Cylindrical balloon

UNIVERSAL DEFORMATIONS OF NEMATIC ELASTOMERS

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

3.6 Cylindrical balloon

to 2 to 3 throughout the deformation. At a stretch of 𝜆_𝑜 = 1, the balloon is again in Case 1, corresponding to being entirely in the 𝐿 region because no deformation has occurred. Immediately upon inflation of the balloon, the balloon jumps to Case 2, where the inner part of the balloon experiences region 𝑀, developing fine-scale microstructure. Then shortly after, the balloon will become entirely in region 𝑀, and the rest of the balloon will develop fine-scale microstructure in response to the pressure. This formation of microstructure by the liquid crystal molecules creates a softer response than the rubber without liquid crystals.

look for a solution of the form

















𝜌 =𝜌(𝑅) 𝜃 = Θ 𝑧 =𝜉 𝑍

. (3.114)

Following the convention found in the paper by Rudykh et al. on electroactive balloons [63], let𝜆= 𝑟/𝑅 be the hoop stretch ratio. Let𝜉 denote the axial stretch ratio. Thus the associated deformation gradient is:

F=

©

­

­

­

«

1

𝜆𝜉 0 0

0 𝜆

0 0 𝜉

ª

®

®

®

¬

, (3.115)

where

𝜆= s

1 𝜉

+ 𝑅²

𝑜

𝑅²

𝜆²

𝑜− 1 𝜉

. (3.116)

For fixed axial ratio 𝜉 = 1, the principal stretches 𝜆₁ ≥ 𝜆₂ ≥ 𝜆₃ are 𝜆₁ = 𝜆, corresponding to ˆe𝜃,𝜆₂=1, corresponding to ˆe𝑧, and𝜆₃= ¹

𝜆, corresponding to ˆe𝜌. Simplifying the regions, we find:

𝐿 : 𝑅 ≥ 𝑅₂ (3.117)

𝑀 : 𝑅₁ ≤ 𝑅 ≤ 𝑅₂ (3.118)

𝑆: 𝑅 ≤ 𝑅₁, (3.119)

where𝑅₁ = q

𝑅²_𝑜(𝜆²𝑜−1)

𝑟−1 and𝑅₂= q

𝑅_𝑜²(𝜆²𝑜−1) 𝑟1/3−1 . This leads to the possibility of the following cases:

• Case 0: 𝑟 > 1 and𝑅₁≤ 𝑅₂ ≤ 𝑅_𝑖: the entire balloon is in region𝐿

• Case 1: 𝑟 > 1 and𝑅₁ ≤ 𝑅_𝑖 < 𝑅₂ < 𝑅_𝑜: the inner portion of the balloon is in region𝑀 and the outer portion is in𝐿

• Case 2: 𝑟 > 1 and𝑅₁≤ 𝑅_𝑖 < 𝑅_𝑜 ≤ 𝑅₂: the entire balloon is in region𝑀

• Case 3: 𝑟 > 1 and𝑅_𝑖 < 𝑅₁ < 𝑅_𝑜 ≤ 𝑅₂: the inner portion of the balloon is in region𝑆and the outer portion is in𝑀

• Case 4: 𝑟 > 1 and𝑅_𝑖 < 𝑅₁ < 𝑅₂ < 𝑅_𝑜: the balloon is in regions𝑆, then 𝑀, then𝐿from inside to outside

• Case 5: 𝑟 ≥ 1 and 𝑅_𝑜 ≤ 𝑅₁ ≤ 𝑅₂: the entire balloon is in region𝑆

Figure 3.7: Diagram of all possible cases in the inflation of a nematic elastomer cylindrical balloon.

A diagram illustrating the various cases can be seen in Figure 3.7.

As in the spherical balloon deformation, we can gain insight from Figure 3.1 re- garding the regions that this deformation will experience. The cylindrical balloon expansion with fixed axial ration𝜉 =1 is the same as a planar extension deformation (the deformation gradient of Equation 3.115 isF =diag(1/𝜆, 𝜆,1)). Recalling that 𝑠 is the largest singular value ofF and𝑡 is the largest singular value of cofF, this means that𝑡 = 𝑠 for this deformation. If we were to follow along the𝑡 = 𝑠 line in Figure 3.1, we see that the deformation will move progressively through region 𝐿 then𝑀 then𝑆.

