§10.1 Mechanical Behaviors: Stress-Strain Regimes, Brittle, Ductile, Elastomeric & Yielding
The mechanical response of solid polymers spans an extraordinary continuum from hard, glass-like brittleness to ductile necking and high-extensibility rubber elasticity, dictated fundamentally by temperature relative to the glass transition ($T_g$) and melting temperature ($T_m$).
Engineering Stress-Strain Definitions
In a uniaxial tensile test, a specimen of initial cross-sectional area $A_0$ and gauge length $L_0$ is pulled at constant deformation rate:
1. Engineering Stress: $\sigma = F / A_0$ (Pascals or $\text{MPa}$).
2. Engineering Strain: $\epsilon = (L - L_0) / L_0 = \Delta L / L_0$ (dimensionless or $\%\text{ strain}$).
3. True Stress: $\sigma_{\text{true}} = F / A = \sigma (1 + \epsilon)$ (assuming constant volume during deformation).
4. True Strain: $\epsilon_{\text{true}} = \ln(L / L_0) = \ln(1 + \epsilon)$.
5. Young's Modulus: $E = \lim_{\epsilon \to 0} (d\sigma / d\epsilon)$ (initial linear elastic slope).
Characteristic Deformation Regimes
1. Brittle Polymeric Glasses ($T \ll T_g$):
- Examples: Atactic polystyrene (PS), poly(methyl methacrylate) (PMMA) at room temperature.
- High Young's modulus ($E = 3.0 - 3.5\text{ GPa}$).
- Fracture occurs prematurely at small strains ($\epsilon_b < 2 - 3\%$) with zero macroscopic plastic yield. Failure is governed by crazing (microvoid formation bridged by drawn polymer fibrils) followed by brittle fracture.
2. Ductile Polymers with Yield and Cold-Drawing ($T < T_g$ but near $T_g$, or semi-crystalline $T_g < T < T_m$):
- Examples: Polycarbonate (PC), high-density polyethylene (HDPE), Nylon 6,6.
- Exhibits an initial linear elastic Hookean regime followed by a prominent yield point (yield stress $\sigma_y$).
- Beyond yield, necking (localized reduction in cross-sectional area) occurs with an engineering stress drop.
- As the neck propagates stably along the gauge length (cold-drawing), polymer chains undergo extensive uncoiling and orientation alignment parallel to the draw axis, resulting in strain hardening until rupture at $\epsilon_b = 50 - 500\%$.
3. Elastomers ($T \gg T_g$, lightly cross-linked):
- Examples: Vulcanized natural rubber, polybutadiene, PDMS silicone.
- Low initial modulus ($E \sim 1 - 10\text{ MPa}$).
- Reversible non-linear deformation up to colossal strains ($\epsilon_b = 500 - 1,000\%$) with instantaneous recovery upon release. Driven purely by conformational entropy.
§10.2 Viscoelastic Phenomena: Creep Compliance, Stress Relaxation & Recovery
Unlike purely elastic Hookean solids (where stress is instantaneous and proportional to strain: $\sigma = E \epsilon$) or purely viscous Newtonian fluids (where stress is proportional to strain rate: $\sigma = \eta \dot{\epsilon}$), polymers exhibit viscoelasticity—a time- and rate-dependent hybrid behavior combining elastic energy storage with viscous energy dissipation.
1. Creep and Creep Compliance $J(t)$
In a creep test, a constant instantaneous engineering stress $\sigma_0$ is applied at $t = 0$ and maintained:
where $H(t)$ is the Heaviside step function. The resulting strain $\epsilon(t)$ increases continuously over time:
- An instantaneous elastic response $\epsilon_0$.
- A time-dependent delayed elastic (anelastic) retarding strain.
- For uncrosslinked polymers, a continuous steady viscous flow.
The creep compliance $J(t)$ is defined as the time-dependent strain per unit applied stress:
(Units: $\text{Pa}^{-1}$ or $\text{MPa}^{-1}$).
2. Stress Relaxation and Relaxation Modulus $E(t)$
In a stress relaxation test, a constant instantaneous strain $\epsilon_0$ is applied at $t = 0$ and held constant:
To maintain this constant strain, the stress required to hold the sample decays continuously over time as macromolecular chains undergo conformational rearrangements, reptation, and disentanglement:
The stress relaxation modulus $E(t)$ is defined as:
(Units: $\text{Pa}$ or $\text{MPa}$). In linear viscoelasticity (Boltzmann superposition principle), $E(t)$ and $J(t)$ are interrelated via the convolution integral:
§10.3 Mechanical Analog Models: Maxwell Fluid, Voigt-Kelvin Solid & Zener Standard Linear Solid
To model viscoelastic constitutive behavior mathematically, physical networks of ideal Hookean springs (elastic storage, modulus $E$) and Newtonian dashpots (viscous dissipation, viscosity $\eta$) are constructed.
1. The Maxwell Model (Viscoelastic Fluid)
A Hookean spring and a Newtonian dashpot connected in series:
- Total strain is the sum of spring and dashpot strains: $\epsilon = \epsilon_s + \epsilon_d$.
- Both elements experience identical stress: $\sigma = \sigma_s = \sigma_d$.
