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

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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.

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 parts, 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.

Solutions for Fundamentals of Modern Manufacturing, 5e
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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.

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 $750,000 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. Applicable 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) = 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 year H = 50(3)(5)(8) = 6000 hr/yr.
Ceq = (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 runs 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. Applicable 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:
Ceq = 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:
Ceq = 125,000(1.30)/(60 x 4 x 4000) = $0.169/min = $10.16/hr
The repaired machine has a cost rate Ceq = 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 20% 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.

Solutions for Fundamentals of Modern Manufacturing, 5e
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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:
Ceq = 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:
Ceq = 125,000(1.30)/(60 x 4 x 4000) = $0.169/min = $10.16/hr
The repaired machine has a cost rate Ceq = 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(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/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 14.4 pc/hr and operating 4000 hr/yr, annual output = 4000(14.4) = 57,600 pc/yr
Total revenue = 57,600(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 costs = $5.00 each. Batch quantity = 40. The actual machining time in the operation =
6.86 min. Time to load and unload each workpiece = 2.0 min. Cost of the cutting tool = $3.00, and
each tool must be changed every 10 pieces. Tool change time = 1.5 min. 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 rate
when setup time is figured in, and (c) cost per piece.
Solution: (a) Processing time To = 6.86 min, part handling time Th = 2.00 min, and tool handling
time Tt = 1.50 min/10 = 0.15 min. Tc = 6.86 + 2.00 + 0.15 = 9.01 min
(b) Average production time per piece including setup time Tp = 1.75(60)/40 + 9.01 = 11.64 min
Hourly production rate Rp = 60/11.64 = 5.16 pc/hr
(c) Equipment cost rate Ceq = $22.00/60 = $0.367/min.
Labor cost rate CL = 16.00(1.50) = $24.00/hr = $0.40/min
Cost of tooling Ct = 3.00/10 = $0.30/pc
Finally, cost per piece Cpc = 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. Labor rate of the worker = $18.00/hr including overhead. However, the worker also

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Solutions for Fundamentals of Modern Manufacturing, 5e
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tends other machines and only spends 20% of his time on this machine. Setup can be ignored
because of the long production run. The molding machine was purchased for $500,000 installed, its
anticipated 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 the
machine, (b) annual quantity of product molded, and (c) cost per piece.
Solution: (a) With a cycle time Tc = 45 sec = 0.75 min, Rp = 60/0.75 = 80 pc/hr. Factoring in the
98% proportion uptime, Rp = 0.98(80) = 78.4 pc/hr
Annual quantity of product = 6000(78.4) = 470,400 pc/yr
(b) Equipment cost rate Ceq = 500,000(1.30)/(60 x 10 x 6000) = $0.1806/min.
Mold cost per piece Ct = 100,000/1,000,000 = $0.10/pc
Labor cost rate CL = 18.00(0.20) = $3.60/hr = $0.06/min
Finally, cost per piece Cpc = 1.20(0.88) + (0.06 + 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 applicable labor overhead rate = 42%. Cost rate of the press =
$22.50/hr and applicable 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, the starting blanks of sheet
metal are manually loaded into the press, which takes 42 sec. The actual press stroke takes only 8
sec. Cost of the starting blanks = $0.43/pc. The press operates 250 days per year, 7.5 hours per day,
but the operator is paid for 8 hours 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 per stamping produced.
Solution: (a) Cycle time Tc = 42 + 8 = 50 sec = 0.833 min
(b) Including setup time, Tp = 75/400 + 0.833 = 1.021 min
Rp = 60/1.021 = 58.78 pc/hr
Excluding setup time, Rc = 60/0.833 = 72.03 pc/hr
(c) Equipment cost rate Ceq = 22.50(1.20)/60 = $0.45/min.
Die cost per piece Ct = 40,000/200,000 = $0.20/pc
Labor cost rate CL = 15.00(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. CL = 0.355(8/7.5) = $0.379
Finally, cost per stamping Cpc = 0.43 + (0.379 + 0.45)(1.021) + 0.20 = $1.48/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 =
$27/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 operation runs 250 days per year, eight hours per day. Assume availability = 100% and scrap
rate = 0. Determine the cost per part produced (a) without the parts 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 Ceq = 18/60 = $0.30/min

