Question 4 of 7: Stress Relaxation of a Nylon Band; Load Sharing in a Glass/Nylon Composite

Question 4: Stress Relaxation of a Nylon Band; Load Sharing in a Glass/Nylon Composite (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) $\sigma_0=1500$ psi at $t=0$; $\sigma=1460$ psi at $t_1=5$ weeks; minimum holding stress $\sigma_{min}=1200$ psi must still be met after $t_2=1$ year $=52$ weeks. (b) $V_{glass}=0.25$, $V_{nylon}=0.75$, $E_{glass}=10.5\times10^6$ psi, $E_{nylon}=0.4\times10^6$ psi.

Find. (a) Initial stress (and, illustratively, load) that must be applied so the band still meets the 1200 psi minimum after 1 year. (b) Fraction of the applied load carried by the glass fibres.

Approach

(a) is a stress-relaxation problem: a polymer held at fixed strain (clamped around the tubes) relaxes stress exponentially with time, $\sigma(t)=\sigma_0e^{-t/\tau}$. The 5-week test data fixes the relaxation time constant $\tau$; the same $\tau$ is then used to find what larger initial stress is needed so the stress has decayed to no less than 1200 psi after a full year. (b) uses the isostrain (parallel, Voigt) composite model appropriate for continuous, aligned, well-bonded fibres sharing a common applied strain.

1. (a) Relaxation time constant from the test data. Using the given relaxation law $\sigma=\sigma_0e^{-t/\tau}$: $$1460=1500\,e^{-5/\tau}\ \Rightarrow\ \tau=\frac{-5}{\ln(1460/1500)}=\frac{-5}{\ln(0.97333)} =\frac{5}{0.02704}=185.0\ \text{weeks}$$
2. (a) Required initial stress for 1-year service. The stress must still equal (at minimum) 1200 psi after $t_2=52$ weeks: $$\sigma_i\,e^{-52/185.0}=1200\ \Rightarrow\ \sigma_i=\frac{1200}{e^{-0.2811}}=\frac{1200}{0.7548}$$ $$\boxed{\sigma_i\approx1590\ \text{psi}}$$ This is the governing, source-independent result: the band must be tightened to about 1590 psi initially so that a full year of relaxation still leaves at least the 1200 psi needed to hold the tubes.
3. (a) Illustrative initial load (assumed thickness). With the assumed cross section $0.5\times0.125=0.0625$ in$^2$ (Check above): $$P_i=\sigma_i\,A=1590(0.0625)\approx\boxed{99.4\ \text{lb}}$$ (scale this directly by the true thickness once known: $P_i=1590\times0.5\times t$).
4. (b) Isostrain load-sharing fraction. Under a common applied strain $e$, each phase carries a load proportional to its own stiffness and volume fraction, $P_i=\sigma_iA_i=E_ie\,A_i\propto E_iV_i$: $$\frac{P_{glass}}{P_{total}}=\frac{E_{glass}V_{glass}}{E_{glass}V_{glass}+E_{nylon}V_{nylon}} =\frac{10.5\times10^6(0.25)}{10.5\times10^6(0.25)+0.4\times10^6(0.75)}$$ $$=\frac{2.625\times10^6}{2.625\times10^6+0.300\times10^6}=\frac{2.625}{2.925}$$ $$\boxed{\frac{P_{glass}}{P_{total}}\approx0.897\ (89.7\%)}$$ Despite being only 25% of the volume, the much stiffer glass fibres carry nearly 90% of the load — a hallmark result of isostrain composite loading.
5. (b) Assumptions. Fibres are continuous, aligned parallel to the load, and perfectly bonded to the nylon matrix (no slip at the interface); both phases remain linear-elastic; strain is uniform across the cross-section (isostrain/Voigt condition), which holds for loading along the fibre direction but would not apply to transverse loading.