上海交通大学：《Design and Manufacturing》课程教学资源（参考文献）Shigley's Mechanical Engineering Design ch5 failures resulting from static loading

Budynas-Nisbett:Shigley's ll.Failure Prevention 5.Failures Resulting from ©The McGraw-Hfl Mechanical Engineering Static Loading Companies,2008 Design,Eighth Edition 5 Failures Resulting from Static Loading Chapter Outline 5-1 Static Strength 208 5-2 Stress Concentration 209 5-3 Failure Theories 211 5-4 Maximum-Shear-Stress Theory for Ductile Materials 211 5-5 Distortion-Energy Theory for Ductile Materials 213 5-6 Coulomb-Mohr Theory for Ductile Materials 219 5-7 Failure of Ductile Materials Summary 222 5-8 Maximum-Normal-Stress Theory for Brittle Materials 226 5-9 Modifications of the Mohr Theory for Brittle Materials 227 5-10 Failure of Brittle Materials Summary 229 5-11 Selection of Failure Criteria 230 5-12 Introduction to Fracture Mechanics 231 5-13 Stochastic Analysis 240 5-14 Important Design Equations 246 205

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition II. Failure Prevention 5. Failures Resulting from Static Loading © The McGraw−Hill 209 Companies, 2008 5Failures Resulting from Static Loading Chapter Outline 5–1 Static Strength 208 5–2 Stress Concentration 209 5–3 Failure Theories 211 5–4 Maximum-Shear-Stress Theory for Ductile Materials 211 5–5 Distortion-Energy Theory for Ductile Materials 213 5–6 Coulomb-Mohr Theory for Ductile Materials 219 5–7 Failure of Ductile Materials Summary 222 5–8 Maximum-Normal-Stress Theory for Brittle Materials 226 5–9 Modifications of the Mohr Theory for Brittle Materials 227 5–10 Failure of Brittle Materials Summary 229 5–11 Selection of Failure Criteria 230 5–12 Introduction to Fracture Mechanics 231 5–13 Stochastic Analysis 240 5–14 Important Design Equations 246 205

210 Budynas-Nisbett:Shigley's ll.Failure Prevention 5.Failures Resulting from T©The McGraw-Hill Mechanical Engineering Static Loading Companies,2008 Design,Eighth Edition 206 Mechanical Engineering Design In Chap.I we learned that strength is a property or characteristic of a mechanical element.This property results from the material identity,the treatment and processing incidental to creating its geometry,and the loading,and it is at the controlling or critical location. In addition to considering the strength of a single part,we must be cognizant that the strengths of the mass-produced parts will all be somewhat different from the others in the collection or ensemble because of variations in dimensions,machining, forming,and composition.Descriptors of strength are necessarily statistical in nature,involving parameters such as mean,standard deviations,and distributional identification. A static load is a stationary force or couple applied to a member.To be stationary, the force or couple must be unchanging in magnitude,point or points of application, and direction.A static load can produce axial tension or compression,a shear load,a bending load,a torsional load,or any combination of these.To be considered static,the load cannot change in any manner. In this chapter we consider the relations between strength and static loading in order to make the decisions concerning material and its treatment,fabrication,and geometry for satisfying the requirements of functionality,safety,reliability,competitiveness, usability,manufacturability,and marketability.How far we go down this list is related to the scope of the examples "Failure"is the first word in the chapter title.Failure can mean a part has sepa- rated into two or more pieces;has become permanently distorted,thus ruining its geometry:has had its reliability downgraded;or has had its function compromised, whatever the reason.A designer speaking of failure can mean any or all of these pos- sibilities.In this chapter our attention is focused on the predictability of permanent distortion or separation.In strength-sensitive situations the designer must separate mean stress and mean strength at the critical location sufficiently to accomplish his or her purposes. Figures 5-1 to 5-5 are photographs of several failed parts.The photographs exem- plify the need of the designer to be well-versed in failure prevention.Toward this end we shall consider one-,two-,and three-dimensional stress states,with and without stress concentrations,for both ductile and brittle materials. Figure 5-1 (a)Failure of a truck drive-shaft spline due to corrosion fatigue.Note that it was necessary to use clear tape to hold the pieces in place. (b)Direct end view of failure. (a)

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition II. Failure Prevention 5. Failures Resulting from Static Loading 210 © The McGraw−Hill Companies, 2008 206 Mechanical Engineering Design In Chap. 1 we learned that strength is a property or characteristic of a mechanical element. This property results from the material identity, the treatment and processing incidental to creating its geometry, and the loading, and it is at the controlling or critical location. In addition to considering the strength of a single part, we must be cognizant that the strengths of the mass-produced parts will all be somewhat different from the others in the collection or ensemble because of variations in dimensions, machining, forming, and composition. Descriptors of strength are necessarily statistical in nature, involving parameters such as mean, standard deviations, and distributional identification. A static load is a stationary force or couple applied to a member. To be stationary, the force or couple must be unchanging in magnitude, point or points of application, and direction. A static load can produce axial tension or compression, a shear load, a bending load, a torsional load, or any combination of these. To be considered static, the load cannot change in any manner. In this chapter we consider the relations between strength and static loading in order to make the decisions concerning material and its treatment, fabrication, and geometry for satisfying the requirements of functionality, safety, reliability, competitiveness, usability, manufacturability, and marketability. How far we go down this list is related to the scope of the examples. “Failure” is the first word in the chapter title. Failure can mean a part has sepa￾rated into two or more pieces; has become permanently distorted, thus ruining its geometry; has had its reliability downgraded; or has had its function compromised, whatever the reason. A designer speaking of failure can mean any or all of these pos￾sibilities. In this chapter our attention is focused on the predictability of permanent distortion or separation. In strength-sensitive situations the designer must separate mean stress and mean strength at the critical location sufficiently to accomplish his or her purposes. Figures 5–1 to 5–5 are photographs of several failed parts. The photographs exem￾plify the need of the designer to be well-versed in failure prevention. Toward this end we shall consider one-, two-, and three-dimensional stress states, with and without stress concentrations, for both ductile and brittle materials. Figure 5–1 (a) Failure of a truck drive-shaft spline due to corrosion fatigue. Note that it was necessary to use clear tape to hold the pieces in place. (b) Direct end view of failure

Budynas-Nisbett:Shigley's ll.Failure Prevention 5.Failures Resulting from ©The McGraw-Hil 21 Mechanical Engineering Static Loading Companies,2008 Design,Eighth Edition Failures Resulting from Static Loading 207 Figure 5-2 Impact failure of a lawn mower blade driver hub.The blade impacted a surveying pipe marker. Figure 5-3 Failure of an overhead-pulley retaining bolt on a weightlifting mochine.A manufacturing error caused a gap that forced the bolt to take the entire moment load. (a) (b) Figure 5-4 Chain test fixture that failed in one cycle.To alleviate complaints of excessive wear,the manufacturer decided to case-harden the material.(a)Two halves showing fracture;this is an excellent example of brittle fracture initiated by stress concentration.(b)Enlarged view of one portion to show cracks induced by stress concentration at the support-pin holes

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition II. Failure Prevention 5. Failures Resulting from Static Loading © The McGraw−Hill 211 Companies, 2008 Figure 5–2 Impact failure of a lawn￾mower blade driver hub. The blade impacted a surveying pipe marker. Figure 5–3 Failure of an overhead-pulley retaining bolt on a weightlifting machine. A manufacturing error caused a gap that forced the bolt to take the entire moment load. Figure 5–4 Chain test fixture that failed in one cycle. To alleviate complaints of excessive wear, the manufacturer decided to case-harden the material. (a) Two halves showing fracture; this is an excellent example of brittle fracture initiated by stress concentration. (b) Enlarged view of one portion to show cracks induced by stress concentration at the support-pin holes. Failures Resulting from Static Loading 207

212 Budynas-Nisbett:Shigley's ll.Failure Prevention 5.Failures Resulting from T©The McGraw-Hil Mechanical Engineering Static Loading Companies,2008 Design,Eighth Edition 208 Mechanical Engineering Design Figure 5-5 Valve-spring failure caused by spring surge in an oversped engine.The fractures exhibit the classic 45 shear failure 5-1 Static Strength Ideally,in designing any machine element,the engineer should have available the results of a great many strength tests of the particular material chosen.These tests should be made on specimens having the same heat treatment,surface finish,and size as the ele- ment the engineer proposes to design;and the tests should be made under exactly the same loading conditions as the part will experience in service.This means that if the part is to experience a bending load,it should be tested with a bending load.If it is to be subjected to combined bending and torsion,it should be tested under combined bending and torsion.If it is made of heat-treated AISI 1040 steel drawn at 500C with a ground finish,the specimens tested should be of the same material prepared in the same manner. Such tests will provide very useful and precise information.Whenever such data are available for design purposes,the engineer can be assured of doing the best possible job of engineering. The cost of gathering such extensive data prior to design is justified if failure of the part may endanger human life or if the part is manufactured in sufficiently large quan- tities.Refrigerators and other appliances,for example,have very good reliabilities because the parts are made in such large quantities that they can be thoroughly tested in advance of manufacture.The cost of making these tests is very low when it is divid- ed by the total number of parts manufactured. You can now appreciate the following four design categories: 1 Failure of the part would endanger human life,or the part is made in extremely large quantities:consequently,an elaborate testing program is justified during design. The part is made in large enough quantities that a moderate series of tests is feasible. 3 The part is made in such small quantities that testing is not justified at all;or the design must be completed so rapidly that there is not enough time for testing

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition II. Failure Prevention 5. Failures Resulting from Static Loading 212 © The McGraw−Hill Companies, 2008 208 Mechanical Engineering Design Figure 5–5 Valve-spring failure caused by spring surge in an oversped engine. The fractures exhibit the classic 45◦ shear failure. 5–1 Static Strength Ideally, in designing any machine element, the engineer should have available the results of a great many strength tests of the particular material chosen. These tests should be made on specimens having the same heat treatment, surface finish, and size as the ele￾ment the engineer proposes to design; and the tests should be made under exactly the same loading conditions as the part will experience in service. This means that if the part is to experience a bending load, it should be tested with a bending load. If it is to be subjected to combined bending and torsion, it should be tested under combined bending and torsion. If it is made of heat-treated AISI 1040 steel drawn at 500◦C with a ground finish, the specimens tested should be of the same material prepared in the same manner. Such tests will provide very useful and precise information. Whenever such data are available for design purposes, the engineer can be assured of doing the best possible job of engineering. The cost of gathering such extensive data prior to design is justified if failure of the part may endanger human life or if the part is manufactured in sufficiently large quan￾tities. Refrigerators and other appliances, for example, have very good reliabilities because the parts are made in such large quantities that they can be thoroughly tested in advance of manufacture. The cost of making these tests is very low when it is divid￾ed by the total number of parts manufactured. You can now appreciate the following four design categories: 1 Failure of the part would endanger human life, or the part is made in extremely large quantities; consequently, an elaborate testing program is justified during design. 2 The part is made in large enough quantities that a moderate series of tests is feasible. 3 The part is made in such small quantities that testing is not justified at all; or the design must be completed so rapidly that there is not enough time for testing