Stress

The stress in region𝐿is

σ^𝐿 =−𝜂^𝐿I. (3.120)

The non-zero components of stress in region 𝑀 are 𝜎^𝑀

𝜌 𝜌 =−𝜂^𝑀+

𝑀

Õ

𝑖=1

𝑐_𝑖𝑝_𝑖|𝐴_𝑀|^𝑝𝑖−12𝑟^1/3 𝜆²

+

𝑁

Õ

𝑗=1

𝑑_𝑗𝑞_𝑗|𝐵_𝑀|^𝑞𝑗−12𝑟^1/6 𝜆

𝜎^𝑀

𝜃 𝜃 =−𝜂^𝑀+

𝑀

Õ

𝑖=1

𝑐_𝑖𝑝_𝑖|𝐴_𝑀|^𝑝𝑖−1 2𝜆 𝑟^1/6

+

𝑁

Õ

𝑗=1

𝑑_𝑗𝑞_𝑗|𝐵_𝑀|^𝑞𝑗−12𝑟^−1/3𝜆² 𝜎^𝑀

𝑧 𝑧 =𝜎^𝑀

𝜃 𝜃,

(3.121)

where

𝐴_𝑀 = 𝑟^1/3 𝜆²

+ 2𝜆 𝑟^1/6

−3 𝐵_𝑀 =𝑟^−1/3𝜆²+ 2𝑟^1/3

𝜆

−3.

(3.122)

The non-zero components of stress in region𝑆are 𝜎^𝑆

𝜌 𝜌 =−𝜂^𝑆+

𝑀

Õ

𝑖=1

𝑐_𝑖𝑝_𝑖|𝐴_𝑆|^𝑝𝑖−12𝑟^1/3 𝜆²

+

𝑁

Õ

𝑗=1

𝑑_𝑗𝑞_𝑗|𝐵_𝑆|^𝑞𝑗−1 2 𝜆²

(𝑟^−1/3𝜆²+𝑟^2/3)

𝜎^𝑆

𝜃 𝜃 =−𝜂^𝑆+

𝑀

Õ

𝑖=1

𝑐_𝑖𝑝_𝑖|𝐴_𝑆|^𝑝𝑖−12𝜆²𝑟^−2/3+

𝑁

Õ

𝑗=1

𝑑_𝑗𝑞_𝑗|𝐵_𝑆|^𝑞𝑗−1(2𝑟^−1/3𝜆²)

1+ 1 𝜆²

𝜎^𝑆

𝑧 𝑧 =−𝜂^𝑆+

𝑀

Õ

𝑖=1

𝑐_𝑖𝑝_𝑖|𝐴_𝑆|^𝑝𝑖−12𝑟^1/3+

𝑁

Õ

𝑗=1

𝑑_𝑗𝑞_𝑗|𝐵_𝑆|^𝑞𝑗−1

2𝑟^−1/3𝜆²+2𝑟^2/3 𝜆²

, (3.123) where

𝐴_𝑆= 𝑟^1/3 𝜆²

+𝑟^1/3+𝑟^−2/3𝜆²−3 𝐵_𝑆=𝑟^−1/3𝜆²+𝑟^−1/3+𝑟^2/3

𝜆²

−3.

