MSE-Askeland Science and Engineering 7e ISM Chapter 04

Chapter i4: iImperfections iin ithe iAtomic iand iIonic iArrangements 4-1 Gold ihas i5.82 i× i108 ivacancies/cm3 iat iequilibrium iat i300 iK. iWhat ifraction iof ithe iatomic isites iis ivacant iat i600 iK? Solution: The inumber iof ivacancies iper icm3 inv iis igiven iby  iQv i nv i in iexp i iRT i i,   where in iis ithe inumber iof iatoms iper icm3, iQv iis ithe ienergy irequired ito iproduce ione imole iof ivacancies, iR iis ithe igas iconstant, iand iT iis ithe iabsolute itemperature. iGold ihas ia idensity iof i19.302 ig/cm3 iand ian iatomic imass iof 196.97 ig/mol. iThus ithe inumber iof iatoms iper icm3 iis 19.302 ig i 1 imol i i i i6.0221023 iatoms i  i 22 3 cm3 196.97 ig 1 imol 5.90 i10 atoms/cm Solving ifor iQv, ithe ienergy iis igiven iby  i  in i  i5.82108 i Qv RT iln i iv i i i8.314 iJ/(mol i iK) i i(300 iK) iln i   in i  i5.901022 i  i80,438 iJ/mol. The ifraction iof ithe ivacant iatomic isites iis igiven iby inv/n. iAt i600 iK, nv i iexp i iQv  i iexp 80,438 iJ/mol  i i9.93108 i. n  RT i  i8.314 iJ/(mol i iK) i i600 iK i     4-2 Calculate ithe inumber iof ivacancies iper im3 ifor igold iat i900 i°C. iThe ienergy ifor ivacancy iformation iis i0.86 ieV/atom. Solution: 4-3 Calculate ithe inumber iof ivacancies iper icm3 iexpected iin icopper iat i1080°C i(just ibelow ithe imelting itemperature). iThe ienergy ifor ivacancy iformation iis i20,000 ical/mol. Solution: n i i(4 iatoms/unit icell) i i8.47 i1022atoms/cm3 i(3.6151108cm)3 n i i8.47 i1022exp[ i i20,000/(1.987)/(1353)]  i8.47 i1022exp( i i7.4393) i i4.981019 ivacancies/cm3 4-4 The ifraction iof ilattice ipoints ioccupied iby ivacancies iin isolid ialuminum iat i660°C iis i10–3. What iis ithe iactivation ienergy irequired ito icreate ivacancies iin ialuminum? Solution: nv/n i= i10–3 i= iexp[–Q/(1.987)/(933)] iln(10–3) i= i–6.9078 i= i–Q/(1.987)/(933) Q i= i12,800 ical/mol 4-5 The idensity iof ia isample iof iFCC ipalladium iis i11.98 ig/cm3, iand iits ilattice iparameter iis i3.8902 iÅ. iCalculate i(a) ithe ifraction iof ithe ilattice ipoints ithat icontain ivacancies; iand i(b) ithe itotal inumber iof ivacancies iin ia icubic icentimeter iof iPd. Solution: 11.98 ig/cm3 i (x) i(106.4 ig/mol) (3.8902 i108cm)3 i(6.022 i1023atoms/mol) (a) (b) x i i3.9918 fraction i i4.0 i i3.9918 i i0.00204 4 number i i0.0082 ivacancies/unit icell i i1.391020 ivacancies/cm3 i(3.8902108cm)3 4-6 The idensity iof ia isample iof iHCP iberyllium iis i1.844 ig/cm3, iand ithe ilattice iparameters iare ia0 i= i0.22858 inm iand ic0 i= i0.35842 inm. iCalculate i(a) ithe ifraction iof ithe ilattice ipoints ithat icontain ivacancies; iand i(b) ithe itotal inumber iof ivacancies iin ia icubic icentimeter iof iBe. Solution: Vunitcell  i(0.22858 inm)2 i(0.35842 inm)cos30 i i0.01622 inm3  i1.622 i1023cm3 (a) From ithe idensity iequation: 1.844 ig/cm3 i (x) i(9.01 ig/mol) (1.622 i1023cm3) i(6.02 i1023atoms/mol) x i i1.9984 fraction i i2 i1.9984 i i0.0008 2 (b) (b) number i i0.0016 ivacancies/unit icell i i0.9861020vacancies/cm3 i1.6221023cm3 4-7 BCC ilithium ihas ia ilattice iparameter iof i3.5089 i× i10–8 icm iand icontains ione ivacancy iper i200 iunit icells. iCalculate i(a) ithe inumber iof ivacancies iper icubic icentimeter; iand i(b) ithe idensity iof iLi. Solution: (a) 1 ivacancy (200) i(3.5089108cm)3  i1.157 i1020vacancies/cm3 (b) iIn i200 iunit icells, ithere iare i399 iLi iatoms. iThe iatoms/cell iare i399/200: (399/200) i(6.94 ig/mol) (3.5089108cm)3(6.021023atoms/mol)  i0.532 ig/cm3 4-8 FCC ilead ihas ia ilattice iparameter iof i0.4949 inm iand icontains ione ivacancy iper i500 iPb iatoms. iCalculate i(a) ithe idensity; iand i(b) ithe inumber iof ivacancies iper igram