Compressed member - DIN Code

[chrede]

Hi all

Hopefully you clever guys can help with this topic.

In the old German code DIN 4114 you were allowed to use a so called w-factor(Omega) on compressed members in order to verify the buckling capacity of a steel member. The w-factor depends of the slenderness (schlankheitsgrad) and the material used (werkstoffe).

eg. If you have a w = 5 you multiply that number on your normal force eg. 100 kN which gives 500 kN.. If 500 kN is less than sigma(zul) x A the bucling capacity should be OK.

I have two questions, I hope you can answer:

1: Is that still allowed now days according to Eurocode (DIN/EN)?

2: If I have a truss member (please see attached) with a normal force (N) and a moment (M) and I need to verify the compression chord (obergurte), do I multiply the normal force only with w-factor or shall I do it with both N + M/h?

Eg. in the attached (eg. w-factor is 5):


1. 20 KN x 5 (w-factor) + 31,3 kN*m / 0,5 m = 163 kN or

2. 5 x (20 KN + 31,3 / 0,5 m) = 813 kN.

Thank

you so much for your help

Best wishes

Chrede

 

[HTURKAK]

Hi chrede (Structural),


Ans 1IN 4114 was the code when Chancellor was Konrad Adenauer era (1952..) and superseded by DIN 18800 and moreover DIN 18800 is superseded by DIN-EN 1993-1-3

ANS 2: When 20kN is axial load for overall truss , the axial load for top chord (obergurte) could be estimated as 20/2= 10kN and axial load for 10 kN/m span loading could be estimated (10*25/8 )/0.5 = 62.5 kN

The members of truss is assumed pin connected and the top chord members will experience bending moment of M=10*1.67**2/8 = 3.47 kN/m

Although the lateral bracing is not clear , the buckling length of truss members could be estimated as the length of the member.

In this case, for ST 37 and λ =105 ω = 2.0

So the top chord members should be checked for axial load N = 2.0*(10+62.5)=145 kN and moment M= 3.47 kN/m (e.g. P/A+M/W ≤ 1.0)

But this calculation SHALL BE performed as per EC-3.

 

[centondollar]

If the member is in both compression and bending, you need to perform a lateral-torsional buckling check for the member. Flexural buckling (Euler buckling, which the "omega-method" is applied to) is the buckling failure mode for purely axially compressed members.

As HTURKAK pointed out, the calculations should be done according to Eurocode 3.

 

[chrede]

Dear HTURKAK and centondollar

Thank you so much for your reply.

It was my expectation, that you have to include the compression force from M/h as well. To exclude it does not give any sense in a truss construction. Yes, the local moment in top chord I have to take care of too by the formula for N+M in Eurocode 3.

The reason why I´m asking is because that I have seen some german static calculations from 2003 where the Omega-method has been used.. when I google it, I can see that it was developed by German engineers of Deutschen Reichsbahn in order to verify the steel bridges..? (

http://www.nadirpoint.de/DIN4114_Das_Omega_Verfahren.pdf)

(seemed a bit old.. but pretty easy to use).. So I was thinking that maybe German engineers still use it?

Thanks again and merry christmas to both of you.

/Chrede

 

[HTURKAK]

Dear chrede , the compression force from M/h already included for the top chord . The compression load for 10 kN/m span loading is estimated (10*25/8 )/0.5 = 62.5 kN ..
I preferred to choose for initial iteration λ =105 so, ω = 2.0..

http://www.nadirpoint.de/DIN4114_Das_Omega_Verfahr...)[/URL] (seemed a bit old.. but pretty easy to use).. So I was thinking that maybe German engineers still use it?]

The Deutsche Reichsbahn was the name of state owned railways in the German Democratic Republic (East Germany), and after German reunification succeeded by the Deutsche Bundesbahn (DB). Probably some old German engineers still using omega method for checking the calculations and feeling safe ..
I remember, ( my generation) in 70's when the calculator era begin we were looking with doubt and were checking the output with slide rule to feel comfort..

Merry christmas to you also..