LECTURE 6 ILLUSRATIVEPROBLEMS: Letus discussfew representativeproblemsdealingwith complexstateofstress to be solvedeitheranalyticallyor graphically. PROB1:Acircularbar40 mm diametercarriesan axialtensileload of 105 kN.Whatis the Valueofshear stress on the planeson whichthe normalstress has a value of 50 MN/m2tensile. Solution: Tensile stress ey= F/A=105 x103 /7c x(0.02)2 = 83.55MN/m2 Nowthe normalskess on an obliqe planeis givenbythe relation 6 9=cysinze 50 x105 = 33.55MN/m2x1O6sin2G e = 50068 Theshearstress on the obliqueplaneis then givenby 19= 1/26ysin20 = 1/2x83.55X106Xsin 101.36 =4o.9e MN/m2 Thereforethe requiredshearstress is 40.96MN/m2 PROB2: Fora givenloadingconditionsthe stateofstress in the wall ofa cylinderis expressedas follows: (a) 85 MN/m2tensile (b) 25 MN/m2tensileat rightanglesto (a) (c) Shearstressesof 60 MN/m2 ontheplanesonwhichthestresses(a)and(b)act;thesheercoupleactingon planescarrying the25 MNlm2 stressis clockwise in effect. Calculatethe principalstressesand the planeson whichtheyact.Whatwould be the effecton these results if owingto a changeofloading (a) becomescompressive while stresses(b) and (c) remainunchanged Solution: The problem may be attempted both analytically as well as graphically. Letus rstobtain the analytical solution The principlestressesare givenbytheformula cqandog =%(o.+o,,):; (o,oy)3+4a1,,, =%(a5+25):% (a5+25)?+(4x5U9) =55:%.5nJ§=55:57 =>01=122iv1Nlm2 02 = -12 Mlwmgcnmpressivej For nding out the planes on which the principle stresses actus the equation ta n2' =[ W] 2r Againthe principalstresseswouldbe givenbythe equation. 1 1 013:0: =Eox +0y]:§(0!-03,): +412 =%(e5 +25) :13(-35 -25)? +(r1}1l3D2) =%(5m 2%(-35 -25)? +(t1xl3D2) =3n:%,l121nn+1.14en = -30 :e1.4 o, = 51.4 lv1NIm2; 02 = 111.4 l1-1Nim2 Again ferfinding out the angles use the following equation. 2 tan2ti'= Tm ax -cry _ 211130H120 e525 -110 :2 11 2E'=tan[-% => e=23.74 Thus,the two principlestressesactingon the two mutuallyperpendicularplanesi.e principleplanesmaybe depictedon the elementas shownbelow: _1_.______ So this is the directionof one principleplane&the principlestressesactingon this would be <51 when is actingnormalto this plane,nowthe directionof otherprincipal planewouldbe900+ 6 because theprincipalplanesarethetwomutually perpendicular plane,hencerotatetheanotherplane9 + 900in thesamedirection to getthe anotherplane,nowcompletethe materialelementife is negativethat meanswe are measuringthe angles in the oppositedirectionto the referenceplaneBC . Thereforethe directionof other principalplaneswould be (~63+ 90} since the angle-9 is alwaysless in magnitudethen 90 hencethe quantity( -9 + 90 )would be positivethereforethe inclinationof otherplanewith referenceplanewould be positivethereforeifjustcomplete the Block.it wouldappearas Retpiarie So wheneverone ofthe anglescomesnegativeto getthe positivevalue, first/ldd90°tothevalueandagainadd900as inthiscase6 =23°74' so 91=23°74' + 90°= 66026./again adding90°alsogivesthedirection ofotherprinciple planes i.e92= 66026 + 90°=156°26' This is howwe can showthe angularpositionof these planesclearly. QRAPHIQAL §QLQTlQN: Mohr's Circle solution:Thesame solutioncan be obtainedusingthe graphicalso utioni.e the Mohr's stress circle,forthe rst part,the blockdiagrambecomes Constructthe graphicalconstructionas perthe steps givenearlier. Takingthe measurementsfrom the Mohr's stress circle,the variousquantitiescomputedare <51 = 120MN/m2 tensile G2=10 MNlm2 compressive 61= 34°counter clockwise fromBC e2= 34°+ 90=124°counter clockwise fromBC Part Second:The requiredcongurationi.ethe blockdiagramfor this case is shownalongwiththe stress circle. Bytakingthe measurements,the variousquantitescomputedare givenas 61= 56.5MN/m2 tensile 62= 106MN/m2 compressive 91= 66015 counter clockwise fromBC 62= 156015 counter clockwise fromBC Salientpoints of Mohr's stress circle: 1. complementaryshearstresses(on planes90° aparton the circle)are equalin magnitude 2. Theprincipalplanesare orthogonal:pointsL and Mare 1800aparton the circle(900apartin material) 3. Thereare no shearstresseson principalplanes:pointL and Mlie on normalstress axis. 4. Theplanesofmaximumshearare45° fromthe principalpointsD and E are 90° , measuredroundthe circlefrom pointsL and M. 5. Themaximumshearstressesare equalin magnitudeand givenbypointsD and E 6. The normalstresses on the planes of maximumshear stress are equal i.e. points D and E both havenormalstress co-ordinatewhich principalstresses. is equal to the two ,3*»?$| \ 2_:'< As we knowthatthe circlerepresentsall possiblestatesof normaland shearstress on anyplanethrougha stresses pointin a material.Furtherwe haveseen thatthe co-ordinatesof the point Qare seen to be the same as those derivedfrom equilibriumof the element.i.e. the normal and shear stress componentson any plane passingthroughthe pointcan befoundusing Mohr's circle.Worthyofnote: 1.ThesidesABandBCoftheelement ABCD, whichare90°apart,arerepresented onthecircleby@ p 3nd E p andtheyare1800apart. 2. it has beenshownthat Mohr's circlerepresentsall possiblestates ata point.Thus,it can be seen ata point.Thus,it, can be seen thattwo planesLP and PM,1800 apartonthediagramandtherefore 900apartinthematerial, onwhichshearstressreis zero.Theseplanesaretermedas principal planesandnormalstressesacting on them are knownas principalstresses. Thus , G1=OL 62 = CM whichtheshearstressis maximum aresituated90°fromtheprincipal planes(on circle),and45°inthematerial. 4.Theminimumnormalstress is justas importantasthe maximum.Thealgebraicminimumstress couldhavea magnitudegreaterthanthatofthe maximumprincipal stress ifthe stateofstress weresuchthatthe centreofthe circleis to the leftof orgin. i.e.if (51= 20MN/m2 (say) «:2=~80MN/m2 (say) Thenxmaxm =(.5'} 2* This can be also understandfrom the circleSinceABand BC are diametricallyoppositethus,whatevermay be their orientation,theywill alwayslie on the diametreor we can saythattheirsum wontchange,it can also be seenfrom analyticalrelations Weknowon = + cus2t?+Tn, sin2' on planeBC;9 = 0 Gm:"5x on planeAB;9 = 270° cng=