Nylon Band Stress Relaxation and Glass/Nylon Composite Load Sharing: 04-BS-11, May 2013, Q4

Question 4: Nylon Band Stress Relaxation and Glass/Nylon Composite Load Sharing (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) Band cross-section $\tfrac12\times\tfrac18$ in; minimum holding stress $\sigma_{min}=1200$ psi; relaxation test: $\sigma_0=1500$ psi $\to$ $\sigma(5\ \text{wk})=1460$ psi; storage time $t=1$ yr $=52$ weeks. (b) Glass/nylon composite, $V_f=25\%$ glass, $E_{glass}=10.5\times10^6$ psi, $E_{nylon}=0.4\times10^6$ psi.

Find. (a) Initial stress/load to apply so the band still exceeds 1200 psi after 1 year. (b) Fraction of load carried by the glass fibres.

Fig. Q4(a) — exponential stress-relaxation decay $\sigma(t)=\sigma_0 e^{-t/\tau}$, calibrated from the 5-week test point and evaluated at 52 weeks.

Approach

Model the polymer's stress relaxation as a single-exponential decay, calibrate the relaxation time $\tau$ from the 5-week test data, then find the initial stress that leaves exactly 1200 psi after 52 weeks; separately, use the isostrain (parallel/Voigt) composite model to split the load between fibre and matrix by stiffness.

Calibrate the relaxation time. Assuming $\sigma(t)=\sigma_0\,e^{-t/\tau}$, $$\tau=\frac{-t}{\ln[\sigma(t)/\sigma_0]}=\frac{-5}{\ln(1460/1500)}=\boxed{185.0\ \text{weeks}}.$$
Required initial stress for 1-year storage. The band must still read $\ge1200$ psi at $t=52$ weeks, so $$\sigma_0'=\frac{\sigma_{min}}{e^{-52/\tau}}=\frac{1200}{e^{-52/185.0}}=\boxed{1589.5\ \text{psi}}.$$
Convert to load. Band area $A=\tfrac12\times\tfrac18=0.0625\ \text{in}^2$: $$P_0=\sigma_0'A=1589.5\times0.0625=\boxed{99.3\ \text{lbf}}.$$
(b) Isostrain load split. With the fibres and matrix bonded and strained equally ($\varepsilon_f=\varepsilon_m$) under load parallel to the fibres, each phase's share of the total force is proportional to its stiffness times its volume fraction: $$\frac{P_f}{P_{total}}=\frac{E_fV_f}{E_fV_f+E_mV_m}=\frac{(10.5\times10^6)(0.25)}{(10.5\times10^6)(0.25)+(0.4\times10^6)(0.75)}=\boxed{89.7\%}.$$

Quantity	Result
Relaxation time, $\tau$	185.0 weeks
Required initial stress	1589.5 psi
Required initial load	99.3 lbf
Fraction of load on glass fibres	89.7%

Check

Part (b) assumes: fibres continuous and aligned parallel to the load, perfect fibre–matrix bonding (no slip), both phases still elastic, and isostrain (Voigt/rule-of-mixtures) loading.

Concept & Theory

Topic. Polymer stress relaxation and isostrain composite load-sharing.

Key relations. $\sigma(t)=\sigma_0e^{-t/\tau}$ (linear-viscoelastic relaxation); isostrain rule of mixtures $P_f/P_{total}=E_fV_f/(E_fV_f+E_mV_m)$.

Why this works. A polymer band under sustained strain relaxes stress as its chains slowly rearrange; the single-exponential model, once calibrated from one short-time data point, extrapolates the same decay law to the long service time. In an aligned continuous-fibre composite loaded along the fibres, compatibility forces equal strain in both phases, so the much stiffer glass fibres carry a disproportionate share of the load despite being only 25% of the volume.

Common pitfall. Extrapolating the exponential decay using only the ratio $\sigma(t)/\sigma_0$ without first solving for $\tau$ explicitly, or (in part b) using volume fraction alone (ignoring the stiffness ratio) to split the load.

Source. Callister & Rethwisch, Materials Science and Engineering, 9th ed., Ch. 15 (polymers, viscoelasticity) and Ch. 16 (composites).

Question 4 of 8: Nylon Band Stress Relaxation and Glass/Nylon Composite Load Sharing

Question 4: Nylon Band Stress Relaxation and Glass/Nylon Composite Load Sharing (20 marks)

Approach