Budynas-Nisbett:Shigley's ll.Failure Prevention 5.Failures Resulting from T©The McGraw-Hill 213 Mechanical Engineering Static Loading Companies,2008 Design,Eighth Edition Failures Resulting from Static Loading 209 4 The part has already been designed,manufactured,and tested and found to be unsatisfactory.Analysis is required to understand why the part is unsatisfactory and what to do to improve it. More often than not it is necessary to design using only published values of yield strength,ultimate strength,percentage reduction in area,and percentage elongation, such as those listed in Appendix A.How can one use such meager data to design against both static and dynamic loads,two-and three-dimensional stress states,high and low temperatures,and very large and very small parts?These and similar questions will be addressed in this chapter and those to follow,but think how much better it would be to have data available that duplicate the actual design situation. 5-2 Stress Concentration Stress concentration (see Sec.3-13)is a highly localized effect.In some instances it may be due to a surface scratch.If the material is ductile and the load static,the design load may cause yielding in the critical location in the notch.This yielding can involve strain strengthening of the material and an increase in yield strength at the small criti- cal notch location.Since the loads are static and the material is ductile,that part can carry the loads satisfactorily with no general yielding.In these cases the designer sets the geometric (theoretical)stress concentration factor K,to unity. The rationale can be expressed as follows.The worst-case scenario is that of an idealized non-strain-strengthening material shown in Fig.5-6.The stress-strain curve rises linearly to the yield strength Sy,then proceeds at constant stress,which is equal to Sy.Consider a filleted rectangular bar as depicted in Fig.A-15-5,where the cross- section area of the small shank is 1 in2.If the material is ductile,with a yield point of 40 kpsi,and the theoretical stress-concentration factor(SCF)K,is 2, A load of 20 kip induces a tensile stress of 20 kpsi in the shank as depicted at point A in Fig.5-6.At the critical location in the fillet the stress is 40 kpsi,and the SCF is K=0max/onom=40/20=2. Figure 5-6 An idealized stress-strain curve.The dashed line depicts a strain-strengthening material. Tensile strain.e

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition II. Failure Prevention 5. Failures Resulting from Static Loading © The McGraw−Hill 213 Companies, 2008 Failures Resulting from Static Loading 209 4 The part has already been designed, manufactured, and tested and found to be unsatisfactory. Analysis is required to understand why the part is unsatisfactory and what to do to improve it. More often than not it is necessary to design using only published values of yield strength, ultimate strength, percentage reduction in area, and percentage elongation, such as those listed in Appendix A. How can one use such meager data to design against both static and dynamic loads, two- and three-dimensional stress states, high and low temperatures, and very large and very small parts? These and similar questions will be addressed in this chapter and those to follow, but think how much better it would be to have data available that duplicate the actual design situation. 5–2 Stress Concentration Stress concentration (see Sec. 3–13) is a highly localized effect. In some instances it may be due to a surface scratch. If the material is ductile and the load static, the design load may cause yielding in the critical location in the notch. This yielding can involve strain strengthening of the material and an increase in yield strength at the small criti￾cal notch location. Since the loads are static and the material is ductile, that part can carry the loads satisfactorily with no general yielding. In these cases the designer sets the geometric (theoretical) stress concentration factor Kt to unity. The rationale can be expressed as follows. The worst-case scenario is that of an idealized non–strain-strengthening material shown in Fig. 5–6. The stress-strain curve rises linearly to the yield strength Sy , then proceeds at constant stress, which is equal to Sy . Consider a filleted rectangular bar as depicted in Fig. A–15–5, where the cross￾section area of the small shank is 1 in2 . If the material is ductile, with a yield point of 40 kpsi, and the theoretical stress-concentration factor (SCF) Kt is 2, • A load of 20 kip induces a tensile stress of 20 kpsi in the shank as depicted at point A in Fig. 5–6. At the critical location in the fillet the stress is 40 kpsi, and the SCF is K = σmax/σnom = 40/20 = 2. 50 0 Sy Tensile strain, Tensile stress , kpsi A B D E C Figure 5–6 An idealized stress-strain curve. The dashed line depicts a strain-strengthening material.

