Spherical balloon

UNIVERSAL DEFORMATIONS OF NEMATIC ELASTOMERS

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

3.5 Spherical balloon

Deformation of spherical balloon expansion

We are interested in the deformation of a balloon, which can be modeled as a spherical shell subjected to an internal pressure. In the undeformed configuration, the balloon has inner radius 𝑅_𝑖 and outer radius 𝑅_𝑜. The internal pressure is denoted by𝑝. The spherical balloon is assumed to remain spherical throughout the deformation. The undeformed sphere has radial coordinate𝑅 ∈ [𝑅_𝑖, 𝑅_𝑜], azimuthal angle Θ ∈ [0,2𝜋), and polar angle Φ ∈ [0, 𝜋), while the deformed sphere has coordinate system 𝜌 ∈ [𝜌_𝑖, 𝜌_𝑜], 𝜃 ∈ [0,2𝜋), and 𝜙 ∈ [0, 𝜋). Following Ericksen, we make the ansatz

















𝜌 =𝜌(𝑅) 𝜃 = Θ 𝜙 = Φ

. (3.49)

The deformation gradient in spherical coordinates is

F =©

­

­

«

𝑑𝜌 𝑑𝑅

𝜌 𝑅

𝜌 𝑅

ª

®

®

®

¬

. (3.50)

With incompressibility, detF =1, we obtain this first-order differential equation 𝑑𝜌

𝑑𝑅 𝜌² 𝑅²

=1. (3.51)

Solving for the deformed radius 𝜌as a function of the undeformed radius𝑅: 𝜌 =

𝑅³+𝑐

1/3

, (3.52)

where𝑐is a constant. Let𝜆denote the azimuthal stretch,𝜆= ^𝜌_𝑅, and let𝜆_𝑜denote the azimuthal stretch at the outer radius,𝜆_𝑜 = ^𝜌(_𝑅^𝑅𝑜)

𝑜 . Then, 𝜆_𝑜 =

𝑅³

𝑜+𝑐1/3

𝑅_𝑜

. (3.53)

Solving for𝑐and plugging𝑐back into Equation 3.52 yields 𝜌 =

𝑅³+𝑅³

𝑜

𝜆³

𝑜−1

1/3

, (3.54)

and the azimuthal stretch is

𝜆= 1+ 𝑅_𝑜

𝑅 3

𝜆³

𝑜−1

!1/3

. (3.55)

Thus, the deformation gradient is

F =©

­

­

«

1 𝜆²

𝜆 𝜆

ª

®

®

®

¬

, (3.56)

and the left Cauchy-Green tensor,b=F F^> is

b=©

­

­

«

1 𝜆⁴

𝜆² 𝜆²

ª

®

®

®

¬

. (3.57)

The principal stretches are𝜆₁ =𝜆₂ = 𝜆(corresponding toe𝜃 ande𝜙) and𝜆₃ = ¹

𝜆²

(corresponding toe𝜌). Thus, 𝑠=𝜆and𝑡 =𝜆². The regions are:

𝐿 : {𝑅 ≥ 𝑅^∗} (3.58)

𝑀 : {𝑅 ≤ 𝑅^∗, 𝑟 ≥ 1} (3.59)

𝑆: {𝑟 ≤ 1}, (3.60)

where

𝑅^∗ =𝑅_𝑜 𝜆³

𝑜−1 𝑟^1/4−1

1/3

. (3.61)

This leads to the possibility of three cases:

• Case 1: 𝑟 > 1 and𝑅^∗ ≤ 𝑅_𝑖: the entire balloon is in region𝐿

• Case 2: 𝑟 >1 and𝑅_𝑖 < 𝑅^∗ < 𝑅_𝑜: the inner region of the balloon𝑅 ∈ [𝑅_𝑖, 𝑅^∗] is in 𝑀, and the outer region𝑅 ∈ [𝑅^∗, 𝑅_𝑜] is in𝐿

Figure 3.3: Diagram of all possible cases in the inflation of a nematic elastomer spherical balloon.

• Case 3: 𝑟 > 1 and𝑅^∗ ≥ 𝑅_𝑜: the entire balloon is in region 𝑀

• Case 4: 𝑟 =1 and the entire balloon is in region𝑆

A diagram illustrating the various cases can be seen in Figure 3.3.