Differentiating with respect to time:
Multiplying by $E$ yields the Maxwell constitutive differential equation:
where $\tau_R \equiv \eta / E$ is the relaxation time. In stress relaxation at constant strain ($\epsilon = \epsilon_0, d\epsilon/dt = 0$):
The stress relaxation modulus is:
The Maxwell model describes uncrosslinked polymer melts: it exhibits instantaneous elasticity and exponential stress relaxation to zero at long times. However, it fails to predict primary creep recovery.
2. The Voigt-Kelvin Model (Viscoelastic Solid)
A Hookean spring and a Newtonian dashpot connected in parallel:
- Both elements experience identical strain: $\epsilon = \epsilon_s = \epsilon_d$.
- Total stress is the sum of spring and dashpot stresses: $\sigma = \sigma_s + \sigma_d = E \epsilon + \eta \frac{d\epsilon}{dt}$.
Rearranging:
where $\tau_C \equiv \eta / E$ is the retardation time. Under constant stress $\sigma_0$ (creep test):
The creep compliance is:
The Voigt-Kelvin model accurately predicts delayed elastic deformation and complete strain recovery upon load removal, but fails in stress relaxation (predicts infinite instantaneous stress).
3. The Standard Linear Solid (Zener Model)
Combines a Maxwell arm in parallel with an equilibrium Hookean spring ($E_2$):
Provides both an instantaneous glassy modulus $E_g = E_1 + E_2$ at $t = 0$, an exponential relaxation decay, and an equilibrium plateau modulus $E_e = E_2$ at $t \to \infty$, accurately capturing crosslinked elastomer behavior.
§10.4 Dynamic Mechanical Analysis (DMA): Storage/Loss Moduli & Loss Factor $\tan \delta$
Dynamic Mechanical Analysis (DMA) is the premier analytical technique for probing the viscoelastic spectrum of polymers across wide frequency and temperature domains.
Oscillatory Harmonic Deformation
A sinusoidal oscillating strain of angular frequency $\omega$ (in $\text{rad/s}$) is applied to the specimen:
Because polymers are viscoelastic, the resulting steady-state stress oscillates at the identical frequency $\omega$ but leads the strain by a phase angle $\delta$ ($0^\circ \le \delta \le 90^\circ$):
Expanding using trigonometric angle addition:
Dividing by maximum strain $\epsilon_0$:
Complex Modulus Components
1. Storage Modulus ($E'$, In-Phase Component):
Represents the elastic energy stored and reversibly recovered per cycle of deformation. Directly proportional to specimen stiffness.
2. Loss Modulus ($E''$, Out-of-Phase Component):
Represents the viscous energy dissipated as internal heat per cycle through molecular friction.
3. Complex Modulus Magnitude ($|E^*|$):
4. Loss Factor (Damping Factor, $\tan \delta$):
The ratio of dissipated energy to stored elastic energy:
- Pure elastic Hookean solid: $\delta = 0^\circ \implies E'' = 0, \tan \delta = 0$.
- Pure viscous Newtonian liquid: $\delta = 90^\circ \implies E' = 0, \tan \delta = \infty$.
- Viscoelastic polymer: $0 < \tan \delta < \infty$.
In a DMA temperature sweep at constant frequency (e.g., $1\text{ Hz}$), the glass transition temperature $T_g$ is detected with unmatched sensitivity as a dramatic multi-decade plunge in storage modulus $E'$ (by 3 orders of magnitude from $\sim 3\text{ GPa}$ to $\sim 1\text{ MPa}$) accompanied by a pronounced peak in $\tan \delta$.
§10.5 The Glass Transition Temperature ($T_g$): Free Volume Theory & Segmental Mobility
The glass transition temperature ($T_g$) is the temperature boundary where an amorphous polymer transitions from a hard, brittle, glassy state ($T < T_g$) into a soft, flexible rubbery or leathery state ($T > T_g$). Unlike crystalline melting ($T_m$), which is a true first-order thermodynamic phase transition with discontinuous changes in enthalpy and volume ($\Delta H_m > 0, \Delta V_m > 0$), the glass transition is a kinetic and pseudo-second-order transition characterized by a step change in heat capacity ($\Delta C_p$) and thermal expansion coefficient ($\Delta \alpha$).
Fox-Flory Free Volume Theory
According to the free volume model formulated by Thomas Fox and Paul Flory: The total macroscopic volume $V$ of a polymer solid consists of:
1. Occupied Volume ($V_{\text{occ}}$): The van der Waals volume occupied by the constituent atoms plus their core vibrational exclusion shells.
2. Free Volume ($V_f$): The interstitial unoccupied empty space between macromolecular chains resulting from packing inefficiencies.
Free volume fraction is defined as:
- Above $T_g$ ($T > T_g$): Thermal energy expands the free volume at thermal expansion rate $\alpha_f$:
where $\alpha_f = \alpha_r - \alpha_g \approx 4.8 \times 10^{-4}\text{ K}^{-1}$. Because $f$ is large ($f > 0.025$), polymer chain segments (typically $20 - 50$ backbone carbon atoms) have ample space to undergo coordinated rotational jumps and crankshaft motions, resulting in rubbery flexibility.
- Cooling toward $T_g$: As temperature decreases, free volume contracts.
- At $T_g$: The free volume fraction drops to a universal critical percolation threshold:
- Below $T_g$ ($T < T_g$): The free volume is too constricted to accommodate coordinated backbone bond rotations. Large-scale segmental motion freezes out. The chains are trapped in a non-equilibrium disordered glassy configuration. The only motions permitted are localized small-amplitude bond vibrations and minor side-group rotations ($eta$- and $\gamma$-relaxations).
