CAESARII_管道应力分析_理论.ppt

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 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 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 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 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 Lam 来计算 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 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 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 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 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 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 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 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 规范要求的载荷工况 膨胀工况说明Whatdoes DS1 DS2 EXP mean Isaloadcasewith T1 EXP thesamething 规范要求的载荷工况 膨胀工况说明Thecodestatesthattheexpansionstressesaretobecomputedfromthe extremedisplacementstressrange Theseareallveryimportantwords Considertheirmeaning EXTREME Inthissenseitmeansthemost orthelargest RANGE Typicallyadifference Whatdifference Thedifferencebetweentheextremes Whatextremes DISPLACEMENT Thisdefineswhatextremestotakethedifferenceof STRESS Whatweareeventuallyafter 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 wedefineitas DS1 DS2 wherethe 1 and 2 refertothedisplacementvector x ofloadcases1and2respectively 规范要求的载荷工况 膨胀工况说明 Obviouslytheloadcasenumbersaresubjecttochangeonajobbyjobbasis Whatdoyougetwhenyoutake DS1 DS2 Well x1 x2 yields x apseudodisplacementvector x isnotarealsetofdisplacementsthatyoucangooutandmeasurewitharuler ratheritisthedifferencebetweentwopositionsofthepipe Oncewehave x wecanusethesameroutinesusedintheOPEorSUScasestocomputeelementforces andfinallyelementstresses 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 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 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 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 规范要求的载荷工况 膨胀工况说明Anotherproofthatthe DS1 DS2 methodisthecorrectwaytogoistoconsiderajobwithtwooperatingtemperatures 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 规范要求的载荷工况 膨胀工况说明Thesecases whilecorrect don taddressthe extreme termofthecoderequirements ThisisbecauseCAESARIIisn tlookingatwhattheloadcomponentsrepresent Tosatisfytherequirementsofthecode theusermustdefineanadditionalloadcase 6 DS1 DS2 EXP Thisloadcasewillbethe extreme thatwilltypicallygoverntheEXPstresscriteria Youcan tdothisatallusingthe T1 onlymethod 规范要求的载荷工况 膨胀工况说明Tosummarize Wetakethedifferencebetweentwoloadcasestodetermineadisplacementrange Fromthisrangewecomputeaforcerangeandthenastressrange Thecoderequirestheextremedisplacementstressrange Theuseronlyhastoworryaboutwhetherornotthe extreme casehasbeenaddressed 线性vs非线性 这个术语指的是边界条件 方程重新被求解 K x f 这是弹簧方程 管系边界条件 例如 约束 指的是刚度或弹簧 可以定义更复杂的边界条件 此时 线性弹簧 的假设将不适用 线性Vs非线性 线性边界条件的一个实例是双向约束 例如 Y 向支撑 线性边界条件的另一个实例是弹簧支吊架 这些约束中力与位移的关系曲线是一条直线 所以这些约束是线性的 直线的斜率为刚度 线性Vs非线性 Y 支撑是非线性支撑 力与位移的关系曲线不是一直线 刚度仅存在于负位移方向 对于正位移 刚度是零 线性Vs非线性 间隙 也是一个非线性支撑 力与位移的关系曲线不是一直线 间隙中没有刚度 LinearvsNon Linear 摩擦使约束成为非线性 大的旋转杆也是非线性约束 文件中的非线性约束意味着 Kope 不等于 Ksus 使用两个其它载荷工况之间的差值来建立 EXP 和 OCC 载荷工况来说明非线性约束 OccasionalLoadCaseSetup Occasionalloadsareconsidered primary sincetheyareforcedriven Occasionalloadsoccurinfrequently Thecodesemployan allowableincrease factorbasedonthefrequencyofoccurrenceinthedeterminationoftheallowable i e k Sh Examplesofoccasionalloadsarewindandearthquake OccasionalLoadCaseSetup ThecodeequationfortheOCCasionalloadcaseis MA Z MB Z kShHere MAisthemomenttermfromtheSUStainedloads andMBisthemomentfromtheOCCasionalloads ThisequationstatesthattheOCCasionalcaseisthesumoftheSUStainedstressesandtheOCCasionalstresses Sowecan trunaloadcasewithjusta WIND loadandsatisfythiscoderequirement Whatabout W P1 WIND asaloadcase OccasionalLoadCaseSetup The W P1 WIND casewillworkfor linear systemsonly For non linear systems thisisnotsufficient forthesamereason T1 isnotsufficientfortheEXPansionloadcase ThebestwaytosetupOCCasionalloadcasesis 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 OccasionalLoadCaseSetup 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 ThisisthenormalOPEratingcaseThisisacombinedOPEratingcasewhichincludestheOCCloadsThisisthestandardSUStainedcaseThisisthestandardEXPansioncaseThisdifferenceyieldstheeffectsoftheOCCasionalloadonthesystem Thisisnotacodecase onlyaconstructioncase therefore OPE Thishandlesnon linearities ThisisourOCCasionalcodecompliancecase stressesfromPrimaryplusOccasionalloads LoadCaseGeneration Maintenance CAESARIIwillrecommendloadcasesfor new jobs By new jobs wemeanjobsthatdonothavea J file For old jobs havinga J file CAESARIIreadsinthedefinedloadcasesandpresentsthemtotheuser Theloadcaseeditingscreenisshownattheright LoadCaseGeneration Maintenance CAESARIIwillrecommendloadcasesfor new jobs By new jobs wemeanjobsthatdonothavea J file For old jobs havinga J file CAESARIIreadsinthedefinedloadcasesandpresentsthemtotheuser Theloadcaseeditingscreenisshownattheright LoadCaseGeneration Maintenance Onthisdialog availableloadtypesarelistedintheupperleftlistbox Availableloadcasetypesarelistedinthelowerleftlistbox Loadcases recommendedorpreviouslydefined areshowninthegridattheright Recommendedloadcasescanalwaysbeobtainedbyclickingonthe Recommend button Theanalysiscommencesbyclickingon therunningman LoadCaseGeneration Maintenance Sayfora new job theloadcasesattherightarerecommended Sayyouacceptandruntheseloadcases Uponreviewingtheoutputyoudiscoverthatpre defineddisplacementsatnode5wereomitted Youreturntoinput addthedisplacements andstarttheStaticAnalysisprocessoragain LoadCaseGeneration Maintenance CAESARIIreadstheseexistingloadcasesandpresentsthem Whatwillyourresultsbeifyouruntheseloadcases Exactlythesameasbefore becausetheseloadcasesdon tincludethepredefineddisplacements Youmustmanuallyadd D1 totheOPEloadcase oraskCAESARIItore recommendtheloadcases LoadCaseGeneration Maintenance Noticetheloadtypelistintheupperleftcontains D1 now Thecorrectedloadcasesareshownattheright LoadCaseGeneration Maintenance Noticetheloadtypelistintheupperleftcontains D1 now Thecorrectedloadcasesareshownattheright LoadCaseGeneration Maintenance Noticetheloadtypelistintheupperleftcontains D1 now Thecorrectedloadcasesareshownattheright Anytimeyouaddorremoveacompleteloadtype theloadcasesareinsufficient Ifyouaddeddisplacementstonode110 wouldtheloadcasesbesufficient InsuringYouAnalyzeWhatYouThinkYou reAnalyzing RememberCAESARIIisafiniteelementprogram RememberCAESARIIusesa3Dbeamelement Rememberyoumusthaveequilibrium ResultantloadsshouldequalappliedloadsGravity weightonly loadcaseshouldequaltheweightofthesystemOtherbasicchecksVerifynodal3DcoordinatesCheckforextremedisplacementsand orloads seehandout 问题解决 当不满意结果时 你应做什么 重新求解方程 K x f 其中我们求解的 x 是位移 由这些位移 我们可以计算单元力 力矩 由这些力 力矩 使用规范方程计算出应力 问题解决 当不满意结果时 你应做什么 如果是应力问题 它可能是由于下面两个问题引起的 与规范有关的问题 SIFs 规范方程等等 极限力和 或力矩如果是力 力矩问题 它可能是由下面两个问题所引起 不正确的单元特性极限位移 ProblemSolving Whatdoyoudowhenyoudon tliketheresults Ifyouhaveadisplacementproblem itcanonlybecausedbytwothings Improperinput density elasticmodulus appliedloads ImproperboundaryconditionsDon tforge