CAESAR_II_管道應力分析_理論

1、CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施緒論緒論3D 梁單元的特征梁單元的特征 無限薄的桿。 描述的所有行為都是根據(jù)端點的位移。 彎曲是粱單元的主要行為。CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施緒論緒論3D 梁單元的特征梁單元的特征僅說明了總體的行為。沒有考慮局部的作用 (表面沒有碰撞)。忽略了二次影響。(使轉角很小)遵循Hooks 定律。CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的

2、實施基本應力基本應力 使用局部坐標系可以將管系應力使用局部坐標系可以將管系應力 (以及產(chǎn)生這些應以及產(chǎn)生這些應力的載荷）力的載荷）the loads that cause them) 分為下面幾種：分為下面幾種： 縱向應力 - SL 環(huán)向應力 - SH 徑向應力 - SR 剪切應力 - CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施縱向應力分量縱向應力分量 沿著管子的軸向。 軸向力 軸向力除以面積 (F/A) 壓力 Pd / 4t or P*di / ( do2 - di2 ) 彎曲力矩 Mc/I 最大應力發(fā)生在圓周的最外面。

3、 I/半徑 Z (抗彎截面系數(shù));使用 M/ZCAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施由于壓力產(chǎn)生的環(huán)向應力由于壓力產(chǎn)生的環(huán)向應力 垂直于半徑 (圓周) Pd / 2t 再一次用薄壁的近似值。 環(huán)向應力很重要，盡管它不是“綜合應力”的一部分。 環(huán)向應力根據(jù)直徑、操作溫度下的許用應力、腐蝕余量，加工偏差和壓力用來定義管子的壁厚。 根據(jù)Barlow, Boardman, Lam來計算。CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施由于壓力產(chǎn)生的徑向應力由于壓

4、力產(chǎn)生的徑向應力 垂直于表面。 內(nèi)表面應力為 -P。 外表面應力通常為 0。 由于最大的彎曲應力發(fā)生在外表面，所以這一項被忽略。CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施剪切應力剪切應力 平面內(nèi)垂直于半徑。 剪切力 這個載荷在外表面最小，因此在管系應力計算中省略了這一項。 在支撐處要求局部考慮。 扭矩 最大的應力發(fā)生在外表面。 MT/2ZCAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施 “綜合應力綜合應力”中的基本應力中的基本應力評價評價 3-D 應力應力

5、S = F / A + Pd / 4t + M / Z 軸向、環(huán)向壓力和縱向彎曲所產(chǎn)生的應力之和。 根據(jù)規(guī)范和載荷工況的不同上式將發(fā)生變化。CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施Basis for “Code Stress Equations”失效理論失效理論 變形能或八面體剪切應力 (根據(jù)米賽斯理論和其它的理論）。 最大剪應力理論 （Columb理論） 。 大多數(shù)理論都根據(jù)這個理論。 由于剪切影響而限制最大主應力 (Rankine理論) 。 CAESAR II 132列輸出應力報告中顯示了米賽斯或最大剪應力強度理論。

6、 應力報告由configuration設置來決定。CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施規(guī)范要求的載荷工況規(guī)范要求的載荷工況 規(guī)范要求使用兩個主要失效方式的失效理論。 一次失效。 二次失效。 (第三種失效方式是偶然失效，它與一次失效相似。）CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施規(guī)范要求的載荷工況規(guī)范要求的載荷工況一次失效情況一次失效情況 力所引起。 非自限性。 重量、壓力和集中力所產(chǎn)生。CAESAR_II_管道應力分析_理論 & CAESAR

7、II 的實施基本應力理論 & CAESAR II 的實施規(guī)范要求的載荷工況規(guī)范要求的載荷工況二次失效情況二次失效情況 位移所引起。 自限性。 溫度、位移和其它變化載荷例如，重力。CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施規(guī)范要求的載荷工況規(guī)范要求的載荷工況 (1) = W + T1 + P1 (OPE) (2) = W + P1 (SUS) (3) = DS1 - DS2 (EXP) 操作工況, 用于: 約束& 設備載荷 最大位移 計算 EXP 工況 持續(xù)工況，用于一次載荷下規(guī)范應力的計算。 膨脹工況，用于 “extrem

