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ME 2151/2151E Principles of Mechanical Engineering Materials Session 2010/2011 SEM1 ME 2151/2151E Fundamentals of Materials Engineering Tutorial 1 Atomic structure and bonding Crystal structures and defects Diffusion

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Page 1: ME2151E Tut 1.pdf

ME 2151/2151E Principles of Mechanical Engineering Materials

Session 2010/2011 SEM1

ME 2151/2151E

Fundamentals of Materials

Engineering

Tutorial 1Atomic structure and bonding

Crystal structures and defects

Diffusion

Page 2: ME2151E Tut 1.pdf

ME 2151/2151E Principles of Mechanical Engineering Materials

2.

1.

Session 2010/2011 SEM1

Page 3: ME2151E Tut 1.pdf

ME 2151/2151E Principles of Mechanical Engineering Materials

3.

Session 2010/2011 SEM1

Page 4: ME2151E Tut 1.pdf

ME 2151/2151E Principles of Mechanical Engineering Materials

4. A copper-zinc alloy has the following properties

Grain diameter (mm) Strength (MPa)

0.015 170

0.025 158

0.035 151

0.050 145

Determine:

(a) The constants in the Hall-Petch equation; and

(b) The grain size required to obtain a strength of 200 MPa

Session 2010/2011 SEM1

Page 5: ME2151E Tut 1.pdf

ME 2151/2151E Principles of Mechanical Engineering Materials

5. A 0.05cm layer of magnesium oxide (MgO) is deposited between layers

of nickel (Ni) and tantalum (Ta) to provide a diffusion barrier that

prevents reactions between the two metals (Figure). At 1400oC, nickel

ions are created and diffuse through the MgO ceramic to the tantalum.

Determine the number if nickel ions that pass through the MgO per

second. The diffusion coefficient of nickel ions in MgO is 9x1012 cm2/s,

and the lattice parameter of nickel at 1400oC is 3.6x10-8cm.

Session 2010/2011 SEM1

Page 6: ME2151E Tut 1.pdf

ME 2151/2151E Principles of Mechanical Engineering Materials

Solutions to Tutorial 1

Session 2010/2011 SEM1

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ME 2151/2151E Principles of Mechanical Engineering Materials

Higher thermal expansion??? Iron or copper

Higher melting point??? Iron or copper Fe: 1538ºC Cu:1084.62ºC

1.

Session 2010/2011 SEM1

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ME 2151/2151E Principles of Mechanical Engineering Materials

Nov 2004, Q2 (6 marks)

Session 2010/2011 SEM1

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ME 2151/2151E Principles of Mechanical Engineering Materials

FCC BCC

4 atoms per FCC unit cell 2 atoms per BCC unit cell

aFCC = 0.3591 aBCC = 0.2863

Close-packed Non close-packed

APF = 0.74 APF = 0.68

CN = 12 CN = 8

2.

Session 2010/2011 SEM1

Page 10: ME2151E Tut 1.pdf

ME 2151/2151E Principles of Mechanical Engineering Materials

x100%Vol

Vol2Vol

FCC

FCCBCC

x100%)3591.0(

)3591.0(2(0.2863)3

33

% Volume change from FCC to BCC =

=

= 1.36%

Session 2010/2011 SEM1

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ME 2151/2151E Principles of Mechanical Engineering Materials

3.

Session 2010/2011 SEM1

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ME 2151/2151E Principles of Mechanical Engineering Materials

Session 2010/2011 SEM1

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ME 2151/2151E Principles of Mechanical Engineering Materials

Session 2010/2011 SEM1

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ME 2151/2151E Principles of Mechanical Engineering Materials

Plane B: Origin at 0,0,0

Intercepts: x = -1/3 ; y = 1/4 ; z =

Reciprocals: = -3; = 4; =0

Miller index = (340)

Alternatively,

Session 2010/2011 SEM1

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ME 2151/2151E Principles of Mechanical Engineering Materials

Session 2010/2011 SEM1

Page 16: ME2151E Tut 1.pdf

ME 2151/2151E Principles of Mechanical Engineering Materials

4. A copper-zinc alloy has the following properties

Grain diameter (mm) Strength (MPa)

0.015 170

0.025 158

0.035 151

0.050 145

Determine:

(a) The constants in the Hall-Petch equation; and

(b) The grain size required to obtain a strength of 200 MPa

Session 2010/2011 SEM1

Page 17: ME2151E Tut 1.pdf

ME 2151/2151E Principles of Mechanical Engineering Materials

Grain diameter (mm) Strength (MPa) d-1/2

0.015 170 8.165

0.025 158 6.325

0.035 151 5.345

0.050 145 4.472

The values of d-1/2 are calculated as shown in above table; one can then plot

the graph of strength versus d-1/2 as:

140

150

160

170

180

4 5 6 7 8 9

d-1/2

Str

en

gth

(M

Pa

)

Session 2010/2011 SEM1

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ME 2151/2151E Principles of Mechanical Engineering Materials

We can determine the constants K and 0 for Hall-Petch equation either

from the graph or by using two of the data points:

(a) 170 = 0 + K (8.165)

145 = 0 + K (4.472)

25 = 3.693 K

K = 6.77 MPa/mm and 0 = 114.7 MPa

(b) To obtain a strength of 200 MPa

200 = 114.7 + 6.77 d

85.3 = 6.77/d

d = 0.0063 m

Session 2010/2011 SEM1

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ME 2151/2151E Principles of Mechanical Engineering Materials

5. A 0.05cm layer of magnesium oxide (MgO) is deposited between layers

of nickel (Ni) and tantalum (Ta) to provide a diffusion barrier that

prevents reactions between the two metals (Figure). At 1400oC, nickel

ions are created and diffuse through the MgO ceramic to the tantalum.

Determine the number if nickel ions that pass through the MgO per

second. The diffusion coefficient of nickel ions in MgO is 9x1012 cm2/s,

and the lattice parameter of nickel at 1400oC is 3.6x10-8cm.

Session 2010/2011 SEM1

Page 20: ME2151E Tut 1.pdf

ME 2151/2151E Principles of Mechanical Engineering Materials

Session 2010/2011 SEM1

The composition of nickel at the Ni/MgO interface is 100% Ni, or

22 3

/ 38

4 / 8.57 10 atoms/cm

3.6 10Ni MgO

Ni atoms unit cellc

cm

The composition of nickel at the Ta/MgO interface is 0% Ni. Thus, the

concentration gradient is:

2224 30 8.57 10 (atoms/cm)

1.71 10 (atoms/cm .cm)0.05 cm

c

x

The flux of nickel atoms through the MgO layer is:

12 2 24 3

13 2

9 10 cm / 1.71 10 atoms/cm .cm

1.54 10 Ni atoms/cm .s

cJ D s

x

J

Page 21: ME2151E Tut 1.pdf

ME 2151/2151E Principles of Mechanical Engineering Materials

Session 2010/2011 SEM1

The total number of nickel atoms crossing the 2cm x 2cm interface per

second is:

13 2

13

Total Ni atoms per second= Area 1.54 10 atoms/cm .s 2 cm 2 cm

6.16 10 (Ni atoms/s)

J

The volume of nickel atoms removed from the Ni/MgO interface per second

is:

13

9 3

22 3

6.16 10 Ni atoms/s0.72 10 (cm /s)

8.57 10 Ni atoms/cm

Or, the thickness by which the nickel layer is reduced each second is:

9 310 6

2

0.72 10 (cm /s)1.8 10 cm/s 1.8 10 m/s

4 cm

For 1m nickel layer to be removed, the time is: 61/1.8 10 556,000 s=154 h