Solution Manual for Fundamentals of Modern Manufacturing: Materials, Processes, and Systems, 5th Edition

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Solutions forFundamentals of Modern Manufacturing, 5e1-11INTRODUCTIONReview Questions1.1About what percentage of the U.S. gross domestic product (GDP) is accounted for by themanufacturing industries?Answer. Although the percentage has varied over the years, currentlythe manufacturing industriesaccount for around 12% of U.S. GDP.1.2Define manufacturing.Answer.Thetextdefinesmanufacturingintwoways:technologicallyandeconomically.Technologically,manufacturingis the application of physical and chemical processes to alter thegeometry, properties, and/or appearance of a given starting material to make parts or products;manufacturingalsoincludesassemblyofmultiplepartstomakeproducts.Economically,manufacturingis the transformation of materials into items of greater value by means of one ormore processing and/or assembly operations. The key point is that manufacturingadds valueto thematerial by changing its shape or properties, or by combining it with other materials.1.3What are the differences between primary, secondary, and tertiary industries? Give an example ofeach 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 and convertsthem to consumer and capital goods. Examples of secondary industries are textiles and electronics.A tertiary industry is in the service sector of the economy. Examples of tertiary industries arebanking and education.1.4What is the difference between a consumer good and a capital good? Give some examples in eachcategory.Answer.Consumer goods are products purchased directly by consumers, such as cars, personalcomputers, TVs, tires, and tennis rackets. Capital goods arethose purchased by companies toproduce goods and/or provide services. Examples of capital goods include aircraft, computers,communication equipment, medical apparatus, trucks and buses, railroad locomotives, machinetools, and construction equipment.1.5What is the difference between soft product variety and hard product variety, as these terms aredefined in the text?Answer.Soft product variety is when there are only small differences among products, such as thedifferences among car models made on the same production line. In an assembled product, softvariety is characterized by a high proportion of common parts among the models. Hard productvariety is whenthe products differ substantially, and there are few common parts, if any. Thedifference between a car and a truck exemplifies hard variety.1.6How are product variety and production quantity related when comparing typical factories?Answer. Generally production quantity is inversely related to product variety. A factory thatproduces a large variety of products will produce a smaller quantity of each. A company thatproduces a single product will produce a large quantity.1.7One of the dimensions of manufacturing capability is technological processing capability. Definetechnological processing capability.

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Solutions forFundamentals of Modern Manufacturing, 5e1-2Answer.Thetechnological processing capabilityof a plant (or company) is its available set ofmanufacturing processes. Certain plants perform machining operations, others roll steel billets intosheet stock, and others build automobiles. The underlying feature that distinguishes these plants isthe processes they can perform. Technological processing capability includes not only the physicalprocesses, but also the expertise possessed by plant personnel in these processing technologies.1.8What are the four categories of engineering materials used in manufacturing?Answer. The four categories of engineering materials are (1) metals, (2) ceramics, (3) polymers,and (4) composite materials, which consists of a non-homogeneous mixture of the other threetypes.1.9What is the definition of steel?Answer.Steelcan be defined as an iron–carbon alloy containing 0.02% to 2.11% carbon. Itscomposition often includes other alloying elements as well, such as manganese, chromium, nickel,and molybdenum, to enhance the properties of the metal.1.10What are some of the typical applications of steel?Answer.Applicationsofsteelincludeconstruction(e.g.,bridges,I-beams,andnails),transportation (trucks, rails, and rolling stock for railroads), and consumer products (automobilesand appliances).1.11What is the difference between a thermoplastic polymer and a thermosetting polymer?Answer.Thermoplastic polymers can be subjected to multiple heating and cooling cycles withoutsubstantially altering the molecular structure of the polymer. Thermosetting polymerschemicallytransform (cure) into a rigid structure on cooling from a heated plastic condition.1.12Manufacturing processes are usually accomplished as unit operations. Define unit operation.Answer. Aunit operation is a single step in the sequence of steps required to transform the startingmaterial into a final product. A unit operation is generally performed on a single piece ofequipment that runs independently of other operations in the plant.1.13In manufacturing processes, what is the difference between a processing operation and an assemblyoperation?Answer. A processing operation transforms a work material from one state of completion to amore advanced state that is closer to the final desired product. It changes the geometry, properties,or appearance of the starting material. In general, processing operations are performed on discretework parts, but certain processing operations are also applicable to assembled items (e.g., paintinga spot-welded car body). An assembly operation joins two or more components to create a newentity, called an assembly, subassembly, or some other term that refers to the joining process (e.g.,a welded assembly is called a weldment).1.14One of the three general types of processing operations is shaping operations, which are used tocreate or alter the geometry of thework part. What are the four categories of shaping operations?Answer. Thefour categories of shaping operations are(1)solidification processes, in which thestarting material is a heated liquid or semifluid that cools and solidifies to form the part geometry;(2) particulate processing, in which the starting material is a powder, and the powders are formedand heated into the desired geometry; (3) deformation processes, in which the starting material is aductile solid (commonly metal) that is deformed to shape the part; and (4) material removalprocesses, in which the starting material is a solid (ductile or brittle), from which material isremoved so that the resulting part has the desired geometry.

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Solutions forFundamentals of Modern Manufacturing, 5e1-31.15What is the difference between net shape processes and near net shape processes?Answer.Net shape processesare manufacturing processes that transform nearly all of the startingmaterial into product and require no subsequent machining to achieve final part geometry. Near netshape processes are ones that require minimum machining to produce the final shape.1.16Identify the four types of permanent joining processes used in assembly.Answer.The four types are welding, brazing, soldering, and adhesive bonding.1.17What is a machine tool?Answer. The term developed during the Industrial Revolution, when it referred to power-drivenmachines used to operate cutting tools previously operated by hand. Modern machine tools aredescribed by the same basic definition, except that the power is electrical rather than water orsteam, and the level of precision and automation is much greater today.1.18What is the difference between special purpose and general purpose production equipment?Answer. General-purpose equipment is more flexible and adaptable to a variety of jobs. It iscommercially available for any manufacturing company to invest in. Special-purpose equipment isusually designed to produce a specific part or product in very large quantities. Another reason maybe because the process is unique and commercial equipment is not available. Some companies withunique processing requirements develop their own special purpose equipment.1.19Define batch production and describe why it is often used for medium-quantity productionproducts.Answer. Batch production is where groups, lots, or batches or materials or parts are processedtogether through the manufacturing operations. All units in the batch are processed at a givenstation before the group proceeds to the next station. In a medium or low quantity productionsituation, the same machines are used to produce many types of products. Whenever a machineswitches from one product to another, a changeover occurs. The changeover requires the machinesetup to be torn down and set up for the new product. Batch production allows the changeover timeto be distributed across a larger number of parts and hence reduce the average operation time perpart.1.20What is the difference between a process layout and a product layout in a production facility?Answer. A process layout is one where the machinery in a plant is arranged based on the type ofprocess it performs. To produce a product it must visit the departments in the order of theoperations that must be performed. This often includes large travel distances within the plant. Aprocess layout is often used when the product variety is large the operation sequences of productsare dissimilar. A product layout is one where the machinery is arranged based on the general flowof the products that will be produced. Travel distance is reduced because products will generallyflow to the next machine in the sequence. A product layout works well when all products tend tofollow the same sequence of production.1.21Name two departments that are typically classified as manufacturing support departments.Answer. A common organizational structure includes the following three manufacturing supportdepartments: (1) manufacturing engineering, (2) production planning and control, and (3) qualitycontrol.1.22What are overhead costs in a manufacturing company?Answer. Overhead costsconsist of all of the expenses of operating the company other thanmaterial, labor, and equipment.

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Solutions forFundamentals of Modern Manufacturing, 5e1-41.23Name and define the two categories of overhead costs in a manufacturing company.Answer. The two categories are (1) factory overhead and (2) corporate overhead.Factory overheadconsists of the costs of running the factory excluding materials, direct labor, and equipment. Thisoverhead category includes plant supervision, maintenance, insurance, heat and light, and so forth.Corporate overhead consists of company expenses not related to the factory, such as sales,marketing, accounting, legal, engineering, research and development, office space, utilities, andhealth benefits.1.24What is the difference between fixed costs and variable costs?Answer. A fixed cost remains constant for any level of production output, whereas variable costsare paid for as they are used. The cost of the factory and equipment are fixed costs. Direct laborand materialsthat are used to produce the productare variable costs.1.25What is meant by the term availability?Answer.Availability is a reliability termwhich is simply the proportion uptime of the equipment.ProblemsAnswers to problems labeled(A)are listed inanAppendix at the back of the book.Manufacturing Economics1.1(A)Acompany invests $750,000 in a piece ofproductionequipment. The cost to install theequipment in the plant =$25,000.Theanticipated lifeof the machine=12years. The machinewillbe usedeight hours per shift,fiveshiftsper week,50 weeks per year.Applicable overhead rate =18%.Assume availability = 100%.Determine theequipmentcost rateif (a) the plant operates oneshift per day and (b) the plant operates three shifts per day.Solution:(a) For a one-shift operation,hours of operation per yearH= 50(1)(5)(8) =2000 hr/yr.Using Eq. (1.8),Ceq= (750,000 +25,000)(1.18)/(60 x 12 x 2000) =$0.635/min= $38.10/hr(b) For a three-shift operation,hours of operation per yearH= 50(3)(5)(8) =6000 hr/yr.Ceq= (750,000 + 25,000)(1.18)/(60 x 12 x 6000) =$0.212/min=$12.70/hrNote the significant advantage the companyhasif it runsthree shifts per day rather than one shift.1.2A production machine was purchased six years ago for an installed price of $530,000. At that timeit was anticipated that the machine would last 10 years and be used 4000 hours per year. However,it is now in need of major repairs that will cost $125,000. If these repairs are made, the machinewill last four more years, operating 4000 hours per year.Applicable overhead rate =30%.Assumeavailability = 100%.Determine the equipmentcost ratefor this machine.Solution:The cost rate under the original conditions was the following:Ceq= 530,000(1.30)/(60 x 10 x 4000) = $0.287/min = $17.23/hrTherepairs will add to that cost rate as follows:Ceq=125,000(1.30)/(60 x 4 x 4000) = $0.169/min = $10.16/hrThe repaired machine has a cost rateCeq= 0.287 + 0.169 =$0.456/min = $27.36/hr1.3Instead of repairing the machine in Problem 1.2, a proposal has been made to purchasea newmachineand scrap the current machine at a zero salvage value. The new machinewillhave aproduction rate that is20% faster than the current equipment,whose production rate=12 parts perhour. Each part has astarting material cost = $1.33and aselling price = $6.40.All parts producedduring the next four years on either machine can be sold at this price.At the end of the four years,the current machine will be scrapped, but the new machinewouldstill be productive for another sixyears.The new machine costs $700,000 installed, has an anticipated life of 10 years, and anapplicable overhead rateof30%.It will be used 4000 hours per year, same as the current machine.

