1.1 2.STRESS STRAINS AND YIELD CRITERIAStressStrain RelationsFig. 1A0 = Original cross section of the specimen.L0 = Original gauge length.Ai = instantaneous cross section of the specimen.Li = instantaneous length of specimen after extension

2. 2Fig. 2: Stress Strain Diagram.P = Proportionality limitE = Elasticity limit Y = Yield pointN = Necking PointF = Fracture PointFig. 3 3. 3Fig. 4 4. 4i.EngineeringFiStress S =Fi = instantaneous loadA0LChange in length L L0 Engineering strain = ==L0 Original length of specimen L0 Fiii. True stress = and Ai LidL L True strain = = log i L L L0 o True stress is defined as load divided by actual cross sectional area (not original cross sectional areaA0) for that particular load. Fi= AiSimilarly, true strain is based on the instantaneous specimen length rather than original length. As suchtrue strain (or incremental strain) is defined asdL d =Where L is length at load F and is the true strain. LThe true strain at load F is then obtained by summing all the increments of equation.Arithmetically, this can be written asdL0 dL1 dL2 dL3 dL = d =+ +++ ...... + n L0L1 L2 L3LnL1 dL L1= L = logL0L0True strain is the sum of each incremental elongation divided by the current length of specimen, whereL0 is original gauge length and Li is the gauge length corresponding to load Fi. The most importantcharacteristics of truestress strain diagram is that true stress increases all the way to fracture. Thus truefracture strength f is greater than the true ultimate strength u in contrast with engineering stresswhere fracture strength is lesser than ultimate strength. 5. 5Relationship between true and engineering stress strainsFrom volume constancy,V = A0 L0 = Ai LiLiA = 0L0 Ai Li L0 Lie= = 1L0 L 0 Li = ( 1 + e)L0FiFi A0 Li= = =S AiA0Ai L0 = S (1 + e) LidLL== log i = log (1 + e) L0 LL0 = log (1 + e)Problems with Engineering StressStrains1. Engineering stressstrain diagram does not give true and accurate picture of deformation characteristics of the material because it takes original cross sectional area for all calculations though it reduces continuously after yield point in extension and markedly after necking. Thats why we get fracture strength of a material less than its ultimate tensile strength is Su > Sf which is not true.2. Total engineering strain is not equal to sum of incremental strains which defies the logic. 6. 6Let us have a specimen with length of 50 mm which then is extended to 66.55 in three steps Length before extension (L0)Length after extensionL L E=L00505015055 55/50 = 0.125560.5 5.55.5/55 = 0.1360.566.556.05 6.05/60.5 = 0.1 5 5.5 6.05Sum of incremental strain =++= 0.1 + 0.1 + 0.1 =0.350 55 60.5Now we will calculate total strain considering original and final length after of extension L3 = 66.55L3 L0 66.55 50 Total engineering strain when extended === 0.331L050 the specimen in one stepThe result is that summation of incremental engineering strain is NOT equal to total engineering strain.Now same procedure is applied to true strain- L L L = 01 + 12 + 2 3 = log 1 + log 2 + log 3 L0L1L2 55 60.5 66.55 = log+ log+ log= 0.286 5055 60.5 L3 66.55But total true strain equals to 0.3 = log= log = 0.286 L050In the case of true strains, sum of incremental strain is equal to the overall strain. Thus true strains areadditive. This is not true for engineering strains.3. 7. 7Fig L0 = length before extension L1 = L0 = length after extensionL1 L 0 L1 L 0 Strain e = I=L0L0 L1 = 0FigTo obtain strain of 1 the cylinder must be squeezed to zero thickness which is only hypothetical andnot true. Moreover, intuitively we expect that strain produced in compression should be equal inmagnitude but opposite in sign.Applying true strain formulation, to extensionL12L 0 = log = log= log 2L0L0To compression; L1 = L0/2L1L0 / 2 = log = log= log 1 / 2 = log 2L0 L0gives consistent results. Thus true strains for equivalent deformation in tension and comprehension are identical except for the sign. Further unlike engineering strains, true strains are consistent with actual phenomenon. 