[ANKU]

Please refer to the jpg for this discussion. I need to determine the compression force and the clamping, or downward force on my part as a result of the gripper fingers "clamping" on my part. You will see that the gripper fingers have a 45 degree chamfer on them and that each finger exerts 10 lbs of force in along the x-direction (the compression force direction). Any help?

Regards,
Anku

[apache405]

I am pretty sure that your part is going to experience around 7.1 pounds for force in the vertical and horizontal directions. I used Force on part=clamp force x sine 45 for the vertical and the same equation but with cosine for the horizontal.

From a pure physics point of view, there is a more complex but more accurate calculation based on friction, but I don't have a piece of paper handy to figure it out (maybe during this evening's physics 2a lecture...).

[ANKU]

I started out this way too, however I could not reconcile the sum of the forces in x and y to equal zero. If the resultant force acting normal to the 45 degree plane is 10 lbs, then of course, the x and y component are 7.1 lbs, but it does not seem that the 10 lbs of clamping force in the x direction can also be considered 10 lbs of force acting normal to the 45 degree plane.

[apache405]

i am going to try and make a force diagram...

[apache405]

the 45 degree plane would exert 10lbs force. the face of the clamping fixture is applying 10 lbs. if the faces were not chamfered, the force is 10 lbs in the x plane. it shouldn't matter that the clamp faces are chamfered. the force will always act in the plane it makes contact in.

also i talked with my physics prof. he said it is basically 7.1 pounds.

If you are still worried, take an extra blank for your part and run it and see what happens.

[ANKU]

So what you are saying is regardless of the clamping face design, the force normal to that plane is always 10 lbs. (i.e. if the claming face is blunt, it is 10 lbs, if it is chamfered at a 10 degree angle, the force normal to that 10 degree face is 10 lbs, and if the face is chamfered at 60 degrees, the force normal to that 60 degree face is still 10 lbs?)

[apache405]

yes.

[ANKU]

Intuitively, it seems that with a blunt face, the force normal to the clamping area is 10 lbs (10 x, 0 y). I buy into this no problem. But it would seem that as a chamfered face is created, starting at 10 degrees from horizontal, the force normal to the clamping face would be slightly less than 10 lbs. since the 10lbs acting in the x direction would be slightly diverted due to the angle into the y direction. And as that angle increases, it seems that even less of the 10lbs is acting normal to the claming face as more of the 10 lbs force is being distributed in the x and y direction. This is my thinking and reasoning for posting this question. However, you are saying that my thought process is flawed and that regardless of the face angle, the force normal to that face is always 10 lbs?

[apache405]

basically. the face of the clamp is going to exert whatever force is applied to whole arm. or at least that what i gather.

[ANKU]

Thank you for your help!

[apache405]

no problem.
good luck with your project!

[cncuser1]

it is 7.1 pounds in each direction IF the part can slide on the WHILE the grippers are applying their force!

If the friction between part and table is high then then the grippers may not apply equal force.

sometimes vaccum tables work well too!

[apache405]

it is 7.1 pounds in each direction IF the part can slide on the WHILE the grippers are applying their force!

If the friction between part and table is high then then the grippers may not apply equal force.

as soon as one of the grippers makes contact the friction force will rise by a maximum of 7.1 pounds because friction is proportional to the normal force from the plane of contact.