iof iPb. Solution: The inumber iof iatoms/cell i= i(499/500)(4 isites/cell) (499/500) i(4) i(207.19 ig/mol) (4.949108cm)3(6.021023atoms/mol)  i11.335 ig/cm3 The i500 iPb iatoms ioccupy i500 i/ i4 i= i125 iunit icells: 1 ivacancy  125 icells [(1/11.335 ig/cm3 i)] i i5.821018vacancies/g   i(4.949108cm)3 i 4-9 Cu iand iNi iform ia isubstitutional isolid isolution. iThis imeans ithat ithe icrystal istructure iof ia iCu-Ni ialloy iconsists iof iNi iatoms isubstituting ifor iCu iatoms iin ithe iregular iatomic ipositions iof ithe iFCC istructure. iFor ia iCu–30% iwt% iNi ialloy, iwhat ifraction iof ithe iatomic isites idoes iNi ioccupy? Solution: In i100 ig iof ia iCu–30% iwt.% iNi ialloy, iCu icomprises i70 ig, iand iNi icomprises i30 ig. iIn i70 ig iof iCu, ithere iare 1 imol 6.022 i1023atoms 23 70 ig iCu i   i6.6310 Cu iatoms. 63.54 ig 1 imol In i30 ig iof iNi, ithere iare 1 imol 6.0221023 iatoms 23 30 ig iCu i   i3.0810 Ni iatoms. 58.71 ig 1 imol Therefore, ithe iNi iatoms ioccupy 3.081023 iNi iatoms 3.081023 iNi iatoms i i6.631023 iCu iatoms or i32% iof ithe iatomic isites.  i0.32 4-10 Au iand iAg iform ia isubstitutional isolid isolution. iThis imeans ithat ithe icrystal istructure iof ia iAu-Ag ialloy iconsists iof iAg iatoms isubstituting ifor iAu iatoms iin ithe iregular iatomic ipositions iof ithe iFCC istructure. iFor ia iAu-50 iat% iAg ialloy, iwhat iis ithe iwt% iAg iin ithe ialloy? Solution: In i1 imol iof ia iAu–50 iat.% iAg ialloy, iAu icomprises i0.5 imol, iand iAg icomprises i0.5 imol. 0.5 imol iAu i i196.97 ig iAu i i98.485 ig iAu. 1 imol 0.5 imol iAg i i107.868 ig iAu i i53.934 ig iAg. 1 imol Thus ithe iwt% iAg iis 53.934 ig 53.934 ig i+98.485 ig  i35.4 iwt% iAg 4-11 A iniobium ialloy iis iproduced iby iintroducing itungsten isubstitutional iatoms iinto ithe iBCC istructure; ieventually ian ialloy iis iproduced ithat ihas ia ilattice iparameter iof i0.32554 inm iand ia idensity iof i11.95 ig/cm3. iCalculate ithe ifraction iof ithe iatoms iin ithe ialloy ithat iare itungsten. Solution: 11.95 ig/cm3 i i(xw i) i(183.85 ig/mol) i i(2 i ixw i) i(92.91 ig/mol) i(3.2554108cm)3 i(6.0221023 iatoms/mol) 248.186 i= i183.85xW i+ i185.82 i– i92.91xW 90.94xW i i62.366 or xW i i0.687 iW iatoms/cell There iare i2 iatoms iper icell iin iBCC imetals. iThus: fw i= i0.687/2 i= i0.344 4-12 Tin iatoms iare iintroduced iinto ian iFCC icopper icrystal, iproducing ian ialloy iwith ia ilattice iparameter iof i3.7589 i× i10–8 icm iand ia idensity iof i8.772 ig/cm3. iCalculate ithe iatomic ipercentage iof itin ipresent iin ithe ialloy. Solution: 8.772 ig/cm3 i i(xSn i) i(118.69 ig/mol) i i(4 i ixSn i) i(63.54 ig/mol) i(3.7589108cm)3 i(6.021023 iatoms/mol) 280.5 i i55.15xSn i i254.16 or xSn i i0.478 iSn iatoms/cell There iare i4 iatoms iper icell iin iFCC imetals; itherefore ithe iat% iSn iis i(0.478/4) i= i11.95% 4-13 We ireplace i7.5 iatomic ipercent iof ithe ichromium iatoms iin iits iBCC icrystal iwith itantalum. iX-ray idiffraction ishows ithat ithe ilattice iparameter iis i0.29158 inm. iCalculate ithe idensity iof ithe ialloy. Solution:  i i(2) i(0.925) i(51.996 ig/mol) i i2(0.075) i(180.95 ig/mol) i i8.262 ig/cm3 i(2.9158108cm)3(6.0221023atoms/mol) 4-14 Suppose iwe iintroduce ione icarbon iatom ifor ievery i100 iiron iatoms iin ian iinterstitial iposition iin iBCC iiron, igiving ia ilattice iparameter iof i0.2867 inm. iFor ithis isteel, ifind ithe idensity iand ithe ipacking ifactor. Solution: There iis ione icarbon iatom iper i100 iiron iatoms, ior i1 iC/50 iunit icells, ior i1/50 iC iper iunit icell: (2) i(55.847 ig/mol) i i(1/50)(12 ig/mol)  i(2.867 i108cm)3(6.021023atoms/mol)  i7.89 ig/cm3 2(4 i/3) i(1.241)3 i i(1/50)(4 i/3) i(0.77)3 Packing iFactor i  i0.681 (2.867)3 