CAESARII-管道应力分析-理论

1、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 & 绪论3D梁单元的的特征仅说明了了总体的的行为。没有考虑虑局部的的作用(表面面没有碰碰撞)。忽略了二二次影响响。(使转角很很小)遵循Hooks定律。基本应力力理论&基本应力力使用局部部坐标系系可以将将管系应应力(以及产产生这些些应力的的载荷）theloadsthatcausethem)分为下面面几种：纵向应力力 -SL环向应力力 -SH径向应力力 -SR剪切应力力 -基本应力力理论&纵向应力力分量沿着管子子的轴向向。轴向力轴向力除除以面积积(F/A)压力Pd/ 4torP*di/ (do

2、2- di2)弯曲力矩矩Mc/I最大应力力发生在在圆周的的最外面面。I/半径Z (抗弯截面面系数);使用用M/Z基本应力力理论&由于压力力产生的的环向应应力垂直于半半径(圆周)Pd/ 2t再一次用用薄壁的的近似值值。环向应力力很重要要，尽管管它不是是“综合合应力”的一部部分。环向应力力根据直直径、操操作温度度下的许许用应力力、腐蚀蚀余量，加工偏偏差和压压力用来来定义管管子的壁壁厚。根据Barlow, Boardman,Lam来计算。基本应力力理论&由于压力力产生的的径向应应力垂直于表表面。内表面应应力为-P。外表面应应力通常常为0。由于最大大的弯曲曲应力发发生在外外表面，所以这这一项被被忽略。

3、基本应力力理论&剪切应力力平面内垂垂直于半半径。剪切力这个载荷荷在外表表面最小小，因此此在管系系应力计计算中省省略了这这一项。在支撑处处要求局局部考虑虑。扭矩最大的应应力发生生在外表表面。MT/2Z基本应力力理论&“综合应力力”中的的基本应应力评价3-D应力S =F/ A+Pd /4t+M/Z轴向、环环向压力力和纵向向弯曲所所产生的的应力之之和。根据规范范和载荷荷工况的的不同上上式将发发生变化化。基本应力力理论&Basisfor“CodeStress Equations”失效理论论变形能或或八面体体剪切应应力(根据米赛赛斯理论论和其它它的理论论）。最大剪应应力理论论（Columb理论）。大多数

4、理理论都根根据这个个理论。由于剪切切影响而而限制最最大主应应力(Rankine理论)。CAESARII132列输出应应力报告告中显示示了米赛赛斯或最最大剪应应力强度度理论。应力报告告由configuration设置来决决定。基本应力力理论&规范要求求的载荷荷工况规范要求求使用两两个主要要失效方方式的失失效理论论。一次失效效。二次失效效。(第三种失失效方式式是偶然然失效，它与一一次失效效相似。）基本应力力理论&规范要求求的载荷荷工况一次失效效情况力所引起起。非自限性性。重量、压压力和集集中力所所产生。基本应力力理论&规范要求求的载荷荷工况二次失效效情况位移所引引起。自限性。温度、位位移和其其它变

5、化化载荷例如如，重力力。基本应力力理论&规范要求求的载荷荷工况(1)=W +T1+P1(OPE)(2)=W +P1(SUS)(3)=DS1-DS2(EXP)操作工况况,用用于:约束&设设备载载荷最大位移移计算EXP工况持续工况况，用于于一次载载荷下规规范应力力的计算算。膨胀工况况，用于于 “extreme displacementstressrange”工况3的的位移是是从工况况1的位位移减去去工况2的位移移而得到到。基本应力力理论&规范要求求的载荷荷工况膨胀工况况说明What does“DS1- DS2(EXP)”mean?Isa loadcasewith “T1(EXP)thesameth

6、ing?基本应力力理论&规范要求求的载荷荷工况膨胀工况况说明Thecodestatesthattheexpansion stresses aretobecomputedfrom theextremedisplacementstress range.These areallveryimportantwords.Considertheirmeaning EXTREME: In thissenseitmeansthe most, or thelargest.RANGE:Typicallyadifference.Whatdifference?The differencebetweentheextrem

