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Khurram's repose is correct however grillage is a term used for steel frame containing beams crossing each other provided under equipment like transformers. tanks etc. Thats because of deformation compatibility. Since both beams have equal geometric stiffness and span, both would be deflecting equally (as what happens in one way slabs where short direction attracts more load). For cases where you have identical beam cross-sections but a shorter beam intersecting a longer beam one, the shorter beam would be governing how much the longer beam would deflect. For cases, where you have a super stiff beam supporting a secondary beam (like a girder or end beam supporting a joist or secondary beam) you can provide "hanger bars". You can check out this thread for more detail: Thanks.

Anf if there is secondary beam and how will they behave ?? there BMD ?? any example provided in engineering book regarding grid system ?

This is relation bw vertical subgrade reaction and bearing capacity not covering horizontal subgrade reaction

Correlation_BC_and_K.pdf

if both beam have same stiffness (same size , same strength) , same span and same loading than , then there will be no secondary beam , both of them will act as primary beam . this system is called grillage system

Assuming a factor of safety of 1.0 (although in reality, you need to decide what FOS you need to achieve...1.5? 2.0?), the weight needed to hold on to the applied load of 100kg is also 100kg. But there is a moment too at the base which is equal to 100kgx1m = 100kg.m. To stabilize this you need to counter this acting moment with 100kg.m stabilizing moment in the opposite direction. In the figure; acting moment is anti-clock wise: 100kg.m You need 100kg.m in clock wise direction (assuming FOS=1.0) which can only be provided by the weight of support block on the right half x length to the middle of right half of support block. So you need to decide either horizontal or vertical dimension of the block too (not just the weight). Note: although the base is just a free standing weight (not fixed) but the pipe in reality is considered fixed into the block. The way you fix the pipe into block will also effect these calculations; either it is firmly fixed or just inserted into an over sized hole for example. There could be many shortcut formulas but you need to fixed some variables first; for example, let's say that for the support block; a) 75% of x dimension is towards right from the center of the pipe 0.75L b) let's assume that height of the block is twice the length so h = 2L c) let's say the dimension into the paper of block is also L (square block on plan); so now the weight of right side of the block = W = density x 0.75L x 2L x L = density x 1.5L³ Now the stabilizing moment will be; 100xFOS = W x 0.75L/2 i.e. W = 2x100xFOS/0.75L if say FOS = 1.5, density of uniformly solid concrete block = 2500 kg/m³ then formula becomes; L = (4M/3750)^0.25 where M = Load x arm (100x1m = 100kg.m in this example).

THANK YOU, SENIORS...

Yes rana bhai.. u r right. i realized that after rechecking the model.. 😀

Special seismic data in ETABS, is used for the enhancement of normally calculated seismic forces, in specific conditions as required by Division IV of Chapter 16, or Chapters 18 through 23 of UBC-97. Inclusion of special seismic data, results in the use of special seismic load combinations of UBC section 1612.4, for the design of structural members. Parameter 'Em', used in these special load combinations, is described in section 1630.1.1 of the code. Generally, special seismic load effects should be included in ETABS models, for all those structural elements for which the code (UBC) requires the use of special seismic load combinations of section 16.12.4. According to Division IV of UBC Chapter 16, special seismic load combinations are to be used in following cases, among others: a. For elements supporting discontinuous systems, as described in Section 1630.8.2. b. For collector elements, splices, and their connections, in accordance with Section 1633.2.6. Regards.