Note that Figure 3.1 can provide us with insight into the regions that this deformation will experience. The spherical balloon expansion is merely equibiaxial stretch (see the deformation gradient of Equation 3.56). Recalling that𝑠 is the largest singular value of F and𝑡 is the largest singular value of cofF, this means that 𝑡 = 𝑠² for this deformation. Following along the 𝑡 = 𝑠² curve in Figure 3.1, we see that the deformation will move progressively through region𝐿then𝑀, never touching𝑆for 𝑟 >1.

Stress

The expressions for stress are as follows: In region𝐿,

σ^𝐿 =−𝜂^𝐿I. (3.62)

In region 𝑀, the non-zero components of the stress are:

𝜎^𝑀

𝜌 𝜌=−𝜂^𝑀 +

𝑀

Õ

𝑖=1

𝑐_𝑖𝑝_𝑖|𝐴_𝑀|^𝑝𝑖−12𝑟^1/3 𝜆⁴

+

𝑁

Õ

𝑗=1

𝑑_𝑗𝑞_𝑗|𝐵_𝑀|^𝑞𝑗−12𝑟^1/6 𝜆²

𝜎^𝑀

𝜃 𝜃 =−𝜂^𝑀 +

𝑀

Õ

𝑖=1

𝑐_𝑖𝑝_𝑖|𝐴_𝑀|^𝑝𝑖−12𝜆² 𝑟^1/6

+

𝑁

Õ

𝑗=1

𝑑_𝑗𝑞_𝑗|𝐵_𝑀|^𝑞𝑗−12𝑟^−1/3𝜆⁴ 𝜎^𝑀

𝜙 𝜙 =𝜎^𝑀

𝜃 𝜃,

(3.63)

where

𝐴_𝑀 = 𝑟^1/3 𝜆⁴

+ 2𝜆² 𝑟^1/6

−3 𝐵_𝑀 =𝑟^−1/3𝜆⁴+ 2𝑟^1/6 𝜆²

−3.

(3.64)

In region𝑆,𝑟 =1 and the non-zero components of the stress are 𝜎^𝑆

𝜌 𝜌 =−𝜂^𝑆+

𝑀

Õ

𝑖=1

𝑐_𝑖𝑝_𝑖|𝐴_𝑆|^𝑝𝑖−1 2 𝜆⁴

+

𝑁

Õ

𝑗=1

𝑑_𝑗𝑞_𝑗|𝐵_𝑆|^𝑞𝑗−1 4 𝜆²

𝜎^𝑆

𝜃 𝜃 =−𝜂^𝑆+

𝑀

Õ

𝑖=1

𝑐_𝑖𝑝_𝑖|𝐴_𝑆|^𝑝𝑖−1(2𝜆²) +

𝑁

Õ

𝑗=1

𝑑_𝑗𝑞_𝑗|𝐵_𝑆|^𝑞𝑗−1(2𝜆²)

𝜆²+ 1 𝜆⁴

𝜎^𝑆

𝜙 𝜙 =𝜎^𝑆

𝜃 𝜃,

(3.65)

where

𝐴_𝑆= 1 𝜆⁴

+2𝜆²−3 𝐵_𝑆=𝜆⁴+ 2

𝜆²

−3.

(3.66)

Solving static equilibrium

The static equilibrium equations (in the absence of body forces) in spherical coor- dinates are

𝜌:

𝜕 𝜎_{𝜌 𝜌}

𝜕 𝜌 +2

𝜎_{𝜌 𝜌} 𝜌

+ 1 𝜌

𝜕 𝜎_{𝜙 𝜌}

𝜕 𝜙

+ cot𝜙 𝜌

𝜎_{𝜙 𝜌}+ 1 𝜌sin𝜙

𝜕 𝜎_{𝜃 𝜌}

𝜕 𝜃

− 1 𝜌

𝜎_{𝜃 𝜃} +𝜎_{𝜙 𝜙}

=0 (3.67) 𝜃 :

