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

Solution Manual for Fundamentals of Modern Manufacturing: Materials, Processes, and Systems, 5th Edition helps you break down challenging textbook sections for easy learning.

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    Solutions for Fundamentals of Modern Manufacturing , 5e 1 - 1 1 INTRODUCTION Review Questions 1.1 About what percentage of the U.S. gross domestic product (GDP) is accounted for by the manufacturing industries? Answer . Although the percentage has varied over the years, currently the manufacturing industries account for around 12% of U.S. GDP. 1.2 Define manufacturing. Answer . The text defines manufacturing in two ways: technologically and economically. Technologically, manufacturing is the application of physical and chemical processes to alter the geometry, properties, and/or appearance of a given starting material to make parts or products; manufacturing also includes assembly of multiple parts to make products. Economically, manufacturing is the transformation of materials into items of greater value by means of one or more processing and/or assembly operations. The key point is that manufacturing adds value to the material by changing its shape or properties, or by combining it with other materials. 1.3 What are the differences between primary, secondary, and tertiary industries? Give an example of each category. Answer . A primary industry is one that cultivates and exploits natural resources, such as agriculture or mining. A secondary industry takes the outputs of primary industries and converts them 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 are banking and education. 1.4 What is the difference between a consumer good and a capital good? Give some examples in each category. Answer . Consumer goods are products purchased directly by consumers, such as cars, personal computers, TVs, tires, and tennis rackets. Capital goods are those purchased by companies to produce goods and/or provide services. Examples of capital goods include aircraft, computers, communication equipment, medical apparatus, trucks and buses, railroad locomotives, machine tools, and construction equipment. 1.5 What is the difference between soft product variety and hard product variety, as these terms are defined in the text? Answer . Soft product variety is when there are only small differences among products, such as the differences among car models made on the same production line. In an assembled product, soft variety is characterized by a high proportion of common parts among the models. Hard product variety is when the products differ substantially, and there are few common parts, if any. The difference between a car and a truck exemplifies hard variety. 1.6 How are product variety and production quantity related when comparing typical factories? Answer . Generally production quantity is inversely related to product variety. A factory that produces a large variety of products will produce a smaller quantity of each. A company that produces a single product will produce a large quantity. 1.7 One of the dimensions of manufacturing capability is technological processing capability. Define technological processing capability.

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    Solutions for Fundamentals of Modern Manufacturing , 5e 1 - 2 Answer . The technological processing capability of a plant (or company) is its available set of manufacturing processes. Certain plants perform machining operations, others roll steel billets into sheet stock, and others build automobiles. The underlying feature that distinguishes these plants is the processes they can perform. Technological processing capability includes not only the physical processes, but also the expertise possessed by plant personnel in these processing technologies. 1.8 What 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 three types. 1.9 What is the definition of steel? Answer . Steel can be defined as an iron – carbon alloy containing 0.02% to 2.11% carbon. Its composition often includes other alloying elements as well, such as manganese, chromium, nickel, and molybdenum, to enhance the properties of the metal. 1.10 What are some of the typical applications of steel? Answer . Applications of steel include construction (e.g., bridges, I - beams, and nails), transportation (trucks, rails, and rolling stock for railroads), and consumer products (automobiles and appliances). 1.11 What is the difference between a thermoplastic polymer and a thermosetting polymer? Answer . Thermoplastic polymers can be subjected to multiple heating and cooling cycles without substantially altering the molecular structure of the polymer. Thermosetting polymers chemically transform (cure) into a rigid structure on cooling from a heated plastic condition. 1.12 Manufacturing processes are usually accomplished as unit operations. Define unit operation. Answer . A unit operation is a single step in the sequence of steps required to transform the starting material into a final product. A unit operation is generally performed on a single piece of equipment that runs independently of other operations in the plant. 1.13 In manufacturing processes, what is the difference between a processing operation and an assembly operation? Answer . A processing operation transforms a work material from one state of completion to a more 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 discrete work part s, but certain processing operations are also applicable to assembled items (e.g., painting a spot - welded car body). An assembly operation joins two or more components to create a new entity, called an assembly, subassembly, or some other term that refers to the joining process (e.g., a welded assembly is called a weldment). 1.14 One of the three general types of processing operations is shaping operations, which are used to create or alter the geometry of the work part . What are the four categories of shaping operations? Answer . The four categories of shaping operations are (1) solidification processes, in which the starting 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 formed and heated into the desired geometry; (3) deformation processes, in which the starting material is a ductile solid (commonly metal) that is deformed to shape the part; and (4) material removal processes, in which the starting material is a solid (ductile or brittle), from which material is removed so that the resulting part has the desired geometry.

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    Solutions for Fundamentals of Modern Manufacturing , 5e 1 - 3 1.15 What is the difference between net shape processes and near net shape processes? Answer . Net shape processes are manufacturing processes that transform nearly all of the starting material into product and require no subsequent machining to achieve final part geometry . Near net shape processes are ones that require minimum machining to produce the final shape. 1.16 Identify the four types of permanent joining processes used in assembly. Answer . The four types are welding, brazing, soldering, and adhesive bonding. 1.17 What is a machine tool? Answer . The term developed during the Industrial Revolution, when it referred to power - driven machines used to operate cutting tools previously operated by hand. Modern machine tools are described by the same basic definition, except that the power is electrical rather than water or steam, and the level of precision and automation is much greater today. 1.18 What 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 is commercially available for any manufacturing company to invest in. Special - purpose equipment 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 not available. Some companies with unique processing requirements develop their own special purpose equipment. 1.19 Define batch production and describe why it is often used for medium - quantity production products. Answer . Batch production is where groups, lots, or batches or materials or parts are processed together through the manufacturing operations. All units in the batch are processed at a given station before the group proceeds to the next station. In a medium or low quantity production situation, the same machines are used to produce many types of products. Whenever a machine switches from one product to another, a changeover occurs. The changeover requires the machine setup to be torn down and set up for the new product. Batch production allows the changeover time to be distributed across a larger number of parts and hence reduce the average operation time per part. 1.20 What 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 of process it performs. To produce a product it must visit the departments in the order of the operations that must be performed. This often includes large travel distances within the plant. A process layout is often used when the product variety is large the operation sequences of products are dissimilar. A product layout is one where the machinery is arranged based on the general flow of the products that will be produced. Travel distance is reduced because products will generally flow to the next machine in the sequence. A product layout works well when all products tend to follow the same sequence of production. 1.21 Name two departments that are typically classified as manufacturing support departments. Answer . A common organizational structure includes the following three manufacturing support departments: (1) manufacturing engineering, (2) production planning and control, and (3) quality control. 1.22 What are overhead costs in a manufacturing company? Answer . Overhead costs consist of all of the expenses of operating the company other than material, labor, and equipment.

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    Solutions for Fundamentals of Modern Manufacturing , 5e 1 - 4 1.23 Name 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 overhead consists of the costs of running the factory excluding materials, direct labor, and equipment. This overhead 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, and health benefits. 1.24 What is the difference between fixed costs and variable costs? Answer . A fixed cost remains constant for any level of production output, whereas variable costs are paid for as they are used. The cost of the factory and equipment are fixed costs. Direct labor and materials that are used to produce the product are variable costs. 1.25 What is meant by the term availability? Answer . Availability is a reliability term which is simply the proportion uptime of the equipment. Problems Answers to problems labeled (A) are listed in an Appendix at the back of the book. Manufacturing Economics 1.1 (A) A company invests $ 7 50,000 in a piece of production equipment. The cost to install the equipment in the plant = $25 ,000 . The anticipated life of the machine = 12 years. The machine will be used eight hours per shift, five shifts per week , 50 weeks per year. A pplicable overhead rate = 18 %. Assume availability = 100%. Determine the equipment cost rate if (a) the plant operates one shift per day and (b) the plant operates three shifts per day. Solution : (a) For a one - shift operation, hours of operation per year H = 50( 1 )(5)(8) = 2 000 hr/yr. Using Eq. (1.8), C eq = ( 7 50,000 + 25 ,000)(1.18)/(60 x 12 x 2000) = $0. 635 /min = $ 3 8 . 10 /hr (b) For a three - shift operation, hours of operation per year H = 50( 3 )(5)(8) = 6 000 hr/yr. C eq = (750,000 + 25,000)(1 . 18)/(60 x 12 x 6000) = $0. 212 /min = $12.70/hr Note the significant advantage the company has if it r u n s three shifts per day rather than one shift. 1.2 A production machine was purchased six years ago for an installed price of $530,000. At that time it 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 machine will last four more years, operating 4000 hours per year. A pplicable overhead rate = 30 %. Assume availability = 100%. Determine the equipment cost rate for this machine. Solution : The cost rate under the original conditions was the following: C eq = 530,000(1.30) /(60 x 10 x 4000) = $0.287/min = $17.23/hr The repairs will add to that cost rate as follows: C eq = 125 ,000(1.30) /(60 x 4 x 4000) = $0.169/min = $10.16/hr The repaired machine has a cost rate C eq = 0.287 + 0.169 = $0.456/min = $27.36/hr 1.3 Instead of repairing the machine in Problem 1.2, a proposal has been made to purchase a new machine and scrap the current machine at a zero salvage value. The new machine will have a production rate that is 2 0% faster than the current equipment , whose production rate = 12 parts per hour. Each part has a starting material cost = $1. 33 and a selling price = $ 6 .40. All parts produced during 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 machine would still be productive for another six years. The new machine costs $700,000 installed, has an anticipated life of 10 years, and an applicable overhead rate of 30 %. It will be used 4000 hours per year , same as the current machine .

