Solution Manual for Manufacturing Processes for Engineering Materials, 6th Edition

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Manufacturing Processes for Engineering Materials, 6th ed.Serope KalpakjianSteven R. Schmid

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Chapter2FundamentalsoftheMechanicalBehaviorofMaterialsQUESTIONS2.1Can you calculate the percent elongation of materialsbased only on the information given in Fig. 2.6?Explain.Recall that the percent elongation is defined by Eq. (2.6)on p. 35 and depends on the original gage length(lo)ofthe specimen. From Fig. 2.6 on p. 39, only the neckingstrain (true and engineering) and true fracture straincan be determined. Thus, we cannot calculate the percent elongation of the specimen; also, note that theelongation is a function of gage length and increaseswith gage length.2.2Explain if it is possible for stress-strain curves in tension tests to reach 0% elongation as the gage length isincreased further.The percent elongation of the specimen is a function ofthe initial and final gage lengths. When the specimen isbeing pulled, regardless of the original gage length, itwill elongate uniformly (and permanently) until necking begins. Therefore, the specimen will always havea certain finite elongation.However, note that as thespecimen’s gage length is increased, the contributionof localized elongation (that is, necking) will decrease,but the total elongation will not approach zero.2.3Explain why the difference between engineering strainand true strain becomes larger as strain increases.Isthis phenomenon true for both tensile and compressivestrains? Explain.Thedifferencebetweentheengineeringandtruestrains becomes larger because of the way the strainsare defined, respectively, as can be seen by inspectingEqs. (2.1) and (2.9).This is true for both tensile andcompressive strains.2.4Using the same scale for stress, the tensile true-stresstrue-strain curve is higher than the engineering stress-strain curve. Explain whether this condition also holdsfor a compression test.During a compression test,the cross-sectional areaof 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 loadto the instantaneous cross-sectional area of the specimen, the true stress in compression will be lower thanthe engineering stress for a given load, assuming thatfriction between the platens and the specimen is negligible.2.5Which of the two tests, tension or compression, requiresahigher capacitytestingmachinethan theother? Explain.The compression test requires a higher capacity machine because the cross-sectional area of the specimenincreases during the test, which is the opposite of a tension test.The increase in area requires a load higherthan that for the tension test to achieve the same stresslevel.Furthermore, note that compression-test specimens generally have a larger original cross-sectionalarea than those for tension tests, thus requiring higherforces.2.6Explain how the modulus of resilience of a materialchanges, if at all, as it is strained: (a) for an elastic, perfectly plastic material, and (b) for an elastic, linearlystrain-hardening material.RecallthatthemodulusofresilienceisgivenbyEq. (2.5) on p. 34 asS2y/(2E).(a) If the material isperfectly plastic, then the yield strength does not increase with strain - see Fig. 2.7c on p. 42. Therefore, themodulus of resilience is unchanged as the material isstrained. (b) For a linear strain hardening material, theyield strength increases with plastic strain. Thereforethe modulus of resilience will increase with strain.1

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2.7If you pull and break a tensile-test specimen rapidly,where would the temperature be the highest? Explainwhy.Since temperature rise is due to the work input, thetemperature 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.2.8Comment on the temperature distribution if the specimen in Question 2.7 is pulled very slowly.If the specimen is pulled very slowly, the temperaturegenerated will be dissipated throughout the specimenand to the environment.Thus, there will be no appreciable temperature rise anywhere, particularly withmaterials with high thermal conductivity.2.9In 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-elongationcurve represents the work done on the specimen.Ifyou divide this latter work by the volume of the specimen between the gage marks, you will determine thework done per unit volume (assuming that all deformation is confined between the gage marks). Will thisspecific work be the same as the area under the true-stress-true-strain curve? Explain. Will your answer bethe same for any value of strain? Explain.If we divide the work done by the total volume of thespecimen between the gage lengths, we obtain the average specific work throughout the specimen.However, the area under the true stress-true strain curverepresents the specific work done at the necked (andfractured) region in the specimen where the strain is amaximum. Thus, the answers will be different. However, up to the onset of necking (instability), the specificwork calculated will be the same. This is because thestrain is uniform throughout the specimen until necking begins.2.10The note at the bottom of Table 2.4 states that as temperature increases,Cdecreases andmincreases.Explain why.The value ofCin Table 2.4 on p. 46 decreases with temperature because it is a measure of the strength of thematerial.The value ofmincreases with temperaturebecause the material becomes more strain-rate sensitive, due to the fact that the higher the strain rate, theless time the material has to recover and recrystallize,hence its strength increases.2.11You are given theKandnvalues of two different materials. Is this information sufficient to determine whichmaterial is tougher?If not, what additional information do you need, and why?Although theKandnvalues may give a good estimate of toughness, the true fracture stress and the truestrain at fracture are required for accurate calculationof toughness. The modulus of elasticity and yield stresswould provide information about the area under theelastic region; however, this region is very small and isthus usually negligible with respect to the rest of thestress-strain curve.2.12Modify the curves in Fig. 2.7 to indicate the effects oftemperature. Explain your changes.These modifications can be made by lowering the slopeof the elastic region and lowering the general height ofthe curves. See, for example, Fig. 2.9 on p. 43.2.13Using a specific example, show why the deformationrate, say in m/s, and the true strain rate are not thesame.The deformation rate is the quantityvin Eqs. (2.16)and (2.17).Thus, whenvis held constant during deformation (hence a constant deformation rate), the truestrain rate will vary (lincreases), whereas the engineering strain rate will remain constant.Hence, the twoquantities are not the same.2.14It has been stated that the higher the value ofm, themore diffuse the neck is, and likewise, the lower thevalue ofm, the more localized the neck is. Explain thereason for this behavior.As discussed in Section 2.2.7, with highmvalues, thematerial stretches to a greater length before it fails; thisbehavior is an indication that necking is delayed withincreasingm.When necking is about to begin, thenecking region’s strength with respect to the rest of thespecimen increases, due to strain hardening. However,the strain rate in the necking region is also higher thanin 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. Theincrease in resistance to necking thus depends on themagnitude ofm. As the tension test progresses, necking becomes morediffuse, and the specimen becomeslonger before fracture; hence, total elongation increaseswith increasing values ofm. As expected, the elongation after necking (postuniform elongation) also increaseswith increasingm. It has been observed that the valueofmdecreases with metals of increasing strength.2.15Explain why materials with highmvalues, such as hotglass and silly putty, when stretched slowly, undergolarge elongations before failure. Consider events taking place in the necked region of the specimen.The answer is similar to Answer 2.14 above.2.16Assume that you are running four-point bending testson a number of identical specimens of the same lengthand 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.2

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As the distance between the upper points of loadingin Fig. 2.19b increases, the magnitude of the bendingmoment decreases. However, the volume of materialsubjected to the maximum bending moment (hence tomaximum stress) increases.Thus, the probability offailure in the four-point test increases as this distanceincreases.2.17Would Eq. (2.10) hold true in the elastic range? Explain.Note that this equation is based on volume constancy,i.e.,Aolo=Al.We know, however, that because thePoisson’s ratioνis less than 0.5 in the elastic range, thevolume is not constant in a tension test; see Eq. (2.47)on p. 71. Therefore, the expression is not valid in theelastic range.2.18Why have different types of hardness tests been developed? How would you measure the hardness of a verylarge object?There are several basic reasons:1.The overall hardness range of the materials2.The hardness of their constituents; see Chapter 3;3.The thickness of the specimen, such as bulk versus foil4.The size of the specimen with respect to that of theindenter5.The surface finish of the part being tested.2.19Which hardness tests and scales would you use forvery thin strips of material, such as aluminum foil?Why?Because aluminum foil is very thin, the indentationson the surface must be very small so as not to affecttest results.Suitable tests would be a microhardnesstest such as Knoop or Vickers under very light loads(see Fig. 2.20 on p. 54).The accuracy of the test canbe validated by observing any changes in the surfaceappearance opposite to the indented side.2.20List and explain the factors that you would consider inselecting an appropriate hardness test and scale for aparticular application.Hardness tests mainly have three differences:1.type of indenter,2.applied load, and3.method of indentation measurement (depth orsurface area of indentation, or rebound of indenter).2.21In a Brinell hardness test, the resulting impression isfound to be an ellipse. Give possible explanations forthis result.Thereareseveralpossiblereasonsforthisphenomenon, but the two most likely are anisotropy in thematerial and the presence of surface residual stressesin the material.2.22Referring to Fig. 2.20, the material for testers are eithersteel, tungsten carbide, or diamond.Why isn’t diamond used for all of the tests?Whilediamondisthehardestmaterialknown,itwould not, for example, be practical to make and usea 10-mm diamond indenter because the costs wouldbe prohibitive. Consequently, a hard material such asthose listed are sufficient for most hardness tests.2.23What role does friction play in a hardness test?Canhigh friction between a material and indenter affect ahardness test? Explain.The effect of friction has been found to be minimal. Ina hardness test, most of the indentation occurs throughplastic deformation, and there is very little sliding atthe indenter-workpiece interface; see Fig. 2.23 on p. 58.2.24Describe the difference between creep and stress relaxation, giving two examples for each as they relate toengineering applications.Creep is the permanent deformation of a part that isunder a load over a period of time, usually occurring atelevated temperatures. Stress relaxation is the decreasein the stress level in a part under a constant strain. Examples of creep include:1.turbine blades operating at high temperatures,and2.high-temperature steam linesand furnace components.Stress relaxation is observed when, for example, a rubber band or a thermoplastic is pulled to a specificlength and held at that length for a period of time. Thisphenomenon is commonly observed in rivets, bolts,and guy wires, as well as thermoplastic components.2.25Referring 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.Note that impacting the specimens shown in Fig. 2.26on p. 61 from the opposite directions would subjectthe roots of the notches to compressive stresses, andthus they would not act as stress raisers. Thus, crackswould not propagate as they would when under tensile stresses. Consequently, the specimens would basically behave as if they were not notched.2.26If you remove the layeradfrom the part shown inFig. 2.27d, such as by machining or grinding, whichway will the specimen curve?(Hint:Assume thatthe part in diagram (d) is composed of four horizontalsprings held at the ends. Thus, from the top down, wehave compression, tension, compression, and tensionsprings.)Since the internal forces will have to achieve a state ofstatic equilibrium, the new part has to bow downward(i.e., it will hold water). Such residual-stress patterns3

