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- 1. 1 2.STRESS STRAINS AND YIELD CRITERIAStress−Strain Relations Fig. 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. 2 Fig. 2: Stress− Strain Diagram.P = Proportionality limit E = Elasticity limit Y = Yield pointN = Necking Point F = Fracture Point Fig. 3
- 3. 3Fig. 4
- 4. 4i. Engineering Fi Stress S = Fi = instantaneous load A0 ∆L Change in length L − L0 Engineering strain = = = L0 Original length of specimen L0 Fiii. True stress σ = and Ai Li dL 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 area A0) for that particular load. Fi σ= AiSimilarly, true strain is based on the instantaneous specimen length rather than original length. As suchtrue strain (or incremental strain) is defined as dL 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 as dL0 dL1 dL2 dL3 dL∈ = ∑d ∈ = + + + + ...... + n L0 L1 L2 L3 Ln L1 dL L1 = ∫ L = log L0 L0True 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 true−stress strain diagram is that true stress increases all the way to fracture. Thus truefracture strength σ f is greater than the true ultimate strength σ 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 Li Li A ∴ = 0 L0 Ai Li − L0 Li e= = − 1 L0 L 0 Li ∴ = ( 1 + e) L0 Fi Fi A0 Li σ= = × =S× Ai A0 Ai L0 σ = S (1 + e) Li dL L ∈= ∫ = log i = log (1 + e) L0 L L0 ∴ ∈ = log (1 + e)Problems with Engineering Stress−Strains1. Engineering stress−strain 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. That’s 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. 6 Let 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 extension ∆L ∆L E= L0 0 50 50 1 50 55 5 5/50 = 0.1 2 55 60.5 5.5 5.5/55 = 0.1 3 60.5 66.55 6.05 6.05/60.5 = 0.1 5 5.5 6.05Sum of incremental strain = + + = 0.1 + 0.1 + 0.1 =0.3 50 55 60.5Now we will calculate total strain considering original and final length after of extension L3 = 66.55 L3 − L0 66.55 − 50∴ Total engineering strain when extended = = = 0.331 L0 50 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 ∈ = ∈0−1 + ∈1−2 + ∈2 − 3 = log 1 + log 2 + log 3 L0 L1 L2 55 60.5 66.55 = log + log + log = 0.286 50 55 60.5 L3 66.55But total true strain equals to ∈0.3 = log = log = 0.286 L0 50In 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. 7 Fig L0 = length before extension L1 = L0 = length after extension L1 − L 0 L1 − L 0 Strain e = ⇒ −I= L0 L0 L1 = 0 FigTo 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 extension L1 2L 0 ∈ = log = log = log 2 L0 L0To compression; L1 = L0/2 L1 L0 / 2 ∈ = log = log = log 1 / 2 = − log 2 L0 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 Load Diameter kN mm kN mm 0 6.40 15.88 5.11 15.30 6.35 15.57 5.08 15.92 6.22 14.90 4.83 16.32 6.10 14.01 4.57 16.5 5.97 13.12 4.32 16.55 5.84 12.45 3.78A. 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 = = π = 617.77 MPa A × 5.84 2 4 P 12.45 × 10 3 2. True fracture stress = = = 1109 MPa A min 11.22 2 2 d 3. True fracture strain = ln 0 = ln 6.4 = 1.053 d 3.78 i 2 2 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 = 0.87 Pmax 16.55 × 10 3 6. Ultimate tensile stress = = π = 514 MPa A max × 6.4 2 4 7. Now, n= log(1+e) = log (1.2) =0.183
- 9. 9 Load Diamete Area True stress True strain = Engg. Stress Engg. Strain KN r mm2 2 Pi d P mm ln 0 = d 2 = d A = 0 −1 Ai i d i 2 (N/mm2) (N/mm ) 0 6.40 32.17 0 0 15.3 6.35 31.67 48.31 0.0156 475.59 0.0158 15.92 6.22 30.39 523.86 0.057 494.87 0.059 16.32 6.10 29.22 558.52 0.096 507.30 0.10 16.5 5.97 27.99 589.50 0.139 512.90 0.149 16.55 5.84 26,79 617.77 0.183 514.45 0.20 15.88 5.11 20.5 774.63 0.45 493.62 0.568 15.57 5.08 20.27 768.13 0.46 484.00 0.587 14.90 4.83 18.32 813.32 0.56 463.16 0.755 14.01 4.57 16.40 854.27 0.67 435.5 0.961 13.12 4.32 14.66 894.95 0.786 407.83 1.19 12.45 3.78 11.22 1109.63 1.053 387.00 1.866Applications 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 stress−strain curves:1. It represents the actual and accurate stress and strain. True strain refers to a length from which that change is produced rather than to original gauge length.The engineering stresse and