§10.6 Factors Governing $T_g$: Chain Stiffness, Steric Bulk, Intermolecular Forces & Cross-Linking
The glass transition temperature of a polymer is governed directly by its chemical constitution and chain architecture:
1. Backbone Chain Flexibility:
- Flexible backbones containing low rotational barrier bonds ($-C-O-C-$, $-Si-O-Si-$, $-C-C-$) exhibit extremely low $T_g$:
- Poly(dimethylsiloxane) (PDMS): $T_g = -123^\circ\text{C}$
- Polyethylene (PE): $T_g \approx -120^\circ\text{C}$ to $-80^\circ\text{C}$
- Poly(ethylene oxide) (PEO): $T_g = -67^\circ\text{C}$
- Rigid backbones containing bulky aromatic rings, heterocyclic units, or conjugated double bonds possess high torsional barriers, elevating $T_g$:
- Poly(ethylene terephthalate) (PET): $T_g = +78^\circ\text{C}$
- Polycarbonate (PC): $T_g = +150^\circ\text{C}$
- Polyetheretherketone (PEEK): $T_g = +143^\circ\text{C}$
- Polyimides (Kapton): $T_g > +380^\circ\text{C}$
2. Steric Bulk of Pendant Substituents:
- Bulky side groups hinder backbone bond rotation, restricting segmental motion and driving $T_g$ upward:
- Polypropylene ($-\text{CH}_3$): $T_g = -10^\circ\text{C}$
- Polystyrene ($-\text{C}_6\text{H}_5$): $T_g = +100^\circ\text{C}$
- Poly(1-vinylnaphthalene): $T_g = +150^\circ\text{C}$
- Conversely, flexible aliphatic side chains act as internal plasticizers, pushing chains apart and increasing free volume (internal plasticization):
- Poly(methyl methacrylate): $T_g = +105^\circ\text{C}$
- Poly(ethyl methacrylate): $T_g = +65^\circ\text{C}$
- Poly(n-butyl methacrylate): $T_g = +20^\circ\text{C}$
- Poly(n-octyl methacrylate): $T_g = -20^\circ\text{C}$
3. Intermolecular Forces (Dipole-Dipole & Hydrogen Bonding):
Strong polar interactions physically anchor neighboring chains, requiring higher thermal energy to unlock segmental motion:
- Polypropylene (non-polar): $T_g = -10^\circ\text{C}$
- Poly(vinyl chloride) (polar $C-Cl$ dipoles): $T_g = +82^\circ\text{C}$
- Polyacrylonitrile (strong $-C\equiv N$ dipoles): $T_g = +105^\circ\text{C}$
- Poly(vinyl alcohol) (interchain hydrogen bonds): $T_g = +85^\circ\text{C}$
- Nylon 6,6 (dense amide hydrogen bond lattice): $T_g = +50^\circ\text{C}$
4. Cross-Link Density:
Covalent cross-links introduce topological constraints that severely restrict chain mobility. As cross-link density increases, $T_g$ increases according to the DiMarzio equation:
Highly cross-linked epoxy or phenolic resins exhibit $T_g > 180 - 250^\circ\text{C}$.
§10.7 The Fox Equation & Plasticization: Copolymer $T_g$ Tuning & Free Volume Addition
The Fox Equation for Random Copolymers and Miscible Blends
In 1956, Thomas G. Fox derived the fundamental thermodynamic equation predicting the glass transition temperature of a homogeneous, single-phase random copolymer or miscible polymer blend:
where:
- $w_1$ and $w_2$ are the mass fractions of components 1 and 2 ($w_1 + w_2 = 1.0$).
- $T_{g, 1}$ and $T_{g, 2}$ are the glass transition temperatures of the corresponding pure homopolymers (expressed strictly in Kelvin).
- $T_g$ is the resulting intermediate glass transition temperature of the copolymer (in Kelvin).
For multicomponent systems containing $N$ comonomers:
Physical Derivation from Free Volume Additivity
Fox derived this relationship by assuming that the free volume of the copolymer mixture at its glass transition ($f_g = 0.025$) is the weighted sum of the free volume contributions of its constituent segments:
Linear expansion above each component's $T_g$ yields the Fox equation.
- Diagnostic Criteria for Phase Homogeneity:
- A single-phase homogeneous random copolymer or completely miscible blend exhibits a single intermediate $T_g$ that obeys the Fox equation.
- In contrast, an immiscible, phase-separated blend (such as HIPS or block copolymers) exhibits two distinct, separate $T_g$ values corresponding to each independent microphase domain.
Plasticization Mechanics
Plasticization is the intentional reduction of polymer $T_g$ through the addition of a compatible, low-volatility small molecule diluent (plasticizer). Because small plasticizer molecules possess vast free volume and low glass transitions ($T_{g, \text{plast}} \approx 120 - 180\text{ K}$), applying the Fox equation demonstrates dramatic $T_g$ suppression:
Adding $30\text{ wt}\%$ DOP ($T_g = 190\text{ K}$) to rigid PVC ($T_g = 355\text{ K}$) drops the glass transition of the plasticized compound to $280\text{ K}$ ($+7^\circ\text{C}$), transforming an unyielding pipe resin into soft, flexible film!
