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Polymer_Science_and_Technology_Fried_Example 1.1
|
Polymer_Science_and_Technology_Fried
|
NUM
|
A polydisperse sample of polystyrene is prepared by mixing three monodisperse samples in the following proportions: $$\begin{array}{ll}1 \mathrm{~g} & 10,000 \text { molecular weight } \\ 2 \mathrm{~g} & 50,000 \text { molecular weight } \\ 2 \mathrm{~g} & 100,000 \text { molecular weight }\end{array}$$ Using this information, calculate the number-average molecular weight, weight-average molecular weight, and PDI of the mixture. Give your answer as a tuple (number-average molecular weight, weight-average molecular weight, PDI).
| null |
We obtain the following:
$$
\begin{aligned}
& \bar{M}_{\mathrm{n}}=\frac{\sum_{i=1}^{3} N_{i} M_{i}}{\sum_{i=1}^{3} N_{i}}=\frac{\sum_{i=1}^{3} W_{i}}{\sum_{i=1}^{3}\left(W_{i} / M_{i}\right)}=\frac{1+2+2}{\frac{1}{10,000}+\frac{2}{50,000}+\frac{2}{100,000}}=31,250 \\
& \bar{M}_{\mathrm{w}}=\frac{\sum_{i=1}^{3} N_{i} M_{i}^{2}}{\sum_{i=1}^{3} N_{i} M_{i}}=\frac{\sum_{i=1}^{3} W_{i} M_{i}}{\sum_{i=1}^{3} W_{i}}=\frac{10,000+2(50,000)+2(100,000)}{5}=62,000 \\
& \mathrm{PDI}=\frac{\bar{M}_{\mathrm{w}}}{\bar{M}_{\mathrm{n}}}=\frac{62,000}{31,250}=1.98
\end{aligned}
$$
|
(31250, 62000, 1.98)
|
(g/mol, g/mol, )
| null |
multiple
|
Example 1.1
| 201,000
| 3
|
hard
|
Materials: Polymers
|
Polymers
| null |
Composites
| null | null | null |
Polymer_Science_and_Technology_Fried_Example 1.2
|
Polymer_Science_and_Technology_Fried
|
NUM
|
A polymer is fractionated and is found to have the continuous molecular-weight distribution shown below as a plot of the weight, $W$, of molecules having molecular weight, $M$, versus $M$. Given this molecular-weight distribution, calculate $\bar{M}_{\mathrm{n}}$ and $\bar{M}_{\mathrm{w}}$. Give your answer as a tuple ($\bar{M}_{\mathrm{n}}$, $\bar{M}_{\mathrm{w}}$).
|
images/Polymer_Science_and_Technology_Fried_Example 1.2_1.jpeg
|
We obtain the following:
$$
\begin{aligned}
& \bar{M}_{\mathrm{n}}=\frac{\int_{10^{3}}^{10^{5}} d M}{\int_{10^{3}}^{10^{5}}(1 / M) d M}=\frac{10^{5}-10^{3}}{\ln \left(10^{5} / 10^{3}\right)}=21,498 \\
& \bar{M}_{\mathrm{w}}=\frac{\int_{10^{3}}^{10^{5}} M d M}{\int_{10^{3}}^{10^{5}} d M}=\frac{\left(M^{2} / 2\right)_{10^{3}}^{10^{5}}}{9.9 \times 10^{4}}=50,500
\end{aligned}
$$
|
(21498, 50500)
| null | null |
multiple
|
Example 1.2
| 201,000
| 1
|
medium
|
Materials: Polymers
|
Polymers
| null |
Composites
| null | null | null |
Polymer_Science_and_Technology_Fried_Example 1.3
|
Polymer_Science_and_Technology_Fried
|
FORMULA
|
The single-parameter Flory distribution is given as
$$
W(X)=X(\ln p)^{2} p^{X}
$$
where $X$ is the degree of polymerization and $p$ is the fractional monomer conversion in a step-growth polymerization. Using this equation, obtain expressions for the number-average and weight-average degrees of polymerization in terms of $X$ and $p$. Give your answer as a tuple (number-average degree of polymerization, weight-average degree of polymerization).
| null |
Using the following geometric series:
\[
\begin{aligned}
& \sum_{X=1}^{\infty} p^{X-1} = 1 + p + p^{2} + p^{3} + \cdots = \frac{1}{1 - p} \quad (-1 < p < 1), \\
& A = \sum_{X=1}^{\infty} X p^{X-1} = 1 + 2p + 3p^{2} + \cdots = \frac{1}{(1 - p)^2}, \\
& B = \sum_{X=1}^{\infty} X^{2} p^{X-1} = 1 + 2^{2}p + 3^{2}p^{2} + \cdots.
\end{aligned}
\]
Since it can be shown that \( B(1 - p) = A(1 + p) \), it follows that
\[
B = \frac{1 + p}{(1 - p)^3},
\]
and then
\[
\begin{aligned}
\bar{X}_{\mathrm{n}} &= \frac{\sum_{X=1}^{\infty} W(X)}{\sum_{X=1}^{\infty} W(X)/X} = \frac{(\ln p)^{2} p \sum_{X=1}^{\infty} X p^{X-1}}{(\ln p)^{2} p \sum_{X=1}^{\infty} p^{X-1}} \\
&= \frac{\sum_{X=1}^{\infty} X p^{X-1}}{\sum_{X=1}^{\infty} p^{X-1}} = \frac{1 / (1 - p)^2}{1 / (1 - p)} = \frac{1}{1 - p}, \\
\bar{X}_{\mathrm{w}} &= \frac{\sum_{X=1}^{\infty} X W(X)}{\sum_{X=1}^{\infty} W(X)} = \frac{(\ln p)^2 p \sum_{X=1}^{\infty} X^2 p^{X-1}}{(\ln p)^2 p \sum_{X=1}^{\infty} X p^{X-1}} \\
&= \frac{\sum_{X=1}^{\infty} X^2 p^{X-1}}{\sum_{X=1}^{\infty} X p^{X-1}} = \frac{(1 + p)/(1 - p)^3}{1/(1 - p)^2} = \frac{1 + p}{1 - p}.
\end{aligned}
\]
|
($\frac{1}{1-p}$, $\frac{1+p}{1-p}$)
| null | null |
multiple
|
Example 1.3
| 201,300
| 1
|
easy
|
Materials: Polymers
|
Polymers
| null |
Composites
|
Phase Diagram
| null | null |
Polymer_Science_and_Technology_Fried_Example 2.1
|
Polymer_Science_and_Technology_Fried
|
NUM
|
Using reactivity ratios of styrene (1) and 4-chlorostyrene (2) calculated from $Q-e$ values given in the table, determine the instantaneous copolymer composition (i.e., mole fraction of styrene) resulting from the free-radical copolymerization of an equimolar mixture of styrene and 4-chlorostyrene.
\begin{tabular}{llr}
\hline Monomer & Q & \multicolumn{1}{l}{ e } \\
\hline Acrylamide & 0.23 & 0.54 \\
Acrylonitrile & 0.48 & 1.23 \\
Butadiene & 1.70 & -0.50 \\
4-Chlorostyrene & 1.33 & -0.64 \\
Ethylene & 0.016 & 0.05 \\
Isobutylene & 0.023 & -1.20 \\
Isoprene & 1.99 & -0.55 \\
Maleic anhydride & 0.86 & 3.69 \\
Methacrylic acid & 0.98 & 0.62 \\
Methyl methacrylate & 0.78 & 0.40 \\
$N$-Vinyl pyrrolidone & 0.088 & -1.62 \\
Styrene & 1.00 & -0.80 \\
Vinyl acetate & 0.026 & -0.88 \\
Vinyl chloride & 0.056 & 0.16 \\
Vinylidene chloride & 0.31 & 0.34 \\
\hline
\end{tabular}
|
images/Polymer_Science_and_Technology_Fried_Example 2.1_1.png
|
From eq. $r_1 = \frac{k_{11}}{k_{12}} = \left( \frac{Q_1}{Q_2} \right) \exp \left[ -e_1 \left( e_1 - e_2 \right) \right]$, we have $r_{i}=\left(\frac{1.0}{1.33}\right) \exp [0.80(-0.80+0.64)]=0.752 \exp (-0.128)=0.662$. and from eq. $r_2 = \frac{k_{22}}{k_{21}} = \left( \frac{Q_2}{Q_1} \right) \exp \left[ -e_2 \left( e_2 - e_1 \right) \right]$, we have $r_{2}=\left(\frac{1.33}{1.0}\right) \exp [0.64(-0.64+0.80)]=1.33 \exp (0.102)=1.47$. These values differ from those given in Table 2-6; however, it is noted that there is typically significant variations in reported reactivity ratios obtained from experiment. For example, the reactivity ratio $\left(r_{2}\right)$ for 4-chlorostyrene in styrene/4-chlorostyrene copolymerization has been reported in the range from 0.76 to 1.76 [2]. Using the $Q-e$ values for $r_{1}$ and $r_{2}$, the instantaneous mole fraction of styrene in the copolymer is: $$\begin{aligned} F_{1} & =\frac{r_{1} f_{1}^{2}+f_{1} f_{2}}{r_{1} f_{1}^{2}+2 f_{1} f_{2}+r_{2} f_{2}^{2}}=\frac{0.662(0.5)^{2}+0.5^{2}}{0.662(0.5)^{2}+2(0.5)^{2}+1.47(0.5)^{2}}=\frac{0.166+0.25}{0.166+0.5+0.368} \\ & =\frac{0.416}{1.03}=0.404 \end{aligned}$$ This result indicates that the copolymer would be expected to be enriched in monomer 2 (i.e., 4-chlorostyrene) as has been observed experimentally.
|
0.404
| null | null |
single
|
Example 2.1
| 220,000
| 3
|
easy
|
Materials: Polymers
|
Polymers
|
Thermal
| null | null | null | null |
Polymer_Science_and_Technology_Fried_Example 2.2
|
Polymer_Science_and_Technology_Fried
|
NUM
|
Calculate the number-average molecular weight of polystyrene obtained at the completion of an anionic (i.e., "living") polymerization in which 0.01 g of $n$-butyllithium and 10 g of styrene monomer are used. The molecular weights of butyllithium and styrene are 64.06 and 104.12 , respectively.
| null |
$$
\begin{gathered}
\bar{X}_{\mathrm{n}} = \frac{[\mathrm{M}]_{\mathrm{o}}}{[\mathrm{I}]_{\mathrm{o}}} = \frac{10 / 104.2}{0.01 / 64.06} = 615.2 \\
\bar{M}_{\mathrm{n}} = \bar{X}_{\mathrm{n}} M_{\mathrm{o}} = 615.2 \times 104.2 = 64{,}055
\end{gathered}
$$
|
64055
| null | null |
single
|
Example 2.2
| 201,000
| 2
|
easy
|
Materials: Polymers
|
Polymers
| null |
Composites
| null | null | null |
Polymer_Science_and_Technology_Fried_Example 2.3
|
Polymer_Science_and_Technology_Fried
|
NUM
|
Consider the RAFT polymerization of 6.55 M methyl methacrylate (100.12 molecular weight) in benzene using a 1,1'-azobis(1-cyclohexanenitrile) $(0.0018 \mathrm{M})$ initiator and $2.48 \times 10^{-2} \mathrm{M}$ of the RAFT agent $S$-dodecyl $S$-(2-cyano-4-carboxy)but-2-yl trithiocarbonate at $90^{\circ} \mathrm{C}$. At a conversion of $92 \%$ of the monomer, estimate the number-average molecular weight.
| null |
The number-average degree of polymerization is first calculated as
$$
\bar{X}_{n} \approx \frac{[\mathrm{M}]_{n}-[\mathrm{M}]_{i}}{[\mathrm{R}]_{n}}=\frac{0.92(6.55) \mathrm{M}}{2.48 \times 10^{-2} \mathrm{M}}=\frac{6.03}{2.48 \times 10^{-2}}=243
$$
The number-average molecular weight is then obtained as
$$
\bar{M}_{n}=243 \times 100.1=24,324
$$
|
24324
| null | null |
single
|
Example 2.3
| 201,000
| 2
|
easy
|
Materials: Polymers
|
Polymers
| null |
Composites
| null | null | null |
Polymer_Science_and_Technology_Fried_Example 13.1
|
Polymer_Science_and_Technology_Fried
|
NUM
|
Calculate the specific density of poly(2,6-dimethyl-1,4-phenylene oxide) using groupcontribution methods.
| null |
The repeating unit of poly(2,6-dimethyl-1,4-phenylene oxide) is an amorphous polymer with a $T_{\mathrm{g}}$ of $214^{\circ} \mathrm{C}$. The molecular weight of the repeating unit is $8(12.011)+15.9994+8(1.00794)=120.151$. Using values from Table 13-1, the molar volume is calculated to be $104+8.0=112.0 \mathrm{~cm}^{3} / \mathrm{mol}$. Specific volume is then calculated as $$V_{\mathrm{a}}=112 \frac{\mathrm{~cm}^{3}}{\mathrm{~mol}} \frac{\mathrm{~mol}}{120.2 \mathrm{~g}}=0.9318 \frac{\mathrm{~cm}^{3}}{\mathrm{~g}}$$ Specific density is then calculated as $1 / 0.9318=1.073 \mathrm{~g} / \mathrm{cm}^{3}$.
|
1.073
|
$g/cm^3$
| null |
single
|
Example 13.1
| 201,200
| 4
|
medium
|
Materials: Polymers
|
Polymers
| null |
Composites
|
Crystal Structure
| null | null |
Polymer_Science_and_Technology_Fried_Example 13.2
|
Polymer_Science_and_Technology_Fried
|
NUM
|
Calculate the fractional free volume (FFV) of atactic polystyrene (aPS) using only group contributions.
|
images/Polymer_Science_and_Technology_Fried_Example 13.2_1.png
|
The repeating unit of PS is $\chemfig{*6(=-=(-(CH_2-[:30]\dot{C})-)=-)}$.
| Group Contribution | $V_{a}$ | $V_{w}$ |
| :--: | :--: | :--: |
| - $\mathrm{CH}_{2}$ - | 16.37 | 10.23 |
| $-\mathrm{CH}\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)-$ | 84.16 | 52.6 |
The molar volume, $V$, of $a$-PS is obtained by summing the group contributions of $V_{\mathrm{a}}$ taken from Table 13-1 as $100.53(16.37+84.16) \mathrm{cm}^{3} / \mathrm{mol}$. Similarly, the van der Waals volume obtained by summing the group contributions, $V_{\mathrm{w}}$ (Table 13-1), is 62.83 $\mathrm{cm}^{3} / \mathrm{mol}(10.23+52.6)$. The occupied volume is then obtained from eq. (13.3) as $V_{\mathrm{o}}=$ $1.3 \times 62.8=81.7 \mathrm{~cm}^{3} / \mathrm{mol}$. Finally, the FFV is obtained from eq. (13.1) as
$$
\mathrm{FFV}=\frac{V-V_{\mathrm{o}}}{V}=\frac{100.5-81.7}{100.5}=\frac{18.8}{100.5}=0.187
$$
|
0.187
| null | null |
single
|
Example 13.2
| 201,000
| 6
|
easy
|
Materials: Polymers
|
Polymers
| null |
Composites
| null | null | null |
Polymer_Science_and_Technology_Fried_Example 13.3
|
Polymer_Science_and_Technology_Fried
|
NUM
|
Calculate the $T_{\mathrm{g}}$ of bisphenol-A polycarbonate (PC) using van Krevelen's parameters. The repeating unit structure of PC is shown in the figure.
|
images/Polymer_Science_and_Technology_Fried_Example 13.3_1.jpeg,images/Polymer_Science_and_Technology_Fried_Example 13.3_2.png
|
The selected structural groups for PC and their corresponding molecular weights and molar glass-transition functions are shown in the table. The molecular weight of the repeating unit is then calculated as $60.0+194.3=254.3$ and the corresponding sum of the molar glass-transition functions is 107 . The $T_{\mathrm{g}}$ is then calculated from eq. $T_g = \frac{\sum\limits_{i} Y_{g,i}}{M} = \frac{Y_g}{M}$ as
$$
T_{\mathrm{g}}=\frac{\sum Y_{\mathrm{g}, i}}{M}=\frac{107 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}}{254.3 \mathrm{~kg} / \mathrm{kmol}} \times \frac{1000 \mathrm{~mol}}{\mathrm{kmol}}=421 \mathrm{~K}
$$
|
421
|
K
| null |
single
|
Example 13.3
| 221,000
| 1
|
easy
|
Materials: Polymers
|
Polymers
|
Thermal
|
Composites
| null | null | null |
Polymer_Science_and_Technology_Fried_Example 13.4
|
Polymer_Science_and_Technology_Fried
|
NUM
|
Calculate the oxygen permeability of polydimethylsiloxane (PDMS) showing in the figure using the molar Permachor values listed in the Table.
|
images/Polymer_Science_and_Technology_Fried_Example 13.4_1.png,images/Polymer_Science_and_Technology_Fried_Example 13.4_2.png
|
The Permachor is calculated using eq. $\pi = \frac{1}{N} \sum_{i} \left( N_i \cdot \Pi_i \right)$ as
$$
\pi=\frac{1}{2}[(1)(-116)+(1)(70)]=-23
$$
Next the nitrogen permeability is obtained from eq. $P(298) = P^*(298) \exp(-s \pi)$ as
$$
P(298)=\left(10^{-12}\right) \exp [(-0.12)(-23)]=1.58 \times 10^{-11} \mathrm{~cm}(\mathrm{STP}) \mathrm{cm} / \mathrm{cm}^{2}-\mathrm{s}-\mathrm{Pa}
$$
Finally, oxygen permeability is obtained from the relative permeability of $\mathrm{O}_{2}$ to $\mathrm{N}_{2}$ (a factor of 3.8 ), giving the following permeability value:
$$
P(298)=3.8 \times 1.58 \times 10^{-11}=6.0 \times 10^{-11} \mathrm{~cm}(\mathrm{STP}) \mathrm{cm} / \mathrm{cm}^{2}-\mathrm{s}-\mathrm{Pa}
$$
An experimental value of the oxygen permeability of PDMS at 308 K is reported as $9.33 \times 10^{-8} \mathrm{~cm}^{3}(\mathrm{STP}) \mathrm{cm} / \mathrm{s} \mathrm{cm}^{2} \mathrm{cmHg}$ [11]. Conversion of pressure units from cmHg to Pa (Appendix D, Table D-2) results in a $P(298)$ value of $7.00 \times 10^{-11} \mathrm{~cm}^{3}(\mathrm{STP}) \mathrm{cm} / \mathrm{cm}^{2}$ -$\mathrm{s}-\mathrm{Pa}$, in good agreement with the experimental value of $6.0 \times 10^{-11} \mathrm{~cm}^{3}(\mathrm{STP}) \mathrm{cm} / \mathrm{cm}^{2}-\mathrm{s}-$ Pa cited above.
|
$6.0 \times 10^{-11}$
|
$\mathrm{cm}^{3}(\mathrm{STP}) \mathrm{cm} / \mathrm{cm}^{2}-\mathrm{s}-\mathrm{Pa}$
| null |
single
|
Example 13.4
| 41,000
| 4
|
easy
|
Properties: Magnetic
| null |
Magnetic
|
Composites
| null | null | null |
Polymer_Science_and_Technology_Fried_Example 13.5
|
Polymer_Science_and_Technology_Fried
|
NUM
|
Estimate the solubility parameters, in units of $(\mathrm{MPa})^{1 / 2}$, for poly(methyl methacrylate) (PMMA) by the method of Small [12]. The density of PMMA is reported to be 1.188 $\mathrm{g} \mathrm{cm}^{-3}$ at $25^{\circ} \mathrm{C}$.
Table 13-4 Molar Attraction Constants at $25^{\circ} \mathrm{C}$
\begin{tabular}{lccc}
\hline & \multicolumn{3}{c}{\begin{tabular}{c}
Molar Attraction Constant, $F$ \\
$\left((\mathbf{M P a})^{1 / 2} \mathbf{c m}^{\mathbf{3}} \mathbf{m o l}^{-1}\right)$
\end{tabular}} \\
Group & Small [12] & Hoy [13] & Van Krevelen [2] \\
\hline$-\mathrm{CH}_3$ & 438 & 303 & 420 \\
$-\mathrm{CH}_2-$ & 272 & 269 & 280 \\
$>\mathrm{CH}-$ & 57 & 176 & 140 \\
$>\mathrm{C}<$ & -190 & 65.5 & 0 \\
$-\mathrm{CH}\left(\mathrm{CH}_3\right)-$ & 495 & $(479)$ & 560 \\
$-\mathrm{C}\left(\mathrm{CH}_3\right)_{2-}$ & 686 & $(672)$ & 840 \\
$-\mathrm{CH}=\mathrm{CH}-$ & 454 & 497 & 444 \\
$>\mathrm{C}=\mathrm{CH}-$ & 266 & 422 & 304 \\
Phenyl & 1504 & 1398 & 1517 \\
$p-\mathrm{Phenylene}$ & 1346 & 1442 & 1377 \\
$-\mathrm{O}-$ (ether) & 143 & 235 & 256 \\
-OH & - & 462 & 754 \\
$-\mathrm{CO}-$ (ketone) & 563 & 538 & 685 \\
$-\mathrm{COO}-$ (ester) & 634 & 668 & 512 \\
$-\mathrm{OCOO}-$ (carbonate) & - & $(904)$ & 767 \\
-CN & 839 & 726 & 982 \\
$-\mathrm{N}=\mathrm{C}=\mathrm{O}$ & - & 734 & - \\
$-\mathrm{NH}-$ & - & 368 & - \\
$-\mathrm{S}-$ (sulfide) & 460 & 428 & 460 \\
-F & $(250)$ & 84.5 & 164 \\
-Cl (primary) & 552 & 420 & 471 \\
-Br (primary) & 696 & 528 & 614 \\
$-\mathrm{CF} \mathrm{F}_3$ (n-fluorocarbon) & 561 & - & - \\
$-\mathrm{Si}-$ & -77 & - & - \\
\hline
\end{tabular}
|
images/Polymer_Science_and_Technology_Fried_Example 13.5_1.jpeg
|
The structure of the PMMA repeat unit is $\chemfig{[-1]CH_2-[::+60]C(-[::+60]CH_3)(-[::-60]C(=[::+60]O)-[::-60]OCH_3)}$
From the available chemical groups listed in the table, the molar-attraction constant for the repeating unit of PMMA can be obtained as follows:
| Group | $\boldsymbol{F}$ | Number of <br> Groups | $\sum \boldsymbol{F}_{\mathbf{i}}$ |
| :-- | :--: | :--: | :--: |
| $-\mathrm{CH}_{3}$ | 438 | 2 | 876 |
| $-\mathrm{CH}_{2^{-}}$ | 272 | 1 | 272 |
| $>\mathrm{C}<$ | -190 | 1 | -190 |
| -COO- (ester) | 634 | 1 | $\underline{634}$ |
| | | | $\underline{1592}$ |
The formula weight of a PMMA repeating unit is calculated from atomic weights as follows:
| C: | $5 \times 12.01115$ | $=$ | 60.06 |
| :-- | :-- | :-- | :-- |
| O: | $2 \times 15.9994$ | $=$ | 32.0 |
| H: | $8 \times 1.00797$ | $=$ | $\frac{8.06}{100.12}$ |
Using the density of PMMA given in the problem statement, the molar volume, $V$, is then calculated as
$$
\frac{200.12}{1.188}=84.28 \mathrm{~cm}^{3} \mathrm{~mol}^{-1}
$$
The solubility parameter is then calculated as
$$
\delta_{i}=\frac{\sum F_{i}}{V_{i}}=\frac{1592}{84.28}=18.9 \mathrm{MPa}^{1 / 2}
$$
This value is within about $17 \%$ of the value of $22.7 \mathrm{MPa}^{1 / 2}$ given in Table 13-5 for PMMA. Recalculation of the solubility parameter using Hoy's and van Krevelen's group contributions given in Table 13-4 gives values of 19.1 and $19.4 \mathrm{MPa}^{1 / 2}$, respectively, which are a little closer to the reported experimental value.
|
18.9
|
$MPa^{1 / 2}$
| null |
single
|
Example 13.5
| 201,000
| 6
|
easy
|
Materials: Polymers
|
Polymers
| null |
Composites
| null | null | null |
Polymer_Science_and_Technology_Fried_Example 13.9
|
Polymer_Science_and_Technology_Fried
|
NUM
|
Using MD solutions, calculate the diffusion coefficient of $\mathrm{O}_{2}(\mathrm{O}=\mathrm{O})$ in polydimethylsiloxane (PDMS) at 298 K.
|
images/Polymer_Science_and_Technology_Fried_Example 13.9_1.jpeg,images/Polymer_Science_and_Technology_Fried_Example 13.9_2.jpeg,images/Polymer_Science_and_Technology_Fried_Example 13.9_3.png
|
Using the COMPASS force field, an amorphous cell was constructed using the same procedures as used in Example 13.8 except four oxygen molecules were built into the cell. As before, 25-ps NVT and 250-ps NPT dynamics were used to equilibrate the cell. The final density at the end of $250-\mathrm{ps}$ NPT dynamics was $1.026 \mathrm{~g} \mathrm{~cm}^{-3}$ at 292.8 K. The amorphous cell, shown in Figure 13-21, was $36.4 \AA$ on a side (density of 1.026 $\mathrm{g} \mathrm{cm}^{-3}$ ). Figure 13-21 shows the four oxygen molecules (in space-filling representation) in the box containing the PDMS chains shown as line representations. Next, 1.0ns NVT dynamics were used to obtain trajectories to determine the diffusion coefficient. Average temperature during the 1.0 -ns NVT dynamics was 298.4 K (s.d. $=3.2$ ) with an average pressure of $0.017 \mathrm{GPa}($ s.d. $=0.084 \mathrm{GPa}$ ).
