Assignment_3_sol(3).doc

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ENGR 2220 Assignment #3

Structure & Properties of Materials

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1. Consider a single crystal of silver oriented such that a tensile stress is applied along a [001] direction. If slip occurs on a

(111)plane and in a [1 01

] direction, and is initiated at an applied tensile stress of 1.1 MPa (160 psi, co!pute the

critical resolved shear stress.

Solution

This problem asks that e compute the critical resol!e" shear stress for sil!er. n or"er to "o this$ e must emplo%

'uation .$ but first it is necessar% to sol!e for the angles an" hich are shon in the sketch belo.

The angle is the angle beteen the tensile a*is+i.e.$ along the ,--1 "irection+an" the slip "irection+i.e.$ [1 -1] .

The angle ma% be "etermine" using 'uation ./ as

= cos1 u1u0 + v1v0 + w1w0

u10 + v1

0 + w10( )u00 + v00 + w00( )

here (for ,--1) u1 -$v

1 -$ "

1 1$ an" (for [1 -1] ) u0

21$ v0 -$ "

0 1. Therefore$ is e'ual to

= cos1(0)(1) + (0)(0) + (1)(1)

(0)2 + (0)2 + (1)2[ ] (1)2 + (0)2 + (1)2[ ]

= cos11

0

= 3

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4urthermore$ is the angle beteen the tensile a*is+the ,--1 "irection+an" the normal to the slip plane+i.e.$ the (111)

plane5 for this case this normal is along a ,111 "irection. Therefore$ again using 'uation ./

= cos1 (-)(1) + (-)(1) + (1)(1)

(-)0 + (-)0 + (1)0[ ] (1)0 + (1)0 + (1)0[ ]

= cos11

3

= 54.7

An"$ finall%$ using 'uation .$ the critical resol!e" shear stress is e'ual to

crss y (cos cos )

= (1.1 MPa) cos(54.7) cos(45)[ ] = (1.1 MPa) 13

1

2

= 0.45 MPa (65.1 psi)

0 of


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0. Consider a single crystal of so!e hypothetical !etal that has the #CC crystal structure and is oriented such that a tensile

stress is applied along a ., 121 direction. If slip occurs on a (111)plane and in a [1 01]direction, co!pute the stress

at "hich the crystal yields if its critical resolved shear stress is $.0 MPa.

Solution

This problem asks for us to "etermine the tensile stress at hich a 466 metal %iel"s hen the stress is applie"

along a ., 121 "irection such that slip occurs on a (111) plane an" in a [1 -1] "irection5 the critical resol!e" shear

stress for this metal is 3.0 MPa. To sol!e this problem e use 'uation .5 hoe!er it is first necessar% to "etermine the

!alues of an" . These "eterminations are possible using 'uation ./. 7o$ is the angle beteen ., 121 an"

[1 -1 ]"irections. Therefore$ relati!e to 'uation ./ let us take u

1 21$ v

1 1$ an" "

1 0$ as ell as u

0 21$ v

0 -$

an" "0 1. This lea"s to

= cos1 u1u0 + v1v0 + w1w0

u10 + v1

0 + w10( )u00 + v00 + w00( )

[ ][ ]

++++

++=

000000

1

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

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

=

10

3cos

7o for the "etermination of $ the normal to the (111) slip plane is the ,111 "irection. Again using 'uation ./$ here

e no take u1 21$ v

1 -$ "

1 0 (for ., 121 )$ an" u0 1$ v0 1$ "0 1 (for ,111). Thus$

[ ][ ]

++++

++= 000000

1

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

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

=

18

0cos

t is no possible to compute the %iel" stress (using 'uation .) as

MPaMPacrss

y 0.10

18

0

10

3

-.

coscos=

==

3 of


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3. %"o previously undefor!ed cylindrical speci!ens of an alloy are to &e strain hardened &y reducing their cross'sectional

areas ("hile !aintaining their circular cross sections. #or one speci!en, the initial and defor!ed radii are 1 !! and )

!!, respectively. %he second speci!en, "ith an initial radius of 1* !!, !ust have the sa!e defor!ed hardness as the first

speci!en+ co!pute the second speci!ens radius after defor!ation.

Solution

n or"er for these to c%lin"rical specimens to ha!e the same "eforme" har"ness$ the% must be "eforme" to the

same percent col" ork. 4or the first specimen

%CW =-0 -d

-0

100 = r

0

2 r

d2

r0

2 100

96:31--)18(

);()18(

0

00

!!

!!!!

4or the secon" specimen$ the "eforme" ra"ius is compute" using the abo!e e'uation an" sol!ing for rdas

rd r

-1

%6:

1--

mm-./1--

931mm)(10 C.

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. - sheet of #CC iron 1.* !! thic/ "as eposed to a car&uriing gas at!osphere on one side and a decar&uriing

at!osphere on the other side at 2*$C. -fter having reached steady state, the iron "as 3uic/ly cooled to roo! te!perature.

%he car&on concentrations at the t"o surfaces of the sheet "ere deter!ined to &e 0.01* and 0.002$ "t4. Co!pute the

diffusion coefficient if the diffusion flu is *.5 10 ')/g!*'s. 7int8 9se :3uation 5.) to convert the concentrations fro!

"eight percent to /ilogra!s of car&on per cu&ic !eter of iron.

Solution

6

(

g=cm0.0

--(.-

--(.-

+

C

-.8; kg 6=m3

7o$ using a rearrange" form of 'uation .3

; 1


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3

3

3

0;@

kg=m8;.-kg=m.;3-

1-0*s@kg=m1--. )0( !

1.3 1-

@11

m

0

=s

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Structure & Properties of Materials. 6onsi"er a metal single cr%stal oriente" such that its slip plane normal makes an angle of 0.8 -ith

the tensile a*is. Three possible slip "irections make angles of /-$ 8.0-$ an" 83.1-ith the same tensile

a*is.(a) :hich slip s%stem(s) ill become acti!e first

(b) f plastic "eformationn begins at a tensile stress of 1.; MPa (08- psi)$ "etermine the critical

resol!e" shear stress for metal.

Solution

:e are aske" to compute the critical resol!e" shear stress for Al. As stipulate" in the problem$ 0.8$ hile

possible !alues for are /-$ 8.0-$ an" 83.1-.

(a) Slip ill occur along that "irection for hich (cos cos ) is a ma*imum$ or$ in this case$ for the largest cos .

6osines for the possible !alues are gi!en belo.

cos(/) -.0

cos(8.0) -.0-

cos(83.1) -.10

Thus$ the slip "irection is at an angle of /0.ith the tensile a*is.

(b) 4rom 'uation .$ the critical resol!e" shear stress is Bust

crss = y (cos cos )max

[ ] MPa-.3)(cos)8.0(cosMPa)(1.; 6

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Assignment_3_sol(3).doc