§10.8 Time-Temperature Superposition (TTS) & The Williams-Landel-Ferry (WLF) Equation
Viscoelastic phenomena are governed by molecular relaxation processes that occur over time scales ranging from microseconds to decades. Measuring relaxation experimentally over 10 decades of time at a single temperature is physically impossible. However, because temperature and time exert mathematically equivalent effects on molecular mobility, Time-Temperature Superposition (TTS) allows short-term measurements taken across multiple temperatures to be shifted into a single universal master curve spanning enormous spans of frequency or time.
The Shift Factor $a_T$
To construct a master curve at an arbitrary reference temperature $T_{\text{ref}}$, relaxation modulus curves $E(t)$ measured at temperature $T$ are shifted horizontally along the logarithmic time axis by a dimensionless shift factor $a_T$:
- If $T > T_{\text{ref}}$: Molecular motions are faster; relaxation occurs at shorter times. The curve must be shifted to the right: $a_T < 1 \implies \log a_T < 0$.
- If $T < T_{\text{ref}}$: Molecular motions are sluggish; relaxation is delayed. The curve must be shifted to the left: $a_T > 1 \implies \log a_T > 0$.
- At $T = T_{\text{ref}}$: $a_T = 1.0 \implies \log a_T = 0$.
The Williams-Landel-Ferry (WLF) Equation
In 1955, Malcolm Williams, Robert Landel, and John D. Ferry demonstrated that for all amorphous polymers in the temperature range from $T_g$ to $T_g + 100^\circ\text{C}$, the shift factor $\log a_T$ obeys a universal empirical equation:
When the polymer's glass transition temperature is chosen as the reference temperature ($T_{\text{ref}} = T_g$), the empirical parameters take on universal values across virtually all amorphous polymers:
where $C_1 = 17.44$ and $C_2 = 51.6\text{ K}$.
Theoretical Derivation from Free Volume Theory
According to the Doolittle viscosity equation, molecular friction factor $\zeta$ depends exponentially on the reciprocal free volume fraction $1/f$:
Substituting the linear expansion $f(T) = f_g + \alpha_f (T - T_g)$:
Dividing numerator and denominator by $\alpha_f$ and converting to base-10 logarithm:
Comparing directly with the WLF equation yields the fundamental physical definitions:
Setting $B \approx 1.0$, $C_1 = 17.44$ yields the universal free volume at $T_g$:
and $C_2 = 51.6\text{ K}$ yields the thermal expansion coefficient of free volume:
This landmark derivation unites empirical polymer rheology with the statistical thermodynamics of free volume!
Worked Practice Problems (9 Challenge Exercises)
Multi-step solved problems covering end-to-end vector statistics, radius of gyration, persistence length, characteristic ratio, and tacticity stereochemistry with line-by-line mathematical proofs.
A dogbone tensile specimen of an engineering thermoplastic with initial gauge dimensions: length $L_0 = 50.0\text{ mm}$, width $w_0 = 10.0\text{ mm}$, thickness $t_0 = 4.00\text{ mm}$ (initial cross-sectional area $A_0 = 40.0\text{ mm}^2 = 4.00 \times 10^{-5}\text{ m}^2$) is tested at room temperature at an elongation rate of $5.0\text{ mm/min}$. The following data points are extracted from the tensile test:
- At $\Delta L = 0.50\text{ mm}$, the applied force is $F = 1,200\text{ N}$ (within the linear elastic limit).
- The maximum tensile yield load is $F_{\text{yield}} = 2,400\text{ N}$ at $\Delta L = 2.50\text{ mm}$.
- Cold drawing extends the specimen until tensile rupture occurs at $F_{\text{rupture}} = 1,800\text{ N}$ and total elongation $\Delta L_{\text{break}} = 35.0\text{ mm}$.
- Total mechanical work of deformation integrated under the load-displacement curve is $W = 68.0\text{ Joules}$.
(a) Calculate the Young's modulus $E$ of the polymer in $\text{GPa}$. (b) Calculate the engineering yield stress $\sigma_y$ and yield strain $\epsilon_y$. (c) Calculate the engineering strain at break $\epsilon_b$ and true strain at break $\epsilon_{\text{true}}$. (d) Calculate the tensile modulus of toughness (energy absorption per unit volume) in $\text{MJ/m}^3$.
Step 1: Young's Modulus $E$
At $\Delta L = 0.50\text{ mm}$:
Engineering stress:
Young's modulus:
Step 2: Yield Stress $\sigma_y$ and Yield Strain $\epsilon_y$
At the yield point ($F = 2,400\text{ N}, \Delta L = 2.50\text{ mm}$):
Step 3: Strain at Break
At break ($\Delta L_{\text{break}} = 35.0\text{ mm}$):
True strain at break:
Step 4: Tensile Toughness (Modulus of Toughness)
Gauge volume of the specimen:
Toughness $U_T$ is total mechanical work per unit volume:
The high toughness ($34\text{ MJ/m}^3$) is characteristic of a ductile engineering thermoplastic capable of extensive plastic cold-drawing.
(a) Young's modulus E = 3.00 GPa; (b) Yield stress sigma_y = 60.0 MPa, Yield strain epsilon_y = 5.00%; (c) Engineering strain at break = 70.0%, True strain = 53.06%; (d) Modulus of toughness = 34.0 MJ/m^3.
