CAESARII官方培训讲义

1、CAESARII官方培训讲义2022-4-25介介 绍绍 培训的目的在于让您了解和掌握 应力分析的基础概念 模型和边界条件的建立 结果的分析和评判 往复压缩机的分析专题 日常遇到的问题和解决方法2022-4-25介介 绍绍3D梁单元的特征梁单元的特征 无限细的杆单元 全部行为靠端点位移决定 弯曲变形是主要的2022-4-25介介 绍绍3D Beam Element Characteristics3D梁单元的特征 仅说明整体行为 无局部作用（表面没有碰撞） 忽略二次影响（小转动） 遵循胡克定律2022-4-25Stress Basics应力基础应力基础局部坐标系下管道应力分类（引发应力的载荷）局

2、部坐标系下管道应力分类（引发应力的载荷）1. 轴向应力轴向应力Longitudinal Stress - SL2. 环向应力Hoop Stress - SH3. 径向应力Radial Stress - SR4. 剪切应力Shear Stress - 2022-4-25轴向应力轴向应力 沿管道轴向Along axis of pipe 轴向力引起Axial Force 轴向力/面积 (F/A) 内压引起Pressure Pd / 4t or P*di / ( do2 - di2 ) 弯矩引起Bending Moment Mc/I 最大应力环向的外表面某点处 I/radius Z (截面模量); u

3、se M/Z2022-4-25压力引起的环向应力压力引起的环向应力 环向（垂直于半径） Pd / 2t 和壁厚紧密相关 环向应力十分重要，但规范应力不考虑它。Hoop is very important, its just not part of the “code stress” 环向应力用来确定壁厚：依据直径、许用应力、腐蚀裕量、加工偏差、压力确定管道壁厚。2022-4-25压力引发的径向应力压力引发的径向应力 沿半径方向向内 内壁的径向应力大小是： -P 外壁的径向应力大小为 0 最大弯曲应力发生在管道的外表面，故该项忽略2022-4-25剪切应力剪切应力Shear Stresses 平

4、面内垂直半径 Shear Force剪力 在外表面剪力很小，应力计算忽略 支架设计有时需要考虑 Torque扭矩 最大应力在外表面 MT/2Z2022-4-253-D 应力评定应力评定 A loaded, 3-D pipe contains a representative infinitesimal stress cube add graphic (Fig 1-13) This stress cube is in equilibrium and can be rotated in space add graphic (rotated cube with loads) This cube ca

5、n be rotated so that shear stresses are zero. This results in the Principal Stresses.2022-4-25Simplifying to a 2-D Stress This plane can be rotated to either eliminate or maximize shear stress by using Mohrs Circle:Since we use the outside surface where radial stress is zero; lets move to a plane el

6、ement:2022-4-25Using Mohrs Circle Cut the square at to calculate S1Cut the square at +90 to calculate S2Cut the square at +45 to calculate max2022-4-25Using Mohrs Circle Brittle material (failure by fracture) - max principal stress Ductile material (failure by general yielding) - max principal stres

7、s is used to set wall thickness Maximum shear stress is a good prediction and errs on the conservative side see p84&85 of Adv. Mech. Of Matls2022-4-25基本应力基本应力 “Code Stress规范应力规范应力”应力评定应力评定Evaluating a 3-D Stress S = F / A + Pd / 4t + M / Z 轴向力、轴向压力，轴向弯矩一起的分量加和 规范不同，上面的算式也不同 那些应力没有包含进来？2022-4-25基本应力基

8、本应力 “Code Stress规范应力规范应力”几个实效理论几个实效理论A Few Failure Theories 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) 。 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。2022-4-25基本应力基本应力 “Code Stress规范应力规范应力”基于最大剪应力实效理论，ASME规范委员会颁布了规范应力方程Based on the ma

9、ximum shear failure theory, the Code Committee developed the “code equations” 目的在于避免管道系统实效Purpose was to reduce system failures这种解决办法很实用，但仍然有问题存在This approach worked well, but there were still problems, even as late as post World War II.研究表明直管道比较符合理论Studies showed systems of straight pipe matched th

10、eory研究表明元件失效比理论发生的早Studies showed systems with fittings failed earlier than theory predicted.ASME规范委员会委托Markl来研究这个问题Code Committee commissioned Markl to study this .2022-4-25基本应力基本应力 “Code Stress规范应力规范应力”Markls 试验和结果试验和结果将试验用的管道充满水，按某个方向和位移反复摇晃管道。Test configurations filled with water and cycled thro

