Statistical models of elasticity in main chain and smectic liquid ...

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108 CHAPTER 4. THE ELASTICITY OF SMECTIC-A ELASTOMERSBefore layer rotation starts λ zx = 0 and so the layer spacing increases asd/d 0 = λ. After layer rotation starts wehave from thesame expression d/d 0 =λ cr . Thus after rotation of the layers starts their spacing remains fixed, andthe only cost in deforming the system is that of shearing the rubber. Thisis because, as the layers rotate, the component of the force along the layersremains constant.The threshold behaviour shown here occurs even for very small values ofB. This is because the way in which the director deforms with the matrixhas been imposed. Physically this is correct for large B. As B is reduced thisconstraint will become less rigidly enforced and the cross-links will be able tomove through one layer to the next. Thus the threshold behaviour predictedfor small B values (∼ µ) is unlikely to be correct in practise.To check the trial solution outlined above and to ensure there are no othersolutions, all the solutions to the minimisation of the free energy are nowcalculated. To factorise the free energy, a change of variables is employed:the variables λ xx and tanφ = − λzxλ xxare used to express the free energy. Theexpression for the free energy density is then given byf = 1 2 µ {λ 2 xx + 1λ 2 xx λ2 +λ2 xxtan 2 φ+(cos 2 φ+rsin 2 φ)λ 2 +b(λcosφ−1) 2 }.Minimising this equation w.r.t. λ xx yields(0 = 2λ xx 1− 1 )λ 2 λ 4 +tan 2 φ . (4.121)xxThis equation has the solutions λ 2 xx = 0,±cosφ λ. The only physical solutionis λ 2 xx = cosφλ . Minimising the free energy w.r.t. φ and substituting for λ xxyields0 = sinφλcos 2 φ +λ2 (r −1)sinφcosφ−bλ(λsinφcosφ−sinφ) (4.122)This equation can be factorised using the solutions previously calculated asfollows(λsinφ cosφ− λ )(crλ 2 (r −1−b)cos 2 φ− λλλ 2 cosφ− 1 )= 0 (4.123)cr λ crprovided that Eq. (4.114) is obeyed. Thus the previous solution (a combinationof the first and second factors) is a minimum. The third factor can beshown to be real only if λ 2 crb < −3/4, and thus is never a relevant solutionin this case. The first solution is thus a minimum in the free energy and theonly solution.
4.3. FINITE DEFORMATION EXAMPLES 109Imposed λ xxA deformation matrix of the form⎛λ = ⎝⎞λ 0 λ xz0 λ yy 0 ⎠ (4.124)0 0 λ zzwill be used here. A λ zx element is not included here because it createsadditional, butphysically uninterestingrotating solutions. Therotation aboutthe y axis would take the layers into the stretch direction (the x direction)and would be energetically unfavourable. A λ xz component is included herebecause from the the equivalent experiment with a nematic elastomer it isexpectedtobenon-zero. However sinceit doesnotfacilitate layer rotation thiselement will turn out to be zero. This deformation tensor can be substitutedinto the free energy expression and the volume conservation constraint usedto eliminate λ zz . The resulting free energy density expression is given by{f = 1 2 µ λ 2 yy +rλ 2 xz +λ 2 + 1 ( ) }12λ 2 +b yy λ2 λ yy λ −1 , (4.125)This free energy density has only a few occurrences of λ xz so it can be easilyminimized w.r.t. λ xz yielding λ xz = 0For this deformation, it is instructive to calculate the Poisson ratios in the(y,z) directions for different values of b. Starting from Eq. (4.125) we makethe substitutions λ zx = 0 and λ xz = 0. The resulting free energy density isgiven by {f = 1 2 µ λ 2 yy +λ2 + 1 ( ) }12λ 2 +b yy λ2 λ yy λ −1 (4.126)Minimisation of this free energy w.r.t. λ yy results in the equationλ 2 λ 4 yy −1 = b(1−λλ yy) (4.127)From this equation it is clear that there are two limits of small and largeb corresponding to λ yy = 1 λ for large b and λ yy = √ 1λfor small b. Thematerial with a small b value is still a smectic in the sense that the directoris constrained to lie along the layer normal. To calculate the Poisson ratiosfrom this expression a small strain expansion is usedλ yy = 1+ǫ (4.128)λ = 1+ω. (4.129)The resulting expression to first order in ω and ǫ is given by4ǫ+2ω +b(ǫ+ω) = 0. (4.130)
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iStatistical models of elasticity i
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AcknowledgementsDuring the course o
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viiiCONTENTS5 Soft elasticity of sm
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2 CHAPTER 1. INTRODUCTION1.2 Neo-cl
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4 CHAPTER 1. INTRODUCTIONticular in
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Chapter TwoThe elasticity of hairpi
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2.1. INTRODUCTION 9Fig. 2.1. This b
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2.2. MODEL OF HAIRPIN CHAINS 11nFig
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2.2. MODEL OF HAIRPIN CHAINS 13za)
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2.2. MODEL OF HAIRPIN CHAINS 15in
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2.2. MODEL OF HAIRPIN CHAINS 17The
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2.2. MODEL OF HAIRPIN CHAINS 1910.8
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2.2. MODEL OF HAIRPIN CHAINS 21This
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2.2. MODEL OF HAIRPIN CHAINS 23f g
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2.3. NON-AFFINE DEFORMATION MODEL 2
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2.4. HAIRPIN CHAIN NETWORK FREE ENE
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2.4. HAIRPIN CHAIN NETWORK FREE ENE
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2.5. CONCLUSIONS 35to their limit
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2.A. GEOMETRY OF A HAIRPIN AT ZERO
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2.B. UNDULATING CHAIN STATISTICS 39
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2.C. COUNTING CONFIGURATIONS BY IND
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2.C. COUNTING CONFIGURATIONS BY IND
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2.D. INTERPRETATION OF FN 452.D Int
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2.E. SPRING CONSTANT OF AN EXTENDED
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50 CHAPTER 3. POLARISATION OF CHIRA
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Page 91 and 92: Chapter FourThe elasticity of smect
Page 93 and 94: 4.1. INTRODUCTION 83wherer 12 is th
Page 95 and 96: 4.1. INTRODUCTION 85HzyxGlobal rota
Page 97 and 98: 4.1. INTRODUCTION 87Figure 4.3: The
Page 99 and 100: 4.1. INTRODUCTION 89−D 2[(v (a)xz
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Page 127: 4.5. CONCLUSIONS 117formation in th
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Page 147 and 148: Bibliography[1] P. J. Flory. Statis
Page 149 and 150: 139[27] J-.C. Meiners and S. R. Qua
Page 151 and 152: 141[58] W. L. McMillan. Measurement
Page 153 and 154: 143[86] L. Golubovic and T. C. Lube
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