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ENGI 7007 Marine Materials Midterm Exam -SOLUTIONS HsiaoSpring 2012

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Name__________________________________________________ Student ID Number ________________

Please write your answers to the following five problems clearly. Each problem is worth 5 points. Note that

each problem may have several parts. You have 90 minutes to complete the exam. The midterm is worth 25%

of your final mark.

Formulas For Your Use on the Midterm (note: not all of them may be used)

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ENGI 7007 Marine Materials Midterm Exam -SOLUTIONS HsiaoSpring 201

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Problem One: Crystallinity and Dislocations (5 points)

(a) (1.5 pts) Figure 1 shows the atomic packing schemes for several different crystallographic directions for ahypothetical metal. For each direction, the circles represent only the atoms contained within a unit cell.

Draw the unit cell given these directions. To what crystal system does the unit cell belong? What would this

crystal structure be called?

Figure 1: Three crystallographic directions in a unit cell

Solution

(a) This unit cell belongs to the tetragonal system since a= b= 0.40 nm, c= 0.50 nm, and = = 90.

(b) This crystal structure would be called face-centered tetragonalsince the unit cell has tetragonal symmetry, and

an atom is located at each of the corners, as well as at the centers of all six unit cell faces. In the figure above, atoms are

only shown at the centers of three faces; however, atoms would also be situated at opposite faces.
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ENGI 7007 Marine Materials Midterm Exam - SOLUTIONS HsiaoSpring 201

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(b) (1.5 pts) Figure 2 shows three different crystallographic planes for a unit cell of a hypothetical metal. Thecircles represent atoms. Draw the unit cell given these three planes. To what crystal system does the unit

cell belong? What would this crystal structure be called? Support your reasoning.

Figure 2: Three crystallographic planes in a unit cell

Solution

The unit cells constructed below show the three crystallographic planes that were provided in the problem

statement.

(a) This unit cell belongs to the orthorhombic crystal system sincea= 0.30 nm, b= 0.40 nm, c= 0.35 nm, and =

= = 90.

(b) This crystal structure would be called body-centered orthorhombic since the unit cell has orthorhombic

symmetry, and an atom is located at each of the corners, as well as at the cell center.
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(c) (2 pts) The magnitude of a Burgers vector can be determined by the following equation: 2

1222

2lkh

ab

Where ais the lattice parameter and [hkl] is the close-packed direction of slip. Determine the value of b

for nickel (Ni). Given that Ni is FCC with an atomic radius of 0.1246 nm. Given that Al is FCC with an atomicradius of 0.1431 nm.

Solution

The close-packed direction would be along the slip direction, which is any of the face diagonals. Thus,

nm.0.4047aandnm0.3524aso,22,FCCFor AlNi ra

nmbNi

249.00112

3524.0 21

222

nmbAl

286.00112

4047.0 21

222

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Problem Two: Mechanical Properties (5 points)

(a) (1 pt) From the tensile stress-strain behavior for the naval brass specimen shown in Figure 3, calculate themodulus of elasticity.

(b) (1 pt) Now determine from the graph the yield strength at a strain offset of 0.002.(c) (1 pt) Calculate the maximum load that can be sustained by a cylindrical specimen having an original

diameter of 12.8 mm (or 10.8 mm).(d) (1 pt) Calculate the change in length of a specimen originally 250 mm (or 200 mm) long that is subjected to a

tensile stress of 345 MPa.

(e) (1 pt) Using Figure 4, estimate the Rockwell B value for the brass at its maximum (tensile) strength.

Figure 3: Stress-Strain Behaviour for Naval Brass

Figure 4: Correlation Between Hardness and Tensile Strength
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Solution

(a) E = (1500) MPa / 0.0016-0 = 93.8 GPa (13.6 x 106psi)(b) The 0.002 strain offset line is constructed as shown in the inset; its intersection with the stress-strain

curve is at approximately 250 MPa, which is the yield strength of the brass.

(c) The maximum load that can be sustained by the specimen is calculated by using as the tensilestrength, which is 450 MPa from the graph:

=57,906 N

=41,224 N

(d) The strain is approximately 0.06, from reading the corresponding strain at point A from the strain axis.

(e) A tensile strength of 450 MPa corresponds to a HRB of ~82.

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Problem Three: Imperfections and Cold-Working (5 points)

(a) (3 pts) Determine whether it is possible to cold work naval brass so as to give a minimum HRB hardness of80 and at the same time have a ductility of at least 20% EL. Justify your decision, using the information from

Figure 4 and Figure 5.