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can be modeled with a set of horizontal tension andcompression springs. Note that the top layer of the materialadin Fig. 2.27d, which is under compression, hasthe tendency to bend the bar upward. When this stressis relieved (such as by removing a layer), the bar willcompensate for it by bending downward.2.27Is it possible to completely remove residual stresses in apiece of material by the technique described in Fig. 2.29if the material is elastic, linearly strain hardening? Explain.ByfollowingthesequenceofeventsdepictedinFig. 2.29 on p. 64, it can be seen that it is not possible to completely remove the residual stresses.Notethat for an elastic, linearly strain hardening material,σc�will never catch up withσt�.2.28Referring to Fig. 2.29, would it be possible to eliminate residual stresses by compression?Assume thatthe piece of material will not buckle under the uniaxialcompressive force.Yes, by the same mechanism described in Fig. 2.29 onp. 64.2.29List and explain the desirable mechanical propertiesfor (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.The following are some basic considerations:(a)Elevator cable: The cable should not elongate elastically to a large extent or undergo yielding as theload is increased. These requirements thus call fora material with a high elastic modulus and yieldstress.(b)Bandage: The bandage material must be compliant, that is, have a low stiffness, but have highstrength 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 sothat it absorbs shock and should have high friction and wear resistance.(d)Fishhook:Afishhookneedstohavehighstrength so that it doesn’t deform permanentlyunder load, and thus maintain its shape. It shouldbe stiff (for better control during its use) andshould be resistant the environment it is used in(such as salt water).(e)Automotive piston: This product must have highstrength at elevated temperatures, high physicaland thermal shock resistance, and low mass.(f)Boat propeller:The material must be stiff (tomaintain its shape) and resistant to corrosion, andalso have abrasion resistance because the propeller encounters sand and other abrasive particles when operated close to shore.(g)Gas turbine blade: A gas turbine blade operatesat high temperatures (depending on its location inthe turbine); thus it should have high-temperaturestrength and resistance to creep, as well as to oxidation and corrosion due to combustion productsduring its use.(h)Staple: The properties should be closely parallelto that of a paper clip.The staple should havehigh ductility to allow it to be deformed withoutfracture, and also have low yield stress so that itcan be bent (as well as unbent when removing it)easily without requiring excessive force.2.30Make a sketch showing the nature and distribution ofthe residual stresses in Figs. 2.28a and b before the partswere cut. Assume that the split parts are free from anystresses. (Hint:Force these parts back to the shape theywere in before they were cut.)As the question states, when we force back the splitportions 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 thesame technique, we find that the specimen in Fig. 2.28bwould have a similar residual stress distribution priorto cutting.2.31It 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 orlower than the actual work done? Explain.If we calculate the heat using a constant specific heatvalue in Eq. (2.62), the work will be higher than it actually is. This is because, by definition, as the specificheat decreases, less work is required to raise the workpiece temperature by one degree.Consequently, thecalculated work will be higher than the actual workdone.2.32Explain whether or not the volume of a metal specimenchanges when the specimen is subjected to a state of(a) uniaxial compressive stress and (b) uniaxial tensilestress, all in the elastic range.For case (a), the quantity in parentheses in Eq. (2.47)on p. 71 will be negative, because of the compressivestress. Since the rest of the terms are positive, the product of these terms is negative and, hence, there will bea decrease in volume (This can also be deduced intuitively.) For case (b), it will be noted that the volumewill increase.2.33It is relatively easy to subject a specimen to hydrostaticcompression, such as by using a chamber filled with aliquid. Devise a means whereby the specimen (say, in4

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the shape of a cube or a round disk) can be subjectedto hydrostatic tension, or one approaching this state ofstress. (Note that a thin-walled, internally pressurizedspherical shell is not a correct answer, because it is subjected only to a state of plane stress.)Two possible answers are the following:1.A solid cube made of a soft metal has all its sixfaces brazed to long square bars (of the same crosssection as the specimen); the bars are made of astronger metal. The six arms are then subjected toequal tension forces, thus subjecting the cube toequal tensile stresses.2.A thin, solid round disk (such as a coin) and madeof a soft material is brazed between the ends oftwo solid round bars of the same diameter as thatof the disk. When subjected to longitudinal tension, the disk will tend to shrink radially. But because it is thin and its flat surfaces are restrainedby the two rods from moving, the disk will be subjected to tensile radial stresses. Thus, a state of triaxial (though not exactly hydrostatic) tension willexist within the thin disk.2.34Referring to Fig. 2.17, make sketches of the state ofstress for an element in the reduced section of the tubewhen it is subjected to (a) torsion only; (b) torsion whilethe tube is internally pressurized; and (c) torsion whilethe tube is externally pressurized.Assume that thetube is a closed-end tube.These states of stress can be represented simply by referring to the contents of this chapter as well as the relevant materials covered in texts on mechanics of solids.2.35A penny-shaped piece of soft metal is brazed to theends of two flat, round steel rods of the same diameteras the piece. The assembly is then subjected to uniaxial tension. What is the state of stress to which the softmetal is subjected? Explain.The penny-shaped soft metal piece will tend to contractradially due to Poisson’s ratio; however, the solid rodsto which it attached will prevent this from happening.Consequently, the state of stress will tend to approachthat of hydrostatic tension.2.36A circular disk of soft metal is being compressed between two flat, hardened circular steel punches of having the same diameter as the disk.Assume that thedisk material is perfectly plastic and that there is nofriction or any temperature effects. Explain the change,if any, in the magnitude of the punch force as the diskis being compressed plastically to, say, a fraction of itsoriginal thickness.Note that as it is compressed plastically,the diskwill expand radially,because of volume constancy.An approximately donut-shaped material will then bepushed radially outward, which will then exert radial compressive stresses on the disk volume underthe punches. The volume of material directly betweenthe punches will now subjected to a triaxial compressive state of stress.According to yield criteria (seeSection 2.11), the compressive stress exerted by thepunches will thus increase, even though the materialis not strain hardening. Therefore, the punch force willincrease as deformation increases.2.37A perfectly plastic metal is yielding under the stressstateσ1,σ2,σ3, whereσ1> σ2> σ3.Explain whathappens ifσ1is increased.Consider Fig. 2.32 on p. 70.Points in the interior ofthe yield locus are in an elastic state, whereas those onthe yield locus are in a plastic state. Points outside theyield locus are not admissible. Therefore, an increaseinσ1while the other stresses remain unchanged wouldrequire an increase in yield stress. This can also be deduced by inspecting either Eq. (2.38) or Eq. (2.39).2.38What is the dilatation of a material with a Poisson’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.It can be seen from Eq. (2.47) on p. 71 that the dilatationof a material withν= 0.5is always zero, regardless ofthe stress state. To examine the case ofν= 0.7, considerthe situation where the stress state is hydrostatic tension.Equation (2.47) would then predict contractionunder a tensile stress, a situation that cannot occur.2.39Can a material have a negative Poisson’s ratio? Give arationale for your answer.Solid material do not have a negative Poisson’s ratio,with the exception of some composite materials (seeChapter 10), where there can be a negative Poisson’sratio in a given direction.2.40As clearly as possible, define plane stress and planestrain.Plane stress is the situation where the stresses in one ofthe direction on an element are zero; plane strain is thesituation where the strains in one of the direction arezero.2.41What test would you use to evaluate the hardness of acoating on a metal surface? Would it matter if the coating was harder or softer than the substrate? Explain.The answer depends on whether the coating is relatively thin or thick. For a relatively thick coating, conventional hardness tests can be conducted, as long asthe deformed region under the indenter is less thanabout one-tenth of the coating thickness.If the coating thickness is less than this threshold, then one musteither rely on nontraditional hardness tests, or elseuse fairly complicated indentation models to extractthe material behavior.As an example of the former,atomic force microscopes using diamond-tipped pyramids have been used to measure the hardness of coatings less than 100 nanometers thick. As an example of5

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the latter, finite-element models of a coated substratebeing indented by an indenter of a known geometrycan be developed and then correlated to experiments.2.42List the advantages and limitations of the stress-strainrelationships given in Fig. 2.7.Several answers that are acceptable, and the student isencouraged to develop as many as possible. Two possible answers are:1.There is a tradeoff between mathematical complexity and accuracy in modeling material behavior2.Some materials may be better suited for certainconstitutive laws than others2.43Plot the data in the inside front cover on a bar chart,showing the range of values, and comment on the results.By the student. An example of a bar chart for the elasticmodulus is shown below.0100200300400500AluminumCopperLeadMagnesiumMolybdenumNickelSteelsStainless steelsTitaniumTungstenElastic modulus (GPa)Metallic materials020040060080010001200CeramicsDiamondGlassRubbersThermoplasticsThermosetsBoron fibersCarbon fibersGlass fibersKevlar fibersSpectra fibersElastic modulus (GPa)Non-metallic materialsTypical comments regarding such a chart are:1.There is a smaller range for metals than for nonmetals;2.Thermoplastics, thermosets and rubbers are orders of magnitude lower than metals and othernon-metals;3.Diamond and ceramics can be superior to others,but ceramics have a large range of values.2.44A hardness test is conducted on as-received metal as aquality check. The results show indicate that the hardness is too high, indicating that the material may nothave sufficient ductility for the intended application.The supplier is reluctant to accept the return of the material, instead claiming that the diamond cone used inthe Rockwell testing was worn and blunt, and hencethe test needed to be recalibrated. Is this explanationplausible? Explain.Refer to Fig. 2.20 on p. 54 and note that if an indenter is blunt, then the penetration,t, under a given loadwill be smaller than that using a sharp indenter. Thisthen translates into a higher hardness. The explanationis plausible, but in practice, hardness tests are fairlyreliable and measurements are consistent if the testing equipment is properly calibrated and routinely serviced.2.45Explain why a 0.2% offset is used to obtain the yieldstrength in a tension test.The value of 0.2% is somewhat arbitrary and is usedto set some standard. A yield stress, representing thetransition point from elastic to plastic deformation, isdifficult to measure.This is because the stress-straincurve is not linearly proportional after the proportional limit, which can be as high as one-half the yieldstrength in some metals. Therefore, a transition fromelastic to plastic behavior in a stress-strain curve is difficult to discern.The use of a 0.2% offset is a convenient way of consistently interpreting a yield pointfrom stress-strain curves.2.46Referring to Question 2.45, would the offset method benecessary for a highly strained hardened material? Explain.The 0.2% offset is still advisable whenever it can beused, because it is a standardized approach for determining yield stress, and thus one should not arbitrarilyabandon standards. However, if the material is highlycold worked, there will be a more noticeable ‘kink’ inthe stress-strain curve, and thus the yield stress is farmore easily discernable than for the same material inthe annealed condition.2.47Explain why the hardness of a material is related to amultiple of the uniaxial compressive stress, since bothinvolve compression of workpiece material.The hardness is related to a multiple of the uniaxialcompressive stress, not just the uniaxial compressivestress, because:6

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1.The volume of material that is stressed is different- in a hardness test, the volume that is under stressis not just a cylinder beneath the inventor.2.The stressed volume is constrained by the elastic material outside of the indentation area. Thisoften requires material to deform laterally andcounter to the indentation direction - see Fig. 2.212.48Without using the words “stress” or “strain”, defineelastic modulus.This is actually quite challenging, but historically sig-nificant, since Thomas Young did not have the benefitof the concept of strain when he first defined modulus of elasticity. Young’s definition satisfies the projectrequirement. In Young’s words:The modulus of the elasticity of any substance is a columnof the same substance, capable of producing a pressure on itsbase which is to the weight causing a certain degree of compression as the length of the substance is to the diminutionof the length.There are many possible other definitions, of course.PROBLEMS2.49A strip of metal is originally 1.0 m long. It is stretchedin three steps: first to a length of 1.5 m, then to 2.5 m,and finally to 3.0 m. Show that the total true strain isthe sum of the true strains in each step, that is, thatthe strains are additive. Show that, using engineeringstrains, the strain for each step cannot be added to obtain the total strain.The true strain is given by Eq. (2.9) on p. 36 asle=lnloTherefore, the true strains for the three steps are:1.5e1= ln= 0.40551.02.5e2= ln= 0.51081.53.0e3= ln= 0.18232.5The sum of these true strains ise= 0.4055+0.5108+0.1823=1.099. The true strain from step 1 to 3 is3e= ln= 1.0991Therefore the true strains are additive. Using the sameapproach for engineering strain as defined by Eq. (2.1),we obtaine1= 0.5,e2= 0.667, ande3= 0.20. The sumof these strains ise1+e2+e3= 1.367. The engineeringstrain from step 1 to 3 isl−lo3−1e=== 2lo1Note that this is not equal to the sum of the engineeringstrains for the individual steps. The following Matlabcode can be used to demonstrate that this is generallytrue, and not just a conclusion for the specific deformations stated in the problem.l0=1;l1=1.5;l2=2.5;l3=3;etot=(l1-l0)/l0+(l2-l0)/l0+(l3-l0)/l0;efin=(l3-l0)/l0;epstot=log(l1/l0)+log(l2/l1)+log(l3/l2);epsfin=log(l3/l0);2.50A paper clip is made of wire 1.00 mm in diameter. Ifthe original material from which the wire is made is arod 15 mm in diameter, calculate the longitudinal anddiametrical engineering and true strains that the wirehas undergone during processing.Assuming volume constancy, we may write22lfdo15=== 225lodf1.00Lettinglobe unity, the longitudinal engineering strainise1=(225−1)/1=224. The diametral engineeringstrain is calculated as1−15ed==−0.93315The longitudinal true strain is given by Eq. (2.9) onp. 36 asle=ln=ln (224)=5.412loThe diametral true strain is1ed= ln=−2.70815Note the large difference between the engineering andtrue strains, even though both describe the same phenomenon.Note also that the sum of the true strains(recognizing that the radial strain isa0.5er= ln=15−2.708)in the three principal directions is zero, indicating volume constancy in plastic deformation.The following Matlab code is useful:7

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d=0.001;d0=0.015;l0=1;lf=(d0/d)ˆ2*l0;e1=(lf-l0)/l0;ed=(d-d0)/d0;epsilon=log(lf/l0);epsilon_d=log(d/d0);2.51A material has the following properties:Sut=350MPaandn= 0.20. Calculate its strength coefficient,K.Note from Eq. (2.11) on p. 36 thatSut,true=Ken=Knnbecause at neckinge=n.From Fig. 2.3,PSut=Ao,wherePis the load at necking. The true ultimate tensile strength would beAoSut,true=P/A=SutA.From Eq. (2.10),Aoln=e= 0.20ATherefore,Ao=exp(0.2)=1.2214ASubstitutingintotheexpressionfortrueultimatestrength,Sut,true=(350MPa)(1.2214)=427MPa.The strength coefficient,K, can then be found as427K==589MPa.0.20.22.52Based on the information given in Fig. 2.6, calculate theultimate tensile strength of 304 stainless steel.From Fig. 2.6 on p. 39, the true stress for 304 stainlesssteel at necking (where the slope changes; see Fig. 2.7e)is found to be about 900 MPa, while the true strain isabout 0.4. We also know that the ratio of the original tonecked areas of the specimen is given byAoln= 0.40AneckorAneck=e−0.40= 0.670AoThus,Sut=(900)(0.670)=603MPa.2.53Calculate the ultimate tensile strength (engineering) ofa material whose strength coefficient is 300 MPa andthat necks at a true strain of 0.25.In this problem,K=300MPaandn= 0.25. Followingthe same procedure as in Example 2.1 on p. 41, the trueultimate tensile strength isσ=(300)(0.25)0.25=212MPaandAneck=Aoe−0.25= 0.779AoConsequently,Sut=(212)(0.779)=165MPa.2.54A material has a strength coefficientK=700MPa. Assuming that a tensile-test specimen made from this material begins to neck at a true strain of 0.20, show thatthe ultimate tensile strength of this material is 415 MPa.The approach is the same as in Example 2.1 on p. 41.Since the necking strain corresponds to the maximumload and the necking strain for this material is givenase=n= 0.20, we have, as the true ultimate tensilestrength:Sut,true=(700)(0.20)0.20=507MPa.The cross-sectional area at the onset of necking is obtained fromAoln=n= 0.20.AneckConsequently,Aneck=Aoexp(−0.20)and the maximum load,P, isP=σA=Sut,trueAoexp(−0.20)=(507)(0.8187)(Ao)=415×106AoSinceSut=P/Ao, we haveSut=415MPa.This isconfirmed with the following Matlab code.K=700e6;n=0.2;Sut_true=K*nˆn;P=Sut_true*exp(-n)2.55A cable is made of four parallel strands of different materials, all behaving according to the equationσ=Ken,wheren= 0.20.The materials, strength coefficientsand cross-sections are as follows:MaterialA:K=450MPa,Ao= 7mm2;MaterialB:K=600MPa,Ao= 2.5mm2;MaterialC:K=300MPa,Ao= 3mm2;MaterialD:K=750MPa,Ao= 2mm2;(a)Calculate the maximum tensile force that this cable can withstand prior to necking.(b)Explain how you would arrive at an answer if thenvalues of the three strands were different fromeach other.8

Solution Manual for Manufacturing Processes for Engineering Materials, 6th Edition - Page 11 preview image

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(a)Necking will occur whene=n=0.20.Atthis point the true stresses in each cable are, fromEq. (2.11) on p. 36,σA=(450)0.20.2=326MPaσB=(600)0.20.2=435MPaσC=(300)0.20.2=217MPaσD=(760)0.20.2=543MPaThe areas at necking are calculated fromAneck=Aoe−n(see Example 2.1 on p. 41):2AA=(7)e−0.2= 5.73mm2AB=(2.5)e−0.2= 2.04mm2AC=(3)e−0.2= 2.46mm2AD=(2)e−0.2= 1.64mmHence the total load that the cable can support isP=(326)(5.73)+(435)(2.04)+(217)(2.46)+(543)(1.64)=4180N.The following Matlab code is helpful:K_A=450e6;K_B=600e6;K_C=300e6;K_D=750e6;A0_A=7/1e6;A0_B=2.5/1e6;A0_C=3/1e6;A0_D=2/1e6;n=0.20;s_A=K_A*nˆn;s_B=K_B*nˆn;s_C=K_C*nˆn;s_D=K_D*nˆn;A_A=A0_A*exp(-1*n);A_B=A0_B*exp(-1*n);A_C=A0_C*exp(-1*n);A_D=A0_D*exp(-1*n);P=s_A*A_A+s_B*A_B+s_C*A_C+s_D*A_D;(b)If thenvalues of the four strands were different, the procedure would consist of plotting theload-elongation curves of the four strands on thesame chart, then obtaining graphically the maximum load. Alternately, a computer program canbe written to determine the maximum load.2.56Using only Fig. 2.6, calculate the maximum load in tension testing of a 304 stainless-steel specimen with anoriginal diameter of 6.0 mm.Observe from Fig. 2.6 on p. 39 that necking begins ata true strain of about 0.4 for 304 stainless steel, andthatSut,trueis about 900 MPa (this is the location ofthe ‘kink’ in the stress-strain curve). The original cross-sectional area isAo=π(0.006m)2/4 = 2.827×10−5m2.Sincen= 0.4, a procedure similar to Example 2.1 onp. 41 demonstrates thatAo=exp(0.4)=1.49AneckThus900Sut==604MPa1.49Hence the maximum load isP= (Sut)(Ao)=(604)(2.827×10−5)orP=17.1kN. The following Matlab code is helpfulto investigate other parameters.n=0.4;Sut_true=900e6;A_0=pi*d0*d0/4;Sut=Sut_true/exp(n);P=Sut*A_0;2.57Using the data given in the inside front cover, calculatethe values of the shear modulusGfor the metals listedin the table.The important equation is Eq. (2.24) which gives theshear modulus asEG=2(1+ν)The following values can be calculated (mid-range values ofνare taken as appropriate):MaterialE(GPa)νG(GPa)Al & alloys69-790.3226-30Cu & alloys105-1500.3439-56Pb & alloys140.434.9Mg & alloys41-450.3215.5-17.0Mo & alloys330-3600.32125-136Ni & alloys180-2140.3169-82Steels190-2000.3073-77Stainless steels190-2000.2974-77Ti & alloys80-1300.3230-49W & alloys350-4000.27138-157Ceramics70-10000.229-417Glass70-800.2428-32Rubbers0.01-0.10.50.0033-0.033Thermoplastics1.4-3.40.360.51-1.25Thermosets3.5-170.341.3-6.342.58Derive an expression for the toughness of a materialrepresented by the equationσ=K(e+ 0.2)nandwhose fracture strain is denoted asef.Recall that toughness is the area under the stress-straincurve, hence the toughness for this material would be9