8. 8Problem:The following data were obtained during the true strain test of nickel specimen. Load Diameter LoadDiameterkN mm kNmm 0 6.4015.885.11 15.30 6.3515.575.08 15.92 6.2214.904.83 16.32 6.1014.014.5716.5 5.9713.124.32 16.55 5.8412.453.78A. Plot the true stress true strain curve:B. Determine the following1. True stress at maximum load.2.True fracture stress.3.True fracture strain.4.True uniform strain .5. True necking strain. 6.Ultimate tensile strength.7.Strain hardening component.Pmax 16.55 10 31. True stress at max load = = = 617.77 MPa A 5.84 24 P12.45 10 32.True fracture stress = = = 1109 MPa A min 11.22 2 2d 3.True fracture strain = ln 0 = ln 6.4 = 1.053d 3.78 i22 d 6.4 4. True uniform strain = ln 0= ln =0.183 d 5.84 i 5. True necking strain = true fracture strain true uniform strain = 1.053 0.183 = 0.87Pmax 16.55 10 3 6. Ultimate tensile stress = = = 514 MPaA max 6.4 24 7. Now, n= log(1+e) = log (1.2) =0.183 9. 9 LoadDiamete Area True stress True strain = Engg. Stress Engg. Strain KNr mm22Pid P mmln 0 =d 2 = d A = 0 1Ai i d i 2(N/mm2)(N/mm ) 0 6.4032.17 00 15.36.3531.6748.31 0.0156475.590.0158 15.92 6.2230.39 523.86 0.057 494.870.059 16.32 6.1029.22 558.52 0.096 507.300.10 16.55.9727.99 589.50 0.139 512.900.149 16.55 5.8426,79 617.77 0.183 514.450.20 15.88 5.1120.5774.63 0.45493.620.568 15.57 5.0820.27 768.13 0.46484.000.587 14.90 4.8318.32 813.32 0.56463.160.755 14.01 4.5716.40 854.27 0.67435.5 0.961 13.12 4.3214.66 894.95 0.786 407.831.19 12.45 3.7811.221109.63 1.053 387.001.866Applications of Engineering Stress and StrainsEngineering stress and strain are useful for many engineering design applications. Computation ofstress and strain is based on initial area or gauge length and therefore engineering stress and strainrepresent only approximations of the real stress and strain in plastic zone.In elastic deformation region (where dimensional changes are small and negligible) the initial andinstantaneous areas are approximately same and hence true stress equals engineering stress. Therefore,in design problems where large dimensional changes do not occur, the use of engineering stress issufficiently accurate and used extensively as it is easier to measure.However, for metal working where large plastic deformations occur and are necessary, theapproximations inherent in engineering stress and strain values are unacceptable. For this reason, thetrue stress and true strains are used.Important advantages of true stressstrain curves:1.It represents the actual and accurate stress and strain.True strain refers to a length fromwhich that change is produced rather than to original gauge length.The engineering stresse and 10. 10strains provides incorrect values after yield point i.e. plastic zone which a main zone of interest formetal working.2. True strains additive i.e. the total overall strain is equal to sum of incremental strains.3. True strains for equivalent deformation in tension and compression are identical except in sign.4. The volume change is related to the sum of the three normal true strains and with volume constancy.5. True stress can be related to true strain. = K () n = (0 + ) n0 = the amount of strain hardening that material received prior to the tension test.6. Truestresstrue strain values are quite sensitive to change in both metallurgical and mechanicalconditions of matter.TruestressstrainEngineering 1. Actual values of gauge length and 1.Original cross sectional areas (A0) iscross sectional