Solution manual for manufacturing processes for engineering materials 6th edition kalpakjian schmid

Solution Manual for Manufacturing Processes for Engineering Materials 6th Edition Kalpakjian Schmid 0134290550 9780134290553

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Chapter 2

Fundamentals of the Mechanical Behavior of Materials © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

QUESTIONS 2.1 Can you calculate the percent elongation of materials based only on the information given in Fig. 2.6? Explain.

strain curve. Explain whether this condition also holds for a compression test.

Recall that the percent elongation is defined by Eq. (2.6) on p. 35 and depends on the original gage length (lo ) of the specimen. From Fig. 2.6 on p. 39, only the necking strain (true and engineering) and true fracture strain can be determined. Thus, we cannot calculate the percent elongation of the specimen; also, note that the elongation is a function of gage length and increases with gage length.

During a compression test, the cross-sectional area of the specimen increases as the specimen height decreases (because of volume constancy) as the load is increased. Since true stress is defined as ratio of the load to the instantaneous cross-sectional area of the specimen, the true stress in compression will be lower than the engineering stress for a given load, assuming that friction between the platens and the specimen is negligible.

2.2 Explain if it is possible for stress-strain curves in tension tests to reach 0% elongation as the gage length is increased further.

2.5 Which of the two tests, tension or compression, requires a higher capacity testing machine than the other? Explain.

The percent elongation of the specimen is a function of the initial and final gage lengths. When the specimen is being pulled, regardless of the original gage length, it will elongate uniformly (and permanently) until necking begins. Therefore, the specimen will always have a certain finite elongation. However, note that as the specimen’s gage length is increased, the contribution of localized elongation (that is, necking) will decrease, but the total elongation will not approach zero.

The compression test requires a higher capacity machine because the cross-sectional area of the specimen increases during the test, which is the opposite of a tension test. The increase in area requires a load higher than that for the tension test to achieve the same stress level. Furthermore, note that compression-test specimens generally have a larger original cross-sectional area than those for tension tests, thus requiring higher forces.

2.3 Explain why the difference between engineering strain and true strain becomes larger as strain increases. Is this phenomenon true for both tensile and compressive strains? Explain.

2.6 Explain how the modulus of resilience of a material changes, if at all, as it is strained: (a) for an elastic, perfectly plastic material, and (b) for an elastic, linearly strain-hardening material.

The difference between the engineering and true strains becomes larger because of the way the strains are defined, respectively, as can be seen by inspecting Eqs. (2.1) and (2.9). This is true for both tensile and compressive strains.

Recall that the modulus of resilience is given by Eq. (2.5) on p. 34 as Sy2 /(2E). (a) If the material is perfectly plastic, then the yield strength does not increase with strain - see Fig. 2.7c on p. 42. Therefore, the modulus of resilience is unchanged as the material is strained. (b) For a linear strain hardening material, the yield strength increases with plastic strain. Therefore the modulus of resilience will increase with strain.

2.4 Using the same scale for stress, the tensile true-stresstrue-strain curve is higher than the engineering stress-

1

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of toughness. The modulus of elasticity and yield stress would provide information about the area under the elastic region; however, this region is very small and is thus usually negligible with respect to the rest of the stress-strain curve.

2.7 If you pull and break a tensile-test specimen rapidly, where would the temperature be the highest? Explain why.

Since temperature rise is due to the work input, the temperature will be highest in the necked region because that is where the strain, hence the energy dissipated per unit volume in plastic deformation, is highest.

2.12 Modify the curves in Fig. 2.7 to indicate the effects of temperature. Explain your changes.

These modifications can be made by lowering the slope of the elastic region and lowering the general height of the curves. See, for example, Fig. 2.9 on p. 43.

2.8 Comment on the temperature distribution if the specimen in Question 2.7 is pulled very slowly.

2.13 Using a specific example, show why the deformation rate, say in m/s, and the true strain rate are not the same.

If the specimen is pulled very slowly, the temperature generated will be dissipated throughout the specimen and to the environment. Thus, there will be no appreciable temperature rise anywhere, particularly with materials with high thermal conductivity.

The deformation rate is the quantity v in Eqs. (2.16) and (2.17). Thus, when v is held constant during deformation (hence a constant deformation rate), the true strain rate will vary (l increases), whereas the engineering strain rate will remain constant. Hence, the two quantities are not the same.

2.9 In a tension test, the area under the true-stress-truestrain curve is the work done per unit volume (the specific work). Also, the area under the load-elongation curve represents the work done on the specimen. If you divide this latter work by the volume of the specimen between the gage marks, you will determine the work done per unit volume (assuming that all deformation is confined between the gage marks). Will this specific work be the same as the area under the truestress-true-strain curve? Explain. Will your answer be the same for any value of strain? Explain.

2.14 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, 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. As the tension test progresses, necking becomes more diffuse, and the specimen becomes longer before fracture; hence, total elongation increases with increasing values of m. As expected, the elongation after necking (postuniform elongation) also increases with increasing m. It has been observed that the value of m decreases with metals of increasing strength.

If we divide the work done by the total volume of the specimen between the gage lengths, we obtain the average specific work throughout the specimen. However, the area under the true stress-true strain curve represents the specific work done at the necked (and fractured) region in the specimen where the strain is a maximum. Thus, the answers will be different. However, up to the onset of necking (instability), the specific work calculated will be the same. This is because the strain is uniform throughout the specimen until necking begins.

2.10 The note at the bottom of Table 2.4 states that as temperature increases, C decreases and m increases. Explain why.

The value of C in Table 2.4 on p. 46 decreases with temperature because it is a measure of the strength of the material. The value of m increases with temperature because the material becomes more strain-rate sensitive, due to the fact that the higher the strain rate, the less time the material has to recover and recrystallize, hence its strength increases.

2.15 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 2.14 above.

2.11 You are given the K and n values of two different materials. Is this information sufficient to determine which material is tougher? If not, what additional information do you need, and why?

2.16 Assume that you are running four-point bending tests on a number of identical specimens of the same length and cross-section, but with increasing distance between the upper points of loading (see Fig. 2.19b).

What changes, if any, would you expect in the test results? Explain.

Although the K and n values may give a good estimate of toughness, the true fracture stress and the true strain at fracture are required for accurate calculation

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2.22 Referring to Fig. 2.20, the material for testers are either steel, tungsten carbide, or diamond. Why isn’t diamond used for all of the tests?

As the distance between the upper points of loading in Fig. 2.19b increases, the magnitude of the bending moment decreases. However, the volume of material subjected to the maximum bending moment (hence to maximum stress) increases. Thus, the probability of failure in the four-point test increases as this distance increases.

While diamond is the hardest material known, it would not, for example, be practical to make and use a 10-mm diamond indenter because the costs would be prohibitive. Consequently, a hard material such as those listed are sufficient for most hardness tests.

2.17 Would Eq. (2.10) hold true in the elastic range? Explain.

2.23 What role does friction play in a hardness test? Can high friction between a material and indenter affect a hardness test? Explain.

Note that this equation is based on volume constancy, i.e., Ao lo = Al. We know, however, that because the Poisson’s ratio ν is less than 0.5 in the elastic range, the volume is not constant in a tension test; see Eq. (2.47) on p. 71. Therefore, the expression is not valid in the elastic range.

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. 2.23 on p. 58.

2.18 Why have different types of hardness tests been developed? How would you measure the hardness of a very large object?

2.24 Describe the difference between creep and stress relaxation, giving two examples for each as they relate to engineering applications.

There are several basic reasons: 1. The overall hardness range of the materials 2. The hardness of their constituents; see Chapter 3; 3. The thickness of the specimen, such as bulk versus foil 4. The size of the specimen with respect to that of the indenter 5. The surface finish of the part being tested.

Creep is the permanent deformation of a part that is under a load over a period of time, usually occurring at elevated temperatures. Stress relaxation is the decrease in the stress level in a part under a constant strain. Examples of creep include: 1. turbine blades operating at high temperatures, and 2. high-temperature steam linesand furnace components.

2.19 Which hardness tests and scales would you use for very thin strips of material, such as aluminum foil?

Why?

Stress relaxation is observed when, for example, a rubber band or a thermoplastic is pulled to a specific length and held at that length for a period of time. This phenomenon is commonly observed in rivets, bolts, and guy wires, as well as thermoplastic components.

Because aluminum foil is very thin, the indentations on the surface must be very small so as not to affect test results. Suitable tests would be a microhardness test such as Knoop or Vickers under very light loads (see Fig. 2.20 on p. 54). The accuracy of the test can be validated by observing any changes in the surface appearance opposite to the indented side.

2.25 Referring to the two impact tests shown in Fig. 2.26, explain how different the results would be if the specimens were impacted from the opposite directions.

2.20 List and explain the factors that you would consider in selecting an appropriate hardness test and scale for a particular application.

Note that impacting the specimens shown in Fig. 2.26 on p. 61 from the opposite directions would subject the roots of the notches to compressive stresses, and thus they would not act as stress raisers. Thus, cracks would not propagate as they would when under tensile stresses. Consequently, the specimens would basically behave as if they were not notched.

Hardness tests mainly have three differences: 1. type of indenter, 2. applied load, and 3. method of indentation measurement (depth or surface area of indentation, or rebound of indenter).

2.26 If you remove the layer ad from the part shown in Fig. 2.27d, such as by machining or grinding, which way will the specimen curve? (Hint: Assume that the part in diagram (d) is composed of four horizontal springs held at the ends. Thus, from the top down, we have compression, tension, compression, and tension springs.)

2.21 In a Brinell hardness test, the resulting impression is found to be an ellipse. Give possible explanations for this result.

There are several possible reasons for this phenomenon, but the two most likely are anisotropy in the material and the presence of surface residual stresses in the material.

Since the internal forces will have to achieve a state of static equilibrium, the new part has to bow downward (i.e., it will hold water). Such residual-stress patterns

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can be modeled with a set of horizontal tension and compression springs. Note that the top layer of the material ad in Fig. 2.27d, which is under compression, has the tendency to bend the bar upward. When this stress is relieved (such as by removing a layer), the bar will compensate for it by bending downward.

(g) Gas turbine blade: A gas turbine blade operates at high temperatures (depending on its location in the turbine); thus it should have high-temperature strength and resistance to creep, as well as to oxidation and corrosion due to combustion products during its use.

2.27 Is it possible to completely remove residual stresses in a piece of material by the technique described in Fig. 2.29 if the material is elastic, linearly strain hardening? Explain.

(h) Staple: The properties should be closely parallel to that of a paper clip. The staple should have high ductility to allow it to be deformed without fracture, and also have low yield stress so that it can be bent (as well as unbent when removing it) easily without requiring excessive force.

By following the sequence of events depicted in Fig. 2.29 on p. 64, it can be seen that it is not possible to completely remove the residual stresses. Note that for an elastic, linearly strain hardening material, σc will never catch up with σt .

0

2.30 Make a sketch showing the nature and distribution of the residual stresses in Figs. 2.28a and b before the parts were cut. Assume that the split parts are free from any stresses. (Hint: Force these parts back to the shape they were in before they were cut.)

0

2.28 Referring to Fig. 2.29, would it be possible to eliminate residual stresses by compression? Assume that the piece of material will not buckle under the uniaxial compressive force.

As the question states, when we force back the split portions in the specimen in Fig. 2.28a on p. 63, we induce tensile stresses on the outer surfaces and compressive on the inner. Thus the original part would, along its total cross section, have a residual stress distribution of tension-compression-tension. Using the same technique, we find that the specimen in Fig. 2.28b would have a similar residual stress distribution prior to cutting.

Yes, by the same mechanism described in Fig. 2.29 on p. 64.

2.29 List and explain the desirable mechanical properties for (a) an elevator cable; (b) a bandage; (c) a shoe sole; (d) a fish hook; (e) an automotive piston; (f) a boat propeller; (g) a gas-turbine blade; and (h) a staple.

2.31 It is possible to calculate 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.

The following are some basic considerations: (a) Elevator cable: The cable should not elongate elastically to a large extent or undergo yielding as the load is increased. These requirements thus call for a material with a high elastic modulus and yield stress.

If we calculate the heat using a constant specific heat value in Eq. (2.62), the work will be higher than it actually is. This is because, by definition, as the specific heat decreases, less work is required to raise the workpiece temperature by one degree. Consequently, the calculated work will be higher than the actual work done.

(b) Bandage: The bandage material must be compliant, that is, have a low stiffness, but have high strength in the membrane direction. Its inner surface must be permeable and outer surface resistant to environmental effects. (c) Shoe sole: The sole should be compliant for comfort, with a high resilience. It should be tough so that it absorbs shock and should have high friction and wear resistance.

2.32 Explain whether or not the volume of a metal specimen changes when the specimen is subjected to a state of (a) uniaxial compressive stress and (b) uniaxial tensile stress, all in the elastic range.

(d) Fish hook: A fish hook needs to have high strength so that it doesn’t deform permanently under load, and thus maintain its shape. It should be stiff (for better control during its use) and should be resistant the environment it is used in (such as salt water).

For case (a), the quantity in parentheses in Eq. (2.47) on p. 71 will be negative, because of the compressive stress. Since the rest of the terms are positive, the product of these terms is negative and, hence, there will be a decrease in volume (This can also be deduced intuitively.) For case (b), it will be noted that the volume will increase.

(e) Automotive piston: This product must have high strength at elevated temperatures, high physical and thermal shock resistance, and low mass. (f) Boat propeller: The material must be stiff (to maintain its shape) and resistant to corrosion, and also have abrasion resistance because the propeller encounters sand and other abrasive particles when operated close to shore.

2.33 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

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