22 PLASTICITY 03,de3 r=YV? Y Figure 2.6.Projection of the Tresca and von Mises yield criteria onto a plane+o2+a3 =a constant. 01,de1 02,dez This can also be expressed as 6=V1/2)[a2+1+(1-a)P]=V1-a+a2o1 (2.10) where a =02/01. For the Tresca criterion, 6=o1-o3 where o1≥o2≥o3. (2.11) 2.5 EFFECTIVE STRAIN The effective strain,E,is defined such that the incremental plastic work per volume is dw =o]ds+o2d82 +o3ds3=dE. (2.12) The von Mises effective strain may be expressed as de-V2[(de2-de)2+(de3-de1)2+(de1-de2)P]/3, (2.13) or more simply as d=V(2/3)(dei+de +de3) (2.14) For proportional straining with a constant ratio of de de2:de3,the total effective strain is E=V(2/3)(e+吃+) (2.15) For the Tresca criterion the effective strain is d =Idsilmax, (2.16) where the subscript,i,refers to the principal strains.Thus for Tresca,the effective strain is the absolutely largest principal strain.The relation provides a simple check when evaluating for the von Mises criterion. IEilmax≤EMises≤1.l5 8ilmax (2.17)
22 PLASTICITY Figure 2.6. Projection of the Tresca and von Mises yield criteria onto a plane σ1 + σ2 + σ3 = a constant. This can also be expressed as σ¯ = (1/2)[α2 + 1 + (1 − α)2]σ1 = 1 − α + α2σ1 (2.10) where α = σ2/σ1. For the Tresca criterion, σ¯ = σ1 − σ3 where σ1 ≥ σ2 ≥ σ3. (2.11) 2.5 EFFECTIVE STRAIN The effective strain, ¯ε, is defined such that the incremental plastic work per volume is dw = σ1dε1 + σ2dε2 + σ3dε3 = σ¯ d¯ε. (2.12) The von Mises effective strain may be expressed as d¯ε = 2[(dε2 − dε3)2 + (dε3 − dε1)2 + (dε1 − dε2)2]/3, (2.13) or more simply as d¯ε = (2/3) dε2 1 + dε2 2 + dε2 3 . (2.14) For proportional straining with a constant ratio of dε1 : dε2 : dε3, the total effective strain is ε¯ = (2/3) ε2 1 + ε2 2 + ε2 3 . (2.15) For the Tresca criterion the effective strain is d¯ε = |dεi|max, (2.16) where the subscript, i, refers to the principal strains. Thus for Tresca, the effective strain is the absolutely largest principal strain. The relation provides a simple check when evaluating ¯ε for the von Mises criterion. |εi|max ≤ ε¯Mises ≤ 1.15|εi|max (2.17)
2.6 FLOW RULES 23 When the von Mises criterion and effective stress is used,the von Mises effective strain must be used.Conversely if the Tresca criterion and effective stress is used,the Tresca effective strain must be used. It should be realized that in both cases the o-g curve in a tension test is the -E curve,since o reduces to o and g reduces to s in a tension test.It is often assumed that strain hardening is described by the o-E curve found in a tension test.However, at large strains there may be deviations from this because of the different changes in crystallographic texture that occur during straining along different paths. 