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Solutions forFundamentals of Modern Manufacturing, 5e1-5The labor rate for either alternative = $24.00/hr which includes applicable overhead costs.Assumeavailability = 100% and scrap rate = 0.Which alternative is more economical using total profit overfour years as the criterion, (a) repairing the current machine or (b) purchasing the new machine?Solution:(a)Thefirst alternative is to repair the current machine. The cost ratewas determined inthe solution to Problem 1.2. Repeating here, the original cost rate is calculated as follows:Ceq= 530,000(1.30)/(60 x 10 x 4000) = $0.287/min = $17.23/hrTherepairs will add to that cost rate as follows:Ceq=125,000(1.30)/(60 x 4 x 4000) = $0.169/min = $10.16/hrThe repaired machine has a cost rateCeq= 0.287 + 0.169 =$0.456/min = $27.36/hrLabor cost = $24.00/hr(given)Given that annual hours of operation = 4000, total cost of production on this machine is calculatedas follows:TC= 4000(24.00 + 27.36) = $205,440/yrAt a production rate of 12 pc/hr and operating 4000 hr/yr, annual output = 4000(12) = 48,000 pc/yrTotal revenue = 48,000(6.40–1.33) = $243,360/yr.Total profit over four years = 4(243,360–205,440)= $151,680(b) The second machine has an equipment cost rate determined as follows:Ceq=700,000(1.30)/(60 x 10 x 4000) = $0.379/min = $22.75/hrLabor cost = $24.00/hr (given)Given that annual hours of operation = 4000, totalannualcost of production on this machine isTC= 4000(24.00 + 22.75) = $187,000/yrProduction rate on the new machine is 20% faster, soproduction rate = 12(1.20) = 14.4 pc/hrAt 14.4pc/hr and operating 4000 hr/yr, annual output = 4000(14.4) =57,600 pc/yrTotal revenue =57,600(6.40–1.33) = $292,032/yr.Total profit over four years = 4(292,032–187,000) = $420,128Conclusion: The new machine should be purchased and the old machine scrapped.1.4(A)Amachinetoolis used tomachinepartsin batches. In one batch of interest, thestartingpiece isa casting thatcosts= $5.00 each. Batch quantity =40. The actualmachiningtime in the operation =6.86min. Time to load and unload each workpiece = 2.0min.Costof the cutting tool= $3.00, andeachtoolmust be changed every10pieces. Tool change time =1.5min.Setup time for the batch =1.75 hr. Hourly wage rate of the operator = $16.00/hr, and the applicable labor overhead rate =50%.Hourly equipment cost rate = $22.00/hr, which includes overhead.Assume availability =100% and scrap rate = 0.Determine (a) the cycle time for the piece, (b) average production ratewhen setup time is figured in, and (c) cost per piece.Solution: (a)Processing timeTo=6.86min,part handling timeTh=2.00 min, and tool handlingtimeTt=1.50min/10 = 0.15min.Tc=6.86+2.00 + 0.15=9.01min(b)Average production time per piece including setup timeTp= 1.75(60)/40 + 9.01 = 11.64 minHourly production rateRp= 60/11.64 =5.16 pc/hr(c)Equipment cost rateCeq=$22.00/60 =$0.367/min.LaborcostrateCL= 16.00(1.50) = $24.00/hr = $0.40/minCost of toolingCt=3.00/10 = $0.30/pcFinally, cost per pieceCpc=5.00+ (0.40+ 0.367)(11.64) + 0.30=$14.23/pc1.5A plastic molding machine produces a product whose annual demand is in the millions.Themachine is automated andused full time just for the production of this product.The molding cycletime = 45 sec. No tooling is required other than the mold, whichcost $100,000 and is expected toproduce 1,000,000 moldings (products).The plastic molding compound costs $1.20/lb. Eachmolding weighs 0.88 lb. The only labor required is for aworkerto periodically retrieve themoldings.Labor rate of theworker=$18.00/hr including overhead.However, the worker also

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Solutions forFundamentals of Modern Manufacturing, 5e1-6tends other machines and only spends20% of his time on this machine.Setup can be ignoredbecause of the long production run. The molding machine was purchased for $500,000 installed, itsanticipated life = 10 years, and it operates 6000 hours per year. Equipment overhead rate = 30%.Assume availability=100%and scrap rate = 0.Determine (a)the hourly production rate of themachine, (b)annual quantity of product molded,and(c) cost per piece.Solution: (a)With a cycle timeTc= 45 sec = 0.75 min,Rp= 60/0.75 = 80 pc/hr. Factoring in the98% proportion uptime,Rp= 0.98(80) =78.4 pc/hrAnnual quantity of product = 6000(78.4) =470,400 pc/yr(b)Equipment cost rateCeq=500,000(1.30)/(60x 10 x 6000)=$0.1806/min.Mold cost per pieceCt= 100,000/1,000,000 = $0.10/pcLaborcostrateCL= 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/pc1.6Astamping press produces sheet-metal stampings in batches. The press is operated by a workerwhose labor rate = $15.00/hrand applicablelaboroverhead rate = 42%.Cost rate of the press =$22.50/hrand applicable equipment overhead rate = 20%.In one job of interest, batch size = 400stampings, and the time to set up the die in the press takes75 min.The die cost $40,000 and isexpected to last for 200,000 stampings.Each cyclein the operation, the starting blanks of sheetmetalare manually loaded into the press, which takes 42 sec.The actual press stroke takes only8sec. Cost of the starting blanks = $0.43/pc. The pressoperates250 days per year, 7.5hours perday,but the operator is paid for8hours per day.Assume availability =100%and scrap rate = 0.Determine (a)cycle time, (b) average production rate with and without setup time included, and(c)cost perstamping produced.Solution: (a)Cycle timeTc= 42 + 8 = 50sec =0.833 min(b)Including setup time,Tp= 75/400 + 0.833 = 1.021 minRp= 60/1.021=58.78 pc/hrExcluding setup time,Rc= 60/0.833 =72.03 pc/hr(c)Equipment cost rateCeq=22.50(1.20)/60 =$0.45/min.Die cost per pieceCt= 40,000/200,000 = $0.20/pcLaborcostrateCL= 15.00(1.42)/60= $0.355/minThis labor cost should be adjusted for the fact that although the press operates 7.5 hr/day, theoperator is paid for 8 hr.CL= 0.355(8/7.5) = $0.379Finally, cost perstampingCpc=0.43+ (0.379+ 0.45)(1.021)+ 0.20=$1.48/pc1.7Aproduction machine operates in a semi-automatic cycle but a worker must tend the machine100% of the time to load parts.Unloading is accomplished automatically.The worker’s cost rate =$27/hr including applicable labor overheadrate. The equipment cost rate of the machine =$18.00/hr including applicable overhead costs.Cost of the starting parts = $0.15/pc. The job runsseveral months so the effect of setup can be ignored.Each cycle, the actual process time = 24 sec,and time to load the part = 6 sec.Automatic unloading takes 3 sec.A proposal has been made toinstall an automatic parts-loading device on the machine. The device would cost $36,000 andwould reduce the part loading time to 3 seceach cycle.Its expected life = 4 years. The devicewould also relieve the worker from full-time attention to the machine. Instead, the worker couldtend four machines, effectively reducing the labor cost to 25% of its current rate for each machine.The operation runs250 days per year,eighthours per day.Assumeavailability= 100%and scraprate = 0.Determinethecost perpartproduced(a) without the parts loading device and (b) with theparts loading device installed.(c) How many days of production are required to pay for theautomatic loading device?In other words find the breakeven point.Solution: (a)Equipment cost rateCeq= 18/60 = $0.30/min

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Solutions forFundamentals of Modern Manufacturing, 5e1-7Labor cost rateCL=27/60 = $0.45/minWithout theloadingdevice,Cpc= 0.15 + (0.45 + 0.30)(24 + 6+ 3)/60 =$0.563/pc(b)Cost rate of the device =36,000/(60 x 4 x 2000) = $0.075/min = $4.50/hrWith theloadingdevice,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 minandCpc= 0.563/pcfrom part (a)Rp=Rc= 60/0.55 = 109.1 pc/hr = 872 pc/dayWith the device,Tc= 24 + 3 + 3 = 33 sec = 0.50 minandCpc= 0.394/pcfrom part (b)At 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(.394)D490.9D= 36,000 + 378.2D(490.9–378.2)D= 112.7D= 36,000D= 319.5 round to320 days1.8(A)In a long-running high-productionoperation, the startingwork partcost=$0.95each, andcycle time = 1.25 min. Equipment cost rate = $28.00/hr, and labor cost rate = $21.00/hr. Both ratesincludeoverhead costs.Tooling cost = $0.05/pc.Availability of the production machine = 96%,and the scrap rate =7%. Determine(a)production rate and(b)finished part cost.Solution: (a) Production rate, including effect of availability(60/1.25)(0.96)=46.08pc/hrHowever, because of the 5% scrap rate, the production rate of acceptable parts isRp=46.08(1-0.07) =42.85pc/hr(b) Factoring inavailability and scrap rate, part cost isCpc= 0.95/0.93 + ((21 + 28)/60)(1.25/(0.93 x 0.96)) + 0.05 =$2.215/pc1.9The starting work part costs$2.25in a batch production operation. Batch quantity = 100parts.Each cycle, part handling time =0.50min, andoperationtime =1.67min.Setup time =50min.Equipment cost rate = $35.00/hr, and labor cost rate = $18.00/hr,includingoverhead costs.There isno tool changeor tool costinthe operation. The machinetoolis100%reliable, and scrap rate =5%. Determine (a) production rate,(b) finished part cost, and(c)number ofhours required tocomplete the batch.Solution: (a)Tc=1.67+0.50=2.17min/pcGivenq=5%, the starting quantity of partsQo=100/0.95=105.3rounded up to106pcDeterminebatchtime, including setup time.Tb=50+106(2.17) =50+230.02=280.02min/batch =4.67hrAverage production rate of partsRp=100/4.67=21.43pc/hrAverage productionrate of acceptable partsRp=21.43(1-0.05) =20.36pc/hr(b)Now determine batch cost, including setup time.Cb=106(2.25) + ((18+35)(4.67)=238.50+247.51= $486.01/batchCpc=486.01/100=$4.86/pcAlternative calculation ofCpc:Cpc=2.25/0.95+((18 + 35)/60)(50/100) +((18 + 35)/60)(2.17/0.95)Cpc=2.368+ 0.442 + 2.018= $4.83/pcThedifference from the previous value of $4.86/pcis duemostlyto the roundup ofQo.(c) The time to complete the batch was computed in part (a) asTb=280min =4.67hr1.10In a batch-productionmachiningoperation, the starting work partis a casting thatcosts$3.50each.Batch quantity = 65 parts. Part handling time each cycle = 2.5 min, and machiningtimeper part=3.44min. 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 ratesinclude overhead costs. The cutting toolinthe operationcosts= $5.75/pcand it must be changed every 18 parts.Tool change time = 3.0 min.Availability of

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Solutions forFundamentals of Modern Manufacturing, 5e1-8the machinetool= 98%, and the scrap rate =4%. Determine (a) production rate and (b) finishedpart cost.(c) How many hours are required to complete the batch?Solution: (a)Tc= 3.44 + 2.5 + 3/15 = 6.14 min/pcGivenq= 4%, the starting quantity of partsQo= 65/0.96 = 67.7 rounded up to 68 pcNow determine batch time, including setup time andavailability, assuming that the availabilityfactor does not apply during setup because the machine is not running.Tb= 75 + 68(6.14)/0.98 = 75 + 426.04 = 501.04 min/batch = 8.35 hrAverage production rate of partsRp= 65/8.35 = 7.784 pc/hrAverage productionrate of acceptable partsRp=7.784(1-0.04) =7.472pc/hr(b) Now determine batch cost, including setup time andavailability, assuming that the availabilityfactor does not apply during setup because the machine is not running. The number of cutting toolsrequired = 68/18 = 3.78 rounded up to 4 tools at $5.75 each = $23.00.Cbatch= 68(3.50) + ((20 + 25)(8.35) + 4(5.75) = 238.00 + 375.75 + 23.00 = $636.75/batchCpc= 636.75/65 =$9.796/pcAlternative calculation ofCpc:Cpc= 3.50/0.96 + ((20 + 25)/60)(75/65) + ((20 + 25)/60)(6.14/(0.98 x 0.96)) + 23.00/65Cpc= 3.646 +0.865 + 4.895 + 0.354 = $9.760/pcThedifference from the previous value of $9.796/pc is due mostly to the roundup ofQo.(c) The time to complete the batch was computed in part (a) asTb=501.04 min = 8.35 hr1.11(A)During a particular 40-hour weekofan automated production operation,336acceptable (non-defective) partsand 22 defective parts were produced. The operation cycle consists of a processingtime of5.73 min,and a part handling time of 0.38 min.Every 60 parts, a tool change is performed,and this takes 7.2min. The machine experienced several breakdowns during the week.Determine(a)hourly production rate of acceptable parts, (b) scrap rate, and (c) availability (proportionuptime) of the machine during this week.Solution: (a)Production rate of acceptable partsRp= 335/40 =8.40 pc/hr(b) Total parts processed during the weekQo= 336 + 22 = 358 pcScrap rateq= 22/358 = 0.0615 =6.15%(c)Cycle time of the unit operationTc=5.73 + 0.38 + 7.2/60 =6.23 minTotal uptime during the week=358(6.23) = 2230.34 min = 37.17 hrProportion uptimeA= 37.17/40 = 0.929 =92.9%1.12Ahigh-production operationwas studied during an80-hrperiod. During that time,a total of sevenequipment breakdowns occurred for a total lost production time of 3.8 hr, andthe operationproduced 38 defective products.No setups were performed during the period.The operation cycleconsists of a processing time of 2.14min, a part handling time of 0.65 min, and a tool change isrequired every 25 parts, which takes 1.50 min.Determine (a) hourly production rate of acceptablepartsand(b) scrap rate duringthe 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 rateof acceptable partsRp= 1566/80 =19.58 pc/hr(b) Scrap rateq=38/1566= 0.0243=2.43%

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Solutions forFundamentals of Modern Manufacturing,5e2-12THE NATURE OF MATERIALSReview Questions2.1The elements listed in the Periodic Table can be divided into three categories. What are thesecategories and give an example of each?Answer. The three types of elements are metals (e.g., aluminum), nonmetals (e.g., oxygen), andsemimetals (e.g., silicon).2.2Which elements are the noble metals?Answer. The noble metals are copper, silver, and gold.2.3What is the difference between primary and secondary bonding in the structure of materials?Answer. Primary bonding is strong bonding between atoms in a material, for example to form amolecule; while secondary bonding is not as strong and is associated with attraction betweenmolecules in the material.2.4Describe how ionic bonding works?Answer. In ionic bonding, atoms of one element give up their outer electron(s) to the atoms ofanother element to form complete outer shells.2.5What is the difference between crystalline and noncrystalline structures in materials?Answer. The atoms in a crystalline structure are located at regular and repeating lattice positions inthree dimensions; thus, the crystal structure possesses a long-range order which allows a high packingdensity. The atoms in a noncrystalline structure are randomly positioned in the material, notpossessing any repeating, regular pattern.2.6What are some common point defects in a crystal lattice structure?Answer. The common point defects are (1) vacancy-a missing atom in the lattice structure; (2)ion-pair vacancy (Schottky defect)-a missing pair of ions of opposite charge in a compound; (3)interstitialcy-a distortion in the lattice caused by an extra atom present; and (4) Frenkel defect-anion is removed from a regular position in the lattice and inserted into an interstitial position notnormally occupied by such an ion.2.7Define the difference between elastic and plastic deformation in terms of the effect on the crystallattice structure.Answer. Elastic deformation involves a temporary distortion of the lattice structure that isproportional to the applied stress. Plastic deformation involves a stress of sufficient magnitude tocause a permanent shift in the relative positions of adjacent atoms in the lattice. Plastic deformationgenerally involves the mechanism of slip-relative movement of atoms on opposite sides of a plane inthe lattice.2.8How 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 to deform; ineffect it becomes stronger.2.9Identify some materials that have a crystalline structure.Answer. Materials typically possessing a crystalline structure are metals and ceramics other thanglass. Some plastics have a partially crystalline structure.

