问题3-14: 马来西亚政府规程中承认ASME II的材料,中国GB6654 16MnR 已列入规范 2506,是否可在马来西亚使用?
Q3-14: For ASME II material have been accepted by Malaysia government, as GB6654 16MnR has been adopted in the code case 2506, is this material can be used in Malaysia?
Answer: It is up to the local jurisdiction to decide whether or not a code case is an acceptable standard.
答复:某个案例是否能被接受,取决于当地权利机构。
问题3-15: 在美国5000m3以上的液化石油气球罐多不多?按VIII-1 或 VIII-2设计时,对地震载荷作用下支柱的应力分析有什么区别、有什么注意事项?
Q3-15: Are there many LPG spherical tanks above 5000m3 in USA? When seismic load is taken into account for the design of support, what is the difference between VIII-1 and VIII-2?
Answer: I do not know the answer to your question regarding the number of LPG spherical tanks above 5000 m³ in the United States. As far as your question concerning consideration of seismic loads, both VIII-1 and VIII-2 will require that the appropriate seismic load be considered in the design of the pressure vessel.
答复:不知道在美国5000 m3以上的液化石油气球罐的数目。关于问题中所提到的地震载荷,在VIII-1册和VIII-2册中都要求压力容器设计中考虑适当的地震载荷。
问题3-16: 常见的球冠封头与筒体连接,是否可以用下图所示的连接,并采用
上图所示同样的计算公式确定球冠的厚度?
Q3-16: For a spherically dish head connecting to cylindrical shell, whether
the structure shown in the lower figure can be used and it can be
calculated as the structure shown in the upper figure?
Answer: Neither head attachment detail shown are addressed in Section VIII,
Division 1. However the upper detail where the spherical cap is inserted into a
cylinder may be designed as a flat head and attached per Fig. UG-34 Sketch (f). The lower detail is not permitted based on the requirements given in UW-13(b)(1) and UW-13(b)(2).
答复:在第VIII卷第1册中没有封头连接的详情规定。但是在图中上面的图,插入筒体的球形管帽根据图UG-34中简图(f),按照平盖设计,下面的图根据UW-13(b)(1) 和 UW-13(b)(2)要求是不允许的。
问题3-17: UG-32(b)中半球形封头的R如何选取,取筒体的一半还是封头球面部分R?
Q3-17: Based on UG-32(b), how to select the R of hemispherical head, is it half of the shell diameter or the radius of head?
Answer: Please see sketch (c) of Fig. 1-4 for a definition of the hemispherical head radius.
答复:请见图1-4中简图(c),半球形封头半径的定义。
问题3-18: 在UW-13.2中,不带支撑的管板(或平盖)或带支撑的管板(或平盖)如何理解?
Q3-18: In figure UW-13.2, how to understand the unstayed tubesheet (or cover) and stayed tubesheet (or cover)?
Answer: An unstayed stayed flat head is one in which there are no stay rods installed to carry the axial load (see UG-47 and UG-48). The weld joint details given in Fig. UW-13.2 are sized to be able to carry the bending moments that will develop at the edge of an un-stayed flat plate.
答复:不带支撑的平盖是指没有用来支撑轴向载荷的支撑棒的平盖(见UG-47 和 UG-48)。其在图UW-13.2中给出的焊接接头详图,用焊缝尺寸来支撑在无支撑平板的周边存在的弯矩。
问题3-19: 凸缘如何计算?
Q3-19: How to calculate the raised flange?
Answer: I am not sure I understand your question. Please see Appendix 2 of VIII-1 for flange design rules.
答复:不理解这个问题。请见VIII-1附录2法兰设计规则。
问题3-20: ASME VIII-1中对膨胀节外压计算长度没有定义,为什么?
Q3-20: There is no definition of the calculation length for expansion joint which bears external pressure, why?
Answer: It is not clear to me if your question is related to a bellows type expansion joint (Appendix 26) or a flanged expansion joint (Appendix 5).
答复:不清楚问题是指波纹管型膨胀节(附录26)还是凸缘型膨胀节。
问题3-21: 在标准中,在确定椭圆封头和碟形封头时,有个限定条件:ts/L ≥ 0.002,请问,该限定值是根据公式来的吗,如果不是,请问是根据什么来的?(上海杨园)
Q3-21:In the code, when determining the thickness of ellipsoidal and torispherical head, there is a limitation which is: ts/L ≥ 0.002, the question is: Is this limitation from Shield & Drucker formula? If not, where is it from?
Answer: This limitation is taken from Welding Research Council Bulletin 444, based on the work of Clarence D. Miller.