2.6 FLOW RULES The strains that result from elastic deformation are described by Hooke's law.There are similar relations for plastic deformation,called the flow rules.In the most general form the flow rule may be written, dsii =d(of/aoij). (2.18) where fis the function of oif that describes yielding (e.g.,the yield criterion.)It is related to what has been called the plastic potential.For the von Mises criterion, differentiation results in de1=d[o1-(1/2)(o2+o3)] de2=d[o2-(1/2)(o3+o1] (2.19) de3=d[o3-(1/2)(o1+o2)】. In these expressions da=d/o which varies with position on the o -E curve.However the ratio of the plastic strains remains constant. de1:d2:de3=[o1-(1/2)(o2+o3)】:[o2-(1/2)(o3+o1]:[o3-(1/2)(a1+o21. (2.20) For Tresca,f=1-03,so the flow rules are simply ds=da,d82 =0 and ds3 =-da. (2.21) EXAMPLE 2.3:Show that for plastic deformation in (a)uniaxial tension and (b) plane-strain compression (82 =0,o3 =0),the incremental work per volume,dw, found from dw =dg is the same as dw =oids +o2ds2 +o3de3... SOLUTION: (a)Substituting o2 =o3 =0 into equation 2.9,G =o1 and substituting ds2=de3 (-1/2)dsi into equation 2.13,so ods =oide (b)Substituting 82=0,03 =0 into dw =o1dg1 +o2ds2+o3ds3 =oids1. Substituting ds2 =0,and de3 =-de into equation 2.13,da V(2/3)+0+(1)2)=(2/3)61.From the flow rules with 82=0anda3 =0,into
2.6 FLOW RULES 23 When the von Mises criterion and effective stress is used, the von Mises effective strain must be used. Conversely if the Tresca criterion and effective stress is used, the Tresca effective strain must be used. It should be realized that in both cases the σ − ε curve in a tension test is the ¯σ − ε¯ curve, since ¯σ reduces to σ and ¯ε reduces to ε in a tension test. It is often assumed that strain hardening is described by the ¯σ − ε¯ curve found in a tension test. However, at large strains there may be deviations from this because of the different changes in crystallographic texture that occur during straining along different paths. 2.6 FLOW RULES The strains that result from elastic deformation are described by Hooke’s law. There are similar relations for plastic deformation, called the flow rules. In the most general form the flow rule may be written, dεi j = dλ(∂ f/∂σi j), (2.18) where f is the function of σi j that describes yielding (e.g., the yield criterion.) It is related to what has been called the plastic potential. For the von Mises criterion, differentiation results in dε1 = dλ[σ1 − (1/2))(σ2 + σ3)] dε2 = dλ[σ2 − (1/2))(σ3 + σ1)] (2.19) dε3 = dλ[σ3 − (1/2))(σ1 + σ2)]. In these expressions dλ = d¯ε/σ¯ which varies with position on the ¯σ − ε¯ curve. However the ratio of the plastic strains remains constant. dε1 : dε2 : dε3=[σ1 −(1/2))(σ2 +σ3)] : [σ2−(1/2))(σ3 +σ1)] : [σ3 − (1/2))(σ1 + σ2)]. (2.20) For Tresca, f = σ1 − σ3, so the flow rules are simply dε1 = dλ, dε2 = 0 and dε3 = −dλ. (2.21) EXAMPLE 2.3: Show that for plastic deformation in (a) uniaxial tension and (b) plane-strain compression (ε2 = 0, σ3 = 0), the incremental work per volume, dw, found from dw = σ¯ d¯ε is the same as dw = σ1dε1 + σ2dε2 + σ3dε3 ... SOLUTION: (a) Substituting σ2 = σ3 = 0 into equation 2.9, ¯σ = σ1 and substituting dε2 = dε3 = (−1/2)dε1 into equation 2.13, so ¯σd¯ε = σ1de1 (b) Substituting ε2 = 0, σ3 = 0 into dw = σ1dε1 + σ2dε2 + σ3dε3 = σ1dε1. Substituting dε2 = 0, and dε3 = −dε1 into equation 2.13, d¯ε = (2/3)(ε2 1 + 0 + (−ε1)2) = (2/ √3)ε1. From the flow rules with ε2 = 0 and σ3 = 0, into