Solution Manual for Fundamentals of Modern Manufacturing: Materials, Processes, and Systems, 5th Edition - Page 11 preview image

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Solutions forFundamentals of Modern Manufacturing,5e2-22.10Identify some materials that possess a noncrystalline structure.Answer. Materials typically having a noncrystalline structure include glass (fused silica), rubber, andcertain plastics (specifically, thermosetting plastics).2.11What is the basic difference in the solidification (or melting) process between crystalline andnoncrystalline structures?Answer. Crystalline structures undergo an abrupt volumetric change as they transform from liquid tosolid state and vice versa. This is accompanied by an amount of energy called the heat of fusion thatmust be added to the material during melting or released during solidification. Noncrystallinematerials melt and solidify without the abrupt volumetric change and heat of fusion.

Solution Manual for Fundamentals of Modern Manufacturing: Materials, Processes, and Systems, 5th Edition - Page 12 preview image

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Solutions forFundamentals of Modern Manufacturing, 5e13MECHANICAL PROPERTIES OF MATERIALSReview Questions3.1What is the dilemma between design and manufacturing in terms of mechanical properties?Answer. To achieve design function and quality, the material must be strong; for ease ofmanufacturing, the material should not be strong, in general.3.2What are the three types of static stresses to which materials are subjected?Answer. tensile, compressive, and shear.3.3State Hooke's law.Answer. Hooke's Law defines the stress-strain relationship for an elastic material:=E, whereE=a constant of proportionality called the modulus of elasticity.3.4What 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;whereastrue stress divides the load by the instantaneous area which decreases as the specimenstretches.3.5Definetensile strengthof a material.Answer. The tensile strength is the maximum load experienced during the tensile test divided by theoriginal area.3.6Defineyield strengthof a material.Answer. The yield strength is the stress at which the material begins to plastically deform. It isusually measured as the 0.2% offset value, which is the point where the stress-strain curve for thematerial intersects a line that is parallel to the straight-line portion of the curve but offset from it by0.2%.3.7Why cannot a direct conversion be made between the ductility measures of elongation and reductionin area using the assumption of constant volume?Answer. Because of necking that occurs in the test specimen.3.8What iswork hardening?Answer. Work hardening, also called strain hardening, is the increase in strength that occurs inmetals when they are strained.3.9Under what circumstancesdoes the strength coefficient have the same value as the yield strength?Answer. When the material is perfectly plastic and does not strain harden.3.10How does the change in cross-sectional area of a test specimen in a compression test differ from itscounterpart in a tensile test specimen?Answer. In a compression test, the specimen cross-sectional area increases as the test progresses;while in a tensile test, the cross-sectional area decreases.3.11What is the complicating factor that occurs in a compression testthat might be considered analogousto necking in a tensile test?Answer. Barreling of the test specimen due to friction at the interfaces with the testing machineplatens.

Solution Manual for Fundamentals of Modern Manufacturing: Materials, Processes, and Systems, 5th Edition - Page 13 preview image

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Solutions forFundamentals of Modern Manufacturing, 5e23.12Tensile testing is not appropriate for hard brittle materials such as ceramics. What is the testcommonly used to determine the strength properties of such materials?Answer. A three-point bending test is commonly used to test the strength of brittle materials. The testprovides a measure called the transverse rupture strength for these materials.3.13How is the shear modulus of elasticityGrelated to the tensile modulus of elasticityE, on average?Answer.G= 0.4E, on average.3.14How is shear strengthSrelated to tensile strengthTS, on average?Answer.S= 0.7TS, on average.3.15What ishardness,and how is it generally tested?Answer. Hardness is defined as the resistance to indentation of a material. It is tested by pressing ahard object (sphere, diamond point) into the test material and measuring the size (depth, area) of theindentation.3.16Why are different hardness tests and scales required?Answer. Different hardness tests and scales are required because different materials possess widelydiffering hardnesses. A test whose measuring range is suited to very hard materials is not sensitive fortesting very soft materials.3.17Define therecrystallization temperaturefor a metal.Answer. The recrystallization temperature is the temperature at which a metal recrystallizes (formsnew grains) rather than work hardens when deformed.3.18Defineviscosityof a fluid.Answer. Viscosity is the resistance to flow of a fluid material; the thicker the fluid, the greater theviscosity.3.19What is the defining characteristic of aNewtonian fluid?Answer. A Newtonian fluid is one for which viscosity is a constant property at a given temperature.Most liquids (water, oils) are Newtonian fluids.3.20What isviscoelasticity, as a material property?Answer. Viscoelasticity refers to the property most commonly exhibited by polymers that defines thestrain of the material as a function of stress and temperature over time. It is a combination of viscosityand elasticity.ProblemsAnswers to problems labeled(A)are listed inanAppendix at the back of the book.Strength and Ductility in Tension3.1(A)(SI Units)A tensile test specimen has a gage length=50 mm andits cross-sectionalarea =100mm2.The specimen yieldsat48,000 N, and the corresponding gage length = 50.23 mm. This is the0.2 percent yield point. The maximum load of87,000 N is reached at a gage length = 64.2 mm.Determine (a) yield strength, (b) modulus of elasticity, and (c) tensile strength.(d) If fracture occursat a gage length of 67.3 mm, determine the percent elongation. (e) If the specimen necked to anarea =53mm2, determine the percent reduction in area.Solution: (a)Y=48,000/100 =480MPa(b)s=E e.Subtracting the 0.2% offset,e= (50.23-50.0)/50.0-0.002 = 0.0026

