Thanks for your response la belle vie, I just struggle to understand why we base the pure concrete punching shear resistance based on the depth "d" to that top reinforcing in the empirical method in the code if we are only relying on concrete tensions.
As the failure plane is on an angle, wouldn’t the top reinforcement (including any other reinforcement within the failure zone) take on vertical (shear friction) and a tension component?
I just find it off how the punching shear formula just asks for depth 'd' of the reinforcement, but we never define what the reinforcement content is. Surely having more reinforcement in the top matt would make a difference?
Also, if we look at section AA in the ACI reference above, can we consider the layer perpendicular to the top most layer to resist in shear friction as the top layer will sort of restrain it from popping out?
Thank you both for your response. So if we had a middle layer of reinforcement (for whatever reason) there will be sufficient concrete to mobilise the bar and provide shear friction to the failure plane?
@harbutmild For the punching shear example, wouldn’t the top bars provide the shear...
Thanks for your response @RabitPete. Would you mind explaining further the mechanism for item 3, with the straight bars? I was thinking if the vertical vector could transfer to a horizontal one and if we have the bar developed past the failure zone it could work?
Hi all,
Just a few questions below:
1. With the 2 U bars shown below on ACI, would that count as 4x Astxfy or 2x Ast x fy for tension capacity of the tension reo?
2. Can I just provide sufficient tension reo on one side of the failure plane (not both sides symmetrically). Using the same...
@Kootk and @strucbells , stupid questions...
1. With the 2 U bars shown below on ACI, would that count as 4x Astxfy or 2x Ast x fy for tension capacity?
2. Can I just provide sufficient tension reo on one side of the failure plane (not both sides symmetrically).
3. Instead of U bars, can we use...