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Opinions questions about using ideCAD Static 7
- Thread starter Ismail Hakki Besler
- Start date
HakanŞahin
Administrator
Hello again, If we go from the existing reinforcement, each of the reinforcement in the left and right support changes the assembly. Let's say we changed the assembly based on the new assembly value... Existing reinforcement in the supports will also change according to the new assembly value. Because the selected will always be larger than necessary. You have to rearrange the assembly as the existing reinforcement in the support changes. Consider also the reinforcement coming from the adjacent opening from the left... That's why we used the reinforcement based on the design moment..."proisa":gnfohnii" said:Yes, you're right, but there may be a difference in interpretation. After all, it is written as the reinforcement ratio in the regulation.
Good afternoon, the type, spacing and coordinates of the additional reinforcement are given in the formwork plan in the attached picture. However, in which direction should this reinforcement be distributed. Is it necessary to distribute this point equally above and below by centering it? Likewise, what should be the distribution direction for iron no. 634? Thanks
HakanŞahin
Administrator
Hello, It is necessary to discard the reinforcement by centering it on Position X, Position Y. Speaking for reinforcement no. 634: Mark the middle of the reinforcement (Position X, Position Y is here)... Left En/2, right [b Go as far as ]Width/2[/b] You will create a rectangular area with Length 5 pieces of reinforcement will be located in this area along Width . mark the middle of it... Go to Width/2 at the top, Width/2 at the bottom. You will create a rectangular area with the length. 24 pieces of reinforcement will be found in this area along the width.."nomask":16vz8ijm" said:Good afternoon, the type spacing and coordinates of the additional reinforcement are given in the formwork plan in the attached picture. But in which direction should this reinforcement be distributed. Is it necessary to distribute this point equally above and below?
Levent Özpak
New Member
Hello, Pattern After taking the plan, the location of the current section line can be moved with the move command, the section line can be selected and the section can be created again with the Create Section command from the right-click menu.Also, Drawing Menu Mold/Basic Plan With the command, a new section line can be defined and a section can be taken according to this line."proisa":3t3ubrd0" said:Hi, Is it possible to change the location of the 1-1 and 2-2 sections taken automatically in the Pattern Plan Drawings? That is, to change the place where the section passes by swiping left or right on the plan.
Thank you very much..."Levent Özpak":2dftd1z0" said:Hello, After taking the mold plan, the location of the existing section line can be moved with the move command, the section line can be selected and re-created with the Create section command from the right-click menu. section can be created. In addition, a new section line can be defined with the Drawing Menu Form/Foundation Plan section command and a section can be taken according to this line."proisa":2dftd1z0" said:Hi, Is it possible to swap 1-1 and 2-2 sections automatically taken in Mold Plan Drawings? In other words, to change the place where the section passes by swiping right left or up and down on the plan.
Levent Özpak
New Member
There is no definition of a stud-bearing beam in the 3D solution. and it is solved together. The effect expressed as the load of the inserted beam is actually the end shear force of the beam. The shear forces of the beams are given together with the beam name as Vshear under the heading "Beam Information and static results" in the Beam Report. The effects expressed as stud beam load, o good work"proisa":315gc4az" said:Where can we see the loads on the main beams due to stud beams in beam reports? Thank you.
Thanks for your answer. I just have one more question. I made some arrangements to type and make the reinforcements continuous in order to get the application sheets related to my project. In order to eliminate the discontinuity caused by the column reinforcements that are less in the lower floors and more in the upper floors, I rearranged the number of reinforcements in the lower floor according to the columns with the most reinforcement, but in the end. I am getting the attached insufficient reinforcement error. But I cannot find this error in the column calculation report. It is not included in the index in the report either.."HakanŞahin":i3roawhl" said:Hello, it is necessary to throw the reinforcement by centering it on Position X, Position Y. Speaking of reinforcement 634: Mark the middle of the reinforcement (This is Position X, Position Y)... Go to the left En/2, to the right En/2. With Length you will create a rectangular area. 5 pieces of reinforcement will be found in this area along Width . Speaking for reinforcement number 150: Mark the middle of the reinforcement... Go as far as Width/2 at the Top, Width/2 at the Bottom. You will create a rectangular area with height. 24 pieces of equipment will be found in this area throughout."nomask":i3roawhl" said:Good afternoon, the type range and coordinate of the additional reinforcement are given in the formwork plan in the attached picture. However, in which direction should this reinforcement be distributed. Is it necessary to distribute this point equally above and below by centering it?
