管道应力分析

1、基本应力理论 使用 M/Z 基本应力理论 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 respectively. 基本应力理论 Is DS1-DS2 t

2、he 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 +Ys, or gaps, or friction), then the answer is

3、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规范要求的载荷工况 膨胀工况说明膨胀工况说明 For this discussion, rearrange the equation to x = f / K, where we know we d

4、ont 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 + P1 / K 基本应力理论 & CAESA

5、R II 的实施 规范要求的载荷工况规范要求的载荷工况 膨胀工况说明膨胀工况说明 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 systems. For non

6、-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规范要求的载荷工况 膨胀工况说明膨胀工况说明 Another proof that the DS1-DS2

7、 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 (EX

8、P) 基本应力理论 & CAESAR II 的实施 规范要求的载荷工况规范要求的载荷工况 膨胀工况说明膨胀工况说明 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

9、 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. 基本应力理论 & CAESAR II 的实施 规范要求的载荷工况规范要求的载荷工况 膨胀工况说明膨胀工况说明 To summarize: We take the difference between two load cases to determine a displ

10、acement 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 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（

11、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。 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

12、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 & CAESAR II 的实施 Occasional Load Case Setup Occasional loads are considered “primary”, since they are force

13、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. 基本应力理论 & CAESAR II 的实施 Occasional Load Case Setup The code equat

14、ion 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 SUStained stresses and the OCCasional stresses. So we cant run a load case wi

15、th 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 not sufficient, for the same reason “T1” is not suffic

16、ient 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 +

17、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 includes the OCC loads This is the standard SUStained case This is the standard EXPansion case This difference yields the effects of the O

18、CCasional 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 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

19、 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 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 & Maint

20、enance 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 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

21、实施 Load Case Generation & Maintenance 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 o

22、btained by clicking on the Recommend button. 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 dis

23、cover that pre-defined displacements at 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

24、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. 基本应力理论 & CAESAR II 的实施 Load Case Generation & Maintenance Notice the load type l

25、ist 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 load cases are shown at the right. 基本应力理论 & CAESAR II 的实施 Load Case Generat

26、ion & 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 be sufficient? 基本应力理论 & CAE

27、SAR II 的实施 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 have equilibrium: Resultant loads should equal applied loads Gravity (weight only) load case should equal the weight of the syst

28、em Other basic checks Verify nodal 3D coordinates 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？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？当不满意结果时，你应做什么？ 如

29、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 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, app

30、lied 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 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 Gather all the data,