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Badar (BAZ)

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Everything posted by Badar (BAZ)

  1. It was a speculation on my part, as I said that I haven't heard/read of it. Encastre beam is synonymous for fixed beam, hence I guessed on its basis. This type of constraint can be envisaged in bridges constructed from structural steel, it can have an application in machines.
  2. It could be a support that restricts rotational and vertical movement, while allowing horizontal movement.
  3. I have'nt heard/read about it. Where have you came across this term?
  4. The statement taken out from ACI document only covers shrinkage cracks in reinforced concrete structures. Hence, it is regarding serviceability limit states. In order to fully understand it, one should know what they mean by 'acceptable level', or 'acceptable design limits'. There must be a paper regarding that, which needs to be studied. You should check latest ACI code; I think. they have not included this provision in the latest code. They are continuing older provision of 0.18/0.2 %. It look like as if they have not taken the recommendation of this committe seriously. It is a confusion.
  5. Reinforced concrete structures, R. Park & T. Paulay Design of Reinforced Concrete, Jack C. McCormac & James K. Nelson Reinforced Concrete, Edward G. Nawy Prestressed Concrete, Edward G. Nawy Design of Prestressed Concrete, Arthur H. Nilson Prestressed Concrete Analysis and Design, Antoine E. Naaman Finite Element Procedures, Klaus-Jurgen Bathe The Finite Element for Solid and Structural Mechanics, Zienkeiwicz & R.L. Taylor Structural Analysis, Aslam Kassimali Structural Analysis, R. C. Hibbeler Theory of Elasticity, S. Timoshenko & J. N. Goodier Seismic Design Of Reinforced Concrete And Masonry Buildings - T.Paulay,M.Priestley FARZAD NAEIM HANDBOOK Wind and EARTHQUAKE Resistant Buildings - Bungale S. Taranath Dynamics of Structures, Anil K. Chopra Structural Dynamics, Mario Paz.
  6. Formulas in the figure titled 'attached thumbnails' gives the equivalent Uniformly distributed load for computing maximum bending moment, where as the other one is the equivalent UDL for maximum shear. Bear in mind that the actual loading is triangular, or trapezoidal, depending upon the location of the beam. Use wlx/3 or W*Lx/6[3-(lx/ly)^2] for computing BM.
  7. WS/3 is for triangular tributary area, while the other one is for trapezoidal tributary area. You can't apply both formulaes for the same beam.
  8. @ Yasir, Is the height of your frame(RCC) about 10m ? I am not able to understand the arrangement made at about 5m; it does not look part of main frame, but an arrangement to support door opening.
  9. @umar I don't think so that checker will ask him to ascertain the effect of story drift on nonstructural member. If his structural system, which, I think, is comprised of RCC frame, satisfy the the drift limits that will limit, or avoid, the damage to cladding/non-structural-members, he does need to do anything else on these non-structural members. Yasir could inquire from him about that aspect, to know if he is worried about the damage to these vertical posts, and attached louvers, due to the drift experienced by RCC frame. In any case you can check the strength of these post against the moment and shear that will result from drift. I think that strength and serviceability limits should be satisfied for these non-structural members, independently of main structural system. It is the mid span deflection due to wind load that also needs attention. The thing you have mentioned refers to the members which are part of a system resisting gravity loads. Code requires us to design these members, meant to resist only gravity loads, for drift inflicted by the members, part of Lateral load resisting systems, on to them . The box-shaped posts, used in current discussion, only supports louvers.
  10. Checker has used the word 'interstory deflection' in his comments. I think he wants you to make sure that these members will not deflect to a extent that will damage those lauvers. So, the values that I have mentioned above are for beam or flexural members. These nonstructural members need to be checked, against wind loads, for mid span deflection. You don't have to model its connects that will resist end moments, it seems unneccesary: model these members as having simple supports, and consequently design the bolts at mid and at top for shear and tension. You can do the same for the bottom connection.
  11. This is what I understood from the drawings: non structural members, about 5 m in length, are called upon to support the lauver. These nonstructural members are supported at bottom, mid height and at top of panel by concrete beams. Now the question is: what should be the 'serviceability requirement' for these non-structural members. According to IBC, floor beams supporting non-plaster ceiling should be limited to a max live load, as well as wind load, deflection of l/240. But I would say that this member should fall in the category of the one's not supporting any ceiling. In that case the deflection for both load cases is limited to l/180 (IBC).
