Comments

• Victor

  February 7

  That's normal but it should converge towards something as time step gets smaller.
  
  In this case, acceleration has such big high frequency components, I don't think 0.01 s fine enough too big to capture it. Maybe add damping to suppress those if they're not important.
  
  Another possibility is that load it too concentrated (500 mm) for the element size (also 500 mm) so there could be artifacts from it straddling element boundaries.
  
  I've found that a rectangular travelling load tends to have this kind of problem and it works better with the unittriangle(w,x) function instead which makes the load taper off at the edges. It might still need higher deck mesh density but not sure.
  
  Here's the formula I used:
  
  -(180/4300/1000) * unittriangle(0.5, (0.9*1.565)*t-40/(0.9*1.565)-x )
  
  It gives broadly similar displacement results to 0.01 s time step with your function but using only a 0.05 s time step.
  
  If you change the first parameter (0.5), it spreads the load out further and can make it smoother by straddling more elements without affecting the total force.
  
  Note that the width of the loaded area is only 3570 mm, not 4300 mm so it's not really 180 N.
• Andrea91

  February 9

  Thanks for your reply. I get better results now.
  With the load formula you proposed, the traction is applied over a length of 1000 mm instead of 500 mm, isn't it?
  
  Secondly, is it possibile to run a similar analysis on a beam model?
  I also have a rougher model of the structure made of beam elements only, but I can't figure out if it supports position dependent loads.
  
  Thanks,
• Victor

  February 9

  The length is still 500 mm (the first argument, 0.5, to the unittriangle function). You can visualize this by refining the mesh enough to see a patch of arrows. I changed 500 to 1000 in the constant term to somehow scale it to agree with your formula but I can't quite understand how the math works.
  
  There isn't a convenient way to do it with beams, sorry. You might have to procedurally generate a separate time-dependent force load at each node.
• Andrea91

  February 10

  I'm not understanding the formula properly. The unit triangular function should have an unit area. Since I am simulating a 180 N travelling load, shouldn't I keep the same length (0.5 m) both in the unit triangular function and in its amplitude (500 mm)?
• Victor

  February 10

  I'll change the traction unit from MPa to Pa because there's a unit conversion factor in there that makes it more confusing to me:
  
  -(180/4.3) * unittriangle(0.5, (0.9*1.565)*t-40/(0.9*1.565)-x )
  
  the unittriangle() function defines a force per unit length whose integral over all x at any t value is 1 N. We then divide by the width of the load area (4.3 m) to get a force per unit area whose integral is also 1 N. Then multiply by -180 to scale that integral from 1 N to -180 N.
• Andrea91

  February 10

  Provided that the function result unit is set correctly, the two functions should provide equal results.
  
  [function in Pa]
  

  -(180/4.3) * unittriangle(0.5, (0.9*1.565)*t-40/(0.9*1.565)-x )

  
  
  [function in MPa]
  

  -(180/4300/1000) * unittriangle(0.5, (0.9*1.565)*t-40/(0.9*1.565)-x )

  
  
  However I get different results in the two cases.
  What am I missing?
• Victor

  February 10

  I get the same displacement at the center deck node at t=20 s. What is the difference you're seeing?
• Andrea91

  February 10

  I've sorted out. Results are equal now.
  
  Can you please clarify the termis of the unittriangle() function?
  
  unittriangle(0.5, (0.9*1.565)*t-40/(0.9*1.565)-x )
  
  0.9*1.565 = velocity
  40/(0.9*1.565) = it looks like space /velocity, but the unit of measure wouldn't be consistent
  
• Victor

  February 10

  40/(0.9*1.565) is half the length of the bridge (unit is meter) so that it starts from the -x end instead of the origin.
  
  unittriangle(w,x) is this with area = 1. You add a constant to x to shift it in space, and add a multiple of t to make it shift linearly with time.