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

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Solution Manual For Fundamentals of Modern Manufacturing: Materials, Processes, and Systems, 7th Edition ensures effective exam preparation with structured textbook notes.

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

Answer. (c).
Manufacturing Economics
1.10 The total time to complete a batch of parts is calculated using which three of the following
terms: (a) actual processing time in the operation, (b) batch quantity, (c) cycle time, (d)
inspection time, (e) setup time, (f) tool change time, and (g) work part handling time?
Answer. (b), (c), and (e).
1.11 The time to load a starting work part into the production machine in a certain unit
production operation = 0.40 min, the actual processing time in the operation = 3.10 min,
and the time to unload the finished part = 0.20 min. In addition, the tool must be changed
every 5 cycles and the tool change time = 1.50 min. The cycle time for the unit operation is
which one of the following: (a) 3.5 min, (b) 3.7 min, (c) 4.0 min, (d) 4.5 min or (e) 5.2
min?
Answer. (c).
1.12 If the initial cost of a production machine including installation is $250,000, and the
planned use of the machine is 2000 hr/yr for 5 years, and the applicable overhead rate on
the equipment is 50%, which one of the following is the corresponding cost rate of this
machine in $/hr: (a) $0.625, (b) $25.00, (c) $37.50, (d) $50.00, or (e) $125.00?
Answer. (c).
Additional Review Questions
1.1 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.
1.2 Annual production quantities made by a factory can be classified into three ranges. Name
the three ranges and the approximate quantities of parts associated with each range.
Answer. The three ranges are (1) low production, with quantities in the range 1 to 100
parts; (2) medium production, with quantities in the range 100 to 10,000 parts; and (3) high
production, with quantities in the range 10,000 to over a million parts.
1.3 Define plant capacity.
Answer. Plant capacity, a.k.a. production capacity, is the maximum rate of production
output that a plant can achieve under assumed operating conditions. Operating conditions
refer to the number of shifts per week, hours per shift, direct labor manning levels in the
plant, and so on.
1.4 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.
1.5 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.
1.6 What are the three types of surface processing operations?
Answer. (1) Cleaning, (2) surface treatments, e.g., sand blasting, and (3) coating and thin
film deposition, e.g., electroplating, painting, physical vapor deposition.
1.7 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.
1.8 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 and 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 operations.
1.9 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.

Additional Problems
1.1 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 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 6,000 hours per year. Equipment overhead rate = 30%. 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 = 6,000(78.4) = 470,400 pc/yr
(b) Equipment cost rate Ceq = 500,000(1.30)/(60  10  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
1.2 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
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4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
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
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(0.394)D
490.9D = 36,000 + 378.2D
(490.9 – 378.2)D = 112.7D = 36,000 D = 319.5 round to 320 days
1.3 In a sequential batch-processing operation, the starting work part costs $4.50 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 the machine tool = 98%, and the scrap rate = 0. Determine
(a) production rate and (b) finished part cost. (c) How many hours are required to complete
the batch?
Solution: (a) Tc = 3.44 + 2.5 + 3/18 = 6.11 min/pc
Given q = 0, the starting quantity of parts Qo = 65 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 + 65(6.11)/0.98 = 75 + 405.26 = 480.26 min/batch = 8.004 hr
Average production rate of parts Rp = 65/8.004 = 8.121 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 = 65(4.50) + ((20 + 25)(8.004) + 4(5.75) = 292.50 + 360.18 + 23.00 = $675.68/batch
Cpc = 675.68/65 = $10.39/pc
Alternative calculation of Cpc:
Cpc = 4.50 + ((20 + 25)/60)(75/65) + ((20 + 25)/60)(6.11/0.98) + 23.00/65
Cpc = 4.50 +0.865 + 4.676 + 0.354 = $10.395/pc (Close enough!)
(c) Time to complete the batch was computed in part (a) as Tb = 480.26 min = 8.004 hr
1.4 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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Additional Multiple-Choice Questions
Atomic Structure and the Elements
2.1 The element with the lowest density and smallest atomic weight is which one of the
following: (a) aluminum, (b) argon, (c) helium, (d) hydrogen, or (e) magnesium?
Answer. (d).
2.2 In the planetary model of atomic structure, the electrons in the outermost shell are called
which one of the following: (a) atomic electrons, (b) orbital electrons, (c) quantum
electrons, or (d) valence electrons?
Answer. (d).
Bonding between Atoms and Molecules
2.3 Which two of the following bond types are called intramolecular bonds: (a) covalent
bonding, (b) electronic bonding, (c) hydrogen bonding, (d) ionic bonding, (e) metallic
bonding, and (f) van der Waals forces?
Answer. (a) and (d).
Crystalline Structures
2.4 How many atoms are there in the hexagonal close-packed (HCP) unit cell: (a) 8, (b) 9, (c)
10, (d) 12, (e) 14, or (f) 17?
Answer. (f).
2.5 Which one of the following metals has a body-centered cubic crystal structure at room
temperature: (a) aluminum, (b) copper, (c) iron, (d) nickel, and (e) zinc?
Answer. (c).
2.6 Twinning can be described as which three of the following: (a) elastic deformation, (b) more
likely at high deformation rates, (c) more likely in metals with HCP structure, (d) plastic
deformation, (e) slip mechanism, and (f) type of dislocation?
2.7 Answer. (b), (c), and (d).
2.8 Grain boundaries in metals interfere with dislocation movement, which contributes to which
one of the following properties that are characteristic of metals: (a) crystal structure, (b)
ductility, (c) electrical conductivity, (d) strain hardening, or (e) thermal conductivity?
Answer. (d).
Noncrystalline (Amorphous) Structures
2.9 As an amorphous material such as glass cools from the molten (liquid) state, it transforms
into the solid state gradually, first going through a transition phase which is called which one
of the following: (a) mushy zone, (b) passing phase, (c) supercooled liquid, (d) superheated
solid, or (e) transformation phase?
Answer. (c).
Engineering Materials

