Plastic center in WSQ/HSQ beams 1

[agps]

Hi everyone!

I want to create a a beam type WSQ (or HSQ c.f picture

http://www.forssellssmide.se/index.php?option=com_content&task=view&id=12&Itemid=30

).

This beam is hollow and its purpose is to support pre-fabricated slabs.

I´m working on a 16.20m span and I´m using steel S275. The loads are not relevant for my question.

When I searched into the supplyer catalog (which I send a copy attached) for a beam that could face my needs, I found none that verified deformations. Therefore I decided to create my own.

After designing a section that verified deformations for this span (L/300) and trying to find its mechanical geometric propreties, my "Plastic Center" (Zpl) had negative values.

I analyzed the problem by assuming initially that, Zpl, would be positioned on the web of the section (50% of the internal forces compressed and the other 50% stressed). I got the value of Zpl= -355mm.

I could re-design my section in order to bring up Zpl values until they become positive. But that´s not my intention. I wanna keep this geometric section.

Just to check, I did the same analyzes to the 1st section of the catalog (attached 1st WSQ180) and Zpl was negative also. I found it realy curiouse after finding Ixx, Wx1, Wx2 values and then not finding "Wpl" (which is not even mencioned in the catalog).

I did the same analysis for the 1st section which has a "Wpl" value mentioned in the catalog (Wpl = 697mm3) and everything went well. I got all the geom. values.

My doubts:

1- Am I analysing the problem correctly assuming initially Zpl in the web? If I analyse it considering Zpl in the Flange I get positive values.
Why and which method is correct?
(Note: I assumed vertical axis (z-z) on the center of the section and the horizontal one (y-y) on level zero of the section under the flange)

2- After getting this results and face them with the catalog, I thought twards the 1st three sections (w/ no Wpl):
Could this sections only work in an elastic analysis and not plastic?

In attached I send a copy of the catalog and a file in Cad to help understanding better the section.

Thank you for taking your time to read my problem.


agps

 

[statiker]

hi agps,

first of all: to calculate Wpl you have to know Mpl.
the plastic center line (where the tension changes to compression) divides the area of the whole cross-section
in two halfes, because of equilibrium...
that means that it is located near or in
the lower flange.

 

[agps]

Hi statiker,

Thank you for ackowledge my problem.

I understand when you say that I´ve to know my Mpl first, but are you answering my first point in a positive way?

"1- Am I analysing the problem correctly assuming initially Zpl in the web? If I analyse it considering Zpl in the Flange I get positive values.
Why and which method is correct?"

Should I assume the plastic center on the flange instead?

Tks

agps

 

[statiker]

first of all we have to satisfy the rule of equilibrium
between inner and outer forces. so if we calculate Mpl, the
sum of horizontal forces Npl of the parts above and below
the plastic center line have to add up to zero.
(pure bending assumed)

if you assume the plastic centerline anywhere adjacent to the correct place some forces remain, which is not allowed.

so vary the location as long as the equilibrium law is not
satisfied and you get the correct solution...

if you need a sketch, let me know !

 

[civilperson]

The sum of T and the sum of C must be equal and with all material being steel the areas are equal on either side of the Plastic Neutral Axis, (thus it is in the lower plate).

 

[agps]

Dear coleagues,

Thank you very much for your help. I´m now sure about my results.

For my specific case de Zpl=33.42mm and is situated on the lower plate.

Before closing this forum, I would like to ask you what´s your considerations facing the following results for my section:

Wel = 381,78mm3
Wpl = 4336648mm3

Therefore the k(shape constant)=11359,12

 

[statiker]

sorry,this result seems impoosible..., k is tooooo high

 

[agps]

Statiker,

Thank you for your feed-back. Obviously the k was tooo high but sometimes It seems like I just needed to hear someone telling me that.

I went back to my excel worksheet and find the mistake.
It was a units problem.

You were very helpful. Thank you "civilperson" as well.

Best regards,

agps

 

[agps]

Just by curiosity:

k = 1.20