4-15 The idensity iof iBCC iiron iis i7.882 ig/cm3, iand ithe ilattice iparameter iis i0.2866 inm iwhen ihydrogen iatoms iare iintroduced iat iinterstitial ipositions. iCalculate i(a) ithe iatomic ifraction iof ihydrogen iatoms; iand i(b) ithe inumber iof iunit icells ion iaverage ithat icontain ihydrogen iatoms. Solution: (a) 7.882 ig/cm3 i 2(55.847 ig/mol) i ix(1.00797 ig/mol) i(2.866108cm)3 i(6.0221023atoms/mol) x i= i0.0449 iH iatoms/cell The itotal iatoms iper icell iinclude i2 iFe iatoms iand i0.0449 iH iatoms. iThus, f i i0.0449 i i0.02195 H 2.0449 (b) iSince ithere iis i0.0449 iH/cell, ithen ithe inumber iof icells icontaining iH iatoms iis cells i i1 i/ i0.0449 i i22.3 or 1 iH iin i22.3 icells 4-16 Suppose ione iSchottky idefect iis ipresent iin ievery itenth iunit icell iof iMgO. iMgO ihas ithe isodium ichloride icrystal istructure iand ia ilattice iparameter iof i0.396 inm. iCalculate i(a) ithe inumber iof ianion ivacancies iper icm3; iand i(b) ithe idensity iof ithe iceramic. Solution: In i10 iunit icells, iwe iexpect i40 iMg i+ i40 iO iions, ibut idue ito ithe idefect: i40 iMg i– i1 i= i39 40 iO i– i1 i= i39 1 ivacancy/(10 icells)(3.96 i× i10–8 icm)3 i= i1.61 i× i1021 ivacancies/cm3  i i(39/40) i(4) i(24.312 ig/mol) i i(39/40) i(4) i(16 ig/mol) i i4.205 ig/cm3 i(3.96108cm)3(6.021023atoms/mol) 4-17 ZnS ihas ithe izinc iblende istructure. iIf ithe idensity iis i3.02 ig/cm3 iand ithe ilattice iparameter iis i0.59583 inm, idetermine ithe inumber iof iSchottky idefects i(a) iper iunit icell; iand i(b) iper icubic icentimeter. Solution: Let ix ibe ithe inumber iof ieach itype iof iion iin ithe iunit icell. iThere inormally iare i4 iof ieach itype. (a) 3.02 ig/cm3 i x(65.38 ig/mol) i ix(32.064 ig/mol) i(5.9583108cm)3(6.0221023ions/mol) x i i3.9478 4 i– i3.9478 i= i0.0522 idefects/unit icell (b) i# iof iunit icells/cm3 i= i1/(5.9583 i× i10–8 icm)3 i= i4.728 i× i1021 iSchottky idefects iper icm3 i= i(4.728 i× i1021)(0.0522) i= i2.466 i× i1020 4-18 Suppose iwe iintroduce ithe ifollowing ipoint idefects. (a) Mg2+ iions isubstitute ifor iyttrium iatoms iin iY2O3; (b) Fe3+ iions isubstitute ifor imagnesium iions iin iMgO; (c) Li1+ iions isubstitute ifor imagnesium iions iin iMgO; iand (d) Fe2+ iions ireplace isodium iions iin iNaCl. What iother ichanges iin ieach istructure imight ibe inecessary ito imaintain ia icharge ibalance? iExplain. Solution: (a) iRemove i2 iY3+ iand iadd i3 iMg2+ i– icreate ication iinterstitial. (b) Remove i3 iMg2+ iand iadd i2 iFe3+ i– icreate ication ivacancy. (c) Remove i1 iMg2+ iand iadd i2 iLi+ i– icreate ication iinterstitial. (d) Remove i2 iNa+ iand iadd i1 iFe2+ i– icreate ication ivacancy. 4-19 Write idown ithe idefect ichemistry iequation ifor iintroduction iof iSrTiO3 iin iBaTiO3 iusing ithe iKröger-Vink inotation. Solution: 4-20 Do iamorphous iand icrystalline imaterials iplastically ideform iby ithe isame imechanisms? Explain. Solution: Without ia icrystalline istructure, ithe iconcepts iof idislocations iand islip isystems idon’t iapply ito iamorphous isystems. iAmorphous isystems itend ito ibe isusceptible ito iflaws iin ithe imaterial. 4-21 What iis ithe iBurger’s ivector iorientation irelationship iwith ithe idislocation iaxis ifor iboth iedge iand iscrew idislocations? Solution: The iedge idislocation iaxis iis iperpendicular ito ithe iBurgers ivector iwhile ithe iscrew idislocation iaxis iis iparallel ito ithe iBurgers ivector. 4-22 What iis ia islip isystem iand iwhat irole idoes iit iplay iin iplastic ideformation? Solution: A islip isystem icomprises ia islip iplane iand islip idirection, iusually iclosed-packed, iin iwhich iline idefects ican imove. iThe imovement iof ithese idefects ior idislocations iis ireferred ito ias islip. iSlip iis iplastic ideformation iin icrystalline isolids. 