Unformatted text preview: 1) You are gm the data at the ﬁght for the tensile
test ofacyhndricalAlrudwithadiametEr at 123 ”Mm ham“)
mm and an initial 1&1ng uf50 8 mm Determine 0 507330
the:
15,100 50.902
2) Modulus ofelasnuty
b) “as“: strength 30.400 51.003
c) Loadatwhiehneckingbeginslooceur
d) Amumldfplarhcmiualalmddflsionn 38"“ 5”“
. 44,800 52.832
You do not need to graph anydnog In complete this prohlan
(15 pm.) 47,300 54.864
2) The most accurate way to determine E is by 463100 56396
graphing the data shown, then obtaining the 42,600 53420
slope of the initial linear pnrtiuu. We will take a
shortcut and just use the ﬁrst two points. This Frrlmne
yields:
(15,100 N)
_ z
A” £(12.8x10 ’ m)
E = _ —“ : 5.04x10‘" Po
A: 50.9027 50.8
50.8
(30,400 N)
A” 5(120x10’3 m)2
E = 7 = 4— = 5.91zr10‘° Po
A: 51.003 — 50.8
50.8 We will lake the average as 1: : 5.07x10“ 1):. h) The tensile strength is simply the maximum engineering stress that the
material can withstand. The maximum engineering stress mnespuuds tn the
maximum load, 47,300 N. Therefore: (47,300 N) TS : : 3.60x10' Po 5 123x104»: z
4( ) c) The load at which necking occurs is simply the maximum load, 47,300 N. d) For the ﬁrst data point given (F : 15,100 N), we are probably in the region of
elastic strain, so the plastil: strain is probably zero. We have already veriﬁed
this, by showing in part a) that the calculated E is the same fur the ﬁrst two
data points. ll'we had reached the nonlinear portion of the stressstrain
curve, where plastic strain becomes signiﬁcant, then the two calculations in
part a) wuuld have yielded different answers. 2) Asingle crystalofCu issubjectedloawnslle stress 050,000 psialoogdle [211]
lhrection. Iftheshpplame and slip dilectiuninthiscrystal are(l ll)and[mi] respectively, delmnine the resolved shear stress on his slip plane. (15 111.5.) Here we need to use equation (8.1) fur the resolved shear stress, but we also need to
use equation (3.5) to determine the cosine of the angles involved: lilo, +1}le + wlw2 cosﬁ:
H1+Nl+wl "1+V1+Wz
2 1+1 1+1 1 2 2
CGS¢=#=7=£=O.4TI
.jl4+1+1i1+1+1i JEJS 3
msr=w_;=ﬁ=nm .fi4+1+1I1+0+1§ _ JEJE 6 mini equation (0.1):
In : 0' (0501 1:052 5, = (50,000 psiXOATlXOJSQ) 6)
TR : 6800 psi 3) The graph 100,00
below “Emmi 90.00 _ 41155.10
stress (or ) against 80.00 _
Crack size (a) his 70 00
been obtained for an ' '
Al sample. Whatlstbe ‘5‘ 60.00 _
plane sumo mini 5 50.00 ,
roughurss ofthis : 40 00 7
material? The upper ' ' 30.00 ,
right point 15
(40,861) (15 p157) 20.00 . .102150 10.00 7 25 530 Assuming Y:1,We 0.00 o . ‘ ‘ ‘ ”'1 "mi“ 0 10 20 30 40
equation (9.11) in a
linear form: 1
a : Klr U—:
7! Thus this problem can be reduoed to ﬁnding the slope of the straight line obtained
by connecting the points above. Note that this goes approximately through the a) ,3 {1110‘5 lieu'1 origin. We will use the point (40,001) 0) determine the slope: ‘3
d)
86.1 mm
40x10" Mm“ 3)
K1 , 5 75 5 ’
he . x10 MP1: —m a) b) C)
L, : 022 Maxim“ d) 4) FrmlheﬁglnebelanemmelbesuessWufordnsaﬂoy. (15m) 2 8 sue(Merl
E S S 3 38 538T ( lwo‘ﬂ swarms psi] «'5 649'CII2OOPFJ 104 10'I 1.0 Studydate creep rate [751le II)
We need In employ equation (9.11), convened In logarithm format: logsl = logKl + .rlloga Sim we are m interested in the .rliurliou energy, we rim simply rhuurr any ofthe three
arms slim. and choose two points mi lhrc rum. w. arbitrarilr Choose the middle Curve
(550°C) and read lun points from the mph. (Iowan) and ao“‘.00), Now We substitute
twice into the above quotient logs}, = llngl + nIoga'l legit” = lugll + "logo; Subtracting thetehroequltinus: 70.14+2.22 = olng E
20 n:4.4 5) Youmgiveu Illaltheﬂexutal snmgth efAersls Zﬂomxavohmle Election
porosity diam and muma atavolume haciidu pomsity arms, What isthe ﬂexuml strength ofnenpclmlls M203? (15 pts.) We want to employ equation (7.22), then solve the 2:2 linear equations based on the
two data [mink given: a}, 2 0° expknP) We should recugnize immediately that nunrpurons M103 corresponds to P : 0,
where as = Ila, so we only need to solve [or 60. Taking the In or both sides [or each
data point: lug)i :luarnl’l lug; =1ucr“ iﬂl’:
Subtracting these two equations: “[53]:H(Pi—Pr) “r:
Substituting:
200 MP
lu[—ll
: "(0.15 — 0.04)
100 MP0 10 = 6.301
Substituting hack into the ﬁrst of the 112 equations: In 200 MP1: : 11w, — (030110.04)
hm,l = 5.550 as = 257 MP0 In the following ﬁgure, the hlnizmtalline is referred to as: (S pls.)
a) Fatigue linril
h) Taughnss
c) Stress lange
d) Stress intensity factor
e) Impact strength m_ 3 § E in F  g 1110' E 103 10‘ 105 los 10’ :05 10° l0m mm IolaHuIe,N
(Insulinl: sale) (of For a polymer Llassilied as an elastnmer, as are strain increases: (5 pts.) The tangent modulus increases
The tensile strength decream
Themuglmess mm The ductility at failure decream which ofthe followmgmatenalshasmecbamral prepmathattypleallymythemosl
mlhlaupmnuemdiemgeumnmmmpmm? (SPE) Metals
Semiconductors
Comes
Mymeis 9) a)
b)
r)
d)
e) 10) I)
b)
I)
0)
EJ During two successive ilexonl sueugth lens of: brittle :emmic. the separation
distauee (L) between support poms was increased by 2: What cocci would this
hm an the measured ﬂmlml straigth(og)? (5 pm) localised by 2x
Demosedby 2x
locreased by 42:
Decreasedhy 4::
No eﬂxt When a mealelo'britde transition is observed with decreasing temperature, the
malainl'scryslnl rum“ is typically: (5 par.) Dlammld FCC BCC HCP None oflhe above ...
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 Spring '08
 Rasmussen

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