Lecture 11 Rubber Elasticity

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online