[guido]

Does anyone have data or can refer to a book or website that has info on cutting forces during milling?

Guido

[pauluk]

hi
do you want the power required to remove metal at a given rate, or the forces that will be put into the job? can you post more infomation

[guido]

Pauluk

Interested in the forces. X,Y and Z direction during milling. Want to design a mini VMC and can't find much info on this subject.

[pauluk]

Ah ok,
I have formulas in a book on calculating the power required for given materials, cutters etc, but I’m not sure they would be much help. I can post them if they are any use to you. You can’t build a machine that is too ridged, but you sure can build one that is not ridged enough, build as ridged as you can

[guido]

I have those formulas also..if you have tabels vor specific cutting forces (Kc) for different materials during milling ...then it would be great if you can post them.

Besides issues on rigidity of the frame..I also need the forces in order to correctly dimension linear guides, ballscrew, motor etc.

I think however that you can build a frame that is to rigid. Mayb more on that later..have to think about it..

Thanx

[HuFlungDung]

Are you planning to mill metals?

I dare say that more power is required of the axis motors to move the machine at rapid, than is required to feed during cutting. IF climb milling procedure is followed, the cutter practically feeds itself, and the servos only regulate this tendency. Of course, there will be instances where the tool must cut full diameter and this takes a bit of muscle, but suffice it to say, if you've got good rapid speed capability, you've got ample power to feed with.

Another factor to consider, is this: how large of a tool will be used, and how dull will it be permitted to get? The dullness (or even edge hone factor) could easily double or quadruple the stiffness requirements of the machine frame to keep vibration within acceptable limits.

There will always be a reaction caused by the tool edge penetrating the work. Worst case, a two flute cutter where the entire cycle of 'edge hits part, spindle and work deflect from each other, metal begins to deform, spindle and work begin to settle back towards one another, the chip completes, and the spindle and work spring free in the opposite direction, then the next edge hits'. This pattern creates the main frequency of cutting vibration (I'm conjecturing), so the real issue, is not the strength of the machine, but how fast will it dampen this vibration.

Dampening is related to material type (how well shock waves propagate through it) and mass. You simply cannot have too much mass while cutting. A heavy machine will cut quietly, while the light 'engineered one' will be ringing like a bell, and next to useless.

My 2 cents. Look at what has already been built. Notice there are no machine tools that look like tinkertoys, they look more like Stonehenge.

[Mcgyver]

who are the engineers out there? isn't the force directly tied to power? i.e. the cutting force of a .100 deep cut with a 1" end mill at .001 per tooth will be less than .200 @ .003 per tooth.

good post Huflung, someone said 'a fly landing on the tailstock will affect the bed shape', - the point is any force on the machine frame will have an effect, you need it massive such that the movement is negligible or insignificant.

I remember reading once that Porsche one year increase unibody rigidity by 50% while also achieving a weight reduction. Some seem to have the view that machine tool designers have let us down by not doing the same, or that with enough intellect and a thoughtful approach convention can be broken and tinker toy marvels will prevail.

Hopefully innovation does bring breakthroughs, but the reality is the Porsche unibody example is a silly analogy to machine tools, the forces, frequency and tolerance of movement are clearly not comparable…try to design something that is removing several cubic inches of steel per minute, with a specified work envelope and holding a tenth or so tolerance (not that we always work to a tenth but how much machine movement is acceptable?) and chances are its not going to be any lighter than what we already have, and perhaps a lot heavier depending on materials

Btw, there usually is correlation between massiveness and quality/accuracy – compare a hardinge or monarch lathe to southbend, so make it as massive as you can

[guido]

I'm an engineer...Do not get your remark about the cutting force...
To calculate the cutting force you need to know the specific cutting force of the material...

I'm not trying to build a tinker toy marvel. I can guarantee you that cutting forces are one of the most relevant issues for commercial machine tool builders.

The goal is to engineer a light construction that is stiff at the same time. Most important is to keep the mass as low as possible near places were vibrations with a big amplitude might occur.

