Breaking a granite surface plate....

[dovidu]

----------------Q 1---------------------
hello it's been two weeks to figure out how much force would it take to break a granite surface plate size of
1) 30x30x8 (cm)
2) 40x60x8 (cm)

literally breaking it in half with a fulcrum underneath and two forces acting on the top

i've posted a number of forums (practical machinist, engineering forum, hobby machinist forum, etc, etc)

all the replies either "refer to your mechanical engineering textbook", "consult a mechanical engineer", use H-beam strength calculation formula, etc.

never a direct answer but beat around the bush type.

i've done my calculation based on the cross section of the granite:

UTS of granite 25~39MPA therefore 25~39 X 240/320 = 6000~9360 / 8000~12480 MPA

which translates to 61182.97~95445.44 / 81577.3~127260.58 kg

seems pretty good not sure if it's correct

----------------Q 2---------------------
also i assumed that as the distance of two forces acting upon the granite increases, proportionally the required force to break the granite decreases
so in order to counter that, using two metallic plates bolted on the surface plate directly beneath the two forces that would direct the force closer to the fulcrum, would it act as a transmitter that brings the two distant forces closer to the fulcrum?

please, i don't want beat around the bush type of answer
thank you :D

 

[Nugatory]

seems pretty good not sure if it's correct

It's not. You've calculated the force required to break a granite block by pulling opposite sides in opposite directions the way you'd break a piece of string. Whoever told you to use the beam strength formula gave you good advice; your problem is equivalent to a beam supported at both ends and subject to a point load at mid-span.

If by "I don't want beat around the bush type answer" you mean "I want someone else to do the work for me" there's a beam strength calculator at engineeringtoolbox.com that will give you your answer.

using two metallic plates bolted on the surface plate directly beneath the two forces that would direct the force closer to the fulcrum, would it act as a transmitter that brings the two distant forces closer to the fulcrum?

Only if the metal plates were very strong and very stiff compared with the granite itself. That seems infeasible for any reasonable thickness of plate.

 

[dovidu]

It's not. You've calculated the force required to break a granite block by pulling opposite sides in opposite directions the way you'd break a piece of string. Whoever told you to use the beam strength formula gave you good advice; your problem is equivalent to a beam supported at both ends and subject to a point load at mid-span.

If by "I don't want beat around the bush type answer" you mean "I want someone else to do the work for me" there's a beam strength calculator at engineeringtoolbox.com that will give you your answer.
Only if the metal plates were very strong and very stiff compared with the granite itself. That seems infeasible for any reasonable thickness of plate.

thank you
good insight on the metal plates

i think i didn't include this, but there will be metal plates bolted top and bottom sides of the granite plate.
wouldn't the granite have to bend the metal plate first to break?
also, in this case wouldn't the plate bring the forces at the edge close to the fulcrum?
once the metal plate has been bent enough to surpass the breakage point of granite, the granite will break

i will check out the beam calc

 

[dovidu]

It's not. You've calculated the force required to break a granite block by pulling opposite sides in opposite directions the way you'd break a piece of string. Whoever told you to use the beam strength formula gave you good advice; your problem is equivalent to a beam supported at both ends and subject to a point load at mid-span.

If by "I don't want beat around the bush type answer" you mean "I want someone else to do the work for me" there's a beam strength calculator at engineeringtoolbox.com that will give you your answer.
Only if the metal plates were very strong and very stiff compared with the granite itself. That seems infeasible for any reasonable thickness of plate.

the shape of the beam is different from a block of rock
unless the granite was shaped was like a beam , it shouldn't applicable
nevertheless i will search for beam calculator that calculates geometry such as the surface plate

it's not that i want someone else to do the work for me, i can't really figure out and seeking for good advice
iam not lazy like that

 

[dovidu]

It's not. You've calculated the force required to break a granite block by pulling opposite sides in opposite directions the way you'd break a piece of string. Whoever told you to use the beam strength formula gave you good advice; your problem is equivalent to a beam supported at both ends and subject to a point load at mid-span.