The MSD plot is shown Figure 13-22. As shown in Figure 13-23, a log-log plot of MSD versus time shows reasonable linearity between 125 and 625 ps (slope $=0.95$, $R^2=0.9988$ ). The good linearity with a slope of the $\log -\log$ plot near unity is a good indication that diffusion is within the Einstein region. The slope in the plot of MSD versus time (ps) over this time range gives a slope of $0.7065 \AA^2 / \mathrm{ps}\left(R^2=0.9974\right)$. The diffusion coefficient at 298 K is then calculated from eq. (13.69) as
$$
D=\frac{\text { slope }}{6}=\frac{0.7065}{6} \frac{\AA^2}{\mathrm{ps}} \frac{\left(10^{-8}\right) \mathrm{cm}^2}{\AA^2} \frac{\mathrm{ps}}{10^{-12} \mathrm{~s}}=0.1178 \times 10^{-4} \frac{\mathrm{~cm}^2}{\mathrm{~s}}=11.8 \times 10^{-6} \frac{\mathrm{~cm}^2}{\mathrm{~s}} .
$$
|
11.8
|
$10^{-6} \frac{\mathrm{cm}^{2}}{\mathrm{~s}}$
| null |
single
|
Example 13.9
| 222,400
| 1
|
hard
|
Materials: Polymers
|
Polymers
|
Thermal
|
Cellular
|
Diffusion & Kinetics
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 2.1
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Chemical analysis in materials science laboratories is frequently done by means of the scanning electron microscope. In this instrument, an electron beam generates characteristic x-rays that can be used to identify chemical elements. This instrument samples a roughly cylindrical volume at the surface of a solid material. Calculate the number of atoms sampled in a $1-\mu \mathrm{m}$-diameter by $1-\mu \mathrm{m}$-deep cylinder in the surface of solid copper. Assuming $\text { density of copper }=8.93 \mathrm{~g} / \mathrm{cm}^{3}$ and $\text { atomic mass of copper }=63.55 \mathrm{amu}$.
| null |
The atomic mass indicates that there are
$$
\frac{63.55 \mathrm{~g} \mathrm{Cu}}{\text { Avogadros number of } \mathrm{Cu} \text { atoms }}
$$
The volume sampled is
$$
\begin{aligned}
V_{\text {sample }} & =\left(\frac{1 \mu \mathrm{~m}}{2}\right)^{2} \times 1 \mu \mathrm{~m} \\
& =0.785 \mu \mathrm{~m}^{3} \times\left(\frac{1 \mathrm{~cm}}{10^{4} \mu \mathrm{~m}}\right)^{3} \\
& =0.785 \times 10^{-12} \mathrm{~cm}^{3}
\end{aligned}
$$
Thus, the number of atoms sampled is
$$
\begin{aligned}
N_{\text {sample }} & =\frac{8.93 \mathrm{~g}}{\mathrm{~cm}^{3}} \times 0.785 \times 10^{-12} \mathrm{~cm}^{3} \times \frac{0.602 \times 10^{24} \text { atoms }}{63.55 \mathrm{~g}} \\
& =6.64 \times 10^{10} \text { atoms. }
\end{aligned}
$$
|
6.64E+10
|
atoms
| null |
single
|
Example 2.1
| 121,400
| 3
|
easy
|
Materials: Metals
|
Metals
|
Thermal
|
Composites
|
Diffusion & Kinetics
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 2.2
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
One mole of solid MgO occupies a cube 22.37 mm on a side. Calculate the density of $\mathrm{MgO}$ (in $\mathrm{g} / \mathrm{cm}^{3}$). $ \text { atomic mass of } \mathrm{Mg}(\text { in } \mathrm{g}) = 24.31g$, $\text { atomic mass of } \mathrm{O}(\text { in } \mathrm{g}) = 16.00g$.
| null |
$$
\begin{aligned}
\text { mass of } 1 \mathrm{~mol} \text { of } \mathrm{MgO}= & \text { atomic mass of } \mathrm{Mg}(\text { in } \mathrm{g}) \\
& + \text { atomic mass of } \mathrm{O}(\text { in } \mathrm{g}) \\
= & 24.31 \mathrm{~g}+16.00 \mathrm{~g}=40.31 \mathrm{~g} . \\
\text { density }= & \frac{\text { mass }}{\text { volume }} \\
= & \frac{40.31 \mathrm{~g}}{(22.37 \mathrm{~mm})^{3} \times 10^{-3} \mathrm{~cm}^{3} / \mathrm{mm}^{3}} \\
= & 3.60 \mathrm{~g} / \mathrm{cm}^{3} .
\end{aligned}
$$
|
3.60
|
g/cm^3
| null |
single
|
Example 2.2
| 1,000
| 2
|
easy
|
Structures: Composites
| null | null |
Composites
| null | null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 2.3
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Calculate the dimensions of a cube containing 1 mol of solid magnesium.
$$
\begin{aligned}
\text { density of } \mathrm{Mg} & =1.74 \mathrm{~g} / \mathrm{cm}^{3} \\
\text { atomic mass of } \mathrm{Mg} & =24.31 \mathrm{amu}
\end{aligned}
$$
Give your answer as a tuple: (volume per mol, cube edge length, cube edge length).
| null |
$$
\begin{aligned}
\text { volume of } 1 \mathrm{~mol} & =\frac{24.31 \mathrm{~g} / \mathrm{mol}}{1.74 \mathrm{~g} / \mathrm{cm}^{3}} \\
& =13.97 \mathrm{~cm}^{3} / \mathrm{mol} \\
\text { edge of cube }= & (13.97)^{1 / 3} \mathrm{~cm} \\
& =2.408 \mathrm{~cm} \times 10 \mathrm{~mm} / \mathrm{cm} \\
& =24.08 \mathrm{~mm}
\end{aligned}
$$
|
(13.97, 2.408, 24.08)
|
(cm³/mol, cm, mm)
| null |
multiple
|
Example 2.3
| 101,000
| 4
|
easy
|
Materials: Metals
|
Metals
| null |
Composites
| null | null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 2.6_1
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Calculate the coulombic force of attraction between $\mathrm{Na}^{+}$ and $\mathrm{O}^{2-}$ in $\mathrm{Na}_{2} \mathrm{O}$ and determine the repulsive force in this case.
$$
r_{\mathrm{Na}^{+}}=0.098 \mathrm{~nm}
$$
$$
r_{\mathrm{O}^{2-}}=0.132 \mathrm{~nm}
$$
Give your answer as a tuple (coulombic force, repulsive force).
| null |
(a)
$$
\begin{aligned}
a_{0} & =r_{\mathrm{Na}^{+}}+r_{\mathrm{O}^{2-}}=0.098 \mathrm{~nm}+0.132 \mathrm{~nm} \\
& =0.231 \mathrm{~nm}
\end{aligned}
$$
Again,
$$
\begin{aligned}
F_{c} & =-\frac{k_{0}\left(Z_{1 q}\right)\left(Z_{2 q}\right)}{a_{0}^{2}} \\
& =-\frac{\left(9 \times 10^{9} \mathrm{~V} \cdot \mathrm{~m} / \mathrm{C}\right)(+1)\left(0.16 \times 10^{-18} \mathrm{C}\right)(-2)\left(0.16 \times 10^{-18} \mathrm{C}\right)}{\left(0.231 \times 10^{-9} \mathrm{~m}\right)^{2}} \\
& =8.64 \times 10^{-9} \mathrm{~N}
\end{aligned}
$$
(b) $F_{R}=-F_{c}=-8.64 \times 10^{-9} \mathrm{~N}$.
|
($8.64 \times 10^{-9}$, $-8.64 \times 10^{-9}$)
|
(N, N)
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multiple
|
Example 2.6_1
| 1,000
| 3
|
easy
|
Structures: Composites
| null | null |
Composites
| null | null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 2.6_2
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
(a) Assuming that the ionic radii are $r_{\mathrm{Na}^{+}}=0.098 \mathrm{~nm}$ and $r_{\mathrm{Cl}^{-}}=0.181 \mathrm{~nm}$., calculate the coulombic force of attraction between $\mathrm{Na}^{+}$ and $\mathrm{Cl}^{-}$ in NaCl. (b) What is the repulsive force in this case?
Give your answer as a tuple (coulombic force of attraction, repulsive force).
| null |
(a) The equilibrium bond length $a_{0}$ is the sum of these radii: $a_{0} = r_{\mathrm{Na}^{+}} + r_{\mathrm{Cl}^{-}} = 0.098 \mathrm{~nm} + 0.181 \mathrm{~nm} = 0.278 \mathrm{~nm}$. The coulombic force of attraction $F_{c}$ is calculated using the formula $F_{c} = -\frac{k_{0}(Z_{1} q)(Z_{2} q)}{a_{0}^{2}}$, where $k_{0} = 9 \times 10^{9} \mathrm{~V} \cdot \mathrm{m} / \mathrm{C}$, $Z_{1} = +1$, $Z_{2} = -1$, and $q = 0.16 \times 10^{-18} \mathrm{C}$. Substituting the values, we get $F_{c} = -\frac{(9 \times 10^{9} \mathrm{~V} \cdot \mathrm{m} / \mathrm{C})(+1)(0.16 \times 10^{-18} \mathrm{C})(-1)(0.16 \times 10^{-18} \mathrm{C})}{(0.278 \times 10^{-9} \mathrm{~m})^{2}} = 2.98 \times 10^{-9} \mathrm{~N}$. (b) The repulsive force $F_{R}$ is equal in magnitude but opposite in direction to the coulombic force, so $F_{R} = -F_{c} = -2.98 \times 10^{-9} \mathrm{~N}$.
|
(2.98 \times 10^{-9}, -2.98 \times 10^{-9})
|
(N, N)
| null |
multiple
|
Example 2.6_2
| 101,100
| 2
|
medium
|
Materials: Metals
|
Metals
| null |
Composites
|
Atomic Bonding
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 2.7
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Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Calculate the minimum radius ratio for a coordination number of 8.
| null |
The ions are touching along a body diagonal. If the cube edge length is termed $l$, then
$$
2 R+2 r=\sqrt{3} l
$$
For the minimum radius ratio coordination, the large ions are also touching each other (along a cube edge), giving
$$
2 R=l
$$
Combining the two equations gives us
$$
2 R+2 r=\sqrt{3}(2 R)
$$
Then
$$
2 r=2 R(\sqrt{3}-1)
$$
and
$$
\begin{aligned}
\frac{r}{R} & =\sqrt{3}-1=1.732-1 \\
& =0.732
\end{aligned}
$$
|
0.732
| null | null |
single
|
Example 2.7
| 100,200
| 6
|
medium
|
Materials: Metals
|
Metals
| null | null |
Crystal Structure
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 2.8
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Estimate the coordination number for the cation in each of these ceramic oxides: $\mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{~B}_{2} \mathrm{O}_{3}, \mathrm{CaO}, \mathrm{MgO}, \mathrm{SiO}_{2}$, and $\mathrm{TiO}_{2}$.
$r_{\mathrm{Al}^{3+}}=0.057 \mathrm{~nm}, r_{\mathrm{B}^{3+}}=0.02 \mathrm{~nm}, r_{\mathrm{Ca}^{2+}}=0.106 \mathrm{~nm}$, $r_{\mathrm{Mg}^{2+}}=0.078 \mathrm{~nm}, r_{\mathrm{Si}^{4+}}=0.039 \mathrm{~nm}, r_{\mathrm{Ti}^{4+}}=0.064 \mathrm{~nm}$, and $r_{\mathrm{O}^{2-}}=$ 0.132 nm .
Give your answer as a tuple (Al2O3, B2O3, CaO, MgO, SiO2, TiO2).
Refer to Table:
\begin{tabular}{|l|l|}
\hline \multicolumn{2}{|c|}{Coordination Numbers for lonic Bonding} \\
\hline Coordination number & Radius ratio, $r / R$ \\
\hline 2 & $0<\frac{r}{R}<0.155$ \\
\hline 3 & $0.155 \leq \frac{r}{R}<0.225$ \\
\hline 4 & $0.225 \leq \frac{r}{R}<0.414$ \\
\hline 6 & $0.414 \leq \frac{r}{R}<0.732$ \\
\hline 8 & $0.732 \leq \frac{r}{R}<1$ \\
\hline 12 & 1 \\
\hline
\end{tabular}
| null |
The ionic radii are: $r_{\mathrm{Al}^{3+}}=0.057 \mathrm{~nm}, r_{\mathrm{B}^{3+}}=0.02 \mathrm{~nm}, r_{\mathrm{Ca}^{2+}}=0.106 \mathrm{~nm}$, $r_{\mathrm{Mg}^{2+}}=0.078 \mathrm{~nm}, r_{\mathrm{Si}^{4+}}=0.039 \mathrm{~nm}, r_{\mathrm{Ti}^{4+}}=0.064 \mathrm{~nm}, \quad$ and $\quad r_{\mathrm{O}^{2-}}=0.132 \mathrm{~nm}$. For $\mathrm{Al}_{2} \mathrm{O}_{3}$, $\frac{r}{R}=\frac{0.057 \mathrm{~nm}}{0.132 \mathrm{~nm}}=0.43$ for which Table gives $\mathrm{CN}=6$. For $\mathrm{B}_{2} \mathrm{O}_{3}$, $\frac{r}{R}=\frac{0.02 \mathrm{~nm}}{0.132 \mathrm{~nm}}=0.15, \text { giving } \mathrm{CN}=2 .^{*}$ *The actual CN for $\mathrm{B}_{2} \mathrm{O}_{3}$ is 3 and for $\mathrm{CaO}$ is 6. Discrepancies are due to a combination of uncertainty in the estimation of ionic radii and bond directionality due to partially covalent character. For $\mathrm{CaO}$, $\frac{r}{R}=\frac{0.106 \mathrm{~nm}}{0.132 \mathrm{~nm}}=0.80, \text { giving } \mathrm{CN}=8$. For $\mathrm{MgO}$, $\frac{r}{R}=\frac{0.078 \mathrm{~nm}}{0.132 \mathrm{~nm}}=0.59, \text { giving } \mathrm{CN}=6$. For $\mathrm{SiO}_{2}$, $\frac{r}{R}=\frac{0.039 \mathrm{~nm}}{0.132 \mathrm{~nm}}=0.30, \text { giving } \mathrm{CN}=4$. For $\mathrm{TiO}_{2}$, $\frac{r}{R}=\frac{0.064 \mathrm{~nm}}{0.132 \mathrm{~nm}}=0.48, \text { giving } \mathrm{CN}=6$.
|
(6, 2, 8, 6, 4, 6)
| null | null |
multiple
|
Example 2.8
| 521,100
| 6
|
medium
|
Materials: Ceramics
|
Ceramics
|
Thermal
|
Composites
|
Atomic Bonding
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 2.9
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Calculate the reaction energy for the polymerization process for polyvinyl chloride (PVC). The vinyl chloride molecule is $\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}$.
Refer to Table:
\begin{tabular}{|l|l|l|l|}
\hline \multicolumn{4}{|l|}{Bond Energies and Bond Lengths for Representative Covalent Bonds} \\
\hline \multirow[b]{2}{*}{Bond} & \multicolumn{2}{|c|}{Bond energy ${ }^{\text {a }}$} & Bond length, nm \\
\hline & kcal/mol & kJ/mol & \\
\hline $\mathrm{C}-\mathrm{C}$ & $88^{\text {b }}$ & 370 & 0.154 \\
\hline $\mathrm{C}=\mathrm{C}$ & 162 & 680 & 0.130 \\
\hline $\mathrm{C} \equiv \mathrm{C}$ & 213 & 890 & 0.120 \\
\hline $\mathrm{C}-\mathrm{H}$ & 104 & 435 & 0.110 \\
\hline $\mathrm{C}-\mathrm{N}$ & 73 & 305 & 0.150 \\
\hline $\mathrm{C}-\mathrm{O}$ & 86 & 360 & 0.140 \\
\hline $\mathrm{C}=\mathrm{O}$ & 128 & 535 & 0.120 \\
\hline C-F & 108 & 450 & 0.140 \\
\hline $\mathrm{C}-\mathrm{Cl}$ & 81 & 340 & 0.180 \\
\hline $\mathrm{O}-\mathrm{H}$ & 119 & 500 & 0.100 \\
\hline $\mathrm{O}-\mathrm{O}$ & 52 & 220 & 0.150 \\
\hline $\mathrm{O}-\mathrm{Si}$ & 90 & 375 & 0.160 \\
\hline $\mathrm{N}-\mathrm{H}$ & 103 & 430 & 0.100 \\
\hline $\mathrm{N}-\mathrm{O}$ & 60 & 250 & 0.120 \\
\hline F-F & 38 & 160 & 0.140 \\
\hline $\mathrm{H}-\mathrm{H}$ & 104 & 435 & 0.074 \\
\hline
\end{tabular}
| null |
The polymerization process for polyvinyl chloride (PVC) involves the connection of several adjacent vinyl chloride molecules, transforming their double bonds into single bonds. The vinyl chloride molecule, $\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}$, undergoes polymerization as follows:
1. The double bond between the carbon atoms in the vinyl chloride molecule is broken.
2. This breaking allows the vinyl chloride molecules to connect, forming single bonds between them.
3. The process repeats, leading to the formation of a long chain polymer, PVC.
The energy associated with this reaction can be calculated using data from bond energies. For each $\mathrm{C}=\mathrm{C}$ bond broken to form two single $\mathrm{C}-\mathrm{C}$ bonds, the reaction energy is calculated as:
$$\mathrm{C}=\mathrm{C} \rightarrow 2 \mathrm{C}-\mathrm{C}$$
Using bond energy data from Table:
$$680 \mathrm{~kJ} / \mathrm{mol} \rightarrow 2(370 \mathrm{~kJ} / \mathrm{mol})=740 \mathrm{~kJ} / \mathrm{mol}$$
The reaction energy is then:
$$(740-680) \mathrm{kJ} / \mathrm{mol}=60 \mathrm{~kJ} / \mathrm{mol}$$
This energy is released during polymerization, making the reaction spontaneous and the product, PVC, stable relative to individual vinyl chloride molecules.
|
60
|
kJ/mol
| null |
single
|
Example 2.9
| 201,123
| 4
|
medium
|
Materials: Polymers
|
Polymers
| null |
Composites
|
Atomic Bonding
|
Shaping
|
Fracture
|
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 2.11
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Calculate the length of a polyethylene molecule, $\left\langle\mathrm{C}_{2} \mathrm{H}_{4}\right\rangle_{n}$, where $n=500$.
| null |
To calculate the length of a polyethylene molecule with $n=500$, we first determine the effective bond length, $l$, using the carbon-carbon bond length and the sine of half the bond angle. The carbon-carbon bond length is $0.154 \mathrm{~nm}$, and the bond angle is $109.5^{\circ}$, leading to an effective bond length calculation as follows:
$$
l = (0.154 \mathrm{~nm}) \times \sin(54.75^{\circ}) = 0.126 \mathrm{~nm}
$$
With two bond lengths per mer and 500 mers, the total molecule length, $L$, is calculated by:
$$
L = 500 \times 2 \times 0.126 \mathrm{~nm} = 126 \mathrm{~nm} = 0.126 \mu \mathrm{m}
$$
|
126
|
nm
| null |
single
|
Example 2.11
| 201,100
| 3
|
easy
|
Materials: Polymers
|
Polymers
| null |
Composites
|
Atomic Bonding
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 2.13
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Given the bonding energy curve for secondary bonding described by the "6-12" potential: $$E=-\frac{K_{A}}{a^{6}}+\frac{K_{R}}{a^{12}}$$ where $K_{A}=10.37 \times 10^{-78} \mathrm{~J} \cdot \mathrm{~m}^{6}$ and $K_{R}=16.16 \times 10^{-135} \mathrm{~J} \cdot \mathrm{~m}^{12}$, calculate the bond energy and bond length for argon. Give your answer as a tuple (bond length, bond energy).
| null |
The equilibrium bond length occurs at $d E / d a=0$: $$\left(\frac{d E}{d a}\right)_{a=a_{0}}=0=\frac{6 K_{A}}{a_{0}^{7}}-\frac{12 K_{R}}{a_{0}^{13}}$$ Rearranging gives us $$a_{0}=\left(2 \frac{K_{R}}{K_{A}}\right)^{1 / 6}=\left(2 \times \frac{16.16 \times 10^{-135}}{10.37 \times 10^{-78}}\right)^{1 / 6} \mathrm{~m}=0.382 \times 10^{-9} \mathrm{~m}=0.382 \mathrm{~nm}$$ The bond energy at $a_{0}$ is $$E(0.382 \mathrm{~nm})=-\frac{K_{A}}{(0.382 \mathrm{~nm})^{6}}+\frac{K_{R}}{(0.382 \mathrm{~nm})^{12}}=-\frac{\left(10.37 \times 10^{-78} \mathrm{~J} \cdot \mathrm{~m}^{6}\right)}{\left(0.382 \times 10^{-9} \mathrm{~m}\right)^{6}}+\frac{\left(16.16 \times 10^{-135} \mathrm{~J} \cdot \mathrm{~m}^{12}\right)}{\left(0.382 \times 10^{-9} \mathrm{~m}\right)^{12}}=-1.66 \times 10^{-21} \mathrm{~J}$$ For 1 mol of Ar, the bonding energy is $$E_{\text{bonding}}=-1.66 \times 10^{-21} \mathrm{~J} / \text{bond} \times 0.602 \times 10^{24} \frac{\text{bonds}}{\text{mole}}=-0.999 \times 10^{3} \mathrm{~J} / \mathrm{mol}=-0.999 \mathrm{~kJ} / \mathrm{mol}$$
|
(0.382, -0.999)
|
(nm, kJ/mol)
| null |
multiple
|
Example 2.13
| 1,100
| 4
|
hard
|
Fundamental: Atomic Bonding
| null | null |
Composites
|
Atomic Bonding
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 3.2
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Calculate the density of copper, given that copper is an fcc metal with an atomic radius of 0.128 nm and a molar mass of 63.55 g/mol.
| null |
The length of the face diagonal in the unit cell for an fcc metal is given by: $$ l = 4 r_{\text{Cu atom}} = \sqrt{2} a $$ Solving for the lattice parameter $a$: $$ a = \frac{4}{\sqrt{2}} r_{\text{Cu atom}} = \frac{4}{\sqrt{2}}(0.128 \text{ nm}) = 0.362 \text{ nm} $$ The density of the unit cell, which contains four atoms, is calculated as: $$ \rho = \frac{4 \text{ atoms}}{(0.362 \text{ nm})^3} \times \frac{63.55 \text{ g}}{0.6023 \times 10^{24} \text{ atoms}} \times \left(\frac{10^7 \text{ nm}}{\text{cm}}\right)^3 = 8.89 \text{ g/cm}^3 $$
|
8.89
|
g/cm^3
| null |
single
|
Example 3.2
| 102,200
| 3
|
hard
|
Materials: Metals
|
Metals
| null |
Cellular
|
Crystal Structure
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 3.3
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Calculate the Ionic Packing Factor (IPF) of MgO, which has the NaCl structure.(Figure 3.9) Ionic radius of \mathrm{Mg}^{2+} is 0.078nm, Ionic radius of \mathrm{O}^{2-} is 0.132nm.
|
images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 3.3_1.jpg
|
To calculate the Ionic Packing Factor (IPF) of MgO with the NaCl structure, follow these steps:
1. Calculate the edge length (a) of the unit cell using the ionic radii of Mg²⁺ and O²⁻:
$$a = 2r_{\mathrm{Mg}^{2+}} + 2r_{\mathrm{O}^{2-}} = 2(0.078 \mathrm{~nm}) + 2(0.132 \mathrm{~nm}) = 0.420 \mathrm{~nm}$$
2. Calculate the volume of the unit cell (V_unit cell):
$$V_{\text{unit cell}} = a^3 = (0.420 \mathrm{~nm})^3 = 0.0741 \mathrm{~nm}^3$$
3. Calculate the total ionic volume of Mg²⁺ and O²⁻ ions in the unit cell. There are four Mg²⁺ ions and four O²⁻ ions per unit cell:
$$4 \times \frac{4}{3} \pi r_{\mathrm{Mg}^{2+}}^3 + 4 \times \frac{4}{3} \pi r_{\mathrm{O}^{2-}}^3 = \frac{16 \pi}{3}[(0.078 \mathrm{~nm})^3 + (0.132 \mathrm{~nm})^3] = 0.0465 \mathrm{~nm}^3$$
4. Calculate the Ionic Packing Factor (IPF):
$$\mathrm{IPF} = \frac{0.0465 \mathrm{~nm}^3}{0.0741 \mathrm{~nm}^3} = 0.627$$
The Ionic Packing Factor for MgO with the NaCl structure is 0.627.
