Ch 02

Chapter 2

Mechanical Behavior, Testing, and Manufacturing Properties of Materials QUALITATIVE PROBLEMS 2.26 On the same scale for stress, the tensile true stress-true strain curve is higher than the engineering stress-engineering strain curve. Explain whether this condition also holds for a compression test. During a compression test, the cross-sectional area of the specimen increases as the load is increased. Since true stress is defined as load divided by the instantaneous crosssectional area of the specimen, the true stress in compression will be lower than the engineering stress for a given load, providing that frictional forces (between the platens and the specimen) are negligible. 2.27 What are the similarities and differences between deformation and strain? The similarities are that they are both a measure of a change in shape; strain is a deformation normalized by initial length, and therefore is dimensionless. 2.28 Can a material have a negative Poisson’s ratio? Give a rationale for your answer. Solid material do not have a negative Poisson’s ratio, with the exception of some composite materials (see Chapter 10), where there can be a negative Poisson’s ratio in a given direction. The rationale is harder to express. It should make sense to students that a material, when stretched, should become narrower in the transverse directions. If Poisson’s ratio were zero, then there would be no lateral deflection. If the Poisson’s ratio were negative, it would expand laterally when stretched longitudinally. Also consider compression - if compressed axially, the material would need to thin laterally. This shouldn’t make sense to students. This can be proven to violate the second law of thermodynamics by calculating all components of strain energy in this case, but this is an advanced proof. 14 © 2014 Pearson Education, Inc. Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction ,storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to : Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Mechanical Behavior and Manufacturing Properties of Materials

15

2.29 It has been stated that the higher the value of m, the more diffuse the neck is, and likewise, the lower the value of m, the more localized the neck is. Explain the reason for this behavior. As discussed in Section 2.2.7 starting on p. 63, with high m values, the material stretches to a greater length before it fails; this behavior is an indication that necking is delayed with increasing m. When necking is about to begin, the necking region’s strength with respect to the rest of the specimen increases, due to strain hardening. However, the strain rate in the necking region is also higher than in the rest of the specimen, because the material is elongating faster there. Since the material in the necked region becomes stronger as it is strained at a higher rate, the region exhibits a greater resistance to necking. The increase in resistance to necking thus depends on the magnitude of m. R 2.30 Explain why materials with high m values, such as hot glass and Silly Putty , when stretched slowly, undergo large elongations before failure. Consider events taking place in the necked region of the specimen. The answer is similar to Answer 2.29 above.

2.31 With a simple sketch, explain whether it is necessary to use the offset method to determine the yield stress, Y , of a material that has been highly cold worked. As can be seen by reviewing Fig. 2.3 on p. 57, a highly cold-worked metal will have a distinct change in slope on its stress-strain curve occurring at the yield point, so that the offset method is not necessary. See also Fig. 2.5 on p. 61. 2.32 Explain why the difference between engineering strain and true strain becomes larger as strain increases. Does this difference occur for both tensile and compressive strains? Explain. The answer lies in the fact that the definitions of engineering strain and true strain are different, the latter being based on the actual or instantaneous dimensions, as can be seen in Eqs. (2.2) and (2.7) on p. 57 and p. 59, respectively. In both cases of tension and compression, the difference increases as strain increases. This is shown quantitatively in Problem 2.74. 2.33 Consider an elastomer, such as a rubber band. This material can undergo a large elastic deformation before failure, but after fracture it recovers completely to its original shape. Is this material brittle or ductile? Explain. This is an interesting question and one that can be answered in a number of ways. From a stress analysis standpoint, the large elastic deformations would lead to blunting of stress concentrations, and the material would be considered ductile. However, in manufacturing, ductility implies an ability to achieve a permanent change in shape; in this case, a rubber band is extremely brittle, as there is essentially no permanent deformation when the rubber band fractures in a tension test, for example. 2.34 If a material (such as aluminum) does not have an endurance limit, how then would you estimate its fatigue life? Materials without endurance limits have their fatigue life defined as a certain number of cycles to failure at a given stress level. For engineering purposes, this definition allows for an estimate of the expected lifetime of a part. The part is then usually taken out of

© 2014 Pearson Education, Inc. Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction ,storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to : Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


16

service before its lifetime is reached. An alternative approach is to use nondestructive test techniques (Section 36.10) to periodically measure the accumulated damage in a part, and then use fracture mechanics approaches to estimate the remaining life. 2.35 What role, if any, does friction play in a hardness test? Explain. The effect of friction has been found to be minimal. In a hardness test, most of the indentation occurs through plastic deformation, and there is very little sliding at the indenter-workpiece interface; see Fig. 2.14 on p. 70. 2.36 Which hardness tests and scales would you use for very thin strips of metal, such as aluminum foil? Explain. A hardness test that produces small indentations would have to be used; also, since aluminum foil is relatively soft, a very light load would be required. Two scales that satisfy these requirements are the Knoop microhardness (HK) and the Vickers hardness (HV) at very light loads (see Fig. 2.13 on p. 69). An area of current research is the use of atomic force microscopy and nanoindenters to obtain the hardness of very thin materials and coatings. The shape of the indenter used is not exactly the same as in Fig. 2.13, and the loads are in the micro- to milli-Newton range. 2.37 Consider the circumstance where aVickers hardness test is conducted on a material. Sketch the resulting indentation shape if there is a residual stress on the surface. There are many possible shapes. Consider the simple sketches below, where the Vickers indentation for a material without residual stresses (the baseline) is shown in red. The green example would be particular to the case where a uniaxial residual stress is aligned with the indenter and is constant through the thickness. The blue example is more typical, and shows a biaxial residual compressive residual stress. If the Vickers inventor is not aligned with the stress, then the shape will be more of a rhomboid than square.

