CHAPTER 6 – ELEMENTS OF GRAIN BOUNDARIES
6.1
(a) Given a small angle tilt boundary, whose angle of tilt of tilt is 0.1 deg., find the spacing between dislocations in the boundary if the if the Burgers vector of the of the dislocations is 0.33 nm. (b) On the assumption that the dislocations conform to the conditions involved in Eq. 4.20, that
⁄2, 8.6 10 and 0.3
, determine an approximate value for the surface
⁄ and ⁄
energy of the of the tilt boundary. Give your answer in
.
Solution:
(a) The angle of tilt of tilt in a small angle tilt boundary,
,
is equal to
b⁄
where b is the Burgers
vector and d is the spacing between the dislocations in the boundary. According to the statement of this of this problem, we are given:
Accordingly,
0.33 3.3 1010 0.1 0180.1 1.75 10 3. 1.3.7351010 1.89 10
(b) Equation 4.20 is for edge dislocations:
41 4 ⁄2 . 9.4545 10, 0.3, 8.6 10 , and 3.3 10. Therefore: 3. 49. 8. 6 10 3 . 3 10 49 . 4 5 10 3.3 10 41 0.3 10 7.5 10 ⁄ Where
is the energy per unit length of one of one of the of the dislocation. However, in a square meter of the of the
boundary there will be 1/d dislocations of this of this length, so that the energy per unit area of the of the
boundary, , will be equal to
1⁄ ⁄.
Therefore,
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7. 5 10 1.89 10 0.0397 ⁄
0.039710 ⁄10 ⁄ 39.7 ⁄ 6.2
According to quantitative metallography,
,
the average number of grain‐boundary intercepts
per unit length of a line laid over a microstructure, is directly related to unit volume, by the relation:
(a) Determine the value of photograph is 350X.
, the surface area per
2 for the microstructure in Fig. 6.1, if the magnification of the
⁄
(b) Assuming that the grain boundary energy of zirconium is about 1
, what would be the
grain‐boundary energy per unit volume in ⁄ of the zirconium in the specimen?
Solution:
(a) The average number of intercepts by a 10 cm long line is approximately 10 or 1/cm. Because of the 350X magnification, this is equivalent to an actual value of grain boundary area per unit volume is
350/.
Thus, the actual
2350 700 / 7 10/ 7 10 ⁄ 0.07 ⁄
However, since each square meter of the surface area is assumed to have an energy of this is equivalent to an energy per unit volume of
6.3
1 ⁄
,
A very fine grained material may have a mean grain intercept of the order of 1 micron or
10. ⁄ .⁄
Assuming the grain boundary energy of the metal is
0.8⁄
, what would be the
approximate value of its grain‐boundary energy per unit volume? Give you answer in both
and
.
Solution:
Given
we may write:
2 2 10 ⁄ © 2009. Cengage Learning, Engineering. All Rights Reserved. 58
and we are given:
0.8 ⁄ Therefore, the total surface energy per unit volume, W, is:
0.82 10 1.6 10 1.6 4.1.1685 0.38⁄ 6.4
(a) Consider Fig. 6.8. If the grain‐boundary energy of a boundary between two iron crystals is
⁄
0.78
angle
, while that between iron and a second phase particle were to be 0.40
should occur at the junction?
⁄
, what
(b) If the surface energy between the iron and the second particle were to be 0.35, what would the angle be? Solution:
(a) This problem considers the case where a boundary between two crystals meets a grain of another phase at a common intersection as illustrated in Fig. 6.17. The equation that corresponds to this case when the boundaries are in static equilibrium is Eq. 6.15. This equation is :
Where
2 cos⁄2 is the surface energy of the boundary between the two iron crystals,
surface energy between an iron crystal and the second phase particle, and
2cos 2 2cos 0.0.7880 25.67° 2cos 0.0.7780 0°
angle at the juncture. Solving the above equation for
(b) If
0.35
, then:
leads to:
is the
is the included
In this case, the second phase would form a thin film lying between the two iron crystals.
6.5
The following data were taken from Jones, R.L. and Conrad, H., TMS‐ AIME , 245 779 (1969), and
give the flow stress , at 4 percent strain, as a function of the gain size of a very high purity
titanium metal. Make a plot of versus parameters k and
. Express k in
⁄
⁄ ⁄
, and from this determine the Hall‐Petch
.
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Solution:
(a)
(b) The straight line drawn through the data points has an intercept on the stress axis of approximately 161 MPa. This makes
161
. The slope of this line, k , may be evaluated
⁄ 321 , 193 √
in terms of the coordinates at its end points. At the right, the coordinates are
⁄ √ . 953⁄ 189⁄ and
.
Accordingly:
and
0.168 ⁄ 0.168 10 ⁄
The Hall Petch equation is:
Where
,and at the left,
is in MPa and d in
.
⁄ 161 0.168 ⁄ ⁄. © 2009. Cengage Learning, Engineering. All Rights Reserved. 60
6.6
Plot the data of the preceding problem, showing
as a function of
⁄
, as well as of
. Do
these plots indicate that there is some justification for Baldwin’s comments on the Hall‐Petch relationship? Solution:
,⁄ The plots in this figure certainly cast a resonable doubt on the Hall‐Petch relationship as a basic law relating the stress to the grain size.
6.7
(a) With regard to the coincident site twist boundary on a {111} fcc metal plane shown in Fig. 6.25, show that using the Ranganathan relationships (Eqs. 6.21, 6.22 and 6.23), that a choice of x = 2 and y = 1 will also give this
Σ 7
boundary.
(b) To what angle of twist does x =2 and y = 1 correspond? (c) Is the fact that x = 2 and y = 1 are able to produce an equivalent coincident site boundary to that in Fig. 6.25 related to the symmetry of the atomic arrangement on the {111} plane? Explain.
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Solution:
(a) The Ranganathan relationship is Eg. 6.22 states that:
Σ Where:
1 1 1 1 3 With x = 2 and y = 1 this yields:
(b) The angle
Σ 2 1 3 7 is determined with the aid of Eq. 6.21
⁄ 2tan 2tan 1 √ 3 2 40.9° 81.8°
2
°
The {111} plane in a fcc metal has a six‐fold symmetry. If we subtract 60 from this value of
°,
°
we obtain 21.8 which in round number equals the 22 angle of rotation of Fig. 6.25.
6.8
,
This problem concerns a coincidence site boundary on a {210} plane of a cubic lattice. Note that the determination does not depend on whether the lattice is simple, face‐centered, or body‐ centered cubic. It holds for all three cases. The basic cell in this plane has x = a and y = coincidence site lattice can be formed with x = 2 and y =1. Assuming a twist boundary,
√ 5
a. A
determine: (a) The twist angle (b)
Σ
for the coincidence structure.
the reciprocal of the density of coincidence sites.
(c) Check you answer for
Σ, and
using two drawings of the atomic structure as revealed on a
{210} plane. Note that this involves an array of rectangular cells, in which x = a and y =
√ 5
a,
where a is the lattice constant of the crystal. In this operation, note that the one drawing of the structure should be made on tracing paper so that it can be place on and rotated over the other so as reveal the coincidence sites.
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Solution:
(a) (b)
2tan ⁄ 2 tan √ 5 96.4° Σ 2 12 12 1 0 4 5 9
(c) The final solution for this part of the problem is shown in the following figure. It corresponds to the arrangement that should be obtained when the two drawings are properly oriented over each other.
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© 2009. Cengage Learning, Engineering. All Rights Reserved. 64
