Lowell D. Outland
IET307
Material Science
IET 307: Materials Science
HW 4 (based on chapter 7),
1) Explain in depth the process of deformation by twinning.
Twinning results from an applied shear stress that can produce atomic
displacements so that on one side of the plane atoms are located in mirror
image positions to the atoms on the other side of that plane. The magnitude
in the twin region is proportional to the distance from the twin plane. Twinning
occurs on a definite crystallographic plane and in a specific direction
depending on the crystal structure.
2) What is recovery, recrystallization and grain growth? Explain this by writing a
paragraph, detailing figure 7.22 that shows the relationship between tensile
strength, annealing temperature and ductility.
Recovery is where some of the strain energy is relieved by dislocation motion.
Causing an absence of external stress. As a result of atomic diffusion at an
elevated temperature. There is a reduction in the number of dislocations and
dislocation configurations are produced that have low strain energies.
Electrical and thermal conductivities are recovered to their precold-worked
status.
Recrystallization is the formation of new grains having approximately equal
dimensions in all directions. That have low dislocation densities and are
characteristic of precold worked condition. Mechanical properties are also
restored to their original state. Recrystallization depends on both time and
temperature.
Grain growth is an increase in the average grain size of a polycrystalline
material. The main force is the reduction in total grain boundary energy.
3) A cylindrical specimen of cold worked copper has ductility (% elongation) of
25%. If its cold worked radius is 10 mm (0.40 in.), what is its radius before
deformation?
This problem calls for us to calculate the precold-worked radius of a cylindrical
specimen of copper that has a cold-worked ductility of 25%EL. From Figure
7.19c, copper that has a ductility of 25%EL will have experienced a deformation
of about 11%CW. For a cylindrical specimen, Equation 7.8 becomes
Lowell D. Outland
IET307
Material Science
 r 2   r 2 
%CW =  0 2 d   100
r0




Since rd = 10 mm (0.40 in.), solving for r0 yields

r0 =
rd
=
%CW
1 
100
10 mm
= 10.6 mm (0.424 in.)
11.0
1 
100
4) What
 is the difference between the slip and twinning mechanisms of plastic
deformation of materials?
In slip mechanisms the atoms on one side of the plane move an equal
distance, while in the twinning mechanism the atoms move proportionally to
the distance of the twin plane.
5) A sheet of aluminum alloy is cold rolled 30%to a thickness of 0.080 in. If the is
then cold rolled to a final thickness of 0.064 in., what is the total percent cold
work done?
First we have to find the original thickness which is 0.1143. To do this I took
0.080 ( total thickness minus 30%)and divided it by 70 (%remaining) which
gave me the original thickness of the aluminum. Then using the formula (Ao –
Ad)/Ao or 0.1143/(0.1143-0.064) =0.0.503/0.1143 = 44%
6) How could you manufacture large propellers for large ships? What factors
would influence the selection of material for this application?
I would manufacture the propeller in separate pieces consisting of the cone and
then the blades. These pieces would be cast from stainless steel, in order to
resist corrosion. The blades could then be welded onto the cone and machined
to a smooth finish to facilitate ease of movement in the water.
Lowell D. Outland
IET307
Material Science
7) Why is it difficult to improve both strength and ductility simultaneously?
Because when strength increases, hardness increases. And as hardness
increases then ductility decreases.
8) A 2 in. diameter copper rod is reduced to a 1.5 in. diameter, and then reduced
again to a final diameter of 1 in. In a second case the 2 in. diameter rod is
reduced in one step from a 2in. to a 1 in. diameter. Calculate the % cold work
for both cases.
Since in both cases the total reduction is equal, we can use the same formula
for both answers.
Using the formula (Ao –Ad)/Ao = (2-1)/2 = .5 or 50%
9) Cold working can’t be used as a strengthening mechanism for materials that
are going to be subject to high temperatures during their use. Explain why.
Because applying high temperatures will remove the effects of cold working.
10) Explain why a tungsten filament used in an incandescent light bulb ultimately
fails, based on what you have learned in this chapter.
The repeated heating and cooling of the filament causes it to become brittle
and eventually fracture.