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Solutions forFundamentals of Modern Manufacturing, 5e3E=s/e= 480/0.0026 =184.6x 103MPa(c)TS=87,000/100 =870MPa(d)EL= (67.3-50)/50 = 17.3/50 = 0.346 =34.6%(e)AR= (100-53)/100 = 0.47=47%3.2(USCS Units)Atensiletest specimen has a gage length=2.0 in anddiameter= 0.798in.Yieldingoccursata load of 31,000 lb. The corresponding gage length = 2.0083 in, whichis the 0.2 percentyield point. The maximum load of58,000 lb is reached at a gage length = 2.47in. Determine (a) yieldstrength, (b) modulus of elasticity, and (c) tensile strength. (d) Iffracture occurs at a gage length of2.62in, determine the percent elongation. (e) If the specimen necked to an area = 0.36in2, determinethe 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 is the elasticstrain to be used in calculating modulus of elasticity.E=s/e=62,000/0.00215 =28.84x 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%3.3(SI Units)During a tensile test in which the starting gage length = 100.0 mm and the cross-sectionalarea =150mm2. During testing,the following force and gage length data are collected:(1) 17,790Nat100.2mm, (2) 23,040N at103.5mm, (3)27,370N at110.5mm, (4) 28,910N at122.0mm, (5)27,400N at130.0mm, and (6) 20,460N at135.1mm.The final data point(6)occurred immediatelyprior to failure. Yielding occurred at a load of 19,390N (0.2% offset value), and the maximum load(4)was28,960N.(a) Plot the engineering stress strain curve. Determine (b) yield strength, (c)modulus of elasticity, (d) tensile strength, and (e) percent elongation.Solution: (a) Student exercise.(b)Yield strength = load at yielding divided by original area; thus,Y=19,390/150 =129.3 MPa(c)The first data point(1)occurredprior to yielding.Modulus of elasticityE=s/eEngineering straine= (100.2-100)/100= 0.002;engineeringstress =17,790/150 = 118.6 MPaE=118.6/0.002=59,300MPa(d)Tensile strength = maximum load divided by original area:TS=28,960/150=193.1MPa(e)Percent elongationEL= (135.10–100)/100 = 0.351 =35.1%Flow Curve3.4(A)(SI Units)In Problem 3.3, determine the strength coefficient and the strain-hardening exponent inthe flow curve equation.Solution: Starting volume of test specimenV=LoAo=100(150) =15,000mm3.We must select two data pointsthat occur after yielding but before necking. Necking occurs afterthe maximum load is reached. The two points selected are (2) and (4): (2)F= 23,040N andL=103.5 mm;and(4)F= 28,910N andL= 122.0mm.(2)A=V/L=15,000/103.5=144.93mm2True stress= 23,040/144.93=159.0MPa.True strain= ln(103.5/100) = 0.0344(4)A=15,000/122.0 =122.95mm2True stress= 28,910/122.95=235.1MPa.True strain= ln(122.0/100) = 0.1989

Solution Manual for Fundamentals of Modern Manufacturing: Materials, Processes, and Systems, 5th Edition - Page 15 preview image

