How to enter the azimuth of the eccentric load on the tower?
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When calculating eccentric loads, the azimuth of the eccentric load is the angle between the positive X-axis direction and the line connecting the coordinate origin to the eccentric load. How is the X-axis determined?
The input for SW6 is the algebraic sum of all the eccentric loads, without considering direction; in other words, the eccentric shaft loads are considered to be in the same direction
Thank you. May I ask, but if multiple eccentric loads are not within 90 degrees of each other in the circumferential direction, does it matter whether their directions are considered or not?
If there is only one load, it’s 0°; if there are multiple loads, it’s the direction angle relative to one of the loads