- 10. 10 strains provides incorrect values after yield point i.e. plastic zone which a main zone of interest for metal 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. σ = K (∈) n σ = (∈0 + ∈) n ∈0 = the amount of strain hardening that material received prior to the tension test.6. True−stress−true strain values are quite sensitive to change in both metallurgical and mechanical conditions of matter. True−stress−strain Engineering 1. Actual values of gauge length and 1. Original cross sectional areas (A0) is cross sectional area is used in used for calculating engineering stress. calculating true stress and true strain. Fi S= Li A0 Fi dL σ= ∈= ∫ Ai L0 L Li − L0 Further strain e = is used. L0 The sum of incremental strains is The sum of incremental strains is not equal to total strain equal to total strain. Unlike load elongation curve, there is no maximum in the true strin curve. The sloppe of the curve in the plastic region decreases with increase in strain 2. The calculated values of stress− 2. The nominal stress (s) defined for the strain are real and very useful in the tensile test in terms of original cross plastic region of the curve. sectional area (A0) is not really stress because the cross sectional area Ai at the instant of load measurement is less than A0 in the evaluation of s. 3. The metal working designers are 3. The structural designers are interested interested in plastic region where in a region where strains are elastic
- 11. 11 difference between Ai and A0 is and difference between Ai and A0 is significant. The true stress−strains negligibly small. But this is not true in give accurate picture and hence it is the plastic region and especially when more useful to metal working designs. maximum load is reached. 4. It is not easy to obtain values of σ 4. It is easy to obtain these values from test since the force Fi and cross through test and convenient−less sectional area (Ai) must be measured costly. These values are widely simultaneously. True stress ( σ ) is available and documented. important in metal working calculation because of its fundamental significance. 5. It is more consistent with the 5. It is less consistent with physical phenomenon 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 hardening Fig. 5The flow curveA true stress–strain 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 σ = k (∈) nwhere: n is strain hardening exponent
- 13. 13 k is strength constant Fig. 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 = 28kN F1 22 × 10 3 ( N) σ1 = = = 3.78 N / mm 2 2 π / 4 d1 π / 4 (8.6) 2 F2 28 × 10 3 T2 = = = 517.5 N / mm 2 π / 4 (8.3) 2 π / 4d2 2 L1 ∈= true strain = log e L0 L1 = Length after deformation L 0 = length before deformation
- 14. 14As volume of specimen remain constant, A 0 × L 0 = A1 × L1 2 π 2 π 2 L1 d d 0 × L 0 = d1 × L1 ∴ = 0 4 4 L 0 d1 2 d d ∴ ∈= log 0 d = 2 log 0 1 d1 ∈1 = true strain for first extension 9 ∴ 2 log = 0.091 8 .6 ∈2 = true strain for second extension d 9 = 2 log 0 = 2 log = 0.1619 d1 8.3 Applying Hollomon equation, n σ1 = K∈1 σ 2 = K ∈n 2 n σ ∈ σ2 ∈ ∴ = 2 or log 2 = n . log 2 σ1 ∈1 σ1 ∈1 σ 517.5 log 2 log n= σ1 = 378 = 0.54 ∴ ∈2 0.1619 log log ∈1 0.091Substituting the value of ‘n’ in equation (1) 378 = k (0.091) 0.54 K = 1385 N / mm 2
- 15. 15∴ strain hardening exponent (n) = 0.54strength coefficient K = 1385 N / mm 2with this information, Hollomon equation can be written as σ = 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 dσ d ∈ = n ∈ . σ 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. 7 F = σ.A dF = σ dA + A dσ = 0 dA dσ ⇒ − = A σFrom constancy of volume, V = A . L A= cross section of spearmen L =length of specimen dV = 0 = A. dL + dA L dA dL∴ − = A L dσ dl∴ = = d∈ σ L dσ ∴ =σ d∈ProblemProve that uniform strain is equal strain hardening exponent (n).Solution:
- 17. 17 Fig.P = load at any instanceA = cross section of specimen. P = A σ = A . k (∈) n σ = k (∈) n (1) Ao A ∈ = log ∴ e∈ = 0 A 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 ] = 0 n ∈u n −1 =∈u n ∴ ∈u = nProblem 1:Hollomon equation for a material is given as σ = 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. σuUltimate tensile strength = = S u = [σ = (1 + e) × S] 1 + σu ∈u = log e (1 + e u )∴ ∈u = (1 + e u ) e = 2.71 (logarithmic base) 1 + eu = en n n σu K . ∈u n ∈ n ∴ Su = = = K u = K n e e 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 2 Engineering elongation strain = e u = 35% = 0.35
- 19. 19uniform true strain ∈u = log (1 + e u ) = log (1.35) = 0.3 σ u = S u (1 + e u ) but n = ∈u = 0.3 = 300 (1 + 0.35) = 405 N / mm 2 . σ u = K (∈u ) n = k (n ) n 405 = K (0.3) 0.3 ∴ K = 581.2 N / mm 2∴ Hollomon equation for given metal is σ = 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. 9 ∈1 u = ∫σ d∈ 0
- 20. 20The true stress–strain curve can be represented by the Hollomon equation σ = K (∈) n . ∈1 ∈1 K ∈n +1 n u = ∫ K (∈) . d ∈ = 0 n +1 0 K ∈1n +1 u= n +1similarly mean flow stress can be found ∈1 ∈1 n ∫σ d∈ ∫ K ∈ .d ∈ K ∈1n +1 K ∈1n 0 0 σm = = = = ∈1 −0 ∈1 (n + 1) ∈1 (n + 1) K ∈1n σm = 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 AA–1100 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. σ = 140 (∈) 0.25 N / mm 2Solutions a. Calculate total strain during deformation
- 21. 21 A0 d ∈ = ln = 2 ln 0 A d 12.7 = 2 ln = 0.199 11.5 Calculate the total volume of bar πd2 π (0.0127 m 2 V= ×l= × 0.1 m = 1.26 × 10 − 5 m 3 4 4 For AA–1100, K = 140 MPa and n = 0.25. Note that, as ∈< n, the deformation is homogeneous ∈1 +1 n Wi = K × ×V n +1 1.25 = 140 × 10 6 N × 0.199 × 1.26 × 10 − 5 m 3 = 187.5 N (J ) m 6 1.25 (b) Mean stress during deformation n ∈1 σm = K × n +1 0.199 0.25 = 140 MPa × = 74.8 MPa 1.25 (c) Peak (maximum) stress applied, from Hollomon equation σ1 = 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 [σ 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 is f (σ x , σ y , σ z, Txy , Tyz , Tzx ) = a constant.or in terms of principal stresses f( f (σ1 , σ 2 , σ 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 poison’s ratio 0.5. σ1 + σ 2 + σ 33. The magnitude of the mean normal stress σ m = does not cause yielding. The 3 assumption that yielding is independent of σm (also called hydrostatic component of the total state of stress) is reasonable if plastic flow depends upon shear mechanism such as slip or twinning. In this context, yield criterion is written as F[(σ1 − σ2), (σ2 − σ3), (σ3 − σ1)] = C which implies that yielding depends upon the size of the Mohr’s circle and not their position. It is shown that if a stress state (σ1, σ2, σ3) will cause yielding, an equivalent stress state (σ1′ , σ 2 ′ , σ 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. σ1 , σ 2 , σ 3 are principal stresses and σ1 > σ 2 > σ 3(algebraically).Then maximum shear stress σ − σ3 Tmax = 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 (σ 0 ).
