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Chapter 11: Materials Engineering
The figures below depict the situations described in Exercises 1 - 6.
Diameter 0.20 mm
L
Cube
Mass M
Figure A)
Plate
1.0
0m
Mass M
1.00 m
Figure B)
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11-1) A mass of M = 1.0 kg is hung from a circular wire of diameter 0.20 mm as shown in the Figure
A) above. What is the stress in the wire?
Need: Stress in wire, σ = ____ GPa.
Know: Stress = force/cross sectional area; g = 9.81 m/s2.
How: Calculate stress from force on wire; divide by area.
Solve: Force on wire = Mg = 1 × 9.81 [kg][m/s2] = 9.81 N
Cross sectional area = πd2/4 = π × 0.000202/4 = 3.14 × 10-8 m2.
∴Stress, σ = 9.81/ 3.14 × 10-8 [N][1/m2] = 3.1 × 108 N/m2 = 0.31 GPa.
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11-2) If the wire shown in Figure A) is stretched from 1.00 m to 1.01 m in length, what is the strain of
the wire?
Need: Strain in wire, ε ___ [0].
Know: Strain is the extension per unit length of wire.
How: Divide extension by the total wire length.
Solve: Strain, ε = 0.010/1.00 = 0.010 or 1.0%. (Notice we used the nominal length of
the wire, 1.00 m – had we used 1.01 m, the result would have been unchanged to our 2
significant figures)
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11-3) In Figure B), when block of metal 1.00 m on a side is placed on a metal plate 1.00 m on a side,
as above, the stress on the plate is 1.00 × 103 N/m2. What is the mass of the metal cube?
Need: Mass of metal cube = ____ kg.
Know: Stress = force/area; g = 9.81 m/s2. σ = 1.00 × 103 N/m2 and A = 1.00 m2.
How: Determine force, then mass.
Solve: σ = 1.00 × 103 N/m2 and A = 1.00 m2, hence force on plate is F = σA = 1.00 ×
103 × 1.00 [N/m2][m2] = 1.00 × 103 N. The corresponding mass is F/g = 1.00 ×
103/9.81 [kg m/s2][s2/m] = 102 kg.
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11-4) Suppose the plate described in exercise 3 and the above figure was 0.011 m thick before the
cube was placed on it. Suppose that placing the cube on it causes a compressive strain of -0.015. Then
how thick will the plate be after the cube is placed on it?
Need: Thickness of plate, T ___ m.
Know - How: Strain, ε = ΔT /T= -0.015 compression/thickness.
Solve: ΔT = -0.015 × 0.011 = -1.65 × 10-4 m. ∴Final thickness of plate = 0.011 - 1.65
× 10-4 m = 1.084 × 10-2 = 0.011 or unchanged to 3 significant figures.
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11-5) Suppose the wire Figure A) is perfectly elastic. When subjected to a stress of 1.00 × 104 Pa, it
shows a strain of 1.00 × 10-5. What is the elastic modulus (i.e., Young’s modulus) of the wire?
Need: Modulus of wire, E = ___ GPa..
Know: Stress, σ = 1.00 × 104 Pa (N/m2) and strain, ε = 1.00 × 10-5.
How: Hooke’s law, σ = Eε
Solve: E = σ/ε = 1.00 × 104/1.00 × 10-5 [Pa]/[0] = 1.00 × 109 = 1.00 GPa.
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11-6) A plate of elastic modulus 1.00 GPa is subjected to a compressive stress of –1.00 × 103 Pa as in
Figure B. What is the strain on the plate?
Need: Plate strain in compression, ε = ___ [0].
Know: Modulus, E = 1.00 GPa = 1.00 × 109 Pa and a compressive stress, σ = -1.00 ×
103.
How: Hooke’s law, σ = Eε
Solve: ε = σ/E = - 1.00 × 103/1.00 × 109 [Pa]/[Pa] = -1.00 × 10-6.
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For Exercises 7- 9, assume that a silicone rubber 3 has the stress-strain diagram shown below.
3. Silicone rubbers are very flexible; their structure consists of polysiloxanes of formula –Si-O- in which the Si atom also
has two CH3 chemical groups per atom, which are nominally orthogonal to the main chain. Incidentally you must
distinguish silicon (a brittle element used in electronics) from silicone, the soft rubber described here.
