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Chapter 15: Problem 11

For thermoplastic polymers, cite five factors that favor brittle fracture.

Short Answer

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Question: List five factors that favor brittle fracture in thermoplastic polymers. Answer: The five factors that favor brittle fracture in thermoplastic polymers are 1) low molecular weight polymers, 2) low temperature, 3) high strain rates, 4) presence of stress concentrators, and 5) environmental factors such as exposure to chemicals, UV radiation, or moisture absorption.

Step by step solution

01

Factor 1: Low molecular weight polymers

Thermoplastic polymers with low molecular weight tend to be more brittle as they have weaker intermolecular interactions and shorter polymer chains, which are less able to absorb and redistribute applied stresses. This results in a greater likelihood of brittle fracture under stress.
02

Factor 2: Low temperature

At low temperatures, the mobility of the polymer chains is reduced, which makes the material more rigid and less ductile. Under these conditions, the thermoplastic polymer is more susceptible to brittle fracture when subjected to stress.
03

Factor 3: High strain rates

Brittle fracture is more likely to occur in thermoplastic polymers when a high strain rate is applied. This is because at high strain rates, the material does not have enough time to deform plastically and redistribute the stress before fracture occurs.
04

Factor 4: Presence of stress concentrators

Stress concentrators, such as notches, cracks, or defects within the thermoplastic polymer, can increase the likelihood of brittle fracture. These concentrators serve as starting points for crack propagation, which can lead to brittle failure of the material when subjected to stress.
05

Factor 5: Environmental factors

Certain environmental factors, such as exposure to specific chemicals or ultraviolet (UV) radiation, can weaken the polymer chains or lead to chain scission, which in turn makes the material more susceptible to brittle fracture under stress. In addition, environmental factors like moisture absorption can change the mechanical properties of thermoplastic polymers and increase the likelihood of brittle fracture.

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Most popular questions from this chapter

The tensile strength and number-average molecular weight for two polyethylene materials are as follows: \begin{tabular}{cc} \hline Tensile Strength (MPa) & Number-Average Molecular Weight \((\mathrm{g} / \mathbf{m o l})\) \\ \hline 85 & 12,700 \\ 150 & 28,500 \\ \hline \end{tabular} Estimate the number-average molecular weight that is required to give a tensile strength of \(195 \mathrm{MPa}\).

Why must fiber materials that are melt-spun and then drawn be thermoplastic? Cite two reasons.

Which of the following polyethylene thin films would have the better mechanical characteristics: (1) formed by blowing, or (2) formed by extrusion and then rolled? Why?

Normal butane and isobutane have boiling temperatures of \(-0.5\) and \(-12.3^{\circ} \mathrm{C}\) (31.1 and \(\left.9.9^{\circ} \mathrm{F}\right)\), respectively. Briefly explain this behavior on the basis of their molecular structures, as presented in Section 14.2.

For some viscoelastic polymers that are subjected to stress relaxation tests, the stress decays with time according to $$ \sigma(t)=\sigma(0) \exp \left(-\frac{t}{\tau}\right) $$ where \(\sigma(t)\) and \(\sigma(0)\) represent the timedependent and initial (i.e., time \(=0\) ) stresses, respectively, and \(t\) and \(\tau\) denote elapsed time and the relaxation time; \(\tau\) is a timeindependent constant characteristic of the material. A specimen of a viscoelastic polymer whose stress relaxation obeys Equation \(15.10\) was suddenly pulled in tension to a measured strain of \(0.6\); the stress necessary to maintain this constant strain was measured as a function of time. Determine \(E_{r}(10)\) for this material if the initial stress level was \(2.76\) MPa (400 psi), which dropped to \(1.72 \mathrm{MPa}\) (250 psi) after \(60 \mathrm{~s}\).
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