|
0.627
| null | null |
single
|
Example 3.3
| 2,200
| 4
|
easy
|
Fundamental: Crystal Structure
| null | null |
Cellular
|
Crystal Structure
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 3.4
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Given $a=0.420 \mathrm{~nm}$, calculate the density of MgO.
| null |
$a=0.420 \mathrm{~nm}$, which gave a unit-cell volume of $0.0741 \mathrm{~nm}^{3}$. The density of the unit cell is
$$
\begin{aligned}
\rho & =\frac{[4(24.31 \mathrm{~g})+4(16.00 \mathrm{~g})] /\left(0.6023 \times 10^{24}\right)}{0.0741 \mathrm{~nm}^{3}} \times\left(\frac{10^{7} \mathrm{~nm}}{\mathrm{~cm}}\right)^{3} \\
& =3.61 \mathrm{~g} / \mathrm{cm}^{3}
\end{aligned}
$$
|
3.61
|
g/cm^3
| null |
single
|
Example 3.4
| 2,200
| 1
|
easy
|
Fundamental: Crystal Structure
| null | null |
Cellular
|
Crystal Structure
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 3.5
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Calculate the number of C and H atoms in the polyethylene unit cell, given a density of $0.9979 \mathrm{~g} / \mathrm{cm}^{3}$ and unit-cell dimensions of $0.741 \mathrm{~nm} \times 0.494 \mathrm{~nm} \times 0.255 \mathrm{~nm}$. Give your answer as a tuple (number of C atoms, number of H atoms).
|
images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 3.5_1.jpeg
|
The volume of the unit cell is calculated as:
$$
V = (0.741 \mathrm{~nm}) \times (0.494 \mathrm{~nm}) \times (0.255 \mathrm{~nm}) = 0.0933 \mathrm{~nm}^{3}
$$
There will be some multiple $(n)$ of $\mathrm{C}_{2} \mathrm{H}_{4}$ units in the unit cell with atomic mass:
$$
m = \frac{n[2(12.01) + 4(1.008)] \mathrm{g}}{0.6023 \times 10^{24}} = (4.66 \times 10^{-23} n) \mathrm{g}
$$
Therefore, the unit-cell density is:
$$
\rho = \frac{(4.66 \times 10^{-23} n) \mathrm{g}}{0.0933 \mathrm{~nm}^{3}} \times \left(\frac{10^{7} \mathrm{~nm}}{\mathrm{~cm}}\right)^{3} = 0.9979 \frac{\mathrm{~g}}{\mathrm{~cm}^{3}}
$$
Solving for $n$ gives:
$$
n = 2.00
$$
As a result, there are:
$$
4 (=2 n) \mathrm{C} \text{ atoms } + 8 (=4 n) \mathrm{H} \text{ atoms per unit cell.}
$$
|
(4, 8)
|
(atoms, atoms)
| null |
multiple
|
Example 3.5
| 202,200
| 5
|
easy
|
Materials: Polymers
|
Polymers
| null |
Cellular
|
Crystal Structure
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 3.6
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Calculate the APF for the diamond cubic structure. (Figure 3.20)
|
images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 3.6_1.jpg
|
Because of the tetrahedral bonding geometry of the diamond cubic structure, the atoms lie along body diagonals. Inspection of Figure 3.20 indicates that this orientation of atoms leads to the equality
$$
2 r_{\mathrm{Si}}=\frac{1}{4}(\text { body diagonal })=\frac{\sqrt{3}}{4} a
$$
or
$$
a=\frac{8}{\sqrt{3}} r_{\mathrm{Si}}
$$
The unit-cell volume then, is,
$$
V_{\text {unit cell }}=a^{3}=(4.62)^{3} r_{\mathrm{Si}}^{3}=98.5 r_{\mathrm{Si}}^{3}
$$
The volume of the eight Si atoms in the unit cell is
$$
V_{\text {atoms }}=8 \times \frac{4}{3} \pi r_{\mathrm{Si}}^{3}=33.5 r_{\mathrm{Si}}^{3}
$$
which gives an atomic packing factor of
$$
\mathrm{APF}=\frac{33.5 r_{\mathrm{Si}}^{3}}{98.5 r_{\mathrm{Si}}^{3}}=0.340
$$
|
0.340
| null | null |
single
|
Example 3.6
| 102,200
| 5
|
medium
|
Materials: Metals
|
Metals
| null |
Cellular
|
Crystal Structure
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 3.7
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Given the atom radius is $0.117 \mathrm{~nm}$, calculate the density of silicon. Given V_{\text {unit cell }} & =98.5 r_{\mathrm{Si}}^{3}.
| null |
$$
\begin{aligned}
V_{\text {unit cell }} & =98.5 r_{\mathrm{Si}}^{3}=98.5(0.117 \mathrm{~nm})^{3} \\
& =0.158 \mathrm{~nm}^{3}
\end{aligned}
$$
giving a density of
$$
\begin{aligned}
\rho & =\frac{8 \text { atoms }}{0.158 \mathrm{~nm}^{3}} \times \frac{28.09 \mathrm{~g}}{0.6023 \times 10^{24} \text { atoms }} \times\left(\frac{10^{7} \mathrm{~nm}}{\mathrm{~cm}}\right)^{3} \\
& =2.36 \mathrm{~g} / \mathrm{cm}^{3} .
\end{aligned}
$$
|
2.36
|
g/cm^3
| null |
single
|
Example 3.7
| 402,200
| 2
|
easy
|
Materials: Semiconductors
|
Semiconductors
| null |
Cellular
|
Crystal Structure
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 3.9
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
FORMULA
|
Which lattice points lie on the [110] direction in the fcc and fco unit cells? Give your answer as a tuple (X, Y, ...) where X, Y, ... are the lattice points in the [110] direction.
| null |
The lattice points on the [110] direction in both fcc (face-centered cubic) and fco (face-centered orthorhombic) unit cells are $000$, $\frac{1}{2} \frac{1}{2} 0$, and $110$. This is consistent for either system, fcc or fco.
|
(000, $\frac{1}{2} \frac{1}{2} 0$, 110)
| null | null |
multiple
|
Example 3.9
| 2,200
| 1
|
medium
|
Fundamental: Crystal Structure
| null | null |
Cellular
|
Crystal Structure
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 3.11
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
What is the angle between the [110] and [111] directions in the cubic system?
| null |
From $\cos \delta=\frac{\mathbf{D} \cdot \mathbf{D}^{\prime}}{|\mathbf{D}|\left|\mathbf{D}^{\prime}\right|}=\frac{u u^{\prime}+v v^{\prime}+w w^{\prime}}{\sqrt{u^2+v^2+w^2} \sqrt{\left(u^{\prime}\right)^2+\left(v^{\prime}\right)^2+\left(w^{\prime}\right)^2}}$.,
$$
\begin{aligned}
\delta & =\arccos \frac{u u^{\prime}+v v^{\prime}+w w^{\prime}}{\sqrt{u^{2}+v^{2}+w^{2}} \sqrt{\left(u^{\prime}\right)^{2}+\left(v^{\prime}\right)^{2}+\left(w^{\prime}\right)^{2}}} \\
& =\arccos \frac{1+1+0}{\sqrt{2} \sqrt{3}} \\
& =\arccos 0.816 \\
& =35.3^{\circ}
\end{aligned}
$$
|
35.3
|
°
| null |
single
|
Example 3.11
| 200
| 1
|
easy
|
Fundamental: Crystal Structure
| null | null | null |
Crystal Structure
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 3.12
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Identify the axial intercepts for the $(3 \overline{1} 1)$ plane. Give your answer as a tuple (X, Y, Z) where X, Y, Z represent the intercepts on the x, y, and z axes respectively.
| null |
For the $a$ axis, intercept $=\frac{1}{3} a$; for the $b$ axis, intercept $=\frac{1}{-1} b=-b$; and for the $c$ axis, intercept $=\frac{1}{1} c=c$.
|
($\frac{1}{3}$, -1, 1)
| null | null |
multiple
|
Example 3.12
| 10,000
| 1
|
medium
|
Properties: Mechanical
| null |
Mechanical
| null | null | null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 3.14
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Given $r_{W atom}=0.137nm$ and $r_{AI atom} = 0.143 nm$, calculate the linear density of atoms along the [111] direction in (a) bcc W and (b) fcc Al.
Give your answer as a tuple (linear density of bcc tungsten, linear density of fcc aluminum).
| null |
(a) For a bcc structure, atoms touch along the [111] direction (a body diagonal). Therefore, the repeat distance is equal to one atomic diameter. Then we find that the repeat distance is
$$
\begin{aligned}
r & =d_{\mathrm{W} \text { atom }}=2 r_{\mathrm{W} \text { atom }} \\
& =2(0.137 \mathrm{~nm})=0.274 \mathrm{~nm}
\end{aligned}
$$
Therefore,
$$
r^{-1}=\frac{1}{0.274 \mathrm{~nm}}=3.65 \text { atoms } / \mathrm{nm}
$$
(b) For an fcc structure, only one atom is intercepted along the body diagonal of a unit cell. To determine the length of the body diagonal, we can note that two atomic diameters equal the length of a face diagonal. And we have
$$
\begin{aligned}
\text { face diagonal length } & =2 d_{\mathrm{Al}} \text { atom } \\
& =4 r_{\mathrm{Al}} \text { atom }=\sqrt{2} a
\end{aligned}
$$
or the lattice parameter is
$$
\begin{aligned}
a & =\frac{4}{\sqrt{2}} r_{\mathrm{Al} \text { atom }} \\
& =\frac{4}{\sqrt{2}}(0.143 \mathrm{~nm})=0.404 \mathrm{~nm}
\end{aligned}
$$
The repeat distance is
$$
\begin{aligned}
r & =\text { body diagonal length }=\sqrt{3} a \\
& =\sqrt{3}(0.404 \mathrm{~nm}) \\
& =0.701 \mathrm{~nm}
\end{aligned}
$$
which gives a linear density of
$$
r^{-1}=\frac{1}{0.701 \mathrm{~nm}}=1.43 \text { atoms } / \mathrm{nm}
$$
|
(3.65, 1.43)
|
(atoms/nm, atoms/nm)
| null |
multiple
|
Example 3.14
| 102,200
| 9
|
easy
|
Materials: Metals
|
Metals
| null |
Cellular
|
Crystal Structure
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 3.15
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Given rW = 0.137 nm for bcc W and rAl = 0.143 nm for fcc Al, calculate the planar density of atoms in the (111) plane of (a) bcc tungsten and (b) fcc aluminum.
Give your answer as a tuple: (bcc_tungsten_planar_density, fcc_aluminum_planar_density).
| null |
(a) For the bcc structure, the (111) plane intersects only corner atoms in the unit cell:
We have
$$
\sqrt{3} a=4 r_{\mathrm{W} \text { atom }}
$$
or
$$
a=\frac{4}{\sqrt{3}} r_{\mathrm{W} \text { atom }}=\frac{4}{\sqrt{3}}(0.137 \mathrm{~nm})=0.316 \mathrm{~nm}
$$
Face diagonal length, $l$, is then
$$
l=\sqrt{2} a=\sqrt{2}(0.316 \mathrm{~nm})=0.447 \mathrm{~nm}
$$
The area of the (111) plane within the unit cell is
$$
\begin{aligned}
A & =\frac{1}{2} b h=\frac{1}{2}(0.447 \mathrm{~nm})\left(\frac{\sqrt{3}}{2} \times 0.447 \mathrm{~nm}\right) \\
& =0.0867 \mathrm{~nm}^2 .
\end{aligned}
$$
There is $\frac{1}{6}$ atom (i.e., $\frac{1}{6}$ of the circumference of a circle) at each corner of the equilateral triangle formed by the (111) plane in the unit cell. Therefore,
$$
\begin{aligned}
\text { atomic density } & =\frac{3 \times \frac{1}{6} \text { atom }}{A} \\
& =\frac{0.5 \text { atom }}{0.0867 \mathrm{~nm}^2}=5.77 \frac{\text { atoms }}{\mathrm{nm}^2}
\end{aligned}
$$
(b) For the fcc structure, the (111) plane intersects three corner atoms plus three face-centered atoms in the unit cell:
We obtain the face diagonal length
$$
l=\sqrt{2} a=\sqrt{2}(0.404 \mathrm{~nm})=0.572 \mathrm{~nm}
$$
The area of the (111) plane within the unit cell is
$$
\begin{aligned}
A & =\frac{1}{2} b h=\frac{1}{2}(0.572 \mathrm{~nm})\left(\frac{\sqrt{3}}{2} 0.572 \mathrm{~nm}\right) \\
& =0.142 \mathrm{~nm}^2
\end{aligned}
$$
There are $3 \times \frac{1}{6}$ corner atoms plus $3 \times \frac{1}{2}$ face-centered atoms within this area, giving
$$
\begin{aligned}
\text { atomic density } & =\frac{3 \times \frac{1}{6}+3 \times \frac{1}{2} \text { atoms }}{0.142 \mathrm{~nm}^2}=\frac{2 \text { atoms }}{0.142 \mathrm{~nm}^2} \\
& =14.1 \text { atoms } / \mathrm{nm}^2
\end{aligned}
$$
|
(5.77, 14.1)
|
(atoms/nm², atoms/nm²)
| null |
multiple
|
Example 3.15
| 102,200
| 9
|
medium
|
Materials: Metals
|
Metals
| null |
Cellular
|
Crystal Structure
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 3.17
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Calculate the planar density of ions in the (111) plane of MgO. Given the lattice constant $a=0.420 \mathrm{~nm}$.
| null |
There are two separate answers to this problem depending on the unit cell's origin. Using the unit cell with its origin on an anion site, the (111) plane has an arrangement comparable to an fcc metal. Alternatively, defining the unit cell with its origin on a cation site results in a comparable arrangement of cations in the (111) plane. In either case, there are two ions per (111) triangle. Given the lattice constant $a=0.420 \mathrm{~nm}$, the length of each (111) triangle side (a unit-cell face diagonal) is $$l=\sqrt{2} a=\sqrt{2}(0.420 \mathrm{~nm})=0.594 \mathrm{~nm}.$$ The planar area is $$A=\frac{1}{2} b h=\frac{1}{2}(0.594 \mathrm{~nm})\left(\frac{\sqrt{3}}{2} 0.594 \mathrm{~nm}\right)=0.153 \mathrm{~nm}^{2},$$ which gives the ionic density as $$\text{ionic density}=\frac{2 \text{ ions }}{0.153 \mathrm{~nm}^{2}}=13.1 \mathrm{~nm}^{-2},$$ or $$13.1\left(\mathrm{Mg}^{2+} \text{ or } \mathrm{O}^{2-}\right) / \mathrm{nm}^{2}.$$
|
13.1
|
nm^{-2}
| null |
single
|
Example 3.17
| 102,200
| 4
|
medium
|
Materials: Metals
|
Metals
| null |
Cellular
|
Crystal Structure
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 3.18
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Given r silicon = 0.117 nm, calculate the linear density of atoms along the [111] direction in silicon.
| null |
To calculate the linear density of atoms along the [111] direction in silicon, we must consider the arrangement of atoms along this direction, which is a body diagonal in the diamond cubic structure. The atoms along the [111] direction are not uniformly spaced, and thus, the calculation requires careful consideration of their positions. Referring to the structure, we find that there are effectively 2 atoms centered along a given body diagonal. The length of the body diagonal in a unit cell, denoted as $l$, can be related to the atomic radius of silicon, $r_{\mathrm{Si}}$, by the equation: $$ 2 r_{\mathrm{Si}} = \frac{1}{4} l $$ Solving for $l$ gives: $$ l = 8 r_{\mathrm{Si}} $$. The atomic radius of silicon is $0.117 \mathrm{~nm}$, thus: $$ l = 8(0.117 \mathrm{~nm}) = 0.936 \mathrm{~nm} $$ Therefore, the linear density is calculated as: $$ \text{linear density} = \frac{2 \text{ atoms}}{0.936 \mathrm{~nm}} = 2.14 \frac{\text{atoms}}{\mathrm{nm}} $$
|
2.14
|
$\frac{\text{atoms}}{\mathrm{nm}}$
| null |
single
|
Example 3.18
| 402,200
| 6
|
easy
|
Materials: Semiconductors
|
Semiconductors
| null |
Cellular
|
Crystal Structure
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 3.19
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Given r silicon = 0.117 nm, calculate the planar density of atoms in the (111) plane of silicon.
| null |
Close observation of the diamond cubic structure shows that the four interior atoms do not lie on the (111) plane. The atom arrangement in this plane is similar to that for the metallic fcc structure. There are two atoms in the equilateral triangle bounded by sides of length $\sqrt{2} a$. From the given data, we see that $a=\frac{8}{\sqrt{3}}(0.117 \mathrm{~nm})=0.540 \mathrm{~nm}$ and $\sqrt{2} a=0.764 \mathrm{~nm}$. The area of the triangle is calculated as $A = \frac{1}{2} b h = \frac{1}{2}(0.764 \mathrm{~nm})\left(\frac{\sqrt{3}}{2} 0.764 \mathrm{~nm}\right) = 0.253 \mathrm{~nm}^{2}$. Therefore, the planar density is $\frac{2 \text{ atoms }}{0.253 \mathrm{~nm}^{2}} = 7.91 \frac{\text{ atoms }}{\mathrm{nm}^{2}}$.
|
7.91
|
$\frac{\text{ atoms }}{\mathrm{nm}^{2}}$
| null |
single
|
Example 3.19
| 101,200
| 6
|
medium
|
Materials: Metals
|
Metals
| null |
Composites
|
Crystal Structure
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 3.20
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Using Bragg's law, calculate the diffraction angles (2θ) for the first three peaks in the aluminum powder pattern, given that the first three peaks correspond to the (111), (200), and (220) planes, the lattice parameter $a=0.404 \mathrm{~nm}$, and the wavelength of the x-ray used is $\lambda=0.1542 \mathrm{~nm}$.
Give your answer as a tuple (diffraction angle for (111) plane, diffraction angle for (200) plane, diffraction angle for (220) plane).
|
images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 3.20_1.jpeg
|
To calculate the diffraction angles (2θ) for the first three peaks in the aluminum powder pattern, we follow these steps using Bragg's law:
1. Calculate the interplanar spacings ($d_{hkl}$) for the (111), (200), and (220) planes using the formula:
$$
d_{h k l}=\frac{a}{\sqrt{h^{2}+k^{2}+l^{2}}}
$$
- For (111):
$$
d_{111}=\frac{0.404 \mathrm{~nm}}{\sqrt{1+1+1}}=\frac{0.404 \mathrm{~nm}}{\sqrt{3}}=0.234 \mathrm{~nm}
$$
- For (200):
$$
d_{200}=\frac{0.404 \mathrm{~nm}}{\sqrt{2^{2}+0+0}}=\frac{0.404 \mathrm{~nm}}{2}=0.202 \mathrm{~nm}
$$
- For (220):
$$
d_{220}=\frac{0.404 \mathrm{~nm}}{\sqrt{2^{2}+2^{2}+0}}=\frac{0.404 \mathrm{~nm}}{\sqrt{8}}=0.143 \mathrm{~nm}
$$
2. Calculate the diffraction angle (θ) for each peak using Bragg's law:
$$
\theta=\arcsin \left(\frac{\lambda}{2 d}\right)
$$
- For (111):
$$
\theta_{111}=\arcsin \left(\frac{0.1542 \mathrm{~nm}}{2 \times 0.234 \mathrm{~nm}}\right)=19.2^{\circ}
$$
Therefore, $(2 \theta)_{111}=38.5^{\circ}$.
- For (200):
$$
\theta_{200}=\arcsin \left(\frac{0.1542 \mathrm{~nm}}{2 \times 0.202 \mathrm{~nm}}\right)=22.4^{\circ}
$$
Therefore, $(2 \theta)_{200}=44.8^{\circ}$.
- For (220):
$$
\theta_{220}=\arcsin \left(\frac{0.1542 \mathrm{~nm}}{2 \times 0.143 \mathrm{~nm}}\right)=32.6^{\circ}
$$
Therefore, $(2 \theta)_{220}=65.3^{\circ}$.
Thus, the diffraction angles (2θ) for the first three peaks in the aluminum powder pattern are $38.5^{\circ}$, $44.8^{\circ}$, and $65.3^{\circ}$ for the (111), (200), and (220) planes, respectively.
|
(38.5, 44.8, 65.3)
|
(degrees, degrees, degrees)
| null |
multiple
|
Example 3.20
| 104,200
| 2
|
hard
|
Materials: Metals
|
Metals
| null |
Surface Texture
|
Crystal Structure
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 4.2
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
How much "oversize" is the C atom in $\alpha-\mathrm{Fe}$ ? Given the radius of the iron item is 0.124 nm and the radius carbon is 0.077 nm. Give your answer as a tuple: (ideal_interstitial_radius, carbon_radius_ratio). Refer to Figure 4.4
|
images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 4.2_1.jpg
|
By inspection of Figure 4.4, it is apparent that an ideal interstitial atom centered at $\frac{1}{2} 0 \frac{1}{2}$ would just touch the surface of the iron atom in the center of the unit-cell cube. The radius of such an ideal interstitial would be
$$ r_{\text {interstitial }}=\frac{1}{2} a-R $$
where $a$ is the length of the unit-cell edge, and $R$ is the radius of an iron atom.
We note that
$$ \begin{aligned} \text { length of unit }- \text { cell body diagonal } & =4 R \\ & =\sqrt{3} a \end{aligned} $$
or
$$ a=\frac{4}{\sqrt{3}} R $$. Then,
$$ r_{\text {interstitial }}=\frac{1}{2}\left(\frac{4}{\sqrt{3}} R\right)-R=0.1547 R $$
$R=0.124 \mathrm{~nm}$, giving
$$ r_{\text {interstitial }}=0.1547(0.124 \mathrm{~nm})=0.0192 \mathrm{~nm} $$
However, $r_{\text {carbon }}=0.077 \mathrm{~nm}$, or
$$ \frac{r_{\text {carbon }}}{r_{\text {interstitial }}}=\frac{0.077 \mathrm{~nm}}{0.0192 \mathrm{~nm}}=4.01 $$
Therefore, the carbon atom is roughly four times too large to fit next to the adjacent iron atoms without strain. The severe local distortion required for this accommodation leads to the low solubility of C in $\alpha-\mathrm{Fe}(<0.1$ at $\%$).
|
(0.0192, 4.01)
|
(nm,)
| null |
multiple
|
Example 4.2
| 112,000
| 7
|
hard
|
Materials: Metals
|
Metals
|
Mechanical
|
Cellular
| null | null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 4.3
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
The fraction of vacant lattice sites in a crystal is typically small. For example, the fraction of aluminum sites vacant at $400^{\circ} \mathrm{C}$ is $2.29 \times 10^{-5}$. Calculate the density of these sites (in units of $\mathrm{m}^{-3}$ ). Given the density of aluminum to be $2.70 \mathrm{Mg} / \mathrm{m}^{3}$ and its atomic mass to be 26.98 amu .
| null |
We find the density of aluminum to be $2.70 \mathrm{Mg} / \mathrm{m}^{3}$ and its atomic mass to be 26.98 amu . The corresponding density of aluminum atoms is then
$$
\begin{aligned}
\text { at. density } & =\frac{\rho}{\text { at. mass }}=\frac{2.70 \times 10^{6} \mathrm{~g} / \mathrm{m}^{3}}{26.98 \mathrm{~g} /\left(0.602 \times 10^{24} \text { atoms }\right)} \\
& =6.02 \times 10^{28} \text { atoms } \cdot \mathrm{m}^{-3} .
\end{aligned}
$$
Then, the density of vacant sites will be
$$
\begin{aligned}
\text { vac. density } & =2.29 \times 10^{-5} \mathrm{atom}^{-1} \times 6.02 \times 10^{28} \text { atoms } \cdot \mathrm{m}^{-3} \\
& =1.38 \times 10^{24} \mathrm{~m}^{-3}
\end{aligned}
$$
|
1.38E+24
|
m^{-3}
| null |
single
|
Example 4.3
| 101,200
| 2
|
easy
|
Materials: Metals
|
Metals
| null |
Composites
|
Crystal Structure
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 4.4
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Using the atomic radii RFe=0.124 nm for bcc Fe and RAl=0.143 nm for fcc Al, calculate the magnitude of the Burgers vector for (a) α-Fe and (b) Al.