24 PLASTICITY 02 =01.Substituting into equation 2.9,=(1/v2)[(a1-01/2)2+(01/2-0)2+ (0-o)2]/2=(3/2)o1.Therefore dw=6de=[(2/W3)de](W3/2)o1=o1de1. EXAMPLE 2.4:A circle 1 cm diameter was printed on a sheet of metal prior to a complex stamping operation.After the stamping,it was found that the circle had become an ellipse with major and minor diameters of 1.300 and 1.100 cm. (a)Determine the effective strain. (b)If a condition of plane stress (o:=0)existed during the stamping,and the ratio a=o2/o1 remained constant what ratio o1/o must have existed? SOLUTION: (a)e1=ln(1.3/1)=0.2624,e2=ln(1.11)=0.0953,e3=-e1-e2=-0.358. E=√(2/3ε7+号+)=[2/3)0.2622+0.09532+0.3582]/2=0.3705 Note that this is larger than 0.358 but not 15%larger. (b)From the flow rules (equation 2.19)with o3 =0,82/81=(202-1)/(201-02). Solving for a,a=o2/o1=(2e2/e1+1)/(e2/e1+2)=[21.1/1.3)+1]/ [(1.1/1.3)+2]=0.946.Now substituting into equation 2.10,G o1√1-0.946+0.946,01/6=1.027. 2.7 NORMALITY PRINCIPLE One interpretation of the flow rules is that the vector sum of the plastic strains is normal to the yield surface*.This is illustrated in three-dimensions in Figure 2.7 and in two-dimensions in Figure 2.8.With isotropic solids,the directions of principal strain and principal stress coincide.As a result the relation de2 (2.22) de 0o2 EXAMPLE 2.5:A thin sheet is subjected to biaxial tension 1=20,o3 =0.The principal strains in the sheet were 82=-(1/4)E1. (a)Using the principle of normality,determine the stress ratio,a =o2/o1,using the von Mises and the Tresca criteria. (b)Show that the normal to the yield locus in both cases corresponds to the answers to (a). SOLUTION: (ae2/e1=-0.25=(o2-0.5o1)/(o-0.5o2).Solving for o2/o1,a=2/7.With the Tresca criterion,82=-(1/4)e1 can occur only at the uniaxial tension corner, s0=0. (b)See Figure 2.9. See D.C.Drucker,Proc.Ist U.S.Nat.Congr.Appl.Mech.(1951),p.487
24 PLASTICITY σ2 = σ1. Substituting into equation 2.9, ¯σ = (1/ √2)[(σ1 − σ1/2)2 + (σ1/2 − 0)2 + (0 − σ1) 2] 1/2 = ( √3/2)σ1. Therefore dw = σ¯ d¯ε = [(2/ √ 3)dε1](√ 3/2)σ1= σ1dε1. EXAMPLE 2.4: A circle 1 cm diameter was printed on a sheet of metal prior to a complex stamping operation. After the stamping, it was found that the circle had become an ellipse with major and minor diameters of 1.300 and 1.100 cm. (a) Determine the effective strain. (b) If a condition of plane stress (σz = 0) existed during the stamping, and the ratio α = σ2/σ1 remained constant what ratio σ1/σ¯ must have existed? SOLUTION: (a) ε1 = ln(1.3/1) = 0.2624, ε2 = ln(1.11) = 0.0953, ε3 = −ε1 −ε2 = −0.358. ε¯ = (2/3)(ε2 1 + ε2 2 + ε2 3) = [(2/3)(0.2622 + 0.09532 + 0.3582)]1/2 = 0.3705 Note that this is larger than 0.358 but not 15% larger. (b) From the flow rules (equation 2.19) with σ3 = 0, ε2/ε1 = (2σ2 − σ1)/(2σ1 − σ2). Solving for α, α = σ2/σ1 = (2ε2/ε1 + 1)/(ε2/ε1 + 2) = [2(1.1/1.3) + 1]/ [(1.1/1.3) + 2] = 0.946. Now substituting into equation 2.10, ¯σ = σ1 √1 − 0.946 + 0.9462, σ1/σ¯ = 1.027. 