Inverted Beams - Shear Wall Meshing Effects

[khalid]

Salam Alaikom Dear professionals!

 

Kindly please help me with the following uncertainties and questions I faced with developing the attached model Etabs model:

 

1. There are some long span (18m) inverted or up-stand beams provided to insure enough headroom in the story bellow. How can I model such kinds of beams in Etabs? How much they are structurally similar to conventional beams?
2. What are the significant effects of shear wall meshing on the structural behavior and the design results? In term of computational efficiency, how much should be the mesh size?
3. What is the proper way of supporting the beam over the shear wall and the grid beam?
4. Does Etabs take care of column’s slenderness?

 

Also if you could take your invaluable time to check the model and give me your professional comments and feedback, which  would be highly appreciated!

 

Thanks and Best Regards

 

Trial I.EDB

[WR1]

Salam Alaikom Dear professionals!

 

Kindly please help me with the following uncertainties and questions I faced with developing the attached model Etabs model:

 

1. There are some long span (18m) inverted or up-stand beams provided to insure enough headroom in the story bellow. How can I model such kinds of beams in Etabs? How much they are structurally similar to conventional beams?
2. What are the significant effects of shear wall meshing on the structural behavior and the design results? In term of computational efficiency, how much should be the mesh size?
3. What is the proper way of supporting the beam over the shear wall and the grid beam?
4. Does Etabs take care of column’s slenderness?

 

Also if you could take your invaluable time to check the model and give me your professional comments and feedback, which  would be highly appreciated!

 

Thanks and Best Regards

 

1. its simple. just study about t and rectangular sections. where you design for T section and where you design as rectangular section for drop beams? for inverted beams just flip these sections and design according to that. another thing to make sure is to design inverted beam shear stress so that they are capable to carry the axial tension.

 

2. important question. meshing in slab or wall or in anything is very important thing in FEA. just read in any finite element book about meshing. it depends on the problem . some times you need refined mesh some times not. google mesh convergence topic. make an example of point load on slab and keep refining mesh at that area and notice the values of stress. does that keep increasing with refined mesh? then ask your self why? and what is the best mesh size that gives accurate results? then study about stress concentration in any good fea book.

 

3. continue the beam atleast to two mesh points inside the wall.

 

4. yes!

[WR1]

what exactly is your question for the model you attached?

[khalid]

Thank you!

 

As per my understanding in the T-Beams the slab serves as the top flange of the beam but in our case, the inverted beam (Rectangular), the slab no longer serve as top flange. Also in inverted Rectangular Beam the stirrups will be additionally subjected to tension transferred from the hanging slab hence this should be considered in the design. 

 

I could not catch your instruction on how to draw the inverted beam or maybe there is no option for them in Etabs 2010 :/   I should design it manually.

 

Does this also affect the hanging slab behavior? 

 

[WR1]

no need to draw it as special inverted beam, just draw it as normal line. you can save as your model and then delete the slabs just to design the inverted beams but put loading on beams from slabs manually as line loads.

 

this way you are not taking t section in stiffness calculations that is conservative. just put the reinforcement that you get from etabs and design shear stirrups manually for additional tension.

 

this is T=Asfy Phi (Phi is 0.9)

 

for example calculate the slab loading including self weight per meter on this beam for example lets say ultimate load on 1.2D+1.6L = 50kn/m (just an assumption)

 

then 50 * 1000 N/m = As (mm²/m) . 420 (N/mm²) . 0.9

 

As = 132 mm²/m .

 

now check how many stirrup legs you have provided per meter. for example (dia 10 4 legs every 200mm)

 

that means you have 5*4=20 legs per meter

 

20 x 78.5 = 1570 mm² / 1m run of beam