Creep of a standard-linear-solid model (Kelvin–Voigt in series with a spring): 23-Chem-B8 Polymer Engineering, December 2014, Q5

Question 5: Creep of a standard-linear-solid model (Kelvin–Voigt in series with a spring) (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. Series combination of a Kelvin–Voigt element ($E_1\parallel\eta_1$) and a lone spring $E_2$, under constant stress $\sigma$. Measured strains: $\epsilon(0^+)=0.002$, $\epsilon(1000\,\text{s})=0.004$, $\epsilon(\infty)=0.006$.

Find. (a) $\epsilon(t)$; (b) the retardation time $\tau_1$; (c) the creep–recovery curve.

Mechanical analogue: a lone spring $E_2$ in series with a Kelvin–Voigt element ($E_1$ parallel to dashpot $\eta_1$). The series spring gives the instantaneous jump; the Kelvin element gives the delayed, fully-recoverable creep.

Approach. Add the strains of series elements: the lone spring responds instantly, the Kelvin element responds with a delayed exponential of time constant $\tau_1=\eta_1/E_1$; pin the two compliances and $\tau_1$ from the three data points.

(a) Total-strain expression

Elements in series carry the same stress $\sigma$ and their strains add. The lone spring gives an instantaneous elastic strain $\sigma/E_2$; the Kelvin–Voigt element under constant stress creeps as $(\sigma/E_1)(1-e^{-t/\tau_1})$. Therefore

$\displaystyle \boxed{\;\epsilon(t)=\frac{\sigma}{E_2} +\frac{\sigma}{E_1}\left(1-e^{-t/\tau_1}\right),\qquad \tau_1=\frac{\eta_1}{E_1}. }$

(b) Retardation time

Instantaneous response fixes $\sigma/E_2$. At $t=0^+$ the Kelvin term is zero, so $\epsilon(0^+)=\sigma/E_2=0.002$.
Long-time plateau fixes $\sigma/E_1$. As $t\to\infty$ the exponential dies and $\epsilon(\infty)=\sigma/E_2+\sigma/E_1=0.006$, hence $\sigma/E_1=0.006-0.002=0.004$.
Intermediate point fixes $\tau_1$. At $t=1000$ s, $$0.004 = 0.002+0.004\left(1-e^{-1000/\tau_1}\right) \;\Rightarrow\; 1-e^{-1000/\tau_1}=0.5 \;\Rightarrow\; e^{-1000/\tau_1}=0.5.$$ Thus $1000/\tau_1=\ln 2$ and $$\tau_1=\frac{1000}{\ln 2}=\boxed{1443\ \text{s}}.$$

(c) Creep and recovery

Creep–recovery response. On loading: an instantaneous jump to $\sigma/E_2=0.002$, then delayed creep toward $\epsilon_\infty=0.006$, passing through 0.004 at 1000 s. On unloading at 1000 s: an instantaneous elastic recovery of $\sigma/E_2=0.002$ (drop to 0.002), then the Kelvin strain decays exponentially with the same $\tau_1$ to zero permanent set.

Because there is no free (series) dashpot, all of the strain is recoverable: the series spring recovers instantly and the Kelvin element relaxes back to zero, leaving no permanent set. At the moment of unloading the Kelvin strain is $\epsilon_K(1000)=0.004(1-e^{-\ln2})=0.002$; it then decays as $0.002\,e^{-(t-1000)/\tau_1}$.

Quantity	Value
Instantaneous compliance $\sigma/E_2$	0.002
Delayed (Kelvin) compliance $\sigma/E_1$	0.004
Retardation time $\tau_1=\eta_1/E_1$	1443 s ($=1000/\ln2$)
Permanent set after recovery	0 (fully recoverable)

Topic: Linear viscoelasticity — the standard linear solid (Zener model) in creep.

Key relations: series strains add; Kelvin creep $(\sigma/E)(1-e^{-t/\tau})$, lone spring $\sigma/E$; retardation time $\tau=\eta/E$; total compliance $J(t)=1/E_2+(1/E_1)(1-e^{-t/\tau_1})$.

Why this works: The lone series spring supplies the instantaneous glassy response; the Kelvin element’s dashpot delays the extension of its parallel spring, giving the characteristic exponential approach to an equilibrium (rubbery) compliance. Because the dashpot is bounded by its parallel spring, it can never flow indefinitely, so the model has no viscous (permanent) flow — unlike a Maxwell or four-element Burgers model. That is exactly why the recovery here returns fully to zero.

Common pitfall: Confusing the retardation time (creep, Kelvin) with the relaxation time (stress relaxation, Maxwell); and assuming a permanent set — there is none unless a free dashpot is placed in series. Numerically, note that $1-e^{-t/\tau}=0.5$ gives $t=\tau\ln2$, not $t=\tau$.

Source: Sperling, Physical Polymer Science 4th ed., Ch. 8 (mechanical models); Rudin & Choi 3rd ed., Ch. on mechanical properties.

Quantity	Value
Instantaneous compliance \(\sigma/E_2\)	0.002
Delayed (Kelvin) compliance \(\sigma/E_1\)	0.004
Retardation time \(\tau_1=\eta_1/E_1\)	1443 s (\(=1000/\ln2\))
Permanent set after recovery	0 (fully recoverable)

Question 5 of 7: Creep of a standard-linear-solid model (Kelvin–Voigt in series with a spring)

Question 5: Creep of a standard-linear-solid model (Kelvin–Voigt in series with a spring) (Part C — 30 marks)

(a) Total-strain expression

(b) Retardation time

(c) Creep and recovery