12.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 extension2Fig. 2: StressStrain Diagram.P = Proportionality limit E = Elasticity limit Y = Yield pointN = Necking PointF = Fracture PointFig. 33Fig. 44i. EngineeringStress 0AFiS Fi = instantaneous loadEngineering strain =

,_

000LL Lspecimen of length Originallength in ChangeLLii. True stress AiFi and True strain

,_

oiiL0LLLlogLdLTrue stress is defined as load divided by actual cross sectional area (not original cross sectional area A0) for that particular load.AiFi Similarly, true strain is based on the instantaneous specimen length rather than original length.As such true strain (or incremental strain) is defined as LdLd Where L is length at load F and is the true strain.The true strain at load F is then obtained by summing all the increments of equation.Arithmetically, this can be written asnnLdLLdLLdLLdLLdLd + + + + + ......33221100 011L0LLLlogLdL True strain is the sum of each incremental elongation divided by the current length of specimen, where L0is original gauge length and Liis the gauge length corresponding to load Fi. The most important characteristics of truestress strain diagram is that true stress increases all the way to fracture.Thus true fracture strength fis greater than the true ultimate strength uin contrast with engineering stress where fracture strength is lesser than ultimate strength.5Relationship between true and engineering stress strains From volume constancy, V = A0 L0 = Ai LiiiAALL00 e =

,_

1LLLL L0i00 i) e 1 (LL0i+ 0ii00iiiLLSAAAFAF ) e 1 ( S + + iL0L0i) e 1 ( logLLlogLdL) e 1 ( log + Problems with Engineering StressStrains1. Engineering stressstrain diagramdoes not give true and accurate picture of deformation characteristicsofthematerial becauseit takesoriginal crosssectional areaforall 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 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 E = 0LL 0 50 501 50 55 5 5/50 = 0.12 55 60.5 5.5 5.5/55 = 0.13 60.5 66.55 6.05 6.05/60.5 = 0.1Sum of incremental strain = 1 . 0 1 . 0 1 . 05 . 6005 . 6555 . 5505+ + + +=0.3Now we will calculate total strain considering original and final length after of extension L3 = 66.55 Total engineering strain when extended = 331 . 05050 55 . 66LL L00 3the 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- 3 2 2 1 1 0 + + =

,_

+ +231201LLlogLLlogLLlog=5 . 6055 . 66log555 . 60log5055log + += 0.286But total true strain equals to286 . 05055 . 66logLLlog033 . 0 In the case of true strains, sum of incremental strain is equal to the overall strain. Thus true strains are additive.This is not true for engineering strains.3.7FigL0 = length before extensionL1 = L0 = length after extensionStraine=00 1LL L00 1LL LI L1= 0FigTo obtain strain of 1 the cylinder must be squeezed to zero thickness which is only hypothetical and not true.Moreover, intuitively we expect that strain produced in compression should be equal in magnitude but opposite in sign.Applying true strain formulation, to extension 2 logLL 2logLLlog0001 To compression; L1 = L0/22 log 2 / 1 logL2 / LlogLLlog0001 gives 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.Problem:8The following data were obtained during the true strain test of nickel specimen.LoadkNDiametermmLoadkNDiametermm0 6.40 15.88 5.1115.30 6.35 15.57 5.0815.92 6.22 14.90 4.8316.32 6.10 14.01 4.5716.5 5.97 13.12 4.3216.55 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.1. True stress at max load = APmax = 2384 . 5410 55 . 16= 617.77 MPa 2. True fracture stress = minAP = 22 . 1110 45 . 123 = 1109 MPa 3. True fracture strain = 2i0ddln

,_

= 278 . 34 . 6ln ,_

= 1.0534. True uniform strain = 2i0ddln

,_

= 284 . 54 . 6ln ,_

=0.183 5. Trueneckingstrain=truefracturestraintrue uniform strain = 1.053 0.183 = 0.876. Ultimate tensile stress = maxmaxAP= 234 . 6410 55 . 16 = 514 MPa7. Now,n= log(1+e) = log (1.2) =0.1839LoadKNDiametermmAreamm2True stress = iiAP(N/mm2)True strain = 2i0ddln

,_

Engg. Stress = AP(N/mm2)Engg. Strain =1dd2i0

,_

0 6.40 32.17 0 015.3 6.35 31.67 48.31 0.0156 475.59 0.015815.92 6.22 30.39 523.86 0.057 494.87 0.05916.32 6.10 29.22 558.52 0.096 507.30 0.1016.5 5.97 27.99 589.50 0.139 512.90 0.14916.55 5.84 26,79 617.77 0.183 514.45 0.2015.88 5.11 20.5 774.63 0.45 493.62 0.56815.57 5.08 20.27 768.13 0.46 484.00 0.58714.90 4.83 18.32 813.32 0.56 463.16 0.75514.01 4.57 16.40 854.27 0.67 435.5 0.96113.12 4.32 14.66 894.95 0.786 407.83 1.1912.45 3.78 11.22 1109.63 1.053 387.00 1.866Applications of Engineering Stress and StrainsEngineeringstressandstrainareuseful formanyengineeringdesignapplications. Computationof stressandstrainisbasedoninitial areaorgaugelengthandthereforeengineeringstressandstrain represent only approximations of the real stress and strain in plastic zone.Inelasticdeformationregion(wheredimensional changes aresmall andnegligible) theinitial and instantaneous areas are approximately same and hence true stress equals engineering stress.Therefore, indesignproblemswherelargedimensional changesdonot occur, theuseofengineeringstressis sufficiently accurate and used extensively as it is easier to measure.However, for metal working where large plastic deformations occur and are necessary, the approximations inherent in engineering stress and strain values are unacceptable. For this reason, the true 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 from which that changeis producedrather thantooriginal gaugelength.Theengineeringstresseandstrains provides incorrect values after yield point i.e. plastic zone which a mainzone of interest for metal working.2. True strains additive i.e. the total overall strain is equal to sum of incremental strains.103. 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.n) ( K n0) ( + 0 = the amount of strain hardening that material received prior to the tension test.6. Truestresstruestrainvaluesarequitesensitivetochangeinbothmetallurgical andmechanical conditions of matter.TruestressstrainEngineering1. Actual values of gauge length and cross sectional area is used in calculating true stress and true strain.iiAF iL0LLdLThe sumof incremental strains is equal to total strainUnlike load elongation curve, there is no maximumin thetrue strincurve. The sloppe of the curve in the plastic region decreases with increase in strain1. Original crosssectional areas(A0) is used for calculating engineering stress.S = 0iAFFurther straine=00 iLL Lisused. The sum of incremental strains is not equal to total strain.2.The calculated values of stressstrain are real and very useful in the plastic region of the curve.2. The nominal stress (s) defined for the tensiletest intermsoforiginal cross sectional area (A0) is not really stress because the cross sectional area Aiat the instant of load measurement is less than A0 in the evaluation of s.3. The metal working designers are interested in plastic region where difference between Aiand A0is significant. The true stressstrains giveaccuratepictureandhenceit is 3. The structural designers are interested inaregionwherestrains areelastic anddifferencebetweenAiandA0is negligibly small.But this is not true in the plastic region and especially when 11more useful to metal working designs.maximum load is reached. 4. It is not easy to obtain values of from test since the force Fiand cross sectional area (Ai) must be measured simultaneously. True stress () is important in metal working calculation because of its fundamental significance.4. It is easy to obtain these values through test and convenientless costly. These values are widely available and documented. 5. It is more consistent with the phenomenon of metal deformation.5. It is less consistent with physical 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 work hardening.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 constitutive relations. 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 the case in structural engineering or for bending. 2.Rigid, perfectly plasticIn most metal forming operations, the permanent strains are much longer than the elastic.One therefore in air no great error by assuming the metal to behave as a rigid body prior to yielding. It is for this reason that one mainly employs perfectly plastic material idealisation. 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 cause the 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 of Holloman equation which is

n) ( k where:n is strain hardening exponent13 k is strength constant Fig. 6In a tension test of stell, a specimen of circular cross section with original diameter 9 mm is used.The loads applied were 22 kN and 28 kN which reduces its diameter to 8.6 mm and8.3 mmrespectively. Determine(i)truestressandtruestrain for given loads (ii) strain hardening exponent and strength coefficient. Solution:0doriginal diameter of specimen = 9 mm1ddiameter of specimen on application of load kN 22 F1 2ddiameter of specimen on application of load kN 28 F2 2232111mm / N 78 . 3) 6 . 8 ( 4 /) N ( 10 22d 4 /F 2223222mm / N 5 . 517d 4 /10 28) 3 . 8 ( 4 /FT

01eLLlog strain true 1LLength after deformation 0Llength before deformationAs volume of specimen remain constant, 1 1 0 0L A L A 14 121020L d4L d4 21001ddLL

,_

10210ddlog 2ddlog

,_

1true strain for first extension091 . 06 . 89log 2 2true strain for second extension 1619 . 03 . 89log 2ddlog 210 Applying Hollomon equation, n11K n22K n1212

,_

or1212log . n log54 . 0091 . 01619 . 0log3785 . 517loglog1logn122 Substituting the value of n in equation (1) 54 . 0) 091 . 0 ( k 378 2mm / N 1385 K strain hardening exponent (n) = 0.54strength coefficientK = 1385 2mm / N15with this information, Hollomon equation can be written as 54 . 0) ( 1385 BothnandKarematerial properties: Thestrainhardeningexponent physicallyreflectstherateat which the material hardens.The derivative of this equation .dnd

,_

In states that fractional change in true stress caused by a fractional change in true strain is determined by the strain hardening exponent (n).Therefore, the stress increases rapidly with strain for a material that hasa largestrainhardening exponent, such as2O 3stainless steel (n = 0.3) compared to a material where n is low such as 4.10 stainless steel (n = 0.1).Plastic InstabilityNeckingor localiseddeformationbeginsat maximumloadwheredecreaseincrosssectional area whichhearstheloadis compensated by increase in strength due in load dF = 0Fig. 7 A . F 16 0 d A dA dF + dAdAFrom constancy of volume, V = A . LA= cross section of spearmenL = length of specimendV = 0 = A. dL + dA L LdLAdA dLdl d

ProblemProve that uniform strain is equal strain hardening exponent (n). Solution: dd17Fig.P = load at any instance A = cross section of specimen. n) ( k . A A P (1) AAlogo AAe0 e A A0(2)Substituting value in equation (1)n0) ( K e A P ] ) ( e [ A K Pn0 At maximum load point on engg.stress strain curve dP = 0 u When true strain 0 ] e ) 1 ( ) ( N e [ A K dPnu1 nu 0 + nu1 nun Problem 1:Hollomon equation for a material is given as. ) ( 140033 . 0 Find the ultimate tensile strength of the material.Solution:n) ( k nu 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 (uTrue strain for uniform elongation is equal to strain hardening exponent. Therefore . nu Ultimate tensile strength = uuu1S + ] S ) e 1 ( [ + ) e 1 ( logu e u+ ) e 1 (u u+ e = 2.71 (logarithmic base) nue e 1 +nnnunnunuuenKeKe. KeS

,_

,_

= 233 . 0mm / N 1 . 69871 . 233 . 01400

,_

UTS=698. 12mm / NThis 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 requires an elongation of 35%. Find strain hardening exponent (n) and strength coefficient (K).Solution:UTS = uS= 300 MPa = 3002mm / NEngineering elongation strain =ue= 35% = 0.35uniform true strain + ) e 1 ( logu u log (1.35) = 0.319 ) e 1 ( Su u u+ but 3 . 0 nu = 300 (1 + 0.35) = 405 . mm / N2

n nu u) n ( k ) ( K 3 . 0) 3 . 0 ( K 405 2mm / N 2 . 581 K Hollomon equation for given metal is3 . 0) ( 2 . 581 Deformation workWork is defined as the product of force and distance.A quantity equivalent to work per unit volume is theproductofstressandstrain. Theareaunder thetruestressstraincurvefor any strain1isthe energy per unit volume (u) or specific energy, of the deformed material.Fig. 9 10d uThe true stressstrain curve can be represented by the Hollomon equation. ) ( Kn 20 101 n10n1 nKd . ) ( K u++ similarly mean flow stress can be found ) 1 n (K) 1 n (Kd . K0dn111 n1110n110m+ + + The work calculated according to above equation assumes that the deformation is homogeneous through out the deforming part. This work is called ideal deformation work.1 nKu1 n1++1 nKn1m+ 21Example:Ideal work of deformationDeformation of fully annealed AA1100 aluminium is governed by the Hollomon equation.Ifa10 cm long bar of this material is pulled in tension from a diameter of 12.7 mm to a diameter of 11.5, calculate the 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. 2 25 . 0mm / N ) ( 140 Solutions a. Calculate total strain during deformation ddln 2AAln0 0 199 . 05 . 117 . 12ln 2 Calculate the total volume of bar

3 52 2m 10 26 . 1 m 1 . 04m 0127 . 0 (l4dV For AA1100, K = 140 MPa and n = 0.25. Note that, as , n > (algebraically).Then maximum shear stress2T3 1max when maxT exceeds a certain value c, specific to that material, yielding will occur.To find the value of c, the material is subjected to uniaxial tensile test and find out yield point strength ). (0For uniaxial tensile test, stress situation is 0 ,3 2 0 1 c2 2T0 3 1max

2 20 3 1 or0 3 1 ii)Material is subjected to pure shear:k1 02 k3 k = shear strength of the material 0 3 1 0k k + 005 . 02k Application:i)Plain stress condition. xy , y xT , 2xy2y x y x1) T (2 2+

,_

+ + + 252xy2y x y x2) T (2 2+

,_

+ + 0 1 when 03 > 0 3 1 + when 03 < ii) Plain strain condition 03 ) ( 22 1 2 + 0 3 1 23 12 + Shortcomings1. An essential short coming of this criterion is that it ignore the effect of intermediate principal stress ). (22. Since pastic flowdepends upon slip phenomenon which is essentially a shearing. Slipis 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 3 2 1 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. 26Von Mises CriterionAccordingtothiscriterion, yieldingwill occurwhenshearstrainenergyperunit volumereachesa critical value.The shear strain energy per unit volume is expressed terms of three principal stresses:[ ]21 323 222 1) ( ) ( ) (G1e + + G = modulus of shear which is a constant. 21 323 222 1) ( ) ( ) ( + + = Constant.(i)For uniaxial tensile test, yielding will occur when 0 ;3 2 0 1 202y2y2 t tan cons ) ( ) ( + Therefore Von Mises criterion can be stated as 2021 323 222 12 ) ( ) ( ) ( + + 20x2z z2y y2x2x z2z y2y x2 ) T T T ( ) ( ) ( ) ( + + + + + i)For plane stress: 02 ii)For plane strain: 23 12 + iii)For pure shear stress condition:k 0 k3 2 1 20213 23 222 12 ) ) ( ) ( + + 2020 020202 ) k k ( ) k 0 ( ) 0 k ( + + 2y202 k 3ky = 0.557 0This is the relationship between shear yield strength and tensile yield strength of the material as per Von Mises criterion. 27 205 . 0 k Tresca criterion0 0577 . 0 k Von Mises criterion. VonMisescriterionsatisfytheexperimental databetterthanTrescaandtherefore3ky valueis normally used. Advantages of Von Mises criterion1. It overcomes major deficiency of Tresca criterion.Von Mises criterion implies that yielding is not dependent onanyparticular normal stressbut instead, dependsonall threeprincipal shearing stresses. 2. VonMises 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: [ ]20zx2yz2xy2 2x z2z y2y x2 ) ( 6 ) ( ) ( ) ( + + + + + T T TEffective stressWith the yield criterion, it is useful to define an effective stress denoted as which is function of the appliesstresses. If themagnitudeof reachesacritical value, thentheappliedstresswill cause yielding. For Von Mises criterion [ ]2 / 121 323 222 1) ( ) ( ) (21 + + For Tresca criterion3 1 280 .For both the criteria.k 3 .Von Misesk 2 TrescaPlane stress condition Plane strain condition1. In plane stress condition, there is no stress in third direction. But there is straininthirddirection. Two principal stresses 1.In plane strain condition, the strain in third direction is absent.

[ ] ) (E11 3 2 2 +

2 ()3 1 + Near yield point and in plastic zone 21 (For plastic defo) ,_

+ 23 12( )2xy2y x y x1T2 2+

,_

+ + + ( )2xy2y x y x2T2)2+

,_

+ + [ ] ) ( E12 1 1 [ ] ) ( E11 2 2 [ ] ) ( 0E12 1 3 + Plane strain conditionIn majority of metal forming operations the problem can be simplified by assuming a condition of plane strain is one.One of the principal strains is zero. [ ] ) ( E13 2 1 1 + [ ] ) ( E13 1 2 2 + 29 [ ] ) ( E12 1 3 3 + let ) ( 03 1 2 2 + for plastic region, Nadai has shown that5 . 0

,_

+ 23 12Thus, for Tresca criterion:33 11,2, + 0 3 1 Von Mises criterion in plane strain: 2021 3233 123 112 ) (2 2 + ,_

+ + ,_

+ ( )2023 12 .23

,_

'0 023 132)0'0155 . 1 = constrained yield strength of the material. 30Yield criterionMaximum shear stress Maximum distortionCriterion (Tresca) energy criterion (Von Misces)Plane Plane PureStress Strain Shear2 min 2, 0 22 1 + 0 k2 3 1 ve .....3 0 1 0 3 1 0 3 1 + ve3 0 3 1 + 2 k = 0k = 20Plane stressPlane strain Pure Shear20 3 12321 + 2312 + 031k 0 3 132 Tresca criterion Von Mises yield criterion1.This criterion is also known asmaximumshearstresscriterionand attributes yielding to slip phenomenon which occurs when maximumshear stress exceeds a value, characteristic tothematerial. Mathematically it can be stated as

0 3 1 1. Van Mises criterion is also known as distortion energy yield criterion.It states that yielding occurs when deformation energy per unit volume of material exceeds certain value which is characteristic of the material. Mathematically, it can be stated as 31where 3 2 , 1, are principal stresses, and .3 2 1 > > 2.Phenomenon of slip is limited to ductile materials and hence application of this criterion is limited to ductile materials.This criterion do not yield good results for brittle materials. 3. Tresca criterion ignores the effect of intermediateprincipal stressandthis is a major draw back of this. []2 / 1 21 323 222 1 0) () ( ) (21 + + Or]2 / 1zx2yz2xy22x z) () (2)zy(2)2)yx(210 + + + +

+ 2. The application of this yield criterion holds goodfor bothductile andbrittle materials. 3. Von Mises criterion take into consideration the intermediate principal stress and hence move realistic. The predications offered by Von Mises criterion conforms empirical data. 4. 5.Locus shown in Figure. It is Hexagonal.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. It is Elliptical.32

Superimposed6. Tresca criterion is preferred in analysis for simplicity. 6. VonMises criterion is preferredwhere more accuracy is desired. Locus of yield as per Tresca criterionBiaxial stress condition is assumed to present locus of yield point on plane paper. 0 , ,2 3 1 yielding will occur if the point plotted is on the boundary or outside. 33Fig. 11 : Tresca yield locus.In the six sectors, the following conditions apply:IY so , 03 1 3+ > > II Y so , 03 1 3+ > > IIIY so , 03 1 3 1+ > > IV Y so , 0 03 3 1 > > > VY so , 01 1 3 > >VI Y so , 01 3 1 3+ > > Locus of yield as per Von Mises criterion1. For a biaxial plane stress condition) 0 (2 the Von Mises criterion can be expressed mathematically, 20 3 12321 + Thistheequationofanellipsewhosemajor semiaxisis02 andwhoseminor semiaxisis .320 The plot of equation is called a yield locus. 34Fig. 12\Comments1. Yielding will occur if the point representing the given stress is plotted and is on the boundary or outside the boundary. 2. Theyieldlocusof maximumshear stresscriterion[Trescacriterion] fall insidethemaximum distortion energy criterion [Von Mises] yield locus. 3. Twoyieldcriteriapredict thesameyieldstressforconditionsofuniaxial stressandbalanced biaxial stress ). (3 1 The greatest divergence between the two criteria occurs for pure shear ). (3 1 4. TheyieldstresspredictedbytheVonMisescriterionis15.5%greater thantheyieldstress, predicted by Tresca criterion. Derive a mathematical expressionfor VonMises yield criterion applicable toplane strainstress condition: Solution:Von Mises yield criterion is stated as 2021 323 222 12 ) ( ) ( ) ( + + where 3 2 1, , are three principal stresses and 0is the yield strength of material.In plane strain stress condition, the intermediate principal stress is arithmetic mean of other two. Assuming ,3 2 1 > > we can write 23 12 + substituting the value of 2 in the above expression( )2021 3233 123 1122 2 + ,_

+ + ,_

+ 2021 323 123 121) (4) (4) ( + + 352023 124) ( 2023 16 / 8 ) ( '0 0 3 132 '0is called constrained strength of material and is 115 times the yield strength under uniaxial tensile test. PROBLEMA stress analysis of a space craft structural member gives the state of stress as below: 111]1

50 0 00 100 300 30 200TijIf the part is made of aluminium alloy with strength 500 MPa, will it exhibit yielding as per Tresca yield criterion and von Mises yield criterion?If not, what is the safety factor?Data given:MPa 200x MPa 100y MPa 50z MPa 30 Ty x(1) Applying von Mises criterion ( ) ( ) ( )2 / 12x2z z2y y2x2x z2z y2y x cT T T 62111]1

,_

+ + + + ( ) ( ) ( ) ( )2 / 122 2 2 2 2c0 0 30 6 200 50 50 100 100 200211]1

+ + + + + + MPa 224c The calculated stress (c) is less than the yield strength of the material ) (0, yielding will not occur as per von Mises criterion36Factor of safety = MPa 224MPa 500c0 = 2.2(ii) Applying Tresca CriterionIn 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. 111]1

111]1

50 0 00 100 300 30 200T TT TT TTz zy zxyz y yxxz xy xij

105 0 00 10 30 3 20111]1

To find the principal stresses 05 0 00 10 30 3 20 I1 = 25 5 10 20z y x + + + y xyxy x2TTI+ z yzyz yTT+ z xzxz xTT= 191 50 100 I2 = 41ij 3T I

0 1010 305 00 335 00 1020 +

= 1000 + 45 I3 =9550 I I I ) ( f3 2213 + 370 955 41 252 3 + + Applying standard method to get cubic roots, 0 955 41 25 ) ( f2 3 + + f(y) = y3 + py2 + qy+ r= 0 p=25q=41r= 955a = ( ) ( ) 625 41 331p q 3312 a = 167.3b = [ ] r 27 pq 9 p 22713+ = [ ] ) 955 ( 27 ) 41 ( ) 25 ( 9 ) 25 ( 22713+ b = 139.25Cos = 3 333 . 167225 . 1393a2b

,_

,_

=99.620100 g=2 ,_

3ag = 14.94y1 = 325362 . 99cos 94 . 143P3cos g + ,_

,_

y1 = 20.8338y2 = 325120362 . 99cos 94 . 143P1203cos g + ,_

+ ,_

+ y1 = 5y3 = 325240362 . 99cos 94 . 143P2493cos g + ,_

+ ,_

+ y3 = 9.161 = 20.83 10= 208.3 MPa2 = 9.16 10 = 91.6 MPa ordered in such a way that 1 > 2 >33 = 5 10 = 50 MPaTo apply Tresca criterion;Tmax= MPa 15 . 1292) 50 ( 3 . 20823 1 Tmax