Application of thermal loads to a slab

[Anonymous]

Hi, I'm a new Robot user who has only recently done a two day course. 

Unfortunately the course did not cover the application of thermal loads to a slab, and as I've never dealt with thermal loads before I'm slightly lost.

If I begin with the simple option of adding a uniform load, I believe that I need to apply the expected difference in temperature to the slab. For example, the temperature may vary between 0 and 40 degrees. I would then apply a value of 40 degrees in dT1. If the top and bottom of the slab varies in temperature by 20 degrees, I would apply this in the gradient tabs. 

Now, how is the gradient affected by the slab thickness? I'm assuming that as the gradient is in K/m, and the slab is 0.25m thick, I would actually apply 20/0.25 = 80K/m in the gradient box.

Am I correct so far or completely barking up the wrong tree?

[Rafal.Gaweda]

It is well described here:

http://forums.autodesk.com/t5/Autodesk-Robot-Structural/Thermal-Load/td-p/3806569

[Anonymous]

Thanks. That was interesting but I'm still not quite following how it would apply in my situation.

We have a large slab and we do not want any movement joints in it.

Three buildings sit on the 450mm thick slab at first floor around the perimeter. 

Underneath there is a car park (unheated) and in the middle a large communal garden.

We want to know how the large slab will be affected by the variations in temperature.

I believe the only parameter in the garden above / car park below area is the annual temperature variation as the temperature either side of the slab should be similar.

In the residential building above / car park below we have the annual temperature variation to the slab (say -10C to 35C), plus the difference in temperature between the upper and lower slab, (say -10C and 25C in the depths of winter).

I assume I would need to enter different surface loads, but what do I actually input in dT1 and gradient? I'm getting more confused by the minute!

[Artur.Kosakowski · Accepted solution]

Ronny,

Try to look at this situation in the following way:

When you erect your structure you have certain temperature e.g. +10 C. This means that the temperature across the slab (bottom, middle, top layers) is the same and it is this + 10C. Then the temperature changes and you have +30C above the slab and -20C below it (Robot assumes the linear change in the temprature across the slab's thickness) . If so the temperature in the middle of the slab is 0.5*(30-20) = + 5 C. This means that you should apply uniform temperature change TX = -5C (the difference between 10C at the assembly stage and 5 C now for the middle layer of the slab). Now you should address the difference in temperature between the top and the bottom. It is + 50 C (difference between the +30 top and - 20 bottom) and this is the additional load dT (gradient) = TZ = + 50 C.

TX is just the 'elongation or shortening' effect whereas TZ is 'bending'.

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.

[Anonymous]

Thanks, it's beginning to become clearer. Will try to implement what you have described.

I'm sure I'll have lots more questions before the project is finished though!

[Anonymous]

OK, I now understand what is going on with respect to dT1 and gradient.

One more thing: Is there a way of modelling temperature variation over a period of time. I'm not even sure what effect this will have, but I have been asked about this as obviously the temperature in the depths of winter will be considerably cooler than at the height of summer. With the slab being so large we feel that this will have an effect on the slab.

Thanks.

[Artur.Kosakowski]

I'm afraid you have to be more specific. Of course you may define a number of temperature loads (e.g. for each month of a year) and use them as components of combinations with other loads but I'm not sure if this is what is needed.

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.

[Anonymous]

Thinking about it, I think your previous reply covers a worst case so I guess my question is answered.

Thanks.