Weld joint factor for the cylinder in hole reinforcement calculations
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If the welding joint coefficient for the cylinder is 0.85, can this value be set to 1 when performing calculations for reinforcement due to openings?
It should be taken based on the welding joint coefficient of the cylinder
When there are no Class A or Class B welds in the reinforcement area, 1.0 can be used.
When there are no Class A or Class B welds in the reinforcement area, 1.0 can be used.
The design factor is set at 0.85, as the new model 150 does not have a requirement prohibiting its use on welds
When there are no Class A or Class B welds in the reinforcement area, 1.0 can be used? If considering actual use (safety margins, etc.), there might be no problem. But from a standard (literal) perspective, I disagree with you; the welding joint coefficient used in the calculation for this opening reinforcement is exactly the same welding joint coefficient used in the calculations for the shell. According to clause 6.3.3.2 a) of GB/T 150.3, for openings in cylinders or spherical shells, δ represents the calculated thickness of the shell at that opening; therefore, the value of φ in the calculated thickness of the shell remains unchanged, regardless of the presence of openings.
When there are no welds in the area reinforced by openings, the joint factor can be set to 1.0 when calculating the theoretical thickness of the shell in that area. As the name implies, a joint is required for a joint factor to exist, as it is the joint that causes a reduction in strength; so why cannot 1.0 be used when there are no joints? This allows full utilization of the excess thickness of the housing.
The new 150 does not have a requirement prohibiting use on welds
The old version of GB150 did not have such a restriction; in the new version of GB150, when reinforcement is not carried out in accordance with clause 6.1.3, openings must not be located on Class A or Class B joints.