11、ugh a predetermined displacement预测失效循环次数Theory should be able to predict “cycles to failure”发现最先失效的管件及其原因Fittings caused early failure because 对管件引入应力集中Stresses concentrations are introduced by fittings分析试验数据，修正轴向应力弯曲项Test data analyzed and a modification to the bending term of the code stress equat

12、ion was introduced:Sbending = i M / z2022-4-25基本应力基本应力 “Code Stress规范应力规范应力”Markls 试验和结果试验和结果 应力增强系数i 和元件的形式有关 对于弯头 “i” 的计算需要如下: 我们需要弯头的几何参数 计算弯头柔性“h” 计算应力增大系数Stress Intensification Factor “i”, 石化规范对平面内、外的SIF取值不同，电力取相同的sif2022-4-25基本应力基本应力 “Code Stress规范应力规范应力”Markls 试验和结果试验和结果 A load “in the plane”

13、 of the fitting causes “in-plane” bending 平面内 A load “out of the plane” of the fitting causes “out-of-plane” bending 平面外2022-4-25基本应力基本应力 “Code Stress规范应力规范应力”Markls 试验和结果试验和结果 规范上的附注十分重要 PetroChem codes modify SIF (and flexibility factor) based on pressure stiffening in a note石化规范规定压力硬化影响和柔性系数 应力算式

14、变化如下S = F / A + Pd / 4t + i M / z 应力增大系数不能小于2022-4-25Basis for “Code Stress Equations”Markls 试验和结果试验和结果The SIF is a “fudge” factorSIF 是个近似的参数The SIF attempts to increase the bending stress computed at the node point, to match the actual higher stress due to the stress concentration caused by the fit

15、ting.引入SIF 在于改变特殊管件应力集中，让他们的应力根实际大小更接近。Markl only tested 4x4 Std fittings !但Markl 仅测试了Additional work is still being done today in the field of SIFs. Results are published in: PVP, WRC, Journal of Pressure Vessel Technology.其余的工作人们仍然在继续进行。2022-4-25规范效验的工况规范效验的工况两种失效： Primary failure一次失效 Secondary fa

16、ilure二次失效 (A third failure mode addressed is Occasional, which is similar to Primary.)2022-4-25规范效验的工况规范效验的工况Primary Failure Case一次失效 力的作用Force Driven 非自限性Not Self-Limiting 重量、压力、集中力Weight, Pressure, Concentrated Forces2022-4-25规范效验的工况规范效验的工况Primary Failure Case一次失效 力的作用Force Driven 非自限性Not Self-Lim

17、iting 重量、压力、集中力, Weight、Pressure, Concentrated Forces2022-4-25规范效验的工况规范效验的工况Secondary Failure Case二次失效 位移作用Displacement Driven 自限性Is Self-Limiting 温度、位移和其他变化载荷引起的 Temperature, Displacement, plus other varying loads - i.e. weight2022-4-25规范效验的工况规范效验的工况 (1) = W + T1 + P1 (OPE) (2) = W + P1 (SUS) (3) =

18、 DS1 - DS2 (EXP) Operating case, used for:热态restraint & equipment loads推力和弯矩maximum displacements最大位移computation of EXP case计算二次应力Sustained case for PRIMARY loads and stress compliance计算一次应力 Expansion case for “extreme displacement stress range”膨胀工况，计算二次应力displacements for case 3 are displacements f

19、rom case 1 minus displacements from case 22022-4-25规范效验的工况规范效验的工况膨胀工况的解释膨胀工况的解释Expansion Case Explained What does “DS1 - DS2 (EXP)” mean? Is a load case with “T1 (EXP) the same thing?2022-4-25Load Cases for Code Compliance膨胀工况的解释膨胀工况的解释Expansion Case Explained The code states that the expansion stre

20、sses are to be computed from the extreme displacement stress range. These are all very important words. Consider their meaning EXTREME极端极端: In this sense it means the most, or the largest. RANGE范围范围: Typically a difference. What difference? The difference between the extremes. What extremes? DISPLAC

21、EMENT位移位移: This defines what extremes to take the difference of. STRESS应力应力: What we are eventually after.2022-4-25Load Cases for Code Compliance膨胀工况的解释膨胀工况的解释Expansion Case Explained Putting everything back together, we are told to compute stresses from the extreme displacement range. How can we do

22、 this?计算最大位移范围的应力 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) of load cases 1 and 2

23、respectively.2022-4-25Load Cases for Code Compliance膨胀工况的解释膨胀工况的解释Expansion Case Explained (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

24、 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.2022-4-25Load Cases for Code Compliance膨胀工况的解释膨胀工况的解释Expansi

25、on Case Explained However, these element forces are also pseudo forces, i.e the difference in forces between two positions of the pipe.力的大小是两个工况力的差值 Similarly, the stresses computed are not real stresses, but stress differences.应力不是真实应力，是应力的差值 This is exactly what the code wants, the stress differen

26、ce, which was computed from a displacement range.二次应力是位移变化量导致的 As to whether or not this stress difference is the extreme, well that depends on the job.2022-4-25Load Cases for Code Compliance膨胀工况的解释膨胀工况的解释Expansion Case Explained DS1-DS2 和 T1“一样吗？. 有可能. 如果是线性系统，答案是一样的。 如果是非线性系统 (如你有 +Ys, or gaps, or

27、 friction), 答案是不一样的。 原因是两个工况应用K x = f。 The reason for this can be found by examining the equation K x = f for the two different methods.2022-4-25Load Cases for Code ComplianceExpansion Case Explained For this discussion, rearrange the equation to x = f / K, where we know we dont really divide by K,

28、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 + P1 / K2022-4-25Load Cases for Code Compliance

29、膨胀工况解释膨胀工况解释Expansion Case Explained 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 问题在于Kope 和 Ksus是否相等. 线性系统相等. 非线性系统不相等2022-4-25Load Cases for Code Compliance膨胀工况解释膨胀工况解释Expansion Case Explained 如

30、果一个系统有两个操作温度。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. 如 T1 = +300, and T2 = -50. CAESAR II 软件自动建立如下工况:(1) W + T1 + P1 (OPE) (2) W + T2 + P1 (OPE)(3) W + P1 (SUS) (4) DS1 - DS3 (EXP)(5

31、) DS2 - DS3 (EXP)2022-4-25Load Cases for Code Compliance膨胀工况解释膨胀工况解释Expansion Case Explained 上述工况正确，但没能说明规范要求的最大应力范围因为CII并不能判断荷载所代表的具体含义 为满足规范的要求，用户必须自己定义:(6) DS1 - DS2 (EXP) 这个工况是最大位移膨胀应力，正是规范所要求的。 您根本不能考虑使用T1来计算膨胀应力.2022-4-25Load Cases for Code ComplianceExpansion Case Explained膨胀工况的解释膨胀工况的解释To su

32、mmarize:概括如下 We take the difference between two load cases to determine a displacement range.两个工况确定位移范围 From this range we compute a force range and then a stress range.由此我们确定力的范围和应力范围 The code requires the extreme displacement stress range.规范要求极端的应力范围 The user only has to worry about whether or not

33、 the “extreme” case has been addressed.用户仅考虑最大应力范围即可2022-4-25Linear vs Non-Linear线性和非线性 Terminology applies to boundary conditions.边界条件的类型 Recall the equation being solved: Kx = f This is the equation of a spring. The piping system boundary conditions (i.e. the restraints) are represented as stiffne

34、sses, or springs.管道边界条件代表刚度或弹簧 More complex boundary conditions can be defined, invalidating the “linear spring” assumption.2022-4-25Linear vs Non-Linear线性和非线性线性约束 boundary condition is a double acting restraint, such as a “Y” support.一种是上下约束 Another example of a linear boundary condition is a sprin

35、g hanger.一种是弹簧 The force versus displacement curve for these restraints is a straight line 力和位移是线性关系 Therefore these restraints are linear.约束是线性的 The slope of the line is the stiffness.斜率是刚度2022-4-25Linear vs Non-Linear线性和非线性非线性约束 A “+Y” support is a non-linear support.支架 Its force vs displacement c

36、urve is not a straight line.力和位移不是直线关系 Stiffness only exists for negative displacements.向下的位移是刚度是存在的 For positive displacements, the stiffness is zero.向上的位移，刚度变为2022-4-25Linear vs Non-Linear线性和非线性 A “gap” is also a non-linear support.间隙的引进 The force vs displacement curve is not a straight line.力和位移不

37、是线性关系 There is no stiffness in the gap.间隙部分没有刚度2022-4-25Linear vs Non-Linear线性和非线性 Friction makes a restraint non-linear摩擦让约束非线性 Large rotation rods are also non-linear restraints大的转动吊杆让约束非线性 Non-linear restraints in a job mean that Kope is not equal to Ksus. 非线性后，热态管道刚度和冷态刚度不一致 (EXP) and (OCC) load

38、 cases must be constructed using the difference between two other load cases to account for non-linear restraints.2022-4-25偶然工况的建立 Occasional loads are considered “primary”, since they are force driven.偶然荷载是主要载荷，力引起的。 Occasional loads occur infrequently.不经常发生 The codes employ an “allowable increase”

39、 factor based on the frequency of occurrence in the determination of the allowable, i.e. k * Sh.基于发生的频率，确定值的大小 Examples of occasional loads are wind and earthquake.偶然载荷是风载荷和地震载荷2022-4-25偶然工况的建立 The code equation for the OCCasional load case is:MA / Z + MB / Z kSh Here, MA is the moment term from the

40、 SUStained loads,冷态荷载引发力矩 and MB is the moment from the OCCasional loads.偶然荷载引发力矩 This equation states that the OCCasional case is the sum of the SUStained stresses and the OCCasional stresses.偶然工况是冷态和偶然的叠加 So we cant run a load case with just a “WIND” load and satisfy this code requirement. What ab

41、out “W + P1 + WIND” as a load case?2022-4-25Occasional Load Case Setup The “W + P1 + WIND” case will work for “linear” 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 i

42、s:(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)2022-4-25Occasional 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

43、normal OPErating case This is a combined OPErating case which includes 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). Th

44、is handles non-linearities. This is our OCCasional code compliance case, stresses from Primary plus Occasional loads.2022-4-25工况的定义和维护 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 reads i

45、n the defined load cases and presents them to the user. The load case editing screen is shown at the right.2022-4-25工况的定义和维护 On this dialog, available load types are listed in the upper left list box.载荷类型 Available load case types are listed in the lower left list box.工况类型 Load cases (recommended or

46、 previously defined) are shown in the grid at the right.推荐生成的工况 Recommended load cases can always be obtained by clicking on the Recommend button. The analysis commences by clicking on “the running man”.2022-4-25Load Case Generation & Maintenance Say for a “new” job, the load cases at the right are

47、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 displacements, and start the Static Analysis processor again.2022-4-25Load Case Generation & Maintenance CAESAR II reads th

48、ese 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 displacements. You must manually add “D1” to the OPE load case, or ask CAESAR II to re-recommend the load cases.2022-4-25

49、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.2022-4-25Load Case Generation & Maintenance Notice the load type list in the upper left contains “D1” now. The corrected load cases are shown at the right

50、.2022-4-25Load 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 insufficient. If you added displacements to node 110, would the load cases

51、be sufficient?2022-4-25确保您分析对象的正确性 Remember CAESAR II is a finite element program.有限元 Remember CAESAR II uses a 3D beam element.3D 梁单元 Remember you must have equilibrium:保持平衡 Resultant loads should equal applied loads作用力等于反作用力 Gravity (weight only) load case should equal the weight of the system重力等于

52、整个系统的总重量 Other basic checks Verify nodal 3D coordinates尺寸输入是否正确 Check for extreme displacements and/or loads最大位移和推力 (see handout)2022-4-25Problem Solving问题的处理What do you do when you dont like the results? Recall the equation being solved:Kx = f where we solve for x, the displacements From these disp

53、lacements, we compute element forces & moments From these forces & moments, the Code equations are applied and we compute stresses.2022-4-25Problem SolvingWhat do you do when you dont like the results? So if you have a stress problem, it can only be caused by two things: A Code related issue (SIFs,

54、code equation, etc.) Extreme forces and/or moments If you have a force/moment problem, they can be caused by only two things: Improper element characteristics Extreme displacements2022-4-25Problem SolvingWhat do you do when you dont like the results? If you have a displacement problem, it can only be caused by two things: Improper input (density, elastic modulus, applied loads) Improper boundary conditions Dont forget to check and recheck the input. Remember that in 3D systems, a load in one location can cause pivoting somewhere else downstream, resulting in excessive forces and mome