cyclic loading - step time

[AleksaBgd94]

Hi everyone,

I want to apply the load as a function of time. The duration of load is 11.84 s. When I define amplitude, should I enter the values for time column from 0-12 for instance, and then define Time period of step in step card as 12?

Or is it possible to leave the time period in step card as 1, but to divide values of time 0-12 by 12 and get the values 0,...,1 and enter them into time column in abaqus?

 

[FEA way]

It depends which type of the analysis you use. For cyclic load usually direct cyclic procedure is used. But I guess you didn't choose it. So there are two possibilities. Either you chose dynamic analysis or static general one. If it's dynamic analysis then time is important and it makes a difference whether you set the time to be 12 or 1 s. But I suppose it's static general step. In this case time has no physical meaning and serves only as a measure for the incrementation of loads. So there's no difference between choosing the step time to be 12 (amplitude from 1 to 12) and leaving default step time equal to 1 (amplitude from 0 to 1).

 

[AleksaBgd94]

Thank you FEA way.

I chose dynamic explicit analysis and set step time to be 12 s and amplitude is 0-12.

 

[FEA way]

No problem. Keep in my mind that in explicit it is often advisable to reduce even time artificially because the higher step time the longer the analysis will take to solve. Thus events are often sped up. However at some point it may influence the results significantly since inertial effects will play role.

 

[AleksaBgd94]

Ok. Thank you very much FEA way.