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I can't figure this out! please help me (regarding surface plate)

D

dovidu

Plastic
Joined
Jun 11, 2017
i am seriously considering starting combination mill lathe machine using surface plate as the base

i will be using through hole not the insert types.

the major question i have on my mind for two weeks now...

how much force would it take to literally break 8cm~10cm surface plate in half

for compressive strength on the net 200MPA, and this source says 2100 MPA (Granite :: MakeItFrom.com)

2100 MPA i really dont believe that... that's very strong

so.... how do i factor in thickness, kg/cm2 i can figure out
maybe use the cross sectional area (thickness) to punch in the numbers?

so for example 40x60x8 (cm) surface plate has 8x40 or 8x60 cross sectional area so.. 320cm2, 480cm2

what force should i use?
tensile strength?
shear?

i am guessing tensile

so tensile strength of granite is 25~39MPA (according to different sources)

320x25 320x39, 480x25, 480x39

8000~12480 MPA, 12000~18720 MPA

815773 ~ 127260.58 kg, 122365.95 ~ 191196.79 kg

the figures seems pretty weird to me.. waaaaaay too high

because i've been utube vids where a dude uses sledge hammer to break boulders in half lol, although it takes lots of hits

one thing though machining forces may locally applied so that may drive down the actual figures...

i'd be happy with even 1/20 the amount indicated above..

it would easily machine 1045 steels, which is my ultimate objective right now


am i calculating right? can some one help me out here
 
dovidu said:
i am seriously considering starting combination mill lathe machine using surface plate as the base

i will be using through hole not the insert types.

the major question i have on my mind for two weeks now...

how much force would it take to literally break 8cm~10cm surface plate in half

for compressive strength on the net 200MPA, and this source says 2100 MPA (Granite :: MakeItFrom.com)

2100 MPA i really dont believe that... that's very strong

so.... how do i factor in thickness, kg/cm2 i can figure out
maybe use the cross sectional area (thickness) to punch in the numbers?

so for example 40x60x8 (cm) surface plate has 8x40 or 8x60 cross sectional area so.. 320cm2, 480cm2

what force should i use?
tensile strength?
shear?

i am guessing tensile

so tensile strength of granite is 25~39MPA (according to different sources)

320x25 320x39, 480x25, 480x39

8000~12480 MPA, 12000~18720 MPA

815773 ~ 127260.58 kg, 122365.95 ~ 191196.79 kg

the figures seems pretty weird to me.. waaaaaay too high

because i've been utube vids where a dude uses sledge hammer to break boulders in half lol, although it takes lots of hits

one thing though machining forces may locally applied so that may drive down the actual figures...

i'd be happy with even 1/20 the amount indicated above..

it would easily machine 1045 steels, which is my ultimate objective right now


am i calculating right? can some one help me out here
Click to expand...

If you want to get an idea of strength, (To break it in half) I would use "simply supported beam equation" assuming the block is supported at the ends and a load is applied in the center, the equation will use tensile strength and beam dimensions. I would design the system to use a maximum of about 1/10 of this force because its a brittle material and loads are not well known.
 
hello, i am aware of EG method
i was thinking solid granite block as the alternative
with correct method, deep drilling is easily done

cost is low, different sizes available, also can cut them to specific sizes
accuracy for big blocks <0.5 microns (except the edges)

the only possible down side might be the tensile strength

which i calculated. a granite with cross section 8x30 or 8x40 = 240cm2 or 320cm2

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
that's actually very good figure right there, kinda hard to believe

i am not so sure if i did the calculation correctly

is this correct?

if so, i am definitely gonna use granite as bed for all my future machine projects

and develop methods to drill through holes effectively and fast
 
I can't figure out why you started yet another thread on the same subject, then posted in the first one to come here.....:nutter:
 
this thread i started to gather people to calculate precisly what kind of load the surface plate can bear

the last post is about washer that supports underside of the granite
 
dovidu said:
this thread i started to gather people to calculate precisly what kind of load the surface plate can bear

the last post is about washer that supports underside of the granite
Click to expand...

STOP THIS !

The whole damn subject is about designing your new machinery base using granite or some other material.

It get's designed as A SYSTEM, so keep it all together in ONE PLACE.

Your starting to be just like "DaveK2" asking separate questions ad naseum for designing
a home shop plasma cutter gantry.
 
ok i am sorry i didn't mean to make you upset
should i delete this right away?

the other post seems to be buried so deep, no one answers back
so i thought it would be nice to focus on this specific topic

if this goes against the policy of this forum i'd happily delete this
 
Your formula is correct if you were loading the granite block like a bolt in tension.

You need to use the formula to calculate bending stress in a beam.

stress= M*c/I

M= Bending Moment at the location in the beam you are interested in.
c= Distance from neutral axis to extreme fiber. for a rectangular cross section that is 1/2 height of beam.
I= Second Moment of Area, sometimes called Moment of Inertia. Value that is calculated from the size and shape of the beam cross section. For a solid rectangular beam I= b*h^3/12 where b is beam width and h is beam height.

Now you have a list of engineering terms to search google for more detailed study.
 
will you be breaking granite to make the plate? if not you shoud be interested in youngs modulus rather than tensile strength. that will give you an idea how stiff your machine base will be.
 
dovidu said:
ok i am sorry i didn't mean to make you upset
should i delete this right away?

the other post seems to be buried so deep, no one answers back
so i thought it would be nice to focus on this specific topic

if this goes against the policy of this forum i'd happily delete this
Click to expand...