8、e displacement stress range” 工況3的位移是從工況1的位移減去工況2的位移而得到。CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施規(guī)范要求的載荷工況規(guī)范要求的載荷工況膨脹工況說明膨脹工況說明 What does “DS1 - DS2 (EXP)” mean? Is a load case with “T1 (EXP) the same thing?CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施規(guī)范要求的載荷工況規(guī)范要求的載荷工況膨脹工

9、況說明膨脹工況說明 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. RANGE: Typically a difference. What difference? The difference bet

10、ween the extremes. What extremes? DISPLACEMENT: This defines what extremes to take the difference of. STRESS: What we are eventually after.CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施規(guī)范要求的載荷工況規(guī)范要求的載荷工況膨脹工況說明膨脹工況說明 Putting everything back together, we are told to compute stresses from th

11、e 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 v

12、ector (x) of load cases 1 and 2 respectively.CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施規(guī)范要求的載荷工況規(guī)范要求的載荷工況膨脹工況說明膨脹工況說明 (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 ve

13、ctor. x is not a real set of displacements that you can go out and measure 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.CAESAR_II_管道應力分析_理

14、論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施規(guī)范要求的載荷工況規(guī)范要求的載荷工況膨脹工況說明膨脹工況說明 However, these element forces are also pseudo forces, i.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 wa

15、nts, the stress difference, which was computed from a displacement range. As to whether or not this stress difference is the extreme, well that depends on the job.CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施規(guī)范要求的載荷工況規(guī)范要求的載荷工況膨脹工況說明膨脹工況說明 Consider the question again; Is DS1-DS2 the same

16、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 results. However, if the system is non-linear (i.e. you have +Ys, or gaps, or friction), then the answer is no. You

17、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.CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施規(guī)范要求的載荷工況規(guī)范要求的載荷工況膨脹工況說明膨脹工況說明 For this discussion, rearrange the equation to x = f

18、 / K, where we know we dont really divide by K, we multiply by its inverse. 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 Can we simplify the above equation as follows? EXP: xexp = W + T1 + P1 / K - W

19、 + P1 / KCAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施規(guī)范要求的載荷工況規(guī)范要求的載荷工況膨脹工況說明膨脹工況說明 Can we simplify the above equation as follows? EXP: xexp = W + T1 + P1 / K - W + P1 / K Canceling like terms (the ones in red) yields: xexp = T1 / K The assumption here is that Kope is the same as Ksus.

20、This assumption is only true for linear systems. For non-linear systems, the stiffness matrix is unique 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.CAESAR_II_管道應力分析_理論 & CAESAR I

21、I 的實施基本應力理論 & CAESAR II 的實施規(guī)范要求的載荷工況規(guī)范要求的載荷工況膨脹工況說明膨脹工況說明 Another proof that the DS1-DS2 method is the correct way to go is to consider a job with two operating 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

22、 + P1 (OPE) (2) W + T2 + P1 (OPE) (3) W + P1 (SUS) (4) DS1 - DS3 (EXP) (5) DS2 - DS3 (EXP)CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施規(guī)范要求的載荷工況規(guī)范要求的載荷工況膨脹工況說明膨脹工況說明 These cases, while correct, dont address the extreme term of the code requirements. This is because CAESAR II isnt looking

23、 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 at all using the T1 only method.CAESAR_II_管道應力分析

24、_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施規(guī)范要求的載荷工況規(guī)范要求的載荷工況膨脹工況說明膨脹工況說明 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 o

25、nly has to worry about whether or not the extreme case has been addressed.CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施線性 vs 非線性 這個術語指的是邊界條件。 方程重新被求解: Kx = f 這是彈簧方程。 管系邊界條件（例如，約束）指的是剛度或彈簧。 可以定義更復雜的邊界條件，此時“線性彈簧”的假設將不適用。CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施線性 Vs 非線性 線性邊

26、界條件的一個實例是雙向約束，例如：“Y”向支撐。 線性邊界條件的另一個實例是彈簧支吊架。 這些約束中力與位移的關系曲線是一條直線。 所以這些約束是線性的。 直線的斜率為剛度。CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施線性 Vs 非線性 “+Y” 支撐是非線性支撐。 力與位移的關系曲線不是一直線。 剛度僅存在于負位移方向。 對于正位移，剛度是零。CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施線性 Vs 非線性 “間隙”也是一個非線性支撐。 力與位移的關系曲線

