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Manual LT buckling length in LRFD steel design module

12 REPLIES 12
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Message 1 of 13
sorgjee
3451 Views, 12 Replies

Manual LT buckling length in LRFD steel design module

Hi is there any way of setting defined lengths for the LT buckling lengths in the LRFD steel design module? The only options that appear show these to be relative lengths as multiples of the bar lengths.

12 REPLIES 12
Message 2 of 13
Rafal.Gaweda
in reply to: sorgjee

If I understood you correctly:

 

lt.jpg



Rafal Gaweda
Message 3 of 13
sorgjee
in reply to: Rafal.Gaweda

Hi Rafal,

 

Yes that was what I was looking for, but it appeared not to be working entirely as hoped. I started setting the LT buckling length, but this was not implented when running the design.

  1. For some reason  the design always sets the Lb as a maximum equal to your Lz. Is this intentional?
  2.  I tried several different combinations, and generally it works assuming you set the Lz manually, the Lb can be set manually, but not implemented as greater than Lz as mentioned in1.
  3. When combining setting Lz through a coeffcient though things get funny. Using Lz = 7xL, and Lb = 7.0m, the design implemented the Lz value correctly, but for some reason set Lb to 4.92m. Any idea about what has happened here? I've tried attaching a screen capture showing the results.

Thanks

Message 4 of 13
Rafal.Gaweda
in reply to: sorgjee

Could you be so kind and either send model or show it on your model (real connections, element , behavior) to let me understand why do you need this coeff or LTB length grater then member lenght?
Only one situation comes to my mind: you are trying to check longer member but consisting of several shorter ones without creating supermember.

 



Rafal Gaweda
Message 5 of 13
sorgjee
in reply to: Rafal.Gaweda

Yes, we got a cranked truss with the chord made up by several members between restraints.

Message 6 of 13
Rafal.Gaweda
in reply to: sorgjee

So replace serie of elements by one or create supermember for design:

 

super.jpg



Rafal Gaweda
Message 7 of 13
sorgjee
in reply to: Rafal.Gaweda

Hi Rafal,

 

Aplogoies for not reply to this one sooner. I've finally got around to doing some screencaps that show my the problems I've been having with superbars.

 

In short, when trying to define lateral restraints for superbars, it looks like the prgramme is only able to identify members that are perpendicular to the z-axis of the members.

 

As you can hopefully see from the illustration below, the restraints from the internal members of the truss are correctly identified for the major axis bukcling and LT buckling. For the minor axis buckling on the other hand, it does not appear to 'see' te slightly angled bracing members that connect to ever second or third chord member.

 

I was hoping there was something wrong with my appraoch, but not been able to find anythign so far.

 

bracing definition.PNG

 

regards,

Even

Message 8 of 13
Rafal.Gaweda
in reply to: sorgjee

Points for buckling lenghts should be automatically recognized, but not for LTB.

 

bucklz.jpg

 

Adjust position for LTB (if you need for buckling also in similar way) as shown below:

 

ltb.jpg

 



Rafal Gaweda
Message 9 of 13
sorgjee
in reply to: Rafal.Gaweda

It is not recognizing anything that is not perfectly in plane with the buckling plane in questions.

 

I made a similar model to yours, and the fist picture shows the beams horizontally from the superbeam correctly picked up as buckling restraint points. The second shows the support point lowered 0.1m out of the plane, and the members are no longer recognised as buckling support points.

 

in plane bracing.PNG

 

 

 

offset bracing.PNG

Message 10 of 13
Rafal.Gaweda
in reply to: sorgjee

Yes, I have forgotten to move them out ouf plane.

 

You are right, from help:

 

"Use the  option to automatically calculate buckling coefficients for the member segments between bracings based on the analysis of rigidity of bracings adjoining the analyzed member in the corresponding buckling plane. Values of calculated buckling coefficients for successive segments are displayed in the field under the bar diagrams."

 

http://docs.autodesk.com/RSA/2012/ENU/filesROBOT/GUID-1177E678-E14E-4C8E-8AB2-0A76B44DF34-955.htm



Rafal Gaweda
Message 11 of 13
sorgjee
in reply to: Rafal.Gaweda

That option appears only to calculate proposed effective lengths of segments based on relative stiffness. It requires the actual bracing points to be predefine, and does not help detecting the non-horizontal bracing members.

 

First picture shows ky<1.0 as calculated, and second picture shows kz = 100 as calculated, probably because no restraints were found.

 

On a separate note, it appears RSA calculates the buckling lengths as <1.0 where there is a crank. What is this based on? As one could expect the effective buckling length to be >1.0 for cranked members. The help link you included also states "For intermediate segments 1.0 is always proposed.", which does not appear to be the case here.