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2.10 Polymers are characterized by which two of the following bonding types: (a) adhesive, (b)
covalent, (c) hydrogen, (d) ionic, (e) metallic, and (f) van der Waals?
Answer. (b) and (f).
2.11 In general, ceramics are characterized by which four of the following properties: (a)
brittleness, (b) chemical inertness, (c) electrically insulating, (d) high electrical conductivity,
(e) high hardness, (f) high thermal conductivity, and (g) low density?
Answer. (a), (b), (c), and (e).
Additional Review Questions
2.1 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.
2.2 Among the common point defects in a crystal lattice structure, what is a Frenkel defect?
Answer. A Frenkel defect in a crystal lattice structure is an ion that is removed from a
regular position in the lattice and inserted into an interstitial position not normally occupied
by such an ion.
2.3 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.

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

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3.10 The shear modulus of elasticity is approximately what percent of Young’s modulus: (a) 10%,
(b) 25%, (c) 40%, (d) 75%, or (e) 100%?
Answer. (c).
Hardness
3.11 Which two of the following hardness tests use a pyramid-shaped indenter: (a) Brinell, (b)
Knoop, (c) Rockwell, (d) scleroscope, and (e) Vickers?
Answer. (b) and (e).
3.12 Which one of the following materials has the highest hardness: (a) alumina ceramic, (b) gray
cast iron, (c) hardened tool steel, (d) high carbon steel, or (e) polystyrene?
Answer. (a).
Effect of Temperature on Properties
3.13 Hot working involves metal forming processes carried out above the recrystallization
temperature: (a) True or (b) false?
Answer. (a).
Additional Review Questions
3.1 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.
3.2 State the flow curve equation, also known as the Holloman-Ludwig equation.
Answer. The flow curve equation is
 = K
n, where
 = true stress,
 = true strain, K =
strength coefficient, and n = strain hardening exponent.
3.3 Which strain always has the greater value, engineering strain or true strain?
Answer. Engineering strain is always greater than true strain.
3.4 A stress-strain relationship that exhibits perfectly elastic behavior is characteristic of what
types of materials?
Answer. Brittle materials such as ceramics, many cast irons, and thermosetting plastics
exhibit perfectly elastic behavior.
3.5 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.
3.6 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. The recrystallization

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temperature is about 0.5 Tm, where Tm is the melting temperature of the metal measured on
an absolute scale (R or K).
3.7 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.
3.8 How does the viscosity of a thermoplastic polymer melt differ from the viscosity of a
Newtonian fluid?
Answer. The viscosity of a thermoplastic polymer melt decreases as flow rate increases,
whereas the viscosity of a Newtonian fluid does not.
Additional Problems
3.1 (USCS Units) A tensile test specimen has a gage length = 2.0 in and diameter = 0.798 in.
Yielding occurs at a load of 31,000 lb. The corresponding gage length = 2.0083 in, which 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 calculate modulus of elasticity.
E = s/e = 62,000/0.00215 = 28.84  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%
3.2 (SI Units) In a tensile test on a steel specimen, true strain = 0.12 at a stress of 250 MPa.
When true stress = 350 MPa, true strain = 0.26. Determine the strength coefficient and the
strain-hardening exponent in the flow curve equation.
Solution: Two equations: (1) 250 = K(0.12)n and (2) 350 = K(0.26)n
350/250 = (0.26/0.12)n 1.4 = (2.1667)n
n ln(2.1667) = ln(1.4) 0.7732 n = 0.3365 n = 0.435
Substituting this value with the data back into the flow curve equation to obtain the value
of the strength coefficient K:
(1) K = 250/(0.12).435 = 629 MPa
Check: (2) K = 350/(0.26).435 = 629 MPa
The flow curve equation is:
 = 629
0.435
3.3 (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

