Mechanical Engineering Materials and Manufacturing Processes
Mechanical Engineering Materials and Manufacturing Processes ME 3322
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ME 3322 Materials and Manufacturing Reading Assignment Chapters 7 amp 8 Please read the sections of Chapters 7 and 8 indicated below and answer the questions for each assigned section Read pages 197 204 1 What caused early materials scientist to postulate the existence of dislocation in crystalline solids When was the existence of dislocations confirmed experimentally by direct observation 2 What type of lattice strain exists above the slip plane Below the slip plane 3 When two edge dislocations of the same sign approach each other what is the nature of their interaction For two dislocations of opposite signs 4 What is an independent slip system How many independent slips systems are found in FCC BCC and HCP Read pp 208 209 5 Is a polycrystalline solid stronger or weaker than a single crystalline solid at ambient temperatures Note at very high temperatures grain boundary sliding actually makes a polycrystalline solid weaker Pay attention to Figures 710 and 711 6 Please write at the bottom of this section in your book a very important fact that is not mentioned namely polycrystalline ductility at very low temperatures lt03 Tm in degrees K requires at least 5 closest packed slip systems Otherwise the material may fail in a very brittle fashion at lower temperatures Read pp 211 225 7 What is the basic strategy used to increase strength and hardness in crystalline metals What are the three approaches to implementing this strategy described in this chapter 8 Give two reasons why grain boundaries impede dislocation motion 9 What is the HallPetch relationship and what does it describe 10 Underline the sentence in your book on pp 189 that indicates the effect of grain size on toughness as well as strength 11 Explain brie y how solid solutions increase the resistance to dislocation motion 12 If the impurity atoms are sufficiently mobile to what region of the crystal will they likely diffuse and what effect does this have on the yield behavior Hint see Fig 610 pp 143 to contrast yielding without dislocation pinning by atoms diffusing to the dislocation line and yielding with dislocation pinning by atoms diffusing to the dislocation line The best example is steel which is iron with less than 1 carbon atoms The small carbon atoms are fairly mobile and can diffuse to the dislocation line pinning the dislocation and giving a sharp yield and yield drop 13 What is strain hardening and what is meant by cold working 14 What happens to YS TS and ductility during strain hardening Hint see Figures 719 and 720 and also Figure 617 pp 154 15 Why does hardness and strength increase with increasing strain 16 What is recovery 17 What happens to crystals during recrystallization What happens to the YS TS and ductility during recrystallization 18 What is the definition of recrystallization temperature What fraction of the melting point in degrees K is the recrystallization temperature book gives a range How does degree of cold work affect Tr Read pages 235 241 19 What is the difference in ductile fracture and brittle fracture 20 Give two reasons ductile fracture is preferred in service 21 How does moderately ductile fracture differ from highly ductile fracture according to Figure 81 Characterize the actual fractures on Figure 83 using the nomenclature in Figure 81 22 Describe the process of ductile fracture as illustrated in Figure 82 23 What are Chevron marks and how do they help to establish the initiation of a fracture 24 What is the difference in trans granular cracking versus intergranular cracking ME 3322 Materials and Manufacturing Chapters 10 and 11 Keys 109 The kinetics of the austenitetopearlite transformation obey the Avrami relationship Using the fraction transformed time data given here determine the total time required for 95 of the austenite to transform to pearlite Fraction T f Time s 02 126 08 28 2 Solution The first thing necessary is to set up two expressions of the form of Equation 1017 and then to solve simultaneously for the values of n and k In order to expedite this process we will rearrange and do some algebraic manipulation of Equation 1017 First of all we rearrange as follows 1 7 y exp 7ktn Now taking natural logarithms ln 1 7 y 7kIquot Or 7111 1 7 y ktquot which may also be expressed as m a Now taking natural logarithms again leads to lnln 71 H Ink n 1m 17 y which is the form of the equation that we will now use Using values cited in the problem statement the two equations are thus ln ln 7 lnk nln126s l 7 02 1 lnln1 7 08 7 lnk nln282 s Solving these two expressions simultaneously for n and kyields n 2453 and k 446 X 104 Now it becomes necessary to solve for the value of I at whichy 095 One of the above equations7viz 7ln 1 7 y ktquot may be rewritten as k I 7 1n 1 7 yjln k Now incorporating into this expression values for n and k determined above the time required for 95 austenite And solving for tleads to transformation is equal to 1 1 09 12453 I H 357 S 464 X 104 1010 The fraction recrystallized time data for the recrystallization at 600 C of a previously deformed steel are tabulated here Assuming that the kinetics of this process obey the Avrami relationship determine the fraction recrystallized after a total time of 22 8 min Fraction Time min Recrystallized 020 131 0 70 29 Solution The first thing necessary is to set up two expressions of the form of Equation 1017 and then to solve simultaneously for the values of n and k In order to expedite this process we will rearrange and do some algebraic manipulation of Equation 1017 First of all we rearrange as follows 17 y exkatquot Now taking natural logarithms ln 17 y iktquot Or iln l 7 y ktquot which may also be expressed as In I 1 y ktquot Now taking natural logarithms again leads to 111 1n1nk nlnt 1 y which is the form of the equation that we will now use The two equations are thus ln ln 4 lnk nlnl3lmin 17 020 ln bl l ln k nln29l min 1 7 070 Solving these two expressions simultaneously for n and kyields n 2112 and k 975 X 104 Now it becomes necessary to solve fory when I 228 min Application oquuation 1017 leads to y l 7 exp ktquot 1 7 expE 975 X 104228 minz39112 051 1015 Suppose that a steel of eutectoid composition is cooled to 550 C 1020 F from 760 C 1400 F in less than 0 5 s and held at this temperature a How long will it take for the austenitetopearlite reaction to go to 50 completion To 100 completion b Estimate the hardness of the