met-2023 (pdf)-sq

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Met-2023: Concepts of Materials Science I

Sample Questions & Answers,(2009)

( Met, PR, FC, MP, CNC, McE )

Q-1.Define the following.

(i) Point Defects (ii) Burgers Vector (iii) Slip and Slip system

(iv) Interplanar spacing (v) Frenkel Defect (vi) Schottky defect

Q-2.Design a heat treatment that will provide 1000 times more vacancies in copper than are

normally present at room temperature. About 20,000 cal/mol are required to produce a

vacancy in copper.The lattice parameter of FCC copper is 0.36151 nm.

Q-3.Determine the number of vacancies needed for a BCC iron lattice to have a

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

(at. Wt. for Fe- 55.847 g/mol )

Q-4.(a) The planar density of the ( 1 1 2 ) plane in BCC iron is 9.94 x 10 1 4 atoms/cm2.

Calculate (i) the planar density of the (110) plane and (ii) the interplanar spacings

for both the (112) and (110) planes. On which plane would s l ip normally occur?

(b) Calculate the length of the Burgers vector in the following materials:

(i) BCC niobium ( a0 – 3.294 Aº )

(ii) FCC silver ( a0 – 4.0862 Aº )

(iii) FCC copper ( a0 – 3.6151 Aº )

Q-5.(a) Define Schmid's law.

(b) An aluminum crystal slips on the (111) plane and in the [110] direction with a 3.5

MPa stress applied in the [111 ] direction. What is the critical resolved shear stress?

Q-6.(a) Calculate the number of vacancies per cm3 expected in copper at 1085°C (just

below the melting temperature). The energy for vacancy formation is 20,000 cal/mol.

( a0 –for Cu - 3.6151 Aº )

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(b) The fraction of lattice points occupied by vacancies in solid aluminum at 660cC is

10-3. What is the energy required to create vacancies in aluminum ?

Q-7. The density of a sample of FCC palladium is 11.98 g/cm3 and its lattice parameter

is 3.8902 A . Atomic mass of Pd is 106.4 g/mol. Calculate

(a) the fraction of the lattice points that contain vacancies and

(b) the total number of vacancies in a cubic centimeter of Pd.

Q-8.(a) Define the rate of Diffusion of “ Fick's First Law”.

(b) Atoms are found to move from one lattice position to another at the rate of 5

x 1 O 5 jumps per second at 400cC when the activation energy for their movement is

30,000 cal/mol. Calculate the jump rate at 750°C.

Q-9.(a) Define (ii) diffusion and (ii) diffusion coefficient.

(b) Consider a diffusion couple set up between pure tungsten and a tungsten-1 at %

thorium alloy. After several minutes of exposure at 2000° C, a transition zone of 0.01 cm

thickness is established. What is the flux of thorium atoms at this time if diffusion is

due to (a) volume diffusion, (b) grain boundary diffusion, and (c) surface diffusion?

The lattice parameter of BCC tungsten is 3.165A°.

Diffusion Coefficient for Thorium in Tungsten

Surface 0.47 exp(-66,400/RT)

Grain boundary 0.74 exp(-90,000/RT)

Volume 1.00exp(-120,000/RT)

Q-10.(a) Define “ Activation energy”.

(b) The diffusion coefficient for Cr in Cr2O3 is 6 x 10 -15 cm2/s at 727°C and is 1x10– 9

cm2/s at 1400°C. Calculate

(i) the activation energy and ( ii ) the constant Do.

Q-11.The surface of a 0.1% C steel is to be strengthened by carburizing. In carburizing, the steel

is placed in an atmosphere that provides 1.2% C at the surface of the steel at a high

temperature. Carbon then diffuses from the surface into the steel. For optimum properties, the

steel must contain 0.45% C at a depth of 0.2 cm below the surface. Design a carburizing heat

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treatment that will produce these optimum properties. Assume that the temperature is high enough

(at least 900°C) so that the iron has the FCC structure.

Q-12.(a) Define “ Fick's Second Law”.

(b) We find that 10 h are required to successfully carburize a batch of 500 steel gears at

900°C, where the iron has the FCC structure. We find that it costs $ 1000 per hour to operate

the carburizing furnace at 900°C and $ 1500 per hour to operate the furnace at 1000° C. Is it

economical to increase the carburizing temperature to 1000°C?

Q-13.A 0.001 in. BCC iron foil is used to separate a high hydrogen gas from a low

hydrogen gas at 650°C. 5 x 108 H atoms/ cm3 are in equilibrium with the hot side of the

foil, and 2 x 103 H atoms/cm3 are in equilibrium with the cold side. Determine

(a) the concentration gradient of hydrogen and

(b) the flux of hydrogen through the foil.

Q-14.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 h, when 1.10% C is present at the surface? Assume

that the iron is FCC.

Q-15.A 0.15% C steel is to be carburized at 1100°C, giving 0.35% C at a distance of

1mm beneath the surface. If the surface composition is maintained at 0.90% C, what

time is required?

Q-16.A 0.02% C steel is to be carburized at 1200°C in 4 h, with a point 0.6 mm beneath

the surface reaching 0.45% C. Calculate the carbon content required at the surface of the

steel.

Q-17.A 1.2% C tool steel held at 1150°C is exposed to oxygen for 48 h. The carbon

content at the steel surface is zero. To what depth will the steel be decarburized to

less than 0.20% C?

Q-18.A BCC steel containing 0.001% N is nitrided at 550°C for 5 h. If the nitrogen content

at the steel surface is 0.08%, determine the nitrogen content at 0.25 mm from the surface.

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(i) % Elongation (ii) Endurance limit (iii) Endurance Ratio

(iv) Fatigue life (v) Elastic deformation

Q-20.(a) Define Engineering Stress, True stress and Engineering Strain, True Strain.

(b) A force of 100,000 N is applied to a 10 mm x 20 mm 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.

Q-21.An aluminum alloy that has a plane strain fracture toughness of 25,000 psi-√in .fails

when a stress of 42,000 psi is applied. Observation of the fracture surface indicates that

fracture began at the surface of the part. Estimate the size of the flaw that initiated

fracture. Assume that f = 1 . 1 .

Q-22.Define Hooke's law and Poisson's ratio.

Q-23.A 0.4-in. diameter, 12-in. long titanium bar has a yield strength of 50,000 psi,

a modulus of elasticity of 16 x 106 psi, and Poisson's ratio of 0.30. Determine the length

and diameter of the bar when a 500-Ib load is applied.

Q-24.A 3-in.diameter rod of copper is to be reduced to a 2-in. diameter rod by being

pushed through an opening. To account for the elastic strain, what should be the

diameter of the opening? The modulus of elasticity for the copper is 17 x 106 psi and the

yield strength is 40,000 psi.

Q-25.Which factors does depend on the ability of a material to resist the growth of a

crack?


(i) Tensile strength (ii) Yield strength (iii) Ductility

Q-27.A ceramic part for a jet engine has a yield strength of 75,000 psi and a plane strain

fracture toughness of 5,000 psi - √in. . To be sure that the part does not fail, we plan to

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assure that the maximum applied stress is only one-third the yield strength. We use a

nondestructive test that will detect any internal flaws greater than 0.05 in. long.

Assuming that f = 1.4, does our nondestructive test have the required sensitivity?

Explain.

Q-28.A large steel plate used in a nuclear reactor has a plane strain fracture toughness of 80,000

psi- √in and is exposed to a stress of 45,000 psi during service. Design a testing or

inspection procedure capable of detecting a crack at the surface of the plate before the crack

is likely to grow at a catastrophic rate.

Q-29.A 850-lb force is applied to a 0.15-in. diameter nickel wire having a yield strength of

45,000 psi and a tensile strength of 55,000 psi. Determine

(a) whether the wire will plastically deform and

(b) whether the wire will experience necking.

Q-30.A solid shaft for a cement kiln produced from the tool steel must be 96 inches long and

must survive continuous operation for one year with an applied load of 12,500 lb. The shaft

makes one revolution per minute during operation. Design a shaft that will satisfy these

requirements.

NOTES;(1) For question no. 11,12,13,14,15,16,17,18, need to provide error function

Table 2-3 and Diffusion Coefficient Table 2-1

(2) For question no. 30, need to provide Figure 3-13. The stress-number of

cycles to failure (S-N) curves for a tool steel and an aluminum alloy.

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Met-02023 Concepts of Materials Science I

Sample Questions and Answers, (2009)


(i) Point Defects (ii) Burgers Vector (iii) Slip and Slip system

(iv) Interplanar spacing (v) Frenkel Defect (vi) Schottky defect

(i) Point Defects

Point defects are localized disruptions of the lattice involving one or, possibly,

several atoms. These imperfections, shown in Figure, may be introduced by

movement of the atoms when they gain energy by heating, during processing of the

material, by introduction of impurities, or intentionally through alloying.

Point defects are (1) vacancy, (2) interstitial atom, (3) substitutional atom, (4) Frenkel

defect, and (5) Schottky defect. All of these defects disrupt the perfect arrangement of the

surrounding atoms.

(ii) Burgers vector ( b )

The direction and the distance that a dislocation moves in each step.

The length of Burgers vector (b) is equal to the repeat distance.

(iii) Slip and slip system

The process by which a dislocation moves and causes a material to deform is

called slip.

The direction, in which the dislocation moves is the s l ip direction and the plane in

which the dislocation moves is the s l ip plane.

The combination of slip direction and slip plane is the slip system.

(iv) Interplanar spacing

The distance between the two adjacent planes of atoms with the same Miller indices is

called the interplanar spacing d(hkl) and the general equation ,

0( ) 2 2 2hk l

adh k l

=+ +

where, a0 = the lattice parameter or lattice constant

h,k,l = Miller indices of adjacent planes of atoms

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(v) Frenkel Defects

A Frenkel defect is a vacancy-interstitial pair formed when an ion jumps from a normal

lattice point to an interstitial site, leaving behind a vacancy.

(vi) Schottky Defects

A Schottky defect is a pair of vacancies in an ionically bonded material; both an

anion and a cation must be missing from the lattice if electrical neutrality is to be

preserved in the crystal. These are common in ceramic materials with the ionic bond.

Q-2. Design a heat treatment that will provide 1000 times more vacancies in copper than are

normally present at room temperature. About 20,000 cal/mol are required to produce a

vacancy in copper. The lattice parameter of FCC copper is 0.36151 nm.

Solution

The lattice parameter of FCC copper is 0.36151 nm.

The number of copper atoms, per cm3 is

228 3

4 / 8.47 10(3.6151 10 )

atoms cell xx cm− = copper atoms/cm3

At room temperature, T = 25 + 273 = 298 K:

n v = (8.47 x 1022) exp [(- 20,000)/ (1.987) (298)]

= 1.815 x 108 vacancies/cm3

We wish to produce 1000 times this number, or nv= 1.815 x 1011 vacancies/cm3.

We could do this by heating the copper to a temperature at which this number of

vacancies forms:

nv = 1 . 8 1 5 x 1011 = (8.47x 1022) exp (-20,000/1.987T)

exp (- 20,000/1.987T ) = 1.85x1011 /(8.47x 1022) = 0.214x10-11

T = 20,000 / (1.987)(26.87) = 375 °K = 102 °C

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By heating the copper slightly above 100°C, then rapidly cooling the copper back to

room temperature, the number of vacancies in the structure may be one thousand times

greater than the equilibrium number of vacancies.

Q-3. Determine the number of vacancies needed for a BCC iron lattice to have a

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

(at. wt. for Fe = 55.847 g/mol )

Solution

Theoretical density of iron can be calculated from the lattice parameter and the atomic

mass.

Since the iron is BCC, two iron atoms are present in each unit cell.

theoretical density of iron ;

3-8 3 23

2 / 55.847 / 7.8814 /(2.866x10 cm ) x (6.02x10 atoms/ mol)

atoms unit cell x g moledensity g cm= =

Let's calculate the number of iron atoms and vacancies that would be present in

each unit cell for the required density of 7.87 g/cm3.

3-8 3 23

( / ) (55.847 / ) 7.8814 /(2.866x10 cm ) x (6.02x10 atoms/ mol)

atoms unit cell x g moledensity g cm= =

3

-8 3 23

7.8814 / (55.847 / )/ 1.997 atoms(2.866x10 cm ) x (6.02x10 atoms/ mol)

g cm x g moleatoms unit cell = =

no. of vacancies per unit cell = 2 – 1.9971 = 0.0029

Or, there should be 0.0029 vacancies per unit cell .

The number of vacancies per cm3 is

Vacancies/cm3 = 0.0029 vacancies per unit cell / (2.866xl0 -8cm)3

= 1.23x1020 vacancies/cm 3

Q-4.(a)The planar density of the ( 1 1 2 ) plane in BCC iron is 9.94 x 10 1 4 atoms/cm2.

Calculate (i) the planar density of the (110) plane and (ii) the interplanar spacings

for both the (112) and (110) planes. On which plane would s l ip normally occur?

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(a)

(b) Calculate the length of the Burgers vector in the following materials:

(i) BCC niobium ( a0 – 3.294 Aº )

(ii) FCC silver ( a0 – 4.0862 Aº )

(iii) FCC copper ( a0 – 3.6151 Aº )

(i) For BCC niobium , a0 = 3.294 Aº

The directions of the Burgers vector, are in [l 11] fo r BCC meta l

The repeat distance is along the [l l 1 ] directions and is equal to one-half of the body

diagonal, since lattice points are located at corners and centers of body.

Body diagonal distance = 3 3 (0.3294 ) 0.5705a x nm nm= =

The length of the Burgers vector, or the repeat distance, is:

1 ( 0.5705 ) 0.28522

b x nm nm= =

(ii) For FCC silver, a0 = 4.0862 Aº

The directions of the Burgers vector, are in [l 10] for FCC meta l

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The repeat distance is along the [l l 0 ] directions and is equal to one-half of the face

diagonal, since lattice points are located at corners and center of the face.

Face diagonal distance = 2 2 (0.40862 ) 0.5778a x nm n= = m


1 ( 0.5778 ) 0.28892

b x nm nm= =

(iii) Copper is FCC , a0 = 0.36151 nm

The directions of the Burgers vector, are of the form(l 10) .

The repeat distance along the ( l l 0 ) directions is one-half the face diagonal, since

lattice points are located at corners and centers of faces.

Face diagonal distance = 2 ( 2) (0.36151) 0.51125a x= = nm


1 ( 0.51125 ) 0.255632

b x nm nm= =

Q-5.(a) Define Schmid's law.

Schmid's law

The relationship between shear stress, the applied stress, and the orientation

of the sl ip system and the resolved shear stress ι in the slip direction is

F Cos CosA

τ φ= λ , Cos Cosτ σ φ= λ

where , τ = the resolved shear stress in the slip direction

σ = the applied stress

φ = the angle between the direction of the force and the normal to the slip plane,

λ = the angle between the direction of force and the sl ip direction.

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Q-5.(b) An aluminum crystal slips on the (111) plane and in the [110] direction with a 3.5

MPa stress applied in the [111 ] direction. What is the critical resolved shear stress?

[111] applied stress direction and normal to slip plane

(111) slip plane [110] slip direction

Solution

Aluminum crystal

slips plane - (111) plane

slip direction - [110] direction

stress applied σ - 3.5 MPa

applied stress direction - [11 1] direction.

the critical resolved shear stress ι = ?

Cos Cosτ σ φ= λ

From the Fig:

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the applied stress direction and the direction of the normal to the slip plane are

the same,

the angle φ = 0 , Cos 0 = 1

23

aCosa

λ = = 2 0.81653=

23.5 1 2.8753

x x Mτ = = Pa

the critical resolved shear stress, 2.875 MPaτ =

Q-6.(a) Calculate the number of vacancies per cm3 expected in copper at 1085°C (just

below the melting temperature). The energy for vacancy formation is 20,000 cal/mol.

( a0 = for Cu - 3.6151 Aº )

Solution

(a) copper metal, FCC structure, 4 atoms/ unit cell ,

80 3.6151 , 3.6151 10 , 20,000 / , 1.987 / .a A x cm Q cal mol R cal mol K−= ° = = =

1085 273 1358 , ?vT C K n= + = =

expvQn n x

RT−

=

.no of atom per unit cellnvolumeof unit cell

=

24 33 8 3

0

4 4 1.106 10 /( ) (3.6151 10 )

atom per unit celln x atom cma x cm−= = =

24

19 3

20,000exp( ) 1.106 10 exp( )1.987 1358

5.12 10 /

vQn n x x x

RT xx vacancies cm

− −= =

=

Q-6.(b) The fraction of lattice points occupied by vacancies in solid aluminum at

660cC is 10-3 .What is the energy required to create vacancies in aluminum ?

Solution

Aluminium metal is FCC structure, 4 atoms/ unit cell ,

3660 273 933 , 10 , ?vnT C K Qn

−= + = = =

expvQn n x

RT−

=

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expvn Qn R

−=

T

310 exp ( )1.987 933

Qx

− −=

12800 /Q cal= mol

Q-7. The density of a sample of FCC palladium is 11.98 g/cm3 and its lattice parameter

is 3.8902 A . Atomic mass of Pd is 106.4 g/mol. Calculate

(a) the fraction of the lattice points that contain vacancies and

(b) the total number of vacancies in a cubic centimeter of Pd.

