CHAPTER 7

DISLOCATIONS AND STRENGTHENING MECHANISMS

PROBLEM SOLUTIONS

7.2 Consider two edge dislocations of opposite sign and having slip planes that are separated by several atomic

distances, as indicated in the following diagram. Briefly describe the defect that results when these two dislocations

become aligned with each other.

Answer

When the two edge dislocations become aligned, a planar region of vacancies will exist between the

dislocations as:

7.6 (a) Compare planar densities (Section 3.11 and Problem 3.60) for the (100), (110), and (111) planes for FCC.

(b) Compare planar densities (Problem 3.61) for the (100), (110), and (111) planes for BCC.

Solution

(a) For the FCC crystal structure, the planar density for the (110) plane is given in Equation 3.19 as

Furthermore, the planar densities of the (100) and (111) planes are calculated in Homework Problem 3.60,

which are as follows:

(b) For the BCC crystal structure, the planar densities of the (100) and (110) planes were determined in

Homework Problem 3.61, which are as follows:

Below is a BCC unit cell, within which is shown a (111) plane.

(a)

The centers of the three corner atoms, denoted by A, B, and C lie on this plane. Furthermore, the (111) plane does

not pass through the center of atom D, which is located at the unit cell center. The atomic packing of this plane is

presented in the following figure; the corresponding atom positions from the Figure (a) are also noted.

(b)

Inasmuch as this plane does not pass through the center of atom D, it is not included in the atom count. One sixth of

each of the three atoms labeled A, B, and C is associated with this plane, which gives an equivalence of one-half

atom.

In Figure (b) the triangle with A, B, and C at its corners is an equilateral triangle. And, from Figure (b), the

area of this triangle is . The triangle edge length, x, is equal to the length of a face diagonal, as indicated in

Figure (a). And its length is related to the unit cell edge length, a, as

or

For BCC, (Equation 3.4), and, therefore,

Also, from Figure (b), with respect to the length y we may write

which leads to . And, substitution for the above expression for x yields

Thus, the area of this triangle is equal to

And, finally, the planar density for this (111) plane is

7.12 Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at

angles of 60° and 35°, respectively, with the tensile axis. If the critical resolved shear stress is 6.2 MPa (900 psi),

will an applied stress of 12 MPa (1750 psi) cause the single crystal to yield? If not, what stress will be necessary?

Solution

This problem calls for us to determine whether or not a metal single crystal having a specific orientation

and of given critical resolved shear stress will yield. We are given that = 60, = 35, and that the values of the

critical resolved shear stress and applied tensile stress are 6.2 MPa (900 psi) and 12 MPa (1750 psi), respectively.

From Equation 7.2

Since the resolved shear stress (4.91 MPa) is less that the critical resolved shear stress (6.2 MPa), the single crystal

will not yield.

However, from Equation 7.4, the stress at which yielding occurs is

7.15 A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied

parallel to the [100] direction. If the critical resolved shear stress for this material is 0.5 MPa, calculate the

magnitude(s) of applied stress(es) necessary to cause slip to occur on the (111) plane in each of the , ,

and directions.

Solution

In order to solve this problem it is necessary to employ Equation 7.4, but first we need to solve for the for

and angles for the three slip systems.

For each of these three slip systems, the will be the same—i.e., the angle between the direction of the

applied stress, [100] and the normal to the (111) plane, that is, the [111] direction. The angle may be determined

using Equation 7.6 as

where (for [100]) u1 = 1, v1 = 0, w1 = 0, and (for [111]) u2 = 1, v2 = 1, w2 = 1. Therefore, is equal to

Let us now determine for the slip direction. Again, using Equation 7.6 where u1 = 1, v1 = 0, w1 = 0 (for

[100]), and u2 = 1, v2 = –1, w2 = 0 (for ). Therefore, is determined as

Now, we solve for the yield strength for this (111)– slip system using Equation 7.4 as

Now, we must determine the value of for the (111)– slip system—that is, the angle between the

[100] and directions. Again using Equation 7.6

λ[ 100 ]−[ 10 {1̄ ]=cos−1 [ (1)(1 )+(0 )(0)+(0 )(−1 )

√ [(1 )2+(0 )2+(0 )2] [(1)2+(0 )2+(−1 )2] ]¿

Thus, since the values of and for this (111)– slip system are the same as for (111)– , so also will y be

the same—viz 1.22 MPa.

And, finally, for the (111)– slip system, is computed using Equation 7.6 as follows:

λ[ 100 ]−[ 01̄ 1]=cos−1 [ (1)(0 )+( 0)(−1)+(0)(1 )

√ [(1 )2+(0 )2+(0 )2] [(0 )2+(−1 )2+(1 )2] ]=cos−1 (0)=90 °

Thus, from Equation 7.4, the yield strength for this slip system is

= 0 .5 MPacos(54 . 7 ° )cos (90 ° )

= 0 .5 MPa(0 .578)( 0)

=¥

which means that slip will not occur on this (111)– slip system.

7.19 The critical resolved shear stress for copper (Cu) is 0.48 MPa (70 psi). Determine the maximum possible yield

strength for a single crystal of Cu pulled in tension.

Solution

In order to determine the maximum possible yield strength for a single crystal of Cu pulled in tension, we

simply employ Equation 7.5 as

7.23 Describe in your own words the three strengthening mechanisms discussed in this chapter (i.e., grain size

reduction, solid-solution strengthening, and strain hardening). Explain how dislocations are involved in each of the

strengthening techniques.

Solution

These three strengthening mechanisms are described in Sections 7.8, 7.9, and 7.10.

7.28 (a) Show, for a tensile test, that

if there is no change in specimen volume during the deformation process (i.e., A0l0 = Adld).

(b) Using the result of part (a), compute the percent cold work experienced by naval brass (for which the

stress–strain behavior is shown in Figure 6.12) when a stress of 415 MPa (60,000 psi) is applied.

Solution

(a) We are asked to show, for a tensile test, that

From Equation 7.8

The following relationship

may be rearranged to read as follows:

Substitution of the right-hand-side of this expression into the above equation for %CW leads to

Now, from the definition of engineering strain (Equation 6.2)

Or, upon rearrangement

Substitution of this expression for l0/ld into the %CW expression above gives

the desired equation

(b) From Figure 6.12, a stress of 415 MPa (60,000 psi) corresponds to a strain of 0.15. Using the above

expression

7.37 (a) What is the driving force for recrystallization?

(b) What is the driving force for grain growth?

Solution

(a) The driving force for recrystallization is the difference in internal energy between the strained and

unstrained material.

(b) The driving force for grain growth is the reduction in grain boundary energy as the total grain boundary

area decreases.

7.41 The average grain diameter for a brass material was measured as a function of time at 650°C, which is shown

in the following table at two different times:

Time (min) Grain Diameter (mm)

40 5.6 × 10–2

100 8.0 × 10–2

(a) What was the original grain diameter?

(b) What grain diameter would you predict after 200 min at 650°C?

Solution

(a) Using the data given and Equation 7.9 (taking n = 2)—that is

we may set up two simultaneous equations with d0 and K as unknowns, as follows:

Solution of these expressions yields a value for d0, the original grain diameter, of

d0 = 0.031 mm,

and a value for K of

(b) At 200 min, the diameter d is computed using a rearranged form of Equation 7.9 (incorporating values

of d0 and K that were just determined) as follows: