What is Vector1, Vector2... in the Displacement input?

Posted by: Tom45

What is Vector1, Vector2... in the Displacement input? - 09/20/19 12:00 PM

What does the Vector1, Vector2, ... Vector 9 represent in the Displacement input screen?
Posted by: Michael_Fletcher

Re: What is Vector1, Vector2... in the Displacement input? - 09/20/19 02:41 PM

You can specify the displacement of a given node on a given element up to 9 different ways. Then you specify when those displacements apply in the load case editor as D1, D2... D9.

Same applies to point loads, F1... F9 and distributed loads, U1... U9.
Posted by: anubis512

Re: What is Vector1, Vector2... in the Displacement input? - 09/23/19 10:29 AM

To add onto the answer above, these are external displacements your piping is seeing in the directions you defined in the vector.
Posted by: Tom45

Re: What is Vector1, Vector2... in the Displacement input? - 11/14/19 06:46 PM

I understand displacements but why call it Vector 1, Vector2 etc.

Is there some significance that I am missing as to why it is called that??
Posted by: Faizal K

Re: What is Vector1, Vector2... in the Displacement input? - 11/14/19 06:57 PM

CAESAR II is just allowing you to define up to 9 sets (vectors) of displacements.

Let's say your line is tied to a much larger header having multiple operating modes. Say case 1, the header goes +1" in the X direction. Case 2 it goes to -1" X and +1" Z. You can define D1 = {1,0,0,0,0,0}*, D2 = {-1,0,1,0,0,0}*. D1 is displacement vector 1, D2 = displacement vector 2 etc.

In your load case setup, you can even combine them too if you want.



*{DX, DY, DZ, RX, RY, RZ}
Posted by: Faizal K

Re: What is Vector1, Vector2... in the Displacement input? - 11/14/19 06:59 PM

also, displacement is a vector since you have magnitude and direction. just like the force input.
Posted by: Michael_Fletcher

Re: What is Vector1, Vector2... in the Displacement input? - 11/15/19 10:01 AM

An additional note. You might be tempted to say D1 = {1,0,0,0,0,0} and D2 = {0,0,1,0,0,0} thus D1+D2 = {1,0,1,0,0,0} but this would be incorrect.

I think CAESAR averages the 2? D1+D2 = {0.5, 0, 0.5, 0, 0, 0}.

(Or it could also do SRSS - I just don't remember.)