Solution

FCC Pd – density = 11.98 g/cm3 , 4 atoms/ unit cell

At. wt. of Pd - 106.4 g/mol

80

3

3.8902 3.8902 10

( ) ?, ( ) ?v v

a A xn na bn cm

−= =

= =

cm

mass of unit celltheoretical densityvolumeof unit cell

=

38 3 23

4 106.4 / 12.0085 /(3.8902 10 ) (6.02 10 / )

atoms per unit cell x g mol g cmx cm x x atom mol

ρ −= =

For theoretical density 12.0085 g/cm3, 4 atoms/ unit cell

For the given density 11.98 g/cm3, ? atoms/ unit cell

For the given density 11.98 g/cm3, 3.9905 atoms/ unit cell

no. of vacancies per unit cell (nv ) = 4 – 3.9905 = 0.0095

nv = 0.0095

n = no. of atom per unit cell = 4

(a) 30.0095 2.375 104

vn xn

−= =

(b)3

20 33 8 3

. 2.375 10 1.6 10 /(3.8902 10 )

vn no of vacancies per unit cell x x vacancies cmcm volumeof unit cell x cm

−

−= = =

Q-8.(a) Define the rate of Diffusion of “ Fick's First Law”.

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The rate at which atoms diffuse in a material can be measured by the flux J, which is

defined as the number of atoms passing through a plane of unit area per unit time.

Fick's first law explains the net flux of atoms:

. cJ Dx

= −

where, J is the flux (atoms/cm2.s),

D is the diffusivity or diffusion coefficient (cm2/s), and

Δc/Δx is the concentration gradient (atoms/cm3.cm).

Q-8.(b) Atoms are found to move from one lattice position to another at the rate of

5 x 1 0 5 jumps per second at 400cC when the activation energy for their movement is

30,000 cal/mol. Calculate the jump rate at 750°C.

Solution

the rate of diffusion = 5 x 1 0 5 jumps per second at 400cC

the activation energy Q = 30,000 cal/mol.

the jump rate at 750°C = ?

1 4000 273 673T C= + = K

2 750 273 1023T C= + = K

By the equation , Rate of diffusion = 0.exp ( )QcRT−

50

1

5 10 exp ( )Qx cRT−

= ----- eq. (1)

02

exp ( )Qx cRT−

= ----- eq. (2)

eq. (2) ÷ eq.(1)

50 0

2 1

exp ( ) / 5 10 exp ( )Q Qx c x cRT R− −

= =T

50 0

30,000 30,000exp ( ) / 5 10 exp ( )1.987 673 1.987 1023

x c x cx x

− −= =

91.08 10 / secx x jump=

Q-9.(a) Define (i) diffusion (ii) diffusion coefficient.

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(a) (i) Diffusion

Diffusion is the movement of atoms within a material. The rate of diffusion is

governed by the Arrhenius relationship — that is, the rate increases exponentially with

temperature.

01

exp ( )QRateof diffusion cRT−

=

where, Q is the activation energy (cal/mol)

R is the gas constant (1.987 cal/mol.K)

T is the absolute temperature in (K)

Co is a constant for a given diffusion system

(ii) Diffusion coefficient (D)

The diffusion coefficient D is related to temperature by an Arrhenius equation.

The diffusion coefficient depends on temperature and activation energy.

0 exp ( )QD DRT−

=

where, D is the diffusion coefficient ( cm2/ s )

Q is the activation energy (cal/mol)

R is the gas constant (1.987 cal/mol.K)

T is the absolute temperature in (K)

Do is a constant for a given diffusion system

Q-9.(b) Consider a diffusion couple set up between pure tungsten and a tungsten-1 at %

thorium alloy. After several minutes of exposure at 2000° C, a transition zone of 0.01 cm

thickness is established. What is the flux of thorium atoms at this time if diffusion is

due to (a) volume diffusion, (b) grain boundary diffusion, and (c) surface diffusion?


Diffusion Coefficient for Thorium in Tungsten

Surface 0.47 exp(-66,400/RT)

Grain boundary 0.74 exp(-90,000/RT)

Volume 1.00exp(-120,000/RT)

Solution

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the number of tungsten atoms/cm3 is:

no. of W atoms per unit cell / vol. of unit cell

= 2 atom per cell / (a0 )3 = 2 / (3.165 x 10-8 cm )3

= 6.3 x 1022 W atoms/cm3

In the tungsten -1 atom % thorium alloy, the number of thorium atoms is:

CTh = ( 0.01 ) ( 6.3 x 1022) = 6.3 x 1020 Th atoms/cm3

In the pure tungsten, the number of thorium atoms is zero. Thus, the

concentration gradient is:

ΔC / Δx = 0 - 6.3 x 1020 / 0.01 cm = - 6.3 x 1022 Tho atom / cm3.cm

T = 2000 C + 273 K = 2273 K

(a) for volume diffusion, D = 1.00exp(-120,000/RT) cm2/ s

J = - D.Δc/Δx

= - 1.00exp(-120,000/1.987 x2273 ) x (- 6.3 x 1022 )

= 1.82 x 1010 Th atoms/cm2 .s

(b) for grain boundary , D = 0.74 exp(-90,000/RT) cm2/ s

J = - D.Δc/Δx

= - 0.74 exp(-90,000/1.987 x2273 ) x (- 6.3 x 1022 )

= 10.3 x 1013 Th atoms/cm2 .s

(c) for surface diffusion , D = 0.47 exp(-66,400/RT) cm2/ s

J = - D.Δc/Δx

= - 0.47 exp(-66,400/1.987 x2273 ) x (- 6.3 x 1022 )

= 12.2 x 1015 Th atoms/cm2 .s

Q-10.(a) Define “ Activation energy”.

