CAESAR II 高级培训讲义 - 基本应力理论&CAESAR II的实施

基本应力理论&CAESARII的实施绪论3D梁单元的特征•无限薄的杆。•描述的所有行为都是根据端点的位移。•弯曲是粱单元的主要行为。基本应力理论&CAESARII的实施绪论3D梁单元的特征•仅说明了总体的行为。•没有考虑局部的作用(表面没有碰撞)。•忽略了二次影响。(使转角很小)•遵循Hook’s定律。基本应力理论&CAESARII的实施基本应力使用局部坐标系可以将管系应力(以及产生这些应力的载荷）theloadsthatcausethem)分为下面几种：纵向应力-SL环向应力-SH径向应力-SR剪切应力-基本应力理论&CAESARII的实施纵向应力分量•沿着管子的轴向。•轴向力–轴向力除以面积(F/A)•压力–Pd/4torP*di/(do2-di2)•弯曲力矩–Mc/I–最大应力发生在圆周的最外面。–I/半径Z(抗弯截面系数);使用M/Z基本应力理论&CAESARII的实施由于压力产生的环向应力•垂直于半径(圆周)•Pd/2t•再一次用薄壁的近似值。•环向应力很重要，尽管它不是“综合应力”的一部分。•环向应力根据直径、操作温度下的许用应力、腐蚀余量，加工偏差和压力用来定义管子的壁厚。•根据Barlow,Boardman,Lamé来计算。基本应力理论&CAESARII的实施由于压力产生的径向应力•垂直于表面。•内表面应力为-P。•外表面应力通常为0。•由于最大的弯曲应力发生在外表面，所以这一项被忽略。基本应力理论&CAESARII的实施剪切应力•平面内垂直于半径。•剪切力–这个载荷在外表面最小，因此在管系应力计算中省略了这一项。–在支撑处要求局部考虑。•扭矩–最大的应力发生在外表面。–MT/2Z基本应力理论&CAESARII的实施“综合应力”中的基本应力评价3-D应力•S=F/A+Pd/4t+M/Z•轴向、环向压力和纵向弯曲所产生的应力之和。•根据规范和载荷工况的不同上式将发生变化。基本应力理论&CAESARII的实施Basisfor“CodeStressEquations”失效理论•变形能或八面体剪切应力(根据米赛斯理论和其它的理论）。•最大剪应力理论（Columb理论）。–大多数理论都根据这个理论。•由于剪切影响而限制最大主应力(Rankine理论)。•CAESARII132列输出应力报告中显示了米赛斯或最大剪应力强度理论。•应力报告由configuration设置来决定。基本应力理论&CAESARII的实施规范要求的载荷工况•规范要求使用两个主要失效方式的失效理论。•一次失效。•二次失效。•(第三种失效方式是偶然失效，它与一次失效相似。）基本应力理论&CAESARII的实施规范要求的载荷工况一次失效情况•力所引起。•非自限性。•重量、压力和集中力所产生。基本应力理论&CAESARII的实施规范要求的载荷工况二次失效情况•位移所引起。•自限性。•温度、位移和其它变化载荷——例如，重力。基本应力理论&CAESARII的实施规范要求的载荷工况•(1)=W+T1+P1(OPE)•(2)=W+P1(SUS)•(3)=DS1-DS2(EXP)•操作工况,用于:–约束&设备载荷–最大位移–计算EXP工况•持续工况，用于一次载荷下规范应力的计算。•膨胀工况，用于“extremedisplacementstressrange”–工况3的位移是从工况1的位移减去工况2的位移而得到。基本应力理论&CAESARII的实施规范要求的载荷工况膨胀工况说明•Whatdoes“DS1-DS2(EXP)”mean?•Isaloadcasewith“T1(EXP)thesamething?基本应力理论&CAESARII的实施规范要求的载荷工况膨胀工况说明•Thecodestatesthattheexpansionstressesaretobecomputedfromtheextremedisplacementstressrange.Theseareallveryimportantwords.Considertheirmeaning…•EXTREME:Inthissenseitmeansthemost,orthelargest.•RANGE:Typicallyadifference.Whatdifference?Thedifferencebetweentheextremes.Whatextremes?•DISPLACEMENT:Thisdefineswhatextremestotakethedifferenceof.•STRESS:Whatweareeventuallyafter.基本应力理论&CAESARII的实施规范要求的载荷工况膨胀工况说明•Puttingeverythingbacktogether,wearetoldtocomputestressesfromtheextremedisplacementrange.Howcanwedothis?•Considertheequationbeingsolved;[K]{x}={f}.•Inthisequation,weknow[K]and{f},andwearesolvingfor{x},thedisplacementvector.•InCAESARII,whenwesetupanexpansioncase,wedefineitasDS1-DS2,wherethe1and2refertothedisplacementvector({x})ofloadcases1and2respectively.基本应力理论&CAESARII的实施规范要求的载荷工况膨胀工况说明•(Obviouslytheloadcasenumbersaresubjecttochangeonajobbyjobbasis.)•WhatdoyougetwhenyoutakeDS1-DS2?•Well{x1}-{x2}yields{x'},apseudodisplacementvector.•{x'}isnotarealsetofdisplacementsthatyoucangooutandmeasurewitharuler,ratheritisthedifferencebetweentwopositionsofthepipe.•Oncewehave{x'},wecanusethesameroutinesusedintheOPEorSUScasestocomputeelementforces,andfinallyelementstresses.基本应力理论&CAESARII的实施规范要求的载荷工况膨胀工况说明•However,theseelementforcesarealsopseudoforces,i.ethedifferenceinforcesbetweentwopositionsofthepipe.•Similarly,thestressescomputedarenotrealstresses,butstressdifferences.•Thisisexactlywhatthecodewants,thestressdifference,whichwascomputedfromadisplacementrange.•Astowhetherornotthisstressdifferenceistheextreme,wellthatdependsonthejob.基本应力理论&CAESARII的实施规范要求的载荷工况膨胀工况说明•Considerthequestionagain;IsDS1-DS2thesameasaloadcasewithjustT1?.•Theanswertothisismaybe.•Ifyouhavealinearsystem(fromaboundaryconditionpointofview),thentheanswerisyes.Youwillgetexactlythesameresults.•However,ifthesystemisnon-linear(i.e.youhave+Ys,orgaps,orfriction),thentheanswerisno.Youwillgetdifferentresults-howdifferentdependsonthejob.•Thereasonforthiscanbefoundbyexaminingtheequation[K]{x}={f}forthetwodifferentmethods.基本应力理论&CAESARII的实施规范要求的载荷工况膨胀工况说明•Forthisdiscussion,rearrangetheequationto{x}={f}/[K],whereweknowwedon'treallydivideby[K],wemultiplybyitsinverse.•OPE:{xope}={fope}/[Kope]={W+T1+P1}/[Kope]•SUS:{xsus}={fsus}/[Ksus]={W+P1}/[Ksus]•EXP:{xexp}={xope}-{xsus}={W+T1+P1}/[Kope]-{W+P1}/[Ksus]•Canwesimplifytheaboveequationasfollows?EXP:{xexp}={W+T1+P1}/[K]-{W+P1}/[K}基本应力理论&CAESARII的实施规范要求的载荷工况膨胀工况说明•Canwesimplifytheaboveequationasfollows?EXP:{xexp}={W+T1+P1}/[K]-{W+P1}/[K]•Cancelingliketerms(theonesinred)yields:{xexp}={T1}/[K]•Theassumptionhereisthat[Kope]isthesameas[Ksus].•Thisassumptionisonlytrueforlinearsystems.•Fornon-linearsystems,thestiffnessmatrixisuniqueforeachloadcaseandtheabovecancellationofloadingtermsisincorrect.•Yougetthewrongstressresultsfortheexpansioncaseifyousetuploadcasesthisway.基本应力理论&CAESARII的实施规范要求的载荷工况膨胀工况说明•AnotherproofthattheDS1-DS2methodisthecorrectwaytogoistoconsiderajobwithtwooperatingtemperatures,oneaboveambientandonebelowambient.•SayT1=+300,andT2=-50.CAESARIIwouldsetuploadcasesasfollows:–(1)W+T1+P1(OPE)–(2)W+T2+P1(OPE)–(3)W+P1(SUS)–(4)DS1-DS3(EXP)–(5)DS2-DS3(EXP)基本应力理论&CAESARII的实施规范要求的载荷工况膨胀工况说明•Thesecases,whilecorrect,don'taddresstheextremetermofthecoderequirements.ThisisbecauseCAESARIIisn'tlookingatwhattheloadcomponentsrepresent.•Tosatisfytherequirementsofthecode,theusermustdefineanadditionalloadcase:–(6)DS1-DS2(EXP)•Thisloadcasewillbetheextreme,thatwilltypicallygoverntheEXPstresscriteria.•Youcan'tdothisatallusingt