Give your answer as a tuple (magnitude for α-Fe, magnitude for Al).
| null |
(a) $|\mathbf{b}|$ is merely the repeat distance between adjacent atoms along the highest atomic density direction. For $\alpha$-Fe, a bcc metal, this distance tends to be along the body diagonal of a unit cell. Fe atoms are in contact along the body diagonal. As a result, the atomic repeat distance is
$$
r=2 R_{F e}
$$
We can then calculate, in a simple way,
$$
|\mathbf{b}|=r=2(0.124 \mathrm{~nm})=0.248 \mathrm{~nm}
$$
(b) Similarly, the highest atomic density direction in fcc metals such as Al tends to be along the face diagonal of a unit cell.This direction is also a line of contact for atoms in an fcc structure. Again,
$$
\begin{aligned}
|\mathbf{b}|=r=2 R_{\mathrm{AI}} & =2(0.143 \mathrm{~nm}) \\
& =0.286 \mathrm{~nm}
\end{aligned}
$$
|
(0.248, 0.286)
|
(nm, nm)
| null |
multiple
|
Example 4.4
| 102,200
| 2
|
easy
|
Materials: Metals
|
Metals
| null |
Cellular
|
Crystal Structure
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 4.5
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Given that Al is fcc with lattice parameter a = 0.405 nm, calculate the separation distance of dislocations in a low-angle $\left(\theta=2^{\circ}\right)$ tilt boundary in aluminum.
|
images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 4.5_1.jpeg
|
$$
|\mathbf{b}| = a / \sqrt{2} =0.286 \mathrm{~nm}
$$
From the figure, we see that
$$
\begin{aligned}
D & =\frac{|\mathbf{b}|}{\theta} \\
& =\frac{0.286 \mathrm{~nm}}{2^{\circ} \times\left(1 \mathrm{rad} / 57.3^{\circ}\right)}=8.19 \mathrm{~nm}
\end{aligned}
$$
|
8.19
|
nm
| null |
single
|
Example 4.5
| 101,200
| 2
|
hard
|
Materials: Metals
|
Metals
| null |
Composites
|
Crystal Structure
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 4.6
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Calculate the grain-size number, $G$, for a micrograph for which you measure $8.04 \frac{\text { grains }}{\text { in. } .^{2}}$ at $100 \times$.
| null |
From Equation,
$$
N=2^{(G-1)}
$$
or
$$
\begin{aligned}
G & =\frac{\ln N}{\ln 2}+1 \\
& =\frac{\ln (8.04)}{\ln 2}+1 \\
& =4.01
\end{aligned}
$$
|
4.01
| null | null |
single
|
Example 4.6
| 5,000
| 3
|
medium
|
Structures: Micro/Nano-structure
| null | null |
Micro/Nano-structure
| null | null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 4.7
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Randomization of atomic packing in amorphous metals (e.g., Figure 4.20) generally causes no more than a $1 \%$ drop in density compared with the crystalline structure of the same composition. Given the normal density for nickel (which would be in the crystalline state) is $8.91 \mathrm{~g} / \mathrm{cm}^{3}$ and the APF for the fcc metal structure is 0.74. Calculate the APF of an amorphous, thin film of nickel whose density is $8.84 \mathrm{~g} / \mathrm{cm}^{3}$.
|
images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 4.7_1.jpg
|
The normal density for nickel (which would be in the crystalline state) is $8.91 \mathrm{~g} / \mathrm{cm}^{3}$. The APF for the fcc metal structure is 0.74 . Therefore, the APF for this amorphous nickel would be
$$
\mathrm{APF}=(0.74) \times \frac{8.84}{8.91}=0.734
$$
|
0.734
| null | null |
single
|
Example 4.7
| 105,200
| 1
|
hard
|
Materials: Metals
|
Metals
| null |
Micro/Nano-structure
|
Crystal Structure
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 5.1
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
The rate at which a metal alloy oxidizes in an oxygen-containing atmosphere is a typical example of the practical utility of the Arrhenius equation (rate $=C e^{-Q / R T}$). For example, the rate of oxidation of a magnesium alloy is represented by a rate constant, $k$. The value of $k$ at $300^{\circ} \mathrm{C}$ is $1.05 \times 10^{-8} \mathrm{~kg} /\left(\mathrm{m}^{4} \cdot \mathrm{~s}\right)$. At $400^{\circ} \mathrm{C}$, the value of $k$ rises to $2.95 \times 10^{-4} \mathrm{~kg} /\left(\mathrm{m}^{4} \cdot \mathrm{~s}\right)$. Calculate the activation energy, $Q$, for this oxidation process (in units of $\mathrm{kJ} / \mathrm{mol}$ ).
| null |
For this specific case, Equation rate $=C e^{-Q / R T}$ has the form
$$
k=C e^{-Q / R T}
$$
Taking the ratio of rate constants at $300^{\circ} \mathrm{C}(=573 \mathrm{~K})$ and $400^{\circ} \mathrm{C}$ $(=673 \mathrm{~K})$, we conveniently cancel out the unknown preexponential constant, $C$, and obtain
$$
\frac{2.95 \times 10^{-4} \mathrm{~kg} /\left[\mathrm{m}^{4} \cdot \mathrm{~s}\right]}{1.05 \times 10^{-8} \mathrm{~kg} /\left[\mathrm{m}^{4} \cdot \mathrm{~s}\right]}=\frac{e^{-Q /(8.314 \mathrm{~J} /[\mathrm{mol} \cdot \mathrm{~K}])(673 \mathrm{~K})}}{e^{-Q /(8.314 \mathrm{~J} /[\mathrm{mol} \cdot \mathrm{~K}])(573 \mathrm{~K})}}
$$
or
$$
2.81 \times 10^{4}=e^{\left\{-Q /(8.314 \mathrm{~J} /[\mathrm{mol} \cdot \mathrm{~K}])\right\}\{1 /(673 \mathrm{~K})-1 /(573 \mathrm{~K})\}}
$$
giving
$$
Q=328 \times 10^{3} \mathrm{~J} / \mathrm{mol}=328 \mathrm{~kJ} / \mathrm{mol}
$$
|
328
|
kJ/mol
| null |
single
|
Example 5.1
| 101,407
| 4
|
easy
|
Materials: Metals
|
Metals
| null |
Composites
|
Diffusion & Kinetics
| null |
Corrosion
|
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 5.2
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
At $400^{\circ} \mathrm{C}$, the fraction of aluminum lattice sites vacant is $2.29 \times 10^{-5}$. Calculate the fraction at $660^{\circ} \mathrm{C}$ (just below its melting point). From the text discussion relative to Figure 5.4, we have $E_{V}=0.76 \mathrm{eV}$.
|
images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 5.2_1.jpg
|
From the text discussion relative to Figure 5.4, we have $E_{V}=0.76 \mathrm{eV}$. Using Equation, we have
$$
\frac{n_{v}}{n_{\text {sites }}}=C e^{-E_{V} / k T}
$$
$$
\begin{aligned}
& \text { At } 400^{\circ} \mathrm{C}(=673 \mathrm{~K}) \text {, we obtain } \\
& \qquad \begin{aligned}
C & =\left(\frac{n_{\mathrm{c}}}{n_{\text {sites }}}\right) e^{+E_{\mathrm{V}} / k T} \\
& =\left(2.29 \times 10^{-5}\right) e^{+0.76 \mathrm{eV} /\left(86.2 \times 10^{-6} \mathrm{eV} / \mathrm{K}\right)(673 \mathrm{~K})}=11.2 \\
& \text { At } 660^{\circ} \mathrm{C}(=933 \mathrm{~K}) \\
& \qquad \frac{n_{\mathrm{c}}}{n_{\text {sites }}}}=(11.2) e^{-0.76 \mathrm{eV} /\left(86.2 \times 10^{-6} \mathrm{eV} / \mathrm{K}\right)(933 \mathrm{~K})}=8.82 \times 10^{-4}
\end{aligned}
\end{aligned}
$$
or roughly nine vacancies occur for every 10,000 lattice sites.
|
8.82E-04
| null | null |
single
|
Example 5.2
| 121,200
| 4
|
hard
|
Materials: Metals
|
Metals
|
Thermal
|
Composites
|
Crystal Structure
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 5.3
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Steel surfaces can be hardened by carburization, as discussed relative to Figure. During one such treatment at $1,000^{\circ} \mathrm{C}$, there is a drop in carbon concentration from 5 to 4 at $\%$ carbon between 1 and 2 mm from the surface of the steel. Estimate the flux of carbon atoms into the steel in this near-surface region. (The density of $\gamma$-Fe at $1,000^{\circ} \mathrm{C}$ is $7.63 \mathrm{~g} / \mathrm{cm}^{3}$.)
Refer to Table:
\begin{tabular}{|l|l|l|l|l|}
\hline \multicolumn{5}{|c|}{Diffusivity Data for a Number of Metallic Systems ${ }^{\text {a }}$} \\
\hline Solute & Solvent & $D_0\left(\mathrm{~m}^2 / \mathrm{s}\right)$ & $Q(\mathrm{~kJ} / \mathrm{mol})$ & $Q(\mathrm{kcal} / \mathrm{mol})$ \\
\hline Carbon & Fcc iron & $20 \times 10^{-6}$ & 142 & 34.0 \\
\hline Carbon & Bcc iron & $220 \times 10^{-6}$ & 122 & 29.3 \\
\hline Iron & Fcc iron & $22 \times 10^{-6}$ & 268 & 64.0 \\
\hline Iron & Bcc iron & $200 \times 10^{-6}$ & 240 & 57.5 \\
\hline Nickel & Fcc iron & $77 \times 10^{-6}$ & 280 & 67.0 \\
\hline Manganese & Fcc iron & $35 \times 10^{-6}$ & 282 & 67.5 \\
\hline Zinc & Copper & $34 \times 10^{-6}$ & 191 & 45.6 \\
\hline Copper & Aluminum & $15 \times 10^{-6}$ & 126 & 30.2 \\
\hline Copper & Copper & $20 \times 10^{-6}$ & 197 & 47.1 \\
\hline Silver & Silver & $40 \times 10^{-6}$ & 184 & 44.1 \\
\hline Carbon & Hcp titanium & $511 \times 10^{-6}$ & 182 & 43.5 \\
\hline
\end{tabular}
|
images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 5.3_1.jpg
|
First, we approximate
$$
\begin{aligned}
\frac{\partial c}{\partial x} \simeq \frac{\Delta c}{\Delta x} & =\frac{5 \text { at }\%-4 \text { at } \%}{1 \mathrm{~mm}-2 \mathrm{~mm}} \\
& =-1 \text { at } \% / \mathrm{mm}
\end{aligned}
$$
To obtain an absolute value for carbon-atom concentration, we must first know the concentration of iron atoms. From the given data,
$$
\rho=7.63 \frac{g}{\mathrm{~cm}^{3}} \times \frac{0.6023 \times 10^{24} \text { atoms }}{55.85 \mathrm{~g}}=8.23 \times 10^{22} \frac{\text { atoms }}{\mathrm{cm}^{3}}
$$
Therefore,
$$
\begin{aligned}
\frac{\Delta c}{\Delta x} & =-\frac{0.01\left(8.23 \times 10^{22} \text { atoms } / \mathrm{cm}^{3}\right)}{1 \mathrm{~mm}} \times \frac{10^{6} \mathrm{~cm}^{3}}{\mathrm{~m}^{3}} \times \frac{10^{3} \mathrm{~mm}}{\mathrm{~m}} \\
& =-8.23 \times 10^{29} \text { atoms } / \mathrm{m}^{4}
\end{aligned}
$$
From Table,
$$
\begin{aligned}
D_{c} \text { in } \gamma-\mathrm{Fe}, 1000^{\circ} \mathrm{C} & =D_{0} e^{-Q / R T} \\
& =\left(20 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right) e^{-(142,000 \mathrm{~J} / \mathrm{mol}) /(8.314 \mathrm{~J} / \mathrm{mol} / \mathrm{K})(1273 \mathrm{~K})} \\
& =2.98 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}
\end{aligned}
$$
Using Equation ($J x=-D \frac{\partial c}{\partial x}$) gives us
$$
\begin{aligned}
J_{x} & =-D \frac{\partial c}{\partial x} \\
& \simeq-D \frac{\Delta c}{\Delta x} \\
& =-\left(2.98 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\right)\left(-8.23 \times 10^{29} \mathrm{atoms} / \mathrm{m}^{4}\right) \\
& =2.45 \times 10^{19} \mathrm{atoms} /\left(\mathrm{m}^{2} \cdot \mathrm{~s}\right)
\end{aligned}
$$
|
2.45 \times 10^{19}
|
atoms/(m^2 \cdot s)
| null |
single
|
Example 5.3
| 101,410
| 5
|
easy
|
Materials: Metals
|
Metals
| null |
Composites
|
Diffusion & Kinetics
|
Joining
| null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 5.4
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
The diffusion result described by Equation ($\frac{c_x-c_0}{c_s-c_0}=1-\operatorname{erf}\left(\frac{x}{2 \sqrt{D t}}\right)$) can apply to the carburization process. The carbon environment (a hydrocarbon gas) is used to set the surface-carbon content $\left(c_{s}\right)$ at $1.0 \mathrm{wt} \%$. The initial carbon content of the steel $\left(c_{0}\right)$ is $0.2 \mathrm{wt} \%$. Using the error-function table, calculate how long it would take at $1,000^{\circ} \mathrm{C}$ to reach a carbon content of $0.6 \mathrm{wt} \%$ [i.e., $\left(c-c_{0}\right) /\left(c_{s}-c_{0}\right)=0.5$ ] at a distance of 1 mm from the surface. Given a diffusivity value ($D = 2.98 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}$).
Refer to Table:
\begin{tabular}{|l|l|l|l|}
\hline \multicolumn{4}{|c|}{The Error Function} \\
\hline $z$ & erf(z) & $z$ & erf(z) \\
\hline 0.00 & 0.0000 & 0.70 & 0.6778 \\
\hline 0.01 & 0.0113 & 0.75 & 0.7112 \\
\hline 0.02 & 0.0226 & 0.80 & 0.7421 \\
\hline 0.03 & 0.0338 & 0.85 & 0.7707 \\
\hline 0.04 & 0.0451 & 0.90 & 0.7969 \\
\hline 0.05 & 0.0564 & 0.95 & 0.8209 \\
\hline 0.10 & 0.1125 & 1.00 & 0.8427 \\
\hline 0.15 & 0.1680 & 1.10 & 0.8802 \\
\hline 0.20 & 0.2227 & 1.20 & 0.9103 \\
\hline 0.25 & 0.2763 & 1.30 & 0.9340 \\
\hline 0.30 & 0.3286 & 1.40 & 0.9523 \\
\hline 0.35 & 0.3794 & 1.50 & 0.9661 \\
\hline 0.40 & 0.4284 & 1.60 & 0.9763 \\
\hline 0.45 & 0.4755 & 1.70 & 0.9838 \\
\hline 0.50 & 0.5205 & 1.80 & 0.9891 \\
\hline 0.55 & 0.5633 & 1.90 & 0.9928 \\
\hline 0.60 & 0.6039 & 2.00 & 0.9953 \\
\hline 0.65 & 0.6420 & & \\
\hline
\end{tabular}
| null |
Using Equation ($\frac{c_x-c_0}{c_s-c_0}=1-\operatorname{erf}\left(\frac{x}{2 \sqrt{D t}}\right)$), we get
$$
\frac{c_{s}-c_{0}}{c_{s}-c_{0}}=0.5=1-\operatorname{erf}\left(\frac{x}{2 \sqrt{D t}}\right)
$$
or
$$
\operatorname{erf}\left(\frac{x}{2 \sqrt{D t}}\right)=1-0.5=0.5
$$
Interpolating from Table gives
$$
\frac{0.5-0.4755}{0.5205-0.4755}=\frac{z-0.45}{0.50-0.45}
$$
or
$$
z=\frac{x}{2 \sqrt{D t}}=0.4772
$$
or
$$
t=\frac{x^{2}}{4(0.4772)^{2} D}
$$
Using the diffusivity calculation, we obtain
$$
\begin{aligned}
t & =\frac{\left(1 \times 10^{-3} \mathrm{~m}\right)^{2}}{4(0.4772)^{2}\left(2.98 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\right)} \\
& =3.68 \times 10^{4} \mathrm{~s} \times \frac{1 \mathrm{~h}}{3.6 \times 10^{3} \mathrm{~s}} \\
& =10.2 \mathrm{~h} .
\end{aligned}
$$
|
10.2
|
h
| null |
single
|
Example 5.4
| 101,400
| 6
|
easy
|
Materials: Metals
|
Metals
| null |
Composites
|
Diffusion & Kinetics
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 5.7
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
A 5-mm-thick sheet of palladium with a cross-sectional area of $0.2 \mathrm{~m}^{2}$ is used as a steady-state diffusional membrane for purifying hydrogen. If the hydrogen concentration on the high-pressure (impure gas) side of the sheet is $0.3 \mathrm{~kg} / \mathrm{m}^{3}$ and the diffusion coefficient for hydrogen in Pd is $1.0 \times 10^{-8} \mathrm{~m}^{2} / \mathrm{s}$, calculate the mass of hydrogen being purified per hour. Assume the hydrogen concentration on the low-pressure side is maintained at 0 kg/m³ (e.g. by vacuum).
| null |
Fick's first law ($J x=-D \frac{\partial c}{\partial x}$) is simplified by the steady-state concentration gradient of Equation ($\frac{\partial c}{\partial x}=\frac{\Delta c}{\Delta x}=\frac{c_h-c_l}{0-x_0}=-\frac{c_h-c_l}{x_0}$.), giving
$$
\begin{aligned}
J x & = -D(\partial c / \partial x) = -D\left[-\left(c_{h}-c_{l}\right) / x_{0}\right] \\
& = -\left(1.0 \times 10^{-8} \mathrm{~m}^{2} / \mathrm{s}\right)\left[-\left(0.3 \mathrm{~kg} / \mathrm{m}^{3}-0 \mathrm{~kg} / \mathrm{m}^{3}\right) /\left(5 \times 10^{-3} \mathrm{~m}\right)\right] \\
& = 0.6 \times 10^{-6} \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{~s} \times 3.6 \times 10^{3} \mathrm{~s} / \mathrm{h} = 2.16 \times 10^{-3} \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{~h}
\end{aligned}
$$
The total mass of hydrogen being purified will then be this flux times the membrane area:
$$
m = J x \times A = 2.16 \times 10^{-3} \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{~h} \times 0.2 \mathrm{~m}^{2} = 0.432 \times 10^{-3} \mathrm{~kg} / \mathrm{h}
$$
|
4.32E-04
|
kg/h
| null |
single
|
Example 5.7
| 153,410
| 2
|
easy
|
Materials: Metals
|
Metals
|
Optical
|
Structural Gradient
|
Diffusion & Kinetics
|
Joining
| null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 5.8
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Using the semi-infinite diffusion expression of Equation ($\frac{c_x-c_0}{c_s-c_0}=1-\operatorname{erf}\left(\frac{x}{2 \sqrt{D t}}\right)$), (a) calculate the penetration of B into A along the grain boundary after 1 hour, defined as the distance, $x$, at which $c_{s}=0.01 c_{s}$ (with $c_{0}=0$ for initially pure A), given $D_{\text {grain boundary }}=1.0 \times 10^{-10} \mathrm{~m}^{2} / \mathrm{s}$. (b) For comparison, calculate the penetration defined in the same way within the bulk grain for which $D_{\text {volume }}=1.0 \times 10^{-14} \mathrm{~m}^{2} / \mathrm{s}$. Give your answer as a tuple (penetration along grain boundary, penetration within bulk grain).
Refer to Table:
\begin{tabular}{|l|l|l|l|}
\hline \multicolumn{4}{|c|}{The Error Function} \\
\hline $z$ & erf(z) & $z$ & erf(z) \\
\hline 0.00 & 0.0000 & 0.70 & 0.6778 \\
\hline 0.01 & 0.0113 & 0.75 & 0.7112 \\
\hline 0.02 & 0.0226 & 0.80 & 0.7421 \\
\hline 0.03 & 0.0338 & 0.85 & 0.7707 \\
\hline 0.04 & 0.0451 & 0.90 & 0.7969 \\
\hline 0.05 & 0.0564 & 0.95 & 0.8209 \\
\hline 0.10 & 0.1125 & 1.00 & 0.8427 \\
\hline 0.15 & 0.1680 & 1.10 & 0.8802 \\
\hline 0.20 & 0.2227 & 1.20 & 0.9103 \\
\hline 0.25 & 0.2763 & 1.30 & 0.9340 \\
\hline 0.30 & 0.3286 & 1.40 & 0.9523 \\
\hline 0.35 & 0.3794 & 1.50 & 0.9661 \\
\hline 0.40 & 0.4284 & 1.60 & 0.9763 \\
\hline 0.45 & 0.4755 & 1.70 & 0.9838 \\
\hline 0.50 & 0.5205 & 1.80 & 0.9891 \\
\hline 0.55 & 0.5633 & 1.90 & 0.9928 \\
\hline 0.60 & 0.6039 & 2.00 & 0.9953 \\
\hline 0.65 & 0.6420 & & \\
\hline
\end{tabular}
|
images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 5.8_1.jpg
|
(a) We can simplify Equation as
$$
\frac{c_{s}-c_{0}}{c_{s}-c_{0}}=1-\operatorname{erf}\left(\frac{x}{2 \sqrt{D t}}\right)=\frac{c_{s}-0}{c_{s}-0}=\frac{0.01 c_{s}}{c_{s}}=0.01
$$
or
$$
\operatorname{erf}\left(\frac{x}{2 \sqrt{D t}}\right)=1-0.01=0.99
$$
Interpolating from Table gives
$$
\frac{0.9928-0.99}{0.9928-0.9891}=\frac{1.90-z}{1.90-1.80}
$$
or
$$
z=\frac{x}{2 \sqrt{D t}}=1.824
$$
and then
$$
\begin{aligned}
x & =2(1.824) \sqrt{D t} \\
& =2(1.824) \sqrt{\left(1.0 \times 10^{-10} \mathrm{~m}^{2} / \mathrm{s}\right)(1 \mathrm{~h})\left(3.6 \times 10^{3} \mathrm{~s} / \mathrm{h}\right)} \\
& =2.19 \times 10^{-3} \mathrm{~m}=2.19 \mathrm{~mm}
\end{aligned}
$$
(b) For comparison,
$$
\begin{aligned}
x & =2(1.824) \sqrt{\left(1.0 \times 10^{-14} \mathrm{~m}^{2} / \mathrm{s}\right)(1 \mathrm{~h})\left(3.6 \times 10^{3} \mathrm{~s} / \mathrm{h}\right)} \\
& =21.9 \times 10^{-6} \mathrm{~m}=21.9 \mu \mathrm{~m}
\end{aligned}
$$
|
(2.19, 21.9)
|
(mm, μm)
| null |
multiple
|
Example 5.8
| 105,406
| 6
|
medium
|
Materials: Metals
|
Metals
| null |
Micro/Nano-structure
|
Diffusion & Kinetics
| null |
Impact
|
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 6.2
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
A 10-mm-diameter bar of 1040 carbon steel is subjected to a tensile load of $50,000 \mathrm{~N}$, taking it beyond its yield point. Calculate the elastic recovery that would occur upon removal of the tensile load.
Refer to Table:
\begin{tabular}{|l|l|l|l|l|}
\hline \multicolumn{5}{|l|}{Tensile Test Data for Some Typical Metal Alloys} \\
\hline Alloy* & E [GPa (psi)] & Y.S. [MPa (ksi)] & T.S. [MPa (ksi)] & Percent elongation at failure \\
\hline 1. 1040 carbon steel & $200\left(29 \times 10^6\right)$ & 600 (87) & 750 (109) & 17 \\
\hline 2. 8630 low-alloy steel & & 680 (99) & 800 (116) & 22 \\
\hline 3. a. 304 stainless steel & $193\left(28 \times 10^6\right)$ & 205 (30) & 515 (75) & 40 \\
\hline b. 410 stainless steel & $200\left(29 \times 10^6\right)$ & 700 (102) & 800 (116) & 22 \\
\hline 4. L2 tool steel & & 1,380 (200) & 1,550 (225) & 12 \\
\hline 5. Ferrous superalloy (410) & $200\left(29 \times 10^6\right)$ & 700 (102) & 800 (116) & 22 \\
\hline 6. a. Ductile iron, quench & $165\left(24 \times 10^6\right)$ & 580 (84) & 750 (108) & 9.4 \\
\hline b. Ductile iron, 60-40-18 & $169\left(24.5 \times 10^6\right)$ & 329 (48) & 461 (67) & 15 \\
\hline 7. a.3003-H14 aluminum & $70\left(10.2 \times 10^6\right)$ & 145 (21) & 150 (22) & 8-16 \\
\hline b. 2048, plate aluminum & $70.3\left(10.2 \times 10^6\right)$ & 416 (60) & 457 (66) & 8 \\
\hline 8. a. AZ31B magnesium & $45\left(6.5 \times 10^6\right)$ & 220 (32) & 290 (42) & 15 \\
\hline b. AM100A casting magnesium & $45\left(6.5 \times 10^6\right)$ & 83 (12) & 150 (22) & 2 \\
\hline 9. a. Ti-5Al-2.5Sn & $107-110\left(15.5-16 \times 10^6\right)$ & 827 (120) & 862 (125) & 15 \\
\hline b. Ti-6Al-4V & $110\left(16 \times 10^6\right)$ & 825 (120) & 895 (130) & 10 \\
\hline 10. Aluminum bronze, $9 \%$ & $110\left(16.1 \times 10^6\right)$ & 320 (46.4) & 652 (94.5) & 34 \\
\hline 11. Monel 400 (nickel alloy) & $179\left(26 \times 10^6\right)$ & 283 (41) & 579 (84) & 39.5 \\
\hline 12. AC41A zinc & & & 328 (47.6) & 7 \\
\hline 13. 50:50 solder (lead alloy) & & 33 (4.8) & 42 (6.0) & 60 \\
\hline 14. $\mathrm{Nb}-1 \mathrm{Zr}$ (refractory metal) & $68.9\left(10 \times 10^6\right)$ & 138 (20) & 241 (35) & 20 \\
\hline 15. Dental gold alloy (precious metal) & & & 310-380 (45-55) & 20-35 \\
\hline
\end{tabular}
| null |
Using Equation ($\sigma=\frac{P}{A_0}$) to calculate engineering stress gives
$$
\begin{aligned}
\sigma & =\frac{P}{A_{u}}=\frac{50,000 \mathrm{~N}}{\pi\left(5 \times 10^{-3} \mathrm{~m}\right)^{2}}=637 \times 10^{6} \frac{\mathrm{~N}}{\mathrm{~m}^{2}} \\
& =637 \mathrm{MPa}
\end{aligned}
$$
which is between the Y.S. ( 600 MPa ) and the T.S. ( 750 MPa ) for this alloy (Table).