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necking, the gage length = 3.0 in, and the corresponding engineering stress = 28,000 lb/in2.
Determine the strength coefficient and the strain-hardening exponent for this metal.
Solution: Starting volume V = LoAo = 2.0(0.5) = 1.0 in3
Area at gage length (1) = 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
Area at gage length (2) = 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
3.4 Prove Equation (3.8): Show that true strain
 = ln(1 + e), where e = engineering strain.
Solution: Starting definitions: (1)
 = ln(L/Lo) and (2) e = (L  Lo)/Lo
Consider definition (2): e = L/Lo  Lo/Lo = L/Lo  1
Rearranging, 1 + e = L/Lo
Substituting this into definition (1),
 = ln(1 + e)
3.5 (SI Units) The following flow curve parameters were obtained for a metal alloy in a tensile
test: strength coefficient = 550 MPa and strain-hardening exponent = 0.25. The same metal is
now tested in a compression test in which the starting height of the specimen = 50 mm and
its diameter = 30 mm. Assuming that the cross section increases uniformly, determine the
load required to compress the specimen to a height of (a) 45 mm and (b) 35 mm.
Solution: The starting volume of test specimen V = hDo2/4 = 50(30)2/4 = 35343 mm3
(a) At h = 45 mm,
 = ln(50/45) = ln(1.1111) = 0.1054
Yf = 550(0.1054).25 = 313.4 MPa
A = V/L = 35343/45 = 785.4 mm2
F = YfA =313.4(785.4) = 246,144 N
(b) At h = 35 mm,
 = ln(50/35) = ln(1.4286) = 0.3567
Yf = 550(0.3567).25 = 425.0 MPa
A = V/L = 35343/35 = 1009.8 mm2
F = YfA = 425.0(1009.8) = 429,165 N
3.6 (USCS Units) A ceramic specimen is tested in a bending test. Its width = 0.50 in and
thickness = 0.25 in. The length of the specimen between supports = 3.0 in. Determine the
transverse rupture strength if failure occurs at a load = 1250 lb.
Solution: TRS = 1.5FL/bt2 = 1.5(1250)(3.0)/(0.5  0.252) = 180,000 lb/in2

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3.7 (USCS Units) A batch of annealed steel has just been received from the vendor. It is
supposed to have a tensile strength in the range 60,000 to 70,000 lb/in2. A Brinell hardness
test in the receiving department yields a value of HB = 118. (a) Does the steel meet the
specification on tensile strength? (b) Estimate the yield strength of the material.
Solution: (a) TS = 500(HB) = 500(118) = 59,000 lb/in2. This lies outside the specified
range of 60,000 to 70,000 lb/in2. However, from a legal standpoint, it is unlikely that the
batch can be rejected on the basis of its measured Brinell hardness number without using an
actual tensile test to measure TS. The formula for converting from Brinell hardness number
to tensile strength is only an approximating equation.
(b) Based on Table 3.2 in the text, the ratio of Y to TS for low carbon steel = 25,000/45,000
= 0.555.
Using this ratio, estimate the yield strength to be Y = 0.555(59,000) = 32,700 lb/in2
3.8 (SI Units) A 125.0-mm-diameter shaft rotates inside a stationary bushing whose inside
diameter = 125.6 mm and length = 50.0 mm. In the clearance between the shaft and the
bushing is a lubricating oil whose viscosity = 0.14 Pa-s. The shaft rotates at a velocity of 400
rev/min; this speed and the action of the oil are sufficient to keep the shaft centered inside the
bushing. Determine the magnitude of the torque due to viscosity that acts to resist the
rotation of the shaft.
Solution: Bushing internal bearing area A = (125.6)2  50/4 = 19729.6 mm2 =
19729.2(10-6) m2
d = (125.6  125)/2 = 0.3 mm
v = (125 mm/rev)(400 rev/min)(1 min/60 sec) = 2618.0 mm/s
Shear rate = 2618/0.3 = 8726.6 s-1
 = (0.14)(8726.6) = 1221.7 Pa = 1221.7 N/mm2
Force on surface between shaft and bushing = (1221.7 N/mm2)(19729.2(10-6)) = 24.1 N
Torque T = 24.1 N  125/2 mm = 1506.4 N-mm = 1.506 N-m

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Additional Multiple-Choice Questions
Volumetric and Melting Properties
4.1 Which one of the following metals has the highest density: (a) copper, (b) iron, (c) lead, (d)
nickel, (e) tungsten, or (f) zinc?
Answer. (e).
4.2 The strength-to-weight ratio of a material is its yield strength divided by its density: (a) True
or (b) false?
Answer. (b). The ratio is based on tensile strength, not yield strength.
Thermal Properties
4.3 The thermal diffusivity of a material is defined as which one of the following: (a) mass
diffusion divided by specific heat, (b) specific heat divided by mass diffusion, (c) specific
heat multiplied by mass diffusion, (d) thermal conductivity divided by specific heat, or (e)
thermal conductivity divided by volumetric specific heat?
Answer. (e).
Electrical Properties
4.4 Copper is generally considered easy to weld because of its high thermal conductivity: (a)
True or (b) false?
Answer. (b). The high thermal conductivity of copper makes it difficult to weld because the
heat flows away from the joint rather than being concentrated to permit melting of the metal.
4.5 Which of the following pure metals is the best conductor of electricity: (a) aluminum, (b)
copper, (c) gold, or (d) silver?
Answer. (d). See Table 4.3.
4.6 Which one of the following materials is classified as a semiconductor: (a) carbon, (b) natural
rubber, (c) polyethylene, or (d) silicon?
Answer. (d).
Electrochemical Processes
4.7 In electrochemical machining, the work part in the electrolytic cell is the (a) anode or (b)
cathode?
Answer. (a).
Additional Review Questions
4.1 Define the coefficient of thermal expansion.
Answer. The coefficient of thermal expansion measures the change in length per degree of
temperature.
4.2 What is the difference in melting characteristics between a pure metal element and a metal
alloy?

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Answer. A pure metal element melts at one temperature (the melting point), while most
metal alloys begin melting at a certain temperature called the solidus and finally completes
the transformation to the molten state at a higher temperature called the liquidus. Between
the solidus and liquidus, the alloy is a mixture of solid and liquid.
4.3 Define specific heat as a material property.
Answer. Specific heat is defined as the quantity of heat required to raise the temperature of a
unit mass of the material by one degree.
4.4 What are the important variables that affect mass diffusion?
Answer. According to Fick's first law, mass diffusion depends on the diffusion coefficient of
the material, which increases rapidly with temperature (so temperature could be listed as an
important variable), concentration gradient, contact area, and time.
4.5 Define electrical resistivity as a material property.
Answer. Resistivity r is the material's capacity to resist the flow of electric current. In a
uniform section of material, r = RA/L, where R = electrical resistance, A = cross-sectional
area of the section, and L = its length.
4.6 What is dielectric strength as a material property?
Answer. Dielectric strength is defined as the electrical potential required to break down the
insulator per unit thickness.
4.7 In an electrolytic cell, what are the names of the positive and negative electrodes?
Answer. The positive electrode is called the anode, and the negative electrode is called the
cathode.
Additional Problems
4.1 (SI units) Determine the length and width of a rectangular nickel plate whose room
temperature (20C) dimensions are 300 mm by 400 mm by 5 mm, if the plate is heated to
250C. The coefficient of thermal expansion
 = 13.3(10-6)/C for nickel.
Solution: Equation (4.1) for length: L2 – L1 = αL1(T2 – T1)
L2 = 300 + 13.3(10-6)(300)(250 – 20) = 300 + 0.918 = 300.918 mm
Equation (4.1) for width: W2 – W1 = αW1(T2 – T1)
W2 = 400 + 13.3(10-6)(400)(250 – 20) = 400 + 1.224 = 401.224 mm
4.2 (USCS units) A zinc die casting has a critical dimension of 10.00 in when it solidifies in the
mold at its melting temperature because 10.00 in is the corresponding mold dimension. What
is the value of this dimension when it cools to room temperature (70°F)? : The melting
temperature of zinc = 787°F, and its coefficient of thermal expansion = 22.2(10-6)/°F.
Solution L2 – L1 =
L1 (T2 – T2)
L2 = 10.00 + 22.2(10-6)(10.00)(70 – 787) = 10.00 – 0.159 = 9.841 in
4.3 (SI units) A steel alloy has a melting point of 1450C. Its specific heat = 0.46 J/gC, and its
heat of fusion = 270 J/g. For a 100-kg block of this steel, determine how much heat is
required to (a) raise its temperature from 25C to its melting point and (b) transform it from
solid to liquid phase.

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

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