(b) (2 pts) A naval brass specimen is a cylindrical rod having an initial diameter of 6 mm that was cold-workedby drawing until the final diameter was reduced to 5 mm. Describe the process that is necessary to meetthe dimensional specifications (dfinal= 5 mm) as well as the mechanical specifications of a yield strength of at

least 340 MPa and a ductility in excess of 20% EL.

Figure 5: Relationship of Cold-Work to Yield Strength, Tensile Strength, and Ductility for Brass, 1040 Steel, and Copper
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(a) A brass sample of HRB 80 has a tensile strength of ~400 MPa, which would require ~18% CW. For brassto have a ductility of at least 20% (EL), it must be cold-worked to at most a percentage of 22% CW.

Thus, yes, it is possible to have a brass sample of HRB 80, with a tensile strength of 400 MPa and a

ductility of 20% (the ductility would be~26% EL actually).

(b) The amount of cold-working performed when a brass sample is changed from a diameter of 6 mm to 5mm is:

This amount of cold-working produces a yield strength of ~390 MPa, which satisfies the first spec but

does not satisfy the second spec of a ductility of at least 20% EL.

For a yield strength of 340 MPa, the brass has to subjected to at least 20% cold-working. For a ductility

of at least 20%, the cold-working can be at most 22%. So we will make it a two-step process, whereby

we will cold-work to an intermediate diameter, anneal and recrystallize to remove the stresses from

the first cold-working, then cold-work from the intermediate diameter to the final diameter of 5 mm in

the amount of 21%CW (which is the average between 20-22%).

r = 2.81 mm

Therefore, the intermediate diameter is 5.6 mm

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Problem Four: Complete Solid Solution Binary Phase Diagram (5 points)

(a) (3 pts) Given the following information, sketch the binary phase diagram for Cu-Ni and label the phasespresent.

Tmeltingof pure Cu = 1085C

Tmeltingof pure Ni = 1455CTliquidusof 20 wt% Ni in Cu = 1200C

Tliquidusof 50 wt% Ni in Cu = 1310C

Tliquidusof 80 wt% Ni in Cu = 1405C

Tsolidusof 30 wt% Ni in Cu = 1200C



(b) (1 pt) Using your binary phase diagram, draw the microstructural development during slow cooling of a 40wt% Ni composition of Cu-Ni alloy.

(c) (1 pt) Calculate the amount of the phase(s) present in 1000 grams of a 50 wt% Ni composition of Cu-Nialloy at 1300C.

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Problem Five: Images and Definitions (5 points)

(a) (1 pt) Identify and describe the black lines in this imageof stainless steel as seen by a transmissionelectron microscopy.

(b) (1 pt) Describe the significance of T = 98C in Figure 6 of 1020 steel.

0

50

100

150

200

250

-150 -100 -50 0 50 100 150

ImpactEnergy(ft-lb

s)

Temperature (C)

Figure 6Test 1

Test 2

~-5C, 98 ft-lbs

These black lines are dislocations movingthrough and blocking other black lines

(dislocations) from moving through the stainlesssteel lattice. We can expect the strength of this

sample to be higher than typical stainless steeldue to this tangle of dislocations. The trade-off

is that ductility is compromised and the samplewill not experience as much plastic deformation

as usual before brittle failure.

This figure shows that the brittle-to-ductile transformation for steel occurs at -5C.

This means that below -5C, steel responds as a brittle material, fracturing in aclean shear, whereas above -5C, 1020 steel responds to stress as a ductile

material, fracturing in a more taffy-like fashion in response to impact from a

Charpy test.

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(c) (2 pts) Given the two-dimensional image of a crystal lattice in Figure 7, identify and label two (2) out of thethree (3) types of point defects clearly.

Solution: only substitutional solute atoms, interstitial solute atoms, and vacancy atoms are point defects.

Figure 7: Two-Dimensional Image of Crystal Lattice Simulation Using Bubbles in a Raft

(d) (1 pt) Choose the best description of the image shown in Figure 8 from the list in the box by circling thecorrect letter.

Figure 8: Microstructural Development from Slow Cooling

a. This image shows two phases present, ofa binary alloy with a hypoeutectic

composition.

b. This image shows two phases present, ofa binary alloy with a hypereutectic

composition.

c. This image shows three phases present,of a binary alloy with a eutecticcomposition.

d. Both a & b are possible.e. None of these descriptions are correct.
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