Activation Energy (Q)

The atom is originally in a low-energy, relatively stable state. In order to move to a new

location, the atom must overcome an energy barrier. This energy barrier is the activation

energy Q. This energy is gained by heat supply.

In diffusion, the activation energy is related to the energy required to move an

atom from one lattice site to another.

The activation energy Q is expressed in (cal/mol).

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Q-10.(b) The diffusion coefficient for Cr in Cr2O3 is 6 x 10 -15 cm2/s at 727°C and is

1x10– 9 cm2/s at 1400°C. Calculate (i) the activation energy and ( ii ) the constant Do.

Solution

The diffusion coefficient for Cr in Cr2O3

D = 6 x 10 -15 cm2/s at T1 = 727°C + 273 K = 1000 K

D = 1 x10– 9 cm2/s at T2 = 1400°C + 273 K = 1673 K

(i) the activation energy Q = ? ( ii ) the constant Do. = ?

0 exp ( )QD DRT−

=

150

1

6 10 exp ( )Qx DRT

− −= ----- eq. (1)

90

2

1 10 exp ( )Qx DRT

− −= ----- eq. (2)

eq. (1) ÷ eq.(2)

15 90 0

1 2

6 10 exp ( ) /1 10 exp ( )Q Qx D x DRT R

− −− −= =

T

15 90 06 10 exp ( ) / 1 10 exp ( )

1.987 1000 1.987 1673Q Qx D x Dx x

− −− −= =

59230 /Q cal= mol

(ii) 90

2

1 10 exp ( )Qx DRT

− −=

90

592301 10 exp ( )1.987 1673

x Dx

− −=

20 0.055 / secD cm=

Q-11.The surface of a 0.1% C steel is to be strengthened by carburizing. In carburizing, the steel is

placed in an atmosphere that provides 1.2% C at the surface of the steel at a high temperature.

Carbon then diffuses from the surface into the steel. For optimum properties, the steel must

contain 0.45% C at a depth of 0.2 cm below the surface. Design a carburizing heat treatment that

will produce these optimum properties. Assume that the temperature is high enough (at least

900°C) so that the iron has the FCC structure.

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Solution

Given: 01.2% , 0.45% , 0.1% , 0.2? , ?

s xC C C C C C xt T

= = = == =

cm

By Fick's second law ,

0

[ ]2

s x

s

C C xerfC C Dt

−=

−

1.2 0.45 0.20.68 [ ]1.2 0.1 2

erfDt

−= =

−

From Table 2-3 , we find that,

0.2 0.712 Dt

= or 0.1 0.71Dt

=

20.1. ( ) 0.01980.71

D t = =

Any combination of D and t whose product is 0.0198 will work.

For carbon diffusing in FCC iron, the diffusion coefficient is related to temperature by:

0 exp ( )QD DRT−

=

From the table 2-1, D0 = 0.23 , Q = 32,900 cal/mol

329000.23 exp ( )1.987

DT

−=

Therefore, the temperature and time of the heat treatment are related by:

0.0198tD

=

( )

0.01980.23exp 16558 /

tT

=−

Some typical combinations of temperatures and times are:

If T = 900°C = 1173 K, then, t = 1 16,174 s = 32.3 hr

If T= 1000°C = 1273 K, then, t = 36,360 s = 10.7 hr

If T = 1100°C = 1373 K, then, t = 14,880 s = 4.13 hr

If T = 1200°C = 1473K, then, t = 6,560 s = 1.82 hr

Q-12.(a) Define “ Fick's Second Law”.

Composition Profile (Fick's Second Law)

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Fick's second law describes the dynamic or non-steady state diffusion of atoms

by the differential equation dc/dt = D (d c/dx2), whose solution depends on the

boundary conditions for a particular situation. One solution is

0

[2

s x

s

C C xerfC C Dt

−=

−]

where,

Cs - concentration of the diffusing atoms at the surface of the material

Cx - the concentration of the diffusing atom at location x below the surface after time t.

C0 - the initial uniform concentration of the diffusing atoms in the material

t - the diffusion time in (s)

x – the depth from the surface of the material ( cm )

D - the diffusion coefficient

Q-12.(b) We find that 10 h are required to successfully carburize a batch of 500 steel gears at

900°C, where the iron has the FCC structure. We find that it costs $ 1000 per hour to operate

the carburizing furnace at 900°C and $ 1500 per hour to operate the furnace at 1000° C. Is it

economical to increase the carburizing temperature to 1000°C ?

Solution

T = 900°C + 273 K = 1 173 K , t1173 = 10 hour

T = 1000°C = 273 K = 1273 K. , t 1273 = ?

For carbon diffusing in FCC iron, from Table,

the activation energy Q = 32,900 cal/mol. for the same carburizing treatment at

1000°C as at 900°C:

D1273 .t 1273 = D1173 .t1173

.t 1273 = D1173 .t1173 / D1273

)1273).(987.1/()900,32exp(

)1173).(987.1/()900,32exp()10(1273 −

−=

ht

t1273 = 3.299 hour

At 900°C, the cost per part is ($ 1000/h) (10h)/500 parts = $ 20/part

At l000°C, the cost per part is ($1500/h) (3.299h)/500 parts=$ 9.90/part

Considering only the cost of operating the furnace, increasing the temperature reduces

the heat-treating cost of the gears and increases the production rate.

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Q-13. A 0.001 in. BCC iron foil is used to separate a high hydrogen gas from a low

hydrogen gas at 650°C. 5 x 108 H atoms/ cm3 are in equilibrium with the hot side of the

foil, and 2 x 103 H atoms/cm3 are in equilibrium with the cold side. Determine

(i)the concentration gradient of hydrogen and

(ii)the flux of hydrogen through the foil.