If by "I don't want beat around the bush type answer" you mean "I want someone else to do the work for me" there's a beam strength calculator at engineeringtoolbox.com that will give you your answer.
Only if the metal plates were very strong and very stiff compared with the granite itself. That seems infeasible for any reasonable thickness of plate.

it's not as simple as "find and use bending beam calculator"
that will undoubtedly result in wrong answer
as matter of fact i have this almost figured out
i think i have to apply both tensile and compression strength to this
on side resists with tensile and the opposite side resists with compression strength
and it will continue until the maximum elongation point, upon which the material breaks

not so beam theory after all

 

[Nidum]

It is beam theory but with some more details needing to be taken into account than is usual in dealing with the more common long slender steel beams .

Simple beam theory becomes unreliable when beams are short and very thick .

Also granite material does not behave quite like steel so special care is needed to identify the actual mechanism of failure involved .

 

[Nugatory]

the shape of the beam is different from a block of rock
unless the granite was shaped was like a beam , it shouldn't applicable
nevertheless i will search for beam calculator that calculates geometry such as the surface plate

Look for a parameter called the "moment" or "cross-sectional moment" or something of that sort; this captures the shape of the beam. But it's not as if a rectangular cross section is exactly an obscure or unusual shape for a beam.

i think i have to apply both tensile and compression strength to this
one side resists with tensile and the opposite side resists with compression strength
and it will continue until the maximum elongation point, upon which the material breaks

And that is how beam theory works. Compression on one side, tension on the other side, both increasing with the deflection.

 

[Chestermiller]

Actually, because the lateral dimension is large compared to the length, the granite should be treated as a plate. The basic analysis for a plate is similar to that of a beam, except for the significant lateral constraint.

This problem should first be solved without any metal reinforcement, and then the effects of the metal reinforcement should be included, treating the combined plate as a laminated composite.

 

[jbriggs444]

Actually, because the lateral dimension is large compared to the length, the granite should be treated as a plate. The basic analysis for a plate is similar to that of a beam, except for the significant lateral constraint.

Ahhh, of course. In a beam, the face under compression will tend to "fatten" out to the side while the face under tension will tend to "suck" in. But in a plate, the surrounding material prevents this. Thank you, Chet, for that insight.

 

[dovidu]

Actually, because the lateral dimension is large compared to the length, the granite should be treated as a plate. The basic analysis for a plate is similar to that of a beam, except for the significant lateral constraint.

This problem should first be solved without any metal reinforcement, and then the effects of the metal reinforcement should be included, treating the combined plate as a laminated composite.

you're right, the reinforcement should be excluded first

 

[dovidu]

Look for a parameter called the "moment" or "cross-sectional moment" or something of that sort; this captures the shape of the beam. But it's not as if a rectangular cross section is exactly an obscure or unusual shape for a beam.

And that is how beam theory works. Compression on one side, tension on the other side, both increasing with the deflection.

this granite will become the base of a machine which requires precision and rigidity

it will not be used as a beam, it's totally different

we're not talking just about pressure applied to extreme ends, there's definitely going to be pressures applied all over the place with all sorts of directions

not to mention the structure of machine parts (linear rail, headstock, mill column, tool post, etc, etc) afftecting this.

i decided to calculate using roughly 1/2 the cross section as tensile and the other 1/2 compression.

i might get rough figures although not perfect.

a physics teacher i know might assist me in doing so, but i doubt it will be all that easy

this is obviously not easy problem i realized, as no one so far in 7 different forums over the last 2 weeks, provide clear answer

i didn't know about any "theories" but used common sense to realize there will be tension and compression. it's quite obvious to me now.

the only variable left is the material property, especially the elongation. granite is considered very brittle therefore you can't just apply any theory and expect it to behave as it is expected to.

 

[dovidu]

Actually, because the lateral dimension is large compared to the length, the granite should be treated as a plate. The basic analysis for a plate is similar to that of a beam, except for the significant lateral constraint.

This problem should first be solved without any metal reinforcement, and then the effects of the metal reinforcement should be included, treating the combined plate as a laminated composite.

when you said "because the lateral dimension is large compared to the length"

the length refers to 8cm length of 40*60*8, and "lateral dimension" as either 40 or 60?

so "the significant lateral constraint" would be related to 40 or 60.

i don't see what's the significant lateral constraint,

or are you saying because this surface granite plate is obviously not shaped like a beam in terms of how wide it is. As a beam is long and slender and this is clearly a plate because of how similar the length and width is. this is obviously not a beam...

 

[Chestermiller]

when you said "because the lateral dimension is large compared to the length"

the length refers to 8cm length of 40*60*8, and "lateral dimension" as either 40 or 60?

so "the significant lateral constraint" would be related to 40 or 60.

i don't see what's the significant lateral constraint,

or are you saying because this surface granite plate is obviously not shaped like a beam in terms of how wide it is. As a beam is long and slender and this is clearly a plate because of how similar the length and width is. this is obviously not a beam...

You're not saying that the plate is short and stubby, are you? Are you really saying that the length of the plate is only 8 cm, compared to the lateral dimensions of 40 and 60 cm?

 

[Fizzy Hicks]

This is easy.

1. By pressing down on each side of the granite, one exerts force on the fulcrum, which then forces the upper surface into tension and the bottom into compression.

2. Once tension in the upper SURFACE of the granite, exceeds it's tensile strength, it breaks by fracture through the mass. Like glass with a score line in the upper surface.

3. Remember it's not the tensile strength of the MASS, or the entire cross sectional area, as in pure tension through out the whole mass, only at the upper surface.

4. Use triangulation and vectors or two upside down right angles triangles, back to back - wide bases at the top, pointy bits on the fulcrum.

 

[Chestermiller]

This is easy.

1. By pressing down on each side of the granite, one exerts force on the fulcrum, which then forces the upper surface into tension and the bottom into compression.

2. Once tension in the upper SURFACE of the granite, exceeds it's tensile strength, it breaks by fracture through the mass. Like glass with a score line in the upper surface.

3. Remember it's not the tensile strength of the MASS, or the entire cross sectional area, as in pure tension through out the whole mass, only at the upper surface.

4. Use triangulation and vectors or two upside down right angles triangles, back to back - wide bases at the top, pointy bits on the fulcrum.

Several other responders have already pointed this out.

 

[dovidu]

when you said "because the lateral dimension is large compared to the length"

the length refers to 8cm length of 40*60*8, and "lateral dimension" as either 40 or 60?

so "the significant lateral constraint" would be related to 40 or 60.

i don't see what's the significant lateral constraint,

You're not saying that the plate is short and stubby, are you? Are you really saying that the length of the plate is only 8 cm, compared to the lateral dimensions of 40 and 60 cm?

no sir, this is a precision granite surface plate. if you google this and look at this images you will see what i meant

 

[dovidu]

This is easy.

1. By pressing down on each side of the granite, one exerts force on the fulcrum, which then forces the upper surface into tension and the bottom into compression.

2. Once tension in the upper SURFACE of the granite, exceeds it's tensile strength, it breaks by fracture through the mass. Like glass with a score line in the upper surface.

3. Remember it's not the tensile strength of the MASS, or the entire cross sectional area, as in pure tension through out the whole mass, only at the upper surface.