4-23 Draw ia iBurgers icircuit iaround ithe idislocation ishown iin iFigure i4-18. iClearly iindicate ithe iBurgers ivector ithat iyou ifind. iWhat itype iof idislocation iis ithis? iIn iwhat idirection iwill ithe idislocation imove idue ito ithe iapplied ishear istress it? iReference iyour ianswers ito ithe icoordinate iaxes ishown. Solution: A inegative iedge idislocation iwith iits iBurgers icircuit iand iBurgers ivector iis ishown iin ithe idiagram. iThe idislocation iwill imove iin ithe idirection idue ito ithe iapplied ishear istress iτ. 4-24 What iare ithe iMiller iindices iof ithe islip idirections: (a) on ithe i(111) iplane iin ian iFCC iunit icell? (b) on ithe i(011) iplane iin ia iBCC iunit icell? Solution: [0 i11], i[01 i1] [1 i11], i[ i11 i1] [ i110], i[1 i10] [ i1 i11], i[11 i1] [ i101], i[10 i1] 4-25 What iare ithe iMiller iindices iof ithe islip iplanes iin iFCC iunit icells ithat iinclude ithe i[101] islip idirection? Solution: (11 i1),(1 i11) (111),(1 i1 i1) 4-26 What iare ithe iMiller iindices iof ithe i{110} islip iplanes iin iBCC iunit icells ithat iinclude ithe [111] islip idirection? Solution: (1 i10),(110) (0 i11),(01 i1) (10 i1),(101) 4-27 Calculate ithe ilength iof ithe iBurgers ivector iin ithe ifollowing imaterials: (a) BCC iniobium; i(b) iFCC isilver; iand i(c) idiamond icubic isilicon. Solution: (a) iThe irepeat idistance, ior iBurgers ivector, iis ihalf ithe ibody idiagonal, ior: b i irepeat idistance i i(½)( i3)(3.294 iÅ) i i2.853 iÅ (b) The irepeat idistance, ior iBurgers ivector, iis ihalf iof ithe iface idiagonal, ior: b i i(½)( i2a0 i) i i(½)( i2)(4.0862 iÅ) i i2.889 iÅ (c) The islip idirection iis i[110], iwhere ithe irepeat idistance iis ihalf iof ithe iface idiagonal: b i i(½) i( i2) i(5.4307 iÅ) i i3.840 iÅ 4-28 Determine ithe iinterplanar ispacing iand ithe ilength iof ithe iBurgers ivector ifor islip ion ithe iexpected islip isystems iin iFCC ialuminum. iRepeat, iassuming ithat ithe islip isystem iis ia i(110) iplane iand ia i[111] idirection. iWhat iis ithe iratio ibetween ithe ishear istresses irequired ifor islip ifor ithe itwo isystems? iAssume ithat ik i= i2 iin iEquation i4-2. Solution: (a) iFor i(111)/[1 i10], b i i(½) i( i2) i(4.04958 iÅ) i i2.863 iÅ (b) iIf i(110)/[1 i11], ithen d111 i 4.04958 iÅ  i2.338 iÅ b i (4.04958 iÅ) i i7.014 iÅ d110 i 4.04958 iÅ  i2.863 iÅ (c) iIf iwe iassume ithat ik i= i2 iin iEquation i4-2, ithen (d i/b)a  i2.338 i i0.8166 (d i/b) 2.863 b  i2.863 i i0.408 7.014  ia b  iexp(2(0.8166)) i i0.44 iexp(2(0.408)) 4-29 Determine ithe iinterplanar ispacing iand ithe ilength iof ithe iBurgers ivector ifor islip ion ithe i(110)[111] islip isystem iin iBCC itantalum. iRepeat, iassuming ithat ithe islip isystem iis ia i(111)/[1 i10] isystem. iWhat iis ithe iratio ibetween ithe ishear istresses irequired ifor islip ifor ithe itwo isystems? iAssume ithat ik i= i2 iin iEquation i4-2. Solution: (a) iFor i(110)/[1 i11]: b i i(½) i( i3) i(3.3026 iÅ) i i2.860 iÅ (b) iIf i(111)/[1 i10] i, ithen: d110 i 3.3026 iÅ  i2.335 iÅ b i 2(3.3026 iÅ) i i4.671 iÅ d111 i 3.3026 iÅ  i1.907 iÅ (c) iIf iwe iassume ithat ik i= i2 iin iEquation i4-2, ithen: (d i/b)a  i2.335 i i0.8166 (d i/b) 2.86 b  i1.907 i i0.408 4.671 a i iexp(2(0.8166)) i i0.44 b exp(2(0.408)) 4-30 The icrystal ishown iin iFigure i4-19 icontains itwo idislocations iA iand iB. iIf ia ishear istress iis iapplied ito ithe icrystal ias ishown, iwhat iwill ihappen ito idislocations iA iand iB? Solution: Under ithe iaction iof ithe iapplied ishear istress ishown, idislocation iA iwill imove ito ithe iright, iand idislocation iB iwill imove ito ithe ileft. iWhen ithe idislocations imeet, ithey iwill iannihilate ibecause ithe icombination iof ia inegative iedge idislocation iand ia ipositive iedge idislocation iwill iform iperfect icrystal. 