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given byfToughness=σdE0f=K(E+0.2)ndE0=K(Ef+0.2)n+1−0.2n+1n+ 12.59A cylindrical specimen made of a brittle material 50mm high and with a diameter of 25 mm is subjected toa compressive force along its axis. It is found that fracture takes place at an angle of45◦under a load of 130kN. Calculate the shear stress and the normal stress,respectively, acting on the fracture surface.Assuming that compression takes place without friction, note that two of the principal stresses will be zero.The third principal stress acting on this specimen isnormal to the specimen and its magnitude is130,000σ3=−=−264MPa(π/4)(0.025)2The Mohr’s circle for this situation is shown below.2=90°The fracture plane is oriented at an angle of 45◦, corresponding to a rotation of 90◦on the Mohr’s circle. Thiscorresponds to a stress state on the fracture plane of−σ=τ=264/2=132MPa.2.60What is the modulus of resilience of a highly cold-worked piece of steel with a hardness of 280 HB? Ofa piece of highly cold-worked copper with a hardnessof 175 HB?Referring to Fig. 2.22 on p. 57, the value ofcin Eq. (2.31)is approximately 3.2 for highly cold-worked steels andaround 3.4 for cold-worked aluminum. Therefore, approximatec=3.3for cold-worked copper.FromEq. (2.31),H280Sy,steel=== 87.5kg/mm2=858MPa3.23.2H175Sy,Cu=== 53.0kg/mm2=520MPa3.33.3From the inside front cover,Esteel=200GPaandECu=124GPa.The modulus of resilience is calculated from Eq. (2.5) on p. 34. For steel:2S2y858×106Modulus of Resilience==2E2(200×109)or a modulus of resilience for steel of1.81 MN-m/m3.For copper,2S2520×106Modulus of Resilience=y=2E2(124×109)or a modulus of resilience for copper of1.09 MNm/m3.Note that these values are slightly different than thevalues given in the text.This is due to the fact that(a) highly cold-worked metals such as these have amuch higher yield stress than the annealed materialsdescribed in the text; and (b) arbitrary property valuesare given in the statement of the problem.2.61Calculate the work done in frictionless compression ofa solid cylinder 40 mm high and 15 mm in diameterto a reduction in height of 50% for the following materials: (a) 1100-O aluminum; (b) annealed copper; (c)annealed 304 stainless steel; and (d) 70-30 brass, annealed.The work done is calculated from Eq. (2.59) where thespecific energy,u, is obtained from Eq. (2.57) on p. 73.Since the reduction in height is 50%, the final height is20 mm and the absolute value of the true strain is40E= ln= 0.693120Kandnare obtained from Table 2.3 as follows:MaterialK(MPa)n1100-O Al1800.20Cu, annealed3150.54304 Stainless, annealed13000.3070-30 brass, annealed8950.49uis then calculated from Eq. (2.57). For example, for1100-O aluminum, whereKis 180 MPa andnis 0.20,uis calculated asKEn+1(180)(0.6931)1.2u=== 96.6MN/m3n+ 11.2The volume is calculated as3V=πr2l=π(0.0075)2(0.04)=7.069×10−6mThe work done is the product of the specific work,u,and the volume,V. Therefore, the results can be tabulated as follows.Materialu(MN/m3)Work(Nm)1100-O Al96.6682Cu, annealed1681186304 Stainless, annealed620438970-30 brass, annealed3482460The following Matlab code can be used to confirm results:10

Solution Manual for Manufacturing Processes for Engineering Materials, 6th Edition - Page 13 preview image

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h0=0.040;d0=0.015;hf=0.020;epsilon=log(h0/hf);K=180e6;n=0.20;u=K*epsilonˆ(n+1)/(n+1);V=pi*d0ˆ2/4*h0;Work=u*V;2.62A tensile-test specimen is made of a material represented by the equationσ=K(E+n)n. (a) Determinethe true strain at which necking will begin. (b) Showthat it is possible for an engineering material to exhibitthis behavior.(a)In Section 2.2.4 on p. 40, it was noted that instability, hence necking, requires the following condition to be fulfilled:dσ=σdEConsequently, for this material we haveKn(E+n)n−1=K(E+n)nThis is solved asn= 0; thus necking begins assoon as the specimen is subjected to tension.(b)Yes, this behavior is possible. Consider a tension-test specimen that has been strained to neckingand then unloaded. Upon loading it again in tension, it will immediately begin to neck.2.63Take two solid cylindrical specimens of equal diameter, but different heights. Assume that both specimensare compressed (frictionless) by the same percent reduction, say 50%. Prove that the final diameters will bethe same.Identify the shorter cylindrical specimen with the subscriptsand the taller one ast, and their original diameter asD. Subscriptsfandoindicate final and original, respectively. Because both specimens undergo thesame percent reduction in height,htfhsf=htohsoand from volume constancy,2htfDto=htoDtfand2hsfDso=hsoDsfBecauseDto=Dso, note from these relationships thatDtf=Dsf.2.64In a disk test performed on a specimen 50 mm in diameter and 2.5 mm thick, the specimen fractures at astress of 500 MPa. What was the load on it at fracture?Equation (2.22) is used to solve this problem. Notingthatσ=500MPa,d=50mm = 0.05 m, andt= 2.5mm = 0.0025 m, we can write2Pσπdtσ=→P=πdt2Therefore(500×106)π(0.05)(0.0025)P== 98kN.2The following Matlab code allows for variation in parameters for this problem.d=0.050;t=0.0025;sigma=500e6;P=sigma*pi*d*t/2;2.65In Fig. 2.29a, let the tensile and compressive residualstresses both be 70 MPa, and the modulus of elasticityof the material be200GPa with a modulus of resilienceof225 kN-m/m3. If the original length in diagram (a)is 500 mm, what should be the stretched length in diagram (b) so that, when unloaded, the strip will be freeof residual stresses?Note that the yield stress can be obtained from Eq. (2.5)on p. 34 asS2Mod. of Resilience=MR=y2EThus,Sy=2(MR)E=2(225×103)(200×109)orSy=300MPa.The strain required to relieve theresidual stress is:σcSy70×106300×106E=+=+= 0.00185EE200×109200×109Therefore,lflfE= ln= ln= 0.00185lo0.500mTherefore,lf= 0.50093m. The following Matlab codeis helpfulsigma_c=70e6;E=200e9;MR=225e3;l0=0.5;Sy=(2*MR*E)ˆ(0.5);epsilon=sigma_c/E+Sy/E;lf=l0*exp(epsilon);11