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11-7) What is the yield strength under compression of the silicone?
Need: Compression yield strength of sample, σ = ____ MPa.
Know - How: Stress-strain relationship known above.
Solve: Reading the graph, the polymer is perfectly elastic, perfectly plastic. Its
compressive yield strength is about -80 MPa
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11-8) A flat saucer made of the above polymer has an initial thickness of 0.0050 m. A ceramic
coffee cup of diameter 0.10 m and mass 0.15 kg is placed on a plate made of the same polymer as
indicated above. What is the final thickness of the plate beneath the cup, assuming that the force of
the cup acts directly downward, and is not spread horizontally by the saucer? Comment on your
answer’s plausibility.
Need: Compressive strain on saucer made of polymer = ___ [0].
Know: Stress-strain relationship known above. Initial thickness of saucer = 0.0050 m.
Contact area from diameter of 0.10 m. Mass = 0.15 kg.
How: 1) Hooke’s Law, ε = σ/E. 2) Given that the applied stress on the saucer acts over
its footprint, σ = Mg/ACup. We’ll need E from the slope of the σ, ε stress-strain
compressive curve.
Solve: Reading the graph, E = -80/-0.4 [MPa]/[0] = 2.0 × 102 MPa (to reading
accuracy)
Further, area of contact of cup = πd2/4 = 7.85 × 10-3 m2; hence σ = force/area = -0.15 ×
9.81/ 7.85 × 10-3 [kg][m/s2][1/m2] = -188 Pa (- since compressive.)
The resulting strain is then ε = σ/E = -188/(2.0 × 102 × 106) [Pa][1/MPa][Mpa/Pa] = 9.4 × 10-7 [0] (dimensionless).
Since the original saucer thickness is 0.0050 m, its compression is εT =
- 9.4 × 10-7 × 0.0050 = - 4.7 × 10-9 m or 4.7 nm. The resulting thickness of the silicone
‘saucer’ under the cup is effectively unchanged (0.0050 - 4.7 × 10-9 = 0.0050 m).
Note that 4.7 nm is in the nano range and a continuum stress-strain model no longer
applies.
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11-9) What is the maximum number of such coffee cups that can be stacked vertically on the saucer
and not cause a permanent dent in the plate?
Need: Maximum load on saucer that does not exceed compressive elastic yield
strength of the silicone = ___ cups.
Know - How: The compressive yield strength (Exercise 7) is -80 MPa; Contact area is
7.85 × 10-3 m2 (Exercise 8) and each cup weighs 0.15 × 9.81= 1.47 N.
Solve: Its compressive yield strength is about σ = F/A = -80 MPa
Hence F = -80 × 7.85 × 10-3 [MPa][m2] = - 0.628 MN = - 6.28 × 105 N.
Number of cups = 6.28 × 105/1.47 [N][cup/N] = 4.3 × 105 cups (a tough balancing act
to follow!)
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11-10) A coat hanger like the one in the illustration below is made from polyvinyl chloride 4. The
“neck” of the coat hanger is 0.01 m in diameter and initially 0.10 m long. A coat hung on the coat
hanger causes the length of the neck to increase by 1.00 × 10-5 m. What is the mass of the coat? The
stress strain diagram is also given below.
Need: Mass of coat = ____ kg.
Know: Stretch of neck = 1.00 × 10-5 m. Nominal length and diameter of neck = 0.10 m
and 0.010 m respectively.
How: The curve above gives Young’s modulus and its yield strength. Then σ = εE and
F = σA.
Solve: Young’s modulus = 48/0.03 [MPa][0] = 1.6 GPa (from slope of graph to
reading accuracy.)
Also ε = 1.00 × 10-5/0.10 [m][1/m] = 1.00 × 10-4 [0].
∴σ = εE = 1.00 × 10-4 × 1.6 × 103 [0][GPa][MPa/GPa] = 0.16 MPa.
Supporting neck area = πD2/4 = 3.14 × 0.0102/4 = 7.85 × 10-5 m2.