4-31 Can iceramic iand ipolymeric imaterials icontain idislocations? Solution: Yes. iAll icrystalline imaterials ican icontain idislocations. 4-32 Why iis iit ithat iceramic imaterials iare ibrittle? Solution: Dislocations ido inot imove ieasily ienough ibecause iof ibonding istrengths, iso ithe imaterials ifail idue ito iflaws isuch ias icracks iand ipores ibefore iany islip ican ioccur. 4-33 What iis imeant iby ithe iterms iplastic iand ielastic ideformation? Solution: Plastic ideformation idescribes ian iirreversible ichange ito ithe ishape iof ian iobject iwhen ia iforce iis iapplied. iElastic ideformation iis ia itemporary ichange iin ishape ithat iis irecovered iwhen ithe iforce iis iremoved. 4-34 Why iis ithe itheoretical istrength iof imetals imuch ihigher ithan ithat iobserved iexperimentally? Solution: Slip iallows imetallic ibonds ito ibe ibroken iindividually irather ithan irequiring iall ibonds iin ia isample ito ibe ibroken iat ionce, ias iis ipredicted iby isimply icounting imetallic ibond istrengths. 4-35 How imany igrams iof ialuminum, iwith ia idislocation idensity iof i1010 icm/cm3, iare irequired ito igive ia itotal idislocation ilength ithat iwould istretch ifrom iNew iYork iCity ito iLos iAngeles i(3000 imiles)? Solution: i i i(3000 imi)(5280 ift/mi)(12 iin./ft)(2.54 icm/in.) i= i4.828 i× i108 icm (4.828108cm) i(2.699 ig/cm3) i (1010cm/cm3) 0.13 ig 4-36 The idistance ifrom iEarth ito ithe iMoon iis i240,000 imiles. iIf ithis iwere ithe itotal ilength iof idislocation iin ia icubic icentimeter iof imaterial, iwhat iwould ibe ithe idislocation idensity? iCompare iyour ianswer ito itypical idislocation idensities ifor imetals. Solution: i(240,000 imi)(5280 ift/mi)(12 iin./ft)(2.54 icm/in.) i= i3.86 i× i1010 icm/cm3 iThis iis ireasonable ias idislocation idensities irange ifrom i106 icm/cm3 ito i1012 icm/cm3. 4-37 Why iwould imetals ibehave ias ibrittle imaterials iwithout idislocations? Solution: The idislocations iallow islip, iwhich iin iturn iallows iWithout idislocations, ithe imetal iwould ibehave ias ia iceramic iand ifail icompletely iwhen iits iultimate istrength iwas ireached. 4-38 Why iis iit ithat idislocations iplay ian iimportant irole iin icontrolling ithe imechanical iproperties iof imetallic imaterials, ihowever, ithey ido inot iplay ia irole iin idetermining ithe imechanical iproperties iof iglasses? Solution: Glasses ido inot icontain idislocations. iGlasses iare iamorphous, iand itherefore, ithey ido inot ihave idefects isuch ias idislocations. 4-39 Suppose iyou iwould ilike ito iintroduce ian iinterstitial ior ilarge isubstitutional iatom iinto ithe icrystal inear ia idislocation. iWould ithe iatom ifit imore ieasily iabove ior ibelow ithe idislocation iline ishown iin iFigure i4-7(c)? iExplain. Solution: The iatom iwould ifit imore ieasily iinto ithe iarea ijust ibelow ithe idislocation idue ito ithe iatoms ibeing ipulled iapart; ithis iallows imore ispace iinto iwhich ithe iatom ican ifit. 4-40 Compare ithe ic/a iratios ifor ithe ifollowing iHCP imetals, idetermine ithe ilikely islip iprocesses iin ieach, iand iestimate ithe iapproximate icritical iresolved ishear istress. iExplain. i(See idata iin iAppendix iA.) (a) Zinc; (b) Magnesium; (c) Titanium; (d) Zirconium; (e) Rhenium; iand (f) Beryllium. Solution: We iexpect imetals iwith ic/a i i1.633 ito ihave ia ilow iτcrss: (a) Zn: i4.9470 i i1.856 i ilow i icrss i2.6648 (b) Mg: i5.209 3.2087  i1.62 i imedium i icrss (c) (d) (e) (f) Ti: i4.6831 i i1.587 i ihigh i icrss i2.9503 Zr: i5.1477 i i1.593 i ihigh i icrss i3.2312 Rh: i4.458 i i1.615 i imedium i icrss i2.760 Be: i3.5842 i i1.568 i ihigh i icrss i2.2858 4-41 Using iSchmid’s iLaw iequation, ithe iresolved ishear istress ioperating ion ia islip idirection/plane iis igiven iby i ir i i icos i icos i. iIf ieither iof ithese iangles iequal i90°, iwhat ihappens ito ithe idislocations iinvolved? Solution: The idislocations iwill inot islip. 