A random copolymer of methyl methacrylate (MMA) and n-butyl acrylate (BA) is synthesized for an architectural coating formulation. The glass transition temperatures of the corresponding pure homopolymers are:
- Poly(methyl methacrylate) (PMMA): $T_{g, 1} = 105.0^\circ\text{C}$
- Poly(n-butyl acrylate) (PBA): $T_{g, 2} = -54.0^\circ\text{C}$
(a) Convert both homopolymer $T_g$ values to Kelvin. (b) Using the Fox equation, calculate the glass transition temperature (in Kelvin and $^\circ\text{C}$) of a random copolymer consisting of $60.0\text{ wt}\%$ MMA and $40.0\text{ wt}\%$ BA. (c) The coating application requires the copolymer to have a glass transition temperature of exactly $T_g = +20.0^\circ\text{C}$ (room temperature balance of hardness and film formation). Calculate the exact mass fraction of MMA comonomer ($w_1$) required in the feed to achieve this target.
Step 1: Convert Temperatures to Kelvin
- PMMA: $T_{g, 1} = 105.0 + 273.15 = 378.15\text{ K}$
- PBA: $T_{g, 2} = -54.0 + 273.15 = 219.15\text{ K}$
Step 2: Calculate $T_g$ for 60/40 Copolymer
Given $w_1 = 0.600$ (MMA) and $w_2 = 0.400$ (BA): The Fox equation is:
Substitute values:
Calculate $T_g$:
Convert to Celsius:
Step 3: Exact Formulation for Target $T_g = +20.0^\circ\text{C}$
Target temperature in Kelvin:
Since $w_2 = 1 - w_1$:
Calculate inverse differences:
Substitute:
The remaining comonomer is $w_2 = 1 - 0.6004 = 39.96\text{ wt}\%$ n-butyl acrylate.
(a) T_g1 = 378.15 K, T_g2 = 219.15 K; (b) 60/40 copolymer T_g = 293.1 K (19.9 °C); (c) Target T_g = 20.0 °C requires 60.04 wt% MMA and 39.96 wt% BA.
A polymer melt is modeled as an ideal Maxwell element consisting of an elastic spring of modulus $E = 5.00 \times 10^6\text{ Pa}$ in series with a viscous dashpot of viscosity $\eta = 2.50 \times 10^8\text{ Pa}\cdot\text{s}$. (a) Calculate the Maxwell relaxation time constant $\tau_R$. (b) An instantaneous tensile strain of $\epsilon_0 = 0.100$ ($10.0\%$) is applied at $t = 0$ and held constant. Calculate the initial stress $\sigma(0)$ immediately following strain application. (c) Calculate the remaining stress $\sigma(t)$ and relaxation modulus $E(t)$ at times $t = 25.0\text{ s}, 50.0\text{ s}, 100.0\text{ s}$, and $250.0\text{ s}$. (d) Calculate the time required for the stress to relax to exactly $1.0\%$ of its initial value.
Step 1: Calculate Relaxation Time $\tau_R$
Step 2: Initial Stress $\sigma(0)$
At $t = 0$, the dashpot has not had time to move. The entire strain is accommodated by the elastic spring:
Step 3: Stress and Modulus Decay Over Time
Constitutive formula:
Given $\tau_R = 50.0\text{ s}$:
1. $t = 25.0\text{ s}$ ($t/\tau_R = 0.50$):
- $\exp(-0.50) = 0.6065$
- $\sigma(25\text{ s}) = 500 \times 0.6065 = 303.3\text{ kPa}$
- $E(25\text{ s}) = 5.00 \times 0.6065 = 3.033\text{ MPa}$
2. $t = 50.0\text{ s}$ ($t = \tau_R$):
- $\exp(-1.00) = 0.3679$
- $\sigma(50\text{ s}) = 500 \times 0.3679 = 183.9\text{ kPa}$
- $E(50\text{ s}) = 5.00 \times 0.3679 = 1.839\text{ MPa}$
3. $t = 100.0\text{ s}$ ($t/\tau_R = 2.00$):
- $\exp(-2.00) = 0.1353$
- $\sigma(100\text{ s}) = 500 \times 0.1353 = 67.7\text{ kPa}$
- $E(100\text{ s}) = 5.00 \times 0.1353 = 0.677\text{ MPa}$
4. $t = 250.0\text{ s}$ ($t/\tau_R = 5.00$):
- $\exp(-5.00) = 0.006738$
- $\sigma(250\text{ s}) = 500 \times 0.006738 = 3.37\text{ kPa}$
- $E(250\text{ s}) = 5.00 \times 0.006738 = 0.0337\text{ MPa}$
Step 4: Time to Relax to $1.0\%$ of Initial Stress
Target:
Taking logarithms:
(a) tau_R = 50.0 s; (b) Initial stress sigma(0) = 500 kPa; (c) sigma(t): 303.3 kPa (25 s), 183.9 kPa (50 s), 67.7 kPa (100 s), 3.37 kPa (250 s); (d) Time to reach 1.0% remaining stress = 230.3 seconds (4.61 * tau_R).
A polymer elastomer is modeled as a Voigt-Kelvin element with an elastic modulus of $E = 2.00 \times 10^7\text{ Pa}$ and a dashpot viscosity of $\eta = 6.00 \times 10^8\text{ Pa}\cdot\text{s}$. (a) Determine the characteristic retardation time $\tau_C$. (b) A constant tensile stress of $\sigma_0 = 1.00 \times 10^6\text{ Pa}$ ($1.00\text{ MPa}$) is applied at $t = 0$. Calculate the instantaneous strain at $t = 0^+$, the strain at $t = \tau_C$, and the equilibrium strain as $t \to \infty$. (c) At $t_1 = 60.0\text{ s}$, the stress is abruptly removed ($\sigma = 0$). Derive the strain recovery equation $\epsilon(t)$ for $t > t_1$ and calculate the residual strain at $t = 90.0\text{ s}$ and $t = 150.0\text{ s}$.
