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Wind load simulations results are incorrect

16 REPLIES 16
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Message 1 of 17
Rahul.JyothulaLR5EC
1287 Views, 16 Replies

Wind load simulations results are incorrect

When a pressure of 0.87 kPa is applied on the structure to calculate wind loads using the wind simulation wizard I am getting results in excess of 0.87kPa which is approx 1.0 to 1.25 kPa

16 REPLIES 16
Message 2 of 17

Hi @Rahul.JyothulaLR5EC ,

 

Did you check the Wind Profile settings?

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Message 3 of 17

Yes, the wind profile setting are 1.0, which is default setting
Message 4 of 17

Hi, it is left as 1.0, which is a default value

Message 5 of 17

Well, it seems something strange here... I have attached the simple test model, the wind load assigned as a wind pressure 1 kPa, area of the wall 15m2, and default wind profile (with factor 1). It is assumed that the horizontal reaction will be 15 kN, however, Robot gave me x2 times more:

Romanich_0-1657022272293.png

Dear @okapawal and @Krzysztof_Wasik  please have a look

 

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Message 6 of 17
Krzysztof_Wasik
in reply to: Romanich

Hi @Romanich 

As far as I understand it wind simulation in Robot is based on CFD analyis (fluid flow simulation).

Defining wind pressure profile means that in space with no obstacles air flows with constant pressure (which can be recalculated to its constant velocity).

 

So pressure is recalculated to air velocity then in that space with constant velocity/ pressure, obstacle is placed. Obstacle geometry changes air flow (initial pressure/speed distribution) and results in modified wind pressure generation on its (obstacle) surfaces.

 

Having constant pressure profile means having constant wind velocity in empty space,  but does not mean that pressure (defined as the profile)  will be applied 1 to 1 to obstacle surfaces.

 

If it would be so easy, it would be enough to apply pressure directly to obstacle surface (as area loads) with no use of wind simulation module.

 



Krzysztof Wasik
Message 7 of 17

Just to think about....generally the wind drag factor to walls is around 2.

Message 8 of 17

Hi @Krzysztof_Wasik ,

 

I have prepared another test model to check the drag coefficients.  

Romanich_0-1657042519469.png

In theory, the ratio between reactions should be close to the ratio between the abovementioned drag coefficients, i.e. 0.8/1.05 = 0.76

 

Romanich_1-1657042789318.pngRomanich_2-1657042842157.png

Comparing reactions (SQR(20.6^2+20.6^2))/39.8=0.73

 

Results are very close to each other and based on this simple test we can conclude that the wind simulation is working fine 🙂

 

 

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Message 9 of 17

Hi @Krzysztof_Wasik @Romanich 

 

I understand that pressure can not be applied 1 to 1 to obstacle surface. when I compare with euro code EN 1991-4-4;2005, there is a external pressure coefficients (Cpe) which will be multiplied to the pressure depending upon windward or leeward side.

 

RahulJyothulaLR5EC_2-1657085040411.png

 

RahulJyothulaLR5EC_0-1657084972473.png

From the table above, for side D which is the windward side the values Cpe is always less than 1. which states that the applied pressure on the side D is always less than the pressure due to wind.

Please clarify.

Message 10 of 17

Hi @Rahul.JyothulaLR5EC 

Robot wind simulation module does not use Eurocode approach (cpe coefficients are not calculated as in the code), but CFD simulation method, so you cannot also expect 1 to 1 results as in the code.

 

Regardless that,  code approach refers to the following obstacle shape.

Krzysztof_Wasik_0-1657088795906.png

 

while in your case obstacle has shape as below (where effect of positive and negative pressure will be bigger that for building shape (d>0) considered in the code).

Krzysztof_Wasik_2-1657089132961.png

 

 

In that case you cannot consider pressure on side D only (positive pressure) but also pressure on side E (negative pressure). This cumulative effect is simulated in Robot

 

 



Krzysztof Wasik
Message 11 of 17

Hi @Krzysztof_Wasik,

 

My point is that, the results from both the methods, CFD (Robot wind simulation) and Codal practices should match with small difference.

Even Euro code considers negative pressure effect,  as shown below

 

RahulJyothulaLR5EC_2-1657091024944.png

There are no openings in the panels so, there will be little to no internal pressure.

My case is a proper building like shown below,

RahulJyothulaLR5EC_3-1657092829281.png

and not this,

RahulJyothulaLR5EC_4-1657093364946.png

 

Your efforts are appreciated. 

 

 

 

Message 12 of 17

@pawelpiechnik 

If only there was some way for Autodesk to communicate how to use this tool properly.

Message 13 of 17

Please, reveal us the final resulted coefficient that should be used in Eurocode to this specific case. Also show us the comparison between that and the response on Robot.

Message 14 of 17

Hi @cerveiramatheus 

 

RSA wind simulation calculates wind pressure on windward side as 1.25kPa  for a Peak velocity pressure of 0.87  kPa, shown in the image below,

RahulJyothulaLR5EC_3-1657174182483.png

 

and As per Eurocode the wind pressure calculated is following,

Dimensions of my building is D x d x h :  25.2 x 8.45 x 6.5

RahulJyothulaLR5EC_2-1657173970439.png

 

 For, h/d = 6.5/8.45 = 0.77

RahulJyothulaLR5EC_1-1657173765101.png

Zone D which is my windward side of the building Cpe is 0.77 from the above table and Cpi of -0.3 

The net wind pressure acting on the windward side D is 0.87 x 0.77 - 0.87 x -0.3 = 0.93 kPa 

 

In conclusion my findings are,

For peak velocity pressure of 0.87 kPa, RSA calculates wind pressure as 1.25 kPa and wind pressure manually calculated as per Euro code is 0.93 kPa.

 

Please find the Robot file attached.

Thanks in advance.

Message 15 of 17

Hi @Rahul.JyothulaLR5EC ,

 

I’ve checked your case based on the formula 5.3 from Eurocode:

Romanich_0-1657180403978.png

CsCd=1,

qp(Ze)=0.87 kPa

 

According to the formula 7.9

Romanich_1-1657180434178.png

From the Figure 7.23 Cf0=2. From figure 7.36 Ѱλ approximately 0.68 (this is a biggest bottleneck how to find the “right” λ value, I've got λ =7.1). Cf = 2 * 1 * 0.68 = 1.36

 

The pressure: 

Cf * qp = 1.36 * 0.87  = 1.18 kPa vs 1.25 kPa in Robot

By the way, changing Ѱλ to 0.71 will give exact 1.25 kPa pressure

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Message 16 of 17

Apologies for delayed response, 

I think, Force coefficient should only be used for structural elements,

RahulJyothulaLR5EC_1-1657531535006.png

 

RahulJyothulaLR5EC_0-1657531297956.png

Let me know your thoughts.

Thank you.

 

Message 17 of 17

Hi @Rahul.JyothulaLR5EC 

 

Romanich_0-1657534564091.png

 

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Have your question been answered successfully? Click 'ACCEPT SOLUTION' button.

Roman Zhelezniak

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