MidTerm solution(1) (3).docx

7/26/2019 MidTerm solution(1) (3).docx

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ENGR 2220U Structure and Properties of

MaterialsMid-Term Examination

Instructor: Ghaus Rizvi

Time: 120 minutes

Date: October 20, 2012

Instructions:

Closed book & closed laptop. A calculator is allowed.

Answer all questions

The questions do not carry equal marks. Spend your time accordingly.

The total mark is 100 and the exam time is 120 minutes.

The marks for each question are given besides the Q# No communication with others at any time. Data sheet is on the back


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Q-1 Multiple choice, each question carries equal marks (15)

1) Ceramics are:

a. Hard and very toughb. Hard but not tough

c. Brittle and softd. malleable and strong

2) Important properties of Metals are:

a. Strong and brittleb. Strong and ductile

c. Hard and non-deformable

d. Stiff and non-deformable

3) Some examples of polymers are:

a. Cotton, Nylon, petroleum, silk and hemp fibers

b. Cotton, Nylon, silk, glass fibers and wool fibersc. Cotton, Nylon, glass fibers, wool, silk and hemp fibers

d. Cotton, Nylon, silk, polyethylene and PVC

4) Which of the following statements is true?a. Biomaterials are materials that mainly contain hydrocarbons

b. Biomaterials have their origins in agricultural products only

c. Biomaterials should not produce toxic substances in human bodyd. Biomaterials easily degrade when exposed to sun light

5) Properties of Iron cannot be changed over a very wide range by imparting proper heat

treatmenta. True

b.False

6) Non-linear stress strain relationship is generally exhibited by

a. Polymeric materials

b. Metallic materialsc. Composites

d. Ceramics

7) Poissons ratio isa. The ratio of applied stress to the resulting strain

b. The ratio of lateral strain to the normal strain

c. The ratio of applied stress to lateral strain

8) True stress on a body deformed by the application of a force is defined as:

a. Force divided by the actual cross sectional area

b. Force multiplied by the actual cross sectional areac. Force multiplied by the original cross sectional area


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d. Force divided by the original cross sectional area

9) A material is isotropic if it:

a. has uniform compositionb. Exhibits uniform properties in all direction

c. Exhibits different properties in different direction

10) The larger the difference in electro negativity between two atoms, the greater is the

tendency to form ionic bonds

a.Tb. F

11) The solubility of a solute in general

a. decreases with increasing pressureb. decreases with increasing temperature

c. is higher when crystal structures of constituents are similar

d. is lower when the atom sizes are similar

12) The equilibrium number of vacancies Nv is a function of the following parameters

a.Total number of atomic sites

b.Temperaturec. Forces between atomsd.Boltzmanns constant

13) In a unit crystal, 110 represents

a. a point within it

b. a direction in it

c. a plane in it

14) Covalent bonds are in general

a. Stronger than ionic bondsb. Weaker than secondary bonds

c. Non directional in nature

d. Directional in nature

15) Polymorphism is defined as

a. the property of a material which enables it to form different crystal structure under

different conditionsb. the property of a material which enables it to form different crystal sizes under

different conditions

c. the property of a material which enables it to form different compositions under

different conditions


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Q1(6 marks)

Sketch the following directions ]210[ and ]121[ in cubic unit cell? (2)

a) ]210[

b) ]121[

1/2

1/2

x

y

z


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Determine the Miller indices for the two planes shown in the following unit cell.(4)

Plane a

Plane b

Q2(8 marks)

If the theoretical density of copper is 8.90 g /cm3, calculate the atomic radius of copper.

First derive the relationship between the lattice parameter a of the crystal structure andthe atomic radius rof copper.

b

a

-1/3

1/2

(121)

a b c

Intercepts 1 -1/2 1

Reciprocal 1/1 -1/1/2 1/1

Lowest integers 1 -2 1

(113)

a b c

Intercepts 1 1 -1/3

Reciprocal 1/1 1/1 -1/1/3

Lowest integers 1 1 -3


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All edges are equal in cubic structure: a =b =c

Close packed direction (use Pythagoras theorem): (a)2+ (a)

2= (4R)

2

2 a2= (4R)

2

a = 4R/2Volume of cell: a

3= (4R/2 )3 = 16R32

Theoretical density eq.:AC

Cu

NV

nA=

A

Cu

NR

nA

)216(=

3

Solving forRfrom the above expression yields

1/3

216=

A

Cu

N

nAR

1amu = 1g/mol

1/3

2atoms/mol106.022g/cm16)(

amu63.55cell)atoms/unit4(=

))( ()(8.9 233

= 1.27 10-8cm = 0.127 nm

Q3(6 marks)

Determine the composition, in weight percent, of an alloy containing 75 atom% nickel

and 25 atom% copper.

