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Using multiply factor on Ig

9 REPLIES 9
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Message 1 of 10
ingolfurj
409 Views, 9 Replies

Using multiply factor on Ig

Hei everybody. I was wondering how you people define a multiply factor on gross moment if inertia during defenition of shell element such as plates and walls. I have done some calculations in a three dimensional modell, non-symetrical and been comparing results as I change this factor and the difference is quite alot according to reaction forces. Can someone give me a comment on how we can interpret cracked moment of inertia into the structure modell so that forces will be devided due to these stiffness changes when concrete cracks or are you guys mabu taking reaction forces from the modell 100% elastic???

Hope you get my point.....and hopfully my point make sense.
9 REPLIES 9
Message 2 of 10
tony.ridley
in reply to: ingolfurj

Untitled.jpg

Message 3 of 10
ingolfurj
in reply to: ingolfurj

Thank you so much for your answer but maby question wasnt clear enough. I am aware how this factor is applied but not sure how to use it. I was more hoping for a recomendation of a way how to define this factor regarding how the program calculates taking into account that the concrete doesnt crack everywhere. Should I for example take the value between Ig and Ic/Ig asauming the reinforcement in the slabs?
Message 4 of 10
ingolfurj
in reply to: ingolfurj

For exaple slab t=200 bars 12mm center to center 100mm gives Ig=6,7e+8 and Ic=1,35e+8 approx. Is it realistic to set the multiply factor 1,35/6,7=0,2 ?? Or maby take the middle value between 1 and 0,2 saying assuming that the slab doesnt crack everywhere......or am I misunderstanding the use of this factor completly? How are you Robot users using this factor?

 

 

Message 5 of 10
tony.ridley
in reply to: ingolfurj

Check ACI code (as an example) for guidance.  Off the top of my head ACI recommends to model slabs as 0.25 x Ig (similar to the number you came up with).

 

 

Message 6 of 10

You may try to use the stiffness map that shows reduction of the stiffness with with the required reinforcement for a guidance.

 

If you find your post answered press the Accept as Solution button please. This will help other users to find solutions much faster. Thank you.

 

 



Artur Kosakowski
Message 7 of 10

Artur,

 

Can you please explain the calculation or method of calculation behind the process that creates this map?

 

Cheers,

Tony

Message 8 of 10

This is the ratio between stiffness calculated based on the properties of the panel (thickness) and 'cracked' stiffness calculated based on selected code provisions with the influence of amount of reinforcement needed for calculations and SLS cases you used for reinforcement calculations.



Artur Kosakowski
Message 9 of 10

Thanks so much for taking time discussing this matter. 

According to your answer Artur then it is possible to assume this factor during the code-check calculation for slabs, as I understand it.  Could you please show this method in a attached picture.  I woud very much appreciate it if you could do that. 

But would it be necessary to repeat the calculations.  What I mean is that first the slab is calculated with this factor set as zero, run the code calculations and estimate this factor, set the factor as assumed and then run code check again??

 

Could be that I am misunderstanding this completly being a new user of Robot.

 

kind regards

 

Ingolfur

Message 10 of 10

According to your answer Artur then it is possible to assume this factor during the code-check calculation for slabs, as I understand it. 

 

The stiffness distribution I indicated is used during the cracked deflection calculations (elastic approach).  

 

If you want to influence the static analysis of the model you should use the method described by Tony. To do so (and use the value from the map) you may need to follow two steps you mentioned below:

 

What I mean is that first the slab is calculated with this factor set as zero, run the code calculations and estimate this factor, set the factor as assumed and then run code check again??

 


If you find your post answered press the Accept as Solution button please. This will help other users to find solutions much faster. Thank you.



Artur Kosakowski

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