- 24. 24For uniaxial tensile test, stress situation is σ1 = σ 0 , σ 2 = σ 3 = 0 σ − σ3 σ 0 ∴ Tmax = 1 = =c 2 2 σ − σ3 σ0 ∴ 1 = or σ1 − σ 3 = σ 0 2 2ii) Material is subjected to pure shear: σ1 = k σ2 = 0 σ 3 = −k k = shear strength of the material σ1 − σ 3 = σ 0 ∴k + k = σ0 σk = 0 = 0.5 σ 0 2Application:i) Plain stress condition. σ x , σ y, Txy 2 σx + σy σx + σy σ1 = + + (T ) 2 2 2 xy 2 σx + σy σx + σy σ2 = − + (T ) 2 2 2 xy ∴ σ1 = σ 0 when σ 3 > 0 σ1 + σ 3 = σ 0 when σ 3 < 0ii) Plain strain condition ∈3 = 0 σ 2 = 2 (σ1 + σ 2 ) σ + σ3 σ1 − σ 3 = σ 0 σ2 = 1 2
- 25. 25Shortcomings1. An essential short coming of this criterion is that it ignore the effect of intermediate principal stress (σ 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 σ1 = σ 2 = σ 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= 1 σG [ (σ1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2 + (σ 3 − σ1 ) 2 ]G = modulus of shear which is a constant. (σ1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2 + (σ 3 − σ1 ) 2 = Constant.(i) For uniaxial tensile test, yielding will occur when σ1 = σ 0 ; σ 2 = σ 3 = 0 (σ y ) 2 + (−σ y ) 2 = cons tan t = 2 σ 0 2 Therefore Von Mises criterion can be stated as (σ1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2 + (σ 3 − σ1 ) 2 = 2 σ 0 2 (σ x − σ y ) 2 + (σ y − σ z ) 2 + (σ z − σ x ) 2 + σ (T 2 + T 2 + T 2 ) = 2σ 0 2 x y y z z xi) For plane stress: σ 2 = 0 σ + σ3ii) For plane strain: σ 2 = 1 2iii) For pure shear stress condition: σ1 = k σ2 = 0 σ 3 = −k (σ1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2 + σ 3 − σ1 ) 2 = 2 σ 0 2 (k 0 − 0) 2 + (0 + k 0 ) 2 (− k 0 − k 0 ) 2 = 2 σ 0 2 σ k 02 = 2 σy2 σy ∴k = = 0.557 σ 0 3
- 27. 27This is the relationship between shear yield strength and tensile yield strength of the material as per VonMises criterion. k = 0.5 σ 0 2 Tresca criterion k 0 = 0.577 σ 0 Von Mises criterion. σyVon Mises criterion satisfy the experimental data better than Tresca and therefore k = value is 3normally 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: [(σ x − σy ) 2 + (σ y − σ z ) 2 + (σ z − σ x ) 2 + 6 (T 2 xy + T 2 yz + T 2 zx ) = 2 σ 0 2 ]Effective stressWith the yield criterion, it is useful to define an effective stress denoted as σ which is function of theapplies stresses. If the magnitude of σ reaches a critical value, then the applied stress will causeyielding.For Von Mises criterion σ= 1 2 [(σ1 − σ2 ) 2 + (σ 2 − σ 3 ) 2 + (σ 3 − σ1 ) 2 ] 1/ 2
