Lecture 11 Rubber Elasticity

Lecture 11 Rubber Elasticity - EMA 6165 – Polymer Physics...

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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 & Department Engineering Engineering EMA 6165 Polymer Physics – AB Brennan EMA 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 EMA 2 Thermodynamics q From our previous lecture: σ0 z G = N v kT (λ z q − λz −2 ) The question was asked, “How can that be if Nv is for a volume? EMA 6165 Polymer Physics – AB Brennan EMA 3 Thermodynamics q The answer lies in the definition of modulus: E y = lim ε →0 ∂ = ∂λ = lim λ →0 ∂ σ0 ∂ ε ∂ σ0 ∂ λ λ − λ −2 ) ) ( N v kT ( = ( N v kT 1 + 2 3 λ q For an extension ratio equal to 1: q Ey = 3( N v kT ) Furthermore, by definition: E y = 2G( 1 + γ ) where γ is Poisson ratio. EMA 6165 Polymer Physics – AB Brennan EMA 4 Thermodynamics Equilibrium Measurements for ∆V = 0 , isotropic materials γ ≅ 0.5 for glassy and semi-crystalline materials thus and E = 3G ∴ G = N v kT Which is given directly by? Hooke’s Law 5 EMA Measurements EMA 6165 Polymer Physics – tAB Brennan too large. Why? yield values of Nv hat are Thermodynamics q The answer: Entanglements. q So how does Nv relate to Mc? q Define Mc? q First count the no. of chains between junctions - Nc q Now consider how the # of junctions relates to the Mn. q If Nc increases, Mn must decrease! EMA 6165 Polymer Physics – AB Brennan EMA 6 Thermodynamics q Therefore: a relationship between density and Avogadro’s number must exist: Nv = q ρN A Mn mass No. * volume mol No. = mass Volume volume = Now, consider the uniaxial stress relationship for an elastomer ( σ0 = N v kT * λ − λ q Substitution yields: ( ρN v kT σ0 = * λ −λ Mc −2 ) where EMA 6165 Polymer Physics – AB Brennan EMA −2 ) N Ak = R 7 Thermodynamics q Hence, one has: ρRT σ0 = * λ −λ Mc q −2 ) Alternatively, this can be written using the following: l λ= l0 q ( Take the derivative of stress w/ respect to l: l −3 dl dσ0 ≅ kTN v + 2l0l l0 l Multiply and divide by l to get: 2 dl dσ 0 = kTN v λ + 2 λ l EMA 6165 Polymer Physics – AB Brennan EMA 8 Thermodynamics q Given the following: dl dλ = E l RTρ 2 E= λ + 2 λ Mc q One can show: q Now examine the significance of these relationships: q E will decrease q Mc increases q q q T increases - λ> 1 > < for λ< 1 for q E will increase E ∝λ E ∝λ EMA 6165 Polymer Physics – AB Brennan EMA − 2 9 Thermodynamics α=1 3RTρ E= Mc q for q Now let’s evaluate the biaxial oriented elastomer q Examples are: – Garbage bag – Packaging film q λy Deform along x and y plane: EMA 6165 Polymer Physics – AB Brennan EMA λx z λy λx 10 Thermodynamics q Consider deformation along the x axis first: σ0x = q N v kT * 2 ) ∂λ x Recalling that we can then show: 1 λz = λ xλ y λ xλ yλz = 1 q ( ∂ λx2 + λ y2 + λz2 − 3 Sustitution gives: σ0x = 2 1 2 ∂ λ x + λ y + 2 2 − 3 λx λy N v kT * 2 ∂ λx EMA 6165 Polymer Physics – AB Brennan EMA 11 Thermodynamics q Then using simple algebra: σ0 = N kT ( λ − λ λ −3 v x x x y σ0 = N v kT ( λx − λx λy −2 −3 y q ) ) −2 True stress is calculated as follows: fz f z A0 A0 σt = = = * = σ0 * A A0 A Area z 1 