the factor m in calculating αm for global initial sway imperfections

[GWZ@CN]

Am I understand the m correctly? My understanding is that when calculate a single span potal frame,m always equals 1,because there alway be a column which Ned is less than the average.And for double span frame,which the inside column NEd is usually much larger than the other two,so m equals 1 most times.
I m confused because I saw a software tutorial vedio,in which the presenter input m=2 for a single span portal frame

m is the number of columns in a row including only those columns which carry a vertical load NEd not less than 50% of the average value of the column in the vertical plane considered​

 

[HTURKAK]

Am I understand the m correctly?

I don't want to disappoint you but your interpretation is wrong.
As you have copy and pasted from EC -1993 ;
m is the number of columns in a row including only those columns which carry a vertical load NEd not less than 50% of the average value of the column in the vertical plane considered and for single span portal frames m=2

.And for double span frame,which the inside column NEd is usually much larger than the other two,so m equals 1 most times.

This is also not true. Consider symmetric double span frame with UDL total 16. The inside mid column will carry 10 units , the side columns will carry 3 units. The average column load is 16/3 =5.33 and 50% of the average makes 2.66 units. All of the tree columns resist more than 50% of the average value . So , m=3

I hope my respond answers your question.

 

[GWZ@CN]

I don't want to disappoint you but your interpretation is wrong.
As you have copy and pasted from EC -1993 ;
m is the number of columns in a row including only those columns which carry a vertical load NEd not less than 50% of the average value of the column in the vertical plane considered and for single span portal frames m=2

This is also not true. Consider symmetric double span frame with UDL total 16. The inside mid column will carry 10 units , the side columns will carry 3 units. The average column load is 16/3 =5.33 and 50% of the average makes 2.66 units. All of the tree columns resist more than 50% of the average value . So , m=3

I hope my respond answers your question.

thanks for your patience.I find I'm error cause I missed the '50%'.maybe I get mistaked by the co-use of '50%' and 'average'.thanks