Crosses in the system when choosing a spoiler and R coefficient

E

eissenhover

New Member
Hello everyone. I have a few questions about the project I uploaded in the attachment. My 1st question; The structure will be used as a windbreaker in a mountainous region. more precisely, a design for trial purposes, to see if it can be used. I got a load of 0.096 tons/m2 by taking the c and q values from the relevant section in ts498. (v=36 m/s, q=0.8 kn/m2, I chose c value as 1.2 for this structure, so w=0.096 ton/m2) normally v value is 28 m/s but I got this upper value. What I want to ask is, how would it be right to give wind load in x and y directions for such a constructor? I defined the coating for both clovers and snatched them. but I made a single load for the y-direction and a single load for the x-direction. I did not add two separate loads in each direction, namely suction and pressure, because it is not like a building. what is the truth of this, especially the value of the load I effect for the x-direction and my way of loading (I defined a coating for each column) is it correct or incomplete? the same way, did I make a mistake in the y-direction? My second question; When choosing the r coefficient in steel structures like this, should I consider the crosses in the system like curtains? that is, we should consider it as a frame or a curtain system. How do I calculate the ace value if we think of the crosses as fretted? In the report, the ace ratio according to the floors appears to be zero. In the table titled “The reason for choosing the r coefficient” in the idecad report, the value of as for the curtain system should be known. in the earthquake regulation (3.4) Buildings where the earthquake loads are carried together with the frames by braced steel shears or cast-in-situ reinforced concrete shears (a) It seems that I should take 6 in the x direction and 5 in the y direction within the scope of the center of the braces. Assuming that the ductility is high because there are crosses in the x direction and normal in the y direction. I think this question has been very soupy. Do I need to articulate the topmost horizontal beams in the 3rd structure? download link:
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Re: windbreaker Hello; 1- a) If your building will be a closed structure; After entering the coating, select the normal type structure from the analysis settings. And you do not enter the wind load. The program will calculate and apply all loads in '+' and '-' directions for X and Y directions in accordance with TS 498. During this process, your columns do not need to be 2 pieces. A total of 4 m of steel element can be assembled in one piece in this way in the field anyway. In this way, you can enter the coating in one piece. If you are going to divide the columns, choose a level where it does not intersect with other elements so that the joint combination can be applied. b) If the structure will be an open structure, select Open Structures from the WIND LOADS SECTION from the analysis settings without any coating. c) If you want to use the calculations you have made by hand, you should not be content with one aspect. If the main goal of the building is to carry the wind load, underestimating the wind loads will not be able to accurately convey the situation to the project. You may experience problems during application. For manual entry, perform calculations for exactly all directions. 2- Central crosses; corresponds to the screen. However, it would be more appropriate to increase the number of diagonals you entered. If you replace two pieces instead of one piece, you can get a more suitable result, especially by aligning the full beams. Regarding the R coefficient, according to DBYBHY 2007 - Table 2.5, you can solve your structure with normal ductility level in both directions and with R=5. 3- It would be a more suitable solution if you connect it with articulation. Good work.
 
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