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    Solutions for Fundamentals of Modern Manufacturing , 5e 1 - 5 The labor rate for either alternative = $24.00/hr which includes applicable overhead costs. Assume availability = 100% and scrap rate = 0. Which alternative is more economical using total profit over four years as the criterion, (a) repairing the current machine or (b) purchasing the new machine? Solution : (a) The first alternative is to repair the current machine. The cost rate was determined in the solution to Problem 1.2. Repeating here, the original cost rate is calculated as follows: C eq = 530,000(1.30) /(60 x 10 x 4000) = $0.287/min = $17.23/hr The repairs will add to that cost rate as follows: C eq = 125 ,000(1.30) /(60 x 4 x 4000) = $0.169/min = $10.16/hr The repaired machine has a cost rate C eq = 0.287 + 0.169 = $0.456/min = $27.36/hr Labor cost = $24.00/hr (given) Given that annual hours of operation = 4000, total cost of production on this machine is calculated as follows: TC = 4000(24.00 + 27.36) = $205,440/yr At a production rate of 12 pc/hr and operating 4000 hr/yr, annual output = 4000(12) = 48,000 pc/yr Total revenue = 48,000( 6 .40 – 1.33 ) = $ 243 , 360 /yr. Total profit over four years = 4(2 43 , 360 – 205,440) = $151,680 (b) The second machine has an equipment cost rate determined as follows: C eq = 700 ,000(1.30) /(60 x 10 x 4000) = $0.379/min = $22.75/hr Labor cost = $24.00/hr (given) Given that annual hours of operation = 4000, total annual cost of production on this machine is TC = 4000(24.00 + 22.75) = $187,000/yr Production rate on the new machine is 20% faster, so production rate = 12(1.20) = 14.4 pc/hr At 1 4.4 pc/hr and operating 4000 hr/yr, annual output = 4000(1 4.4 ) = 57 , 6 00 pc/yr Total revenue = 57 , 6 00(6.40 – 1.33) = $ 292 , 032 /yr. Total profit over four years = 4( 292 , 032 – 187 , 000 ) = $ 420 , 128 Conclusion: The new machine should be purchased and the old machine scrapped. 1.4 (A) A machine tool is used to machine parts in batches. In one batch of interest, the starting piece is a casting that cost s = $ 5 . 00 each . Batch quantity = 40 . The actual machining time in the operation = 6 . 86 min. Time to loa d and unload each workpiece = 2.0 min. C ost of the cutting tool = $ 3 . 0 0, and each tool must be changed every 10 pieces . Tool change time = 1 . 5 min. Setup time for the batch = 1 . 7 5 hr. Hourly wage rate of the operator = $16. 0 0/hr, and the applicable labor overhead rate = 5 0%. H ourly 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 rate when setup time is figured in, and (c) cost per piece. Solution : (a) P rocessing time T o = 6 . 86 min, p art handling time T h = 2 . 0 0 min, and tool handling time T t = 1 . 50 min/ 1 0 = 0.1 5 min. T c = 6 . 86 + 2 . 0 0 + 0.1 5 = 9 . 01 min (b) A verage production time per piece including setup time T p = 1.75(60)/40 + 9.01 = 11.64 min Hourly pro duction rate R p = 60/11.64 = 5.16 pc/hr (c) E quipment cost rate C eq = $22.00/60 = $0. 367 /min. L abor cost rate C L = 16.00(1.50) = $24.00/hr = $0.40/min Cost of tooling C t = 3 . 0 0/ 1 0 = $0. 30 /pc Finally, cost per piece C pc = 5 . 00 + (0. 40 + 0. 367 )( 11.64 ) + 0. 30 = $ 14 . 23 /pc 1.5 A 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. The molding 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 molding compound costs $1.20/lb. Each molding weighs 0.88 lb. The only labor required is for a worker to periodically retrieve the moldings. L abor rate of the worker = $18.00/hr including overhead. However, the worker also

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    Solutions for Fundamentals of Modern Manufacturing , 5e 1 - 6 tends other machines and only spends 2 0% of his time on this machine. Setup can be ignored because of the long production run. The molding machine was purchased for $ 5 00,000 installed, its anticipated life = 10 years, and it operates 6000 hours per year. Equipment overhead rate = 30%. Assume a vailability = 100 % and scrap rate = 0 . Determine (a) the hourly production rate of the machine , (b) annual quantity of product molded , and ( c ) cost per piece. Solution : (a) With a cycle time T c = 45 sec = 0.75 min, R p = 60/0.75 = 80 pc/hr. Factoring in the 98% proportion uptime, R p = 0.98(80) = 78.4 pc/hr Annual quantity of product = 6000(78.4) = 470,400 pc/yr (b) E quipment cost rate C eq = 5 00,000(1.30) / ( 60 x 10 x 6000) = $0. 18 06 /min. Mold cost per piece C t = 100,000/1,000,000 = $0.10/pc L abor cost rate C L = 1 8 .00 (0. 2 0 ) = $ 3 . 6 0/hr = $0. 0 6 /min Finally, cost per piece C pc = 1.20(0.88) + (0. 0 6 + 0. 1806 )( 0 . 75 ) + 0.10 = $ 1 . 34 /pc 1.6 A stamping press produces sheet - metal stampings in batches. The press is operated by a worker whose labor rate = $15.00/hr and a pplicable labor overhead rate = 4 2 %. C ost rate of the press = $22.50/hr and a pplicable equipment overhead rate = 20%. In one job of interest, batch size = 400 stampings, and the time to set up the die in the press takes 75 min . The die cost $40,000 and is expected to last for 200,000 stampings. Each cycle in the operation, t he starting blanks of sheet metal ar e manually loaded into the press , which takes 42 sec. The actual press stroke take s only 8 sec. Cost of the starting blanks = $0. 43 /pc. The press operates 250 days per year, 7.5 hours per day , but t he operator is paid for 8 hours per day. Assume a vailability = 100 % and scrap rate = 0 . Determine (a) cycle time, (b) average production rate with and without setup time included , and ( c ) cost per stamping produced . Solution : (a) C ycle time T c = 4 2 + 8 = 50 sec = 0. 833 min (b) Including setup time, T p = 75/400 + 0.833 = 1.021 min R p = 60/ 1 . 021 = 58.78 pc/hr Excluding setup time, R c = 60/0.833 = 72.03 pc/hr ( c ) E quipment cost rate C eq = 22.50(1.20)/60 = $0. 45 /min. Die cost per piece C t = 40,000/200,000 = $0.20/pc L abor cost rate C L = 1 5 .00( 1 . 42 ) /60 = $0. 355 /min This labor cost should be adjusted for the fact that although the press operates 7.5 hr/day, the operator is paid for 8 hr. C L = 0.355(8/7.5) = $0.379 Finally, cost per stamping C pc = 0 . 43 + (0. 3 79 + 0. 45 )( 1 . 021 ) + 0.20 = $ 1 . 4 8 /pc 1.7 A production machine operates in a semi - automatic cycle but a worker must tend the machine 100% of the time to load parts. Unloading is accomplished automatically. The worker’s cost rate = $2 7 /hr including applicable labor overhead rate . The equipment cost rate of the machine = $18.00/hr including applicable overhead costs. Cost of the starting parts = $0.15/pc. The job runs several 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 to install an automatic parts - loading device on the machine. The device would cost $ 36 ,000 and would reduce the part loading time to 3 sec each cycle . Its expected life = 4 years. The device would also relieve the worker from full - time attention to the machine. Instead, the worker could tend four machines, effectively reducing the labor cost to 25% of its current rate for each machine. The operat ion 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 loading device and (b) with the parts loading device installed. (c) How many days of production are required to pay for the automatic loading device? In other words find the breakeven point. Solution : (a) Equipment cost rate C eq = 18/60 = $0.30/min