𝜕 𝜎_{𝜌 𝜃}

𝜕 𝜌 +2

𝜎_{𝜌 𝜃} 𝜌

+ 1 𝜌

𝜕 𝜎_𝜙𝜃

𝜕 𝜙

+ 1

𝜌sin𝜙

𝜕 𝜎_{𝜃 𝜃}

𝜕 𝜃 +

𝜎_{𝜃 𝜌} 𝜌

+ cot𝜙 𝜌

𝜎_𝜙𝜃 +𝜎_{𝜃 𝜙}

=0 (3.68) 𝜙:

𝜕 𝜎_{𝜌 𝜙}

𝜕 𝜌 +2

𝜎_{𝜌 𝜙} 𝜌

+ 1 𝜌

𝜕 𝜎_{𝜙 𝜙}

𝜕 𝜙

+ 1

𝜌sin𝜙

𝜕 𝜎_{𝜃 𝜙}

𝜕 𝜃 +

𝜎_{𝜙 𝜌} 𝜌

+cot𝜙 𝜌

𝜎_{𝜙 𝜙}−𝜎_{𝜃 𝜃}

=0, (3.69) and the boundary conditions for this problem are as follows:

𝜎_{𝜌 𝜌}|𝜌=𝜌𝑖 =−𝑝 (3.70)

𝜎_{𝜌 𝜌}|𝜌=𝜌_𝑜 =0. (3.71)

In all cases, solving the 𝜙 equation yields 𝜂 = 𝜂(𝜌, 𝜃). Similarly, solving the 𝜃 equation, we find that 𝜂 = 𝜂(𝜌). Finally, to solve the 𝜌 equation, we will rewrite the stress expressions as

𝜎_{𝜌 𝜌} =−𝜂+𝜎ˆ_{𝜌 𝜌} (3.72)

𝜎_{𝜃 𝜃} =𝜎_{𝜙 𝜙} =−𝜂+𝜎ˆ_{𝜃 𝜃}, (3.73) and so the boundary conditions can be rewritten as follows

−𝜂(𝜌 =𝜌_𝑖) +𝜎ˆ_{𝜌 𝜌}(𝜌= 𝜌_𝑖) =−𝑝 (3.74)

−𝜂(𝜌 = 𝜌_𝑜) +𝜎ˆ_{𝜌 𝜌}(𝜌 = 𝜌_𝑜) =0. (3.75)

Case 1

In Case 1, 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 (3.76)

−𝜂^𝐿(𝜌 =𝜌_𝑜) +𝜎ˆ^𝐿

𝜌 𝜌(𝜌 =𝜌_𝑜)=0. (3.77) The stress in this region is found in Equation 3.62. Thus, the𝜌equilibrium equation yields

𝑑𝜎^𝐿

𝜌 𝜌

𝑑𝜌 +2

𝜎^𝐿

𝜌 𝜌

𝜌

− 1 𝜌

𝜎^𝐿

𝜃 𝜃 +𝜎^𝐿

𝜙 𝜙

=0 (3.78) 𝑑(−𝜂^𝐿+𝜎ˆ^𝐿

𝜌 𝜌) 𝑑𝜌

+2

(−𝜂^𝐿+𝜎ˆ^𝐿

𝜌 𝜌) 𝜌

− 1 𝜌

−𝜂^𝐿 +𝜎ˆ ^𝐿

𝜃 𝜃−𝜂^𝐿 +𝜎ˆ^𝐿

𝜙 𝜙

=0 (3.79)

∫ ^𝜌𝑜

𝜌_𝑖

𝑑

−𝜂^𝐿 +𝜎ˆ^𝐿

𝜌 𝜌

=

∫ ^𝜌𝑜

𝜌_𝑖

0𝑑𝜌 (3.80) 𝑝=0 (3.81) Thus, for Case 1, the inner pressure is:

𝑝=0. (3.82)

Case 2

In Case 2, 𝑅 ∈ [𝑅_𝑖, 𝑅^∗] is in region 𝑀, and the 𝑅 ∈ [𝑅^∗, 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 (3.83)

−𝜂^𝐿(𝜌 =𝜌_𝑜) +𝜎ˆ^𝐿

𝜌 𝜌(𝜌 =𝜌_𝑜)=0, (3.84) 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^∗). (3.85)

