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jammer3425

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Everything posted by jammer3425

  1. can anyone tell me from which code does the following table belongs to for info.xlsx
  2. for sure I will upload it in a time of two to three days
  3. sir you are superb thanx alot for such genoerosity you really rock
  4. jammer3425

    Quote

    There comes lot of juices and spices in mans life but a man made his own destiny. (and After some years he says cant I undo it)
  5. jammer3425

    Quote

    Architects may interpret design as being the production of drawings and models to show what a new building will actually look like. To civil and structural engineers design is taken to mean the entire planning process for a new building structure, bridge, tunnel, road, etc., from outline concepts and feasibility studies through mathematical calculations to working drawings which could show every last nut and bolt in the project that is why it is said that from where the vision of architects ends the vision of engineer begins zubaida apa k totkay safha paintalees line no 1 or yeh he aik akaam ki baat aj tak us nay ki hai
  6. jammer3425

    Quote

    life is all grey till u put some colors in it
  7. sir I am working on it as per the guidelines you people have given to me and soon will upload my results here please if anyone of you get sometime to check that will be an honour for me
  8. per fwo ki polician yahan na discuss kereen bhai
  9. its an honor to be a part of of a community where we have ppl like u ali welcome will surely benefit from your skills
  10. one of my senior has done it but i knw it was done by rule of thumb i am onto it and have almost done it only want to show my calculations to some structural engineer
  11. sir UGWT Data length 84 feet width 24 feet height 12 feet and take the centre line in x y same for ohwt as of ugwt i.e both have common 0,0 the data for ohwt dia 16 feet height 10 feet and ohwt is 60 feet above ff lvl
  12. actually sir the case is like that i got a task to design a tank the tank situation is an ohwt is coming out from the ugwt at its top and i have to bring out a column from the ugwt both have same foundation ao if i bring out series of columns to support my ohwt the situation seems awkward to me waht u suggest @sir BAZ as per your view point in that particular should I set my dynamics analysis aparameter b4 analysis in ETABS if so give a hint to do that
  13. maybe they are every where but wo kehtay hain na jb apnay sir pay perti hai na tb daikhi hui cheezain b bhool jati hai one think should i do it or shall i not
  14. CAN WE REST 35000 GALLONS CIRCULAR WATER TANK AT HEIGHT OF 60 FEET ON SINGLE COLUMN OR IS IT NECESSARY TO PROVEIDE BARCING ?
  15. I Saw presentation based on safe In the presentaion i Am unable to clear myself with the following items or concepts 1. define --- slab property data a box showing slab property data appears where we enter modulus of elasticity and type column and the last thing we enter is thickness to be short which thickness is being asked here because when i put the value of thickness my 3 d shows a part of lenth of column under the footing what type and what value is required here 2. In point object information the value of applied loads are required to be entered or the value that it shows is that exported from ETABS to safe?
  16. when the egg is good it is denser and a denser egg is said to be a good egg and once it is rotten the matter inside gets lighter so it floats the concept behind transformation can be taken from the example that when a hen sits on the egg a chick comes out after 21 days which is even more denser lolz u can weigh that urself too so if a chick which is totally different from a fluid inside can hatch from the egg then y cant a matter inside change from denser to lighter so here is where i saved ur money from the x ray procedure recommnded by sir makhzumi lollz
  17. dont worry dnt give me a chance to jam lolz
  18. see the curved behaviour of the columns under different end conditions
  19. I agree with sir umar the end behaviour of the columns depends on the type ot the member it is supporting and the pinned end column can be taken for the reference to deduce effective length of the cantilever column.
  20. sir no doubdt this site is a wonderful way of sharing knowledge and getting your knowledge enhanced here is wot i thought and i hope you wont mind that the thing is when we spent on somethng we also get bound to it by virtue of time like when we go to school we pay fees and then abide by the timing not only student do that but also the teachers because one is spending and the other is gaining. But in doing that a relationship is also developed a relation of student and a teacher. i hope you are getting wot i m saying so let come straight to the point sir make an arrangement for online classes in which students will be given access through passwords and the one teaching the particular subject is also monitored by the experienced moderators. Sir here in pakistan we are lacking such coarses and a liitle touch to the software knowhow is given at the university level so taking an opertunity to shout in the shout box I would like to ask you to arrange for classes on ETABS, staad, safe, etc in a proper way to benefit us and the members who have already joined will be given premium membership free of cost this is my thought hope you will manage in a better way
  21. ANYONE HAVING SPREADSHEET FOR DESIGN OF STEEL COLUMNS AND CHANNELS ANCHORED TO COLUMNS TO SUPPORT DECK? STAIR DETAIL.dwg
