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, Lamé来计算。 • • • •
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Hook’s 定律。
基本应力理论 & 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
规范要求的载荷工况
• • • 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 have +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
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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 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
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。
基本应力理论 & CAESAR II 的实施
••Biblioteka 规范要求的载荷工况• 膨胀工况说明 For this discussion, rearrange the equation to {x} = {f} / [K], where we know we don't 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}
• (2) = W + P1 (SUS) • (3) = DS1 - DS2 (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 • What does “DS1 - DS2 (EXP)” mean? • Is a load case with “T1 (EXP) the same thing?
• 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 综合应力” 综合应力
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。 • 根据规范和载荷工况的不同上式将发生变化。
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 以及产生这些应 力的载荷） 分为下面几种： 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 - τ
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) • 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 (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 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.
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 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 between the extremes. What extremes? • DISPLACEMENT: This defines what 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 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 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