alloy that has completely transformed to pearlite Solution We are called upon to consider the isothermal transformation of an ironcarbon alloy of eutectoid composition a From Figure 1022 a horizontal line at 550 C intersects the 50 and reaction completion curves at about 25 and 6 seconds respectively these are the times asked for in the problem statement b The pearlite formed will be fine pearlite From Figure 103011 the hardness of an alloy of composition 076 Wt C that consists of fine pearlite is about 265 BB 27 HRC 1016 Brie y cite the differences between pearlite bainite and spheroidite relative to microstructure and mechanical properties Solution The microstructures of pearlite bainite and spheroidite all consist of ocferrite and cementite phases For pearlite the two phases exist as layers which alternate with one another Bainite consists of very fine and parallel needleshaped particles of cementite that are surrounded an ocferrite matrix For spheroidite the matrix is ferrite and the cementite phase is in the shape of sphereshaped particles Bainite is harder and stronger than pearlite which in turn is harder and stronger than spheroidite 1019 Make a copy of the isothermal transformation diagram for an iron carbon alloy of eutectoid composition Figure 1022 and then sketch and label time temperature paths on this diagram to produce the following microstructures a 100 fine pearlite b 100 tempered martensite c 50 coarse pearlite 25 bainite and 25 martensite Solution Below is shown the isothermal transformation diagram for a eutectoid ironcarbon alloy with time temperature paths that will yield a 100 fine pearlite b 100 tempered martensite and c 50 coarse pearlite 25 bainite and 25 martensite Tempevature 5 Tempevalure F Tlme 5 1023 Name the microstructural products of eutectoid ironwarbon alloy 0 76 nt 0 specimens thm are rst w J mstenite 39 39 4 t mu mm a 200 Cs D 10000 and c 20 Us m n n 4 r 101 specimens ofan ironcarbon alloy of eutectoid composition that are continuously cooled to room temperature at a Variety ofrates Figure 1027 is used in these determinations a At a rate of 200 Cs only martensite forms b At a rate of 100 Cs both martensite and pearlite form c At a rate of 20 Cs only ne pearlite forms 1024 Figure 1040 shows the continuous cooling transformation diagram for a 113 1 C ironcarbon alloy Make a copy of this gure and then sketch and label continuous cooling cuwes to yield the following microstmcmms a Fine pearlite andproeutectoid cementite b Martensite c Martensite andproeutectoid cementite 0 Come pearlite andproeutectmu cementt39te e Mmenxite tw pearlite andproeutectoid cementite Solution Below is shown a continuous cooling transformation diagram for a 113 wt C ironcarbon alloy with continuous cooling paths that will produce a ne pearlite and proeutectoid cementite b martensite c martensite and proeutectoid cementite d coarse pearlite and proeutectoid cementite and e martensite fine pearlite and proeutectoid cementite Temperature Hz Tim 5 1025 Citetwo 39 I 39 39 39 39 39 39 quot 391 alloy steels Solution Two important differences between continuous cooling transformation diagrams for plain carbon and alloy steels are 1 for an alloy steel a bainite nose will be present which nose will be absent for plain carbon alloys and 2 the pearliteproeutectoid noses for plain carbon steel alloys are positioned at shorter times than for the alloy steels 1027 Name the 39 urnlurk 0f4340 alloy rt I 39 that are 39 L auxtem39te then cooled to room temperm um at the following rmer q 10 Cs b 1 Cs c 0 1 Cx and w 0 01 Cs Solution This problem asks for the microstructural products that form when specimens of a 4340 steel are continuously cooled to room temperature at several rates Figure 1028 is used for these determinations a At a cooling rate of lOOCs only martensite forms b At a cooling rate of lOCs both martensite and bainite form c At a cooling rate of 0l Cs martensite proeutectoid ferrite and bainite form d At a cooling rate of 00l Cs martensite proeutectoid ferrite pearlite and bainite form 1030 Brie y explain why ne pearlite is harder and stronger than coarse pearlite which in turn is harder and stronger than sph eroidite Solution The hardness and strength of ironcarbon alloys that have microstructures consisting of ocferrite and cementite phases depend on the boundary area between the two phases The greater this area the harder and stronger the alloy inasmuch as 1 these boundaries impede the motion of dislocations and 2 the cementite phase restricts the deformation of the ferrite phase in regions adjacent to the phase boundaries Fine pearlite is harder and stronger than coarse pearlite because the alternating ferritecementite layers are thinner for fine and therefore there is more phase boundary area The phase boundary area between the spherelike cementite particles and the ferrite matrix is less in spheroidite than for the alternating layered microstructure found in coarse pearlite 1031 Cite two reasons why martensite is so hard and brittle So ution Two reasons why martensite is so hard and brittle are 1 there are relatively few operable slip systems for the bodycentered tetragonal crystal structure and 2 virtually all of the carbon is in solid solution which produces a solidsolution hardening effect 1033 Brie y explain why the hardness of tempered martensite diminishes with tempering time at constant temperature and with in creasing temperature at constant tempering time This question asks for an explanation as to why the hardness of tempered martensite diminishes with tempering time at constant temperature and with increasing temperature at constant tempering time The hardness of tempered martensite depends on the ferritecementite phase boundary area39 since these phase boundaries are barriers to dislocation motion the greater the area the harder the alloy The microstructure of tempered martensite consists of small spherelike particles of cementite embedded within a ferrite matrix As the size of the cementite particles increases the phase boundary area diminishes and the alloy becomes softer Therefore with increasing tempering time the cementite particles grow the phase boundary area decreases and the hardness diminishes As the tempering temperature is increased the rate of cementite particle growth also increases and the alloy softens again because of the decrease in phase boundary area 1034 Brie y describe the simplest heat treatment procedure that would be used in converting a 0 76 wt C steel from one microstructure to the other as follows a Spheroidite t0 tempered martensite b Tempered martensite t0 pearlite e Bainite t0 martensite d Martensite t0 pearlite e Pearlite t0 tempered martensite f Tempered martensite t0 pearlite g Bainite t0 tempered martensite h Tempered martensite t0 spheroidite Solution In this problem we are asked to describe the simplest heat treatment that would be required to convert a eutectoid steel from one microstructure to another Figure 1027 is used to solve the several parts of this problem a For spheroidite to tempered martensite austenitize at a temperature of about 760 C quench to room temperature at a rate greater than about l40 Cs then isothermally heat at a temperature between 250 and 650 C b For tempered martensite to pearlite austenitize at a temperature of about 760 C then cool to room temperature at a rate less than about 35 Cs c For bainite to martensite first austenitize at a temperature of about 760 C then quench to room temperature at a rate greater than about l40 Cs d For martensite to pearlite first austenitize at a temperature of about 760 C then cool to room temperature at a rate less than about 35 Cs e For pearlite to tempered martensite first austenitize at a temperature of about 760 C then rapidly quench to room temperature at a rate greater than about l40 Cs then isothermally heat treat temper at a temperature between 250 and 650 C f For tempered martensite to pearlite first austenitize at a temperature of about 760 C then cool to room temperature at a rate less than about 35 Cs g For bainite to tempered martensite first austenitize at a temperature of about 760 C then rapidly quench to room temperature at a rate greater than about l40 Cs then isothermally heat treat temper at a temperature between 250 and 650 C h For tempered martensite to spheroidite simply heat at about 700 C for approximately 20 h 1039 F or a eutectoid steel describe isothermal heat treatments that would be required to yield specimens having the following Rockwell hardn esses39 a 93 HRB b 40 HRC and c 27 HRC Solution For this problem we are asked to describe isothermal heat treatments required to yield specimens having several Brinell hardnesses a From Figure 103011 in order for a 076 wt C alloy to have a Rockwell hardness of 93 HRB the microstructure must be coarse pearlite Thus utilizing the isothermal transformation diagram for this alloy Figure 1022 we must rapidly cool to a temperature at which coarse pearlite forms ie to about 675 C allow the specimen to isothermally and completely transform to coarse pearlite At this temperature an isothermal heat treatment for at least 200 s is required b This portion of the problem asks for a hardness of 40 HRC the microstructure could consist of either 1 about 75 fine pearlite and 25 martensite Figure 1032 or 2 tempered martensite Figure 1035 For case 1 after austenitizing rapidly cool to about 580 C Figure 1022 hold at this temperature for about 4 s to obtain 75 fine pearlite and then rapidly quench to room temperature For case 2 after austenitizing rapidly cool to room temperature in order to achieve 100 martensite Then temper this martensite for about 2000 s at 535 C Figure 1035 c From Figure 103011 in order for a 076 wt C alloy to have a Rockwell hardness of 27 HRC the microstructure must be fine pearlite Thus utilizing the isothermal transformation diagram for this alloy Figure 1022 we must rapidly cool to a temperature at which fine pearlite forms ie at about 580 C allow the specimen to isothermally and completely transform to fine pearlite At this temperature an isothermal heat treatment for at least 7 s is required Chapter 11Key 113 What is thefunction ofalloying elements in tool steels olution The alloying elements in tool steels eg Cr V W and Mo combine with the carbon to form very hard and wearresistant carbide compounds 115 On the basis of microstructure brie y explain why gray iron is brittle and weak in tension olution Gray iron is weak and brittle in tension because the tips of the graphite akes act as points of stress concentration 1111 What is the chief difference between heattreatable and nonh eattreatable alloys 0 ution The chief difference between heattreatable and nonheattreatable alloys is that heattreatable alloys may be strengthened by a heat treatment wherein a precipitate phase is formed precipitation hardening or a martensitic transformation occurs Nonheattreatable alloys are not amenable to strengthening by such treatments 1113 Cite advantages and disadvantages of hot working and cold working Solution The advantages of cold working are 1 A high quality surface finish 2 The mechanical properties may be varied 3 Close dimensional tolerances The disadvantages of cold working are 1 High deformation energy requirements 2 Large deformations must be accomplished in steps which may be expensive 3 A loss of ductility The advantages of hot working are 1 Large deformations are possible which may be repeated 2 Deformation energy requirements are relatively low The disadvantages of hot working are 1 A poor surface finish 2 A variety of mechanical properties is not possible 1114 a Cite advantages of forming metals by extrusion as opposed to rolling b Cite some disadvantages Solution a The advantages of extrusion as opposed to rolling are as follows 1 Pieces having more complicated crosssectional geometries may be formed 2 Seamless tubing may be produced b The disadvantages of extrusion over rolling are as follows 1 Nonuniform deformation over the crosssection 2 A variation in properties may result over a crosssection of an extruded piece 1115 List four situations in which castin g is the preferred fabrication technique Solution Four situations in which casting is the preferred fabrication technique are 1 For large pieces andor complicated shapes 2 When mechanical strength is not an important consideration 3 For alloys having low ductilities 4 When it is the most economical fabrication technique 1119 In your own words describe the following heat treatment procedures for steels and for each the intended nal microstructure full annealing normalizing quenching and tempering Solution Full annealingHeat to about 50 C above the A3 line Figure lllO if the concentration of carbon is less than the eutectoid or above the A1 line if the concentration of carbon is greater than the eutectoid until the alloy comes to equilibrium then furnace cool to room temperature The final microstructure is coarse pearlite NormalizingHeat to at least 55 C above the A3 line Figure lllO if the concentration of carbon is less than the eutectoid or above the A cm line if the concentration of carbon is greater than the eutectoid until the alloy completely transforms to austenite then cool in air The final microstructure is fine pearlite QuenchingnHeat to a temperature within the austenite phase region and allow the specimen to fully austenitize then quench to room temperature in oil or water The final microstructure is martensite TemperingHeat a quenched martensitic specimen to a temperature between 450 and 650 C for the time necessary to achieve the desired hardness The final microstructure is tempered martensite 1120 Cite three sources of internal residual stresses in metal components What are two possible adverse consequences of these stresses Solution Three sources of residual stresses in metal components are plastic deformation processes nonuniform cooling of a piece that was cooled from an elevated temperature and a phase transformation in which parent and product phases have different densities Two adverse consequences of these stresses are distortion or warpage and fracture 1121 Give the approximate minimum temperature at which it is possible to austenitize each of the following iron carbon alloys during a normalizing heat treatment a 020 wt C b 076 wt C and c 095 wt C Solution a For 020 Wt C heat to at least 905 C 1660 F since the A3 temperature is 850 C 1560 F b For 076 Wt C heat to at least 782 C l440 F since the A3 temperature is 727 C l340 F c For 095 Wt C heat to at least 840 C 1545 F since theAcm temperature is 785 C l445 F 1124 Brie y explain the difference between hardness and hardenability Hardness is a measure of a materials resistance to localized surface deformation whereas hardenability is a measure of the depth to which a ferrous alloy may be hardened by the formation of martensite Hardenability is determined from hardness tests 1126 How would you expect a decrease in the austenite grain size to affect the hardenability of a steel alloy Why Solution A decrease of austenite grain size will decrease the hardenability Pearlite normally nucleates at grain boundaries and the smaller the grain size the greater the grain boundary area and consequently the easier it is for pearlite to form 1128 Construct radial hardness pro les for the following a A 50mrn 2in diameter cylindrical specimen of an 8640 steel alloy that has been quenched in moderately agitated oil Solution In the manner of Example Problem 111 the equivalent distances and hardnesses tabulated below were determined from Figures 1114 and 11171 Radial Equivalent HRC W Distance mm Hardness Surface 7 54 34 R l l 5 0 Midradius l 4 45 Center 1 6 44 The resulting hardness profile is plotted below g m 8 r u x N I 42 50 Position b A 75min 3in diameter cylinc 39ical specimen ofa 5140 steel alloy that ha been quenched in moderately agitated oil Solution In the manner of Example Problem 111 the equivalent distances and hardnesses tabulated below were determined from Figures 1114 and 1117b Radial Equivalent HRC m istance mrn Hardness Surface 13 41 34 R 1 7 5 3 7 Midradius 22 33 Center 25 3 2 The resulting hardness pro le is plotted below Hardness HRC 75 mm Position 511 65min Z in diameter cylindi39ical specimen of an 8620 steel alloy that liar been quenched in moderately agitated water Solution In the manner of Example Problem 111 the equivalent distances and hardnesses tabulated below were determined l romFigures 1115 and 111711 Radial Equivalent HRC W Distance mrn Hardness Surface 25 42 34 R 7 3 1 Midradius 1 1 25 Center 13 24 The resulting hardness pro le is plotted below Hardness H RC k 65 mm I Position A 70min 2 31in diameter cylinc 39ical specimen of a 1040 steel alloy that hm been quenched in moderately agitated water Solution In the manner of Example Problem 111 the equivalent distances and hardnesses tabulated below were determined from Figures 1114 and 111711 i Equivalent m m ELIE655 Surface 3 48 34 R 8 30 Midradius 13 23 Center 15 22 The resulting hardness pro le is plotted below Hardness HRC l 7ll rum i Position 1130 Compm39e precipitation hardening Section 11 9 and the hardening 0f39teel by quenching and tempering Sections 105 106 and 108 with regm39d to are a The total hem treatment procedure 1 The microxtmcmrex thm develop 6 39 39 during th Solution 39 follows L the steps for a 39 39 39 39 39 l Austenitize above the upper critical temperature 2 Quench to a relatively low temperature 3 Temper at a temperature below the eutectoid 4 Cool to room temperature With regard to precipitation hardening the steps are as follows 1 2 Quench to a relatively low temperature 3 neatin to a 4 Cool to room temperature b For the hardening of steel the microstructnres that form at the Various heat treating stages in part a l Austenite 2 Martensite 3 Tempered martensite 4 Tempered martensite For h J 1 Single phase 2 Single phasesupersaturated 3 Small platelike particles of a new phase within a matrix of the original phase 4 Same as 3 c For the hardening afsleel the mechanical characteristics for the various steps in part a are as follows 1 Not important 2 The steel becomes hard and brittle upon quenching 3 During tempering the alloy softens slightly and becomes more ductile 4 No significant changes upon cooling to or maintaining at room temperature For precipitation hardening the mechanical characteristics for the various steps in part a are as follows 1 Not important 2 The alloy is relatively soft 3 The alloy hardens with increasing time initially and becomes more brittle39 it may soften with overaging 4 The alloy may continue to harden or overage at room temperature ME 3322 Chapter 9 Homework Name l Derive Lever s rule Note it is based on conservation of mass as the fundamental principle on which the rule is based hi g cmmmw ng Cd 0ch 3 00 SEre LVL VCOJCQ CAD V 06 39 r 2 Using the c ta 1111551 288 co f a cgp p rnickel alloy with 40 wt Ni at 1250C A What phases are present Q 1 LIPQ M B What is t e chemical composition of eac hase in wt nickel C What is the weight of each phase present in the system m 01 I 72 7 3 How would the coppernickel alloy s microstructure change with faster non 325 MWWZ equilibrium cooling compared to slower equilibrium cooling M177 4 Why does the strength of the 60 Ni 40 wt Cu exceed the strengt of pure copper or pure nickel j ali gal djm 2i 5 The four Hume Rothery rules for limited solubility rules 1 and 3 and limited solubility 14 are given on page pp 9495 Apply these rules to the nickel copper system and see if they predict an unlimited solid solution which is obviously what the phase diagram indicates Note a difference in valence of l is acceptable for unlimited 7 solubility but a difference of 2 is not Electronegativity data is on pp 27 A difference of 01 is okay for unlimited solubility Greater difference than 01 will diminish the solubility preventing unlimited solubility One can achieve limited solubility with ifferences in atomic size of up to 15 but unlimited solubility requires difference less m an 10WF 39 s are listLel 39 iig e the front apbf gbook Rules 1 I 2 Mining munimikngmkani 6 For coppersilver phase diagram on pp 298 answer the following questions QS F06 A What is the solubility limit of silver in copper Hint what is the greatest a cunt of silver you can put into solution in the copper rich alpha phase Q Ma 2 8 B What is the solubility limit of copper in silver Hint what is the great st mount of copper you can put into solution in silver rich beta phase m g C Which of the Hume Rothery factors indicates that copper and silver will not form an unlimited solid solution like copper and nickel do 1 Hint review my notes in problem 5 which are more complete than the book Wk gQW wa 293 D The eutectic structure is composed of alpha and eta phases How much of each phase do you have at a temperature of 778C at the eutectic composition of 719 M 375 a awt a39 H B What mechanical propei39ties woulayou expect for alpha phase and beta a I alpha gaffquot garaif a K mmmw phases Strong with low ductility or soft with high ductile Beta arm dyo i dg m B F What mechanical propertie W0 ld you expect forwthgeutecticstrugture of the silvercopper alloy qualitative om ductility orsoftlow strength duitil e I Cf Explain your answer brie y Circle one 6 37 w 1555 W 7C G Sterling silver has 75 wt Cu and 925 wt Ag At 779C what would be the phases present What would be the composition of the phases At this temperature and composition where is the Cu 9 7 w 755 409 i 9 9 HM H If you cooled it slowly to 200C what phases wodld be present Where is the copper after slow cooling from 779C to 200C 77167 at 3 A 44 4 96 M I If you cooled it very rapidly from 790C to 200C little or no diffusion where would the copper be 7 4quot 9701 7 f pi tgt f V7 J Pure silver annealed and cooled rapidlyrto ioom temperature has a yield strength of 7800 psi Sterling silver annealed and cooled rapidly to room temperature has a yield strength of 18000 psi What is the strengthening mechanism for sterling silver with this thermal history hint choices are strain hardening grain size re nement g substitutional solid solution strennth inB interstitial solid solution strengthening or hard second phase strengthening Circle one of choices Would it have good ductility 7 or the lead tin phase diagram on pp 309 what are the microstructural constituents for a 5050 composition 3 nn Figure 917 Choices ar primary alpha any of these microstru ural costituent that appear in Figure 917 A Which microstructural constituent is continuous 3g 2 m B Compare Figures 917 for a eutectic structure and 933 for a eutectoid structure Which microstructural constituent is continuous Figure 917 and which ignicrostructural constituent is continuous in Figure 933 917 a 66 C and 9 1 W 933 Elquot It dWit F63 CWhy this difference 018 IS 96 39 H C 54 I74M0 3 Pp C Wha kind of properties would you expect the eutectic of this alloy to have 66 8 Look at the phase diagram for copper and zinc on pp 312 Copper alloyed with zinc is called brass A Any eutectic reactions Any peritectic reactions V85 Q B What is the maximum solubility of zinc in copper rich phase alpha N 32g 2 C Zinc obviously has a high solubility in copper but not unlimited solubility as does nickel in copper Which Hume Rothery rule is violated that prevents unlimited W505 5 solubility of zinc in copper er WS JLQM SULY cXQILUte 1 91 W D The yield strength of 7030 brass is 15000 psi Pure copper has a yield 539 60w 6quot stren h of 10000 psi What is the strengthening mechanism for annealed 7030 brass W Why is the strengthening not more signi cant 0quot 12 tquot is 5 11103 Which base in the MgPb phase diagram pp 313 is an intermetallic compound ncfi a39 o 3 C 9 A What mechanical properties should the intermetallic compound h ve d g B What mechanical properties should the alpha phase have 4 U C What phases would be present for the eutectic structure that forms at 67 Pb 0amp2 MiZPb 30 D How much of each phase would you have in the eutectic just after it forms just below 465C Note that for normal cooling rates these percentages Mia will remain the same as you 0001 from 465C to room temperature since there will be insufficient time for diffusion to occur unless very very slow cooling takes place B What mechanical properties would this eutectic microstructural constituent have Note that if eutectic microstructural constituents h ve more than 1215 wt brittle phase they will have very poor ductility 5 a 02 6 OF46 3 2 10 For the iron carbon phase diagram on pp 319 answer the following questions A For a eutectoid 076 wt carbon steel what will be the phases present at room temperature d t 33 B Whatxwill be the wt 0 0 carbon in each of these phases 3 and 24 m 763C WOW in 04 C 02W 7 72A 2 C What with be the wt of each phase in this eutectoid microstructure 6mg aft 35 322 a a a Page D What mechanical pro erties would you expect for this microst ctural 3G constituent called pearlite o mq Apr On a 10me p LQULU 4 Note the elongation in pearlite is 10 YS7000d and UTSl30000 Pure iron is 45 elongation YS 40900 psi UTS 45000 psi 11 Using the phase diagam o p 319 determine what is the solubility limit of carbon 37 in alpha iron 0W 0 AWhat is the olubility limit of carbon in gamma iron y 33 3 Lib 1 BWhy is the solubility of carbon in gamma iron 100X greater than the solubility of carbon in alpha iron Hint the ratio of the radius of the interstitial site in FCC to the radius of the atoms in FCC is rR 0414 and in BCC it is rRO29l Calculate the radius of the interstitial site for each type of site by substituting R0 124 nm for iron to 39 determine the radius r of the interstitial Site and then comparing this to the radius of v 03 atomic carbon to see how easily it can be tted into either of these sites aw 5 l waan 46 m ace 39 034073 I3 H as 238 WWW C For iron with 43 wt carbon at1146C determine the phases that are prese CZ p M 49 24 4quot their respective chemical compositions w 1 6 E 77 t e 5 7 L61 expected mechanical properties for each phase k gill 57740 3 the wt of each phase in the eutectic structure r a F7 and 7 S the mecha ical properties you would expect for the eutectic stru ture for ironcarbon t 5 CH system 34469 a D For a 018 wt carbon steel determine the amount of proeutectoid alpha M 4 6 phase and the amount of gamma phase at 728C M 9 W B What is t e composition of eac of these phases any Y 40 o m yawv EC F What happ ns to the proeutectoid alpha as the temperature is cooled from 728C 4 to 7250 my Wt 422M rampa6 G Wha happens todthe gamma ase s it is cooled from 728C to 725C szarms ow army gs lt4 15ch ankl H which of tese giigstrptzctural constituents will be continuous I Hdlw much cementite and how much alpha phase is in the eutectoid pearlite 64 50 Mt 5 J What are the mechanical properties of thge alpha phase a Mk r t 51 and the cementite phase 396 8 K What then should be the mechanical properties of the pearlite eutectoid microstructural constituent and the proeAtectoid microstrucgral const tuent Hint see i Si Wage17117 LOILHOVZ br 8 63 10D above gala yf 0t Sta b L What then would you expect the properties to be of the noncarbon all y w1th 461261 la wt Ma 018 wt carbon if it was air cooled to room temperature from 900C 12 For steel with 14 wt carbon answer the following uestions 5 A What phases will be present at 728C YI 61C d B What will be the chemical compositions of each phase 1 L7 E E251 67 63 C As you cool from 728C to 725C What hap ens to each phase 1 Q 5 an 717 W VAalfF zC Qatarf a s D What will be the wt of each phase that forms from gamma as gamma is w cooled from 728C to 725C zCL B What would be the name of this microstructural constituent that forms from a gamma F W at would be the name of the microstructural constituent that forms as Q Q proeutectoid cementite is cooled from 728 to 725 WWJ39FBE G W at kind of echan39cal properties should it have M at 8 golft6 H Which of these two microstructural constitue ts ill be continuous Hint s e 49 Fig 933 MM Why pm orig 3216 01 81mm IQ lff r 1 ME 3322 Materials and Manufacturing Processes Homework Chapters 10 and 11 Chapter 10 1 09 1010 1015 1016 1019 Hint part c will involve a series of quenches and isothermal transformations 1023 1024 attached 1025 1027 1030 1031 39 1034 1039 Chapter 11 1 13 115 1111 1113 1114 1115 1119 1120 1121 1124 1126 1128 1130 Figure 1040 Continuousz transformation diagram wt C iron carbon allo Temperature C Time s 8P3 14 L v l39 ME 3322 Materials and Manufacturing Homework Key Chapter 17 l71 a Brie r explain the di izrence between oxidation and reduction electrochemical reactions 17 ll39hich reaction occurs at the anode and which at the cathode Solution to Oxidation is the process by which an atom gives up an electron or electrons to become a cation Reduction is the process by which an atom acquires an extra electron or electrons and becomes an anion b Oxidation occurs at the anode reduction at the cathode 175 A Zn39Zn39 concentration cell is constructed in which both electrodat are pure im The Zn concentrationfor one cell halfi l 0 ill for the other I 039 39 M t Is a voltage generated between the mo cell halves I39m what is its magnitude and which electrode will be oxidized If no voltage is praduced extplain this result Solution This problem calls for us to determine whether or not a voltage is generated in a Znan39 concentration cell and if so its magnitude Let us label the Zn cell having a LG M an solution as cell 1 and the other as cell 2 Fluthennore assume that oxidation occurs within cell 2 wherein an 10 2 M Hence Zn an Zn an and employing Equation I720 leads to I 00592 ani AV T lo T an 00592 102 l oos 2 2 l LOAIJ 9 V Therefore a voltage of 00592 V is genemted when oxidation occurs in the cell having the 2113 concennau39on of 1041 177 An electrochemical cell is constructed such that on one side a pure nickel electrode is in contact with a solution containing NF ions at a concentration of x 0quot M The other cell half consists of a pure F c electrode that is immersed in a solution af F 2 ion having a concentration of 0 M At what ternqu will the potential between the two electrodes to 0 HO V Solution This problem asks for us to calculate the temperature for a nickeliron electrochemical cell when the potential between the Ni and Fe electrodes is 0140 V On the basis of their relative positions in the standard emf series Table 17l assume that F is oxidized and Ni is reduced Thus the electrochemical reaction that occurs within this cell isjust V Ni2 Fe Ni Fe3 Generated by CamScanner Thus Equation l720 is written in the form RT rah V V 39 1 A N l e quotF quotNi2 Solving this expression for Tgives T E AV I39 i VF R Fain Nili In The standard potentials from Table l7l are Vc 0440 V and V i 0250 V Therefore 296500 Cmolio14o v 0250 v o44o wi 83 lmolK In 01 M 3 x lo3 M T 33 K 58 C 178 For the following pairs of alloys that are coupled in seawater predict the possibility of corrosion if corrosion is probable note which metalalloy will corrode a Aluminum and magnesium b Zinc and a lowcarbon steel c Brass 60Cu 402n and Martel 70Ni 30Cu d Titanium and 304 stainless steel 2 Cast iron and 316 stainless steel Solution This problem asks for several pairs of alloys that are immersed in seawater to predict whether or not corrosion is possible and if it is possible to note which alloy will conode In order to make these predictions it is necessary to usa the galvanic series Table 172 If both of the alloys in the pair reside within the same set of brackets in this table then galvanic corrosion is unlikely However if the two alloys do not lie within the same set of brackets then that alloy appearing lower in the table will experience corrosion a For the aluminummagnesium couple corrosion is possible and a nesium will corrode b For the zinclow carbon steel couple corrosion is possible andW c For the brassmane couple corrosion is unlikely inasmuch as o a oys appear within the same set of brackets d For the titanium304 stainless steel pair the stainless steel will corrode inasmuch as it is below titanium in both its active and passive states e For the cast iron3 6 stainless steel couple the cast iron will corrode since it is below stainless steel in both active and passive states I 79 a From the galvanic series Table 172 cite three metals or alloys that may be used to galvanically protect 304 stainless steel in the active state As Concept Check 174b notes galvanic corrosion is prevented by making an electrical contact between the two metals in the couple and a third metal that is anodic to the other two Using the galvanic series name one metal that could be used to protect a copper aluminum galvanic couple Solution Generated by CamScanner I It The Following metals and alloys may be used 0 galvanically protect 304 stainless steel in the native slate cusijn iron eck aanLalumjnum a gys cadmium zine mugncsiummylgncsium glioyi lltem metalsInlays appear below cast iron in the galvanic series Table 172 x 39 Ulwnd mggnesnum may be ttsed to protect a copperaluminum gnlvamc couple these metals urc unodlc to aluminum m the galvamc series Generated by CamScanner l722 For each farm of corrosion other Ihan uniform do rhefollmt39ing a cribs why where and the conditions under which the corrosion occurs b C Ila three measures Ihal maI be taken to prevent or control it For each of the forms of corrosion the conditions under which it occurs and measures that may be taken to prevent or control it are outlined in Section l77 1724 Briefly explain why for a small anodeIo caihade area ratio the corrosion rate will be higher than for a large ratio Solution For a small anodetocathode area ratio the corrosion rate will be higher than for a large ratio The reason for this is that for some given current flow associated with the corrosion reaction for a small area ratio the current density at the anode will be greater than for a large ratio The corrosion rate is proportional to the current density 1 according to Equation 1724 727 Brie y describe the two techniques that are usedfor galvanic protection Solution Descriptions of the two techniques used for galvanic protection are as follows A sacri cial anode is elecm cally coupled to the metal piece to be protected which anode is also situated in the corrosion environment The sacri cial anode is a metal or alloy that is chemically more reactive in the particular environment It the anode preferentially oxidizes and upon giving up electrons to the other metal protects it from electrochemical corrosion 2 An im ressed current from an external dc power source provides excess electrons to the metallic structureW Generated by CamScanner Chapters 8 Solutions 85 A specimen ofa 4340 steel alloy having a plane strain fracture toughness of 45Wa1E 41 shE is exposed to a stress of1000 MPa 145 000 psi Will this specimen experiencefracture ifit is known that the largest surface crack is 0 75 mm 003 in long Why or why not Assume that the parameter Yhas a value of10 Solution This problem asks us to determine whether or not the 4340 steel alloy specimen will fracture when exposed to a stress of 1000 MPa given the values of K18 Y and the largest value of a in the material This requires that we solve for 08 from Equation 86 Thus 45 MPa4 m 927 MPa 133500 psi 5 X 103 m K1c Y M 1011r07 Gc Therefore fracture will most likely occur because this specimen will tolerate a stress of 927 MPa 133500 psi before fracture which is less than the applied stress of 1000 MPa 145000 psi 86 Some aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 35Wa1E 319sAE It has been determined that fracture results at a stress of 250 MPa 36250 ps1 when the maximum or critical internal crack length is 20 mm 008 in For this same component and alloy will fracture occur at a stress level of 325 MPa 47125 psi when the maximum internal crack length is 10 mm 004 in Why or why not Solution We are asked to determine if an aircraft component will fracture for a given fracture toughness 35 MPan stress level 325 MPa and maximum internal crack length 10 mm given that fracture occurs for the same component using the same alloy for another stress level and internal crack length It first becomes necessary to solve for the parameter Y using Equation 85 for the conditions under which fracture occurred ie o 250 MPa and 2a 20 mm Therefore Y K16 35 MPa 250 CW 250 MPa 1r w Now we will solve for the product Y 011ra for the other set of conditions so as to ascertain whether or not this value is greater than the K18 for the alloy Thus Excerpts from thiswork may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Yo 1ra 250325 MPa 1r 322 MPaJE 295 ksi Therefore fracture will not occur since this value 323 is less than the K18 of the material 35 MPa 810 A structural component in theform ofa wide plate is to befabricatedfrom a steel alloy that has a plane strain fracture toughness of 770Wa1E 70 I laidH and a yield strength of1400 MPa 205000 psi The flaw size resolution limit of the flaw detection apparatus is 40 mm 016 in Ifthe design stress is one half of the yield strength and the value on is I 0 determine whether or not a criticalflawfor thisplate is subject to detection Solution This problem asks that we determine Whether or not a critical aw in a Wide plate is subject to detection given the limit of the aw detection apparatus 40 mm the value of K18 77 the design stress 02 in which oy 1400 MPa and Y 10 We first need to compute the value of ac using Equation 87 thus 12 Z 77 MP a iKIc 1 WE 00039 In 39 mm 015 in C 7 1 Yo 1r 1014002MPa Therefore the critical aw is not subject to detection since this value of aC 39 mm is less than the 40 mm resolution limit 813 Following is tabulated data that were gatheredfrom a series of Charpy impact tests on a tempered 4140 steel alloy T 2 C lmnact Fnergv J 100 89 3 75 88 6 50 87 6 25 85 4 0 82 9 725 78 9 75 0 731 765 66 0 775 59 3 785 4 7 9 71 00 34 3 Excerpts from thiswork may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 7125 293 7150 27 1 71 75 25 0 a Plot the data as impact energy versus temperature b Determine a ductiletobrittle transition temperature as that temperature corresponding to the average of the maximum and minimum impact energies 5 Determine a ductiletobrittle transition temperature as that temperature at which the impact energy is 70 1 Solution The plot of impact energy Versus temperature is shown below 00 I I I I oo o I Impact Energy J ON 0 J o I I I I 20 I I I 200 150 gt100 50 0 50 100 Temperature C b The average of the maximum and minimum impact energies from the data is Average 893 I 25 572 I As indicated on the plot by the one set of dashed lines the ductiletobrittle transition temperature according to this criterion is about 775 C c Also as noted on the plot by the other set of dashed lines the ductiletobrittle transition temperature for an impact energy of 70 J is about 755 C Excerpts from rm work may be reproduced by instructors for distribution on a noteforeprofit basis for Iesung or instructional purposes only to Any by Sections 107 01108 afthe 816 An 80 mm 031 in diameter cylindrical rodfabricatedfrom a red brass alloy Figure 834 is subjected to reversed tensioncompression load cycling along its axis If the maximum tensile and compressive loads are 7500 N1700 y and 7500 N 1700 lbf respectively determine itsfatigue life Assume that the stress plotted in Figure 834 is stress amplitude Solution We are asked to determine the