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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 short-cut 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 stress-strain 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 Study-date 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 lu-n 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 mum-a 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 (Insulin-l: sale) (of For a polymer L-lassilied 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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