(3.124)

Solving static equilibrium

Static equilibrium (in the absence of body forces) is obtained when divσ = 0. In cylindrical coordinates and for a symmetric tensorσ,

𝜕 𝜎_{𝜌 𝜌}

𝜕 𝜌 + 1

𝜌

𝜕 𝜎_{𝜌 𝜃}

𝜕 𝜃

+ 𝜎_{𝜌 𝜌}−𝜎_{𝜃 𝜃} 𝜌

+ 𝜕 𝜎_{𝜌 𝑧}

𝜕 𝑧

=0,

𝜕 𝜎_{𝜌 𝜃}

𝜕 𝜌 + 1

𝜌

𝜕 𝜎_{𝜃 𝜃}

𝜕 𝜃

+ 2𝜎_{𝜌 𝜃} 𝜌

+ 𝜕 𝜎_{𝜃 𝑧}

𝜕 𝑧

=0,

𝜕 𝜎_{𝜌 𝑧}

𝜕 𝜌 + 1

𝜌

𝜕 𝜎_{𝜃 𝑧}

𝜕 𝜃

+ 𝜎_{𝜌 𝑧} 𝜌

+ 𝜕 𝜎_{𝑧 𝑧}

𝜕 𝑧

=0.

(3.125)

The boundary conditions for this problem are as follows:

𝜎_{𝜌 𝜌}|𝜌=𝜌_𝑖 =−𝑝 (3.126)

𝜎_{𝜌 𝜌}|_{𝜌=𝜌𝑜} =0. (3.127)

For every case, the𝜃 and 𝑧equilibrium equations yield that 𝜂 =𝜂(𝜌). This leaves only the𝜌equilibrium equation to be solved. We will rewrite the stress expressions as

𝜎_{𝜌 𝜌} =−𝜂+𝜎ˆ_{𝜌 𝜌} (3.128)

𝜎_{𝜃 𝜃} =−𝜂+𝜎ˆ_{𝜃 𝜃}. (3.129)

Case 0

In Case 0, the entire balloon is in region 𝐿. The two boundary conditions for this case are:

−𝜂^𝐿(𝜌= 𝜌_𝑖) +𝜎ˆ ^𝐿

𝜌 𝜌(𝜌 = 𝜌_𝑖)=−𝑝 (3.130)

−𝜂^𝐿(𝜌 =𝜌_𝑜) +𝜎ˆ^𝐿

𝜌 𝜌(𝜌 =𝜌_𝑜)=0. (3.131) The𝜌 equilibrium equation yields

𝑑𝜎^𝐿

𝜌 𝜌

𝑑𝜌 + 1

𝜌

𝜎^𝐿

𝜌 𝜌−𝜎^𝐿

𝜃 𝜃

=0 (3.132)

𝑑(−𝜂^𝐿+𝜎ˆ^𝐿

𝜌 𝜌) 𝑑𝜌

+ 1 𝜌

ˆ 𝜎^𝐿

𝜌 𝜌−𝜎ˆ^𝐿

𝜃 𝜃

=0 (3.133)

∫ ^𝜌𝑜

𝜌𝑖

𝑑

−𝜂^𝐿 +𝜎ˆ^𝐿

𝜌 𝜌

=

∫ ^𝜌𝑜

𝜌𝑖

0𝑑𝜌 (3.134)

𝑝 =0, (3.135)

and thus, for Case 1, the inner pressure is:

𝑝=0. (3.136)

Case 1

In Case 1, 𝑅 ∈ [𝑅_𝑖, 𝑅₂] is in region 𝑀, and the 𝑅 ∈ [𝑅₂, 𝑅_𝑜] is in 𝐿. The two boundary conditions can be rewritten specific to the region:

−𝜂^𝑀(𝜌= 𝜌_𝑖) +𝜎ˆ ^𝑀

𝜌 𝜌(𝜌 = 𝜌_𝑖)=−𝑝 (3.137)

−𝜂^𝐿(𝜌 =𝜌_𝑜) +𝜎ˆ^𝐿

𝜌 𝜌(𝜌 =𝜌_𝑜)=0, (3.138) and there is an additional boundary condition for continuity between the two regions:

−𝜂^𝐿(𝜌 =𝜌₂) +𝜎ˆ^𝐿

𝜌 𝜌(𝜌 =𝜌₂) =−𝜂^𝑀(𝜌 =𝜌₂) +𝜎ˆ^𝑀

𝜌 𝜌(𝜌= 𝜌₂). (3.139) First, we solve the equilibrium equations in region 𝐿. The𝜌equation yields:

𝑑𝜎^𝐿

𝜌 𝜌

𝑑𝜌 + 1

𝜌

𝜎^𝐿

𝜌 𝜌−𝜎^𝐿

𝜃 𝜃

=0 (3.140)

𝑑(−𝜂^𝐿 +𝜎ˆ^𝐿

𝜌 𝜌) 𝑑𝜌

=0 (3.141)

∫ ^𝜌𝑜

𝜌₂

𝑑

−𝜂^𝐿 +𝜎ˆ^𝐿

𝜌 𝜌

=

∫ ^𝜌𝑜

𝜌₂

0𝑑𝜌 (3.142)

−𝜂^𝐿+𝜎ˆ^𝐿

𝜌 𝜌

|^𝜌𝜌^𝑜₂ =0 (3.143)

𝜂^𝑀(𝜌₂) −𝜎ˆ^𝑀

𝜌 𝜌(𝜌₂) =0. (3.144)

In region M, the𝜌equation yields:

𝑑𝜎^𝑀

𝜌 𝜌

𝑑𝜌 + 1

𝜌

𝜎^𝑀

𝜌 𝜌−𝜎^𝑀

𝜃 𝜃

=0 (3.145)

𝑑(−𝜂^𝑀 +𝜎ˆ ^𝑀

𝜌 𝜌) 𝑑𝜌

+ 1 𝜌

ˆ 𝜎^𝑀

𝜌 𝜌−𝜎ˆ^𝑀

𝜃 𝜃

=0 (3.146)

∫ 𝜌₂ 𝜌𝑖

𝑑

−𝜂^𝑀 +𝜎ˆ^𝑀

𝜌 𝜌

=

∫ 𝜌₂ 𝜌𝑖

1 𝜌

ˆ 𝜎^𝑀

𝜃 𝜃 −𝜎ˆ^𝑀

𝜌 𝜌

𝑑𝜌 (3.147)

𝑝 =

∫ ^𝜌₂

𝜌_𝑖

1 𝜌

ˆ 𝜎^𝑀

𝜃 𝜃 −𝜎ˆ^𝑀

𝜌 𝜌

𝑑𝜌 . (3.148)

The right-hand side requires a change of integration variable from 𝜌 to 𝑅. Noting that𝑑𝜌= _𝜆(¹_𝑅)2𝑑𝑅, we obtain an expression for the inner pressure:

𝑝=

∫ ^𝑅₂

𝑅𝑖

1 𝜆(𝑅)³𝑅

𝜎ˆ ^𝑀

𝜃 𝜃(𝜌=𝜆(𝑅)𝑅) −𝜎ˆ ^𝑀

𝜌 𝜌(𝜌 =𝜆(𝑅)𝑅)

𝑑𝑅 . (3.149) Case 2

In Case 2,𝑅 ∈ [𝑅_𝑖, 𝑅_𝑜]is in region𝑀. The boundary conditions are:

−𝜂(𝜌_𝑖) +𝜎ˆ^𝑀

𝜌 𝜌(𝜌_𝑖) =−𝑝 (3.150)

−𝜂(𝜌_𝑜) +𝜎ˆ^𝑀

𝜌 𝜌(𝜌_𝑜) =0. (3.151)

Similar to the previous cases, the𝜌equilibrium equation yields the pressure:

𝑝 =

∫ ^𝑅𝑜

𝑅_𝑖

1 𝜆(𝑅)³𝑅

𝜎ˆ^𝑀

𝜃 𝜃(𝜌 =𝜆(𝑅)𝑅) −𝜎ˆ^𝑀

𝜌 𝜌(𝜌 =𝜆(𝑅)𝑅)

𝑑𝑅 . (3.152) Case 3

In Case 3, 𝑅 ∈ [𝑅_𝑖, 𝑅₁] is in region 𝑆 and 𝑅 ∈ [𝑅₁, 𝑅_𝑜] is in region 𝑀. The boundary conditions are as follows:

−𝜂^𝑀(𝜌 =𝜌_𝑜) +𝜎ˆ^𝑀

𝜌 𝜌(𝜌= 𝜌_𝑜) =0 (3.153)

