1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

B Breaking a granite surface plate...

  1. Jun 23, 2017 #1
    ----------------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 text book", "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
     
  2. jcsd
  3. Jun 23, 2017 #2

    Nugatory

    User Avatar

    Staff: Mentor

    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.
     
  4. Jun 23, 2017 #3
    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
     
  5. Jun 23, 2017 #4
    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
     
  6. Jun 23, 2017 #5

    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
     
  7. Jun 23, 2017 #6

    Nidum

    User Avatar
    Science Advisor
    Gold Member

    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 .
     
  8. Jun 23, 2017 #7

    Nugatory

    User Avatar

    Staff: Mentor

    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.
     
  9. Jun 23, 2017 #8
    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.
     
  10. Jun 23, 2017 #9

    jbriggs444

    User Avatar
    Science Advisor

    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.
     
  11. Jun 23, 2017 #10
    you're right, the reinforcement should be excluded first
     
  12. Jun 23, 2017 #11
    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.
     
  13. Jun 23, 2017 #12
    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...
     
  14. Jun 23, 2017 #13
    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?
     
  15. Jun 24, 2017 #14
    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.
     
  16. Jun 24, 2017 #15
    Several other responders have already pointed this out.
     
  17. Jun 24, 2017 #16

    no sir, this is a precision granite surface plate. if you google this and look at this images you will see what i meant
     
  18. Jun 24, 2017 #17

    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/logic


    it'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)
     
    Last edited: Jun 24, 2017
  19. Jun 24, 2017 #18
    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.
     
  20. Jun 25, 2017 #19

    Nidum

    User Avatar
    Science Advisor
    Gold Member

    Granite_surface_plate_breaking_load_2017-Jun-25_08-58-29AM-000_CustomizedView3750219237.png

    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 ?
     
    Last edited: Jun 25, 2017
  21. Jun 25, 2017 #20

    Nidum

    User Avatar
    Science Advisor
    Gold Member

    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 .
     
  22. Jun 25, 2017 #21

    Nidum

    User Avatar
    Science Advisor
    Gold Member

    Granite machine base with feet and pad 40kN_corrected geometry ..png
    40 kN applied load .

    Worst displacement 0,003 mm at central point .

    Least safety factor 8.
     
    Last edited: Jun 25, 2017
  23. Jun 25, 2017 #22
    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
     
  24. Jun 25, 2017 #23

    Nidum

    User Avatar
    Science Advisor
    Gold Member

    Ok . That could be interesting . Back later .
     
  25. Jun 25, 2017 #24
    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
     
  26. Jun 25, 2017 #25
    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" machine


    and 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]
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted