Cutting Technology
Introduction
This introduction will help you use information provided by tool manufacturers in the application of their tools. Estimating cutting forces being transferred into the workpiece is just one tool to use in a competitive workholding environment.
The information presented here is only to be a guideline and not the final decision. Use this information with a cutting tool brochure you get from your cutting tool supplier as an aid in determining your cutting forces. Much of the calculations presented here are readily available from many sources. Your cutting supplier may even have a slide chart you can obtain to do equations for you.
The operations described here include boring, drilling, end milling and face milling.

Drilling involves using a multi-fl uted tool with a helix spiral. The tool is driven in as it is rotated to create a round hole.
End Milling uses a multi-fluted rotary tool with or without removable (inserts) teeth to remove material along the edge of the workpiece. The cut is usually very shallow and the depth is many times the thickness of the cut.
Face Milling involves a very shallow depth, but a very wide cut. Cutters can range up 12 inches (300 mm) or more in diameter and can have many replaceable teeth (inserts).
These examples are only a very small sample of operations that can use hydraulic workholding. Cutting force determinations These cutting force examples involve face milling. The largest use of hydraulic workholding is by far for some sort of milling operations. 1 Imperial system Cutting Force (Pounds) = Spindle Horsepower x 26400 (Horsepower to foot pounds per minute at 80% efficiency)/Cutting Speed (In tool surface feet per minute) Spindle Horsepower = Unit Power (Horsepower per cubic inches of material removed per minute) x Material removal rate (Cubic Inches per Minute)
Material removal rate (Cubic inches per minute) = Width of the cut (Inches) x Depth of the Cut (Inches) x Feed per cutter tooth (Inches) x Number of cutter teeth x Spindle RPM Example An 8-inch diameter cutter with 10 teeth (inserts) is machining low silicon aluminum at 3000 SFM (surface feet per minute).
First, you must convert surface feet/minute into tool RPM/Solving Tool RPM= SFMDiameter (Inch) x .2618 = 1432 Tool RPM
Now you can determine your material removal rate. An independent tool catalog lists a feed per tooth of 0.008" maximum at 3000 SFM at cut depth of 0.1".
This gives 8" (diameter cutter) x 0.100" (cut depth) x 0.008" (feed per tooth) x 10 (number of teeth) x 1432 (spindle RPM)= 91.6 cubic inches per minute material removal rate
Next, spindle horsepower is found using unit HP from the table Spindle Horsepower = 91.6 x 0.4 (Unit Horsepower for Aluminum with a dull tool) = 36.6 HP.
Note this Horsepower is for fixture design and not for machine tool horsepower requirements.
For example a true 40 HP machine can remove aluminum well over 200 cubic inches per minute.
Using the original formula:
36.6 hp x 26,400/3000 SFM = 322 lbs. 3000 SFM of force being transmitted into the work.
Force is transmitted in the same direction as the cutter movement. In other words, if the cutter moves right to left in the diagram below, the cutter force is transmitted from right to left.
Using a safety factor of 2 for rigid clamping gives 644 pounds in line parallel to the line force and 483 pounds using an elastic medium such as hydraulics with a safety factor of 1.5. Note this force does not take into account any sort of friction factors if you plan on relying on friction force between a swing cylinder and the workpiece. For example: The coefficient of friction for lubricated aluminum is .12 (flooded with coolant) this same 483 pounds of force becomes 483/.12 = 4025 pounds. This uses clamp force only and does not take into account any direct forces that may be developed by the cylinders that located the workpiece against fixed locators. Unit Power for dull tools [imperial system]
Material | Unit Power hp/In3/min | |||
Hardness | Turning HSS & Carbide Tools |
Drilling HSS Drills |
Milling HSS & Carbide Tools |
|
STEELS | 85-200 Bhn | 1.4 | 1.3 | 1.4 |
Plain Carbon | 35-40 Rc | 1.7 | 1.7 | 1.9 |
Alloy Steels | 40-50 Rc | 1.9 | 2.1 | 2.2 |
Tool Steels | 50-55 Rc 55-58 Rc |
2.5 4.2 |
2.6 3.2 |
2.6 3.2 |
CAST IRONS | 110-190 Bhn | 0.9 | 1.2 | 0.8 |
Gray, ductile and malleable | 190-320 Bhn | 1.7 | 2.0 | 1.4 |
STAINLESS STEELS | 135-275 Bhn | 1.6 | 1.4 | 1.7 |
30-45 Rc | 1.7 | 1.5 | 1.9 | |
TITANIUM | 250-375 Bhn | 1.5 | 1.4 | 1.4 |
NICKEL ALLOYS | 80-360 Bhn | 2.5 | 2.2 | 2.4 |
ALIMINUM ALLOYS | 30-150 Bhn | 0.3 | 0.2 | 0.4 |
MAGNESIUM ALLOYS | 40-90 | 0.3 | 0.2 | 0.2 |
COPPER ALLOYS | 10-80 RB 80-100 RB |
0.8 1.2 |
0.6 1.0 |
0.8 1.2 |
Metric System
Cutting Force (Newtons) = Spindle Power (kW) x 48000 (80% effi ciency) / Cutting Speed (Meters per minute).