  22. A column is essentially a vertical member designed to transmit a compressive load. Being a compression member, it is reasonable to suppose that a column would fail by crushing of the material when the load reached a high enough value, but for most columns failure occurs at a lower load than the crushing strength; this is because most columns are relatively slender, i.e. they are long in relation to their lateral dimensions which are termed as stocky columns. It is generally seen that when a slender member is loaded in compression, as for example when a slender garden cane is leaned on rather heavily, it will bow sideways or buckle, and if the load is then increased further the cane will eventually fail in bending. If, on the other hand, a stocky column is used, one with a low length to breadth ratio, then a crushing mode of failure is more likely than a buckling mode. For example, if a block of timber 50 mm x 50 mm x 100 mm high were loaded in compression, one could not imagine it failing by buckling. Thus the normal compression elements, length and lateral dimension play a part in determining the mode of failure that will result. Also, for a given section, there will be a critical length of the compression member below which it will be crushed and above which it will buckle. The shape of a column is also very important. For example, a sheet of cardboard has practically no strength as a column, but if bent to form an angle section or other shapes as shown below, it is capable of supporting a load It follows that by intelligent use of available material, economical columns can be constructed. Buckling of slender columns and struts If a long thin flexible rod is loaded longitudinally in compression, it is noticeable that it deflects readily near the mid point of its length with a considerable amount of displacement. The phenomenon is called BUCKLING and occurs when the stresses in the rod are still well below those required to cause a shearing type failure. However, if the length of the rod is gradually reduced, whilst still applying the axial load, a length is eventually reached below which the rod will not buckle. Its failure will be internal. I.e. material failure rather than structural distortion. Columns and struts may therefore be described as either SHORT(stocky) or SLENDER depending on its mode of failure. A short column or strut will fail internally by yielding in the case of ductile materials, such as mild steel, or by shearing in the case of brittle materials such as concrete. A slender column or strut will fail by buckling, where a relatively large bending distortion will develop along its length. The member does not collapse immediately but remains in bent equilibrium unless the yield strength of the material has been exceeded. The buckling phenomenon is an example of unstable equilibrium, whereas the behaviour of a short strut is that of stable equilibrium. The axial load to cause buckling is called the critical load (P). For a given load, a critical length may also be deduced. In the case of slender structural columns or struts, the critical buckling load and the critical length depend upon a number of factors, such as the shape and size of the cross-section, the relationship between the length of the column and its lateral dimensions and the degree of fixity at both ends. For a strut of given length which is pinned at both ends, the minimum load at which buckling will occur may be determined using a mathematical analysis which produces what is known as the EULER FORMULA. The Swiss mathematician, Leonhard Euler (1707 – 83), calculated the load at which a column would buckle if it were axially loaded and pinned at its ends. Thus, the Euler buckling load for an axially loaded pin ended column is given by: PE = the Euler buckling load E = Young’s modulus for the material I = the least second moment of area of the section L = the length of the strut between the pinned ends The magnitude of the buckling load given by this formula is the appropriate value for initially straight struts which are pinned at both ends and are subject to an axial load only. In the cases where one end is fixed and the other end is pinned, or where both ends are fixed, the effective length has to be determined by multiplying the length between supports by an effective length factor. 1. Both ends pinned Effective length = actual length x 1.0 2. Both ends fixed Effective length = actual length x 0.5 3. One end pinned other end fixed Effective length = actual length x 0.7 4. One end fixed, other end completely free Effective length = actual length x 2.0 Effective Length Constant Table Column Boundary Effective Length Constant **C Free - Free 1.0 Fixed - Free 2.0 Fixed - Pinned .707 Fixed - Guided 1.0 Fixed - Fixed .5 Guide - Free 2.0 Guided - Pinned 2.0 Guided - Guided 1.0 Pinned - Free 1.0 Pinned - Pinned 1.0 In practice, the assumptions of the Euler formula rarely hold good and some bending occurs in the strut, causing bending stresses. Also the initial straightness and eccentricity of loading are almost impossible to produce in practice with any reasonable degree of accuracy. Therefore several empirical formulae have been developed for practical use. The Rankine-Gordon formula and the Perry-Robertson formula. Perry-Robertson is the one most used but it is rather complex and so tables have been produced to assist the designer and these can be found in the respective codes of practice. The safe axial load depends on: A = area of cross-section r = least radius of gyration of the section Le = effective length (takes into account length and end conditions). The safe stress depends upon: Slenderness ratio = Le/r Where r = (I/A)1/2 I= moment of inertia A= cross sectional area If the slenderness ratio > (greater than) critical slenderness ratio, then the column is treated as a long column and the Euler buckling formula is applicable. If slenderness ratio is < (less than) the critical slenderness ratio, the column is treated as a short column. In short columns, failure may occur by compression without significant buckling and at stresses exceeding the proportional limit. For this condition, Johnson's formula is applicable: For columns that fail subsequent to the onset of inelastic behavior, the constant of proportionality must be used rather than the modulus of elasticity (Engesser formula). The constant of proportionality, Et, is the slope of the stress-strain diagram beyond the proportional limit, termed the tangent modulus. Note within the linearly elastic range, E = Et. Because of the problems previously mentioned with the Euler equation, the effective length factors are modified to take into account these practical limitations. 1. Both ends pinned Effective length = actual length x 1.0 2. Both ends fixed Effective length = actual length x 0.7 3. One end pinned, other end fixed Effective length = actual length x 0.85 4. One end fixed, other end completely free Effective length = actual length x 2.0 Note: These values for effective length factor apply to design to the current BS but may well vary in the new Euro Codes and thus reference should be made to the appropriate Euro Code for actual design values.
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