Clamping Force
When properly designed and controlled, a hydraulic clamping mechanism prevents the workpiece from moving when machining forces are applied, yet will not cause permanent distortion to occur in the workpiece. As already mentioned, it is important to ensure that the clamping cylinders themselves do not move or distort the positioned part.
The clamping force needed to keep a part in place is determined by the machining forces to be applied to the part. Cutting tool suppliers provide information about the forces produced by each type of tool when machining various materials. Following are example calculations for a face milling operation.
Example
An 8inch diameter cutter with 10 teeth (inserts) is to be used to machine low silicon aluminum at 3000 SFM (surface feet per minute). How much force does it apply to the part, and how much clamping force is required to restrain the part?
Step 1
Convert SFM to RPM using this formula:




where
RPM = spindle speed
SFM = tool surface feet per minute
D = tool diameter in inches


Inserting the numbers, 




Step 2
Determine the material removal rate using this formula:



MRR = (W)(D)(F)(N)(RPM) 


where
MRR = material removal rate in in3/minute
W = width of cut in inches
D = depth of cut in inches
F = feed per tooth in inches
N = number of cutter teeth
RPM = spindle speed


For the 3000 SFM specified in this example, a typical tool catalog lists a feed per tooth of 0.008” maximum at a cut depth of 0.1”. Inserting the numbers, 


MRR = (8”)(0.1”)(0.008”)(10)(1432) = 91.6 in3/minute 

Step 3
Calculate spindle hp:
For milling aluminum with a dull tool, a typical fixturing reference table lists a spindle horsepowertoMRR ratio (generally referred to as “unit power”) of 0.4. Therefore, spindle hp = (0.4)(91.6) = 36.6 hp.
It should be noted that this horsepower calculation is for fixture design and not for machine tool horsepower requirements. (A true 40 hp machine, for example, can remove aluminum at more than 200 in3/minute.)
Step 4
Calculate force transmitted from the tool to the workpiece using this formula:






where
Cutting force = force transmitted to the workpiece, in lb
hp = the spindle hp calculated in step 3
SFM = tool surface feet per minute
26,406 = 33,000 ftlb/min conversion factor, with 80% efficiency


Inserting the numbers, 




Using a safety factor of 1.5 for hydraulic clamping, the allowance for cutting force becomes 483 lb.
Step 5
Calculate the required clamping force using this formula:






where
Fclamp = required clamping force to resist cutting force, in lb
Ftool = cutting force, in lb
k = coefficient of friction


The coefficient of friction for lubricated aluminum (flooded with coolant) is 0.12. Inserting the numbers, 




This is the clamp force required to keep the workpiece immobile while subject to the cutting tool force if clamping alone is holding the workpiece in place. Other elements, such as positioning cylinders and locating stops, may contribute forces that reduce the required clamping force.
The foregoing calculations determine the required hydraulic system parameters. Recall that the force produced by a cylinder equals the product of cylinder crosssectional area and hydraulic fluid pressure. Cylinder catalogs list cylinders in terms of their maximum clamping force when maximum rated hydraulic pressure is applied.
In practice, the hydraulic system should be operated at 5075% of its rated pressure, so clamping cylinders selected for the above example should have a total capacity that is 1.332.0 times the required force. That means the total maximum rated clamping force of all cylinders in the example should be within the range 53538050 lb. (e.g. three cylinders rated 2600 lb each.)
Force is not the only cylinder parameter of concern. The next section describes additional factors in cylinder selection.
