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Lecture11

# Lecture11 - EMA 6165 Polymer Physics Rubber Elasticity...

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Unformatted text preview: EMA 6165 Polymer Physics Rubber Elasticity Lecture 11 Dr. Anthony Brennan University of Florida Department of Materials Science & Engineering EMA 6165 Polymer Physics AB Brennan 1 Agenda Introduction Thermo-Elastic Behavior Statistical Mechanical Approach Uniaxial Deformation Biaxial Deformation Swollen State Deviations from Classical Theories Affine Deformation Phantom Behavior Mooney-Rivlin Model EMA 6165 Polymer Physics AB Brennan 2 Thermodynamics q From our previous lecture: z 0 G = N v kT ( q z - z -2 ) 3 The question was asked, "How can that be if Nv is for a volume? EMA 6165 Polymer Physics AB Brennan Thermodynamics q The answer lies in the definition of modulus: E y = lim 0 = - -2 ) ) ( N v kT ( 1 + 2 3 0 = lim 0 0 = ( N v kT q q For an extension ratio equal to 1: Furthermore, by definition: E y = 3 N v kT ( ) 4 E y = 2G( 1 + ) where is Poisson ratio. EMA 6165 Polymer Physics AB Brennan Thermodynamics Equilibrium Measurements for V = 0 , isotropic materials 0.5 for glassy and semi-crystalline materials thus and E = 3G G = N v kT Hooke's Law 5 Which is given directly by? Measurements yield values of Nv that are too large. Why? EMA 6165 Polymer Physics AB Brennan Thermodynamics q q q The answer: Entanglements. So how does Nv relate to Mc? Define Mc? q First count the no. of chain segments between junctions - Nc Now consider how the # of junctions relates to the Mn. If Nc increases, Mn must decrease! EMA 6165 Polymer Physics AB Brennan 6 q q Thermodynamics q Therefore: a relationship between density and Avogadro's number must exist: Nv = q N A Mn = mass No. * volume mol No. = mass Volume volume Now, consider the uniaxial stress relationship for an elastomer 0 = N v kT * - q Substitution yields: ( -2 ) N v kT 0 = * - Mc ( -2 ) where N Ak = R 7 EMA 6165 Polymer Physics AB Brennan Thermodynamics q Hence, one has: RT 0 = * - Mc ( -2 ) q Alternatively, this can be written using the following: l = l0 Take the derivative of stress w/ respect to l: l -3 dl d0 kTN v + 2l0l l0 l q Multiply and divide by l to get: 2 dl d 0 = kTN v + 2 l EMA 6165 Polymer Physics AB Brennan 8 Thermodynamics q Given the following: dl d = E l q One can show: RT 2 E= + 2 Mc q Now examine the significance of these relationships: q E will decrease q Mc increases q T increases - q E will increase q q for > 1 > < for < 1 EMA 6165 Polymer Physics AB Brennan E E - 2 9 Thermodynamics 3RT E= Mc q for =1 q Now let's evaluate the biaxial oriented elastomer Examples are: Garbage bag Packaging film Deform along x and y plane: EMA 6165 Polymer Physics AB Brennan x z q y y q x 10 Thermodynamics q Consider deformation along the x axis first: 0x = q N v kT * 2 x2 + y2 + z2 - 3 x ( ) Recalling that we can then show: x yz = 1 q 1 z = x y 2 1 2 x + y + 2 2 - 3 x y N v kT * 2 x 11 Sustitution gives: 0x = EMA 6165 Polymer Physics AB Brennan Thermodynamics q Then using simple algebra: x 0 = N kT ( - -3 v x x -2 y 0 = N v kT ( x - x y -2 y z -3 ) ) q True stress is calculated as follows: fz f z A0 A0 t = = = * = 0 * A A0 A Area 1 1 0 * * = 0z * * lx l y x y l0 x l0 y z EMA 6165 Polymer Physics AB Brennan 12 Thermodynamics q And 1 z = t z = 0z z x y q Conversely: tx = 0x x t y = 0 y y t z = 0z z 13 q Which gives us: EMA 6165 Polymer Physics AB Brennan Thermodynamics q Biaxial Stress in x direction t x = N v kT x + 2 - x - 2 y - 2 q ( ) Biaxial Stress in y direction t y = N v kT y + 2 - x - 2 y - 2 q ( ) 2 Differences in Principal Stresses: t x - t y = N v kT x -y ( 2 ) 14 EMA 6165 Polymer Physics AB Brennan Thermodynamics Let's apply this to a simple example: What happens when one tries? Difficult to start How do you help start it? You stretch it. Why? EMA 6165 Polymer Physics AB Brennan 15 Thermodynamics Why is it easier after the first difficult expansion? Consider the stresses: 0i = x2 + y2 + z2 - 3 N v kT * 2 i ( ) R t = 0 = f Area dR EMA 6165 Polymer Physics AB Brennan 16 Thermodynamics Express force in terms of pressure P= f f = 2 Area R f = P* R 2 PR P R t = 0 = = 2d 2 R d 2 EMA 6165 Polymer Physics AB Brennan Area = 2Rd 17 Thermodynamics Assume biaxial orientation: R = R0 x yz = 1 x = z 1 z = 2 Thus: And: z = 1 2 2 2 1 + + 4 - 3 0i = N v kT * 2 EMA 6165 Polymer Physics AB Brennan 18 Thermodynamics Which leads to: 0i = 2 1 2 + 4 - 3 N v kT * 2 x 0i = 0i = 2 1 PR t = 0 = 2 N v kT * - 4 = 2d EMA 6165 Polymer Physics AB Brennan N v kT -5 * ( 4 - 4 ) 2 1 N v kT * - 5 19 Thermodynamics Rearranging, yields: 4dN v kT 2 1 P= * - 4 R Since it is known: and 2 R = R0 or V = 0 = RR 2 o then 4 R0 d 0 = 4 R d R0 d0 d = 2 * d0 = R 2 EMA 6165 Polymer Physics AB Brennan 20 Thermodynamics Substitution gives: P 4d 0 N v kT - 1 -7 = * - R0 ( ) for a spherical balloon, where the thin film exhibits V = 0 Plot P versus Where is the maximum P? EMA 6165 Polymer Physics AB Brennan 21 Summary Relationships for uniaxial and biaxial orientation of elastomers were developed. Relationships were based upon statistical thermodynamics. Both Shear and Young's Modulus were defined for uniaxial deformation EMA 6165 Polymer Physics AB Brennan 22 References Introduction to Physical Polymer Science, 3rd Edition, Lesley H. Sperling, Wiley Interscience (2001) ISBN 0-471-32921-5 Principles of Polymer Chemistry, P.J. Flory (1953) Cornell University Press, Inc., New York. The Physics of Polymers, Gert Strobl (1996) Springer-Verlag, New York. Some figures were reproduced from Principles of Polymer Chemistry, P.J. Flory (1953) Cornell University Press, Inc., New York. EMA 6165 Polymer Physics AB Brennan 23 ...
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