For composition, in weight percent, of a 75 at% Ni-25 at% Cu alloy, employ mentioned

below 1amu = 1gmol

100=Ni CuCuNiNi

NiNi

ACAC

ACC

100)55.63)(25()69.58)(75(

)69.58)(75(=

amuamu

amu

= 73.47 wt%

Similarly use equation for wt.% of Cu


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100=Ni CuCuNiNi

CuCu

ACAC

ACC

100

)55.63)(25()69.58)(75(

)55.63)(25(=

amuamu

amu

= 26.53 %

Q4(15)

To case harden a steel part, carbon environment is used to set the surface carbon contentat 0.9 wt%. The initial carbon content of steel is 0.2 wt%. Calculate the time it would,

take at 1000C, to reach a carbon content of 0.5 wt% at a distance of 1.5 mm from the

surface. (Hint: use the diffusion coefficient of gamma () iron)

Co = 0.2 wt.% T = 1000oC = 1000 + 273 = 1273

ok

Cs = 0.9 wt.%

Cx = 0.5 wt.%x = 1.55 mm

t=

Cx C0Cs C0

= 1 erfx

2 Dt

Dt

x

CC

CC

s

x

2erf1=0.428=

20.09.0

20.05.0=

0

0

0.572=0.4281=2

erf

Dt

x

By linear interpolation using data from Table 5.1

z erf(z)

0.55 0.563

z 0.572

0.60 0.6039


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563.06039.0

563.0572.0=

563.060.0

55.0

z

Dt

x

z 2=0.5503=

Now, from Table 5.2, at 1000oC (1273 K)

RT

QDD dexp=0

D = (2.3 10-5 m2/s)exp 148,000 J/mol

(8.31 J/mol- K)(1273 K)

= 1.93 10-11m2/s

Thus,

)()/sm1093.1()2(

105.1=0.5503

211

3

t

m

Solving for tyields

t= 9.6 104s = 26.73 h

Q5 (15)

For a bronze alloy, the stress at which plastic deformation begins is 267 MPa and themodulus of elasticity is 115 GPa.

(a) What is the maximum load (in N) that may be applied to a specimen having a circular

cross-sectional area of 300 mm2 without plastic deformation?

(b) If the original length of the specimen in part a) is 137 mm, what is the maximum

length (in mm) to which it may be stretched without causing plastic deformation?

Calculate the new diameter corresponding to stretched length. (v=0.33)

(a) Determination of max. load at which plastic deformation begins

)300()( 2-6260 m10N/m10267== AF yy

= 80100 N


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(b)The maximum length at which the sample may be deformed without plastic

deformation is determined by combining two equations

li = l0 1 E

mm137.32=

10115

2671mm)(137=

3

MPa

MPa

Determination of new diameter, di (v= 0.34)

z= L/Li= (Lf-Li)/Li= (137.32-137)/137 = 137.32/137 = 0.0023

z = - x/vRearrange the formula

x= - zv = - 0.0023 * 0.34 = - 0.000794x= d/do= (di-do)/do-0.000794 = (di-19.54)/19.54

di = 19.524 mm2

Q6 (10)

A suspension bridge is to be supported by a series of steel wires. It is estimated that theload on each wire will be 11,100 N. Determine the minimum required wire diameter

assuming a factor of safety of 3 and yield strength of 1030 MPa.

Determination of working stress

MPa33.433=

3

1030=

3

= MPay

w

Since the force is given, the area may be determined from Equation 6.1, and subsequently

the original diameter d0may be calculated as

Eo

oi

l

ll

A

F


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Rearrange the equation:

A0 =F

w=

d0

2

2

Finally

)33.343( 260

/10

)100,11)(4(=

4=

mN

NFd

w

do= 6.41 10-3m = 6.41 mm

Q7(15)

Consider a single crystal of BCC iron oriented such that a tensile stress is applied along[010] direction.Compute the resolved shear stress along a (110) plane and in a [111]

direction when a tensile stress of 52 MPa is applied.

The angle is the angle between the tensile axisi.e., along the [010] directionand theslip directioni.e. ]111[ . The angle may be determined using Equation

cos1 u1u2 v1v2 w1w2

u12 v1

2 w12 u22 v22 w22

where (for [010]) u1 = 0, v1 = 0, w1 = 1, and (for ]111[ ) u2 = 1, v2 = 0, w2 = 1.

Therefore, where (for [111]) u1= 0, v1= 0, w1= 1, and (for [1 01]) u2=1, v2= 0, w2=

1. Therefore, is equal to

222222

1

)1()1()1()0()1()0(

)1)(0()1(1)()1)(0(cos


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7.54

3

1cos 1

Furthermore, is the angle between the tensile axisthe [010] directionand the

normal to the slip planei.e., the (110) plane; Therefore,

222222

1

)0()1()1()0()1()0(

)0)(0()1(1)()1)(0(cos

45

2

1cos 1

Finally, the critical resolved shear stress is equal to

crss =y (cos cos )

MPa24.12=)45cos()7.54cos(MPa)(52=

Q8(10)

Give short answers to the following:

1. Write down the names of at least two zero dimensional defects. (1)

VacanciesSubstitutional atoms

Interstitial atoms

Self-interstitial atoms

2. What is the difference between in critical resolved shear stress & resolved shear

stress (2)

Critical resolved shear stress is the threshold value of shear stress at which slip

initiates in the system while resolved shear stress is the shear component in the slipdirection, of applied tensile stress

3. Why interstitial diffusion is faster than vacancy diffusion? (3)Due to small atomic sizes and availability of more interstitial sites in atomic structure

4. Deform a metal and it gets harderwhy? (3)


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After deformation metal becomes harder due to increase in dislocation density that

generates the strain field in the vicinity of dislocation and hinders their motion.

5. What is a necessary condition for substitution diffusion to occur? (1)There should be a vacancy in the crystal structure for substitution diffusion

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MidTerm solution(1) (3).docx