Assignment 2 Sol(1)

8/10/2019 Assignment 2 Sol(1)

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ENGR 2220 Assignment #2Structure & Properties of Materials

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1. Calculate the concentration of vacancies in aluminum at room temperature (25 0C . Assume that20!000 cal are re"uire to pro uce a mole of vacancies in aluminum.

Solution$%he lattice parameter of CC aluminum is 0.'0'' 51 nm. %herefore! the num)er of aluminumatoms or lattice points per cm * is

*22*+ ,100'5.100'' 51.'(

,' - cmatoms Al x

cm xcell atoms

n =

At room temperature! %-25 2/* -2 +

( )

=

K xk mol cal mol cal cmatoms x

RT Qn vv 2 +.,+/.1

,20000e p,100'5.(e pn- *22

- 1.2 10+

vacancies,cm*

2. Calculate the composition! in 3eight percent an atom percent! of an allo4 that contains 15+.0g titanium! 20.* g of aluminum! an +. g of vana ium.

Solution

%he concentration! in 3eight percent! of an element in an allo4 ma4 )e compute using amo ifie form of 6"uation '.*. or this allo4! the concentration of titanium ( C %i is 7ust

C Ti =mTi

mTi + mAl + mV 100

3t8+'.5'-100 .+*.205+1

5+1 - ++ kg kg kg

kg

Similarl4! for aluminum

3t810.+-100 .+*.205+1

0.*2 -Al ++ kg kg kg

kg C

An for vana ium

1 of


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3t8'.-100 .+*.205+1

.+ -9 ++ kg kg kg

kg C

Atom percent:

Masses must ne t )e converte into moles (6"uation '.' ! as

mol**00.(-,/.+/' 5+0001

--%i mol g

g Am

nTi

Tim

mol/52.'-,. +2 0*002

-Al mol g

g n m

mol1 +.+-,0. '5 00+

-9 mol g

g n m

:o3! emplo4ment of a mo ifie form of 6"uation '.5! gives

100

-

%i ++V Al Ti

Ti

mmm

m

nnn

nC

at8/+.1+-100 +.+1'./52*00.*

*00.* -

++ mol mol mol mol

100

-

Al ++V Al Ti

Al

mmm

m

nnn

nC

at81/.+2-100 +.+1'./52*00.*

52.'/ -

++ mol mol mol mol

100

-

9 ++V Al Ti

V

mmm

m

nnn

nC

at8'.0-100 +.+1'./52*00.*

+.+1 -

++ mol mol mol

mol

*. The outer surface of a steel gear is to be hardened by increasing its carbon content. The carbon isto be supplied from an external carbon rich atmosphere! "hich is maintained at an elevatedtemperature. A diffusion heat treatment at #$% C &''#( K) for '* min increases the carbonconcentration to *.#* "t+ at a position '.* mm belo" the surface. ,stimate the diffusion time re-uired at % % C &(/( K) to achieve this same concentration also at a '.* mm position. Assume that the

2 of


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'. %he iffusion coefficient for Cr * in Cr 2= * is 10 >15 cm 2,s at /2/ 0C an is 1 10 > cm 2,s at 1'00 0C.Calculate (a the activation energ4 an () the constant @ 0.

Solution

(a

(1 /*>?,mol(+.*1e p

(1000>?,mol(+.*1e p

-,101,10

0

0

2

215

Q 2

Q 2

scm x scm x

( )[ ] [ ]Q x 0000'+0/.0e p0.0000/1>0.00012>e p-10 =

[ ]Q0000'+0/.0-02'.12 -2.5 10 5 ?,mol

() [ ] 0+o5

o 1055.1+.1/e p@(1 /*>?,mol(+.*1,2.5 10>

e p@-101 2 xmol 3

x ==

@o - 0.0 '5 cm 2,s

%. An 1CC iron carbon alloy initially containing *.$( "t+ C is carburi4ed at an elevated temperatureand in an atmosphere "herein the surface carbon concentration is maintained at '.* "t+. 5f after /#.%h the concentration of carbon is *./6 "t+ at a position / mm belo" the surface! determine the

temperature at "hich the treatment "as carried out.