−𝜂^𝑆(𝜌 = 𝜌_𝑖) +𝜎ˆ^𝑆

𝜌 𝜌(𝜌 =𝜌_𝑖) =−𝑝, (3.154) and there is an additional boundary condition for continuity between the two regions:

−𝜂^𝑀(𝜌 = 𝜌₁) +𝜎ˆ^𝑀

𝜌 𝜌(𝜌= 𝜌₁) =−𝜂^𝑆(𝜌= 𝜌₁) +𝜎ˆ^𝑆

𝜌 𝜌(𝜌 =𝜌₁). (3.155) In region 𝑀, the𝜌equilibrium equation yields

𝜂^𝑆(𝜌₁) −𝜎ˆ^𝑆

𝜌 𝜌(𝜌₁) =int₁ (3.156)

=

∫ 𝑅𝑜

𝑅₁

1 𝜆(𝑅)³𝑅

𝜎ˆ^𝑀

𝜃 𝜃(𝜌 =𝜆(𝑅)𝑅) −𝜎ˆ^𝑀

𝜌 𝜌(𝜌 =𝜆(𝑅)𝑅) 𝑑𝑅 . (3.157)

In region𝑆, we have

∫ ^𝜌₁

𝜌_𝑖

𝑑

−𝜂^𝑆+𝜎ˆ^𝑆

𝜌 𝜌

=

∫ ^𝜌₁

𝜌_𝑖

1 𝜌

ˆ 𝜎^𝑆

𝜃 𝜃 −𝜎ˆ^𝑆

𝜌 𝜌

𝑑𝜌 (3.158)

−𝜂^𝑆(𝜌₁) +𝜎ˆ^𝑆

𝜌 𝜌(𝜌₁) −𝜎ˆ^𝑆

𝜌 𝜌(𝜌_𝑖) +𝜂^𝑆(𝜌_𝑖)=

∫ 𝜌₁ 𝜌𝑖

1 𝜌

ˆ 𝜎^𝑆

𝜃 𝜃 −𝜎ˆ^𝑆

𝜌 𝜌

𝑑𝜌 (3.159)

−int₁+ 𝑝=int₂, (3.160)

where

int₂=𝜎ˆ^𝑆

𝜌 𝜌(𝜌₁) =

∫ ^𝑅₁

𝑅𝑖

1 𝜆(𝑅)³𝑅

𝜎ˆ^𝑆

𝜃 𝜃(𝜌 =𝜆(𝑅)𝑅) −𝜎ˆ^𝑆

𝜌 𝜌(𝜌 =𝜆(𝑅)𝑅) 𝑑𝑅 .

(3.161) This yields the result

𝑝 =int₁+int₂, (3.162)

or

𝑝 =

∫ ^𝑅𝑜

𝑅₁

1 𝜆(𝑅)³𝑅

𝜎ˆ^𝑀

𝜃 𝜃(𝜌=𝜆(𝑅)𝑅) −𝜎ˆ ^𝑀

𝜌 𝜌(𝜌 =𝜆(𝑅)𝑅) 𝑑𝑅

+

∫ ^𝑅₁

𝑅𝑖

1 𝜆(𝑅)³𝑅

𝜎ˆ^𝑆

𝜃 𝜃(𝜌 =𝜆(𝑅)𝑅) −𝜎ˆ^𝑆

𝜌 𝜌(𝜌=𝜆(𝑅)𝑅) 𝑑𝑅 .

