[gabi68]

Hi to all,

I try to find (maybe that not exist) a formula for calculating power (in N) for an end mill to cut or profile some material (Al, wood, plastic, ...). I need that for another calculation for deflection of a aluminum profile. Formula for deflection is:
d2 = F x L*power3 / 48E x I x 10*power4
Where
F = load(N);
L = Unsupported lenght (mm);
E = 70 000N/mm2 (modulus of elasticity);
I = Moment of Inertia (cm4);
D = Deflection of profile (mm);
W = Section Modulus (cm3);

I don't know value for that "F".

My question is: it is somewhere a table with values for a known dia of tool, DOC and RPM for diferent materials.

Thank you
Gabi

[engkit]

Hi Gabi68
It is a long time since I have had to do this but from what I remember is the following:-

It takes 1hp to remove 1cu inch/min or 22mm2/min per watt spindle power with sharp tools on steel but that is not what you need to check deflection you need to know feed force. that is dependant on cutter geometry
To cut aluminium you would more than likely need one third to a quarter the power (be able to remove three/ four times the material per watt) and wood could be half again
at least. I could be wrong as it has been thirty odd years.
I will try to find out if we still have the cutting tool manuals which deal with machinability of different materials.
I will say that feed force is dependant also on the type of cut, ie. full width of cutter, one half cutter, climb or conventional cutting ect. It might be difficult to explain in words all the actions going on. Remember that the cutter torque will be applied to the material at the half dia of the cutter (partial cut) you should be able to work an approximation to the feed force from that.
cutting speeds for the materials also change dependant on the material and cutter material. Plain carbide tools cut steel at 100 metres/min alumin at 300 metres/min wood at 600metres?/min. coated carbides are completely out of of the question we cut alloy steels at 300 to 400m/min on 40 Kw spindle machines with 2.5 Kw servo feed motors

[rowbare]

I googled "feed speed horsepower calculator" and this is one of the hits I got.

http://www.custompartnet.com/calcula...ing-horsepower

[gabi68]

Thank you engkit and rowbare. Engkit maybe you will find that book and post some figures.

Thanks again
Gabi

[skiingman]

Doing design for my router so I know where you are coming from...If you use the "google custom search" there are some good threads from years past on this specific subject.

First thought:

You ask for power in N. Newtons are the SI measure of force. Power is the Watt.

Spindle power: The power required at the spinning tool to cut material in a given set of circumstances. Tables and calculators to determine spindle power are available on the web, some are more detailed than others. Ensure the calculator you are using is intended for the machining operation you intend (as in, milling rather than drilling) as the differences in power required can be substantial.

What you really want, and is more difficult to come by, is the resultant force in a worst case cutting condition for the toughest material you intend to machine. There is no simple equation or table for this, because it depends very strongly on the geometry of the tool, the work, and the toolpath. As the tool is rotating and shearing away chips, the resulting force on the machine varies as the angular position of the cutter contact with the work changes. See "climb" vs. "conventional" milling and "self feeding". Depending on cutter geometry, there may well be a Z component as well as the X and Y components of the force.

For the design process, you need a worst case scenario. AFAIK, this would be in "conventional" milling where the line of action of the cutter tooth is parallel to the direction of feed. In this case, you can use the basic relation of force to power to calculate the resultant force on the machine.

An excellent description and illustration is available at:
http://www.carrlane.com/Catalog/inde...3B2853524B5A59

Another thought:

You are concerned with the deflection of your beam, but remember that this force is acting through the cutter, which isn't coincident with your beam. Your stiffness calculations need to consider the moments imposed by the arm from the cutter to the beam, and of course they also need to include all the other elements in your machine beyond just the aluminum structural beam. When I complete my machine, I plan on doing as much work as possible on tall fixtures that limit Z travel and are mounted as near to the edge of the table (support of the gantry) as possible.

[TheMillMan]

Hi to all,

I try to find (maybe that not exist) a formula for calculating power (in N) for an end mill to cut or profile some material (Al, wood, plastic, ...). I need that for another calculation for deflection of a aluminum profile. Formula for deflection is:
d2 = F x L*power3 / 48E x I x 10*power4
Where
F = load(N);
L = Unsupported lenght (mm);
E = 70 000N/mm2 (modulus of elasticity);
I = Moment of Inertia (cm4);
D = Deflection of profile (mm);
W = Section Modulus (cm3);

I don't know value for that "F".

My question is: it is somewhere a table with values for a known dia of tool, DOC and RPM for different materials.

Thank you
Gabi

i don't know of any tables but you could enter your equation into excel and graph the different materials having different modulus. so set up the equation, deflection with respect to force at a given E and L.

D2 = (F*L^3) / (48*E*I*10^4)

or force with respect to deflection (you list W, but i don't see it in the equation)

F=(D2(E*I*10^4))/L^3


you know the force because the machine can only produce so much before it stalls and if not find it by experimentation (the machine will force the tool to the max until the tool brakes or the machine brakes). so i would say the tool force is dependent on the machine that drives it. what you don't know is deflection so that is unknown and a graphing calculator will tell you every value of deflection at a given force at a set length

i am not sure about the equations so please review them so we can help you with your problem. later i will look up this information myself.

i could be wrong but the machine can only push so hard then something gives, so in my opinion you know the force. its the linear force exerted by the ball screw.

it might be possible to contact the tool manufacturer and ask them what the max linear force it takes to brake their tool and that could be a good stating point or you could, if the e-mill is not to big clamp the tool somehow and by experimentation hang weight from the tool end till it brakes then weigh it, that will be the max force you could apply to the tool.

i hope this helps

the mill man

[davereagan]

The correct formula is Deflection = F*L*L*L/48*E*I

There is no 10 to the 4th power. It is very important to watch the units for each quantity entered into the formula. This applies to a cantilevered beam, with load on one end, support at the other end, and L being the length in between. E is the modulus of Elasticity for the material. I i the polar moment of inertia, which raises the height of the beam to the 4th power, which is probably where the 10 to the 4th power got into the formula given in a previous post. The Machinery's handbook explains this thoroughly with examples and different formulae for different bending applications. If you are trying to find out how much you can cut based on how much bending stress the cutter can handle, it will be even more complicated. What helix, how sharp, runout (uneven flute loading). The rules of thumb are the place to start, like a chipload of 1% of the cutter diameter in slotting aluminum with a rigid setup.

[pst]

hi defelection farmula for cantilever beams........check out the link
http://www.clag.org.uk/beam.html#equations

i think it will surly help you out

[davereagan]

Those formulas are some of the most powerful I learned in engineering school. Notice the deflection being a function of the cube of the length. So for any given endmill, if you hang it out twice as far with the same side load, it will deflect 8 times more, not twice or even four times. This can be used to your advantage when reaming, because runout is proportional to length. If you have a 1/4" reamer that is 6" long, it is better to leave it out there all the way, rather than choke up on it and have 3 inches hanging out. Yes, it will have twice the runout as the shorter setup, but it will only be 1/8 as stiff and will follow the hole rather than cut a bigger hole. I think that's why jobber reamers are longer than most holes you'll be reaming.

[veteq]

So, you dont know what the force is that is acting on the beam as a result off the milling proces, you will need to know the tangial force that occurs between the mill and the material. This force will make the beam deform.

To get that force you will need to do some extra calculations, it aint hard but it is a longer calculation with parameters that can be found in some machining tables.

You cann find a nice book about it on this website:
http://www.alliedchennai.com/31748ip.html

You should buy it, its including a example off the needed calculation.

regards,

Roy