## Robot Structural Analysis

- Subscribe to RSS Feed
- Mark Topic as New
- Mark Topic as Read
- Float this Topic to the Top
- Bookmark
- Subscribe
- Printer Friendly Page

# EN 1990:2002 code combinations

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

I'm doing the code combinations with the automatic combinations option (EN 1990:2002 code).

First, I have to do them with the last option called "manual combinations-generate" because the previous options are not able to performance a non linear second order analysis. Is there any way to solve this, in order not to have a huge list of combinations and just have the ULS envelope?

Second, I only see the availability of making ULS combinations for the STR combinations of the code, what about the GEO and EQU combinations of this code??

Third, In my strucrure, I've already done some manual combinations, now I want to have the whole code combinations...I use this tool but, in the tab "cases" i just select the simple cases, and I ensure that the manual combinations are not selected, so then, why when I go to the "groups" tab I still have the combinations? I've opened the file in repair mode but this still happens.

Solved! Go to Solution.

# Re: EN 1990:2002 code combinations

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

See the attached picture. You can generate combinations using two regulations of the same type and you can select the load cases you want to include as well as the combination types you want to generate. ACC includes seismic.

*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**

# Re: EN 1990:2002 code combinations

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

Which nature should I choose for the combinations? Does this nature affects in somewhere in a certain way?

# Re: EN 1990:2002 code combinations

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

This is irrelevant unless you want to use one of these combinations in the definition of another combination and you care about the default load factor it should get.

*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**

# Re: EN 1990:2002 code combinations

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

Hi again,

Going over the combinations created automatically, I've noticed that something is not going correctly.

In the attached image you can see the EN 1990:2002 coeficientes, where for the cathegory H are empty, if empty box means 0 it would be ok...but I think it's working as if it would be 1.

when make the combinations, this loads of the H category have to be 0 when are acommpanying the leading action.

# Re: EN 1990:2002 code combinations

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

what I mean is as shown in the following image.

LL E: Live load category E.

LL H: Live load category H.

When leading the LLE, the LLH souldn't be incluided in the combination as its acompanying factor is 0.

So the combination should be 1,35*DL1+1,5*LLE+**0*1,5*LLH**+0,6*1,5*W+0,5*1,5*S

# Re: EN 1990:2002 code combinations

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

What is the difference of this two ways? The combinations are different.

# Re: EN 1990:2002 code combinations

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

LL E: Live load category E.

LL H: Live load category H.

When leading the LLE, the LLH souldn't be incluided in the combination as its acompanying factor is 0.

So the combination should be 1,35*DL1+1,5*LLE+0*1,5*LLH+0,6*1,5*W+0,5*1,5*S

In fact there should be two sets of ciombinatios and as far as I can see Robot creates them corerctly:

1. E is main load so we have 0 factor for H

1,35*DL1+1,5*LLE+0*1.5*LLH

2. H is main load and then E get 1

1,35*DL1+1,5*LLH+1*1,5*LLE

**Artur Kosakowski**

# Re: EN 1990:2002 code combinations

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

The point is that Robot don't do what you and me agree:

Robot is doing this:

1. E is main load so we have 0 factor for H

1,35*DL1+1,5*LLE+**1***1.5*LLH

2. H is main load and then E get 1

1,35*DL1+1,5*LLH+1*1,5*LLE

# Re: EN 1990:2002 code combinations

- Mark as New
- Bookmark
- Subscribe
- Subscribe to RSS Feed
- Highlight
- Email to a Friend
- Report Inappropriate Content

Send me your file please.

**Artur Kosakowski**