You can "bump" an older post to bring it back to attention. However, doing this too frequently is considered rude. You should have permanent links to any threads you start where you want to check back for answers. These can be in your internet browser Bookmarks or Favorites or even copied and pasted into a text document. Simply right click the address bar to highlight the full address, Copy, and then Paste into wherever you want it stored.
 
dian said:
will you be breaking granite to make the plate? if not you shoud be interested in youngs modulus rather than tensile strength. that will give you an idea how stiff your machine base will be.
Click to expand...

thank you so much for your input i will re do my calculations

to make a smaller sized or miniture sized machine with base of 20x20 or smaller, i can cut the granite in half, linearly

the left over pieces can be made to granite cubes or squares

i am just trying to find the "stiffness" of the base you mentioned, thank you
 
Erich said:
Your formula is correct if you were loading the granite block like a bolt in tension.

You need to use the formula to calculate bending stress in a beam.

stress= M*c/I

M= Bending Moment at the location in the beam you are interested in.
c= Distance from neutral axis to extreme fiber. for a rectangular cross section that is 1/2 height of beam.
I= Second Moment of Area, sometimes called Moment of Inertia. Value that is calculated from the size and shape of the beam cross section. For a solid rectangular beam I= b*h^3/12 where b is beam width and h is beam height.

Now you have a list of engineering terms to search google for more detailed study.
Click to expand...


it seems that i can't apply this to a block of granite, which is basically a rectagular 6 dimensional object

it is not H/I shaped material
 
dian said:
will you be breaking granite to make the plate? if not you shoud be interested in youngs modulus rather than tensile strength. that will give you an idea how stiff your machine base will be.
Click to expand...

the thing with young's modulus is that it seems to be irrelevant actually..

for example tungsten carbide and diamond has higher young's modulus than steel (many many times)

but we all know that a tungsten carbide or diamond machine base will crack

as opposed to a cast iron base which will withstand far greater bending stress

so i guess we need to know how to calculate bending load for granite

i think what i have done above may be related to that

what do you think?
 
Erich said:
Your formula is correct if you were loading the granite block like a bolt in tension.

You need to use the formula to calculate bending stress in a beam.

stress= M*c/I

M= Bending Moment at the location in the beam you are interested in.
c= Distance from neutral axis to extreme fiber. for a rectangular cross section that is 1/2 height of beam.
I= Second Moment of Area, sometimes called Moment of Inertia. Value that is calculated from the size and shape of the beam cross section. For a solid rectangular beam I= b*h^3/12 where b is beam width and h is beam height.

Now you have a list of engineering terms to search google for more detailed study.
Click to expand...



i have several theories that i came up with over the days

(1) the forces that will act on the granite surface plate based machine would be..

the forces interacting between 2 linear guide ways on the surface plate and the headstock(lathe) or column(mill)

so for the rails, forces will be distributed out more evenly

for the headstock/column the distribution of force is dependant on the surface area on which it is installed on


(2) the surface area of linear rails and headstock/column really doesn't matter as in (1).

in the end if the granite is gonna break, it's gonna break at the weakest point in between guide way / headstock-column


(3) a sort of lever action caused by the linear rails may create a variable that makes this calculation not so simple

the rigidity will increase with the rails and larger headstock/column base


i'd really appreciate your participation in this debate

now that i see it, this is not just about the surface plate

but fundamental questions of mechanical engineering/designing
 
You are correct. Designing a machine tool is mechanical engineering.
ME's begin by drawing a Free Body Diagram (FBD). This is a simplified picture of the parts of your machine with all applied forces. That means gravity, cutting forces on the tool bit etc.
Each individual part gets its own FBD. Where one part touches or is attached to another part you have what are called reaction forces. Those come in equal and opposite pairs (Newton and his third law)

You Sum up the forces acting in each of the X,Y,Z directions and set them equal to zero (Newtons first law).

Now that you know the forces acting on each part, You calculate the stresses inside each part and size them so the stresses are with in the safe limits for your materials.

If all this makes no sense, you need to go get a degree in Mechanical Engineering or hire one to design your machine.
 
Get a good mechanical engineering text. And read it. Continue reading it until you understand why your question can not be answered as it was stated. Then you will probably come to the realization that you need to read it more. And others too.

My thoughts include the fact that virtually all machine tools are made with cast iron for the base parts. Think about that for a while.
 
Granite is sold every day in many forms

Find a source that provides something that you feel is suited to you tool build. If it breaks in service, Purchase another block only thicker!

Inventing is not following "engineering guidelines", but it helps.

Bottom line, THIS thread has no end.
 
this cannot be calculated using beam theories

as matter of fact i just realized i did my early calculation wrong, but no one pointed out

in order to calculate this i have to know the granite's maximum allowable elongation point (+ the X force required to do that)

consider both tensile and compression strength

i am still working on this, probably won't take long
 
20170623_210111.jpg

explained in the picture

need more opinions and inputs on this


i think tensile strength mainly determines the breakage point

but some compression has to occur at the bottom side

so tensile ~ compression in a gradient fashion
 
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