MASSACHUSES INSIUE OF ECHNOLOGY DEPARMEN OF MAERIALS SCIENCE AND ENGINEERING CAMBRIDGE, MASSACHUSES 02139 3.22 MECHANICAL PROPERIES OF MAERIALS PROBLEM SE 5 SOLUIONS 1. (Hertzber 6.2) If it takes 300 seconds for the relaxation modulus to decay to a particular value at, to what temperature must the material have been raised to effect the same decay in seconds? Usin the time-temperature equivalence relationship for amorphous polymers, t 17.44( ) lo = t 51.6 + 17.44( ) lo =. 300 51.6 + = + 4.77 K 2. (Hertzber 6.3) Calculate the relaxation time for lass and comment on its propensity for stress relaxation at room temperature. E 70GPa and η 1 GPa-s. Relaxation time is defined as the ratio of viscosity to modulus. GPa-s = η 1.4 sec E = 70 GPa = he relaxation time is a measure of the time it takes for viscous behavior to play a sinificant role in deformation. he relaxation time calculated for lass is over 400 years, indicatin that there is limited stress relaxation. hus, lass behaves essentially completely elastically. 3. Your Irish post doc asked you to determine the time for the relaxation modulus to decay to a particular value at 75 C by testin the polymer at its lass transition temperature, 0 C. She said a simple calculation usin the empirical time-temperature relationship for amorphous polymers would ive you the desired result. You find experimentally that the relaxation time is 217.2 seconds for the relaxation modulus to decay to the particular value. However, the refrieration unit that was supposed to keep the temperature at 0 C (lass transition temperature) was not functionin and the test was carried out at 20 C, the ambient temperature in the laboratory. Can you use the data from this test to determine the relaxation time to decay to the specified value at 75 C? If so, what is the value? It is possible to determine the relaxation time to decay at 75 C. he relaxation time is 0.0007527 sec. o find the relaxation time at some arbitrary temperature, we need to find the relaxation time at the lass transition temperature or the lass transition temperature plus 50 K. We are iven the lass transition temperature, = 0 K. We find the time to the particular relaxation modulus at by the empirical timetemperature relation for amorphous polymers.
t 17.44( ) lo = t 51.6 + 217.2 sec 17.44(20 C 0 C) lo = t 51.6 + 20 C 0 C 7 1.616 sec t = Now we can use the time to the relaxation modulus at to compute the time to a particular relaxation modulus at a iven temperature. We do this at 75 C. t 17.44( ) lo = t 51.6 + t 17.44(75 C 0 C) + C C lo 1.616 7 = sec 51.6 75 0 4 7.528 sec t = 4. (Hertzber 6.5) he deformation response of a certain polymer can be described by the Voit model. If E = 400 MPa and η = 2 MPa-s, compute the relaxation time. Compute ε () t for times to 5τ when the steady state stress is MPa. How much creep strain takes place when t = τ and when t =? he relaxation time can be calculated by 9 τ = η E = 2 MPa sec 400MPa = 5 sec. he strain experienced by a Voit element can be expressed by the expression σ t τ ε = ( 1 e E ). he strain at the indicated times is 2 ε( t = τ) = 1.58 ε( t = 5 τ) = 2.48 ε ( t = ) = 2.5 2 2. (Hertzber 6.6) Compare the fractional amount of the total deformation that would occur if t = τ when η = 2 MPa-s and η = 8 MPa-s, respectively. he relaxation time, τ, by definition is the time to reach a particular strain. herefore, for each material, at t =τ, the strains will be the same. 9 However, if one uses a time of 5 sec for the η = 8 MPa-smaterial, we can calculate the strain. τ = η E = = 9 8 MPa sec 400MPa 2 sec t τ = 0.25 ε ( t = 5 sec) = 5.53 9 3 his makes intuitive sense, as the strain at a iven time for a more viscous material is lower than the low viscosity material.
5. o improve the description of polymer behavior, Maxwell and Voit models can be combined in series. For the four-element viscoelastic model shown below, derive an expression for the strain as a function of time for a iven applied stress. Discuss the advantaes of the four-element viscoelastic model over the Maxwell and Voit models. o find the total strain as a function of time, we can sum the elastic, viscoelastic, and viscous strain components. ε () t = ε() t + ε() t + ε() t η V E V sprin Voit dashpot σ σ t σ ε () t = + (1 e ) + t E E η M V M is defined as the relaxation time for the Voit element. he four-element viscoelastic model accounts for instantaneous strain and creep strain. Upon unloadin, the Voit element accounts for creep recovery and the sinle dashpot accounts for permanent deformation that is irrecoverable. hus, this simplistic model is much more accurate than the Maxwell or Voit elements alone. he Maxwell element is unable to account for creep recovery and the Voit element does not account for instantaneous strain or permanent deformation. 6. Sketch a lo-lo plot of relaxation modulus versus temperature for an amorphous polymer with crosslinks. a. Identify the various reions and characteristic relaxation modulus values on the plot.. CORRECION: here is no viscous portion in a cross-linked polymer. he rubbery reime should extend to the riht.
