Solutions Manual for Manufacturing Processes for EngineeringMaterials 5th edition by Kalpakjian and Schmid Link download: https://digitalcontentmarket.org/download/solutions-manual-for- manufacturing-processes-for-engineering-materials-5th-edition-by-kalpakjian-and- schmid/ Chapter 2 Fundamentals of the Mechanical Behavior of Materials Questions 2.1 Can you calculate the percent elongation of increases. Is this phenomenon true for both ma-terials based only on the information ten-sile and compressive strains? Explain. given in Fig. 2.6? Explain. The difference between the engineering and Recall that the percent elongation is defined by true strains becomes larger because of the Eq. (2.6) on p. 33 and depends on the original way the strains are defined, respectively, as gage length (lo) of the specimen. From Fig. 2.6 can be seen by inspecting Eqs. (2.1) on p. 30 on p. 37 only the necking strain (true and engi- and (2.9) on p. 35. This is true for both tensile neering) and true fracture strain can be deter- and com-pressive strains. mined. Thus, we cannot calculate the percent 2.4 Using the same scale for stress, we note that elongation of the specimen; also, note that the the tensile true-stress-true-strain curve is elongation is a function of gage length and in- higher than the engineering stress-strain creases with gage length. curve. Ex-plain whether this condition also 2.2 Explain if it is possible for the curves in Fig. holds for a compression test. 2.4 to reach 0% elongation as the gage length During a compression test, the cross-sectional is in-creased further. area of the specimen increases as the specimen The percent elongation of the specimen is a height decreases (because of volume function of the initial and final gage lengths. constancy) as the load is increased. Since true When the specimen is being pulled, regardless stress is de-fined as ratio of the load to the of the original gage length, it will elongate uni- instantaneous cross-sectional area of the formly (and permanently) until necking begins. specimen, the true stress in compression will be Therefore, the specimen will always have a cer- lower than the en-gineering stress for a given tain finite elongation. However, note that as the load, assuming that friction between the platens specimen’s gage length is increased, the contri- and the specimen is negligible. bution of localized elongation (that is, necking) 2.5 Which of the two tests, tension or will decrease, but the total elongation will not compression, requires a higher capacity approach zero. testing machine than the other? Explain. 2.3 Explain why the difference between engineering The compression test requires a higher capacity strain and true strain becomes larger as strain machine because the cross-sectional area of the 1 specimen increases during the test, which is stress-true strain curve represents the specific the opposite of a tension test. The increase in work done at the necked (and fractured) region area requires a load higher than that for the in the specimen where the strain is a maximum. ten-sion test to achieve the same stress level. Thus, the answers will be different. However, up Fur-thermore, note that compression-test to the onset of necking (instability), the spe-cific specimens generally have a larger original work calculated will be the same. This is cross-sectional area than those for tension because the strain is uniform throughout the tests, thus requiring higher forces. specimen until necking begins. 2.10 The note at the bottom of Table 2.5 states that 2.6 Explain how the modulus of resilience of a ma- as temperature increases, C decreases and terial changes, if at all, as it is strained: (1) for m increases. Explain why. an elastic, perfectly plastic material, and (2) for an elastic, linearly strain-hardening material. The value of C in Table 2.5 on p. 43 decreases with temperature because it is a measure of the 2.7 If you pull and break a tension-test specimen strength of the material. The value of m in- rapidly, where would the temperature be the creases with temperature because the material highest? Explain why. becomes more strain-rate sensitive, due to the fact that the higher the strain rate, the less time Since temperature rise is due to the work the material has to recover and recrystallize, input, the temperature will be highest in the hence its strength increases. necked region because that is where the strain, hence the energy dissipated per unit 2.11 You are given the K and n values of two dif- volume in plastic deformation, is highest. ferent materials. Is this information sufficient to determine which material is tougher? If not, 2.8 Comment on the temperature distribution if the what additional information do you need, and specimen in Question 2.7 is pulled very slowly. why? If the specimen is pulled very slowly, the tem- Although the K and n values may give a good perature generated will be dissipated through- estimate of toughness, the true fracture stress out the specimen and to the environment. and the true strain at fracture are required for Thus, there will be no appreciable accurate calculation of toughness. The modu- temperature rise anywhere, particularly with lus of elasticity and yield stress would provide materials with high thermal conductivity. information about the area under the elastic re-gion; however, this region is very small and 2.9 In a tension test, the area under the true-stress- is thus usually negligible with respect to the true-strain curve is the work done per unit vol- rest of the stress-strain curve. ume (the specific work). We also know that the area under the load-elongation curve rep- 2.12 Modify the curves in Fig. 2.7 to indicate the resents the work done on the specimen. If you effects of temperature. Explain the reasons for divide this latter work by the volume of the your changes. specimen between the gage marks, you will de- These modifications can be made by lowering termine the work done per unit volume (assum- the slope of the elastic region and lowering ing that all deformation is confined between the the general height of the curves. See, for gage marks). Will this specific work be the same example, Fig. 2.10 on p. 42. as the area under the true-stress-true-strain curve? Explain. Will your answer be the same 2.13 Using a specific example, show why the defor- for any value of strain? Explain. mation rate, say in m/s, and the true strain rate are not the same. If we divide the work done by the total volume of the specimen between the gage lengths, we The deformation rate is the quantity v in Eqs. obtain the average specific work throughout the (2.14), (2.15), (2.17), and (2.18) on pp. 41-46. specimen. However, the area under the true Thus, when v is held constant during de- 2 formation (hence a constant deformation rate), However, the volume of material subjected to the true strain rate will vary, whereas the engi- the maximum bending moment (hence to neering strain rate will remain constant. Hence, max-imum stress) increases. Thus, the the two quantities are not the same. probability of failure in the four-point test increases as this distance increases. 2.14 It has been stated that the higher the value of m, the more diffuse the neck is, and likewise, the 2.17 Would Eq. (2.10) hold true in the elastic lower the value of m, the more localized the range? Explain. neck is. Explain the reason for this behavior. Note that this equation is based on volume As discussed in Section 2.2.7 starting on p. 41, con-stancy, i.e., Aolo = Al. We know, however, with high m values, the material stretches to a that because the Poisson’s ratio ν is less than greater length before it fails; this behavior is an 0.5 in the elastic range, the volume is not indication that necking is delayed with increasing constant in a tension test; see Eq. (2.47) on p. m. When necking is about to be-gin, the necking 69. There-fore, the expression is not valid in region’s strength with respect to the rest of the the elastic range. specimen increases, due to strain hardening. However, the strain rate in the necking region is 2.18 Why have different types of hardness tests also higher than in the rest of the specimen, been developed? How would you measure because the material is elon-gating faster there. the hard-ness of a very large object? Since the material in the necked region becomes There are several basic reasons: (a) The overall stronger as it is strained at a higher rate, the region hardness range of the materials; (b) the hard- exhibits a greater re-sistance to necking. The ness of their constituents; see Chapter 3; (c) the increase in resistance to necking thus depends on thickness of the specimen, such as bulk versus the magnitude of m. As the tension test foil; (d) the size of the specimen with respect to progresses, necking be-comes more diffuse, and that of the indenter; and (e) the surface finish of the specimen becomes longer before fracture; the part being tested. hence, total elongation increases with increasing values of m (Fig. 2.13 on p. 45). As expected, the 2.19 Which hardness tests and scales would you elongation after necking (postuniform elongation) use for very thin strips of material, such as also increases with increasing m. It has been alu-minum foil? Why? observed that the value of m decreases with metals of increas-ing strength. Because aluminum foil is very thin, the indenta- tions on the surface must be very small so as not to affect test results. Suitable tests would be 2.15 Explain why materials with high m values a microhardness test such as Knoop or Vickers (such as hot glass and silly putty) when under very light loads (see Fig. 2.22 on p. 52). stretched slowly, undergo large elongations The accuracy of the test can be validated by ob- before failure. Consider events taking place in serving any changes in the surface appearance the necked re-gion of the specimen. opposite to the indented side. The answer is similar to Answer 2.14 above. 2.20 List and explain the factors that you would con- sider in selecting an appropriate hardness test 2.16 Assume that you are running four-point bend-ing and scale for a particular application. tests on a number of identical specimens of the same length and cross-section, but with in- Hardness tests mainly have three differences: creasing distance between the upper points of loading (see Fig. 2.21b). What changes, if any, (a) type of indenter, would you expect in the test results? Explain. (b) applied load, and As the distance between the upper points of (c) method of indentation measurement loading in Fig. 2.21b on p. 51 increases, the (depth or surface area of indentation, or magnitude of the bending moment decreases. rebound of indenter). 3 The hardness test selected would depend on 2.23 Describe the difference between creep and the estimated hardness of the workpiece, its stress-relaxation phenomena, giving two size and thickness, and if an average exam-ples for each as they relate to hardness or the hardness of individual engineering ap-plications. microstructural compo-nents is desired. For instance, the scleroscope, which is portable, Creep is the permanent deformation of a part is capable of measuring the hardness of large that is under a load over a period of time, usu- pieces which otherwise would be difficult or ally occurring at elevated temperatures. impossible to measure by other techniques. Stress relaxation is the decrease in the stress level in a part under a constant strain. The Brinell hardness measurement leaves a fairly Examples of creep include: large indentation which provides a good measure of average hardness, while the Knoop test leaves (a) turbine blades operating at high a small indentation that allows, for example, the tempera-tures, and determination of the hardness of individual phases in a two-phase alloy, as well as inclusions. The (b) high-temperature steam linesand small indentation of the Knoop test also allows it to furnace components. be useful in measuring the hardness of very thin Stress relaxation is observed when, for layers on parts, such as plating or coatings. Recall example, a rubber band or a thermoplastic is that the depth of in-dentation should be small pulled to a specific length and held at that length relative to part thick-ness, and that any change on for a period of time. This phenomenon is the bottom sur-face appearance makes the test commonly observed in rivets, bolts, and guy results invalid. wires, as well as thermoplastic components. 2.21 In a Brinell hardness test, the resulting impres- 2.24 Referring to the two impact tests shown in Fig. sion is found to be an ellipse. Give possible 2.31, explain how different the results would explanations for this phenomenon. be if the specimens were impacted from the There are several possible reasons for this opposite directions. phenomenon, but the two most likely are Note that impacting the specimens shown in anisotropy in the material and the presence of Fig. 2.31 on p. 60 from the opposite directions surface residual stresses in the material. would subject the roots of the notches to com- 2.21 Referring to Fig. 2.22 on p. 52, note that the pressive stresses, and thus they would not act material for indenters are either steel, as stress raisers. Thus, cracks would not propa- tungsten carbide, or diamond. Why isn’t gate as they would when under tensile stresses. diamond used for all of the tests? Consequently, the specimens would basically behave as if they were not notched. While diamond is the hardest material known, it would not, for example, be practical to make 2.25 If you remove layer ad from the part shown in and use a 10-mm diamond indenter because Fig. 2.30d, such as by machining or grinding, the costs would be prohibitive. Consequently, which way will the specimen curve? (Hint: As- a hard material such as those listed are sume that the part in diagram (d) can be mod- sufficient for most hardness tests. eled as consisting of four horizontal springs held at the ends. Thus, from the top down, we 2.22 What effect, if any, does friction have in a have compression, tension, compression, and hard-ness test? Explain. tension springs.) The effect of friction has been found to be mini- Since the internal forces will have to achieve a mal. In a hardness test, most of the indentation state of static equilibrium, the new part has to occurs through plastic deformation, and there is bow downward (i.e., it will hold water). Such very little sliding at the indenter-workpiece residual-stress patterns can be modeled with a interface; see Fig. 2.25 on p. 55. set of horizontal tension and compression 4 springs. Note that the top layer of the mate- (d) Fish hook: A fish hook needs to have rial ad in Fig. 2.30d on p. 60, which is under high strength so that it doesn’t deform compression, has the tendency to bend the perma-nently under load, and thus bar upward. When this stress is relieved maintain its shape. It should be stiff (for (such as by removing a layer), the bar will better con-trol during its use) and should compensate for it by bending downward. be resistant the environment it is used in (such as salt water). 2.26 Is it possible to completely remove residual (e) Automotive piston: This product must stresses in a piece of material by the technique have high strength at elevated tempera- described in Fig. 2.32 if the material is elastic, tures, high physical and thermal shock linearly strain hardening? Explain. re-sistance, and low mass. By following the sequence of events depicted in (f) Boat propeller: The material must be stiff Fig. 2.32 on p. 61, it can be seen that it is not (to maintain its shape) and resistant to possible to completely remove the residual corrosion, and also have abrasion re- stresses. Note that for an elastic, linearly strain sistance because the propeller 0 hardening material, σc will never catch up with encounters sand and other abrasive 0 σt . particles when op-erated close to shore. (g) Gas turbine blade: A gas turbine blade 2.27 Referring to Fig. 2.32, would it be possible to op-erates at high temperatures eliminate residual stresses by compression (depending on its location in the turbine); in-stead of tension? Assume that the piece of thus it should have high-temperature ma-terial will not buckle under the uniaxial strength and resis-tance to creep, as com-pressive force. well as to oxidation and corrosion due to Yes, by the same mechanism described in combustion products dur-ing its use. Fig. 2.32 on p. 61. (h) Staple: The properties should be closely parallel to that of a paper clip. The staple 2.28 List and explain the desirable mechanical prop- should have high ductility to allow it to be erties for the following: (1) elevator cable, (2) deformed without fracture, and also have bandage, (3) shoe sole, (4) fish hook, (5) au- low yield stress so that it can be bent (as tomotive piston, (6) boat propeller, (7) gas- well as unbent when removing it) easily turbine blade, and (8) staple. without requiring excessive force. The following are some basic considerations: 2.29 Make a sketch showing the nature and distribu-tion of the residual stresses in Figs. (a) Elevator cable: The cable should not 2.31a and b before the parts were split (cut). elon-gate elastically to a large extent or Assume that the split parts are free from any un-dergo yielding as the load is stresses. (Hint: Force these parts back to the increased. These requirements thus call shape they were in before they were cut.) for a mate-rial with a high elastic modulus and yield stress. As the question states, when we force back (b) Bandage: The bandage material must be the split portions in the specimen in Fig. 2.31a compliant, that is, have a low stiffness, on p. 60, we induce tensile stresses on the but have high strength in the membrane outer surfaces and compressive on the inner. direc-tion. Its inner surface must be Thus the original part would, along its total permeable and outer surface resistant to cross section, have a residual stress distribu- environmen-tal effects. tion of tension-compression-tension. Using the same technique, we find that the (c) Shoe sole: The sole should be compliant specimen in Fig. 2.31b would have a similar for comfort, with a high resilience. It residual stress distribution prior to cutting. should be tough so that it absorbs shock and should have high friction and wear 2.30 It is possible to calculate the work of plastic re-sistance. deformation by measuring the temperature rise 5 in a workpiece, assuming that there is no heat (b) A thin, solid round disk (such as a coin) loss and that the temperature distribution is and made of a soft material is brazed be- uniform throughout. If the specific heat of the tween the ends of two solid round bars material decreases with increasing temperature, of the same diameter as that of the disk. will the work of deformation calculated using the When subjected to longitudinal tension, specific heat at room temperature be higher or the disk will tend to shrink radially. But lower than the actual work done? Explain. because it is thin and its flat surfaces are restrained by the two rods from moving, If we calculate the heat using a constant the disk will be subjected to tensile radial specific heat value in Eq. (2.65) on p. 73, the stresses. Thus, a state of triaxial (though work will be higher than it actually is. This is not exactly hydrostatic) tension will exist because, by definition, as the specific heat within the thin disk. decreases, less work is required to raise the workpiece temper-ature by one degree. 2.33 Referring to Fig. 2.19, make sketches of the Consequently, the calcu-lated work will be state of stress for an element in the reduced higher than the actual work done. section of the tube when it is subjected to (1) torsion only, (2) torsion while the tube is in- 2.31 Explain whether or not the volume of a metal ternally pressurized, and (3) torsion while the specimen changes when the specimen is tube is externally pressurized. Assume that sub-jected to a state of (a) uniaxial the tube is closed end. compressive stress and (b) uniaxial tensile stress, all in the elastic range. These states of stress can be represented simply by referring to the contents of this For case (a), the quantity in parentheses in chapter as well as the relevant materials Eq. (2.47) on p. 69 will be negative, because covered in texts on mechanics of solids. of the compressive stress. Since the rest of 2.34 A penny-shaped piece of soft metal is brazed the terms are positive, the product of these to the ends of two flat, round steel rods of the terms is negative and, hence, there will be a same diameter as the piece. The assembly is decrease in volume (This can also be then subjected to uniaxial tension. What is the deduced intuitively.) For case (b), it will be state of stress to which the soft metal is sub- noted that the volume will increase. jected? Explain. 2.32 We know that it is relatively easy to subject a The penny-shaped soft metal piece will tend specimen to hydrostatic compression, such as to contract radially due to the Poisson’s ratio; by using a chamber filled with a liquid. Devise a however, the solid rods to which it attached means whereby the specimen (say, in the shape will prevent this from happening. of a cube or a thin round disk) can be subjected Consequently, the state of stress will tend to to hydrostatic tension, or one approaching this approach that of hy-drostatic tension. state of stress. (Note that a thin-walled, inter- nally pressurized spherical shell is not a correct2.35 A circular disk of soft metal is being com- answer, because it is subjected only to a state pressed between two flat, hardened circular of plane stress.) steel punches having the same diameter as the disk. Assume that the disk material is Two possible answers are the following: perfectly plastic and that there is no friction or any tem-perature effects. Explain the change, (a) A solid cube made of a soft metal has all if any, in the magnitude of the punch force as its six faces brazed to long square bars (of the disk is being compressed plastically to, the same cross section as the specimen); say, a fraction of its original thickness. the bars are made of a stronger metal. The six arms are then subjected to equal Note that as it is compressed plastically, the tension forces, thus subjecting the cube to disk will expand radially, because of volume equal tensile stresses. constancy. An approximately donut-shaped 6 material will then be pushed radially out-ward, 2.40 What test would you use to evaluate the hard- which will then exert radial compressive ness of a coating on a metal surface? Would stresses on the disk volume under the punches. it matter if the coating was harder or softer The volume of material directly between the than the substrate? Explain. punches will now subjected to a triaxial com- pressive state of stress. According to yield cri- teria (see Section 2.11), the compressive stress The answer depends on whether the coating is exerted by the punches will thus increase, even relatively thin or thick. For a relatively thick though the material is not strain hardening. coating, conventional hardness tests can be Therefore, the punch force will increase as de- con-ducted, as long as the deformed region formation increases. under the indenter is less than about one-tenth 2.36 A perfectly plastic metal is yielding under the of the coating thickness. If the coating thickness stress state σ1, σ2, σ3, where σ1 > σ2 > σ3. is less than this threshold, then one must ei-ther Explain what happens if σ1 is increased. rely on nontraditional hardness tests, or else use fairly complicated indentation models to Consider Fig. 2.36 on p. 67. Points in the in- extract the material behavior. As an exam-ple of terior of the yield locus are in an elastic state, the former, atomic force microscopes us-ing whereas those on the yield locus are in a plas- diamond-tipped pyramids have been used to tic state. Points outside the yield locus are not measure the hardness of coatings less than 100 admissible. Therefore, an increase in σ 1 while nanometers thick. As an example of the lat-ter, the other stresses remain unchanged would re- finite-element models of a coated substrate quire an increase in yield stress. This can also being indented by an indenter of a known ge- be deduced by inspecting either Eq. (2.36) or ometry can be developed and then correlated to Eq. (2.37) on p. 64. experiments. 2.37 What is the dilatation of a material with a Pois- son’s ratio of 0.5? Is it possible for a material to have a Poisson’s ratio of 0.7? Give a rationale for your answer. 2.41 List the advantages and limitations of the It can be seen from Eq. (2.47) on p. 69 that the stress-strain relationships given in Fig. 2.7. dilatation of a material with ν = 0.5 is always zero, regardless of the stress state. To examine the case of ν = 0.7, consider the situation where Several answers that are acceptable, and the the stress state is hydrostatic tension. Equation student is encouraged to develop as many as (2.47) would then predict contraction under a possible. Two possible answers are: (1) there is tensile stress, a situation that cannot occur. a tradeoff between mathematical complex-ity 2.38 Can a material have a negative Poisson’s and accuracy in modeling material behavior and ratio? Explain. (2) some materials may be better suited for certain constitutive laws than others. Solid material do not have a negative Poisson’s ratio, with the exception of some composite ma- terials (see Chapter 10), where there can be a negative Poisson’s ratio in a given direction. 2.42 Plot the data in Table 2.1 on a bar chart, 2.39 As clearly as possible, define plane stress and show-ing the range of values, and comment plane strain. on the results. Plane stress is the situation where the stresses in one of the direction on an element are zero; plane strain is the situation where By the student. An example of a bar chart for the strains in one of the direction are zero. the elastic modulus is shown below. 7 Metallic materials that the hardness is too high, thus the mate- Tungsten rial may not have sufficient ductility for the in- tended application. The supplier is reluctant to Titanium accept the return of the material, instead Stainless steels claim-ing that the diamond cone used in the Rockwell testing was worn and blunt, and Steels hence the test needed to be recalibrated. Is Nickel this explanation plausible? Explain. Molybdenum Refer to Fig. 2.22 on p. 52 and note that if an Magnesium indenter is blunt, then the penetration, t, un- Lead der a given load will be smaller than that Copper using a sharp indenter. This then translates into a higher hardness. The explanation is Aluminum plausible, but in practice, hardness tests are 0 100 200 300 400 500 fairly reliable and measurements are Elastic modulus (GPa) consistent if the testing equipment is properly calibrated and routinely serviced. Non-metallic materials Spectra fibers 2.44 Explain why a 0.2% offset is used to determine Kevlar fibers the yield strength in a tension test. Glass fibers The value of 0.2% is somewhat arbitrary and is Carbon fibers used to set some standard. A yield stress, Boron fibers repre-senting the transition point from elastic to Thermosets plas-tic deformation, is difficult to measure. This Thermoplastics is because the stress-strain curve is not linearly proportional after the proportional limit, which Rubbers can be as high as one-half the yield strength in Glass some metals. Therefore, a transition from elas- Diamond tic to plastic behavior in a stress-strain curve is Ceramics difficult to discern. The use of a 0.2% o ffset is a 0 200 400 600 800 1000 1200 convenient way of consistently interpreting a yield point from stress-strain curves. Elastic modulus (GPa) 2.45 Referring to Question 2.44, would the o ff-set Typical comments regarding such a chart are: method be necessary for a highly-strained- (a) There is a smaller range for metals than hardened material? Explain. for non-metals; The 0.2% offset is still advisable whenever it (b) Thermoplastics, thermosets and rubbers can be used, because it is a standardized ap- are orders of magnitude lower than met- proach for determining yield stress, and thus als and other non-metals; one should not arbitrarily abandon standards. (c) Diamond and ceramics can be superior However, if the material is highly cold worked, to others, but ceramics have a large there will be a more noticeable ‘kink’ in the range of values. stress-strain curve, and thus the yield stress 2.43 A hardness test is conducted on as-received is far more easily discernable than for the metal as a quality check. The results indicate same material in the annealed condition. 8 Problems 2.46 A strip of metal is originally 1.5 m long. It is Assuming volume constancy, we may write stretched in three steps: first to a length of lo = df 2 = 1.20 2 1.75 m, then to 2.0 m, and finally to 3.0 m. lf do 15 = 156.25 ≈ 156 Show that the total true strain is the sum of the true strains in each step, that is, that the strains are additive. Show that, using Letting l0 be unity, the longitudinal engineering engineering strains, the strain for each step strain is e1 = (156 −1)/1 = 155. The diametral cannot be added to ob-tain the total strain. engineering strain is calculated as The true strain is given by Eq. (2.9) on p. 35 as e = 1.2 − 15 = 0.92 d 15 − = ln l The longitudinal true strain is given by lo Eq. (2.9) on p. 35 as Therefore, the true strains for the three steps are: 1 = ln 75 1 . 1 .5 = 0.1541 l = ln l o = ln (155) = 5.043 2. 2 = ln 1 . 75 = 0.1335 The diametral true strain is 0 20 3 = ln 2.0 = 0.4055 1 . 3.0 d = ln 15 = −2.526 Note the large difference between the engineer- The sum of these true strains is = 0.1541 + ing and true strains, even though both describe 0.1335 + 0.4055 = 0.6931. The true strain from the same phenomenon. Note also that the sum of the true strains (recognizing that the radial step 1 to 3 is strain is r = ln 0.60 = −2.526) in the three 7.5 3 principal directions is zero, indicating volume constancy in plastic deformation. = ln 1.5 = 0.6931 Therefore the true strains are additive. Us-ing 2.48 A material has the following properties: UTS = the same approach for engineering strain as 50, 000 psi and n = 0.25 Calculate its strength defined by Eq. (2.1), we obtain e 1 = 0.1667, coefficient K. e2 = 0.1429, and e3 = 0.5. The sum of these Let us first note that the true UTS of this ma- strains is e1 +e2 +e3 = 0.8096. The n engineering strain from step 1 to 3 is terial is given by UTStrue = Kn (because at necking = n). We can then determine the l l 3 1.5 1.5 value of this stress from the UTS by follow-ing e= − o= − = =1 a procedure similar to Example 2.1. Since n = lo 1.5 1.5 0.25, we can write Note that this is not equal to the sum of the A o engineering strains for the individual steps. A UTStrue = UTS neck = UTS e0.25 2.47 A paper clip is made of wire 1.20-mm in di- = (50, 000)(1.28) = 64, 200 psi ameter. If the original material from which the wire is made is a rod 15-mm in diameter, Therefore, since UTStrue = Kn , n calcu-late the longitudinal and diametrical UTS 64, 200 engineer-ing and true strains that the wire has K= true = = 90, 800 psi under-gone. n 0.25 n 0.25 9 2.49 Based on the information given in Fig. 2.6, cal- (a) Calculate the maximum tensile load that culate the ultimate tensile strength of this cable can withstand prior to necking. annealed 70-30 brass. (b) Explain how you would arrive at an an- swer if the n values of the three strands From Fig. 2.6 on p. 37, the true stress for an- were different from each other. nealed 70-30 brass at necking (where the slope becomes constant; see Fig. 2.7a on p. 40) is found to be about 60,000 psi, while the (a) Necking will occur when = n = 0.3. At true strain is about 0.2. We also know that the this point the true stresses in each cable ratio of the original to necked areas of the n are (from σ = K ), respectively, specimen is given by 0.3 σA = (450)0.3 = 314 MPa Ao 0.3 σB = (600)0.3 = 418 MPa ln Aneck = 0.20 σC = (300)0.3 0.3 = 209 MPa or 0.3 Aneck σD = (760)0.3 = 530 MPa = e−0.20 = 0.819 The areas at necking are calculated as A o −n fol-lows (from Aneck = Aoe ): Thus, −0.3 2 AA = (7)e = 5.18 mm UTS = (60, 000)(0.819) = 49, 100 psi −0.3 2 AB = (2.5)e = 1.85 mm −0.3 2 2.50 Calculate the ultimate tensile strength (engi- AC = (3)e = 2.22 mm −0.3 2 neering) of a material whose strength AD = (2)e = 1.48 mm coefficient is 400 MPa and of a tensile-test Hence the total load that the cable can specimen that necks at a true strain of 0.20. support is In this problem we have K = 400 MPa and n = P = (314)(5.18) + (418)(1.85) 0.20. Following the same procedure as in +(209)(2.22) + (530)(1.48) Example 2.1, we find the true ultimate tensile = 3650 N strength is (b) If the n values of the four strands were 0.20 σ = (400)(0.20) = 290 MPa dif-ferent, the procedure would consist of plot-ting the load-elongation curves of and the four strands on the same chart, then −0.20 Aneck = Aoe = 0.81Ao obtain-ing graphically the maximum load. Consequently, Alter-nately, a computer program can be written to determine the maximum load. UTS = (290)(0.81) = 237 MPa 2.52 Using only Fig. 2.6, calculate the maximum load in tension testing of a 304 stainless-steel 2.51 A cable is made of four parallel strands of dif- round specimen with an original diameter of ferent materials, all behaving according to the 0.5 in. n equation σ = K , where n = 0.3 The materi- als, strength coefficients, and cross sections We observe from Fig. 2.6 on p. 37 that necking are as follows: begins at a true strain of about 0.1, and that the true UTS is about 110,000 psi. The origi-nal 2 2 cross-sectional area is Ao = π(0.25 in) = 0.196 Material A: K = 450 MPa, Ao = 7 mm ; 2 2 in . Since n = 0.1, we follow a procedure similar Material B: K = 600 MPa, Ao = 2.5 mm ; to Example 2.1 and show that 2 Material C: K = 300 MPa, Ao = 3 mm ; Ao 2 = e0.1 = 1.1 Material D: K = 760 MPa, Ao = 2 mm ; A neck 10 Thus 2.55 A cylindrical specimen made of a brittle mate- 110, 000 rial 1 in. high and with a diameter of 1 in. is UTS = = 100, 000 subjected to a compressive force along its psi 1.1 axis. It is found that fracture takes place at an ◦ Hence the maximum load is angle of 45 under a load of 30,000 lb. Calculate the shear stress and the normal F = (UTS)(Ao) = (100, 000)(0.196) stress acting on the fracture surface. or F = 19, 600 lb. Assuming that compression takes place without friction, note that two of the principal 2.53 Using the data given in Table 2.1, calculate stresses will be zero. The third principal the values of the shear modulus G for the stress acting on this specimen is normal to metals listed in the table. the specimen and its magnitude is The important equation is Eq. (2.24) on p. 49 30, 000 which gives the shear modulus as σ = = 38, 200 psi 3 π(0.5)2 E G= The Mohr’s circle for this situation is shown 2(1 + ν) below. The following values can be calculated (mid- range values of ν are taken as appropriate): Material E (GPa) ν G (GPa) Al & alloys 69-79 0.32 26-30 Cu & alloys 105-150 0.34 39-56 2 =90° Pb & alloys 14 0.43 4.9 Mg & alloys 41-45 0.32 15.5-17.0 Mo & alloys 330-360 0.32 125-136 Ni & alloys 180-214 0.31 69-82 Steels 190-200 0.30 73-77 Stainless steels 190-200 0.29 74-77 ◦ Ti & alloys 80-130 0.32 30-49 The fracture plane is oriented at an angle of 45 , ◦ W & alloys 350-400 0.27 138-157 corresponding to a rotation of 90 on the Mohr’s Ceramics 70-1000 0.2 29-417 circle. This corresponds to a stress state on the Glass 70-80 0.24 28-32 fracture plane of σ = −19, 100 psi and Rubbers 0.01-0.1 0.5 0.0033-0.033 τ = 19, 100 psi. Thermoplastics 1.4-3.4 0.36 0.51-1.25 Thermosets 3.5-17 0.34 1.3-6.34 2.56 What is the modulus of resilience of a highly cold-worked piece of steel with a hardness of 2.54 Derive an expression for the toughness of a material whose behavior is represented by 300 HB? Of a piece of highly cold-worked n the equation σ = K ( + 0.2) and whose cop-per with a hardness of 150 HB? fracture strain is denoted as f . Referring to Fig. 2.24 on p. 55, the value of c in Eq. (2.29) on p. 54 is approximately 3.2 for Recall that toughness is the area under the highly cold-worked steels and around 3.4 for stress-strain curve, hence the toughness for cold-worked aluminum. Therefore, we can this material would be given by approximate c = 3.3 for cold-worked copper. Z f Toughness = σd However, since the Brinell hardness is in units 2 0 of kg/mm , from Eq. (2.29) we can write = Z f n K ( + 0.2) d T steel = 3.2 = 3.2 = 93.75 kg/mm 2 = 133 ksi H 300 0 h T = = 2 = n + 1 ( f + 0.2) − 0.2 i Cu 3.3 3.3 = 45.5 kg/mm = 64.6 ksi K n+1 n+1 H 150 11 6 2 From Table 2.1, Esteel = 30 × 10 psi and ECu The volume is calculated as V = πr l = 6 2 −6 3 = 15 × 10 psi. The modulus of resilience is π(0.0075) (0.04) = 7.069 × 10 m . The calculated from Eq. (2.5). For steel: work done is the product of the specific work, Y2 (133, 000)2 u, and the volume, V . Therefore, the results Modulus of Resilience = = can be tabulated as follows. 6 2E 2(30 × 10 ) or a modulus of resilience for steel of 295 in- 3 u Work lb/in . For copper, 3 Y 2 (62, 200) 2 Material (MN/m ) (Nm) Modulus of Resilience = = 1100-O Al 222 1562 2E 2(15 × 10 ) 6 Cu, annealed 338 2391 or a modulus of resilience for copper of 129 304 Stainless, annealed 1529 10,808 3 70-30 brass, annealed 977 6908 in-lb/in . Note that these values are slightly different than the values given in the text; this is due to the 2.58 A material has a strength coefficient K = 100, fact that (a) highly cold-worked metals such as 000 psi Assuming that a tensile-test spec- these have a much higher yield stress than the imen made from this material begins to neck annealed materials described in the text, and at a true strain of 0.17, show that the ultimate (b) arbitrary property values are given in the tensile strength of this material is 62,400 psi. statement of the problem. The approach is the same as in Example 2.1. 