4. Use triangulation and vectors or two upside down right angles triangles, back to back - wide bases at the top, pointy bits on the fulcrum.

actually, i realized the compression and tensile stress occurs "AT THE SAME TIME"

this applies to all materials

this you can realize by pure common sense and intuition/logicit's more like the two stresses occur at the same time. as the force increases the bottom side compresses

to a point where the bend incurred will be sufficient enough to cause break at the top surface

(this is so because compression strength of granite is way greater than tensile, breakage ALWAYS OCCURS AT THE TOP SURFACE)

(at the very moment the crack occurs at the top, we can determine the "breakage point" of this material, as it will be the only useful structural strength of this specific material for mechanical applications)

therefore i strongly believe compression strength of this material is a very important aspect of this

calculation

this is purely based on my experience/logic/intuition. i don't have any knowledge of theories.

i make the theories on my own (although it may be invented before by someone else)

 

[Fizzy Hicks]

There is no compression in the lateral sense, only compression is in a thin line on the contact area over the fulcrum, and that is acting vertically.

The mass is parting by cleavage and fracture - a product of tension alone.

Anyway by the sounds of it, your wanting to plant a little machine on a little block of granite, why stress about this issue for? Sounds like a partial inquiry, an exploration of the issue and a partial wasting of everyone's time to me.

 

[Nidum]

I've inverted the arrangement of supports and load that you described initially . Seems more logical for a load coming from machine weight to act downwards and centrally .

Is this now a reasonable interpretation of the geometry of your problem ?

 

[Nidum]

Estimating the actual breaking load is going to be problematic . A better approach would be to determine a load at which we can be reasonably certain that the block will not break .

Quick FEA showed that with 40 kN on the central bar the bar itself failed first due to local crushing . Granite block still had large reserves of strength remaining .

With this load vertical deflection of top surface central point of block is 0. 02 mm relative to the outer ends .

If the actual machine is centrally mounted on the block with a large footprint area and similarly if the block is mounted on the support structure with well spaced pads with a large total footprint area I actually think the block is going to be so resistant to breakage and deflection that you need not consider this aspect of your design study any further .

Fitting steel plates is both not necessary and undesirable . The granite block is adequately stable , strong and accurate as supplied . Steel plates will just add complexity for no advantage .

 

[Nidum]

40 kN applied load .

Worst displacement 0,003 mm at central point .

Least safety factor 8.

 

[Chestermiller]

Estimating the actual breaking load is going to be problematic . A better approach would be to determine a load at which we can be reasonably certain that the block will not break .

Quick FEA showed that with 40 kN on the central bar the bar itself failed first due to local crushing . Granite block still had large reserves of strength remaining .

With this load vertical deflection of top surface central point of block is 0. 02 mm relative to the outer ends .

If the actual machine is centrally mounted on the block with a large footprint area and similarly if the block is mounted on the support structure with well spaced pads with a large total footprint area I actually think the block is going to be so resistant to breakage and deflection that you need not consider this aspect of your design study any further .

Fitting steel plates is both not necessary and undesirable . The granite block is adequately stable , strong and accurate as supplied . Steel plates will just add complexity for no advantage .

The usual (tacit) assumption is that the footprint areas are relatively large (say, on the order of a few plate thicknesses), so that the loading is not really close to the knife-edge type loading implied by Fizzy Hicks in post #18, and local crushing will not be a problem. Under these circumstances, ordinary plate bending analysis applies; I'm curious to know what does your FEM analysis predicts for such circumstances and, by comparison, what ordinary plate theory would predict for the failure load. If you're willing to do the FEM, I'd be willing to do the standard bending analysis for comparison.

Chet

 

[Nidum]

Ok . That could be interesting . Back later .

 

[dovidu]

Estimating the actual breaking load is going to be problematic . A better approach would be to determine a load at which we can be reasonably certain that the block will not break .

Quick FEA showed that with 40 kN on the central bar the bar itself failed first due to local crushing . Granite block still had large reserves of strength remaining .

With this load vertical deflection of top surface central point of block is 0. 02 mm relative to the outer ends .

If the actual machine is centrally mounted on the block with a large footprint area and similarly if the block is mounted on the support structure with well spaced pads with a large total footprint area I actually think the block is going to be so resistant to breakage and deflection that you need not consider this aspect of your design study any further .

