3.40 Sketch within a cubic unit cell the following planes ...

3.40 Sketch within a cubic unit cell the following planes:

(a) (01 1 )

(b) (112 )

(c) (102 )

(d) (13 1)

Solution

The planes called for are plotted in the cubic unit cells shown below.

3.41 Determine the Miller indices for the planes shown in the following unit cell:

Solution

For plane A we will leave the origin at the unit cell as shown; this is a (403) plane, as summarized below.

Intercepts

Intercepts in terms of a, b, and c

x

y

z

a

¡Þb

2c

2

1

2

¡Þ

Reciprocals of intercepts

2

0

Reduction

4

0

Enclosure

3

2

3

3

2

3

(403)

For plane B we will move the origin of the unit cell one unit cell distance to the right along the y axis, and

one unit cell distance parallel to the x axis; thus, this is a (1 1 2) plane, as summarized below.

Intercepts

x

y

¨Ca

¨Cb

z

c

2

Intercepts in terms of a, b, and c

¨C1

¨C1

Reciprocals of intercepts

¨C1

¨C1

Reduction

(not necessary)

Enclosure

(1 1 2)

1

2

2

4.4 Below, atomic radius, crystal structure, electronegativity, and the most common valence are tabulated, for

several elements; for those that are nonmetals, only atomic radii are indicated.

Cu

Atomic Radius

(nm)

0.1278

C

0.071

H

0.046

O

0.060

Ag

0.1445

FCC

1.9

+1

Al

0.1431

FCC

1.5

+3

Co

0.1253

HCP

1.8

+2

Cr

0.1249

BCC

1.6

+3

Fe

0.1241

BCC

1.8

+2

Ni

0.1246

FCC

1.8

+2

Pd

0.1376

FCC

2.2

+2

Pt

0.1387

FCC

2.2

+2

Zn

0.1332

HCP

1.6

+2

Element

Crystal Structure

Electronegativity

Valence

FCC

1.9

+2

Which of these elements would you expect to form the following with copper:

(a) A substitutional solid solution having complete solubility

(b) A substitutional solid solution of incomplete solubility

(c) An interstitial solid solution

Solution

In this problem we are asked to cite which of the elements listed form with Cu the three possible solid

solution types. For complete substitutional solubility the following criteria must be met: 1) the difference in atomic

radii between Cu and the other element (?R%) must be less than ¡À15%, 2) the crystal structures must be the same,

3) the electronegativities must be similar, and 4) the valences should be the same, or nearly the same. Below are

tabulated, for the various elements, these criteria.

Element

?R%

Cu

C

H

O

Ag

¨C44

¨C64

¨C53

+13

Crystal

Structure

?Electronegativity

FCC

FCC

Valence

2+

0

1+

Al

Co

Cr

Fe

Ni

Pd

Pt

Zn

+12

-2

-2

-3

-3

+8

+9

+4

FCC

HCP

BCC

BCC

FCC

FCC

FCC

HCP

-0.4

-0.1

-0.3

-0.1

-0.1

+0.3

+0.3

-0.3

3+

2+

3+

2+

2+

2+

2+

2+

(a) Ni, Pd, and Pt meet all of the criteria and thus form substitutional solid solutions having complete

solubility. At elevated temperatures Co and Fe experience allotropic transformations to the FCC crystal structure,

and thus display complete solid solubility at these temperatures.

(b) Ag, Al, Co, Cr, Fe, and Zn form substitutional solid solutions of incomplete solubility. All these metals

have either BCC or HCP crystal structures, and/or the difference between their atomic radii and that for Cu are

greater than ¡À15%, and/or have a valence different than 2+.

(c) C, H, and O form interstitial solid solutions. These elements have atomic radii that are significantly

smaller than the atomic radius of Cu.

4.9 Calculate the composition, in weight percent, of an alloy that contains 218.0 kg titanium, 14.6 kg of aluminum,

and 9.7 kg of vanadium.

Solution

The concentration, in weight percent, of an element in an alloy may be computed using a modified form of

Equation 4.3. For this alloy, the concentration of titanium (CTi) is just

CTi =

=

mTi

¡Á 100

mTi + mAl + mV

218 kg

¡Á 100 = 89.97 wt%

218 kg + 14.6 kg + 9.7 kg

Similarly, for aluminum

C Al =

And for vanadium

14.6 kg

¡Á 100 = 6.03 wt%

218 kg + 14.6 kg + 9.7 kg

CV =

9.7 kg

¡Á 100 = 4.00 wt%

218 kg + 14.6 kg + 9.7 kg

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