214 Budynas-Nisbett:Shigley's ll.Failure Prevention 5.Failures Resulting from T©The McGraw-Hill Mechanical Engineering Static Loading Companies,2008 Design,Eighth Edition 210 Mechanical Engineering Design A load of 30 kip induces a tensile stress of 30 kpsi in the shank at point B.The fillet stress is still 40 kpsi(point D),and the SCFK Omax/onom Sy/o =40/30 1.33. At a load of 40 kip the induced tensile stress (point C)is 40 kpsi in the shank. At the critical location in the fillet,the stress (at point E)is 40 kpsi.The SCF K =Omax /onom Sy/=40/40 =1. For materials that strain-strengthen,the critical location in the notch has a higher Sy. The shank area is at a stress level a little below 40 kpsi,is carrying load,and is very near its failure-by-general-yielding condition.This is the reason designers do not apply K:in static loading of a ductile material loaded elastically,instead setting K:=1. When using this rule for ductile materials with static loads,be careful to assure yourself that the material is not susceptible to brittle fracture (see Sec.5-12)in the environment of use.The usual definition of geometric (theoretical)stress-concentration factor for normal stress K,and shear stress Kis is Omax K1Onom (a) Tmax =Kistnom (6 Since your attention is on the stress-concentration factor,and the definition of onom or Thom is given in the graph caption or from a computer program,be sure the value of nominal stress is appropriate for the section carrying the load. Brittle materials do not exhibit a plastic range.A brittle material"feels"the stress concentration factor K,or Ks,which is applied by using Eq.(a)or(b). An exception to this rule is a brittle material that inherently contains microdiscon- tinuity stress concentration,worse than the macrodiscontinuity that the designer has in mind.Sand molding introduces sand particles,air,and water vapor bubbles.The grain structure of cast iron contains graphite flakes(with little strength),which are literally cracks introduced during the solidification process.When a tensile test on a cast iron is performed,the strength reported in the literature includes this stress concentration.In such cases K,or Kis need not be applied. An important source of stress-concentration factors is R.E.Peterson,who com- piled them from his own work and that of others.'Peterson developed the style of pre- sentation in which the stress-concentration factor K,is multiplied by the nominal stress nom to estimate the magnitude of the largest stress in the locality.His approximations were based on photoelastic studies of two-dimensional strips(Hartman and Levan, 1951:Wilson and White,1973),with some limited data from three-dimensional photoelastic tests of Hartman and Levan.A contoured graph was included in the pre- sentation of each case.Filleted shafts in tension were based on two-dimensional strips. Table A-15 provides many charts for the theoretical stress-concentration factors for several fundamental load conditions and geometry.Additional charts are also available from Peterson. Finite element analysis(FEA)can also be applied to obtain stress-concentration factors.Improvements on K,and Kis for filleted shafts were reported by Tipton,Sorem, and Rolovic.3 R.E.Peterson,"Design Factors for Stress Concentration,"Machine Design,vol.23.no.2.February 1951; no.3.March 1951;no.5.May 1951;no.6.June 1951;no.7,July 1951. 2Walter D.Pilkey,Peterson's Stress Concentration Factors.2nd ed.John Wiley&Sons,New York.1997. 3S.M.Tipton,J.R.Sorem Jr.,and R.D.Rolovic."Updated Stress-Concentration Factors for Filleted Shafts in Bending and Tension,"Trans.ASME,Journal of Mechanical Design,vol.118.September 1996,pp.321-327

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition II. Failure Prevention 5. Failures Resulting from Static Loading 214 © The McGraw−Hill Companies, 2008 210 Mechanical Engineering Design • A load of 30 kip induces a tensile stress of 30 kpsi in the shank at point B. The fillet stress is still 40 kpsi (point D), and the SCF K = σmax/σnom = Sy/σ = 40/30 = 1.33. • At a load of 40 kip the induced tensile stress (point C) is 40 kpsi in the shank. At the critical location in the fillet, the stress (at point E) is 40 kpsi. The SCF K = σmax/σnom = Sy/σ = 40/40 = 1. For materials that strain-strengthen, the critical location in the notch has a higher Sy . The shank area is at a stress level a little below 40 kpsi, is carrying load, and is very near its failure-by-general-yielding condition. This is the reason designers do not apply Kt in static loading of a ductile material loaded elastically, instead setting Kt = 1. When using this rule for ductile materials with static loads, be careful to assure yourself that the material is not susceptible to brittle fracture (see Sec. 5–12) in the environment of use. The usual definition of geometric (theoretical) stress-concentration factor for normal stress Kt and shear stress Kts is σmax = Ktσnom (a) τmax = Ktsτnom (b) Since your attention is on the stress-concentration factor, and the definition of σnom or τnom is given in the graph caption or from a computer program, be sure the value of nominal stress is appropriate for the section carrying the load. Brittle materials do not exhibit a plastic range. A brittle material “feels” the stress concentration factor Kt or Kts, which is applied by using Eq. (a) or (b). An exception to this rule is a brittle material that inherently contains microdiscon￾tinuity stress concentration, worse than the macrodiscontinuity that the designer has in mind. Sand molding introduces sand particles, air, and water vapor bubbles. The grain structure of cast iron contains graphite flakes (with little strength), which are literally cracks introduced during the solidification process. When a tensile test on a cast iron is performed, the strength reported in the literature includes this stress concentration. In such cases Kt or Kts need not be applied. An important source of stress-concentration factors is R. E. Peterson, who com￾piled them from his own work and that of others.1 Peterson developed the style of pre￾sentation in which the stress-concentration factor Kt is multiplied by the nominal stress σnom to estimate the magnitude of the largest stress in the locality. His approximations were based on photoelastic studies of two-dimensional strips (Hartman and Levan, 1951; Wilson and White, 1973), with some limited data from three-dimensional photoelastic tests of Hartman and Levan. A contoured graph was included in the pre￾sentation of each case. Filleted shafts in tension were based on two-dimensional strips. Table A–15 provides many charts for the theoretical stress-concentration factors for several fundamental load conditions and geometry. Additional charts are also available from Peterson.2 Finite element analysis (FEA) can also be applied to obtain stress-concentration factors. Improvements on Kt and Kts for filleted shafts were reported by Tipton, Sorem, and Rolovic.3 1 R. E. Peterson, “Design Factors for Stress Concentration,” Machine Design, vol. 23, no. 2, February 1951; no. 3, March 1951; no. 5, May 1951; no. 6, June 1951; no. 7, July 1951. 2 Walter D. Pilkey, Peterson’s Stress Concentration Factors, 2nd ed, John Wiley & Sons, New York, 1997. 3 S. M. Tipton, J. R. Sorem Jr., and R. D. Rolovic, “Updated Stress-Concentration Factors for Filleted Shafts in Bending and Tension,” Trans. ASME, Journal of Mechanical Design, vol. 118, September 1996, pp. 321–327

Budynas-Nisbett:Shigley's ll.Failure Prevention 5.Failures Resulting from T©The McGraw-Hill 215 Mechanical Engineering Static Loading Companies,2008 Design,Eighth Edition Failures Resulting from Static Loading 211 5-3 Failure Theories Section 5-1 illustrated some ways that loss of function is manifested.Events such as distortion,permanent set,cracking,and rupturing are among the ways that a machine element fails.Testing machines appeared in the 1700s,and specimens were pulled,bent, and twisted in simple loading processes. If the failure mechanism is simple,then simple tests can give clues.Just what is simple?The tension test is uniaxial (that's simple)and elongations are largest in the axial direction,so strains can be measured and stresses inferred up to"failure."Just what is important:a critical stress,a critical strain,a critical energy?In the next several sections, we shall show failure theories that have helped answer some of these questions. Unfortunately,there is no universal theory of failure for the general case of mate- rial properties and stress state.Instead,over the years several hypotheses have been formulated and tested,leading to today's accepted practices.Being accepted,we will characterize these "practices"as theories as most designers do. Structural metal behavior is typically classified as being ductile or brittle,although under special situations,a material normally considered ductile can fail in a brittle manner(see Sec.5-12).Ductile materials are normally classified such that 0.05 and have an identifiable yield strength that is often the same in compression as in ten- sion (Syr=Syc=Sy).Brittle materials,<0.05,do not exhibit an identifiable yield strength,and are typically classified by ultimate tensile and compressive strengths,Sr and Suc,respectively (where Se is given as a positive quantity).The generally accepted theories are: Ductile materials(yield criteria) Maximum shear stress (MSS),Sec.5-4 Distortion energy (DE),Sec.5-5 Ductile Coulomb-Mohr (DCM).Sec.5-6 Brittle materials (fracture criteria) Maximum normal stress (MNS).Sec.5-8 Brittle Coulomb-Mohr (BCM).Sec.5-9 Modified Mohr(MM).Sec.5-9 It would be inviting if we had one universally accepted theory for each material type,but for one reason or another,they are all used.Later,we will provide rationales for selecting a particular theory.First,we will describe the bases of these theories and apply them to some examples. 5-4 Maximum-Shear-Stress Theory for Ductile Materials The maximum-shear-stress theory predicts that yielding begins whenever the maximum shear stress in any element equals or exceeds the maximum shear stress in a tension- test specimen of the same material when that specimen begins to yield.The MSS theory is also referred to as the Tresca or Guest theory Many theories are postulated on the basis of the consequences seen from tensile tests.As a strip of a ductile material is subjected to tension,slip lines (called Liider lines)form at approximately 45 with the axis of the strip.These slip lines are the