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    Solutions for Fundamentals of Modern Manufacturing , 5e 1 - 7 Labor cost rate C L = 27/60 = $0.45/min Without the loading device, C pc = 0.15 + (0.45 + 0.30)(24 + 6 + 3 )/60 = $0.5 63 /pc (b) Cost rate of the device = 3 6,000/(60 x 4 x 2000) = $0.0 7 5/min = $ 4 .5 0 /hr With the loading device, C pc = 0.15 + (0.45 /4 + 0.30 + 0.075 )(24 + 3 + 3)/60 = $0. 394 /pc ( c ) Without the device, T c = 24 + 6 + 3 = 33 sec = 0.55 min and C pc = 0.563/pc from part (a) R p = R c = 60/0.55 = 109.1 pc/hr = 872 pc/day With the device, T c = 24 + 3 + 3 = 33 sec = 0.50 min and C pc = 0.394/pc from part (b) At 100% reliability and no setup time, R p = R c = 60/0.50 = 120.0 pc/hr = 960 pc/day Let D = number of days of production at which the two alternatives are equivalent. 872(0.563) D = 36,000 + 960(.394) D 490.9 D = 36,000 + 378.2 D (490.9 – 378.2) D = 112.7 D = 36,000 D = 319.5 round to 320 days 1.8 (A) In a long - running high - production operation, the s tarting work part cost = $0.95 each , and cycle time = 1.25 min. Equipment cost rate = $ 28 .00/hr, and labor cost rate = $2 1 .00/hr . Both rates include overhead costs. Tooling cost = $ 0. 05/pc. Availability of the production machine = 9 6 %, 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) = 4 6. 08 pc/hr However, because of the 5% scrap rate, the production rate of acceptable parts is R p = 4 6. 08 (1 - 0.0 7 ) = 42 . 85 pc/hr (b) Factoring in availability and scrap rate, part cost is C pc = 0.95/0.93 + ((21 + 28)/60)(1.25/(0.93 x 0.96)) + 0.05 = $2.215/pc 1.9 T he starting work part cost s $ 2 . 25 in a batch production operation . B atch quantity = 100 parts. Each cycle, p art handling time = 0 .5 0 min , and operation time = 1 . 67 min . Setup time = 50 min. Equipment cost rate = $ 3 5 .00/hr, and labor cost rate = $ 18 .00/hr , includ ing overhead costs. There is no tool change or tool cost in the operation. T he machine tool is 100 % reliable , and scrap rate = 5 %. Determine (a) production rate , (b) finished part cost , and (c) number of hours required to complete the batch . Solution : (a) T c = 1 . 67 + 0 .5 0 = 2 . 17 min/pc Given q = 5 %, the starting quantity of parts Q o = 100 /0.9 5 = 105.3 rounded up to 106 pc D etermine batch time, i ncluding setup time . T b = 50 + 106 ( 2 .1 7 ) = 50 + 230 . 0 2 = 280.02 min/batch = 4 . 67 hr Average production rate of parts R p = 100 / 4 . 67 = 21 . 43 pc/hr Average production rate of acceptable parts R p = 21 . 43 (1 - 0.0 5 ) = 20 . 36 pc/hr (b) Now determine batch cost, including setup time. C b = 106 ( 2 . 25 ) + (( 18 + 35 )( 4 . 67 ) = 238. 50 + 247 . 51 = $ 486 . 01 /batch C pc = 486.01 / 100 = $ 4 . 86 /pc Alternative calculation of C pc : C pc = 2.25 /0.95 + (( 18 + 35 )/60)( 50 / 100 ) + (( 18 + 35 )/60)( 2 . 17 /0.9 5 ) C pc = 2. 368 + 0.442 + 2.018 = $ 4 . 83 /pc The difference from the previous value of $ 4 . 86 /pc is due mostly to the roundup of Q o . (c) The time to complete the batch was computed in part (a) as T b = 280 min = 4 . 67 hr 1.10 In a batch - production machining operation, the starting work part is a casting that cost s $ 3 . 50 each . Batch quantity = 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 = $2 5 .00/hr, and labor cost rate = $2 0 .00/hr . Both rates include overhead costs. T he cutting t ool in the operation costs = $ 5.75 /pc and it must be changed every 18 parts . Tool change time = 3.0 min. Availability of

