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This year Structural PE question 1

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Structural Mindset

Structural
Aug 3, 2020
8
Hello. Need help with one of the PE questions my friend got last friday. It is basically about critical perimeter of punching shear. We are told that we have a loaded forklift. each tire gets 8000 pounds of load. Tire is inflated to 120psi pressure. Tires are 8 inch wide. d is also given. Formula is straight forward if you know the contact area of the tire. That is what the problem comes down to. Is there a relationship between tire air pressure and contact area?
Thank you
 
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I think you can get the contact area from the 8000lb load and 120psi. Divide the 8000/120.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
 
Contact area is equal to the FORCE divided by the PRESSURE.
 
Have a look in any engineering mechanics book for pressure

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
 
Not a consensus relationship. There have been several studies over the years and there is disagreement between researchers as to the actual relationship....some think it is the tire pressure plus 10 percent....others think it is the tire pressure minus 10 or 20 percent. Here is a paper that addresses the concept pretty well and in my opinion has some credible data and equations:



 
 https://files.engineering.com/getfile.aspx?folder=e0977863-b304-429e-972f-f4d0504bbfd7&file=Tire_Contact_Pressure.pdf
It is unevenly distributed over the contact patch, and that distribution varies with tire construction and pressure and load and the phase of the moon. I'd use load/pressure, in context. Worst case is a pointload, and a tire running over a nut and bolt on concrete probably approaches that, but of course that would give you infinite shear stress. Another option would be to look at Hertzian contact stress distribution. That's in Timoshenko.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376
 
Also, don't forget that the effective perimeter for calculating punching shear capacity will be rather bigger than the perimeter of the presumed tyre contact patch on the top of the slab. It is commonly asusmed that the load spreads at say 45 degrees, so at the mid-plane of a slab with thickness "d", a contact patch of a x b will have an effective size of (a + d) x (b + d).

 
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