First, we solve the equilibrium equations in region 𝐿. Again, the stress in region𝐿 is found in Equation 3.62. The𝜌equation yields:

𝑑𝜎^𝐿

𝜌 𝜌

𝑑𝜌 +2

𝜎^𝐿

𝜌 𝜌

𝜌

− 1 𝜌

𝜎^𝐿

𝜃 𝜃 +𝜎^𝐿

𝜙 𝜙

=0 (3.86) 𝑑(−𝜂^𝐿+𝜎ˆ^𝐿

𝜌 𝜌) 𝑑𝜌

+2(−𝜂^𝐿+𝜎ˆ^𝐿

𝜌 𝜌) 𝜌

− 1 𝜌

−𝜂^𝐿 +𝜎ˆ ^𝐿

𝜃 𝜃−𝜂^𝐿 +𝜎ˆ^𝐿

𝜙 𝜙

=0 (3.87) 𝑑

−𝜂^𝐿 +𝜎ˆ^𝐿

𝜌 𝜌

𝑑𝜌

=0 (3.88)

−𝜂^𝐿+𝜎ˆ^𝐿

𝜌 𝜌

|^𝜌𝑜

𝜌^∗ =0 (3.89)

−𝜂^𝐿(𝜌_𝑜) +𝜎ˆ ^𝐿

𝜌 𝜌(𝜌_𝑜)

−

−𝜂^𝐿(𝜌^∗) +𝜎ˆ^𝐿

𝜌 𝜌(𝜌^∗)

=0 (3.90)

−𝜂^𝐿(𝜌^∗) +𝜎ˆ^𝐿

𝜌 𝜌(𝜌^∗)=0 (3.91) In region M, the stress is that of Eqns. 3.63 and 3.64. The𝜌equation yields:

𝑑𝜎^𝑀

𝜌 𝜌

𝑑𝜌 +2

𝜎^𝑀

𝜌 𝜌

𝜌

− 1 𝜌

𝜎^𝑀

𝜃 𝜃 +𝜎^𝑀

𝜙 𝜙

=0 (3.92) 𝑑(−𝜂^𝑀 +𝜎ˆ ^𝑀

𝜌 𝜌) 𝑑𝜌

+ 1 𝜌

2 ˆ𝜎^𝑀

𝜌 𝜌−𝜎ˆ ^𝑀

𝜃 𝜃 −𝜎ˆ^𝑀

𝜙 𝜙

=0 (3.93)

∫ ^𝜌∗

𝜌𝑖

𝑑

−𝜂^𝑀+𝜎ˆ^𝑀

𝜌 𝜌

=−

∫ ^𝜌∗

𝜌𝑖

2 𝜌

ˆ 𝜎^𝑀

𝜌 𝜌−𝜎ˆ ^𝑀

𝜃 𝜃

𝑑𝜌 (3.94)

−𝜂^𝑀(𝜌^∗) +𝜎ˆ ^𝑀

𝜌 𝜌(𝜌^∗)

−

−𝜂^𝑀(𝜌_𝑖) +𝜎ˆ^𝑀

𝜌 𝜌(𝜌_𝑖)

=

∫ 𝜌^∗ 𝜌𝑖

2 𝜌

ˆ 𝜎^𝑀

𝜃 𝜃 −𝜎ˆ^𝑀

𝜌 𝜌

𝑑𝜌 (3.95) 𝑝 =

∫ 𝜌^∗ 𝜌𝑖

2 𝜌

ˆ 𝜎^𝑀

𝜃 𝜃 −𝜎ˆ^𝑀

𝜌 𝜌

𝑑𝜌 . (3.96) The right-hand side requires a change of integration variable from 𝜌 to 𝑅. Noting that𝑑𝜌= ¹

𝜆(𝑅)²𝑑𝑅, we obtain an expression for the inner pressure:

𝑝 =

∫ ^𝑅∗

𝑅_𝑖

2 𝜆(𝑅)³𝑅

𝜎ˆ^𝑀

𝜃 𝜃(𝜌 =𝜆(𝑅)𝑅) −𝜎ˆ^𝑀

𝜌 𝜌(𝜌 =𝜆(𝑅)𝑅)

𝑑𝑅 . (3.97)