2.38 Which of the two tests, tension or compression, would require a higher capacity of testing machine, and why? The compression test requires a higher capacity machine since the cross-sectional area of the specimen increases as the test progresses. The increase in area requires a load higher than that for the tension test to achieve the same stress level. Also, there is friction between the flat dies (platens) and the workpiece surfaces in a compression test (see Sections 2.3 on p. 65 and 14.2 on p. 339) which results in higher pressures than in tension; this higher pressure then requires larger forces for the same cross-sectional area. In addition, there is more redundant work in compression testing than in tension testing, so the material will work harden more (unless the test is conducted at elevated temperatures). 2.39 List and explain briefly the conditions that induce brittle fracture in an otherwise ductile metal.



17

Brittle fracture can be induced by high deformation rates, lower temperatures (particularly those with bcc structure), the presence of stress concentration (notches and cracks), state of stress, radiation damage, corrosion (including hydrogen embrittlement). In each case, the stress needed to cause yielding is raised above the stress needed to cause failure, or the stress needed for a crack to propagate is below the yield stress of the material (as with stress concentration). 2.40 List the factors that you would consider in selecting a hardness test. Explain why. Hardness tests mainly have three differences: (a) type of indenter, (b) applied load, and (c) method of indentation measurement, i.e., depth or surface area of indentation, or rebound of indenter. The hardness test selected would depend on the estimated hardness of the workpiece, its size and thickness, and if average hardness or the hardness of individual microstructural components is desired. For instance, the scleroscope, which is portable, is capable of measuring the hardness of large pieces that cannot be used for measurement by other techniques. The Brinell hardness test leaves a fairly large indentation, thus providing a good measure of average hardness, while the Knoop test leaves a small indentation that allows for determination of the hardness of the individual phases in a two-phase alloy. The small indentation of the Knoop test also allows it to be useful in measuring the hardness of very thin layers or plated layers on parts. Note that the depth of indentation should be small relative to part thickness, and that any change in the appearance of the bottom surface the part will make the test results invalid. Figure 2.15 on p. 72 is a useful guide for determining which hardness test is valid for a class of material. Note that often numerous hardness tests are suitable for a material. In these cases, the best hardness test is the one that has one or more of the following characteristics: • The best hardness test is often one that can be performed quickly; thus, it may be desirable to also select a hardness test based on available equipment. • Hardness tests are often specified by customers as part of a quality control requirement. Whatever form of hardness test is specified by the customer is the appropriate one to use. • A hardness test that is most commonly used in a plant may be the best choice since technicians will be most familiar with the test protocol and the equipment is most likely to be in good calibration. • Experimental error can be minimized by selecting a hardness test that gives the largest penetration or indentation size. 2.41 On the basis of Fig. 2.5, can you calculate the percent elongation of the materials listed? Explain. Recall that elongation (total) is defined by Eq. (2.4) on p. 60 and depends on the original gage length (lo ) of the specimen. Note that if the gage length encompasses a necked region only, it will register a larger percent elongation than if the gage length is four or five times as long as the necked region, for example. From Fig. 2.5 on p. 61, the true necking and fracture strains can be determined. Since this is a true strain, it is



18

not dependent on the gage length. That is, regardless of the gage length, the same true stress will be measured. Since there is no way of incorporating gage length, one cannot obtain the percent elongation from Fig. 2.5. (See also the answer to Problem 2.57.) 2.42 If a metal tension-test specimen is rapidly pulled and broken, where would the temperature be highest, and why? Since temperature rise is due to work input, it is obvious that the temperature will be highest in the necked region because that is where the strain is highest and, hence, the energy dissipated per unit volume in plastic deformation is highest. 2.43 Comment on your observations regarding the contents of Table 2.2. By the student. There are a large number of acceptable answers to this problem. A student may compare the values in the table between different materials or material classes. Alternatively, the students may comment on the size of the range in properties. Students should be encouraged to develop well thought-out answers to this question. 2.44 Will the disk test be applicable to a ductile material? Why or why not? With a ductile material, a point load on a disk results in the circular disk being flattened at the platens and attaining elliptical shape of the originally round specimen. The flattening converts the point load to a distributed load, completely changing the stress state in the piece. Therefore, Eq. (2.10) on p. 66 is not valid, and the usefulness of the test is compromised. 2.45 What hardness test is suitable for determining the hardness of a thin ceramic coating on a piece of metal? For a thin ceramic coating, it is still important that the hardness of the coating and not the substrate be measured. Most ceramics have limited ductility (Section 8.3 starting on p. 200), so that Knoop or Vickers tests are suitable, although the Mohs test can also be used to obtain a qualitative value. Because of the increasing importance of coatings, special microhardness tests have been developed for their hardness measurement. 2.46 Wire rope consists of many wires that bend and unbend as the rope is run over a sheave. A wire-rope failure is investigated, and it is found that some of the wires, when examined under a scanning electron microscope, display cup-and-cone failure surfaces, while others display transgranular fracture surfaces. Comment on these observations. There are a large number of potential reasons for this behavior. However, a likely explanation is that when the wire rope was in use, it was run over a sheave or drum repeatedly. As a result, some of the wires in the rope failed due to fatigue, so they display brittle fracture surfaces. At some point, enough wires have failed so that the remaining wires fail due to static overload, and display cup-and-cone failure surfaces as a result. 2.47 A statistical sampling of Rockwell C hardness tests are conducted on a material, and it is determined that the material is defective because of insufficient hardness. The supplier claims that the tests are flawed because the diamond cone indenter was probably dull. Is this a valid claim? Explain. Refer to Fig. 2.13 on p. 69 and note that if an indenter is blunt, then the penetration, t, under a given load will be smaller than that using a sharp indenter. This then translates