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition II. Failure Prevention 5. Failures Resulting from Static Loading © The McGraw−Hill 215 Companies, 2008 Failures Resulting from Static Loading 211 5–3 Failure Theories Section 5–1 illustrated some ways that loss of function is manifested. Events such as distortion, permanent set, cracking, and rupturing are among the ways that a machine element fails. Testing machines appeared in the 1700s, and specimens were pulled, bent, and twisted in simple loading processes. If the failure mechanism is simple, then simple tests can give clues. Just what is simple? The tension test is uniaxial (that’s simple) and elongations are largest in the axial direction, so strains can be measured and stresses inferred up to “failure.” Just what is important: a critical stress, a critical strain, a critical energy? In the next several sections, we shall show failure theories that have helped answer some of these questions. Unfortunately, there is no universal theory of failure for the general case of mate￾rial properties and stress state. Instead, over the years several hypotheses have been formulated and tested, leading to today’s accepted practices. Being accepted, we will characterize these “practices” as theories as most designers do. Structural metal behavior is typically classified as being ductile or brittle, although under special situations, a material normally considered ductile can fail in a brittle manner (see Sec. 5–12). Ductile materials are normally classified such that εf ≥ 0.05 and have an identifiable yield strength that is often the same in compression as in ten￾sion (Syt = Syc = Sy ). Brittle materials, εf < 0.05, do not exhibit an identifiable yield strength, and are typically classified by ultimate tensile and compressive strengths, Sut and Suc, respectively (where Suc is given as a positive quantity). The generally accepted theories are: Ductile materials (yield criteria) • Maximum shear stress (MSS), Sec. 5–4 • Distortion energy (DE), Sec. 5–5 • Ductile Coulomb-Mohr (DCM), Sec. 5–6 Brittle materials (fracture criteria) • Maximum normal stress (MNS), Sec. 5–8 • Brittle Coulomb-Mohr (BCM), Sec. 5–9 • Modified Mohr (MM), Sec. 5–9 It would be inviting if we had one universally accepted theory for each material type, but for one reason or another, they are all used. Later, we will provide rationales for selecting a particular theory. First, we will describe the bases of these theories and apply them to some examples. 5–4 Maximum-Shear-Stress Theory for Ductile Materials The maximum-shear-stress theory predicts that yielding begins whenever the maximum shear stress in any element equals or exceeds the maximum shear stress in a tension￾test specimen of the same material when that specimen begins to yield. The MSS theory is also referred to as the Tresca or Guest theory. Many theories are postulated on the basis of the consequences seen from tensile tests. As a strip of a ductile material is subjected to tension, slip lines (called Lüder lines) form at approximately 45° with the axis of the strip. These slip lines are the

216 Budynas-Nisbett:Shigley's ll.Failure Prevention 5.Failures Resulting from T©The McGraw-Hil Mechanical Engineering Static Loading Companies,2008 Design,Eighth Edition 212 Mechanical Engineering Design beginning of yield,and when loaded to fracture,fracture lines are also seen at angles approximately 45 with the axis of tension.Since the shear stress is maximum at 45 from the axis of tension,it makes sense to think that this is the mechanism of failure.It will be shown in the next section,that there is a little more going on than this.However, it turns out the MSS theory is an acceptable but conservative predictor of failure;and since engineers are conservative by nature,it is quite often used. Recall that for simple tensile stress,o P/A,and the maximum shear stress occurs on a surface 45 from the tensile surface with a magnitude of mx=/2.So the maximum shear stress at yield ismax=Sy/2.For a general state of stress,three prin- cipal stresses can be determined and ordered such that o1>02>03.The maximum shear stress is then tmax =(a1-03)/2(see Fig.3-12).Thus,for a general state of stress,the maximum-shear-stress theory predicts yielding when ax=-a、S, .22 0r01-3≥S (5-1) 2 Note that this implies that the yield strength in shear is given by Ssy =0.5Sy (5-21 which,as we will see later is about 15 percent low (conservative). For design purposes,Eq.(5-1)can be modified to incorporate a factor of safety,n. Thus, S Tmax or1-03=S (5-3) 2n Plane stress problems are very common where one of the principal stresses is zero, and the other two,A and og,are determined from Eq.(3-13).Assuming that A >og. there are three cases to consider in using Eq.(5-1)for plane stress: Case 1:A g0.For this case,o1 =A and o3 =0.Equation (5-1) reduces to a yield condition of OA≥S, (5-41 Case 2:A 202OB.Here,a1=A and o3=og,and Eq.(5-1)becomes A-OB Sy (5-51 Case 3:02A zog.For this case,o1=0 and o3 =aB,and Eq.(5-1)gives OB≤-S, (5-61 Equations (5-4)to (5-6)are represented in Fig.5-7 by the three lines indicated in the OA,og plane.The remaining unmarked lines are cases for og oA,which are not nor- mally used.Equations(5-4)to(5-6)can also be converted to design equations by sub- stituting equality for the equal to or greater sign and dividing S,by n. Note that the first part of Eq.(5-3).Tmx=Sy/2n.is sufficient for design purposes provided the designer is careful in determining Tmax.For plane stress,Eq.(3-14)does not always predict mx However,consider the special case when one normal stress is zero in the plane,say ox and xy have values andoy=0.It can be easily shown that this is a Case 2 problem,and the shear stress determined by Eg.(3-14)is tmax.Shaft design problems typically fall into this category where a normal stress exists from bending and/or axial loading,and a shear stress arises from torsion