答复:这一限制是根据Clarence D. Miller的工作,由焊接研究委员会第444通报而来。
问题3-22: 在标准中,L-9.2.3中似乎有错误:
标准中,L-9.2.3中第5步,调整的MDMT=9°F, 高于壳体调整后的MDMT。假如欲保持由壳体确定的2°FMDMT, 使封头直边段A类对接接头调整后的MDMT为2°F,封头直边段的最小厚度能很容易地由下式确定:
tn = (100trE*) / (100-DR) + C
但,事实上是不容易的!
首先,我们来推导上面公式:
其中,tn—新的厚度值, in.
Tn—新的厚度tn所对应的未调整的MDMT值,°F
2°F—期望达到的调整后的 MDMT
步骤 3: 比值 (trE*) / (tr-C)
步骤 4: (1- Radio)100 = 100[1- (trE*) / (tr-C)]
步骤 5: Tn-100[1- (trE*) / (tr-C)]= 2
因此, Tn-2 = 100-(100trE*) / (tr-C)
令 DR = 在步骤2确定的满应力MDMT状态多期望的降低值,即, DR = Tn-2,
所以: tn = (100trE*) / (100-DR) + C, 也就是上面的公式
从公式的推导中可以看到, DR = Tn-2, tn 是未知数,Tn对应于tn,也是未知数,那么, DR就同样是未知数。但在标准中,DR = 18-2 = 16°F. 其中18°F 是根据表UCS-66采用直线内插法得到控制厚度为0.804in的未调整的MDMT.,不是新的厚度值tn. 所对应的未调整的MDMT值,所以,这是错误的。
在标准中, tn = 0.855in. 根据表UCS-66, 控制厚度为0.855in的未调整的MDMT 是 22°F,
调弄整后的 MDMT = 22-16 = 6°F, 不是期望值 2°F.
可见,在公式中, tn = (100trE*) / (100-DR) + C, 同时存在两个未知数,对确定tn是不容易的。
假设封头直边段的最小厚度为0.953in, 根据表UCS-66, 控制厚度为0.9535in的未调整的MDMT 是28°F.
步骤3: 比值(trE*) / (tr-C) =0.74
步骤4: (1- Radio)100 = 100[1- (trE*) / (tr-C)] = 26°F
步骤5: 调整后的MDMT = 28-26 =2°F 为期望值,因此假设正确, tn = 0.953in(上海杨园)
Q3-22: There seems something wrong in L-9.2.3 in the code:
In the code, L-9.2.3: Step 5. Adjusted MDMT=9°F, this is warmer than the adjusted MDMT determined for the shell. Assuming it is desired that the 2°F MDMT established by the shell be maintained, the minimum head skirt thickness that will result in a 2°F adjusted MDMT for the Category A butt joint in the head skirt can be easily determined by the following formula:
tn = (100trE*) / (100-DR) + C
But, it is not easy!
First, let’s derivate the formula:
Where, tn—new nominal thickness, in.
Tn—unadjusted MDMT for governing thickness of tn,°F
2°F—the finally adjusted MDMT
Step 3: Radio (trE*) / (tr-C)
Step 4: (1- Radio)100 = 100[1- (trE*) / (tr-C)]
Step 5: Tn-100[1- (trE*) / (tr-C)]= 2
So, Tn-2 = 100-(100trE*) / (tr-C)
Where, DR = desired reduction in the full-stress MDMT determined in Step 2, that is, DR = Tn-2, So: tn = (100trE*) / (100-DR) + C, that is the formula.
From the derivation, DR = Tn-2, DR is the difference between adjusted MDMT for governing thickness of tn and the finally adjusted MDMT. tn is an unknown value, so DR is also unknown. But in the code, DR = 18-2 = 16°F. 18°F is the unadjusted MDMT for governing thickness of 0.804in., not of tn. So, it is wrong.
In the code, tn = 0.855in. From Table UCS-66, the adjusted MDMT for governing thickness of 0.855in. is 22°F,
Adjusted MDMT = 22-16 = 6°F, not 2°F.
Sum up, in the formula, tn = (100trE*) / (100-DR) + C, there are two unknown value, it is not easy to determine tn,
Suppose: the minimum head skirt thickness is 0.953in, from Table UCS-66,the adjusted MDMT for governing thickness of 0.953in is 28°F.
Step 3: Radio (trE*) / (tr-C) =0.74
Step 4: (1- Radio)100 = 100[1- (trE*) / (tr-C)] = 26°F
Step 5: Adjusted MDMT = 28-26 =2°F So, tn = 0.953in is right.
Answer: I have reviewed the example given in L-9.2.3 and find it to be correct.
答复:我看了在L-9.2.3中给出的例子,发现有误需要纠正。
问题3-23: ASME规范正文内容是强制性的,是否案例也是强制性的?
Q3-23: The formal content in ASME code is mandatory. Whether is any code case mandatory?
Answer: Code Cases are not mandatory unless local jurisdiction dictates them to be.
答复:案例不是强制性的,除非当地有这方面的强制性规定。
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