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    Solutions for Fundamentals of Modern Manufacturing , 5e 1 - 8 the machine tool = 9 8 %, and the scrap rate = 4 %. Determine (a) production rate and (b) finished part cost. (c) How many hours are required to complete the batch? Solution : (a) T c = 3.44 + 2.5 + 3/15 = 6.14 min/pc Given q = 4%, the starting quantity of parts Q o = 65/0.96 = 67.7 rounded up to 68 pc Now determine batch time, including setup time and availability , assuming that the availability factor does not apply during setup because the machine is not running. T b = 75 + 68(6.14)/0.98 = 75 + 426.04 = 501.04 min/batch = 8.35 hr Average production rate of parts R p = 65/8.35 = 7.784 pc/hr Average production rate of acceptable parts R p = 7.784 (1 - 0.0 4 ) = 7 . 472 pc/hr (b) Now determine batch cost, including setup time and availability , assuming that the availability factor does not apply during setup because the machine is not running. The number of cutting tools required = 68/18 = 3.78 rounded up to 4 tools at $5.75 each = $23.00. C batch = 68(3.50) + ((20 + 25)(8.35) + 4(5.75) = 238.00 + 375.75 + 23.00 = $636.75/batch C pc = 636.75/65 = $9.796/pc Alternative calculation of C pc : C pc = 3.50/0.96 + ((20 + 25)/60)(75/65) + ((20 + 25)/60)(6.14/(0.98 x 0.96)) + 23.00/65 C pc = 3.646 +0.865 + 4.895 + 0.354 = $9.760/pc The difference from the previous value of $9.796/pc is due mostly to the roundup of Q o . (c) The time to complete the batch was computed in part (a) as T b = 501.04 min = 8.35 hr 1.11 (A) During a particular 40 - hour week of an automated production operation , 33 6 acceptable (non - defective) parts and 22 defective parts were produced . The operation cycle consists of a processing time of 5 . 7 3 min , and a part handling time of 0.38 min. Every 60 parts, a tool change is performed, and this takes 7. 2 min. T he machine experienced several breakdowns during the week. Determine (a) hourly production rate of acceptable parts, (b) scrap rate, and ( c ) availability (proportion uptime) of the mac hine during this week. Solution : (a) Production rate of acceptable parts R p = 335/40 = 8.40 pc/hr (b) Total parts processed during the week Q o = 336 + 22 = 358 pc Scrap rate q = 22/358 = 0.0615 = 6.15% (c) Cycle time of the unit operation T c = 5 . 7 3 + 0.38 + 7. 2 /60 = 6.23 min Total uptime during the week = 358(6.23) = 2230.34 min = 37.17 hr Proportion uptime A = 37.17/40 = 0.929 = 92.9% 1.12 A high - production operation was studied during an 80 - hr period . During that time, a total of seven equipment breakdowns occurred for a total lost production time of 3.8 hr , and the operation produced 38 defective products . No setups were performed during the period. The operation cycle consists of a processing time of 2.1 4 min, a part handling time of 0.65 min, 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 operation T c = 2.14 + 0.65 + 1.50/25 = 2.85 min Hours of production during 80 hours = 80 – 3.8 = 76.2 hr Total number of parts produced = 76.2(60)/2.85 = 1604 pc Number of acceptable parts produced = 1604 – 38 = 1566 pc Production rate of acceptable parts R p = 1566/80 = 19.58 pc/hr (b) Scrap rate q = 38 / 1566 = 0. 0243 = 2 . 43 %

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    Solutions for Fundamentals of Modern Manufacturing , 5 e 2 - 1 2 THE NATURE OF MATERIALS Review Questions 2.1 The elements listed in the Periodic Table can be divided into three categories. What are these categories and give an example of each? Answer . The three types of elements are metals (e.g., aluminum), nonmetals (e.g., oxygen), and semimetals (e.g., silicon). 2.2 Which elements are the noble metals? Answer . The noble metals are copper, silver, and gold. 2.3 What 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 a molecule; while secondary bonding is not as strong and is associated with attraction between molecules in the material. 2.4 Describe how ionic bonding works? Answer . In ionic bonding, atoms of one element give up their outer electron(s) to the atoms of another element to form complete outer shells. 2.5 What 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 in three dimensions; thus, the crystal structure possesses a long - range order which allows a high packing density. The atoms in a noncrystalline structure are randomly positioned in the material, not possessing any repeating, regular pattern. 2.6 What 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 - an ion is removed from a regular position in the lattice and inserted into an interstitial position not normally occupied by such an ion. 2.7 Define the difference between elastic and plastic deformation in terms of the effect on the crystal lattice structure. Answer . Elastic deformation involves a temporary distortion of the lattice structure that is proportional to the applied stress. Plastic deformation involves a stress of sufficient magnitude to cause a permanent shift in the relative positions of adjacent atoms in the lattice. Plastic deformation generally involves the mechanism of slip - relative movement of atoms on opposite sides of a plane in the lattice. 2.8 How do grain boundaries contribute to the strain hardening phenomenon in metals? Answer . Grain boundaries block the continued movement of dislocations in the metal during straining. As more dislocations become blocked, the metal becomes more difficult to deform; in effect it becomes stronger. 2.9 Identify some materials that have a crystalline structure. Answer . Materials typically possessing a crystalline structure are metals and ceramics other than glass. Some plastics have a partially crystalline structure.

