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This is a good discussion because of the fact that it reminds us how the use terminology/ terms can lead to difference of approach. For example, when we talk about column punching shear, actually the limit state in that case is that the slab is punching through the column. Its not the column that requires checking but the slab which we always do. Similarly, the slab would be punching through the shear wall too, not the other way around as summarized above. Thanks.

Dear mhdhamood, I am happy to help............If you still not clear just let me know.....I ll send you some literature regarding this. What Rana Waseem summarized is absolutely perfect.........I just wanted to add in his Point # 1 that during this out of plane design we need to satisfy ACI 21.13.6 for Flat Slabs. In case of Beam supported slab no need to consider ACI 21.13.6. Thanks Muneeb

Very confusing discussion to me! Some notes: mhdhamood; please be clear and easy in your questions! Let me say what I think about slab system for this issue; 3 different cases 1. Only gravity load is present Slab is checked against out-of-plane bending Punching is checked against P and M arising from these loading 2. Only E loading (actually assume E as horizontal load because E and gravity loads are always accompanies together). Slab will resist this by axial stiffness (in-plane or diaphragm design). And columns will resist this load by bending stiffness. So in slabs, no out-of-plane bending..right? No slab will also have bending moment out-of-plane. Model a one-bay frame and apply a horizontal load of 100KN at LEFT-TOP joint and see the moments in beam. In this case, you will check slab design against axial load and this out-of-plane bending. For punching in this case, you will use this same M. And for axial force (P) P = 0 for one story frame (refer to frame model) P = P from column above for multi-story 3. When E and gravity both acts. For example in combinations including D+L+E There will be additional component of moment here called, P-Delta So you will use M = moment from p-delta + moment from step 1 + moment from step 2 and use P = P from step 1 + P from step 2 of the column above the slab you will use this P and M for calculating punching anywhere in slab location. If you are calculating punching for shear wall, it will follow the same procedure. Having said all this, please correct me if my understanding of this is not upto the highest standard. Thanks. Simplicity is sexy

Thank you so much Mr Muneeb , I get your point. Thank you for your patience.

Dear mhdhamood, My dear as I told you before that you are 100% right about this.... .....But these are inplane forces not out of plane forces. The picture you showed me is for inplane forces. As I told you before that you need to design for gravity for out of plane by satisfying 21.13.6 and then inplane forces (which you showed me) you nedd to design as well, which itself is a very complex kind of design. You not only need to design for collector forces as you showed but also for Shear Friction, Shear Reinforcement and Tension Reinforcement.......BUT ALL OF THIS DESIGN IS FOR INPLANE FORCES NOT FOR OUT OF PLANE FORCES You are confusing inplane and out of plane design. I ll summarize again: 1- Out of Plane: You need to check the thickness of slab (for flexure and shear )by satisfying 21.13.6 using gravity loads 2- In Plane : You need to check the thickness and provide shear reinforcement and flexural reinforcement for inplane design uding Lateral Loads. These reinforcement includes Tension (Chord) Reinforcement, Shear Friction Reinforcement, Shear Reinforcement, Retaining Wall Distribution Reinforcement: Retaining Wall Vertical Reinforcement The second type of design (for inplane forces) is not easy to handle. It involves a lot of knowledge and engineering judgement and is not common at all. If you want to design for this I can provide you guidelines. Hope it will be more clear now Thanks Muneeb

Dear mhdhamood, Sorry I forgot to add one thing. You can design the Flat Slab based on the gravity loading and then design the diaphragm for lateral forces. BUT there is one condition you need to check. In case of Flat Slab, if you are designing it on gravity loading then you need to check the story drift ratios. Please check ACT 21.13.6 (b ). We need to satisfy this condition if we want Flat Slab to be designed as gravity member. Thanks Muneeb

yes i am putting point load on node. Now after applying load as pressure and taking Poisson ratio as zero, there is no shear stress. Thank you for your response.

and how did u apply the axial load? as a point load on a node? if as a point load, then yeah the column will exp some (fake) shear stresses because in reality there is no point load, it should be uniformly distributed on the top surface of this column mesh (remember to reduce load to half on edge nodes, read about consistent vs lumped loads in FEA). Are you using 3d brick elements for the mesh? After doing this, you also need to turn of poissons ratio effect if there is any. Make v=0 and then all shear stresses should be zero.

Muneeb Badar Rana Waseem Thank you so much dears for this amazing discussion. because previously I was checking the slab in shear wall punching form the earthquake load combinations (it was wrong and over design) and because of high moment's values due to earthquake I was encountering punching shear problems ..... So from now I will just check the punching from the gravity loads

Dear mhdhamood, in your next comment you are asking which axial force to use, top shear wall or bottom shear wall whcih will cause rotation. Dear these shear forces can not be used to check the slab thickness as these are inplane shear forces when transferred from shaer wall to slab or from slab to shear wall. Even IF YOU PROVIDE THE THICKNESS FOR THIS SHEAR, IT WILL NOT BE EFFECTIVE IN CASE OF EARTHQUAKE AS YOU PROVIDED THE THICKNESS WHERE IT IS NOT REQUIRED. You have to increase the thickness in the basis of inplane bending not on the basis of out out of plane bending. Thanks Muneeb

Sir what difference would it make if we dont assign diaphragm in ETABS?

Diaphragms connect all individual members into a seismic force resisting system. If you have concrete floors, it is imperative that your floor will do the above mentioned job plus distribute the seismic load to all the connecting members too. Diaphragms can be rigid or flexible depending upon their stiffness. For practise in Pakistan, if your flooring is type conc slab > 5" consider your diaphragm to be rigid. A rigid diaphragm distributes forces to connecting members based on connecting members stiffness, where as a flexible diaphragm distributes loading based on tributary areas of supporting members to diaphragm. Take diaphragm as a frame that connects all members, and makes them act together.