# diagonal shear + overturning



## McEngr (Aug 3, 2012)

My boss asked me to check a vessel/hopper footing for shear and overturning at a 45 degree angle to the square footing. I tried to derive this, but it involves integration. Anyone have a quickie formula that I can use?

Thanks.

btw~I hate to use RISA Foundation for something so little like this, so any help is appreciated.


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## ipswitch (Aug 4, 2012)

I wonder if you can check the diagonal shear the same way you check the shear in the web of a bulb tee. I know that too is an iterative process but a professor I had used a technique (or assumption) that circumvented that interative process. I'll have to check my notes.


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## ipswitch (Aug 4, 2012)

I looked for my notes and they might me at the office. I do believe if you can google the diagonal shear check in bulb tee design you can stop after one iteration.


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## ipswitch (Aug 4, 2012)

Start at section 9.4.11.1.1 You may be able to modify this equation and do as little as 2 iterations.

/&gt;http://www.pci.org/view_file.cfm?file=MNL-133-97_ch9_4.pdf


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## McEngr (Aug 4, 2012)

ipswitch,

I'm only checking the bearing capacity. I'm not checking the internal shear (flexural or two-way).


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## ipswitch (Aug 5, 2012)

and section 10.14 of chap 6 of ACI Notes is no good?


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## kevo_55 (Aug 6, 2012)

Are you using a mat foundation?

I've never done a mat foundation for this but I did use a pile foundation once. I simply took the polar moment of inertia for the pile group and calculated the maximum compression/uplift.

With the mat foundation, I'd most likely bust out RISA. I have no idea how to do that by hand, so I'd also be asking around if what RISA gives actually makes sense too.


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## McEngr (Aug 6, 2012)

We have RISA 3D+Foundation here at the office. I've been thinking I could set up a template, but I'm against using analysis programs that I can easily do in a half-page of hand calcs. I believe this is possible. The question is multi-faceted because the bearing equation changes after you have reached the bearing length beyond the cross-axis of the square footing. It gets even more complicated with a rectangular (B and L different) footing. I think this would be similar to solving gradients with integrals in a three dimensional calculus class. I'll admit, I could whip this out without much problem in college, but as the saying goes - "Don't use it, and you lose it."


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## ipswitch (Aug 6, 2012)

Maybe I'm out of my league here. You can't use a punching shear formula for footings?


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## McEngr (Aug 6, 2012)

The punching shear would be the same for a circular cross-section above the slab. I would neglect the diagonal directional force for punching shear since there isn't any code-prescribed way of addressing it. I believe it also doesn't control, but I could be wrong.


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## ipswitch (Aug 6, 2012)

All I can tell you is this 'Notes on ACI' is pretty heavy duty.


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## McEngr (Aug 6, 2012)

ipswitch said:


> All I can tell you is this 'Notes on ACI' is pretty heavy duty.


It's heavy duty, and you'll think it's way overkill for the exam when it's all said and done. However, it will prepare you for almost anything that's thrown at you. I can't tell you what I had on the exam, but I know of at least 2 or 3 problems that were out of left field that I was able to finagle(sp?) my way to solve. Good luck, ipswitch. I'm trying to pay it forward anyway I can, my man.


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## dakota_79 (Aug 6, 2012)

I'm thinking you're over-complicating it, which leads me to believe I may be missing what you're trying to do. But to check bearing stresses under an odd-shaped footing (or for non-orthogonal lateral loads on a rectangular footing - same thing really), you just calc the moment of inertia of the footing plan and do your P/A + Mc/I to get the stress at any point, assuming you don't have negative (tension) stress at any point.

For overturning, similarly it doesn't matter what shape/orthogonality you have, you're just summing your forces at their c.g.'s at an arm about a point.

But again, I gather I'm missing your point! Any sketches you can provide?


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## McEngr (Aug 7, 2012)

The eccentricity is outside of the 'kern', therefore, it's not that simple.


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## dakota_79 (Aug 7, 2012)

Should've figured, my bad.

See if this helps:

http://www.scribd.com/doc/35268258/Wilson-Bearing-Pressures-for-Rectangular-Footings-With-Biaxial-Uplift


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