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    Solutions for Fundamentals of Modern Manufacturing , 5 e 2 - 2 2.10 Identify some materials that possess a noncrystalline structure. Answer . Materials typically having a noncrystalline structure include glass (fused silica), rubber, and certain plastics (specifically, thermosetting plastics). 2.11 What is the basic difference in the solidification (or melting) process between crystalline and noncrystalline structures? Answer . Crystalline structures undergo an abrupt volumetric change as they transform from liquid to solid state and vice versa. This is accompanied by an amount of energy called the heat of fusion that must be added to the material during melting or released during solidification. Noncrystalline materials melt and solidify without the abrupt volumetric change and heat of fusion.

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    Solutions for Fundamentals of Modern Manufacturing , 5e 1 3 MECHANICAL PROPERTIES OF MATERIALS Review Questions 3.1 What 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 of manufacturing, the material should not be strong, in general. 3.2 What are the three types of static stresses to which materials are subjected? Answer . tensile, compressive, and shear. 3.3 State Hooke's law. Answer . Hooke's Law defines the stress - strain relationship for an elastic material:  = E  , where E = a constant of proportionality called the modulus of elasticity. 3.4 What 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 true stress divides the load by the instantaneous area which decreases as the specimen stretches. 3.5 Define tensile strength of a material. Answer . The tensile strength is the maximum load experienced during the tensile test divided by the original area. 3.6 Define yield strength of a material. Answer . The yield strength is the stress at which the material begins to plastically deform. It is usually measured as the 0.2% offset value, which is the point where the stress - strain curve for the material intersects a line that is parallel to the straight - line portion of the curve but offset from it by 0.2%. 3.7 Why cannot a direct conversion be made between the ductility measures of elongation and reduction in area using the assumption of constant volume? Answer . Because of necking that occurs in the test specimen. 3.8 What is work hardening ? Answer . Work hardening, also called strain hardening, is the increase in strength that occurs in metals when they are strained. 3.9 Under what circumstances does the strength coefficient have the same value as the yield strength? Answer . When the material is perfectly plastic and does not strain harden. 3.10 How does the change in cross - sectional area of a test specimen in a compression test differ from its counterpart 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. 3.11 What is the complicating factor that occurs in a compression test that might be considered analogous to necking in a tensile test ? Answer . Barreling of the test specimen due to friction at the interfaces with the testing machine platens.

