管道应力分析.ppt

1、基本应力理论&CAESAR II 的实施绪论绪论3D 梁单元的特征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 的实施绪论绪论3D 梁单元的特征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 的实施基本应力基本应力 使用局部坐标系可以将管系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）力的载荷）the loads that cause them)分为下面几种：分为下面几种：纵向应力-SL环向应力

2、-SH径向应力-SR剪切应力-基本应力理论&CAESAR II 的实施纵向应力分量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。基本应力

3、理论&CAESAR II 的实施由于压力产生的径向应力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“综合应力综合应力”中的基本应力中的基本应力评价评价 3-D 应力应力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

4、。基本应力理论&CAESAR II 的实施Basis for“Code Stress Equations”失效理论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。(第三种失效方式是偶然失效，它与一次失效相似。）基

5、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一次失效情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规范要求的载荷工况二次失效情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，用于一次载荷下规范应力的计算。膨胀工况，用于“extreme dis

6、placement 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规范要求的载荷工况膨胀工况说明膨胀工况说明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规范要求的载荷工况膨胀工况说明膨胀工况说明The code states that the expansion stresses are to be computed from the extreme d

7、isplacement 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.S

8、TRESS: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规范要求的载荷工况膨胀工况说明膨胀工况说明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 know K and f,and we are solving fo

9、r 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规范要求的载荷工况膨胀工况说明膨胀工况说明(Obviously the load case numbers are subject to change on a job b

10、y 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

11、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规范要求的载荷工况膨胀工况说明膨胀工况说明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 stres

12、s 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规范要求的载荷工况膨胀工况说明膨胀工况说明Consider the question again;Is DS1-DS2 the sa

13、me 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 di

14、fferent 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 dont really divide by K,we multiply b

15、y its inverse.OPE:xope=fope/Kope=W+T1+P1/KopeSUS:xsus=fsus/Ksus=W+P1/KsusEXP:xexp=xope-xsus=W+T1+P1/Kope-W+P1/KsusCan we simplify the above equation as follows?EXP:xexp=W+T1+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-

16、W+P1/KCanceling like terms(the ones in red)yields:xexp=T1/KThe 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&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 method is the correct way to go is to co