Cylinder opening calculation problem

The software cannot calculate it, because the software will treat the rectangular shape as a round hole, so the diameter of the shell will be larger when the hole is opened, so it can only be calculated by hand. The aperture diameter is the long axis size of the hole.

Thanks for the reply upstairs. However, when opening a 200X40 rectangular hole on a DN150 cylinder, it is impossible to assume that the hole will be circular based on the long axis size.

You can refer to the calculation method of equal area reinforcement above GB150! !

This post was last edited by watermind at 2010-1-1 20:15 and returned to the 3rd floor. 1. The basic assumption of equal-area reinforcement is to open small holes in a large flat plate, and you can just treat the cylinder as a large flat plate. 2. Why SW6 cannot calculate this is because the reinforcement calculation in the program is based on round holes. Of course, it is not possible to open a 200 round hole on a 150 tube, so it is not calculated. In fact, the holes we opened are long holes, not round holes, so it may be calculated by hand, but not by calculation. 3. From the perspective of film stress, the longitudinal joint stress is the largest, so we only consider the long axis direction, without considering the circumferential direction. 4. From the essence of reinforcement, reinforcement solves the problem of local film stress. In this case, the local stress in the long axis direction is the largest, so only the long axis direction needs to be considered. 5. Specifically, for example, the outer size of the liquid level gauge is 200X40, the inner size is 100X20, the height is 30, and the pipe thickness is 10, then the required area is less than 100 10) 6. Some people say that this example is a large opening and the pressure area method should be used. In fact, the pressure area method and the equal area method are essentially the same. They both solve the problem of local stress. However, the effective reinforcement width of the pressure area method is the local stress when the end of the cylinder is subjected to a uniform load. The attenuation range. The effective reinforcement width of the equal area method is the attenuation range of local stress when a small hole is opened in a large flat plate. Relatively speaking, when a large diameter cylinder is opened, the pressure area method is often used, which is more conservative. When a small diameter cylinder is opened, the equal area method is often used, which is more conservative. 7. No matter which method is used, the peak stress is solved by limiting the size of the major and minor axes of the opening. At the same time, square holes are not recommended for the opening. Therefore, the case mentioned by the poster is not good in terms of size and shape, so it should not be used under harsh working conditions. It should at least be rounded when necessary.

Open a 200X40 rectangular hole on the DN150 cylinder. The equal area method can be used.

Stress analysis is not difficult to calculate intuitively using finite element.

The equal area reinforcement method can be calculated by referring to page 175 of the second edition of "Process Equipment Design"

Calculated using WRC107 Bulletin, (local stress module in SW6)

If the size in the long axis direction is too long, it will increase the local stress at the connection between the end of the long axis and the cylinder. In this case, it is appropriate to use finite element analysis.