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We can use that model if you are happy with it .
If I did the math correctly, for the force at failure at the fulcrum R, I get: $$R=\frac{2}{3}\frac{wh^2}{L}\frac{\sigma_{ult}}{\sqrt{1-\nu+\nu^2}}$$Nidum said:
In this analysis, I assumed pure bending. This was probably not too good an approximation because of the stubbiness of the plate. Probably shear bending was a factor too. For a failure criterion, I used von Mises.Nidum said:Interesting . I'll do the FEA with that load later today .
Chestermiller said:In this analysis, I assumed pure bending. This was probably not too good an approximation because of the stubbiness of the plate. Probably shear bending was a factor too. For a failure criterion, I used von Mises.
The gold standard for this will be the calculation that Nidum is doing. If the equation that I presented is anywhere close to being in the right ballpark, it predicts that doubling the thickness would quadruple the allowable load. You can work out for yourself (from my equation) the rest of the proportionalities involved.dovidu said:sir, it turns out this is way more complicated than i initially predicted
the physics teacher i know said he couldn't figure it out, he had never encountered such problem ever in his life
he just kept saying sorry...
i actually might do a real life test with small pieces but... would it just be multiplying the figure to thicker plates...
not sure it would be purely proportional
for example, if i tested 10x10x2 (cm) piece and get required force to break x
would it take 2x force for granite piece 10x10x4
that's only thinking aboout thickness alone, what about 20x20x4
would it apply the same?? just simply by multiplying the force x proportional to the size?
if i try two different tests with different dimension and observe that the value increased is proportional, then hm... i guess.. that's actually good idea :D
Chestermiller said:The gold standard for this will be the calculation that Nidum is doing. If the equation that I presented is anywhere close to being in the right ballpark, it predicts that doubling the thickness would quadruple the allowable load. You can work out for yourself (from my equation) the rest of the proportionalities involved.
If you want to learn more about all this, Google "three point bending test." The solution in Wiki confirms my 2/3 factor. But, as I said, the analysis applies most accurately for long thin plates, which is not the case for your samples.
what is your problem with this?dovidu said:quadruple...![]()
Are you sure about the colors in the figure? Shouldn't the bottom be red (high tensile stress) and the top be blue (high compressive stress), rather than the colors being oriented vertically.?Nidum said:View attachment 206202
Load : 158 kN uniformly distributed on top of central wedge .
Top surface :
Max fibre stress : 33.9 MPa compressive .
Deflection : 0.1 MM
Bottom surface :
Max fibre stress : 21.2 MPa tensile .
Deflection : 0.097 MM
Values sampled at centre of block as seen in plan view .
Picture shows deflected shape .
Stress concentration effects supressed .
There already seems to be reasonable agreement between the analytic and the FEA results .
I want to try another run now with the applied and reaction loads acting on the plate as distributed loads acting on small definite areas instead of on the almost line contact areas of the wedges .
Something like 25% of plate thickness times the plate width for each contact .
nothing!Chestermiller said:what is your problem with this?