(3.163)

Case 4

In Case 4,𝑅 ∈ [𝑅_𝑖, 𝑅₁]is in region𝑆,𝑅 ∈ [𝑅₁, 𝑅₂]is in region𝑀, and𝑅 ∈ [𝑅₂, 𝑅_𝑜] is in region𝐿. The boundary conditions are

−𝜂^𝐿(𝜌 =𝜌_𝑜) +𝜎ˆ^𝐿

𝜌 𝜌(𝜌= 𝜌_𝑜) =0 (3.164)

−𝜂^𝑆(𝜌 = 𝜌_𝑖) +𝜎ˆ^𝑆

𝜌 𝜌(𝜌 =𝜌_𝑖) =−𝑝, (3.165) and there are additional boundary conditions for continuity between the regions:

−𝜂^𝑀(𝜌 =𝜌₁) +𝜎ˆ ^𝑀

𝜌 𝜌(𝜌 =𝜌₁) =−𝜂^𝑆(𝜌 =𝜌₁) +𝜎ˆ^𝑆

𝜌 𝜌(𝜌 = 𝜌₁) (3.166)

−𝜂^𝑀(𝜌 =𝜌₂) +𝜎ˆ ^𝑀

𝜌 𝜌(𝜌 =𝜌₂) =−𝜂^𝐿(𝜌 = 𝜌₂) +𝜎ˆ^𝐿

𝜌 𝜌(𝜌= 𝜌₂). (3.167) Region 𝐿yields the result

𝜂^𝐿(𝜌₂) −𝜎ˆ ^𝐿

𝜌 𝜌(𝜌₂) =0. (3.168)

Region 𝑀 yields

−𝜂^𝑀(𝜌₂) +𝜎ˆ^𝑀

𝜌 𝜌(𝜌₂) −𝜎ˆ^𝑀

𝜌 𝜌(𝜌₁) +𝜂^𝑀(𝜌₁) =

∫ ^𝜌₂

𝜌₁

1 𝜌

ˆ 𝜎^𝑀

𝜃 𝜃 −𝜎ˆ^𝑀

𝜌 𝜌

𝑑𝜌 (3.169)

−𝜂^𝐿(𝜌₂) +𝜎ˆ^𝐿

𝜌 𝜌(𝜌₂) −𝜎ˆ^𝑆

𝜌 𝜌(𝜌₁) +𝜂^𝑆(𝜌₁) =

∫ ^𝜌₂

𝜌₁

1 𝜌

ˆ 𝜎^𝑀

𝜃 𝜃 −𝜎ˆ^𝑀

𝜌 𝜌

𝑑𝜌 (3.170) 𝜂^𝑆(𝜌₁) −𝜎ˆ^𝑆

𝜌 𝜌(𝜌₁) =

∫ ^𝜌₂

𝜌₁

1 𝜌

ˆ 𝜎^𝑀

𝜃 𝜃 −𝜎ˆ^𝑀

𝜌 𝜌

𝑑𝜌, (3.171) so

𝜂^𝑆(𝜌₁) −𝜎ˆ^𝑆

𝜌 𝜌(𝜌₁)=int₁ (3.172)

=

∫ ^𝑅₂

𝑅₁

1 𝜆(𝑅)³𝑅

𝜎ˆ ^𝑀

𝜃 𝜃(𝜌 =𝜆(𝑅)𝑅) −𝜎ˆ^𝑀

𝜌 𝜌(𝜌 =𝜆(𝑅)𝑅) 𝑑𝑅 . (3.173) Region𝑆yields

−𝜂^𝑆(𝜌₁) +𝜎ˆ^𝑆

𝜌 𝜌(𝜌₁) −𝜎ˆ^𝑆

𝜌 𝜌(𝜌_𝑖) +𝜂^𝑆(𝜌_𝑖) =

∫ 𝜌₁ 𝜌𝑖

1 𝜌

ˆ 𝜎^𝑀

𝜃 𝜃 −𝜎ˆ ^𝑀

𝜌 𝜌

𝑑𝜌 (3.174)

−int₁+𝑝 =

∫ ^𝜌₁

𝜌_𝑖

1 𝜌

ˆ 𝜎^𝑀

𝜃 𝜃 −𝜎ˆ ^𝑀

𝜌 𝜌

𝑑𝜌 (3.175)

−int₁+𝑝 =int₂, (3.176)

where int₂=

∫ 𝑅₁ 𝑅𝑖

1 𝜆(𝑅)³𝑅

𝜎ˆ^𝑆

𝜃 𝜃(𝜌 =𝜆(𝑅)𝑅) −𝜎ˆ^𝑆

𝜌 𝜌(𝜌 =𝜆(𝑅)𝑅)

