Chapter 2

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.

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.

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.

List and explain briefly the conditions that induce brittle fracture in an otherwise ductile metal.

Brittle fracture can be induced by high deformation rates, lower temperatures (particu- larly 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).

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.

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 cross- sectional 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.

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= √2 2 (πD) D− D −dTherefore, solving for d, 2P 2 2(500) 2 d= D2− D−(πD)(HB) = 102− 10−[π(10)](4) =9.79mm 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: 1.854P 1.854P 1.854(1) HV= L2 → L= HX = 4 =0.68mm This is much more reasonable, suggesting that the Vickers test is the best alternative for lead. iii. 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. iv. 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. v. 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. vi. Granite. The hardness of granite varies according to the source, but it is approx- imately 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.

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.

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 require- ment. 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.

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 ofservice 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.

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 isnot 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.)

A statistical sampling of Rockwell C hardness tests are conducted on a mate- rial, 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 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.

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.

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 com- posite 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.

Explain why materials with high m values, such as hot glass and Silly Putty⃝R , 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.

Explain why the difference between engineering strain and true strain be- comes 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.

Which of the two tests, tension or compression, would require a higher ca- pacity 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).

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.

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.

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.

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.

Consider an elastomer, such as a rubber band. This material can undergo a large elastic deformation before failure, but after fracture it recovers com- pletely 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.

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.

Some coatings are extremely thin - some as thin as a few nanometers. Ex- plain why even the Knoop test is not able to give reliable results for such coatings. Recent investigations have attempted to use highly polished dia- monds (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 one- tenth 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).

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.