Moving Load Lane Question

Creation date: 2/10/2021 6:20 AM    Updated: 2/10/2021 6:20 AM
Question:

Dear Support Team,

I have a question on the moving load - I made a very simple test model as attached to show my question.  

On a simple supported system, I defined the max. and min. number of loaded lanes = 1.  The goal is to fix the number of lanes to 1, so that I can get the results for 1 loaded lane only.  However, I am still getting a result of 0's as shown, representing a case of where there is no moving load (0 loaded lanes).

The reason I would like to have with 1 loaded lanes only is because in the later load combination, I'd like to combine 1 loaded moving load case with other static load case that associated with moving load (say, longitudinal train loads due to acceleration).  This static load case occur concurrently with the moving load; while there is no moving lane, there's no such static case.  For now it seems that moving load includes both 1 loaded lane and non-loaded lane, making it hard to separate the two cases and combine with other static load.

Please advise how to fix the # of loaded lane for the analysis - for example in the test model, how to fix the lane to 1 only.

Thanks,






Answer:

Hi DK LEE,

Thank you for the quick response.  Yes, I think your suggestion works, although it would be tedious for all the concurrent loadings in the 2 track cases.  The typical routine I would do is to get the concurrent demands as shown.  As there will be Service Load Combination for 2 track cases, Strength Load combination for 2 track cases, and so on.  It would be tedious to determine which case below is the "true 2track case" so I can add LF2, and which one is the "actual 1track case" so I can add LF1.  But still workable. 



The reason I want to bring this up is that for light rail structures, there seems always some load cases that are combined with the moving load, and the number of the lanes would matter.  I understand that Midas cannot change the current rule of the program, but just want to bring this issue to your awareness.

Thanks again for all the emails.  I think we resolved the issue. 

Thanks,



Hi,

Thank you very much for your detailed explanation and understanding the algorithm of the program..
How about this idea? 
For the two lanes loaded, find the maximum compressive stress of the footing (element no. 8), which would be the critical stress at the footing caused by concurrent FZ, MY, and MX. As shown below, two lanes are loaded. 

And then convert this train loads to equivalent static loads from the Moving Load Tracer.

Run analysis and you can find the concurrent axial force, moment-y, and moment-z of the footing.


Finally, combine this static load case with LF2 in the load combination.

Please check and let me know.

Regards,
DK Lee
Technical Manager

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Hi DaeKeun Lee,

Thank you for the email.  Let me explain what I need to do first and see if you have better suggestions.


First of all, the model consists of 2 spans, and these 2 spans are supported by the abutment as shown.  The abutment consists of backwall, stem, and footing at the bottom.  I’d like to find the stress at the footing calculating from the concurrent vertical reaction FZ, longitudinal moment MY, and transverse moment MX.    


Per design criteria, we have to consider train longitudinal acceleration & deceleration load (call it LF) along with the live loads.  If there’s no live load, LF = 0; if there’s one lane loaded, there will be a constant value LF1; if there’re two lanes loaded, there will be a constant value LF2.  Values 0, LF1, and LF2 differ by quite a bit and thus affect the longitudinal moment MY. 


From modeling standpoint for the best compliant with the code, in the load combination I would define 0track (no moving load, no LF), 1track (1 lane moving load + LF1), 2tracks (2 lanes moving load + LF2).  



Below shows the result at the bottom of the footing from 2 loaded lanes.  In this model in the lane definition, I defined the max. and min. lane number as 2.  And I used the load combination as 2tracks (2 lanes moving load + LF2).  


From max. MX standpoint, 1 lane loaded creates max. torsion as shown, even I defined max. and min. as 2.  If strictly following the code, this max. MX from 1 lane loaded would only need to add LF1 for the footing design, instead of LF2 as defined in the load combination.    


Your second question brings up a very interesting point.  Consider max. MX case specifically:

1)      In my opinion. two lanes loaded: 

a.      MX + LF2

b.      MX is a bit smaller than max. MX from one lane loaded

c.       MX combined with LF2 may still control the stress at footing


2)      One lane loaded:  Max. MX + LF1

a.       This case can be obtained from one lane loaded only



This is the reason I am trying to separate the # of lanes into 0 loaded, 1 loaded, and 2 loaded cases, so that I can combine them with the corresponding LFs.  From what I have now I think it is conservative but acceptable. But I'd like to hear your thoughts on this - any better way to do this?