2.7 NORMALITY PRINCIPLE One interpretation of the flow rules is that the vector sum of the plastic strains is normal to the yield surface∗. This is illustrated in three-dimensions in Figure 2.7 and in two-dimensions in Figure 2.8. With isotropic solids, the directions of principal strain and principal stress coincide. As a result the relation dε2 dε1 = −∂σ1 ∂σ2 . (2.22) EXAMPLE 2.5: A thin sheet is subjected to biaxial tension σ1 = σ2 = 0, σ3 = 0. The principal strains in the sheet were ε2 = −(1/4)ε1. (a) Using the principle of normality, determine the stress ratio, α = σ2/σ1, using the von Mises and the Tresca criteria. (b) Show that the normal to the yield locus in both cases corresponds to the answers to (a). SOLUTION: (a) ε2/ε1 = −0.25 = (σ2 − 0.5σ1)/(σ1 − 0.5σ2). Solving for σ2/σ1, α = 2/7. With the Tresca criterion, ε2 = −(1/4)ε1 can occur only at the uniaxial tension corner, so α = 0. (b) See Figure 2.9. ∗ See D. C. Drucker, Proc. 1st U.S. Nat. Congr. Appl. Mech. (1951), p. 487
2.7 NORMALITY PRINCIPLE 25 02 3 dEv yield surface dE1 two-dimensional 9 yield surface 6 loading line Figure 2.7.Three-dimensional yield surfaces.If a material is loaded to yielding at A,the resulting plastic strain is represented by a vector,dey normal to the yield surface and which is the vector sum of ds1,de2 and ds3. 03 (Biaxial tension: dev dey 01 01 0 dev (Uniaxial tension) -03 Figure 2.8.lllustration of normality.Note that the ratio of de3/ds is given by the normal to the yield surface at the point of yielding
2.7 NORMALITY PRINCIPLE 25 dεv dε1 dε2 dε3 σ1 σ2 σ3 yield surface two-dimensional yield surface loading line Figure 2.7. Three-dimensional yield surfaces. If a material is loaded to yielding at A, the resulting plastic strain is represented by a vector, dεv normal to the yield surface and which is the vector sum of dε1, dε2 and dε3. Figure 2.8. Illustration of normality. Note that the ratio of dε3/dε1 is given by the normal to the yield surface at the point of yielding
26 PLASTICITY 02 02 2=0 along this line =号 01 det de2 0 01 014 +01 de ea61 must occur at corner c2=一e along this line 02 (a) (b) Figure 2.9.Normals to yield loci for s2/s1=-0.25. 2.8 DERIVATION OF THE VON MISES EFFECTIVE STRAIN The effective stress-strain function is defined such that the incremental work per volume is dw =o de+o2ds2+o3ds3=dE.For simplicity consider a stress state with o3=0.Then ad =de+o2ds2 o1ds1(1+ap), (2.23) where a =o2/o1 and p=ds2/ds1.Then de=de1(o1/6)1+ap). (2.24) From the flow rules,p ds2/ds1 [o2+(1/2)01]/[o1+(1/2)02]=(2a-1)/ (2-a)or a=(2p+1)/2+p). (2.25) Combining equations 2.24 and 2.25, de=de1(o1/G)[2(1+p+p)/(2+p]. (2.26) With o3=0,the von Mises expression for is 11/2 a=(oi+a-g1o2)=(1-a+a2)R. (2.27) Combining equations 2.25 and 2.26, -（）+p+e (2.28) Since p de2/ds1, d=(2)(a+dd+d) (2.29)
26 PLASTICITY Figure 2.9. Normals to yield loci for ε2/ε1 = −0.25. 2.8 DERIVATION OF THE VON MISES EFFECTIVE STRAIN The effective stress-strain function is defined such that the incremental work per volume is dw = σ1dε1 + σ2dε2 + σ3dε3 = σ¯ d¯ε. For simplicity consider a stress state with σ3 = 0. Then σ¯ d¯ε = σ1dε1 + σ2dε2 = σ1dε1(1 + αρ), (2.23) where α = σ2/σ1 and ρ = dε2/dε1. Then d¯ε = dε1(σ1/σ¯ )(1 + αρ). (2.24) From the flow rules, ρ = dε2/dε1 = [σ2 + (1/2)σ1]/[σ1 + (1/2)σ2] = (2α − 1)/ (2 − α) or α = (2ρ + 1)/(2 + ρ). (2.25) Combining equations 2.24 and 2.25, d¯ε = dε1(σ1/σ¯ )[2(1 + ρ + ρ)/(2 + ρ]. (2.26) With σ3 = 0, the von Mises expression for ¯σ is σ¯ = (σ2 1 + σ 2 2 − σ1σ2) 1/2 = (1 − α + α2 ) 1/2 . (2.27) Combining equations 2.25 and 2.26, σ1 σ¯ = 2 + ρ √3 (1 + ρ + ρ2 ) 1/2 . (2.28) Since ρ = dε2/dε1, d¯ε = 2 √3 dε2 1 + dε1dε2 + dε2 2 1/2 (2.29)