B Breaking a granite surface plate....

[Nidum]

We can use that model if you are happy with it .

 

[Nidum]

 

[Chestermiller]

View attachment 206138

If I did the math correctly, for the force at failure at the fulcrum R, I get: $$R=\frac{2}{3}\frac{wh^2}{L}\frac{\sigma_{ult}}{\sqrt{1-\nu+\nu^2}}$$
where h=0.08m, L=0.35m, and W=0.3m. This gives a value for R of 158 kN.

 

[Nidum]

Interesting . I'll do the FEA with that load later today .

 

[Chestermiller]

Interesting . I'll do the FEA with that load later today .

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.

 

[dovidu]

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.

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]

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

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.

 

[dovidu]

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.

quadruple...

 

[Nidum]

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 .

 

[Chestermiller]

quadruple...

what is your problem with this?

 

[Chestermiller]

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 .

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]

That view shows the deflected shape and the colours represent local vertical deflection values .

I'll post the stress distribution views asap .

 

[dovidu]

what is your problem with this?

nothing!
i am just amused :D

 

[Nidum]

I have not forgotten about the FEA . Other work had to have priority last couple of days .