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  1. 2 likes
    A couple of reasons... thicker material cools more slowly and the steel normalises a little and also thinner material has a greater distortion in rolling and is 'cold worked' a little more. There may be other reasons, and someone with a metallurgical background can fill in the blanks. Dik
  2. 2 likes
    For single members restrained by bracing members which are supported by almost rigid "foundations" (bracing systems etc), which will not deform themselves under the axial load in the bracing member: 1) Derive minimum axial stiffness from elastic theory Cmin=2(1+cos(180/(L/a))*NEd/a, where "L" is the span of the restrained member between its outermost supports, "a" is the uniform spacing of bracing members along the restrained member, "NEd" is the average compression force in the restrained compression member or the maximum compressive force resultant of the member in bending. *If bracing members are not spaced uniformly one must use the smallest spacing on both sides of the bracing member for which the minimum stiffness requirement is calculated. In general one should take into account the reduction in stiffness of the bracing member due to it's connections and adequately increase the required actual axial stiffness of the bracing member (bar) itself (won't go into detail about this here, sorry, that is just to detailed of a topic). The only thing I will add is that the cumulative axial stiffness of connections on both ends of the bracing member must be at least twice the "Cmin" calculated above. For members in bending one can use this approximate equation to derive compression force: - by EN 1993-1-1 (will provide conservative results): NEd=MEd/h, where "MEd" is the maximum bending moment in the member; "h" is the depth of the member in the relevant bending plane; - by EN 1995-1-1 (to me this seems more reasonable and logic for economic design): NEd=(1-kcrit)MEd/h, where "kcrit" is the lateral-torsional buckling capacity reduction factor which is obtained analysing restrained member as if it would not be restrained, thus with it's full span L ["kcrit" is the symbol used for timber but the meaning is the same as "X" for steel], everything else as before. Comparing this equation to the one in EN 1993-1-1 one can see that this equation actually takes into account the amount to which the restrained member wants to buckle, so if you use really stiff beam with low bending moment you would not need to restrain it as it's capacity would be adequate with full span L, whereas using EN 1993-1-1 approach you should still design restraint for the member to maximum forces. 2) Derive axial force in the bracing member: FEd=a0/(a/(2NEd)-1/C), where "FEd" is the axial force in the bracing member, "a0" is the local bow imperfection for the restrained member (see table 5.1. of EN 1993-1-1 for steel members, for example), and "C" is the actual axial stiffness of the bracing member. Everything else is as before. I won't go into details about designing bracing members for systems of restrained members that are not supported by rigid "foundations" (thus design of bracing trusses, etc). P.S. I usually conservatively assume 2% of the compression force in the restrained member which is determined as described above. P.S.S. One must exert caution when using the above method when designing bracing members for a bracing system that is supported on rigid "foundation" if it restraints multiple members, as the requirements for stiffness are quite different then as well as the restraint forces in each subsequent part of the bracing member between adjacent restrained members.
  3. 1 like
    Hallo. Read some literature on load path method in strut-and-tie design. Also, check information on correct strut-and-tie models for pile caps - you shouldn't assume that the horizontal strut is lying on the interface of the column/pile cap - it should be positioned inside the pile cap. Cheers
  4. 1 like
    I would say you have summarized our discussion in this post and cleared almost all doubts. Pleasure to discuss with you. Hope you won't mind for any more doubts. Thank you.
  5. 1 like
    I'm not sure I completely understand what you're asking in your last post. I'll try to answer though: the regulation (not the standard), which equals the law, says that when you declare a single characteristic, it's enough. This does not include "NPD" ofcourse. So, a single actual value is sufficient. As a reply to your previous post: CE and DoP needs to exist twice: one copy for the client, another for your archives. You're legally bound to keep a copy for 10 years. You do not have to submit it to the government/...
  6. 1 like
    Thanks for replies guys, I looked up the clause you suggested reinis but it didn't really cover what I was looking for. Edu90 this etag publication seems to cover what I was looking for although I have not had a chance to study it in detail. I have actually came across it before but I never thought of using it for this situation. This wasn't for a certain job or anything it was more just for self teaching as I had seen this situation and was wondering how to deal with it. I am going to do an example calculation to add to my reference material for future. Thanks again
  7. 1 like
    Read through this book - http://www.steelconstruction.info/index.php?title=Special:ImagePage&t=Sci+p360.pdf - there's a lot of good information on the matter. Anyway, lateral and torsional restraint is always recommended although you can only have lateral restraint. In such a situation elastic critical moment Mcr will be smaller compared to one in similar situation with torsional restraint added. In such situations, you have to calculate Mcr with appropriate "analytical" or empirical methods. Although I would always recommend using freeware "LTBeamN" where you can precisely model support/restraint condition in a specialized FEA package. Cheers
  8. 1 like
    for lateral TORSIONAL buckling you must have aside of lateral restrain a torsional restrain. Hope pic attached explains... Hence, once you put the "hands" you have point LTB restrains not continuous
  9. 1 like
    An excerpt of the regulation 305/2011: Where applicable, provisions for an intended use or uses of a construction product in a Member State, aimed at fulfilling basic requirements for construction works, determine the essential characteristics the performance of which should be declared. In order to avoid an empty declaration of performance, at least one of the essential characteristics of a construction product which are relevant for the declared use or uses should be declared. So, declaring a single prestation is enough to fulfill all legal requirements.