b. Explain the trends in relaxation modulus in the various reions in terms of the bondin in polymers. For low temperatures times, the relaxation approaches a maximum limitin value where the material exhibits lassy behavior associated with neliible molecule semental motions. In this reion, the primary carbon-carbon bonds are stretched. At intermediate temperatures, the material transitions to a reion of leathery behavior associated with short-rane molecule semental motion. At this elevated temperature, enouh thermal enery is present to break weak secondary bonds between the polymer chains that permits semental motion. At still hiher temperatures, thermal enery is reat enouh to permit complete molecule movements in the rubbery reion. Note that as the temperature is increased into the rubbery reion, the modulus increases due to entropic effects. c. How would each reime chane if the polymer had no cross-linkin? A polymer without cross-linkin would have a limited rubbery reion and experience visous flow as a liquid. Physical entanlement of polymer chains causes networks to form and restrict molecular flow. he restriction is responsible for rubbers (heavily cross-linked) havin a stable rubbery reion. See sketch in part (a). 7. Creep compliance values for polyethylene are iven. (hours) J(t) (psi -1-4 ) 0 0.600 0 0.700 200 0.720 300 0.730 400 0.740 500 0.770 Consider a sample of polyethylene (cross-section 0.5 in. 0.1 in.) for the iven load history at the same test conditions as the data above. Load (lbs.) Duration (hours) 20 0 5 200 50 0 0 0 a. Calculate the strain at 200 and 500 hours. 20lbs σ (0 hours) = = 400 psi 0.5in 0.1in -15lbs σ (0hours) = = 300psi 0.5in 0.1in 45lbs σ (300hours) = = 900psi 0.5in 0.1in -50 lbs σ (400 hours) = = 00 psi 0.5in 0.1in Usin the Boltzmann superposition principle, we calculate the strains. ε ( t= 200 hours) = σ(0 hours) Jt ( 0 hours) + σ(0 hours) Jt ( 0 hours) ε ( t = 200 hours) = σ(0 hours) J (200 hours) + σ(0 hours) J (0 hours) ε t = hours = + 4 1 4 1 ( 200 ) (400 psi)(0.72 psi ) ( 300 psi)(0.70 psi ) ε ( t = 200 hours) = 0.0078
ε ( t= 500 hours) = σ(0hours) Jt ( 0hours) + σ(0hours) Jt ( 0hours) + σ(300 hours) J( t 300 hours) + σ(400 hours) J( t 400 hours) ε ( t = 500 hours) = σ(0hours) J (500hours) + σ(0hours) J (400hours) + σ(300 hours) J(200 hours) + σ(400 hours) J(0 hours) 4 1 4 1 ε ( t = 500 hours) = (400 psi)(0.77 psi ) + ( 300 psi)(0.74 psi ) + + 4 1 4 1 (900psi)(0.72 psi ) ( 00psi)(0.70 psi ) ε ( t = 500 hours) = 0.0034 b. Sketch a qualitative strain-time plot. 8. Derive the constitutive relation for the viscoelastic model shown below. Based on the confiuration, we know that the strain in the sprin k 1 is equal to the combined strain in the sprin k 2 and the dashpot η, and that the total stress is equal to the sum of the stress in the sprin k 1 and the sprin k 2. Because the sprin k 2 is in series with the dashpot η, they bear the same stress. ε = ε = ε + ε k1 k2 η σ = σ + σ = σ + σ k1 k2 k1 η Lookin at the elastic response of the sprin k 1 ives σ k1 = k1εk1 = k1ε σ = k1ε + σk2 or σk2 = σ k1ε. akin the time derivative of the first equation yields ε = ε + ε = σ k + σ η. k2 η k2 2 k2 Combinin the last two equations leads to the desired result. ε = σ k + σ η k2 2 k2
( k ) k ( k ) ε = σ ε + σ ε 1 2 1 dε 1 dσ k1 dε σ k1 = + ε dt k dt k dt η η 2 2 σ 1 dσ k 1 k 1 dε + = ε + 1+ η k2 dt η k2 dt η 9. Evaluate the viscosity of the simple lass shown below by the followin approximate procedure. Assume that half of the atoms pairs in the lass are in a position permittin an activated shear to the left (state A) and that the remainin half of the atom pairs are in the complementary position (state B). In the absence of a stress, states A and B have the same enery, F, and are separated by an activation enery of * manitude, F. he stress σ s raises the enery of atoms in state A, and lowers that of atoms in state B. he difference in enery between the two states is F = 2σ s γ Ω, where γ is the shear strain that occurs when an atom pair stretch from A to B, (which you may take to be unity) and 2Ω is the volume of two atoms. he vibration frequency (attempt frequency) of atom pairs is ν. Calculate the rate of shear, γ, of the unit volume of the liquid, subjected to a shear stress, σ s, by calculatin the nu,ber of atom pairs jumpin, per second, from A to B and from B to A. Assume that σ sω k (this means that exp( σ sω k ) 1 + σ sω k ) where k, is Boltzmann s constant and is the absolute temperature and derive an equation for the viscosity of the liquid. (You may assume that a switchin event, althouh it converts a pair of atoms in the state A into a pair in state B, also creates with the atoms surroundin it new pairs in state A, so that the fraction of atoms in state A remains constant and equal to one half.) State A State B