7、es.What extremes?DISPLACEMENT:This defineswhatextremestotake thedifference of.STRESS:What we areeventually after.基本应力力理论&规范要求求的载荷荷工况膨胀工况况说明Putting everythingbacktogether,weare toldtocomputestressesfrom theextreme displacementrange.Howcanwedothis?Considertheequationbeingsolved;Kx =f.Inthis equation,w

8、eknow Kandf,andweare solvingfor x,the displacementvector.InCAESARII, whenwesetupanexpansion case, we defineitasDS1 -DS2,wherethe 1and2 refer to thedisplacementvector (x) of loadcases1and2respectively.基本应力力理论&规范要求求的载荷荷工况膨胀工况况说明(Obviously theload casenumbersaresubjecttochangeonajobbyjob basis.)What do

9、 yougetwhenyoutakeDS1 -DS2?Well x1-x2 yieldsx,a pseudodisplacementvector.x is nota realset of displacementsthatyoucan go outandmeasurewith aruler, ratheritisthedifferencebetween twopositionsofthe pipe.Once we havex,wecan usethesameroutinesused in theOPEorSUS cases to computeelementforces, andfinally

10、 elementstresses.基本应力力理论&规范要求求的载荷荷工况膨胀工况况说明However,theseelement forcesarealsopseudoforces,i.ethe differenceinforces betweentwo positionsofthepipe.Similarly, thestressescomputedarenot realstresses, butstressdifferences.This is exactlywhatthecodewants,the stressdifference,whichwascomputedfroma displac

11、ementrange.Astowhether or notthis stressdifference is theextreme,well thatdependsonthejob.基本应力力理论&规范要求求的载荷荷工况膨胀工况况说明Considerthequestionagain; IsDS1-DS2 thesame as aloadcase withjustT1?.Theanswer to thisismaybe.Ifyouhavealinearsystem(fromaboundaryconditionpointofview), thenthe answerisyes.Youwillgete

12、xactlythesameresults.However,ifthesystem isnon-linear(i.e.youhave+Ys, or gaps, or friction),thentheanswer is no.You willget differentresults -how differentdepends on thejob.Thereason forthis canbefoundbyexaminingthe equation Kx=ffor thetwodifferent methods.基本应力力理论&规范要求求的载荷荷工况膨胀工况况说明Forthisdiscussion

13、,rearrangethe equation to x= f/ K,whereweknowwedontreallydivide by K,wemultiplybyits inverse.OPE:xope =fope /Kope =W+T1+ P1/ KopeSUS:xsus =fsus /Ksus =W+P1/KsusEXP:xexp =xope -xsus =W+T1+ P1/ Kope-W+P1/KsusCanwesimplifythe above equation as follows?EXP:xexp= W +T1+P1/K-W+ P1/ K基本应力力理论&规范要求求的载荷荷工况膨胀工

14、况况说明Canwesimplifythe above equation as follows?EXP:xexp= W+ T1 +P1 /K -W+P1 /KCancelingliketerms(the onesinred)yields:xexp =T1/ KTheassumptionhere is thatKope is thesame as Ksus.This assumptionisonlytrue forlinearsystems.Fornon-linearsystems,thestiffness matrixisuniquefor eachloadcase andtheabovecan

15、cellation of loadingtermsisincorrect.Youget thewrongstressresultsforthe expansioncase if yousetupload cases thisway.基本应力力理论&规范要求求的载荷荷工况膨胀工况况说明Another proof thatthe DS1-DS2methodisthe correctway to go is to consider ajob withtwo operatingtemperatures, oneaboveambient andonebelowambient.SayT1=+300,and

16、T2=-50.CAESARIIwouldsetuploadcasesasfollows:(1)W+ T1 +P1(OPE)(2)W+ T2 +P1(OPE)(3)W+ P1 (SUS)(4)DS1 -DS3 (EXP)(5)DS2 -DS3 (EXP)基本应力力理论&规范要求求的载荷荷工况膨胀工况况说明Thesecases,whilecorrect,dontaddresstheextreme termofthe coderequirements.Thisisbecause CAESARIIisntlooking at whatthe loadcomponentsrepresent.Tosati

17、sfy therequirementsofthe code, theuser mustdefine an additionalloadcase:(6)DS1 -DS2 (EXP)This loadcasewill be theextreme, thatwilltypicallygovern theEXPstress criteria.Youcantdothisatallusingthe T1onlymethod.基本应力力理论&规范要求求的载荷荷工况膨胀工况况说明Tosummarize:Wetake thedifference betweentwo loadcasestodetermine a

18、displacement range.From thisrangewecomputea force range andthen astress range.Thecoderequirestheextremedisplacementstress range.Theuseronly hastoworryaboutwhether or nottheextreme casehas beenaddressed.基本应力力理论&线性vs非非线性这个术语语指的是是边界条条件。方程重新新被求解解:Kx= f这是弹簧簧方程。管系边界界条件（例如，约束）指的是是刚度或或弹簧。可以定义义更复杂杂的边界界条件，此时“

19、线性弹弹簧”的的假设将将不适用用。基本应力力理论&线性Vs非非线性线性边界界条件的的一个实实例是双双向约束束，例如如：“Y”向支撑。线性边界界条件的的另一个个实例是是弹簧支支吊架。这些约束束中力与与位移的的关系曲曲线是一一条直线线。所以这些些约束是是线性的的。直线的斜斜率为刚刚度。基本应力力理论&线性Vs非非线性“+Y”支撑是非非线性支支撑。力与位移移的关系系曲线不不是一直直线。刚度仅存存在于负负位移方方向。对于正位位移，刚刚度是零零。基本应力力理论&线性Vs非非线性“间隙”也也是一个个非线性性支撑。力与位移移的关系系曲线不不是一直直线。间隙中没没有刚度度。基本应力力理论&LinearvsNo

20、n-Linear摩擦使约约束成为为非线性性。大的旋转转杆也是是非线性性约束。文件中的的非线性性约束意意味着Kope不等于Ksus。使用两个个其它载载荷工况况之间的的差值来来建立(EXP)和 (OCC)载荷工况况来说明明非线性性约束。基本应力力理论&Occasional LoadCaseSetupOccasional loads areconsidered “primary”,sincetheyareforcedriven.Occasional loads occur infrequently.Thecodesemploy an “allowableincrease” factorbasedon

21、thefrequency of occurrenceinthe determinationofthe allowable,i.e.k* Sh.Examplesofoccasional loads arewind andearthquake.基本应力力理论&Occasional LoadCaseSetupThecodeequationforthe OCCasionalloadcase is:MA/ Z+MB/ ZkShHere,MAisthemoment termfromtheSUStained loads,andMBisthemoment fromthe OCCasionalloads.Thi

22、s equation statesthat theOCCasional caseisthe sumoftheSUStained stresses andtheOCCasionalstresses.Sowecantrunaload casewithjust a“WIND” loadand satisfythiscode requirement.What about “W +P1+WIND”asa loadcase?基本应力力理论&Occasional LoadCaseSetupThe“W+P1+ WIND” casewillwork for“linear”systems only.For“non

23、-linear” systems, thisisnot sufficient, forthesamereason“T1”isnotsufficientforthe EXPansionload case.ThebestwaytosetupOCCasionalload cases is:(1)W+ P1 +T1(OPE)(2)W+ P1 +T1+WIND (OPE)(3)W+ P1 (SUS)(4)DS1 -DS3 (EXP)(5)DS2 -DS1 (OPE)(6)ST5 +ST3 (OCC)基本应力力理论&Occasional LoadCaseSetup(1)W +P1+T1(OPE)(2)W+

24、 P1 +T1+WIND (OPE)(3)W+ P1 (SUS)(4)DS1 -DS3 (EXP)(5)DS2 -DS1 (OPE)(6)ST5 +ST3 (OCC)This is thenormalOPErating caseThis is acombinedOPErating casewhichincludesthe OCCloadsThis is thestandardSUStainedcaseThis is thestandardEXPansioncaseThis differenceyields theeffects of theOCCasional loadonthe system

25、.Thisisnot acodecase,only aconstruction case, therefore(OPE).Thishandles non-linearities.This is ourOCCasional codecompliancecase,stressesfrom PrimaryplusOccasional loads.基本应力力理论&Load CaseGeneration& MaintenanceCAESARIIwillrecommendloadcasesfor“new” jobs.By“new” jobs, we meanjobsthat do nothave a“._

26、J”file.For“old”jobs,having a“._J”file,CAESAR II reads in thedefined loadcasesand presents themtothe user.Theloadcase editingscreen is shown at theright.基本应力力理论&Load CaseGeneration& MaintenanceCAESARIIwillrecommendloadcasesfor“new” jobs.By“new” jobs, we meanjobsthat do nothave a“._J”file.For“old”jobs

27、,having a“._J”file,CAESAR II reads in thedefined loadcasesand presents themtothe user.Theloadcase editingscreen is shown at theright.基本应力力理论&Load CaseGeneration& MaintenanceOnthis dialog,available loadtypesare listedintheupperleftlist box.Availableloadcase types arelistedinthe lower leftlistbox.Load

28、 cases (recommendedorpreviously defined) areshowninthegridattheright.Recommendedload cases canalwaysbeobtainedbyclickingonthe Recommend button.Theanalysiscommences by clicking on “therunningman”.基本应力力理论&Load CaseGeneration& MaintenanceSayfor a“new”job,theloadcasesattherightare recommended.Sayyou acc

29、eptandrun these loadcases.Upon reviewingtheoutput youdiscoverthat pre-defined displacementsatnode5 wereomitted.Youreturn to input,addthe displacements, andstarttheStatic Analysis processoragain.基本应力力理论&Load CaseGeneration& MaintenanceCAESARIIreadstheseexistingloadcasesandpresentsthem.What willyourre

30、sults be if youruntheseloadcases?Exactly thesame as before,becausetheseload cases dont includethe predefineddisplacements.Youmustmanuallyadd“D1”totheOPE loadcase,orask CAESARIItore-recommendthe loadcases.基本应力力理论&Load CaseGeneration& MaintenanceNoticethe loadtypelist in theupperleft contains “D1”now.

31、Thecorrected loadcasesare shown at theright.基本应力力理论&Load CaseGeneration& MaintenanceNoticethe loadtypelist in theupperleft contains “D1”now.Thecorrected loadcasesare shown at theright.基本应力力理论&Load CaseGeneration& MaintenanceNoticethe loadtypelist in theupperleft contains “D1”now.Thecorrected loadcas

32、esare shown at theright.Anytimeyouadd or removea complete loadtype,the loadcasesare insufficient.Ifyouaddeddisplacementstonode 110,wouldthe loadcasesbesufficient?基本应力力理论&InsuringYouAnalyzeWhat YouThinkYoureAnalyzingRememberCAESARIIisafiniteelementprogram.RememberCAESARIIusesa 3D beamelement.Remember

33、youmusthave equilibrium:Resultantloadsshould equal appliedloadsGravity (weightonly)loadcase shouldequaltheweight of thesystemOtherbasicchecksVerifynodal3DcoordinatesCheckforextremedisplacements and/orloads(see handout)基本应力力理论&问题解决决当不满意意结果时时，你应应做什么么？重新求解解方程：Kx= f其中我们们求解的的 x是位移。由这些位位移，我我们可以以计算单单元力&力矩。

34、由这些力力&力矩，使使用规范范方程计计算出应应力。基本应力力理论&问题解决决当不满意意结果时时，你应应做什么么？如果是应应力问题题，它可可能是由由于下面面两个问问题引起起的：与规范有有关的问问题(SIFs、规范方程程等等）极限力和和/或力力矩如果是力力/力矩矩问题，它可能能是由下下面两个个问题所所引起：不正确的的单元特特性极限位移移基本应力力理论&Problem SolvingWhat do youdowhen youdontlike theresults?Ifyouhavea displacementproblem,itcanonlybecausedbytwo things:Improperinput(density,elasticmodulus,applied loads)ImproperboundaryconditionsDontforgettocheckand recheckthe input.Remembert