In my project, Beam is connected to the column from two sides and another beam is connected to it in the middle. Again, we could not see the effects of the connected beam. The report states the columns at the two ends of the beam."Levent Özpak":1fb04kzn" said:There is no definition of a stuck-bearing beam in the 3D solution. The beams, which are expressed as stuck and bearing, are solved at once and together in the analysis. The effect expressed as the load of the inserted beam is actually the end shear force of the beam. The shear forces of the beams are given in the Beam Report under the heading "Beam Information and static results", with the name of the beam as Vshear. The influences expressed as stud beam load are those influences. Good work"proisa":1fb04kzn" said:Where can we see the loads on main beams due to stud beams in beam reports? Thank you.
HAVE YOU TRIED TO TIGHTEN THE STRIPES OF THE COLUMNS THAT FAILLED?"nomask":1hv81apc" said:Thanks for your answer. I have one more question. I made some precautions. In order to eliminate the discontinuity caused by the column reinforcements that are few in the lower floors and more in the upper floors, I rearranged the number of reinforcements in the lower floors according to the columns with the most reinforcement, but as a result, I get the error of insufficient reinforcement in the appendix. But I cannot find this error in the column calculation report. .It is not included in the index in the report either.."HakanŞahin":1hv81apc" said:Hello, it is necessary to discard the reinforcement by centering it on Position X, Position Y. Talking about reinforcement no. 634: Mark the middle of the reinforcement (Position X, Position Y). here)... Move to the left Width/2, to the right Width/2 You will create a rectangular area with Length. b]Wide [/b] will be located in this area. Speaking for the reinforcement number 150: Mark the middle of the reinforcement... Go as far as Width/2 at the Top, Width/2 at the Bottom. You will form a rectangular area with the Length. 24 pieces of reinforcement, Width throughout.."nomask":1hv81apc" said:Good afternoon, the type range and coordinate of the additional reinforcement are given in the formwork plan in the attached picture. In which direction is it necessary to distribute it. Is it necessary to distribute this point equally above and below by centering it?
Levent Özpak
New Member
According to the picture you attached, the right end V shear value of the KZ06 beam is the beam effect at the point connected to the KZ22. also, the static information and results of the KZ22 beam are not shown in the report."proisa":23cijb36" said:In my project, Beam is connected to the column from its two sides and another beam is connected to it from the middle. Again, we could not see the effects of the connected beam. The report states the columns at both ends of the report."Levent Özpak":23cijb36" said:3 In the multidimensional solution, there is no definition of a inserted-bearing beam. In the analysis, the beams expressed as inserted-bearing beams are solved at once and together. The effect expressed as the load of the inserted beam is actually the end shear force of the beam. Shear forces of beams are described in the Beam Report, "Beam Information and Static It is given together with the name of the beam as Vshear under the heading "Results of the Beam". The effects expressed as stud beam load are those influences. Good work"proisa":23cijb36" said:Where can we see the loads on main beams due to stud beams in beam reports? Thank you.
windtalker
New Member
Hello, floors are not listed on the floor parameters screen in the attached project. Is this because a rigid diaphragm has not been created? The project is used as a warehouse and the live load participation coefficient should be taken as 0.8. Since I can't see the floor parameters, I can't enter this value. Do you have any suggestions about this? If the analysis is done with the default settings, that is, n = 0.3, will it cause any problems? Thanks for your help.
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HakanŞahin
Administrator
"windtalker":1d6zw8re" said:Hello, floors are not listed on the floor parameters screen in the attached project. Is it because the rigid diaphragm has not been created? The project is considered a warehouse and the live load participation factor should be taken as 0.8. I cannot enter this value because I cannot see the floor parameters. Do you have any suggestions on this subject.Is it a problem if analysis is done with default settings, that is, n=0.3 is taken?Thanks for your help.You do not have permission to view link Log in or register now.
This effect is not related to the rigid diaphragm, but to the modeling of the floor mass as point or diffuse masses. The program becomes visible in the semi-rigid diaphragm solution, as it cannot reduce the system to point mass at the diaphragm level and automatically calculates with distributed masses. If the floor mass is modeled as a point, the xy translation of the floor and its rotation about the z axis are reduced to the point mass with the appropriate value. Here, assuming that the floor will move rigidly, a single RZ rotational mass is also modeled, which takes into account the effect of resisting rotation of the masses dispersed on the floor, as a result, when the equation is solved, results in RZ degrees of freedom. When calculations are made with scattered masses in the elements, only the masses that resist translation are modeled as scattered in the system. Although the movement of these masses separately produces torsion-like displacement modes in the system, it is not seen in the results since a mass in the RZ degree of freedom is not recognized mathematically. The RZ values obtained in the solution with the point mass are actually an approximation value calculated as a result of simplifying this complex many linear motions with the rigid assumption and reducing them to a single value.
windtalker
New Member
Hello. Is there any problem in terms of data entry in the attached file? Especially regarding the definition of slanted floors. Thanks for your help... download link:
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