figure a) figure b) · 2012-08-16 · ... a flat saucer made of the above polymer has an initial...

Kosky, Balmer, Keat and Wise: Exploring Engineering, Third Edition

Chapter 11, Materials Engineering The figures below depict the situations described in Exercises 11.1 – 11.6.

Mass M

LDiameter 0.20 mm

Mass MMass M

LDiameter 0.20 mm

1.00 m

1.00 m

Mass MCube

Plate 1.00 m

1.00 m

Mass MCube

Plate

Figure A) Figure B)

Mass M

LDiameter 0.20 mm

Mass MMass M

LDiameter 0.20 mm

1.00 m

1.00 m

Mass MCube

Plate 1.00 m

1.00 m

Mass MCube

Plate

Figure A) Figure B)

Copyright ©2012, Elsevier, Inc 1


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.



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)



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.



11-4) Suppose the plate described in Exercise 11.3 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.



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.



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.



For Exercises 11.7- 11.9, assume that a silicone rubber1 has the stress-strain diagram shown below.

1. 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.



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



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.



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!)



11-10) A coat hanger like the one in the illustration below is made from polyvinyl chloride2. 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.

2. PVC is a very inexpensive polymer of the monomer, CH2=CHCl with repeating unit -CH2(CHCl)-.



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 11.12 - 11.16, 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.



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



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.



11-14) Which of the following will happen as a result of the collision in Exercise 11.13? 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.



11-15) In Exercise 11.14 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 mv2

max < 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 mv2

max = 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.



Problems 11.16 - 11.19 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.



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. Mass3 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.

3. πD3/6 is volume of a sphere of diameter; again diameters are preferred to radii.



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.



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.



11-19) Suppose a shield exactly 0.01 m thick of the material in Exercise 11.18 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.

0,0ε = -0.10

Toughness

σ, yield

0,0ε = -0.10

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



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. 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. 1) Engineering Ethics Matrix:

Options Canons

a) Release the design

b) Release with warning

c) Try to find a stronger plastic

d) Tell boss, prepare to go to higher level

Hold paramount the safety, health and welfare of the public.

Does not meet canon

Meets canon Meets canon Meets canon

Perform services only in the area of your competence

Does not apply

Meets canon Meets canon only if you are competent to make materials switch with safety

Meets canon

Issue public statements only in an objective and truthful manner

Silence here is an untruthful public statement


Act for each employer or client as faithful agents or trustees

Does not meet canon, Puts employer at risk


Avoid deceptive acts

Is deceptive Meets canon Meets canon Meets canon

Conduct themselves honorably …

Dishonorable Meets canon Meets canon Meets canon



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) and c), while better than the clearly unacceptable option a) are inferior to d) because their hasty nature compromises ethical engineering design.



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 Options Canons

a) Reject the shipment

b) Retest c) Accept d) Ask boss for advice

Hold paramount the safety, health and welfare of the public.

Does not meet canon

Meets canon Does not meets canon

Meets canon

Perform services only in the area of your competence

Meets canon if you accept the competence of a superior

Meets canon Meets canon You may not be competent as a public spokesman

Issue public statements only in an objective and truthful manner

Silence here is an untruthful public statement

Meets canon Meets canon Meets canon if letter is objective and truthful

Act for each employer or client as faithful agents or trustees

Meets canon, Meets canon Meets canon Going over the heads of management is not being a faithful agent

Avoid deceptive acts

Meets canon Meets canon May be deceptive if done without giving reasons

Meets canon

Conduct themselves honorably …

Meets canon Meets canon May be dishonorable if it results in safety being swept under the rug.

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.


figure a) figure b) · 2012-08-16 · ... a flat saucer made of the above polymer has an initial...

Documents