1、基本应力理论&CAESAR II 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&CAESAR II 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Hooks 定律。,基本应力理论&CAESAR II 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）the loads that cause them)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&CAESAR II 的实施,纵向应力分量,沿着
2、管子的轴向。轴向力轴向力除以面积(F/A)压力Pd/4t or P*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&CAESAR II 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,Lam来计算。,基本应力理论&CAESAR II 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
3、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&CAESAR II 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&CAESAR II 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&CAESAR II 的实施,Basis for“Code Stress Equations”,失效理论变形能或八面体剪切应力(根据米赛斯理论
4、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)。CAESAR II 132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&CAESAR II 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&CAESAR II 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&CAESAR II 的实施,规范要求
5、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&CAESAR II 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“extreme displacement stress range”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&CAESAR II 的实施,规范要求的载荷工况,膨胀工况说明What does“DS1-DS2(EXP)”mean?
6、Is a load case with“T1(EXP)the same thing?,基本应力理论&CAESAR II 的实施,规范要求的载荷工况,膨胀工况说明The code states that the expansion stresses are to be computed from the extreme displacement stress range.These are all very important words.Consider their meaning EXTREME:In this sense it means the most,or the largest.R
7、ANGE:Typically a difference.What difference?The difference between the extremes.What extremes?DISPLACEMENT:This defines what extremes to take the difference of.STRESS:What we are eventually after.,基本应力理论&CAESAR II 的实施,规范要求的载荷工况,膨胀工况说明Putting everything back together,we are told to compute stresses f
8、rom the extreme displacement range.How can we do this?Consider the equation being solved;K x=f.In this equation,we know K and f,and we are solving for x,the displacement vector.In CAESAR II,when we setup an expansion case,we define it as DS1-DS2,where the 1 and 2 refer to the displacement vector(x)o
9、f load cases 1 and 2 respectively.,基本应力理论&CAESAR II 的实施,规范要求的载荷工况,膨胀工况说明(Obviously the load case numbers are subject to change on a job by job basis.)What do you get when you take DS1-DS2?Well x1-x2 yields x,a pseudo displacement vector.x is not a real set of displacements that you can go out and me
10、asure with a ruler,rather it is the difference between two positions of the pipe.Once we have x,we can use the same routines used in the OPE or SUS cases to compute element forces,and finally element stresses.,基本应力理论&CAESAR II 的实施,规范要求的载荷工况,膨胀工况说明However,these element forces are also pseudo forces,i
11、.e the difference in forces between two positions of the pipe.Similarly,the stresses computed are not real stresses,but stress differences.This is exactly what the code wants,the stress difference,which was computed from a displacement range.As to whether or not this stress difference is the extreme
12、,well that depends on the job.,基本应力理论&CAESAR II 的实施,规范要求的载荷工况,膨胀工况说明Consider the question again;Is DS1-DS2 the same as a load case with just T1?.The answer to this is maybe.If you have a linear system(from a boundary condition point of view),then the answer is yes.You will get exactly the same resul
13、ts.However,if the system is non-linear(i.e.you have+Ys,or gaps,or friction),then the answer is no.You will get different results-how different depends on the job.The reason for this can be found by examining the equation K x=f for the two different methods.,基本应力理论&CAESAR II 的实施,规范要求的载荷工况,膨胀工况说明For t
14、his discussion,rearrange the equation to x=f/K,where we know we dont really divide by K,we multiply by its inverse.OPE:xope=fope/Kope=W+T1+P1/KopeSUS:xsus=fsus/Ksus=W+P1/KsusEXP:xexp=xope-xsus=W+T1+P1/Kope-W+P1/KsusCan we simplify the above equation as follows?EXP:xexp=W+T1+P1/K-W+P1/K,基本应力理论&CAESAR
15、 II 的实施,规范要求的载荷工况,膨胀工况说明Can we simplify the above equation as follows?EXP:xexp=W+T1+P1/K-W+P1/KCanceling like terms(the ones in red)yields:xexp=T1/KThe assumption here is that Kope is the same as Ksus.This assumption is only true for linear systems.For non-linear systems,the stiffness matrix is uniq
16、ue for each load case and the above cancellation of loading terms is incorrect.You get the wrong stress results for the expansion case if you setup load cases this way.,基本应力理论&CAESAR II 的实施,规范要求的载荷工况,膨胀工况说明Another proof that the DS1-DS2 method is the correct way to go is to consider a job with two o
17、perating temperatures,one above ambient and one below ambient.Say T1=+300,and T2=-50.CAESAR II would setup load cases as follows:(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&CAESAR II 的实施,规范要求的载荷工况,膨胀工况说明These cases,while correct,dont address the extreme term of th
18、e code requirements.This is because CAESAR II isnt looking at what the load components represent.To satisfy the requirements of the code,the user must define an additional load case:(6)DS1-DS2(EXP)This load case will be the extreme,that will typically govern the EXP stress criteria.You cant do this
19、at all using the T1 only method.,基本应力理论&CAESAR II 的实施,规范要求的载荷工况,膨胀工况说明To summarize:We take the difference between two load cases to determine a displacement range.From this range we compute a force range and then a stress range.The code requires the extreme displacement stress range.The user only ha
20、s to worry about whether or not the extreme case has been addressed.,基本应力理论&CAESAR II 的实施,线性 vs 非线性,这个术语指的是边界条件。方程重新被求解:Kx=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。,基本应力理论&CAESAR II 的实施,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。
21、,基本应力理论&CAESAR II 的实施,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。,基本应力理论&CAESAR II 的实施,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。,基本应力理论&CAESAR II 的实施,Linear vs Non-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。,基本应力理论&CAES
22、AR II 的实施,Occasional Load Case Setup,Occasional loads are considered“primary”,since they are force driven.Occasional loads occur infrequently.The codes employ an“allowable increase”factor based on the frequency of occurrence in the determination of the allowable,i.e.k*Sh.Examples of occasional loads
23、 are wind and earthquake.,基本应力理论&CAESAR II 的实施,Occasional Load Case Setup,The code equation for the OCCasional load case is:MA/Z+MB/Z kShHere,MA is the moment term from the SUStained loads,and MB is the moment from the OCCasional loads.This equation states that the OCCasional case is the sum of the
24、SUStained stresses and the OCCasional stresses.So we cant run a load case with just a“WIND”load and satisfy this code requirement.What about“W+P1+WIND”as a load case?,基本应力理论&CAESAR II 的实施,Occasional Load Case Setup,The“W+P1+WIND”case will work for“linear”systems only.For“non-linear”systems,this is n
25、ot sufficient,for the same reason“T1”is not sufficient for the EXPansion load case.The best way to setup OCCasional load 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),基本应力理论&CAESAR II 的实施,Occasional Load Case Setup,(1)W+P1+T1(OPE)(2)W+P1+T1+WIN
26、D(OPE)(3)W+P1(SUS)(4)DS1-DS3(EXP)(5)DS2-DS1(OPE)(6)ST5+ST3(OCC),This is the normal OPErating caseThis is a combined OPErating case which includes the OCC loadsThis is the standard SUStained caseThis is the standard EXPansion caseThis difference yields the effects of the OCCasional load on the system
27、.This is not a code case,only a construction case,therefore(OPE).This handles non-linearities.This is our OCCasional code compliance case,stresses from Primary plus Occasional loads.,基本应力理论&CAESAR II 的实施,Load Case Generation&Maintenance,CAESAR II will recommend load cases for“new”jobs.By“new”jobs,we
28、 mean jobs that do not have a“._J”file.For“old”jobs,having a“._J”file,CAESAR II reads in the defined load cases and presents them to the user.The load case editing screen is shown at the right.,基本应力理论&CAESAR II 的实施,Load Case Generation&Maintenance,CAESAR II will recommend load cases for“new”jobs.By“
29、new”jobs,we mean jobs that do not have a“._J”file.For“old”jobs,having a“._J”file,CAESAR II reads in the defined load cases and presents them to the user.The load case editing screen is shown at the right.,基本应力理论&CAESAR II 的实施,Load Case Generation&Maintenance,On this dialog,available load types are l
30、isted in the upper left list box.Available load case types are listed in the lower left list box.Load cases(recommended or previously defined)are shown in the grid at the right.Recommended load cases can always be obtained by clicking on the Recommend button.The analysis commences by clicking on“the
31、 running man”.,基本应力理论&CAESAR II 的实施,Load Case Generation&Maintenance,Say for a“new”job,the load cases at the right are recommended.Say you accept and run these load cases.Upon reviewing the output you discover that pre-defined displacements at node 5 were omitted.You return to input,add the displace
32、ments,and start the Static Analysis processor again.,基本应力理论&CAESAR II 的实施,Load Case Generation&Maintenance,CAESAR II reads these existing load cases and presents them.What will your results be if you run these load cases?Exactly the same as before,because these load cases dont include the predefined
33、 displacements.You must manually add“D1”to the OPE load case,or ask CAESAR II to re-recommend the load cases.,基本应力理论&CAESAR II 的实施,Load Case Generation&Maintenance,Notice the load type list in the upper left contains“D1”now.The corrected load cases are shown at the right.,基本应力理论&CAESAR II 的实施,Load Case Generation&Maintenance,Notice the load type list in the upper left contains“D1”now.The corrected