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Solutions for Fundamentals of Modern Manufacturing, 5e
1-7
Labor cost rate CL = 27/60 = $0.45/min
Without the loading device, Cpc = 0.15 + (0.45 + 0.30)(24 + 6 + 3)/60 = $0.563/pc
(b) Cost rate of the device = 36,000/(60 x 4 x 2000) = $0.075/min = $4.50/hr
With the loading device, Cpc = 0.15 + (0.45/4 + 0.30 + 0.075)(24 + 3 + 3)/60 = $0.394/pc
(c) Without the device, Tc = 24 + 6 + 3 = 33 sec = 0.55 min and Cpc = 0.563/pc from part (a)
Rp = Rc = 60/0.55 = 109.1 pc/hr = 872 pc/day
With the device, Tc = 24 + 3 + 3 = 33 sec = 0.50 min and Cpc = 0.394/pc from part (b)
At 100% reliability and no setup time, Rp = Rc = 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
490.9D = 36,000 + 378.2D
(490.9 – 378.2)D = 112.7D = 36,000 D = 319.5 round to 320 days
1.8 (A) In a long-running high-production operation, the starting work part cost = $0.95 each, and
cycle time = 1.25 min. Equipment cost rate = $28.00/hr, and labor cost rate = $21.00/hr. Both rates
include overhead 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) = 46.08 pc/hr
However, because of the 5% scrap rate, the production rate of acceptable parts is
Rp = 46.08(1-0.07) = 42.85 pc/hr
(b) Factoring in availability and scrap rate, part cost is
Cpc = 0.95/0.93 + ((21 + 28)/60)(1.25/(0.93 x 0.96)) + 0.05 = $2.215/pc
1.9 The starting work part costs $2.25 in a batch production operation. Batch quantity = 100 parts.
Each cycle, part handling time = 0.50 min, and operation time = 1.67 min. Setup time = 50 min.
Equipment cost rate = $35.00/hr, and labor cost rate = $18.00/hr, including overhead costs. There is
no tool change or tool cost in the operation. The 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) Tc = 1.67 + 0.50 = 2.17 min/pc
Given q = 5%, the starting quantity of parts Qo = 100/0.95 = 105.3 rounded up to 106 pc
Determine batch time, including setup time.
Tb = 50 + 106(2.17) = 50 + 230.02 = 280.02 min/batch = 4.67 hr
Average production rate of parts Rp = 100/4.67 = 21.43 pc/hr
Average production rate of acceptable parts Rp = 21.43(1 - 0.05) = 20.36 pc/hr
(b) Now determine batch cost, including setup time.
Cb = 106(2.25) + ((18 + 35)(4.67) = 238.50 + 247.51 = $486.01/batch
Cpc = 486.01/100 = $4.86/pc
Alternative calculation of Cpc:
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/pc
The difference from the previous value of $4.86/pc is due mostly to the roundup of Qo.
(c) The time to complete the batch was computed in part (a) as Tb = 280 min = 4.67 hr
1.10 In a batch-production machining operation, the starting work part is a casting that costs $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 = $25.00/hr,
and labor cost rate = $20.00/hr. Both rates include overhead costs. The cutting tool 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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1-8
the machine tool = 98%, 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) Tc = 3.44 + 2.5 + 3/15 = 6.14 min/pc
Given q = 4%, the starting quantity of parts Qo = 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.
Tb = 75 + 68(6.14)/0.98 = 75 + 426.04 = 501.04 min/batch = 8.35 hr
Average production rate of parts Rp = 65/8.35 = 7.784 pc/hr
Average production rate of acceptable parts Rp = 7.784(1 - 0.04) = 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.
Cbatch = 68(3.50) + ((20 + 25)(8.35) + 4(5.75) = 238.00 + 375.75 + 23.00 = $636.75/batch
Cpc = 636.75/65 = $9.796/pc
Alternative calculation of Cpc:
Cpc = 3.50/0.96 + ((20 + 25)/60)(75/65) + ((20 + 25)/60)(6.14/(0.98 x 0.96)) + 23.00/65
Cpc = 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 Qo.
(c) The time to complete the batch was computed in part (a) as Tb = 501.04 min = 8.35 hr
1.11 (A) During a particular 40-hour week of an automated production operation, 336 acceptable (non-
defective) parts and 22 defective parts were produced. The operation cycle consists of a processing
time of 5.73 min, and a part handling time of 0.38 min. Every 60 parts, a tool change is performed,
and this takes 7.2 min. The 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 machine during this week.
Solution: (a) Production rate of acceptable parts Rp = 335/40 = 8.40 pc/hr
(b) Total parts processed during the week Qo = 336 + 22 = 358 pc
Scrap rate q = 22/358 = 0.0615 = 6.15%
(c) Cycle time of the unit operation Tc = 5.73 + 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 2.14 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 Tc = 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 Rp = 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, 5e
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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, 5e
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 = 100
mm2. The specimen yields at 48,000 N, and the 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 mm2, determine the percent reduction in area.
Solution: (a) Y = 48,000/100 = 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 = 480/0.0026 = 184.6 x 103 MPa
(c) TS = 87,000/100 = 870 MPa
(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) A tensile test specimen has a gage length = 2.0 in and diameter = 0.798 in. Yielding
occurs at a load of 31,000 lb. The corresponding gage length = 2.0083 in, which 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, determine the percent elongation. (e) If the specimen necked to an area = 0.36 in2, determine
the percent reduction in area.
Solution: (a) Area A = π(0.798)2/4 = 0.50 in2. Yield strength Y = 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 elastic
strain to be used in calculating modulus of elasticity.
E = s/e = 62,000/0.00215 = 28.84 x 106 lb/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-sectional
area = 150 mm2. During testing, the following force and gage length data are collected: (1) 17,790 N
at 100.2 mm, (2) 23,040 N at 103.5 mm, (3) 27,370 N at 110.5 mm, (4) 28,910 N at 122.0 mm, (5)
27,400 N at 130.0 mm, and (6) 20,460 N at 135.1 mm. The final data point (6) occurred immediately
prior to failure. Yielding occurred at a load of 19,390 N (0.2% offset value), and the maximum load
(4) was 28,960 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 first data point (1) occurred prior to yielding. Modulus of elasticity E = s/e
Engineering strain e = (100.2 - 100)/100 = 0.002; engineering stress = 17,790/150 = 118.6 MPa
E = 118.6/0.002 = 59,300 MPa
(d) Tensile strength = maximum load divided by original area: TS = 28,960/150 = 193.1 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 = LoAo = 100(150) = 15,000 mm3.
We must select two data points that occur after yielding but before necking. Necking occurs after
the maximum load is reached. The two points selected are (2) and (4): (2) F = 23,040 N and L =
103.5 mm; and (4) F = 28,910 N and L = 122.0 mm.
(2) A = V/L = 15,000/103.5 = 144.93 mm2
True stress
 = 23,040/144.93 = 159.0 MPa. True strain
 = ln(103.5/100) = 0.0344
(4) A = 15,000/122.0 = 122.95 mm2
True stress
 = 28,910/122.95 = 235.1 MPa. True strain
 = ln(122.0/100) = 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) 159.0 = K(0.0344)n and (4) 235.1 = K(0.1989)n
235.1/159.0 = (0.1989/0.0344)n
1.479 = (5.782)n
ln(1.479) = 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 = 159.0/(0.0344).223 = 337.1 MPa
K = 235.1/(0.1989).223 = 337.2 MPa
The flow curve equation is
 = 337.15 0.223
3.5 (SI Units) In a tensile test on a steel specimen, true strain = 0.11 at a stress of 245 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) 245 = K(0.11)n and (2) 340 = K(0.31)n
340/245 = (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 = 245/(0.11).316 = 492.1 MPa
Check: (2) K = 340/(0.31).316 = 492.3 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/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)n and (2) 55,000 = K(0.24)n
55,000/37,000 = (0.24/0.08)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/in2
Check: (2) K = 55,000/(0.24)0.3608 = 92,041 lb/in2
Averaging the calculated values of K, the flow curve equation is:
 = 92,040
0.3608
3.7 (SI Units) A tensile test specimen begins to neck at a true strain = 0.22 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 = 0.22.
Using this value in the flow curve equation, we have K = 257/(0.22).22 = 358.6 MPa
The flow curve equation is:
 = 358.6
0.22
3.8 (SI Units) A tensile test provides the following flow curve parameters: strain-hardening exponent =
0.25 and strength coefficient = 500 MPa. Determine (a) flow stress at a true strain = 1.0 and (b) true
strain at a flow stress = 500 MPa.
Solution: (a) Yf = 500(1.0).25 = 500 MPa