area is used in used for calculating engineering stress.calculating true stress and true strain.FiS=Li A0FidL== AiL0 L Li L0Further strain e = is used.L0The sum of incremental strains isThe sum of incremental strains is notequal to total strainequal to total strain.Unlike load elongation curve, there isno maximum in the true strin curve.The sloppe of the curve in the plasticregion decreases with increaseinstrain 2. The calculated values of stress 2. The nominal stress (s) defined for thestrain are real and very useful in thetensile test in terms of original crossplastic region of the curve.sectional area (A0) is not really stressbecause the cross sectional area Ai atthe instant of load measurement is lessthan A0 in the evaluation of s. 3. The metal working designers are 3.The structural designers are interestedinterested in plastic region wherein a region where strains are elastic 11. 11difference between Ai and A0 is and difference between Ai and A0 issignificant. The true stressstrainsnegligibly small. But this is not true ingive accurate picture and hence it is the plastic region and especially whenmore useful to metal working designs. maximum load is reached. 4. It is not easy to obtain values of 4. It is easy to obtain these valuesfrom test since the force Fi and crossthroughtest and convenientlesssectional area (Ai) must be measuredcostly. These values are widelysimultaneously. True stress ( ) isavailable and documented.importantinmetalworkingcalculation becauseofitsfundamental significance. 5. It is more consistent with the 5. It is less consistent with physicalphenomenon of metal deformation.phenomenon of metal deformation.Idealisation of stress strain curvesThe solutions to the plasticity problems are quite complex. To obtain solution to these problems,stress strain curves are idealized by [i] neglecting elastic strains and/or [ii] ignoring the effect workhardening. Idealization and simplification restrict its field of application.1. Elastic perfectly plasticIt considers elastic strains and neglects effects of work hardening; it yields more difficult constitutiverelations. As a consequence, it also leads to greater mathematical difficulties in practical applications.It must be used for those processes in which elastic and plastic strains are of the same order. This is thecase in structural engineering or for bending.2. Rigid, perfectly plasticIn most metal forming operations, the permanent strains are much longer than the elastic. One thereforein air no great error by assuming the metal to behave as a rigid body prior to yielding. It is for thisreason that one mainly employs perfectly plastic material idealisation. 12. 12(a) Perfectly elastic, brittle (b) Perfectly rigid plastic (c) Rigid, linear strain hardening(d) Elastic perfectly plastic (e) Elastic linear strain hardeningFig. 5The flow curveA true stressstrain curve is frequently called a flow curve because it gives the stress required to causethe metal to flow physically to given strain.The plastic region of a true stress strain curve for many materials has a general form in the form ofHolloman equation which is = k () nwhere: n is strain hardening exponent 13. 13 k is strength constantFig. 6In a tension test of stell, a specimen of circular cross section with original diameter 9 mm is used. Theloads applied were 22 kN and 28 kN which reduces its diameter to 8.6 mm and 8.3 mm respectively.Determine (i) true stress and true strain for given loads (ii) strain hardening exponent and strengthcoefficient.Solution:d 0 = original diameter of specimen = 9 mmd1 = diameter of specimen on application of load F1 = 22kN d 2 = diameter of specimen on application of load F2 = 28kNF122 10 3 ( N)1 == = 3.78 N / mm 22 / 4 d1 / 4 (8.6) 2F228 10 3T2 == = 517.5 N / mm 2 / 4 (8.3) 2 / 4d22L1= true strain = log e L0L1 = Length after deformationL 0 = length before deformation 14. 