supports

T

tuğra

New Member
take it easy, I have a question about supports in idecad. I attached a beam to two columns and assigned a fixed support from the analysis model without combining it, but the project gave an error. In the analysis model, you said solid support to the red dot, but isn't the solid support a connection that does not transfer moment. There are situations that do not fit my logic. Could you please explain that to me?
 
Hello there; You do not need to assign any supports after two columns have entered 1 beam. The red dot symbolizes the node and if only the red dot is visible, the node is fixed. So it transfers torque. Connections that do not transmit torque are hinged, they are not fixed, you have wrong information about that. If you want to create a fixed joint point, do not make any intervention by entering 2 columns and 1 beam. Analyze. As a result of the analysis, if you switch to the right-click analysis view section effects tab on the perspective screen, you will see moments at the nodal points where the beam is connected in the moment diagram under vertical loads. If you want to make a hinged connection, you can enter the properties of the beam and mark the i and j ends for M2 and M3 from the freedoms sub-tab. If you look at the cross-section effects after the analysis, you can see that the moment at the beam nodal points is 0.
 
I just had one more question. Can I combine two profiles and assign a column or beam as a single profile. I want to put two 60x120x3mm profiles side by side and make them as columns. Will the column inertia mononet be like a single element. Or should I connect them together?
 
Hello there; The program does not combine the moments of inertia when you put them together or join them together. For this, you need to combine 2*-60*120*3 profiles in the building tree with an appropriate method. You should do the analysis and design with this model. 20-30 cm etc. between the two column pieces in the field. If there will be gaps, you should set up a different model in the drawing and connect the two cross-sections with an artificial column joint in the joints tab. Good work.
 
take it easy, Ms. Nurgul, I shared a picture from a slide you told, here the column is held in the weak direction (x direction) buckling length is taken as lx=4500mm ly=9000 mm, why were the buckling lengths taken this way. axis) buckling length is taken, can you explain the logic of this?
 
I'm sorry, I'm pretty new, I have another question. In the question in the pictures I sent above, how do I introduce the loads (G and Q) to the program, will I enter them as a single load. Because there are combinations about earthquake loads in it, how can I do that. Isn't there? So we can't calculate as a single combination using only G + Q load OR 1.4G load?
 
Hello there; Select the element to be loaded. Right click the mouse while on it, go to the add load - add point load section in the floating tab that opens. From here, you can apply the G and Q loads in the question in the program. The program automatically creates combinations according to the selected regulation. There is no way to turn them off. You can evaluate the controls at the design stage only by looking at the results in the combination of G and Q. If you switch to the top toolbar analysis and design in the program, you can see which regulation is selected in the section in the image below.
 
the column is held in the weak direction (x direction) buckling length is taken as lx=4500mm ly=9000 mm, why were the buckling lengths taken this way :(
 
Is there a sample project file on the page where you are looking at this example? You can see it when you check the model. lx=4500 is taken because there is a beam connected to the column in the x direction from 4500 mm, that is, it is held in the x direction. In the Y direction, no element is connected to the column for 9000 mm. In this case, the retained length of the column becomes 9000 mm.
 
I'm sorry, I said the opposite. I asked for lcy=4500 mm when he should have taken lcx=4500 mm, he got 9000mm, lcy=9000mm, but I wrote it wrong.
 
Can you let me know which project is wrong? I only see the images, if the direction is entered differently in the model, the results change.
 
The picture I posted above, not on the project (the example you explained in the webinar for bending and buckling in pressure elements) In this example, according to the x and y axis sets on the figure, while the column is held in the x axis direction, lcx=9000mm ; You have taken lcy = 4500 mm. If the figure is held in the x-axis direction, why did we not take it as lcx=4500 ;lcy:9000mm. Does it have a special case?
 
In the picture I took, the strong axis (y-axis) is given as the weak axis (x-axis), of course, if I don't have the wrong information, that's what I see in the picture.
 
To better understand the example, download the Steel Structures Application Guide from the website of the Ministry of Environment and Urbanization. Go to Axial Pressure section. You can also look at the steel structures lecture notes. When calculating under axial pressure, its lateral stiffness in the perpendicular direction to the relevant direction is checked. It will be clearer when you see the whole example.
 
yes, it is taken in the vertical direction, I understand. Well, in the idecad program, I connected it in the major (2) direction and it was split in two in the major direction, I threw the picture down, without looking. I am confused here.
 
Hello there; The program makes this calculation automatically. You do not need to interfere. If you want to test the program, you can set the geometry in the example to be exactly the same and compare the results of the calculation and program manually. Actually, you have to work as a subject.
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link will also help you.
 
Hello there; For detailed information on the subject
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link. If you are a member of the portal, you will also be able to see technical information. You can activate local axes in the program from the perspective screen right click analysis display tab.
 
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