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Solutions for Fundamentals of Modern Manufacturing, 5e
5
(b)
 = (500/500)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 = 0.20 and its
strength coefficient = 26,000 lb/in2. Determine (a) flow stress at a true strain = 0.15 and (b) true strain
at a flow stress = 20,000 lb/in2. (c) What is the likely metal in this problem?
Solution: (a) Yf = 26,000(0.15).20 = 17,791 lb/in2
(b)
 = (20,000/26,000)1/.20 = (0.7692)5.0 = 0.269
(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 tensile test. The starting specimen had a gage
length = 2.0 in and an area = 0.50 in2. At one point during the test, the gage length = 2.3 in, and the
corresponding engineering stress = 25,000 lb/in2. At another point prior to necking, the gage length =
3.0 in, and the corresponding engineering stress = 28,000 lb/in2. Determine the strength coefficient
and the strain-hardening exponent for this metal.
Solution: Starting volume V = LoAo = 2.0(0.5) = 1.0 in3
(1) Area at gage length = 2.3 in: A = V/L = 1.0/2.3 = 0.4348 in2
Given engineering stress = 25,000 lb/in2, force F = 25,000(0.5) = 12,500 lb
True stress
 = 12,500/0.4348 = 28,749 lb/in2 and true strain
 = ln(2.3/2.0) = 0.1398
(2) Area at gage length = 3.0 in: A = 1.0/3.0 = 0.3333 in2
Given engineering stress = 28,000 lb/in2, force F = 28,000(0.5) = 14,000 lb
True stress
 = 14,000/0.3333 = 42,000 lb/in2 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(0.4055)n
42,000/28,749 = (0.4055/0.1398)n
1.4609 = (2.9006)n
ln(1.4609) = n ln(2.9006)
0.3791 = 1.0649 n n = 0.356
(1) K = 28,749/(0.1398).356 = 57,915 lb/in2
Check: (2) K = 42,000/(0.4055).356 = 57,915 lb/in2
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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