The elastic recovery can be calculated from Hooke's law ($\sigma=E \epsilon$) using the elastic modulus of Table:
$$
\begin{aligned}
\epsilon & =\frac{\sigma}{E} \\
& =\frac{637 \times 10^{6} \mathrm{~Pa}}{200 \times 10^{9} \mathrm{~Pa}} \\
& =3.18 \times 10^{-3}
\end{aligned}
$$
|
3.18E-03
| null | null |
single
|
Example 6.2
| 111,321
| 2
|
medium
|
Materials: Metals
|
Metals
|
Mechanical
|
Composites
|
Phase Diagram
|
Shaping
|
Elastic
|
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 6.3
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
(a) A 10-mm-diameter rod of 3003-H14 aluminum alloy is subjected to a $6-\mathrm{kN}$ tensile load. Calculate the resulting rod diameter. (b) Calculate the diameter if this rod is subjected to a $6-\mathrm{kN}$ compressive load. Give your answer as a tuple (resulting diameter under tensile load, resulting diameter under compressive load).
Refer to Table 6.1:
\begin{tabular}{|l|l|l|l|l|}
\hline \multicolumn{5}{|l|}{Tensile Test Data for Some Typical Metal Alloys} \\
\hline Alloy* & E [GPa (psi)] & Y.S. [MPa (ksi)] & T.S. [MPa (ksi)] & Percent elongation at failure \\
\hline 1. 1040 carbon steel & $200\left(29 \times 10^6\right)$ & 600 (87) & 750 (109) & 17 \\
\hline 2. 8630 low-alloy steel & & 680 (99) & 800 (116) & 22 \\
\hline 3. a. 304 stainless steel & $193\left(28 \times 10^6\right)$ & 205 (30) & 515 (75) & 40 \\
\hline b. 410 stainless steel & $200\left(29 \times 10^6\right)$ & 700 (102) & 800 (116) & 22 \\
\hline 4. L2 tool steel & & 1,380 (200) & 1,550 (225) & 12 \\
\hline 5. Ferrous superalloy (410) & $200\left(29 \times 10^6\right)$ & 700 (102) & 800 (116) & 22 \\
\hline 6. a. Ductile iron, quench & $165\left(24 \times 10^6\right)$ & 580 (84) & 750 (108) & 9.4 \\
\hline b. Ductile iron, 60-40-18 & $169\left(24.5 \times 10^6\right)$ & 329 (48) & 461 (67) & 15 \\
\hline 7. a.3003-H14 aluminum & $70\left(10.2 \times 10^6\right)$ & 145 (21) & 150 (22) & 8-16 \\
\hline b. 2048, plate aluminum & $70.3\left(10.2 \times 10^6\right)$ & 416 (60) & 457 (66) & 8 \\
\hline 8. a. AZ31B magnesium & $45\left(6.5 \times 10^6\right)$ & 220 (32) & 290 (42) & 15 \\
\hline b. AM100A casting magnesium & $45\left(6.5 \times 10^6\right)$ & 83 (12) & 150 (22) & 2 \\
\hline 9. a. Ti-5Al-2.5Sn & $107-110\left(15.5-16 \times 10^6\right)$ & 827 (120) & 862 (125) & 15 \\
\hline b. Ti-6Al-4V & $110\left(16 \times 10^6\right)$ & 825 (120) & 895 (130) & 10 \\
\hline 10. Aluminum bronze, $9 \%$ & $110\left(16.1 \times 10^6\right)$ & 320 (46.4) & 652 (94.5) & 34 \\
\hline 11. Monel 400 (nickel alloy) & $179\left(26 \times 10^6\right)$ & 283 (41) & 579 (84) & 39.5 \\
\hline 12. AC41A zinc & & & 328 (47.6) & 7 \\
\hline 13. 50:50 solder (lead alloy) & & 33 (4.8) & 42 (6.0) & 60 \\
\hline 14. $\mathrm{Nb}-1 \mathrm{Zr}$ (refractory metal) & $68.9\left(10 \times 10^6\right)$ & 138 (20) & 241 (35) & 20 \\
\hline 15. Dental gold alloy (precious metal) & & & 310-380 (45-55) & 20-35 \\
\hline
\end{tabular}
Refer to Table 6.3:
\begin{tabular}{|l|l|l|l|}
\hline \multicolumn{4}{|l|}{Poisson's Ratio and Shear Modulus for the Alloys of Table 6.1} \\
\hline Alloy & $\nu$ & $G(\mathrm{GPa})$ & $G / E$ \\
\hline 1. 1040 carbon steel & 0.30 & & \\
\hline 2. 8630 carbon steel & 0.30 & & \\
\hline 3. a. 304 stainless steel & 0.29 & & \\
\hline 6. b. Ductile iron, 60-40-18 & 0.29 & & \\
\hline 7. a. 3003-H14 aluminum & 0.33 & 25 & 0.36 \\
\hline 8. a. AZ31B magnesium & 0.35 & 17 & 0.38 \\
\hline b. AM100A casting magnesium & 0.35 & & \\
\hline 9. a. $\mathrm{Ti}-5 \mathrm{Al}-2.5 \mathrm{Sn}$ & 0.35 & 48 & 0.44 \\
\hline b. Ti-6Al-4V & 0.33 & 41 & 0.38 \\
\hline 10. Aluminum bronze, $9 \%$ (copper alloy) & 0.33 & 44 & 0.40 \\
\hline 11. Monel 400 (nickel alloy) & 0.32 & & \\
\hline
\end{tabular}
| null |
(a) From Equation ($\sigma=\frac{P}{A_0}$), the engineering stress is $$\sigma =\frac{P}{A_{0}} =\frac{6 \times 10^{3} \mathrm{~N}}{\pi\left(\frac{10}{2} \times 10^{-3} \mathrm{~m}\right)^{2}}=76.4 \times 10^{6} \frac{\mathrm{~N}}{\mathrm{~m}^{2}}=76.4 \mathrm{MPa}$$ From Table 6.1, we see that this stress is well below the yield strength (145 MPa) and, as a result, the deformation is elastic. From Equation ($\sigma=E \epsilon$), we can calculate the tensile strain using the elastic modulus from Table 6.1: $$\epsilon=\frac{\sigma}{E}=\frac{76.4 \mathrm{MPa}}{70 \times 10^{3} \mathrm{MPa}}=1.09 \times 10^{-3}$$ If we use Equation ($\nu=-\frac{\epsilon_x}{\epsilon_z}$) and the value for $v$ from Table 6.3, the strain for the diameter can be calculated as $$\epsilon_{\text {diameter }} =-\nu \epsilon_{z}=-(0.33)\left(1.09 \times 10^{-3}\right) =-3.60 \times 10^{-4}$$ The resulting diameter can then be determined (analogous to Equation $\epsilon=\frac{l-l_0}{l_0}=\frac{\Delta l}{l_0}$) from $$\epsilon_{\text {diameter }}=\frac{d_{f}-d_{0}}{d_{0}}$$ or $$d_{f} =d_{0}\left(\epsilon_{\text {diameter }}+1\right)=10 \mathrm{~mm}\left(-3.60 \times 10^{-4}+1\right) =9.9964 \mathrm{~mm}$$ (b) For a compressive stress, the diameter strain will be of equal magnitude but of opposite sign; that is, $$\epsilon_{\text {diameter }}=+3.60 \times 10^{-4}$$ As a result, the final diameter will be $$d_{f} =d_{0}\left(\epsilon_{\text {diameter }}+1\right)=10 \mathrm{~mm}\left(+3.60 \times 10^{-4}+1\right) =10.0036 \mathrm{~mm}$$
|
(9.9964, 10.0036)
|
(mm, mm)
| null |
multiple
|
Example 6.3
| 111,001
| 9
|
hard
|
Materials: Metals
|
Metals
|
Mechanical
|
Composites
| null | null |
Elastic
|
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 6.4
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
A glass plate contains an atomic-scale surface crack. (Take the cracktip radius $\simeq$ diameter of an $\mathrm{O}^{2-}$ ion.) Given that the crack is $1-\mu \mathrm{m}$ long, the theoretical strength of the defect-free glass is 7.0 GPa , and $r_{\mathrm{O}^{2-}}=0.132 \mathrm{~nm}$. Calculate the breaking strength of the plate.
| null |
This is an application of Equation ($\sigma_m \simeq 2 \sigma\left(\frac{c}{\rho}\right)^{1 / 2}$):
$$
\sigma_{m}=2 \sigma\left(\frac{c}{\rho}\right)^{1 / 2}
$$
Rearranging gives us
$$
\sigma=\frac{1}{2} \sigma_{m}\left(\frac{\rho}{c}\right)^{1 / 2}
$$
Using given data, we have
$$
\begin{aligned}
\rho & =2 r_{\mathrm{O}^{2-}}=2(0.132 \mathrm{~nm}) \\
& =0.264 \mathrm{~nm}
\end{aligned}
$$
Then,
$$
\begin{aligned}
\sigma & =\frac{1}{2}\left(7.0 \times 10^{9} \mathrm{~Pa}\right)\left(\frac{0.264 \times 10^{-9} \mathrm{~m}}{1 \times 10^{-6} \mathrm{~m}}\right)^{1 / 2} \\
& =57 \mathrm{MPa} .
\end{aligned}
$$
|
57
|
MPa
| null |
single
|
Example 6.4
| 611,003
| 5
|
easy
|
Materials: Glasses
|
Glasses
|
Mechanical
|
Composites
| null | null |
Fracture
|
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 6.5
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
The following data are collected in a flexural test of a nylon to be used in the fabrication of lightweight gears:
Test piece geometry: $7 \mathrm{~mm} \times 13 \mathrm{~mm} \times 100 \mathrm{~mm}$,
Distance between supports $=L=50 \mathrm{~mm}$,
and
Initial slope of load-deflection curve $=404 \times 10^{3} \mathrm{~N} / \mathrm{m}$.
Calculate the flexural modulus for this engineering polymer.
|
images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 6.5_1.jpg
|
Referring to Figure 6.14 and Equation ($E_{\mathrm{flex}}=\frac{L^3 m}{4 b h^3}$), we find that
$$
\begin{aligned}
E_{\text{flex }} & =\frac{L^{3} m}{4 b h^{3}} \\
& =\frac{\left(50 \times 10^{-3} \mathrm{~m}\right)^{3}\left(404 \times 10^{3} \mathrm{~N} / \mathrm{m}\right)}{4\left(13 \times 10^{-3} \mathrm{~m}\right)\left(7 \times 10^{-3} \mathrm{~m}\right)^{3}} \\
& =2.83 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}=2,830 \mathrm{MPa} .
\end{aligned}
$$.
|
2.83E+09
|
N/m^2
| null |
single
|
Example 6.5
| 211,000
| 1
|
easy
|
Materials: Polymers
|
Polymers
|
Mechanical
|
Composites
| null | null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 6.6
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
A small, uniaxial stress of $1 \mathrm{MPa}(145 \mathrm{psi})$ is applied to a rod of high-density polyethylene. (a) Given $E=830 \mathrm{MPa}$, what is the resulting strain? (b) Given $E=1.3 \mathrm{MPa}$, repeat for a rod of vulcanized isoprene. (c) Given $E=2 \times 10^5 \mathrm{MPa}$, repeat for a rod of 1040 steel. Give your answer as a tuple (strain for polyethylene, strain for isoprene, strain for steel).
| null |
(a) For this modest stress level, we can assume Hooke's law behavior: $$\epsilon=\frac{s}{E}$$ Given $E=830 \mathrm{MPa}$. So, $$\epsilon=\frac{1 \mathrm{MPa}}{830 \mathrm{MPa}}=1.2 \times 10^{-3}$$ (b) Given $E=1.3 \mathrm{MPa}$, or $$\epsilon=\frac{1 \mathrm{MPa}}{1.3 \mathrm{MPa}}=0.77$$ (c) Given $E=200 \mathrm{GPa}=2 \times 10^{5} \mathrm{MPa}$, or $$\begin{aligned}\epsilon & =\frac{1 \mathrm{MPa}}{2 \times 10^{5} \mathrm{MPa}} \\ & =5.0 \times 10^{-6}\end{aligned}$$
|
(1.2 \times 10^{-3}, 0.77, 5.0 \times 10^{-6})
| null | null |
multiple
|
Example 6.6
| 111,001
| 3
|
medium
|
Materials: Metals
|
Metals
|
Mechanical
|
Composites
| null | null |
Elastic
|
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 6.7
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
In the absence of stress, the center-to-center atomic separation distance of two Fe atoms is 0.2480 nm (along a $\langle 111\rangle$ direction). Under a tensile stress of $1,000 \mathrm{MPa}$ along this direction, the atomic separation distance increases to 0.2489 nm . Calculate the modulus of elasticity along the $\langle 111\rangle$ directions.
| null |
From Hooke's law ($\sigma=E \epsilon$),
$$
E=\frac{\sigma}{\epsilon}
$$
with
$$
\epsilon=\frac{(0.2489-0.2480) \mathrm{nm}}{0.2480 \mathrm{~nm}}=0.00363
$$
giving
$$
E=\frac{1,000 \mathrm{MPa}}{0.00363}=280 \mathrm{GPa}
$$
|
280
|
GPa
| null |
single
|
Example 6.7
| 115,201
| 3
|
easy
|
Materials: Metals
|
Metals
|
Mechanical
|
Micro/Nano-structure
|
Crystal Structure
| null |
Elastic
|
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 6.8
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
A zinc single crystal is being pulled in tension, with the normal to its basal plane (0001) at $60^{\circ}$ to the tensile axis and with the slip direction [1120] at $40^{\circ}$ to the tensile axis.
(a) What is the resolved shear stress, $\tau$, acting in the slip direction when a tensile stress of $0.690 \mathrm{MPa}(100 \mathrm{psi})$ is applied?
(b) What tensile stress is necessary to reach the critical resolved shear stress, $\tau_{c}$, of $0.94 \mathrm{MPa}(136 \mathrm{psi})$?
Give your answers as a tuple: (resolved shear stress, tensile stress)
| null |
(a) From Equation ($\tau=\frac{F \cos \lambda}{A / \cos \varphi}=\frac{F}{A} \cos \lambda \cos \varphi=\sigma \cos \lambda \cos \varphi$),
$$
\begin{aligned}
\tau & =\sigma \cos \lambda \cos \varphi \\
& =(0.690 \mathrm{MPa}) \cos 40^{\circ} \cos 60^{\circ} \\
& =0.264 \mathrm{MPa}(38.3 \mathrm{psi})
\end{aligned}
$$
(b) From Equation ($\tau_c=\sigma_c \cos \lambda \cos \varphi$),
$$
\tau_{\mathrm{c}}=\sigma_{\mathrm{c}} \cos \lambda \cos \varphi
$$
or
$$
\begin{aligned}
\sigma_{c} & =\frac{\tau_{c}}{\cos \lambda \cos \varphi}=\frac{0.94 \mathrm{MPa}}{\cos 40^{\circ} \cos 60^{\circ}} \\
& =2.45 \mathrm{MPa}(356 \mathrm{psi})
\end{aligned}
$$
|
(0.264, 2.45)
|
(MPa,MPa)
| null |
multiple
|
Example 6.8
| 115,202
| 2
|
easy
|
Materials: Metals
|
Metals
|
Mechanical
|
Micro/Nano-structure
|
Crystal Structure
| null |
Plastic
|
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 6.9
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
A Brinell hardness measurement is made on a ductile iron (100-70-03, air-quenched) using a 10-mm-diameter sphere of tungsten carbide. A load of $3,000 \mathrm{~kg}$ produces a $3.91-\mathrm{mm}$-diameter impression in the iron surface. Calculate the BHN of this alloy. (The correct units for the Brinell equation of Table are kilograms for load and millimeters for diameters.)
Refer to Table:
\begin{tabular}{|c|c|c|}
\hline \multicolumn{3}{|l|}{Common Hardness Tests} \\
\hline Test & Indenter & Hardness formula \\
\hline \multirow[t]{3}{*}{Brinell} & \multirow[t]{3}{*}{Steel or tungsten carbide sphere} & $$
\mathrm{BHN}=\frac{2 P}{\pi D\left[D-\sqrt{D^2-d^2}\right]}
$$ \\
\hline & & where $P=$ load \\
\hline & & $$
\begin{aligned}
& D=\text { sphere diameter }=10 \mathrm{~mm} \\
& d=\text { indent diameter (in } \mathrm{mm} \text { ) }
\end{aligned}
$$ \\
\hline \multicolumn{3}{|l|}{Rockwell type} \\
\hline A & Diamond cone under 60 kg load & $R_A=100-500 t$ \\
\hline B & $1 / 16 \mathrm{in}$. dia. steel sphere under 100 kg load & $R_B=130-500 t$ \\
\hline C & Diamond cone under 150 kg load & $R_C=100-500 t$ \\
\hline D & Diamond cone under 100 kg load & $R_D=100-500 t$ \\
\hline E & $1 / 8 \mathrm{in}$. dia. steel sphere under 100 kg load & $R_E=130-500 t$ \\
\hline F & $1 / 16 \mathrm{in}$. dia. steel sphere under 60 kg load & $R_F=130-500 t$ \\
\hline G & $1 / 16 \mathrm{in}$. dia. steel sphere under 150 kg load & $R_G=130-500 t$ \\
\hline \multicolumn{3}{|l|}{with, in all cases, $t=$ depth of indentation (in mm )} \\
\hline
\end{tabular}
| null |
From Table,
$$
\begin{aligned}
\mathrm{BHN} & =\frac{2 P}{\pi D\left(D-\sqrt{D^{2}-d^{2}}\right)} \\
& =\frac{2(3,000)}{\pi(10)\left(10-\sqrt{10^{2}-3.91^{2}}\right)}=240
\end{aligned}
$$
|
240
| null | null |
single
|
Example 6.9
| 111,000
| 2
|
medium
|
Materials: Metals
|
Metals
|
Mechanical
|
Composites
| null | null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 6.10
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
In a laboratory creep experiment at $1,000^{\circ} \mathrm{C}$, a steady-state creep rate of $5 \times 10^{-1} \%$ per hour is obtained in a metal alloy. The creep mechanism for this alloy is known to be dislocation climb with an activation energy of $200 \mathrm{~kJ} / \mathrm{mol}$. Predict the creep rate at a service temperature of $600^{\circ} \mathrm{C}$. (Assume that the laboratory experiment duplicated the service stress.)
| null |
Using the laboratory experiment to determine the preexponential constant in Equation ($\dot{\epsilon}=C e^{-Q / R T}$), we obtain
$$
\begin{aligned}
C & =\dot{\epsilon} e^{+Q / R T} \\
& =\left(5 \times 10^{-1} \% \text { per hour }\right) e^{+\left(2 \times 10^{5} \mathrm{~J} / \mathrm{mol}\right) /[8.314 \mathrm{~J} / \mathrm{molK})](1,273 \mathrm{~K})} \\
& =80.5 \times 10^{6} \% \text { per hour. }
\end{aligned}
$$
Applying this amount to the service temperature yields
$$
\begin{aligned}
\dot{\epsilon} & =\left(80.5 \times 10^{6} \% \text { per hour }\right) e^{-\left(2 \times 10^{5}\right) /(8.314)(873)} \\
& =8.68 \times 10^{-5} \% \text { per hour. }
\end{aligned}
$$
Note. We have assumed that the creep mechanism remains the same between 1,000 and $600^{\circ} \mathrm{C}$.
|
8.68E-05
|
% per hour
| null |
single
|
Example 6.10
| 111,405
| 2
|
easy
|
Materials: Metals
|
Metals
|
Mechanical
|
Composites
|
Diffusion & Kinetics
| null |
Creep
|
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 6.12
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
The relaxation time for a rubber band at $25^{\circ} \mathrm{C}$ is 60 days. (a) If it is stressed to 2 MPa initially, how many days will be required before the stress relaxes to 1 MPa ? (b) If the activation energy for the relaxation process is $30 \mathrm{~kJ} / \mathrm{mol}$, what is the relaxation time at $35^{\circ} \mathrm{C}$ ? Give your answer as a tuple (days for stress to relax to 1 MPa, relaxation time at $35^{\circ} \mathrm{C}$).
| null |
(a) From Equation ($\sigma=\sigma_0 e^{-t / t}$), $$\sigma=\sigma_{0} e^{-t / \tau}$$ and $$1 \mathrm{MPa}=2 \mathrm{MPa} e^{-t /(60 \mathrm{~d})}$$ Rearranging the terms yields $$t=-(60 \text { days })\left(\ln \frac{1}{2}\right)=41.5 \text { days. }$$ (b) From Equation ($\frac{1}{\tau}=C e^{-Q / R T}$), $$\frac{1}{\tau}=C e^{-Q / R T}$$ or $$\frac{1 / \tau_{25^{\circ} \mathrm{C}}}{1 / \tau_{35^{\circ} \mathrm{C}}}=\frac{e^{-Q / R(298 \mathrm{~K})}}{e^{-Q / R(308 \mathrm{~K})}}$$ or $$\tau_{35^{\circ} \mathrm{C}}=\tau_{25^{\circ} \mathrm{C}} \exp \left[\frac{Q}{R}\left(\frac{1}{308 \mathrm{~K}}-\frac{1}{298 \mathrm{~K}}\right)\right]$$ giving, finally, $$\begin{aligned} \tau_{35^{\circ} \mathrm{C}} & =(60 \text { days }) \exp \left[\frac{30 \times 10^{3} \mathrm{~J} / \mathrm{mol}}{8.314 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{~K})}\left(\frac{1}{308 \mathrm{~K}}-\frac{1}{298 \mathrm{~K}}\right)\right] \\ & =40.5 \text { days. } \end{aligned}$$
|
(41.5, 40.5)
|
(days, days)
| null |
multiple
|
Example 6.12
| 311,400
| 6
|
medium
|
Materials: Elastomers
|
Elastomers
|
Mechanical
|
Composites
|
Diffusion & Kinetics
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 6.13
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
A soda-lime-silica glass used to make lamp bulbs has an annealing point of $514^{\circ} \mathrm{C}$ and a softening point of $696^{\circ} \mathrm{C}$. Calculate the working range and the melting range for this glass.
Assuming:
– Annealing point η = 10^13.4 P at 514 °C
– Softening point η = 10^7.6 P at 696 °C
– Working range bounded by η = 10^8 P and 10^4 P
– Melting range bounded by η = 500 P and 50 P
Give your answer as a tuple: (working range lower, working range upper, melting range lower, melting range upper).
| null |
The following is an application of Equation ($\eta=\eta_0 e^{+Q / R T}$):
$$
\eta=\eta_0 e^{+Q / R T} .
$$
Given
$$
\text { annealing point }=514+273=787 \mathrm{~K} \text { for } \eta=10^{13.4} \mathrm{P}
$$
and
$$
\begin{aligned}
\text { softening point } & =696+273=969 \mathrm{~K} \text { for } \eta=10^{7.6} \mathrm{P}, \\
10^{13.4} \mathrm{P} & =\eta_0 e^{+Q /[8.314 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{~K})[(787 \mathrm{~K})}, \\
10^{7.6} \mathrm{P} & =\eta_0 e^{+Q /[8.314 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{~K})](969 \mathrm{~K})},
\end{aligned}
$$
and
$$
\frac{10^{13.4}}{10^{7.6}}=e^{+Q\left[[8.314 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{~K})](1 / 787-1 / 969) K^{-1}\right.},
$$
or
$$
Q=465 \mathrm{~kJ} / \mathrm{mol}
$$
and
$$
\begin{aligned}
\eta_0 & =\left(10^{13.4} \mathrm{P}\right) e^{-\left(465 \times 10^3 \mathrm{~J} / \mathrm{mol}\right) /[8.314 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{~K})(787 \mathrm{~K})} \\
& =3.31 \times 10^{-18} \mathrm{P}
\end{aligned}
$$
The working range is bounded by $\eta=10^4 \mathrm{P}$ and $\eta=10^8 \mathrm{P}$. In general,
$$
T=\frac{Q}{R \ln \left(\eta / \eta_0\right)}
$$
For $\eta=10^4 \mathrm{P}$,
$$
\begin{aligned}
T & =\frac{465 \times 10^3 \mathrm{~J} / \mathrm{mol}}{[8.314 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{~K})] \ln \left(10^4 /\left[3.31 \times 10^{-18}\right]\right)} \\
& =1,130 \mathrm{~K}=858^{\circ} \mathrm{C} .