4-42 A isingle icrystal iof ian iFCC imetal iis ioriented iso ithat ithe i[001] idirection iis iparallel ito ian iapplied istress iof i5000 ipsi. iCalculate ithe iresolved ishear istress iacting ion ithe i(111) islip iplane iin ithe i[110], i[0 i11] i, iand i[101] islip idirections. iWhich islip isystem(s) iwill ibecome iactive ifirst? Solution:  i= i54.76° τ i= i5000 icos i54.76° icos iλ λ110 i= i90° λ011 i= i45° λ101 i= i45° τ i= i0 τ i= i2040 ipsi iactive τ i= i2040 ipsi iactive 4-43 A isingle icrystal iof ia iBCC imetal iis ioriented iso ithat ithe i[001] idirection iis iparallel ito ithe iapplied istress. iIf ithe icritical iresolved ishear istress irequired ifor islip iis i12,000 ipsi, icalculate ithe imagnitude iof ithe iapplied istress irequired ito icause islip ito ibegin iin ithe i[111] direction ion ithe i(110), i(011), iand i(101) islip iplanes. Solution: CRSS i= i12,000 ipsi i= iσ icos icosλ iλ i= i54.76° 12, i000 ipsi i i cos icos i 110 i= i90o σ i= i∞ 011 i= i45° σ i= i29,412 ipsi 101 i= i45° σ i= i29,412 ipsi 4-44 A isingle icrystal iof isilver iis ioriented iso ithat ithe i(111) islip iplane iis iperpendicular ito ian iapplied istress iof i50 iMPa. iList ithe islip isystems icomposed iof iclose-packed iplanes iand idirections ithat imay ibe iactivated idue ito ithis iapplied istress. Solution: i i iThere iare ifour ipossible islip iplanes iin ian iFCC icrystal. iThey iare ithe i(111), (11 i1),(1 i11) i, iand i(11 i1) i. iEach islip iplane icontains ithree islip idirections. iThus ithere iare ia itotal iof itwelve ipotential islip isystems iin ian iFCC isingle icrystal. iSome iof ithese islip isystems iwill inot ibe iactivated, ihowever, ibecause ieither ithe islip iplane ior ithe islip idirection iis iperpendicular ito ithe iapplied istress. iThe itwelve ipotential islip isystems iare listed ibelow, iand ithose ithat iare inot iactivated iare icrossed iout. iIf ithe i(111) islip iplane is iperpendicular ito ithe iapplied istress iof i50 iMPa, ithen ithe iplane inormal i– ithe i[111] idirection i– iis ithe idirection iof ithe iapplied istress. iThus ithere ican ibe ino ishear istress iacting ion ithe i(111) iplane. iAll islip isystems ithat iinclude ithe i(111) iplane iare ieliminated. iIn ithe i(111) isystem, ithe i[01 i1] iis iperpendicular ito ithe i[111] idirection, and itherefore, iis ieliminated. iIn ithe i(1 i11) isystem, ithe i[101] iis iperpendicular ito ithe [111] idirection, iand itherefore, iis ieliminated. iIn ithe i(11 i1) i, ithe i[110] iis iperpendicular ito ithe i[111] idirection, iand itherefore, iis ieliminated. (11 i1)[011] (111)[110] (111)[110] (111)[101] (111)[101] (111)[011] 4-45 Why iis iit ithat isingle icrystal iand ipolycrystalline icopper iare iboth iductile; ihowever, ionly isingle icrystal, ibut inot ipolycrystalline, izinc ican iexhibit iconsiderable iductility? Solution: Single icrystal imetals iwith ithe iFCC istructure iare iinherently iductile ibecause iFCC ihas ia ilarge inumber iof islip isystems i(12). iThus iregardless iof ithe iorientation iof ithe iapplied istress, ia inumber iof islip isystems iwill ibe iactive. iThe iHCP istructure ihas ia irelatively ismall inumber iof islip isystems i(3), iand ithese islip isystems iare iparallel ito ieach iother. iWhen ithe iapplied istress iis ioriented ifavorably ifor ishear, ithe islip isystems iwill ibe iactive iand ithe isingle icrystal iwill ideform iin ia iductile ifashion. iWhen ithe iHCP imaterial iis ipolycrystalline, iit iwill ibehave iin ia ibrittle ifashion ibecause icross islip idoes inot ioccur. 