Applied Science 278 Midterm Examination - October 21 51 ,2013

THE UNIVERSITY OF BRITISH COLUMBIA

Department of Materials Engineering

APSC 278 Engineering Materials

MID-TERM EXAMINATION, October 21, 2013

This is a closed book examination.

Use of relevant material stored in programmable calculators prohibited.

Answer all questions. Show all work and units.

The complete exam is 8 pages in length

Name: __________________________

Student #: ____________________

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Applied Science 278 Midterm Examination - October 21 S\ 2013

A cylindrical rod 750 mm in length, having a diameter of 12.5 mm is to be subjected to a tensile load. The rod should experience neither plastic deformation nor an elongation of more than 2 mm when the applied load is 45 KN.

Tensile Strength / Poisson RatioModulus of Yield Strength / Material ~~Jm) J?Wcj~ !MPaElasticity/ GPa MPa '!,.. qj , . 70 350 450 033Aluminum alloy

.X'1es :x. .

100 400 500 034Brass )(Ok .

1:11._-'01

Applied Science 278 Midterm Examination - October 21 51,2013

(4) Show that the true stress in a tensile test can be expr sed as: "o~J J ivA;." -fv~~ >te,~1 - 0"T =0"(1 + cy

PIII)k Je(O!J\-~ ~ Q i)J~~IC1WH>/~ Ao~ ~ AI.(fT; ~:: .Il; ~ (fi' ,~i :: (f., . .'''-(0 t ~ :; (] (Iff'

Af A010 (,0 to .

(3) Calculate the strain hardening exponent for an alloy in which a true stress of 450 MPa produces a true strain of 0.15. Assume a value of 950 MPa for the strength coefficient.

f\ fJT:: K~

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Applied Science 278 Midtenn Examination - October 21st, 2013

(3) sketch the FCC unit cell and then calculate the atomic packing factor assuming the atoms to be hard spheres.

o 'c;v

~/~~ZCl'-I 0 Q~~hR

APF~~tA~fr'S )(v"J~ ~c~~/tc ') "" t)( ~l(t< ICdi{

Applied Science 278 Midterm Examination - October 21 st , 2013

(2) define what is meant by the Burgers vector of a dislocation.

Mlt~{\ifwJ~ ~~ dA(w,'Q-v\ of 7ti l~irl.t ~l f-Orh~ C{\)O~ltfJ VVlt p.. &t'>(~"n

(4) compare and contrast the microstructure and dislocation density of cold and hot worked materials.

Cdd fVorf.ul - de~(~ ~~ , l1AJh ~1Io~ ~j~

ho+NO t'kJ - (U'aJ Mid er ""'- ....lot J afilA j I luw rk foCi Vh~ ~tJ~ ~ etWf'1 -ft; fhtJ~ .

11t!.t .Ir~ : &)"lV (b) - e~pt~ >t*~ fov atu~ kJ f1I (Jet, lttfu", Ih Vfv~flU. ~ o-r /h01-t- is ~ fWe-~ ~ ~t. ~J C::InJvf)k..., ~ l~{wk;~ :' v&l~~ct, h~ ~M~ ~/lRAt, N~,fhvA/6~l ~~W OK ~ ~i/v~ b~. 74 ~O'f I~ 14vJe, )0 c4~1+ ~ ~ ~ MCtM '~ ~1? ~!J ~;~~~ ~. )c~(,;- ~I~(f)\y\ C1 I"Sr/affJr\ W N 11 ho~ j~ -'~e(~.~ ~e11 A. I - 11. . - _, , r".. ,,1;r' f::::... r ~ .l.-v.J. 1\".1.,.. (-I)" _LFI' j,~ L"lA. J

Applied Science 278 Midterm Examination - October 21 51 , 20 I 3

(4) Steady-state creep data taken for iron at a stress level of 140 MPa gives the following data. If it is known that the value of the stress exponent for this alloy is 8.5 , compute the steady state creep rate at 875 C

}J;.eady state creep rate (h' I) Stress (MPa) Temperature (0C) Ce) 6.6x I 0-4 140 820 f7hJ 8.8x I O'l 140 925 \:7

(4) The aluminum alloy, AA2024-TJ, has a plane strain fracture toughness of 44 M Parm and a yield stress of 345 MPa. For the geometry shown below, determine the maximum load which can be applied to the plate such that it does not fai I under the following circumstances: thickness of 5 mm, a width of 100 mm and a crack length, a, of 20 mm.

F

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Applied Science 278 Midtenn Examination - October 21 " , 2013

1~-(J~ ~

rJ) frAVh.. Vt,

(3) Can you get fatigue failures without applying oscillating loads? Why?

f\;~1 .-{tJ1~~ re11~rt~ I$p O>ttt(tt;h ~cb.J Osc",/(cd)J lotu,{~ ~V1lv tri ~Uw,.iJ/AI1 t~ (JVVJc.J ptt-jh'&t~ fo tdt ihihtitJtt ~ ~ C-r&\tk!, . No o~{./)'{a;h~ I.x~, rto \~~.J

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Applied Science 278 Midterm Examination - October 21 s', 2013

Useful Formulae

;r: y v=--=-

0 :

E = 2G(1+ v) 1 0"y 2 U =-0" X=

r 2 y 2

0" y + 0""

toughness ~ 2 x & J

O"r =0"(1+) &r =In(l + &) aT =K;

nA/ NA p= Vc

nA =2d sine

&s = K 2(yn exp ( ~;) Ny = N exp( -~ )

E () theoretical =10 Til = (yCOS>COSA

am ~a,rl+{;Jl

0" =(2EY s) }i c 7rQ

a , ~ ( 2E(r,;,,+ relf (Y y = fz;; i y (e)2Jrr K=YaJ;;

B~2tJJ R = (Y min

a max

M ="xMn ~ I I

M ="wMI ,w ~

MwDPw =nw =~ m

L =Nd sinCe)2

r=dFn or r = AFn i.e. d:Ed

EwJ. = E\(.E.] 2 p,

Ee =EJVJ + EI//Vm EmErE = .

C EmVr + ErVm Ff = EfVf Fm EmVm

Ff = Ef / Em

Fe Ef / Em + V m / Vf

a~d 1=

c 2Tc

Ec=KENf + EmVm

v = IR pl

R=-;;: 1

(5=P

J = (5E V

E=l

(5 = nleltle + pleltlh Va

tle or tlh = E -Egap/

ni ex: e kT Vs

n=C NA=6.023 x 1023 atoms/mol

R = 8.314 Jmol-1K-1

Charge on electron = 1.6 x 10.19 C

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