∴Force = σA = 0.16 × 106 × 7.85 × 10-5 [MN/m2][N/MN][m2] = 13 N.
Hence mass of coat = 13/9.81 [kg m/s2][s2/m] = 1.3 kg.
4. PVC is a very inexpensive polymer of the monomer, CH2=CHCl with repeating unit -CH2(CHCl)-.
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11-11) What’s the minimum of such coats as per the previous Exercise must be hung below the neck
of coat hanger to cause the coat hanger neck to remain stretched after the coats are removed?
Need: # coats to cause the hanger neck to yield ____ .
Know: From previous exercise, one coat is 13 N and stress per coat = 0.16 MPa. Also
from stress-strain diagram, yield occurs at 48 MPa.
How: # coats × 0.16 MPa = Yield in MPa
Solve: # coats × 0.16 MPa = 48 MPa. ∴# coats = 300.
(This is surely too large – failure would have occurred somewhere else before the
hanger’s neck failed.)
For Exercises 12-16, imagine the idealized situation as shown in diagram A) below for an artillery
shell striking an armor plate made of an (imaginary) metal called “armories”. Diagram B is the
stress/strain diagram for armories. Assume the stress/strain diagram is to the armories’ failure.
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11-12) Suppose the shell has mass 1.00 kg, and is traveling at 3.0 × 102 m/s. How much TKE does it
carry?
Need: TKE of shell = ___ J.
Know: Mass = 1.00 kg and speed v = 3.0 × 102 m/s.
How: TKE = ½ mv2
Solve: TKE = ½ × 1.00 × (3.0 × 102)2 [kg][m/s]2 = 4.5 × 104 [kg m/s2][m] = 4.5 × 104
[N m] = 4.5 × 104 J
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11-13) Assume that the energy transferred in the collision between the shell and the armor plate in
the previous Exercise affects only the shaded area of the armor plate beneath contact with the flat tip
of the shell (a circle of diameter 0.010 m) and does not spread out to affect the rest of the plate. What
is the energy density delivered within that shaded volume by the shell?
Need: Energy density in armor ____ J/m3.
Know: Energy delivered by shell = 4.5 × 104 J and volume of “armory” affected =
diameter of 0.010 m to depth of 0.10 m.
How: Volume affected = π d2T/4 [m]2[m] and energy known.
Solve: Volume affected = π × 0.0102 × 0.10/4 [m]2[m] = 7.85 × 10-6 m3.
∴Energy density delivered = 4.5 × 104/7.85 × 10-6 [J][1/m3] = 5.7 × 109 J/m3.
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11-14) Which of the following will happen as a result of the collision in Exercise 13? Support your
answer with numbers.
a) The shell will bounce off without denting the armor plate.
b) The armor plate will be damaged, but will protect the region beyond it from the collision.
c) The shell will destroy the armor plate and retain sufficient kinetic energy with which to
harm the region beyond the plate.
Need: Predict whether armor will survive, ____ Yes/no? What margin?
Know: Energy deposited = 5.7 × 109 J/m3 and σ,ε diagram for the armor.
How: Compare energy deposited with capability to absorb energy, compressive area
under σ,ε diagram to yield at ε = 0.2%.
Note: It’s an assumption that the armor will fail in compression since the far side of
the armor from the impact point is being stretched into tension. In fact, this is better
assessed by advanced multidimensional methods of analysis.
Solve: At yield, the armor can elastically absorb ½ × 1.00 × 104 × 0.2/100
[GPa][%][fraction/%] = 10. GPa. But 10. GPa = 10. × 109 = 1.0 × 1010 N/m3.
Compare this to energy deposited of 5.7 × 109 J/m3.
The energy deposited is less than the armor’s elastic yield capability, so it will not take a
permanent set. Answer is thus a) The shell will bounce off without denting the armor plate.
However, it is uncomfortably close to yielding and a permanent set to the armor; the armored
tank crew will at least get quite a headache.
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11-15) In Exercise 14 what is the highest speed the artillery shell can have, yet still bounce off the
armor plate without damaging it?
Need: Max speed of shell, v, = ___ m/s allowing the armor (and tank crew) to survive.
Know: Mass of shell = 1.00 kg. Elastic energy absorption capability of armor from the
previous Exercise, is 1.0 × 1010 J/m3. Volume V of armor engaged = 7.85 × 10-6 m3.
How: ½ mv2max < Energy absorbed to failure × V [J/m3][m3]. Need to calculate the
energy /volume absorbed in plastic yield from ε = 0.20% to failure at ε = 0.70% at a
compressive stress of 1.00 × 104 GPa.
Solve: Total plastic energy absorbed at failure = 1.00 × 104 × (0.7 - 0.2)/100
[GPa][%][fraction/%] = 50. GPa.
∴ total energy absorbed to failure is 10. (Elastic) + 50. (Plastic) = 60. GPa.
∴equating ½ mv2max = 60 [GPa] × V [m3] and sorting out the units,
vmax = √(2 × 6.0 × 1010 × 7.85 × 10-6/1.00) √{[N m/m3][m3][1/kg]} = 970 m/s.
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Problems 16 -19 involve the situation depicted below. Consider a “micrometeorite” to be a piece of
mineral that is approximately a sphere of diameter 1. × 10-6 m and of density 2.00 × 103 kg/m3. It
travels through outer space at a speed of about 5.0 × 103 m/s relative to a spacecraft. Your job as an
engineer is to provide a micrometeorite shield for the spacecraft. Assume that if the micrometeorite
strikes the shield, it affects only a volume of the shield 1.0 × 10-6 m in diameter and extending through
the entire thickness of the shield.
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11-16) Using the stress-strain diagram for steel in Figure 11.11, determine the minimum thickness a
steel micrometeorite shield would have to be to protect the spacecraft from destruction (even though
the shield itself might be dented, cracked or even destroyed in the process).
Need: Thickness of shield, T = ____ mm.
Know: KE of micrometeorite and volume of steel affected; also the steel with
properties shown in Figure 11.11 is symmetric with respect to tension and
compression. Its compressive yield is -2.0 × 102 MPa at ε = -0.15% and its fracture
occurs at ε = -1.0%. Micrometeorite density 2.00 × 103 kg/m3 and relative speed = 5.0
× 103 m/s.
How: Compare the shield’s toughness with the KE/volume material. If contact area is
A, 1/2 mv2/AT = toughness gives T, where toughness is area to failure under σ,ε
diagram.
Solve: Steel toughness = ½ × (-2.0 × 102) × (-0.0015) + (-2.0 × 102) × (-0.01 –
(-0.0015)) [MPa][0] = 0.15 (elastic) + 1.7 (plastic) [MN/m2] = 1.85 MN/m2 = 1.85 ×
106 N/m2.
Impact area of micrometeorite = πD2/4 = π × (1.0 × 10-6)2/4 = 7.85 × 10-13 m2.
Mass 5 of micrometeorite = ρ × πD3/6 = 2.00 × 103 × π ×
(1.0 × 10-6)3/6 [kg/m3][m3] = 1.05 × 10-15 kg.
KE deposited = ½ mv2 = ½ × 1.05 × 10-15 × (5.0 × 103)2 [kg][m/s]2 = 1.31 × 10-8 J.
For T in m, 1.31 × 10-8/(7.85 × 10-13 × T) [N m][1/m3] = 1.85 × 106 N/m2 or T = 9.0
mm.
This implies a pretty hefty mass of steel to be orbited. Also much faster
micrometeorites can be imagined than 5,000 m/s relative to the space craft.
5. πD3/6 is volume of a sphere of diameter; again diameters are preferred to radii.
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11-17) Using the stress/strain properties for a polymer (Figure 11.13 and Table 11.1), determine
whether a sheet of this polymer 0.10 m thick could serve as a micrometeorite shield, if this time the
shield must survive a micrometeorite strike without being permanently dented or damaged. Assume
the properties of the polymer are symmetric to tension and to compression.
Need: Polymer shield will survive undamaged ___ Yes/No?
Know: KE of micrometeorite = 1.31 × 10-8 J; impact area = 7.85 × 10-13 m and σ = 55
MPa and ε = 0.30 at yield. T = 0.10 m.
How: Compare the shield’s toughness with the KE/volume material.
Solve: Toughness at yield for polymer = ½ × 55 × 106 × 0.30 [N/m2][0] = 8.3 × 106
J/m3.
KE deposited/volume material = 1.31 × 10-8 /(7.85 × 10-13 × 0.10) [J][1/m3] = 1.67 ×
105 J/m3 < 8.3 × 106 J/m3.
Hence 0.1 m of this polymer will survive micrometeorite unscathed.
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11-18) Suppose one was required to use a micrometeorite shield no more than 0.01 meters thick.
What would be the required toughness of the material from which that shield was made, if the shield
must survive a micrometeorite strike without being permanently dented or damaged.
Need: Toughness of micrometeorite shield ___ MPa.
Know: KE of micrometeorite = 1.31 × 10-8 J; impact volume =
7.85 × 10-15 m3 (i.e., 7.85 × 10-13 × 0.01).
How: Toughness = KE/volume material affected.
Solve: KE/volume = 1.31 × 10-8/7.85 × 10-15 = 1.67 × 106 J/m3 = 1.67 MPa.
∴Toughness required = 1.67 MPa.
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11-19) Suppose a shield exactly 0.01 m thick of the material in Exercise 18 exactly met the
requirement of surviving without denting or damage at a strain of -0.10, yet any thinner layer would
not survive.
a) What is the Young’s Modulus (elasticity) of the material?
b) What is the yield strength of the material?
Need: E = ____ GPa and yield strength = ___ MPa..
Know: Toughness = 1.67 MPa and εmax = -0.10. Also there is no plastic zone in this
material.
How: Draw σ,ε diagram and deduce what is required from it.
ε = -0.10
0,0
Toughness
σ, yield
Solve: The area under the σ,ε diagram = 1.67 MPa = ½ × σYield × 0. 10.
∴σYield = -33 MPa
E is the slope of the graph.
∴E = -33/(-0.10) [MPa][0] = 0.33 GPa
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11-20) Your company wants to enter a new market by reverse engineering a popular folding kitchen
step stool whose patent has recently expired. Your analysis shows that by simplifying the design and
making all the components from injection molded PVC plastic you can product a similar product at a
substantially reduced cost.
However, when you make a plastic prototype and test its performance you find that it is not as
strong and does not work as smoothly as your competitor’s original stool. Your boss is anxious to get
your design into production because he has promised the company president a new high profit item
by the end of this quarter. What do you do?
a) Release the design. Nearly everything is made from plastic today, and people don’t expect
plastic items to work well. You get what you pay for in the commercial market.
b) Release the design, but put a warning label on it limiting its use to people weighing less than
150 pounds.
c) Quickly try to find a different, stronger plastic that can still be injection molded and adjust the
production cost estimate upward.
d) Tell your boss that your tests show the final product to be substandard and ask if he wants to
put the company’s reputation at risk. If he/she presses you to release your design, make an
appointment to meet with the company president.
1) Apply the Fundamental Canons: Engineers, in the fulfillment of their professional duties, shall:
1) Hold paramount the safety, health, and welfare of the public -rules out option a).
Deliberately releasing an inferior product is not holding the welfare of the public
paramount
2) Perform services only in the area of their competence - again, rules out option a).
A design engineer is not competent to judge what people expect, and what “you
get what you pay for” means in terms of product quality.
3) Issue public statements only in an objective and truthful manner - this argues
against option a). Release of a product amounts to a public statement that the
product is good, in this case a deceptive statement.
4) Act for each employer or client as faithful agents or trustees - this argues in favor
of option d). It is a better choice than options b) or c) because b) and c) require a
hasty fix that is likely not to work, and might put the employer at risk if the
product failed.
5) Avoid deceptive acts - this argues against options b) and c), as it would be
deceptive to pass off a quick fix as responsible engineering design.
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6) Conduct themselves honorably, responsibly, ethically, and lawfully so as to enhance
the honor, reputation and usefulness of the profession - this argues in favor of option
d), the most honorable way of pursuing your concerns about the product.
4) Engineering Ethics Matrix:
Options
Canons
Hold
paramount
the safety,
health and
welfare of
the public.
Perform
services only
in the area of
your
competence
Issue public
statements
only in an
objective and
truthful
manner
Act for each
employer or
client as
faithful
agents or
trustees
Avoid
deceptive
acts
Conduct
themselves
honorably …
c) Try to find
a stronger
plastic
d) Tell boss,
prepare to go
to higher level
a) Release
the design
b) Release
with
warning
Does not
meet canon
Meets canon Meets canon
Meets canon
Does not
apply
Meets canon Meets canon
only if you
are competent
to make
materials
switch with
safety
Meets canon Meets canon
Meets canon
Meets canon
Does not
meet canon,
Puts
employer at
risk
Meets canon Meets canon
Meets canon
Is deceptive
Meets canon Meets canon
Meets canon
Dishonorable
Meets canon Meets canon
Meets canon
Silence here
is an
untruthful
public
statement
Solution: An ethical engineer must do option d) or some improvement on it: halting
production and assembling a team to improve the product before release. Options b)
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and c), while better than the clearly unacceptable option a) are inferior to d) because
their hasty nature compromises ethical engineering design.
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362
11-21) As quality control engineer for your company you must approve all material shipments for
your suppliers. Part of this job involves testing random samples from each delivery and make sure
they meet your company’s specifications. Your tests of a new shipment of carbon steel rods
produced yield strengths 10% below specification. When you contact the supplier they claim their
tests show the yield strength for this shipment is within specifications. What do you do?
a) Reject the shipment and get on with your other work.
b) Retest samples of this shipment to see if new data will meet the specifications.
c) Accept the shipment since the supplier probably have better test equipment and has been
reliable in the past.
d) Ask your boss for advice.
1) Apply the Fundamental Canons: Engineers, in the fulfillment of their professional duties, shall:
1) Hold paramount the safety, health, and welfare of the public - rules out option c). If an
engineer has doubts about the safety or quality of a product, he is ethically obliged to
explore, not suppress, those doubts.
2) Perform services only in the area of their competence - applies equally to all options
3) Issue public statements only in an objective and truthful manner - does not yet apply, as
no public statements have been made
4) Act for each employer or client as faithful agents or trustees - this argues in favor of
option d). In case of doubt, it is better to stop and get others involved than to try a quick
fix by yourself. Option b), retesting the samples, does not get at the real problems, since
your tests may be defective, so repeating them will not solve the problem. When tests
such as yours and the customers disagree, the differences should be explained before
proceeding further.
5) Avoid deceptive acts - this argues against options a) and c), as sweeping under the rug
the disagreement of customer and supplier by accepting one test or the other would be a
deceptive conclusion.
6) Conduct themselves honorably, responsibly, ethically, and lawfully so as to enhance the
honor, reputation and usefulness of the profession - this argues in favor of option d).
Disagreements involving engineering decisions should be aired, not suppressed.
2) Engineering Ethics Matrix
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Options
Canons
Hold
paramount
the safety,
health and
welfare of
the public.
Perform
services only
in the area of
your
competence
Issue public
statements
only in an
objective and
truthful
manner
Act for each
employer or
client as
faithful
agents or
trustees
Avoid
deceptive
acts
a) Reject the
shipment
b) Retest
Does not
meet canon
Meets canon Does not
meets canon
Meets canon
Meets canon
if you accept
the
competence
of a superior
Silence here
is an
untruthful
public
statement
Meets canon Meets canon
You may
not be
competent
as a public
spokesman
Meets canon
if letter is
objective
and truthful
Meets canon,
Meets canon Meets canon
Meets canon
Meets canon May be
deceptive if
done
without
giving
reasons
Meets canon
Meets canon May be
dishonorable
if it results
in safety
being swept
under the
rug.
Meets canon
Conduct
themselves
honorably …
c) Accept
363
Meets canon Meets canon
d) Ask boss
for advice
Going over
the heads of
management
is not being
a faithful
agent
Meets canon
Solution: An ethical engineer must do option d) or some improvement on it. As in the
preceding problem, this is not really an ethical dilemma but simple engineering best practice.
When a serious problem potentially threatening product safety or quality arises, stop and fix
the problem. Don’t try to hide it and move on.
Copyright ©2010, Elsevier, Inc