4-46 Explain iwhy ihexagonal iclose-packed imetals itend ito ihave ia ilimited iability ito ibe istrain ihardened. Solution: HCP imetals iwith ic/a iratios i i1.633 ihave ilimited iability ito istrain iharden ibecause iof ithe ilimited inumber iof iparallel islip isystems ithese imetals ipossess. iSince ithere iare ionly ithree iparallel islip isystems, ithe iopportunities ifor idislocations ito iinteract iand ibecome itangled iare ilow. iIn iaddition, isince ithese imetallic isystems ido inot ihave ithe iability ito icross-slip, iductility iis ialso ilimited. 4-47 Why iis iit ithat icross islip iin iBCC iand iFCC imetals iis ieasier ithan iin iHCP imetals? iHow idoes ithis iinfluence ithe iductility iof iBCC, iFCC, iand iHCP imetals? Solution: Cross islip iis ieasier iin iBCC iand iFCC imetals ithan iin iHCP imetals ibecause ithere iare imultiple inon-parallel iand iintersecting islip isystems iin ithe iBCC iand iFCC istructures. iThis imakes iit ipossible ifor idislocations ito imove ifrom ione islip iplane ito ianother. iAll iof ithe ifavored islip isystems iin iHCP imetals iare iparallel ito ieach iother imaking icross islip idifficult. 4-48 Arrange ithe ifollowing imetals iin ithe iexpected iorder iof iincreasing iductility: iCu, iTi, iand iFe. Explain. Solution: From ileast iductile ito imost iductile, ithe iorder iis iTi, iFe, iand iCu. iTi iis iHCP iand ithus iwill ibe ithe imost ibrittle idue ito ithe ilimited inumber iof islip isystems. iBCC imetals i(such ias iFe) itend ito ihave ihigher istrengths iand ilower iductilities ithan iFCC imetals i(such ias iCu) idue ito itheir ihigher icritical iresolved ishear istrengths. 4-49 What iare ithe iimperfections iin ithe iatomic iarrangements ithat ihave ia isignificant ieffect ion ithe imaterial ibehavior? iGive ian iexample iof ieach. Solution: Imperfections iinclude ipoint i(vacancies), iline i(dislocations) iand isurface idefects i(grain iboundaries). 4-50 The istrength iof ititanium iis ifound ito ibe i65,000 ipsi iwhen ithe igrain isize iis i6.7 i104 iin. iand i82,000 ipsi iwhen ithe igrain isize iis i3.15 i10 in. iDetermine i(a) ithe iconstants iin ithe iHall-Petch iequation; iand i(b) ithe istrength iof ithe ititanium iwhen ithe igrain isize iis ireduced to i8.00 i10–6 iin. i. Solution: 65,000 i i  iK 1  i i i38.633 iK 82,000 i i  iK 1  i i i178.17 iK (a) By isolving ithe itwo isimultaneous iequations: K i i121.8 ipsi (b) i i i i60,293121.8/  i i60,293 ipsi  i103,400 ipsi i i103.4 iksi 4-51 A icopper-zinc ialloy ihas ithe iproperties ishown iin ithe itable ibelow: iGrain idiameter i(mm) Strength i(MPa) d–½ 0.015 170 iMPa 8.165 0.025 158 iMPa 6.325 0.035 151 iMPa 5.345 0.050 145 iMPa 4.472 Determine i(a) ithe iconstants iin ithe iHall-Petch iequation; iand i(b) ithe igrain isize irequired ito iobtain ia istrength iof i200 iMPa. Solution: The ivalues iof id–½ iare iincluded iin ithe itable; ithe igraph ishows ithe irelationship. iWe ican idetermine iK iand iσo ieither ifrom ithe igraph ior iby iusing itwo iof ithe idata ipoints. 