14As volume of specimen remain constant, A 0 L 0 = A1 L12 2 2L1d d 0 L 0 = d1 L1 = 0 44L 0 d1 2d d = log 0d = 2 log 0 1d1 1 = true strain for first extension 9 2 log = 0.0918 .6 2 = true strain for second extension d 9= 2 log 0 = 2 log = 0.1619 d1 8.3 Applying Hollomon equation, n1 = K1 2 = K n 2 n 2 = 2 orlog 2 = n . log 2 1 1 11 517.5log 2log n=1 = 378 = 0.54 2 0.1619 loglog 10.091Substituting the value of n in equation (1) 378 = k (0.091) 0.54 K = 1385 N / mm 2 15. 15strain hardening exponent (n) = 0.54strength coefficientK = 1385 N / mm 2with this information, Hollomon equation can be written as = 1385 () 0.54Both n and K are material properties: The strain hardening exponent physically reflects the rate atwhich the material hardens. The derivative of this equation d d = n .In states that fractional change in true stress caused by a fractional change in true strain is determined bythe strain hardening exponent (n). Therefore, the stress increases rapidly with strain for a material thathas a large strain hardening exponent, such as 3O 2 stainless steel (n = 0.3) compared to a materialwhere n is low such as 4.10 stainless steel (n = 0.1).Plastic InstabilityNecking or localised deformation begins at maximum load where decrease in cross sectional areawhich hears the load is compensated by increase in strength due in loaddF = 0 16. 16 Fig. 7F = .AdF = dA + A d = 0 dA d =AFrom constancy of volume, V = A . L A= cross section of spearmen L =length of specimendV = 0 = A. dL + dA LdA dL =ALd dl== dL d =dProblemProve that uniform strain is equal strain hardening exponent (n).Solution: 17. 17Fig.P = load at any instanceA = cross section of specimen.P = A = A . k () n = k () n (1)Ao A = log e = 0A A A = A 0 e (2)Substituting value in equation (1) P = A 0 e K () n P = K A 0 [e () n ]At maximum load point on engg. stress strain curve dP = 0 = uWhen true strain dP = K A 0 [e N (u ) n 1 + (1) e u n ] = 0n u n 1 =u n u = nProblem 1:Hollomon equation for a material is given as = 1400 () 0.33 . Find the ultimate tensile strength of thematerial.Solution: 18. 18Ultimate tensile strength of a material is measured at maximum load point and where necking begins.Upto the necking point, deformation is uniform throughout its gauge length. It is a engineering stress(S u ).True strain for uniform elongation is equal to strain hardening exponent. Therefore u = n.uUltimate tensile strength = = S u = [ = (1 + e) S]1 + uu = log e (1 + e u ) u = (1 + e u )e = 2.71 (logarithmic base)1 + eu = enn n u K . u n n Su === K u = K n ee en en =0.33 0.33 1400 2.71 = 698.1 N / mm 2 UTS = 698. 1 N / mm 2This shows that ultimate strength of a material can be calculated from the value of K and n.Problem 2:A metal obeys Hollomon relationship and has a UTS of 300 MPa. To reach the maximum load requiresan elongation of 35%. Find strain hardening exponent (n) and strength coefficient (K).Solution:UTS = S u = 300 MPa = 300 N / mm 2Engineering elongation strain = e u = 35% = 0.35 19. 19uniform true strain u = log (1 + e u ) = log (1.35) = 0.3 u = S u (1 + e u )but n = u = 0.3= 300 (1 + 0.35) = 405 N / mm 2 . u = K (u ) n = k (n ) n405 = K (0.3) 0.3 K = 581.2 N / mm 2 Hollomon equation for given metal is = 581.2 () 0.3Deformation workWork is defined as the product of force and distance. A quantity equivalent to work per unit volume isthe product of stress and strain. The area under the true stress strain curve for any strain 1 is the energyper unit volume (u) or specific energy, of the deformed material. Fig. 91u = d0 20. 20The true stressstrain curve can be represented by the Hollomon equation = K () n . 