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Article 4.3.4.6 Control * IMPORTANT EXPLANATION *
- Thread starter Ismail Hakki Besler
- Start date
Ismail Hakki Besler
Administrator
You are right. That's what happens on the cell phone in a hurry. I fixed it as follows by going to the computer and copy-pasting the relevant icons from the regulation. For example, the X direction is n. for mode More precisely, the modal floor shear force for each mode is obtained by multiplying the following quantities; Vtxn = mtxn x SaR(Tn) Vtxn = Base shear modal effective mass x reduced design spectral acceleration (Vibration period) See Fig. 2018 Earthquake Code APPENDIX 4B MODAL ACCOUNT METHODS 4B0 ICONS
Ismail Hakki Besler
Administrator
at 10.05 In the account I explained above We have improved it by making a change. The current is version 10.05. We have improved the Mdev calculation for shear walls with tie beams. Mdev calculation has been changed for shear walls without tie beams."ibrahim93":2hp2qps1" said:ismail bey The project, which exceeded 75% in the versions before 10.05, is now around 0.25. I guess it was made on 10.03. What changed in 10.05 v. This rate made such a difference
Ismail Hakki Besler
Administrator
"ibrahim93":1dima5sq" said:The projects we delivered on 10.03v are now wrong then? and the projects we have finished are garbage because it is impossible to achieve this ratio
Current version is 10.05."Ismail Hakki Besler":1dima5sq" said:In 10.05, we developed it by making a change in the account I explained above. The current version is 10.05. We have developed the Mdev calculation for shear walls. Mdev calculation has been changed for shear walls without tie beams.
Ismail Hakki Besler
Administrator
It does not matter if the projects licensed in the old version are wrong. These projects are probably low-rise ribbed. We think that the drafters of the regulation tested these rules for multi-storey buildings. In low-rise flat structures, this control may be should be done in the logic of alphas. If you say it is wrong, we can ask whether the projects made with the 2007 regulation are also wrong. Also, this article is probably in the regulation, which may change. For this version of the regulation, we are currently doing it as the regulation says do. ."ibrahim93":3tx4ntix" said:is it wrong
yusuf15hira17
New Member
Re: ideCAD Static IDS v10.05 Published (14.03.2019) hello, the hollow wall system smaller than 17.50 m, whose images I have shared in the appendix, is a full curtain system in a plan. We could not provide the 0.75 ratio, although it was not.
Re: ideCAD Static IDS v10.05 Released (14.03.2019) yusuf15hira17, The building does not have a screen in the X direction, so it is expected that Mdev > 0.75M0 should not be provided anyway, but the Y direction is thought-provoking. It takes very little shear force when it is alone on the curtain (non-core) and outer axles. For this reason, Mdev is probably low. Regardless of the regulation, the rectangular (I fret) often takes little cutting. If you share your model, we will look at it in more detail and reply.
Ismail Hakki Besler
Administrator
Re: ideCAD Static IDS v10.05 Released (14.03.2019)
This project is the most typical example. Although there are much less shears in the X direction, a reasonable ratio is 0.56, since there is a core in the very center, while reducing the axial forces of all walls in the Y direction. It's impossible for you to provide because we couldn't take it into account. if you wish, you can solve it by manually entering the R coefficients in the 2018 regulation. If you were solving this project with a constructive program, you would do it like that z."yusuf15hira17":g7mdzzfd" said:hello, the hollow floor system smaller than 17.50m, whose images I have shared in the attached, is a full curtain system in a plan, omega value in 2 corners We could not achieve the ratio of 0.75 even though there are shear walls with a connecting beam of 0.46 and there are no columns.
Hi İsmail, I agree with you in the calculation of the total overturning moment, but in the definition made for the tie beam wall in the regulation, it refers to the sum of the Tie Beam shear forces at the wall base as tension/pressure instead of the moment contribution of the Wall Axial Forces. In other words, axial forces are not taken into account in the hollow curtain, neither in the regulation nor in its applications. Also, is it possible to provide a basis for the calculation of the Total Rolling Moment, which you describe below and which I think should be the same? I couldn't find anything. Although my licensing information says it should be as you say.
A simple model "curtain without tie beams" has been checked in the program. The results are given below. (It can be controlled by creating a similar model) In this model, everything is chosen symmetrically. The sum of the moments to be created by the wall axial forces N1 and N2 with respect to the center of gravity of the system should cancel each other out. Therefore, in this system, the value of alphaM should be 1, but it was calculated as 0.80 in the program. Based on the analysis results; In the process performed, it was not found by which distance the N1=N2=8.005 tf value was multiplied (arm of force). The calculation of M0 as you described is given below. I'd appreciate it if you let me know if I missed something. good work, best regards...Story total overturning moment = MDev of columns + Mdev of walls + Normal forces of columns x distance from center of gravity + Normal forces of walls x distance from center of gravity Earthquake The regulation takes into account the part shown in red only for curtains with tie beams that meet the 1/3 requirement. It never takes into account the part shown in blue
Ismail Hakki Besler
Administrator
Nv is the sum of the ligament shear forces. Check that the 40/50 beam meets the 1/3 requirement. If it doesn't, it won't be 1.00. The force arm is printed in the report with 1/3 control. It is difficult to meet the requirement of coupling with 40/50 beam at 4.60 m. Click on the link below and read.
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Nv is the curtain axial force.Nv is the sum of the axial forces of the tie beam.
I think this expression must have been written inadvertently? In the example I shared above, I specifically stated that "gapless shear" is designed, so I am aware that beam shears cannot create a "force couple".Nv is the curtain axial force.
The 1/3 condition is not met and according to your definition for Mo, I say that the ratio should be 1 in this system. I need to refresh the question.Check that 40/50 beam satisfies the 1/3 requirement. If it does not, it will not be 1.00
If the program does as you describe, is there anything I missed while verifying it? (details in the previous message) M0 = 248.92tfm, but 312.76 tfm reports are printed. My acceptance: Since both of the 8.005 tf Curtain Axial forces are in the same direction, the system will not be able to generate a rotation with respect to the center of gravity. What the program does: M1+M2+N1.d1+N2.d2.... M1+M2=248.92 If M0=312.76 tfm, the difference should consist of Shear Axial Forces. 312.76 - 248.92=63.84 tfm divide this result by N1, N2 or N1+N2 (which is not meaningful to me), the distance you will find will not be a consistent distance according to the system geometry. The distances of N1N2 to the System Center of Gravity are d1=d2= 3.80 m. You can rest assured that the bond beam (hollow) model has been worked on. I will share its details during the day. good work, best regards.Story total overturning moment = MDev of columns + Mdev of walls + Normal forces of columns x distance to center of gravity + Normal forces of walls x distance to center of gravity
Ismail Hakki Besler
Administrator
Yes, I wrote it accidentally. Indeed; Nv will be the sum of the shear forces of the tie beam. It is already written like that in the image below. Since the Mo equivalent that ideCAD calculates is not calculated, the equivalent you calculated will not be the same as the tipping. Mo is calculated modally. If Mo is calculated as equivalent, the wall end forces must also be found with the equivalent earthquake calculation. If Mo is calculated modally, the curtain end forces must also be found modally. Mo is calculated modally in the program. Curtain end forces are calculated modally. Modal results are unsigned results, so they are not + -, they are just calculated as absolute values. Total tipping moment of storey = MDev of columns + Mdev of walls + Normal forces of columns x distance from center of gravity + Normal forces of walls x distance from center of gravity My expression in this above paragraph is the entire total tipping stability of the structure. Earthquake regulation does not say to calculate Mdev/Mo by providing this balance. In the statement I wrote above, I do not say that it should be done like that. I say offering; Even if there are only curtains in the system, if all of these curtains do not meet the requirement of stringed curtains and 1/3, I say it is almost impossible to meet the 0.75 requirement of the regulation. While writing this statement, I emphasize that it cannot be caused by the normal force effects that I have shown in blue and red.
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condition 1/3 This balance is achieved in systems that provide curtains only. And the ratio is 1. Because the regulation says to consider Nv in curtains with tie beams that meet the 1/3 requirement. This ratio will not be 1 in systems that do not meet the 1/3 requirement and have only curtains. Because the regulation does not say to consider Nv in curtains that do not meet this requirement. Add the ideCAD data for which you have given the numerical values. Express it without the need for long words; In this system, the water value is found by the program in the X direction of the tipping ratio, whereas the water value should rise. One project and two values please. The first value is what the program found, the second value is what you found. I will add the overturning moment with Etabs to modal calculations during the week.Hello Mr. Muhammed, Nv is the shear force in the tie beam. The shear force in the beam becomes the axial force at the curtain end with respect to the curtain. We can say that shear force at the end of the tie beam = shear axial force, since no additional or self-loading is added to the results in pure earthquake loading. Another issue is that due to earthquake loading in the pair of walls (connector beam shear), the axial force should be tension on one wall and pressure on the other wall. In this example, the reason you see the same direction (pressure) on both curtains is due to the solution without a diaphragm (combining element-specific results in the settings), loads are shown as absolute values. That is, if the tie-beam rule was provided in your calculation, c*Nv should be taken into account.
I checked the calculations Mr. Ismail, I have a mistake about the axial forces. thank you for the information. Erhan, thank you for your contribution.
Mr. İsmail, I don't know if this was necessary... We are trying to understand something here. I think that's the purpose of the forum. Why are you guessing? Got a style problem? I think how many words I need to use to describe the problem, let us decide so that we can express ourselves more easily. I noticed this expression in your message later and I am sharing the question I asked in detail below without retracting it. (Just keep it like this for once, it won't happen again, you can be sure.) Corrected; Mo = M1+M2+N1d1+N2d2 Mdev = M1+M2 These values for the project; Mo = 124.462+124.462+8.505*3.8+8.505*3.8 = 313.565 Mdev = 124.462*2 = 248.924 AlphaM= 248.924/313.565 = 0.793 there is only a slight difference in the value read from the program. ( Mo = 312.76 / 313.56) For Shears with Holes: We collect the beam shear forces and carry them to the curtain floor in the form of tension/pressure. In the control of the loyalty level report; Nv value and M1 M2 moments are also printed as quite different results. When we add the sum of shear forces taken from the program to the Mo calculation as a force couple at the wall base, we get the following result. Calculation results taken from the program: In Application Examples Handbook BA2/10, it is said that the porous wall is handled independently of the structure and analyzed under triangular fictitious load. Is this difference in program results related to the calculation method? Does the program perform structure-independent control under fictitious load? good work, best regards...Add the ideCAD data for which you have given the numerical values. Express it without the need for long words; In this system, the water value is found by the program in the X direction of the tipping ratio, whereas the water value should rise. One project and two values please. The first value is what the program finds, the second value is yours.
Ismail Hakki Besler
Administrator
Before seeing the project There is only 2 curtains and 1 tie beam between them. If there is no other column and curtain; So like the project in this title
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