MECH2419 Test 2015 Solution

Test for MECH2419  –  Properties  Properties of Materials

Date: 22nd October, 2015

Time: 12:30pm-1:10pm

 Name of student: ___________________________ (Write clearly in block letters) University Number: ________________________ Instructions: Answer ALL questions in the spaces provided below. You may use a calculator.

(1) Sketch the unit cells for the ferrite and austenite phases of iron. In the unit cell for austenite, indicate the (111) plane, and in a separate drawing, sketch the atomic  positions on this this plane. Repeat these these steps for the (110) plane in the unit cell for ferrite. ferrite. (8 marks) Answer: Ferrite has BCC structure

(1 mark)

(BCC structure 1 mark, (110) plane 2 marks) Austenite has FCC structure

(1 mark)

(FCC structure 1 mark, (111) plane 2 marks)

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(2) Given that the lattice constant of austenite Fe is 3.52Å, estimate the lattice constant of ferrite Fe. The atomic mass of Fe is 55.85gmol -1. Calculate the densities (mass per unit volume) of both the ferrite and austenite forms of Fe. [Hint: Avogadro’s number =

10.]

6.02×

(6 marks)

Answer:

√ 3  =  √ 2 = 4 × (atomic radius) So,   =  2 /3×  =  2 /3 × 3.52 = 2.87 Å

(2 marks)

Densities: In bcc (ferrite), there are 2 atoms, so density =

 ×.×/(.× ) = 7848.9 kgm-3. (.×)

(2 marks)

In fcc (austenite), there are 4 atoms, so density =

 ×.×/(.× ) = 8508.6 kgm-3. (.×)

2

(2 marks)

(3) The energy U  of a solid crystal is represented by the following equation:  A

U  



 B

m

r 



n

r 

where r  is the separation between atoms, m = 2 and n = 10. Given that the equilibrium atomic spacing occurs at r   3 10 m and U   4 eV, calculate the values of  A and  B. Calculate also the stiffness of an atomic bond at its equilibrium length, and estimate the Young modulus of the crystal. (8 marks) 10

Answer: The energy function now reads:

At equilibrium at

U o

8

 A  

2

r o

r   r o ,



 Ar o 5r o

 

2

or

 A 



4

8

 B



 Ar o 5

force  F  

 A 

2

r 

dU 

.

10

r 

2 A 

dr 

 B 

3

r 

10 B 

11

r 



0,

so

2 A 3

r o

10 B 

11

r o

8

 or  B 

 Ar o 5

.

4 A

10

5r o U o

U  



2

5r o

5  0.3 4

2

4



 0.45 eV nm2 =

7.2 10

 J nm2

(2 marks)

 J nm10.

(2 marks)

20

8



0.45  0.3 5

  

Stiffness  =   | 



0.59 10

5

eV nm10 =

9.4  10





| =     +  =  ×.× + ×.× =     (.) (.)        

= 1.41 × 10− J nm-2 = 141 N/m Young’s modulus

25

 =  =   .  = 470 GPa

(2 marks) (2 marks)

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Questions 4 to 7 concern the Al-rich side of the Al-Cu phase diagram shown below:

   %    1    2   ~    %    3   ~

(4) Which of the following statements are true? (A) 10 wt.% of Cu can dissolve homogeneously into Al at 500ºC. (B) The   phase is a substitutional alloy with the FCC structure. (C) The  -CuAl2 phase contains roughly 1 Al atom out of every 3 atoms. (D) An Al-Cu alloy with 4 wt.% Cu well equilibrated at room temperature sho uld consist of two solid phases. (E) The solubility of Cu into Al decreases with increasing temperature from room temperature to ~540ºC. (F) There is no eutectic reaction in the Al-Cu phase diagram. Answer (e.g. A and B, etc.): __B and D__________________ (2 marks) As shown in the phase diagram above, an Al-Cu alloy containing 10 wt % of Cu overall would be partially melted when maintained at 600ºC for sufficient time. Determine the following from the phase diagram: (5) Weight % of Cu in the liquid phase at 600ºC: _~21%_(see red line in phase diagram above) (2 marks) (6) Weight % of Cu in the solid phase at 600ºC: __~3%_(see blue line)__ (2 marks) (7) % by weight of the liquid phase present in the alloy at 600ºC: By lever rule, (10-3)/(213) = 38.9% (2 marks) 4

Total mark of paper = 30 Report mark as xx/30, e.g. 25/30

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