11K n +1nu = K () . d =0 n +1 0 K 1n +1u= n +1similarly mean flow stress can be found11 n d K .d K 1n +1 K 1n0 0 m = = ==1 0 1(n + 1) 1 (n + 1) K 1nm =n +1The work calculated according to above equation assumes that the deformation is homogeneous throughout the deforming part. This work is called ideal deformation work.Example: Ideal work of deformationDeformation of fully annealed AA1100 aluminium is governed by the Hollomon equation. If a 10 cmlong bar of this material is pulled in tension from a diameter of 12.7 mm to a diameter of 11.5, calculatethe following:a. the ideal work per unit volume of aluminium required;b. the mean stress in the aluminium during deformation;c. the peak stress applied to the aluminium. = 140 () 0.25 N / mm 2Solutionsa.Calculate total strain during deformation 21. 21 A0d = ln = 2 ln 0 Ad 12.7= 2 ln= 0.199 11.5 Calculate the total volume of bard2 (0.0127 m 2 V= l= 0.1 m = 1.26 10 5 m 3 4 4 For AA1100, K = 140 MPa and n = 0.25. Note that, as < n, the deformation is homogeneous1 +1 n Wi = K Vn +1 1.25= 140 10 6 N 0.199 1.26 10 5 m 3 = 187.5 N (J ) m 61.25 (b)Mean stress during deformation n1 m = K n +1 0.199 0.25 = 140 MPa = 74.8 MPa 1.25 (c)Peak (maximum) stress applied, from Hollomon equation 1 = K 1 = 140 MPa 0.199 0.25 = 93.5 MPa nYield CriterionYield point under simplified condition of uniaxial tension is widely known and documented. But suchsimplified conditions [1 Pure uniaxial tension 2 Pure shear] are rare in reality. In many situationscomplex and multiaxial stresses are present and in this situation it is necessary to know when a materialwill yield. Mathematically and empirically, the relationships between the yield point under uniaxialtensile test and yield strength under complex situations have been found out. These relationships areknown as yield criteria. Thus yield criterion is defined as mathematical and empirically derived 22. 22relationship between yield strength under uniaxial tensile load and yielding under multiaxial complexstress situation.Yield Criterion is a law defining the limit of elastic behaviour under any possible combination ofstresses is called yield criterion. Yield criterion is a mathematical expression which unites experimentalobservations with mathematical expressions n a phenomenological manner. Yield criteria is primarilyused to predict if or when yieldingwill occur under combined stress states in terms of particular properties of the metal being stressed [ 0 , K] .Any yield criterion is a postulated mathematical expression of the stress that will induce yielding or theonset of plastic deformation. The most general form isf ( x , y , z, Txy , Tyz , Tzx ) = a constant.orin terms of principal stresses f( f (1 , 2 , 3 ) = CFor most ductile metals that are isotropic, the following assumptions are invoked:1.There is no Bauschinger effect, thus the yield strengths in tension and compression are equivalent.Bauschinger effectThe lowering of yield stress for a material when deformation in one direction is followed by deformationin the opposite direction, is called Bauschinger effect. 23. 23 Fig. 102.The constancy of volume prevails so that plastic equivalent of poisons ratio 0.5.1 + 2 + 33.The magnitude of the mean normal stress m =does not cause yielding. The3assumption that yielding is independent of m (also called hydrostatic component of the total stateof stress) is reasonable if plastic flow depends upon shear mechanism such as slip or twinning. Inthis context, yield criterion is written asF[(1 2), (2 3), (3 1)] = Cwhich implies that yielding depends upon the size of the Mohrs circle and not their position. It isshown that if a stress state (1, 2, 3) will cause yielding, an equivalent stress state(1 , 2 , 3 ) will cause yielding, if,Two widely used yield criterion:1. Tresca criterion or maximum shear stress criterion.2. Von Mises criterion or distortion energy criterion.1.Tresca criterionTresca found that plastic flow in a metal begins when tangential stress attains a value.Assume that a body is subjected to triaxial stresses. 