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
2、应力理论 & CAESAR II 的实施,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 / 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。,
3、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
4、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
5、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
6、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? 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 compute
7、d 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 between the extremes. What extremes? DISPLACEMENT: This defines what ext
8、remes 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 from the extreme displacement range. How can we do this? Consider the equation being solved; K x = f. In this equation, we
9、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 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
10、 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 measure with a ruler, rather it is the difference between two positions of the pipe. Onc
11、e 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.e the difference in forces between two positions of the pipe. Similarly, the
12、 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, 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
13、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 results. However, if the system is non-linear (i.e. you have +Y
14、s, 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 this discussion, rearrange the equation to
15、 x = f / 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 /
16、 K - W + P1 / K,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 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. This assumption is only true for linear s
17、ystems. 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.,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-DS
18、2“ 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 + P1 (OPE) (2) W + T2 + P1 (OPE) (3) W + P1 (SUS) (4) DS1 - DS3 (EXP) (5) DS2 - DS3 (
19、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 the 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)
20、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.,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 th
21、is range we compute a force range and then a stress range. The code requires the extreme displacement stress range. The user only has 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
22、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。,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
23、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 & CAESAR II 的实施,Occasional Load Case Setup,Occasional loads are considered “primary”, since they are force driven. Occasional loads occur
24、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.,基本应力理论 & CAESAR II 的实施,Occasional Load Case Setup,The code equation for the OCCasional load cas
25、e 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 SUStained stresses and the OCCasional stresses.So we cant run a load case with just a “WIND” load and satisf
26、y 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 not sufficient, for the same reason “T1” is not sufficient for the EXPansion load case
27、. 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 + WIND (OPE)(3) W + P1 (SUS) (4) D
28、S1 - DS3 (EXP) (5) DS2 - DS1 (OPE)(6) ST5 + ST3 (OCC),This is the 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
29、 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 “ne
30、w” 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,CAESAR II will recommend load
31、 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 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,
32、On this dialog, available load 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 b
33、utton. The analysis commences by clicking on “the 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 a
34、t node 5 were omitted. You return to input, add the displacements, 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 s
35、ame 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.,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
36、.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 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 l
37、ist 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 be sufficient?,基本应力理论 & CAESAR II 的实施,Insuring You Analyze What You
38、 Think Youre Analyzing,Remember CAESAR II is a finite element program. Remember CAESAR II uses a 3D beam element. Remember you must 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 co
39、ordinates Check for extreme displacements and/or loads (see handout),基本应力理论 & CAESAR II 的实施,问题解决,当不满意结果时，你应做什么？重新求解方程： Kx = 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。,基本应力理论 & CAESAR II 的实施,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
40、两个问题所引起： 不正确的单元特性 极限位移,基本应力理论 & CAESAR II 的实施,Problem Solving,What 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
41、the input. Remember that in 3D systems, a load in one location can cause pivoting somewhere else downstream, resulting in excessive forces and moments. Try to isolate the load causing the problem, and trace its origin.,基本应力理论 & CAESAR II 的实施,Problem Solving,Design by Analysis - The Design Cycle Gath
42、er all the data, with assumptions Generate the model and load sets Perform the analysis Check the results and assumptions Diagnose any problems Make corrections and refinements as necessary Re-run the modified model/load sets Document the analysis,基本应力理论 & CAESAR II 的实施,其它主题,关联节点 - CNODEsCNODEs 是 CA
43、ESAR II 中一个很灵活、很有用的功能。 如何使用CNODEs最简单的方法是记注短语：自由度关系 把CNODE看作一个“球绞”，则一些自由度是相关的（例如平动），而一些自由度不是（例如转动）。,基本应力理论 & CAESAR II 的实施,其它主题,CNODE 实例 放在钢结构上的管子容器壳体上的管嘴,基本应力理论 & CAESAR II 的实施,其它主题,CNODE 实例 膨胀节拉杆弹簧杆连接,基本应力理论 & CAESAR II 的实施,其它主题,CNODE 实例 假设在一工字梁上有一管子。 假设用一个 U形螺栓将管子与梁连接起来。 假设我们在分析中要包括梁，则模型该如何来模拟？第1步
44、定义哪个自由度相关联，哪个自由度相互没有关系。Y & X 是相关的，而 Z, RX, RY和 RZ 是互相独立的。,基本应力理论 & CAESAR II 的实施,其它主题,CNODE 实例CAESAR II 数据表显示了“约束辅助区”。 如果我们在CNODE区定义一个节点号，则我们已经定义了一个CNODE连接。 这里我们定义了两个约束。 由于定义了 CNODE ，则约束不是与空间中的一个固定点相连，而是与 CNODE相连。,基本应力理论 & CAESAR II 的实施,其它主题,关联节点 - CNODES CAESAR II 膨胀节模块自动设置拉杆的 CNODEs 。 管嘴的辅助项(WRC-297, API-650, & BS-5500)也自动设置管嘴 / 壳体连接的CNODEs 。 CNODEs 的手动定义可以归纳为下面的几个连接类型： 钢结构上的管子 吊架杆 夹套管,