Spindle Power = Unit Power (kilowatts per cubic centimeters of material remove per minute) x Material removal rate (cubic centimeters per minute).
Material removal rate (Cubic centimeters per minute) = Width of cut (mm) x depth of cut (mm) x feed per tooth (mm) x number of teeth x spindle RPM/1000.
Example:
A 200 mm cutter with 10 teeth is machining low silicon aluminum at 1000 MPM (meters per minute).
Solving Tool RPM = MPM x 1000 Diameter (mm) x p (= 1592 Tool RPM
The same tool catalog lists a feed per tooth of 0.2 mm at 1000 MPM and a cutting depth of 2.5 mm. This gives an 200 mm cutter x 2.5 mm depth x 0.2 mm feed x 10 teeth x 1592 Tool RPM/1000 = 1592 cm3/min.
Spindle power = 1592 x 0.018 = 28.7 kW This too is power from a fixture design standpoint; the actual operation will use less power than indicated here.
Using the original formula transposed is: Cutting Force 1378 (Newtons) = 28.7 (kW) x 48000 (80% effi ciency) / 1000 (MPM cutting speed)
Multiply by a safety factor of 2 for rigid clamping and by 1.5 for elastic clamping (hydraulic).
This calculation does not take into account coefficients of friction when using clamp cylinders. For example, if the aluminum has a coefficient of .12 (flooded with coolant), the clamping force becomes 1378/.12 = 11483 Newtons of force. This calculation does not take into account forces being generated by the fixture positioning cylinders.
Use these numbers and set up your hydraulic system to run at about 50 to 75% of its rated pressure. This leaves some reserve for a later date when the process is optimized and you need more holding/ clampforce for higher speeds and feeds. If you design to the maximum now, you have nothing in reserve.
Unit Power for dull tools [metric system]
Material | Hardness | TURNING P1 HSS AND CARBIDE TOOLS |
DRILLING P HSS DRILLS |
MILLING P d HSS AND CARBIDE TOOLS |
STEELS, WROUGHT AND CAST | 85-200 Bhn | 0.064 | 0.059 | 0.064 |
Plain Carbon | 35-40 Rc | .077 | .077 | .086 |
Alloy Steels | 40-50 Rc | .086 | .096 | .100 |
Tool Steels | 50-55 Rc 55-58 Rc |
.114 .191 |
.118 .146 |
.118 .146 |
CAST IRONS | 110-190 Bhn | .41 | .055 | .036 |
Gray, ductile and malleable | 190-320 Bhn | .077 | .091 | .064 |
STAINLESS STEELS, WROUGHT AND CAST |
135-275 Bhn | .073 | .064 | .077 |
Ferritic, austenitic and martensitic | 30-45 Rc | .077 | .068 | .086 |
TITANIUM | 250-375 Bhn | .068 | .064 | .064 |
NICKEL ALLOYS | 80-360 Bhn | .114 | .100 | .109 |
ALIMINUM ALLOYS | 30-150 | .014 | .009 | .018 |
MAGNESIUM ALLOYS | 40-90 | .009 | .009 | .009 |
COPPER ALLOYS | 10-80 RB 80-100 RB |
.036 .055 |
.027 .046 |
.036 .055 |