19

into a higher hardness. The explanation is plausible, but in practice, hardness tests are fairly reliable and measurements are consistent if the testing equipment is properly calibrated and routinely serviced. 2.48 In a Brinell hardness test, the resulting impression is found to be elliptical. Give possible explanations for this result. Two possible explanations for an elliptical impression after a Brinell test are: (a) An obvious reason is the possible presence of asymmetric residual stresses in the surface layers of the material before the test. (b) The material itself may be highly anisotropic, such as a fiber-reinforced composite material, or due to severe cold working. 2.49 Some coatings are extremely thin – some as thin as a few nanometers. Explain why even the Knoop test is not able to give reliable results for such coatings. Recent investigations have attempted to use highly polished diamonds (with a tip radius around 5 nm) to indent such coatings in atomic force microscopes. What concerns would you have regarding the appropriateness of the test results? With a coating of thickness of 5 nm, the stressed volume has to be approximately onetenth this depth, which begins to approach the size of individual atoms. Thus, a knoop indentor would need to have a tip radius that was atomically sharp in order to get results. Even with highly polished diamond tips in atomic force microscopes, this scale problem is unavoidable. However, there are additional concerns in that the diamond indenter may not be symmetric, there are large adhesive forces at the small scales, there are complicated elastic and viscoelastic recovery at small length scales, there may be residual stresses at the surface, and the stressed volume may or may not contain a dislocation (whereas with Knoop tests, there is always a number of dislocations). 2.50 Select an appropriate hardness test for each of the following materials, and justify your answer: i. Cubic boron nitride ii. Lead iii. Cold-drawn 0.5%C steel iv. Diamond v. Caramel candy vi. Granite Figure 2.15 on p. 72 is a useful guide for selecting hardness tests. i. Cubic boron nitride is very hard, and useful data can be obtained only from the Knoop and Mohs tests. The Mohs scale is qualitative and does not give numerical values for hardness, so the Knoop test is preferable. ii. Lead. As shown in Fig. 2.15, lead is so soft that only the Brinell and Vickers tests yield useful data. Recognizing that lead is very soft, the lightest loads in these tests should be used. Consider the expected results in this test if a typical value of hardness is 4 HB or 4 HV. For the Brinell test, Fig. 2.13 suggests that the expected indentation for a 500 kg load is: 2P HB = √ (πD) D − D2 − d2



Therefore, solving for d, s d=

2P D2 − D − (πD)(HB)

s

2 =

2(500) 102 − 10 − [π(10)] (4)

20

2 = 9.79 mm

Note that this dimension is almost the same as the diameter of the indentor, and makes the usefulness of the test highly questionable. For the Vickers test, the expected indentation test, using the lowest allowable load of 1 kg, is: r r 1.854P 1.854P 1.854(1) HV = → L= = = 0.68 mm L2 HX 4

iii.

iv.

v.

vi.

This is much more reasonable, suggesting that the Vickers test is the best alternative for lead. Cold-drawn 0.5% steel. From Fig. 2.15, all of the hardness tests are suitable for this material. As discussed in Problem 2.40, the best choice for this material will depend on a number of factors. Diamond. The hardness of diamond is difficult to obtain. The hardness of diamond is really determined by extrapolating the hardness on the Mohs curve to another scale in Fig. 2.15. The hardness of diamond is usually quoted as 8000 to 10,000 HK. Caramel (candy). This would be an interesting experiment to perform, but the result will be that none of the hardness tests can be used for this material because it is far too soft. Also, the hardness of caramel is strongly temperature-dependent and that it creeps, so that hardness measurement may be meaningless. Granite. The hardness of granite varies according to the source, but it is approximately around apatite on the Mohs scale. Thus, various hardness tests can give valuable information on granite. Note, however, that in inspecting granite surfaces, one can see various regions within which there would be hardness variations. The particular hardness test selected will depend on various factors, as discussed in part (c) above.

QUANTITATIVE PROBLEMS 2.51 A paper clip is made of wire 0.5 mm in diameter. If the original material from which the wire is made is a rod 25 mm in diameter, calculate the longitudinal engineering and true strains that the wire has undergone during processing. Engineering strain is defined by Eq. (2.2) on p. 58. Thus, because of volume constancy in plastic deformation, we may write 2 2 lf do 25 = = = 2500 lo df 0.5 or lf = 2500lo . Therefore, the engineering strain is, from Eq. (2.2) on p. 57, e=

lf − lo 2500lo − lo = = 2499 lo lo



21

True strain is defined by Eq. (2.7) on p. 59. Hence 2500lo l = ln = 7.824 = ln lo lo Note the large difference in strains, even though they both describe the same phenomenon. (See also the last paragraph in Sec. 2.2.3 on p. 59.) 2.52 A 250-mm-long strip of metal is stretched in two steps, first to 300 mm and then to 400 mm. Show that the total true strain is the sum of the true strains in each step; in other words, the true strains are additive. Show that, in the case of engineering strains, the strains cannot be added to obtain the total strain. We first calculate the true strains for each step: Step 1:

1 = ln(300/250) = 0.182

Step 2:

2 = ln(400/300) = 0.288

Thus the total true strain is total = 0.182 + 0.288 = 0.470 The total strain may also be calculated from the final and initial dimensions as total = ln(400/250) = 0.470 thus giving the same answer. Engineering strains are calculated likewise. Thus, Step 1:

e1 = (300 − 250)/250 = 0.200

Step 2:

e2 = (400 − 300)/300 = 0.333

Thus the total engineering strain is 0.200+0.333=0.533. However, when calculated from the initial and final dimensions, the total engineering strain is etotal = (400 − 250)/250 = 0.6 which is higher than the sum of the engineering strains. Consequently, engineering strains are not additive, whereas true strains are. 2.53 Identify the two materials in Fig. 2.5 that have the lowest and the highest uniform elongations. Calculate these quantities as percentages of the original gage lengths. The magnitude of uniform elongation is directly related to the true strain at the onset of necking. As we see in Fig. 2.4c on p. 60, the necking strain on a true stress-true strain curve corresponds to the beginning of the straight portion of the curve. Thus,



