Shear Safety Status on Columns (Tape window)

[harkan92]

In an existing architectural project, there are two strip band windows on one side of the shop floor. I have three column elements respectively affected by this band window. I have taken the lnet value of 0.5 m for my columns, which will exhibit short column behavior due to the rigidity of the wall. (band window height 50 cm) After the analysis, I left the lnet value to the "Find the Program" option as the amount of stirrups thrown out was not enough for me. My column height is 4 meters. As a result of the analysis I made again, I determined from the reports that the amount of stirrups thrown and the shear force value (''And'&#39 coming to the column for both cases were the same. In such a case, short column behavior is observed, but I could not understand how to take precautions against this situation in our program. How can I prevent this behavior? Thank you ... Civil Eng. Hakan ARKAN

 

[HakanŞahin]

In an existing architectural project, there are two strip band windows on one side of the shop floor. I have three column elements respectively affected by this band window. I have taken the lnet value of 0.5 m for my columns, which will exhibit short column behavior due to the wall rigidity. (band window height 50 cm) After the analysis, as the amount of stirrups thrown into my eyes was low, I left the lnet value to "Find the Program" option. My column height is 4 meters. ''And'&#39 program is the same. quote] Hello, For the columns that you think will affect the short column behavior of the wall rigidity, make sure to adjust the column settings. Enter the Ln value to be based on the right column behavior.

 

[harkan92]

I entered this value manually as lnet = 0.5 m, then I had it analyzed. However, when I press the Let the program find option for the lnet value, I still see that the same Ve value is found in the column break safety verifications in the report. The program finds the same result whether the lnet value is 0.5 m or 4 m.

 

[HakanŞahin]

I manually entered this value as lnet = 0.5 m, then I had the analysis done. But then when I clicked the program find the lnet value option, I still see that the same Ve value is found in the column shear safety verifications in the report. The program lnet value is 0.5 m Even if it is 4 m, it still finds the same result.

Hello again, in the regulation column shear Ve is limited to the value Ve found with R=2. In columns, "Ve shear force" is derived from "Ve (R=2)" If it is greater than the found shear force value, And (R=2) limit value is used. If the column you are looking at gets stuck at the limit value, you may have thought that the results were not different. I am adding a small example to clarify the subject:[/ b] If we look for the column major direction: S01 column Mu=0, Ma=156.16, Ln=0.5(defined) Ve=(Mu+Ma)/ln= (0+156.16)/0.5=312.32 > Ve(R =2)=180.50 tf Ve=180.50 tf SO2 column that appears in the report Mu=0, Ma=227.77, Ln=1.5(defined) Ve=(Mu+Ma)/ln=( 0+227.77)/1.5=151.85 < V e(R=2)=235.01 appears in the report Ve=151.85 tf SO3 column Mu=0, Ma=156.16, Ln=2.5(program find) Ve=(Mu+Ma)/ln=(0+ 156.16)/2.5=62.46 < Ve(R=2)=180.5 Ve=62.46 tf appearing in the report

 

[harkan92]

I entered this value manually as lnet = 0.5 m and then analyzed it. However, when I press the Let the program find option for the lnet value, I still see that the same Ve value is found in the column break safety verifications in the report. The program finds the same result whether the lnet value is 0.5 m or 4 m.

Hello again, in the regulation column shear force Ve is limited to the value Ve found with R=2. If "Ve shear force" is greater than the shear force value found from "Ve (R=2)" in the columns, the limit value Ve (R=2) is used. If the column you are looking at is stuck at the limit value, you may have thought that the results were not different. I'm adding a small example for clarity: For the column major aspect: S01 column Mu=0, Ma=156.16, Ln=0.5(defined) And=(Mu+ Ma)/ln= (0+156.16)/0.5=312.32 > Ve(R=2)=180.50 tf Ve=180.50 tf SO2 column Mu=0, Ma=227.77, Ln=1.5(defined) Ve=(Mu+Ma)/ln=(0+227.77)/1.5=151.85 < Ve(R=2)=235.01 appearing in the report Ve=151.85 tf SO3 column Mu=0, Ma=156.16, Ln=2.5(program find) Ve=(Mu+Ma)/ln=(0+156,16)/2.5=62.46 [/ b]< Ve(R=2)=180.5 Ve=62.46 tf

appearing in the report The subject is understood. Thank you very much for your understanding and effort.