MINISTRY OF SCIENCE AND TECHNOLOGY

DEPARTMENT OF TECHNICAL AND VOCATIONAL EDUCATION

GOVERNMENT TECHNOLOGICAL COLLEGES / INSTITUTES

AGTI Year II (Metallurgy)

Date : 21-9-2006 (Thursday) Time : 8:30 –

11:30 AM

MET-02023 CONCEPTS OF MATERIALS SCIENCE

1(a) What are the main classes of engineering materials? What are some of the

important properties of each of these engineering materials?

(b) Determine the number of vacancies needed for a BCC iron lattice to have a

density of 7.87 g/cm3. The lattice parameter of the iron is 2.866 x 10-8 cm.

(c) Tungsten has an unusually high melting temperature of 3410C, making it difficult

to process into useful shapes. Design a process by which small diameter filaments

can be produced from tungsten.

Ans: (a)There are five main classes of engineering materials. They are

(1) metal

(2) ceramic

(3) polymers

(4) electronic materials

(5) composites

Metal: have good strength, good ductility and formability, good electrical and

thermal conductivity, and moderate temperature resistance.

ceramics are strong, serve as good electrical and thermal insulators; are often

resistance to damage by high temp and corrosive environments, but are brittle.

Polymers have relatively low strength, are not suitable for us at high temp,

Have good corrosion resistance, good electrical and thermal insulation.

Electronic materials posses unique electrical and optical properties that make

Them essential components in electric and communication devices.

Composites are mixture of materials that provide unique combinations of

Mechanical and physical properties that cannot be found in any single material.

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(c) Tungsten has an unusually high melting nature of 3410˚C, making it difficult to

Into useful shapes. Design a process by which small diameter filaments can be

Produced from tungsten.

Because of tungsten’s high melting temperature, most costing processes cannot be

used for it. A common method for producing tungsten is by a powder metallurgy

process. Powder particles of tungsten oxide (WO3), a ceramic, are heated in a

hydrogen atmosphere, the subsequent react in produces metallic tungsten

particles and H2O. The tungsten powder particles are consolidated at high

temperature and pressure into simple rods. The rods can then be wires-drawn

(a forming process) to progressively smaller diameters until the correct size is

produced.

II.(a) Sketch the following planes and directions within each cubic unit cell:

(10 ),(012),(211)(3 ),[ 1 ],[0 2],[11 ],[ 01]

(b) A metal having a cubic structure has a density of 2.6 g/cm3. An atomic weight of

87.62g/mole, and a lattice parameter is 6.0849A˚. One atom is associated with

Each lattice point. Determine the crystal structure of the metal and the packing

Factor of this meal.

(c) Determine the repeat distance, linear density and picking fraction for BCC

Lithium, which has a lattice parameter of 0.3508nm in the [111],[011]and [010].

Ans: (a)

( 1 0 ) ( 0 1 2 )

2

( 2 1 1 ) ( 3 )

[ 1 ] [ 0 2 ]

[1 1 ] [ 0 1]

(b) =2.6g/cm3

Atomic weight =87.62 g/mole

a0 =6.0849 A˚ =6.0849 x 10-8 cm

crystal structure =?

Packing factor =?

=

3

No: of atom/cell =

=

= 4.026

= 4 atoms /cell

crystal structure is FCC.

For FCC structure,

a0 =

packing factor =

=

=

= = 0.74

(c) r =1.519 A˚ = 1.519 x 10-8 cm ----------FIG------

Repeat distance =?

Linear density =?

Packing fraction=?

a0 =0.3508 nm= 0.3508 x 10-7cm

for BCC lithium,

Repeat distance = =

= 30.38 x10-9 cm

Linear density = =

= 32.92 x 106 l.p/cm

Packing fraction = linear density x 2

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= 32.92 x 106 x 2 x1.519 x 10-8

= 0.9999 (1)

Repeat distance = = x 0.3508 x 10-7

= 49.61 x 10-9 cm

Linear density = =

= 20.15 x 106 l.p/cm

Packing fraction = linear density x 2

= 20.15 x 106 x 2 x1.519 x 10-8

= 0.612

Repeat distance = = 0.3508 x 10-7

Linear density = =

= 28.51 x 106 l.p/cm

Packing fraction = linear density x 2

=28.51 x 106 x 2 x1.519 x 10-8

= 0.866

III.(a) A single crystal of a BCC metal is oriented so that the [001] direction is parallel to

the applied stress. If the critical resolved shear stress required for slip is 12000psi,

Calculate the magnitude of the applied stress required to cause slip to begin [111]

Direction on the (011),(101),( 01).

(b) Determine the interplanner spacing and the length of Burgers vector for slip on

The expected slip system in BCC tantalum.(lattice parameter is 3.3026 A˚).

Repeat, assuming that the slip system is a (111)/[1 0]. What is the ratio between

the shear stress required for slip for the two systems? Assume the k=2

Ans: (a): = 12000 psi

= ?