Case 3

In Case 3, all of the balloon is in region 𝑀. The two boundary conditions for this case are:

−𝜂^𝑀(𝜌= 𝜌_𝑖) +𝜎ˆ ^𝑀

𝜌 𝜌(𝜌 = 𝜌_𝑖) =−𝑝 (3.98)

−𝜂^𝑀(𝜌 =𝜌_𝑜) +𝜎ˆ^𝑀

𝜌 𝜌(𝜌 =𝜌_𝑜) =0. (3.99)

The stress in region 𝑀 can be found in Eqns. 3.63 and 3.64. The 𝜌 equilibrium equation yields:

𝑑𝜎^𝑀

𝜌 𝜌

𝑑𝜌 +2

𝜎^𝑀

𝜌 𝜌

𝜌

− 1 𝜌

𝜎^𝑀

𝜃 𝜃 +𝜎^𝑀

𝜙 𝜙

=0 (3.100) 𝑑(−𝜂^𝑀 +𝜎ˆ ^𝑀

𝜌 𝜌) 𝑑𝜌

+ 1 𝜌

2 ˆ𝜎^𝑀

𝜌 𝜌−𝜎ˆ ^𝑀

𝜃 𝜃 −𝜎ˆ^𝑀

𝜙 𝜙

=0 (3.101)

∫ ^𝜌𝑜

𝜌_𝑖

𝑑

−𝜂^𝑀 +𝜎ˆ^𝑀

𝜌 𝜌

=−

∫ ^𝜌𝑜

𝜌_𝑖

2 𝜌

ˆ 𝜎^𝑀

𝜌 𝜌−𝜎ˆ ^𝑀

𝜃 𝜃

𝑑𝜌 (3.102)

−𝜂^𝑀(𝜌_𝑜) +𝜎ˆ^𝑀

𝜌 𝜌(𝜌_𝑜)

−

−𝜂^𝑀(𝜌_𝑖) +𝜎ˆ^𝑀

𝜌 𝜌(𝜌_𝑖)

=

∫ ^𝜌𝑜

𝜌_𝑖

2 𝜌

ˆ 𝜎^𝑀

𝜃 𝜃 −𝜎ˆ^𝑀

𝜌 𝜌

𝑑𝜌 (3.103) 𝑝=

∫ ^𝜌𝑜

𝜌_𝑖

2 𝜌

ˆ 𝜎^𝑀

𝜃 𝜃 −𝜎ˆ^𝑀

𝜌 𝜌

𝑑𝜌 .

(3.104) As in Case 2, we employ a change of integration variable from 𝜌 to 𝑅 and obtain the inner pressure as:

𝑝 =

∫ 𝑅𝑜

𝑅𝑖

2 𝜆(𝑅)³𝑅

𝜎ˆ^𝑀

𝜃 𝜃(𝜌 =𝜆(𝑅)𝑅) −𝜎ˆ^𝑀

𝜌 𝜌(𝜌 =𝜆(𝑅)𝑅)

𝑑𝑅 . (3.105) Case 4

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

−𝜂^𝑆(𝜌 =𝜌_𝑖) +𝜎ˆ^𝑆

𝜌 𝜌(𝜌= 𝜌_𝑖) =−𝑝 (3.106)

−𝜂^𝑆(𝜌 = 𝜌_𝑜) +𝜎ˆ^𝑆

𝜌 𝜌(𝜌 = 𝜌_𝑜) =0. (3.107)

The stress in region 𝑆 can be found in Eqns. 3.65 and 3.66. The 𝜌 equilibrium equation yields:

𝑑𝜎^𝑀

𝜌 𝜌

𝑑𝜌 +2

𝜎^𝑀

𝜌 𝜌

𝜌

− 1 𝜌

𝜎^𝑀

𝜃 𝜃 +𝜎^𝑀

𝜙 𝜙

=0 (3.108) 𝑑(−𝜂^𝑀 +𝜎ˆ ^𝑀

𝜌 𝜌) 𝑑𝜌

+ 1 𝜌

2 ˆ𝜎^𝑀

𝜌 𝜌−𝜎ˆ ^𝑀

𝜃 𝜃 −𝜎ˆ^𝑀

𝜙 𝜙

=0 (3.109)

