Andrea91

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Andrea91
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  • I’ve noticed that, as the elastic modulus of the rigid elements increases, they seem to provide an alternative load path that bypasses the pier. When I consider an axial stiffness of the rigid elements approximately 1E10 times greater than that of t…
  • @disla I've just gone through your model. Since the pier is modelled using brick elements and the constraint equations link the forces between the beam nodes and the pier nodes, the model introduces the moment I was looking for. Unfortunately, I nee…
  • The deck of the bridge consists of three longitudinal beams and four cross beams. To simplify the model, the slab is not explicitly modelled, and the loads are applied directly to the beams. This is a common simplification in bridge analysis. Each …
  • Probably I’m not understanding how rigid elements are supposed to be used properly. From the attached model, it looks like the rigid elements prevent the pier from actually carrying the loads. The axial load in the pier is much lower than expected, …
  • @Victor I also tried to introduce flexible joint on beams: - For the two outer bearings, that provide vertical support only, vertical truss elements should work. - For the central bearing, which provides both vertical and transverse restraint, how …
  • I have tried using constraint equations, but they seem to work only partially. In fact, constraint equations appear to transfer forces, but not moments. To clarify the issue, consider the bridge with one of the two spans removed. Because of the ecc…
  • Yes, this is the only way I have found out. But in my specific case I have many nodes and it becomes coumbersome to deal with. I hoped there would be another method.
  • 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 mea…
  • 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…
  • 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…
  • 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 …
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