 

buck1.PNG

 

buck2.PNG

Message 12 of 13
Rafal.Gaweda
in reply to: sorgjee

Yes you are write. It is written in help.

 

"Buckling coefficients of component segments - Defines values of buckling (lateral buckling) coefficients for the member segments between bracings. After selecting an icon representing the analytical model (bar diagram) of the analyzed member,Robot proposes values of coefficients for extreme segments. For intermediate segments 1.0 is always proposed."

 

I found description of algorithm:

 

A. GENERAL INFORMATION

Calculations of the column buckling length according to the automatic procedure (commonly called Automatic Buckling Length – ABL) in the ROBOT program consist in analyzing geometry of bars located in the column’s neighborhood and attempting to adjust it to code formulas for regular rectangular frames.

The option is obtainable only for 2D structures.

B. ALGORITHM OF OPERATION

The program analyzes separately both column end nodes, calculating for each of them their stiffness values according to code regulations. In order to apply code formulas, a user should know stiffness of the analyzed column (which is known from definition), stiffness values of transversal beams that meet in a node as well as stiffness of the adjoining column. The last two ones, called further - by convention - ”beam” stiffness and ”column” stiffness, are calculated in the following manner.

  1. A bar adjoining the node is analyzed considering all its further connections (in other words, together with the entire bar chain). A user calculates the stiffness of the entire bar chain, which either affects beam stiffness or column stiffness of the node depending on the bar chain direction.
  2. First bar in a bar chain determines its direction:        
          - column direction,        
          - beam direction,          
          - intermediate direction.

The column direction indicates direction contained within the area of ±15° from the direction determined by the initial analyzed column.        
The beam direction indicates direction contained within the area of ±15° from the direction perpendicular to the initial analyzed column.        
All bars that are not included in the above-listed classification, belong to the ”intermediate” group.           
The division presented is illustrated in figure 1.

  1. Stiffness of the ”intermediate” bar chain (equal to J/L) is replaced by equivalent stiffness values: column stiffness Jc (J/Lc) and beam stiffness Jb (J/Lb), assuming for fictitious column and beam the same moment of inertia J as for the inclined bar chain as well as modified length values Lc = k*L*cosa, Lb  = k*L*sina (k is a multiplier, whereas a is an angle between a column and direction of the vector connecting beginning and end of a bar chain).     
          Based on the condition J = Jc + Jb., the following is obtained     
                                  1/L = 1/Lc + 1/Lb            
          and this allows calculating multiplier k = (sin*cos)/(sin+cos).
  2. An end of a bar chain is determined by:

a)     a node in which at least 3 bars meet

b)    support

c)     release in node or element (hinge).

d)    change of direction by the angle greater than ±30° with respect to the initial position

e)     too large number of changes in bar stiffness (more than 10)       
A change in stiffness reaching the order of 1.0e-12 is considered insignificant and is not included in calculations. Since version 12.5, the substitute stiffness is calculated based on the following formula  (J1*L1+J2*L2)/(L1+L2).

  1. A chain of bars with an unconstrained ending is not included in stiffness calculations, similarly as a chain of bars beginning with a hinge (release in element at the bar chain beginning).
  2. The program takes account of the support method (ends) applied to bar chains (rotational release, fixed support, fixed elastic support).         
  3. Influence of a longitudinal force on stiffness values is not considered. This is the analysis of purely geometrical character.

Column and beam stiffness values (calculated as a relation of moment of inertia to length) for individual chains of bars are added up, which enables determining the ultimate beam stiffness and column stiffness of a node after analyzing all the bars that meet in a node. These values are substituted in the appropriate code formulas.

If there is a support or hinge in a node, analysis of a bar chain is not performed and the support pattern implies the appropriate equivalent stiffness of the node. If both nodes are supported, then the buckling length coefficients corresponding to those known from the material strength theory are adopted.

C. COMPARISON WITH CALCULATIONS ACCORDING TO MODELS WITH ADJOINING BARS

For simple rectangular frames, results obtained on the basis of ABL correspond to results obtained due to applying the appropriate model with adjoining bars. In case of composed bars of different stiffness values (e.g. bars with brackets), the results will be identical only when ”superbars” have been used in the model (only then the method of calculating averaged branch stiffness will be identical as for ABL). Application of the list of adjoining bars provides minimally different results.

Another situation in which differences between the methods may occur, concerns the inclined spandrel beams exceeding the area of ±15° (see fig.1). Then it is necessary to calculate manually equivalent length values of columns and beams following the rules specified in point.

 

abl.jpg


Figure 1.




Rafal Gaweda
Message 13 of 13
sorgjee
in reply to: Rafal.Gaweda

Thanks. That goes a long way towards clarifying the details of how it is set up.

 

Back to the original issue, it looks like this limitation in the program prevents superbars from being an option, and that all buckilng lenghts will have to be manually calculated and entered.

 

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