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition II. Failure Prevention 5. Failures Resulting from Static Loading 216 © The McGraw−Hill Companies, 2008 212 Mechanical Engineering Design beginning of yield, and when loaded to fracture, fracture lines are also seen at angles approximately 45° with the axis of tension. Since the shear stress is maximum at 45° from the axis of tension, it makes sense to think that this is the mechanism of failure. It will be shown in the next section, that there is a little more going on than this. However, it turns out the MSS theory is an acceptable but conservative predictor of failure; and since engineers are conservative by nature, it is quite often used. Recall that for simple tensile stress, σ = P/A, and the maximum shear stress occurs on a surface 45° from the tensile surface with a magnitude of τmax = σ/2. So the maximum shear stress at yield is τmax = Sy/2. For a general state of stress, three prin￾cipal stresses can be determined and ordered such that σ1 ≥ σ2 ≥ σ3. The maximum shear stress is then τmax = (σ1 − σ3)/2 (see Fig. 3–12). Thus, for a general state of stress, the maximum-shear-stress theory predicts yielding when τmax = σ1 − σ3 2 ≥ Sy 2 or σ1 − σ3 ≥ Sy (5–1) Note that this implies that the yield strength in shear is given by Ssy = 0.5Sy (5–2) which, as we will see later is about 15 percent low (conservative). For design purposes, Eq. (5–1) can be modified to incorporate a factor of safety, n. Thus, τmax = Sy 2n or σ1 − σ3 = Sy n (5–3) Plane stress problems are very common where one of the principal stresses is zero, and the other two, σA and σB, are determined from Eq. (3–13). Assuming that σA ≥ σB, there are three cases to consider in using Eq. (5–1) for plane stress: Case 1: σA ≥ σB ≥ 0. For this case, σ1 = σA and σ3 = 0. Equation (5–1) reduces to a yield condition of σA ≥ Sy (5–4) Case 2: σA ≥ 0 ≥ σB . Here, σ1 = σA and σ3 = σB , and Eq. (5–1) becomes σA − σB ≥ Sy (5–5) Case 3: 0 ≥ σA ≥ σB . For this case, σ1 = 0 and σ3 = σB , and Eq. (5–1) gives σB ≤ −Sy (5–6) Equations (5–4) to (5–6) are represented in Fig. 5–7 by the three lines indicated in the σA, σB plane. The remaining unmarked lines are cases for σB ≥ σA, which are not nor￾mally used. Equations (5–4) to (5–6) can also be converted to design equations by sub￾stituting equality for the equal to or greater sign and dividing Sy by n. Note that the first part of Eq. (5-3), τmax = Sy/2n, is sufficient for design purposes provided the designer is careful in determining τmax. For plane stress, Eq. (3–14) does not always predict τmax. However, consider the special case when one normal stress is zero in the plane, say σx and τxy have values and σy = 0. It can be easily shown that this is a Case 2 problem, and the shear stress determined by Eq. (3–14) is τmax. Shaft design problems typically fall into this category where a normal stress exists from bending and/or axial loading, and a shear stress arises from torsion

Budynas-Nisbett:Shigley's ll.Failure Prevention 5.Failures Resulting from T©The McGraw-Hill 21m Mechanical Engineering Static Loading Companies,2008 Design,Eighth Edition Failures Resulting from Static Loading 213 Figure 5-7 B The maximum-shear-stress MSS)theory for plane stress, where aA and ag are the two nonzero principal stresses. Case 2 Case 3 5-5 Distortion-Energy Theory for Ductile Materials The distortion-energy theory predicts that yielding occurs when the distortion strain energy per unit volume reaches or exceeds the distortion strain energy per unit volume for yield in simple tension or compression of the same material. The distortion-energy (DE)theory originated from the observation that ductile materials stressed hydrostatically exhibited yield strengths greatly in excess of the val- ues given by the simple tension test.Therefore it was postulated that yielding was not a simple tensile or compressive phenomenon at all,but,rather,that it was related some- how to the angular distortion of the stressed element.To develop the theory,note,in Fig. 5-8a,the unit volume subjected to any three-dimensional stress state designated by the stresses o1.02,and o3.The stress state shown in Fig.5-8b is one of hydrostatic tension due to the stresses ov acting in each of the same principal directions as in Fig.5-8a. The formula for oav is simply oav =1+02+03 (a) 3 Thus the element in Fig.5-8b undergoes pure volume change,that is,no angular dis- tortion.If we regard oav as a component of o1,02,and o3,then this component can be I>02>0 (a Triaxial stresses (b)Hydrostatic component (c)Distortional componen Figure 5-8 (a)Element with triaxial stresses;this element undergoes both volume change and angular distortion.(b)Element under hydrostatic tension undergoes only volume change.(c]Element has angular distortion without volume change

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition II. Failure Prevention 5. Failures Resulting from Static Loading © The McGraw−Hill 217 Companies, 2008 Failures Resulting from Static Loading 213 5–5 Distortion-Energy Theory for Ductile Materials The distortion-energy theory predicts that yielding occurs when the distortion strain energy per unit volume reaches or exceeds the distortion strain energy per unit volume for yield in simple tension or compression of the same material. The distortion-energy (DE) theory originated from the observation that ductile materials stressed hydrostatically exhibited yield strengths greatly in excess of the val￾ues given by the simple tension test. Therefore it was postulated that yielding was not a simple tensile or compressive phenomenon at all, but, rather, that it was related some￾how to the angular distortion of the stressed element. To develop the theory, note, in Fig. 5–8a, the unit volume subjected to any three-dimensional stress state designated by the stresses σ1, σ2, and σ3. The stress state shown in Fig. 5–8b is one of hydrostatic tension due to the stresses σav acting in each of the same principal directions as in Fig. 5–8a. The formula for σav is simply σav = σ1 + σ2 + σ3 3 (a) Thus the element in Fig. 5–8b undergoes pure volume change, that is, no angular dis￾tortion. If we regard σav as a component of σ1, σ2, and σ3, then this component can be B A –Sy Sy Sy –Sy Case 2 Case 3 Case 1 Figure 5–7 The maximum-shear-stress (MSS) theory for plane stress, where σA and σB are the two nonzero principal stresses. Figure 5–8 (a) Element with triaxial stresses; this element undergoes both volume change and angular distortion. (b) Element under hydrostatic tension undergoes only volume change. (c) Element has angular distortion without volume change. 3 1 > 2 > 3 2 1 (a) Triaxial stresses (b) Hydrostatic component 3 – av 2 – av 1 – av (c) Distortional component av av av = +