𝑑𝑅 . (3.177) Thus,

𝑝 =int₁+int₂, (3.178)

or

𝑝 =

∫ ^𝑅₂

𝑅₁

1 𝜆(𝑅)³𝑅

𝜎ˆ^𝑀

𝜃 𝜃(𝜌 =𝜆(𝑅)𝑅) −𝜎ˆ^𝑀

𝜌 𝜌(𝜌 =𝜆(𝑅)𝑅) 𝑑𝑅

+

∫ 𝑅₁ 𝑅𝑖

1 𝜆(𝑅)³𝑅

𝜎ˆ^𝑆

𝜃 𝜃(𝜌 =𝜆(𝑅)𝑅) −𝜎ˆ^𝑆

𝜌 𝜌(𝜌=𝜆(𝑅)𝑅) 𝑑𝑅 .

(3.179)

Case 5

In Case 5,𝑅 ∈ [𝑅_𝑖, 𝑅_𝑜]is in region𝑆. The boundary conditions are as follows:

−𝜂^𝑆(𝜌 =𝜌_𝑜) +𝜎ˆ^𝑆

𝜌 𝜌(𝜌 =𝜌_𝑜) =0 (3.180)

−𝜂^𝑆(𝜌 = 𝜌_𝑖) +𝜎ˆ^𝑆

𝜌 𝜌(𝜌 =𝜌_𝑖) =−𝑝 . (3.181) The𝜌 equilibrium equation yields the result

𝑝=

∫ ^𝑅𝑜

𝑅_𝑖

1 𝜆(𝑅)³𝑅

𝜎ˆ^𝑆

𝜃 𝜃(𝜌 =𝜆(𝑅)𝑅) −𝜎ˆ^𝑆

𝜌 𝜌(𝜌 =𝜆(𝑅)𝑅)

𝑑𝑅 . (3.182)

Results

The calculations were performed inMATLABwith inner radius 𝑅_𝑖 =1 cm and outer radius 𝑅_𝑜 = 1.1 cm. The parameters that were used for the generalized Mooney- Rivlin model are the same as for the spherical balloon except with 𝑝₂ = 5, which yields an effective shear modulus 𝜇 = ¹₂

Í𝑁

𝑖=1𝑐_𝑖𝑝₁+Í𝑀 𝑗=1𝑑_𝑗𝑞_𝑗

= 6.55·10⁴ Pa.

The results for the inflation a cylindrical balloon with fixed axial stretch can be seen in Figures 3.8–3.9b.

Figure 3.8: Comparison of this work’s generalized Mooney-Rivlin model with the trace formula model of [23] for inflation pressure of a cylindrical balloon.

Figure 3.8 compares the results from the generalized Mooney-Rivlin model with that of the BTW theory (which features a neo-Hookean energy structure). Similarly to the spherical balloon, the BTW model is unable to capture the correct effect of elasticity in the material at very high values of stretch, so the two curves deviate starting around a stretch of𝜆_𝑜 ≈ 3. This work’s Mooney-Rivlin model captures the physics of the rubber extension at high stretch correctly.

Figure 3.9a and 3.9b shows the results for the balloon inflation at varying anisotropy parameter𝑟. The response of the balloon is stiffest in the isotropic state (𝑟 = 1), and gets correspondingly softer as 𝑟 increases, as expected. We also note that for 𝑟 = 1, the deformation of the balloon passes through (𝜆_𝑜, 𝑝/𝜇) = (1,0) in its undeformed state and then the pressure begins immediately rising upon𝜆_𝑜becoming greater than 1. However, for𝑟 > 1, the fact that the balloon moves through Case 0 for finite values of 𝜆_𝑜 means that the balloon starts inflating at points that are not at (𝜆_𝑜, 𝑝/𝜇) = (1,0), as mentioned in Chapter 2. This is due to the fact that there is a spontaneous deformation associated with the nematic state, because the liquid crystal molecules spontaneously orient along a preferred direction and

(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