(011) / [1 0]

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C C

Cos = =

Cos =

=

cross = Cos Cos

= = =29393.87691 psi

(101)/[111]

Cos = , Cos = =

cross = Cos Cos

= = = 29393.87691 psi

( 01) / [111]

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A

B

B

A

A

C B

ao

03a

C

A

ao

ao B

02aA

Cos = , Cos = =

cross = Cos Cos

= = = 29393.87691 psi

(b) for expected slip system for BCC =(110)/ [111],

(1 0)/ [1 1]

a0 = 3.3026 A˚ =3.3026x 10-8 =0.33026 nm

dhkl = = = 0.23353 nm

= 2.33 A˚

b = = 0.28601 nm = 2.8601 A˚

BCC =C exp (- )

= C exp (- 2 )

= 0.19534 C

Assume slip system and direction;

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B

CA

C C

A

ao

aoB

A02a

A

B A

A A

A

B

C

ao

ao

02a

(111) / [1 0]

dhkl = = = 0.1907 nm =1.9067 A˚

b = = x 0.33026 = 0.4671 nm 4.6705 A˚

assume = C exp (- )

= C exp (- 2 )

= 0.44198 C

= = 0.44196

IV.(a) A force of 100000 N is applied to a 10 mm x 20mm iron bar having a yield

Strength of 400 MPa and a tensile strength of 480 MPa. Determine

(i) whether the bar will plastically deform and

(ii) whether the bar will experience necking

(b) A cyclical load of 1500 lb is to be exerted at the end of a 10 in long aluminum

Beam. The bar must survive for at least 106 cycles. What is the minimum

diameter of the bar?

Ans: (a) F =100,000 N

A = 10 x20 mm2

= = = 500 N/mm2

T = 480 MPa = 480 N/mm2

y = 400 MPa = 400 N/mm2

(i) the bar will plastically deform

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02a

A B

C

a0

a0

= 500 N/mm2 = 400 N/mm2

(ii) the bar will experience necking

= 500 N/mm2 = 480 N/mm2

(b) N = 106 cycles

L = 10 in

= 14 MPa (from graph)

= 14 MPa x =2030 psi

F = 1500 lb

=

d3 = = = 75.22

d = 4.22 in

V(a) The diffusion coefficient for Cr in Cr2O3 is 6 x10-15 cm2/s at 1400˚C. Calculate

(i) the activation energy (ii) the constant D0.

Ans:(a) What temperature is required to obtain 0.50% C at a distance of 0.5 mm

beneath

The surface of a 0.20% C steel in 2 hr, when 1.1%C is present at the surface?

Assume the iron is FCC, activation energy is 32900 cal/mole, D0 = 0.23 cm2/s.

erf

0 0

0.10 0.1125

0.2 0 0.227

0.30 0.3286

0.40 0.4284

0.50 0.5205

0.60 0.6039

0.70 0.6778

0.80 0.7421

0.90 0.7970

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1.00 0.8427

1.50 0.9661

2.00 0.9953

(a) D = 6 x10-15 ,T =727˚C = 1000K

D = 1 x 10-9 , T =1400˚C = 1673K

(i) Q =?

D1000 = D0 exp (- )

D0 = ….(1)

D1673= D0 exp (- )

D0 = …..(2)

From equation 1& 2

=

=

=

6 x 10-6 =

6 x 10-6 = exp [(-5.027 +3.0082) x10-4]

6 x 10-6 =exp (-2.0188 x 10-4 Q)

Ln 6 x 10-6 =-2.0188 x 10-4Q

-12.0238 = -2.0188 x 10-4Q

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Q = =5.9559 cal/mol

= 59.559 x1013 cal/mol

(ii) D0 =?

D0 = = = = 0.0624

(a) T =?

Cx = 0.5%

C0 = 0.2%

Cs = 1.1%

x = 0.5 mm =0.05cm

t = 2 hr = 7200s

the iron is FCC structure

= erf ( )

= erf ( )

= erf ( )

0.67 = erf ( )

0.7 =

=

=4.2 x 10-4

D = 1.772 x 10-7

Q = 32900 cal/mol

D0 =0.23cm2/s

D = D0

1.772 x 10-7 =0.23

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=

170.236 x 10-9 =

T =

T =1176.2543 K

T = 903.2543˚C

VI Define the following:

(i) coefficient of thermal expansion (ii) allotropy

(iii) hardness test (iv) modulus of resilience

(v) Fick’s first law (vi) interstitial diffusion

(vii) Hall Petch equation (viii) strain hardening coefficient

(ix) cold working (x) diffusion coefficient

(i) coefficient of thermal expansion

The amount by which a material changes its dimensions when the

Temperature changes. A material with a low efficient of thermal

Expansion tends to retain its dimensions when temperature changes.

(ii) allotropy

The characteristics of a material being able to exist in more than one crystal

Structure depending on temperature and pressure.

(iii) hardness test

Measures the resistance of a material to penetration by a sharp object.

Common hardness tests included the Brinell test, Rockwell test, kno test,

And Vickers test.

(iv) Modulus of resilience

The maximum elastic energy absorbed a material when a load is applied.

(v) Fick’s first law

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The equation relating the flux of atoms by diffusion to the diffusion

coefficient and the concentration gradient. [J =-D ]

J = flux (atoms/cm2.s)

D = diffuse coefficient (cm2/s)

= concentration gradient (atoms/cm3 .cm )

(vi) Interstitial diffusion

Diffusion of small atoms from one into rstitial position to another in the

Crystal structure.

(vii) Hall Petch equation

[ ]

= yield strength

d = average diameter

and k are the constant for the metal

(viii) Strain hardening coefficient

The effect that strain has on the result strength of the materials. A material

With a high strain hardening coefficient obtains higher strength with only

Small amounts of deform atm or strain.

(ix) Cold working

Deformation of a metal below the crystallization temperature . During cold

Working the number of dislocations increase, causing the

(x) Diffusion coefficient

A temperature depended coefficient related to the rate at which atoms diffuse .

The diffusion coefficient depends on temperature and activation energy.

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