1 , 2 , 3 are principal stresses and 1 > 2 > 3(algebraically).Then maximum shear stress 3Tmax = 1 2when Tmax exceeds a certain value c, specific to that material, yielding will occur. To find the valueof c, the material is subjected to uniaxial tensile test and find out yield point strength ( 0 ). 24. 24For uniaxial tensile test, stress situation is1 = 0 , 2 = 3 = 0 3 0 Tmax = 1==c 22 3 0 1= or1 3 = 022ii) Material is subjected to pure shear: 1 = k2 = 0 3 = k k = shear strength of the material 1 3 = 0 k + k = 0 k = 0 = 0.5 02Application:i) Plain stress condition. x , y, Txy 2 x + y x + y 1 = + + (T ) 222 xy 2 x + y x + y 2 = + (T ) 222 xy 1 = 0when 3 > 01 + 3 = 0when 3 < 0ii)Plain strain condition 3 = 0 2 = 2 (1 + 2 ) + 31 3 = 0 2 = 12 25. 25Shortcomings1. An essential short coming of this criterion is that it ignore the effect of intermediate principal stress( 2 ).2. Since pastic flow depends upon slip phenomenon which is essentially a shearing.Slip is practically absent in brittle materials. Therefore application of this criterion is limited to ductile materials. This criterion is not applicable to crystalline brittle material which cannot be brought into plastic state under tension but yield a little before compress fracture in compression.3. Failure of/ yielding of a material under triaxial pure tension condition where 1 = 2 = 3 can not be explained by this criterion.4. It suffers from a major difficulty that it is necessary to know in advance which are maximum and minimum stresses.5. Moreover, the general form of this criterion is far more complicated than the Von Mises criterion. Therefore Von Mises criterion is preferred in most theoretical (not practical) work.For sake of simplicity, in analysis, this criterion is widely used in practice. 26. 26Von Mises CriterionAccording to this criterion, yielding will occur when shear strain energy per unit volume reaches acritical value. The shear strain energy per unit volume is expressed terms of three principal stresses: e= 1G[ (1 2 ) 2 + ( 2 3 ) 2 + ( 3 1 ) 2 ]G = modulus of shear which is a constant. (1 2 ) 2 + ( 2 3 ) 2 + ( 3 1 ) 2 = Constant.(i) For uniaxial tensile test, yielding will occur when 1 = 0 ; 2 = 3 = 0 ( y ) 2 + ( y ) 2 = cons tan t = 2 0 2 Therefore Von Mises criterion can be stated as (1 2 ) 2 + ( 2 3 ) 2 + ( 3 1 ) 2 = 2 0 2 ( x y ) 2 + ( y z ) 2 + ( z x ) 2 + (T 2 + T 2 + T 2 ) = 2 0 2x y y z z xi) For plane stress: 2 = 0 + 3ii) For plane strain: 2 = 12iii) For pure shear stress condition: 1 = k 2 = 0 3 = k (1 2 ) 2 + ( 2 3 ) 2 + 3 1 ) 2 = 2 0 2 (k 0 0) 2 + (0 + k 0 ) 2 ( k 0 k 0 ) 2 = 2 0 2 k 02 = 2 y2yk == 0.557 03 27. 27This is the relationship between shear yield strength and tensile yield strength of the material as per VonMises criterion.k = 0.5 0 2 Tresca criterionk 0 = 0.577 0 Von Mises criterion. yVon Mises criterion satisfy the experimental data better than Tresca and therefore k =value is3normally used.Advantages of Von Mises criterion1. It overcomes major deficiency of Tresca criterion. Von Mises criterion implies that yielding is not dependent on any particular normal stress but instead, depends on all three principal shearing stresses.2. Von Mises criterion conforms the experimental data better than Tresca and therefore more realistic.3. Since it involves squared terms, the result is independent of sign of individual stresses. This is an important since it is not necessary to know which is the largest and the smallest principal stress in order to use this criterion.Von Mises yield criteria:[( x y ) 2 + ( y z ) 2 + ( z x ) 2 + 6 (T 2 xy + T 2 yz + T 2 zx ) = 2 0 2]Effective stressWith the yield criterion, it is useful to define an effective stress denoted as which is function of theapplies stresses. If the magnitude of reaches a critical value, then the applied stress will causeyielding.For Von Mises criterion = 12[(1 2 )2+ ( 2 3 ) 2 + ( 3 1 ) 2 ] 1/ 2 28. 28For Tresca criterion = 1 3 0 . For both the criteria. 3 k . Von Mises 2 k Tresca Plane stress condition Plane strain condition1. In plane stress condition, there is no stress 1. In plane strain condition, the strain in third direction is absent.in third direction.12 = [ 2 ( 3 + 1 )]E 2 = ( 1 + 3 ) Near yield point and in plastic zone 1 = (For plastic defo) 2 + 3 2 = 1But there is strain in third direction.Two2principal stresses2 x + y x + y 1 =2 + 2 + Txy 2 ( ) 2x + y x + y ) 2 =2 2( ) + Txy 2 11 = [ 1 ( 2 )] E 12 = [ 2 (1 )] E 13 = [ 0 (1 + 2 )] EPlane strain condition 29. 29In majority of metal forming operations the problem can be simplified by assuming a condition of planestrain is one. One of the principal strains is zero.1 1 = [ 1 ( 2 + 3 )]E1 2 = [ 2 (1 + 3 )]E1 3 = [ 3 (1 + 2 )]Elet 2 = 0 2 = (1 + 3 )for plastic region, Nadai has shown that = 0.5 + 3 2 = 1 2 + 3Thus, for Tresca criterion: 1 , 1 , 3 2 1 3 = 0Von Mises criterion in plane strain:2 2 + 3 + 3 1 1 + 1 3 + ( 3 1 ) 2 = 2 0 2 2 23( 1 . 3 ) 2 = 2 0 22 2 1 3 ) 2 = 0 = 0 3 0 = 1.155 0 = constrained yield strength of the material. 30. 30Yield criterionMaximum shear stressMaximum distortionCriterion (Tresca)energy criterion (Von Misces)PlanePlane PureStress StrainShear + 2 2 0, min = 2= 1 1 = 3 = k 2 = 0 21 = 0 ..... 3 = ve1 3 = 01 + 3 = 01 + 3 = 0 3 = ve2k= 0 0k=2 Plane stress Plane strainPure Shear1+3 112 + 3 2 1 3 = 0 22 =k=0232 1 3 = 03Tresca criterionVon Mises yield criterion 1. This criterion is also known as1. Van Mises criterion is also known as maximum shear stress criterion anddistortion energy yield criterion. It statesattributesyieldingto slipthat yielding occurs when deformationphenomenonwhich occurs whenenergy per unit volume of materialmaximum shear stress exceeds a exceedscertainvaluewhichisvalue, characteristic to the material. characteristic ofthe material.Mathematically it can be stated as Mathematically, it can be stated as 31. 31 1 3 = 00 =1 [(1 2 ) 2 + ( 2 3 ) 2] 2 where 1, 2 , 3 are principal+ ( 3 1 ) 2 1 / 2 stresses, and 1 > 2 > 3 .Or1 0 =( y ) 2 ) 2 + ( y z ) 2 x 2+ ( z x ) 2 + ( 2 xy + 2 yz + 2 zx ) ] 1/ 22. Phenomenon of slip is limited to 2. The application of this yield criterion ductilematerialsandhenceholds good for both ductile and brittle application of this criterion is limitedmaterials. to ductile materials. This criterion do not yield good results for brittle materials.3. Tresca criterion ignores the effect of 3. Von Mises criterion take into considera intermediate principal stress and thistion the intermediate principal stress and is a major draw back of this. hence move realistic. The predications offered by Von Mises criterion conforms empirical data.4.4. The yield stress predicted by Von Mises criterion is 15. 5% greater than the yield stress predicted by Tresca criterion.5. Locus shown in Figure. 5. . Locus shown in Figure. It is Hexagonal.It is Elliptical. 32. 32 Superimposed6. Von Mises criterion is preferred where 6. Tresca criterionis preferred in more accuracy is desired.analysis for simplicity.Locus of yield as per Tresca criterionBiaxial stress condition is assumed to present locus of yield point on plane paper. 1 , 3 , 2 = 0yielding will occur if the point plotted is on the boundary or outside. 33. 33 Fig. 11 : Tresca yield locus. In the six sectors, the following conditions apply:I 3 > 1 > 0, so 3 = + Y II 3 > 1 > 0, so 3 = + YIII 1 > 0 > 3 , so 1 3 = + Y IV 0 > 1 > 3 > 0, so 3 = Y V 0 > 3 > 1 , so 1 = YVI 3 > 0 > 1 , so 3 1 = + YLocus of yield as per Von Mises criterion1.For a biaxial plane stress condition ( 2 = 0) the Von Mises criterion can be expressed mathematically, 12 + 3 2 1 3 = 0 2 This the equation of an ellipse whose major semiaxis is2 0 and whose minor semiaxis is 2 0 . The plot of equation is called a yield locus. 3 34. 34 Fig. 12Comments1. Yielding will occur if the point representing the given stress is plotted and is on the boundary or outside the boundary.2. The yield locus of maximum shear stress criterion [Tresca criterion] fall inside the maximum distortion energy criterion [Von Mises] yield locus.3. Two yield criteria predict the same yield stress for conditions of uniaxial stress and balanced biaxial stress (1 = 3 ). The greatest divergence between the two criteria occurs for pure shear(1 = 3 ).4. The yield stress predicted by the Von Mises criterion is 15.5% greater than the yield stress, predicted by Tresca criterion.Derive a mathematical expression for Von Mises yield criterion applicable to plane strain stresscondition:Solution:Von Mises yield criterion is stated as(1 2 ) 2 + ( 2 3 ) 2 + ( 3 1 ) 2 = 2 0 2where 1 , 2 , 3 are three principal stresses and 0 is the yield strength of material. In plane strainstress condition, the intermediate principal stress is arithmetic mean of other two. Assuming1 > 2 > 3 , we can write + 32 = 1 2substituting the value of 2 in the above expression 2 2 + 3 + 3 1 1 + 1 3 + ( 3 1 ) 2 = 2 0 22 2 35. 35(1 3 ) 2 (1 3 ) 2 ( 3 1 ) 2 + += 2 02 4 41 (1 3 ) 2= 2 024 (1 3 ) 2 = 8 / 6 0 2 2 1 3 = 0 = 03 0is called constrained strength of material and is 115 times the yield strength under uniaxial tensiletest.PROBLEMA stress analysis of a space craft structural member gives the state of stress as below: 200 300 Tij = 30 100 0 0 0 50 If the part is made of aluminium alloy with strength 500 MPa, will it exhibit yielding as per Tresca yieldcriterion and von Mises yield criterion? If not, what is the safety factor?Data given: x = 200 MPa y = 100 MPa z = 50 MPaTx y = 30 MPa(1)Applying von Mises criterion1/ 21 2 c = 2( ) () x y 2 + y z 2 + ( z x ) 2 + 6 T 2 T 2 + T 2 x yy z z x 36. 36( ) 1/ 21 ( 200 100) 2 + (100 + 50) 2 + ( 50 200) 2 + 6 30 2 + 0 2 + 0 2 c = 2 c = 224 MPaThe calculated stress ( c ) is less than the yield strength of the material ( 0 ) , yielding will not occur asper von Mises criterion 0 500 MPaFactor of safety == = 2.2 c 224 MPa(ii) Applying Tresca Criterion In order to apply this criterion, it is necessary to know the magnitude and sign of three principal stresses stress situation can be written in matrix form. Txy Txz 200 300 x Tij = Tyx yTyz = 30 100 0 T Tzy z 00 50 zx 20 3 0 = 3 10 0 10 0 0 5 To find the principal stresses 20 3 0310 0=000 5 I1 = x + y + z = 20 + 10 5 = 25 xTxy y Tyz xTxzI2 = ++ Txy yTyzzTxz z = 191 50 100 37. 37 I2 = 41 I 3 = Tij 10 0 3 0 3 10 = 203+00 50 510 0 = 1000 + 45 I3 = 955 f () = 3 I1 2 + I 2 I 3 = 0 3 25 2 + 41 + 955 = 0Applying standard method to get cubic roots,f () = 3 25 2 + 41 + 955 = 0f(y) = y3 + py2 + qy + r = 0 p = 25q = 41r =955a=13( 1 )3q p 2 = ( 3 41 625)3a = 167.3b=1 [27 2p 3 9 pq + 27r ]=127 [2(25) 3 9(25) ( 41) + 27(955) ] b = 139.25 38. 38b 139.25 = 3 3 Cos =a 167.3 2 2 3 3 = 99.620 100 a g = 2 3 g = 14.94 P 99.62 25 y1 = g cos = 14.94 cos +3 3 3 3 y1 = 20.83 P 99.62 25 y2 = g cos + 120 = 14.94 cos + 120 +3 3 3 3 y1 = 5 P 99.62 25 y3 = g cos + 249 = 14.94 cos + 240 +3 3 3 3 y3 = 9.161 = 20.83 10 = 208.3 MPa 2 = 9.16 10 = 91.6 MPa ordered in such a way that 1 > 2 > 3 3 = 5 10 = 50 MPaTo apply Tresca criterion; 1 3208.3 ( 50) Tmax === 129.15 MPa220 Tmax