[pauluk]

Hi All
Here are the power constants (Kp) for different materials, which you probably have anyway, but there is probably a direct comparison since the cutting forces are directly related to the power used to create the forces.

Power Constants Kp
Material Power Constant (inch units) Power Constant (metric units)
Aluminium
Cast 0.25 0.68
Rolled (Hard) 0.33 0.90
Bronze
Hard 0.91 2.48
medium 0.50 1.36
Soft 0.33 0.90
Brass
hard 0.83 2.27
medium 0.50 1.36
soft 0.25 0.68
leaded 0.30 0.82
Grey cast iron
from 0.28 0.76
to 0.91 2.48
plain carbon steel
from 0.63 1.72
to 1.14 3.11
Free machining steels
From 0.41 1.12
To 0.62 1.69
Tool steel (unhardened)
from 0.75 2.05
to 1.30 3.55
Cast steel
from 0.62 1.69
to 0.86 2.35


To echo what Mcgyver was talking about, cars, be they Porsche or any other flex in reaction to loads created by driving this is how they absorb shock loads (imagine hitting a deep pot hole at high speed). If the car was so rigid that it did not flex to any degree your high performance Porsche would probably weigh about 65 tons, have about 1200 BHP and do 40 MPH. Vehicles like this do exist, they are called tanks and they flex very little. So if you want to build a machine which takes very light cuts with a high-speed spindle at relatively low feed rates build a Porsche. If you want to take heavier cuts with bigger cutters at higher feed rates build a tank. Me i would take the tank any day.

I fear I may be talking rubbish so I’ll stop there

[guido]

huflungdung,

Yes planning to mill metals.

At this point I tend to say that I'm not interested in rapids because I won’t use it as a high volume production machine. So my philosophy is that my machine need enough power to cut metals...and at the same time I want to keep the cost down as much as possible...meaning small motors, linear guides ball screw etc. Probably after selecting the motors based on cutting forces I will be left with acceptable rapids.

The machine will be relatively small...tool size of max 10mm D.

[Mcgyver]

guido, not that you were doing so, but in general there are a lot of threads where it seems like people harbor this notion that machine tools are as massive as they are because the engineers/designers really haven't done their jobs as well as they could. I agree with you, other than supporting its own mass, the cutting force is what is at issue, but that is only 1/2 of it - we could agree on what the cutting force is and still build radically different machines depending on what deflection is acceptable.
I'm not an engineer so you'll quickly be able to run circles around me on technical points, how ever having a 2000lb cast iron mill wanting to hop around the shop is also compelling towards the belief that big bloody cast iron pieces make for a good machine.

In theory, there should be no discrepancy. If we plugged in the forces and deflection tolerances into a model, we should get the same answer – my point is that there is no free lunch and the answer will be the type and massiveness we see on existing tools.

Re cutting forces and power, I’m not sure of this so I presented it as a question. it’s been a long time since high school physics, but isn’t the magnitude of the force relative to the power? Common sense says there is a correlation between the power, or material removal rate (the calculation of which requires a constant for each material as you mention) and the force. iirc force is a vector having both a magnitude and direction. If the magnitude either is, or is tied to the power, then what we’re missing is the direction. Unless it’s a shaper or planer, the direction is constantly changing as the cutter rotates, hence the machine must be built to take the loads in all directions. Hence the idea that the cutting forces involved depend on the power used, which depends on the material removal rate. The max material removal rate does change for materials, but at the max, won’t the power be the same?

Just to convince myself I’m only partially crazy, here’s some material I found that ties together material constants and power to calculate the force magnitude
http://www.hytec.com/Designin.pdf

[HuFlungDung]

Probably after selecting the motors based on cutting forces I will be left with acceptable rapids.

If you like The net effect will be the same: determine how fast you want the machine to move, and you will have lots of power left over to do the feeding. If the machine cannot rapid any faster than it feeds, you will soon put it in the dumpster

Typically, the table and workpiece have a sizable mass. You can apply your formula to determine exactly how much power it takes to accelerate this load to full rapid speed. This is the most work that the feed motors have to do.

The amount of power it takes to cut the material is readily available information, but that only applies to spindle horsepower. So you could have a 10hp spindle (if you like) chomping away on this part. That 10hp could take a good bite and the force that the spindle/tool is putting on the machine has little to do with the XYZ motor power.

Feed horsepower could be miniscule, especially if climb milling.

[mactec54]

I have some infomation that mite help a little The( machinery's Handbook) has all
the imformation you are looking for. I found this as well in some other software
Aluminum at 10 IPM Depth of cut .100 Dia. of cutter 10mm Machining HP .24
Aluminum at 20 IPM Depth of cut .250 Dia. of cutter 10mm Machining HP .59
Tool Steel at 10 IPM Depth of cut .100 Dia. of cutter 10mm Machining HP 1.02
Tool Steel at 20 IPM Depth of cut .250 Dia. of cutter 10mm Machining HP 2.55
mactec54

[guido]

I agree that the net result might turn out the same..
However I cannot ignore cutting forces if I want to design something..
There is a difference between building somehing for speed/acceleration or something capable of resisting forces..

If I design something based on rapid acceleration and speed I will a light construction with matching screw and motors...
This construction will nog be very good at handling cutting forces..So first I need to have an idea about the cutting forces select components that match that and calculate the motors I need to get the desired/acceptable acceleration and speed using these components...

I doubt feed horsepower will be miniscule...the power you need for the spindle is directly related to the cutting force...cutting force on the spindle means the same cutting force on the workpiece and machine...these forces are in te kN range..

[HuFlungDung]

What do we know? Unit horsepower tells us how much energy is consumed, on average, by reducing a certain volume of solid material to chips. Most of the energy goes into heat and sound.

Spindle torque required for milling depends on the depth of cut, the shear strength of the material being cut and the diameter of the tool. But, tool edge geometry can have a tremendous effect on how difficult it is for the tool edge to begin to shear the material. This determines the actual cutting force. But the force can only be transmitted properly if the machine is rigid enough to not deflect, so that the tool edge practically follows the intended path through the material.

Does it matter if the material moves into the cutter (under constant feed) as the chip is being sheared? Efficiency of chip flow takes major precedence in feed force. Again, this would seem to me to be a highly variable amount of force.

Spindle thrust force for opening a hole by drilling has much to do with tool geometry as well. The force is high considering the inefficiency of the drill point at penetrating the part, since it must extrude the material at the center web...essentially rubbing it out of the way. If you were to drill a pilot hole first, to permit the tool to enter the cut without the extrusion effect, the limitation on the rate of feed has little to do with force, but is limited more by spindle torque, (and tool edge strength) than penetrating power of the feed mechanism.

Once the tool edge has begun to shear the material, an increase in feedrate changes the angle of entry of the tool, but does it proportionately increase the feed pressure? No, my experience tells me it does not. When climb milling a part with 1/2 tool engagement, the tool shears the material and applies sufficient force to pull the work into the tool. The use of screws to regulate table feed causes the nut and the machine base to absorb most of the force of the cut, ie., the feed motor is not doing an equivalent amount of work compared to the spindle.

The tool geometry has the greatest effect (that I can imagine) on the force required to shear the material. The shear strength of the material should be readily available information.

I suspect the best you can come up with is a certain performance specification based on the usage of some kind of standard test tool. The thrust for drilling would probably be a good all round indicator of how much feed pressure you will need to create with your feed motors. You can then determine the bending moment on the column of your machine, and work back from that, how stout the components need to be.

[Geof]

Guido;

This quote from Hu (Post #12) in my opinion encapsulates what you need to consider and it also pulls in McGyver's comments about power.

"The amount of power it takes to cut the material is readily available information, but that only applies to spindle horsepower. So you could have a 10hp spindle (if you like) chomping away on this part. That 10hp could take a good bite and the force that the spindle/tool is putting on the machine has little to do with the XYZ motor power."