fatigue life for a cylindrical red brass rod given its diameter 80 mm and the maximum tensile and compressive loads 7500 N and 7500 N respectively The first thing that is necessary is to calculate values of omax and 0min using Equation 61 Thus 6 Fmax Fmax max A0 d j i0 2 LNZ 150 X 106 Nm2 150 MPa 22500 psi 80 X 103 In W f F 0min 1 A 2 7150 gtlt 106 Nm2 7150 MPa 722500psi n80 X 103 mjz 2 Now it becomes necessary to compute the stress amplitude using Equation 816 as ea W mi 150 MR 728150 ma 150 MPa 22500 psi From Figure 834 f for the red brass the number of cycles to failure at this stress amplitude is about l X 105 cycles 8 l 8 The fatigue data for a brass alloy are given as follows Stress Amplitude MPaZ Cycles to Failure 310 2 x105 223 1x105 Excerpts from thiswork may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 191 3 x 10 168 1 x 107 153 3 x 107 143 1 x 10 134 3 x 10 127 1 x 109 a JWIke an S39erlot stress amplitude versux logarithm cycles to zilure using these data b Determine the fatigue strength at 5 X 10 eyees 5 Determine the fatigue life for 200 MPa Solution a The fatigue data for this alloy are plotted below Stress amplitude MPa Lug cycles to failure b As indicated by the A set of dashed lines on the plot the fatigue strength at 5 x 105 cycles log 5 x 105 57 is about 250 MPa c As noted by the B set of dashed lines the fatigue life for 200 MPa is about 2 gtlt 106 cycles ie the log ofthe lifetime is about 63 Exeapts from this work may be reproduced by instructors for distribution on a notrforrprofit basis for testing or instructional purposes only to Any by Sectiam107 01108 ofthe 824 Brie y explain the di erence between fatigue striations and beachmarks both in terms of a size and b origin Solution a With regard to size beachmarks are normally of macroscopic dimensions and may be observed with the naked eye fatigue striations are of microscopic size and it is necessary to observe them using electron microscopy b With regard to origin beachmarks result from interruptions in the stress cycles each fatigue striation is corresponds to the advance of a fatigue crack during a single load cycle Excerpts from thiswork may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 825 List four measures that may be taken to increase the resistance to fatigue of a metal alloy Solution Four measures that may be taken to increase the fatigue resistance of a metal alloy are 1 Polish the surface to remove stress amplification sites 2 Reduce the number of internal defects pores etc by means of altering processing and fabrication techniques 3 Modify the design to eliminate notches and sudden contour changes 4 Harden the outer surface of the structure by case hardening carburizing nitriding or shot peening Excerpts from thiswork may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful 827 The following creep data were taken on an aluminum alloy at 400 IC 7509 and a constant stress of 25 MPa 3660psi Plot the data as strain versus time then determine the steadystate or minimum creep rate Note The initial and instantaneous strain is not included Time min Strm39n Time min Strm39n 0 0000 16 0135 2 0025 18 0153 4 0043 20 0172 6 0065 22 0193 8 0078 24 0218 10 0092 26 0255 12 0109 28 0307 14 0120 30 0368 m These creep data are plotted below 04 r I 2 Time min The steadystate creep rate AUAt is the slope of the linear region ie the straight line that has been superimposed on the curve as 5 0230 e 009 70 x 10393 min391 At 30 min710 min Exca pts from this work may be reproduced by instructors for distribution on a notrforrprofit basis for testing or instructional purposes only to Any by Sections 107 01108 afthe W 828 A specimen 750 mm 30 in long ofan S 590 alloy Figure 831 is to be exposed to a tensile stress of 80 MPa 11600 psi at 815 C 1500 17 Determine its elongation after 5000 h Assume that the total ofboth instantaneous and primary creep elongations is 15 mm 0 06 in Solution From the 815 C line in Figure 831 the steady state creep rate gig is about 55 X 10396 h391 at 80 MPa The steady state creep strain cs therefore is just the product of 3S and time as cs Eng x time 55 X 106 h15000 h 00275 Strain and elongation are related as in Equation 6239 solving for the steady state elongation Alf leads to Al log 750 mm 00275 206 mm 081 in Finally the total elongation is just the sum of this Als and the total of both instantaneous and primary creep elongations ie 15 mm 006 in Therefore the total elongation is 206 mm 15 mm 221 mm 087 in 830 Ifa componentfabricatedfrom an S 590 alloy Figure 830 is to be exposed to a tensile stress of300 MPa 43500 psi at 650 0C 1200 17 estimate its rupture lifetime Solution This problem asks us to calculate the rupture lifetime of a component fabricated from an S590 alloy exposed to a tensile stress of 300 MPa at 650 C All that we need do is read from the 650 C line in Figure 830 the rupture lifetime at 300 MPa this value is about 600 h 834 Steadystate creep rate data are given belowfor nickel at 1000 0C 1273 K 3 S4 47 MPa psi 10 4 15 2175 10 5 45 650 Excerpts from thiswork may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful If it is known that the activation energy for creep is 272000 Jmol compute the steadystate creep rate at a temperature of850 0C 1123 K and a stress level of25 MPa 3625 psi Solution Taking natural logarithms of both sides of Equation 820 yields 7 Q8 ln gi an2 nlno 77 RT With the given data there are two unknowns in this equationnamely K2 and n Using the data provided in the problem statement we can set up two independent equations as follows 272 000 J mol ln1 gtlt10 4 s l anz nln15 MPa 7 831 JmolK1273 K 272 000 Imol m1 gtlt10 6 54 anz nln45 MPa 7 831 JmolK1273 K Now solving simultaneously for n and K2 leads to n 3825 and K2 466 s39l Thus it is now possible to solve for dg at 25 MPa and 1123 K using Equation 820 as 39 7 Q Eng 7 K20nexp 272 000 Jmol 466 s 125 MPa3825expi 831 Imol K1123 K 228 X 10395 s391 836 Cite three metallurgicanrocessing techniques that are employed to enhance the creep resistance of metal alloys Solution Three metallurgicalprocessing techniques that are employed to enhance the creep resistance of metal alloys are 1 solid solution alloying 2 dispersion strengthening by using an insoluble second phase and 3 increasing the grain size or producing a grain structure with a preferred orientation Excerpts from thiswork may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful Excerpts from thiswork may be reproduced by instructors for distribution on a notforprofit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful