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Turkish Hello, when I analyze the building that I modeled as a single storey and basement (rigid floor), the program closes. If I do not choose the solid as a rigid floor, I can analyze it without any problems. How should I analyze this structure, which will be completely underground?
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Analysis in single storey and basement (rigid storey) structures
- Thread starter Muhpen
- Start date
Re: Rigid floor analysis Hello My Suggestions --Semi-rigid diaphragm, --Basement floor rigidities from floor general settings, --Flood precipitate on curtains, (fixed so far) --Removing wall load over beams, - - By arranging the stair supports, - If there is a soil load on the outer curtains, give a solution by giving a horizontal soil load. Unver ÖZCAN
HakanŞahin
Administrator
Re: Rigid floor analysis
Hello, according to TDY, floor force will be applied to the rigid basement floor as Fbk= Ao*I*Wbk/1.5.Your building weight Wbk=359.05 tons, earthquake zone coefficient 1st zone Ao=0.4, building importance factor I =1.2.Fbk=0.4*1.2*356.78/1.5= 114.89 tf. Select "Defined" in the floor parameters and enter 115 tf as Fx, Fy. If solid, do not mark as rigid basement In the analysis of a single-storey building with a rigid basement, the closure of the program is among our notes."Muhpen":3dq5qyqd" said:Hi, the program closes when I analyze the building that I modeled as a single storey and basement (rigid storey). How should I analyze this structure that will be under the ground?
sahinsukru
New Member
Re: Rigid floor analysis
The definition of a rigid basement floor is meaningful if there are floors or floors other than the rigid floor.In my opinion, since your building has three floors and all three floors have the same characteristics, the definition of a rigid floor is the same. 3A.6.4 - Especially for transition floors, which are located in the transition from normally rigid floors to very rigid basement floors, and transfer the majority of the earthquake forces formed in the upper floors to the perimeter curtains in the basement floors, there is no need to do this. It is essential to provide in-plane rigidity and strength."MaFiAMaX":2lqp99t8" said:Greetings, Mr. Hakan, Will there be a regulation on this subject soon? Load defined in v10 cannot be entered. I remove the rigid floor signs and take R=1.5 and get modal analysis.
Re: Rigid floor analysis Mr. Şükrü, Thank you for your comment. I think differently. In your opinion, basement floors that cannot be a normal building above are not covered by the regulation's basement floors. However, I think that what is meant by basement is the structure that is under the ground. The free vibration modes of the structures in the soil are compatible with the ground. Like the superstructure, the free vibration mode does not depend on the weight and stiffness of the structure. In other words, the horizontal forces that occur in the structure are not inertial forces, as in the superstructure. The earthquake load reduction coefficient is taken as 1.5, since the structure receives maximum accelerations in the ground. As you stated, a mathematical error occurs when the calculation is made. Namely; If I do not specify any floor of the building as a rigid basement in ideCAD, the articles 4.7.5.3 and 4.8.5.3 of the regulation will not be applied in both equivalent and modal analysis. Suppose the period of the structure is close to zero. Let's take T=0 and substitute it in formula 4.1b, we get Ra = D. Generally, D=3 or 2,5 for the superstructure. As a result, the Ra value is taken higher than it should be. Moreover, sometimes in a building with a very large mass, even if there are basement curtains around it, the period value is not close to zero, as expected. In a normal structural analysis (basement floor unmarked), the Ra value will be even higher and less earthquake force will be calculated. However, when this structure is underground, it will behave differently from the T= 2pi*In Root (m/k) value you calculated. This is how I interpret it. I hope the software team will fix our mistake if there is one.
Ismail Hakki Besler
Administrator
Re: Rigid floor analysis
You can do this. Alternatively, you can mark all floors as rigid and enter another non-rigid floor on the top floor. You can enter a very small column on this floor and solve it."MaFiAMaX":10ctufxa" said:Greetings, Mr. Hakan, Will there be a regulation on this subject soon? Load defined in v10 cannot be entered. I remove the rigid floor signs and take R=1.5 and get modal analysis.
sahinsukru
New Member
Re: Rigid floor analysis
Hello Mr. Erhan; First of all, thanks for your contribution. In my opinion, a floor does not need to be underground to be a rigid floor. There is no such statement in the regulation that I know of. In order for the items you wrote to be valid, a situation like 3A.6.4 must be present. In addition, this article also imposes the condition of 'providing sufficient in-plane rigidity and strength in transition floors'. During the construction of basement perimeter curtains, the surrounding ground will be disturbed and this disturbed ground will act as a dilatation between the ground and the building. I maintain my opinion that if there is no stiffness difference between floors as in 3A.6.4, the definition of rigid floor should not be applied. Note: In my opinion, the following article is about this issue. 4.5.7.1 - As stated in 3A.6.4, adequate in-plane stiffness and in-plane stiffness in the transition floors slabs, which are involved in the transition from normal floors to very rigid basement floors and have to transfer all or most of the inertial forces formed in the upper floors to the perimeter walls of the basement floors suddenly. Ensuring resilience is essential. This condition is also valid for other transition floors where sudden stiffness changes are made for other reasons. I would like to read it if it will contribute to this issue at an academic level."MaFiAMaX":vkfvimj8" said:Sükrü Bey, Thank you for your comment. I think differently. According to your opinion, basement floors that cannot be above a normal building are not within the scope of basement floors in the regulation. However, I think that basement means the structure that is under the ground. The free vibration modes of the structures in the soil are compatible with the ground. The free vibration mode, like the superstructure, does not depend on the weight and rigidity of the structure. That is, the horizontal forces in the structure as in the superstructure are not inertial forces. The structure in the ground is maximum The earthquake load reduction coefficient is taken as 1.5. As you stated, there is a mathematical error when the calculation is made. Namely, if I do not specify any floor of the building as a rigid basement floor in ideCAD, in both equivalent and modal analysis, the 4.7.5.3 and 4.8 of the regulation will be taken. Items .5.3 will not be applied as they are. Let's assume the period of the structure is close to zero. Let's take T=0 and in formula 4.1b we get If we put it in, we get Ra = D. Generally, D=3 or 2,5 for the superstructure. As a result, the Ra value is taken higher than it should be. Moreover, sometimes in a building with a very large mass, even if there are basement curtains around it, the period value is not close to zero, as expected. In a normal structural analysis (basement floor unmarked), the Ra value will be even higher and less earthquake force will be calculated. However, when this structure is underground, it will behave differently from the T= 2pi*In Root (m/k) value you calculated. This is how I interpret it. I hope the software team will correct our mistake.