- 28. 28For Tresca criterion σ = σ1 − σ 3 σ ≥ σ 0 ……………………. For both the criteria. σ ≥ 3 k …………………. Von Mises σ ≥ 2 k …………………… Tresca Plane stress condition Plane strain condition 1. In plane stress condition, there is no stress 1. In plane strain condition, the strain in third direction is absent. in third direction. 1 ∈2 = [ σ 2 − υ(σ 3 + σ1 )] E ∴ σ 2 = υ ( σ1 + σ 3 ) Near yield point and in plastic zone 1 υ = (For plastic defo) 2 σ + σ3 σ2 = 1 But there is strain in third direction. Two 2 principal stresses 2 σx + σy σx + σy σ1 = 2 + 2 + Txy 2 ( ) 2 σx + σy σx + σy ) σ2 = 2 − 2 ( ) + Txy 2 1∈1 = [ σ1 − υ (σ 2 )] E 1∈2 = [ σ 2 − υ (σ1 )] E 1∈3 = [ 0 − υ (σ1 + σ 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 = [ σ1 − υ (σ 2 + σ 3 )] E 1 ∈2 = [ σ 2 − υ (σ1 + σ 3 )] E 1 ∈3 = [ σ 3 − υ (σ1 + σ 2 )] Elet ∈2 = 0 ⇒ σ 2 = υ (σ1 + σ 3 ) for plastic region, Nadai has shown that υ = 0.5 σ + σ3 σ2 = 1 2 σ + σ3Thus, for Tresca criterion: σ1 , 1 , σ3 2 σ1 − σ 3 = σ 0Von Mises criterion in plane strain: 2 2 σ + σ3 σ + σ3 σ1 − 1 + 1 − σ 3 + (σ 3 − σ1 ) 2 = 2 σ 0 2 2 2 3 ( σ1 . σ 3 ) 2 = 2 σ 0 22 2 σ1 − σ 3 ) 2 = σ 0 = σ 0 3 σ 0 = 1.155 σ 0 = constrained yield strength of the material.
- 30. 30 Yield criterionMaximum shear stress Maximum distortionCriterion (Tresca) energy criterion (Von Misces)Plane Plane PureStress Strain Shear σ + σ2σ 2 − 0, σ min = σ 2 σ= 1 σ1 = −σ 3 = k σ 2 = 0 2σ1 = σ 0 .....σ 3 = ⊕ ve σ1 − σ 3 = σ 0 σ1 + σ 3 = σ 0σ1 + σ 3 = σ 0 σ 3 = Οve 2k= σ0 σ0 k= 2 Plane stress Plane strain Pure Shear σ1+σ3 1 σ12 + σ 3 2 − σ1σ 3 = σ 0 2 σ2 = k= σ0 2 3 2 σ1 − σ 3 = σ0 3 Tresca criterion Von Mises yield criterion 1. This criterion is also known as 1. Van Mises criterion is also known as maximum shear stress criterion and distortion energy yield criterion. It states attributes yielding to slip that yielding occurs when deformation phenomenon which occurs when energy per unit volume of material maximum shear stress exceeds a exceeds certain value which is value, characteristic to the material. characteristic of the material. Mathematically it can be stated as Mathematically, it can be stated as
- 31. 31 σ1 − σ 3 = σ 0 σ0 = 1 [(σ 1 − σ2 ) 2 + (σ 2 − σ 3 ) 2 ] 2 where σ1, σ 2 , σ 3 are principal + (σ 3 − σ1 ) 2 1 / 2 stresses, and σ1 > σ 2 > σ 3 . Or 1 σ0 = (σ − σ y ) 2 ) 2 + (σ y − σ z ) 2 x 2 + (σ z − σ x ) 2 + σ (τ 2 xy + τ 2 yz + τ 2 zx ) ] 1/ 22. Phenomenon of slip is limited to 2. The application of this yield criterion ductile materials and hence holds good for both ductile and brittle application of this criterion is limited materials. 