How to calculate punching shear stress?

[NTCONLINE]

Hi,

I've found this document from CSI SAFE software, set out the procedure to calculate
punching shear stress that involves unbalanced moments.

https://files.engineering.com/getfi...2127-45bf-b59a-e6c762f73fff&file=ftn01-03.pdf

but it has limitations that the shape of the critical punching sections has to be parallel to X or Y axis.

What is the procedure that you would follow to calculate the maximum punching shear stress for a circular shape or a shape that has an edge not parallel to X or y-axis?

Thank you in advance.
NTCOnline.

 

[jayrod12]

If it was me, I would probably make a conservative assumption of a square column that provides similar punching shear area.

For example
600 diameter column and 200 thick slab,
Circular punching shear area = pi*(0.6+0.2)*0.2 = 0.502m^2
Equivalent square column (0.502 m^2/0.2)/4)-0.2= .427m therefore I would run the calcs based on a 400 square column to be conservative.

 

[BridgeSmith]

You should be able to back out the angle and area of the failure planes that are being used in the calculation. Then, you would substitute your calculated surface area. What it appears to be calculating is a truncated pyramid (typical for punching shear calcs), some parts of which are interrupted by the edges of the slab. In the case of a circular loading, you'll have a cone instead of a pyramid. For a square or rectangle oriented at something other than parallel or perpendicular to an edge that is within the pyramid, you'll have to do some fancy geometry to figure out the surface area of the remaining portion of the pyramid.

 

[BridgeSmith]

I'm not sure the equivalent square column would be conservative, jayrod (maybe it is - I haven't done the math). An inscribed square area would definitely be conservative. My 'eyeball' guess is that the equivalent square column would have very close to an equivalent failure surface area and therefore very close to an equivalent capacity. I believe a square with an equivalent perimeter length would have the same failure surface area, assuming no slab edges intersect the failure surface. In that case, I think it would depend on the orientation of the assumed equivalent square. At that point it's probably just easier to calculate the surface area of the cone directly.

 

[jayrod12]

That's what I was doing, calculating the surface area of the cone, converting it to an equivalent square failure plane. I wasn't comparing bearing areas otherwise that would've resulted in a 525x525 as opposed to a 400x400. I feel with the shorter lever arm for much of the square failure plane as opposed to the circular, that we're being doubly or triply conservative. But to be honest, I don't have any hard numbers to back that up. Just my guy feel.

 

[BridgeSmith]

I see now that you were doing a surface area, jayrod, but it appears to be the surface area of a cylinder and a converting to a rectangular shape, rather than a cone to a pyramid, although you should get a similar answers. It just looked too simple to be the surface area calc I was expecting, at first glance.

 

[jayrod12]

The surface area of the cone would be partially more, same for the pyramid. But to me, it's conservative to ignore the 1.414*t when doing the area calc. and it makes my life so much easier.

 

[BridgeSmith]

I agree the ratio of the area of a cone to a pyramid is very similar to that of cylinder to a cuboid, so your simplification is easily close enough to approximate punching shear capacity (especially considering the huge amount of approximation involved otherwise). I was just trying to explain away my erroneous assumption.

 

[rapt]

You should be able to calculate the Polar Moment of Inertia of the shear plane for any shape to a reasonable degree of accuracy. I remember doing it for circles and combined circles and rectangles for edge circular columns many years ago.I think an old copy of the PCI Manual used to define the methodology fairly well. Not sure about the current one.

 

[Agent666]

In NZS3101 (NZ concrete standard) it specifically says a circular column may be replaced with a square column of equal area (for what its worth).

Also fib 12 guide also has quite a bit of stuff relating to review of various standards and in particular some stuff on circular columns and how they are treated by various standards around the world.

Otherwise do what Rapt noted as this is the "code approach" for any irregular section/punching shear perimeter, I can't imagine its that hard and is somewhat analogous to finding the polar moment of inertia for weld groups (I think...famous last words perhaps...). Report back if you go this way to compare with other assumptions.

 

[Agent666]

For a comparison point as far as how alternative software might treat the scenario, RISA also note that they convert a circular column to an equivalent square for design checks

 

[Agent666]

One other point that makes the assumption of an equivalent square member easier to digest in terms of adopting the code equations directly, is that the equations for the transfer of unbalanced moment via direct flexure and the remainder by eccentricity of shear are all obviously in terms of the width of a 'rectangular' critical section (defined as b1 and b2 in ACI), how well this relationship translates to circular sections I have no idea (as my code says circular sections should be equated as square because there is little research/testing on how these relationships are applied to circular columns for the mechanism of transferring the moment via flexure or eccentricity of shear).

But i would assume you can use it directly for any irregular section taking the critical width equal to the widest(?) part of the critical section. For a circle or a square the ratio of b1/b2 is obviously the same anyway if taking the widest location stands true, so the ratio of moment being transferred by flexure or by eccentricity of shear is the same whether you take a circle or equivalent square.

 

[NTCONLINE]

Rapt,

I agree that we should be able to calculate the Polar Moment of Inertia of the shear plane for any shape. However, the textbook calculations usually assume the axis of torsion is perpendicular to the section. So I am quite unsure for the case where the axis of rotation is not perpendicular to the section, which is quite often the case in punching shear (one shear plane might not be perpendicular or parallel to the bending axes)

Maybe I should project the shear plane to the plane that perpendicular to the rotational axis, and go ahead and calculate the torsional constant J of the projected plane? Does this sound reasonable?


NTC

 

[Agent666]

Check this thread