1 σ0 * * = σ 0z * * lx l y λx λ y l0 x l0 y z EMA 6165 Polymer Physics – AB Brennan EMA 12 Thermodynamics q And q 1 λz = → σ t z = σ 0z λ z λ xλ y Conversely: σtx = σ0x λ x σt y = σ0 y λ y σ t z = σ 0z λ z q Which gives us: EMA 6165 Polymer Physics – AB Brennan EMA 13 Thermodynamics Biaxial Stress in x direction q ( σ t x = N v kT λ x + 2 − λ x − 2 λ y − 2 ) Biaxial Stress in y direction q ( σ t y = N v kT λ y + 2 − λ x − 2 λ y − 2 q ) Differences in Principal Stresses: σt x − σt y = [ N v kT λ x λ y 2 2 EMA 6165 Polymer Physics – AB Brennan EMA 14 Thermodynamics Let’s apply this to a simple example: Difficult to start How do you help start it? What happens when one tries? You stretch it. Why? EMA 6165 Polymer Physics – AB Brennan EMA 15 Thermodynamics Why is it easier after the first difficult expansion? Consider the stresses: σ 0i = ( ) ∂ λ x2 + λ y2 + λz2 − 3 N v kT * 2 ∂ λi σ t = σ0λ = f Area R dR EMA 6165 Polymer Physics – AB Brennan EMA 16 Thermodynamics Express force in terms of pressure P= f f = 2 Area πR f= P* π R 2 Area = 2πRd PR Pπ R σ t = λσ 0 = = 2d 2π R d 2 EMA 6165 Polymer Physics – AB Brennan EMA 17 Thermodynamics Assume biaxial orientation: R λ= R0 Thus: λ xλ yλz = 1 λ λz = 1 2 λx = λz 1 λz = 2 λ 2 2 1 ∂ λ + λ + 4 − 3 λ σ 0i = N v kT * 2 ∂λ And: EMA 6165 Polymer Physics – AB Brennan EMA 18 Thermodynamics Which leads to: σ 0i = 21 ∂ 2λ + 4 − 3 N v kT λ * 2 ∂x σ0i = σ0i = N v kT −5 * ( 4λ − 4λ ) 2 1 N v kT * λ − 5 λ 2 1 PR σt = σ0 λ = 2 N v kT * λ − 4 = λ 2d EMA 6165 Polymer Physics – AB Brennan EMA 19 Thermodynamics Rearranging, yields: 4dN v kT 2 1 P= * λ − 4 R λ Since it is known: ∆V = 0 and then R = R0 λ or λ = RR o 4π R0 d 0 = 4π R d 2 2 2 R0 d0 d = 2 * d0 = R λ EMA 6165 Polymer Physics – AB Brennan EMA 20 Thermodynamics Substitution gives: P= 4d 0 N v kT 1 * λ − 7 3 R0 λ λ for a spherical balloon, where the thin film exhibits ∆V = 0 Plot P versus λ Where is the maximum P? EMA 6165 Polymer Physics – AB Brennan EMA 21 Summary • Relationships for uniaxial and biaxial Relationships orientation of elastomers were developed. developed. • Relationships were based upon Relationships statistical thermodynamics. statistical • Both Shear and Young’s Modulus Both were defined for uniaxial deformation were EMA 6165 Polymer Physics – AB Brennan EMA 22 References • Introduction to Physical Polymer Science, 3rd Introduction Edition, Lesley H. Sperling, Wiley Interscience (2001) ISBN 0-471-32921-5 (2001) • Principles of Polymer Chemistry, P.J. Flory (1953) Principles Cornell University Press, Inc., New York. Cornell • The Physics of Polymers, Gert Strobl (1996) The Springer-Verlag, New York. Springer-Verlag, • Some figures were reproduced from Some • Principles of Polymer Chemistry, P.J. Flory (1953) Principles Cornell University Press, Inc., New York. Cornell EMA 6165 Polymer Physics – AB Brennan EMA 23 ...
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This note was uploaded on 07/20/2011 for the course EMA 6165 taught by Professor Brennan during the Spring '08 term at University of Florida.

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