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Solutions forFundamentals of Modern Manufacturing, 5e4Substituting these values into the flow curve equation, we have(2)159.0=K(0.0344)nand (4)235.1=K(0.1989)n235.1/159.0= (0.1989/0.0344)n1.479= (5.782)nln(1.479) =nln(5.782)0.3914= 1.7547nn= 0.223Substituting this value with the data back into the flow curve equation, we obtain the value of thestrength coefficientK:K=159.0/(0.0344).223=337.1MPaK=235.1/(0.1989).223=337.2MPaThe flow curve equation is=337.150.2233.5(SI Units)In a tensile test on asteelspecimen, true strain = 0.11at a stressof245MPa. When truestress =340MPa, true strain = 0.31. Determine the strength coefficient and the strain-hardeningexponent in the flow curve equation.Solution:Two equations:(1)245=K(0.11)nand (2)340=K(0.31)n340/245= (0.31/0.11)n1.3878= (2.8182)nnln(2.8182) = ln(1.3333)1.0361n= 0.3277n= 0.316Substituting this value with the data back into the flow curve equation, we obtain the value of thestrength coefficientK:(1)K=245/(0.11).316=492.1MPaCheck:(2)K=340/(0.31).316=492.3MPaThe flow curve equation is:=492.20.3163.6(USCS Units)During a tensile test, a metal has a true strain = 0.08at a true stress = 37,000 lb/in2.Later, at a true stress = 55,000 lb/in2, true strain = 0.24. Determine the strength coefficient and strain-hardening exponent in the flow curve equation.Solution: (1) 37,000 =K(0.08)nand (2) 55,000 =K(0.24)n55,000/37,000 = (0.24/0.08)n1.4865 = (3.0)nnln(3.0) = ln(1.4865)1.0986n= 0.3964n= 0.3608Substituting this value with the data back into the flow curve equation, we obtain the value of thestrength coefficientK:(1)K= 37,000/(0.08)0.3608=92,038lb/in2Check:(2)K= 55,000/(0.24)0.3608=92,041lb/in2Averaging the calculated values ofK, the flow curve equation is:=92,0400.36083.7(SI Units)Atensile testspecimenbegins to neck at a true strain = 0.22andtrue stress =257MPa.Without knowing any more about the test,determinethe strength coefficient and the strain-hardeningexponent in the flow curve equation?Solution: If we assume thatn=when necking starts, thenn= 0.22.Using this value in the flow curve equation, we haveK=257/(0.22).22=358.6MPaThe flow curve equation is:=358.60.223.8(SI Units)A tensile test providesthe followingflow curve parameters: strain-hardening exponent=0.25and strength coefficient=500 MPa. Determine (a) flow stress at a true strain = 1.0 and (b) truestrain at a flow stress =500 MPa.Solution: (a)Yf=500(1.0).25=500 MPa

Solution Manual for Fundamentals of Modern Manufacturing: Materials, Processes, and Systems, 5th Edition - Page 16 preview image

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Solutions forFundamentals of Modern Manufacturing, 5e5(b)= (500/500)1/.25= (1.0)4.0=1.003.9(A)(USCS Units)The flow curve for a certain metal has a strain-hardening exponent=0.20anditsstrength coefficient=26,000 lb/in2. Determine (a) flow stress at a true strain = 0.15 and (b) true strainat a flow stress =20,000 lb/in2.(c) What is the likely metal in this problem?Solution: (a)Yf=26,000(0.15).20=17,791lb/in2(b)= (20,000/26,000)1/.20= (0.7692)5.0=0.269(c) The flow curve parameters indicate that the metal isprobablyannealed pure aluminum,according to Table 3.4.3.10(USCS Units)A metal isplasticallydeformed in a tensile test. The starting specimen had a gagelength = 2.0 in and an area = 0.50 in2. At one pointduring thetest,thegage length = 2.3in, andthecorresponding engineering stress = 25,000 lb/in2. At another point prior to necking, the gage length =3.0in, and the corresponding engineering stress = 28,000 lb/in2. Determine the strength coefficientand the strain-hardening exponent for this metal.Solution: Starting volumeV=LoAo= 2.0(0.5) = 1.0 in3(1)Area at gage length = 2.3 in:A=V/L= 1.0/2.3= 0.4348in2Given engineering stress = 25,000 lb/in2,forceF= 25,000(0.5) = 12,500 lbTrue stress=12,500/0.4348=28,749lb/in2andtrue strain= ln(2.3/2.0) = 0.1398(2)Area at gage length = 3.0 in:A= 1.0/3.0= 0.3333in2Given engineering stress = 28,000 lb/in2,forceF= 28,000(0.5) = 14,000 lbTrue stress=14,000/0.3333= 42,000 lb/in2andtrue strain= ln(3.0/2.0) = 0.4055These two data pointscan be usedto 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=57,915lb/in2Check:(2)K= 42,000/(0.4055).356=57,915lb/in2The flow curve equation is:=57,9150.3563.11(SI Units)A tensile test specimen has a starting gage length = 75.0 mm. It is elongated during the testto a length = 110.0 mm before necking occurs. Determine (a) the engineering strain and (b) the truestrain. (c) Compute and sum the engineering strains as the specimen elongates from: (1) 75.0 to 80.0mm, (2) 80.0 to 85.0 mm, (3) 85.0 to 90.0 mm, (4) 90.0 to 95.0 mm, (5) 95.0 to 100.0 mm, (6) 100.0to 105.0 mm, and (7) 105.0 to 110.0 mm. (d) Is the result closer to the answer to part (a) or part (b)?Does this help to show what is meant by the term true strain?Solution: (a) Engineering straine= (110-75)/75 = 35/75 =0.4667(b) True strain= ln(110/75) = ln(1.4667) =0.383(c) (1)L= 75 to 80 mm:e= (80-75)/75 = 5/75 = 0.0667(2)L= 80 to 85 mm:e= (85-80)/80 = 5/80 = 0.0625(3)L= 85 to 90 mm:e= (90-85)/85 = 5/85 = 0.0588(4)L= 90 to 95 mm:e= (95-90)/90 = 5/90 = 0.0556(5)L= 95 to 100 mm:e= (100-95)/95 = 5/95 = 0.0526(6)L= 100 to 105 mm:e= (105-100)/100 = 5/100 = 0.0500(7)L= 105 to 110 mm:e= (110-105)/105 = 5/105 = 0.0476_____________________________________________

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