Step 1: Retardation Time $\tau_C$
Step 2: Creep Strains under Load
The Voigt-Kelvin creep equation is:
Equilibrium strain:
1. At $t = 0^+$:
The dashpot cannot deform instantaneously ($d\epsilon/dt = \sigma / \eta$), so:
2. At $t = \tau_C = 30.0\text{ s}$:
3. As $t \to \infty$:
Step 3: Strain Recovery for $t > t_1 = 60.0\text{ s}$
At $t_1 = 60.0\text{ s}$ ($t_1 / \tau_C = 60.0 / 30.0 = 2.00$): Strain achieved prior to unloading:
When the load is removed at $t_1$, the stress drops to zero: $\sigma = E \epsilon + \eta \frac{d\epsilon}{dt} = 0$. Rearranging:
Integrating for $t > t_1$:
Calculate residual strains:
1. At $t = 90.0\text{ s}$ ($t - t_1 = 30.0\text{ s} = 1.00 \tau_C$):
2. At $t = 150.0\text{ s}$ ($t - t_1 = 90.0\text{ s} = 3.00 \tau_C$):
As $t \to \infty$, $\epsilon(t) \to 0$. The Voigt-Kelvin element undergoes $100\%$ complete elastic strain recovery.
(a) tau_C = 30.0 s; (b) epsilon(0) = 0%, epsilon(30 s) = 3.16%, equilibrium epsilon_inf = 5.00%; (c) epsilon(t) = epsilon(t_1) * exp(-(t - t_1) / tau_C); Residual strain: 1.59% at 90 s, 0.22% at 150 s (fully recovers to 0%).
A Dynamic Mechanical Analysis (DMA) test in tension is conducted on a viscoelastic polymer at frequency $f = 1.00\text{ Hz}$ (angular frequency $\omega = 2 \pi f = 6.283\text{ rad/s}$) at $T = 25.0^\circ\text{C}$. The applied oscillatory strain is $\epsilon(t) = 0.0050 \sin(\omega t)$ (peak amplitude $\epsilon_0 = 0.0050$). The resulting steady-state stress is measured to be:
(a) Determine the peak stress amplitude $\sigma_0$ and phase angle $\delta$ in degrees. (b) Calculate the complex modulus magnitude $|E^*|$, storage modulus $E'$, and loss modulus $E''$ in $\text{MPa}$. (c) Calculate the loss factor $\tan \delta$. (d) Calculate the mechanical energy dissipated per cycle per unit volume $\Delta U_{\text{cycle}} = \pi \sigma_0 \epsilon_0 \sin\delta$ in $\text{kJ/m}^3$.
Step 1: Stress Amplitude and Phase Angle
From the equation $\sigma(t) = \sigma_0 \sin(\omega t + \delta)$:
- $\sigma_0 = 8.50 \times 10^6\text{ Pa} = 8.50\text{ MPa}$
- $\delta = 0.350\text{ rad}$
Convert phase angle to degrees:
Step 2: Complex, Storage, and Loss Moduli
1. Complex Modulus $|E^*|$:
2. Storage Modulus $E'$:
3. Loss Modulus $E''$:
Step 3: Loss Factor $\tan \delta$
(Alternatively: $\tan(20.054^\circ) = 0.36506$). A $\tan \delta$ value of $0.365$ indicates significant viscoelastic damping, characteristic of a polymer operating in its glass-rubber transition zone.
Step 4: Dissipated Energy per Cycle $\Delta U_{\text{cycle}}$
The energy dissipated as heat per unit volume during one complete cycle is:
Given:
- $\sigma_0 = 8.50 \times 10^6\text{ Pa}$
- $\epsilon_0 = 0.0050$
- $\sin\delta = 0.34292$
Substitute:
This dissipated mechanical energy ($45.8\text{ kJ/m}^3$) is converted entirely into internal thermal heating, demonstrating why high-frequency cyclic loading causes self-heating in viscoelastic dampers and tires.
(a) sigma_0 = 8.50 MPa, delta = 20.05°; (b) |E*| = 1,700 MPa, E' = 1,597 MPa, E'' = 583 MPa; (c) tan delta = 0.365; (d) Energy dissipated per cycle = 45.79 kJ/m^3.
Stress relaxation measurements are carried out on an amorphous poly(vinyl acetate) (PVAc) elastomer with glass transition temperature $T_g = 32.0^\circ\text{C}$ ($305.15\text{ K}$). The universal WLF parameters referenced to $T_g$ are $C_1 = 17.44$ and $C_2 = 51.6\text{ K}$:
(a) Calculate the shift factors $\log a_T$ and $a_T$ for temperatures $T = 42.0^\circ\text{C}, 52.0^\circ\text{C}, 72.0^\circ\text{C}$, and $82.0^\circ\text{C}$ relative to the reference temperature $T_{\text{ref}} = T_g = 32.0^\circ\text{C}$. (b) At $T = 72.0^\circ\text{C}$, the shear relaxation modulus is measured to be $G(t) = 1.00 \times 10^5\text{ Pa}$ at time $t = 10.0\text{ seconds}$. Using the principle of Time-Temperature Superposition, calculate the equivalent time $t_{\text{ref}}$ required for the polymer to relax to this identical modulus value at room temperature $T = 32.0^\circ\text{C}$. (c) Express this equivalent time in hours, days, or years, and comment on the power of TTS for predictive accelerated aging.