170 i i io i iK(8.165) (a) 145 i i io i iK(4.472) 25 i i3.693K K i i6.77 iMPa/  io i i114.7 iMPa (b) iTo iobtain ia istrength iof i200 iMPa: 200 i i114.7 i i6.77/ 85.3 i i6.77/ d i= i0.0063 imm 4-52 If ithere iwere i50 igrains iper iin2 iin ia iphotograph iof ia imetal itaken iat i100× imagnification, icalculate ithe iASTM igrain isize inumber i(n). Solution: N i= i2 i(n-1) 50 i= i2(n-1) ln i50 i= i(n–1)*ln i2 3.912 i= i(n–1) i(0.6931) 4.605 i= i0.6931n n i= i6.64 4-53 If ithe iarea iof ia iphotograph imeasured i7.812 iin2 iand i23 igrains iwere idocumented, iwhat iwould ithe iASTM igrain isize inumber i(n) ibe? Solution: 4-54 You ihave ithe ichoice ito ieither ipurchase ia icopper ialloy ithat ihas ian iASTM igrain isize iof i5 ior iASTM igrain isize iof i8. iYou ican’t idecide iif ithere iis imuch iof ia idifference ibetween ithese itwo. iDetermine ihow imany igrains/in2 iwould iappear ion ia iphotograph itaken iat i100× ifor ia imetal igiven ithese itwo iASTM igrain isize ichoices. iDoes ithis iseem ilike ia isignificant idifference? Solution: This iis ia ivery isignificant idifference ibecause ithere iis iare i8 itimes imore igrains iper iunit iarea iwith ian iASTM igrain isize iof i8 icompared ito ian iASTM igrain isize iof i5. 4-55 The imill itest ireport i(MTR) ion ia istructural isteel iyou iare iconsidering iusing ion ian ielevated ipedestrian ibridge iover ia ibusy iroadway ishows ian iASTM igrain isize iof i1.5. iWould iyou iaccept ithis isteel ifor ithis iparticular iproject? iExplain. Solution: An iASTM igrain isize iof i1.5 imeans ivery ilarge igrains. iFor ithe isake iof ithis iapplication, ivery ilarge igrains iwill iprovide ilow istrength iand itoughness iwhich iare inot idesirable iin istructural iapplications. 4-56 For ian iASTM igrain isize inumber iof i8, icalculate ithe inumber iof igrains iper isquare iinch i(a) iat ia imagnification iof i100 iand i(b) iwith ino imagnification. Solution: N i= i2n–1 iN i= i28–1 i= i27 i= i128 igrains/in.2 No imagnification imeans ithat ithe imagnification iis i“1”: i(27)(100/1)2 i= i1.28 i× i106 igrains/in.2 4-57 Determine ithe iASTM igrain isize inumber iif i20 igrains/square iinch iare iobserved iat ia imagnification iof i400. Solution: (20)(400/100)2 i= i2n–1 log(320) i= i(n–1)log(2) 2.505 i(n–1)(0.301) ior in i= i9.3 4-58 Determine ithe iASTM igrain isize inumber iif i25 igrains/square iinch iare iobserved iat ia imagnification iof i50. Solution: 25(50/100)2 i= i2n–1 log(6.25) i= i(n–1)log(2) i0.796 i(n–1)(0.301) ior in i= i3.6 4-59 Determine ithe iASTM igrain isize inumber ifor ithe imaterials iin iFigure i4-14 iand iFigure i4-20. Solution: (a) iThere iare iabout i26 igrains iin ithe iphotomicrograph, iwhich ihas ithe idimensions 2.375 iin. i× i2 iin. iThe imagnification iis i100, ithus 26 (2.375)(2)  i2n1 log(5.47) i i0.738 i i(n i1)log(2) n i i3.5 (b) iThere iare iabout i59 igrains iin ithe iphotomicrograph, iwhich ihas ithe idimensions 2.25 iin. i× i2 iin. iThe imagnification iis i500, ithus 59(500/100)2 i (2.25)(2) 2n1 log(328) i i2.516 i i(n i1) ilog(2) n i i9.4 There iare iabout i28 igrains iin ithe iphotomicrograph, iwhich ihas ithe idimensions i2 iin. i× 2.25 iin. iThe imagnification iis i200, ithus 28(200/100)2 i (2.25)(2) 2n1 log(24.889) i i1.396 i i(n i1)log(2) in i i5.6 4-61 The iangle iθ iof ia itilt iboundary iis igiven iby isin(θ/2) i= ib/2D i[see iFigure i4-15(a)]. iVerify ithe icorrectness iof ithis iequation. Solution: From ithe ifigure, iwe inote ithat ithe igrains iare ioffset ione iBurgers ivector, ib, ionly ifor itwo ispacings iD. iThen iit iis iapparent ithat isin(θ/2) imust ibe ib idivided iby itwo iD. 4-62 Calculate ithe iangle iθ iof ia ismall-angle igrain iboundary iin iFCC ialuminum iwhen ithe idislocations iare i5000 iÅ iapart. i(See iFigure i4-15 iand ithe iequation iin iProblem i4-61.) Solution: b i i(½)( i2)(4.04958) i i2.8635 iÅ iand iD i i5000 iÅ sin( i/2) i 2.8635 (2)(5000)  i0. θ/2 i= i0.0164 θ i= i0.0328° 4-63 For iBCC iiron, icalculate ithe iaverage idistance ibetween idislocations iin ia ismall-angle igrain iboundary itilted i0.50°. i[See iFigure i4-15(a).] Solution: sin(0.5/2) i i½( i i i3)(2.866) 2D 0. i= i1.241/D iD i= i284 iÅ