∫ ^𝜌𝑜

𝜌_𝑖

𝑑

−𝜂^𝑀 +𝜎ˆ^𝑀

𝜌 𝜌

=−

∫ ^𝜌𝑜

𝜌_𝑖

2 𝜌

ˆ 𝜎^𝑀

𝜌 𝜌−𝜎ˆ ^𝑀

𝜃 𝜃

𝑑𝜌 (3.110)

−𝜂^𝑀(𝜌_𝑜) +𝜎ˆ^𝑀

𝜌 𝜌(𝜌_𝑜)

−

−𝜂^𝑀(𝜌_𝑖) +𝜎ˆ^𝑀

𝜌 𝜌(𝜌_𝑖)

=

∫ ^𝜌𝑜

𝜌_𝑖

2 𝜌

ˆ 𝜎^𝑀

𝜃 𝜃 −𝜎ˆ^𝑀

𝜌 𝜌

𝑑𝜌 (3.111) 𝑝=

∫ ^𝜌𝑜

𝜌_𝑖

2 𝜌

ˆ 𝜎^𝑀

𝜃 𝜃 −𝜎ˆ^𝑀

𝜌 𝜌

𝑑𝜌 .

(3.112) As in Case 2, we employ a change of integration variable from 𝜌 to 𝑅 and obtain the inner pressure as:

𝑝 =

∫ ^𝑅𝑜

𝑅𝑖

2 𝜆(𝑅)³𝑅

𝜎ˆ^𝑀

𝜃 𝜃(𝜌 =𝜆(𝑅)𝑅) −𝜎ˆ^𝑀

𝜌 𝜌(𝜌 =𝜆(𝑅)𝑅)

𝑑𝑅 . (3.113) Results

The calculations were performed inMATLAB. The solutions are plotted in Figures 3.4–

3.5b with the inner radius 𝑅_𝑖 =1 cm and outer radius 𝑅_𝑜 =1.1 cm. The following parameters for the generalized Mooney-Rivlin model were used: 𝑀 = 2, 𝑁 = 1, 𝑐₁=1.0·10⁵Pa,𝑐₂=1.90·10²Pa,𝑑₁ =1.59·10⁻²Pa,𝑝₁=1.3,𝑝₂=5, and𝑞₁=2.

This yields an effective shear modulus 𝜇= ¹₂ Í𝑁

𝑖=1𝑐_𝑖𝑝_𝑖+Í𝑀 𝑗=1𝑑_𝑗𝑞_𝑗

=6.52·10⁴ Pa.

The pressure𝑝, normalized by the effective shear modulus𝜇, is plotted as a function of the azimuthal stretch at the outer radius 𝜆_𝑜. Figure 3.4 shows the comparison between the generalized Mooney-Rivlin model of this work and the BTW (neo- Hookean) model of previous work in the field [23] for an anisotropy parameter of 𝑟 =8. The neo-Hookean model is unable to capture the correct effect of elasticity in the material at very high values of stretch (e.g. for𝜆_𝑜greater than about 6). As seen in experiments of rubber balloons [73], the balloon experiences a subsequent

Figure 3.4: Comparison of this work’s generalized Mooney-Rivlin model with the trace formula model of [23] for inflation pressure of a spherical balloon.

stiffening due to the further stretching of the polymer, or “effects of the limited extensibility of the network", as indicated by the increase in pressure at high stretch, which is correctly captured by our generalized Mooney-Rivlin model.

(a) (b)

Figure 3.5: Spherical balloon results: (a) Pressure-stretch curves at varying anisotropy parameter. (b) Progression of the spherical balloon solution through individual case numbers.

Figure 3.5a shows the results for 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. As the pressure 𝑝 increases, the balloon undergoes deformation according to the various cases, as seen in Figure 3.5b. For 𝑟 = 1 (rubber), the balloon is in Case 1 (region𝐿) at𝜆_𝑜=1 when there is no deformation, but for the rest of the deformation the balloon is entirely in region𝑆, corresponding to a solid-like response with no microstructure formation, since the material is a rubber with no liquid crystals. For𝑟 > 1, the solutions move progressively from Case 1

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.