Solution

In BCC iron foil, the flux of hydrogen through the foil, from hot side to cold side.

at 650°C., T = 650 + 273 = 923 K

Cinitial = 5 x 108 H atoms/ cm3

Cfinal = 2 x 103 H atoms/cm3

thickness - Δx = 0.001 in. = 0.001 in x 2.54 cm

(i) the concentration gradient (Δc/Δx) = ?

(ii) the flux of hydrogen through the foil = ?

(i) the concentration gradient (Δc/Δx) = )54.2()001.0(

105102 83

xxx −

= - 1969x108 H atoms/ cm3

(ii) the flux of hydrogen through the foil , J = - D.Δc/Δx

0 exp ( )QD DRT−

=

From Table 2-1, H in BCC iron, D0= 0.0012 , Q = 3600

= )923.987.1/00,36exp(0012.0 −=D

J = - D.Δc/Δx

= - x (- 1969x10)923.987.1/00,36exp(0012.0 − 8 )

= 0.33 x108 H atoms/ cm2.s

Q-14. 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 h, when 1.10% C is present at the surface? Assume

that the iron is FCC.

Solution

Given: 01.1% , 0.5% , 0.2% , 0.5 0.052 3600 7200sec , ?

s xC C C C C C x mmt hr x T

= = = = == = =

cm

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By Fick's second law

0

[2

s x

s

C C xerfC C Dt

−=

−]

1.1 0.5 0.050.667 [ ]1.1 0.2 2

erfDt

−= =

−


0.05 0.6852 Dt

=

20.05. ( 0.685) 0.001332

D t x= =

0 exp ( )QD DRT−

=

From the table 2-1, C in FCC iron, D0 = 0.23 , Q = 32,900 cal/mol

).987.1/900,32exp(23.0 TD −=

D.t = 0.00133 , D = 0.00133/ t , D = 0.00133/ 7200

and D = 0.00133/ 7200 ).987.1/900,32exp(23.0 TD −=

0.00133/ 7200 ).987.1/900,32exp(23.0 T−=

T = 1180 K = 1180 – 273 = 907 C

Q-15. A 0.15% C steel is to be carburized at 1100°C, giving 0.35% C at a distance of

1mm beneath the surface. If the surface composition is maintained at 0.90% C, what

time is required?

Solution

Given; 00.9% , 0.35% , 0.15% , 1 0.1

1100 273 1373 , ( ) ?s xC C C C C C x mm

T C K time t

= = = = =

= + = =

cm


0

[2

s x

s

C C xerfC C Dt

−=

−]

]2

1.0[733.015.09.035.09.0

tDerf==

−−


0.1 0.7862 Dt

=

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D.t = [ 0.1/2 x 0.786 ]2 = 0.00405

0 exp ( )QD DRT−

=


-6 20.23exp(-32,900 /1.987.1373) 1.332 10 /D x= = cm s

D.t = 0.00405 ,

t = 0.00405 / D = 0.00405 / 1.332 x 10-6 = 3040 s

time ( t ) = 3040 sec = 51 min

Q-16. A 0.02% C steel is to be carburized at 1200°C in 4 h, with a point 0.6 mm beneath

the surface reaching 0.45% C. Calculate the carbon content required at the surface of the

steel.

Solution

Given; 0?, 0.45% , 0.02% , 0.6 0.06

1200 273 1473 , ( ) 4 4 3600 4400secs xC C C C C x mm cm

T C K time t hr x

= = = = =

= + = = = =


0

[2

s x

s

C C xerfC C Dt

−=

−]

0.45 0.06[ ]0.02 2

s

s

C erfC Dt−

=−

0.06[2

erfDt

] = ?

0 exp ( )QD DRT−

=


= 3.019 x 10)1473.987.1/900,32exp(23.0 −=D -6 cm2 /s

6

0.06[ ] [0.144]2 3.019 10 .4. 3600.

erf erfx x x−

=


0.1442

xDt

= , [ ] 0.1612

xerfDt

=

161.002.045.0

=−−

s

s

CC

Cs = 0.53 C %

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Q-17. A 1.2% C tool steel held at 1150°C is exposed to oxygen for 48 h. The carbon

content at the steel surface is zero. To what depth will the steel be decarburized to

less than 0.20% C ?

Solution

Given; 00.% , 0.2% , 1.2% , ( ) ?

1150 273 1423 , ( ) 4 4 3600sec 4400secs xC C C C C C x cm

T C K time t hr x

= = = =

= + = = = =


0

[2

s x

s

C C xerfC C Dt

−=

−]

]2

[1667.02.102.00

tDxerf==

−−

1667.0]2

[ =tD

xerf , From Table 2-3 , we find that,

0.1492

xDt

= ,

0 exp ( )QD DRT−

=


= 2.034 x 10)1423.987.1/900,32exp(23.0 −=D -6 cm2 /s

.D t = 62.034 10 .48. .3600x x x− = 0.5929 ,

149.02

=tD

x , 149.05929.02

=x

x , x = 0.177 cm

Q-18. A BCC steel containing 0.001% N is nitrided at 550°C for 5 h. If the nitrogen content

at the steel surface is 0.08%, determine the nitrogen content at 0.25 mm from the surface.

Solution

Given; 00.08 %, ? %, 0.001 %, 0.25 0.25

550 273 823 , ( ) 5 5 3600sec 8000secs xC N C N C N x mm c

T C K time t hr x

= = = = =

= + = = = =

m


tD

xerfCCCC

s

xs

2[

0

=−− ]

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]2

[001.008.0

08.0tD

xerfCx =−−

0 exp ( )QD DRT−

=

From the table 2-1, N in BCC iron, D0 = 0.0047 , Q = 18,300 cal/mol

= 6.488 x 10)823..987.1/300,18exp(0047.0 xD −= -8 cm2 /s

D .t = 86.488. .10 . .5. .3600x x x− = 0.0342 ,

=]2

[tD

xerf ]0342.02

025.0[x

erf

3655.02

=tD

x , =]2

[tD

xerf 0.394

]2

[001.008.0

08.0tD

xerfCx =−− = 0.394

Cx = 0.049 N%

Q-19. Define the following.