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 this tion 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 Superimposed 6. Von Mises criterion is preferred where 6. Tresca criterion is 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. σ1 , σ 3 , σ2 = 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 σ 3 > σ1 > 0, so σ 3 = + Y II σ 3 > σ1 > 0, so σ 3 = + Y III σ1 > 0 > σ 3 , so σ1 − σ 3 = + Y IV 0 > σ1 > σ 3 > 0, so σ 3 = − Y V 0 > σ 3 > σ1 , so σ1 = − Y VI σ 3 > 0 > σ1 , so σ 3 − σ1 = + YLocus of yield as per Von Mises criterion1. For a biaxial plane stress condition (σ 2 = 0) the Von Mises criterion can be expressed mathematically, σ12 + σ 3 2 − σ1σ 3 = σ 0 2 This the equation of an ellipse whose major semiaxis is 2 σ 0 and whose minor semiaxis is 2 σ 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 (σ1 = σ 3 ). The greatest divergence between the two criteria occurs for pure shear (σ1 = −σ 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 (σ1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2 + (σ 3 − σ1 ) 2 = 2 σ 0 2where σ1 , σ 2 , σ 3 are three principal stresses and σ 0 is the yield strength of material. In plane strainstress condition, the intermediate principal stress is arithmetic mean of other two. Assumingσ1 > σ 2 > σ 3 , we can write σ + σ3 σ2 = 1 2substituting the value of σ 2 in the above expression 2 2 σ + σ3 σ + σ3 σ1 − 1 + 1 − σ 3 + ( σ 3 − σ1 ) 2 = 2 σ 0 2 2 2
- 35. 35 (σ1 − σ 3 ) 2 (σ1 − σ 3 ) 2 (σ 3 − σ1 ) 2 + + = 2 σ02 4 4 1 σ (σ1 − σ 3 ) 2 = 2 σ02 4 (σ1 − σ 3 ) 2 = 8 / 6 σ 0 2 2 ∴ σ1 − σ 3 = σ0 = σ0 3σ 0 is called constrained strength of material and is 1−15 times the yield strength under uniaxial tensiletest.PROBLEMA stress analysis of a space craft structural member gives the state of stress as below: 200 30 0 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:σ x = 200 MPaσ y = 100 MPaσ z = − 50 MPaTx y = 30 MPa(1) Applying von Mises criterion 1/ 2 1 2 σc = 2 ( ) ( ) σ x − σ y 2 + σ y − σ z 2 + (σ z − σ x ) 2 + 6 T 2 − T 2 + T 2 x y y z z x
- 36. 36 ( ) 1/ 2 1 ( 200 − 100) 2 + (100 + 50) 2 + ( − 50 − 200) 2 + 6 30 2 + 0 2 + 0 2 σc = 2 ∴ σ c = 224 MPaThe calculated stress ( σ c ) is less than the yield strength of the material (σ 0 ) , yielding will not occur asper von Mises criterion σ0 500 MPaFactor of safety = = = 2.2 σc 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. σ Txy Txz 200 30 0 x Tij = Tyx σy Tyz = 30 100 0 T Tzy σz 0 0 − 50 zx 20 3 0 = 3 10 0 × 10 0 0 − 5 To find the principal stresses σ 20 − σ 3 0 3 10 − σ 0 =0 0 0 − 5−σ ∴ I1 = σ x + σ y + σ z = 20 + 10 − 5 = 25 σx Txy σy Tyz σx Txz I2 = + + Txy σy Tyz σz Txz σz = 191 − 50 − 100
- 37. 37 ∴ I2 = 41 I 3 = Tij 10 0 3 0 3 10 = 20 −3 +0 0 −5 0 −5 10 0 = − 1000 + 45 ∴ I3 = − 955 f (σ) = σ 3 − I1σ 2 + I 2 σ − I 3 = 0 σ 3 − 25 σ 2 + 41 σ + 955 = 0Applying standard method to get cubic roots, f (σ) = σ 3 − 25 σ 2 + 41 σ + 955 = 0 f(y) = y3 + py2 + qy + r = 0 ∴ p = − 25 q = 41 r = 955 a= 1 3 ( 1 ) 3q − p 2 = ( 3 × 41 − 625) 3 a = − 167.3 b= 1 [ 27 2p 3 − 9 pq + 27r ] = 1 27 [2(−25) 3 − 9(−25) ( 41) + 27(955) ] ∴ b = 139.25
- 38. 38 −b − 139.25 = 3 3 Cos φ = a − 167.3 2 × − 2× − 3 3 ∴ φ = 99.620 ∴ φ ≈ 100 −a g = 2× 3 g = 14.94 φ P 99.62 25 y1 = g cos − = 14.94 cos + 3 3 3 3 ∴ y1 = 20.83 φ P 99.62 25 y2 = g cos + 120 − = 14.94 cos + 120 + 3 3 3 3 ∴ y1 = −5 φ P 99.62 25 y3 = g cos + 249 − = 14.94 cos + 240 + 3 3 3 3 ∴ y3 = 9.16σ1 = 20.83 × 10 = 208.3 MPaσ 2 = 9.16 × 10 = 91.6 MPa ordered in such a way that σ1 > σ 2 > σ 3σ 3 = −5 × 10 = − 50 MPaTo apply Tresca criterion; σ1 − σ 3 208.3 − ( −50) Tmax = = = 129.15 MPa 2 2 σ0 Tmax < = 250 MPa 2Hence, yielding will not occur as per Tresca criterion
- 39. 39 σ0 / 2 250Factor of safety = = = 1.94 Tmax 129.15PROBLEM1. A thin walled tube with closed ends is to be subjected to maximum internal stress pressure of 0.35 N/mm2 in service. The mean radius of the tube is to be 304.8 mm and it is not to yield in any region. a. If the material has σ 0 = 7 N/mm2, what minimum thickness ‘t’ should be specified according to Tresca and von Mises criterion. b. If the shear yield strength ‘K’ , were specified as 2.8 N/mm2, find ‘t’.Soln: pr a. σ1 = −−−−−−−−−−−−−− hoop stress 2t pr σ2 = −−−−−−−−−−−−−− axial stress 2t σ3 = 0 −−−−−−−−−−−−−− radial stress → Using Tresca criterion σ1 − σ 3 = σ 0 pr = σ0 t 0.35 × 304. 8 =7 t t = 15.24 mm → Using von Mises criterion
- 40. 40 σ1 σ2 = 2 ∴ (σ1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2 + (σ 3 − σ1 ) 2 = 2 σ 0 2 2 2 σ σ σ1 − 1 + 1 − 0 + ( 0 − σ1 ) 2 = 2 σ 0 2 2 2 σ12 σ12 + + σ12 = 2 σ 0 2 4 4 6 σ12 ∴ = 2σ 0 2 4 2 ∴ σ1 = σ0 3 p×r 2 = × 60 t 3 0.35 × 304.8 2 = ×7 t 3 ∴ t = 13.19 mm→ when σ 0 is the specified property and ‘t’ is the unknown, the Tresca criterion is more conservative. b. K = 2.8 N/mm2 → Using Tresca criterion σ1 − σ 3 = 2K pr = 2K t 0.35 × 304.8 = 2 × 2.8 t t = 19.05 mm → Using von Mises criterion
- 41. 41 (σ1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2 + (σ 3 − σ1 ) 2 = 6K 2 6σ12 = 6K 2 4 ∴ σ1 = 2K p×r = 2K t 0.35 × 304.8 = 2 × 2.8 t ∴ t = 19.05 mmNOTE:When ‘K’ is the specified property, both criteria predict the same value for ‘t’.2. Consider the same problem as above except that ‘t’ is specified as being 25.4 mm and the values of σ 0 and K are unknown using both yield criteria. a. Determine the value of σ 0 to prevent yielding. b. Determine the value of K to prevent yielding. a. Tresca criterion: σ1 − σ 3 = σ 0 pr = σ0 t pr pr σ1 = ; σ2 = ; σ 3 = 0 t t 0.35 × 304. 8 = σ0 25.4 ∴ σ 0 = 4.2 N/mm2 → von Mises criterion
- 42. 42 2 σ1 = σ0 3 2 3 0.35 × 304.8 σ0 = 3 × σ1 = × 2 2 25.4 ∴ σ 0 = 3.64 N / mm 2b. σ1 = 2K 4.2 ∴ K= 2 ∴ K = 2.1 N/mm2 6 σ12 ∴ = 6K2 4 ∴ σ1 = 2K ∴ K = 2.1 N/mm2 ****************

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