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About Gurudev

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  1. Hello forum, Which countries National annex (NA) for EN 1993-1-1 refers to reliability class for selection of Execution class(EXC). I have seen couple of NA which address consequence class for EXC selection.
  2. Hello, See EN 1998-1, 3.2.1(5)P. It gives the threshold values for very low seismicity. Though i dont have experience for this kind of structure, In my opinion type of structure do not decide the requirement.
  3. In case of fixed column base plate in SCI 398, design for moment is only given for major axis. Is it presuming that, for minor axis there will be bracing system always transferring load directly to foundation? so no moment exists in minor direction
  4. I went through Part 5. But couldnt find joint (see attached image)
  5. Hello forum, Does any one have idea, how to design column splice with cap end plate for open section column? I went through SCI P358, but it only covers Hollow section column for this connection.
  6. Sorry wrong interpretation. This is applicable to compression members only.
  7. As per EN 1993-1-1, 3.2.3(3)B NOTE B: The use of table 2.1 of EN 1993-1-10 for σEd = 0.25 f(y) is recommended. (even some NA also recommend it) Don't you think stress state of 0.25f(y) value is too conservative? OR it is just recommended, one may use upto 0.75 f(y). I mean if it is mandatory, then toughness selection will govern, rather than c/s design as per EN 1993-1-1 (may not be for compression members). Isn't it?
  8. Hello forum once again, Below I have attached procedure for selecting reference temperature for toughness selection in case of fatigue load. ************************************************************************************ Tmd = 10 deg C, Lowest air temperature Variabe load Q2 = 1000 Kg @ 90 RPM Max applied reference stress shall be calculated for combination as per Eq. 2.1: TEd = Reference temperature as per Eq. 2.2 = Tmd + ΔTɛ' (Considering recommended values for adjustment factors) ΔTɛ' = Adjustment for a strain rate other than the reference strain rate ɛ'0 = - [ (1440-fy)/550 ] * [ ln(ɛ'/ɛ'0)]^1.5 where, fy = yield strength = 355 N/mm2 (for S355 grade) ɛ' = Actual strain rate = 90 /min. (since speed of shaft is 90 RPM) = 1.5 /sec ɛ'0 = Reference strain rate = 4e-4 /sec. [from 2.3.1(2)] ΔTɛ' = - [ (1440-355)/550 ] * [ ln(1.5/4e-4)]^1.5 = - 46.57 deg. C Tmd = 10 deg C, Lowest air temperature TEd = 10 - 46.57 = -36.57 deg. C ***************************************************************************************** Am I using right procedure? If no, please let me know my mistakes. If yes, in case the rotor speed producing dynamic load rotates @ 1500 RPM i.e., strain rate of 1500/60 = 25 /sec. Adjustment for strain rate would have been, 79.73 deg. C. TEd = 10-79.73 = 69.73 deg.C It means in such case, we will not be able to select thickness from Table 2.1 and will need to use other method for toughness selection. eg. fracture mechanics Am I right?
  9. Hello Reinis, may be your PMI is full, limit is only 50. Need some assistance.
  10. Thanks for the confirmation. Have a good day.
  11. No doubt about this. But my que is regarding need of designing the butt weld joining P2 to P3 (see c/s diagram) along the length of beam (its location coincides with neutral axis), when web satisfies one of the criteria (6.18 or 6.19) and it is not slender according to 6.22 ? Since butt welds with full penetration may be assumed to have same capacity as parent metal (EN 1993-1-8). Yes, need to revise basics.
  12. In EN1993-1-1, 6.2.6 shear, what do we actually check, transverse shear stress or average shear stress (Ved/shear area)? Now I think we check transverse shear stresses, am I right? If the web is not slender and we have qualified (checked) that it passes 6.2.6 shear criteria, then what is the need of designing full penetration butt weld at neutral axis. (off course made of higher strength weld filler). Since butt welds with full penetration may be assumed to have same capacity as parent metal (EN 1993-1-8) What do you think?
  13. Actually I went too fast without noticing the use of right words. In above post transverse shear stress means average shear stress = (SF / total c/s area). And longitudinal shear stress (mistakenly I thought as they are parallel axis of beam hence long. shear strs, but they are also perpendicular to cross sect hence they are called as transverse shear stresses) means transverse shear stress (which is max at NA and 0 at extreme fibres of beam c/s). I have attached cross section diagram of hypothetical case I was thing about.
  14. Hello forum, Lets consider a beam with simply support and point load. Now if we have to design this beam (not with reference to any code) by the basics of strength of machine elements; we check it for bending stresses and transverse shear force (am i right?) Why dont we consider longitudinal shear? OR It is not governing or pressumed as compared to bending stresses and transverse shear stresses. If lets say it is built up I section (web is not slender), and web is made of 2 plates butt welded longitudinally. If the beam is designed for BM and trans. SF stresses, and it is safe, still we have to design longitudinal weld for long shear stresses? I know this may be very basic que, but i got stuck.
  15. We are doing this as for improving our capability to produce steel structures with CE marking. It is project specific, we get layout from clients and hence loading (wind, seismic category....etc)