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2.66A horizontal rigid barc-cis subjecting specimenato tension and specimenbto frictionless compressionsuch that the bar remains horizontal. (See the accompanying figure.) The forceFis located at a distance ratioof 2:1. Both specimens have an original cross-sectionalarea of0.0001 m2and the original lengths area=200mmandb=115mm. The material for specimenahasa true-stress-true-strain curve ofσ=(700MPa)E0.5.Plot the true-stress-true-strain curve that the materialfor specimenbshould have for the bar to remain horizontal.F21abccxFrom the equilibrium of vertical forces and to keep thebar horizontal, we note that2Fa=Fb. Hence, in termsof true stresses and instantaneous areas, we have2σaAa=σbAbFrom volume constancy we also have, in terms of original and final dimensionsAoaLoa=AaLaandAobLob=AbLbwhereLoa=(0.200/0.115)Lob= 1.73Lob. From theserelationships we can show that0.2Lbσb= 2Kσa0.115LaSinceσa=KE0.5awhereK=700MPa, we can nowwrite0.4KLb√σb=Ea0.115LaHence, for a deflection ofx,0.4K0.115−x0.2 +xσb=ln0.1150.2 +x0.2The true strain in specimen b is given by0.115−xEb=ln0.115By inspecting the figure in the problem statement, wenote that while specimenagets longer, it will continue exerting some forceFa. However, specimenbwilleventually acquire a cross-sectional area that will become infinite asxapproaches 115 mm, thus its strengthmust approach zero.This observation suggests thatspecimenbcannot have a true stress-true strain curvetypical of metals, and that it will have a maximum atsome strain. This is seen in the plot ofσbshown below.350280210140700 00.51.01.52.02.5True stress (MPa)Absolute value of true strain2.67Inspect the curve that you obtained in Problem 2.66.Does a typical strain-hardening material behave in thatmanner? Explain.Based on the discussions in Section 2.2.3, it is obviousthat ordinary metals would not normally behave in thismanner.However, under certain conditions, the following could explain such behavior:•When specimenbis heated to higher and highertemperatures as deformation progresses, with itsstrength decreasing asxis increased further afterthe maximum value of stress.•In compression testing of brittle materials, such asceramics, when the specimen begins to fracture.•If the material is susceptible to thermal softening, then it can display such behavior with a sufficiently high strain rate.2.68Show that you can take a bent bar made of an elastic,perfectly plastic material and straighten it by stretching it into the plastic range. (Hint:Observe the eventsshown in Fig. 2.29.)The series of events that takes place in straighteninga bent bar by stretching it can be visualized by starting with a stress distribution as in Fig. 2.29a on p. 64,which would represent the unbending of a bent section. As we apply tension, we algebraically add a uniform tensile stress to this stress distribution. Note thatthe change in the stresses is the same as that depictedin Fig. 2.29d, namely, the tensile stress increases andreaches the yield stress,Sy.The compressive stressis first reduced in magnitude, then becomes tensile.Eventually, the whole cross section reaches the constant yield stress,Sy. Because we now have a uniformstress distribution throughout its thickness, the bar becomes straight and remains straight upon unloading.12

