Homework Solution 6

Homework Solution 6 - Chapter 1-1 Saluticn E First...

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Unformatted text preview: Chapter 1-1, Saluticn E? First determine an , the intrinsic electrical cenductiyity, tbr Gafiis at ETC [30D K}: a =50 = qumfl +flp} = [1 41-11121” m'3)(1-scs1e'” 1:)[11aae in2 1ty-sl+ 11-11211 in2 His-st] =1.s:i~s:111'T [c.ml'l Neat, calculate a at 125T: 1393 K} for the energygap cf 1.4T e111": —E 511:1" c' =c'ne ' 1:3,, I, _ ssp{—1-4i sy1[ta}{s-sesia-5 syisltsss Kfl} IFssss sisp{—1.4’1 sy1[ta}{s-sesie'5 syisltsce afl} 1:3,“, :53“, ifs-=tggfll [1 seam—7111.1 sst'l]t11194- a} D1 =1.sr:1114trs. 1'1 Chapter 1-1, Saluticn 59 1s} Fssisiaisb. 13-511, an = Less-1:1 11‘T (1:1. nil—1 siisi a, =1-4? as is: asses at rec {31111 1c- Thus, for use {see In, the intrinsic electrical ccnductiyity is: amp: —1.4i'sy [ 1 1 ] errp _ __ 53111111; Elflfifl 1-1142!"5 eWFill 343 3m} 1:1,.“=15,“sssptssstsl=[Latissie‘F Iii’i-isil'1]t51:1sia}=s.;1.'1:s111ti tn-ml'l Chapter 1-1, Salutinn "1"? —5 r: :1 =w= 535x104 F 1" 1111012111" Asa" 5s1e'5m [Sflfixlfl'leimllilflfll s1: “ =%=L59Klfl4m=fl.lfiflmm c 5.33K1D'g F Chapter 1-1, Saluticn E-tl sass Bane, is cabled tc 120°C, the central Ti4+ sstisii and the sumttufldhlg 02‘ ssisiis siiis slightly in cppcsite directicns and thereby create a small electric dipele ml". At the critical temperature cflEtIF'C, the shifting is accempanied by a crystalline structural change fi'cnr cubic tc tetragcnal- This transfcrniaticn temperature is called the Curie temperature. Chapter 14, Salutien EH} 'When ccmpriessise fcrces are applied acrcss a sample cfa farm-electric material, the sample length is reduced, and ccnsequently, the distance between dipcles is reduced As a result, the cyerall dipele irairnent cfthe material is decreased, which in turn, changes the charge density at the sample ends and causes a yeltage pctential te deselcp- Similarly, ifan electrical field is applied acr‘css the sample, the resulting charge densityr alteraticn at the sample ends willcause dirreensicnal changes inthedirecticn cftheappliedfield. The sarrple tlnispa'ndueesthemecbanical respcnae cf physical elengaticn er shrinkage. Chapter E, 5nl1rticn 1E First,thelcadmustbeecrrsertedtuafcrce, F =mia =[13EHJ kg][9-31tfli52}=12,i'r53 N. Thfimgiflfifliflg stress isthen, a is ass 11 a = —=’—=4c.ss1at Pa = mamas A11 grass is)2 Chapter '3, SnILrti-nn 13 i—i'u _ 2-35 in-—2.Dfl i11- _ _ = 11.1 T5 1, 2.1111 111. engineering strain E = Chapter 6, Snluticn 25 Frcb- EEG: Stress 11-5. Strain fifllllll 5MB g acne SEMI 15 men 1EH1|J [LEI [LIII [LIE llLlEl llLlE I120 sup-1 1mm1mrn} Theultimate tensile strength, based entire engineering stress-staincnrve, is 514 Ma. Chapter 6, 5nl1rticn 4t} Theprineipal slip planes andslip directicns fchCC metalsare {111:1}snd (ill), respectively. Chapter 6, Enlirticn 41 Thcprineipal slip planes anddinecticnafcrHCPnietalsare [11 u 1:11] and (11E 11). Chapter 6, Snlmicn fill [at [11 111] [1 1:11] Eh:- [11111] .,,, A???“ The direclien cftbe applied stress, the slip plane and each cfthe three slip directicns are shewn inFig.1[a]| abcye while Fig- {b} depicts the {110] plane with the directicns cfthe applied stress and the ncrmal tn the slip plane. Fer all three slip systems, (it can be calculated based en the gecnren'y depicted inFig- {b}: ccsrlrr=ccs sf[afin] A, = 4 5 n {a} Fcr slip system ¢=4§ \f3— fl' the gecnletry chig. {cl verifies that it is equal tc 43”. Thus, [ 11 11 1 ] {b} Fcr slip system [1 1 1] J3— 11 the geometry efFig- {c} verifies that his equal ten 43'"- Thus, erase [ 11 1 1 ] again, '."-.' e r In a 1 [c] Fcr slip system [:1 ll] [1 l l] fi'cnrthegecmen'y depicted withinthe [GD 1] plane chig. {e}, his equal tn 911°. Thus, [ CI 12] 1 ] Chapter 1|]l Saluticn ass During elastic defermaticn, the cct-‘alent bends cf the mclecular' chains stretch. Subsequertly, the chains unceil in a precess that inychres bcth elastic and plastic defermaticn. Finally, plach defermatien cccurs as seccnda.r_I..r dipule bends are breken, allen-ing the chains tc slide relative tc each ether and establish new dipele bending threes. Chapter 11, Salutien fit} The lack cfplastieity in crystalline ceramics is attributed tc- their icnic and ecsalent chemical bands. In ceyalent crystals and cc'salentlybcnded ceramics, atcms bend thrcugh the exchange cfelectr‘cn charge between pairs cf electrons, in a specific and directienal manner. Ccnsequently, if the ceramic is sufficiently stressed, electIen-pair' bends undergc irreparable separaticn, causing the material to experience brittle fiacture ...
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