Solution

%his pro)lem as s us to compute the temperature at 3hich a nonstea 4>state ' .5 h iffusion

anneal 3as carrie out in or er to give a car)on concentration of 0.*5 3t8 C in CC e at a position

'.0 mm )elo3 the surface. rom 6"uation 5.5

2t

x

C C

C C

s

x

2 erf 1-0.20+*-

2+.00.1

2+.0'*.0 -

0

0

=r

0./ 1-2

erf

2t x

' of


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. A 1.01 cm iameter! *0.'+ cm long titanium )ar has a 4iel strength of *''./ MPa! a mo ulus ofelasticit4 of 110.* BPa! an Poisson s ratio of 0.*0. @etermine the length an iameter of the )ar 3hena 2225 : loa is applie .

Solution

The stress is D = F/A = 2225 N/( /4)(1.016 x 10 -2 m.) 2 = 27.4 MPa The applied stress is much less tha the !ield stre "th# there$%re &%%'e sla ca *e used.

The strai is + = D/E = 27.4 MPa / (110., x 10 , MPa) = 0.00024 m./m.

0.0002'++*0'+.0

*0'+.0 - =

m

ml

l

l l f

o

o f

lf = 0.,04 76 i .

r%m P%iss% s rati% = +lat / +l% " = 0.,+lat = (0.,)(0.00024 ) = 0.0000746 m./m.

0.0000/'(>0101(.0

0101(.0 - =

m

md

d

d d f

o

o f

df = 0.01015 m.

. A cylindrical rod '** mm long and having a diameter of '*.* mm is to be deformed using a tensile

load of $'!*** 7. 5t must not experience either plastic deformation or a diameter reduction of morethan .% '* 6 mm. 8f the materials listed as follo"s! "hich are possible candidates9 3ustify your

choice&s).

Material

Modulus of Elasticity

(GPa)Yield Strength

(MPa)

Poissons

Ratio

Aluminum alloy * $** *.66

:rass alloy '*' 6** *.6/

;teel alloy $* /** *.6*

Titanium alloy '*


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ENGR 2220 Assignment #2Structure & Properties of Materialsapplie < this means that the stress correspon ing to this loa not e cee the 4iel strength of the

material. Epon computing the stress

M2 /.*+- :,m102 /.*+-

21010

000!21-

2

-- 22*200

m

7

d

1

A

1

=f the allo4s liste ! the %i! Frass an steel allo4s have 4iel strengths greater than 2 /.*+ MPa.

Gelative to the secon criterion (i.e.! that d )e less than /.5 10>* mm ! it is necessar4 to

calculate the change in iameter H d for these three allo4s. rom 6"uation .+

- x

4

-

d d 0

,

= , d d 0

:o3! solving for H d from this e pression!

d -

d 0

,

or the steel allo4

mm10+/.*-1020/

mm(10MPa*+./2(*0.0(- *>* =>a

d

%herefore! the steel is a can i ate.

or the %i allo4

mm10'.+-1010/

mm10(MPa*+./2(*'.0(-

*>* =>ad

Ience! the titanium allo4 is not a can i ate.

or the Frass allo4

mm100.-10101

mm10(MPa*+./2(*'.0(- *>*

=>a

d

+ of


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Ience! the Frass allo4 is not a can i ate.

(. A cylindrical specimen of a hypothetical metal alloy is stressed in compression. 5f its original and

final diameters are '$.*** and '$.*$% mm! respectively! and its final length is '** mm! compute itsoriginal length if the deformation is totally elastic. The elastic and shear moduli for this alloy are '*%?>a and 6#. ?>a! respectively.

Solution

%his pro)lem as s that 3e compute the original length of a c4lin rical specimen that is stresse

in compression. ;t is first convenient to compute the lateral strain x as

*>

0

102.0+-mm12.000

mm2.0001mm12.025-- d

d x

;n or er to etermine the longitu inal strain 4 3e nee PoissonJs ratio! 3hich ma4 )e compute using

6"uation . < solving for 4iel s

=

, 2 ?

1 =105 10 3 MPa

(2) (39.7 10 3 MPa ) 1 = 0.322

:o3 4 ma4 )e compute from 6"uation .+ as

*>*

10'/.(-*22.0

100+.2--

x

4

:o3 solving for l 0 using 6"uation .2

l 0 -

l i1 + 4

mm 100.(5-10'/.(1

mm001 -

*

of

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Assignment 2 Sol(1)