Hu has considered the force parallel to the direction of movement and correctly points out that climb milling will pull the work into the cutter; but McGyver points out that with a milling cutter the direction of the force rotates.

I am glad Hu used 10 hp; I did some heavy straight X axis roughing on aluminum with a 3/4" two flute cutter running at 10,000 rpm with 0.4" depth of cut both radial and length and adjusted the feed until I was running at 100% to 120% load with a 10 hp spindle. If you do the calculation you find the load on the cutting edge is around 160-200 lb so the side load on the spindle in the Y direction peaked at this twice per revolution. The tool deflection in this direction was in the region of 0.003" and this includes everything; bending of the tool itself (which is probably negligible), deflection of the entire spindle assembly due to play in the spindle bearing and the Z axis linear guides, deflection of the table sideways on the X axis guides and axial deflection of the Y ball screw (which is probably negligible also).

This deflection more or less correlates with a crude experiment I did with a dial gauge between the table and spindle. Bracing against the table and pulling as hard as possible on the spindle I am able to get a deflection of a bit less than 0.001".

You need to figure out the hp you need based on the rate at which you want to remove metal. From this for different tool diameters you can calculate the tool load, i.e. the side load on your machine, and then aim to build a structure that keeps this deflection within acceptable limits.

[Mcgyver]

hi Geof, haven't seen you around in awhile, if you follow that link I posted, a few pages in they give material constants for hp/cubic inch/minute, as per your last paragraph

does the dia really matter? it along with depth and feed determines the removal rate, if we're removing at the same but with different dia cutters the power's still the same, right?

[ImanCarrot]

This was posted elsewhere, so Credits to the chap who originaly posted it (can't remember who), but if you go to www.lovejoytool.com and fill in some stuff, they'll send you a nifty slide rule type calculator for free that's got all sorts of brilliant info on it including Horsepower, DOC, WOC, RPM, IPM, FPT, #inserts, SFM, Diameter, HP Factor vs various alloys etc etc. It really is excellent it even has hardness conversions from Shore to Rockwell to Brinell! well handy

[Geof]

McGyver;

I have never gone into the hp versus material removal very deeply because I think it really only applies when you have nice straight predictable cuts. I doubt very much that you would find the same relationship between all combinations of dia, rpm and feed because you are making different size chips. More power will be needed to convert the same volume of removed material into smaller chips. I have also read some articles (can't find them now!) that suggested the power needed was dependent on a combination of both cutting speed and depth of cut; at a very high speed such as 25,000 rpm with a 1/2" cutter running at .025" per tooth (I forget the depth of cut) it was claimed the fracture mode of the chip changed to a brittle mode failure. This was in 7xxx series aluminum alloys so it does seem reasonable. But these conditions rarely apply even in commercial CNC work let alone a hobby machine situation.

[loves80z]

http://www.carrlane.com/Catalog/inde...3B2853524B5A59

Workholdong places have the info U are after. Based on HP of the spindle. Below is copied pasted from the link above. There are pics with it on the site I didn't get

Kevin

An important step in most fixture designs is looking at the planned machining operations to estimate cutting forces on the workpiece, both magnitude and direction. The "estimate" can be a rough guess based on experience, or a calculation based on machining data. One simple formula for force magnitude, shown in Figure 3-10, is based on the physical relationship:

Please note: "heaviest-cut horsepower" is not total machine horsepower; rather it is the maximum horsepower actually used during the machining cycle. Typical machine efficiency is roughly 75% (.75). The number 33,000 is a units-conversion factor.
Figure 3-10. A simple formula to estimate the magnitude of cutting forces on the workpiece.

The above formula only calculates force magnitude, not direction. Cutting force can have x-, y-, and/or z-axis components. Force direction (and magnitude) can vary drastically from the beginning, to the middle, to the end of the cut. Figure 3-11 shows a typical calculation. Intuitively, force direction is virtually all horizontal in this example (negligible z-axis component). Direction varies between the x and y axes as the cut progresses.
Figure 3-11. Example of a cutting force calculation.

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