user277418

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About user277418

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  • Birthday 05/16/87

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    Tallinn, Estonia
  1. Only one link active. Can you upload this books somewhere or send to email? nem0v@mail.ru
  2. page 103 http://www.eurocodes.fi/1997/1997-1/background/Evaluation of Eurocode 7.pdf
  3. If I understand your question correct. If internal supports designed as pinned movable. Al least one or both of the edge supports should be pinned non-movable in the plane of the section you posted. So the roof structure not slide away from wind load action.
  4. Thank You for the replay ReinisGailitis. Because You so often answer on my questions I have feeling that only You remained on this forum ))
  5. In which case necessary to use CsCd for buildings? Is it not easier to use wind pressure on cladding to spread wind pressure force to building's frame?
  6. You can look in this paper. Page 20-21. Article 11 This confused me too. One more confusing thing is that load category F accounts vehicles up to 30 kN, but maximum authorized mass for standard cars (European driving licence category B ) is 3500kg=34.4kN. So, if strictly adhere to Eurocode rules, all parkings necessary to design for load by category G May be it will be revised in new edition of Eurocode.
  7. Hi This forum about Eurocodes, so you can look in EN 1993-6 for crane girder design
  8. Hi According to annex H, EN 1994-1-2, to account bending in CFT column it is neccessery to calculate equivalent axial force Nequ=Nfi,Ed/(fis*fidelta) fis - reinforcement ratio coefficient and choosed by picture H.1 and assume fis=0.4=no reinforcement fidelta - eccentricity coefficient and choosed by picture H.2. It depends on ratio of eccentricity DeltaSmall=Mfi,Ed/Nfi,Ed to column cross-section width D and ratio of equivalent column length in fire L to column cross-section width D. Assume DeltaSmall/D=0.5 as maximum possible for this method. I calculated 2 columns: 1. L=1000 mm, D=100 mm > fidelta=0.35 > Nequ=1000/(0.4*0.35)=7143kN 2. L=4000 mm, D=100 mm > fidelta=0.55 > Nequ=1000/(0.4*0.55)=4546kN So, equvalent axial force for slender column lower than for more stiff one. Is there any misstake in annex H or my calculations? I expected opposite situation. Best regards! PS: I cannot attach pictures, because of some error in forum
  9. You need to model flexible connection. Calibrate connection stiffness to get 150kNm moment in the beam to column joint.
  10. Hi. From EN 1994-1-1 6.6.1.3(4) If the plastic resistance moment exceeds 2,5 times the plastic resistance moment of the steel member alone, additional checks on the adequacy of the shear connection should be made at intermediate points approximately mid-way between adjacent critical cross-sections. What additional checks mentioned in this article?
  11. Hello. The question is in the title. I tried to compare results from different software and little bit confused. I thought that results will be very similar, but in my opnion they are not. Description of the models and results below in DETAILS. Maybe something did wrong. Waiting for criticism and clues. Archives with model files and calculations can be downloaded here >> https://www.dropbox.com/sh/o9f8upf3u1pvy6v/AAAH5CFhyjd8JdA7L9aCQ_r0a?dl=0 There everything is divided according to the programs.
  12. I don't see it. Some of descriptions actually have some pdf articles but they are not full AD. Look attachment.
  13. Thank You Reinis. It helped. But there only brief description of each AD without possibility to download full AD. So my request still valid.
  14. When I go by your link only gray screen appeared and that is all. Look attachment. I tried at work and home, the result is the same. If you can download it from website, please share.