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    Solutions for Fundamentals of Modern Manufacturing , 5e 2 3.12 Tensile testing is not appropriate for hard brittle materials such as ceramics. What is the test commonly 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 test provides a measure called the transverse rupture strength for these materials. 3.13 How is the shear modulus of elasticity G related to the tensile modulus of elasticity E , on average? Answer . G = 0.4 E , on average. 3.14 How is shear strength S related to tensile strength TS , on average? Answer . S = 0.7 TS , on average. 3.15 What is hardness, and how is it generally tested? Answer . Hardness is defined as the resistance to indentation of a material. It is tested by pressing a hard object (sphere, diamond point) into the test material and measuring the size (depth, area) of the indentation. 3.16 Why are different hardness tests and scales required? Answer . Different hardness tests and scales are required because different materials possess widely differing hardnesses. A test whose measuring range is suited to very hard materials is not sensitive for testing very soft materials. 3.17 Define the recrystallization temperature for a metal. Answer . The recrystallization temperature is the temperature at which a metal recrystallizes (forms new grains) rather than work hardens when deformed. 3.18 Define viscosity of a fluid. Answer . Viscosity is the resistance to flow of a fluid material; the thicker the fluid, the greater the viscosity. 3.19 What is the defining characteristic of a Newtonian 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. 3.20 What is viscoelasticity , as a material property? Answer . Viscoelasticity refers to the property most commonly exhibited by polymers that defines the strain of the material as a function of stress and temperature over time. It is a combination of viscosity and elasticity. Problems Answers to problems labeled (A) are listed in an Appendix at the back of the book. Strength and Ductility in Tension 3.1 (A) (SI Units) A tensile test specimen has a gage length = 50 mm and its cross - sectional area = 1 00 mm 2 . T he specimen yields at 4 8,000 N , and t he corresponding gage length = 50.23 mm. This is the 0.2 percent yield point. The maximum load of 87 ,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 occurs at a gage length of 67.3 mm, determine the percent elongation. (e) If the specimen necked to an area = 53 mm 2 , determine the percent reduction in area. Solution : (a) Y = 4 8,000/ 1 00 = 480 MPa (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 for Fundamentals of Modern Manufacturing , 5e 3 E = s / e = 4 8 0/0.0026 = 18 4 . 6 x 10 3 MPa (c) TS = 87 ,000/ 1 00 = 870 MPa (d) EL = (67.3 - 50)/50 = 17.3/50 = 0.346 = 34.6% (e) AR = ( 1 00 - 53 )/ 1 00 = 0. 47 = 47 % 3.2 (USCS Units) A tensile test specimen has a gage length = 2.0 in and diameter = 0. 798 in. Y ield ing occurs at a load of 3 1 ,000 lb. The corresponding gage length = 2.0083 in , which is the 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) If fracture occurs at a gage length of 2. 62 in, d etermine the percent elongation. (e) If the specimen necked to an area = 0. 3 6 in 2 , determine the percent reduction in area. Solution : (a) Area A = π( 0. 798) 2 /4 = 0.50 in 2 . Yield strength Y = 3 1 ,000/0.5 0 = 6 2 ,000 lb/in 2 (b) s = E e . Subtracting the 0.2% offset, e = (2.0083 - 2.0)/2.0 - 0.002 = 0.00215 . This is the elastic strain to be used in calculating modulus of elasticity. E = s / e = 62 ,000/0.002 15 = 28 . 84 x 10 6 lb/in 2 (c) TS = 58 ,000/0.5 = 116 ,000 lb/in 2 (d) EL = (2. 62 - 2.0)/2.0 = 0. 62 /2.0 = 0. 31 = 31 % (e) AR = (0.5 - 0. 3 6 )/0.5 = 0. 28 = 28 % 3.3 (SI Units) During a tensile test in which the starting gage length = 1 00 .0 mm and the cross - sectional area = 150 mm 2 . During testing, the following force and gage length data are collected : (1) 17,79 0 N at 1 00 .2 mm, (2) 23,04 0 N at 1 0 3 .5 mm, (3) 27, 37 0 N at 1 1 0.5 mm, (4) 28,91 0 N at 1 22 .0 mm, (5) 27, 40 0 N at 1 30 .0 mm, and (6) 20,46 0 N at 1 35 .1 mm. The final data point (6) occurred immediately prior to failure. Yielding occurred at a load of 19,39 0 N (0.2% offset value), and t h e maximum load (4) was 2 8 , 96 0 N . (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 f irst data point (1) occurred prior to yielding. Modulus of elasticity E = s / e Engineering s train e = (1 00 .2 - 1 00 )/1 00 = 0.00 2; engineering stress = 17,790/150 = 118.6 MPa E = 118.6 /0.00 2 = 5 9 , 300 MPa (d) Tensile strength = maximum load divided by original area: TS = 2 8 , 960 / 150 = 19 3 . 1 MPa (e) Percent elongation EL = ( 135.10 – 100)/100 = 0.351 = 35.1% Flow Curve 3.4 (A) (SI Units) In Problem 3.3, determine the strength coefficient and the strain - hardening exponent in the flow curve equation. Solution : Starting volume of test specimen V = L o A o = 1 00 ( 150 ) = 15,000 mm 3 . We must s elect two data points that occur after yielding but before necking. Necking occurs after the maxi m um load is reached. The two points selected are (2) and (4) : ( 2 ) F = 23 , 04 0 N and L = 1 0 3 .5 mm; and ( 4 ) F = 28 , 91 0 N and L = 1 22 . 0 mm. ( 2 ) A = V / L = 15,000 / 10 3 .5 = 14 4 . 93 mm 2 True s tress  = 23 , 04 0 / 14 4 . 93 = 1 59 . 0 MPa. True s train  = ln(1 0 3 .5/1 00 ) = 0. 0 344 ( 4 ) A = 15,000 /1 22 .0 = 122 . 9 5 mm 2 True stress  = 28 , 91 0 / 122.9 5 = 235 . 1 MPa. True strain  = ln(1 22 .0/1 00 ) = 0. 1989