22

from Fig. 2.5 on p. 61, we note that the lowest uniform elongation is for 1112 cold-rolled steel, with a necking strain of about 0.05. The highest is for 304 stainless steel, although 70-30 annealed brass is close to it, with a necking strain of about 0.45. To relate these values to a percentage of the gage length (lo ), we convert true strain to percent elongation as follows: = ln(l/lo ) Or, l/lo = e . Hence, Elongation = (e − 1) × 100 Thus, for 1112 cold-rolled steel Elongation = (e0.05 − 1) × 100 = 5% And for 70-30 brass, the specific value is Elongation = (e0.45 − 1) × 100 = 57% 2.54 Plot the ultimate strength vs. stiffness for the materials listed in Table 2.2, and prepare a three-dimensional plot for these materials where the third axis is their maximum elongation in 50 mm. The graphs are constructed using the minimum values in Table 2.2 on p. 58 when ranges of values exist. They are as follows: 800

Titanium

UTS (MPa)

UTS (MPa)

500

Steel

400 Nickel

Magnesium

300

Zinc

100

Copper Aluminum

Lead 0

50

100

2500

Spectra fibers Carbon fibers

2000 1500 1000

200

150

200

E (GPa)

250

300

Thermoplastics Thermoplastics, reinforced Thermosets Silicon carbide Ceramics, glass Silicon nitride

500

Molybdenum 350

Boron fibers

Kevlar fibers

3000

Tungsten

600

0

Glass fibers

3500

700

0

0

50

100

150

200

250

300

350

400

E (GPa)

2.55 If you remove the layer of material ad from the part shown in Fig. 2.30d (for instance, by machining or grinding), which way will the specimen curve? ( Hint: Assume that the part shown in sketch d in the figure is composed of four horizontal springs held at the ends. Thus, from the top down, you have compression, tension, compression, and tension springs.) The part will bow downward, with the center of curvature below the part (that is, it will hold water). The internal forces will have to achieve a state of static equilibrium. The top layer of material ad, which is under compression, has the tendency to bend the bar



23

upward. When this stress is relieved, such as by removing the top layer by machining or grinding, the bar will compensate by bending downward. Such residual stress systems can be modeled with a set of horizontal tension and compression springs. Thus, starting at the top, we have a compression, tension, compression, and tension spring. 2.56 Prove that the true strain at necking equals the strain hardening exponent. Note that at necking, dσ =σ d Therefore, using a power law constitutive model, σ = Kn dσ = Knn−1 d Therefore, since dσ/d = σ, Kn = Knn−1 which is solved as = n. 2.57 Percent elongation is always defined in terms of the original gage length, such as 50 mm or 2 in. Explain how percent elongation would vary as the gage length of the tensile-test specimen increases. (Hint: Recall that necking is a local phenomenon.) As gage length increases, the percent elongation (total) will decrease to a value representing uniform elongation. For instance, consider taking a gage length that just encompasses the necking region (thus approaching zero gage length); the percent elongation in this region will be very high because the effect of localized deformation will tend to greatly increase the elongation. As gage length increases, however, the localized deformation zone has less and less effect on the total elongation. 2.58 Make a sketch showing the nature and distribution of residual stresses in Fig. 2.31a and b, prior to the material’s being cut. (Hint: Assume that the split parts are free from any stresses; then force these parts back to the shape they originally had.) As the problem states, we assume that the split parts shown in Fig. 2.31a and b on p. 81 are free of stresses. After unbending the two halves (that is, closing the gap by pushing the two halves together), you will note that the inner (mating) surfaces will now be subjected to longitudinal compressive stresses, while the outer surfaces are in tension. Consequently, this represents the state of stress prior to slitting. 2.59 You are given the K and n values of two different metals. Is this information sufficient to determine which metal is tougher? If not, what additional information do you need? Although the K and n values may give a good estimate of toughness, true fracture stress and true strain at fracture are also required for accurate calculation of the toughness of the metals. The modulus of elasticity and yield strength would provide information about the area under the elastic region (known as modulus of resilience; see Example 2.2 on p. 71), but this region is very small and is negligible compared to the rest of the curve.



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2.60 A cable is made of two strands of different materials, A and B, and cross sections, as follows: For material A, K = 60, 000 psi, n = 0.5, Ao = 0.6 in2 ; for material B, K = 30, 000 psi, n = 0.5, Ao = 0.3 in2 . Calculate the maximum tensile force that this cable can withstand prior to necking. Note that necking will occur when = n = 0.5. At this point, the true stresses in each cable are (using σ = Kn ), respectively, σA = 60, 0000.5 = 42, 400 psi and σB = 30, 0000.5 = 21, 200 psi The true areas at necking can be calculated as follows: AA = 0.6e−0.5 = 0.364 in2 and AB = 0.3e−0.5 = 0.182 in2 Thus the total load that the cable can support is Ptotal = (42, 400)(0.364) + (21, 200)(0.182) = 19, 300 lb 2.61 On the basis of the information given in Fig. 2.5, calculate the ultimate tensile strength (engineering) of 304 stainless steel. Recall that the definition of engineering stress is σeng = P/Ao , and that for true stress it is σ = P/A. From Fig. 2.5 on p. 61, the true stress at necking for 304 stainless steel is found to be about 130,000 psi, while the true strain is about 0.45. We also know that the ratio of the original to the necked areas of the specimen is given by Ao ln = 0.45 Aneck or

Aneck = e−0.45 = 0.63 Ao

Thus, the engineering stress is calculated as σeng = (130, 000)(0.63) = 81, 900 psi 2.62 In a disk test performed on a specimen 1.00 in. in diameter and 1/4 in. thick, the specimen fractures at a stress of 40,000 psi. What was the load on it? For this problem, we use Eq. (2.10) on p. 66. Thus, σ=

2P ; πdt

P =

σπdt (40, 000 psi)π(1 in.)(0.25 in.) = = 15, 700 lb 2 2



25

2.63 A piece of steel has a hardness of 300 HB. Calculate its tensile strength, in MPa and in psi. Using Eqs. (2.14) and (2.15) on p. 71 we find that UTS = 3.5(HB) = 3.5(300) = 1050 MPa and in psi the similar calculation yields UTS = 500(300) = 150, 000 psi = 150 ksi 2.64 A metal has the following properties: UTS = 70, 000 psi and n = 0.20. Calculate its strength coefficient, K. Let’s first note that the true UTS of this material is given by σ = Knn . We can then determine the value of this stress from the UTS by following a procedure similar to the Example 2.1 on p. 62. Since UTS = P/Ao , P = (UTS)Ao = 70, 000Ao Also, P = σAneck = σAo e−0.20 = σ(0.8187)Ao Therefore, the true stress at necking is 70, 000Ao = σ(0.8187)Ao

→

σ = 85, 500 psi

Therefore, from Eq. (2.8) and realizing that at necking = n, σ = 85, 500 psi = Knn

→

K=

85, 500 = 118, 000 psi 0.200 .20

2.65 Using only Fig. 2.5, calculate the maximum load in tension testing of an annealed copper specimen with an original diameter of 5 mm. We note from the figure that the necking strain for annealed copper is 0.40 and that the true stress at necking is about 300 MPa. The true area at necking is therefore hπ i Aneck = Ao e−n = 52 e−0.40 = 13.16 mm2 4 Hence the maximum load is Pmax = (300 MN/m2 )(13.16 mm2 ) = 3950 N 2.66 Estimate the modulus of resilience for a highly cold worked piece of steel having a hardness of 250 HB, and for a piece of highly cold-worked copper with a hardness of 100 HRB. Refer to Example 2.2 on p. 71. For steel, the yield strength is roughly one-third the hardness, or 250 H = = 83.33 kg/mm2 = 118 ksi Ysteel = 3 3



26

For copper, the same relationship cannot be used, but an Internet search for highly cold-worked copper, Ycopper =

100 H = = 33.3 kg/mm2 = 47.4 ksi 3.3 3.3

From Table 2.1, Esteel = 30 × 106 psi and Ecopper = 15 × 106 psi. Therefore, from Example 2.2 for steel, Modulus of Resilience =

Y2 118, 000 = = 232 in-lb/in3 2E 2(30 × 106 )

and for copper, Modulus of Resilience =

47, 400 Y2 = = 74.9 in-lb/in3 2E 2(15 × 106 )

2.67 A metal has a strength coefficient K = 100, 000 psi and n = 0.25. Assuming that a tensile-test specimen made from this metal begins to neck at a true strain of 0.25, show that the ultimate tensile strength is 59,340 psi. The true UTS is given by True UTS = Knn = 100, 000(0.25)0.25 = 70, 710 psi The true UTS is also related to the UTS by: True UTS = UTS(Ao /Aneck ) = UTSen = 1.28 UTS Hence, UTS=70,710/1.28=55,200 psi. 2.68 Plot the true stress–true strain curves for the materials listed in Table 2.3. Note that these curves can be plotted using Eq. (2.8) on p. 61, but the strain at failure is unknown. Some of the materials have their strain at failure indicated in Fig. 2.5 on p. 61. Otherwise, the data must be obtained from an external source such as a handbook or internet search. The following data summarizes the plot below:



Material Aluminum 1100-O 2024-T4 6061-O 6061-T6 7075-O Brass 70-30, annealed 85-15, cold-rolled Cobalt-base alloy Copper, annealed Steel Low-C, annealed 4135, annealed 4135, cold-rolled 4340, annealed 304 stainless, annealed 410 stainless, annealed

27

K (MPa)

n

f

180 690 205 410 400

0.20 0.16 0.20 0.05 0.17

1.9 0.13 0.15 0.077 0.19

900 580 2070 315

0.49 0.34 0.50 0.54

1.3 0.49 0.14 1.5

530 1015 1100 640 1275 960

0.26 0.17 0.14 0.15 0.45 0.10

0.18 0.16 0.14 0.11 1.3 0.15

The plot is as follows: 1000

304 stainless

Cobalt-base alloy 4135 CR

Stress (MPa)

800

70-30 Brass

410 stainless 4135 annealed

600

2024-T4 Al 4340 annealed 85-15 Brass

400

6061-O Al

Copper, annealed

7075-O Al 6061-O Al

200

1100-O Al

0 0

0.5

1.0

1.5

2.0

Strain



28

2.69 The design specification for a metal requires a minimum hardness of 80 HRA. If a Rockwell test is performed and the depth of penetration is 60 µm, is the material acceptable? Refer to Fig. 2.13 on p. 69. The Rockwell hardness is defined as HRA = 100 − 500t = 100 − 500(0.06) = 70 Note that it is important that the depth of penetration be converted to millimeters before calculating hardness. Using the correct units, it is seen that the hardness is too low, and the material is not acceptable. 2.70 Calculate the major and minor pyramid angles for a Knoop indenter, and compare your results with those obtained from Vickers and Rockwell A indenters. Refer to Fig. 2.13 on p. 69 for the geometry of the indenters. For a Knoop indenter with b/t = 4, the pyramid’s minor angle is given by β = tan−1 (2) = 63◦ Also, since L/b = 7.11 and b/t = 4, then L Lb = = (7.11)(4) = 28.44 t bt Therefore, the major pyramid angle is α = tan−1 (14.22) = 86◦ 2.71 If a material has a target hardness of 300 HB, what is the expected indentation diameter? From Fig. 2.13 on p. 69 the indentation diameter depends on the load and the Brinell hardness according to the equation HB =

2P √ (πD) D − D2 − d2

where P is the load (either 500, 1500 or 3000 kg) and D is the diameter of the indenting ball (10 mm). This equation can be solved for d as: s 2 2P 2 d= D − D− πD(HB) For a hardness of 300 HB, this results in the following for the different loads used in a Brinell test: P (kg) 500 1500 3000

d (mm) 1.45 2.50 3.51



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2.72 A Rockwell A test was conducted on a material and a penetration depth of 0.15 mm was recorded. What is the hardness of the material? What material would typically have such a hardness value? If a Brinell hardness test were to be conducted on this material, give an estimate of the indentation diameter if the load used was 1500 kg. From Fig. 2.13 on p. 69, the Rockwell hardness is calculated from HRA = 100 − 500t = 100 − 500(0.15) = 25 HRA From Fig. 2.15 on p. 72, a material with a 25 HRA should have a Brinell hardness of around 70 HB (note that extrapolation is needed to obtain this estimate). Therefore, from Fig. 2.12, 2P HB = √ (πD) D − D2 − d2 or, solving for d, s 2 d= D − D−

2P (πD)(HB)

s

2

102

=

2(1500) − 10 − π(10)(70)

2 = 5.04 mm

2.73 For a cold-drawn 0.5% carbon steel, will a Rockwell C test or a Brinell test at 500 kg result in a deeper penetration? Referring to Fig. 2.15 on p. 72, such a material has a hardness of around 39 HRC and 350 HB. Now referring to Fig. 2.13, for the Rockwell C test, 39 = 100 − 500t

→

t = 0.122 mm

For the Brinnel test (see Problem 2.72), assuming the lightest load s 2P 2 d= D − = 1.34 mm (πD)(HB) Refer to the sketch below for the geometry.

θ

5 mm

5 mm x d/2



30

Note that sin θ = (d/2)/5 and x = 5 − 5 cos θ. Therefore, for d = 1.34 mm, the depth of penetration is x = 0.045 mm, which is smaller than the Rockwell C hardness test. However, if 1500 kg is used, the depth of penetration is found to be x = 0.135 mm, which is larger than the Rockwell C test. Therefore, the penetration can be larger for the Brinell test, depending on the load used. 2.74 A material is tested in tension. Over a 1-in. gage length, the engineering strain measurements are 0.01, 0.02, 0.03, 0.04, 0.05, 0.1, 0.15, 0.2, 0.5, and 1.0. Plot the true strain versus engineering strain for these readings. The engineering strain is easy to calculate from Eq. (2.2) on p. 58, and the true strain is given by Eq. (2.7) on p. 60 as l = ln lo The following chart can then be made of the data points: e

0.01 0.01

0.02 0.0198

0.03 0.0296

0.04 0.0392

0.05 0.0488

0.1 0.0953

0.15 0.1397

0.2 0.182

0.5 0.405

1.0 0.69

The plot is as follows: 1.0

True strain, ε

0.8

Slope = 1

0.6 0.4 0.2 0 0

0.2

0.4 0.6 Engineering strain, e

0.8

1.0

2.75 A horizontal rigid bar c-c is subjecting specimen a to tension and specimen b to frictionless compression such that the bar remains horizontal. (See Fig. P2.75.) The force F is located at a distance ratio of 2 : 1. Both specimens have an original cross-sectional area of 1 in2 and the original lengths are a = 8 in. and b = 4.5 in. The material for specimen a has a true stress-true strain curve of σ = 100, 0000.5 . Plot the true stress-true strain curve that the material for specimen b should have for the bar to remain horizontal.



31

a F 2

1

c

c b

x

From the equilibrium of vertical forces and to keep the bar horizontal, we note that 2Fa = Fb . Hence, in terms of true stresses and instantaneous areas, we have 2σa Aa = σb Ab From volume constancy we also have, in terms of original and final dimensions Aoa Loa = Aa La and Aob Lob = Ab Lb where Loa = (8/4.5)Lob = 1.78Lob . From these relationships we can show that 8 Lb σb = 2 Kσa 4.5 La Since σa = K0.5 a where K = 100, 000 psi, we can now write Lb √ 16K a σb = 4.5 La Hence, for a deflection of x, σb =

16K 4.5

4.5 − x 8+x

s

ln

8+x 8

The true strain in specimen b is given by b = ln

4.5 − x 4.5

By inspecting the figure in the problem statement, we note that while specimen a gets longer, it will continue exerting some force Fa . However, specimen b will eventually acquire a cross-sectional area that will become infinite as x approaches 4.5 in., thus its strength must approach zero. This observation suggests that specimen b cannot have a true stress-true strain curve typical of metals, and that it will have a maximum at some strain. This is seen in the plot of σb shown below.



32

50,000

True stress (psi)

40,000 30,000 20,000 10,000 0

0

0.5 1.0 1.5 2.0 Absolute value of true strain

2.5

SYNTHESIS, DESIGN, AND PROJECTS

2.76 List and explain the desirable mechanical properties of (a) an elevator cable, (b) a paper clip, (c) a leaf spring for a truck, (d) a bracket for a bookshelf, (e) piano wire, (f ) a wire coat hanger, (g) the clip for a pen, and (h) a staple. i. Elevator cable: The cable should not elongate elastically to a great extent or undergo yielding under the load applied. These requirements call for a material with a high elastic modulus and yield stress. The cable should also be sufficiently flexible and ductile to be wrapped around the drums during its use. ii. Paper clip: The clip material must possess high ductility to allow it to be formed without fracture. In its normal use, the clip should recover elastically when removed as well as apply sufficient force to hold papers together (thus should possess high elastic modulus). iii. Leaf spring: The function of the leaf spring is to absorb energy elastically upon static or dynamic loading. The energy should be absorbed in an elastic manner because after absorbing the energy, the spring should return to its original shape. This requires a high yield stress and high elastic modulus that maximizes the area under the elastic portion of the stress-strain curve (modulus of resilience). iv. Bracket: Much like an elevator cable, but not as critical, a material with a high yield stress and elastic modulus would be required. These properties would keep the shelf from sagging excessively under load. v. Piano wire: A piano wire is under high tension (hence high stress) to achieve the desired tone. The wire should be able to maintain this stress level for a period of time, or the piano would become out of tune. Thus, its yield stress should be high in order to sustain the tension required to produce the tone, and should not be susceptible to stress relaxation. Furthermore, it should be sufficiently ductile to be able to be wound around the tightening mechanisms.



33

vi. Wire coat hanger: Much like the paper clip, the hanger requires a large amount of deformation in its manufacturing, so the material must possess high ductility. The hanger should also be able to maintain its shape when clothes are hung on it, hence it should possess sufficient yield strength and elastic modulus. vii. Clip for a pen. A clip for a pen is intended to affix the pen to a pocket or other cloth or paper; it is important that the clip material have the proper stiffness so that the force needed to make it attach is neither too high nor too low. The material should have a high resilience, in that it should recover strains elastically, and it should do so fairly rapidly. viii. Staple: The properties should closely parallel a paper clip. The staple should have high ductility to allow it to be deformed without fracture, and also a low yield stress so the staple can be bent (as well as removed) easily without requiring the user to exert a great amount of force. 2.77 When making a hamburger, you may have observed the type of cracks shown in Fig. 2.20d. What would you do to avoid such cracks? [Note: Test hamburger patties by compressing them at different temperatures, and observe the crack path (i.e., the path through the fat particles, the meat particles, or their interface).] These cracks do occur, and they can be eliminated by making sure that the chunk of ground meat is not too cold. In experiments, which can be performed by the student, it can be observed that when cold, cracks propagate through the meat particles rather than through the fat. As temperature increases, cracks would propagate through the fat, which eventually softens and melts as higher temperatures are reached (which is easily observed when frying ground meat in a pan to make, for example, meat sauce). R 2.78 An inexpensive claylike material called Silly Putty is generally available in stores that sell toys and games. Obtain a sample and perform the following experiments: (a) Shape it into a ball, and drop it onto a flat surface. (b) Reround the ball and place a heavy book on it for one minute. (c) Shape the putty into a long rod, and pull on it–first slowly, then very quickly. Describe your observations, referring to the specific sections in this chapter where each particular observation is relevant. Silly putty is an extremely strain-rate sensitive material. When shaped into a ball and bounced, it undergoes a high strain rate and acts as very rigid, elastic material; i.e., it bounces. When placed on a table and subjected to a static force such as placing a book in it, it deforms as a very soft, plastic material and there is very little elastic recovery. When stretched slowly, it is very ductile; when stretched quickly, it can be seen that the ductility is drastically reduced, and the stiffness and strength are also much higher.

2.79 Make individual sketches of the mechanisms of testing machines that, in your opinion, would be appropriate for tension, for torsion, and for compression testing of specimens at different rates of deformation. What modifications would you make on these machines to include the effects of temperature on material properties? By the student. The important features are that the jaws of the machine must move at constant speed; this can be simply done using a power screw drive, with a load cell



34

to measure force. More elaborate machines can use hydraulic power sources and control systems to achieve any desired mode of loading. For testing at elevated temperatures, the machine must be protected, and the specimen must be insulated from the environment to maintain constant temperature during the test. This clearly requires a test chamber that is thermally insulated. 2.80 In tension testing of specimens, mechanical and electronic instruments are typically used to measure elongation. Make sketches of instruments that would be suitable for this purpose, commenting on their accuracy. What modifications would you make to these instruments to include the use of specimens at elevated temperatures? By the student. There are a number of ways to measure elongation. Commercial extensometers usually have a means to clip the ends of the extensometer to a specimen, and these clips are initially 25 or 50 mm (or a standard length) apart. This allows more direct measurement of elongation [see Eq. (2.2); the quantity lo is pre-defined by the spacing of clips]. Measurement of elongation can take place with potentiometers or strain gages; the students should be encouraged to develop creative solutions to this problem. 2.81 Obtain small pieces of different metallic and nonmetallic materials, including stones. Rub them against each other, observe the scratches made, and order them in a manner similar to the Mohs hardness numbering system. By the student. Some subtle points are that scratches may be difficult to detect and measure when the hardness of the specimen and that of the harder body are close to each other. Also, the materials to be tested may have hardness variations along their surfaces. 2.82 Demonstrate the stress-relaxation phenomenon by tightly stretching thin plastic strings between two nails placed at the ends of a long piece of wood. Pluck the strings frequently, to test the tension as a function of time. Repeat the test at a higher temperature by placing the fixture in an oven set on low. By the student. This experiment can be easily performed with a fishing line, but one must be careful with a wooden support in an oven. The string tone decreases in frequency over time, demonstrating stress relaxation. 2.83 Demonstrate the impact toughness of a piece of round chalk by first using a triangular file to produce a V-notch on the cylindrical surface (as shown in Fig. 2.19a) and then bending the chalk to break it. By the student. Chalk is relatively easy to break by hand. With a carefully filed, sharp notch, a dramatic reduction in the required force can be easily observed. With a simple fixture (such as using a vise), the chalk can be cantilevered and small weights dropped from ever higher distances to demonstrate the level of energy required to fracture the chalk. 2.84 Using a large rubber band and a set of weights, obtain the force–displacement curve for the rubber band. Is the result different from the stress–strain curves shown in Fig. 2.4? Comment. By the student. See also Fig. 7.14 on p. 191 and note that the behavior of rubber is highly nonlinear elastic. Two qualitative examples of load-elongation or stress-strain



35

curves are given below. Some rubber bands have a prestress, so that a non-zero force is needed to obtain a small deflection, but these are rare. Most have section where the slope of the stress-strain curve is low, but the curve is very steep near fracture. This can be easily felt with a rubber band. Occasionally, there will be an initial stiff section as shown. Fracture

Load or stress

Finite load needed for incremental strain

*

Low modulus extension

*

High modulus extension at high strain

Initial high modulus may be absent

Extension or strain 2.85 Design a test protocol to obtain the work of plastic deformation by measuring the temperature rise in a workpiece, assuming that there is no heat loss and that the temperature distribution is uniform throughout. If the specific heat of the material decreases with increasing temperature, will the work of deformation calculated using the specific heat at room temperature be higher or lower than the actual work done? Explain. By the student. The design should involve some form of plastic deformation, and the incorporation of some means of measuring temperature, such as thermocouples. If temperatures are measured, and the specific heat decreases with increasing temperature, then for a given temperature rise, the work of deformation will be higher than actually occurs. 2.86 Find or prepare some solid circular pieces of brittle materials, such as chalk, ceramics, etc. and subject them to the type of test shown in Fig. 2.9 by using the jaws of a simple vise. Describe your observations as to how the materials fracture. Repeat the tests, using ductile materials, such as clay, soft metals, etc., and describe your observations. By the student. When carefully performed, one can demonstrate the failure mode as described in the disc test on p. 67, namely, that the brittle material fractures in tension along the midplane of the specimen. Note that most vises have a knurled, rough surface finish which can influence the results; a hard and flat spacer may need to be inserted into the jaws of the vise on opposite ends of the disk. Also, for a ductile material, the disk will flatten, and it may be very difficult to actually cause fracture if a soft specimen is being used in the experiment.



36

2.87 Take several rubber bands and pull them at different temperatures, including from a frozen state. Comment on their behavior, such as ductile or brittle. Note that the behavior will depend on the specific rubber in the rubber band. However, students should be able to demonstrate changes in material behavior. The stiffness should increase as the temperature decreases, and the ductility will decrease. Students can perform such investigations fairly easily using simple weights to apply loads and rulers to measure deflection. Rubber bands can be frozen or the tests can be conducted outside in cold climates. 2.88 Devise a simple fixture for conducting the bend tests shown in Fig. 2.11. Test sticks of various brittle materials by loading them with dead weights until they break. Verify the statement in the text that the specimens on the right in the figure will fracture sooner than the ones on the left. By the student. This test can be performed using specially fabricated fixtures on a tension test machine. However, it can also be done very simply by using a small-diameter rod, suspended between two desks and the load applied by a basket with two strings. Three-point bending can be simulated by placing both strings at the same location, and four-point bending by spacing the strings along the rod. The load in the basket can be increased until the rod fails. Other alternatives are also possible. 2.89 By pressing a small ball bearing against the top surfaces of various materials, such as clay and dough, observe the shape of the indentation with a magnifier, referring to those shapes shown in Fig. 2.14a and b. By the student. Note that with simple materials, the indentation shape (when viewed from above) will always be circular. An interesting specimen can be carefully prepared by reinforcing threads placed in the material so that a non-circular patch can be created. A side view of a deformed zone can be obtained by laminating materials, such as using clays with different colors. 2.90 Describe your observations regarding Fig. 2.14c. There are many potential observations and students should be encouraged to develop their own well-stated insights. Examples of observations are the following: • The penetration depth is very small compared to the depth at which material is still affected by the hardness test. • The aspect ratio of a typical hardness test can be described, note the large diameter of the indent compared to its depth. • There doesn’t seem to be evidence of shear at the indentor, although it is admittedly difficult to discern from this image. • A ridge on the hardness test as shown in Fig. 2.14b doesn’t appear to be present, but from the image it looks like material may have encountered flow in the thickness direction (i.e., out of the page). 2.91 Embed a small steel ball in a soft block of material such as clay, and compress the clay as shown in Fig. 2.24a. Then slice the clay carefully along the center plane and observe the deformation of the material. Repeat the experiment by embedding a small round jelly bean in the clay and deforming the material. Comment on your observations.



37

By the student. The steel ball will not deform, just as is shown in the middle sketch of Fig. 2.24b on p. 78. The jelly bean has to be soft in order to simulate the behavior R in a carefully shown in Fig. 2.24a. An alternative is to use two colors of Play-Doh prepared specimen, with one material chilled to increase its stiffness. 2.92 Devise a simple experiment, and perform tests on materials commonly found around the house by bending them at different temperatures, for a qualitative assessment of their transition temperature, as shown in Fig. 2.25. By the student. This experiment can be performed fairly easily on materials such as butter, chocolate, or cheese. The temperature can be varied and controlled by putting the materials in a freezer or increasing it by immersion into water heated to different temperatures. A qualitative assessment of transition temperature can thus be done; also, a number of experiments can be performed to assess ductility (such as shearing the material and observing the strains that can be achieved - shearing can include smearing with a knife). 2.93 Obtain some solid and some tubular metal pieces, and slit them as shown in Fig. 2.31. Comment on whether there were any residual stresses in the parts prior to slitting them. By the student. The answers can vary widely depending on the materials and their processing history. Most tubes have been extruded and drawn and will have some residual stresses notable. Some tubes may be annealed and therefore be residual stress free. 2.94 Explain how you would obtain an estimate of the hardness for a carbon nanotube. (See Section 8.6.2.) This is an open-ended problem, and the student should be encouraged to consider a wide range of solutions. The hardness of a nanotube would be very difficult to measure directly. It may be more practical to measure properties that depend on hardness. For example, one could evaluate a carbon nanotube by conducting a wear test in an atomic force microscope and then comparing to wear tests on other materials to infer the hardness. One could measure the breaking stress of a carbon nanotube and relating this to hardness by H = 3Y (see p. 72). Alternatively, one could conduct molecular dynamics simulations of carbon nanotubes being penetrated by a flat indentor. These approaches are not perfect, but the student should be encouraged to develop ideas. 2.95 Without using the words “stress” or “strain,” define elastic modulus. This is actually quite challenging, but historically significant, since Thomas Young did not have the benefit of the concept of strain when he first defined modulus of elasticity. Young’s definition satisfies the project requirement. In Young’s words: The modulus of the elasticity of any substance is a column of the same substance, capable of producing a pressure on its base which is to the weight causing a certain degree of compression as the length of the substance is to the diminution of the length. There are many possible other definitions, of course. 2.96 We know that it is relatively easy to subject a specimen to hydrostatic compression, such as by using a chamber filled with a liquid. Devise a means whereby the specimen (say, in the shape of a cube or a round disk) can



38

be subjected to hydrostatic tension, or one approaching this state of stress. (Note that a thin-walled, internally pressurized spherical shell is not a correct answer, because it is subjected only to a state of plane stress.) By the student. Students should be encouraged to design test specimens and means for applying stresses. One solution is to have a cube with a fixture adhesively bonded to each of the sides of the cube, and the same load applied to each face. Another alternative is to place a sphere in a mold of a stronger material and then reduce temperature (if the coefficient of thermal expansion is the same). Students or groups should be encouraged to describe more options.


Ch 02

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