2.57 Calculate the work done in frictionless compres- Since the necking strain corresponds to the sion of a solid cylinder 40 mm high and 15 mm maximum load and the necking strain for this in diameter to a reduction in height of 75% for material is given as = n = 0.17, we have, as the following materials: (1) 1100-O aluminum, the true ultimate tensile strength: (2) annealed copper, (3) annealed 304 stainless 0.17 steel, and (4) 70-30 brass, annealed. UTStrue = (100, 000)(0.17) = 74, 000 psi. The work done is calculated from Eq. (2.62) The cross-sectional area at the onset of on p. 71 where the specific energy, u, is necking is obtained from obtained from Eq. (2.60). Since the reduction ln Aneck = n = 0.17. in height is 75%, the final height is 10 mm and Ao the absolute value of the true strain is = ln 10 = 1.386 40 Consequently, −0.17 K and n are obtained from Table 2.3 as follows: Aneck = Aoe Material K (MPa) n and the maximum load, P , is 1100-O Al 180 0.20 −0.17 Cu, annealed 315 0.54 P = σA = (UTStrue)Aoe 304 Stainless, annealed 1300 0.30 = (74, 000)(0.844)(Ao) = 62, 400Ao lb. 70-30 brass, annealed 895 0.49 Since UTS= P/Ao, we have UTS = 62,400 psi. The u values are then calculated from Eq. (2.60). For example, for 1100-O 2.59 A tensile-test specimen is made of a material aluminum, where K is 180 MPa and n is 0.20, represented by the equation σ = K ( + n) . n u is calcu-lated as (a) Determine the true strain at which necking 1.2 K n+1 (180)(1.386) 3 will begin. (b) Show that it is possible for an u= = = 222 MN/m engineering material to exhibit this behavior. n+1 1.2 12 (a) In Section 2.2.4 on p. 38 we noted that 2 in and the original lengths are a = 8 in. and b = instability, hence necking, requires the 4.5 in. The material for specimen a has a true- fol-lowing condition to be fulfilled: 0.5 stress-true-strain curve of σ = 100, 000 . Plot the true-stress-true-strain curve that the material dσ for specimen b should have for the bar to d=σ remain horizontal during the experiment. Consequently, for this material we have n−1 n a Kn ( + n) = K ( + n) This is solved as n = 0; thus necking be- F gins as soon as the specimen is 2 1 subjected to tension. c c (b) Yes, this behavior is possible. Consider b a tension-test specimen that has been x strained to necking and then unloaded. Upon loading it again in tension, it will immediately begin to neck. From the equilibrium of vertical forces and to 2.60 Take two solid cylindrical specimens of equal di- keep the bar horizontal, we note that 2F a = ameter but different heights. Assume that both Fb. Hence, in terms of true stresses and specimens are compressed (frictionless) by the instanta-neous areas, we have same percent reduction, say 50%. Prove that 2σaAa = σbAb the final diameters will be the same. From volume constancy we also have, in Let’s identify the shorter cylindrical specimen terms of original and final dimensions with the subscript s and the taller one as t, and their original diameter as D. Subscripts f A L =A L oa oa a a and o indicate final and original, respectively. and Be-cause both specimens undergo the same A L =A L percent reduction in height, we can write ob ob b b h h where Loa = (8/4.5)Lob = 1.78Lob. From these tf = sf relationships we can show that h h to so 8 Lb σ =2 Kσ and from volume constancy, 4.5 b a La 2 0 .5 h D Since σa = K a where K = 100, 000 psi, we tf = to can now write h D σb = 4.5 La L √ a to tf 16K b and 2 h D sf = so Hence, for a deflection of x, h D s so sf b 4.5 8+x 8 4.5 − x σ = 16K Because Dto = Dso, we note from these rela- 8+x ln tionships that Dtf = Dsf . 2.61 A horizontal rigid bar c-c is subjecting specimen The true strain in specimen b is given by a to tension and specimen b to frictionless com- pression such that the bar remains horizontal. = ln 4.5 − x (See the accompanying figure.) The force F is b 4.5 located at a distance ratio of 2:1. Both speci- By inspecting the figure in the problem state- mens have an original cross-sectional area of 1 ment, we note that while specimen a gets 13 longer, it will continue exerting some force F a. Equation (2.20) is used to solve this problem. However, specimen b will eventually acquire Noting that σ = 500 MPa, d = 40 mm = 0.04 a cross-sectional area that will become infinite m, and t = 5 mm = 0.005 m, we can write as x approaches 4.5 in., thus its strength must 2P σπdt approach zero. This observation suggests that specimen b cannot have a true stress- σ = πdt → P= 2 true strain curve typical of metals, and that it Therefore will have a maximum at some strain. This is 6 seen in the plot of σb shown below. P = (500 × 10 )π(0.04)(0.005) = 157 kN. 2 2.64 In Fig. 2.32a, let the tensile and compressive 50,000 residual stresses both be 10,000 psi and the 6 modulus of elasticity of the material be 30×10 40,000 3 psi, with a modulus of resilience of 30 in.-lb/in . If the original length in diagram (a) is 20 in., True stress (psi) 30,000 what should be the stretched length in diagram (b) so that, when unloaded, the strip will be free of residual stresses? 20,000 Note that the yield stress can be obtained from Eq. (2.5) on p. 31 as 10,000 Y2 Mod. of Resilience = MR = 0 2E 0 0.5 1.0 1.5 2.0 2.5 Absolute value of true strain Thus, p p 6 Y= 2(MR)E = 2(30)(30 × 10 ) 2.62 Inspect the curve that you obtained in Problem 2.61. Does a typical strain-hardening material or Y = 42, 430 psi. Using Eq. (2.32), the strain behave in that manner? Explain. required to relieve the residual stress is: = σc +Y = 10, 000 + 42, 430 = 0.00175 Based on the discussions in Section 2.2.3 6 6 start-ing on p. 35, it is obvious that ordinary E E 30 × 10 30 × 10 met-als would not normally behave in this Therefore, manner. However, under certain conditions, = ln lo = ln 20 in. = 0.00175 the follow-ing could explain such behavior: lf lf • When specimen b is heated to higher Therefore, lf = 20.035 in. and higher temperatures as deformation pro-gresses, with its strength decreasing 2.65 Show that you can take a bent bar made of an as x is increased further after the elastic, perfectly plastic material and straighten maximum value of stress. it by stretching it into the plastic range. (Hint: • In compression testing of brittle Observe the events shown in Fig. 2.32.) materials, such as ceramics, when the The series of events that takes place in straight- specimen be-gins to fracture. ening a bent bar by stretching it can be visu- • If the material is susceptible to thermal alized by starting with a stress distribution as in softening, then it can display such behav- Fig. 2.32a on p. 61, which would represent the ior with a sufficiently high strain rate. unbending of a bent section. As we apply tension, we algebraically add a uniform tensile 2.63 In a disk test performed on a specimen 40-mm stress to this stress distribution. Note that the in diameter and 5 m thick, the specimen frac- change in the stresses is the same as that de- tures at a stress of 500 MPa. What was the picted in Fig. 2.32d, namely, the tensile stress load on the disk at fracture? 14 increases and reaches the yield stress, Y . affect yielding. In other words, the material will The compressive stress is first reduced in still yield according to yield criteria. magnitude, then becomes tensile. Eventually, the whole cross section reaches the constant Let’s consider the distortion-energy criterion, yield stress, Y . Because we now have a although the same derivation could be per- uniform stress dis-tribution throughout its formed with the maximum shear stress criterion thickness, the bar be-comes straight and as well. Equation (2.37) on p. 64 gives remains straight upon un-loading. 2 2 2 2 (σ1 − σ2) + (σ2 − σ3) + (σ3 − σ1) = 2Y 2.66 A bar 1 m long is bent and then stress re- lieved. The radius of curvature to the neutral Now consider a new stress state where the axis is 0.50 m. The bar is 30 mm thick and is prin-cipal stresses are made of an elastic, perfectly plastic material 0 with Y = 600 MPa and E = 200 GPa. Cal- σ1 = σ 1 + p culate the length to which this bar should be 0 stretched so that, after unloading, it will be- σ2 = σ 2 + p come and remain straight. 0 σ3 = σ3 + p When the curved bar becomes straight, the which represents a new loading with an addi- en-gineering strain it undergoes is given by tional hydrostatic pressure, p. The distortion- the ex-pression energy criterion for this stress state is t e=2ρ 0 0 2 0 0 2 0 0 2 2 (σ1 − σ2 ) + (σ2 − σ3 ) + (σ3 − σ1 ) = 2Y where t is the thickness and ρ is the radius to the neutral axis. Hence in this case, or (0.030) 2 2 e= = 0.03 2Y = [(σ1 + p) − (σ2 + p)] 2 + [(σ2 + p) − (σ3 + p)] Since Y = 600 MPa and E = 200 GPa, we find + [(σ3 + p) − (σ1 + p)] 2 that the elastic limit for this material is at an elastic strain of which can be simplified as Y 600 MPa 2 2 2 2 e= E= 200 GPa = 0.003 (σ1 − σ2) + (σ2 − σ3) + (σ3 − σ1) = 2Y which is much smaller than 0.05. Following which is the original yield criterion. Hence, the the description in Answer 2.65 above, we find yield criterion is unaffected by the superposi- that the strain required to straighten the bar is tion of a hydrostatic pressure. e = (2)(0.003) = 0.006 2.68 Give two different and specific examples in which the maximum-shear-stress and the distortion- or energy criteria give the same answer. lf − lo = 0.006 l = 0.006l + l In order to obtain the same answer for the two lo → f o o yield criteria, we refer to Fig. 2.36 on p. 67 for plane stress and note the coordinates at which or lf = 1.006 m. the two diagrams meet. Examples are: simple 2.67 Assume that a material with a uniaxial yield tension, simple compression, equal biaxial ten- stress Y yields under a stress state of principal sion, and equal biaxial compression. Thus, ac- stresses σ1, σ2, σ3, where σ1 > σ2 > σ3. Show ceptable answers would include (a) wire rope, that the superposition of a hydrostatic stress, p, as used on a crane to lift loads; (b) spherical on this system (such as placing the specimen in pres-sure vessels, including balloons and gas a chamber pressurized with a liquid) does not storage tanks, and (c) shrink fits. 15 2.69 A thin-walled spherical shell with a yield stress 2.71 What would be the answer to Problem 2.70 if the Y is subjected to an internal pressure p. With maximum-shear-stress criterion were used? appropriate equations, show whether or not the pressure required to yield this shell Because σ2 is an intermediate stress and depends on the particular yield criterion used. using Eq. (2.36), the answer would be Here we have a state of plane stress with equal σ1 − 0 = Y biaxial tension. The answer to Problem 2.68 hence the yield stress in plane strain will be leads one to immediately conclude that both the equal to the uniaxial yield stress, Y . maximum shear stress and distortion energy 2.72 A closed-end, thin-walled cylinder of original criteria will give the same results. We will now length l, thickness t, and internal radius r is demonstrate this more rigorously. The princi-pal subjected to an internal pressure p. Using the membrane stresses are given by generalized Hooke’s law equations, show the pr σ =σ = 1 2 change, if any, that occurs in the length of this 2t cylinder when it is pressurized. Let ν = 0.33. and A closed-end, thin-walled cylinder under inter- σ3 = 0 nal pressure is subjected to the following prin- Using the maximum shear-stress criterion, we cipal stresses: find that pr pr σ1 − 0 = Y σ1 = 2t ; σ2 = t ; σ3 = 0 hence where the subscript 1 is the longitudinal di- 2tY rection, 2 is the hoop direction, and 3 is the p= r thickness direction. From Hooke’s law given Using the distortion-energy criterion, we have by Eq. (2.33) on p. 63, 2 2 2 2 1 (0 − 0) + (σ2 − 0) + (0 − σ1) = 2Y 1 = E [σ1 − ν (σ2 + σ3)] Since σ1 = σ2, then this gives σ1 = σ2 = Y , and = 1 pr − 1 pr + 0 the same expression is obtained for pressure. E 2t 3 t pr 2.70 Show that, according to the distortion-energy = 6tE criterion, the yield stress in plane strain is 1.15Y where Y is the uniaxial yield stress of Since all the quantities are positive (note that in the material. order to produce a tensile membrane stress, the pressure is positive as well), the longitudinal A plane-strain condition is shown in Fig. 2.35d strain is finite and positive. Thus the cylinder on p. 67, where σ1 is the yield stress of the becomes longer when pressurized, as it can 0 material in plane strain (Y ), σ3 is zero, and 2 also be deduced intuitively. = 0. From Eq. 2.43b on p. 68, we find that σ 2 2.73 A round, thin-walled tube is subjected to ten- = σ1/2. Substituting these into the distortion- sion in the elastic range. Show that both the energy criterion given by Eq. (2.37) on p.64, thickness and the diameter of the tube σ 2 σ 2 decrease as tension increases. 1 1 2 2 σ1 − 2 + 2 −0 + (0 − σ1) = 2Y The stress state in this case is σ 1, σ2 = σ3 = and 2 0. From the generalized Hooke’s law 3σ 1 = 2Y 2 equations given by Eq. (2.33) on p. 63, and 2 denoting the axial direction as 1, the hoop hence direction as 2, and the radial direction as 3, 2 we have for the hoop strain: σ1 = √ 3 Y ≈ 1.15Y 1 νσ 1 2 = E [σ2 − ν (σ1 + σ3)] = − E 16 Therefore, the diameter is negative for a Taking the natural log of both sides, tensile (positive) value of σ1. For the radial strain, the generalized Hooke’s law gives l1l2l3 1 νσ 1 ln lololo = ln(1) = 0 3 = E [σ3 − ν (σ1 + σ2)] = − E since ln(AB) = ln(A) + ln(B), ln lo + ln lo + ln lo =0 Therefore, the radial strain is also negative l1 l2 l3 and the wall becomes thinner for a positive value of σ1. From the definition of true strain given by 2.74 Take a long cylindrical balloon and, with a thin l1 felt-tip pen, mark a small square on it. What will l be the shape of this square after you blow up Eq. (2.9) on p. 35, ln 0 = 1, etc., so that the balloon: (1) a larger square, (2) a rectan-gle, 1 + 2 + 3 = 0. with its long axis in the circumferential di- rections, (3) a rectangle, with its long axis in the 2.76 What is the diameter of an originally 30-mm- longitudinal direction, or (4) an ellipse? Per-form diameter solid steel ball when it is subjected this experiment and, based on your obser- to a hydrostatic pressure of 5 GPa? vations, explain the results, using appropriate equations. Assume that the material the bal-loon From Eq. (2.46) on p. 68 and noting that, for is made of is perfectly elastic and isotropic, and this case, all three strains are equal and all that this situation represents a thin-walled three stresses are equal in magnitude, closed-end cylinder under internal pressure. E − 3= 1− 2ν ( 3p) This is a simple graphic way of illustrating the generalized Hooke’s law equations. A balloon is a readily available and economical method of where p is the hydrostatic pressure. Thus, from demonstrating these stress states. It is also en- Table 2.1 on p. 32 we take values for steel of couraged to assign the students the task of pre- ν = 0.3 and E = 200 GPa, so that dicting the shape numerically; an example of a E − 200 − valuable experiment involves partially inflating 2ν 1 = 1 − ( p) = − 0.6 ( 5) the balloon, drawing the square, then expand- ing it further and having the students predict the dimensions of the square. or = −0.01. Therefore Although not as readily available, a rubber Df tube can be used to demonstrate the effects of tor-sion in a similar manner. ln Do = −0.01 Solving for Df , 2.75 Take a cubic piece of metal with a side length lo and deform it plastically to the shape of a −0.01 −0.01 D f = Do e = (20)e = 19.8 mm rectangular parallelepiped of dimensions l1, l2, and l3. Assuming that the material is rigid and perfectly plastic, show that volume constancy 2.77 Determine the effective stress and effective requires that the following expression be strain in plane-strain compression according satis-fied: 1 + 2 + 3 = 0. to the distortion-energy criterion. The initial volume and the final volume are Referring to Fig. 2.35d on p. 67 we note that, con-stant, so that for this case, σ3 = 0 and σ2 = σ1/2, as can be seen from Eq. (2.44) on p. 68. According to l1l2l3 the distortion-energy criterion and referring to lololo = l1l2l3 → lololo = 1 Eq. (2.52) on p. 69 for effective stress, we find 17 that where V is the volume of the sphere. We inte- σ¯ = √ 2 σ1 −2 1 + 21 + (σ1)2 1/2 grate this equation between the limits V o and 1 σ 2 σ 2 Vf , noting that = √24 + 4 + 1 1/2 σ1 2tY 1 1 1 p= r and 1 √3 √3 3 ! σ V = 4πr 3 = √2 √ 2 σ1 = 2 1 so that Note that for this case 3 = 0. Since volume 2 dV = 4πr dr constancy is maintained during plastic defor- Also, from volume constancy, we have mation, we also have 3 = − 1. Substitut-ing 2 these into Eq. (2.54), the effective strain is ro to t= found to be r2 Combining these expressions, we obtain 2 ¯= √ 1 rf dr rf 3 2 Z 2 W = 8πY ro to ro r = 8πY ro to ln ro 2.78 (a) Calculate the work done in expanding a 2- which is the same expression obtained earlier. mm-thick spherical shell from a diameter of 100 To obtain a numerical answer to this prob-lem, mm to 140 mm, where the shell is made of a note that Y should be replaced with an 0.5 ¯ ma-terial for which σ = 200+50 MPa. (b) average value Y . Also note that 1 = 2 = Does your answer depend on the particular ln(140/100) = 0.336. Thus, yield cri-terion used? Explain. 1.5 ¯ 50(0.336) For this case, the membrane stresses are Y = 200 + 1.5 = 206 MPa given by Hence ¯the work done is pt W = 8πY ro2to ln ro σ =σ = 1 2 r f 2t and the strains are =2 = ln fo 6 2 1 = 8π(206 × 10 )(0.1) (0.001) ln(70/50) fr = 17.4kN-m The yield criterion used does not matter be- Note that we have a balanced (or equal) biaxial cause this is equal biaxial tension; see the an- state of plane stress. Thus, the specific energy swer to Problem 2.68. (for a perfectly-plastic material) will, according to either yield criteria, be 2.79 A cylindrical slug that has a diameter of 1 in. u = 2σ1 1 = 2Y ln ro and is 1 in. high is placed at the center of a 2- rf in.-diameter cavity in a rigid die. (See the accompanying figure.) The slug is surrounded The work done will be by a compressible matrix, the pressure of which is given by the relation W = (Volume)(u) V = 4πro2to 2Y ln ro pm = 40, 000 psi rf V om 2 = 8πY ro to ln ro r where m denotes the matrix and V om is the orig- f inal volume of the compressible matrix. Both the slug and the matrix are being compressed by a Using the pressure-volume method of work, piston and without any friction. The ini-tial we begin with the formula pressure on the matrix is zero, and the slug Z material has the true-stress-true-strain curve of 0.4 W= p dV σ = 15, 000 . 