\end{aligned}
$$
For $\eta=10^8 \mathrm{P}$,
$$
\begin{aligned}
T & =\frac{465 \times 10^3 \mathrm{~J} / \mathrm{mol}}{[8.314 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{~K})] \ln \left(10^8 /\left[3.31 \times 10^{-18}\right]\right)} \\
& =953 \mathrm{~K}=680^{\circ} \mathrm{C} .
\end{aligned}
$$
Therefore,
$$
\text { working range }=680 \text { to } 858^{\circ} \mathrm{C} \text {. }
$$
For the melting range, $\eta=50$ to 500 P . For $\eta=50 \mathrm{P}$,
$$
\begin{aligned}
T & =\frac{465 \times 10^3 \mathrm{~J} / \mathrm{mol}}{[8.314 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{~K})] \ln \left(50 /\left[3.31 \times 10^{-18}\right]\right)} \\
& =1,266 \mathrm{~K}=993^{\circ} \mathrm{C}
\end{aligned}
$$
For $\eta=500 \mathrm{P}$,
$$
\begin{aligned}
T & =\frac{465 \times 10^3 \mathrm{~J} / \mathrm{mol}}{[8.314 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{~K})] \ln \left(500 /\left[3.31 \times 10^{-18}\right]\right)} \\
& =1,204 \mathrm{~K}=931^{\circ} \mathrm{C} .
\end{aligned}
$$
Therefore,
$$
\text { melting range }=931 \text { to } 993^{\circ} \mathrm{C} .
$$
|
(680, 858, 931, 993)
|
(°C, °C, °C, °C)
| null |
multiple
|
Example 6.13
| 601,000
| 9
|
medium
|
Materials: Glasses
|
Glasses
| null |
Composites
| null | null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 7.2
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
A $0.1-\mathrm{m}$-long $\mathrm{Al}_{2} \mathrm{O}_{3}$ furnace tube is heated from room temperature $\left(25^{\circ} \mathrm{C}\right)$ to $1,000^{\circ} \mathrm{C}$. Assuming the tube is not mechanically constrained, calculate the increase in length produced by this heating. Given the thermal-expansion coefficient for $\mathrm{Al}_{2} \mathrm{O}_{3}$ as $8.8 \times 10^{-6} \mathrm{~mm} /\left(\mathrm{mm} \cdot{ }^{\circ} \mathrm{C}\right)$.
| null |
Rearranging Equation ($\alpha=\frac{d L}{L d T}$),
$$
d L=\alpha L d T
$$
We can assume linear thermal expansion using the overall thermal-expansion coefficient for this temperature range given in Table 7.2. Then,
$$
\begin{aligned}
\Delta L & =\alpha L_{0} \Delta T \\
& =\left[8.8 \times 10^{-6} \mathrm{~mm} /\left(\mathrm{mm} \cdot{ }^{\circ} \mathrm{C}\right)\right](0.1 \mathrm{~m})(1,000-25)^{\circ} \mathrm{C} \\
& =0.858 \times 10^{-3} \mathrm{~m} \\
& =0.858 \mathrm{~mm}
\end{aligned}
$$
|
0.858
|
mm
| null |
single
|
Example 7.2
| 121,000
| 1
|
easy
|
Materials: Metals
|
Metals
|
Thermal
|
Composites
| null | null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 7.3
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Calculate the steady-state heat transfer rate (in $\mathrm{J} / \mathrm{m}^{2} \cdot \mathrm{~s}$ ) through a sheet of copper 10 mm thick if there is a $50^{\circ} \mathrm{C}$ temperature drop (from $50^{\circ} \mathrm{C}$ to $0^{\circ} \mathrm{C}$ ) across the sheet. Given the thermal conductivity for copper at 300 K as $398 \mathrm{~J} / \mathrm{s} \cdot \mathrm{m} \cdot \mathrm{K}$
| null |
Rearranging Equation ($k=-\frac{\Delta Q / \Delta t}{A(\Delta T / \Delta x)}$),
$$(\Delta Q / \Delta t) / A=-k(\Delta T / \Delta x)$$
Over this temperature range (average $T=25^{\circ} \mathrm{C}=298 \mathrm{~K}$ ), we can use the thermal conductivity for copper at 300 K given in Table 7.4, giving
$$
\begin{aligned}
(\Delta Q / \Delta t) / A & =-\left(398 \mathrm{~J} / \mathrm{s} \cdot \mathrm{~m} \cdot \mathrm{~K}\right)\left(\left[0^{\circ} \mathrm{C}-50^{\circ} \mathrm{C}\right]\right) /\left[10 \times 10^{-3} \mathrm{~m}\right] \\
& =-\left(398 \mathrm{~J} / \mathrm{s} \cdot \mathrm{~m} \cdot \mathrm{~K}\right)\left(-5 \times 10^{-3}{ }^{\circ} \mathrm{C} / \mathrm{m}\right)
\end{aligned}
$$
The K and ${ }^{\circ} \mathrm{C}$ units cancel each other out given that we are dealing with an incremental change in temperature, so
$$(\Delta Q / \Delta t) / A=1.99 \times 10^{6} \mathrm{~J} / \mathrm{m}^{2} \cdot \mathrm{~s}$$
|
1.99 \times 10^{6}
|
\mathrm{J} / \mathrm{m}^{2} \cdot \mathrm{s}
| null |
single
|
Example 7.3
| 121,400
| 1
|
medium
|
Materials: Metals
|
Metals
|
Thermal
|
Composites
|
Diffusion & Kinetics
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 7.4
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Estimate the amount of heat (in J) required to raise 2 kg of (a) $\alpha$ - iron, (b) graphite, and (c) polypropylene from room temperature ($25^{\circ} \mathrm{C}$) to $100^{\circ} \mathrm{C}$. Give your answer as a tuple (heat for $\alpha$ - iron, heat for graphite, heat for polypropylene).
| null |
To calculate the amount of heat required to raise the temperature of a substance, we use the formula:
$$
Q = m \cdot C_p \cdot \Delta T
$$
where:
- $Q$ is the heat in joules (J),
- $m$ is the mass in kilograms (kg),
- $C_p$ is the specific heat capacity in J/(kg·°C),
- $\Delta T$ is the change in temperature in °C.
For all three materials, the change in temperature $\Delta T$ is $100^{\circ} \mathrm{C} - 25^{\circ} \mathrm{C} = 75^{\circ} \mathrm{C}$.
(a) For $\alpha$ - iron, the specific heat capacity $C_p$ is approximately $450 \mathrm{J}/(\mathrm{kg} \cdot {}^{\circ} \mathrm{C})$.
$$
Q_{\text{iron}} = 2 \mathrm{kg} \times 450 \mathrm{J}/(\mathrm{kg} \cdot {}^{\circ} \mathrm{C}) \times 75^{\circ} \mathrm{C} = 67,500 \mathrm{J}
$$
(b) For graphite, the specific heat capacity $C_p$ is approximately $710 \mathrm{J}/(\mathrm{kg} \cdot {}^{\circ} \mathrm{C})$.
$$
Q_{\text{graphite}} = 2 \mathrm{kg} \times 710 \mathrm{J}/(\mathrm{kg} \cdot {}^{\circ} \mathrm{C}) \times 75^{\circ} \mathrm{C} = 106,500 \mathrm{J}
$$
(c) For polypropylene, the specific heat capacity $C_p$ is approximately $1,880 \mathrm{J}/(\mathrm{kg} \cdot {}^{\circ} \mathrm{C})$.
$$
Q_{\text{polypropylene}} = 2 \mathrm{kg} \times 1,880 \mathrm{J}/(\mathrm{kg} \cdot {}^{\circ} \mathrm{C}) \times 75^{\circ} \mathrm{C} = 282,000 \mathrm{J}
$$
Therefore, the amount of heat required is:
(a) $67,500 \mathrm{J}$ for $\alpha$ - iron,
(b) $106,500 \mathrm{J}$ for graphite, and
(c) $282,000 \mathrm{J}$ for polypropylene.
|
(67500, 106500, 282000)
|
(J, J, J)
| null |
multiple
|
Example 7.4
| 121,000
| 4
|
easy
|
Materials: Metals
|
Metals
|
Thermal
|
Composites
| null | null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 8.1
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
You are required to use a furnace-cooled $\mathrm{Fe}-\mathrm{Mn}-0.05 \mathrm{C}$ alloy in a structural design that may see service temperatures as low as $0^{\circ} \mathrm{C}$. Suggest an appropriate Mn content for the alloy.
|
images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 8.1_1.jpeg
|
Figure provides the specific guidance we need. A $1 \% \mathrm{Mn}$ alloy is relatively brittle at $0^{\circ} \mathrm{C}$, whereas a $2 \% \mathrm{Mn}$ alloy is highly ductile. Therefore, a secure choice (based on notch-toughness considerations only) would be
$$
\text { Mn content }=2 \% \text {. }
$$
|
2
|
%
| null |
single
|
Example 8.1
| 111,000
| 1
|
easy
|
Materials: Metals
|
Metals
|
Mechanical
|
Composites
| null | null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 8.2
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
A high‐strength steel has a yield strength of 1 460 MPa and a fracture toughness K_IC of 98 MPa·√m. Assuming a through‐thickness center crack (Y = 1), determine the half‐crack length a that will cause catastrophic failure when the applied tensile stress reaches ½ Y.S.
| null |
We may use Equation ($K_{\mathrm{IC}}=Y \sigma_f \sqrt{p a}$) with the realization that we are assuming an ideal case of plane strain conditions. In lieu of specific geometrical information, we are forced to take $Y=1$. Within these limitations, we can calculate
$$
K_{\mathrm{IC}}=Y \sigma_{f} \sqrt{\pi a}
$$
With $Y=1$ and $\sigma_{f}=0.5$ Y.S.,
$$
K_{\mathrm{IC}}=0.5 \text { Y.S. } \sqrt{\pi a}
$$
or
$$
\begin{aligned}
a & =\frac{1}{\pi} \frac{K_{\mathrm{IC}}^{2}}{(0.5 \text { Y.S. })^{2}} \\
& =\frac{1}{\pi} \frac{(98 \mathrm{MPa} \sqrt{\mathrm{~m}})^{2}}{[0.5(1,460 \mathrm{MPa})]^{2}} \\
& =5.74 \times 10^{-3} \mathrm{~m} \\
& =5.74 \mathrm{~mm}
\end{aligned}
$$
|
5.74
|
mm
| null |
single
|
Example 8.2
| 111,003
| 1
|
easy
|
Materials: Metals
|
Metals
|
Mechanical
|
Composites
| null | null |
Fracture
|
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 8.3
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Given that a quality-control inspection can ensure that a structural ceramic part will have no flaws greater than $25 \mu \mathrm{~m}$ in size, calculate the maximum service stress available with (a) SiC and (b) partially stabilized zirconia.
Given K_IC(SiC)=3 MPa·√m and K_IC(PSZ)=9 MPa·√m.
Give your answer as a tuple (SiC, partially stabilized zirconia).
| null |
In lieu of more specific information, we can treat this problem as a general fracture mechanics problem using Equation ($K_{\mathrm{IC}}=Y \sigma_f \sqrt{p a}$) with $Y=1$, in which case
$$
\sigma_{f}=\frac{K_{\mathrm{IC}}}{\sqrt{\pi a}}
$$
This problem assumes that the maximum service stress will be the fracture stress for a part with flaw size $=a=25 \mu \mathrm{~m}$.
(a) For SiC,
$$
\sigma_{f}=\frac{3 \mathrm{MPa} \sqrt{\mathrm{~m}}}{\sqrt{\pi \times 25 \times 10^{-6} \mathrm{~m}}}=339 \mathrm{MPa}
$$
(b) For PSZ,
$$
\sigma_{f}=\frac{9 \mathrm{MPa} \sqrt{\mathrm{~m}}}{\sqrt{\pi \times 25 \times 10^{-6} \mathrm{~m}}}=1,020 \mathrm{MPa}
$$
|
(339, 1020)
|
(MPa, MPa)
| null |
multiple
|
Example 8.3
| 511,003
| 1
|
easy
|
Materials: Ceramics
|
Ceramics
|
Mechanical
|
Composites
| null | null |
Fracture
|
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 8.4
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Given only that the alloy for a structural member has a tensile strength of 800 MPa, estimate a maximum permissible service stress knowing that the loading will be cyclic in nature and that a safety factor of 2 is required. Use Figure 8.16 as a guide.
|
images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 8.4_1.jpg
|
If we use Figure 8.16 as a guide, a conservative estimate of the fatigue strength will be $$\text { F.S. }=\frac{1}{4} \text { T.S. }=\frac{1}{4}(800 \mathrm{MPa})=200 \mathrm{MPa} \text {. }$$ Using a safety factor of 2 will give a permissible service stress of $$\text { service stress }=\frac{\text { F.S. }}{2}=\frac{200 \mathrm{MPa}}{2}=100 \mathrm{MPa} \text {. }$$ Note. The safety factor helps to account for, among other things, the approximate nature of the relationship between F.S. and T.S.
|
100
|
MPa
| null |
single
|
Example 8.4
| 111,004
| 2
|
medium
|
Materials: Metals
|
Metals
|
Mechanical
|
Composites
| null | null |
Fatigue
|
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 8.5
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Static fatigue depends on a chemical reaction (Figure) and, as a result, is another example of Arrhenius behavior. Specifically, at a given load, the inverse of time to fracture has been shown to increase exponentially with temperature. The activation energy associated with the mechanism of Figure 8.19 is $78.6 \mathrm{~kJ} / \mathrm{mol}$. If the time to fracture for a soda-lime-silica glass is 1 second at $+50^{\circ} \mathrm{C}$ at a given load, what is the time to fracture at $-50^{\circ} \mathrm{C}$ at the same load?
|
images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 8.5_1.jpg
|
As stated, we can apply the Arrhenius equation (rate $=C e^{-Q / R T}$). In this case,
$$
t^{-1}=C e^{-Q / R T}
$$
where $t$ is the time to fracture.
At $50^{\circ} \mathrm{C}(323 \mathrm{~K})$,
$$
t^{-1}=1 \mathrm{~s}^{-1}=C e^{-\left(78.6 \times 10^{3} \mathrm{~J} / \mathrm{mol}\right) /\left[8.314 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{~K}\right](323 \mathrm{~K})}
$$
giving
$$
C=5.15 \times 10^{12} \mathrm{~s}^{-1}
$$
Then,
$$
\begin{aligned}
t_{-50^{\circ} \mathrm{C}}^{-1} & =\left(5.15 \times 10^{12} \mathrm{~s}^{-1}\right) e^{-\left(78.6 \times 10^{3} \mathrm{~J} / \mathrm{mol}\right) /\left[8.314 \mathrm{~J} /\left(\mathrm{mol}^{\circ} \mathrm{K}\right)\right](223 \mathrm{~K})} \\
& =1.99 \times 10^{-6} \mathrm{~s}^{-1}
\end{aligned}
$$
or
$$
\begin{aligned}
t & =5.0 \times 10^{5} \mathrm{~s} \\
& =5.0 \times 10^{5} \mathrm{~s} \times \frac{1 \mathrm{~h}}{3.6 \times 10^{3} \mathrm{~s}} \\
& =140 \mathrm{~h} \\
& =5 \text { days, } 20 \mathrm{~h} .
\end{aligned}
$$
|
140
|
h
| null |
single
|
Example 8.5
| 621,403
| 5
|
easy
|
Materials: Glasses
|
Glasses
|
Thermal
|
Composites
|
Diffusion & Kinetics
| null |
Fracture
|
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 8.6
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Calculate the fraction of x-ray beam intensity transmitted through a 10 -mm-thick plate of low-carbon steel. Take the beam energy to be 100 keV . Because of the small amount of carbon and its inherently low absorption of x-rays, the steel can be approximated as elemental iron.
Refer to Table:
\begin{tabular}{|l|l|l|}
\hline \multicolumn{3}{|l|}{Linear Absorption Coefficient of Various Elements for an X-Ray Beam with Energy $=100 \mathrm{keV}(=0.1 \mathrm{MeV})$} \\
\hline Element & Atomic number & $\mu\left(\mathrm{mm}^{-1}\right)$ \\
\hline Aluminum & 13 & 0.0459 \\
\hline Titanium & 22 & 0.124 \\
\hline Iron & 26 & 0.293 \\
\hline Nickel & 28 & 0.396 \\
\hline Copper & 29 & 0.410 \\
\hline Zinc & 30 & 0.356 \\
\hline Tungsten & 74 & 8.15a \\
\hline Lead & 82 & 6.20 \\
\hline
\end{tabular}
| null |
Using Equation ($I=I_0 e^{-\mu x}$) and the attenuation coefficient from Table,
$$
I=I_{0} e^{-\mu x}
$$
or
$$
\begin{aligned}
I / I_{0} & =e^{-\mu x} \\
& =e^{-\left(0.293 \mathrm{~mm}^{-1}\right)(10 \mathrm{~mm})} \\
& =e^{-2.93}=0.0534
\end{aligned}
$$
|
0.0534
| null | null |
single
|
Example 8.6
| 151,000
| 1
|
hard
|
Materials: Metals
|
Metals
|
Optical
|
Composites
| null | null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 9.1
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
At $200^{\circ} \mathrm{C}$, a $50: 50 \mathrm{~Pb}-\mathrm{Sn}$ solder alloy exists as two phases, a lead-rich solid and a tin-rich liquid. Calculate the degrees of freedom for this alloy and comment on its practical significance.
| null |
Using Equation 9.2 (i.e., assuming a constant pressure of 1 atm above the alloy), we obtain
$$
F=C-P+1
$$
There are two components ( Pb and Sn ) and two phases (solid and liquid), giving
$$
F=2-2+1=1
$$
As a practical matter, we may retain this two-phase microstructure upon heating or cooling. However, such a temperature change exhausts the "freedom" of the system and must be accompanied by changes in composition.
|
1
| null | null |
single
|
Example 9.1
| 125,300
| 1
|
medium
|
Materials: Metals
|
Metals
|
Thermal
|
Micro/Nano-structure
|
Phase Diagram
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 9.3
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
An alloy of overall composition x = 50 wt% B is slowly cooled to temperature T₁, at which the liquid‐solution composition is 18 wt% B and the solid‐solution composition is 66 wt% B (see the revised phase diagram). If you start with 1 kg of this alloy and equilibrate at T₁, calculate the masses of the liquid and solid phases present.
Give your answer as a tuple (amount of liquid phase, amount of solid phase).
| null |
Using Equations ($\frac{m_\alpha}{m_\alpha+m_\beta}=\frac{x_\beta-x}{x_\beta-x_\alpha}$) and ($\frac{m_\beta}{m_\alpha+m_\beta}=\frac{x-x_\alpha}{x_\beta-x_\alpha}$), we obtain
$$
\begin{aligned}
m_{\mathrm{L}} & =\frac{x_{\mathrm{SS}}-x}{x_{\mathrm{SS}}-x_{\mathrm{L}}}(1 \mathrm{~kg})=\frac{66-50}{66-18}(1 \mathrm{~kg}) \\
& =0.333 \mathrm{~kg}=333 \mathrm{~g}
\end{aligned}
$$
and
$$
\begin{aligned}
m_{\mathrm{SS}} & =\frac{x-x_{\mathrm{L}}}{x_{\mathrm{SS}}-x_{\mathrm{L}}}(1 \mathrm{~kg})=\frac{50-18}{66-18}(1 \mathrm{~kg}) \\
& =0.667 \mathrm{~kg}=667 \mathrm{~g} .
\end{aligned}
$$
|
(333, 667)
|
(g, g)
| null |
multiple
|
Example 9.3
| 121,300
| 2
|
easy
|
Materials: Metals
|
Metals
|
Thermal
|
Composites
|
Phase Diagram
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 9.4
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
For 1 kg of eutectoid steel at room temperature, calculate the amount of each phase $\left(\alpha\right.$ and $\left.\mathrm{Fe}_{3} \mathrm{C}\right)$ present. Give your answer as a tuple (amount of $\alpha$, amount of $\mathrm{Fe}_{3} \mathrm{C}$).
|
images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 9.4_1.jpg
|
Using Equations ($\frac{m_\alpha}{m_\alpha+m_\beta}=\frac{x_\beta-x}{x_\beta-x_\alpha}$) and ($\frac{m_\beta}{m_\alpha+m_\beta}=\frac{x-x_\alpha}{x_\beta-x_\alpha}$) and Figure 9.19, we have
$$
\begin{aligned}
m_{\alpha} & =\frac{x_{\mathrm{Fe}_{3} \mathrm{C}}-x}{x_{\mathrm{Fe}_{3} \mathrm{C}}-x_{\alpha}}(1 \mathrm{~kg})=\frac{6.69-0.77}{6.69-0}(1 \mathrm{~kg}) \\
& =0.885 \mathrm{~kg}=885 \mathrm{~g}
\end{aligned}
$$
and
$$
\begin{aligned}
m_{\mathrm{Fe}_{3} \mathrm{C}} & =\frac{x-x_{\alpha}}{x_{\mathrm{Fe}_{3} \mathrm{C}}-x_{\alpha}}(1 \mathrm{~kg})=\frac{0.77-0}{6.69-0}(1 \mathrm{~kg}) \\
& =0.115 \mathrm{~kg}=115 \mathrm{~g} .
\end{aligned}
$$
|
(885, 115)
|
(g, g)
| null |
multiple
|
Example 9.4
| 111,300
| 2
|
medium
|
Materials: Metals
|
Metals
|
Mechanical
|
Composites
|
Phase Diagram
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 9.5
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
A partially stabilized zirconia is composed of $4 \mathrm{wt} \% \mathrm{CaO}$. This product contains some monoclinic phase together with the cubic phase, which is the basis of fully stabilized zirconia. Estimate the mole percent of each phase present at room temperature. Give your answer as a tuple: (mol_percent_monoclinic, mol_percent_cubic).
Noting that $4 \mathrm{wt} \% \mathrm{CaO}=8 \mathrm{~mol} \% \mathrm{CaO}$ and assuming that the solubility limits shown in Figure 9.29 do not change significantly below $500^{\circ} \mathrm{C}$.
|
images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 9.5_1.jpg
|
Noting that $4 \mathrm{wt} \% \mathrm{CaO}=8 \mathrm{~mol} \% \mathrm{CaO}$ and assuming that the solubility limits shown in Figure 9.29 do not change significantly below $500^{\circ} \mathrm{C}$, we can use Equations ($\frac{m_\alpha}{m_\alpha+m_\beta}=\frac{x_\beta-x}{x_\beta-x_\alpha}$) and ($\frac{m_\beta}{m_\alpha+m_\beta}=\frac{x-x_\alpha}{x_\beta-x_\alpha}$):
$$
\begin{aligned}
\text { mol \% monoclinic } & =\frac{x_{\mathrm{cub}}-x}{x_{\mathrm{cub}}-x_{\mathrm{mono}}} \times 100 \% \\
& =\frac{15-8}{15-2} \times 100 \%=53.8 \mathrm{~mol} \%
\end{aligned}
$$
and
$$
\begin{aligned}
\mathrm{mol} \% \text { cubic } & =\frac{x-x_{\text {mono }}}{x_{\text {cub }}-x_{\text {mono }}} \times 100 \% \\
& =\frac{8-2}{15-2} \times 100 \%=46.2 \mathrm{~mol} \%
\end{aligned}
$$
|
(53.8, 46.2)
|
(mol%, mol%)
| null |
multiple
|
Example 9.5
| 521,200
| 1
|
medium
|
Materials: Ceramics
|
Ceramics
|
Thermal
|
Composites
|
Crystal Structure
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 9.6
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Figure shows the microstructural development for an $80 \mathrm{wt} \% \mathrm{~B}$ alloy. Consider instead 1 kg of a $70 \mathrm{wt} \% \mathrm{~B}$ alloy.
(a) Calculate the amount of $\beta$ phase at $T_{3}$.
(b) Calculate what weight fraction of this $\beta$ phase at $T_{3}$ is proeutectic.
