Constant-rate extension of Maxwell and Kelvin–Voigt materials: 23-Chem-B8 Polymer Engineering, December 2014, Q6

Question 6: Constant-rate extension of Maxwell and Kelvin–Voigt materials (Part C — 30 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Constant strain-rate ($\dot\epsilon$) extension. Polymer A = Kelvin–Voigt, $E_A=1.0\times10^{8}\ \text{dyn/cm}^2$ ($=10\ \text{MPa}$), $\tau_A=1.0\times10^{5}\ \text{s}$. Polymer B = Maxwell, $E_B=1.0\times10^{10}\ \text{dyn/cm}^2$ ($=1\ \text{GPa}$), $\tau_B=2.0\ \text{s}$. Each element viscosity is $\eta=E\tau$.

Find. (a) maximum extension (rupture strain) and (b) tensile strength for each polymer.

The two rheological models: Polymer B as a Maxwell element (spring and dashpot in series — flows viscously, stress saturates) and Polymer A as a Kelvin–Voigt element (spring parallel to dashpot — stress climbs without bound as strain accumulates).

Approach. Write each model’s stress response to a constant strain rate $\epsilon=\dot\epsilon\,t$; the qualitative shapes then dictate which quantity (a limiting stress or a limiting strain) each model can supply, and a numerical value follows once a strain rate and a rupture criterion are fixed.

Check — data not fully specified

A unique number for “maximum extension” and “tensile strength” requires two inputs the question does not give: the applied strain rate $\dot\epsilon$ and a rupture criterion (a break stress or break strain). Following the exam’s own Note 1 (“state any assumptions”) and Note 2 (“provide a solution methodology even without all data”), the closed-form responses are derived below and illustrated with an assumed laboratory rate $\dot\epsilon=1.0\times10^{-2}\ \text{s}^{-1}$. The method and the qualitative contrast are the graded content.

Model responses at constant strain rate

Maxwell (Polymer B). With spring and dashpot in series, $\dot\epsilon=\dot\sigma/E+\sigma/\eta$. For constant $\dot\epsilon$, $$\sigma(t)=\eta\dot\epsilon\left(1-e^{-t/\tau}\right)=E\tau\dot\epsilon\left(1-e^{-\epsilon/(\tau\dot\epsilon)}\right).$$ The stress saturates at $\sigma_{\max}=\eta\dot\epsilon=E\tau\dot\epsilon$: the dashpot flows as fast as it is stretched, so the material yields/draws at essentially constant stress and can extend to large strains. Maxwell therefore supplies a well-defined tensile strength (the plateau) but an unbounded extension (limited only by the true break criterion).
Kelvin–Voigt (Polymer A). With spring parallel to dashpot, $\sigma=E\epsilon+\eta\dot\epsilon=E\epsilon+E\tau\dot\epsilon$. The stress climbs without bound as strain accumulates; there is no plateau. Kelvin–Voigt therefore supplies a recoverable, retarded-elastic response with a well-defined limiting strain (it cannot flow), and its “strength” is whatever stress corresponds to the rupture strain.

Illustrative stress–strain curves at $\dot\epsilon=10^{-2}\ \text{s}^{-1}$: the Maxwell material (B) plateaus at $\eta\dot\epsilon$, while the Kelvin–Voigt material (A) rises linearly (offset by the viscous term) with no plateau.

Illustrative numbers (assumed $\dot\epsilon=10^{-2}\,\text{s}^{-1}$)

Polymer B tensile strength (plateau). $\eta_B=E_B\tau_B=(10^{10})(2.0)=2\times10^{10}\ \text{dyn s/cm}^2$, so $$\sigma_{B,\max}=\eta_B\dot\epsilon=(2\times10^{10})(10^{-2}) =\boxed{2\times10^{8}\ \text{dyn/cm}^2\;(=20\ \text{MPa})}.$$ Extension is set by the draw/rupture criterion, not the model — B is the ductile, flow-dominated material.
Polymer A stress at a reference strain $\epsilon=1$. $\eta_A=E_A\tau_A=(10^{8})(10^{5})=10^{13}\ \text{dyn s/cm}^2$, so the viscous offset $\eta_A\dot\epsilon=(10^{13})(10^{-2})=10^{11}\ \text{dyn/cm}^2$ already dwarfs the elastic term $E_A\epsilon\le10^{8}$. Because $\tau_A$ ($10^5$ s) is enormous compared with any test duration, Polymer A behaves essentially as a stiff, retarded solid that reaches very high stress at small strain — the brittle, elasticity-dominated material.

Polymer	Model	$\eta=E\tau$	Character	Illustrative result ($\dot\epsilon=10^{-2}\,\text{s}^{-1}$)
B	Maxwell	$2\times10^{10}$	ductile / draws (stress plateaus)	tensile strength $\approx\eta\dot\epsilon = 2\times10^{8}$ dyn/cm² (20 MPa); large extension
A	Kelvin–Voigt	$10^{13}$	brittle / retarded-elastic (stress climbs)	small rupture strain; high stress (viscous term $\sim10^{11}$ dyn/cm² dominates)

Topic: Constant strain-rate response of the two elementary viscoelastic models; the Deborah number and ductile-vs-brittle behaviour.

Key relations: Maxwell $\sigma=\eta\dot\epsilon(1-e^{-t/\tau})$, plateau $\eta\dot\epsilon$; Kelvin–Voigt $\sigma=E\epsilon+\eta\dot\epsilon$; $\eta=E\tau$; Deborah number $De=\tau\dot\epsilon$ (or $\tau/t_{\text{test}}$).

Why this works: Whether a viscoelastic solid behaves as a liquid or a solid on the test time-scale is governed by $De$. Polymer B has a short $\tau=2$ s, so during any normal test $De\ll1$: it relaxes as fast as it is loaded, stress saturates and it draws (ductile). Polymer A has a huge $\tau=10^5$ s, so $De\gg1$: the dashpot is essentially locked, the material responds elastically and reaches high stress at small strain (brittle). The two models are complementary — Maxwell defines a limiting stress, Kelvin defines a limiting strain.

Common pitfall: Trying to extract both a rupture strain and a tensile strength from a single element without a strain rate or a break criterion — the models supply only one natural limit each. Also, confusing the Kelvin “relaxation time” wording in the question with a true relaxation time; for a Kelvin element $\tau=\eta/E$ is a retardation time.

Source: Sperling 4th ed., Ch. 8; Rudin & Choi 3rd ed., mechanical-properties chapter (Maxwell & Voigt models).

Question 6 of 7: Constant-rate extension of Maxwell and Kelvin–Voigt materials

Question 6: Constant-rate extension of Maxwell and Kelvin–Voigt materials (Part C — 30 marks)

Model responses at constant strain rate

Illustrative numbers (assumed \(\dot\epsilon=10^{-2}\,\text{s}^{-1}\))

Polymer	Model	\(\eta=E\tau\)	Character	Illustrative result (\(\dot\epsilon=10^{-2}\,\text{s}^{-1}\))
B	Maxwell	\(2\times10^{10}\)	ductile / draws (stress plateaus)	tensile strength \(\approx\eta\dot\epsilon = 2\times10^{8}\) dyn/cm² (20 MPa); large extension
A	Kelvin–Voigt	\(10^{13}\)	brittle / retarded-elastic (stress climbs)	small rupture strain; high stress (viscous term \(\sim10^{11}\) dyn/cm² dominates)