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2.69A bar 1 m long is bent and then stress relieved.Theradius of curvature to the neutral axis is 0.50 m. Thebar is 25 mm thick and is made of an elastic, perfectlyplastic material withSy=500MPaandE=207GPa. Calculate the length to which this bar should bestretched so that, after unloading, it will become andremain straight.A review of bending theory from a solid mechanicstextbook is necessary for this problem.In particular,it should be recognized that when the curved bar becomes straight, the engineering strain it undergoes isgiven by the expressionte=2ρwheretis the thickness andρis the radius to the neutral axis. Hence in this case,(0.025)e== 0.0252(0.50)SinceSy=500MPaandE=207GPa, we find that theelastic limit for this material is at an elastic strain ofSy500MPae=== 0.00242E207GPawhich is smaller than 0.025. Therefore, we know thatthe bar must be loaded in the plastic range.Following the description in Answer 2.68 above, the strain required to straighten the bar is twice the elastic limit, ore=(2)(0.00242)=0.0048orlf−lo= 0.0048→lf= 0.005lo+loloorlf= 1.0049m.The following Matlab code is helpful.l=1;rho=0.5;Sy=500e6;E=207e9;t=0.025;epsilon=t/2/rho;epsilon_y=Sy/E;epsilon_b=2*epsilon_y;lf=l/(1-epsilon_b);2.70Assume that a material with a uniaxial yield strengthSyyields under a stress state of principal stressesσ1,σ2,σ3, whereσ1> σ2> σ3. Show that the superposition of a hydrostatic stresspon this system (such asplacing the specimen in a chamber pressurized with aliquid) does not affect yielding. In other words, the material will still yield according to yield criteria.This solution considers the distortion-energy criterion,although the same derivation could be performed withthe maximum shear stress criterion as well. Equation(2.39) on p. 67 gives222(σ1−σ2)+ (σ2−σ3)+ (σ3−σ1)= 2Sy2Now consider a new stress state where the principalstresses areσI=σ1+p1σI=σ2+p2Iσ=σ3+p3which represents a new loading with an additional hydrostatic pressure,p.The distortion-energy criterionfor this stress state is222(σ1I−σ2I)+ (σ2I−σ3I)+ (σ3I−σ1I)= 2Sy2or22S2=[(σ1+p)−(σ2+p)]y2+[(σ2+p)−(σ3+p)]2+[(σ3+p)−(σ1+p)]which can be simplified as222(σ1−σ2)+ (σ2−σ3)+ (σ3−σ1)= 2S2ywhich is the original yield criterion. Hence, the yieldcriterion is unaffected by the superposition of a hydrostatic pressure.2.71Give two different and specific examples in which themaximum-shear-stress and the distortion-energy criteria give the same answer.In order to obtain the same answer for the two yieldcriteria, we refer to Fig. 2.32 on p. 70 for plane stressand note the coordinates at which the two diagramsmeet. Examples are: simple tension, simple compression, equal biaxial tension, and equal biaxial compression. Thus, acceptable answers would include (a) wirerope, as used on a crane to lift loads; (b) spherical pressure vessels, including balloons and gas storage tanks;and (c) shrink fits.2.72A thin-walled spherical shell with a yield strengthSyis subjected to an internal pressurep. With appropriateequations, show whether or not the pressure requiredto yield this shell depends on the particular yield criterion used.Here we have a state of plane stress with equal biaxial tension.The answer to Problem 2.71 leads one toimmediately conclude that both the maximum shearstress and distortion energy criteria will give the sameresults. We will now demonstrate this more rigorously.The principal membrane stresses are given byprσ1=σ2=2tandσ3= 013

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Using the maximum shear-stress criterion, we find thatσ1−0 =Sy.Hence2tSyp=rUsing the distortion-energy criterion, we have(0−0)2+ (σ2−0)2+(0−σ1)2= 2S2ySinceσ1=σ2, then this givesσ1=σ2=Sy, and thesame expression is obtained for pressure.2.73Show that, according to the distortion-energy criterion,the yield strength in plane strain is1.15Sy, whereSyisthe uniaxial yield strength of the material.A plane-strain condition is shown in Fig. 2.35d, whereσ1is the yield stress of the material in plane strain(S�)y,σ3is zero, andE2= 0. From Eq. 2.43b, we findthatσ2=σ1/2. Substituting these into the distortion-energy criterion given by Eq. (2.37),22σ1σ1σ1−+−0+ (0−σ1)2= 2S2y22and3σ12= 2Sy22hence2σ1=√Sy≈1.15Sy32.74What would be the answer to Problem 2.73 if the maximum shear stress criterion were used?Becauseσ2isanintermediatestressandusingEq. (2.38) on p. 67, the answer would beσ1−0 =Sy.Hence, the yield stress in plane strain will be equal tothe uniaxial yield stress,Sy.2.75A closed-end, thin-walled cylinder of original lengthlthicknesst, and internal radiusris subjected to aninternal pressurep.Using the generalized Hooke’slaw equations, show the change, if any, that occurs inthe length of this cylinder when it is pressurized. Letν= 0.25.A closed-end, thin-walled cylinder under internal pressure is subjected to the following principal stresses:prσ1= 2t;prσ2=t;σ3= 0where the subscript 1 is the longitudinal direction, 2is the hoop direction, and 3 is the thickness direction.From Hooke’s law given by Eq. (2.34) on p. 66,1E1=[σ1−ν(σ2+σ3)]E1pr1pr=−+ 0E2t3tpr=6tESince all the quantities are positive (note that in orderto produce a tensile membrane stress, the pressure ispositive as well), the longitudinal strain is finite andpositive. Thus the cylinder becomes longer when pressurized, as it can also be deduced intuitively.2.76A round, thin-walled tube is subjected to tension in theelastic range. Show that both the thickness and the diameter decrease as tension increases.The stress state in this case isσ1,σ2=σ3= 0. From thegeneralized Hooke’s law equations given by Eq. (2.34),and denoting the axial direction as 1, the hoop direction as 2, and the radial direction as 3, we have for thehoop strain:1νσ1E2=[σ2−ν(σ1+σ3)]=−EETherefore, the diameter is negative for a tensile (positive) value ofσ1. For the radial strain, the generalizedHooke’s law gives1νσ1E3=[σ3−ν(σ1+σ2)]=−EETherefore, the radial strain is also negative and the wallbecomes thinner for a positive value ofσ1.2.77Take a long cylindrical balloon and, with a thin felt-tippen, mark a small square on it. What will be the shapeof this square after you blow up the balloon, (a) a largersquare; (b) a rectangle with its long axis in the circumferential direction; (c) a rectangle with its long axis inthe longitudinal direction; or (d) an ellipse?Performthis experiment, and, based on your observations, explain the results, using appropriate equations. Assumethat the material the balloon is made up of is perfectlyelastic and isotropic and that this situation representsa thin-walled closed-end cylinder under internal pressure.This is a simple graphic way of illustrating the generalized Hooke’s law equations.A balloon is a readily available and economical method of demonstratingthese stress states. It is also encouraged to assign thestudents the task of predicting the shape numerically;an example of a valuable experiment involves partiallyinflating the balloon, drawing the square, then expanding it further and having the students predict the dimensions of the square.Although not as readily available, a rubber tube can beused to demonstrate the effects of torsion in a similarmanner.2.78Take a cubic piece of metal with a side lengthloanddeform it plastically to the shape of a rectangular parallelepiped of dimensionsl1,l2, andl3. Assuming thatthe material is rigid and perfectly plastic, show thatvolume constancy requires that the following expression be satisfied:E1+E2+E3= 0.14

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