  12. Columns also take part in resistance od lateral loads; they do so by rotation of beam column joint; and due to confinement of column core, confinement beam-column joint and by accounting for extra shear and bending requirement in plastic range. In moment-resisting-frame structures, they are only source; but in dual system, or a system consisting of large number of shear walls, their effect is less.
  13. I think it depends on the lateral load resisting system, which depends on seismic zone, and which also depends on the arrangment and size of remaining vertical members. Ductility demands a specially reinforced boundary regions in shear walls, and it must of that much stiffness to attract that much forces, which will depend on the size, and detailing of others members of lateral load resisting system.
  14. I have'nt understood your question. What do you mean by interpret; ETABS have clearly defined the symbols/acronyms that they use in design report. It is a common practice to add torsion reinforcement in flexural reinforcement, and divide it over the depth of beam: top bottom and middle. Stirrups are also able to arrest the cracks due to torsion. But the important point is to ascertain the importance of torsion reinforcement in a beam. The beam that transfers load from a slab which is supported on four /three sides don't need all the torsion reinforcement reported by ETABS, as it can redistribute the stressed in the form of flexure and shear in beam and slab. Only minium torsion reinforcement is required in this case. In these cases you can run the analysis by reducing tosion stiffness of beam so that beam does not fail in shear+torsion in ETABS. The Beam part of a cantilever slab has to transfer the moment through torsion as no other load path is available in this case, and redistribution is not possible.
  15. There is more confidence on tension carrying abilities of reinforcement than there is on ability of concrete to resist compression; ductility is also a factor, shear failure is less ductile and more catastrophic.
  16. you should compare results with analytical/empirical methods i.e manual calculations. You need to read manuals of these soft wares, use google.
  17. ETABS is great tool for dealing with RCC frame and steel frame structures according to the established building codes. You cannot perform nonlinear analysis in ETABS; it has limited ability e.g. for pushover . You can model masonry but it is not the best tool for masonry because you can only remain in elastic region, and you do not have control on element type. Likewise, it is not suitable for water tanks as well, especially underground water tanks. You cannot perform soil-structure interaction, which is the on of the reason that its is not suitable for underground water tanks. In finite element analysis, forces and displacements are computed at nodes and then distributed inside the element. So every FEM software does the same.
  18. what are those numbers in options? Do they represent multiple of spacing required for grade 40?
  19. I would not say anything about best structural design firms, but I would name the structural engineer, who impressed most: Rizwan Mirza. He runs the firm by the name of Rizwan Mirza Consulting Engineers, Lahore.
  20. Is it a one storey structure ? What is angle of inclination of the wall? I don't understand why are you referring to M11 and M22 as out of plane; one of them can be roughly considered as in-plane. What is source of torsion in your wall?
  21. Use elements with shell behavior to calculate shear and bending stresses, and then you can compute the required reinforcement accordingly. Mmax and Mmin will give max stressed areas but you will need M11 and M22 to get economical distribution of reinforcement. You are saying 2 out of plane bending moments because of moment generated gravity and out of plane shear ? If that is the case, then the two contributions should be added.
  22. It might not be irrational, he has conducted tests and then reported the values. Depending upon the method a preparation of test samples, results can vary hugely. It depends upon many factors, amount of heat a clay received in kiln, mix ratio of mortar, ratio of of area between two bricks and mortar area, amount of water absorbed by before laying of brick. There may be other issues as well; construction activity needs to be monitored to justify higher values, and it not possible, economically, in those regions.
  23. The values, I given, in my earlier post are from American code. After your comment, I went through a thesis of a Pakistani researcher (doing research in Pakistan), and he recommended to use 350f'm (perpendicular to bed joints) for unreinforced masonry in northern regions of Pakistan.. The value for parallel to bed joint is half of its value in perpendicular direction. It is very conservative estimate and should not used in areas where construction practice is better. I am attaching his thesis,other Pakistani researchers have also reported similar values. doct thesis pakistani masonry.pdf
  24. consult Building Code Requirements for Masonry Structures ACI 530-88 - ASCE 5-88. Em = 700(f'm) can also be used for clay masonry, where f'm is compressive strength of masonry. .
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