝜙i ⊗EhΦi (3.184)

= 𝑑𝜌 𝑑𝑅

eh𝜌i ⊗Eh𝑅i + 𝜌 𝑅

eh𝜃i ⊗EhΘi + 𝜌 𝑅

eh𝜙i ⊗EhΦi. (3.185)

Thus,

Fh𝑖 𝑗i=©

­

­

«

𝑑𝜌 𝑑𝑅

𝜌 𝑅

𝜌 𝑅

ª

®

®

®

¬

. (3.186)

Since nematic elastomers are incompressible, we constrain detF=1:

𝑑𝜌 𝑑𝑅

𝜌² 𝑅²

=1 (3.187)

𝜌²𝑑𝜌=𝑅²𝑑𝑅 (3.188)

𝜌³=𝑅³+𝑐 . (3.189)

We solve for the constant𝑐 in terms of 𝜌_𝑖 (the deformed radius at the undeformed radius𝑅 =𝑅_𝑖) and obtain

𝑐= 𝜌³

𝑖 −𝑅³

𝑖, (3.190)

so we have a new expression for the deformed radius,𝑟 ℎ𝑜: 𝜌 =

𝑅³+𝜌³

𝑖 −𝑅³

𝑖

1/3

. (3.191)

Let the non-dimensional azimuthal stretch at the inner radius be denoted by 𝜆_𝑖 = 𝜌_𝑖

𝑅_𝑖

. (3.192)

Then in terms of𝜆_𝑖, the deformed radius is 𝜌 =

h

𝑅³+𝑅³

𝑖

𝜆³

𝑖 −1 i1/3

, (3.193)

and the ratio of the deformed radius to undeformed radius is 𝜆= 𝜌

𝑅

=

"

1+ 𝑅³

𝑖

𝑅³

𝜆³

𝑖 −1

#1/3

. (3.194)

Thus, the deformation gradient is

Fh𝑖 𝑗i=©

­

­

«

1 𝜆²

𝜆 𝜆

ª

®

®

®

¬

. (3.195)

The principal values ofFare𝜆₁ ≥ 𝜆₂ ≥ 𝜆₃: 𝜆₁ =𝜆₂=𝜆(which correspond to ˆe𝜃

and ˆe𝜙) and𝜆₃= ¹

𝜆² (corresponding to ˆe𝜌).

There are the same possible cases as presented in the spherical balloon, except that for cavitation we have a new definition for𝑅^∗:

𝑅^∗ =𝑅_𝑖 𝜆³

𝑖 −1 𝑟^1/4−1

!1/3

. (3.196)

Stress

The stress state is the same as that for the spherical balloon, seen in Section 3.5.

Solving static equilibrium

This section is the same as that of the spherical balloon, but with the following boundary conditions imposed:

𝜎_{𝜌 𝜌}(𝜌 =𝑜) = 𝑝 (3.197)

𝜎_{𝜌 𝜌}(𝜌 =𝑖) =0. (3.198)

We solve static equilibrium in the absence of body forces.

Case 1

In Case 1,𝑅 ∈ [𝑅_𝑖, 𝑅_𝑜]is in region𝐿. The outer pressure then is

𝑝=0. (3.199)

Case 2

In Case 2, 𝑅 ∈ [𝑅_𝑖, 𝑅^∗] is in region 𝑀 and 𝑅 ∈ [𝑅^∗, 𝑅_𝑜] is in region 𝐿. The two boundary conditions can be rewritten specific to the region:

−𝜂^𝑀(𝜌 =𝜌_𝑖) +𝜎ˆ^𝑀

𝜌 𝜌(𝜌= 𝜌_𝑖) =0 (3.200)

−𝜂^𝐿(𝜌= 𝜌_𝑜) +𝜎ˆ^𝐿

𝜌 𝜌(𝜌 = 𝜌_𝑜) = 𝑝, (3.201) and there is an additional boundary condition for continuity between the two regions:

−𝜂^𝐿(𝜌 =𝜌^∗) +𝜎ˆ ^𝐿

𝜌 𝜌(𝜌 =𝜌^∗) =−𝜂^𝑀(𝜌 =𝜌^∗) +𝜎ˆ^𝑀

𝜌 𝜌(𝜌 = 𝜌^∗). (3.202) The result from region𝐿 is

−𝜂^𝐿(𝜌^∗) +𝜎ˆ ^𝐿

𝜌 𝜌(𝜌^∗) = 𝑝 . (3.203)