Give your answer as a tuple: (beta_phase_amount, proeutectic_fraction).
|
images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 9.6_1.jpg
|
(a) Using Equation ($\frac{m_\beta}{m_\alpha+m_\beta}=\frac{x-x_\alpha}{x_\beta-x_\alpha}$.) gives us
$$
\begin{aligned}
m_{\beta, T_j} & =\frac{x-x_\alpha}{x_\beta-x_\alpha}(1 \mathrm{~kg})=\frac{70-30}{90-30}(1 \mathrm{~kg}) \\
& =0.667 \mathrm{~kg}=667 \mathrm{~g}
\end{aligned}
$$
(b) The proeutectic $\beta$ was that which was present in the microstructure at $T_2$ :
$$
\begin{aligned}
m_{\beta, T_2} & =\frac{x-x_{\mathrm{L}}}{x_\beta-x_{\mathrm{L}}}(1 \mathrm{~kg})=\frac{70-60}{90-60}(1 \mathrm{~kg}) \\
& =0.333 \mathrm{~kg}=333 \mathrm{~g}
\end{aligned}
$$
This portion of the microstructure is retained upon cooling through the eutectic temperature, giving
$$
\begin{aligned}
\text { fraction proeutectic } & =\frac{\text { proeutectic } \beta}{\operatorname{total} \beta} \\
& =\frac{333 \mathrm{~g}}{667 \mathrm{~g}}=0.50
\end{aligned}
$$
|
(0.667, 0.50)
|
(kg,)
| null |
multiple
|
Example 9.6
| 125,300
| 6
|
easy
|
Materials: Metals
|
Metals
|
Thermal
|
Micro/Nano-structure
|
Phase Diagram
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 9.7
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
For 1 kg of $0.5 \mathrm{wt} \% \mathrm{C}$ steel, given at T = 727 °C (≈εutectoid): Cγ≈0.77 wt %, Cα≈0.02 wt %, calculate the amount of proeutectoid $\alpha$ at the grain boundaries.
|
images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 9.7_1.jpg,images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 9.7_2.jpg
|
Using Figure 9.40 for illustration and Figure 9.19 for calculation, we essentially need to calculate the equilibrium amount of $\alpha$ at $728^{\circ} \mathrm{C}$ (i.e., 1 degree above the eutectoid temperature). Using Equation ($\frac{m_\alpha}{m_\alpha+m_\beta}=\frac{x_\beta-x}{x_\beta-x_\alpha}$), we have
$$
\begin{aligned}
m_{\alpha} & =\frac{x_{\gamma}-x}{x_{\gamma}-x_{\alpha}}(1 \mathrm{~kg})=\frac{0.77-0.50}{0.77-0.02}(1 \mathrm{~kg}) \\
& =0.360 \mathrm{~kg}=360 \mathrm{~g} .
\end{aligned}
$$
Note. You might have noticed that this calculation near the eutectoid composition used a value of $x_{\alpha}$ representative of the maximum solubility of carbon in $\alpha-\mathrm{Fe}(0.02 \mathrm{wt} \%)$. At room temperature, this solubility goes to nearly zero.
|
0.360
|
kg
| null |
single
|
Example 9.7
| 125,300
| 1
|
medium
|
Materials: Metals
|
Metals
|
Thermal
|
Micro/Nano-structure
|
Phase Diagram
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 9.8
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
For 1 kg of $3 \mathrm{wt} \% \mathrm{C}$ gray iron, calculate the amount of graphite flakes present in the microstructure (a) at $1,153^{\circ} \mathrm{C}$ and (b) at room temperature. Give your answer as a tuple (amount at $1,153^{\circ} \mathrm{C}$, amount at room temperature).
|
images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 9.8_1.jpg,images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 9.8_2.jpg
|
(a) Using Figures 9.20 and 9.41, we note that $1,153^{\circ} \mathrm{C}$ is just below the eutectic temperature. Using Equation ($\frac{m_\beta}{m_\alpha+m_\beta}=\frac{x-x_\alpha}{x_\beta-x_\alpha}$.) gives us
$$
\begin{aligned}
m_{\mathrm{C}} & =\frac{x-x_{\mathrm{y}}}{x_{\mathrm{C}}-x_{\mathrm{y}}}(1 \mathrm{~kg})=\frac{3.00-2.08}{100-2.08}(1 \mathrm{~kg}) \\
& =0.00940 \mathrm{~kg}=9.40 \mathrm{~g}
\end{aligned}
$$
(b) At room temperature, we obtain
$$
\begin{aligned}
m_{\mathrm{C}} & =\frac{x-x_{\alpha}}{x_{\mathrm{C}}-x_{\alpha}}(1 \mathrm{~kg})=\frac{3.00-0}{100-0}(1 \mathrm{~kg}) \\
& =0.030 \mathrm{~kg}=30.0 \mathrm{~g}
\end{aligned}
$$
|
(9.40, 30.0)
|
(g, g)
| null |
multiple
|
Example 9.8
| 125,300
| 2
|
easy
|
Materials: Metals
|
Metals
|
Thermal
|
Micro/Nano-structure
|
Phase Diagram
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 9.9
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Consider 1 kg of an aluminum casting alloy with $10 \mathrm{wt} \%$ silicon. (a) Upon cooling, at what temperature would the first solid appear? (b) What is the first solid phase, and what is its composition? (c) At what temperature will the alloy completely solidify? (d) How much proeutectic phase will be found in the microstructure? (e) How is the silicon distributed in the microstructure at $576^{\circ} \mathrm{C}$? Give your answer as a tuple (first solid appearance temperature, first solid phase composition wt%, complete solidification temperature, proeutectic phase weight in g, mass of silicon in proeutectic alpha in g, mass of silicon in eutectic alpha in g, mass of silicon in eutectic beta in g).
We follow this microstructural development with the aid of Figure 9.13. The following values are needed from the phase diagram: liquidus temperature for 10 wt% Si is 595°C, the composition of the alpha phase at the liquidus is 1 wt% Si, the eutectic temperature is 577°C, the eutectic composition is 12.6 wt% Si, the composition of the alpha phase at the eutectic is 1.6 wt% Si, and the composition of the beta phase is 100 wt% Si.
|
images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 9.9_1.jpg
|
(a) For this composition, the liquidus is at $\sim 595^{\circ} \mathrm{C}$.
(b) It is solid solution $\alpha$ with a composition of $\sim 1 \mathrm{wt} \% \mathrm{Si}$.
(c) At the eutectic temperature, $577^{\circ} \mathrm{C}$.
(d) Practically all of the proeutectic $\alpha$ will have developed by $578^{\circ} \mathrm{C}$. Using Equation ($\frac{m_\alpha}{m_\alpha+m_\beta}=\frac{x_\beta-x}{x_\beta-x_\alpha}$), we obtain
$$
\begin{aligned}
m_{\alpha} & =\frac{x_{\mathrm{L}}-x}{x_{\mathrm{L}}-x_{\alpha}}(1 \mathrm{~kg})=\frac{12.6-10}{12.6-1.6}(1 \mathrm{~kg}) \\
& =0.236 \mathrm{~kg}=236 \mathrm{~g}
\end{aligned}
$$
(e) At $576^{\circ} \mathrm{C}$, the overall microstructure is $\alpha+\beta$. The amounts of each are
$$
\begin{aligned}
m_{\alpha} & =\frac{x_{\beta}-x}{x_{\beta}-x_{\alpha}}(1 \mathrm{~kg})=\frac{100-10}{100-1.6}(1 \mathrm{~kg}) \\
& =0.915 \mathrm{~kg}=915 \mathrm{~g}
\end{aligned}
$$
and
$$
\begin{aligned}
m_{\beta} & =\frac{x-x_{\alpha}}{x_{\beta}-x_{\alpha}}(1 \mathrm{~kg})=\frac{10-1.6}{100-1.6}(1 \mathrm{~kg}) \\
& =0.085 \mathrm{~kg}=85 \mathrm{~g} .
\end{aligned}
$$
However, we found in (d) that 236 g of the $\alpha$ is in the form of relatively large grains of proeutectic phase, giving
$$
\begin{aligned}
\alpha_{\text {eutectic }} & =\alpha_{\text {total }}-\alpha_{\text {proeutectic }} \\
& =915 \mathrm{~g}-236 \mathrm{~g}=679 \mathrm{~g}
\end{aligned}
$$
The silicon distribution is then given by multiplying its weight fraction in each microstructural region by the amount of that region:
$$
\begin{aligned}
& \text { Si in proeutectic } \alpha=(0.016)(236 \mathrm{~g})=3.8 \mathrm{~g} \\
& \text { Si in eutectic } \alpha=(0.016)(679 \mathrm{~g})=10.9 \mathrm{~g}
\end{aligned}
$$
and
$$
\text { Si in eutectic } \beta=(1.000)(85 \mathrm{~g})=85.0 \mathrm{~g}
$$
Finally, note that the total mass of silicon in the three regions sums to 99.7 g rather than $100.0 \mathrm{~g}(=10 \mathrm{wt} \%$ of the total alloy) due to roundoff errors.
|
(595, 1, 577, 236, 3.8, 10.9, 85.0)
|
(°C, wt%, °C, g, g, g, g)
| null |
multiple
|
Example 9.9
| 125,320
| 7
|
easy
|
Materials: Metals
|
Metals
|
Thermal
|
Micro/Nano-structure
|
Phase Diagram
|
Shaping
| null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 9.10
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
The solubility of copper in aluminum drops to nearly zero at $100^{\circ} \mathrm{C}$. What is the maximum amount of $\theta$ phase that will precipitate out in a $4.5 \mathrm{wt} \%$ copper alloy quenched and aged at $100^{\circ} \mathrm{C}$ ? Express your answer in weight percent.
|
images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 9.10_1.png
|
As indicated in Figure, the solubility limit of the $\theta$ phase is essentially unchanging with temperature below $\sim 400^{\circ} \mathrm{C}$ and is near a composition of $53 \mathrm{wt} \%$ copper. Using Equation ($\frac{m_\beta}{m_\alpha+m_\beta}=\frac{x-x_\alpha}{x_\beta-x_\alpha}$) gives us
$$
\begin{aligned}
\mathrm{wt} \% \theta & =\frac{x-x_{\kappa}}{x_{\theta}-x_{\kappa}} \times 100 \%=\frac{4.5-0}{53-0} \times 100 \% \\
& =8.49 \%
\end{aligned}
$$
|
8.49
|
%
| null |
single
|
Example 9.10
| 125,300
| 1
|
easy
|
Materials: Metals
|
Metals
|
Thermal
|
Micro/Nano-structure
|
Phase Diagram
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 9.11
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
For a 50:50 $\mathrm{Pb}-\mathrm{Sn}$ solder at a temperature of $200^{\circ} \mathrm{C}$, determine (i) the phases present, (ii) their compositions, and (iii) their relative amounts (expressed in weight percent). Give your answer as a tuple (phases present, compositions, relative amounts in weight percent).
|
images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 9.11_1.jpg
|
(a) Using Figure 9.16, we find the following results at $200^{\circ} \mathrm{C}$: i. The phases are $\alpha$ and liquid. ii. The composition of $\alpha$ is $\sim 18 \mathrm{wt} \% \mathrm{Sn}$ and of L is $\sim 54 \mathrm{wt} \% \mathrm{Sn}$. iii. Using Equations ($\frac{m_\alpha}{m_\alpha+m_\beta}=\frac{x_\beta-x}{x_\beta-x_\alpha}$) and ($\frac{m_\beta}{m_\alpha+m_\beta}=\frac{x-x_\alpha}{x_\beta-x_\alpha}$), we have $$\begin{aligned} \mathrm{wt} \% \alpha & =\frac{x_{\mathrm{L}}-x}{x_{\mathrm{L}}-x_{\alpha}} \times 100 \%=\frac{54-50}{54-18} \times 100 \% \\ & =11.1 \% \end{aligned}$$ and $$\begin{aligned} \mathrm{wt} \% \mathrm{~L} & =\frac{x-x_{\alpha}}{x_{\mathrm{L}}-x_{\alpha}} \times 100 \%=\frac{50-18}{54-18} \times 100 \% \\ & =88.9 \% \end{aligned}$$ (b) Similarly, at $100^{\circ} \mathrm{C}$, we obtain i. $\alpha$ and $\beta$. ii. $\alpha$ is $\sim 5 \mathrm{wt} \% \mathrm{Sn}$ and $\beta$ is $\sim 99 \mathrm{wt} \% \mathrm{Sn}$. iii. wt $\% \alpha=\frac{x_{\beta}-x}{x_{\beta}-x_{\alpha}} \times 100 \%=\frac{99-50}{99-5} \times 100 \%=52.1 \%$ and $$\mathrm{wt} \% \beta=\frac{x-x_{\alpha}}{x_{\beta}-x_{\alpha}} \times 100 \%=\frac{50-5}{99-5} \times 100 \%=47.9 \%$$
|
(11.1, 88.9, 52.1, 47.9)
|
(%, %, %, %)
| null |
multiple
|
Example 9.11
| 21,000
| 2
|
medium
|
Properties: Thermal
| null |
Thermal
|
Composites
| null | null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 9.12
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
A fireclay refractory ceramic can be made by heating the raw material kaolinite, $\mathrm{Al}_{2}\left(\mathrm{Si}_{2} \mathrm{O}_{5}\right)(\mathrm{OH})_{4}$, thus driving off the waters of hydration. Determine the phases present, their compositions, and their amounts for the resulting microstructure (below the eutectic temperature). Give your answer as a tuple (Amount of SiO₂ in mol%, Amount of Mullite in mol%).
|
images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 9.12_1.jpg
|
A modest rearrangement of the kaolinite formula helps to clarify the production of this ceramic product:
$$\mathrm{Al}_{2}\left(\mathrm{Si}_{2} \mathrm{O}_{5}\right)(\mathrm{OH})_{4}=\mathrm{Al}_{2} \mathrm{O}_{3} \cdot 2 \mathrm{SiO}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}$$
The firing operation yields
$$\mathrm{Al}_{2} \mathrm{O}_{3} \cdot 2 \mathrm{SiO}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O} \xrightarrow{\text { heat }} \mathrm{Al}_{2} \mathrm{O}_{3} \cdot 2 \mathrm{SiO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \uparrow$$
The remaining solid, then, has an overall composition of
$$\begin{aligned}
\operatorname{mol} \% \mathrm{Al}_{2} \mathrm{O}_{3} & =\frac{\operatorname{mol} \mathrm{Al}_{2} \mathrm{O}_{3}}{\operatorname{mol} \mathrm{Al}_{2} \mathrm{O}_{3}+\operatorname{molSiO}_{2}} \times 100 \% \\
& =\frac{1}{1+2} \times 100 \%=33.3 \%
\end{aligned}$$
Using Figure 9.23, we see that the overall composition falls in the $\mathrm{SiO}_{2}+$ mullite two-phase region below the eutectic temperature. The $\mathrm{SiO}_{2}$ composition is $0 \mathrm{~mol} \% \mathrm{Al}_{2} \mathrm{O}_{3}$ (i.e., $100 \% \mathrm{SiO}_{2}$). The composition of mullite is $60 \mathrm{~mol} \% \mathrm{Al}_{2} \mathrm{O}_{3}$.
Using Equations $\frac{m_\alpha}{m_\alpha+m_\beta}=\frac{x_\beta-x}{x_\beta-x_\alpha}$ and ($\frac{m_\beta}{m_\alpha+m_\beta}=\frac{x-x_\alpha}{x_\beta-x_\alpha}$) yields
$$\begin{aligned}
\operatorname{mol} \% \mathrm{SiO}_{2} & =\frac{x_{\text {mullite }}-x}{x_{\text {mullite }}-x_{\mathrm{SiO}_{2}}} \times 100 \%=\frac{60-33.3}{60-0} \times 100 \% \\
& =44.5 \mathrm{~mol} \%
\end{aligned}$$
and
$$\begin{aligned}
\operatorname{mol} \% \text { mullite } & =\frac{x-x_{\mathrm{SiO}_{2}}}{x_{\text {mullite }}-x_{\mathrm{SiO}_{2}}} \times 100 \%=\frac{33.3-0}{60-0} \times 100 \% \\
& =55.5 \mathrm{~mol} \%
\end{aligned}$$
|
(44.5, 55.5)
|
(mol %, mol %)
| null |
multiple
|
Example 9.12
| 525,300
| 5
|
hard
|
Materials: Ceramics
|
Ceramics
|
Thermal
|
Micro/Nano-structure
|
Phase Diagram
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 10.1
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
At $900^{\circ} \mathrm{C}$, growth rate, $\dot{G}$, is a dominant term in the crystallization of a copper alloy. By dropping the system temperature to $400^{\circ} \mathrm{C}$, the growth rate drops six orders of magnitude and effectively reduces the crystallization rate to zero. Calculate the activation energy for selfdiffusion in this alloy system.
| null |
The following is a direct application of Equation ($\dot{G}=C e^{-Q / R T}$):
$$
\dot{G}=C e^{-Q / R T}
$$
Considering two different temperatures yields
$$
\begin{aligned}
\frac{\dot{G}_{900^{\circ} \mathrm{C}}}{\dot{G}_{400^{\circ} \mathrm{C}}} & =\frac{C e^{-Q / R(900+273) \mathrm{K}}}{C e^{-Q / R(400+273) \mathrm{K}}} \\
& =e^{-Q / R(1 / 1,173-1 / 673) \mathrm{K}^{-1}}
\end{aligned}
$$
which gives
$$
\begin{aligned}
Q & =-\frac{R \ln \left(\dot{G}_{900^{\circ} \mathrm{C}} / \dot{G}_{400^{\circ} \mathrm{C}}\right)}{(1 / 1,173-1 / 673) \mathrm{K}^{-1}} \\
& =-\frac{[8.314 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{~K})] \ln 10^6}{(1 / 1,173-1 / 673) \mathrm{K}^{-1}}=181 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}
$$
|
181
|
kJ/mol
| null |
single
|
Example 10.1
| 121,400
| 1
|
hard
|
Materials: Metals
|
Metals
|
Thermal
|
Composites
|
Diffusion & Kinetics
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 10.2
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
(a) How much time is required for austenite to transform to $50 \%$ pearlite at $600^{\circ} \mathrm{C}$ ? (b) How much time is required for austenite to transform to $50 \%$ bainite at $300^{\circ} \mathrm{C}$ ? Give your answer as a tuple (time for pearlite transformation, time for bainite transformation).
|
images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 10.2_1.png
|
(a) This problem is a direct application of Figure 10.7. The dotted line denotes the halfway point in the $\gamma \rightarrow \alpha+\mathrm{Fe}_{3} \mathrm{C}$ transformation. At $600^{\circ} \mathrm{C}$, the time to reach that line is $\sim 3 \frac{1}{2} \mathrm{~s}$.
(b) At $300^{\circ} \mathrm{C}$, the time is $\sim 480 \mathrm{~s}$, or 8 min.
|
(3.5, 480)
|
(s, s)
| null |
multiple
|
Example 10.2
| 105,000
| 2
|
medium
|
Materials: Metals
|
Metals
| null |
Micro/Nano-structure
| null | null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 10.3
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
FORMULA
|
(a) Calculate the microstructure of a 0.77 wt $\% \mathrm{C}$ steel that has the following heat treatment: (i) instantly quenched from the $\gamma$ region to $500^{\circ} \mathrm{C}$, (ii) held for 5 s , and (iii) quenched instantly to $250^{\circ} \mathrm{C}$. (b) What will happen if the resulting microstructure is held for 1 day at $250^{\circ} \mathrm{C}$ and then cooled to room temperature? (c) What will happen if the resulting microstructure from part (a) is quenched directly to room temperature? Give your answer as a tuple (microstructure after part (a), microstructure after part (b), microstructure after part (c)).
|
images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 10.3_1.jpg,images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 10.3_2.jpg
|
(a) By having ideally fast quenches, we can solve this problem precisely in terms of Figure 10.7. The first two parts of the heat treatment lead to $\sim 70 \%$ transformation to fine pearlite. The final quench will retain this state: $$ 30 \% \gamma+70 \% \text { fine pearlite }\left(\alpha+\mathrm{Fe}_{3} \mathrm{C}\right) $$ (b) The pearlite remains stable, but the retained $\gamma$ will have time to transform to bainite, giving a final state: $$ 30 \% \text { bainite }\left(\alpha+\mathrm{Fe}_{3} \mathrm{C}\right)+70 \% \text { fine pearlite }\left(\alpha+\mathrm{Fe}_{3} \mathrm{C}\right) $$ (c) Again, the pearlite remains stable, but most of the retained $\gamma$ will become unstable. For this case, we must consider the martensitic transformation data in Figure 10.11. The resulting microstructure will be $$ 70 \% \text { fine pearlite }\left(\alpha+\mathrm{Fe}_{3} \mathrm{C}\right)+\sim 30 \% \text { martensite. } $$ (Because the martensitic transformation is not complete until $\sim 46^{\circ} \mathrm{C}$, a small amount of untransformed $\gamma$ will remain at room temperature.)
|
(30 \% \gamma + 70 \% \text{ fine pearlite }(\alpha + \mathrm{Fe}_{3} \mathrm{C}), 30 \% \text{ bainite }(\alpha + \mathrm{Fe}_{3} \mathrm{C}) + 70 \% \text{ fine pearlite }(\alpha + \mathrm{Fe}_{3} \mathrm{C}), 70 \% \text{ fine pearlite }(\alpha + \mathrm{Fe}_{3} \mathrm{C}) + \sim 30 \% \text{ martensite})
| null | null |
multiple
|
Example 10.3
| 125,300
| 3
|
easy
|
Materials: Metals
|
Metals
|
Thermal
|
Micro/Nano-structure
|
Phase Diagram
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 10.4
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Estimate the quench rate needed to avoid pearlite formation in (a) $0.5 \mathrm{wt} \% \mathrm{C}$ steel, (b) $0.77 \mathrm{wt} \% \mathrm{C}$ steel, and (c) $1.13 \mathrm{wt} \% \mathrm{C}$ steel. Give your answer as a tuple (quench rate for 0.5 wt% C steel, quench rate for 0.77 wt% C steel, quench rate for 1.13 wt% C steel).
Reference:
From Figure 10.16 for a $0.5 \mathrm{wt} \% \mathrm{C}$ steel, we must quench from the austenite boundary $\left(770^{\circ} \mathrm{C}\right)$ to $\sim 520^{\circ} \mathrm{C}$ in $\sim 0.6 \mathrm{~s}$. From Figure 10.11 for a $0.77 \mathrm{wt} \% \mathrm{C}$ steel, we quench from the eutectoid temperature $\left(727^{\circ} \mathrm{C}\right)$ to $\sim 550^{\circ} \mathrm{C}$ in $\sim 0.7 \mathrm{~s}$. From Figure 10.15 for a $1.13 \mathrm{wt} \% \mathrm{C}$ steel, we quench from the austenite boundary $\left(880^{\circ} \mathrm{C}\right)$ to $\sim 550^{\circ} \mathrm{C}$ in $\sim 0.35 \mathrm{~s}$.
|
images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 10.4_1.jpg,images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 10.4_2.jpg,images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 10.4_3.jpg
|
In each case, we are looking at the rate of temperature drop needed to avoid the pearlite "knee":\n(a) From Figure 10.16 for a $0.5 \mathrm{wt} \% \mathrm{C}$ steel, we must quench from the austenite boundary $\left(770^{\circ} \mathrm{C}\right)$ to $\sim 520^{\circ} \mathrm{C}$ in $\sim 0.6 \mathrm{~s}$, giving\n$$\frac{\Delta T}{t}=\frac{(770-520)^{\circ} \mathrm{C}}{0.6 \mathrm{~s}}=420^{\circ} \mathrm{C} / \mathrm{s}$$\n(b) From Figure 10.11 for a $0.77 \mathrm{wt} \% \mathrm{C}$ steel, we quench from the eutectoid temperature $\left(727^{\circ} \mathrm{C}\right)$ to $\sim 550^{\circ} \mathrm{C}$ in $\sim 0.7 \mathrm{~s}$, giving\n$$\frac{\Delta T}{t}=\frac{(727-550)^{\circ} \mathrm{C}}{0.7 \mathrm{~s}}=250^{\circ} \mathrm{C} / \mathrm{s}$$\n(c) From Figure 10.15 for a 1.13 wt $\% \mathrm{C}$ steel, we quench from the austenite boundary $\left(880^{\circ} \mathrm{C}\right)$ to $\sim 550^{\circ} \mathrm{C}$ in $\sim 0.35 \mathrm{~s}$, giving\n$$\frac{\Delta T}{t}=\frac{(880-550)^{\circ} \mathrm{C}}{0.35 \mathrm{~s}}=940^{\circ} \mathrm{C} / \mathrm{s}$$
|
(420, 250, 940)
|
(°C/s, °C/s, °C/s)
| null |
multiple
|
Example 10.4
| 125,300
| 3
|
easy
|
Materials: Metals
|
Metals
|
Thermal
|
Micro/Nano-structure
|
Phase Diagram
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 10.8
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
(a) Calculate the amount of $\theta$ phase that would precipitate at the grain boundaries in the equilibrium microstructure shown in Figure 10.25.