218 Budynas-Nisbett:Shigley's ll.Failure Prevention 5.Failures Resulting from T©The McGraw-Hil Mechanical Engineering Static Loading Companies,2008 Design,Eighth Edition 214 Mechanical Engineering Design subtracted from them,resulting in the stress state shown in Fig.5-8c.This element is subjected to pure angular distortion,that is,no volume change. The strain energy per unit volume for simple tension is u=eo.For the element of Fig.5-8a the strain energy per unit volume is u=e101+e202+e303]. Substituting Eq.(3-19)for the principal strains gives =2+听+G-2a+oa+oo】 (6 The strain energy for producing only volume change u can be obtained by substitut- ing oav for 01,o2,and o3 in Eq.(b).The result is 3蓝1-2w) v=2 (d) If we now substitute the square of Eq.(a)in Eq.(c)and simplify the expression,we get ,=6E(o+号+听+2012+20203+2030)） 1-2v (5-7刀 Then the distortion energy is obtained by subtracting Eq.(5-7)from Eq.(b).This gives 1+v「(o1-02)2+(o2-03)2+(a3-0)2 ld u-lv (5-8) 3E 2 Note that the distortion energy is zero if o1 =o2 =03. For the simple tensile test,at yield,o1=Sy and o2 =03 =0,and from Eq.(5-8) the distortion energy is 1+v =3E号 (5-9 So for the general state of stress given by Eq.(5-8),yield is predicted if Eq.(5-8) equals or exceeds Eg.(5-9).This gives T1-22+@2-o2+o-n2]p ≥S (5-10) 2 If we had a simple case of tension o,then yield would occur when S.Thus,the left of Eq.(5-10)can be thought of as a single,equivalent,or effective stress for the entire general state of stress given by o,o2,and o3.This effective stress is usually called the von Mises stress,o',named after Dr.R.von Mises,who contributed to the theory.Thus Eq.(5-10),for yield,can be written as a'≥S, (5-11) where the von Mises stress is 0= 「o1-22+(@2-32+(a3-m)27p 2 (5-12) For plane stress,let oA and og be the two nonzero principal stresses.Then from Eq.(5-12),we get '=(oi-aaag+ag)i (5-13)

Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition II. Failure Prevention 5. Failures Resulting from Static Loading 218 © The McGraw−Hill Companies, 2008 214 Mechanical Engineering Design subtracted from them, resulting in the stress state shown in Fig. 5–8c. This element is subjected to pure angular distortion, that is, no volume change. The strain energy per unit volume for simple tension is u = 1 2 σ . For the element of Fig. 5–8a the strain energy per unit volume is u = 1 2 [1σ1 + 2σ2 + 3σ3]. Substituting Eq. (3–19) for the principal strains gives u = 1 2E σ2 1 + σ2 2 + σ2 3 − 2ν(σ1σ2 + σ2σ3 + σ3σ1) (b) The strain energy for producing only volume change uv can be obtained by substitut￾ing σav for σ1, σ2, and σ3 in Eq. (b). The result is uv = 3σ2 av 2E (1 − 2ν) (c) If we now substitute the square of Eq. (a) in Eq. (c) and simplify the expression, we get uv = 1 − 2ν 6E  σ2 1 + σ2 2 + σ2 3 + 2σ1σ2 + 2σ2σ3 + 2σ3σ1  (5–7) Then the distortion energy is obtained by subtracting Eq. (5–7) from Eq. (b). This gives ud = u − uv = 1 + ν 3E  (σ1 − σ2) 2 + (σ2 − σ3) 2 + (σ3 − σ1) 2 2  (5–8) Note that the distortion energy is zero if σ1 = σ2 = σ3. For the simple tensile test, at yield, σ1 = Sy and σ2 = σ3 = 0, and from Eq. (5–8) the distortion energy is ud = 1 + ν 3E S2 y (5–9) So for the general state of stress given by Eq. (5–8), yield is predicted if Eq. (5–8) equals or exceeds Eq. (5–9). This gives  (σ1 − σ2) 2 + (σ2 − σ3) 2 + (σ3 − σ1) 2 2 1/2 ≥ Sy (5–10) If we had a simple case of tension σ , then yield would occur when σ ≥ Sy . Thus, the left of Eq. (5–10) can be thought of as a single, equivalent, or effective stress for the entire general state of stress given by σ1, σ2, and σ3. This effective stress is usually called the von Mises stress, σ , named after Dr. R. von Mises, who contributed to the theory. Thus Eq. (5–10), for yield, can be written as σ ≥ Sy (5–11) where the von Mises stress is σ =  (σ1 − σ2) 2 + (σ2 − σ3) 2 + (σ3 − σ1) 2 2 1/2 (5–12) For plane stress, let σA and σB be the two nonzero principal stresses. Then from Eq. (5–12), we get σ =  σ2 A − σAσB + σ2 B 1/2 (5–13)