27.9: Questions
- Page ID
- 32812
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Quick questions
You should be able to answer these questions without too much difficulty after studying this TLP. If not, then you should go through it again!
Which of these methods of characterisation would be helpful in identifying a ceramic?
- Answer
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Correct. Ceramics are very hard to identify by eye, using EDS can give information about what elements are present and the proportions in which they are present. SEM will give an idea of microstructure and may help to identify processing methods, but does not give information about the actual composition. IR spectroscopy and DSC are commonly used in characterising polymers: IR identifies materials by their bonds, DSC by the relationship between heat capacity and temperature.
Deeper questions
The following questions require some thought and reaching the answer may require you to think beyond the contents of this TLP.
Match the material to the properties, try to think of reasons why the material may have been chosen.
Material 1 has a shiny 'spangled' appearance due to large dendritic grains on the surface, it is a structural component within the article and must withstand relatively high stresses, and may experience wear in service, it must not corrode in warm air and is in a low cost article.
- Answer
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D. Galvanised Steel: The spangled appearance is due to large zinc grains on the surface, the metal is galvanised to prevent corrosion. Steel is a better structural material than pure aluminium, and it is much cheaper. Stainless steel might also fulfil the requirements of the material, but it is more expensive. Coating the steel with tin provides corrosion protection, but only if the coating is not damaged.
Tinned steel usually has a fairly uniform appearance; it may be shiny, but does not have visible grains on the surface. It is not the best choice due to the stresses and wear the component must undergo. The corrosion protection comes from a layer of tin, if this is damaged the steel may corrode.
Stainless steel offers very good corrosion resistance due to a layer of chromium oxide forming on the surface. Steel is a very good structural material, but stainless steel is reasonably expensive (some of the other choices are more economical).
Unalloyed aluminium is not a good structural material, and it is more expensive than steels.
Match the material to the properties, try to think of reasons why the material may have been chosen.
Material 2 is a brightly coloured, low-density component, it must be tough, rigid and non-toxic, the recycling mark is number 7.
- Answer
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B. ABS: the recycling mark immediately rules out PE (identified by recycling marks 2-HDPE and 4-LDPE) and PET (identified by recycling mark 1). PTFE is more commonly known as Teflon, and it is usually used as a very low-friction coating. ABS is the only material that satisfies the evidence and requirements: it is a very tough, easily coloured polymer.
The recycling mark for PE is 2 for High Density Polyethylene and 4 for Low Density Polyethylene.
The recycling mark for PET is 1.
PTFE is more commonly known as Teflon, and it is usually used as a very low-friction coating. It is unsuitable for this application.
Match the material to the properties, try to think of reasons why the material may have been chosen.
Material 3 is an electrical component, it forms a contact that is subjected to a reasonable amount of wear, it must not corrode in air.
- Answer
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C. Low carbon steel and stainless steel create non-conducting oxides, so without some other coating they are not good choices for electrical contacts. Copper has a very high conductivity, but the addition of zinc greatly increases the wear resistance.
Low carbon steel may corrode in air and forms a non-conducting oxide so without some other coating this is not a good choice for an electrical contact.
Copper has a very high conductivity, but not very good mechanical properties.
Stainless steel forms a non-conducting oxide so without some other coating this is not a good choice for an electrical contact.
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