Polyacrylonitrile Molecular Weight Averages; Stress-Relaxation Life of a Fastener: 04-BS-11, December 2019, Q6

Question 6: Polyacrylonitrile Molecular Weight Averages; Stress-Relaxation Life of a Fastener (20 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. (a) Six $(N_i,\,\overline M_i)$ chain-count/mean-molecular-weight groups as tabulated, for polyacrylonitrile, repeat unit (CH$_2$CHCN)$_n$. (b) Stress-relaxation fastener: $\sigma_0=4000$ psi, $\sigma(100\text{ h})=3500$ psi, minimum functional stress $\sigma_{min}=2500$ psi.

Find. (a) $\overline M_n$, $\overline M_w$, and degree of polymerization DP. (b) Time at which the stress falls to $2500$ psi.

Approach

(a) The table gives chain counts $N_i$ at each mean molecular weight $\overline M_i$, so $\overline M_n=\Sigma N_i\overline M_i/\Sigma N_i$ and $\overline M_w=\Sigma N_i\overline M_i^2/\Sigma N_i\overline M_i$; DP $=\overline M_w/M_{repeat}$ using the given weight-average. (b) Polymer stress relaxation follows $\sigma(t)=\sigma_0e^{-t/\lambda}$; fit the one data point to find the relaxation time $\lambda$, then solve for the time at which $\sigma=2500$ psi.

(a) Number-average molecular weight. $$\overline M_n = \frac{\Sigma N_i\overline M_i}{\Sigma N_i} = \frac{10000(3000)+18000(6000)+17000(9000)+15000(12000)+9000(15000)+4000(18000)}{73{,}000}$$ $$\boxed{\overline M_n \approx 9290\ \text{g/mol}}$$
(a) Weight-average molecular weight. $$\overline M_w = \frac{\Sigma N_i\overline M_i^2}{\Sigma N_i\overline M_i}$$ Using the same six terms, now weighted by $\overline M_i^2$ in the numerator and by $\Sigma N_i\overline M_i = 678.0\times10^{6}$ in the denominator: $$\boxed{\overline M_w \approx 11{,}200\ \text{g/mol}}$$
(a) Degree of polymerization. The repeat unit is CH$_2$–CH(CN)–, formula C$_3$H$_3$N: $$M_{repeat} = 3(12.01)+3(1.01)+14.01 = 53.07\ \text{g/mol}$$ $$DP = \frac{\overline M_w}{M_{repeat}} = \frac{11{,}200}{53.07}$$ $$\boxed{DP \approx 211}$$
(b) Relaxation time from the one data point. Stress relaxation at constant strain follows $\sigma(t)=\sigma_0 e^{-t/\lambda}$. Using $\sigma(100\text{ h})=3500$ psi: $$\lambda = \frac{-t}{\ln(\sigma/\sigma_0)} = \frac{-100}{\ln(3500/4000)}$$ $$\boxed{\lambda \approx 749\ \text{h}}$$
(b) Time to reach the minimum functional stress. Solving the same relation for $t$ at $\sigma=2500$ psi: $$t = -\lambda\ln\!\left(\frac{\sigma_{min}}{\sigma_0}\right) = -749\ln\!\left(\frac{2500}{4000}\right)$$ $$\boxed{t \approx 352\ \text{h} \approx 14.7\ \text{days}}$$

Quantity	Result
Number-average MW, $\overline M_n$	≈ 9290 g/mol
Weight-average MW, $\overline M_w$	≈ 11,200 g/mol
Degree of polymerization, DP	≈ 211
Relaxation time, $\lambda$	≈ 749 h
Fastener life (to 2500 psi)	≈ 352 h (≈ 14.7 days)

Concept & Theory

Topic. Polymer molecular-weight statistics; polymer viscoelastic stress relaxation.

Key relations. $\overline M_n=\Sigma N_i\overline M_i/\Sigma N_i$; $\overline M_w=\Sigma N_i\overline M_i^2/\Sigma N_i\overline M_i$; $DP=\overline M_w/M_{repeat}$ (or $\overline M_n/M_{repeat}$, depending which average is requested); stress relaxation $\sigma(t)=\sigma_0e^{-t/\lambda}$.

Why this works. $\overline M_w$ weights each chain-length fraction by its own mass (not just its count), so it is more sensitive to the presence of a few very long chains than $\overline M_n$ — this is why $\overline M_w > \overline M_n$ always, and their ratio (the polydispersity index) measures how broad the chain-length distribution is. Stress relaxation is the signature viscoelastic response of a polymer held at constant strain: molecular chains gradually disentangle/reorient to relieve stress, and the exponential decay with a single relaxation time is the simplest (Maxwell-element) model of that process.

Common pitfall. Using $\overline M_n$ instead of $\overline M_w$ for DP when the question specifically asks for the weight-average-based value; and, for the relaxation problem, trying to use two different $(t,\sigma)$ points as if fitting a line on linear axes instead of recognizing that only $\ln(\sigma/\sigma_0)$ vs. $t$ is linear (a single point plus the known $\sigma_0$ is enough to fix $\lambda$ exactly).

Source. Callister & Rethwisch, Materials Science and Engineering, 9th ed., Ch. 14–15 (polymer molecular weight, degree of polymerization, viscoelasticity and stress relaxation).

Question 6 of 7: Polyacrylonitrile Molecular Weight Averages; Stress-Relaxation Life of a Fastener

Question 6: Polyacrylonitrile Molecular Weight Averages; Stress-Relaxation Life of a Fastener (20 marks)

Approach