27、不是一直線。 間隙中沒有剛度。CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施Linear vs Non-Linear 摩擦使約束成為非線性。 大的旋轉桿也是非線性約束。 文件中的非線性約束意味著 Kope 不等于 Ksus。 使用兩個其它載荷工況之間的差值來建立(EXP) 和 (OCC) 載荷工況來說明非線性約束。CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施Occasional Load Case Setup Occasional loads are con

28、sidered “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 are wind and earthquake.CAESAR_II_管道應力分析_理論 & CAE

29、SAR II 的實施基本應力理論 & CAESAR II 的實施Occasional Load Case Setup The code equation for the OCCasional load case is:MA / Z + MB / Z kSh Here, 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 SUSta

30、ined 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?CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施Occasional Load Case Setup The “W + P1 + WIND” case will work for “l(fā)inea

31、r” systems only. For “non-linear” systems, this is not 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 + S

32、T3 (OCC)CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施Occasional Load Case Setup (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) This is the normal OPErating case This is a combined OPErating case which includ

33、es the OCC loads This is the standard SUStained case This is the standard EXPansion case This difference yields the effects of the OCCasional load on the system. This is not a code case, only a construction case, therefore (OPE). This handles non-linearities. This is our OCCasional code compliance c

34、ase, stresses from Primary plus Occasional loads.CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施Load Case Generation & Maintenance CAESAR II will recommend load cases for “new” jobs. By “new” jobs, we mean jobs that do not have a “._J” file. For “old” jobs, having a “._J” file, CAESAR II r

35、eads in the defined load cases and presents them to the user. The load case editing screen is shown at the right.CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施Load Case Generation & Maintenance CAESAR II will recommend load cases for “new” jobs. By “new” jobs, we mean jobs that do not hav

36、e 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.CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施Load Case Generation & Maintenance On this dialog, available load

37、types are listed 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 b

38、y clicking on “the running man”.CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施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

39、 at node 5 were omitted. You return to input, add the displacements, and start the Static Analysis processor again.CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施Load Case Generation & Maintenance CAESAR II reads these existing load cases and presents them. What will your results be if you

40、 run these load cases? Exactly the same as before, because these load cases dont include the predefined displacements. You must manually add “D1” to the OPE load case, or ask CAESAR II to re-recommend the load cases.CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施Load Case Generation & Main

41、tenance Notice the load type list in the upper left contains “D1” now. The corrected load cases are shown at the right.CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施Load Case Generation & Maintenance Notice the load type list in the upper left contains “D1” now. The corrected load cases a

42、re shown at the right.CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施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. Any time you add or remove a complete load type, the load cases are insuff

43、icient. If you added displacements to node 110, would the load cases be sufficient?CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施Insuring You Analyze What You Think Youre Analyzing Remember CAESAR II is a finite element program. Remember CAESAR II uses a 3D beam element. Remember you must

44、 have equilibrium: Resultant loads should equal applied loads Gravity (weight only) load case should equal the weight of the system Other basic checks Verify nodal 3D coordinates Check for extreme displacements and/or loads (see handout)CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施問題解決當不

45、滿意結果時，你應做什么？當不滿意結果時，你應做什么？ 重新求解方程：Kx = f 其中我們求解的 x是位移。 由這些位移，我們可以計算單元力& 力矩。 由這些力 & 力矩，使用規(guī)范方程計算出應力。CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施問題解決當不滿意結果時，你應做什么？當不滿意結果時，你應做什么？ 如果是應力問題，它可能是由于下面兩個問題引起的： 與規(guī)范有關的問題 (SIFs、規(guī)范方程等等） 極限力和/或力矩 如果是力/力矩問題，它可能是由下面兩個問題所引起： 不正確的單元特性 極限位移CAESAR_II_管道應力分析_理論 & CAESAR II 的實施基本應力理論 & CAESAR II 的實施Problem SolvingWhat do you do when you dont like the results? If y