Office: Dearborn Hall 303E

Phone and Voice Mail: (541) 737-7016

FAX (541) 737-2600

This page last updated April 12, 2010

SOLUTIONS POSTED DATE: Monday April 12, 2010.

1) In class we discussed how two edge dislocations
of opposite sign on the same slip plane will annihilate each other,
leaving no crystal defect and a perfect crystal behind. Look at the case
in the figure to the right. The two edge dislocations are on parallel
slip planes a few units cells apart, but they overlap in the vertical
direction. When the two dislocations come together and are aligned,
they do not annihilate. Explain why not and what defect is left behind.
Making a sketch may be helpful to illustrate your description. |

a) FCC (110).

b) BCC (110)

c) For each plane, calculate the planar packing fraction (area occupied by atoms on plane / area of plane) in percent.

3) A single crystal of a metal that has the FCC crystal structure is oriented
such that a tensile stress is applied parallel to the [110] direction. If the
critical resolved shear stress for this material is 1.75 [MPa], calculate
the magnitude(s) of applied stress(es) necessary to cause slip to occur on
the (111) plane in each of the [1 -1 0], [1 0 -1], and [0 1 -1] directions.

4) In a copper wiring drawing process, a 150mm diameter billet of Cu is drawn
down to a final wire diameter of 2mm.

a) What is the ductility measured in terms of area reduction?

b) If the original billet is 2 meters long, how long is the final wire?

c) What is is final engineering strain of the wire?

d) What is the final true strain of the wire?

5) Table 7.1 lists the slip systems for Cu (an FCC metal)
as {111}<1 -1 0>,
and says that there are 12 slip systems for FCC.

a) Why does slip in FCC Cu happen on the combination of planes and directions given as {111}<1 -1 0>?

b) What does it mean to say that there are 12 slip systems for FCC? Explain with words and sketches.

SOLUTIONS

1) *...they do not annihilate. Explain
why not..*.

If the two (opposite sign) dislocations were
on the same slip plane, they would come together and annihilate,
leaving a perfect crystal. In this case, when they overlap, there
will be a small extra set of atoms left in the region between the
two slip planes. If the slip planes are separated by only a single
unit cell, this would essentially be one extra atom, which is essentially
an interstitial. Since the dislocations are line defects, the interstitial
atom will actually be a If the slip planes of the two dislocations are farther apart, they will leave a "strip" of extra atoms along the line of the dislocations. Eventually the extra atoms in this strip will diffuse away, becoming distributed interstitials or annihilating with vacancies, thereby lowering the total strain energy in the crystal. |

2) *Show with a drawing the atom positions...*

2a) |
2c) The planar packing fraction (PPF) is defined as the area of the atoms on the plane divided by the total area of the plane. For the FCC (110) plane shown to the left we can use the red area to make our calculation. The total number of atom cross sections in the red area is 2 atoms (4 quarter-atom pieces at the corners and 2 half-atom pieces on the edges). This gives a total atom area of The area of the red plane is found from the dimensions of the rectangle. The rectangle is the lattice parameter, a, high, and the width is a face diagonal, 1.41a. We know the relationship between the lattice parameter, a, and the atomic radius from the hard sphere model: |

2b) BCC (110)

2c) Planar density of (110)-BCC:
For the BCC (110) plane shown to the right we can use the pink area to make our calculation. The total number of atom cross sections in the red area is 2 atoms (4 quarter-atom pieces at the corners and 1 whole-atom in the center). This gives a total atom area of The area of the pink plane is found from the dimensions of the rectangle. The rectangle is the lattice parameter, a, high, and the width is a face diagonal, 1.41a. We know the relationship between the lattice parameter, a, and the atomic radius from the hard sphere model: |
2b) |

3) *A single crystal of a metal that has the FCC crystal
structure is oriented such that...*

To solve this problem, we need to be able to calculate the resolved shear stress in each of these planes and directions. To do that, we need to know the angles between the applied stress and the slip directions. The easiest way to determine these (in a CUBIC crystal, such as FCC) is to use the dot product rule. The angle we are looking for first is the angle between the applied stress and the slip direction, which is often given the symbol of LAMBDA:

Secondly, we need to know the angle between the applied stress and the NORMAL to the slip plane (called PHI). For cubic crystals (and cubic crystals ONLY), the normal direction to the plane (h, k, l) is the direction [h, k, l]. Thus, the plane normal for the (111) plane in FCC is the direction [111], and the dot product rule gives us the angle PHI as:

OKAY...now we're ready to apply the critical resolved shear stress equation (equation 7.4) to this problem to find the applied stress for which slip occurs (the YIELD strength) for each of these three slip systems.

For slip system (111)[1 -1 0]:

For slip system (111)[1 0 -1]:

For slip system (111)[0 1 -1]:

CONCLUSION? There are two slip systems that are oriented equally to produce slip at 4.29 [MPa]. We could expect the yield strength to be about 4.3 [MPa], and that several slip systems will be operating simultaneously.

4a) *What is the ductility measured in terms of area reduction?*

4b) *If the original billet is 2 meters long, how long is the final wire?*

In plastic deformation, the VOLUME is conserved, so

4c) *What is is final engineering strain of the wire?*

Engineering strain definition is in chapter 6:

4d) *What is the final true strain of the wire?*

True strain definition is also in chapter 6:

5a) *Why does slip in FCC Cu happen on the combination
of planes and directions given as {111}<1 -1 0>?*

The {111}-family of planes are the closest packed planes in the FCC crystal structure. These planes are the farthest apart, so that the bonds that must be broken for slip to occur are of lowest energy, making slip easiest on this plane.

The <1 -1 0>-family of directions are the close-packed directions in the FCC
crystal structure, and are also in the {111} planes. When slip occurs in a
close-packed direction, the elastic energy needed to create a dislocation line
is the smallest, because the burger's vector is the smallest in these directions.

5b)* What does it mean to say that there are 12 slip systems for FCC? Explain
with words and sketches.*

There are four different (111) planes in the family of {111} FCC planes. These are shown in the figure below as the teal, blue, green and red planes. (NOTE that the plane (111) is the same plane as the (-1 -1 -1) plane, so only one is needed.)

For each of the close packed planes, there are three close-packed directions of the family <1 -1 0>. The three close-packed directions on the (111) plane are shown in the figure. Each of the four close-packed planes has three separate close-packed directions, giving a total of 12 slip systems.