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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
2、本应力理论 & 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
4、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。 (
5、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
6、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? 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 c
7、omputed 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 wh
8、at 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 from the extreme displacement range. How can we do this? Consider the equation being solved; K x = f. In this equatio
9、n, 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 & CAESAR II 的实施,规范要求的载荷工况,膨胀工况说明 (Obviously the loa
10、d 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 measure with a ruler, rather it is the difference between two positions of the pip
11、e. 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.e the difference in forces between two positions of the pipe. Similarl
12、y, 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, 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
13、,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 results. However, if the system is non-linear (i.e. you h
14、ave +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 this discussion, rearrange the equat
15、ion 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 / 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
16、+ P1 / 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 systems. For non-linear systems, the stiffness matrix is unique for e