18 F The absolute value of the true strain in the slug is given by 1 = ln 1 − d , with which we can determine the value of σ d for any d. The cross-sectional area of the 1" workpiece at any d is Aw = π in 2 1" Compressible 2" 4(1 − d) matrix and that of the matrix is Obtain an expression for the force F versus π 2 Am = π − 4(1 d) in pis-ton travel d up to d = 0.5 in. − The total force, F , on the piston will be The required compressive stress on the slug is 160, 000 F = F w + F m, σ1 = σ + σ2 = σ + d. where the subscript w denotes the workpiece and m the matrix. As d increases, the matrix pressure increases, thus subjecting the slug We may now write the total force on the to transverse compressive stresses on its piston as circum-ference. Hence the slug will be subjected to tri-axial compressive stresses, 160, 000 160, 000 F = Aw σ+ d + Am d lb. with σ2 = σ3. Using the maximum shear- 3 3 stress criterion for simplic-ity, we have σ1 = σ + σ 2 The following data gives some numerical re- sults: where σ1 is the required compressive stress on the slug, σ is the flow stress of the slug mate-rial corresponding to a given strain, and given as σ d Aw σ F 0.4 2 = 15, 000 , and σ2 is the compressive stress (in.) (in ) (psi) (lb) due to matrix pressure. Lets now deter-mine the 0.1 0.872 0.105 6089 22,070 matrix pressure in terms of d. 0.2 0.98 0.223 8230 41,590 The volume of the slug is equal to π/4 and the 0.3 1.121 0.357 9934 61,410 volume of the cavity when d = 0 is π. Hence 0.4 1.31 0.510 11,460 82,030 the original volume of the matrix is V om = 4 3 0.5 1.571 0.692 12,950 104,200 π. The volume of the matrix at any value of d is then And the following plot shows the desired re- V = π(1 d) m − −4 =π 3 4 −d in , π 3 sults. from which we obtain 120 Force(kip) V V V 4 40 = om − m = d. 80 V V 3 om om 3 Note that when d = in., the volume of the 4 ma-trix becomes zero. The matrix pressure, hence σ2, is now given by 00 0.1 0.2 0.3 0.4 0.5 σ2 = 4(40, 000) d = 160, 000 d (psi) Displacement (in.) 3 3 19 2.80 A specimen in the shape of a cube 20 mm on (a) For a perfectly-elastic material as shown in each side is being compressed without friction in Fig 2.7a on p. 40, this expression becomes a die cavity, as shown in Fig. 2.35d, where the 2 1 2 width of the groove is 15 mm. Assume that the E1 linearly strain-hardening material has the true- Z 1 u= 0 Ed=E 2 0 = 2 stress-true-strain curve given by σ = 70 + 30 MPa. Calculate the compressive force required (b) For a rigid, perfectly-plastic material as when the height of the specimen is at 3 mm, shown in Fig. 2.7b, this is according to both yield criteria. Z 1 u= Y d = Y ( )0 1 = Y 1 0 We note that the volume of the specimen is con-stant and can be expressed as (c) For an elastic, perfectly plastic material, this is identical to an elastic material for 1 (20)(20)(20) = (h)(x)(x) < Y /E, and for 1 > Y /E it is where x is the lateral dimensions assuming the Y /E 1 specimen expands uniformly during compres- sion. Since h = 3 mm, we have x = 51.6 Z Z u = Z0 σ d = 0 1 Ed+ Y /E Yd mm. Thus, the specimen touches the walls E Y 2 Y and hence this becomes a plane-strain − = 2 E +Y 1 E problem (see Fig. 2.35d on p. 67). The 2 2 absolute value of the true strain is Y Y Y = ln 3 = 1.90 20 = 2E + Y 1 − E =Y 1 − 2E (d) For a rigid, linearly strain hardening ma- terial, the specific energy is We can now determine the flow stress, Yf , of the material at this strain as 2 1 Ep Z 1 Yf = 70 + 30(1.90) = 127 MPa u= 0 (Y + Ep ) d = Y 1 + 2 The cross-sectional area on which the force is (e) For an elastic, linear strain hardening acting is ma-terial, the specific energy is identical to an elastic material for 1 < Y /E and for 2 Area = (20)(20)(20)/3 = 2667 mm 1 > Y /E it is According to the maximum shear-stress crite- u = Z0 1 Y + Ep − E d Y rion, we have σ1 = Yf , and thus = 1 Y 1 − Ep + Ep d Force = (127)(2667) = 338 kN Z0 E According to the distortion energy criterion, E 2 p Ep we have σ1 = 1.15Yf , or 1 Force = (1.15)(338) = 389 kN. = Y1 − E 1 + 2 2.82 A material with a yield stress of 70 MPa is sub-jected to three principal (normal) stresses 2.81 Obtain expressions for the specific energy for of σ1, σ2 = 0, and σ3 = −σ1/2. What is the a material for each of the stress-strain curves value of σ1 when the metal yields according shown in Fig. 2.7, similar to those shown in to the von Mises criterion? What if σ2 = σ1/3? Section 2.12. The distortion-energy criterion, given by Eq. Equation (2.59) on p. 71 gives the specific en- ergy as (2.37) on p. 64, is Z 1 2 2 2 2 u= σd (σ1 − σ2) + (σ2 − σ3) + (σ3 − σ1) = 2Y 0 20 Substituting Y = 70 MPa and σ 1, σ2 = 0 and 2.84 A 50-mm-wide, 1-mm-thick strip is rolled to a final thickness of 0.5 mm. It is noted that the σ3 = −σ1/2, we have σ 2 σ strip has increased in width to 52 mm. What is 2 1 the strain in the rolling direction? 1 2 2 2(70) = (σ1) + − 2 + − 2 − σ1 thus, The thickness strain is σ1 = 52.9 MPa l mm 0.5 If Y = 70 MPa and σ1, σ2 = σ1/3 and σ3 = t = ln lo = ln 1 mm = −0.693 −σ1/2 is the stress state, then The width strain is σ1 2 σ1 σ1 2 2 σ l 52 mm + 3 σ1 2(70) = 1−3 −2 2 w = ln lo = ln 50 mm = 0.0392 2 Therefore, from Eq. (2.48), the strain in the + − 2 − σ1 = 2.72σ1 rolling (or longitudinal) direction is l = 0 − Thus, σ1 = 60.0 MPa. Therefore, the stress level to initiate yielding actually increases 0.0392 + 0.693 = 0.654. when σ2 is increased. 2.85 An aluminum alloy yields at a stress of 50 MPa in 2.83 A steel plate has the dimensions 100 mm × 100 uniaxial tension. If this material is subjected to mm × 5 mm thick. It is subjected to biaxial the stresses σ1 = 25 MPa, σ2 = 15 MPa and σ3 tension of σ1 = σ2, with the stress in the thick- = −26 MPa, will it yield? Explain. ness direction of σ3 = 0. What is the largest According to the maximum shear-stress crite- possible change in volume at yielding, using the rion, the effective stress is given by Eq. (2.51) von Mises criterion? What would this change in on p. 69 as: volume be if the plate were made of copper? From Table 2.1 on p. 32, it is noted that for σ¯ = σ1 − σ3 = 25 − (−26) = 51 MPa steel we can use E = 200 GPa and ν = 0.30. However, according to the distortion-energy For a stress state of σ 1 = σ2 and σ3 = 0, the cri-terion, the effective stress is given by Eq. von Mises criterion predicts that at yielding, (2.52) on p. 69 as: 2 2 2 2 1 (σ1 − σ2) + (σ2 − σ3) + (σ3 − σ1) = 2Y q 2 2 2 or σ¯ = √ 2 (σ1 − σ2) + (σ2 − σ3) + (σ3 − σ1) or 2 2 2 2 (σ1 − σ1) + (σ1 − 0) + (0 − σ1) = 2Y 2 2 2 (25 − 15) + (15 + 26) + (−26 − 25) Resulting in σ1 = Y . Equation (2.47) gives: r σ¯ = 2 1 − 2ν = (σx + σy + σz) or σ¯ = 46.8 MPa. Therefore, the effective E stress is higher than the yield stress for the 1 − 2(0.3) maximum shear-stress criterion, and lower = [(350 MPa) + (350 MPa] 200 GPa than the yield stress for the distortion-energy = = 0.0014 criterion. It is impossible to state whether or not the mate-rial will yield at this stress state. Since the original volume is (100)(100)(5) = 3 An accurate statement would be that yielding 50,000 mm , the stressed volume is 50,070 3 3 is imminent, if it is not already occurring. mm , or the volume change is 70 mm . For copper, we have E = 125 GPa and ν = 2.86 A cylindrical specimen 1-in. in diameter and 1-in. 0.34. Following the same derivation, the high is being compressed by dropping a weight dilatation for copper is 0.0006144; the of 200 lb on it from a certain height. After 3 stressed volume is 50,031 mm and thus the deformation, it is found that the temper-ature 3 ◦ change in volume is 31 mm . rise in the specimen is 300 F. Assuming 21 no heat loss and no friction, calculate the fi- Similarly, for the second step where h1 = 70 nal height of the specimen, using the mm and h2 = 40 mm, following data for the material: K = 30, 000 h −h 40 − 70 psi, n = 0.5, density = 0.1 lb/in , and specific 3 e2 = 2 1 = = −0.429 ◦ h1 70 heat = 0.3 BTU/lb· F. = ln h1 = ln 70 = −0.560 2 This problem uses the same approach as in h2 40 Ex-ample 2.8. The volume of the specimen is πd2h π(1)2(1) Note that if the operation were conducted in V= = = 0.785 in3 44 one step, the following would result: The expression for heat is given by h h 40 100 e= 2− o = − = −0.6 Heat = cpρV T ho 100 = ln ho = ln 100 = −0.916 = (0.3)(0.1)(0.785)(300)(778) h2 40 = 5500ft-lb = 66, 000 in-lb. where the unit conversion 778 ft-lb = 1 BTU As was shown in Problem 2.46, this indicates has been applied. Since, ideally, that the true strains are additive while the en- gineering strains are not. K n+1 Heat = Work = Vu=Vn+1 2.88 Assume that the specimen in Problem 2.87 has an initial diameter of 80 mm and is made = (0.785) (30, 1.5 000) of 1100-O aluminum. Determine the load 1.5 required for each step. Solving for , From volume constancy, we calculate 1.5 = 1.5(66, 000) = 4.20 r d1 = do r = 80 = 95.6 mm (0.785)(30, 000) h h1 70 o 100 Therefore, = 2.60. Using absolute values, we d2 = do r h2 = 80 r 40 = 126.5 mm have ho 100 ln hf = ln h f = 2.60 ho 1 in. Based on these diameters the cross-sectional area at the steps is calculated as: Solving for hf gives hf = 0.074 in. π 2 2 A1 = 4d 1 = 7181 mm 2.87 A solid cylindrical specimen 100-mm high is compressed to a final height of 40 mm in two π 2 2 steps between frictionless platens; after the A2 = 4d 2 = 12, 566 mm first step the cylinder is 70 mm high. Calculate As calculated in Problem 2.87, 1 = 0.357 and the engineering strain and the true strain for total = 0.916. Note that for 1100-O aluminum, both steps, compare them, and comment on K = 180 MPa and n = 0.20 (see Table 2.3 on your ob-servations. p. 37) so that Eq. (2.11) on p. 35 yields In the first step, we note that ho = 100 mm and σ1 = 180(0.357) 0.20 = 146.5 MPa h1 = 70 mm, so that from Eq. (2.1) on p. 30, 0.20 σ2 = 180(0.916) = 176.9 Mpa e = h1 − ho = 70 − 100 = 0.300 Therefore, the loads are calculated as: 1 ho 100 − P1 = σ1A1 = (146.5)(7181) = 1050 kN and from Eq. (2.9) on p. 35, 1= ln ho = ln 100 = −0.357 P2 = (176.9)(12, 566) = 2223 kN h1 70 22 2.89 Determine the specific energy and actual 2.91 The area of each face of a metal cube is 400 energy expended for the entire process 2 m , and the metal has a shear yield stress, k, described in the preceding two problems. of 140 MPa. Compressive loads of 40 kN and 80 kN are applied at different faces (say in the From Eq. (2.60) on p. 71 and using total = x- and y-directions). What must be the 0.916, K = 180 MPa and n = 0.20, we have compressive load applied to the z-direction to 1.2 K n+1 (180)(0.916) cause yield-ing according to the Tresca u= = = 135 MPa criterion? Assume a frictionless condition. n+1 1.2 2 2.90 A metal has a strain hardening exponent of Since the area of each face is 400 mm , the stresses in the x- and y- directions are 0.22. At a true strain of 0.2, the true stress is 20,000 psi. (a) Determine the stress-strain 40, 000 = −100 MPa relationship for this material. (b) Determine σx = − 400 the ultimate tensile strength for this material. 