Solution Manual For Fundamentals of Modern Manufacturing: Materials, Processes, and Systems, 7th Edition

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Additional Multiple-Choice QuestionsWhat Is Manufacturing?1.1The term plant capacity refers to the maximum rate of production that a factory canachieve under assumed levels of labor manning, hours of operation per week, and otheroperating conditions: (a) True or (b) false?Answer. (a).Materials in Manufacturing1.2A metal alloy is composed of two or more elements, all of which are metallic: (a) True or(b) false?Answer. (b). A metal alloy is composed of two or more elements, at least one of which ismetallic. An example is steel which is an iron-carbon alloy.1.3Which one of the following ceramics is not crystalline: (a) alumina, (b) clay, (c) glass, (d)titanium nitride, or (e) tungsten carbide?Answer. (c).Manufacturing Processes1.4Which two of the following are referred to as thin film deposition processes: (a) chemicalvapor deposition, (b) electroplating, (c) extrusion, (d) painting, and (e) physical vapordeposition?Answer. (a) and (e).1.5Which one of the following is a property-enhancing operation: (a) drilling, (b) extrusion,(c) forging, (d) grinding, (e) heat treatment, or (f) painting?Answer. (e).1.6Which two of the following processes provide a permanent joint: (a) expansion fits, (b)nuts and bolts, (c) screws, (d)threaded fasteners, and (e) welding?Answer. (a) and (e).1.7Fixtures are commonly used as work-holding devices in which two of the followingprocesses: (a) casting, (b) extrusion, (c) machining, (d) rolling, (e) stamping, and (f)welding?Answer. (c) and (f).Production Systems1.8Low-quantity production of large complex products is most closely associated with whichone of the following layout types: (a) cellular layout, (b) fixed-position layout, (c) processlayout, or (d) product layout?Answer. (b).1.9Batch production is traditionally associated with which one of the following layout types:(a) cellular layout, (b) fixed-position layout, (c) process layout, or (d) product layout?

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Answer. (c).Manufacturing Economics1.10The total time to complete a batch of parts is calculated using which three of the followingterms: (a) actual processing time in the operation, (b) batch quantity, (c) cycle time, (d)inspection time, (e) setup time, (f) tool change time, and (g) work part handling time?Answer. (b),(c), and (e).1.11The time to load a starting work part into the production machine in a certain unitproduction operation = 0.40 min, the actual processing time in the operation = 3.10 min,and the time to unload the finished part = 0.20 min. In addition, the tool must be changedevery 5 cycles and the tool change time = 1.50 min. The cycle time for the unit operation iswhich one of the following: (a) 3.5 min, (b) 3.7 min, (c) 4.0 min, (d) 4.5 min or (e) 5.2min?Answer. (c).1.12If the initial cost of a production machine including installation is $250,000, and theplanned use of the machine is 2000 hr/yr for 5 years, and the applicable overhead rate onthe equipment is 50%, which one of the following is the corresponding cost rate of thismachine in $/hr: (a) $0.625, (b) $25.00, (c) $37.50, (d) $50.00, or (e) $125.00?Answer. (c).AdditionalReviewQuestions1.1What are the differences between primary, secondary, and tertiary industries? Give anexample of each category.Answer. A primary industry is one that cultivates and exploits natural resources, such asagriculture or mining. A secondary industry takes the outputs of primary industries andconverts them to consumer and capital goods. Examples of secondary industries aretextiles and electronics. A tertiary industry is in the service sector of the economy.Examples of tertiary industries are banking and education.1.2Annual production quantities made by a factory can be classified into three ranges. Namethe three ranges and the approximate quantities of parts associated with each range.Answer. The three ranges are (1) low production, with quantities in the range 1 to 100parts; (2) medium production, with quantities in the range 100 to 10,000 parts; and (3) highproduction, with quantities in the range 10,000 to over a million parts.1.3Define plant capacity.Answer. Plant capacity, a.k.a. production capacity, is the maximum rate of productionoutput that a plant can achieve under assumed operating conditions. Operating conditionsrefer to thenumber of shifts per week, hours per shift, direct labor manning levels in theplant, and so on.1.4One of the three general types of processing operations is shaping operations, which areused to create or alter the geometry of the work part. What are the four categories ofshaping operations?

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Answer. The four categories of shaping operations are (1)solidification processes, inwhich the starting material is a heated liquid or semifluid that cools and solidifies to formthe part geometry; (2) particulate processing, in which the starting material is a powder,and the powders are formed and heated into the desired geometry; (3) deformationprocesses, in which the starting material is a ductile solid (commonly metal) that isdeformed to shape the part; and (4) material removal processes, in which the startingmaterial is a solid (ductile or brittle), from which material is removed so that the resultingpart has the desired geometry.1.5What is the difference between net shape processes and near net shape processes?Answer.Net shape processesare manufacturing processes that transform nearly all of thestarting material into product and require no subsequent machining to achieve final partgeometry. Near net shape processes are ones that require minimum machining to producethe final shape.1.6What are the three types of surface processing operations?Answer. (1) Cleaning, (2) surface treatments, e.g., sand blasting, and (3) coating and thinfilm deposition, e.g., electroplating, painting, physical vapor deposition.1.7What is the difference between special purpose and general purpose productionequipment?Answer. General-purpose equipment is more flexible and adaptable to a variety of jobs. Itis commercially available for any manufacturing company to invest in. Special-purposeequipment is usually designed to produce a specific part or product in very large quantities.Another reason may be because the process is unique and commercial equipment is notavailable. Some companies with unique processing requirements develop their own specialpurpose equipment.1.8What is the difference between a process layout and a product layout in a productionfacility?Answer. A process layout is one where the machinery in a plant is arranged based on thetype of process it performs. To produce a product it must visit the departments in the orderof the operations that must be performed. This often includes large travel distances withinthe plant. A process layout is often used when the product variety is large and the operationsequences of products are dissimilar. A product layout is one where the machinery isarranged based on the general flow of the products that will be produced. Travel distance isreduced because products will generally flow to the next machine in the sequence. Aproduct layout works well when all products tend to follow the same sequence ofproduction operations.1.9What is the difference between fixed costs and variable costs?Answer. A fixed cost remains constant for any level of production output, whereasvariable costs are paid for as they are used. The cost of the factory and equipment are fixedcosts. Direct labor and materials that are used to produce the product are variable costs.

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Additional Problems1.1A plastic molding machine produces a product whose annual demand is in the millions.The machine is automated and used full time just for the production of this product. Themolding cycle time = 45 sec. No tooling is required other than the mold, which cost$100,000 and is expected to produce 1,000,000 moldings (products). The plastic moldingcompound costs $1.20/lb. Each molding weighs 0.88 lb. The only labor required is for aworker to periodically retrieve the moldings. Labor rate of the worker = $18.00/hrincluding overhead. However, the worker also tends other machines and only spends 20%of his time on this machine. Setup can be ignored because of the long production run. Themolding machine was purchased for $500,000 installed, its anticipated life = 10 years, andit operates 6,000 hours per year. Equipment overhead rate = 30%. Availability = 100% andscrap rate = 0. Determine (a) the hourly production rate of the machine, (b) annual quantityof product molded, and (c) cost per piece.Solution: (a) With a cycle timeTc= 45 sec = 0.75 min,Rp= 60/0.75 = 80 pc/hrFactoring in the 98% proportion uptime,Rp= 0.98(80) =78.4 pc/hrAnnual quantity of product = 6,000(78.4) =470,400 pc/yr(b) Equipment cost rateCeq= 500,000(1.30)/(60106000) = $0.1806/minMold cost per pieceCt= 100,000/1,000,000 = $0.10/pcLabor cost rateCL= 18.00(0.20) = $3.60/hr = $0.06/minFinally, cost per pieceCpc= 1.20(0.88) + (0.06 + 0.1806)(0.75) + 0.10 =$1.34/pc1.2A production machine operates in a semi-automatic cycle but a worker must tend themachine 100% of the time to load parts. Unloading is accomplished automatically. Theworker’s cost rate = $27/hr including applicable labor overhead rate. The equipment costrate of the machine = $18.00/hr including applicable overhead costs. Cost of the startingparts = $0.15/pc. The job runs several months so the effect of setup can be ignored. Eachcycle, the actual process time = 24 sec, and time to load the part = 6 sec. Automaticunloading takes 3 sec. A proposal has been made to install an automatic parts-loadingdevice on the machine. The device would cost $36,000 and would reduce the part loadingtime to 3 sec each cycle. Its expected life = 4 years. The device would also relieve theworker from full-time attention to the machine. Instead, the worker could tend fourmachines, effectively reducing the labor cost to 25% of its current rate for each machine.The operation runs 250 days per year, eight hours per day. Assume availability = 100%and scrap rate = 0. Determine the cost per part produced (a) without the parts loadingdevice and (b) with the parts loading device installed. (c) How many days of productionare required to pay for the automatic loading device? In other words find the breakevenpoint.Solution: (a) Equipment cost rateCeq= 18/60 = $0.30/minLabor cost rateCL= 27/60 = $0.45/minWithout the loading device,Cpc= 0.15 + (0.45 + 0.30)(24 + 6 + 3)/60 =$0.563/pc(b) Cost rate of the device = 36,000/(6042000) = $0.075/min = $4.50/hrWith the loading device,Cpc= 0.15 + (0.45/4 + 0.30 + 0.075)(24 + 3 + 3)/60 =$0.394/pc(c) Without the device,Tc= 24 + 6 + 3 = 33 sec = 0.55 min andCpc= 0.563/pcRp=Rc= 60/0.55 = 109.1 pc/hr = 872 pc/day

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With the device,Tc= 24 + 3 + 3 = 33 sec = 0.50 min andCpc= 0.394/pcAt 100% reliability and no setup time,Rp=Rc= 60/0.50 = 120.0 pc/hr = 960 pc/dayLetD= number of days of production at which the two alternatives are equivalent.872(0.563)D= 36,000 + 960(0.394)D490.9D= 36,000 + 378.2D(490.9–378.2)D= 112.7D= 36,000D= 319.5 round to320 days1.3In a sequential batch-processingoperation, the starting work part costs $4.50 each. Batchquantity = 65 parts. Part handling time each cycle = 2.5 min, and machining time per part =3.44 min. It takes 75 min to set up the machine for production. Equipment cost rate =$25.00/hr, and labor cost rate = $20.00/hr. Both rates include overhead costs. The cuttingtool in the operation costs = $5.75/pc and it must be changed every 18 parts. Tool changetime = 3.0 min. Availability of the machine tool = 98%, and the scrap rate = 0. Determine(a) production rate and (b) finished part cost. (c) How many hours are required to completethe batch?Solution: (a)Tc= 3.44 + 2.5 + 3/18 = 6.11 min/pcGivenq= 0, the starting quantity of partsQo= 65 pcNow determine batch time, including setup time and availability, assuming that theavailability factor does not apply during setup because the machine is not running.Tb= 75 + 65(6.11)/0.98 = 75 + 405.26 = 480.26 min/batch = 8.004 hrAverage production rate of partsRp= 65/8.004 =8.121 pc/hr(b) Now determine batch cost, including setup time and availability, assuming that theavailability factor does not apply during setup because the machine is not running. Thenumber of cutting tools required = 68/18 = 3.78 rounded up to 4 tools at $5.75 each =$23.00.Cbatch= 65(4.50) + ((20 + 25)(8.004) + 4(5.75) = 292.50 + 360.18 + 23.00 = $675.68/batchCpc= 675.68/65 =$10.39/pcAlternative calculation ofCpc:Cpc= 4.50 + ((20 + 25)/60)(75/65) + ((20 + 25)/60)(6.11/0.98) + 23.00/65Cpc= 4.50 +0.865 + 4.676 + 0.354 = $10.395/pc (Close enough!)(c) Time to complete the batch was computed in part (a) asTb=480.26 min = 8.004 hr1.4A high-production operation was studied during an 80-hr period. During that time, a totalof seven equipment breakdowns occurred for a total lost production time of 3.8 hr, and theoperation produced 38 defective products. No setups were performed during the period.The operation cycle consists of a processing time of 2.14 min, a part handling time of 0.65min, and a tool change is required every 25 parts, which takes 1.50 min. Determine (a)hourly production rate of acceptable parts and (b) scrap rate during the period.Solution: (a) Cycle time of the unit operationTc= 2.14 + 0.65 + 1.50/25 = 2.85 minHours of production during 80 hours = 80–3.8 = 76.2 hrTotal number of parts produced = 76.2(60)/2.85 = 1604 pcNumber of acceptable parts produced = 1604–38 = 1566 pcProduction rate of acceptable partsRp= 1566/80 =19.58 pc/hr(b) Scrap rateq= 38/1566 = 0.0243 =2.43%

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Additional Multiple-Choice QuestionsAtomic Structure and the Elements2.1The element with the lowest density and smallest atomic weight is which one of thefollowing: (a) aluminum, (b) argon, (c) helium, (d) hydrogen, or (e)magnesium?Answer. (d).2.2In the planetary model of atomic structure, the electrons in the outermost shell are calledwhich one of the following: (a) atomic electrons, (b) orbital electrons, (c) quantumelectrons, or (d) valence electrons?Answer. (d).Bonding between Atoms and Molecules2.3Which two of the following bond types are called intramolecular bonds: (a) covalentbonding, (b) electronic bonding, (c) hydrogen bonding, (d) ionic bonding, (e) metallicbonding, and (f) van der Waals forces?Answer. (a) and (d).Crystalline Structures2.4How many atoms are there in the hexagonal close-packed (HCP) unit cell: (a) 8, (b) 9, (c)10, (d) 12, (e) 14, or (f) 17?Answer. (f).2.5Which one of the following metals has a body-centered cubic crystal structure at roomtemperature: (a) aluminum, (b) copper, (c) iron, (d) nickel, and (e) zinc?Answer. (c).2.6Twinning can be described as which three of the following: (a) elastic deformation, (b) morelikely at high deformation rates, (c) more likely in metals with HCP structure, (d) plasticdeformation, (e) slip mechanism, and (f) type of dislocation?2.7Answer. (b), (c), and (d).2.8Grain boundaries in metals interfere with dislocation movement, which contributes to whichone of the following properties that arecharacteristic of metals: (a) crystal structure, (b)ductility, (c) electrical conductivity, (d) strain hardening, or (e) thermal conductivity?Answer. (d).Noncrystalline (Amorphous) Structures2.9As an amorphous material such as glass cools from the molten (liquid) state, it transformsinto the solid state gradually, first going through a transition phase which is called which oneof the following: (a) mushy zone, (b) passing phase, (c) supercooled liquid, (d) superheatedsolid, or (e) transformation phase?Answer. (c).Engineering Materials

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2.10Polymers are characterized by which two of the following bonding types: (a) adhesive, (b)covalent, (c) hydrogen, (d) ionic, (e) metallic, and (f) van der Waals?Answer. (b) and (f).2.11In general, ceramics arecharacterized by which four of the following properties: (a)brittleness, (b) chemical inertness, (c) electrically insulating, (d) high electrical conductivity,(e) high hardness, (f) high thermal conductivity, and (g) low density?Answer. (a), (b), (c), and (e).AdditionalReviewQuestions2.1Describe how ionic bonding works?Answer. In ionic bonding, atoms of one element give up their outer electron(s) to the atomsof another element to form complete outer shells.2.2Among the common point defects in a crystal lattice structure, what is a Frenkel defect?Answer. A Frenkel defect in a crystal lattice structure is an ion that is removed from aregular position in the lattice and inserted into an interstitial position not normally occupiedby such an ion.2.3How do grain boundaries contribute to the strain hardening phenomenon in metals?Answer. Grain boundaries block the continued movement of dislocations in the metal duringstraining. As more dislocations become blocked, the metal becomes more difficult todeform; in effect it becomes stronger.

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AdditionalMultiple-ChoiceQuestionsStress–Strain Relationships3.1Which of the following are the three basic types of static stress to which a material can besubjected: (a) compression, (b) hardness, (c) reduction in area, (d)shear, (e) tensile, (f) truestress, and (f) yield?Answer. (a), (d), and (e).3.2The plastic region of the stress-strain curve for a metal is characterized by a proportionalrelationship between stress and strain: (a) True or (b) false?Answer. (b). It isthe elastic region that is characterized by a proportional relationshipbetween stress and strain. The plastic region is characterized by the flow curve, which is apower function.3.3The modulus of elasticity is also known as which one of the following: (a) elastic limit, (b)strain hardening exponent, (c) strength coefficient, or (d) Young’s modulus?Answer. (d).3.4If engineering stress and true stress were measured simultaneously during a tensile test,which would have the higher value: (a) engineering stress or (b) true stress?Answer. (b).3.5Which one of the following materials has the highest modulus of elasticity: (a) aluminum,(b) diamond, (c) steel, (d) titanium, or (e) tungsten?Answer. (b).3.6Which of the following types of stress-strain relationship best describes the behavior ofbrittle materials such as ceramics and thermosetting plastics: (a) elastic and perfectly plastic,(b) elastic and strain hardening, (c) perfectly elastic, or (d) none of the above?Answer. (c).3.7Which of the following types of stress-strain relationship best describes the behavior ofmetals at temperatures above their respective recrystallization points: (a) elastic and perfectlyplastic, (b) elastic and strain hardening, (c) perfectly elastic, or (d) none of the above?Answer. (a).3.8The strength coefficient and strain hardening exponent measured in a tensile test for a givenmetal are nearly identical to those measured in a compression test: (a) True or (b) false?Answer. (a).3.9Cleavage is a failure mode associated with ceramics and metakls operating at lowtemperatures, in which slip occurs along certain crystallographic planes: (a) True or (b)false?Answer. (b). Cleavage is a failure mode in which separation rather than slip occurs alongcertain crystallographic planes.

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3.10Theshear modulus of elasticity is approximately what percent of Young’s modulus: (a) 10%,(b) 25%, (c) 40%, (d) 75%, or (e) 100%?Answer. (c).Hardness3.11Which two of the following hardness tests use a pyramid-shaped indenter: (a) Brinell, (b)Knoop, (c) Rockwell, (d) scleroscope, and (e) Vickers?Answer. (b) and (e).3.12Which one of the following materials has the highest hardness: (a) alumina ceramic, (b) graycast iron, (c) hardened tool steel, (d) high carbon steel, or (e) polystyrene?Answer. (a).Effect of Temperature on Properties3.13Hot working involves metal forming processes carried out above the recrystallizationtemperature: (a) True or (b) false?Answer. (a).AdditionalReviewQuestions3.1What is the difference between engineering stress and true stress in a tensile test?Answer. Engineering stress divides the load (force) on the test specimen by the original area;whereas truestress divides the load by the instantaneous area which decreases as thespecimen stretches.3.2State the flow curve equation, also known as the Holloman-Ludwig equation.Answer. The flow curve equation is=Kn, where= true stress,= true strain,K=strength coefficient, andn= strain hardening exponent.3.3Which strain always has the greater value, engineering strain or true strain?Answer. Engineering strain is always greater than true strain.3.4A stress-strain relationship that exhibitsperfectly elastic behavior is characteristic of whattypes of materials?Answer. Brittle materials such as ceramics, many cast irons, and thermosetting plasticsexhibit perfectly elastic behavior.3.5How does the change in cross-sectional area of a test specimen in a compression test differfrom its counterpart in a tensile test specimen?Answer. In a compression test, the specimen cross-sectional area increases as the testprogresses; while in a tensile test, the cross-sectional area decreases.3.6Define therecrystallization temperaturefor a metal.Answer. The recrystallization temperature is the temperature at which a metal recrystallizes(forms new grains) rather than work hardens when deformed. The recrystallization

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temperature is about 0.5Tm, whereTmis the melting temperature of the metal measured onan absolute scale (R or K).3.7Why are different hardness tests and scales required?Answer. Different hardness tests and scales are required because different materials possesswidely differing hardnesses. A test whose measuring range is suited to very hard materials isnot sensitive for testing very soft materials.3.8How does the viscosity of a thermoplastic polymer melt differ from the viscosity of aNewtonian fluid?Answer. The viscosity of a thermoplastic polymer melt decreases as flow rate increases,whereas the viscosity of a Newtonian fluid does not.Additional Problems3.1(USCS Units) A tensile test specimen has a gage length = 2.0 in and diameter = 0.798 in.Yielding occurs at a load of 31,000 lb. The corresponding gage length = 2.0083 in, which isthe 0.2 percent yield point. The maximum load of 58,000 lb is reached at a gage length =2.47 in. Determine (a) yield strength, (b) modulus of elasticity, and (c) tensile strength. (d) Iffracture occurs at a gage length of 2.62 in, determine the percent elongation. (e) If thespecimen necked to an area = 0.36 in2, determine the percent reduction in area.Solution: (a) AreaA= π(0.798)2/4 = 0.50 in2.Yield strengthY= 31,000/0.50 =62,000 lb/in2(b)s=E e. Subtracting the 0.2% offset,e= (2.00832.0)/2.00.002 = 0.00215. This isthe elastic strain to calculate modulus of elasticity.E=s/e= 62,000/0.00215 =28.84106lb/in2(c)TS= 58,000/0.5 =116,000 lb/in2(d)EL= (2.622.0)/2.0 = 0.62/2.0 = 0.31 =31%(e)AR= (0.50.36)/0.5 = 0.28 =28%3.2(SI Units) In a tensile test on a steel specimen, true strain = 0.12 at a stress of 250 MPa.When true stress = 350 MPa, true strain = 0.26. Determine the strength coefficient and thestrain-hardening exponent in the flow curve equation.Solution: Two equations: (1) 250 =K(0.12)nand (2) 350 =K(0.26)n350/250 = (0.26/0.12)n1.4 = (2.1667)nnln(2.1667) = ln(1.4)0.7732n= 0.3365n= 0.435Substituting this value with the data back into the flow curve equation to obtain the valueof the strength coefficientK:(1)K= 250/(0.12).435=629 MPaCheck: (2)K= 350/(0.26).435= 629 MPaThe flow curve equation is:= 6290.4353.3(USCSUnits) A metal is plastically deformed in a tensile test. The starting specimen had agage length = 2.0 in and an area = 0.50 in2. At one point during the test, the gage length = 2.3in, and the corresponding engineering stress = 25,000 lb/in2. At another point prior to

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necking, the gage length = 3.0 in, and the corresponding engineering stress = 28,000 lb/in2.Determine the strength coefficient and the strain-hardening exponent for this metal.Solution: Starting volumeV=LoAo= 2.0(0.5) = 1.0 in3Area at gage length(1)= 2.3 in:A=V/L= 1.0/2.3 = 0.4348 in2Given engineering stress = 25,000 lb/in2, forceF= 25,000(0.5) = 12,500 lbTrue stress= 12,500/0.4348 = 28,749 lb/in2and true strain= ln(2.3/2.0) = 0.1398Area at gage length(2)= 3.0 in:A= 1.0/3.0 = 0.3333 in2Given engineering stress = 28,000 lb/in2, forceF= 28,000(0.5) = 14,000 lbTrue stress= 14,000/0.3333 = 42,000 lb/in2and true strain= ln(3.0/2.0) = 0.4055These two data points can be used to determine the parameters of the flow curve equation.(1) 28,749 =K(0.1398)nand (2) 42,000 =K(0.4055)n42,000/28,749 = (0.4055/0.1398)n1.4609 = (2.9006)nln(1.4609) =nln(2.9006)0.3791 = 1.0649nn= 0.356(1)K= 28,749/(0.1398).356= 57,915 lb/in2Check: (2)K= 42,000/(0.4055).356= 57,915 lb/in2The flow curve equation is:= 57,9150.3563.4Prove Equation (3.8): Show that true strain= ln(1 +e), wheree= engineering strain.Solution: Starting definitions: (1)= ln(L/Lo) and (2)e= (LLo)/LoConsider definition (2):e=L/LoLo/Lo=L/Lo1Rearranging, 1 +e=L/LoSubstituting this into definition (1),= ln(1 +e)3.5(SI Units) The following flow curve parameters were obtained for a metal alloy in a tensiletest:strength coefficient= 550 MPa andstrain-hardening exponent= 0.25. The same metal isnow tested in a compression test in which the starting height of the specimen = 50 mm andits diameter = 30 mm. Assuming that the cross section increases uniformly, determine theload required to compress the specimen to a height of (a) 45 mm and (b) 35 mm.Solution: The starting volume of test specimenV=hDo2/4 = 50(30)2/4 = 35343 mm3(a) Ath= 45 mm,= ln(50/45) = ln(1.1111) = 0.1054Yf= 550(0.1054).25= 313.4 MPaA=V/L= 35343/45 = 785.4 mm2F=YfA=313.4(785.4) =246,144 N(b) Ath= 35 mm,= ln(50/35) = ln(1.4286) = 0.3567Yf= 550(0.3567).25= 425.0 MPaA=V/L= 35343/35 = 1009.8 mm2F=YfA= 425.0(1009.8) =429,165 N3.6(USCS Units) A ceramic specimen is tested in a bending test. Its width = 0.50 in andthickness = 0.25 in. The length of the specimen between supports = 3.0 in. Determine thetransverse rupture strength if failure occurs at a load = 1250 lb.Solution:TRS= 1.5FL/bt2= 1.5(1250)(3.0)/(0.50.252) =180,000 lb/in2

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3.7(USCS Units) A batch of annealed steel has just been received from the vendor. It issupposed to have a tensile strength in the range 60,000 to 70,000 lb/in2. A Brinell hardnesstest in the receiving department yields a value ofHB= 118. (a) Does the steel meet thespecification on tensile strength? (b) Estimate the yield strength of the material.Solution: (a)TS= 500(HB) = 500(118) =59,000 lb/in2.This lies outside the specifiedrange of 60,000 to 70,000 lb/in2. However, from a legal standpoint, it is unlikely that thebatch can be rejected on the basis of its measuredBrinell hardness numberwithout using anactual tensile test to measure TS. The formula for converting fromBrinell hardness numberto tensile strength is only an approximating equation.(b) Based on Table 3.2 in the text, the ratio ofYtoTSfor low carbon steel = 25,000/45,000= 0.555.Using this ratio, estimate the yield strength to beY= 0.555(59,000) =32,700 lb/in23.8(SI Units) A 125.0-mm-diameter shaft rotates inside a stationary bushing whose insidediameter = 125.6 mm and length = 50.0 mm. In the clearance between the shaft and thebushing is a lubricating oil whose viscosity = 0.14 Pa-s. The shaft rotates at a velocity of 400rev/min; this speed and the action of the oil are sufficient to keep the shaft centered inside thebushing. Determine the magnitude of the torque due to viscosity that acts to resist therotation of the shaft.Solution: Bushing internal bearing areaA=(125.6)250/4 = 19729.6 mm2=19729.2(10-6) m2d= (125.6125)/2 = 0.3 mmv= (125mm/rev)(400 rev/min)(1 min/60 sec) = 2618.0 mm/sShear rate = 2618/0.3 = 8726.6 s-1= (0.14)(8726.6) =1221.7 Pa = 1221.7 N/mm2Force on surface between shaft and bushing = (1221.7 N/mm2)(19729.2(10-6)) = 24.1 NTorqueT= 24.1 N125/2 mm =1506.4 N-mm = 1.506 N-m

Solution Manual For Fundamentals of Modern Manufacturing: Materials, Processes, and Systems, 7th Edition - Page 14 preview image

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AdditionalMultiple-ChoiceQuestionsVolumetric and Melting Properties4.1Which one of the following metals has the highest density: (a) copper, (b) iron, (c) lead, (d)nickel, (e) tungsten, or (f) zinc?Answer. (e).4.2Thestrength-to-weight ratio of a material is its yield strength divided by its density: (a) Trueor (b) false?Answer. (b). The ratio is based on tensile strength, not yield strength.Thermal Properties4.3The thermal diffusivity of a material is defined aswhich one of the following: (a) massdiffusion divided by specific heat, (b) specific heat divided by mass diffusion, (c) specificheat multiplied by mass diffusion, (d) thermal conductivity divided by specific heat, or (e)thermal conductivity divided by volumetric specific heat?Answer. (e).Electrical Properties4.4Copper is generally considered easy to weld because of its high thermal conductivity: (a)True or (b) false?Answer. (b). The high thermal conductivity of copper makes it difficult to weld because theheat flows away from the joint rather than being concentrated to permit melting of the metal.4.5Which of the following pure metals is the best conductor of electricity: (a) aluminum, (b)copper, (c) gold, or (d) silver?Answer. (d). See Table 4.3.4.6Which one of the following materials is classified as a semiconductor: (a) carbon, (b) naturalrubber, (c) polyethylene, or (d) silicon?Answer. (d).Electrochemical Processes4.7In electrochemical machining, the work part in the electrolytic cell is the(a) anode or (b)cathode?Answer. (a).AdditionalReviewQuestions4.1Define the coefficient of thermal expansion.Answer. The coefficient of thermal expansion measures the change in length per degree oftemperature.4.2What is thedifference in melting characteristics between a pure metal element and a metalalloy?

Solution Manual For Fundamentals of Modern Manufacturing: Materials, Processes, and Systems, 7th Edition - Page 15 preview image

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Answer. A pure metal element melts at one temperature (the melting point), while mostmetal alloys begin melting at a certain temperature called the solidus and finally completesthe transformation to the molten state at a higher temperature called the liquidus. Betweenthe solidus and liquidus, the alloy is a mixture of solid and liquid.4.3Definespecific heatas a material property.Answer. Specific heat is defined as the quantity of heat required to raise the temperature of aunit mass of the material by one degree.4.4What are the important variables that affect mass diffusion?Answer. According to Fick's first law, mass diffusion depends on the diffusion coefficient ofthe material, which increases rapidly with temperature (so temperature could be listed as animportant variable), concentration gradient, contact area, and time.4.5Define electricalresistivityas a material property.Answer. Resistivityris the material's capacity to resist the flow of electric current. In auniform section of material,r=RA/L, whereR= electrical resistance,A= cross-sectionalarea of the section, andL= its length.4.6What isdielectric strengthas a material property?Answer. Dielectric strength is defined as the electrical potential required to break down theinsulator per unit thickness.4.7In anelectrolytic cell, what are the names of the positive and negative electrodes?Answer. The positive electrode is called the anode, and the negative electrode is called thecathode.Additional Problems4.1(SI units) Determine the length and width of a rectangular nickel plate whose roomtemperature (20C) dimensions are 300 mm by 400 mmby 5 mm, if the plate is heated to250C. The coefficient of thermal expansion= 13.3(10-6)/C for nickel.Solution:Equation (4.1) for length:L2–L1=αL1(T2–T1)L2= 300 + 13.3(10-6)(300)(250–20) = 300 + 0.918 =300.918 mmEquation (4.1) for width:W2–W1=αW1(T2–T1)W2= 400 + 13.3(10-6)(400)(250–20) = 400 + 1.224 =401.224 mm4.2(USCS units) A zinc die casting has a critical dimension of 10.00 in when it solidifies in themold at its melting temperature because10.00 in is the corresponding mold dimension. Whatis the value of this dimension when it cools to room temperature (70°F)? : The meltingtemperature of zinc = 787°F, and its coefficient of thermal expansion = 22.2(10-6)/°F.SolutionL2–L1=L1(T2–T2)L2= 10.00 + 22.2(10-6)(10.00)(70–787) = 10.00–0.159 =9.841 in4.3(SI units) A steel alloy has a melting point of 1450C. Its specific heat = 0.46 J/gC, and itsheat of fusion = 270 J/g. For a 100-kg block of this steel, determine how much heat isrequired to (a) raise its temperature from 25C to its melting point and (b) transform it fromsolid to liquid phase.

Solution Manual For Fundamentals of Modern Manufacturing: Materials, Processes, and Systems, 7th Edition - Page 16 preview image

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Solution. (a) Required heatH=mC(Tm–Ta),m= 100 kg = 100(103) gH= 100(103g)(0.46 J/gC)(1450C–25C) =65,550,000 J(b) Required heat for meltingH=mHf=100(103g)(270J/g) =27,000,000 J4.4(SI units) What is the resistanceRof a length of copper wire whose length = 20 m andwhose diameter = 0.40 mm?Solution:R=rL/A,A=(0.40)2/4 = 0.1257 mm2= 0.1257(10-6) m2From Table 4.3,resistivityr= 1.710-8-m2/mR= (1.710-8-m2/m)(20 m)/( 0.1257(10-6) m2) =2.705

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