Moving Load over a bridge deck

AndreaC91

Hi,

I'm trying to model the effect of a uniformly distributed load crossing the deck of a bridge, in order to assess how the structural response of the bridge depends on the crossing velocity.
In short, the load should be a pressure distributed over a rectangular area shorter than the bridge main span length.
I have already tried to introduce the load through a pressure over each deck element as a function of time, but at the end of the day I just get the static result that I was expecting by considering all the static configurations with the load moving forward (see also attached file).
Probably I should model the load as a mass, but I can't understand if this is possible with Mecway.
Do you have any advice?

Thanks,
Andrea

Victor

The solution isn't static, but the dynamic response is at too high a frequency to see much. Reduce time step size.

A quicker way to make a moving load is using a function of both position and time. I've done that in the attached file using the heaviside() step function. See also https://mecway.com/forum/discussion/1116/moving-loads-analysis-bridge-crane

If you're OK with a non-uniform load, there's a function intended for moving point loads unittriangle(w,x) which avoids some discretization error from the sharp edge of a uniform load.

As you pointed out, this neglects the inertial mass of the vehicle. I'm not sure if you can do that, at least with the UI and internal solver. With CCX, you might be able to model the vehicle as a solid element with contact and self-weight.

AndreaC91

Thanks for your suggestion, Victor.

Can you please explain how to set up the two proposed functions (heaviside(), unitriangle() ) or at least point out where I can find some detailed information?

With reference to heaviside(), how did you set up the arguments? I mean, what do "20" and "8" represent?

52327 * heaviside(20-8*t-z) * heaviside(-(20-8*t-z-10))

Thanks

Victor

I don't have a clear idea of why it's like that. Though 20 and -20 are z at the ends of the path. 8 could be velocity - 20m/(8m/s)=2.5s to cover 20 m?.

Here's an example of unittriangle. Now the constant in front means force instead of pressure and width and positions of the two paths are described by the formula so you can apply it to all the deck's upper faces.

1234 * unittriangle(5,20-8*t-z) * (unittriangle(1.04, x-1.04) + unittriangle(1.04, x+1.04))/2

The last two terms are averaged so as not to double the load with two tracks.