80, 000 This solution follows the same approach as in σy = − 400 = −200 MPa Example 2.1. From Eq. (2.11) on p. 35, and where the negative sign indicates that the recognizing that n = 0.22 and σ = 20, 000 psi stresses are compressive. If the Tresca criterion for = 0.20, is used, then Eq. (2.36) on p. 64 gives n 0.22 σ=K → 20, 000 = K(0.20) σmax − σmin = Y = 2k = 280 MPa or K = 28, 500 psi. Therefore, the stress- It is stated that σ3 is compressive, and is strain relationship for this material is there-fore negative. Note that if σ 3 is zero, 0.22 then the material does not yield because σ max σ = 28, 500 psi − σmin = 0 − (−200) = 200 MPa < 280 MPa. To determine the ultimate tensile strength for There-fore, σ3 must be lower than σ2, and is the material, realize that the strain at necking calculated from: is equal to the strain hardening exponent, or = n. Therefore, σmax − σmin = σ1 − σ3 = 280 MPa n 0.22 σult = K(n) = 28, 500(0.22) = 20, 400 psi or The cross-sectional area at the onset of σ3 = σ1 − 280 = −100 − 280 = −380 MPa necking is obtained from ln Aneck = n = 0.22 Ao 2.92 A tensile force of 9 kN is applied to the ends of a solid bar of 6.35 mm diameter. Under load, the diameter reduces to 5.00 mm. Assuming Consequently, uniform deformation and volume constancy, (a) −0.22 determine the engineering stress and strain, (b) Aneck = Aoe determine the true stress and strain, (c) if the and the maximum load is original bar had been subjected to a true stress of 345 MPa and the resulting diameter was 5.60 P = σA = σultAneck. mm, what are the engineering stress and engi- neering strain for this condition? Hence, −0.22 First note that, in this case, do = 6.35 mm, df P = (20, 400)(Ao)e = 16, 370Ao = 5.00 mm, P =9000 N, and from volume Since UTS= P/Ao, we have con-stancy, 16, 370Ao lf d 2 6.35 2 UTS = = 16, 370 psi 2 2 o Ao lodo = lf d f → lo = d2 = 5.002 = 1.613 f 23 (a) The engineering stress is calculated This problem uses a similar approach as for from Eq. (2.3) on p. 30 as: Ex-ample 2.1. First, we note from Table 2.3 on p. 37 that for cold-rolled 1112 steel, K = P 9000 760 MPa and n = 0.08. Also, the initial cross- σ= =π = 284 MPa π 2 2 Ao 4 (6.35)2 sectional area is Ao = 4 (10) = 78.5 mm . For annealed 1112 steel, K = 760 MPa and n and the engineering strain is calculated = 0.19. At necking, = n, so that the strain will from Eq. (2.1) on p. 30 as: be = 0.08 for the cold-rolled steel and l − lo lf = 0.19 for the annealed steel. For the cold- e= = − 1 = 1.613 − 1 = 0.613 rolled steel, the final length is given by Eq. l l (2.9) on p. 35 as o o (b) The true stress is calculated from Eq. (2.8) l on p. 34 as: = n = ln lo P 9000 σ= = Solving for l, = 458 MPa A π4 (5.00)2 n 0.08 l = e lo = e (25) = 27.08 mm and the true strain is calculated from Eq. (2.9) on p. 35 as: The elongation is, from Eq. (2.6), = ln lo = ln 1.613 = 0.478 lf − lo 27.08 − 25 l Elongation = × 100 = × 100 f lo 25 or 8.32 %. To calculate the ultimate strength, (c) If the final diameter is d f = 5.60 mm, then we can write, for the cold-rolled steel, π 2 2 the final area is Af = 4 d f = 24.63 mm . n 0.08 If the true stress is 345 MPa, then UTStrue = Kn = 760(0.08) = 621 MPa P = σA = (345)(24.63) = 8497 ≈ 8500 N As in Example 2.1, we calculate the load at necking as: Therefore, the engineering stress is −n P = UTStrueAoe calcu-lated as before as P 8500 So that σ= =π = 268 MPa P UTStrueAoe−n Ao 4 (6.35)2 UTS = = = UTStruee−n Similarly, from volume constancy, Ao Ao l d2 6.352 This expression is evaluated as f= o= = 1.286 −0.08 UTS = (621)e = 573 MPa Therefore, the engineering strain is Repeating these calculations for the annealed specimen yields l = 30.23 mm, elongation = lf 20.9%, and UTS= 458 MPa. e= − 1 = 1.286 − 1 = 0.286 l 2.94 During the production of a part, a metal with a o yield strength of 110 MPa is subjected to a 2.93 Two identical specimens 10-mm in diameter stress state σ1, σ2 = σ1/3, σ3 = 0. Sketch the and with test sections 25 mm long are made Mohr’s circle diagram for this stress state. De- of 1112 steel. One is in the as-received condi- tion and the other is annealed. What will be termine the stress σ1 necessary to cause the true strain when necking begins, and what yielding by the maximum shear stress and the will be the elongation of these samples at that von Mises criteria. instant? What is the ultimate tensile strength For the stress state of σ1, σ1/3, 0 the following for these samples? figure the three-dimensional Mohr’s circle: 24 Because the radius is 5 mm and one-half the penetration diameter is 1.5 mm, we can obtain α as 1.5 α = sin−1 = 17.5◦ 3 2 1 5 The depth of penetration, t, can be obtained from ◦ t = 5 − 5 cos α = 5 − 5 cos 17.5 = 0.23 mm For the von Mises criterion, Eq. (2.37) on p. 2.96 The following data are taken from a stainless 64 gives: steel tension-test specimen: 2 2 2 2 Load, P (lb) Extension, l (in.) 2Y = (σ1 − σ2) + (σ2 − σ3) + (σ3 − σ1) σ 2 σ 2 1600 0 1 1 2 2500 0.02 = σ1 − 3 + 3−0 + (0 − σ1) 3000 0.08 4 2 1 2 2 14 2 3600 0.20 = 9 σ1 + 9 σ1 + σ1 = 9 σ1 4200 0.40 Solving for σ1 gives σ1 = 125 MPa. According 4500 0.60 to the Tresca criterion, Eq. (2.36) on p. 64 on 4600 (max) 0.86 p. 64 gives 4586 (fracture) 0.98 2 2 σ1 − σ3 = σ1 = 0 = Y Also, Ao = 0.056 in , Af = 0.016 in , lo = 2 in. Plot the true stress-true strain curve for the or σ1 = 110 MPa. material. 2.95 Estimate the depth of penetration in a Brinell The following are calculated from Eqs. (2.6), hardness test using 500-kg load, when the (2.9), (2.10), and (2.8) on pp. 33-35: sam-ple is a cold-worked aluminum with a yield stress of 200 MPa. A σ 2 Note from Fig. 2.24 on p. 55 that for cold- l l (in ) (ksi) worked aluminum with a yield stress of 200 0 2.0 0 0.056 28.5 MPa, the Brinell hardness is around 65 0.02 2.02 0.00995 0.0554 45.1 2 kg/mm . From Fig. 2.22 on p. 52, we can esti- 0.08 2.08 0.0392 0.0538 55.7 mate the diameter of the indentation from the 0.2 2.2 0.0953 0.0509 70.7 expression: 0.4 2.4 0.182 0.0467 90. 0.6 2.6 0.262 0.0431 104 2P 0.86 2.86 0.357 0.0392 117 HB = √ 2 2 0.98 2.98 0.399 0.0376 120 (πD)(D − D − d ) from which we find that d = 3.091 mm for D = The true stress-true strain curve is then 10mm. To calculate the depth of pene-tration, plotted as follows: consider the following sketch: 160 (ksi) 120 stress, 80 5 mm 40 True 3 mm 00 0.2 0.4 True strain, 25 2.97 A metal is yielding plastically under the stress 2.98 It has been proposed to modify the von Mises state shown in the accompanying figure. yield criterion as: a a a 20 MPa (σ1 − σ2) + (σ2 − σ3) + (σ3 − σ1) = C where C is a constant and a is an even inte- ger larger than 2. Plot this yield criterion for a = 4 and a = 12, along with the Tresca and von 40 MPa Mises criteria, in plane stress. (Hint: See Fig. 2.36 on p. 67). For plane stress, one of the stresses, say σ 3, is zero, and the other stresses are σA and σB. 50 MPa The yield criterion is then a a a (a) Label the principal axes according to their (σA − σB) + (σB) + (σA) = C proper numerical convention (1, 2, 3). For uniaxial tension, σA = Y and σB = 0 so a (b) What is the yield stress using the Tresca that C = 2Y . These equations are difficult to criterion? solve by hand; the following solution was (c) What if the von Mises criterion is used? obtained using a mathematical programming package: (d) The stress state causes measured strains of 1 = 0.4 and 2 = 0.2, with 3 not B being measured. What is the value of 3? von Mises a=4 Y a=12 (a) Since σ1 ≥ σ2 ≥ σ3, then from the figure Tresca σ1 = 50 MPa, σ2 = 20 MPa and σ 3 = −40 MPa. A (b) The yield stress using the Tresca Y criterion is given by Eq. (2.36) as σ −σ =Y max min So that Y = 50 MPa − (−40 MPa) = 90 MPa (c) If the von Mises criterion is used, then Note that the solution for a = 2 (von Mises) Eq. (2.37) on p. 64 gives and a = 4 are so close that they cannot be 2 2 2 distinguished in the plot. When zoomed into a (σ1 − σ2) + (σ2 − σ3) + (σ3 − σ1) = 2Y portion of the curve, one would see that the a 2 or = 4 curve lies between the von Mises curve and the a = 12 curve. 2 2 2 2Y = (50 − 20) + (20 + 40) + (50 + 2 2.99 Assume that you are asked to give a quiz to stu- 40) or dents on the contents of this chapter. Prepare 2 2Y = 12, 600 three quantitative problems and three qualita- which is solved as Y = 79.4 MPa. tive questions, and supply the answers. (d) If the material is deforming plastically, By the student. This is a challenging, open- then from Eq. (2.48) on p. 69, ended question that requires considerable focus and understanding on the part of the 1 + 2 + 3 = 0.4 + 0.2 + 3 = 0 or student, and has been found to be a very = −0.6. valuable home-work problem. 3 26
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