Step 1: Calculate WLF Shift Factors
Formula: $\log a_T = \frac{-17.44 (T - T_g)}{51.6 + (T - T_g)}$ where $T_g = 32.0^\circ\text{C}$:
1. $T = 42.0^\circ\text{C}$ ($T - T_g = +10.0^\circ\text{C}$):
2. $T = 52.0^\circ\text{C}$ ($T - T_g = +20.0^\circ\text{C}$):
3. $T = 72.0^\circ\text{C}$ ($T - T_g = +40.0^\circ\text{C}$):
4. $T = 82.0^\circ\text{C}$ ($T - T_g = +50.0^\circ\text{C}$):
Step 2: Equivalent Relaxation Time at $T = 32.0^\circ\text{C}$
According to TTS:
Given:
- Test temperature $T = 72.0^\circ\text{C}$
- Measurement time $t = 10.0\text{ s}$
- Shift factor $a_T = 2.423 \times 10^{-8}$
Step 3: Conversion and Physical Interpretation
Convert $t_{\text{ref}}$:
- In hours:
- In days:
- In years ($365.25\text{ days/year}$):
A test lasting just 10 seconds at $72^\circ\text{C}$ directly predicts the stress relaxation behavior after 13 years of continuous service at room temperature ($32^\circ\text{C}$)! This illustrates the incredible utility of Time-Temperature Superposition in aerospace, civil infrastructure, and polymer lifetime prediction.
(a) log(a_T): -2.83 (42 °C), -4.87 (52 °C), -7.62 (72 °C), -8.58 (82 °C); (b) t_ref = 4.13 x 10^8 seconds; (c) Equivalent time = 13.1 years (demonstrates 10-second high-T test predicting 13-year ambient relaxation).
Starting from the Doolittle empirical equation for liquid viscosity in terms of fractional free volume $f$:
where $A$ and $B$ are constants ($B \approx 1.0$), and the definition of the shift factor $a_T = \eta(T) / \eta(T_g)$: (a) Show that:
(b) Assuming the fractional free volume expands linearly above $T_g$:
where $\alpha_f$ is the thermal expansion coefficient of free volume, derive algebraically the WLF equation:
and express the constants $C_1$ and $C_2$ in terms of $B, f_g$, and $\alpha_f$. (c) Given the experimental values $C_1 = 17.44$ and $C_2 = 51.6\text{ K}$, calculate the numerical values of the fractional free volume at $T_g$ ($f_g$) and the thermal expansion coefficient $\alpha_f$ (assuming $B = 1.00$).
Step 1: Derivation of $\ln a_T$ from Doolittle Equation
The Doolittle viscosity equation is:
At the reference temperature $T_g$:
Subtracting the two equations:
Step 2: Substitution of Linear Free Volume Expansion
Substitute $f(T) = f_g + \alpha_f (T - T_g)$:
Multiply by $B$:
Divide numerator and denominator of the fraction by $\alpha_f$:
Convert natural logarithm to common base-10 logarithm ($\log x = \ln x / \ln(10) = \ln x / 2.302585$):
Comparing with the standard WLF form $\log a_T = \frac{-C_1 (T - T_g)}{C_2 + (T - T_g)}$:
This completes the exact theoretical derivation!
Step 3: Calculate $f_g$ and $\alpha_f$
Given $C_1 = 17.44$, $C_2 = 51.6\text{ K}$, and $B = 1.00$:
- Solve for $f_g$:
- Solve for $\alpha_f$:
This proves that at the glass transition, all amorphous polymers collapse to a universal fractional free volume of $2.5\%$, and free volume expands above $T_g$ with a universal thermal expansion rate of $4.8 \times 10^{-4}\text{ K}^{-1}$!
(a) ln(a_T) = B (1/f(T) - 1/f_g); (b) WLF form proved with C_1 = B / (2.303 f_g) and C_2 = f_g / alpha_f; (c) f_g = 0.0249 (2.49% free volume at T_g); alpha_f = 4.83 x 10^-4 K^-1.
The Standard Linear Solid (Zener model) consists of a Maxwell element (spring $E_1$ in series with dashpot $\eta$) connected in parallel with an equilibrium spring $E_2$. (a) Derive the differential constitutive equation relating stress $\sigma(t)$ and strain $\epsilon(t)$:
where $\tau_R = \eta / E_1$. (b) For harmonic oscillatory strain $\epsilon(t) = \epsilon_0 e^{i \omega t}$ and stress $\sigma(t) = \sigma_0 e^{i(\omega t + \delta)}$, show that the complex modulus $E^*(\omega) = \sigma(t) / \epsilon(t)$ is:
(c) Identify the storage modulus $E'(\omega)$, loss modulus $E''(\omega)$, and loss tangent $\tan \delta(\omega)$. (d) Find the angular frequency $\omega_{\text{max}}$ at which the loss modulus $E''$ achieves its peak maximum, and calculate $E''_{\text{max}}$.
Step 1: Derivation of the Constitutive Equation
In the parallel arrangement:
- Total stress is the sum of both arms: $\sigma = \sigma_1 + \sigma_2$.