(i) % Elongation (ii) Endurance limit (iii) Endurance Ratio

(iv) Fatigue life (v) Elastic deformation

(i) % Elongation - The total percentage increases in the length of a specimen during

a tensile test.

0

0

% 100%fl lElongation x

l−

=

where lf – final length of the specimen

l0 – initial length of the specimen

(ii) Endurance limit - The stress below which a material w i l l not fail in a fatigue test.

(iii) Endurance Ratio - The endurance l imi t divided by the tensile strength of the

material. The ratio is about 0.5 for many ferrous metals.

lim 0.5Endurance itEndurance ratiotrnsile strength

= ≅

The endurance ratio allows us to estimate fatigue properties from the tensile test.

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(iv) Fatigue life - The number of cycles permitted at a particular stress before a material

fails by fatigue. Fatigue life tells us how long a component survives at a particular stress.

(v) Elastic deformation - Deformation of the material that is recovered when the applied

load is removed.

Q-20. (a) Define Engineering Stress, True stress and Engineering Strain, True Strain.

(b) A force of 100,000 N is applied to a 10 mm x 20 mm 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.

Solution

(a) Engineering Stress - The applied load, or force, divided by the original cross

sectional area of the material.

0

. sec .

force FEngineering stressinitial cross tional area A

σ= = =

True stress - The load divided by the actual cross-sectional area of the

specimen at that load.

True stress = σ 1 = 1

1

AF

, σ 2 = 2

2

AF

Engineering Strain - The amount that a material deforms per unit length in a tensile test.

0

0

l lEngineering strainl

ε −= =

where l – final length of the specimen


True Strain - The strain, given by ε t = ln (l/l0), produced in a material.

True Strain ε t = ⎥⎦

⎤⎢⎣

⎡

0

lnll

where l – final length of the specimen


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Solution

Q-20.(b) Applied force = 100,000 N

(i) area of iron bar = 10 mm x 20 mm = 200 mm2

yield strength - 400 MPa

tensile strength - 480 MPa.

0

. sec .

force FEngineering stressinitial cross tional area A

σ= = =

σ = 100,000 N / 200 mm2 = 500 N/mm2

1 MPa = 1 N/mm2 , 500 MPa

applied stress σ is greater than yield strength ( 500 MPa > 400 MPa ),therefore,

the iron bar will plastically deform.

(ii) applied stress σ is greater than tensile strength ( 500 MPa > 480 MPa ),therefore,

the bar will occur necking.

Q-21. An aluminum alloy that has a plane strain fracture toughness of 25,000 psi-√in .fails

when a stress of 42,000 psi is applied. Observation of the fracture surface indicates that

fracture began at the surface of the part. Estimate the size of the flaw that initiated

fracture. Assume that f = 1 . 1 .

Solution

plane strain fracture toughness Kc = 25,000 psi-√in .

stress σ = 42,000 psi is applied.

the fracture began at the surface of the part.

the size of the flaw = ?. Assume that f = 1 . 1 .

KIC = f .σ. a.π

25,000 = 1.1 x 42,000 x a.π

a = 0.093 in

the initial flaw size on the surface – 0.093 in

Q-22. Define Hooke's law and Poisson's ratio.

Hooke's law

The relationship between stress and strain in the elastic portion of the

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stress-strain curve. The modulus of elasticity, or Young's modulus, E, is the slope of

the stress-strain curve in the elastic region. This relationship is Hooke's law :

E = εσ

where E - Young's modulus

σ - engineering stress

ε - engineering strain

Poisson's ratio

The ratio between the lateral and longitudinal strains in the elastic

region. Poisson's ratio, μ, relates the longitudinal elastic deformation produced by a

simple tensile or compressive stress to the lateral deformation that occurs simultaneously:

μ = nallongitudio

lateral(

)(ε

ε− , ( μ is about 0.3)

Q-23. A 0.4-in. diameter,12-in. long titanium bar has a yield strength of 50,000 psi,

a modulus of elasticity of 16 x 106 psi, and Poisson's ratio of 0.30. Determine the length

and diameter of the bar when a 500-Ib load is applied.

Solution

a titanium bar ,

diameter = 0.4-in., length = 12-in.

yield strength = 50,000 psi,

modulus of elasticity E = 16 x 106 psi,

Poisson's ratio μ = 0.30.

the length = ? and the diameter of the bar = ? when a 500-Ib load is applied.

F = 500 lb

the stress σ = F/A = 500 lb / (π/4 )(0.4 in.)2 = 3979 psi

By Hook's Law, E σε

=

the strain Eσε = = 3979 psi / 16 x 106 psi = 0.00024868 in/in

Engineering strain = 0

0

120.00024868

12f fl l ll

ε− −

= == =

12.00298fl i= n

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Poisson's ratio: ( ) 0.3(

laterallongitudionalεμ

ε−

= = −

ε(lateral) = - (μ )x ε(longitudinal) = - (0.3 ) (0.00024868) = -0.0000746 in/in

0

0

0.4- 0.0000746 /

0.4f fD D D

in inD− −

= =

0.39997fD i= n

the length of the bar = 12.00298 in

the diameter of the bar = 0.39997 in

Q-24. A 3-in.diameter rod of copper is to be reduced to a 2-in. diameter rod by being

pushed through an opening. To account for the elastic strain, what should be the

diameter of the opening? The modulus of elasticity for the copper is 17 x 106 psi and the

yield strength is 40,000 psi.

Solution

A copper rod is to be reduced in diameter,

D0 = 3-in. , D1 = 2-in. dia ,

the diameter of the opening ?

modulus of elasticity for the copper, E = 17 x 106 psi

yield strength σ = 40,000 psi.

To get 2 in diameter bar, the diameter of the opening must be smaller than the

final dia.

E σε

= , ε = σ / E = 40,000 / 17 x 106 = 0.00235

ε = 0.00235, Engineering strain = 0 0

0 0

l l D Dl D

ε − −= =

D = 2 in. , D0 = ?

0

0

20.00235 DD−

= , D0 = 1.995 in

the diameter of the opening die = 1.995 in.

Q-25.Which factors does depend on the ability of a material to resist the growth of a crack?

Solution

The ability of a material to resist the growth of a crack depends on a large number

of factors:

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1). the larger the flaws size, the smaller the permitted stress.

2). Increasing the strength of a given metal usually decreases ductility and gives a

lower fracture toughness.

3).Thicker, more rigid materials have lower fracture toughness than thin materials.

4). Increasing the rate of application of the load, such as in an impact test, typically

reduces the fracture toughness of the material.

5). Increasing the temperature normally increases the fracture toughness, just as in the

impact test.

6). A small grain size normally improves fracture toughness, more point defects and

dislocations reduce fracture toughness.


(i) Tensile strength (ii) Yield strength (iii) Ductility

Solution

(i) Tensile strength

The stress obtained at the highest applied force is the tensile strength, which is the

maximum stress on the engineering stress-strain curve.

(ii) Yield strength

The stress applied to a material that just causes permanent plastic

deformation.

(iii) Ductility

The ability of a material to be permanently deformed without breaking when a force is

applied. Ductility measures the amount of deformation that a material can withstand

without breaking.

Q-27. A ceramic part for a jet engine has a yield strength of 75,000 psi and a plane strain

fracture toughness of 5,000 psi - √in. . To be sure that the part does not fail, we plan to

assure that the maximum applied stress is only one-third the yield strength. We use a

nondestructive test that will detect any internal flaws greater than 0.05 in. long.

Assuming that f = 1.4, does our nondestructive test have the required sensitivity ?

Explain.

Solution

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A ceramic part for a jet engine ,

yield strength σ = 75,000 psi

plane strain fracture toughness K IC = 5,000 psi - √in. .

the maximum applied stress = one-third the yield strength. = 1/3 x 75,000 psi

internal flaw size = 2.a , f = 1.4,

K IC = f.σ. a.π

5000 = 1.4 x 1/3 x 75,000 . a.π

a = 0.0065 in

the length of internal flaw = 2.a = 2 x 0.0065 in = 0.013 in

Non- destructive test can detect the flaw size of length = 0.05 in.

Now, the flaw size is 0.013 in. and therefore, our nondestructive test have no the

required sensitivity.

Q-28. A large steel plate used in a nuclear reactor has a plane strain fracture toughness of 80,000

psi- √in and is exposed to a stress of 45,000 psi during service.Design a testing or

inspection procedure capable of detecting a crack at the surface of the plate before the crack

is likely to grow at a catastrophic rate.

Solution

steel plate used in a nuclear reactor,

applied stress σ = 45,000 psi

plane strain fracture toughness K IC = 80,000 psi - √in.

surface crack = a = ? , f = 1,

Under these condition, to determine the minimum size of crack.

K IC = Kc = f.σ. a.π

80,000 = 1 x 45,000 x a.π

a = 1 in.

This minimum crack size ( 1 in.) on the surface can be observed visually.

If the growth rate of the crack is slow, the inspection is performed by regular

method.

Q-29. A 850-lb force is applied to a 0.15-in. diameter nickel wire having a yield strength of

45,000 psi and a tensile strength of 55,000 psi. Determine

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(a) whether the wire will plastically deform and

(b) whether the wire will experience necking.

Solution

nickel wire,

yield strength = 45,000 psi , tensile strength = 55,000 psi.

applied force = 850-lb , initial diameter D0 = 0.15-in.

(i)whether the wire will plastically deform and

(ii)whether the wire will experience necking.

0

. sec .

force FEngineering stress on the wireinitial cross tional area A

σ= = =

σ = 850 lb / (π/4) (d )2 = 48,100 Psi

(i) Applied stress σ is greater than yield strength ( 48,100 Psi > 45,000 Psi ),

therefore, the nickel wire will plastically deform.

(ii)Applied stress σ is less than tensile strength ( 48,100 Psi < 55,000 Psi ),

therefore, the nickel wire will no necking occur.

Q-30. A solid shaft for a cement kiln produced from the tool steel must be 96 inches long and

must survive continuous operation for one year with an applied load of 12,500 lb. The shaft

makes one revolution per minute during operation. Design a shaft that will satisfy these

requirements.

Solution

a tool steel solid shaft for a cement kiln,

length L = 96 in. , applied load ( F ) = 12,500 lb.

continuous operation for one year, and one revolution per minute during operation.

Design a shaft that will satisfy these requirements.

It means that minimum diameter of the shaft (d ) = ?

Number of cycle / year = N = 1 . .365. . 24. . .60.min1.min . .1. . .1. . .1. .

cycle x days x hr xx year x day x hr

N = 5.265 x 105 cycles/year

From Fig. 3-13 , S-N curve for tool steel,

N = 5.265 x 105 ,,, applied stress σ = 72,000 Psi or 72 Ksi.

applied stress σ must be less than 72,000 Psi or 72 Ksi.

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By the equation, 3

..18.10d

FL=σ

3

10.18. .96. .1250072,000 x xd

=

5.54d i= n

For these conditions, the diameter of the shaft 5.54 in. will operate for one year.

But, safety is required in the design without failure.

In Fig. endurance limit 60,000 Psi, < 72,000 Psi., this condition is minimum diameter required

to prevent failure.

3

10.18. .96. .1250060,000 x xd

=

5.88d i= n

The condition that will operate for more than one year.

without failure, the minimum diameter of the shaft (d ) is 5.88 in

* * * * * * * * * * * * *END* * * * * * * * * * * *

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