Fitting steel plates is both not necessary and undesirable . The granite block is adequately stable , strong and accurate as supplied . Steel plates will just add complexity for no advantage .

thank you for your input.

i currently have 3 surface plates in my room and when you looks and it and touch it, it really is a solid material

can't imagine this thing breaking so eaily

also 10cm versions are available, but cost and weight also increases

i think if 8cm fails 10cm is a next thing to try

if i am not going to use underside 1inch thick steel plate reinforcement as a washer, i am going to use washers 1cm thick made of 12L14 material with 5~8mm thick rim

to disperse force away from the drilled hole area to prevent cracks

currently i have drilled 4x 8cm holes 10mm diameter and mounted a hiwin hgh30ca series linear guide

the top of the rail has 2 micron flatness and so far it feels pretty solid

this machine is going to be a precision surface grinder

my next projects that i haven't designed in cad yet are lathe and mill

i will be machining steel in CNC and possibly use toolchagers

 

[dovidu]

good cad design, if you created that just for this discussion i give a plus to you. good job!

some things to note though, your design is inverted from what I've been talking abought throughout this thread (it really doesn't matter though)

also, i don't consider 40x60x8 a small size for a base. this machine's base alone is 80kgs, that's not a "small" machineand i don't know what are you talking about?

if you think harder you will realize when you break any material, compression always occur one side and tension another

since almost all material have stronger compression than tension, the tension side always fail first, you've seen it all your life[/QUOTE]

 

[dovidu]

View attachment 206037
40 kN applied load .

Worst displacement 0,003 mm at central point .

Least safety factor 8.

good work!

thank you for your dedication!

gradient dispersion of force from center to edge as expected

if linear rails and reinforcement plates are installed, this could be quite a different story

however the result you've shown gives me hope

4 ton pressure and 0.003mm displacement hm... this could definitely be improved by reinforcement

i need figure out maximum displacement of granite. will 0.003mm displacement create cracks?

it's only 3 microns but i guess it could happen

 

[dovidu]

for those wondering what my current project looks like here's the picture

it's a small machine using 30x30x8(cm) granite surface plate

a small but powerful variable speed grinder with vertical type stone will mounted on the Z axis

diamond dresser to true up the stone

i am expecting quite accurate surface grinding

one thing to note though, as the height of the head increases, rigidity decreases

so 1~15cm tall parts preferred for my operation

the maximum is up to 28 cm thick

currently drilled four holes and mounted bottom rail

i know it's just single rail but i used 30 series rail rated for tons of pressure in all direction

dual rail should be better but obviously more space will be sacrificed and more complexity

what do you guys think of this?

 

[dovidu]

here's actual pictures of it

currently under construction

have many projects running at the same time so progress is a bit slow

but slowly will be finished eventually

i designed in such way so rigidity loss is miminum with minimal deflection

low part grinding should be a breeze

very tall parts probably have to take very small depth of cut

the washers i made are 9mm thick, 25mm diameter, 12L14 material, with a small contact side relief around center with 5~8mm rim(contact point)

the bolts are M8 X 150 (4 of them) 6cms apart

measured the top side of the rail with a test indicator

2 micron flatness from end to end

very likely error from the rail not the plate

 

[Nidum]

...these circumstances, ordinary plate bending analysis applies; I'm curious to know what does your FEM analysis predicts for such circumstances and, by comparison, what ordinary plate theory would predict for the failure load. If you're willing to do the FEM, I'd be willing to do the standard bending analysis for comparison.

A massive granite slab supported and loaded by steel wedges is not very representative of normal engineering practice . We need a different model to explore your question properly . Would you like to suggest something suitable ?

 

[Chestermiller]

A massive granite slab supported and loaded by steel wedges is not very representative of normal engineering practice . We need a different model to explore your question properly . Would you like to suggest something suitable ?

I don't follow the diagrams the OP provided. Is your diagram in post #19 not adequate?

 

[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.