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 microdiscontinuity 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 compiled them from his own work and that of others.1 Peterson developed the style of presentation 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 presentation 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 material 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 tension (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 tensiontest 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 principal 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 normally used. Equations (5–4) to (5–6) can also be converted to design equations by substituting 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 values 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 somehow 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 distortion. 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 substituting 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)