- Strain across both arms is identical: $\epsilon = \epsilon_1 = \epsilon_2$.
For the equilibrium spring arm:
For the Maxwell arm:
Multiply by $\eta$:
where $\tau_R = \eta / E_1$. Substitute $\sigma_1 = \sigma - \sigma_2 = \sigma - E_2 \epsilon$:
Rearranging terms:
Since $\eta = E_1 \tau_R$:
Therefore:
Step 2: Complex Modulus $E^*(\omega)$
Substitute $\epsilon(t) = \epsilon_0 e^{i \omega t}$ and $\sigma(t) = \sigma_0 e^{i \omega t}$:
Substitute into the differential equation:
Solve for $E^*(\omega) = \sigma / \epsilon$:
Separate $E_2$:
Multiply numerator and denominator by the complex conjugate $(1 - i \omega \tau_R)$:
Adding $E_2$:
Step 3: Storage, Loss Moduli and $\tan \delta$
1. Storage Modulus:
- At $\omega \to 0$ (low frequency / equilibrium): $E' \to E_2$.
- At $\omega \to \infty$ (high frequency / glassy): $E' \to E_1 + E_2$.
2. Loss Modulus:
3. Loss Tangent:
Step 4: Maximum Loss Modulus $E''_{\text{max}}$
To find the maximum of $E''(\omega)$, differentiate with respect to $\omega$:
Setting numerator to zero:
Substitute $\omega_{\text{max}} = 1/\tau_R$ into $E''$:
The loss modulus achieves its symmetrical peak at frequency $\omega = 1/\tau_R$ with maximum height equal to half the relaxing modulus $E_1$.
(a) Differential equation derived; (b) E(omega) derived with real and imaginary parts; (c) E'(omega) = E_2 + E_1omega^2tau_R^2 / (1 + omega^2tau_R^2), E''(omega) = E_1omegatau_R / (1 + omega^2*tau_R^2); (d) Peak loss occurs at omega_max = 1 / tau_R with E''_max = E_1 / 2.
According to the statistical thermodynamic theory of rubber elasticity, the free energy of an ideal elastomeric network arises from conformational entropy loss upon network deformation. For an affine network of $N$ network subchains per unit volume:
where $\lambda_i = L_i / L_{i, 0}$ are the extension ratios. (a) For incompressible uniaxial extension ($\lambda_x = \lambda, \lambda_y = \lambda_z = 1/\sqrt{\lambda}$), derive the true stress $\sigma_{\text{true}}$ and engineering stress $\sigma_{\text{eng}}$:
where $G = N k_B T = \frac{\rho R T}{M_c}$ is the shear modulus, and $M_c$ is the number-average molecular weight between cross-links. (b) A vulcanized polyisoprene rubber (density $\rho = 0.920\text{ g/cm}^3$) is tested at $T = 25.0^\circ\text{C}$ ($298.15\text{ K}$). At an extension ratio of $\lambda = 2.00$ ($100\%$ elongation), the engineering tensile stress is measured to be $\sigma_{\text{eng}} = 1.050\text{ MPa}$. Calculate:
- The shear modulus $G$ of the rubber.
- The number of active network subchains per cubic meter $N$.
- The average molecular weight between cross-links $M_c$.
- The average number of isoprene repeat units ($M_0 = 68.12\text{ g/mol}$) per network subchain.
Step 1: Derivation of the Stress-Strain Relation
For incompressible deformation, volume is constant: $V = L_x L_y L_z = V_0 \implies \lambda_x \lambda_y \lambda_z = 1$. Under uniaxial tension along the x-axis:
Substitute into the Helmholtz free energy per unit volume:
The engineering stress is the derivative with respect to $\lambda$:
The true stress is $\sigma_{\text{true}} = \lambda \sigma_{\text{eng}} = G (\lambda^2 - 1/\lambda)$.
Step 2: Calculate Shear Modulus $G$
Given $\lambda = 2.00$ and $\sigma_{\text{eng}} = 1.050\text{ MPa} = 1.050 \times 10^6\text{ Pa}$:
Rearranging for $G$:
Step 3: Calculate Subchain Density $N$
Since $G = N k_B T$:
Given:
- $G = 6.000 \times 10^5\text{ Pa}$
- $k_B = 1.38065 \times 10^{-23}\text{ J/K}$
- $T = 298.15\text{ K} \implies k_B T = 4.1164 \times 10^{-21}\text{ J}$
Step 4: Calculate Molecular Weight Between Cross-Links $M_c$
Since $G = \frac{\rho R T}{M_c}$:
Given:
- $\rho = 0.920\text{ g/cm}^3 = 920\text{ kg/m}^3$
- $R = 8.31446\text{ J/(mol K)}$
- $T = 298.15\text{ K} \implies R T = 2,478.96\text{ J/mol}$
- $G = 6.000 \times 10^5\text{ Pa}$
Step 5: Isoprene Units per Network Chain
Given isoprene repeating unit $M_0 = 68.12\text{ g/mol}$:
Each network strand between cross-link junction points contains approximately 56 isoprene monomer units, providing sufficient conformational degrees of freedom to support reversible entropy elasticity.
(a) sigma_eng = G * (lambda - 1/lambda^2) derived; (b) Shear modulus G = 600 kPa (0.600 MPa); (c) Subchain density N = 1.46 x 10^26 chains/m^3; (d) M_c = 3,801 g/mol; Average 56 isoprene units per cross-link strand.