(b) What is the maximum amount of Guinier-Preston zones to be expected in a $4.5 \mathrm{wt} \% \mathrm{Cu}$ alloy.
Give your answer as a tuple: (theta_phase_wt_percent, guinier_preston_zones_wt_percent).
|
images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 10.8_1.jpg,images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 10.8_2.jpg
|
(a) Using the $\mathrm{Al}-\mathrm{Cu}$ phase diagram (Figure 9.27) and Equation ($\frac{m_\beta}{m_\alpha+m_\beta}=\frac{x-x_\alpha}{x_\beta-x_\alpha}$), we obtain
$$
\text { wt } \begin{aligned}
\% \theta & =\frac{x-x_\kappa}{x_\theta-x_\kappa} \times 100 \%=\frac{4.5-0}{53-0} \times 100 \% \\
& =8.49 \%
\end{aligned}
$$
(b) As the G.P. zones are precursors to the equilibrium precipitation, the maximum amount would be $8.49 \%$.
|
(8.49,8.49)
|
(%, %)
| null |
multiple
|
Example 10.8
| 125,300
| 1
|
hard
|
Materials: Metals
|
Metals
|
Thermal
|
Micro/Nano-structure
|
Phase Diagram
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 10.10
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Calculate the maximum amount of monoclinic phase you would expect to find in the microstructure of a PSZ with an overall composition of $3.4 \mathrm{wt} \% \mathrm{CaO}$. An alloy with 3.4 wt% CaO corresponds to approximately 7 mol% CaO. In the CaO-ZrO2 system, at room temperature, the monoclinic phase contains approximately 2 mol% CaO and the cubic phase contains approximately 15 mol% CaO.
|
images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 10.10_1.jpg
|
This is another example of the important relationship between phase diagrams and kinetics. We can make this prediction by calculating the equilibrium concentration of monoclinic phase at room temperature in the $\mathrm{CaO}-\mathrm{ZrO}_{2}$ system (Figure 9.29). A composition of $3.4 \mathrm{wt} \%$ CaO is approximately $7 \mathrm{~mol} \% \mathrm{CaO}$ (as shown by the upper and lower composition scales in Figure 9.29). Extrapolating the phase boundaries in the monoclinic + cubic two-phase region to room temperature gives a monoclinic phase composition of $\sim 2 \mathrm{~mol} \% \mathrm{CaO}$ and a cubic phase composition of $\sim 15 \mathrm{~mol} \% \mathrm{CaO}$. Using Equation ($\frac{m_\alpha}{m_\alpha+m_\beta}=\frac{x_\beta-x}{x_\beta-x_\alpha}$), we get
$$
\begin{aligned}
\operatorname{mol} \% \text { monoclinic } & =\frac{x_{\text {cubic }}-x}{x_{\text {cubic }}-x_{\text {mono }}} \times 100 \% \\
& =\frac{15-7}{15-2} \times 100 \% \\
& =62 \mathrm{~mol} \%
\end{aligned}
$$
Note. We have made this calculation in terms of mole percent because of the nature of the plot in Figure 9.29. It would be a straightforward matter to convert the results to weight percent.
|
62
|
mol %
| null |
single
|
Example 10.10
| 125,200
| 1
|
easy
|
Materials: Metals
|
Metals
|
Thermal
|
Micro/Nano-structure
|
Crystal Structure
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 11.1
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
In redesigning an automobile for a new model year, 25 kg of conventional steel parts are replaced by aluminum alloys of the same dimensions. Calculate the resulting mass savings for the new model, approximating the alloy densities by those for pure Fe and Al , respectively. Using $\rho_{\mathrm{Fe}}=7.87 \mathrm{Mg} / \mathrm{m}^{3}$ and $\rho_{\mathrm{Al}}=2.70 \mathrm{Mg} / \mathrm{m}^{3}$.
| null |
Using $\rho_{\mathrm{Fe}}=7.87 \mathrm{Mg} / \mathrm{m}^{3}$ and $\rho_{\mathrm{Al}}=2.70 \mathrm{Mg} / \mathrm{m}^{3}$. The volume of steel parts replaced would be
$$
V=\frac{m_{\mathrm{Fe}}}{\rho_{\mathrm{Fe}}}=\frac{25 \mathrm{~kg}}{7.87 \mathrm{Mg} / \mathrm{m}^{3}} \times \frac{1 \mathrm{Mg}}{10^{3} \mathrm{~kg}}=3.21 \times 10^{-3} \mathrm{~m}^{3}
$$
The mass of new aluminum parts would be
$$
m_{\mathrm{Al}}=\rho_{\mathrm{Al}} V_{\mathrm{Al}}=2.70 \mathrm{Mg} / \mathrm{m}^3 \times 3.21 \times 10^{-3} \mathrm{~m}^3 \times \frac{10^3 \mathrm{~kg}}{1 \mathrm{Mg}}=8.65 \mathrm{~kg}
$$
The resulting mass savings is then
$$
m_{\mathrm{Fe}}-m_{\mathrm{Al}}=25 \mathrm{~kg}-8.65 \mathrm{~kg}=16.3 \mathrm{~kg}
$$
|
16.3
|
kg
| null |
single
|
Example 11.1
| 101,000
| 3
|
easy
|
Materials: Metals
|
Metals
| null |
Composites
| null | null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 11.2
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
For $1,000 \mathrm{~kg}$ of container glass $\left(15 \mathrm{wt} \% \mathrm{Na}_{2} \mathrm{O}, 10 \mathrm{wt} \% \mathrm{CaO}, 75 \mathrm{wt} \% \mathrm{SiO}_{2}\right)$, what is the raw material batch formula (weight percent of $\mathrm{Na}_{2} \mathrm{CO}_{3}, \mathrm{CaCO}_{3}$, and $\mathrm{SiO}_{2}$ )? Give your answer as a tuple (weight percent of $\mathrm{Na}_{2} \mathrm{CO}_{3}$, weight percent of $\mathrm{CaCO}_{3}$, weight percent of $\mathrm{SiO}_{2}$).
Assume the following atomic weights (in amu): Na = 22.99, C = 12.00, O = 16.00, Ca = 40.08.
| null |
$1,000 \mathrm{~kg}$ of glass consists of 150 kg of $\mathrm{Na}_{2} \mathrm{O}, 100 \mathrm{~kg}$ of CaO , and 750 kg of $\mathrm{SiO}_{2}$. From given data, $$\begin{aligned} \operatorname{mol} \mathrm{wt} \mathrm{Na}_{2} \mathrm{O} & =2(22.99)+16.00 \\ & =61.98 \mathrm{amu} \\ \operatorname{mol} \mathrm{wt} \mathrm{Na}_{2} \mathrm{CO}_{3} & =2(22.99)+12.00+3(16.00) \\ & =105.98 \mathrm{amu} \\ \operatorname{mol} \mathrm{wt} \mathrm{CaO} & =40.08+16.00 \\ & =56.08 \mathrm{amu} \end{aligned}$$ and $$\begin{aligned} \operatorname{mol} \mathrm{wt} \mathrm{CaCO}_{3} & =40.08+12.00+3(16.00) \\ & =100.08 \mathrm{amu} \\ \mathrm{Na}_{2} \mathrm{CO}_{3} \text { required } & =150 \mathrm{~kg} \times \frac{105.98}{61.98}=256 \mathrm{~kg} \\ \mathrm{CaCO}_{3} \text { required } & =100 \mathrm{~kg} \times \frac{100.08}{56.08}=178 \mathrm{~kg} \end{aligned}$$ and $$\mathrm{SiO}_{2} \text { required }=750 \mathrm{~kg}$$ The batch formula is $$\begin{aligned} & \frac{256 \mathrm{~kg}}{(256+178+750) \mathrm{kg}} \times 100=21.6 \mathrm{wt} \% \mathrm{Na}_{2} \mathrm{CO}_{3} \\ & \frac{178 \mathrm{~kg}}{(256+178+750) \mathrm{kg}} \times 100=15.0 \mathrm{wt} \% \mathrm{CaCO}_{3} \end{aligned}$$ and $$\frac{750 \mathrm{~kg}}{(256+178+750) \mathrm{kg}} \times 100=63.3 \mathrm{wt} \% \mathrm{SiO}_{2}$$
|
(21.6, 15.0, 63.3)
|
(wt%, wt%, wt%)
| null |
multiple
|
Example 11.2
| 601,000
| 7
|
easy
|
Materials: Glasses
|
Glasses
| null |
Composites
| null | null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 11.4
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
In firing 5 kg of kaolinite, $\mathrm{Al}_{2}\left(\mathrm{Si}_{2} \mathrm{O}_{5}\right)(\mathrm{OH})_{4}$, in a laboratory furnace to produce an aluminosilicate ceramic, how much $\mathrm{H}_{2} \mathrm{O}$ is driven off?
| null |
To determine the amount of $\mathrm{H}_{2} \mathrm{O}$ driven off during the firing of kaolinite, we first need to calculate the molar mass of kaolinite and the molar mass of the water that is released. The chemical formula of kaolinite is $\mathrm{Al}_{2}\left(\mathrm{Si}_{2} \mathrm{O}_{5}\right)(\mathrm{OH})_{4}$, which can be simplified to $\mathrm{Al}_{2}\mathrm{Si}_{2}\mathrm{O}_{5}(\mathrm{OH})_{4}$.
The molar mass of kaolinite is calculated as follows:
\[ \text{Molar mass of } \mathrm{Al}_{2}\mathrm{Si}_{2}\mathrm{O}_{5}(\mathrm{OH})_{4} = (2 \times 26.98) + (2 \times 28.09) + (5 \times 16.00) + (4 \times (1.01 + 16.00)) = 258.16 \text{ g/mol} \]
The molar mass of the water ($\mathrm{H}_{2} \mathrm{O}$) that is driven off is:
\[ \text{Molar mass of } \mathrm{H}_{2} \mathrm{O} = 2 \times 1.01 + 16.00 = 18.02 \text{ g/mol} \]
From the chemical formula, we can see that for every mole of kaolinite, 2 moles of water are released. Therefore, the mass of water released per mole of kaolinite is:
\[ 2 \times 18.02 = 36.04 \text{ g} \]
Now, to find out how much water is driven off from 5 kg of kaolinite, we first convert the mass of kaolinite to moles:
\[ \text{Moles of kaolinite} = \frac{5000 \text{ g}}{258.16 \text{ g/mol}} \approx 19.37 \text{ moles} \]
Then, the mass of water driven off is:
\[ 19.37 \text{ moles} \times 36.04 \text{ g/mole} \approx 698.0 \text{ g} \]
Therefore, approximately 698.0 g of $\mathrm{H}_{2} \mathrm{O}$ is driven off when firing 5 kg of kaolinite.
|
698.0
|
g
| null |
single
|
Example 11.4
| 501,000
| 7
|
medium
|
Materials: Ceramics
|
Ceramics
| null |
Composites
| null | null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 11.5
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
In firing 5 kg of kaolinite, $\mathrm{Al}_{2}\left(\mathrm{Si}_{2} \mathrm{O}_{5}\right)(\mathrm{OH})_{4}$, in a laboratory furnace to produce an aluminosilicate ceramic, how much $\mathrm{H}_{2} \mathrm{O}$ is driven off?
| null |
Note that
$$
\mathrm{Al}_{2}\left(\mathrm{Si}_{2} \mathrm{O}_{5}\right)(\mathrm{OH})_{4}=\mathrm{Al}_{2} \mathrm{O}_{3} \cdot 2 \mathrm{SiO}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}
$$
and
$$
\mathrm{Al}_{2} \mathrm{O}_{3} \cdot 2 \mathrm{SiO}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O} \xrightarrow{\text { heat }} \mathrm{Al}_{2} \mathrm{O}_{3} \cdot 2 \mathrm{SiO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \uparrow
$$
Then,
$$
\begin{aligned}
1 \mathrm{~mol} \mathrm{Al}_{2} \mathrm{O}_{3} \cdot 2 \mathrm{SiO}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}= & {[2(26.98)+3(16.00)] \mathrm{amu} } \\
& +2[28.09+2(16.00)] \mathrm{amu} \\
& +2[2(1.008)+16.00] \mathrm{amu} \\
= & 258.2 \mathrm{amu},
\end{aligned}
$$
and
$$
2 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}=2[2(1.008)+16.00] \mathrm{amu}=36.03 \mathrm{amu} .
$$
As a result, the mass of $\mathrm{H}_{2} \mathrm{O}$ driven off will be
$$
\mathrm{m}_{\mathrm{H}_{2} \mathrm{O}}=\frac{36.03 \mathrm{amu}}{258.2 \mathrm{amu}} \times 5 \mathrm{~kg}=0.698 \mathrm{~kg}=698 \mathrm{~g}
$$
|
698
|
g
| null |
single
|
Example 11.5
| 521,000
| 5
|
hard
|
Materials: Ceramics
|
Ceramics
|
Thermal
|
Composites
| null | null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 11.6
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Assume a glass bottle is formed at 680 °C with a viscosity of 10^7 P. The activation energy for viscous deformation is Q = 460 kJ/mol and the pre‐exponential viscosity factor is η₀ = 6.11×10^–19 P. Calculate the annealing temperature range, defined as the temperatures at which the viscosity lies between 10^12.5 P and 10^13.5 P.
Give your answer as a tuple (lower temperature, upper temperature).
| null |
We have an application for Equation ($\eta=\eta_0 e^{+Q / R T}$):
$$
\eta=\eta_{0} e^{+Q / R T}
$$
For $680^{\circ} \mathrm{C}=953 \mathrm{~K}$,
$$
10^{7} \mathrm{P}=\eta_{0} e^{+\left(460 \times 10^{3} \mathrm{~J} / \mathrm{mol}\right) /\left[8.314 \mathrm{~J} /\left(\mathrm{mol}^{\prime} \mathrm{K}\right)\right](953 \mathrm{~K})}
$$
or
$$
\eta_{0}=6.11 \times 10^{-19} \mathrm{P}
$$
For the annealing range, $\eta=10^{12.5}$ to $10^{13.5} \mathrm{P}$. For $\eta=10^{12.5} \mathrm{P}$,
$$
\begin{aligned}
\mathrm{T} & =\frac{460 \times 10^{3} \mathrm{~J} / \mathrm{mol}}{[8.314 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{~K})] \ln \left(10^{12.5} / 6.11 \times 10^{-19}\right)} \\
& =782 \mathrm{~K}=509^{\circ} \mathrm{C}
\end{aligned}
$$
For $\eta=10^{13.5} \mathrm{P}$,
$$
\begin{aligned}
\mathrm{T} & =\frac{460 \times 10^{3} \mathrm{~J} / \mathrm{mol}}{[8.314 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{~K})] \ln \left(10^{13.5} / 6.11 \times 10^{-19}\right)} \\
& =758 \mathrm{~K}=485^{\circ} \mathrm{C}
\end{aligned}
$$
Therefore,
$$
\text { annealing range }=485^{\circ} \mathrm{C} \text { to } 509^{\circ} \mathrm{C}
$$
|
(485, 509)
|
(^{\circ}C, ^{\circ}C)
| null |
multiple
|
Example 11.6
| 611,400
| 7
|
easy
|
Materials: Glasses
|
Glasses
|
Mechanical
|
Composites
|
Diffusion & Kinetics
| null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 11.25
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
We can use the $\mathrm{Al}-\mathrm{Si}$ phase diagram (Figure) to illustrate the principle of zone refining. Assuming that we have a bar of silicon with aluminum as its only impurity, (a) calculate the segregation coefficient, $K$, in the Si-rich region and (b) calculate the purity of a $99 \mathrm{wt} \%$ Si bar after a single pass of the molten zone. (Note that the solidus can be taken as a straight line between a composition of $99.985 \mathrm{wt} \% \mathrm{Si}$ at $1,190^{\circ} \mathrm{C}$ and $100 \mathrm{wt} \% \mathrm{Si}$ at $1,414^{\circ} \mathrm{C}$.)
Give your answer as a tuple: (segregation_coefficient, purity_silicon).
|
images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 11.25_1.jpg
|
(a) Close inspection of Figure indicates that the liquidus curve crosses the $90 \% \mathrm{Si}$ composition line at a temperature of $1,360^{\circ} \mathrm{C}$. The solidus line can be expressed in the form $y=m x+b$ with $y$ being the temperature and $x$ being the silicon composition (in wt \%). For the conditions stated, $1,190=m(99.985)+b$ and $1,414=m(100)+b$. Solving the equations gives $m=1.493 \times 10^{4}$ and $b=-1.492 \times 10^{6}$. At $1,360^{\circ} \mathrm{C}$ (where the liquidus composition is $90 \% \mathrm{Si}$ ), the solid composition is given by $1,360=1.493 \times 10^{4} x-1.492 \times 10^{6}$ or $x=\frac{1,360+1.492 \times 10^{6}}{1.493 \times 10^{4}}=99.99638$. The segregation coefficient is calculated in terms of impurity levels; that is, $c_{s}=100-99.99638=0.00362 \mathrm{wt} \% \mathrm{Al}$ and $c_{l}=100-90=10 \mathrm{wt} \% \mathrm{Al}$ yielding $K=\frac{c_{s}}{c_{l}}=\frac{0.00362}{10}=3.62 \times 10^{-4}$. (b) For the liquidus line, a similar straight-line expression takes on the values $1,360=m(90)+b$ and $1,414=m(100)+b$ yielding $m=5.40$ and $b=874$. A 99 wt \% Si bar will have a liquidus temperature $T=5.40(99)+874=1,408.6^{\circ} \mathrm{C}$. The corresponding solidus composition is given by $1,408.6=1.493 \times 10^{4} x-1.492 \times 10^{6}$ or $x=\frac{1,408.6+1.492 \times 10^{6}}{4,924}=99.999638 \mathrm{wt} \% \mathrm{Si}$. An alternate composition expression is $\frac{(100-99.999638) \% \mathrm{Al}}{100 \%}=3.62 \times 10^{-6} \mathrm{Al}$ or 3.62 parts per million Al. Note. These calculations are susceptible to round-off errors. Values for $m$ and $b$ in the solidus-line equation must be carried to several places.
|
(3.62e-4, 99.999638)
|
(, wt%)
| null |
multiple
|
Example 11.25
| 121,302
| 1
|
medium
|
Materials: Metals
|
Metals
|
Thermal
|
Composites
|
Phase Diagram
| null |
Plastic
|
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 12.1
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
How much $\mathrm{H}_{2} \mathrm{O}_{2}$ must be added to ethylene to yield an average degree of polymerization of 750 ? Assume that all $\mathrm{H}_{2} \mathrm{O}_{2}$ dissociates to OH groups that serve as terminals for the molecules, and express the answer in weight percent.
Assume the following atomic weights (in g/mol or amu):
H = 1.008
O = 16.00
C = 12.01
|
images/Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 12.1_1.jpeg
|
Referring to Figure, we note that there is one $\mathrm{H}_{2} \mathrm{O}_{2}$ molecule ( $=$ two OH groups) per polyethylene molecule. Thus,
$$
\mathrm{wt} \% \mathrm{H}_{2} \mathrm{O}_{2}=\frac{\mathrm{mol} \mathrm{wt} \mathrm{H}_{2} \mathrm{O}_{2}}{750 \times\left(\mathrm{mol} \mathrm{wt} \mathrm{C}_{2} \mathrm{H}_{4}\right)} \times 100
$$
Using the given data yields
$$
\begin{aligned}
\mathrm{wt} \% \mathrm{H}_{2} \mathrm{O}_{2} & =\frac{2(1.008)+2(16.00)}{750[2(12.01)+4(1.008)]} \times 100 \\
& =0.162 \mathrm{wt} \%
\end{aligned}
$$
|
0.162
|
wt %
| null |
single
|
Example 12.1
| 201,000
| 1
|
hard
|
Materials: Polymers
|
Polymers
| null |
Composites
| null | null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 12.2
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
Calculate the molecular weight of a polyacetal molecule with a degree of polymerization of 500.
Assume the following atomic weights (in g/mol or amu):
C = 12.01
H = 1.008
O = 16.00
| null |
The molecular weight of a polyacetal molecule with a degree of polymerization of 500 is calculated using the formula: $$\text{mol wt}\left(\mathrm{CH}_{2} \mathrm{O}\right)_{n}=n\left(\text{mol wt } \mathrm{CH}_{2} \mathrm{O}\right)$$ Using the data, we obtain $$\begin{aligned} \operatorname{mol} \text{ wt }\left(\mathrm{CH}_{2} \mathrm{O}\right)_{n} & =500[12.01+2(1.008)+16.00] \mathrm{amu} \\ & =15,010 \mathrm{amu} \end{aligned}$$
|
15010
|
amu
| null |
single
|
Example 12.2
| 201,000
| 1
|
easy
|
Materials: Polymers
|
Polymers
| null |
Composites
| null | null | null |
Introduction_to_Materials_Science_for_Engineers_Shackelford_Example 12.3
|
Introduction_to_Materials_Science_for_Engineers_Shackelford
|
NUM
|
A sample of polyethylene has an average degree of polymerization of 750. (a) What is the coiled length? (b) Determine the extended length of an average molecule. Give your answer as a tuple: (coiled_length, extended_length).
Refer to Table:
\begin{tabular}{|l|l|l|l|}
\hline \multicolumn{4}{|l|}{Bond Energies and Bond Lengths for Representative Covalent Bonds} \\
\hline \multirow[b]{2}{*}{Bond} & \multicolumn{2}{|c|}{Bond energy ${ }^{\text {a }}$} & \multirow{2}{*}{Bond length, nm} \\
\hline & kcal/mol & kJ/mol & \\
\hline $\mathrm{C}-\mathrm{C}$ & $88{ }^{\text {b }}$ & 370 & 0.154 \\
\hline $\mathrm{C}=\mathrm{C}$ & 162 & 680 & 0.130 \\
\hline $\mathrm{C} \equiv \mathrm{C}$ & 213 & 890 & 0.120 \\
\hline $\mathrm{C}-\mathrm{H}$ & 104 & 435 & 0.110 \\
\hline $\mathrm{C}-\mathrm{N}$ & 73 & 305 & 0.150 \\
\hline $\mathrm{C}-\mathrm{O}$ & 86 & 360 & 0.140 \\
\hline $\mathrm{C}=\mathrm{O}$ & 128 & 535 & 0.120 \\
\hline C-F & 108 & 450 & 0.140 \\
\hline $\mathrm{C}-\mathrm{Cl}$ & 81 & 340 & 0.180 \\
\hline $\mathrm{O}-\mathrm{H}$ & 119 & 500 & 0.100 \\
\hline O-O & 52 & 220 & 0.150 \\
\hline $\mathrm{O}-\mathrm{Si}$ & 90 & 375 & 0.160 \\
\hline $\mathrm{N}-\mathrm{H}$ & 103 & 430 & 0.100 \\
\hline $\mathrm{N}-\mathrm{O}$ & 60 & 250 & 0.120 \\
\hline F-F & 38 & 160 & 0.140 \\
\hline $\mathrm{H}-\mathrm{H}$ & 104 & 435 & 0.074 \\
\hline
\end{tabular}
| null |
(a):
Using Equations ($\bar{L}=l \sqrt{m}$) and ($m=2 n$), we have
$$\bar{L}=l \sqrt{2 n}$$
From Table , $l=0.154 \mathrm{~nm}$, giving
$$\begin{aligned}
\bar{L} & =(0.154 \mathrm{~nm}) \sqrt{2(750)} \\
& =5.96 \mathrm{~nm}
\end{aligned}$$
(b):
Using Equations ($L_{\mathrm{ext}}=m l \sin \frac{109.5^{\circ}}{2}$) and ($m=2 n$,), we obtain
$$\begin{aligned}
L_{\text{ext}} & =2 n l \sin \frac{109.5^{\circ}}{2} \\
& =2(750)(0.154 \mathrm{~nm}) \sin \frac{109.5^{\circ}}{2} \\
& =189 \mathrm{~nm}
\end{aligned}$$
|
(5.96, 189)
|
(nm, nm)
| null |
multiple
|
Example 12.3
| 201,000
| 2
|
easy
|
Materials: Polymers
|
Polymers
| null |
Composites
| null | null | null |
End of preview. Expand
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MatSciBench Dataset
MatSciBench is a comprehensive benchmark dataset for materials science question answering. The dataset contains questions spanning various domains within materials science, including polymer science, crystallography, thermodynamics, and more.
Dataset Structure
The dataset is provided as a CSV file (qa.csv) with the following columns:
qid: Question identifiersource: Source of the questiondomain: Scientific domaintype: Question type (NUM for numerical, etc.)question: The question textimage: Associated image file (if any)solution: Step-by-step solutionanswer: Final answerunit: Units for numerical answersnotes: Additional notesnumber_of_answers: Number of possible answersoriginal_qid: Original question ID from sourcecategory_vector: Categorical classificationsteps_count: Number of solution stepsdifficulty_level: Difficulty ratingprimary_category: Main category classification- Additional category columns for Materials, Properties, Structures, etc.
Usage
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- Materials science question answering research
- Benchmarking AI models on scientific reasoning
- Educational applications in materials science
Citation
If you use this dataset in your research, please cite the original MatSciBench paper.
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