How do you calculate req clamping power?

[Shotout]

I was working on a fixture design for a friend that has a one man shop, kind of an educational exercise, and one thing that struck me was just how much force will be placed on the clamps I picked out. The clamp is rated to max 1500lbs but there will be a lot of force above that clamp so I was wondering if someone could offer me some help. Below is a jpg of the part located on the fixture plate. can anyone give me an idea on how to calculate the machine forces placed on the hardware? The blank material is 1X1X1in 304SS for scale. Thanks

[Geof]

I suppose it would be possible to get some idea of the forces from the spindle horsepower needed for a particular set of cutting conditions, but I get an estimate by a bit of a hokey way. The maximum servo force on the X and Y axes of most of my machines is around 2000 lbs, I rarely see the servos operating at more than 10% to 20% load on a straight milling cut on aluminum, steel might go up to 40% sometimes. So I conclude that I rarely put a force of more than 400 lbs on a part when it is being machined. I should mention the corresponding spindle load might be 80% or more so I am taking good cuts.

Drilling can put much more force and I often push the servos over 100%, often drills are pushing the part down into the clamp or vise and things will move during drilling that will never move during milling.

Your point about force above the clamp is a good one; you need to consider leverage which can multiply to tool force many times down at the clamping point.

[Shotout]

Glad to know I'm not to far off the mark. I always assumed that if the cutter isn't loaded past the feed/tooth then the servo could only put so much load on a part but it is nice to hear that from someone wiser than myself. My friend likes the fixture design so if he gets the bid he plans on making two of them and I appreciate the help, makes me feel more confident about the whole thing. He can program to counter some of the leverage, maybe an extra roughing pass at the first DOC, or whatever his preference is.

On the drilling he will probably go with an inserted drill so he should be able to control the forces somewhat better than a carbide tipped drill. If it starts loading the spindle to much just rotate and run on. Shame he doesn't have a 4th axis to drill the side hole, but that is another fixture for another day.

Thanks
Scott

Originally Posted by Geof I suppose it would be possible to get some idea of the forces from the spindle horsepower needed for a particular set of cutting conditions, but I get an estimate by a bit of a hokey way. The maximum servo force on the X and Y axes of most of my machines is around 2000 lbs, I rarely see the servos operating at more than 10% to 20% load on a straight milling cut on aluminum, steel might go up to 40% sometimes. So I conclude that I rarely put a force of more than 400 lbs on a part when it is being machined. I should mention the corresponding spindle load might be 80% or more so I am taking good cuts. Drilling can put much more force and I often push the servos over 100%, often drills are pushing the part down into the clamp or vise and things will move during drilling that will never move during milling. Your point about force above the clamp is a good one; you need to consider leverage which can multiply to tool force many times down at the clamping point.