Thanks,






Hi,

I am DaeKeun Lee, a technical director of MIDAS IT. I have discussed with the development team about this issue Please let me reply to this ticket.
 

First of all, in your model, the whole span consists of just one long element. If we divide it into 10 elements and view the vehicle position for the span center, we can see that two lanes are loaded as shown below and the positions of the two vehicles will cause the maximum torsion.



This is the case for all elements except for the start/end point of the span for which one vehicle is located inside the span and the other vehicle is located outside the span so that maximum torsion can be found. We cannot say that this algorithm is incorrect. Rather, it is the widely accepted algorithm among a lot of practical engineers for a very long time. 


Secondly, we thought about your comment that two lanes must be loaded because the user entered 2 for both maximum and minimum number of lanes. Then, the following question arises. How should two vehicles be positioned on the two lanes inside the span? Is it case 1 or case 2 or case 3 and on what basis? We have not found the appropriate answer between ourselves.





Please understand that we cannot change the current rule of the program. If you let us know why you need to do this, we will try to find any workaround for you or we can consider this for the future development if there is no workaround.


Regards,
DK Lee
Technical Manager

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Hi Anuj,

So being that, even though I forced the software to apply 2 lanes, it still can apply only one lane for max. torsion.. I don't think this is correct.  This is the case when only one lane is applied.

Can you get back to the developers and see if they can fix it?

Thanks,



Hello,

Greetings!

I have got an update from the developers regarding the issue. When maximum and minimum number of loaded lanes is assigned as 2, the program applies the vehicle on the two lanes but some wheel loads are removed when their effects are favorable. 
                 As in the case of attached model for torsion, single loaded lane is more critical as compared to 2 loaded lanes. That is the reason why only 1 loaded lane is shown in the moving load tracer. 
                 However in the case of My 2 loaded case is more critical as compared to single loaded lane. Hence 2 loaded lanes are shown in moving load tracer results.

Regards,
Anuj Asati
Sr. Technical Support Engineer

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Thank you Anuj!

Hello,

I have asked the developers regarding this. I will update you once I receive a response from the developers.

Regards,
Anuj Asati
Sr. Technical Support Engineer

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Hi Anuj,

Thank you for the email. It seems that the beam end release is causing the confusion.  Let me make another model - please see attached. 


In this model, I set 2-lanes of live loads and max/min number of lanes = 2.



This time, let's find the results of max. torsion MX in the support at Node 20.  Since the beam is restrained by torsion MX.

The results of the max. torsion MX, due to MVmax: only the NB Track lane is loaded from the moving tracer 

Similarly, MVmin is when the SB Track lane is loaded

I wondered since two lanes are fixed, why there's still one lane that causes max. MX?  Am I still missing something here?

Thanks,



Hello,

Greetings!

As mentioned in the earlier response that there will be no hogging moment generated in the structure due to beam end release in element no. 1 and moment release in all the supports. This is the reason we are getting 0 moment  for MVmin case as shown by you in the image. If the beam end release is removed the scenario changes and hogging moment will come near internal support.


After removing this beam end release:


Hope this helps.

Regards,
Anuj Asati
Sr. Technical Support Engineer

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Hi Anuj,

Thanks for the response.  It appears that it does not only occur at the support (released moment) end.  Here using the same model, I tried element 5 - about the middle of the span.  The max. moment makes sense.  However, for the min. moment, the moment = 0 kip-ft and there is no loaded train on the system from Moving Tracer.

Please let me know if you see the same thing.

Thanks,





Hello,

Greetings!

Thanks for getting in touch.
Your approach is absolutely correct in order to get the results for only one lane loaded. The results are also correct. Moment (Ry) is released for all the supports. Element number 1 also has a beam end release for moment at i end. Due to this there will be no hogging moment in the beam. For MV_max case the results are as shown below:


This seems to be correct. In the query the results are shown for element 1 at I end. This is support location. This is the reason for 0  moment at that location.  Only one lane is loaded with the given moving load definition.
Hope this helps. Kindly let us know if any further assistance is needed.

Regards,
Anuj Asati
Sr. Technical Support Engineer

Click here to give FEEDBACK
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