  10. 1 like
    In essence, yes, it does behave differently, although, in reality, the difference should be smaller than the one prescribed by design codes, as there should be at least some margin for error in the assumption. Yes, use the largest thickness to determine strength of steel member. Cheers
  11. 1 like
    Yes, in a simply supported slab you usually check ULS bending and shear, whereas in SLS you check stress limitations, crack widths (or the simplified requirements) and deflections. If you have particular situations which require design by strut and tie models, then that's a different story. I don't use 7.4.2. as it has been shown previously that the method is unconservative in some cases, especially if you don't limit the factor by which you take into account actual stress in the tension steel to 1,5, as per UK NA to EN 1992-1-1. Anyone designing flexural members to EN 1992-1-1 should read "Comparison of deflection calculations and span-to-depth ratios in BS 8110 and Eurocode 2" by R. L. Vollum. To note, I have never ever found a situation where actual calculation by simplified method (using the bending moment at the span for curvature and using factor K to account for shape of bending moment diagram instead of integration) also taking into account curvature due to shrinkage would have yielded a thinner slab than the span-to-effective depth ratios, so I don't believe that method is conservative. And I'm never even taking K>1, only K=1. Also, I'm not sure it takes into account curvature due to shrinkage. I don't use that span-to-eff.depth method for mainstream design, only for preliminary, and also with caution. There are many books and examples on many different topics related to RC structure design to EC, but not many specifically about creep and shrinkage effects beyond deflection calculations. Cheers
  12. 1 like
    Hallo, and welcome to the forum! In general, it is accepted that one should use the thickest part of a rolled section to determine yield strength of the member. For welded sections of course one can assume different strengths, although again, it is usual to assume the lowest one. Cheers
  13. 1 like
    I think my first response perfectly answered you question. I may have added that EN 1992-1-1 is where you should look for specifics to Eurocode. But beyond that, I don't think that anyone should have quoted code requirements in this case - if internet is available to you there should be no problem for you to read them on your own.
  14. 1 like
    Karuba, there's a section in EC2 which I think gives you what you're after. That section of the code is what you need and if you can't meet the minimums in the table and need a tighter bend there are more checks required. Another option in this case is to provide a welded transverse bar. Anyway, hope it helps
  15. 1 like
    Declare NPD if you question its use. You only have to declare one characteristic.
  16. 1 like
    Hallo. There are many workshop presentations at the bottom of the page: http://eurocodes.jrc.ec.europa.eu/showpage.php?id=2014_07_WS_Steel Cheers
  17. 1 like
    One thing I forgot to add is that from my understanding the provided methodology for deriving uniformly distributed load for design of bracing systems as presented in both EC3 and EC5 is only applicable for horizontal diaphragms, or more precisely - diaphragms that are simply supported between two supports (vertical bracing systems). As one can see vertical bracing systems are not such systems, as they're basically a cantilever beam, not simply supported one. The reason behind this is the way of derivation of the uniformly distribute load, as it assumes sinus shape deflection for the bracing system between two supports. Cheers
  18. 1 like
    EN 1993-1-1 section 5.3.3 only provides info concerning bracing systems in the form of uniformly distributed load, although one can derive the force in one of the bracing members from this UDL by multiplying it with the spacing between adjacent bracing members. In my opinion the most thorough information is provided in EN 1995-1-1 section 9.2.5. This section deals with bracing members for single members as well as groups of members by bracing systems as in EN 1993-1-1. This section provide a bit more insight into the matter. For real understanding one must read through some literature. As I've spent some time understanding this topic to a degree which is necessary for practical design I'll provide some of my findings in the next replay to this topic.
  19. 1 like
    Hi everyone Which system of prestressed bolts do you prefer (and why), which is more popular - HV (10.9) or HR (8.8 10.9) ? Second question is where can I find clamp lengths of bolts of HR system (HV is described in EN 14399-4)?