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    Solutions for Fundamentals of Modern Manufacturing , 5e 4 Substituting these values into the flow curve equation, we have ( 2 ) 1 59 . 0 = K (0. 0 344 ) n and ( 4 ) 235. 1 = K (0. 1989 ) n 235. 1 / 1 59 . 0 = (0. 1989 /0. 0 344 ) n 1. 4 79 = ( 5.782 ) n ln(1. 4 79 ) = n ln( 5.782 ) 0. 3914 = 1. 7547 n n = 0. 223 Substituting this value with the data back into the flow curve equation, we obtain the value of the strength coefficient K : K = 1 59 . 0 /(0. 0 344 ) . 223 = 337 . 1 MPa K = 235. 1 /(0. 1989 ) . 223 = 3 37 . 2 MPa The flow curve equation is  = 337 . 1 5  0. 223 3.5 (SI Units) In a tensile test on a steel specimen, true strain = 0. 11 at a stress of 2 4 5 MPa. When true stress = 340 MPa, true strain = 0. 31 . Determine the strength coefficient and the strain - hardening exponent in the flow curve equation. Solution : Two equations: (1) 2 4 5 = K (0. 11 ) n and (2) 340 = K (0. 31 ) n 340 / 2 4 5 = (0. 31 /0. 11 ) n 1. 3878 = ( 2.8182 ) n n ln( 2.8182 ) = ln(1. 3333 ) 1. 0361 n = 0. 3277 n = 0. 316 Substituting this value with the data back into the flow curve equation, we obtain the value of the strength coefficient K : (1) K = 2 4 5 /(0. 11 ) . 316 = 492.1 MPa Check: (2) K = 340 /(0. 31 ) . 316 = 4 92 . 3 MPa The flow curve equation is:  = 492 . 2  0. 316 3.6 (USCS Units) During a tensile test, a metal has a true strain = 0. 08 at a true stress = 37,000 lb/in 2 . Later, at a true stress = 55,000 lb/in 2 , true strain = 0.2 4 . Determine the strength coefficient and strain - hardening exponent in the flow curve equation. Solution : (1) 37,000 = K (0. 08 ) n and (2) 55,000 = K (0.2 4 ) n 55,000/37,000 = (0.2 4 /0. 08 ) n 1.4865 = ( 3.0 ) n n ln( 3.0) = ln(1.4865) 1 . 0986 n = 0.3964 n = 0. 3608 Substituting this value with the data back into the flow curve equation, we obtain the value of the strength coefficient K : (1) K = 37,000/(0. 08 ) 0. 3608 = 92 , 038 lb/in 2 Check: (2) K = 55,000/(0. 24 ) 0. 3608 = 92 , 041 lb/in 2 Averaging the calculated values of K , t he flow curve equation is:  = 92 , 040  0. 3608 3.7 (SI Units) A tensile test specimen begins to neck at a true strain = 0.2 2 and true stress = 257 MPa. Without knowing any more about the test, determine the strength coefficient and the strain - hardening exponent in the flow curve equation? Solution : If we assume that n =  when necking starts, then n = 0.2 2 . Using this value in the flow curve equation, we have K = 257 /(0.2 2 ) .2 2 = 358.6 MPa The flow curve equation is:  = 358.6  0.2 2 3.8 (SI Units) A tensile test provides the following flow curve parameters: strain - hardening exponent = 0. 2 5 and strength coefficient = 5 00 MPa. Determine (a) flow stress at a true strain = 1.0 and (b) true strain at a flow stress = 5 00 MPa. Solution : (a) Y f = 5 00(1.0) . 25 = 5 00 MPa

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    Solutions for Fundamentals of Modern Manufacturing , 5e 5 (b)  = ( 5 00/ 5 00) 1/. 25 = (1.0) 4.0 = 1.00 3.9 (A) (USCS Units) The flow curve for a certain metal has a strain - hardening exponent = 0.2 0 and its strength coefficient = 26 ,000 lb/in 2 . Determine (a) flow stress at a true strain = 0. 1 5 and (b) true strain at a flow stress = 2 0,000 lb/in 2 . (c) What is the likely metal in this problem? Solution : (a) Y f = 26 ,000(0. 1 5) .2 0 = 17 , 791 lb/in 2 (b)  = ( 2 0,000/ 26 ,000) 1/.2 0 = (0. 7692 ) 5 .0 = 0.26 9 (c) The flow curve parameters indicate that the metal is probably annealed pure aluminum, according to Table 3.4. 3.10 (USCS Units) A metal is plastically deformed in a tensi le test . The starting specimen had a gage length = 2.0 in and an area = 0.50 in 2 . At one point during the test, the gage length = 2. 3 in, and the corresponding engineering stress = 2 5 ,000 lb/in 2 . A t another point prior to necking, the gage length = 3. 0 in, and the corresponding engineering stress = 28,000 lb/in 2 . Determine the strength coefficient and the strain - hardening exponent for this metal. Solution : Starting volume V = L o A o = 2.0(0.5) = 1.0 in 3 (1) Area at gage length = 2.3 in: A = V / L = 1.0/2. 3 = 0.4 348 in 2 Given engineering stress = 25,000 lb/in 2 , force F = 25,000(0.5) = 12,500 lb T rue stress  = 12,500 /0.4 348 = 28 , 749 lb/in 2 and true strain  = ln(2. 3 /2.0) = 0. 1398 (2) Area at gage length = 3.0 in: A = 1.0/3. 0 = 0.3 333 in 2 Given engineering stress = 28,000 lb/in 2 , force F = 28,000(0.5) = 14,000 lb T rue stress  = 14,000 /0.3 333 = 4 2 , 0 00 lb/in 2 and true strain  = ln(3. 0 /2.0) = 0. 4055 These two data points can be used to determine the parameters of the flow curve equation. (1) 28,749 = K (0. 1398 ) n and (2) 42,000 = K (0.4 055 ) n 42,000 / 28,749 = (0.4 0 55 /0. 1398 ) n 1. 4609 = (2. 9006 ) n ln(1. 4609 ) = n ln(2. 9006 ) 0. 3791 = 1 . 0649 n n = 0. 3 56 (1) K = 28,749 /(0. 1398 ) . 356 = 57 , 915 lb/in 2 Check: (2) K = 4 2 , 0 00/(0. 4055 ) . 356 = 57,915 lb/in 2 The flow curve equation is:  = 57 , 915  0. 356 3.11 (SI Units) A tensile test specimen has a starting gage length = 75.0 mm. It is elongated during the test to a length = 110.0 mm before necking occurs. Determine (a) the engineering strain and (b) the true strain. (c) Compute and sum the engineering strains as the specimen elongates from: (1) 75.0 to 80.0 mm, (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) 100.0 to 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 strain e = (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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