The result from region𝑀 is

−𝜂^𝑀(𝜌^∗) +𝜎ˆ^𝑀

𝜌 𝜌(𝜌^∗) −𝜎ˆ^𝑀

𝜌 𝜌(𝜌_𝑖) +𝜂^𝑀(𝜌_𝑖) =

∫ ^𝜌∗

𝜌_𝑖

2 𝜌

ˆ 𝜎^𝑀

𝜃 𝜃 −𝜎ˆ ^𝑀

𝜌 𝜌

𝑑𝜌, (3.204) or

𝑝=

∫ ^𝑅∗

𝑅_𝑖

2 𝜆(𝑅)𝑅

𝜎ˆ^𝑀

𝜃 𝜃(𝜌 =𝜆(𝑅)𝑅) −𝜎ˆ^𝑀

𝜌 𝜌(𝜌 =𝜆(𝑅)𝑅) 1

𝜆(𝑅)²𝑑𝑅 . (3.205)

Case 3

In Case 3, all of the balloon is in region𝑀. Note that the isotropic state (𝑟 =1 and 𝜆≥ 1) falls under this case. The two boundary conditions for this case are:

−𝜂^𝑀(𝜌 =𝜌_𝑖) +𝜎ˆ^𝑀

𝜌 𝜌(𝜌 = 𝜌_𝑖) =0 (3.206)

−𝜂^𝑀(𝜌 = 𝜌_𝑜) +𝜎ˆ ^𝑀

𝜌 𝜌(𝜌 =𝜌_𝑜) = 𝑝 . (3.207) The result of solving static equilibrium is

𝑝 =

∫ ^𝑅𝑜

𝑅_𝑖

2 𝜆(𝑅)𝑅

𝜎ˆ ^𝑀

𝜃 𝜃(𝜌 =𝜆(𝑅)𝑅) −𝜎ˆ^𝑀

𝜌 𝜌(𝜌 =𝜆(𝑅)𝑅) 1

𝜆(𝑅)²𝑑𝑅 . (3.208) Case 4

In Case 4, all of the balloon is in region𝑆. Note that the isotropic state (𝑟 =1) falls under this case. The two boundary conditions for this case are:

−𝜂^𝑆(𝜌 =𝜌_𝑖) +𝜎ˆ^𝑆

𝜌 𝜌(𝜌 =𝜌_𝑖) =0 (3.209)

−𝜂^𝑆(𝜌= 𝜌_𝑜) +𝜎ˆ^𝑆

𝜌 𝜌(𝜌= 𝜌_𝑜) = 𝑝 . (3.210) The result of solving static equilibrium is

𝑝=

∫ ^𝑅𝑜

𝑅𝑖

2 𝜆(𝑅)𝑅

𝜎ˆ^𝑆

𝜃 𝜃(𝜌 =𝜆(𝑅)𝑅) −𝜎ˆ^𝑆

𝜌 𝜌(𝜌 =𝜆(𝑅)𝑅) 1

𝜆(𝑅)²𝑑𝑅 . (3.211) Results

The calculations were performed in MATLAB and the solutions are plotted below.

The outer radius is 𝑅_𝑜 = 1 cm, and we take the inner radius to be much smaller than the outer radius with a value of𝑅_𝑖 =1·10⁻⁸m. The parameters used for the generalized Mooney-Rivlin model were the same as for the spherical balloon.

The pressure 𝑝, normalized by the shear modulus 𝜇, is plotted as a function of the azimuthal stretch at the inner radius𝜆_𝑖. Figure 3.12 shows the comparison between the generalized Mooney-Rivlin model of this work and the neo-Hookean model of previous work [23] for an anisotropy parameter of 𝑟 = 8. The BTW theory and generalized Mooney-Rivlin theory deviate starting around a stretch of𝜆_𝑖 ≈15. The BTW model predicts that the cavitation pressure plateaus, whereas the Extended Mooney-Rivlin model does not.

Figure 3.13a shows the cavitation results for varying anisotropy parameter 𝑟. As expected, the isotropic case,𝑟 =1, is the stiffest response, and the response is softer as𝑟 increases and becomes more nematic.

Figure 3.12: Comparison of this work’s generalized Mooney-Rivlin model with the trace formula model of [23] for pressure of a growing spherical cavity inside a bulk disk.

(a) (b)

Figure 3.13: Cavitation results: (a) Results for cavitation at varying anisotropy pa- rameter. (b) Progression of the cavitation solution through individual case numbers.

Figure 3.13b shows the solution’s case at various values of𝜆_𝑖. The disk is in Case 1 where the entire structure is undeformed and in region 𝐿. Then, for𝑟 = 1, the rest of the deformation belongs to Case 4, where the disk is entirely in region 𝑆, having a purely elastomer response with no liquid crystal effects. For𝑟 >1, the disk develops microstructure in the area immediately surrounding the void that forms during cavitation. Because the void is so small compared to the length scale of the disk, the structure never moves into Case 3 (where the entire disk would be in region 𝑀).